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NOTES ON BUILDING CONSTRUCTION
PART IV.
CALCULATIONS FOR BUILDING STRUCTURES— COURSE
FOR HONOURS
NOTES ON BUILDING CONSTRUCTION
Arranged to meet the requirements of the Syllabus of the Science and Art
Department of the Committee of Council on Education, South
Kensington.
In Four Parts. Medium %vo. Sold separately.
Part I. — First Stage, or Elementary Course. With Illustratiom,
los. 6d.
Part II. — Second Stage, or Advanced Course. With 479 Illustra-
tions. \os. 6d.
Part III. — Materials. Course for Honours. With 188 Illustrations.
21s.
Part IV. — Calculations for Building Structures. Course for
Honours. With 597 Illustrations.
Official Report on the Examination in Building Construction, held by the Science
AND Art Department, South Kensington, in May 1875.— "The want of a text-book in this
subject, arranged in accordance with the pubHshed Syllabus, and therefore limiting the students and
teachers to the prescribed course, has lately been well met by a work published by Messrs. Rivingtons,
entitled ' Notes on Building Construction.' "
"Those things which writers of elementary books generally pass over are here explained with
minuteness. . . . Altogether the book is one which it is a pleasure to recommend. Its primary
object may be to support the Science and Art Department, but it will be found to be of wider use ; and
if the parts which are to follow are prepared as carefully as this is, the ' Notes on Building Construc-
tion' will far surpass any work of the kind hitherto published." — Architect.
" Something of the sort was verj' much needed. . . . The whole series when published will
be a great boon to young students." — Builder.
" One of the most sensible and really reliable aids to students of construction we have seen for a
long time. If the remaining Parts are up to the standard successfully aimed at in Part I., the work
cannot fail to become the standard text-book for students." — Building News.
" It very rarely happens that explanations are given with such clearness as those in ' Notes on
Building Construction,' and the dullest student cannot fail to grasp the idea intended to be conveyed.
... As a work of reference it will at once take a leading place."— Builder's Weekly Reporter.
" Certainly the four parts will, judging from the first, form the besl text-book on the subject ex-
tant." — English Mechanic.
"The work throughout is got up in the most admirable style, and is profusely illustrated with
well-drawn engravings." — Timier Trades' lournal.
" The whole will form a compendious series of volumes of very great value to 'practical' men.
The text is prepared in an extremely simple and consecutive manner, advancing from rudimental
and general statements to those which are comparatively advanced ; it is a thoroughly coherent self-
sustained account. . . . We can testify that its contents justify the promises of the title, that we
have missed nothing which we looked for" and had a right to expect would be included in the
volume." — Athenceimi.
LONDON : LONGMANS, GREEN & CO.
RIVINGTON'S SERIES OF NOTES ON BUILDING CONSTRUCTION
BUILDING CONSTRUCTION
ARRANGED TO MEET THE REQUIREMENTS OF
THE SYLLABUS OF THE SCIENCE ART DEPARTMENT
OF THE COMMITTEE OF COUNCIL ON EDUCATION,
SOUTH KENSINGTON
PART IV.
CALCULATIONS FOR BUILDING STRUCTURES-
COURSE FOR HONOURS
LONDON
LONGMANS, GREEN, AND CO.
AND NEW YORK : 15 EAST i6th STREET
1891
N
ON
A II rights reserved
PREFACE TO PART IV.
rpHE writer desires to express to the Publishers, and to
those who have done him the honour of readinaf the
p)revious Parts of this work, his regret for the great delay
tJhat has occurred in the preparation of this volume for the
Press.
More than one serious illness and the pressure of his
dluties have obliged him again and again to lay this work
emtirely aside. Finding that he was not likely to have
Leisure or strength to complete the volume, he has, with
t;he consent of the Publishers, entrusted most of it to
ai friend, by which arrangement he feels that he has con-
fi'erred a great benefit upon his readers.
The object in view has been to explain fully and
c^learly the simplest and best methods for calculating the
p)roper forms and dimensions to be given to Timber and
Eron Beams, Cantilevers, Plate, Box, and Trussed Girders,
Eloofs, Walls, Arches, Water pipes, etc. etc., showing by
eixamples the calculations required even for the details.
In carrying out this object the use of Advanced Mathe-
rmatics has been as far as possible avoided. The methods
e3xplained are chiefly those that can be effected almost
e3ntirely by drawing and measuring lines, and the calcu-
kations given require but a very slight knowledge of
IMathematics. Short rules are added for the use of prac-
vi PREFACE
tical men, and also tables by which many calculations
are altogether avoided.
The great advantage of using simple graphic methods
instead of elaborate calculations is that the former, if in-
correctly carried out, at once proclaim the fact by the
polygons of forces refusing to " close," so that an error has
to be faced at once and rectified ; whereas an error in a
figure, or even of a decimal point, in calculations quite
correct in other respects, may escape notice, and lead to an
utterly erroneous result, which, if acted upon, may en-
danger the stability of the structure to which it is applied.
It is hoped that this volume will be found to explain
not only the calculations that may be called for in the
Honours Examination at South Kensington, but also all
that can possibly be required in connection with ordinary
buildings. Upon the more complicated engineering struc-
tures it does not profess to enter.
PREFACE
vii
NOTE.
The following is an extract from the Syllabus of the
Science and Art Department of the Committee of Council
on Education, South Kensington.
It shows the heads of the examination in connection
with the calculation of structures for Honours, and
opposite to each subject the portion of this volume in
which the information required is to be found.
FIRST STAGE, OR ELEMENTARY COURSE.
No examination as to the calculation of structures.
SECOND STAGE, OR ADVANCED COURSE.
All that is required Avill be found in Part II.
IRequirements of Syllabus.
He must he able to solve
siimple problems in the theory
ojf construction, and to deter-
miine the safe dimensions of
irron or luooden beams subjected
tio dead loads.
EXAMINATION FOR HONOURS.
Where dealt with in this Volume.
Simple problems in the theory of con-
struction, equilibrium, chap. ii. ; beams,
chap. iii. ; dimensions of timber beams,
pp. 54, 75-80 ; wrought iron beams, chap,
iv. ; cast iron beams, chap. v.
In ordinary roof trusses and
firamed structures of a similar
dlescription, he must be able to
tirace the stresses, brought into
aiction by the loads, from the
20i>oints of application to the
ptoints of support, as well as
too determine the nature and
aimount of the stresses on the
different members of the truss,
amd, consequently, the quantify
ojf material required in each
poart.
Stresses on frames in general, chap,
ix.; open-webbed girders, chap. x. ; trussed
beams, chap. xi. ; roofs, chap. xii. ; deter-
mination of the quantity of material re-
quired in each part, i.e. of the dimensions
of the parts, tension and compression
bars, chap, vi.; joints and connections,
chap. vii.
viii
PREFACE
In ordinary ivalls and re- Stability of walls, piers, chimneys
taining walls, he must he able (house), and enclosure walls, cha^). xiii. ;
to ascertain the conditions retaining walls for water and eartli, chap.
necessary to stability, neglect- xiv.
ing the strength of the mortar.
In addition to the above-mentioned chapters, which
give all the information required for the examinations,
other chapters have been added upon Plate Girders,
Arches, and Hydraulics, in order that the volume may
be practically useful as an aid in solving all the calcula-
tions that can be required in connection with ordinary
Building Construction.
NOTATION AND WORKING STRESSES.
N.B. — In order to prevent constant repetitions in the body of the work, and
to assist the student, the meaning of tlie letters used in the notation, and
the working stresses used in the calculations in this book, are given in a
collected and complete form in Tables XXV. and XXVI. at the end of the
volume (pp. 348-351), so that they can easily be referred to at once without
consulting the index.
LIST OF BOOKS CONSULTED.
Adam's Designing Cast and Wrought Iron Structures.
Aide Memoire for the use of Officers of the Eoyal Engineers.
Bow's Economics of Construction.
Clark's Manual of Rules, Tables, and Data.
Cunningham's Applied Mechanics.
Fidler's Practical Treatise on Bridge Construction.
Hurst's Architectural Surveyor's Hand-Book.
Molesworth's Pocket-book of Engineering Eormulse.
Proceedings of the Institute of Civil Engineers. ^
Eankine's Civil Engineering.
Rankine's Useful Rules and Tables.
Ritter's Elementary Theory and Calculation of Iron Bridges and Roofs.
Stoney on Strains.
Transactions of the Society of Engineers.
TJnwin's Elements of Machine Design — Part I.
Unwin's Wrought-Iron Bridges and Roofs.
AVray's Application of Theory to the Practice of Construction (revised by
Seddon).
CONTENTS OF PART IV.
Chapter I.
introductory-
Terms IN Use. — Load — Dead Load — Live Loads — Comparative effect of
Dead and Live Load — Breaking Load — Working Load or Safe Load —
Strain — Stress — Tension — Compression — Transverse Stress — Shearing
Stress — Bearing Stress — Torsion — Stresses classified as regards Descrip-
tion — Stresses classified as regards Intensity — Limiting Stress — Safe
Working Stress — Intensity of Stress — Factor of Safety — Fracture —
Table of Stresses and Mode of Fracture — Elasticity — Elastic Limit — Re-
peated Stresses — Elongation and Shortening — Modulus of Elasticity
— Deflection — Stiffness .... Pages 1-13
Chapter II.
EQUILIBRIUM.
External Forces acting on a Structure. — Reactions — Rules for find-
ing Reactions at Supports- — Examples.
Stresses produced by the External Forces . Pages 14-23
Chapter III.
BEAMS.
Bending Moments. — Rule for finding Bending Moment — Bending Mo-
ment under various conditions.
Moment of Resistance. — Method of ascertaining by reasoning — by
Graphic Method — by Experiment — by Calculation — General Formulae
— Practical Formulte — Neutral Layer — Neutral Axis — Equivalent
Area of Resistance — Modulus of Rupture.
Shearing Stress. — Vertical — Horizontal — Rules for finding amount of
Vertical Shearing Stress — Graphic Representations, various cases — Dis-
tribution of Shearing Stress.
Beams of Uniform Strength.
xii
CONTENTS
Deflection. — Formulae — Simpler Formulee.
Fixed Beams. — Various Cases — Continuous Girders.
Examples. — Timber Cantilevers — Joist — Useful Notes . Pages 24-80
Chapter IV.
WROUGHT IRON ROLLED BEAMS.
Finding Moment of Resistance — Graphic Method — Mathematical Method
— Moment of Inertia — Practical Formulae — Examples — Remarks on
Rolled Beams — Girders built up with the aid of X Beams.
Pages 81-93
Chapter V.
CAST IRON GIRDERS.
Finding Moment of Resistance — Approximate Formulae — Graphic Method
— Mathematical Method — Practical Points connected with the form of
Cast Iron Girders — Examples — Cantilever — Girder . Pages 94-104
Chapter VI.
TENSION AND COMPRESSION BARS.
Tension Bars. — Symmetrical Stress — Effective Area — Unsymmetrical
Stress — Examples.
Compression Bars or Struts. — Fidler's Rules for Struts — Long and
Short Compression Bars— Gordon's Formula — Formulae and Table for
Wooden Struts — Examples — Cast Iron Columns — Timber Struts —
Wrought Iron Strut — X Iron Principal . . Pages 105-121
Chapter VII.
JOINTS AND CONNECTIONS.
Points to be observed in making Joints.
Riveted Joints. — Riveted Joints in Tension — Lap Joint — Butt Joint
with Single Cover Plate— Butt Joint with Double Cover Plate —
General Formulae for Riveted Joints other than Grouped Joints —
Formulae for Grouped Joints — Examples — Oblique Riveted Joints —
Riveted Joints in Compression — Dimensions for Riveted Joints — Note
upon Figured and actual Sizes of Rivets and Rivet Holes.
Pin Joints. — Example — Joint in Tie-rod Iron Roof — Joint at Apex and
Feet of Truss.
Screws. — Practical Formulae — Example — Tie Rod.
CONTENTS
xm
CoTTKR Joints — Practical Formulae — Example.
Joints in Wooden Structures.— i;xamj9Zes— Fished Joint— Scarfed
Joint— Joint at foot of Principal Kafter of a Roof . Pages 122-153
Chapter VIII.
PLATE GIRDERS.
General Form — Practical points to be attended to.
Flanges.— Graphic Method of finding length of Plates— Example.
Web.— Shearing Stress— Example— Joints in the Web— Example— Con-
nection of the Web with the Flanges— Example.
Example op Plate Girder with Single Web.
Example of Box Girder.
Hog-backed Girders . - ■ • Pages 154-176
Chapter IX.
BRACED OR FRAMED STRUCTURES.
General Remarks.— Simple Frames— Triangle of Forces— Polygon of
Forces— Different Forms of Framed Structures— Perfectly Braced
Frame — Overbraced Frame.
Maxwell's Diagrams— Method of Drawing— Bow's System of Notation.
Method of Sections.— Rule— Definitions . • Pages 177-186
Chapter X.
BRACED OR OPEN WEBBED GIRDERS.
Warren Girders.
Lattice Girders.
Whipple-Murphy Girder.— Bow-string Girder.
Examples.— Practical Formula— Graphic Method . Pages 187-194
i
Chapter XI.
TRUSSED BEAMS.
ExAMVh-E— Graphic Method— Method of Sections . . Pages 195-198
Chapter XII.
BOOFS.
Loads to be borne by Roofs.— Permanent Load — Occasional Loads —
Snow — Wind Pressure — Stresses in Roofs of Special Shape — in
Ordinary Roofs.
xiv CONTENTS
Distribution of the Loads.— Example— J^eaciions at AUtMs—Wmdi
on one side of Eoof.
Example— Iron Eoof Truss.— Finding Stresses by Maxwell Diagrams
— Calculation of PrinciiDal when carrying a number of Smal Purlins
— as Jointed — as Continued.
Diagrams for various Eoof Trusses . , Pages 199-217
Chapter XIII.
STABILITY OF BEICKWOEK AND MASONEY STEUCTUEES.
Single Block — Centre of Pressure — Distribution of Pressure.
Uncemented Block Structures.— Conditions of Stability— ^mm_?jfe—
Brick Pier — House Cbimney — Enclosure Walls . Pages 218-229
Chapter XIV.
EETAINING WALLS.
Usual Cross Sections for Eetaining Walls.
Eetaining Walls for Water.— Example by Graphic Method— Practical
Eules.
Eetaining Walls for Earth. — Mathematical Formuh^ — Graphic
Method— Practical Formulae— Examples— Baker's Practical Eules—
Foundations ..... Pages 230-245
Chapter XV.
AECHES.
Graphic Method of Determining the Stability of an Arch Line
of Eesistance— Moseley's Principle of Least Eesistance— Thickness of
Arch Eing — Abutment — Table of Thickness of Arches.
Example ...... Pages 246-259
»
Chapter XVI.
HYDEAULICS.
Definitions.— Pressure— Head of Pressure and Elevation— Loss of
Head — Wetted Perimeter — Hydraulic Mean Depth.
Motion op Liquids in Pipes. — Discharge from a Pipe running full
Formula— Examples — Practical Formulaj — Losses of Head — Due to
Orifice of Entry — Due to Velocity— Due to Bends and Elbows — Ex-
amples — Water Supply to a House — Intermittent — Constant — Examples
CONTENTS
XV
— Ditto to a Street — Examples — Discharge from a Pipe running partially
full — Formulse — Example — Drain Pipes — Example of Simple System
of Drains — Egg-shaped Sewer — Example.
Jets. — Height and Eange — Nozzles — Examples . . Pages 260-294
APPENDICES.
NUMBER PAGE
I. Factors of Safety ......... 295
II. Note on Equilibrium . . . . . . . 296
III. To Draw a Parabola . . . . . , . 297
IV. To show that the Neutral Axis in a Beam passes through
the Centre of Gravity of the Cross section . . . 298
V. Shearing Stress Diagrams . . . . . . 298
VI. Graphic Method of finding Stresses in Beam . . . 299
VII. Stresses upon Beams with various Distributions of Load . 302
VIII. Comparison of Strength and Stiffness of Supported and
Fixed Beams . . . . . . • • 303
IX, Deflection of Rectangular Beams ..... 304
X. Strength of Beams of Different Sections . . . . 304
XI. Continuous Beams, Eeactions and Bending Moments . . 305
XII. Centre of Gravity 308
XIII. Graphic Method of finding Centre of Gravity ... 309
XIV. Moment of Inertia of various Sections .... 310
XV. Formulse for finding approximately the Weight of Girders 311
XVI. Rules for Drawing Maxwell's Diagrams . . . . 311
XVII. Bow's System of Lettering Maxwell's Diagrams . , 313
XVIII. Calculation of Stresses on Iron Roof by the Method of
Sections ......... 314
XIX. Centre of Pressure in Masonry Joints . . . . 321
XX. Arch with Unsymmetrical Load ..... 322
XXI. Formulse to be remembered . . . . . . 324
XVI
CONTENTS
TABLES.
NUMBER PAGE
A. Factors of Safety ......... 9
B. Stresses and Modes of Fracture ...... 9
C. Factors for Deflection of Beams — given Load . . . . 67
D. „ „ „ — given Limiting Stress . . 68
E. Distance of Points of Contra-flexure . . . . . 70
F. Fixed Beams — Bending Moments for various Cases . . . 74
G. Gordon's Formula for Long Columns — Values of a . . . 114
H. Iron Roof — Tabulation of Stresses . . . . . . 213
J. Flovp- of Liquids in Pipes — Darcy's Formula — Value of C . . 265
L Ultimate Resistance of Materials . . . . . 327
la. Safe Resistance of Materials . . . . . . 328
II. Safe Loads for Rectangular Beams of Northern Pine or
Baltic Fir 329
III. Deflection of Rectangular Beams of Northern Pine or
Baltic Fir 330
IV. Deflection of Wrought Iron Rolled Beams . . . . 331
V. Strength of Struts (Wrought Iron, Cast Iron, and Steel) . 332
VL „ Wooden Struts 334
VIL „ Timber Columns 334
VIIL „ Rivets 335
IX. Dimensions of Eyes of Wrought Iron Tension Bars . . 335
X. Weight of Angle Iron and Tee Iron . . . . 336
XL Web Plates— Thickness of 337
XII. Weight of Roof Framing, ceilings, etc. . . . . 337
XIII. „ Coverings 339
XIV. Wind Pressure 339
XV. Scantlings for Wooden Roofs 340
XVa. Coefiicients of Friction . . . . . . . 341
XVI. Retaining Walls (Angle of Repose of Earths and Value of K) 342
XVII. Weights of Earths, Stone, etc 342
XVIIa. Thickness required for Arches . . . . . . 343
XVIII. Flow of Water in Pipes running full (gallons per min.) . 343
XIX. „ „ „ (cubic feet per min.) . 344
XX. Cast Iron Pipes — Thickness, Weight, and Strength . . 345
CONTENTS xvii
NUMBER PAGE
XXI. Loss of Head due to Bends ...... 346
XXII. „ „ Elbows 346
XXIII. Pipes flowing partially full 347
XXIV. Jets — Factors to find Initial Velocity . . . . 347
Notation and Working Stresses.
XXV. Notation employed in this Volume ..... 348
XXVI. Working Stresses „ ...... 350
EXAMPLES.
1. External Forces acting on a Girder . . . . . 21
2. External Forces acting on a Cantilever . . . . . 22
3. „ a Roof 23
4. Bending Moment for a Beam . . . . . . 27
5. Timber Cantilever for Balcony, Uniform Cross Section . . 76
5a. ,, ,, „ Strength and Width . 77
6. Timber Beam Loaded in Centre . : . . . . 78
7. Fir Joist for Given Load 79
8. Safe Distributed Load for an I Wrought Iron Joist of given Dimensions 8 8
9. „ „ „ „ weight, etc. 89
Qa. „ „ „ section 89
10. IE Joist required for a given Distributed Load ... 89
11. Cast Iron Cantilever . . . . . . . . 101
11a. „ Girder ... 103
12. Eectangular Tension Bar . . •. . . . . 108
13. Tension Rod - . . . 108
14. Rectangular Timber Strut . . . . . . . 115
15. Cast Iron Column . . . . . . . . 116
16. „ „ 117
17. Timber Strut . 118
18. „ „ .118
19. Wrought Iron Strut 119
20. T Iron Principal of a Roof 120
21. Lap Joint (Riveted) 124
22. Double-Cover Riveted Joint 128
23. „ Grouped Joint : . 131
24. „ „ for Plate Girder . . . . 132
h
xviii CONTENTS
NUMBER PAGE
25. Oblique Riveted Joint 134
26. Pin Joint 140
26a. Joints at Apex and Feet of Roof Truss . . . • 144
27. Screw of Tie-Rod 147
28. Cotter Joint 1^8
29. " Fished " Joint for Wooden Beam 149
30. "Scarfed" „ „ 151
31. Joint at foot of Principal Rafter (Wood) 152
32. Flanges for Plate Girder of 20 feet Span .... 156
33. Web for Plate Girder 159
34. Joint in the Web of a Plate Girder 161
35. Pitch of Rivets to connect the Web to the Flanges . . 161
36. Plate Girder Avith Single Web 162
37. Box Girder I'J'O
38. Stresses in Warren Girder . . . . • • • 187
38a. „ „ hy Graphic Method . . . 194
39. „ Whipple- Murphy Girder 190
39a. „ „ „ M Graphic Method . . 194
40. Trussed AVooden Beam . ...... 196
41. Loads on an Iron Roof .... ... 204
42. Stresses in an Iron Roof Truss 209
43. Pressure on a Masonry Joint . . . • • • 219
44. Brick Pier 222
45. Chimney 225
46. Enclosure Wall 227
47. Reservoir Wall 234
48. Retaining Wall for Light Vegetable Earth .... 241
49. „ ,, London Clay 243
50. Surcharged Retaining Wall, Loamy Earth .... 243
50a. Foundations 244
51. Semicircular Brick Arch 254
51a. Abutment for Arch 259
52. Velocity of Water flowing through a Pipe . . . . 265
53. „ „ „ „ .... 266
54. „ „ „ „ .... 266
55. „ „ „ „ .... 266
56. Loss of Head due to Orifice of Entry 269
57. „ „ Velocity 270
CONTENTS xix
NUMBEIl p^^Qj,
58. Loss of Head due to a Bend 270
59. „ an Elbow ...... 270
60. „ the Resistance of the Pipe . . . 271
61. Water Supply for a House, Intermittent Service . . . 271
62. „ „ Constant Service . . . . 274
63. "Water Supply to eight Houses in a Street . . . . 277
64. Velocity of Sewage in a Drain Pipe 282
65. Simple System of Drains . . . . . . , 283
66. Egg-shaped Sewer . . 9 3g
67. Jet of Water 289
68. Vertical Jet of Water, effect of Nozzle . . . ■. . 291
69. Water Supply for House and Jet 292
IN APPENDICES.
70. Graphic Method of Finding Stresses in Loaded Beam . . 300
71. Drawing Maxwell's Diagrams 312
72. Bow's System of Lettering Maxwell's Diagrams . . . 313
73. Stresses on Roof found by Method of Sections . . . 314
Chapter I.
INTRODUCTORY.
THE object of this Part of the Course is to teach a student
how to design those parts of buildings which require to be
carefully proportioned, in order that the cost of material and labour
may be a minimum, consistently with the structure being of
sufficient strength and permanence ; but it does not deal in any
way with the question of artistic design or appearance. It is
assumed that the student is familiar with Parts I., II., and III. of
this Course, in fact, that he understands the usual forms of parts
of ordinary buildings, and the nature of the various materials in
use for the construction of buildings.
In designing any structure care should be taken that each jpart
is strong enough to resist the forces that may act upon it. To
insure this result it is necessary to know —
1. The nature and direction of the stresses ^ that each part may
be called upon to resist.
In ordinary buildings these stresses are produced by : (a) The force of
gravity {i.e. the weight of the parts of the structure and of any load they
may have to carry). (6) The force of the wind.
2. The most suitable description of material.
3. The lest form and dimensions to he given to each part in
order that it may be able to resist the stress that may act upon it.
When the student has ascertained these particulars, and de-
signed the structure accordingly, he may be sure not only that
each part will be strong enough, but that no more material will
be used than is necessary to give the strength required ; in fact,
that the materials will be used without waste.
It should, however, be remembered that in practice cases
frequently occur, especially where the parts are small, in which
the minimum dimensions that can practically be allowed are
greater than the maximum required by theory.
^ See p. 6 for definition of this term.
B.C. IV. B
2
NOTES ON BUILDING CONSTRUCTION
It is of course easy in many cases to make a structure strong
enough by using plenty of material; but an engineer or archi-
tect who understands his work will use only sufticient material
to make sure of a safe and permanent structure, taking care to
dispose it so as to obtain a maximum of strength at a mini-
mum cost.
The calculations required for designing the different parts of
buildings are not difficult, and they require very little knowledge
of mathematics. In many books on the subject formulae are given
without any explanation as to how they are obtained, and are
therefore rather bewildering, and tend to wrap the subject in a
veil of mystery which does not properly belong to it.
An attempt will be made in this chapter to explain the
principles upon which the calculations depend, and then, in
subsequent chapters, to show how these principles are applied to
ordinary cases which occur in practice. If the student under-
stands the principles, he will always be able to apply them to
any unusual case v/hich may arise in his own experience.
TEEMS IN USE.
In order to clear the ground, it is desirable 'first to explain
the meanings of the various terms that have to be used in con-
sidering the subject. Many of these definitions have been given
in Part III., but they are now repeated in a somewhat different
form.
Load The external forces that act upon any structure,
tofyether with the weight of the structure itself, are called the
"load."
Thus in the case of a beam AB (Fig. i) supported at the ends, and
carrying a load W at its middle point, the external forces acting are (1)
the weight of the beam itself, (2) the weight of the load. The direction
of these forces is shown by the arrows. In Fig. 2 the beam is uniformly
TERMS IN USE
3
loaded througliout part of its length, the forces acting being (1) its own
weight, w w tv, etc. (2) the weight of the tank and its contents, ttt, etc.
On an inclined rafter (Fig. 3) there may be the weight of the rafter and
roof-covering acting vertically, and the force of the wind acting normally,
i.e. at right angles to the slope (air, being a fluid, exerts a normal pressure).
Fig. 4 shows two men carrying a ladder with a boy sitting upon it, the load
they have to bear is (1) the weight of the ladder, (2) the weight of the boy.
Again, in the case of a column carrying a girder (Fig. 5) which supports
a building, the weight (10 tons) on the head of the column is the load to
which, it is subjected. The load on the foundation of the column, will be 10
tons + the weight of the column.
Or in the case of a wire from which a lamp is suspended (Fig. 6), th.e
weight of the lamp, say 50 lbs., is the load upon the wire.
Tlie load in each case is the total of the external forces, i.e.
Fig. 4.
of the weight of beam + the weight of tank and its contents ; the
weight of rafter + force of wind, etc.
The weight of the structure itself is in some cases small com-
pared with the load upon it and can in practice be ignored, as for
instance that of the beam supporting a tank of water, or the wire
supporting the lamp. In others the weight of the structure itself
is important, as in the case of the ladder, which forms a consider-
able part of the total load.
Distribution of Load. — The load may be concentrated at the
centre of the beam (Fig. i), or concentrated at any point (Fig. 18),
4 NOTES ON BUILDING CONSTRUCTION
or uniformly distributed over the whole of the beam (Fig. i6), or
over a ^portion (Figs. 2 and 20).
Dead Load is that which is very gradually applied, and which
remains steady.
Thus the water might be poured into the tank very gradually, and when
once poured in would remain quiet, forming a dead load.
In the same way the weight quietly placed and remaining steadily on
the column, and the lamp on the rod, would be dead loads.
Of course the weight of a structure itself is a dead load.
yri \ ■•" ;___,,_,,.C__J Fig. 6.
Fig. 5.
*LiVE Loads are those which are suddenly applied or are ac-
companied by shocks or vibration.
Thus the boy might jump suddenly upon the ladder (Fig. 4), causing a
shock. An excited crowd upon a balcony or floor might make sudden and
fitful movements causing jars and vibration — in either case the load would
be called a " live load," not a dead or steady load.
There are but few cases in which the parts of ordinary buildings
are subject to live loads — the chief instance being in floors subject,
as before mentioned, to suddenly moving loads. In engineering
structures, however, railway bridges are subject to the live loads
caused by swiftly passing trains, breakwaters and sea walls are
liable to the sudden impact of the waves ; and so in parts of
TERMS IN USE
5
machines, and in other cases, the loads acting upon a structure
are often suddenly applied.
Comparative Effect of Dead and Live Loads. — It has been
found by experiment that a live load produces nearly twice the
effect that a dead load of the same weight would produce.
Therefore to find the dead load which would produce the same effect as
a given live load, the latter must be multiplied by 2.
This operation is called converting the live load into an equivalent dead
load.
Thus a floor girder may weigh 30 lbs. per foot run ; if the load upon it
of 250 lbs. per foot run be a live load, the total equivalent dead load will be
(30 lbs. + 2 X 250 lbs.) = 530 lbs. per foot run.
The Breaking Load for any structure or piece of material is
that dead load which will just produce fracture in the structure
or material.
The Working Load or Safe Load is the greatest dead load
which the structure or material can safely be permitted to bear in
practice.
It may be useful to know the load that would cause rupture in the
structure, i.e. the breaking load, but the load that should actually be applied
to it, i.e. the working load, must be so much smaller as to put all danger of
rupture out of the question.
The breaking load or the working load may be either live load
or dead load, or a combination of both ; but for convenience it is
usual to reduce it all to an equivalent dead load, by doubling the
live load and adding it to the dead load, as in the example
given above.
Strain is the alteration in the shape of a body produced by
a stress.
Thus a stress of tension, or tensile stress, produces a stretching
or strain of elongation, a compressive stress leads to a shortening
or squeezing strain, a transverse stress to a bending strain, and
so on.
At one time the word " strain " was generally used instead of the word
"stress" to denote the forces of tension, compression, etc., and it is still so used
in many treatises on applied mechanics ; but under the high authority and
guidance of the late Professor Rankine, the best writers on the subject use
the word " stress " to signify the forces acting upon a body,^ and the word
"strain" to mean the alteration of figure that takes place under the action of
those stresses.
^ "The word 'stress' has been adopted as a general term to comprehend various
forces which are exerted between contiguous bodies or parts of bodies, and which are
distributed over the surface of contact of the masses between which they act." —
Civil Engineering, by Professor Rankine, p. 161.
6
NOTES ON BUILDING CONSTRUCTION
It should be noticed that this use of the word " strain " is different from
the more usual one, in which it is implied that some injury has been done
to the material of the body which has been strained.
Stress. — When a force acts upon a structure or piece of material
it produces an alteration of form or strain which may be merely
temporary, lasting only while the force is applied, or it may be
permanent, or may eventually end in rupture. This alteration in
form calls out a resistance in the internal structure of the material
which is called the stress upon the body.-^
Thus the weight of the lamp (Fig. 6) tends to alter the form of the rod
which supports it by making it elongate, tending to tear it in two ; but the
strength of the fibres composing the rod resists this tendency to alter its form ;
and it is subject to a stress in the direction of its length. ^
Again, the weight on the girder in Fig. 5 tends to shorten the column,
and would, if sufficiently great, crush it. The weight (Fig. i) tends to bend
the beam, and would, if sufficiently great, break it (Fig. 31). The tendency,
however, in each case is resisted by the internal strength of the body.
The various kinds of stress are as follows : —
Tension. — This stress elongates the body upon which it acts,
and tends to cause rupture by tearing asunder.
There are many parts of ordinary structures subject to a tensile stress ;
e.g. the tie beam TT of a roof (Fig. 7) is in tension, because the rafters tend to
spread out at the feet and thus pull uj)on the ends of the tie beam.
CoMPEESSiON. — This stress shortens the body upon which it
acts and tends to cause rupture by crushing.
Many parts of ordinary buildings constantly undergo compression, for
example, columns and story posts, also the struts of roofs, which are com-
pressed in the direction of their length by the weight of the roof that they
support.
^ Theoretically the smallest force acting upon a body produces a permanent altera-
tion in its form (see p. 10). Practically, however, the permanent change of form
produced by forces which are very small in proportion to the strength of the struc-
ture may be ignored.
2 Strictly, the stress in the rod increases towards the supporting beam, owing to
the weight of the rod itself, and in the same way the stress in the column is greater
near the ground than at the top, on account of the weight of the column itself, but
the increase is so small that it may practically be neglected.
Fig. 7.
TERMS IN USE
7
TiRANSVEESE STRESS. — This stress bends the body on which
it acts and tends to break it across.
Instances of this stress in connection witli buildings will at once occur to
the student, e.g. in lintels or bressumers carrying walls, joists and girders of
floors, etc., tlie rafters of a roof under the influence of wind and load.
Shearing Stress is that produced when one part of a body
is forcibly pressed or pulled so as to tend to make it slide over
another part.
As, for example, when two plates riveted together (Fig. 8), are severed
A
Fig. 9.
by pulling or pushing in opposite directions, the rivet r is sheared — one plate
sliding upon the other (Fig. 9).
Bearing Stress is that which occurs when one body presses
against another, so as to tend to produce indentation or cutting.^
As, for example, when a rivet holding a plate cuts into the plate, making
the hole larger: thus in Fig. 10 the plates A, B, being pulled in opposite
Tiff. 10.
Fig. 11.
directions, the rivet c, being of harder iron than the plate B, has borne upon it
and cut into it, making the hole larger as shown at d, Fig. 11.
Torsion is the stress produced by twisting the parts of a body
in opposite directions, or, what comes to the same thing, by fixing
one part and twisting another around the axis of the fixed part.^
This stress is common in machinery, but not in ordinary building struc-
tures, and it will therefore not be further considered.
Stresses classified as regards Description. — Summing up
therefore, it appears that there are six different kinds of stress,
namely —
Tensile Shearing Transverse
Compressive Bearing Torsional
^ Bearing stress resolves itself into stresses of compression and shear, i.e., in Fig.
1 1, compression of the metal in front of the bolt, and shearing of the metal at the sides.
2 It can be shown that this stress resolves itself into tension, compression, and
shear.
8
NOTES ON BUILDING CONSTRUCTION
The three latter can, however, be resolved into combinations
of the three former.
The various stresses are also classified as follows : —
Stresses classified as regards Intensity. — Beeaking Stkess
is that stress at which a material will just give way.
Limiting Steess is the term applied to a stress which is pur-
posely limited to less than the breaking stress of the material.
Thus it may be known that a bar of iron will bear a stress of 20 tons
per square inch before breaking, but yet it may be determined, in order to be
quite sure that the bar will be safe, not to load it with more than 5 tons per
square inch. In this case, 5 tons would be the limiting stress ; any less stress
might be applied, but not a greater stress than the limiting stress, i.e. 5
tons per square inch.
The limiting stress should in all cases be less than the elastic limit of
the material (see p. 10).
Safe Woeking Steess is the stress that may in practice be
safely allowed upon the parts of a structure.
The amount of this stress depends upon the material, the nature of the
stress, or the nature of load, whether live or dead, to which the structure is
subjected. It is ascertained by dividing the breaking stress by a factor of
safety, which varies under the several conditions mentioned above, and is
determined by experience (see p. 9 and Appendix I.)
Intensity of Stress is the amount of the stress, expressed in units
of weight (such as tons), divided by the area of the surface over
which it acts, expressed in units of area (such as inches); provided
of course that the stress is uniformly distributed over the surface.
If it is not uniformly distributed, the intensity of stress will of
necessity be greater at some points than at others. It is generally
expressed in tons per square inch, sometimes in lbs. per square inch.
Thus in the case of the lamp hanging by a rod, or rather wire (p. 4), the
stress is uniformly distributed over the cross section of the rod. Assuming
that this cross section has an area of -wi-jth. square inch,
T . c . weight of lamp 50 lbs.
Intensity of stress = ^ — = —
area of cross section -^^-^ square inch
= 50 X 224 lbs. per square inch,
or 5 tons per square inch.
Again, if a bar 2 inches square, that is, with a sectional area of 4 square
inches, has 40 tons hanging from it, the total tensile stress on the bar is 40
tons, but the intensity of stress is 10 tons per square inch.
The cast iron column shown in Fig. 5 may be taken as another example.
The load is lO'O tons, and assuming that the area of the cross section is 8
square inches, the average intensity of stress is ^= 1-25 ton per square inch.^
^ As will be explained in Chapter VI. , unless the column is very short, it cannot
be assumed that the stress is uniformly distributed over the cross section, and there-
fore the intensity of stress will vary from point to point, and the maximum intensity
may be far greater than the average.
TERMS IN USE
9
The Factor of Safety is the ratio in which the breaking load
exceeds the working load.
This ratio depends upon the nature of the load and that of
the material, and it is found by experience.
The following Table shows the Factors of Safety given by Professor
Kankine in bis Useful Mules and Tables.
TABLE A.
Factors of Safety.
Dead Load.
Live Load.
For perfect materials and work-
manship .....
For good ordinary ( Metals .
materials and I Timber
workmanship ( Masonry
2
3
4 to 5
4
4
6
8 to 10
8
It will be seen that, for the reasons given above, the factor of safety for
a live load is taken at double that for a dead load.
When a load is mixed, i.e. partly live and partly dead, the live portion
may be converted into an equivalent amount of dead load, and the factor
of safety for dead load then applied to the whole.
The factors of safety shown in Table A are lower than can be safely used
for the work ordinarily met with.
Factors of safety which have been used in practice for different special
structures are given in Appendix I.
Fracture. — When a body is subjected to a stress, and the
stress at any point is greater than the material can withstand,
fracture ensues.
The nature of the fracture depends of course upon the kind of stress to
which the body has been subjected.
Thus a tensile stress carried far enough produces fracture by tearing, a
compressive stress by crushing, a transverse stress by cross breaking, etc.
Table of Stresses and Strains and Modes of Fracture.
The following Table shows the various stresses and the strains,
and description of fracture to which they lead when carried
suffi-ciently far.
TABLE B.
stresses.
Strains.
Modes of Fracture.
Tensile or pulling .
Compressive or thrusting
Transverse ....
Shearing ....
Bearing .....
Torsional ....
Extension .
Compression
Bending
Distortion .
Indentation .
Twisting
Tearing.
Crushing.
Breaking across.
Cutting asunder.
Cutting and crushing.
"Wrenching asunder.
lo NOTES ON BUILDING CONSTRUCTION
Elasticity is the property which all bodies have (in a greater
or less degree of perfection) of returning to their original figure
after being strained by the application of any kind of stress.-^
When the original figure is completely and quickly recove:ed,
the elasticity is said to be perfect.
Thus if an ivory billiard ball be dropped upon a hard smooth sbne,
smeared with black, a black spot of considerable size will be found on the
ball, showing that the ball was to some extent flattened at the momer.t of
impact.
Upon examination of the ball, however, it will be found that all flataess
has disappeared, showing that it has returned to its spherical shape ; oj, at
any rate, that no deviation from that shape can be detected by the eye o: by
ordinary modes of measurement.
When the original figure is not completely recovered, but remains per-
manently altered, the elasticity is said to be imperfect, and the disfigvirenent
produced is called a permanent set or set.
Although there is reason to believe that the elasticity of every
solid is imperfect, so that the slightest strain produces a set,
the elasticity of most building materials is practically perfect, up
to a certain point; that is to say, that stresses up to a certain
limit can be applied and removed, and the resulting disfigurement,
or change of figure, is only temporary; there is no appreciable
permanent set — that is, no set that can be measured with any
ordinary instruments. Stresses above this limit, however, cause
permanent sets.
The Elastic Limit of any material is the maximum stress per
square inch that can be applied to it without causing any appre-
ciable permanent set.
An illustration will perhaps make the meaning of the preceding terms
more clear than the definitions alone would do.
If a weight be suspended from a rod so as to cause a tensile stress in the
direction of its length, the rod will at once be elongated.
It will stretch a certain proportion of its own length. This proportion
will vary according to the description and quality of the material and to the
amount of weight applied.
Taking the case of wrought iron as an illustration :
If a weight of 1 ton be hung from the end of a wrought-iron rod of aver-
age quality, having a cross-sectional area of 1 square inch, the bar will
stretch about x^^Fo^^ P^^^ original length.
If the weight be removed, the bar will immediately recover itself, that is,
it will return to its original length. If measured by any ordinary means of
measurement, it will be found to be of the same length that it was before
the weight was applied.
^ This must not be confounded with the popular meaning of the word "elasticity,"
i. e. the property of being easily stretched.
TERMS IN USE
II
This recovery of the bar occurs, however, only up to a certain point. If
the load amounts to a considerable proportion of the breaking weight, the
result produced is very different.
For example, if instead of 1 ton, a weight of 1 2 tons be applied to the
bar just mentioned, the iron will stretch about xoV (1*1^ length. Upon
removal of the weight, however, it will not entirely recover itself, and upon
measurement it will be found to be a minute portion longer than it was before.
This slight increase in the length of the bar is called the permanent set.
It is evident, then, that there is a very important line to be drawn — on
one side of it are the stresses the application of which will produce no appre-
ciable permanent set, on the other side are the stresses which do produce an
appreciable permanent set. This line of demarcation is the Elastic Limit or
limit of elasticity, and its value is generally stated in lbs. or tons per square
inch, and it is important that materials should not be subjected to stresses
exceeding the elastic limit.
The above remarks have been made with regard to a tensile stress, but
the same thing occurs with a body under compression — weights placed upon the
end of a strut produce no permanent shortening, or set, up to a certain limit ;
weights greater than this permanently shorten the strxit to an appreciable
extent. This point is called, as before, the elastic limit or limit of elasticity.
Repeated Stresses — Variations of Stress in amount and kind. — It was shown long
ago by experiment that a stress not much exceeding half the breaking stress for
iron or steel would cause rupture after being applied and removed a great number
of times.
This used to be explained by pointing out that if the stress exceeded the elastic
limit each application of the load would cause a slight increase in the permanent
set, i.e., the length of the bar would be slightly but permanently increased each
time the load was applied until at last rupture would take place.
More recent experiments on iron and steel have shown, however, that the action
of repeated stresses is not so simple. Wohler has shown that if a bar be subjected
to repeated stresses subject to variation in amount or in kind, the load nuist be
much less than that which the bar will bear when it is steadily applied.
Thus if t will just break a bar when steadily applied in tension, a tensile stress
varying from 0 to | will break it if applied thousands of times, and | will break it
if repeatedly applied as compression and tension alternately.
The cases in which repeated stresses vary in kind are rare and unimportant in
building construction. They do, however, frequently vary in amount, and the
abov-e shows that the working stress for repeated loads should on no accoimt exceed
half the breaking stress — in practice it is much less.
Elongation and Shoetening. — Another point to be noticed
is that in wrought iron, and in most building materials, the
temporary elongations or shortenings produced before the limit of
elasticity is reached are proportional to the loads which produce
thoise elongations or shortenings.
Thus in the bar above referred to — if a load of 1 ton produce an elonga-
tion, of 12 ^0 0^^ length, 2 tons will produce j-gWo"^^^' ^ t^Io o^^^^'
and so on until 12 tons produce an elongation of ^--^^(j^ths = ^ ^ ths the
length.
At this point, however, the permanent set becomes appreciable, and beyond
it the elongations are not in proportion to the load, but increase more
12
NOTES ON BUILDING CONSTRUCTION
rapidly than the load. Thus 13 tons will produce an elongation slightly-
greater than j^'^^Q-^hs of the length, and so on in an increasing ratio.
The Modulus of Elasticity is a number representing the
ratio of the intensity of stress (of any kind) to the intensity of
strain ^ (of the same kind) produced by that stress, so long as the
elastic limit is not passed.
Thus the modulus of tensile elasticity of any material is found by
dividing the tensile stress in lbs. per square inch of sectional area by the
elongation (produced by that stress) expressed as a portion of the length of
the body.
For instance, if a weight of 1 ton hung from an iron bar produce an
elongation of xto-tto^^ length of the bar, the modulus of elasticity of
the bar will be 2240 Ibs.-h 3-2^00 = 26,880,000 lbs. This is rather lower
than the modulus of average wrought iron.
Similarly, the modulus of compressive elasticity is found by dividing the
compressive stress in lbs. per square inch of section by the shortening (produced
by that stress) expressed as a fraction of the length of the body.
In most building materials the modulus of tensile elasticity and that of
compressive elasticity are practically equal to one another so long as the
stresses do not exceed the elastic limit.
In advanced works on applied mechanics other moduli are used which,
however, are not required in ordinary calculations, except the modulus of
rupture, see p. 52, and need not be further referred to in these notes.
Moduli of elasticity for different materials are given in Table I.
Deflection is the bending of any body caused by a transverse
stress.
Thus the ladder shown in Fig. 14 is bent by the weight of the
boy sitting upon it.
If the intensity of the stress be below the elastic limit, the
deflection will disappear when the stress is removed ; but if the
intensity of stress be in excess of the elastic limit, a permanent
set will remain.
Stiffness in a material is the power that it may possess to
resist being strained out of its proper shape ; this is a very
different thing from the strength, or power to resist rupture.
Compare, for instance, glass and wrought iron.
The stiffness of the material depends upon its elasticity;
the greater the value of the modulus of elasticity, the greater the
stiffness of the body.
^ By "intensity of strain" is meant the amount of alteration of figure per unit
of length ; thus if a bar be elongated under tensile stress, the intensity of strain
would be the amount of elongation (expressed as a fraction of a foot) per foot of the
bar. Expressing the total elongation as a fraction of the length of the bar amounts
to the same thing.
TERMS IN USE
13
T]u stiffness of a structure depends not only upon the elasticity
of the material of which it is composed, bnt upon its arrangement.
Thus a floor with deep narrow joists is much stiffer than one of
the same strength with wide and shallow joists.
This quality has an importance only second to that of strength
in fact, strength and stiffness must be considered together.
In a floor, for example, although the joists may be strong enough to
resist breaking, if they are not stiff enough, the floor will be springy and
uncomfortable, and if they have a ceiling attached to them it will be cracked,
and may be rendered dangerous by the deflection or bending of the joists.
In roofs, the rafters, if not stiff enough, will bend or sag, causing ugly
hollows, and in some cases lodgment of wet upon the roof-covering.
Other Physical Properties of bodies do not affect the calcula-
tions about to be entered upon, a discussion of them would tend
only to confuse the student, and they will therefore not be alluded
to in this volume.
Chapter II,
EQUILIBRIUM.
HEIST external forces act upon a body, its powers of resistance
T t are called out until they are just sufficient to balance the
external forces ; when this balance is maintained, the external
forces and those of resistance are said to be in equilibrium.
When the external or active forces require, for equilibrium,
greater resistances than the body can offer, movement ensues in
the form of rupture of the material if the internal resistances are
ineffectual, or of movement of the body itself if the external
resistances are insufficient.
Thus if a rope be fastened to a man, and a boy pull at it with a force of
30 lbs., the man passively resists until there is a stress upon him of 30 lbs.,
and equilibrium is established. If, however, the man begins to pull the rope
with a force of 80 lbs., equilibrium is disturbed and movement commences —
the boy is dragged forward ; or if he ties the rope to a wall, and the man
pulls with sufficient force, the rope is broken.
Every structure is under the action of two distinct sets of
forces, namely —
1. The external forces. 2. The internal forces, or the
stresses.
And in order that the structure may not move, or, as it is
termed, may be " in equilibrium," it is necessary in the first place
that the external forces be in equilibrium amongst themselves;
and secondly, that the internal forces, or stresses, be not greater
than the resistance the material of the body is able to offer.
The external forces on a structure consist of the various loads
it has to bear, and of the reactions at the points of support.
It will thus be seen that, to design a structure as regards
strength, three distinct operations are required, namely —
1. To find the external forces acting on the structure.
2. To find the stresses produced by those forces.
3. To calculate the dimensions of the various parts of the
structure so that the stresses (or, more strictly speaking, the inten-
EXTERNAL FORCES ACTING ON A STRUCTURE 15
sity of stress) shall nowhere be greater than the material is safely-
able to resist.
The resistance is always exactly equal and opposite to the stress up to
the point of rupture ; and the resistance offered at that moment is called the
ultimate resistance. This introduces the important subject of resistance of
materials on which ultimately depend all calculations made in connection
with the strength of structures, and unfortunately it is a subject of which our
knowledge is by no means as full as might be wished. At the end of this
book will be found a Table giving the ultimate resistance to the various stresses
of tension, compression, shearing, etc., of some of the materials usually employed
in building construction, ^ but more detailed information is given in Part III.
These operations involve the application of the conditions of
equilibrium of a body, the general consideration of which requires
a greater knowledge of mathematics than is assumed for this
Course ; the large majority of cases that occur in building practice
can, however, be treated in an elementary manner, which we now
proceed to do in the order mentioned above.
EXTEENAL FOECES ACTING ON A STEUCTUEE.
The loads a structure has to bear may be given, but usually
they have to be ascertained in each case, and it will be shown in
the sequel how to do this.
For instance, it might be given that a certain beam has to bear a weight
of 10 tons ; but in designing a roof, the weight of the roof-covering, the loads
caused by the wind and snow, etc., would have to be ascertained or estimated.
The reactions at the points of support are determined, as soon
as the loads are known, by the application of the following
principle :
Whenever a force acts upon a body tending to move it in a
particular direction, this force must be opposed by an equal and
opposite force, or else the body will move — that is, it will not be in
equilibrium.
This is Newton's third law, namely, that to every action there
is always an equal and contrary reaction ; or, the mutual actions
of any two bodies are always equal and oppositely directed in the
same straight line. Or, putting it another way,
If a body is in equilibrium under the action of two forces, it
is self-evident that the two forces must be equal and oppositely
directed.
1 See Table I.
i6 NOTES ON BUILDING CONSTRUCTION
Taking, for instance, the. case of two men pulling at a rope (Fig. 1 2).
Fig. 12.
Supposing the man A pulls with a force of 30 lbs., B must pull in the
opposite direction also with a force of
30 lbs. or the rope would move to-
ward A.
Now, supposing B ties his end of
the rope to a ring in the wall (Fig. 1 3)
and goes away, while A goes on pulling
with a force of 30 lbs. The wall will
now supply a force of 30 lbs. to oppose
the pull of 30 lbs. The force supplied
by the wall is called the reaction of the
wall.
The next case to consider is
that of a body in equilibrium under the action of parallel forces,
as in Fig. 14, showing two men carrying a boy on a ladder.
'lbs. The boy's weight and
the weight of the ladder act
vertically downwards (Fig.
1 4) ; the men therefore have
to supply vertical forces act-
ing upwards to maintain
equilibrium.
Supposing the boy to
weigh 60 lbs. and the ladder
90 lbs., the whole weight to
^'^S- be carried by the men is
150 lbs. That is, there are forces amounting to 150 lbs. acting vertically
downwards, and the men have to supply the reactions to resist these forces,
and prevent the ladder from moving downwards. If the boy is in the
middle, it is clear that each man will have to bear half the total weight,
amounting for each to 75 lbs.
In the same way, when a horizontal beam supporting a single load placed
at its centre (Fig. i 5), or loaded symmetrically throughout its length (Fig.
16), rests upon two walls, half the total weight is borne by each wall, or,
in other words, each wall has to supply a reaction equal to half the weight.
When the loads to be carried are symmetrically placed with
regard to the supporting bodies, and are symmetrical in them-
selves, the reaction afforded by each of these supporting bodies
is equal, as in the illustrations given above.
EXTERNAL FORCES ACTING ON A STRUCTURE 17
When, however, the weights
with regard to the supporting b
w
A ^
2 '
' 2
Fig. 15.
afforded hy one support differs from that to be afforded by the
other support.
Eeverting to the illustration of the boy carried on the ladder, if the boy
does not sit in the centre of the ladder (Fig. 17), then the man A, nearer to
whom he sits, will bear more of his weight than the other man — that is, he has
to supply a greater reaction.
In order to ascertain the 1- so zz*
amount of weight each man
will have to bear, a simple
calculation is necessary,
founded upon the principle
of the lever, which no doubt
is familiar to the student.
By the principle of the
lever we know that to pro-
duce equilibrium the up-
ward force or reaction act-
ing at A, multiplied by the leverage AL, must equal the reaction at B
multiplied by BL.
We have, therefore, considering now only the weight of the boy,
Eeaction at A x AL = reaction B x BL,
Eeaction A x 5 = reaction B x 1 5,
Eeaction A = ^ x reaction B,
= 3 x reaction B.
Eeaction A -H reaction B = 4 x reaction B.
But we know that the two reactions together must be equal to the weight.^
.'. 4 X Eeaction B = 60 lbs.
.•. Eeaction B = 1 5 lbs.
And Eeaction A = 45 lbs.
Therefore we see that, because the boy is three times as far from B as from
A, A has three times as much of his weight to carry as B.
^ From the conditions of equilibrium (see Appendix IL) it appears that —
by taking moments about A,
Wx5-Ebx20 = 0.
B.C. IV.
C
i8 NOTES ON BUILDING CONSTRUCTION
We have, however, ignored the weight of the ladder, which, being a
uniform load of 90 lbs., produces a reaction of 45 lbs. at A and 45 lbs, at B.
Eeaction caused by
Boy. Ladder.
The total reaction at A= 45 + 45 = 90
B= 15 + 45 = 60
Total reactions caused by weight of boy and ladder = 150
The general rule for finding the reaction at either support,
caused by a load, placed anywhere on a beam supported at both
ends, may be stated as follows : —
Rules for finding the Reactions at Supports.
jluU I. — Tlie reaction at either support, caused hy a single load
(W) placed anywhere upon a heam supported at both ends,is equal to the
load multiplied by the distance from its centre of gravity to the other
support, and divided by the length of the beam between the supports.
Single Load. — Thus with refer-
ence to Fig, 1 8
WxDB
Eeaction at A = R,
AB
Fig. 18.
Wliere W is in the centre, we have
DB = |AB, and
Reaction at A = = — ^ — = W-
AB
General Case of Single Load.
tance of W from A = a, we have
-If the load = W, the span = I, the dis-
R.
I — a
W
(1).
Rb = 7W
(1 A).
R^t?W^
If TF he in the centre,
then I
(1 B).
I 2
R. = R,
Fig. 19.
Two OR MORE Loads. —
When a number of weights
rest upon a beam the reac-
tions can be found either by
calculating the reactions produced by each weight separately, and adding
them together, or by considering that they produce the same effects as would
be produced by a weight equal to all of them put together, acting through
their common centre of gravity.
The second method is to be adopted in preference when the position of
the centre of gravity is self-evident. Thus in Fig. 20 it is clear that the four
weights, each of 100 lbs., produce the same reactions that would be produced
1 n
A i I
tOOtbi f 00 lbs
< — 5>--*
tOOtb, 100 Ibt P
^ iOlo'. >f R=|x4
io;,o -
EXTERNAL FORCES ACTING ON A STRUCTURE 19
one weight of 400 lbs. placed midway between them, that is, at the
centre of gravity of the weights (Fig. 21). ^
Uniform Load. — Again
the reactions produced by
the uniformly distributed
load Sw (Fig. 16) will be
equal to those of a con-
centrated load W = ^10 act-
ing through the centre of
gravity of 8 w.
When the load is uni-
formly distributed over a por-
tion or the whole of the beam,
the reactions produced by it
will be the same as those
! produced by a single weight
equal to the load and placed
at the centre of gravity of
the load. Thus in Fig. 2 1
the reactions produced by
the load of 400 lbs. distri-
buted along EF would be
; exactly the same as those R=^*>loo
I produced by a single weight
of 400 lbs. acting at D, the
e.g. of EF (see Fig. 21).
JVhen the weights are unequal and placed unsymmetricaUy, their reactions
are best found by the first method, that is, by taking each in turn and finding
the reactions produced by it. The reactions produced at either support will
be the sum of the reactions produced by each weight at that support.
W2
Fig. 21.
w
•s:,o-->< io'„(fi-
ItOO lbs
Fig. 22.
I
t—5'„ O'—Xk — 5;, 0-->U 10'„0"--
^100 lbs
^300 lbs
Fig. 23,
Thus in Fig. 23 the reactions produced would be as follows : —
At A.
At B.
Reactions produced by Wj ....
Reactions produced by W2 ....
Total reactions caused by Wi + "VV,
Or substituting the values of Wj and "Wg, we
have
Total reactions caused by Wj, + Wj
i.e. by 300 lbs. +100 lbs.
20
20 ^2
i^^ + 2>^
= 150 + 75
= 225
20
20 "2
20^1 + 20^2
= 150 + 25
= 175
20 NOTES ON BUILDING CONSTRUCTION
The general case of two weights unsymmetrically placed, as shown in Fig. 24.
W2
::!:dr^::t::::^!:::z^
-I-
riff. 24.
Suppose a to be the distance from the abutment A to the point of applica-
tion of W^, 6 the similar distance for "Wg, and I the span or distance between
the supports. Then
I - a I — h.
Keaction at A = — ^ — x Wj -\ j-
Eeaction at B = ? + jW^
Wo
I
Several loads placed ^msymrfietrically.—^'h2i't&\&i: the number of weights,
the reactions are found in a similar manner.
Thus in Fig. 25 there are five weights, Wg, Wg, W^, W5, the distances
Fig. 25.
of which from A are respectively a, h, c, d, e, and I being the span or distance
between the supports.
l-a l-h ^„ ^ - C ,Tr ^ - ^ -.TT ^ ~ ^ -rtr
The reaction at A = ^ + + W3 + + W,.
Q/ 1) c d ^
The reaction at B = - + ^ + ^ W3 + + ^-^5.
The weights W3, W^, W5 being known, as also the lengths of
a, b, c, d, e, and I also known, it is easy to substitute the values and calculate
the reactions.
We arrive then at the following rule : —
Bule II. — The reaction at either support caused hy any numher
of loads placed upon a beam supported at each end is equal to the
sum of the reactions at that support caused hy each load taken
independently.
EXTERNAL FORCES ACTING ON A STRUCTURE 21
Or, stated in another way, the reactions at the supports A and B caused
by a number of loads Wp Wg, Wg, W^, W5, Wg, etc., will be as follows : —
Total reaction at A = reaction at A of 4- reaction at A of Wg + reaction
at A of W3 + reaction at A of "W4 + reaction at A of W5 + reaction at A
of Wg, etc.
Similarly, the total reaction at B = reaction at B of W^ + reaction at B of
Wg + reaction at B of W3 + reaction at B of W^ + reaction at B of W5 + re-
action at B of Wg, etc.
The reactions of the individual weights at A and B are of course found
by Rule 1.
Load on the Supports. — It is evident that the load on a support is equal
and opposite to the reaction the support exerts on the beam. Thus in the
case of the men carrying the boy on a ladder, when the boy is sitting in the
centre of the ladder the downward pressure on each man is equal to the
reaction, or to 75 lbs. ; when, however, the boy is not sitting at the centre
of the ladder, but at the position shown in Fig. 1 7, the man A has to bear
a downward pressure of 90 lbs. and the man B of 60 lbs.
In order to familiarise the student with the method of finding
the reactions caused by various forms of load, it will be well to
take as examples a few cases such as occur in practice.
External Forces acting on a Girder.
Example 1.— A girder AB of 30 feet span (Fig. 26) supports a
c
A
B
^ 2o'o" >
SO'O- M
so'o"- — ■ g
Fig. 26.
beam C at a point 10 feet from the column at A. The load
on the beam C is uniformly distributed, and amounts, including
the weight of this beam, to 46 tons. The girder AB itself weighs
2'6 tons. Find the proportion of weight to be borne by the
column and the wall.
This weight will be equal to the reaction at each point of support. Since
46
the beam C is uniformly loaded, it will impose a load of — = 23 tons on the
girder AB at the point C. The girder AB is therefore subject to two loads,
namely — •
1 . Its own weight = 2 '6 tons.
2. The load upon it = 23 tons.
The former is a uniformly distributed dead load, the reactions of which
NOTES ON BUILDING CONSTRUCTION
can be found as at p. 17. The latter is a single load placed unsymmetric-
ally iipori the girder, and its reactions can be found by Rule I.^
At A ; tons.
At B ; tons.
The reaction of the weight of girder itself will be
1-3
15-3
1-3
7-7
Total reactions
16-6
9-0
If the reactions have been correctly worked out, their sum must be equal
to the sum of the loads.
In this case the sum of the weights is 23 + 2-6 = 25-6 tons.
The sum of the reactions as found above is 16-6 + 9*0 = 25-6 tons.
The result is the same in both cases — which shows the working to have
been correct.
If in this example the beam C had been supported in the centre of AB,
23
the reactions due to it at A and B would have been equal each to — = 11-5
tons, and the total reactions would have been altered accordingly — each being
equal to 1-3 + 11*5 = 12-8 tons. The column A would thus have to bear
less, and the wall B more, than in the former case.
External Forces acting on a Cantilever.
Example 2.— A wrouglit-iron cantilever, as shown in Fig. 27, is
bolted to a wall and sustains
a weight of 1 cwt. Find
the other external forces
acting on the cantilever.
To oppose the downward
tendency of W, the two bolts AB
must supply an upward reaction
W, but the proportion in
E,
which is divided between
the two bolts depends upon
the method of fixing. For in-
stance, if the lower bolt were
working in a slot, the top bolt
would have to supply the whole
■^^S" of R and vice versa.
"W also tends to turn the cantilever around the point B in the direction
of the hands of a watch, and to oppose this the bolt at A must supply a
horizontal reaction R^. From the principle of the lever, or, as it is termed,
by taking moments about B (see Appendix II.) —
6xW-4xRi = 0,
or R^ = ^W = l-5 cwt.
1 Reaction at A
Reaction at
15-3,
7-7.
STRESSES PRODUCED BY THE EXTERNAL FORCES 23
The bolt at A must therefore be of sufficient strength to withstand a pull
of 1'5 ton, or in other words 1-5 ton is the stress in the top bolt.
The reaction Eg ^ clearly equal to the reaction at A^^, but acting
in the opposite direction.
External Forces acting on a Boof.
Example 3.— The roof shown in Fig. 28 is loaded with equal
w
/
^\
v*^-— — ; 1
\ A\
...8>-
Fig. 28.
loads. W at the points D, C, and E. Find the reactions at the
supports A and B.
It can be seen at once from symmetry that the reactions must each be
W
equal to W + — , However, applying Rule II., p. 20,
24 W 8 „^
^^ = 32^ + ¥+r2^^
8 W 24
w
:W +
w
The preceding must suffice to give a general idea of the manner of finding
the reactions at the supports, and of thus determining all the external forces
acting on the body or structure. Other cases, such for example as the re-
actions due to the pressure of the wind, acting on one side of a roof, will be
considered in subsequent chapters.
The next step (see p. 14) is to find the internal stresses pro-
duced hy the external forces, and for this we must consider the
various structures separately, commencing with beams (Chap. III.)
Chapter III.
BEAMS/
EIGS. 3 1 and 3 2 show the manner in which a rectangular
wooden beam, supported at the ends, breaks when subjected
to a concentrated load greater than it can bear. The beam bends.
Fig. 31.
sinking most at the point under the weight, and the fibres oi the
upper portion of the beam are crushed, and those of the lower
portion torn asunder, as shown on a larger scale in Eig. 3 2.
Fig. 32,
There are therefore certain external forces at work tending to
break the beam across, and certain internal forces which are resist-
ing this tendency to rupture.
In order to ascertain the strength of the beam, as compared
with any load it may be called upon to bear, it is necessary to
ascertain the nature and extent —
1. Of the external forces tending to break the beam across.
2. Of the internal forces tending to resist rupture,
^ The formulce for practical use are given in Equation 30, p. 54, and in
Appendices VII. and XXL
BEAMS
A uniform load of sufficient magnitude would also produce
rupture in the same way, but with a slight difference as to the
form assumed by the beam just before the time of rupture.
When a beam is supported at both ends and subject to a safe
load, it will bend to a certain extent, and the upper fibres will be
in compression and the lower fibres in tension.
Now if we examine the external forces acting upon the beam in
I'ig- 3 3j we see that the weight acting downwards causes a reaction
at each support acting upwards, just as in Fig. 1 7 the weight of the
boy on the ladder rendered it necessary for an upward force to be
exerted by the man at each end.
The downward vertical force
caused by the weight acting to-
wards the centre of the earth is
balanced by the upward forces ^
caused by the reaction at each end,
and the whole structure is in a state
of equilibrium {i.e. the beam supports the weight).
Conversely, we may say that the reactions acting vertically
upwards are balanced by the weight acting vertically downwards.
They are in the position of vertical forces each acting at the
end of a lever under the weight as a fulcrum, and the action is
the same as in bending a stick across one's knee.
When we consider the effect of these reactions upon the beam
we see that each acts with a leverage equal to the distance from
the support to the point immediately under the weight, about
which the beam tends to break.
The power of these reactions to bend or break the beam, there-
fore, consists of -D /.-U ^ A\ Kf^
(the reaction at A) x AC,
or of Ej, (the reaction at B) x BC.
Bending moment. — This is called the lending moment or, in
some books, the moment of flexure.
Since the bending moment of the reaction at C (the section of the
beam under the weight) is proportional not only to the magnitude of R^ but
also to the lever arm AC with which it acts, it is therefore equal. to
AC X R^,
therefore the bending moment due to R^ at C
CB
AB
r\x>
AC.-^.W .... (2).
26 NOTES ON BUILDING CONSTRUCTION
In the same way the bending moment due to the reaction R3 at C is
CB X R3
AO
= CB.^.W .... (3),
which shows that the bending moment at C due to is equal to the bending
moment due to R^, a manifestly correct result.^
Let it now be required to find the. bending moment at any
point P. Considering the left-hand part of the beam, the bend-
ing moment is seen to be
^ AP.CB
^P^^^ = -AB-^-
Now considering the right-hand part of the beam ; the reaction Rj, tends
to turn that part of the beam against the direction of the hands of a watch,
but the weight itself opposes and tends to turn it in the contrary direction.
Thus the bending moment will be equal to the bending power of R3, less
that of W, that is
= PB . R3 - PC . W,
AO
= (PB. — -PC)W,
PB.AC-PC.AB „^
- . W,
AB
(PO + CB)AC - PC(AC + CB)
AB
W,
CB.AC-PC.CB (AC-PC)CB AP.CB
AB AB AB
The same result as before.
The bending moment at any point of a beam can therefore
be found by the following
Eule for finding the Bending Moment at any given Point.
Consider either of the portions into which the beam is divided
at the given point, and multiply each force acting on that portion
by the distance of its point of application from the given point ;
these products can be taken as positive when the force tends to
turn the portion of the beam under consideration in the direction
of the hands of a watch, and negative when in the opposite
direction. Distributed loads are to be reckoned as acting at
their centre of gravity (see p. 19).
The algebraic^ sum of these products is the bending moment
required.
1 Strictly, the distances AC and CB should be measured horizontally, but in
practice the bending of a beam is so small that these distances can be measured
without any appreciable error along the beam, or in other words may be taken to
be, after bending, equal to what they were before bending.
2 That is, the positive products are to be added and the negative products
deducted.
BEAMS— BENDING MOMENT
27
In practice that portion of the beam would be chosen on
which, the fewest forces are acting.
Example 4. — As a simple numerical example let
W= 100 lbs.
AC = 6 feet
CB;= 14 feet
|ab =
20 feet.
4R8
'^^^^ K. = iixlOO = 70lbs.
^ 20
E3 = — X 100 = 30 lbs.
° 20
And tlie bending moment at C
= CBxE,
It is also equal to
ACxE.
^100 Ihs
Fie. 34.
: 14 feet X 30 lbs.,
420 foot-lbs.
6 feet X 70 lbs.,
420 foot-lbs.
As will be seen subsequently, it is sometimes better to express lengths
in inclies, and so expressed the bending moment at C would be
= 420 X 12 = 5040 inch-lbs.
Further, if the loads are expressed in cwts. the bending moments will be
expressed in inch-cwts., and in inch-tons if the loads are expressed in tons.
BENDING MOMENT UNDEE VAEIOUS CONDITIONS.^
We proceed now to consider the Bending Moment or Moment
of Flexure caused in a beam fixed or supported in various ways,
by the different distributions of load that are most likely to occur
in practice.
The following will be the notation used throughout : —
M = Bending moment or moment of flexure in inch-pounds, inch-cwts., or
inch-tons, according to the denomination in which W and w are expressed.
Mg = Bending moment at centre.
My = Moment at point P.
= Moment at point S, and so on.
A and B are used for the points of support.
'W = Total weight on the beam in lbs., cwts., or tons.
w = Weight per inch run of the beam in lbs., cwts., or tons.
I = length \ of the beam,
6 = breadth > all in
d = depth ) inches.
= Reaction at support A.
= Reaction at support B.
^ The formulae for practical use will be found numbered 4 to 27 on pp. 28 to 40,
also in a condensed form in Appendix VII.
28 NOTES ON BUILDING CONSTRUCTION
Case 1. BEAM FIXED AT ONE END AND LOADED AT THE OTHER END.
To find the bending moment Mp at any point P, consider the portion
PE of the beam as shown in Fig. 35, then by the rule (p. 26) the bending
moment is equal to the weight W multiplied by the leverage with which it
acts, namely PE.
Now if X be the distance of the point P from the wall, the leverage
with which W acts at P is - x), therefore
Mp-W(Z-a;)
(4).
<— X— »k-
•< l-x ^
Ficr. 35.
Fig. 35a.
At the wall A the weight will be acting with the leverage of the full
length of the beam Z, hence Mj^ = WZ .... (5).
At the point E the leverage is nothing, and the bending moment
M^ = 0
Graphic Representation. — The above
can be shown graphically thus : —
Draw AE (Fig. 36) as before.^ Plot
A/ vertically to represent WZ (the bend-
ing moment at A) to some convenient
scale of inch-tons to an inch.^ Join E/
and draw P^ parallel to A/; then P</
will represent the bending moment at
P, for, by similar triangles,
so that Pgr and A/ are in the same
-^^S- proportion as the respective bending
moments, and the value in inch-tons of can be ascertained by scaling
the length of Vg.
Case 2. — beam fixed at one end, loaded uniformly.
Here if w is the load for unit of span the whole load is wl, and at the
^ This and the similar figures following are merely graphic representations of the
results obtained algebraically. Appendix YI. shows a graphic process applied to
obtain the results themselves.
2 For instance, supposing M^i is 100 inch-tons, then if a scale of 100 inch-tons to
the inch is chosen {i.e. one inch represents 100 inch-tons), A/ will have to be made
1 inch long. Supposing Vg is measured and found to be 0-65 inch, then Mr = 65
inch-tons.
BEAMS— BENDING MOMENT
29
point A (Fig. 37) it acts with a mean leverage ) equal to the distance from
its centre of gravity to the point A.
Therefore
(6).
The load tending to bend the beam about any section P is that upon PE,
A IP E|
I I
< X >K l-MC M
Fig. 37.
P El
!
t< l-X >1
Fig. 38.
= w(Z — £c], and it acts with a leverage equal to the distance of its centre of
l — x
gravity from P, i.e. — — -, therefore
(I - X)2
(7).
Fig. 39.
A
1 1 1
1 1 1
1 1 1
1 1
/ i
r
Fig. 40. \
Graphic Representation. — This case may be shown graphically, as in Fig, 39.
30 , NOTES ON BUILDING CONSTRUCTION
Draw AE as before, and A/ — — to represent the bending moment at A.
Draw /E a parabola with its vertex at E ; then the ordinate dropped
w(l — cc)^
from any point P, distant x from A, will be equal to — , which is the
bending moment at that point. ^
Simpler Construction. — To be of ready practical use, a simpler construc-
AE
tion is needed. Now if A/ is made less than — — a circular arc drawn
through /, and tangent to AE at E, will practically coincide with the required
parabola, and can be used instead of it.
This is shown in Fig. 40, the circle being in full and the parabola in
dotted lines. The value of the above limit as to the proportion of A/ to AE
AE
is shown on the same figure, where A/ is made = and the divergence
between the circle and parabola is seen to be not sufficiently great as to lead
to erroneous results, but in the case of A/' which is equal to — the error is
seen to be considerable.
An easy method of finding the centre of the required circle in such cases
is also shown on this figure. EC is drawn perpendicular to AE, E/^ is joined
and bisected at b, and 60 is drawn perpendicular to E/^^. The point of its
intersection with EC is obviously the required centre.
Case 3. — beam fixed at one end and loaded with a unifoemly
DISTRIBUTED LOAD, AND ALSO WITH A CONCENTRATED LOAD AT
ITS EXTREMITY.
This case (Fig. 41) is a combination of Cases 1 and 2.
Fig. 41.
Fig. 42.
Graphic Representation. — This case is shown graphically in Fig. 42 ; the
^ For a proof of this, reference may be made to any work on conic sections.
BEAMS— BENDING MOMENT
31
diagram from Fig. 36 being plotted above, and that from Fig. 39 below
the beam AE.^
The bending moment at A will be the sum of the bending moments
caused by W and by wZ, so that, combining Equation 5 with Equation 6,
M, = WZ+— (8),
= A/n-A/,
= /i/(less the thickness of the beam in Fig. 42).
Combining Equation 4 with Equation 7,
M, = W(Z-.) + ^fc^' . . . (9),
= Pi + = (less the thickness of the beam).
It is not absolutely correct to say that ln,f— the bending moment at A, and
that ig = the bending moment at P, for it is obvious that the depth of the beam
must be deducted from these ordinates, or the beam must be represented by
a single line when drawing the diagrams.
This case would apply to a balcony crowded with people and having a
very heavy railing.
GcLSt 4 (Fig. 43). BEAM FIXED AT ONE END AND LOADED WITH
SEVERAL CONCENTRATED WEIGHTS, W^, Wg, Wg, ETC.
The bending moment at any point produced by all the weights is equal
to the sum of the bending moments produced by each at that point.
W3
3
— /, — ^
Fiff. 43.
Fig. 44.
Graphic Representation (Fig. 44).
Draw Ae = 'Wjl and join Ee,
ef =^¥"2^2 and join F (the point on Ee under W^) with/,
fg = WgZg and join G (the point on F/ under Wg) with g.
It is evident that
M, = W,l + W,l, + W,l„ .... (10),
= Ae + ef+fg = Ag.
At any point P distant x from A
M, = W^{1 -x) + W,il, -x) + W,(l, - X), . (11),
= 'Pi + ij+jk = 'Pk.
^ Both diagrams must be plotted to the same scale— that is, the same number of
inch-tons or inch-lbs. to an inch must be taken for both.
NOTES ON BUILDING CONSTRUCTION
Case 5 (Fig. 45). — beam fixed at one end and suppoeting a
LOAD UNIFOKMLY DISTKIBUTED OYER A PAET OF ITS LENGTH.
Tf Z being the loaded portion of the heam
At the point A M.^ = wz{\z + y) .... (12).
At any point Q in AB distant x' from A
llL^ = wz{lz + y-x') . . . (12 A).
At any point P in BD distant x from A
'M.^ = w{z + ij-x)y.\{z + y- x),
= lw{z + y-x)^ . . . (12 B).
At the points = loa + 13 = |w22 . . . (12 C).
At the point D M„ = 0.
At any point in DE M = 0.
Fig. 46.
Graphic Representation (Fig. 46).
Draw AG = = + ^/) and join G and the middle point between
B and D.
From B draw the vertical line BF,^ cutting the line from G in F.
Between the points F and D (the points immediately under the extremi-
ties of the load) draw a semi-parabola as for a beam BD uniformly loaded
(Case 2).
The vertical distance between GFD and ABD at any point gives the
bending moment at that point.
Case 6 (Fig. 47). — beam supported at both ends and loaded
IN THE centre.
Fig. 47. Fig. 48.
^ If BF be made less than -r--, the parabola can be drawn as a circle.
BEAMS— BENDING MOMENT
Mo = Reaction at A or B x leverage,
33
Reaction at A or B =
Hence
At any point P
leverage =
,r W I Wl
W
2'
I
W/l
■■ — I ;
2 \2
(13) .
(14) ,
M, = 0,
Graphic Representation. — Fig. 48 shows this Case graphically.
Draw CD to represent to some convenient scale M(,=
Join
AD and BD, then the vertical distance from AB to AD or DB at any point
is the bending moment at that point.
Thus PF is the bending moment at P and is equal V2 ~ ^)*
Case 7 (Fig. 49). — beam suppoeted at both ends and
UNIFOKMLY LOADED,
According to the rule given at p. 26 the reaction R^^^ produces a
positive bending moment at C = ^ x ^, and the load on AC produces a nega-
id I
tive bending moment in the opposite direction at C = — x -.
2i 4
Hence the bending moment at C is
wl I wl I
M„ = — X X-
' 2 2 2 4
(15).
To find the bending moment at any point P distant x from the centre of
the beam, consider the portion AP of the beam, then by the rule —
I
I /I
2 \2
w/l
+ x
2V2
w/l
^2(2- + ^
%u /P
2V4
B.C.-
-IV.
(16).
D
NOTES ON BUILDING CONSTRUCTION
Graphic Representation (Fig. 50). — As in Case 2 the bending moments are
graphically represented by a portion of a parabola ADB as shown in Fig.
50, and if a suitable scale is chosen so that CD is less than ^ AC or -J AB,
the arc of a circle can be used instead of the parabola without appreciable
error. The method of drawing a parabola is given in Appendix III.
To obtain, therefore, the bending moment at any point in AB, plot CD to
represent — to some convenient scale, so that CD is less than ^ AB. Draw
a circular arc through the points A, D, and B, then PQ will represent the
bending moment at the point P; P^Qi the bending moment at P^, and so on.
The bending moment at P can also be obtained by considering the portion PB of
the beam. It will then be found that i
The same value as before but with a negative sign. This difference in signs
expresses the fact that the forces acting on the portion AP of the beam tend to turn
that portion in the positive direction, i.e. in the same direction as the hands ot a
watch, whereas the forces acting on the portion PB tend to turn that portion in the
negative direction.
The maximmn bending moment occurs at the centre of the beam, and is
— . Now wl is the total load, which is written W, so that M„ = — .
Comparison between Central and Distributed Load.— On referring to^Case 6
it will be seen that a central load W produces a bending moment — at the
centre of the beam, that is, double the bending moment produced by a uni-
formly distributed load of the same amount. In other words, the safe dis-
tributed load on a beam is double the safe concentrated load at the centre.
Case 8 (Fig. 51). — beam suppokted at both ends, loaded in
CENTRE AND ALSO UNIFORMLY.
This is a combination of Cases 6 and 7.
Graphic Representation.— In the graphic representation. Fig. 52, the
--a;
A
B
A'
J
Pig. 51. Fig. 52.
diagram of bending moments for the distributed load wl is below, and that for
BEAMS— BENDING MOMENT
35
the single load W is above the line AB. It will be easily seen that
the total bending moment at the centre = for the central load + for
the distributed load. Therefore, combining Equation 13 with Equation 15,
M, = M„ forW + M„fori(;/,
And further, by combining Equation 14 with Equation 16,
= for W + Mp for wl.
= 1W -f A^hr
(17).
. . (18).
It will be observed that the bending moment at any point of the beam
can be obtained by adding together the ordinate of the triangle and of the
curve at that point.
Case 9 (Fig. 53). — beam suppoeted at both ends loaded
AT any point.
In this case the maximum bending moment is at D, the point of applica-
tion of the load. With regard to the reactions at A and B see p. 17.
l-m
^
m ^
-l-m-x —
Fig. 53.
Here the bending moment immediately under W is :
Bending moment at D = M„ = reaction at A x leverage, AD,
m
= y W x{l- m),
m{l — m)
I
W
(19).
Or
M^, = - reaction at B x leverage, BD,
'I — m\
-J— jW xm,
m{l — in)
I
W.
At any point P in AD to the left of and at a distance x from W
Mp = X leverage,
m
= J W X (Z - m - a;) . . , (20).
At any point Q in DB to the right of and at a distance y from W
= - X leverage, BQ,
l — m
= ^ ' • • (21)-
At A and B and each = 0.
36
NOTES ON BUILDING CONSTRUCTION
Graphic Representation. — The bending moments are represented hj a tri-
angle, as shown in Fig. 54.
Case 10 (Fig. 55). — beam supported at the ends and
PARTIALLY LOADED WriH A UNIFORM LOAD.
Fig. 55 shows the manner in which the beam is loaded, and the values of
the reactions and R^j are as follows :
I — m + n
1 (l — m — n){l — m + n)
R^ = (i — m — u)w X
I
l + m — n
21
2 (l — m- n)(l + m-n)
R^ = (Z - m - n)io X ^ = ^ . w.
1+ m-n
-M-^QQQQ)^
~ 2 ■
'(l-in -n) w
Fig. 55.
The distance of any point (P, P', or P") from A will be called x, and from
B the distance will be called y.
The bending moment at any point P situated between A and E is
Mj. = R^ X X,
(Z - m + n)(Z — m — n)
21
(22).
Similarly the bending moment at any point P" situated between F and B at
a distance y from B will be found to be (considering the portion P"B of the
beam and neglecting the negative sign)
(l + m — n)(l — m — n)
ll,>, = li,xy = ^-^ ^\ \wy
(23).
Lastly, the bending moment at any point P' situated between E and F
is equal to
Mj., = 'Rj^x X - {x- m)w X — ^ ,
(I — m + n)(l - m — n) (x-m)"^
= ^ '.'wx-'^—^.w (24).
It can also be shown by the application of the differential calculus that
Mp' is a maximum when
^ = — SZ— • • • • (25).
Graphic Representation. — It will be found that if the distributed load over
BEAMS— BENDING MOMENT
37
EF were concentrated, one half at E and the other half at F, that neither
nor R3 would be altered in value, and hence M^. and Mp" would also have the
same value. But the bending moment at P' would be changed and would
be equal to
•,r/ (^-m + 7i)(Z-m-TC) (l-m-n)w
M . ivx - (x - m),
therefore
21
(x — nn)w
I — m — n — {x — m)
Fig. 56 represents the part EF of the beam by itself, and it is clear that
he— z
I — m — n = l^; and if z is the distance of P' from the centre of EF,
I.
Hence
Mp' - M' ' =
+ 2
+ z
It will therefore be seen that the bending moment at P' is greater when
the load is distributed than when it is concentrated at E and F. And further,
Ly referring to Equation 16, it will be seen that the increase can be represented
graphically by the ordinate of a parabola as shown in Fig. 57, or practically
wl ^ I
by the ordinate of an arc of a circle, if — ^ is represented by less than J.
Now the diagram of bending moments, when the loads are concentrated
at E and F, is as shown in Fig 58, where
and
EQ = R^ X m =
FQi = E3xn =
(l-m + n)(l — m ■
n
(Z + m - n){l — in — n)
21
■.vm,
38 NOTES ON BUILDING CONSTRUCTION
so that the diagram, when the load is distributed, can he found by adding the
diagrams in Figs. 57 and 58 together, as shown in Fig. 59.
Instructive lessons can be learnt from this case by varying the values of m
and n.
Supposing, for instance, that m=n, then the maximum bending moment will be
at the point for which (Equation 25)
that is, at the centre of the beam, and the bending moment at that point, from
E(,;[uation 24, will be
n ■ '2'
V7 ^ \
-\l-^-\-vx\w,
(26).
If now m = 0, so that n is also 0, that is, the load is distributed uniformly over
the whole beam, then Mc = -g •
the result previously obtained when considering Case 7 (Equation 15).
Again, if m = ^, that is, the central half of the beam is uniformly loaded,
M -4 16.
w.
BEAMS— BENDING MOMENT
39
from which it appears that if the two outer quarters of the beam were loaded and
the central half unloaded,
32
So that the load on the central half produces a three times greater bending moment
than the remainder of the load.
Equation 26 can be written
M^=|^^ + m^(Z- 2m)M;.
Now (Z - 1m)w is the total load on the beam, that is W, so that
Let m = \ and also m=hi the load will therefore be concentrated at the centre of the
Z 2
beam (Case 1), then M — "W
0 4 * '
the result previously obtained (Equation 13).
The case of a beam loaded with a concentrated load placed anywhere upon it can
be deduced in a similar manner to the above from Equation 24, an exercise which
is, however, left to the student.
It will thus be seen that the case under consideration includes all the previous
cases.
Case 11 (Fig. 6o). — beam suppoeted at both ends, loaded
WITH ANY NUMBER OF LOADS.
The total reaction at each abutment is equal to the sum of the reactions
w,
N 0
1^ \0i) V osy
\ \
\ \
\ \
1
/ /
'^3 /
t
h
1 '
Z = the span
OT = the distance of "Wi from A
,, "W2 from A
mz— „ "W3 from A
Fig. 60.
caused by each load separately, thus (Fig. 60)
I
I
I
40
NOTES ON BUILDING CONSTRUCTION
The moment at any point P distant x from A is (if P be between aad O3)
Mp = Reaction A x leverage — weights between P and A x leverage,
f^Jl-m.) y^Jl-m^ Wo(Z-m„)\ , , , ,
= {-^l +-i + I ) xa;-W^(x-m)-W2(x-m2) (27).
Graphic Representation^ (Fig. 60). — In the graphic representation the cotted
lines show the triangle of bending moments due to each load separately (see
^^^^ Tlius Ar^B is the triangle for W^^,
Ap.^B is the triangle for Wg,
AggB is the triangle for "Wg.
The bending moments caused by all the loads are represented graphically
by a polygon, which can be obtained as follows : —
The bending moment caused by immediately under itself is o^r,, the
bending moment at this same point caused by the weight W2 is o^p^, and the
bending moment caused by Wg is o^g^ ; the total bending moment at this
point caused by W^, W^, and Wg is equal to o^r^ + o-^^p-^ + o-^q^, and o^t^ is set off
equal to this.
In the same way the total bending moment at Wg is found and set off as
o^t^, and the total bending moment at Wg as Og^g.
Join A^j^, t^t^, t^t^, and ^gB, and the polygon of moments is obtained.
MOMENT OF EESISTANCE.
The application of the preceding rules enables us to ascertain
the effect produced by external forces acting transversely upon a
beam — that is, in other words, to calculate the bending moments
produced by loads or other external forces.
The next step is to investigate the nature of the resistance
offered to these forces.
This resistance is afforded by the internal structure of the
beam, which, for instance, consists in timber and wrought iron
of a number of fibres lying closely together and running in the
direction of the length of the beam, but in cast iron of closely
adhering grains.
There are two methods by which the value of this resistance
can be ascertained —
1. By reasoning upon certain assumptions,
2. By experiment.
Of these two methods the first is not practically used in the
calculation of beams of rectangular cross section, but its applica-
tion to them is described somewhat fully in detail, in order to
clear the ground for the better understanding of the calculation
of beams of more complex section.
Method of Ascertaining the Moment of Resistance by Reasoning.
■ — The effects produced upon the fibres of a beam by external forces acting
upon the beam may be shown as follows : —
^ See Appendix VI. where a similar ease is worked entirely by a graphic method.
BEAMS— MOMENT OF RESISTANCE
41
Let Fig. 61 represent a straight rectangular beam supported at A and B
and Fig. 6 1 a a cross section of the same beam. Mark upon tlie beam two vertical
straight lines op, qr, and also a line NL drawn parallel to the length of the
beam and midway of its depth, and intersecting op in n and qr in L
c
d
e
i
/
Fig. 61a.
P
Fiff. 61.
The beam can be considered as made up of horizontal layers of fibres, so
that oq will represent a portion of the top layer and pr of the bottom layer ;
NL will represent the central layer, nl a portion of it, and cd and ef portions
of intermediate layers.
If the beam be now slightly bent as in Fig. 62 (where, however, the bend-
C
b-
e'
1 f'
P
r'
Fig. 62.
Coin
.pressioit'
Tension
ing is exaggerated to make the case clearer), it will be found that the lines
op and qr, originally vertical, are inclined inwards, and that nl remains
unchanged in length. It therefore follows at once that o'q' and c'd' are
shorter than oq and cd, but that p'r' and
e'f are longer than pr and ef.
Fig. 63 is an enlargement of the
central part of Fig. 62, and o^p^ has
been drawn through n parallel to q'r';
clearly 0^0' and c^c are the amounts by
which oq and cd have been shortened
by the bending, and similarly e'e-^ and
p'p.^ are the amounts by which ef and
pr have been lengthened.
It is evident, therefore, that all the
fibres above nl have been compressed,
and those below nl extended, by the
stress put upon them by the bending
action.
Only the fibres in the layer repre-
sented by NL are uninfluenced by the
bending action ; they are neutral, hav-
ing neither compression nor tension to resist,
the neutral layer.
P'
Fig. 63.
This layer of fibres is called
42 NOTES ON BUILDING CONSTRUCTION
If a cross section of the beam be made, the neutral layer ^ will be
intersected in a straight line (00, Fig. 6ia), and this straight line is called the
neutral axis.
As already mentioned at p. 24, if the bending of the beam were carried
far enough to produce cross breaking, it would be found that the fibres at
the cross section of maximum stress above the neutral layer were crushed,
and the fibres below were torn asunder.
It will therefore be seen that the resistance of the beam to bending is
afforded by the resistance of the fibres above NL to compression and of those
below NL to tension, and the question is. How can we arrive at the value of
this resistance ?
Figs. 64 and 65 may make us understand more clearly exactly what it is
we wish to find out.
Fig. €4.
Taking a simple case of a beam supported at both ends and loaded in the
centre.
Fig. 64 shows the general direction of the forces which produce the
bending moment and of those which offer resistance at the cross section CD
under the load.
and are the reactions at A and B respectively which act with a
leverage equal to half the span of the opening, i.e. AC or BC, and
R^ X AC or
R3 X BC
Supposing now that the beam is cut across at CD and separated, the load
being also divided equally ; and that by some means the stress on each fibre
is maintained exactly at what it was before the beam was cut ; the two halves
of the beam will clearly remain in equilibrium, and the state of things is
represented in Fig. 65.
Considering the left-hand half of the beam, it will be seen that the forces
W
R^ and — tend to turn that portion of the beam in the positive direction, and
that the internal stresses tend to resist it in the negative direction. The forces R^
: the bending moment.
1 The neutral layer is sometimes called the neutral surface and in some works
the neutral axis.
BEAMS— MOMENT OF RESISTANCE
43
and — , being equal and parallel, form what is called a coufle, and the pro-
duct of one of the forces into the distance between them (i.e. x AC) is
called the moment of the couple. It will be seen that the moment of this
couple is equal to the bending moment at the section CD. To ensure equi-
librium it is clear that the resisting stresses must have a. turning moment
equal and opposite to the bending moment. This moment is called the
moment of resistance of the section.
The same reasoning applies to the right half of the beam, and can also be
extended to any section of the beam.
A A
B
W W
Fig. 65.
We have therefore established the important relation that at any section
of a beam
Bending Moment = Moment of Resistance.
The moment of resistance at a section P will be denoted by Mp, and the
next step is to find an expression for its value.
From Figs. 64 and 65 we gain a general idea of the nature of the
resistance which is offered by the fibres of the cross section and of the way
in which this resistance counteracts the effects of the bending moment.
We wish now to ascertain more exactly the amount of the resistance
offered by the fibres of a cross section, and how the resistance is distributed
among those fibres ; whether they all resist equally, or whether more resist-
ance is offered by some than by others.
In order to do this, certain assumptions must be made.
These assumptions are based on experiment, and are practically true
when the stresses do not exceed the elastic limit ; and as the working stress
should not even approach, far less exceed, the elastic limit, these assumptions
«iiui be accepted as giving practically reliable results. They are, however,
not true so soon as the elastic limit is overstepped, and the divergence
becomes greater and greater as the point of rupture is approached. It is clear,
therefore, that the breaking load cannot be deduced by working on these
assumptions, which are as follows : —
1. That the stretching or shortening in any fibre is proportional to the
stress of tension or compression that comes upon it.
44 NOTES ON BUILDING CONSTRUCTION
For example — If a compression of 1 ton per square iiicli shortens a fibre
by xttVtt^^ i*"^ length, a compression of 2 tons per square inch on the fibre
will shorten it yo^o o^-'^^ ^'^ length, 3 tons Y^^QQ-ths, and so on, up to the
elastic limit ; and in the same way, if a tension of 1 ton per sqnare inch
stretches a fibre j^Vo^^j ^ ^^^^^ '^'^ stretch it yQ^Q-yths, 3 tons y-^^^-^ths, and
5 tons y-Q^Q-g-ths.
2. That the amoixnt of shortening produced by a given stress causing
compression will be equal to the amount of stretching produced by an equal
stress causing tension.
Thus if a compression of 1 ton per square inch ujion a fibre shortens it
y^^jyth of its length, a tension of 1 ton per square inch will stretch it j qVo^-'^
of its length, and so on.
3. That the amount of stretching or shortening in a fibre is proportional
to its distance from the neutral axis of the section under consideration.^
It follows from assumptions 2 and 3 (see Appendix IV.) that the neutral
axis passes through the centre of gravity of the section.
Thus if CDEF (Fig. 66) represents the cross section of a beam 12" deep,
then 00, the neutral axis, is to be drawn
through the centre of gravity of the cross
section, that is in this case midway between
CE and DF.
p
o
s —
Q-^r i
If PQ be 1 inch from 00,
RS „ 4 inches ,,
TU „ 4 inches ,,
VX ,, 5 inches
X—
F—
Fig. 66.
then the shortening of each fibre in the
layer ES will be four times that of the
fibres at PQ ; the shortening at CE six
times that at PQ. The stretching at VX will be |- of that at TU, and the
stretching at DF = |- that at TU.
In the case under consideration, it is to be observed that, owing to the
neutral axis being midway between CE and DF, the amotmt of shortening
of the fibres at CE is equal to the lengthening of the fibres at DF.
It will be seen that in Fig. 63, o^^ti^j, and o'np' are drawn as straight lines,
which expresses graphically assumption 3.
Combining assumptions 1 and 3, it will be
seen that the stress in any fibre is proportional to
its distance from the neutral axis, and further,
from assumption 2, that the numerical value of
the tension in a fibre is equal to the numerical
value of the compression in a fibre at the same
distance on the other side of the neutral axis.
Thus in Fig. 66, if the tension on the fibres at TU
is 8 cwts. per square inch, the compression on the
fibres at RS is also 8 cwts. per square inch.
The amount of stress is shown graphically in
Fig. 67 ; that is, if CE represents the compression
at CE, will represent the compression at PQ, and so on.
^ This assumption has been proved by experiment. See A Manual of Civil
Engineering, by W. T. M. Rankine, F.R.S.S., etc., p. 250, 10th edition.
— O
BEAMS— MOMENT OF RESISTANCE
45
Stresses in beam 1" wide. — Applying these assumptions in the first instance
to a simple case. Suppose A^B (Fig. 68) to be a rectangular beam 1" wide
and 12" deep, supported at the ends and loaded in the centre, then each inch
in depth of the beam will contain one square inch of fibres.
Suppose also that the " elastic limit " of the material of which the beam
is composed is 1 0 tons ^ per square inch. We know that the fibres will bear
that stress without injury, but being determined to be well under that limit,
we wish to arrange that the stress upon the fibres shall nowhere exceed 6
tons per square inch. The limiting stress then is to be 6 tons per square inch.
On reference to Case 6, p. 32, it will be seen that the maximum bending
moment occurs at the centre of the beam. If, therefore, the stress on the
extreme fibres at the centre of the length of the beam does not exceed 6 tons
per square inch, we can be quite sure that it will not exceed it anywhere else.
Again, by assumptions 1 and 3, the stress upon the fibres is in proportion
to their distance from the neutral axis. The stress at the neutral axis is
nothing, but the compression upon the fibres above the neutral axis becomes
w
greater and greater as they approach the extreme fibre A^Bj, and the tension
upon the fibres below the neutral axis becomes greater and greater as they
approach the extreme fibre AB.
Considering, therefore, the central cross section CD of the beam ; if c'd' be
drawn so that Cc' = Bd' = 6 tons on a scale of 10 tons to one inch (Figs. 68 and
69a), the stress at any other point will be represented by the line drawn across
the triangles of stress BOd' and COc' at that point parallel to the neutral layer.
Thus mm' represents the stress at the point m =' 5 tons per square inch,
as measured on the scale, nn' represents the compression at the point n = 2
tons per square inch, and pp' represents the tension at ^ = on the scale 4 J tons
per square inch.
^ These are taken merely as simple convenient figures, not with reference to any
particular material.
NOTES ON BUILDING CONSTRUCTION
Neutral
D
Fig. 69a.
C tons
6 tons
The mean compression on the fibres above the neutral axis will be
that represented by 22 ( = 3 tons) at a point q, half-way between O and the
extreme fibre.
In the same way the mean
tension on the fibres below
the neutral axis is rr' = 3 tons.
This mean compression
acts over a surface 6" x 1", the
total compression is therefore
6" X 1" X 3 tons = 18 tons.
In the same way, below
the neutral axis the total ten-
sion is = 3 tons X 6" x 1" = 1 8
tons.
We have then in Fig. 69 a cross section of the beam, and in Fig. 6ga a
diagram which shows the resistance offered by the fibres (equal to the stress
upon them) at any point in the cross section, and also the total resistance of the
cross section both in compression and tension. The diagram also shows us
how this resistance is distributed, i.e. that the fibres at the neutral axis
contribute nothing to the resistance, but that the resistance of the fibres
increases in proportion to their distance from the neutral axis until it reaches
a maximum in the extreme layers of fibres CE and DF, where each undergoes a
stress equal to the limiting stress of 6 tons per square inch.
Graphic Method of Finding the Moment of Eesistance.^ — It is,
however, convenient to combine the cross section
of the beam with the diagram as follows : —
Draw the cross section of a beam CEDF as
before (Fig. 70), and make a new scale of tons
in which the length CE = 6 tons, the limiting
stress per square inch, and draw the stress dia-
gram (Fig. 71) similar to Fig 69a above.
Fig. 7 2 is merely a modification of Fig. 7 i
and clearly gives the same results, and can be
inserted in Fig. 70 as shown. Now the area of
the triangle CEn is half the area of the rectangle
CEOO, and, therefore, the total compression is — Tig- 70.
= area of rect. CEOO x 3 tons,
= area of triangle CEn x 6 tons,
6x1 ^ ^
== X 6 = 1 8 tons,
the same value as before. In other words, a compression of 6 tons uniformly
distributed over the triangle CEu is equivalent to the actual compression
existing at the section of the beam. And in the same way a tension of 6
tons uniformly distributed over the triangle nDF is equivalent to the actual
tension. The areas of the triangles CEn and nI>F can therefore be called the
equivalent areas of resistance.
In the case of beams more than 1 inch wide, the application of this method
of showing the quantity and distribution of the resistances is very simple.
Stresses in Beam 4 inches wide. — Thus a beam 4 inches wide and 12
^ This method is described as it illustrates the principles and for the sake of those
who prefer drawing to calculatious, but it is easier to use the formulae, p. 54.
Fig. 71. Fig. 72.
BEAMS— MOMENT OF RESISTANCE
47
inches deep is the same as four beams each an inch wide and 12 inches deep
(such as that we have been considering) placed side by side.
The diagram showing the intensity of stress on the fibres of such a beam
will be the same as before (Fig. 69a).
In the diagram (Fig. 73) showing the quantity and distribution of stress,
as there are 4 inches in CE, each acted upon by a stress of 6 tons per square
inch, CE will represent 4x6 = 24 tons. CEw will be the equivalent area for
compression, and nDF the equivalent area for tension.
The area CEw = ^ x 4" x 6",
= 12 square inches.
The resistance to compression of CEr!,= 6 tons x 12 square inches,
= 72 tons.
The tensile resistance of ?iDF = x 4" x 6" x 6 tons,
= 72 tons :
or in each case four times as much as the resistance (18 tons) of the beam 1"
wide and 12" deep.
In the same way the resistance of any beam may be
found, whatever its breadth, depth, and strength of fibre. c
In all these diagrams it will be noticed that the
equivalent area representing tension is equal to that
representing compression. It is evident that this must
be the case, for if the total compression above the
neutral axis were not equal to the tension below, there ^
would not be equilibrium.
Value of Moment of Resistance. — Having thus drawn
the triangles showing the amount and distribution of
the resistance of the cross section, it will be easy to ^ f '
ascertain the value of the moment of resistance, which,
as explained at p. 43, has to be made equal to the "
bending moment produced by the load.
It will make the general rule more clear if we deal first with a particular case.
Let us take the case we have already been considering, that of a beam 4"
wide and 1 2" deep, having a limiting stress upon the extreme fibres of 6 tons
per square inch.
We know from p. 46 that the amount and distribution of the resistance in
compression is shown by the triangle CEn, that in tension by nDF, Fig. 73.
This diagram tells us then that if 6 tons per square inch according to the
scale were acting uniformly all over the triangle CEn, it would ofi"er in
quantity and distribution exactly the same resistance as is offered by the vary-
ing stress which actually occurs over the rectangle CEOO. Again, the actual
varying tension on CODE is represented in quantity and distribution by 6
tons (on the scale) per square inch acting uniformly all over the triangle nDF.
Now if 6 tons were acting on each square inch of CE?i, we know by the
elementary rules of mechanics that it will be the same thing if we consider
the total number of tons acting over CE», as being concentrated at the centre
of gravit}^ and acting there.
The centre of gravity of CEu is at the point c, and that of iiDF at (/.^
It has been shown above that the total compression is 72 tons, and that
f the total tension is the same.
^ c is 2 inches or J of cji from e, ^ is 2 inches from/.
48
NOTES ON BUILDING CONSTRUCTION
The resistance of the cross section is therefore equivalent to two forces,
each equal to 72 tons, and acting at c and g.
On reference to p. 43 it will be seen that these forces form a cou'ple,
whose moment can be found by multiplying one of the forces by the dis-
tance between the forces, or, as it is called, by the arm of the couple, in this case
eg. The moment of this couple is clearly the moment of resistance required.
The moment of resistance of the cross section under consideration is
therefore = 7 2 x c£f.
These centres of gravity are found by bisecting CE at e and DF at/ —
joining ef and taking ec and gf equal to 1 of en and nf respectively.
Now eg is evidently equal to cn + ng Fig. 74, and
cn = fen = -I x 6 = 4"
ng-
or the moment of resistance
.•• cg =
X 6 = 4
Fig. 74.2
Thus, as in Fig. 73,
so that the distance eg
= 72 tons X 8 inches,
= 576 inch-tons.^
Safe resistance Area — General Case. — Now
to take a general case (Fig. 74). Suppose the
breadth of the beam in inches is called b, the
depth in inches d, and the limiting stress to
be allowed on the outside fibres tons per
square inch. As in the previous case, the
triangles CEw and nD¥ are the equivalent
areas of resistance each ^bd, the stress being
?•„ tons per square inch. The total compres-
sion is therefore r^ x ^bd, and the total
tension has the same value.
cn = ng = i-~,
or the length of the arm of the couple, will be
ld + ^d = |d
Eesulting Formula. — -The moment of the couple, i.e. the moment of
resistance required, is therefore
r^x^bdx^d = ^ rj)d^.
We have now foiind a general formula for the moment of resistance of a
rectangular beam, in terms of its breadth, depth, and of the limiting stress
allowed upon the extreme fibres.
The method by which this formula has been arrived at is shown step by
step in the preceding paragraphs, and as the formula has been deduced from
^ Another way of looking at this is : The fibres in CEn act with an average leverage
e(iual to tlie distance from their centre of gravity to the neutral axis -on— 4 inches ;
in the same way the fibres of ?iDF act with an average leverage of 4".
Thus we have total resistance
tons tons
CEnx6x4" + 7iDFx6x4",
= 72x4 +72x4,
= 576 inch-tons.
2 The shaded portions of this figure represent in magnitude and distribution the
stresses over the section of the beam— just when the extreme fibres on each side are
subjected to the limiting stress.
BEAMS— MOMENT OF RESISTANCE
49
the geometrical demonstration, it is in agreement with it, and may be used
instead of working out each case graphically.
Two conclusions are self-evident from the formula, viz. : (1) That the
moment of resistance in a rectangular beam varies directly as the breadth of
the beam, i.e. that a beam 12x4 will have a moment of resistance twice as
great as one 12x2; and (2) That it varies as the square of the depth of the
beam, i.e. that a beam 12x2 will have a moment of resistance four times
as great as one 6x2.
It will be understood that the limiting stress should not be greater than
the worlcing stress that can safely be allowed upon the particular material
used and never greater than the elastic limit.
When such a stress is used, the result gives the worlcing resistance of the
beam ; but it should be remembered that if the breaking stress were used, the
resistance found would not be the ultimate resistance, or actual resistance to
breaking, because when the breaking point is approached the assumptions
made (p. 43) no longer hold good.
The value of the moment of resistance of any beam of rectangular section
with any limiting stress can be obtained either by drawing the resistance
diagrams and calculating the amount of the moment from them, or by
substituting the values of h, d, and r„ in the formula.
Effect of depth upon strength. — To illustrate the self-evident conclusions
mentioned above, suppose we wish to ascer-
tain the relative strength of three beams
A, B, 0, of the same length, but differing ^
in depth — A being 4" deep, B 6" deep, C 8"
deep, and the width of each being 2". Sup-
pose also that the limiting stress to be D
allowed upon the extreme fibres is to be ^
taken in each case at 3 tons.
We will calculate these beams first from
the diagrams showing quantity and distri-
bution of resistance, then by the formula.
The diagrams are as in Fig. 75.
Calculation from Diagrams or Resistance.
Calculation erom Formula
To
1
Beam.
2
Dimen-
sions.
3
Resistance of section
= area of GE?!, or
DwP X worlcing
stress.
4
Lengtli of
arm of couple,
i.e. from
c to g.
= id
5
Moment of
Resistance =
Resistance of
section x arm
of couple.
6
Value of To, h,
d.
7
Value of Moment of
Resistance
TO-.
A
^ p.
2" X 4"
tons tons
ix2x2x3=: 6
fx 4"= 1"
inch-
tons tons
6x f" =16
To = 3 tons
b = 2" c^ = 4"
inch-
tons
0 2x4x4 ,
3 X =lb
6
B
2" X 6"
ix2x3x3= 9
|x6" = V-"
9xi/"=36
ro = 3 tons
6 = 2" d = 6"
„ 2x6x6
3x — - — =36
0
2" X 8"
4 X 2 X 4 X 3 = 12
I X 8"=-V-"
12x J/"=64
ro = S tons
6 = 2" d=8"
„ 2x8x8
3 X — - — = 64
6
We see from the above that the
B.C. IV.
moments of
resistance found in column
E
5°
NOTES ON BUILDING CONSTRUCTION
5 by calculation from the diagrams are tlie same as the moments of resistance
found in column 7 from the formula.
Moreover, we see that the width, etc., being the same, the moment of
resistance, i.e. the strength of the beams, varies as :
16 -.36 : 64, i.e. as
4^ : 6^ : 8^, i.e. as
the squares of the depths of the beams.
Effect of width upon strength. — Let us see the effect of varying the width
upon the strength of a beam.
Take beam C, and instead of 2", give it first a width of 3", then of 4",
then of 6", calculating by the formula we have :
Size of Beam.
8" X 2"
8" X 3"
8" X 4"
8" X 6"
Value of Moment of Besistance
3 X
3 X
3 X
3 X
2x8x8
3x8x8
6
4x8x8
6
6x8x8
= 64
= 96
= 128
= 192
We see from the above that when the depth remains the same the resist-
ance or strength of the beam varies as the width.
Thus the resistance of a beam 3" wide is one and a half times that of a
beam 4" wide, twice that of a beam 6" wide, three times that of a beam
2" wide.
General rule. — We see then, as already mentioned at p. 49, that the strength
of rectangular teams of the same material vary directly as their breadth and as
the square of their depth.
Tliis indeed is apparent from an examination of the formula, for the
resistance of the beam is shown by that formula to vary as bd^.
General Formula.— This formula,
Moment of resistance = r^, — ■ . (28),
6
is therefore, as has been mentioned above, a general ex-
pression for the value of the moment of resistance of a beam of
rectangular section, when it is subjected to such a load that
the limiting stress on the fibres nowhere exceeds the elastic
limit of the material.^
If the assumptions mentioned at p. 43 were true for all conditions of stress,
^ This formula is practically the same as that given by Kankine and other
writers, who have approached the subject from the side of advanced mathematics. They
r I . . »
show the moment of resistance to be equal to where I is the moment of inertia of
BEAMS— MOMENT OF RESISTANCE 51
;]iat is, if they were true for every degree of stress from very small ones up to
;he breaking point, then would be that value of the stress either of tension
)r compression which was known to cause rupture by tearing or crushing in
;he fibres of the material, and could be ascertained by direct experiments
ipon the tensile strength and resistance to compression of the material.
Experiment has shown, however, that these assumptions are true only for
stresses such as are well within the elastic limit of the material, the shorten-
ing or lengthening of the fibres varying regularly with the compressive or
tensile stresses (see p. 11) so long as the elastic limit is not exceeded. Beyond
that, the ratios of the shortening and lengthening to the corresponding stresses
producing them advance more and more rapidly until, on approaching
fracture, they become irregular, but always considerably in excess of what they
were under the lower stresses.
The effect of stresses in tension often differs also from the effect of the
same stresses in compression, and the result of all this is that the position of
the neutral axis is altered — it no longer passes through the centre of gravity
of the cross section ; e.g. if the material is at the point of rupture stronger to
resist compression than tension, the neutral axis will then be nearer the outer-
most fibre in compression than to the outermost fibre in tension. Thus in a
beam of such material, supported at both ends and loaded, the neutral axis
will be nearer the top than the bottom edge.
This difference between facts and the assumptions, at the time when the
breaking stress is approached, renders the theory founded upon them inappli-
cable for the purpose of ascertaining the resistance of beams to actual 'breaking.
It should be noticed that in these examples the weight of the beam itself is left
out of the consideration. When it is large enough to require attention it must, of
course, be treated as a distributed load, and its effect be added to those of the load
proper.
METHOD OF ASCERTAINING THE MOMENT OF RESISTANCE
BY EXPERIMENT AND CALCULATION.
When we liave to deal with heams loaded so that they are
more than slightly bent, or when the intensity of stress upon their
fibres exceeds the elastic limit, the theory and the method just
described no longer apply ; there is no sufficiently simple theory
which is applicable, and we must have recourse to experiment.
Careful experiments have accordingly been made, from which
it has been found that the resistance of rectangular beams of the
same length and of similar section similarly loaded up to the
ireahing point varies directly as the breadth and as the square (/'
the depth, i.e. as hd'^.
This seems to agree with the result arrived at by reasoning in
the case of beams only slightly bent by a load.
the cross section, and yo is the distance from the neutral axis to the extreme fibre
subject to a stress To. In the case of a rectangular timber beam 2/o=^ and
hd? r I IxJP'
1 = — , therefore — =ro — ; and the formula reduces to that given above.
NOTES ON BUILDING CONSTRUCTION
If tlie agreement were complete, tlie formiila given at p. 50, i.e.
M = r„ — ,
0 g ,
might be applied even to cases of beams loaded up to the breaking point, by-
taking r„ as the ultimate resistance of the fibres per square incli to tension or
compression.
This formula may be and is used for beams so loaded, when an important
alteration has been made in it. The coefficient r„ no longer represents the
resistance of the material to tension or compression up to the elastic limit ;
but experiment shows that it has an entirely new value — depending upon the
material, the form of beam, and disposition of the load— and that this new
value can only be ascertained by experiment, and will be denoted by /„.
Value of f„ for timber. — The method in which this value was ascertained
was as follows : —
Small experimental beams each 1" square and 1 2" long between the sup-
ports were supported at the ends and loaded in the centre with a weight W.
From Equation 13 we know that in such beams
the bending moment = .
Substituting/, for r„ in Equation 28, Ave have the moment of resistance
= f!^ (29).
Hence = f — ,
r 6
or /o=-— -Xy-,
4 0(1-
The weight W was gradually increased until the beam broke, and then
the value of W, I, h, and d in the above equation being known, / could be
found.
Modulus of rupture of timber. — The value of /„ thus found is called the
modulus of rupture ^ of the material. It varies of course for different materials
— the stronger the material the higher the modulus. Thus a beam of oak
would require a greater weight W to break it than a beam of pine, and /„
would therefore be greater.
It is not accurately known why the modulus of rupture thus found by
experiment differs so widely, as it does in many instances, from the resistance
of the fibres to either tension or compression.
It must be remembered, however, that the fibres of beams are not in-
1 The modulus of rupture thus found is always equal to 18 times the weight W
that will break the beam across. The reason for this is easily seen if the values of
I, b, and d for the experimental beam are substituted in Equation 29 above —
1 = 12",
6 = 1",
d = l".
4 ^bd^'
^Wxl_2 6
4 1x12
BEAMS— MOMENT OF RESISTANCE
53
dependent filaments, but are united together laterally with, an adhesion less
or greater according to the nature of the material.
This lateral adhesion has been shown by Mr. Barlow to lead to stresses of
a very complicated character, which so far have not been dealt with by any
simple mathematical theory, and its effect is practically to alter the value of
the coefficient/,.
There is also another consideration. The strength of the individual
fibres cannot be uniformly the same, as it is assumed to be. So long as the
limit of elasticity is not passed, this difference of strength is so small as not to
affect the result ; but after that point the weaker fibres begin to yield more
than the others, and finally, just before rupture, these weaker fibres break
first and leave the whole strain to come successively upon fewer and fewer
fibres, uutil complete rupture takes place. Under these conditions, /„ cannot
be a constant coefficient.
The value of /„ obtained by the experiments above referred to is gener-
ally used in practice, though it is in most cases too high, for the following reason.
The small beams (speaking of timber) experimented upon were as a rule
selected specimens perfectly dry, free from defects, such as large or dead
knots, etc., which tend to interrupt the continuity of the fibres and thus
weaken the beam.
Large scantlings cannot be obtained so well seasoned or free from defects
as the small experimental beams, and therefore in calculating the strengtli
of large beams the value of /, should be taken considerably lower than the
value found for the experimental selected beam of the same material.
For example, in the case of Red pine, the modulus of rupture found by
experiments upon small selected beams, varies from 7100 to 9540 Ibs.^ ;
whereas the average modulus found by breaking large beams is 45 cwts. or
5040 lbs. -per square inch.^
It would be expensive to ascertain the modulus by breaking very large
beams, and after all, even if this were done, the timber of the beams used
Avould probably not be exactly similar to that experimented upon, and would
therefore have a different modulus.
It will be seen, then, that the method of ascertaining and applying the
modulus of rupture is only a rough way of correcting the imperfections
of the theory upon which the formula is based.
The method answers, however, sufficiently well in practice, and it has the
advantage of being very simple. An engineer in a distant country having to
deal with an entirely new material can at once ascertain its modulus of
rupture by experimenting upon small bars as above described ; and having
found the modulus he can use the ordinary formula given above.
It is far easier for him to find the modulus of rupture in this way than
to find the amount of extension and compression of the material under
different stresses — which would be necessary if the perfect mathematical theory
could be worked upon.
Modulus of rupture for different sections of beam. — In using the modulus of rupture
thus obtained by experiment, it must be remembered that it is only apphcable in
the calculation of beams of similar section to the experimental beam from which the
modulus was deduced.
In cases where the modulus obtained from a rectangular beam of cast iron placed
thus D was applied to the calculation of the resistance of a beam with a circular sec-
^ Rankine's Useful Rules and Tables, p. 199. ^ Seddon's Builders' Work.
NOTES ON BUILDING CONSTRUCTION
tion, the error caused was found to be 174 per cent, and in a case where it was applied
to a square beam placed on edge, thus, 0> the error was 190 per cent.
In practice, however, beams of these sections are not used for building construc-
tion in any material, and they need not be further considered.
Even a difference in the method of loading the beam used, as compared with the
method in which the experimental beam was loaded to find the modulus, has been
found to cause errors in results ; but these errors are comparatively small, and may
be neglected in using a method which after all is merely a rough and ready way of
arriving at the result.
The moduli of rupture given in Table I. are those obtained by breaking beams
by means of weights placed on their centres, but these moduli may, without material
error, be used in calculating beams with any distribution of load. ^
Practical Formula. — In practice, the rough, and ready way of using the
formula (Equation 29) in conjunction with the modulus of rupture is as
follows : —
To find the weight that will break a beam. — Find from pp. 27 to 40 the
bending moment, M, for the given conditions of load — equate this with the
moment of resistance, M (Equation 29, p. 52), and we have
M='^-^ .... (30).
Take the case of a beam supported at the ends and loaded at the centre,^
W"^ = breaking weight in centre
I = length j
h = breadth I all in the same dimensions, either feet or inches,
d = depth J
= modulus of rupture for the material (see Table I.)
M = — (see Equation 13, p. 33)
To find the load that a learn can safely hear. — is, however, the hreaJang
load. To find the safe load the factor of safety F must be introduced. Taking
W as the safe load with that factor.
FW:
W =
" 61
fo^.
To find the section of a beam to hear safely a given weight. — Take the
same case as before — a beam supported at ends and loaded in centre.
fjbd^
From this find hd^, and assuming a value for either b or d, find the other,
Avhich will give the sectional dimensions of the beam.
Examples of the calculation of beams are given at p. 75 and seq.
^ For moduli of rupture of other materials, see Part III.
It must be clearly understood that M will vary according to the arrangement
of the beam and loads. Its different values are given at pp. 27 to 40, also in a
concise form in Appendix VII. (see also Appendix XXI.)
BEAMS— SHEARING STRESS
55
SHEAEING STEESS.
We have hitherto considered only the effect of the direct
stresses upon a beam, that is, the stresses which eventually act
upon the fibres by extending or compressing them in the direction
of their length.
There are, however, other stresses acting upon a beam the
existence of which will be rendered evident upon considering the
next three diagrams.
Vertical Shearing Stress. — If a beam AB, Fig. 76, supported
at the ends, be bent, the fibres above
the neutral layer are evidently com-
g pressed and those below it extended,
but no other stresses in the beam
are apparent to the observer.
If, however, the continuity of the fibres be interrupted by
cutting the beam into vertical slices, these slices, when the beam
is loaded, will slip one past the other, as shown in Eig. 77.
In the uncut beam there must be the same tendency for each
section to slide vertically upon the next, although actual sliding
is prevented by the continuity of the fibre or cohesion of the
Fig. 77. Fig. 78.
particles. The force causing this tendency is called the vertical
shearing stress.
Horizontal Shearing Stress. — In the same way, if the beam
be divided into different layers of fibres by cutting it into planks,
the planks or layers of fibres will slide upon each other as in
Fig. 78, when loaded.
In the uncut beam there must be a tendency to slide in the
same way, and the force causing this tendency is called the
horizontal shearing stress.
Bules for finding Amount of Vertical Shearing Stress.^
The amount of the vertical shearing stress is easily ascertained
by the application of two simple rules —
^ The practical formulce for finding the shearing stresses are on pp. 57 to 60,
numbered (31) to (43c), each being in connection with figures showing the distribution
of loads, diagrams of shearing stresses, etc.
S6 NOTES ON BUILDING CONSTRUCTION
Fig. 79
is a
cantilever,
or
a beam
/ k
h
/
1
L
til
A t
S
Fig. 79.
e
B
1. Vertical Shearing Stress in Cantilevers. — The
shearing stress at any section is equal to the weight on the beam
between that section and the outer end.
2. Vertical Shearing Stress in Beams supported at each
End. — The shearing stress at any section is equal to the difference
between the reaction at either support and the weight between that
support and the section in question.
These rules hardly require any proof, but it may make them more clear
to examine a case of each.
SheariTU) stress in cantilevers. — AB,
loaded at the end with a weight W.
We may divide the beam into ima-
ginary vertical sections at ki, hg, fe;
these are only a few out of an infinite
number of sections infinitely near to
one another.
Now "W acting downwards tends to
shear off the piece B/, with a force W,
acting, as shown by arrow thus, on the section fe ; this is resisted by the
opposite surface of the section fe, the resisting force acting as shown by the
arrow -f- upwards.
Supposing then that the resistance prevents the shearing at surface fe,
consider the next surface hg. Here the load W still acts with a force W
downwards, tending to shear off the piece Hh.
Again in the same way at hi, W the load acts downwards, and a
resistance = W acts upwards, as shown by the arrows. Nothing has intervened
to increase or diminish the force W caused by the load.^ The same takes
place at any intermediate section, right back to Al.
The shearing stress at any section from B to A is therefore always the
same, i.e. = W as shown graphically in Fig. 82.
Shearing stress in supported learns. — As regards the second rule, take the
case of a beam (Fig 80) supported at the ends and uniformly loaded.
SupjDosing the beam to be 8' span and loaded with tu lbs. per foot run :
and each = 4u\
At the section marked the reaction 'Rj^ = 4w acts upwards ^, and is
W/ lUj Wg
Fiir. 80,
Tm. 81.
resisted by the four weights 4w acting downwards, so that the shearing force
at section 1 = R..
^ The weight of the beam itself does really increase it, but we have arranged to
neglect it as insignificant. If it is not insignificant it would cause a shearing stress
similar to that caused by a distributed load, as described on the next page.
BEAMS— SHEARING STRESS
57
At the section marked one-fourth of the upward force of the
reaction = 4i« is cancelled by the weight acting downwards, so that the
upward force at the section 2 is Zw, which is resisted by Zw acting down-
wards between it and the centre.
Similarly at section 3 the shearing force is (R_^ - 2w) = 4w-2w = 2iu,
and at section 4 it is Iw.
At section 5 (the centre) the upward force of the reaction is quite
neutralised by the four weights iw^, w^, w^, acting downwards and the shear-
ing force = 0.
Thus at each section the shearing force = — weights between the
section and support A, as stated in Rule 2 above. The same reasoning
applies to the part CB of the beam.
The shearing stress is evidently decreasing from Aiv at the supports to 0
at the centre, and may be graphically shown as in Fig. 8 1 .
The truth of the rules is so apparent, and their application
to different cases so simple, that it will be sufficient to show a few
of the most useful cases graphically, giving at the same time the
value of the shearing stress in terms of the weights and lengths
of the beams.
The shearing stress at any point P is denoted by Sp, and at
any other points A, B, C by S^, S^, S^, etc., respectively.
Graphic Representations ^ of Vertical Shearing Stress,
and Values for the Same.
Case 1 (Fig. 82). — beam fixed at one end and loaded at the
OTHEK.
Shearing stress S = W throughout . . (31).
/ _>1
--/-AT ^—
/
/-X
n ip
)) w
Fig. 83.
AND LOADED
Fig. 82.
Case 2 (Fig. 83). — beam fixed at one end
TJNIFOEMLY THROUGHOUT ITS LENGTH.
Sp = wx. The point P being distant x from B . (32).
8^ = wl (32 A).
Case 3 (Fig. 84). — beam fixed at one end, loaded at the
OTHER END AND ALSO UNIFORMLY.
A combination of Cases 1 and 2. The loads are omitted in the figure
for clearness, but are as in Figs. 82 and 83.
^ These are merely graphic representations of results obtained algebraically, but
in Appendix VI. will be seen a graphic method of obtaining the shearing stresses.
-rmi,'. j„^X..m ..4ta:>fl«1^
58
NOTES ON BUILDING CONSTRUCTION
Sj. = W + wx
(33).
(33 A).
Fig. 84.
Case 4 (Fig. 85). — beam fixed at one end and loaded with
SEVERAL concentrated WEIGHTS W^, Wg, Wg.
S, = Wi + W2 + W3 . . . (34).
S at any section = the sum of the loads between that section and the
outer end of the beam. ....... (34 A).
Case 5 (Fig. 86). — beam fixed at one end and loaded
UNIFORMLY OVER A PART OF ITS LENGTH.
The shearing stress at any point P under the load distant x from A is
^^ = w{z + y -x), . . . (35),
and Sj^ = 'wz, . . . . . (35 A).
S^^wz (35 B).
w
n lut it
It 1
wlliilllll
Fig. 87.
Fig. 86
Case 6 (Fig. 87). — beam supported at both ends and loaded
IN the centre.^
w
S throughout =
(36),
except at C where the stress changes direction and
S, = 0. . . .(36 A).
The arrows indicate the different directions of the shearing stress 011 the
opposite side of the centre. Thus
S anywhere in CA the stress is
thus tending to shear the
^^S- 88. Fig. 89. ^g^j^ -pig. 88, whereas in
the half BO the stress is thus tending to shear the beam as in Fig. 89. At
C the stresses in opposite directions cancel one another, and there is no shear.
^ See Appendix V.
BEAMS— SHEARING STRESS
59
Case 7 (Fig. 90)! — beam suppokted at the ends and uniformly
LOADED.
See the remarks on p. 56 with, regard to this case.
Case 8 (Fig. 91). — beam supported at both ends, loaded in the
CENTRE AND ALSO UNIFORMLY.
This case is a comhination, of 6 and 7. The load is oraitted for clearness.
wl W
^ wl W
(38).
. (38 A).
S(,= 0 (stress changing direction) (38 B).
Case 9 (Fig. 92). — beam supported at both ends and loaded at
Between A and D
Between B and D
ANY POINT.
(39).
(40).
At D
Sj, = 0 (stresses changing direction) (40 A).
6o
NOTES ON BUILDING CONSTRUCTION
Case 10 (Fig. 93). — beam supported at both ends and
PARTIALLY LOADED WITH A UNIFORM LOAD.
Between A and E
Between B and F
(I — mf —
At any section P under the partial load distant x from A
(I — lllf — TC^
21
(x-m) > V)
(41) .
(41 A).
(42) .
Referring to p. 38 it will be a good exercise for the student to deduce the previous
cases of beams supported at both ends from this case.
Case 11 (Fig. 94). — beam supported at both ends and loaded
"WITH ANY NUMBER OF CONCENTRATED LOADS.
Between A and D
Between D and E
Between E and F
Between F and B
^ — — I — + I
S = R,-W
S = - W - W,
S = R =
Wm, W„m„ Worn,
I
+
3"''3
I
(43).
(43 A).
(43 B).
(43 C).
U— -
t<-
U—
.i.vt2 A
.f/ia !-
Horizontal Shearing Stress. — We have hitherto considered
only the vertical shearing stress, which tends to make sections of
a loaded beam slide vertically npon one another, as in Fig. 77.
We have now to consider the horizontal shearing stress, which
tends to make the layers slide horizontally upon one another, as
in Fig. 78.
BEAMS— SHEARING STRESS
6i
Fig. 95.
This horizontal shearing stress is at every point in the learn
equal to the vertical shearing stress at that point.
Tliis is proved thus : — On or near the neutral layer of an unloaded rect-
angular beam draw a little square, Fig. 95.
If the beam is then loaded until it bends
slightly, this square will be distorted into
a rhombus, Fig. 96. Now we know that
at the neutral layer of a rectangular beam
there are no direct stresses.
The distortion must therefore have been
produced by the vertical shearing stresses
V, V acting as shown by the arrows, and it is opposed by the horizontal
shearing stresses H, H. Now since there is equilibrium, H and H must evi-
dently be equal to V and Y, or the horizontal shearing stress must be equal
to the vertical shearing stress at the point. ^
We know then the total amount of shearing stress at each section — the
vertical is found as before explained and the horizontal is equal to it.
Distribution of Shearing Stress. — It is necessary to know
further how these shearing stresses are distributed over the section.
It can be proved mathematically that the shearing stress is
distributed over the section, as shown in Tig. 97.
The curve is a parabola,
the relative proportion of
whose ordinates is shown in
the diagram. The area^ of
the parabola represents to scale
the total shearing stress at the
section.
From this the horizontal
shearing stress at any point
can easily be ascertained. Thus
suppose the total shearing
stress S at any section of a
beam 1 2" deep is 2 tons. Then
I NL X CD = S,
f X NL X 12" = 2 tons,
NL == 2 tons X 1^ X y 5
_ 6
— 2 4'
= i ton.
Having thus found the stress on NL to be ^ ton, we make a scale on
which the length NL = ^ ton, and we can find the stress at any other point
by measuring the ordinate of the parabola at that point.
It is unimportant in most cases to know the exact distribution of the
shearing stress over the section, but we can see from the diagram that it is
greatest at the centre and vanishes at the top and bottom layers of the
beams.
^ Cunningham, p. 244.
- The area of a parabola = | the circumscribed rectangle. The area = | x CD x NL.
62 NOTES ON BUILDING CONSTRUCTION
This is exactly the converse of the distribution of the direct stresses.
The difference is clearly shown by the diagrams, Figs, 98 and 99.
Direct Stresses
Neutral
u
iMyer
U F
Fig. 99.
We see from these diagrams that the shearing stress is greatest along the neutral
layer where the direct stress is the least, and least at the upper and lower fibres where
the direct stress is greatest.
In Fig. 99 the amount of the shearing stress is exaggerated in order that
it may be clearly shown, but its amount is generally very small in com-
parison with, the direct stresses ; so small that in rectangular beams whicli
have so much, more substance near the centre than is required to meet the
direct stresses, there is sure to be plenty to meet the small shearing stress,
and its consideration may therefore safely be neglected.
In built-up iron beams, however, it is different ; in those the shearing stress plays
an important part (see p. 157).
BEAMS OF UNIFOEM STEENGTH.
If we glance at the graphic representation of the bending
3
moments at the different points of
a beam supported at the ends and
uniformly loaded throughout its
length. Fig. 100, we see that the
bending moment is greatest at the
^^g- centre and gradually diminishes to-
wards the ends, and that in fact exactly at the points of support
there is no bending stress whatever.
If, therefore, the beam be made of the same section through-
out, it is evident that it is unnecessarily strong at every point
except the centre, and that it may be reduced in section as it
approaches the points of support, because the bending moment
becomes less and less as those points are approached.
There are three ways in which the beam may be reduced so
as to make it of equal strength throughout its length —
1. By varying the depth, keeping the breadth the same ; as in
Figs. 102-105.
2. By varying the breadth, keeping the depth the same ; as in
Figs. 106-109.
BEAMS OF UNIFORM STRENGTH 63
3. By varying both breadth and depth.
The same observations apply to other cases of loading. Thus,
for example, in a cantilever of rectangular section, with either a
single or a uniform load, there is superfluous material except near
the point of fixing (see Figs. 102, 103), and in a supported
beam carrying an isolated load there is superfluous material except
under the load.
In all such beams then a certain portion of the material is
superfluous, and can be cut away without injuring their strength.
In timber beams, as a rule, it does not pay to cut away this
superfluous material, the cost of labour being greater than the
value of the material saved ; but in some cases for the sake of
appearance, or in large beams to reduce the weight, it may be
desirable.
It can, however, never be necessary to cut timbers to the
exact curves shown. These curves are the forms which are theo-
retically correct, but practically they are valuable only as dividing
the parts of the beam which have no excess of strength from
those parts which are unnecessarily strong and can therefore be
cut away, thus making a beam of approximate uniform strength.
The following figures show the shapes of beams of equal strength through-
out their length so far as the direct stresses only are concerned, and in each
case ignoring the shearing stress and the weight of the beam itself.
The shearing stresses would necessitate some substance being left on at the
ends of the supported beams.
It is not difficult to prove mathematically that beams of equal strength
throughout their length are of the forms shown, but such mathematical proof
would here take i;p more space than can be afforded for it, and the question
is not one of much practical importance. Case 1 (p. 28) is, however, worked
out as follows, and the other cases can be worked out in a similar manner.
Calculation to ascertain the form of a cantilever of equal strength throughout
its length, loaded at one end and of the same breadth throughout.
Bending moment at any point P (Fig. loi) distant x from the end W = Wx.
Moment of resistance at P = ^fy^.
Fig. 101.
64 NOTES ON BUILDING CONSTRUCTION
Wl'
X
V
Then since at any point bending moment = moment of resistance, we have —
At P : Wx = i/;%2. At A : = ^M^-
t.
which is the equation to a parabola with its vertex at B ; the under side of
the beam would therefore be shaped to this curve.
Shapes of Beams of uniform Strength.
a. WHEN THE BREADTH IS CONSTANT THROUGHOUT THE LENGTH OF THE
BEAM, BUT THE DEPTH VARIED TO SUIT THE VARYING STRESS.
Figs. 102-105 show the disposition of load and the shape of the beam
in elevation, the dotted line being the beam of constant depth and breadth
as calculated.
In Fig. 102, AB is a parabola with the vertex at B.
In Fig. 103, AB is a straight line.
Fig. 102. K ' Fig. 103.
In Fig. 104, AB and PB are parabolas with their vertices at A and B.
Fig. 104.
In Fig. 105, ABC is a semi-ellipse, AB being the major axis.
BEAMS OF UNIFORM STRENGTH
65
A '^-^-^^"^ ---.-.^c^-- -
1 ✓ ^ — ■ — ^ ^
A
< 1
Jf- .X.
f ^ \
r
Fig. 105.
I. WHEN THE DEPTH IS KEPT CONSTANT THROUGHOUT THE LENGTH OF THE
BEAM, AND THE BREADTH VARIED TO SUIT THE VARYING STRESS.
Figs. 106-109 show the disposition of load and the shape of the beam
in plan.
In Fig. 106, ABD is a triangle.
In Fig. 107, AB, DB are both parabolas with vertices at B.
In Fig. 108, DAC and CBD are two triangles with their bases at DC.
In Fig. 109, ACB and ADB are parabolas with their vertices at the centre
points C, D.
66
NOTES ON BUILDING CONSTRUCTION
DErLECTION.i
In many structures it is necessary that the beams, whether of
wood or iron, forming part of the structure, should be not only
strong enough, but stiff enough ; that is, not only must rupture
be prevented, but the beams must not bend too much.
For example, it would never do if tlie joists supporting a ceiling were to
bend beyond a certain small amount under the load, although they might be
strong enough, because a very slight bending is sufficient to cause cracks in the
ceiling below. Such joists must therefore be stiff as well as strong.
It is necessary, therefore, to be able to calculate how stiff a given
beam is, that is, how much it wiU bend under its load, in order to
be able to judge whether the amount of bending is likely to cause
inconvenience ; and if so, to increase the size of the beam in such
a manner as to prevent such excessive and inconvenient bending.
The amount of deflection that may be expected in a beam of
known form and material, with a given amount and distribution
of load, can easily be ascertained by the use of the formulae given
below. The investigation of the formulae, however, involves the
use of the Calculus ; their proof will therefore not be attempted
in this work.
The student must be content to accept these formulas, which
are based upon those given by Professor Eankine, after investiga-
tions made by himself and other mathematicians.
Formulae for ascertaining the Deflection of Beams of any Kind.
General Formula. — Th e general formula applicable to all kinds of beams,
of any material and with various distributions of load, is
Fig. 110.
In this formula
A = the maximum deflection in inches. Thus in Fig. iio, if the beam
AB deflect as shown by the arc Ai/B, the line xy is denoted in the above
formula by A.
W = the total load on the beam.
^ The formulae for practical use are Epilations 44, 45, 46, 47, pp. 66, 67, 68, and
in Appendices VIII., IX., XXI. Examples are given, pp. 76 to 80.
BEAMS— DEFLECTION 67
I = the length of the beam in inches.
E = the value of the modulus of elasticity of the material expressed in
the same units as W.
I = the moment of inertia about the neutral axis of that section of the
beam where the greatest stress occurs with the given distribution of load.
n IS a coefficient, the value of which varies for each class of beam and for
each distribution of the load.
The following Table shows the different values of n for three classes of
beams and for four distributions of load.
TABLE C.
Arrangement of Beam and Load.
Class of Beam.
1
Uniform cross
section
throughout its
length. See
Figs. 61 and
127.
2
Uniform
strength. Same
depth through-
out its length.
See Figs. 106 to
T09.
3
Uniform
strength.
Same breadth
throughout
its length.
See Figs. 102
to 105.
Fixed at one end, loaded at the other .
j; loaded uniformly
Supported at both ends, loaded in the middle
) ) , . loaded uniformly .
-
1
a
1
■S"
6
1
\
1
1
1
1
1
1
Maximum deflection under any load.— To find the maximum deflection of
any beani under a given load it is merely necessary to substitute the values
ot the different letters, and the result gives the value of A.
The values of W and I are of course known ; E is given in Table I •
and I, the moment of inertia, can be found by means of the information
given m Appendix XIV.
The deflection under proof load, or under loads which produce a known
stress, may be ascertained from the following formulae
n'r^l^
A = -
for all beams, and
^_ n'.(rc + rt)l^
(45)
for beams with cross sections of equal strength. 1
In these formulae r,, is the greatest stress on the weakest side of the
beam ; ijo is the distance of the neutral axis from the extreme fibre of the
beam on that weakest side at the section of greatest stress ; r^, are the
limiting stresses in compression and tension respectively ; dg the depth of the
beam at the section of greatest stress. The other letters have the same signi-
fication as in Equation 44.
The following Table shows the different values of n' for three classes of
beams and for four distributions of load.
1 These are beams of such a form that the limiting stress is reached on the upper-
most and lowermost fibre of the beam at the same instant.
68 NOTES ON BUILDING CONSTRUCTION
TABLE D.
Arrangement of Beam and Load.
Class of Beam.
1
Uniform cross
section
throughout its
length. See
Figs. 6ia and
126.
2
Uniform
strength and
same depth
throughout its
length. See
Figs. 106 to
109.
3
Uniform
strength and
same breadth
throughout
its length.
See Figs. 102
to 105.
Fixed at one end, loaded at the other .
,, uniformly loaded .
Supported at both ends, loaded in the middle
,, uniformly loaded .
1
4
1
5
1
8
1
?
1
2
!
Simpler Formula. — For ordinary rectangular beams of the same section
throughout their length, such as wooden beams nearly always are, the for-
mula 44 can be much simplified by su.bstituting for the moment of inertia
I its value for a rectangular section, namely y--. (See App. XIV. and IX.)
1 'in'Wl^
Equation 44 becomes ^~~E6d^ .... (47).
Comparison hetiveen strength and stiffness. — We found that the strength of
bd^
rectangular timber beams varied as fo-f clii'ectly as the breadth, as the
square of the depth, and as the modulus of rupture of the material, but
inversely as the length.
By examining Equation 47 we see that the amount of deflection of
P
rectangular beams varies as ^-^3 : that is, the greater the length the greater the
deflection ; the greater the modulus of elasticity, the breadth, or the depth, the
less the deflection.
The deflection arises from want of stiffness, therefore the stiff7iess or resist-
ance to deflection will vary according to an exactly opposite set of condi-
EM^
tions— in fact, it varies as —^3- . It will be greater directly as E is greater
(thus an oak beam, for which E is greater than for a similar fir beam, will be
stiffer than the fir beam), also directly as the breadth and cube of the depth
are greater ; but, on the other hand, it will be less as the cube of the length is
greater : that is, of two beams of the same section and material, the longer
will be less stift" than the other, not in proportion to its length, but of the
cube of its length.
Recapitulating, we see that for beams of rectangular cross section
The strength of beams varies as — ^ — ,
„ stiffness „ „
Further, since the strength varies as bd^ and the stiffness as bd^, we see that
by reducing the breadth and increasing the depth in such a Avay that the strength
remains constant, we can obtain stiffer beams. Thus a beam 4 inches broad
and 6 inches deep has the same strength as one 1 inch broad and 1 2 inches deep
FIXED BEAMS
69
(4 X 62 = 1 X 122), ijut 4 X 6^ = 864 and 1 x 12^ = 1728, so that the second
beam is twice as stiff as the first, though it only contains half the material.
Amount of deflection allowable. — This depends greatly upon the nature of
the structure. It is generally considered that for floors, or roof-timbers sup-
porting ceilings, the maximum deflection allowable is -^jy" per foot of span.
For roofs or floors without ceilings a somewhat greater deflection, say per
foot run, and for rough or temporary structures per foot of span may
be allowed in the case of supported beams.
In cantilevers supporting ceilings the deflection allowable should be
per foot of length, without ceilings and for rough structures per foot
of length.
It will be seen that these allowances make the depression at the end of
the cantilever the same as that in the centre of the supported beam.
FIXED BEAMS.i
The general investigation of the formulse for the calculation
of beams rigidly fixed instead of being merely supported, at one
or both ends, involves mathematical knowledge quite beyond the
province of these Notes.
The important points of difference between fixed and sup-
ported beams can, however, be explained in an elementary manner,
and the student will be enabled intelligently to use the formulae
connected with the calculation of such beams, and indeed to make
the calculations in an approximate manner without having recourse
to the formulae at all.
Nature of stresses upon a beam fixed at the ends. — The stresses produced by
loads upon a beam, whose ends are rigidly fixed, are different both in distribu-
tion and amount from those produced by the same loads upon a beam whose
ends are merely supported.
Thus in a beam supported at the ends and loaded in the centre, as in
Fig. 1 1 1, we know that the beam deflects, forming an approximately circular
YW
Fig. 111.
arc,2 the part of the beam above the neutral layer is under compression (ccc),
the part below in tension (ttt).
If, however, the ends be rigidly fixed and prevented from tilting up, as
in Fig. 112, the curve of deflection caused by the weight is quite different.
1 Theformulce for practical use are given in Equations 48 to 52, pp. 71, 72, and
in Appendix VIII.
2 For beams of uniform section the curve of flexure is nearer an hyperbola when
the load is at the centre.
70
NOTES ON BUILDING CONSTRUCTION
From A to / and from B to i tlie bottom surface of tlie beam is concave ;
from / to i it is convex.
Fig. 112.
Moreover, the stresses upon the beam are different from those o; the
supported beam. From A to /and from B to t the upper surface is in teision
(itt), the lower surface is in compression {ccc). From f to i the upper surface
is in compression (ccc), and the lower surface is in tension (tU).
It will be seen that at the points / and i, where the curvature changes^
the nature of the stress also changes ; the upper surface, which from A to /
and from B to i was in tension, changes to compression, and the lower sur-
face, which was under compression, to tension.
These points, where the curvature and the stress change, are called the
points of contra-flexure.
Their distance from A and B varies according to the shape of the beam
and the nature of the load. When that distance is known, the calculation of
the bending moments, etc., becomes a simple matter.
In fact, the beam thus fixed and loaded is exactly in the condition of two
cantilevers A/ and Bi, carrying a beam fi between them, which is supported
at its ends / and i by hanging from the ends / and i of the cantilevers.
The cantilevers may be calculated as shown in Case 1, p. 28, and the
supported portion as in Case 6, p. 32.
The only difficulty, therefore, is to ascertain the exact distances of the
points of contra-flexure from A and B.
Position of points of contra-flexure in fixed teams with different
distributions of loads. — Without making any attempt to describe
the mathematical investigations by which the positions of the
points of contra-flexure are ascertained, we will give their positions
for different classes of beams with various distributions of loads.
TABLE E.
Distance or Points of Contra-Flexure.
Arrangement of Beam and Load.
Class of Beam.
No. of Fig.
Uniform cross
section.
Uniform strength,
with
uniform depth.
Distance of points of contra-flexure.
Fixed at both ends. Load in centre .
„ ,, Loaded uniformly
Fixed at one end andl ^oad in centre .
supported at the other )
Fixed at one end and ^ r j j
, , , , , . , \ Loaded uniformly
supported at the other 1
0-25 I from A and B
0-211 I „
0-273 I from A
0-267 I from A
0-25 Z from A and B
0-25 I i,
0-33 I from A
0-33 I from A
Fig. 113
Fig. 115
Fig. 117
FIXED BEAMS
71
It will be seen that in beams fixed at both ends the points of
contra-fiexure vary from '2111 to -25 Z from the ends; and in
beams fixed at one end and supported at the other the point of
contra-fiexure is from -267 ^ to -83 ^ from the fixed end; so that
the distance of the point of contra-fiexure from the fixed end of
any beam may, for approximate calculations, be taken without
much error at ^ Z.
Various Cases of Fixed Beams.
One or two cases may now be calculated as examples : —
Case 1. BEAM OF UNIFOKM CEOSS SECTION FIXED AT BOTH
ENDS AND LOADED AT THE CENTKE.
The points of contra-fiexure will be 0-25 I from the walls A and B (see
Table E).
It is evident that A/ and Bi, Fig. 1 1 3, are in the condition of canti-
W
levers ; each loaded at the end with a weight besides the weight of tlie
beam fi which may be ignored.
. •. = weight X leverage,
= 17x7=^ ■ . • ■ (48).
W I Wl
(49).
I
The portion fi is like an ordinary supported beam of a length equal to -
and loaded with W in the centre.
W
The reaction at / and at i will be — .
. •. = Keaction x leverage,
(50)
Graphically the moments of flexure may be shown as in Fig. 1 1 4.
72
NOTES ON BUILDING CONSTRUCTION
Shearing stresses. — In this case the shearing stresses are the same as in a
beam supported at both ends and loaded in the centre (see Appendix VIII..)
Deflection.— It can be shown that the maximum deflection is ^th of tlhe
maximum of a similar beam supported at both ends (see Appendix VIII.)
Case 2. — beam of uniform cross section fixed at both ends
AND UNIFORMLY LOADED.
In this case the points of contra-flexure are distant -211? from A and IE
Fig. 115.
1 —
1
'■Unifoj'vi load "
■>■
C
— 0-5781-
->T<— 0-311 /—>.^
■■0-083wl^ =
Fig. 115.
= weight distributed on cantilever x leverage + weight at end
X length of cantilever,
= wx-2lUx—^ + X 21H,
= 0-022w;Z2 + 0-061^^2^
12 (51)-
= Reaction x leverage - weight between C and / x leverage,
•5781^1 -5781 -blSwl -5781
~-~2~''—2
_ 0-334?/7Z2 wl^
8 -24 • • (52).
12 24 values of and M„ as can be proved bpy
another method; the fraction 0-211 is only approximate.
FIXED BEAMS
73
Graphic Representation. — Graphically the tending moments may be shown
in the shaded portions of Fig. 1 1 6.
Fig. 116.
The curve is a parabola whose centre ordinate
^d^ wl^ rvl'^
As in the preceding case, the shearing stresses are the same as in a beam
supported at both ends and uniformly loaded (see Appendix VIII.)
Deflection. — It can be shown that the maximum deflection is i-th of
the maximum deflection of a similar beam supported at both ends (see
Ap>]3endix VIII.)
Case 3. — beam of uniform cross section fixed at one end,
SUPPORTED AT THE OTHER, AND LOADED UNIFORMLY.
In this case the beam is like a cantilever A/ and a supported beam /B, as
shown in the portion Ai of Fig. 115.
The point of contra-flexure / is at -211 the span ^ from A in Fig. 1 1 5 and
Ai =-789Z, therefore in this case Af=^ll = •261l.
The upper portion of A/ is evidently in tension and the lower portion in
coimpression ; the upper portion of B/ is in compression, the lower portion
in tension.
In the figure the portions in compression are shown in thick lines, those
in tension in thin lines. They will be practically the same for a beam
loaded in the centre.
Fig. 117.
Bending moments for fixed beams with, different distribution of loads. —
It is unnecessary to work out any more cases. The bending moments at
dilfferent points for three classes of beams, with one or both ends fixed and
wiith different distributions of load, are given i n the following Table.
^ Approximately (see p. 70).
74
NOTES ON BUILDING CONSTRUCTION
TABLE F.
Class of Beam.
Arrangement of Beam and Load.
Uniform cross section.
Bending moment
Uniform strength, with
uniform depth.
Bending moment
At fixed ends.
In centre.
At fixed ends.
In centre.
Fixed both ends. Load in centre
!, ,, Loaded uniformly
» Loaded uniformly
and in centre
Fixed one end and "1 t i •
supported at other /^"^^^ centre
if", IUZ2
J W/
IS
In centre of
supported
portion
Loaded uniformly
J wl-
In centre of
supported
portion
6
— ""^'-'v^ >JV.^>-- J.i-pp^jJ.±>_liA Y XXJ..
Oljections to fixed learns in practice. — Fixing the ends of beams
is generally objectionable in practice. In the case of timber, con-
fining the ends keeps them from the air and causes them to rot ;
in the case of iron beams it prevents them from expanding and
contracting freely. ^
^ There are other reasons against fixing special forms of beams,
which will be mentioned in treating of those forms.
As fixed beams are rarely required in practice, it is not
necessary to go farther into the subject to consider their deflection;
shearing stresses, etc.
Enough has been said to show the student that the stresses
in a beam fixed at the ends are very different from those in a
supported beam, not only in amount but in distribution.
Continuous Girders. When a beam or girder extends withouit
break m itself over two spans or more it is said to be continuouts.
Fig. 1 1 8 shows to an exaggerated degree the curves formed by
Fig. 118.
a uniformly loaded girder supported over two spans, AB, BC, ancd
fixed at the ends.
It will be seen that the general arrangement resembles twfo
pairs of cantilevers, A/, Bt and B/, Ck, and between them two beams?,
fi, jk, supported at the ends.
^ The fornudce for practical use are given in Apjiendix XI.
BEAMS
75
The upper portions of A/, Bi, B/, and C/c are evidently in
tension, the lower in compression; whereas in fi, jh the upper
P'ortions are in compression and the lower in tension.
The points of contra-flexure are at /, i, j, and k.
Again, if the ends of the girder are not fixed, but merely sup-
p.orted, the curves will be as shown in an exaggerated form in Fig.
f tt B t I
Fig. 119.
1. 1 9, and the portions in compression and tension respectively
a.re shown by the thick and thin lines as before.
It will be seen that the general curves of each span in Fig. 1 1 8
r'esemble those of a beam fixed at both ends (see Fig. 1 1 2), and
tihe curves of each span in Fig. 1 1 9 resemble those of a beam
fiixed at one end and loaded at the other, as shown in Fig. 1 1 7.
The distances of the points of contra-flexure from the
aibutments, and the value of the bending moments, vary according
t'-o the section of the girder, the distribution of the load, etc.
The calculations connected with continuous girders are very
cjomplicated, and not suited for a work of this land. Enough
imformation has been given above to indicate which portions are
i:n compression and which in tension, and that is all that is
r-equired by the syllabus ; but as cases occur of bressummers con-
tinuous over two spans and of rafters continuous over many spans,
s5ome further information on the subject is given in Appendix XI.
EXAMPLES.
In order to illustrate the application of the rules given in
t:he previous pages, it will be well to give in some detail the
c3alculations necessary for a few examples of such timber beams
ais are used in practice.
Before making calculations it is always desirable to state the
^preliminaries with great care. These include all known par-
tticulars relating to the case, such as the span, the nature and dis-
tbribution of the load, also the moduli of rupture and elasticity of
tthe materials, etc. etc.
Great care must be taken also that the dimensions, weights,
eetc, are all expressed in the same units, i.e. that feet and inches,
76 NOTES ON BUILDING CONSTRUCTION
tons and pounds, etc., are not mixed up in the formulae ; for if so,
grave errors will ensue.
Timber Cantilever for Balcony (uniform cross section).
Example 5.- — To find the dimensions for oak cantilevers to support a balcony
to carry a dead load.
Conditions. — A temporary balcony 30 feet long and 6 feet wide is sup-
ported by 11 cantilevers of English oak, built into a wall and projecting 6
feet from it.
The weight of the balcony platform is about 20 lbs. per foot superficial,
and it is liable to a dead load of about 100 lbs. per foot superficial.
Find the scantling necessary for the cantilevers if they are made of
uniform rectangular section throughout : (1) for strength ; (2) for stiffness ;
the d'eflection being limited to -^fj" per foot.
Preliminaries. — Here each cantilever, except the one at each end of the
balcony, has to support an area of 6' x 3' = 1 8 feet superficial.
The load on this area is
Platform 20x18= 360 lbs.
Load 100 X 18 = 1800 „
Total = 2160 lbs.
Let h = the breadth of the beam. [A breadth convenient for fixing the
platform to, 3" or 4", may be assumed, say 4".]
Z = the length of the beam = 72".
2160
w = load per inch run of the beam = ■ = 30 lbs.
/ ^
/o = an ordinary modulus of rupture for English oak= 10,000 Ibs.^
E = an average modulus of elasticity for English oak= 1,200,000 lbs.
F = the factor of safety for a temporary timber structure with a dead
load = 5.
Calculation for strength. — By referring to Fig. 39, Case 2, p. 29, we see
that the bending moment is greatest at A — the point of fixing. As the
cantilever is to be of uniform section throughout its length, if it is strong
enough to resist this bending moment at A, it will be more than strong
enough in all other parts.
Now we know that at any point the bending moment = the moment of
resistance when the forces are in equilibrium. Introducing the factor of
safety F, by multiplying it into iv, we have (Equations 6, p. 29, and 29, p. 52)
(Fw)P_ bd^
— a Jo '
rf2 =
6 '
_6 X 5 X 30 lbs. X 722
~ 2 X 10,000 lbs. X 4"'
= 58-3,
(Z= 7'6 inches. 2
^ It is to be understood that moduli of rupture, elasticity, resistances to tension
and compression, etc. etc., used in this and all other examples, are taken from Table I.
2 If & had been taken at 3 inches, d would have been 8 '8 inches, a stiffer beam
with less timber in it.
BEAMS
77
This is the depth necessary when strength only is considered.
We must now see if it is sufficient for stiffness.
Calculation for stiffness— To ascertain the deflection of this cantilever
UnWl^
when fully loaded, the general formula (equation 47) is A= -gyg • -l-iie
beam is of uniform cross section fixed at one end and uniformly loaded, there-
fore w=i (see Table C, p. 67, col. 1, line 2).
12 x|x 2160x 723
•'• ' 1,200,000 lbs. X 4" X 7-6^'
= 0-57 inch.
Since the deflection is to be limited to ^" per foot run of the beam (see p.
69), le. for the whole, it is clear that a beam 4" x 7-6" is not stiff enough.
To find the necessary depth put A =
We have
g „ 12nWP
Transposing,
, 40 X 12 X i X 2160 X 723
6 X 1,200,000 X 4
= 1680,
fZ= 11-9, say 12".
This depth would be adopted, being greater than the depth (7-6") required
for strength only. Therefore, to obtain the necessary stiffness, a beam
4"x 12" is required, supposing that it is decided the breadth shall be 4".
If a breadth of 3" is sufficient, then it will be found by repeating the above
calculations that the dimensions are
(1) For strength, 3" x 8f",
(2) For stiffness, 3" x 13".
Timber Cantilever for Balcony (uniform strength,
uniform breadth).
Example 5 A. — To find the dimensions for oak cantilevers to support a
balcony to carry a dead load.
Conditions and Preliminaries. — Same as in Example 5; but supposing that
for some reason it is desirable to have the cantilevers for the balcony, described
in Example 5, as light as possible, or that it is thought that their appearance
would be improved by making them of uniform strength, keeping the breadth
the same but varying the depth.
For strength the depth of the beam at the wall would be as before 7 -6".
As regards stiffness, to find the depth for the allowed deflection ^V', we use
the same formula as before, but the value of » is ^ instead of | (see Table C,
p. 67, col. 3, line 2).
We have
^ „_12nW;3
^ Ebd^ '
40 X 12 x-|x 2160 x 723
6 x 1,200,000 x 4 '
= 6718,
d=l8-9,
= 19 inches nearly.
78
NOTES ON BUILDING CONSTRUCTION
Referring to Fig. 103 we see that the theoretical form of the beam would
be as shown in Fig. 120.
The calculated deptJi is
given at AC, where the beam
is built into the wall, and
the line joining CB is the
under side of the beam.
In practice the end
would not* be brought to a
feather edge as shown at B,
but would have a thickness
there of an inch or two, as
shown in Fig. 121. It will
be seen that in most cases
there would be no advan-
tage in using a cantilever
of uniform strength, for
though the theoretical form
would contain less timber,
still, practically, it would
contain nearly as much, and
would involve more labour
than the beam of uniform
depth throughout.
It would be better,
Fig, 121. therefore, to iise the latter
in ordinary cases.
Balcony with live load. — It will be noticed that the load on the balcony
has been taken as a dead load.
If, however, the balcony is liable to be occupied by a crowd of excited
people moving about, it would be safer to consider them as causing a live
load. This live load should be reduced to a corresponding dead load by
doubling it. The total weight would then be as follows
Weight of platform as before = 360 lbs.
Weight of people 2 X 1800 =3600 „
Total weight = 3960 lbs.
Taking this value for the weight instead of 2160 lbs., the calculation would
proceed as before with the following results : —
Uniform cross section
(1) For strength, 4" x 10-4",
(2) For stiffness, 4" x 14-5".
Uniform strength
(1) For strength, 4" x 10-4",
(2) For stiffness, 4" x 23-1".
In practice it would be preferable to adopt a framed cantilever either of
wood or iron. See also Examples 2 and 10.
Example 6 -Timber Beam loaded in Centre. — To find the load a given
baulk will carry. A pitch-pine baulk of 20 feet span 12" wide 12" deep is
BEAMS
79
to be loaded in the centre with a dead load. It is to be used for temporary-
purposes. What weight will it safely carry ?
Preliminaries. —
Z = 120" 6=12" d=12",
/„= 1100,
F = 4. Factor of safety.
Introducing F we have from Equation 13, p. 33, and Equation 29, p.
52, or from Appendix VII.
_ 11,000 X 12 X 122 X 4
~ 6 X 120 X 4 '
= 3 tons (about).
Fir Joist to carry given Load.
Example l.—To find the scantling for a fir joist in a single floor.
Conditions. — A joist of Baltic fir has a span of 20 feet, and is to carry a
uniformly distributed dead load of 2400 lbs. Find the scantling it should
have for (1) strength and (2) stiffness.
This case will not be worked out in full, as it is a very simple one.
Preliminaries. ■ —
1=120 inches
for Baltic fir = 6600 lbs.
E „ =1,444,000 lbs.
2400 lbs.
w = — — = 10 lbs. per inch.
F = factor of safety = 8.
Calculation for strength. — Calculating for a beam of uniform rectangular
section throughout its length
(Fw)P _ hd^
8
6 X (Fw)l^
= 524.
Taking 6 = 3. d^=ll5, and d=l 3",
as 13" would be inconvenient. Take 6 = 4. rf^ = 131, and d = 1 1-4,
Calculation for stiffness. — To calculate deflection of the beam if 4" wide
and 11-4" deep. By Equation 47, p. 68
. , l2n'Wf
we have A = .
Ebd^
Here n = -^l^ (see Table C, col. 1 , line 4).
12 X 5 X 2400 X 2403
A =
384 X 1,444,000 x 4 x 1 1*43'
= -1 inch.
8o
NOTES ON BUILDING CONSTRUCTION
If, however, the beam is to carry a ceiling below, the deflection should
not exceed -^q" per foot of span, or = |" altogether.
To find the depth that the beam should have, so that the deflection
shall not exceed |", we have by Equation 47
1" =
2
d-.
Eb
1800,
: 1 2 inches (about).
Fig. 122.
TJseful Notes.
Strongest rectangular beam that can he cut from a round log (Fig. 122).
Trisect a diameter "AB of the log. From the points of tri-
section C, D raise perpendiculars CE and DF, cutting the cir-
cumference at E and F. Join AE, AF, FB, and BE ; then
AEBF is the strongest beam that can be cut out of the log.
It can be shown that
FB 1 „ ,
AF = V2~'^"^ """^"^y-
Stiffest rectangular beam that can be cut out of a round log
(Fig. 123).
Divide a diameter AB of the log into four equal parts at
the points C, O, and D. From C and D draw perpendiculars
cutting the circumference at E and F. Join AE, AF, FB,
and BE. Then AEBF is the stiffest beam that can be cut
out of the log, and it can be shown that
FB^ J_^4
AF-V3 7-
Proportion of breadth to depth in a beam. —
We have seen above that the ratio of breadth
to depth is :
For the strongest rectangular beam cut out
of a round log, 1 to fj^.
For the stiffest rectangular beam cut out of a round log,
1 to J2>.
When, therefore, a single rectangular beam has to be cut from
a round log, its proportions will be regulated by one of the
above rules, according to whether its strength or stiffness is more
important.
In practice, however, the beams used are not generally cut
simply out of round logs, and the proportions are governed by
other considerations.
The width is made sufficient for lateral stiffness and for fixing
the superstructure, and the depth as great as convenient under
the circumstances, taking care not to exceed the market size of the
timber.
Chaptek IV
WROUGHT IRON ROLLED BEAMS.^
rriHE best section for a beam made of wrought iron is very
different from that for a timber beam.
We have seen that the direct stresses produced upon the cross
section of a timber beam may be graphically shown as in Fig. 1 24.
They are greatest on the layers most remote c e
from the neutral axis, gradually decrease as they p
approach the axis, and dwindle to nothing at all
at the axis itself ~'
In a rectangular solid section, however, just
as much material is provided to resist the stress d f
at PQ or VX, where this stress is smaU com- -^^S- i^i.
paratively, as at CE or DE where it is largest, and indeed just
as much at the neutral axis where there is no direct stress at all,
but merely the maximum part of a comparatively small shearing
stress (see p. 61),
It is evident then that, so far as the direct stresses are con-
cerned, the portions COD and EOE are useless, and might be
removed. To resist the shearing stress a small portion must be
left near the neutral axis.
Nevertheless it would be undesirable to cut away these por-
tions in the case of timber beams, as in most cases it would destroy
the continuity of the fibres, and would lessen the resistance of
the beam to lateral forces, that is, to those forces acting hori-
zontally, which, although not as a rule calculated for (their amount
being uncertain), should be allowed for in practice. Moreover,
the wood obtained by thus cutting the beam would not pay for
the labour of cutting. But in iron, which has in any case to be
formed to a particular shape, it is evidently desirable for the sake
of economy of material and weight to make beams of the shape
best adapted to resist the stresses that will come upon them.
^ The. formulce for practical use are Equations 54, 55, 56, p. 87, 88, and in
Appendix XXI. Examples are given at pp. 88 to 91.
B.C. — IV. a
utral
82 NOTES ON BUILDING CONSTRUCTION
Wrought iron beams are therefore rolled of the form shown
in rig. 125.
c E CE being called the upper flange
Ki DF „ lower
I KM „ web
iv, I By this arrangement, as will be seen, the bulk of
^emmmx^ the metal is placed in the flanges where the greatest
Fig. 125. direct stresses exist, and these flanges are connected
by the web KM. If the thickness of the web were regulated by
the shearing stress only, the amount of which is comparatively
small, it would be made very thm.
Practically, however, for reasons which will soon be men-
tioned, the web is made thicker than is necessary to resist the
shearing stress.
Let us now take a section of an ordinary rolled iron beam
(Fig. 126), such as is kept in stock by merchants, and ascertain
the nature and distribution of the direct stress upon it.
Resistance of Wrought Iron.
Ultimate Resistance. — We know from Parts II. and III. that the breaking
stresses for average wrought iron may be taken as follows : —
Tension . . . 25 tons per square inch.
Compression . . . 16 to 20 „ ,,
Shearing ... 20 „
But as rolled beams are often of inferior iron the breaking stress in tension
should not be taken higher than 20 tons.
Working Resistance. — The working or limiting stresses that may with
safety be allowed in the case of rolled beams ^ are
In Tension ... 5 tons per square inch.
Compression . . . 4 „ „
Shearing . . . 4 „ „
These are lower than those sometimes allowed in the case of iron for roof
trusses.
To find the Moment of Resistance of an I Beam.
There are several different ways, more or less accurate, of finding the
moment of resistance of a rolled iron beam.
A simple, and at the same time accurate method, is founded upon exactly
1 In consequence of the iron in rolled beams being free from joints, welds, and
other sources of defective workmanship, it is the practice of some engineers to take
the working stresses upon them as follows : —
In Tension . . . . 6 tons per square inch.
Compression . . . 5 ,, ,,
Shearing . . . 5 ,, ,,
As such beams are, however, often made from inferior iron, it is better and safer to
adhere to the limiting stresses given above.
WROUGHT IRON ROLLED BEAMS
83
the same principles, and worked out in almost the same manner, as the cal-
culation described at p. 43 et seg. for a timber beam.
Graphic Method, inchiding Web. — The internal direct stresses produced
111 a wrought iron beam, slightly bent by a central load, are similar to those
produced m a timber beam.
The fibres above the neutral axis are in compression ; those below it in
tension.
_ The intensity of stress in the different layers of fibres is greatest in those
which are most remote from the neutral axis, and becomes smaller and smaller
as they approach the neutral axis, until, at the axis itself, there is no direct
stress at all.
The amount and distribution of tlie stresses may be graphically shown,
just as they are for a timber beam in Fig. 74.
Fig. 126 is the section of a rolled iron beam 10" deep, with flanges 4"
wide and |" thick, the web being also J" thick.
I
t... 1 .- k^'l- / .
V— j-A ions per sq.in.
h
3.G tons per sq.in.
Fig. 126.
Linear scale
J inch= 1 foot.
M' P' F 7/ /
Fig. 128. Fig. 127.
Linear scale I inch = 1 foot. Scale of stresses
Scale of stresses 1 inch— 16 to7is. 1 inch—l& tons.
The centre of gravity of this section will be at the central point n, and
the neutral axis will pass through it (see p. 44).
The limiting or working stress for wrought iron in compression being
4 tons, and in tension 5 tons per square inch, it is evident that the limit will
be first reached in the flange which is under compression, i.e. in the present
case (a beam supported at the ends and loaded) the upper flange.
To put this in other words : When the stress upon each flange is 4 tons
per square inch, the compression flange will be undergoing the limiting stress
allowed for it, but the tension flange will be undergoing only four-fifths of
the stress that it could rightly bear.
Fig. 1 27 is a diagram showing the intensity of stress on the different layers
NOTES ON BUILDING CONSTRUCTION
of fibres at the moment when the extreme fibres under compression at CE
are subjected to the limiting stress of 4 tons per square inch at ce.
When this intensity is called out, a similar stress df of 4 tons per square
inch occurs on the lowermost layer DF. This layer would safely withstand
5 tons per inch, but that stress cannot be called out without exciting a similar
stress of 5 tons per square inch on the uppermost compression layer CE,
i.e. a stress in excess of the limit of 4 tons per square inch.
A cross section of the rolled iron beam is shown in Fig. 128, with the
angles square for the sake of simplicity, and having marked upon it the
amount and distribution of the stress upon the section.
At CE the compression flange is 4" wide, the stress at CE is therefore four
times ce = 4 X 4 tons =16 tons.
At GH the flange is also 4" wide, and the stress four times gh = 4" x 3-6
tons = 14-4 tons.
At KL the web is | inch wide and the stress upon it at that point is =
^xgh = ^ ■xZ-Q = 1-8 ton. Two points k and I are thus obtained on the
diagram of stress, as shown in Fig. 129,
!^ which is an enlargement of part of Fig. 128.
|\ ;! The diagram of stress is completed (the
\\ l\ web being rectangular) by joining kn and
'■ (Fig. 128). A moment's consideration will
show that the same two points can be obtained
by dropping perpendiculars KK' and LL' on
to CE, and joining K'n and L'ti, the inter-
sections with KL are the points required.
Thus the hatched portion above the
neutral axis represents the amount and dis-
tribution of the compression on the fibres of
Fig- 129. the beam.
In the same manner may be constructed the hatched diagram of tensile
stresses below the neutral axis.
It will be seen at once that the result we have obtained amounts to
drawing Cn, En as far as their intersection with GH, and Bn, Fn as far as
their intersection with I, J, and then completing the diagram by joining K'n,
L'n, M.'n, and T'n. So that in practice Fig. 127 is not required.
Thus CoknlpE is the equivalent area of compressive stress. This area,
multiplied by 4 tons per square inch, will give the actual amount of resist-
ance to compressive stress.
Similarly the hatched figure DrmnpsF below the neutral axis, multiplied
by the intensity of stress 4 tons per square inch, gives the amount of resist-
ance to tension.
Now we know that in the case of the timber beam the moment of
resistance is equal to the stress of either kind multiplied by the distance between
the centres of gravity of the equivalent areas shown on the diagram.
It is, however, simpler in this case, and quicker, to take the parts of the
rolled beam separately.
Thus, in Fig. 128, the moment of resistance of the flanges is equal to
stress area CoknlpE x distance between the centre of gravity of CoknlpE and
of DrmnpsF.
The moment of resistance of the web = stress kin x distance between the
centres of gravity of kin and mpn.
WROUGHT IRON ROLLED BEAMS
Putting tliis into figures we have —
M of Flanges = x
85
Limiting
Stress
Distance between Centres of
Gravity of Areas CofcnZpE
and DmnpsF
tons
X 4 X
:|x||-x4x 9-53,
= 72-43.
9 "5 3 inches,
M of Web =
Stress Area
kin
^ Limiting Distance between Centres of
Stress Gravity of Areas
^ I 215"" ^ ^2" ^ i I X 4 tons X 6 inches,
= |ix4x6,
_ =24-3.
Therefore, M of Flanges and Web = 72-4 + 24-3,
= 96*7 inch- tons.
STRESS DIAGRAMS.
In cases where the flanges of the beam diff'er very greatly from rect-
Fig. 130. Fig. 131.
angles, the nearest equivalent rectangles may be sketched in and the cal-
culations carried out as if the flanges were of this shape.
Thus in Fig. 1 30 abed may be taken as the shape of the flange ; and in Fig.
131 the flange may be considered as made up of opqj- and stuv. The moment
of resistance of each rectangle should then be found, as previously explained,
and the sum of these moments will be the moment of resistance required.
Alternative Method. — The following is another method, which may
occasionally be employed with advantage.
In Fig. 132 the stress diagram for the shallow rectangle sv is obtained as
explained for Fig. 128. Now it is evident that the depth su of the rectangle
may be diminished until it becomes nothing, without altering the position of
the points u-^ and The rectangle su would then coincide with the line itv.
It thus appears that u-^v^ represents the stress in the layer of fibres uv ;
and in the same way cc^i/^ represents the stress in the layer xy. Hence u-^,
x^, y^, and are points in the stress diagram of the flange.
And similarly any number of points can be found on the stress diagram,
as shown in Fig. 132.
Area and centre of gravity of stress. — The area of the stress diagram and
the distance of its centre of gravity from n must now be found. The former
can be obtained by reducing in the usual way to a triangle, or better still,
by means of a planimeter, should one be available ; the latter can also be
determined by a graphic method, which is too complicated for these Notes.
86 NOTES ON BUILDING CONSTRUCTION
Both, however, can be found by the following simple artifice with ample
accuracy for all practical purposes.
n Draw the stress dia-
gram on a thick piece
of Bristol board, and cut
it carefully out with a
sharp penknife.
To find the area
weigh the cut-out dia-
gram, and then weigh
one square inch of the
same Bristol board ; the
first weight divided by
the second will give the
area required in square
inches.
To find the centre of
gravity. — Make a small
hole at each of the
corners D and F, so that
when a needle is passed
through, the piece of
cardboard will swing
perfectly freely. Attach
a small weight to a fine thread and secure the thread to the needle as shown
in Fig. 133. Mark the position of the thread on the diagram, both when sus-
pended from D and from F ; the intersection of these two lines is the centre
of gravity required. In the particular case under consideration the centre of
gravity lies on the centre line through n, owing
to the symmetry of the diagram.
Mathematical Method of finding the moment
of resistance of a wrought iron I beam. — The
graphic method described, pp. 83 to 85, gives exactly
the same result as the accurate mathematical formula
given in Rankine's and other works, and is simpler
to understand.
The mathematical formula is —
■ (53),
where ro = limiting stress per square inch,
I = moment of inertia (see Appendix XIV.),
d
I is worked out for the present example in App. Fig. 133.
XIV., and found to be 120*9. Substituting in Equation 53, we have
To—i tons for compression,
2/0 = 5 inches,
M:
yo
4x120-9
M = 96-7 as before.
When the beam with its load is in equilibrium, the moment of resistance thus
found is of course equal to the bending moment found in the manner alreaidy
explained.
WROUGHT IRON ROLLED BEAMS
87
The calculations necessary to ascertain the weight a given rolled iron beam will
carry, or to find the dimensions of such a beam to carry a load of particular weight
and distribution, are similar to those already described for timber beams.
Examples of some of the cases likely to occur in practice are given as illustrations
at p. 88 et seq.
Approximate and Practical Pormulse for Rolled
Iron Beams.
Approximate Method, No. 1 {ignoring web), — In most books of formulas,
and in many other works describing the calculation of rolled beams, the resist-
ance to direct stresses afforded by the web is ignored ; the resistance of the
flanges alone is calculated.
Thus in Fig. 128, p. 83, the resistance indicated by the triangle Mn would
be omitted.
Instead, however, of taking the stress area CoklplS, as representing the
varying resistance of the flange, the whole area of the flange is taken.
Again, instead of taking the distance between the centres of gravity of
the stress areas as the length of the arm of the couple, the full depth of the
beam is taken.
The resulting formula stands thus —
M = area of flange x limiting stress x depth,
= AxroXd . . . . . . (54).
This formula gives defective results, and the thicker the web (in propor-
tion) the greater the error. The error will therefore be greater in small
rolled beams (as in the above instance) than in large ones. Taking the beam
just dealt with, and ro = 4: tons
M = 4x|x4xl0,
= 80 inch-tons.
We know from p. 85 that its actual value, as there calculated, is 96*7
inch-tons.
This formula should therefore be employed only as a trial-method to
pick out a suitable section, as shown in Example 9, and to be followed up
by the accurate method already explained.
Approximate Method, No. 2 {including web). — In the method just described
the flanges only are considered, the web ignored.
From a glance at Fig. 128, p. 83, it will be seen that the moment of the
triangle of stress thus ignored, i.e. Jcln, is approximately
= area Mn x §dl x r„,
= \ area of web x x r^,
= ^ area of web x ^ x r„.
If, therefore, we add ^ area of web to the area of the flange and multiply
by d, we shall have another formula for the moment of resistance, namely —
M = (A + 1- area web) xr^y.d . . . (55).
Taking the same beam as before,
M = (4 X 1 4- ^ X 10 X I) X 4 X 10,
= 113-inch tons.
This formula gives too large a result, and is therefore not a safe one
to use; the true value of M lies between the values found by the two approxi-
mate formulae.
88 NOTES ON BUILDING CONSTRUCTION
Practical Formula for Rolled Iron Beams, whose Length,
Breadth, Depth, and Weight are given.
The student should observe that in the case of rolled iron
beams the choice is practically limited to the sizes which are
usually rolled by manufacturers, and in many cases to what happens
to be in stock. Most manufacturers publish a list of sections, and
many also supply illustrations of full-sized sections. The process
is therefore to select from such a list that section which best fulfils
the requirements of the case. For such a purpose the approxi-
mate formulae given above are very useful, but it sometimes happens
that only the depth, the width, and the weight per foot are
obtainable, and it is to be observed that this is the information
usually given when specifying for a rolled iron beam. The follow-
ing formula based on these data is therefore useful, namely —
Approximate Method, No. 3.— Safe distributed load in tons
= 0-55(^^- 0-3&(^)j^. . . (56),
where w — weight of rolled iron beam in lbs. per foot,
& = breadth in inches,
d — depth in inches,
L=rspan in feet.
This formula is constructed on the supposition that the
maximum safe stress is 5 tons per square inch, and it will be
found on trial to give fairly good results.
Distributed Load that can be safely borne by a Boiled
Iron Beam of given Dimensions.
Example 8. — Conditions. — Suppose an X rolled iron beam to be of the
section sliown in Fig. 126 and to be 1 2 feet between supports, what load uni-
formly distributed over its length will it safely carry ?
Calculation by mathematical Method. — Preliminaries. — Taking the
safe limit of stress in compression at 4 tons per square inch, we know from p.
86 that the moment of resistance of this beam = 96'7 inch-tons.
The maximum bending moment (see Case 7, p. 33) Mg = — .
But Mc = M,
wl^ = 96-7 inch-tons,
or since , , ^ ^ , -, . w/
1=12 feet = 144
, 96-7 X 8
■••^ = «*=-T44-'
= 5-37 tons.
Calculation by approximate Method No. 1. — The value of M found by the
first approximate method which ignores the web was 80 inch-tous (see p. 87).
WROUGHT IRON ROLLED BEAMS
89
Hence ^^=80,
8
, 80 X 8
wl=
144 '
= 4 '44 tons.
Thus this method gives as the distributed weight that can safely be carried by
the beam 4*44 tons, whereas by the more accurate method it was found to be 5*37
tons.
Distributed Load that can safely be borne by a Rolled Iron
Beam of given Length, Width, Depth, and Weight.
Example 9.— A rolled iron beam 12 inclies deep and 6 inches broad is
found from a Table of Sections to weigh 56 lbs. to the foot. Hence, from
the a;pproximate formula No. 3 (56), for a span of 18 feet we have
Safe distributed load = 0-55(56 - 0-3 x 6 x 12)i|,
= 12-6 tons.
The average thickness of the flange of such an iron beam is 1 inch.
Hence, using approximate formula No. 1,
— = A + r„ + c?.
OP-,. .-, , , 8x6x1x5x12
Saie distributed load wl = — — — ,
18 X 12 '
= 13-3 tons.
The moment of inertia of this section will be found to be 400. Hence
by the mathematical formula (see Equation 53, p. 86)
wl^ rj.
Safe distributed load wl = ^ ^ , ^ —
6x18x12'
= 12-3 tons.
Example 9 A.— Again, take the case of the rolled iron beam 16" x 6" of
which the moment of inertia is 804, and let us find the safe distributed load
for a span of 20 feet, when the beam is supported at the ends.
On referring to a Table of Sections such as is published by manufacturers
it will be seen that this section weighs 63 lbs. to the foot. Hence by
approximate formula No. 3
Safe distributed load = 0-55 (63 - 0-3 x 6 x 16)i|,
= 15 tons ;
and by the accurate mathematical formula
5 X 804 X 8
Safe distributed load :
8 X 20 X 12'
16-8 tons.
Section for a Joist of a specified Span to carry a given
Distributed Load.
Example 10. — Conditions. — A rolled iron floor girder of 20 feet span
and fixed at the ends has to carry a uniformly distributed dead load of 1
ton per foot run of its length.
Find the dimensions required
(1) For strength,
(2) For stiffness.
NOTES ON BUILDING CONSTRUCTION
Preliminaries. —
10 = \ ton per foot rim = -J^ ton per inch =186 lbs.
1 = 20 feet = 240 inches.
i;' = 29,000,000 lbs. for wrought iron bars.
Resistance to direct Stresses. — This being a case of a beam fixed at the ends,
the greatest bending moment is at the point of fixing (see Case 2, p. 72),
and is (Equation 51, p. 72)
M.=A-(210)^
Therefore
12
= 400 inch-tons,
and hence the rolled beam must have a moment of resistance of 400 inch-
tons.
It is practically necessary to select a section from those kept by manu-
facturers. Supposing, therefore, we have a sheet of sections before us, and,
as a first trial, select a section 12" deep, with flanges 6" wide, and 1" thick.
Using the approximate formula No 1 given at p. 87 (Equation 54) we find
M=6xlx4xl2,
= 288 inch- tons,
w^hich is a great deal too small.
Trying next a section 17" x 6" x we find
M = 6xl|x4x 17,
= 510 inch-tons,
which is more than necessary. A section 1 6" x 6"
X 1" gives M = 384 inch- tons. This section will
therefore probably do, since the value of M found
by the approximate formula is too low.
The moment of resistance of this section may
be found by the graphic method thus :
The upper and lower flanges may be taken as
having a mean thickness of 1 inch, and the web
^ ^ inch thick.
The equivalent areas are drawn as described at
p. 83, and we have the following moments of
resistance (Fig. 134) : —
For flanges
For web
tons inch-tons
1 (6" + |6") X 1" X 15" X 4 = 337-5
TV'x|xix7"xf Xl4"x4 = 64-3
Total moment of resistance = 401 "8
This section will therefore be suitable in point of strength.
^ 15", the distance taken between the centres of gravity of the stress diagrams
for the flanges, is a little less than the true distance, because the stress diagram for
the flange is a trapezium, not a rectangle. The true distance is
+ 3 6-1-16 V'
= 2(7-F0-51),
= 15-02,
and the moment of resistance for the flanges becomes 342 inch-tons.
WROUGHT IRON ROLLED BEAMS
91
It must of course be understood that no holes are to be made in either
flange. If any are necessary, the breadth or thickness of the top and bottom
flanges must be increased accordingly.
Besistayice to Shearing.- — We know then that the beam selected is able to
bear the direct stresses that will come upon it ; the next question is whether
the web will bear the shearing stresses.
The greatest shearing stress is at the points of fixing, and it there amounts
^ wl . . 20 tons
to — (see Case 7, p. 59) = — ^ — =10 tons.
This has to be resisted by the sectional area of the web. This area is
14" X Y^g-" = nearly 8 square inches,
. 10 tons
the shearing stress will therefore be ^ = Ij ton per square inch.
Now the safe working shearing stress may be taken as 4 tons per sqiiare
inch, so that the web is amply strong enough.
Deflection. — From Case 2, p. 72, and Appendix VIII., we see that the maxi-
mum deflection of a beam of uniform section fixed at both ends and uniformly
loaded throughout its length is at the centre, and is
wP I
^= El'' 384-
We know (Equation 53, p. 86) that
- d
I = Mx— ,
16
= ^^^X2-74'
A = -
= 804.
186 x 2403 1
29,000,000 x 804 384'
= 0-06 inch,
or much less than the specified limit for floor girders (see p. 69) of per
foot of span, viz. in this case, = 1".
The section of girder chosen is therefore both strong enough and stiff
enough for the purpose, if care is taken in setting it that the ends are
properly fixed.
Remarks on Rolled Beams.
Large rolled beams not theoretically economical. Rolled heams
much used because cheap and useful. Can be fixed at ends. EoUed
iron beams when more than 12 or 14 inches deep are not theoreti-
cally economical, because their section is uniform throughout their
length, and their flanges are equal to one another. They are
therefore not of uniform strength throughout their length, and thus
more material is used in them than is theoretically required.
A glance at pp. 34 and 59 will show that in a section of uniform strength
both web and flanges would vary in dimensions at different parts according to
the stress that comes upon them.
NOTES ON BUILDING CONSTRUCTION
Thus for a beam with a uniformly distributed load and supported at the
ends the area of the flanges required at the centre to meet the maximum direct
stress becomes unnecessarily great as the ends are approached, because the
direct stress diminishes (see Fig. 50) towards the ends.
This is still more the case with a load at the centre (Fig. 48).
The web, on the contrary, would theoretically require to be greater at the
supported ends to resist the shearing stress, and would gradually diminish
towards the centre.
The flange in tension might be smaller than that in compression, as it
can be safely subjected to a greater stress per square inch.
In the same way, beams of uniform strength for other dispositions of load
would theoretically require to have flanges and web with varying sections to
suit the stresses at diff'erent points.
The method in which rolled beams are manufactured (see Part III.)
renders it impossible that their section can be varied to suit the stress that
comes upon them at different points.
It results, therefore, that these beams rolled of uniform section throughout
must contain more material than is theoretically required. This is not of
much importance in small beams ; when, however, the beam is large, the
waste of material becomes important, and it is better to build up such a beam,
or girder, with plates and angle irons so disposed as to have at each point, as
nearly as may be, the exact amount of material which is required to meet the
stress at that point.
The method of doing this is described in Chapter VIII.
EoUed beams are cheap and convenient, and are therefore used
for a great many useful purposes in spite of the fact that they
contain more metal than is theoretically necessary to meet the
stresses upon them.
One advantage of rolled iron beams is that they can be placed
with either flange uppermost, thus preventing all chance of mis-
takes which occur with other girders put up by ignorant men.
Moreover, as the flanges are equally strong, these beams can be
fixed at the ends, thus greatly adding to their strength — giving
them a great advantage over cast iron girders, which cannot so
readily be fixed.
Of course as compared with cast iron girders they possess a
still greater advantage in being made of a sound and reliable
material which will not give way suddenly.
Steel rolled joists or beams are similar in form to those of
wrought iron, and can be calculated in exactly the same way,
using the following working stresses : tension 6-|- tons, compres-
sion 6-|- tons per square inch.
Market Sections. — EoUed I joists can readily be obtained
in either iron or steel, from 3 to 30 inches in depth, with flanges
up to 8" wide ; and in iron they may be obtained of much larger
WROUGHT IRON ROLLED BEAMS
93
dimensions. The largest sizes are not economical to use, as
explained at p. 92.
Girders huilt up with the aid of I iron beams are sometimes
used. The following are some of the forms : —
Fig. 135.
Fig. 136.
Fig. 137.
rig. 135 may be useful in some cases, but a beam built up
with plate and angle iron would probably be better.
Figs. 136 and 137 are distinctly faulty owing to the waste
of material at the centre of the section.
Chapter V.
CAST IRON GIRDERS.^
THE calculation for cast iron girders may be based upon the
same simple methods that have already been explained for
timber beams and wrought iron I girders.
There is, however, one principal point of difference, which
makes the calculation at first sight not quite so straightforward
and simple as in the cases previously explained.
This difference is caused by the nature of the material. In
the timber and wrought iron beams the resistance to tension per
square inch was the same or not very different from the resistance
to compression ; but in cast iron the resistance to tension is very
small compared with the resistance to compression, as will be seen
at once by a glance at the following figures.
Resistance of Cast Iron.
Ultimate Resistance. — The ultimate resistance, i.e. the resistance to actual
rupture of cast iron obtained under prosper specifications for girder -work, may-
be taken as follows : —
Tension ... 9 tons per square inch.
Compression. . . 48 „ „ ^
Shearing . . . 8^ „ „
TForking Resistance. — Working or safe limiting stress to be put upon cast
iron per square inch : —
Tension . . . ij tons per square inch.
Compression. . .8 ,, ,,
Shearing . . . 2"4 „
It will be seen from the above that the ultimate resistance
to tension is about one-sixth of the resistance to compression.
The limiting stress to be allowed in tension is therefore one-sixth
of that allowed in compression.
Manges. — If the tension flange is not made equal in strength
to the compression flange it will tear across before the compressive
1 The formula for practical use will be found at p. 95, Equation 57, and in App.
XXI.
2 The above figures are rather higher than those given at p. 315, Part III., for
the average ultimate strength of ordinary varieties found in the market.
CAST IRON GIRDERS
95
stress upon the other flange is equal to what the flange is able to
bear. It will be seen that approximately, if the web be ignored
altogether, the tension flange should, in order to be of equal
strength with the compression flange, have an area six times as large.
Approximate Formula for Calculation of Cast Iron Girders.
The formulse ordinarily used for the calculation of the strength
of cast iron girders are based upon the valuable experiments of
the late Mr. Eaton Hodgkinson.
Those experiments were made with girders having webs so
thin that the webs may practically be ignored, because they afforded
very little assistance to the flanges in resisting the direct stress.
Mr. Hodgkinson arrived at the conclusion that Fig. 138 is the best form
of section for a beam required to bear an ultimate or breaking strain : —
"The section of the flanges being in the ratio of
6 to 1, or nearly in that of the mean crushing and
tensile strength of cast iron. "
He also inferred from his experiments that " when
the length, depth, and top flange in different cast iron
beams with very large flanges are the same, and the
thickness of the vertical part between the flanges is small
and invariable, the strength is nearly in proportion to
the size of the bottom flange." Also that "in beams
which vary only in depth, every other dimension being ^^^^^ ^^MWMX Mm ~d'-'nft
the same, the strength is nearly as the depth."
From these data^ he found the moment of Fio-. 138.
resistance = C x a x c^.
a being the area of the tension flange in inches.
d being the depth of the girder.
C being a constant or modulus found, by breaking several experimental
beams, to be 6|- tons.
This value of C gives the resistance to actual breaking; the working
modulus may be taken at one-fifth of this, i.e. 1^ ton.
As the bending moment equals the moment of resistance we have
M = Oxaxd . . . . (57).
For instance, in the case of a beam supported at the ends and loaded in
the centre
= Cad,
W =
4Cad
"IT'
4 X tons X ad
26 tons X ad
' I ■
^ These results are not here stated in the exact form given by Mr. Hodgkinson,
but are modified to make them in accordance with the formulse given above for timber
and wrought iron beams.
96
NOTES ON BUILDING CONSTRUCTION
If I be reduced from inches to feet W = ^ ; and this is tlie shape in
which the formula is generally given in books, where W is the breaking weight
in tons.
It will be seen that this formula roughly gives the moment of
resistance of the tension flange.
In practice it is found that when cast iron girders fail, it is
generally by the rupture of the flange which is in tension — the
flange in compression is generally more than strong enough.
If, therefore, the tension flange is strong enough, the girder is
likely to be safe.
The strength of the compression flange may, however, be
approximately found by the same formula by substituting 48 tons,
the breaking stress in compression, for the value of C.
In recent practice the area of the top flange is raised from
one-sixth of the bottom flange to one-third or one-fourth, so that
by increased breadth sufficient resistance shall be offered to lateral
bending.
This formula is very simply applied as shown in Examples
11 and 11 A, and though it is only approximate, it is quite accurate
enough for the calculation of cast iron girders. The material of
which they are made is so treacherous and uncertain that it is
always necessary to use a large factor of safety, or (what amounts
to the same thing) a small limiting stress in calculating them.
Any minute accuracy in the method used for ascertaining their
resistance would therefore be useless.
Graphic Method of ascertaining the Section
for a Cast Iron Girder.^
The distribution of stress over a cast iron girder under a safe
load, and subject only to a small limiting stress, can be shown in a
manner similar to that already described for timber and for wrought
iron beams, and its strength determined from the diagram thus
obtained.
Such a diagram will show the actual distribution of the
stresses over both flanges and the web at the moment when
the extreme fibres under tension are subjected to the working or
limiting stress, and can easily be constructed as follows.
^ This is described as it illustrates principles, and for the sake of those who
prefer drawing to calculations.
CAST IRON GIRDERS
97
Take the case of a girder of the section shown in Fig. 139, supported at
the ends and loaded in the centre.
The lower flange will be in tension, and
the centre of gravity will be at a point 7"
from the bottom edge.^
The neutral axis NA will pass through the
centre of gravity.
The diagram of intensity of stress (Fig. 140)
and the diagram of distribution of stress (Fig.
141) are drawn in the same way as at p. 83.
If we limit the tensile stress on the lower-
most fibre to li ton (the working tensile stress),
we see that the maximum compressive stress per
square inch d'a' upon the uppermost fibre will
be (in proportion to its distance from the neutral
. , 13 ^, , 13 3 39
axis) ~ X Vc' = y X - tons = = tons.
Now the fibres in compression could safely bear 8 tons per square inch,
so that in this girder they are not called upon to bear more than about one-
third of the safe working stress.
—^''/itons—->\
Neutral Axis
'i-^l'Atonpi 1-6 KV
Fig. 140. Fig. 141.
Scale of Stresses 1 inch = 4 tons. Linear scale ^ inch = 1 foot.
Scale of stresses 1 inch= 12 tons.
So far as mere compressive stress goes, the upper flange is therefore un-
necessarily large, though it may nevertheless be desirable to make it of the
dimensions shown, for the reason mentioned at p. 96.
The figure showing the distribution of stress is obtained as before by
multiplying the number of inches of fibres at each point by the intensity of
stress at that point obtained from Fig. 140.
See Appendix XII.
B.C.
-IV.
H
98 NOTES ON BUILDING CONSTRUCTION
Thus at CD there are 4" of fibres, each acted upon by a stress equal to
13 13
d'a' or ^ X the limiting stress, so that C-^D^ = x 4 = 7-| inches. At gh there
12
is 1 inch of fibres acted upon by — of the limiting stress, so that the width
12
of the stress area at this point is = -;^ inch and reduces gradually until at
the neutral axis the direct stress is nothing.
The equivalent shaded area in tension below the neutral axis is of course
equal to the equivalent shaded area in compression above the neutral axis,
being in each case about 17y square inches.
The centres of gravity of these two equivalent areas are 15*97 inches
apart, and the moment of the couple, i.e. the moment of resistance of the
girder, is equal to the equivalent stress area of either kind in square inches x
the limiting stress per square inch x the distance between centres of gravity =
17y square inches x 1|- ton x 15-97 inches = 417*5 inch-tons.
Comparison with Result from approximate Formula. — Let us compare
this result with that obtained from the approximate formula given at p. 95, namely —
Taking C — ton, as at p. 95, we find
Mr=lx(16xlJ)x20,
— 400 inch-tons.
But the limiting stress has been taken above at 1^ ton per square inch. If C be
taken =1^ ton, M = 480 tons.
Fig. 142. Fig. 143.
Linear scale |" = 1 foot. Scale of stresses
Scale of stresses 1 inch— 12 tons. 1 inch—^ tons.
CAST IRON GIRDERS
99
Section with tapering Weh. — In a girder of tlie section shown in Fig. 142,
and at p. 169, Part I., where the web tapers in thickness from the upper to
the lower flange, supj)orted at the ends and loaded, the stresses would be as
shown in Fig. 143.
It will also be noticed that the maximum stress in compression, i.e. that
on the extreme upper fibre, is only 3|- tons.
Mathematical Method for calculating a Cast Iron Girder.
The mathematical method may be described, though the graphic method just
described is equally accurate, and safer to use.
The formula used for M is — I, or — I, in which
Vc yt
rc = limiting stress in compression.
,, tension.
2/c = distance from NA to extreme fibre of compression flange.
i/j= ,, tension flange.
I = moment of inertia.
The former of these two formulse is used when the compression flange is weaker
than the other ; the latter formula when (as is usually the case) the tension flange
is the ^'eaker.
A case worked out by this method is shown at p. 104.
Figs. 141 and 142 give a clear idea of the nature and distribution of
the direct stresses on a cast iron girder, but for the reasons given at p. 96 it
is seldom necessary to go into any accurate calculations for such girders.
The shearing stress is distributed in the same way as described at p. 61
for timber beams. It is nearly all borne by the web, being a maximum at
the neutral axis and diminishing as the flanges are approached, until at the
flanges themselves the shearing stress is almost nothing.
Practical Points connected with the rorm of Cast Iron
Girders.^
Proportions of Flanges. — In both the sections given it will
be seen that the compression flange is never called upon to
undergo anything like the working stress that it can safely bear.
Thus in Fig. 140 the stress on the extreme fibre in compression
is only 2^1 tons instead of 8 tons per sc[uare inch, and in Fig. 143
it is only 3^ tons per square inch.
It appears, therefore, that so far as resistance to crushing is
concerned, the compression flange might be made smaller ; practi-
cally, however, other points have to be considered.
The flange which undergoes compression must not be made
too narrow, or (though safe against crushing) it will fail by buck-
ling sideways.
The greater the span the wider the flange should be. No rule
has been laid down for this, but it is well not to make the com-
pression flange narrower than one-sixtieth of the span.
^ See also Part II.
NOTES ON BUILDING CONSTRUCTION
It will be further seen that in some cases, when from construction of the
whole building it is impossible for the cast girder to buckle, the compression
flange need not have an area exceeding one-sixth of the area of the tension
flange, and this proportion is sometimes taken as the universal rule.
When, however, the compression flange has to be widened to resist cross-
buckling, or when it is required to be wide to support a wide form of load
such as a wall above, then it is often necessary to make it of greater area
than one-sixth of that of the tension flange. The consequence is that some
writers propose one-fourth and others one-third as the proportion (see p. 96).
Each case must, however, depend upon circumstances. As a
rule a compression flange having one-sixth the area of the tension
flange is large enough, if it is wide enough to resist cross-buckling,
and to carry the load upon it.
The web must of course contain sufficient area to withstand the
shearing stress that comes upon it.
In order to prevent it from "buckling" sideways it is customary to
strengthen it by feathers or stiffeners, as shown at /in Fig. 144 and in Part
1. Figs. 247, 248.
These stiffeners should, however, be avoided as far as possible, because
the angles formed by their junction with the web and flanges tend to produce
weak points in the castings, for the reasons explained in Part III.
The depth of a cast iron girder must of course in most cases
be governed by circumstances, but when possible it is desirable
to give it a depth of at least one-twelfth the span, or one-tenth
when considerable stiffness is required. The depth is, however,
very often made much less than this in practice.
Camber. — Cast iron girders should be constructed with a rise in the centre equal
to -j-Itj to -2-^(5^ of their span, so that they may not when loaded sag or droop below
the horizontal line.
Points connected loith casting. — So far as the requirements of good casting
go,^ for flanges 2' wide it is not wise to have a less thickness than 1 1", and
for flanges 18" wide not less than 1", but as a rule 2 the thickness of any part
should not exceed of the width of the part.
When one flange is to be thinner than the other, as in Fig. 142, the web
may taper from one to the other, each end being equal in thickness to the
flange it joins.
Some founders prefer that the thickness of the metal should be the same
throughout web and flanges, as in Fig. 141.
In any case, there should be no sudden variation in the thickness at any
part, and no sharp angles.
Girders of uniform Strength. — The method by which cast iron girders
are made is such that they can without any difficulty be cast so as to
be of uniform strength, the form varying according to the load they have
to carry.
Thus for a girder with a uniformly distributed load the shape would be
1 Wray.
^ Adams.
CAST IRON GIRDERS
lOI
approximately that shown in Figs. 144 or 145, the material being reduced
where the stress is less, so as to form curves.^
The alteration in section may be obtained by altering the depth, as in
Fig. 144, and leaving the flanges of uniform width throughout, or by
altering the width of the flanges at different points according to the stress 'upon
them, keeping the depth of the girder constant throughout, as in Figs 146
147.
ELEVA TION.
f— I]
TO
Fig. 146.
PLAN
Fig. 147.
A cast iron girder should never be fixed at the ends, because of the
change from compression to tension which takes place (as shown in Fig. 112)
in the top flange as the fixed end is approached. This is not desirable even
when the girder is designed accordingly, and is unsafe if the ordinary design
is followed, for then the top flange at the point of fixture will be unduly
strained in tension.
EXAMPLES OF CAST IRON BEAMS.
Cast Iron Cantilever.
Example 11. — Conditions. — A cast iron
cantilever for a balcony (Fig. 148) has a
projection of 6 feet, and has to carry a
distributed dead load (including its own
weight) of I ton per foot run. The depth
at the wall must not exceed 3 feet. Find s'o'l
the dimensions for the cantilever. '
Preliminaries.
w = ^ ton = weight per foot run on the
cantilever including its own weight,
Fig. 148.
^ Theoretically the curve for Fig. 144 should be something between a parabola
and an ellipse. In practice an arc of a circle with ends rounded is used, the ends
being half the depth of the centre. The curves on plan Fig. 147 are parabolas.
^ The cantilevers are assumed to be 5 feet apart.
I02 NOTES ON BUILDING CONSTRUCTION
I = length. = 72 inches.
C = 1 ton per square inch = coefficient for Hodgkinson's formula.
a = area of top (or tension) flange in inches (to he found).
d = depth of cantilever at wall =36 inches.
E = modulus elasticity of cast iron = 17,000,000 Ihs. per square inch.
Now since M = M,
hence (Case 2, p. 28) '^^'^^
that is, ^ijx 72 X 36 = 1 xax 36.
Hence area of tension flange is a = 3 square inches, and the flange
may be 4" x f "
area of compression flange = ^ x 3,
= i square inch.
Thus the web is thick enough to act as a compression flange, or
an enlargement may be made for appearance, as shown in Fig.
149.
The shearing stress at the wall will be = 'wZ = 3 tons (Fig. 83,
p. 57).
The thickness of web t required to resist shearing
shearing stress
~ depth of web x working resistance to shearing per square inch'
3 tons
t = ,
36 x 2'
the actual thickness f" is therefore 18 times as strong as it need be.
Great waste of metal and sacrifice of appearance would result from
making the cantilever of uniform depth throughout. The depth required by
theory at different points may be found by equating the moment of flexure
at each point with the moment of resistance.
At any point P distant x from the wall
W{1 - £C)2
2
We have then — — = Cad,
, ^v(l — 1
2 Ga
Taking Z = 6 feet, When £c=l A=^^^^ ^ I^'
= |.2 X 1 = II = 25 inches
2 cZ = |2xi= i|=16 „
■ 3 = X ^ = =9 „
.4 _ 2 2 y i _ JL = 4
: 5 cZ = i2 X i = • T-V = 1 inch
X
X
X
x = Q d = 0.
Setting off these values as ordinates at the different points, and drawing
the curve of the bottom flange through the extremities of the ordinates,
we have Fig. 1 50.
CAST IRON GIRDERS
103
The curve might have been ascertained by drawing a parabola with its
apex at B through CB.
It has also been shown that area of
the web is largely in excess of the require-
ments. It is evident, therefore, that large
holes may be made in the web, as in Fig.
148, which will lighten the cantilever and
improve its appearance without making
it too weak. The end of the cantilever
may with advantage be thickened as
shown by the dotted line, Fig. 148.
The deflection of the cantilever cannot be accurately ascertained, because
it has been designed by an approximate formula, in which the resistance
afforded by the web has not been taken into account.
Assuming, however, that the stress on the tension flange will not exceed
the limit of 1 ton per square inch, the probable deflection may be ascertained
by Equation 45, p. 67.
A:
in which n' ■-
Hence
1-.
A =
1 (for a cantilever of uniform strength and breadth),
■ 1 ton,
:72",
: 1 6 inches nearly.
1 x 2240 x 722
17,000,000x 16'
= 0'04 inch = inch.
The allowable deflection, even if there were to be a ceiling below, is
per foot run of cantilever = -^V', so that this cantilever is amply stiff
enough.
If any holes are mads in the upper flange to aid in securing the platform,
a sufiicient width must be added to the upper flange to make up for the loss
in area caused by the holes.
Clearly, also, it is not strictly accurate to take the load (including weight
of cantilever) uniformly at ^ ton per foot, as the weight of the cantilever
itself varies from 8 lbs. per inch run at the wall to almost nil at the end.
In this small cantilever this is of no consequence, but in larger work it would
become important.
Cast Iron Girder.
Example 11 A. — To find the weight that a given girder will bear.
Conditions. — A number of old cast iron girders in stock have their cross
section throughout of the form and dimensions shown in Fig. 1 39. Span
= 20 feet. What distributed load per foot run will they safely bear when
supported at the ends ?
1. By the approximate rule (Equation 57), taking limiting stress in tension
at 1|- ton per square inch,
— = Qad,
8
Cad X 8
NOTES ON BUILDING CONSTRUCTION
l|x 16x20x8
^ 2402 '
= 0-067 ton per inch,
= 0"8 ton per foot run.
If C is taken at 1^ ton, as recommended at p. 95, the safe distributed load'
■would be |- 0*8 = 0*66 ton per foot run.
2. By the graphic method, taking the limiting stress in tension as 1^ ton.
From p. 98 we know M to be 417*5 inch-tons. Hence
-==417-5.
=0-058 ton per inch,
= 0-70 ton per foot.
3. By the mathematical method. — Finding I as in Appendix XIV. to be
1950, and knowing the tension flange to be the weakest, we have, since
wl^ 1950
"8"" 7 '
= 417*8 inch-tons,
and i<; = 0-058 ton per inch, as by the graphic method.
Deflection. — To find the deflection of this girder when loaded with a
distributed load of *7 ton per foot as found above by the graphic and mathe-
matical methods.
From Equation 44, p. 66, and Table C, we have A= -^v-,
5
n = for a girder of uniform cross section supported at the ends and
loaded uniformly.
I = 1950, see Appendix XIV. E = 17,000,000 lbs. per square
inch.
5 14 X 2240 X (20 x 12)3
A = V i iripli
384 17,000,000 x 1950 6
The deflection allowable if there is to be a ceiling below would be per
foot of span, or = The girders would therefore be amply stiff enough.
Chapter VI.
TENSION AND COMPRESSION BARS.
"Tlsr the previous chapters we have considered various methods of
J- calculating the strength of beams made of one piece of material
— either wood, wrought or cast iron. Such beams can only be
made of comparatively small dimensions, and in the case of wood
or wrought iron it is, as already pointed out, uneconomical to make
them beyond a certain size, because, in a uniform section, which does
not accommodate itself to the varying stresses, much material is
thrown away.
Large beams or girders are therefore built up of smaller pieces
suitably connected together, thus forming either plate girders or
open-work girders in wrought iron, or trusses in timber. More-
over, roof trusses are composed of a number of pieces or members
jointed together (see Parts I.-II.) The various members of such
structures will be subject to tension or compression or shear
according to the position of the member in the girder or roof
truss. The first step, after designing such a structure, is to
ascertain the amount and nature of the stress each member has
to bear ; and the next to design each member so as to safely
resist the stress in it, as will be explained in the present chapter.
Lastly, the various connections must be of adequate strength to
transmit the stress from one member to another. This forms the
subject of "joints" and will be considered in the next chapter.
It will, however, be best to leave for the present the question
of finding the stresses in the various members ; we will therefore
deal first with the design of separate members, assuming the stress
upon each as known.
Tension Bars.
Symmetrical Stress. — Let AB, Fig. 151, represent a tension
bar stretched by the forces E^, E^. The dotted line shows the line
of action of these forces, and we will suppose that it is exactly
in the centre of the bar. By cutting the bar across at C, as
io6 NOTES ON BUILDING CONSTRUCTION
shown in Fig. 152, the tension forces excited in the fibres of the
bar are, so to speak, brought to view ; and since the line of
Fig. 151.
Fig. 152.
action of the force is central, these forces are equal, or, in other
words, the stress is uniformly distributed over the cross section of
the bar. Clearly therefore : Safe resistance to tension of the bar
= safe resistance to tension of the material per unit of area x area
of cross section.
Thus a wrought iron bar 2'5" x 1" could resist safely 5 tons
per square inch or 5x2'5=:12-5 tons.
Or, again, a round iron bar 1" diameter could resist safely
5 X 0-78 = 3-9 tons.
Effective Area.— Let us now inquire into the effect of making
holes in the bar, such as would be made for rivets.
In the first place let there be one hole at the centre line of
the bar, as at D, Fig. 153. The cross section at D is clearly the
Fig. 153.
R;
Fig. 154.
weakest in the bar, and it is there where rupture would take
place. The stress at the section is shown in Fig. 154, and it will
be noticed that, as before, the stress is uniformly distributed, but
that the area over which it acts is reduced. This reduced area
is called the effective area of the tension bar and : Safe resistance
to tension of bar = safe resistance to tension of the material X effective
area. Or in symbols
R = nx A
(58).
The same formula would clearly be applicable if two holes
TENSION BARS
107
symmetrically placed were made in the bar, as shown in Figs. 155
and 156.
0
0
5-
Fig. 155.
Fig. 156.
If now a third hole be made, as shown in Fig. 1 5 7, it becomes
a question whether the bar will break as shown in Fig. i 5 8, or as
-^5-
Fig. 157.
-ei^ ;
Eg
Fig. 158.
Fig. 159.
shown in Fig. 159. It has been found by experiment on ordinary
fibrous iron that a bar is most likely to break along such a line
that the area of the section along the line of fracture is a mini-
mum. Therefore, if a + c + c + a is less than a + & + a, the bar
will tend to break, as shown in Fig. 159, and the effective area will
be 2(a, + c) X thickness of the bar.
Unsymmetrical Stress. — So far it has been supposed that the
holes are placed symmetrically with regard to the centre line of
R
0
>-
Fig. 160.
Fig. 161.
the bar. Fig, 160 shows a tension bar with a hole made in it
nearer to one edge than to the other, and Fig. 161 exhibits the
stresses in the fibres. It is clear that the intensity of stress will
no longer be uniform, as in the previous cases, but will be greater
at the edge near the hole than elsewhere ; the distribution of the
stress can be found by calculation, but by a method rather beyond
the scope of these Notes. The student should, however, observe
?o8 NOTES ON BUILDING CONSTRUCTION
that an unsymmetrically placed hole weakens a tension bar more
than a symmetrically placed hole does, and in the same way it
will be seen that if the tension in the bar is not transmitted along
the centre line, the bar will be weakened as shown in Tigs. 162,
Fig. 162.
Fig. 163.
163; for the intensity of stress being greatest at one edge, the
bar will begin to yield first at that edge, and its whole area will
never be opposed to the full working stress.
It should be an aim, therefore, in designing tension bars to
arrange the holes in them symmetrically, so that the stress shall
be transmitted along the centre line. When this cannot be done,
it is necessary to slightly increase the cross section accordingly.
The formula Rj = x A is applicable to tension rods. The
ends of such rods are often screwed with a minus thread, in which
case the effective area is the area at the bottom of the threads
(see p. 143). In fact, this formula is of general application when
the stress is uniformly distributed over the cross section, and to
use it the weakest section of the tension bar or rod must be found,
and the area of that section is the effective area.
Example 12. — A rectangular iron tension bar is required to resist a stress
of 1 0 tons. The thickness of the metal is 1", and allowance is to be made for
one f ' rivet hole. Find the width.
If a is the total width the effective width is a — f , and the effective area
(a — -I)!" square inches.
Assuming a safe resistance to tension of 5 tons per square inch, we have
5(a-f)i=10,
whence a = 4| inches.
Example 13. — The stress in a wrought iron tension rod is V'SS tons.
The ends are formed into links for pin joints (see Example 26, p. 140).
Find the necessary diameter.
The links at the ends of the rod w^ould be made of such dimensions as to
be of at least equal strength with the rod. The strength of the rod is there-
fore not reduced in any way, and consequently the effective area is the full
area of the cross-section. Allowing a safe stress of 5 tons per -square inch,
we have clearly 7-83 = 5xA, orA=l-56 square inch, which corresponds to
a diameter of 1'4 inch, or a rod if" diameter would do.
COMPRESSION BARS
109
Compression Bars or Struts.^
It might, at first sight, be thought that compression bars or
struts could be treated in the same manner as tension bars, and that
the resistance of a compression bar could be obtained by
multiplying the area of the cross section by the resist-
ance per square inch. But a moment's consideration
will show that this cannot be the case, for in a tension
bar the forces acting on it tend to straighten it, whereas C
the forces acting on a compression bar practically tend
to bend it,^ as shown in Fig. 1 64, thus causing a bend-
ing moment which has to be resisted in addition to the
compression proper. Fig. 165 represents the portion of
-<
\
- 1
a
■3— ^~
Fig. 165.
the bar at AB to an enlarged scale, the arrows of length
a show the compression proper, and the other arrows the
bending stresses, which are distributed as in the case
of a beam (see p. 43). The maximum intensity of
compression will be at B and equal to a + h. At A
it will be a minimum and equal a — h, and if & > a (as Yig. 164.
in the figure), the fibres at A will be in tension.
It is clear that a + h must not exceed the safe resistance to
compression of the material.
Fidler's rules for Struts.- — Unfortunately, however, there are many
difficulties in the way of finding the value of b. In the first place it depends
on the amount to which the resultant stress deviates from the centre of
gravity of the cross-section of the strut, and it further also depends, as shown
in a paper ^ by Professor Fidler, M. Inst. C.E,, on inequalities in the modulus
^ The formulae, for practical use will be found for iron struts in Table V. , in
Equation 60, p. 113, and for timber in p. 115, and Table VI. As the question of
long struts has hitherto been in an unsatisfactory condition three or four methods
of dealing with them are given. Prof. Fidler's being preferred.
^ Theoretically, with the stress passing absolutely through the centre line of the
strut, there is no tendency to bend, but practically this condition is never fulfilled,
and the slightest divergence causes a bending stress ; so that, at best, the state of
the strut is one of unstable equilibrium.
^ Published in Minutes of Proceedings, Inst. C.E., vol. Ixxxvi., 1886.
I lO
NOTES ON BUILDING CONSTRUCTION
of elasticity whidi may exist in different parts of the column ; for instance,
a variation of 2 per cent in the modulus of elasticity in a solid cylindrical
column will reduce the strength by 26 per cent.
Struts of rectangular section. — It can be shown, by an investigation too
difficult for these Notes, that the average ultimate stress per square inch for an
ideal column of rectangular section, in which there is no variation of the
modulus of elasticity, is given by the formula
r, = 7r2E.^V^' .... (59),
where d is the shortest side of the rectangular cross section and I the length of
1,0,000
A
20,000
10,000
^ JVROUGHT
IRON.
Q 100 200 300
Length divided by radius of gyration.
Fig. 166.
the column. This formula is given vnBer Gonstructeur, by Eouleaux; it is based
on Euler's theory, and is plotted graphically in Fig. i66 in the curve BHC, on
the assumption that the material is wrought iron, that is, taking E = 1200 tons.
In this figure the abscissae measured along OP are the ratios ^ Vl2 • ^)
the ordinates are the corresponding breaking weights in lbs. per square inch.
AB is drawn at the ultimate resistance to crushing of the material.
Now Prof. Fidler shows in the paper above alluded to, that under the most
unfavourable conditions which are all likely to occur in practice, the column
may be weakened to the extent shown by the curve ATC. This curve therefore
gives the minimum strength of wrought iron columns of rectangular cross-
_ I
section for varying ratios of . ^. The actual strength of any particular
column may therefore lie anywhere between these two curves, and Prof Fidler
comes to this conclusion : " that the strength of columns cannot be defined by any
^ This ratio is length divided by radius of gyration (see top of next page). If /c be the
,. , ,. . ,■ ,^ o moment of inertia of the cross section
radius of gyration oi a cross section then = — , -.
area oi the cross section
= for rectangular sections (see Appendix XIV.) ^ ^ " ^^12
AOO
P
COMPRESSION BARS
III
hard, and fast line, even when the modulus for the whole column and the
ultimate strength of the material are accurately known ; but, on the contrary,
the strength may have any value less than that of the ideal column within
certain limits. The strength of columns must, therefore, be represented by
an area, as shown in Fig. i66, within which the results of individual
experiments may be expected to place themselves at haphazard,"
The advanced student is strongly recommended to read Prof. Fidler's
paper,! in which will be found the formula he arrives at — a formula which
is, however, too complicated for these Notes.
Struts of any cross section. — "We have confined ourselves so far to columns
of rectangular cross section ; the same reasoning, however, applies, whatever
may be the cross section, but we must substitute for — =d (which is the
radius of gyration for a rectangular section) (see Equation 59) the expression
for the radius of gyration of any cross section, which can be found from
,r, J. p X- \o Moment of inertia of the cross section
(Eadius of gyration)^ — ■ r .
Area oi cross section
The value of the moment of inertia of the cross section varies according to
the axis about which it is taken, but when taken about an axis through the
CG of the cross section, perpendicular to the direction in which the column
tends to bend, it is a minimum, and the radius of gyration found from this
moment of inertia is the one to insert in the formula.
Practical Value of d. — It is, however, somewhat laborious to obtain this
radius of gyration, and for practical purposes, at any rate for building con-
y'^h^ ,\ i*--d — >!
1 •{"■
Fig. 167.
struction, the value of can be taken as shown in Fig. 167, multiplying by a
factor n as shown in Table V., which has been compiled from Fidler's formula
and from similar tables given in the paper, and is recommended for use.
Many other formulae have been proposed for finding the strength of
columns, and some of them will be mentioned later on.
Short and Long Compression Bars.
It is often the practice to divide compression bars into two
classes, viz. short compression bars and long compression bars.
Short compression bars are those which fail by direct crushing,
and their strengths are calculated as if the pressure were uniformly
distributed over the cross section. Long compression bars are those
^ Also Bridge Construction, by T. Claxton Fidler.
112
NOTES ON BUILDING CONSTRUCTION
which fail by bending, so that some of the fibres are torn. A
sudden change in the manner of failing has therefore been assumed.
Such an assumption is contrary to all experience and common sense,
and it is evident that these two causes of failure lie at the extreme
ends of a series, and that between them failure arises from a
gradually varying combination of the two causes.
Eeferring to Fig. 165 a compression bar may be said to
commence to fail by cross-breaking when & = a ; but in fact the
bending moment has begun to act from the first, and has been gradu-
ally increasing the compression on one side of the bar and lessening
it at the other, and it is therefore not right to assume that the
pressure is uniformly distributed. It may be mentioned that some
experiments by Mr. Kirkaldy point to the same conclusion.
The following Table gives the proportion between length and least
dimension of the cross section at which compression bars have been assumed
to cease to become short and become long.^
Wood ....
20
Wrought Iron
10
Cast Iron ....
5
Of course there is in reality no abrupt transition from short to long com-
pression bars as assumed in this
Table, but the figures are given,
as they are much used with the
ordinary formulae.
Securing ends of long
struts. — The manner in
which the ends of a long
compression bar are secured
has a great influence on its
powers of resistance. If
the ends are hinged or
rounded the bar will tend to
bend as indicated in Fig. 168,
but if they are fixed or flat
the bar will be stiffer and will
tend to bend as shown in Figs.
169 and 170. Manifestly
a compression bar with the
Fig. 168. Fig. 169. Fig. 170. Fig. 171.
1 That is, it lias been assumed that a wooden strut whose length is more than 20
times the least dimension of its cross section will fail by bending ; if less than 20
times that it will fail by crushing.
COMPRESSION BARS
113
ends fixed is considerably stronger than one with the ends
rounded. Again, one end may be rounded and the other fixed,
in which case the strength will have an intermediate value
(see Fig. 171). In any particular case, before the ends of
a compression bar are assumed to be fixed, the matter should
be carefully considered, because it often happens that what may
appear at first sight to be fixed is not fixed at all. The criterion
is whether the ends are so connected as to constrain the bar to
bend as shown in Fig. 169. Thus the end of a bar ^
fixed to a T iron as in Fig. 1 7 2 is not " fixed," because
the bar can bend as shown in Fig. 1 7 3 by twisting the
T iron. There is, however, a certain amount of con-
straint due to the resistance of the
T iron to bending, so that the bar
is stronger than if its ends were
purely hinged. These last remarks
apply to the case when the ends are
flat (see Fig. 174).
Gordon's Formula. — The formula
principally used in this country for long
compression bars is that known as Gor-
don's formula, namely —
Ar.
Fiff. 172. Fig. 173.
1 +a
(60),
where a is a constant determined by experiment, d is the least dimension of
W ROUGH
T IRON.
L
Red. colui
ins- ends
'ounded
20 30 UO 50 00
Length divided by lesser side of section.
Fig. 175.
B.C. IV.
114
NOTES ON BUILDING CONSTRUCTION
the cross section, as indicated in Fig. 167, A the area of the cross section, and
the safe resistance to compression of the material per square inch.
The value of the constant a depends not only on the material but also
orv the form of the compression bar. The following Table (G) gives some
oi these values.
To compare this formula with Fidler's formula, when the material is
wrought iron, the curves shown in Fig. 175 have been dra\^'n. It will be
seen that the agreement is fairly good, but that Gordon's formula gives lower
results for short columns and higher results for the longer columns.
TABLE G.
Material.
Timber
AVrought iron
Cast iron ■
Form of Cross-section.
Values of a when
Rectangular )_
Circular J
Rectangular
Circular (solid)
Circular (hollow)
LT+n
]XIi-i
Circular (solid)
Circular (hollow)
Rectangular
Cross-shaped
Both ends
One end is
are
Both ends
rounded,
rounded or
are fixed.
the other
hinged.
fixed.
4
1
1
250
250
100
4
1
1
2500
2500
1000
\ *
1
1
( 900
900
360
1
1
1
100
400
160
1
1
1
2"00
800
320
3
3
3
400
1(500
640
3
3
3
200
800
320
13 cwts.
}- 4 tons.
>■ 8 tons.
N.B. — In using this Table it must be remembered that it is calculated iviih a
factor of safety of 4- In many structures, such as those of cast iron or timber, a
factor of safety of 6 to 8 or 10 is necessary for safety, and care must be taken in
such cases to modify the value of r^ accordingly.
Formulse for Wooden Struts — The formulae in use for wooden struts; are
more at variance than those for iron, and it will be useful to compare som.e of
them. For instance : —
Professor Rouleaux gives this formula in Der Constructei
,IE
/2 ■
• factor of safety
(61)),
in which I is the moment of inertia, E the modulus of elasticity which redmces
to the following, taking E= 1,300,000 Ibs.^ and a factor of safety betwee;n 4
and 5, ^ bd^
R, = 2000-3- cwts.
^ The letters have been changed to agree with the notation in this book. See
also Equation 59.
^ This modulus must be converted into cwts. before reducing the formula, because
the safe load is to be expressed in cwts.
COMPRESSION BARS 115
Professor Ritter gives the following —
1 + 0-0027
. . . (62),
whicli is the same as Gordon's, but with a different constant. In both these
formulee the ends are supposed to be rounded.
Example 14, — To compare the formulae let us find by each of them the
safe resistance to compression of a wooden strut 10 feet long and 4" x 3"
cross section, ends rounded.
By Gordon's formula :
By Rouleaux's formula :
4x3x13 _
4 /120\
^+25oV"3-;
5 '8 cwts.
4 X 33
R, = 2000
By Ritter's formula :
R.
1202'
15 cwts.
4 X 3 X 13
1 + 0-0027 X 402'
29-3 cwts.
These results vary so, that it is difficult to make use of them, and we
must resort to some other method.
Table for Wooden Struts. — Table VI. was deduced from some
experiments by Mr. Kirkaldy,^ but modified to agree with Mr.
Fidler's views (see p. 110).
The above example works out as follows by means of this Table : —
I
Ratio = ^ = 40 (col. R in Table),
Safe intensity of compression per square inch= 1-06 cwt..
Total safe resistance = 4 x 3 x 1-06 = 12-7 cwts.
Again, from Table VII. (Stoney's Table) we find for a ratio of 40, 80
tons as the breaking load per square foot, or taking a factor of safety of 4 the
safe load = 20 tons per square foot. Hence in the case of the above example
20 X 20 X 4 X 3
^ 12 X 12 '
= 33-3 cwts.
This Table is, however, deduced from experiments with large timbers with
flat ends adjusted as in ordinary practice, so that the ends might be con-
sidered as partially fixed. Working out the example again for fixed ends by
means of Table VI. we find
Safe intensity of compression for ratio 40
= 2-75 cwts. per sqiiare inch,
. •. R<. = 1 2 X 3-60 = 43-2 cwts.
Table for practical use. — It is clear from the above tbat the subject of
1 Corps Papers, Royal Engineers (Vols. XXII. and XXIII.)
ii6
NOTES ON BUILDING CONSTRUCTION
long wooden struts is in an unsatisfactory condition. On the whole, Table
VI. is recommended for use, and it moreover saves a considerable amount
of numerical work. As a further comparison this Table is shown graphic-
ally in Fig. 176 by the curve ATC ; likewise Rouleaux's formula, which,
as already mentioned at p. 11 0, is the formula for an ideal column, is shown
by the curve BHC. It will be observed that the curve ATC agrees well
with the corresponding curve ATC in Fig. 166, which can be looked upon
as an indication of the reliability of the Table.
A B
\ \h
WOOD.
-
Red, Coht
77ms- ends
rounded.
%
\ V
—
0 10 so so UO 50 GO 70 80
Length divided by lesser side of section.
Fig. 176.
We now proceed to work out some examples.
Cast Iron Column.
Example 15. — Find the dimensions of the cast iron column supporting
the end A of the rolled iron girder of Example 1, p. 21. The cross section
to be circular and hollow.
The load on the top of the pillar is equal to the reaction on the girder,
which has already been found to be 16"6 tons. The length of the pillar is
10 feet. The practice with cast iron pillars is to make the ends large and
flat, and such pillars can be considered as fixed. In this example it will be
assumed that the ends of the pillar are " fixed." ^
Assume, as a first trial, that the external diameter is 4 inches, and tliat the
thickness (i) of metal is ^ inch. Then
Z = 120 inches,
^ It is, however, though usual, not good practice to have a large bearing surface
on the top of a column, because the deflection of the girder causes nearly all the
stress to pass down one side of the column, thus producing a bending moment. This
may be avoided by having a small bearing surface at the top of the column, but
then it would be in the condition of being fixed at the lower end and roundec at the
top, and the value of a in Gordon's formula would be 3^^.
COMPRESSION BARS
117
d = 4 inches,
t = 0"5 inch,
= 8 tons.
4 4
= 7ri(cZ-0-7ri(4-i),
= 5"5 square inches.
Or using a table of areas of circles
A = area of circle 4" diameter - area of circle 3" diameter,
= 12-58 - 7-08 = 5-5.
Gordon's Formula. — Hence, using Gordon's formula, since
(1= (for fixed ends),
8 X 5'5
I^c=i , 1 ./i20N2=19-6 tons.
^ "f" 'Sou
Fidler's Formula. — Again, referring to Table V., p. 332, we find the
value of n for a circular hollow column to be 3-1. The ratio of ^ to is
I
therefore - = 3-lxi|^ = 93 (see p. 306). From Table V. we find the
I
safe stress in tons per square inch. When — = 95 (the nearest to 93) the safe
K
stress is 3'66 tons per square inch.
Hence
Safe load = 5-5 X 3-66,
= 20 tons.
This is rather more strength than required, but looking to the uncertain
nature of cast iron, and the possibility of flaws, this section is not too large.
Since the actual load is 16*6, and according to Gordon's formula the safe
load is 19"6 tons with a maximum intensity of stress of 8 tons per square
inch, the actual maximum intensity of stress will be
16-6
X 8 = 6*8 tons per square inch.
Cast Iron Column.
Example 16. — Find the safe load that can be placed on a cast iron
column of M section of the following dimensions : Length 1 5 feet, width of
flanges 6", width across the flanges 8", thickness of metal (flanges and
web) 1".
It will be found that the area of the web is half the area of the flanges
together. Hence, referring to Table V. and p. 333, the proper value to take
for n is 4-2, so that
I 15x12
Ratio - = X 4-2 = 126.
K 6
If, therefore, the ends are supposed to be rounded, the safe stress per square
inch is 0-86 ton, and the safe load
= (2x6 + 6xl)x0-86 = 15-5 tons.
But if the ends are supposed to be fixed, the safe stress would be 2-19 tons
per square inch, Table V., and the safe load 39-4 tons. In practice it
would not, however, be safe to rely on the ends being fixed unless special
precautions are taken.
ii8
NOTES ON BUILDING CONSTRUCTION
Timber Strut.
Example 17. — Find the cross section of a rectangular strut of Riga
fir 13' 6" long, and capable of withstanding safely a stress of 25 cwts., the
ends being considered rounded.
There are clearly any number of cross sections that will fulfil the given
conditions. If it were required that the strut should contain the minimum
quantity of wood, then the cross section would be a square and the side of
the square could be found by means of Table VI. by a series of approxima-
tions. But in practice one side of the scantling is generally determined to
enable the strut to fit in with other work. In this example let it be assumed
that one side of the cross section is to be 4" ; then
^ . 13' 6"
Ratio = — TTT- = 40,
4
and, referring to Table VI., the safe stress per square inch
= 1-06 cwt.
Hence the other side required
_ 25 _ „
~ 4^1^ ~ ^'^ •
Or a cross section of 4" x 6" would be suitable.
Timber Strut.
Example 18. — Find the cross section of a rectangular strut of Riga fir
13' 6" long, and capable of withstanding safely a stress of 12 cwts. One side
of the cross section is fixed at 4", the ratio ^ will be 40, and the ends are
supposed to be rounded.
If the side to be found is greater than 4", the safe stress per square inch
will be as in the last example = 1*06 cwt., the side required is
12
~4 X 1-06 ""^"^ ■
The side to be found is therefore smaller than the given side, and the
ratio is therefore greater than 40, so that the safe stress must be less than
1"06 cwt. per square inch. In other words, the required side must be greater
than 2'8" and must also be less than 4".
In a case like the present we must proceed by a series of approximations.
Assume that the required side is 3|", then
^ . 13' 6"
Ratio = , „ =50,
safe stress = 0'66 cwt. per square inch, and therefore the safe stress the strut
can withstand
= 0-66 X 3| X 4 = 8-6 cwts.
The side is therefore still too small, therefore assume 3-|" as the required
side, then
^ . 13' 6"
Ratio = —-3 — = 43,
and safe stress = 0'92 cwt. per square inch.
Hence total safe stress
0-92 X 3| X 4 = 13-8 cwts.
Therefore, according to the calculations, a cross section of 3|" x 4" is
rather stronger than necessary, but in practice it would be a suitable se^ction.
COMPRESSION BARS
119
Wrought Iron Strut.
Example 19, — Design a wrought iron strut of tlie form shown in Fig.
177 to bear a stress of 1'69 ton, the length being 2' 11".
In struts of this form it can always be arranged to make the tendency of
the strut to bend as a whole less than the tendency of the sides to bend
between the distance pieces,^ or in other words, such a strut will tend to fail
as indicated in Fig. 178. The strength of such struts will therefore to a large
extent depend on the number of distance pieces used. Clearly the portions
of each side bar between two distance pieces (such as ah, Fig. 178) can be
looked upon as a long compression bar hinged at the ends.
A simple way of designing such struts is to assume some convenient cross
section for the flat bar iron forming the sides, and then to calculate how far
apart the distance pieces can safely be placed.
Fig. 178.
In this example the stress is small, and therefore very small side bars
will probably suffice.
For instance, assume x |-" as the cross section of each side bar. Then,
as each has to bear half the stress or 0'85 ton, the intensity of stress will be
0-85 . ,
~ 1| X ^ ~ ■'■'^ square inch.
On referring to Table V. it will be seen that this intensity of stress
corresponds to a ratio of 135. Hence, since « = 3'5 for a rectangular cross
section, maximum safe distance between distance pieces
-3 X — = 14-5"
-8 X 3-5 -^^^ •
One distance piece is therefore hardly sufficient, and it would be better
to use two, as shown in Fig. 179.
^ The student is recommended to test this statement in the present case by
taking the value of 7i = 3"0 from Table V.
NOTES ON BUILDING CONSTRUCTION
2.
■11% ^ uVs'--
11% ■>
I
Fig. 179.
The cross section of the side bars might, however, with advantage be
increased being rather narrow) say to 2", then intensity of stress
A.Q K
— = 1'13 ton per square inch.
0-85
2^ I
Whence ratio = 152, and maximum distance between distance pieces
1S2
= |x-^ = 16-5".
T Iron Principal of a Roof.
Example 20.— The T iron principal of a roof is subject to direct compres-
sion only. The portions between the supports are 1 0-8 feet long, and the stress
is T-Sa tons. Find a suitable section for the T iron (see Example 42, 209).
The purlins are supposed to be placed at the joints of the roof ; the
principal is therefore only subject to direct compression and not to any bend-
ing stress, as would be the case were the purlins distributed along its length.
In this case, therefore, a section of J iron in which the stem is about equal
to the table will be the best.
Now in a roof of this size the principal would be made of a single T iron
from wall plate to ridge, and on referring to Table H, p. 213, it will be seen
that the upper part (DC) of the principal has less stress to bear than the lower
part AD (Fig. 373), so that the upper part assists the lower part to resist
buckling. Moreover, the purlin tends to " fix " the principal at D. The
lower end (A) of the principal can, however, only be regarded as "rounded."
On the whole, therefore, the principal from A to D can be regarded as a long
column with one end "fixed" and the other "rounded," and the safe stress
taken as the mean.
Try in the first instance a section 4y x 4" x From p. 333
obtain 4-9 as a suflaciently accurate value for n. Then from Table V.
we
Ratio 7
k
I 10-8x12
X 4-9 = 162.
Safe stress = 1(0-98 x 2*14) =
Total safe stress = 1-56 x (4 + 4)|,
= 6-3 tons.
Next try the section 4^" x 41" x Then
I
1-56 ton per square inch.
Ratio -
Jc
10-8 X 12
— ^ — X 4-9 = 141 nearly.
Safe stress = 1-86 ton per square inch.
Total safe stress = 1-62 x (4 + 4 1)1,
= 7-9 tons.
which will just do
Comparison with Molesworth's Tables
On referring to the Table for
trussed rafter roofs in Molesworth's Pocket-Booh of Engineering Formulce., it
COMPRESSION BARS
121
will be found that tlie section of the principal is given as 3i" x 3 x \'. The
roofs referred to in the above Table are of the same kind and shape as the
one in which the stresses are found in Example 42, and the rise span)
and camber of tie-rod are the same as assumed in the example. On the
other hand, however, as stated in Molesworth's Pocket-Booh, in the Table
the trusses are supposed to be 6 feet 8 inches apart, whereas in the example
they were taken as being 8 feet apart ; and although not stated, it is probable
that the covering on the roofs in the Table is corrugated iron or zinc, whereas
in the example the roof-covering is countess slates on |" boarding.
It will be interesting to investigate this a little more closely.
Now, on referring to Example 42, Table H, it will be seen that the stress
in the principal (AD) due to roof-covering, etc., and snow is 4-45 tons, and
on referring to p. 209 it will be further seen that this stress is due to a load
of 20-2 lbs. per square foot covered by the roof. If the roof-covering is altered
to corrugated iron the load will be reduced by about 8 lbs. per square foot (see
Table XIII.), or say to 12-2 lbs. per square foot. The stress of 4-45 tons
will therefore be reduced to
1 2*2
4-45 X ^q:^ = 2"68 tons,
and accordingly the total stress in the principal will be reduced from 7 "7 8
tons to
7-78 - 4-45 -1- 2-68 = 6-01 tons,
and diminishing the distance apart of the trusses from 8 feet to 6 feet 8 inches
will further reduce the stress to
6'8"
6-01 X -7-77= 5-01 tons.
8 0
We have now to find what stress a T iron principal 10-8 feet long and
X 3 X ^ cross section can bear.
As before,
10-8x12
Ratio = ^x 4-9 = 212.
Safe stress per square inch = 1-01 ton.
Total safe stress = I'Ol x (3 -i- 3) J,
= 3 tons.
From the above it appears that it would be wiser to use the stronger
section obtained by the calculations on p. 120, at any rate for important roofs
and in exposed situations, where full allowance for wind pressure ought to
be made.
Comparison with Hurst's Table. — In Hurst's Pocket-Booh 4i" x 3|-" x i"
are given as the dimensions for the principals for a trussed rafter roof 40 feet
span, rise | span, and camber of tie-rod span, covering countess slates on
boards. Distance apart of trusses, 6 feet.
Now in this case the length of the principal rafter from A to D (Fig. 360)
is 11-7 feet, so that
11-7x12 „
Eatio = — ^ X 4-9 = 196.
Safe stress per square inch= 1'12.
Total safe stress = (4-5 + 3)i x 1-12,
= 4-8 tons nearly,
which would be almost enough for a corrugated iron covering.
Chapter VII.
JOINTS AND CONNECTIONS.
A GEEAT variety of joints for connecting the various members
of wooden and iron structures have been described in Parts I.
and II. It is proposed to show in this Part how to determine the
dimensions of the various portions of a joint when the stress to be
borne is known.
Eiveted joints are principally used for built-up wrought iron
beams and roofs, but occasionally pin joints are employed. In
roofs, screw connections, cotter joints, and knuckle joints are used,
and these will also be described in this chapter.
Points to be observed in making Joints.
These are as follows ; —
1. The members of the structure connected together should be
weakened as little as possible. They are liable to be weakened,
as will be seen in the sequel, either by cutting away material (for
instance, a minus screw thread), or by the design of the joint being
such as not to transmit the stress uniformly over the cross section.
2. Each part of the joint should be proportioned to the stre;ss
it has to bear.
3. Abutting surfaces in compression should be placed jas
nearly as possible perpendicular to the stress.
EIVETED JOINTS.
The manner in which these joints are made and the technic;al
names given to them were dealt with in Part I., and it ther'e-
fore only remains to show how the strength of such joints can
be calculated.
We will first consider riveted joints in tension.
Riveted Joints in Tension.
Lap Joint. — Starting with a single lap joint with only ome
RIVETED JOINTS
123
Fig. 181.
rivet, as shown in Fig. 180, it is clear that this joint is liable
to fail —
1. By one or both bars
tearing across, Tig. 181.
2. By the rivet cutting
into the metal of the bar, or by
the bar cutting into the rivet,
according to which metal is the
softer. Fig. 182.
3. By the rivet "being
sheared, Fig. i 8 3 (single shear
in this case).
4. By the rivet shearing
a piece out of the bar, Fig.
184.
Let t be the thickness, h
the breadth of the bar, and cl
the diameter of the clenched
rivet and of the rivet-hole,^ also
the tension in the bar, the
working resistance per
square inch in tension = I—
5 tons.
1. To guard against
the first manner of failing
Y., = Q)-d)tr, (63).
2. The resistance to
bearing Tj, has been found
by experiment to vary as
the thickness and as the diameter of the rivet, that is, as td ; so
that if is the safe resistance to bearing per square inch
Fig. 182.
Fig. 183.
(64).
The value of of course depends on the quality of the iron
employed, but it is usually safe to take it as 8 tons per square inch.
3. The resistance of the rivet to shearing T^ varies as the area
of the cross section of the rivet, namely 0-l8d^, therefore
T^= 0-7 8d^ XT, . . . (65).
Tg can be taken at 4 tons per square inch.
4. To guard against the fourth manner of failing, the lap
should be sufficient to prevent the piece of the bar shearing out.
1 See p. 136.
124
NOTES ON; BUILDING CONSTRUCTION
Two surfaces have to be sheared, each havin^ an area oi tx ~
2
(where I is the lap) and the resistance to shearing is therefore
= ilr^ .... (66).
The joint should be so arranged that T^,, T„ and T/ are each
greater than T, so that the metal of the tension bar may be sub-
jected to the full, safe, tensile stress it is capable of resisting ; that
is, the joint itself must not be less strong but a little stronger
than the tension bar to be connected.
Example 21. — As an example assume
6=1-4", t = l", d=f.
Then
Ef = (1-4 - 0-75)i X 5 = 1 -62 ton.
T6 = ix fx 8 = 2-99 tons.
T, = 0-78 x (1)2 x 4 = 1-76 ton.
By putting T/= 1-62 ton we get the least value of /, namely-
l-62_ 1-62
'~^~~2~'
= 0-81 inch.
Practically, however, a lap of only 0-81 inch is quite insufficient, because
the rivet holes would be much too close to the end of the bar, from a manufac-
turing point of view. In the present case the lap should be
2 X IlxcZ
2^ inches
(see p. 136).
Single riveted lap joint. — The equations already deduced can be easily
applied to the case of the joint shown in Fig. 185, thus
R, = (6-3(^)ir^,
Ts = 3(0-78tZ2)r,.
Fig. 185.
Double riveted lap joint-
Fig. 186.
-And similarly for the joints shown in Fig. 186
Rt = ib-3d)trt,
Tb = t{6d)rb,
T,= 6(078d2)r,.
Butt joint with single cover plate. — The joint shown in Fig.
I 87, with a single cover plate, can be looked upon as two joints
like that shown in Fig. 185 put close together, and the equations
RIVETED JOINTS
125
for that joint will therefore be applicable. Clearly the cover plate
should have at least the same thickness as the plates to be
joined.'-
The double-riveted single
cover joint shown in Fig. 1 88
can be calciilated from the
equations for the joint shown
in Fig. I 86.
In the above it has been
assumed, in finding the resist-
ance to tearing of the bars,
that the tensile stress was
uniformly distributed over the cross section, A glance at Figs.
o ■
1
Fig. 188.
189 and 190 will, however, show that this assumption is not
Fig. 189.
warranted, and that the strength of the bar is less than cal-
culated, the tendency of the stress being to draw the bars into
the position shown. This is one of the cases mentioned at p.
108, in which the stress is not transmitted down the centre of
the bar, and an allowance should be made ; but the values of T^
and Tj are not affected.
Butt joint with double cover plate. — This reduction in the
strength of a tie-bar can be avoided by using two cover plates, as
shown in Fig. 191; which arrangement clearly allows of the stress
being uniformly distributed. As regards the thickness of the cover
plates, each should have at least half ^ the thickness of the plates
^ Prof. Unwin says that a single cover plate should be 1| the thickness of the
pLates joined, at least in boiler work. — Machine Design.
" Prof. Unwin says §, at least in boiler work.
126
NOTES ON BUILDING CONSTRUCTION
to be joined ; the resistances of the covers to tearing and to bear-
ing will then be at least
equal to those of the
plates. It is to be ob-
served that the shearing-
resistance of the rivets is
doubled by the use of two
cover plates, because each
rivet has to shear along
tivo sections instead of only
along one, as shown in Tig.
double shear." ^
191.
192. These rivets are said to be in
Joint, Fig. 191.-
GO
Q Q
00
QQ
QO
Fig. 193.
Fig. 192.
-The equations to find the strength of this joint are therefore
-Rt = (b~Sd)trt,
Tg=6(0-78(^2)r«.
Comparing the joint shown in Fig. 193 with that shown in
Fig. 1 94 it will be seen that,
other things being equal, the
first weakens the plates more
than the second does, but
that they have the same
bearing resistance and the
same resistance to shearing.
The joint shown in Fig.
195 has the same beariujg
and shearing resistance a.s
the two above, but weaken.s
the plates by only one rive t
hole.
Considering a sectiom
AA, the plate is weakened
by two rivet holes he, but be-
fore it can tear across thiis
section, the first or leading
rivet a must fail by shearing
or bearing, which practically makes up the difference : so that thits
1 Some experiments show that the resistance to sliearing on two sections is abouit
§ less than double the resistance on one section.
Fig. 195.
RIVETED JOINTS
127
section is as strong as the one through the leading rivet. Similarly,
the section through the three rivets c?e/is stronger than the two
first, there being three rivets ahc to fail as above, to set against the
weakening due to two rivet holes. It should, however, be noticed
that the cover plates should be made thicker than half the thick-
ness of the plates, since their section is reduced by three rivet holes.
Joint, Fig, 195. — The equations for this joint are therefore
= (6 - d)trt,
Ts = 2 X 6{0-78dys,
and the thickness t' of each of the cover plates can be found from
or
Rt
* ~ 2(6 - 3d)rt
Joint, Fig. 196. — These equations also apply to the joint shown in Fig.
\^\^ — vyc \y; O
Fig. 196.
1 96 except as regards the cover plates. The upper plate, in which there is
no joint, acts simply as a distance piece, and does not affect the resistance to
shearing, or to bearing, of the rivets. But the lower cover plate now trans-
mits more stress than the upper one, as shown by the arrows, and should
therefore be made thicker ; in the present case very nearly in the proportion
1 2
j = -, so that and ^t' are the proper thicknesses to give to the top and
bot/tom cover plates respectively. There is, however, a danger, when carry-
ing out the work, of transposing the cover plates, so that in practice both are
maide of equal thickness, and usually of the same thickness as the plates to
be jointed, so as to be on the side of safety.
Gfeneral Formulas for riveted Joints other than grouped Joints.
The student ought to have no difficulty in deducing from the
foregoing the following general formula for riveted joints (other
than grouped joints).
Let Ej be the tensile stress to be borne by the tension bar
(denoted by Ej in the previous equations).
a the number of rivets required for resistance to hearing.
^ the number of rivets required for resistance to shearing.
128 NOTES ON BUILDING CONSTRUCTION
s the number of sections at which each rivet tends to shear
(i.e. 5=1 for lap and single cover joints, and s = 2 for double
cover joints).
7c the number of rivets in the first or outermost row, that is,
the number of rivet holes by which the tension bar is weakened.
h the breadth of the bar.
t the thickness of the bar and also of the cover plates.
d the diameter of rivet holes.
'^^"'^ E, = r,.(6-M)< .... (67).
rt b — M
= n-~d- • ' ' ■ (68).
_ X sx 0-78^2'
rt {b - kd) t
(69).
Tg ' 0-78sd'^ ■
Whichever is greater, a or /3, must be adopted as the number
of rivets in the joint.
The calculations connected with the above are much shortened
by the use of Table VIII.
Double-cover Riveted Joint.
Example 22.— Design a double-cover riveted joint similar to Fig. 195,
the plates to be joined being 9" broad and ^" thick.
We have
s = 2,
1=1,
b = 9,
/ _ 5"
— 8 '
and from the practical rules for Eiveted Joints, p. 136,
9 - 1 X i-
Hence from (68) a = f • ts"^'
16"
= 6 nearly.
(9-il)f
(0-78 X 2 x(i|)2'
= 4 nearly.
Each side of the joint must therefore contain six rivets. On reference
to p. 136, it will be seen that the minimum pitch ought to be 2|-" ; thus only
three rivets can be placed in one row across the width of the bar. Hence
the required joint will be as shown in Fig. 197.
With the aid of Table VIII. the result can be obtained more quickly as
follows : —
The resistance to bearing of one rivet in |" plates is 4*7 tons, and
RIVETED JOINTS
129
the resistance to double shear is 2 x 3"45 = 6'9 tons. The number required
for bearing must therefore be adopted. This number is
^R, ^ 5(9-1 xlpl
4.7 4.7
= 6, nearly, as before.
1
L U/^ 1
0® )
Fig. 197.
Formulae for Grouped Joints.
We have next to consider " grouped " joints (see p. 9 3,
Part I.)
Equations for Joints with three Plates. For Shearing. — Taking a
1 — 1 1 — !• '1 — r — 1
r-H M 1 1 1 1 i
M 1 1 1 1 i ' i 1
' ! ' i j j i i '
A BCD
Fig. 198.
^ ^ r\ r\ 00
?
■ VLV O O O KJKJ
Fig. 199.
simple case of three plates joined together as shown in Fig. 198, we see that
the joint can fail by the shearing of all the rivets, as indicated in Fig. 199.
Let TCj be the number of rivets in each end group (i.e. between A and B
and D and E) ; be the number of rivets between the joints (i.e. between
B and C and C and D) ; t the thickness of the plates, and t.^ the thickness
of the covers.
Then it is clear that the number of rivets sheared through, if the joint
were to fail as above, is
27ij + 2n.2,
and therefore
R, = (2ni + 2?i2)0-78cZ2rs . . . (70).
For bearing. — The joint might also fail by the rivets cutting into the
plates, as shown in Fig. 200. From A to B the resistance to bearing is
tt-^ X dtdr^,
and from B to C it is
and from C to D
So that altogether
B.C. — IV.
n^ X 2tdrfi,
n^ X tdr;,.
(71).
K
NOTES ON BUILDING CONSTRUCTION
From Equations 70 and 71 the total number of rivets required can be
obtained. Two values will be found for 2 {n^ + n^, one being the number of
y Ongtnally tn
! contact with rivet.
Separation between
End of plates.
Fig. 200,
rivets required for shearing, and the other the number required for bearing ;
the larger value is to be taken.
So far we have only found the sum of + In^, and to obtain n-^ and
separately we must consider two other ways in which the joint may fail.
^ ^ ^
The value of n._
— CT-cr
Fig. 201.
Failing as in Fig. 201. — In the first place it may fail as shown in Fig.
201, by the rivets between B and D shearing and the cover plates tearing
across at B and D'.
The resistance to tearing of the cover plates is, if is the number of
rivets to be deducted,
2 X (6 - n^d)t^rt.
depends on the width of the plates and on the pitch
of the rivets ; for instance, with 10" plates and 3" pitch,
ng would be 3, as shown in Fig. 202.
The resistance to shearing of the rivets between B
and D' is
27).2 X 0-78(/2rs,
so that
= 2(6 - n^d)t^rt + x 0-7 Sfi^r, . (72).
It must now be determined wliat value should be
given to t^. On referring to p. 127 it will be seen that
when two plates have to be joined, the cover plate
nearest the joint should be made f i thick, and by the same reasoning it will
appear that for three plates
We can therefore find the value of from equation 72, and hence the
value of riy
Practically, as already mentioned, it is usual to make t^ = t, but this will
not affect the above calculations.
Gutting into covers and plates. — The joint can also fail by the rivets cutting
into the covers between A and B, and into the plates between B and D.
The resistance to bearing of the covers between A and B is
2n.^t^dri),
1.
Fig. 202.
.-it.
RIVETED JOINTS
and the resistance of the plates to bearing between B and D is
2n2*(^rj, + ^^2^<^r^, (see p. 129).
So that, assuming as before = t,
% = i2n^ + 37i^)tdri, . . . (73).
It is only necessary to see that the values already obtained for and 71^
are sufficient to give the bearing resistance required by this equation.
On the whole, therefore, the strength of a grouped joint with three
plates can be calculated from the four equations 70, 71, 72, and 73.
Similar equations would be obtained whatever the number of plates to be
joined, and the following are the general equations which the student should
check by working out several cases.
Ge7ieral Equations. — Let N be the number of plates on each side of the
joint which are to be connected, then
for shearing (see 70)
Rt= (271^ + (N- 1)»2} 0-7 8(^2 . . (74),
for bearing (see 71)
N r )
R,= -| 2«i + (N-l)n2 j-rfirs ■ . . . (75),
for tearing of covers and shearing (see 72)
R, = 2(6-'H3(Z)i2?-t + (N-l)n2xO-78rf2rs . (76),
for bearing on plates and covers (see 73)
R, = I 2??,^ + |(N - 1)^2 I dtrt, . . (77).
The total number of rivets is evidently
+ (N - l)n2 (77«).
These equations can be solved very simply by the use of
Table VIII., as will appear from the following example.
Double-cover Gro7iped Joint.
Example 23. — Design a double-cover grouped joint to connect four plates,
each 9" x |", the end group of rivets being so arranged as to only weaken
the plates by one rivet hole.
Preliminaries. — Let cZ= l|- inch, which is the usual size of rivet for |"
plates, and let r^ = 5 tons, r& = 8 tons, and = 4 tons.
Then
R, = 4(9-li)f x5,
= 118 tons.
Now from Table VIII.
0-78(^2^^ = 3-98 tons,
dtrjj = Q-l4: „
.-. -(Z<j-6= 13-48 „
Clearly, therefore, a greater number of rivets are required for shearing
than for bearing.
Hence, from Equation 74
2?i, + 3ft2 = 3-g^g = 29-6 . . . (78).
132 NOTES ON BUILDING CONSTRUCTION
We must now apply Equation 76. Following the same reasoning as at
p. 127 it will be seen that for four plates
4
i, = -i.
Hence assuming 11^^ = 3 we have (76)
118 = 2 (9-3x 1|) fxfx 5 + 3n2x4,
whence
118-34 ^
The arrangement shown in Fig. 203 only gives Wg^^j '^^^^ enough,
QQ
QQ
@Q
QQ
QQ
QQ
QQ
QQ
QQ
Q^ \
•.Q^ /
/-V /-\
^
^
^ /-^
Fig. 203.
considering that the safe resistances assumed for bearing and shearing are
rather low. Moreover, in practice would be made equal to t. Substituting
this value (6) for n.-^ in equation 78,
2ftj = 29-6 - 3 X 6,
= 11-6.
n-^= 5-8.
Six rivets therefore will do.
Substituting these values of n-^ and n^in Equation 77 we find
R, = (2 X 6 + |x 3 X 6)6-74,
• =323-5 tons,
so that there is no fear of the joint failing in the fourth manner.
The joint is shown in Fig. 203.
Douhle-cover Grouped Joint for Plate Girder.
Example 24. — Design the right-hand joint for the lower boom of the
plate girder shown in Fig. 259 and Plate A.
On referring to Example 36 it will be seen that the plates to be joined
are two in number, and that they are 9" wide and f" thick ; and further
that f " rivets are used.
Owing to the angle irons connecting the booms to the web, only two rows
of rivets can be used, so that = 2.
Take
rt = 5 tons per square inch,
''6 = 8 „ „
= 4 „ „
From Table VIII. it appears that for f " rivets
resistance to shearing = 1-77 ton,
resistance to bearing in f" plate = 2*2 tons.
so that
RIVETED JOINTS
N
— d<rb = 2-2 tons.
133
16.
Hence a less number of rivets are required for bearing tlian for shearing.
Now since two f " rivets have to be deducted
K,= 2x{9-2x|)f x5,
= 28-2 tons.
Hence total number of rivets required for shearing
_28-2
~ T?
Applying Equation 7 6 we find, remembering that for two plates = ^t,
28-2 = 2(9 - 2 X f)f X f X 5 + W2 X 1-77,
28-2 -18-75
. •. Too = -— — = 5 "3 nearly.
And since = 2, three rows of rivets will be required in the length of half
the joint.
Lastly, since from (77a)
2n^ + ng = 1 6,
. •. = 5 "35,
so that three rows must be used.
In practice, however, would be made equal to t, and the above calcula-
tions would be modified as follows —
28-2 = 2(9 -2 X Dfx S + jigX 1-77,
28-2 - 28-2
»o = = 0.
1-77
This appears to be an absurd result, but since there are only two plates
to be joined, and the cover plates are made of equal thickness to the plates,
^—i —
U3
Fig. 204.
'L. I. Wrapper 3x3x^/s
_ Q _ \© ©
Fig. 205.
it follows that the covers are able to transmit the stress withou.t the help of
any rivets between the joints. In practice, however, this would be a bad
construction, and a couple of rows of rivets would probably be used, as shown
in Figs. 204 and 205.
On referring to Example 36 it will be seen that one of the covers can be
134
NOTES ON BUILDING CONSTRUCTION
formed by extending one of the plates of the booms ; the other cover can
be added in the usual way, and the angle irons can be cranked over it as
shown in Fig. 259 (an objectionable arrangement), or the upper cover can be
formed by angle wrappers as shown in Fig. 206.
Oblique riveted joints. — So far we have
only considered riveted joints connecting plates
placed in prolongation one of another, but
clearly the same rules are applicable when one
plate, or set of plates, is placed at an angle to
the others as shown in Fig. 208.
Ys "wcb
k 3"xs'x%
Fig.
206.
Fig. 207.
Fig. 208.
Example 25. — Thus, taking the dimensions given in Fig. 207 and
deducting two rivets, we have, assuming rj = 5 tons — ri = 8 tons — and Vg = 4
tons, R, = (6-2x 1)1x5,
= 11-25 tons.
If a is the number of rivets for bearing —
axlx|x8 = ll-25,
a = 4 nearly.
And if fi is the number of rivets for shearing —
^x 0-78(1)2 x 4 = 11-25,
/8 = 6-3 nearly.
So that the joint is weaker in shearing than in bearing.
Using Table VIII. we find resistance to bearing of one rivet
= 2-99 tons.
Resistance to shearing of one rivet
= 1-77 ton.
Therefore number of rivets required
11"25
= = 6-3
1-77
as before, and six rivets will do.
It might be objected that, in calculating R^,, only one rivet should be
deducted to obtain the effective cross section of the tension bar, but it will
RIVETED JOINTS
135
be observed tliat the leading rivet is miicli nearer to one edge of the bar than
to the other, and therefore weakens the bar by more than one rivet hole, as
explained at p. 126. Professor Eeilly has treated this subject at some length
in a paper published in Vol. XXIX. Minutes Proceedings of the Institution
of Civil Engineers, and shows that a similar faulty arrangement of rivets,
though of frequent occurrence in practice, will increase the intensity of stress
on one side of the bar by as much as 26 per cent. Professor Eeilly not only
recommends that the leading rivet should be placed on the centre line of the
bar, but also that all the rivets be placed symmetrically about the central line, or
at any rate that the centre of gravity of the rivets lie on the central line, which
he calls the mean fibre. Thus arranged, the joint we have just calculated
would be as shown in Fig. 209. Only one rivet need now be deducted,
so that
E, = (6-1)^x5,
= 13-1 tons,
The rivets are rather numerous, and it would be better therefore to use
larger ones. Ee-calculating with |" rivets we find
Rx = (6-|)ix5,
= 12-8 tons.
And from Table VIII.
Eesistance to bearing = 3-49 tons;
Eesistance to shearing = 2-4 „
Therefore
12-8
a = — -— = 5'3.
2-4
Five or six rivets ought therefore to be used, and with six rivets the
arrangement of the joint would be as shown in Fig. 210.
Riveted Joints in Compression. ■
At one time it was considered that the compression stress
was transmitted across the joint by the plates abutting against
each other, and such would be the case if the joints were perfectly
136 NOTES ON BUILDING CONSTRUCTION
fitted ; the rivets would then transmit no stress, and their only
function would be to hold the plates together. Such fitting
cannot, however, be ensured in practice, and it is therefore
assumed that the whole stress is transmitted by the rivets, and
the calculations are in no wise different from those required for
tension joints. The following points, however, should be ob-
served —
1. No deduction is to be made for rivet holes, as the metal of
rivet completely fills up the hole and passes on the stress.
2. The longitudinal pitch of the rivets sometimes requires to
be less than in tension joints, or the plates might buckle between
the rivets as shown in Fig. 211.
Fig. 211.
Dimensions for Riveted Joints.
Diameter of rivet holes.— II t is the thickness of the thickest plate to be
jointed —
Fairbairn's rule : 2t for plates under i" thick,
~ 5) ^" thick and over.
Un win's rule : d = 1-2 Jt.
Pitch of rivets. — Minimum pitch of rivets =2xd; minimum distance 1 of
centre of rivet hole from edge of plate = 1 1. c7.
The pitch of rivets in girder work usually varies from 3 to 5 inchies,
but it should not exceed 10 to 12 times the thickness of a single plate, as
otherwise damp may get in between the plates and cause rust, which in tinne
swells and bursts them asunder 2; 4" is a common pitch for girder wonk.
The pitch is sometimes made greater in the tension than in the compressicon
flange.
Note upon the Difference between the Figured and Actual
Sizes of Rivets and Rivet Holes.
In the calculations given above, the rivets and the rivet holes are assumted
to be of exactly the sizes shown or figured upon the drawings.
In the actual work, however, neither the rivets nor the rivet holes aire
of the sizes shown upon the drawings.
Iron rivets.~To begin with, the rivet as purchased is generally ^ incch
smaller diameter than its nominal size— that is, a so-called i inch rivet woulld
be ^4 inch in diameter.
When, however, the rivet is heated and clenched it will, if the work Ibe
^ tV" to should be added in the case of thick plates and rivets. - Stoney.
RIVETED JOINTS 137
properly done, very nearly i fill the hole, and the hole will be found to be
considerably larger than the nominal size of the rivet, especially m the case
of punched holes. , , •, . t x xi,
FwiM holes in iron plates.— ^Yhen a hole is punched m a plate the
diameter of the punch is generally j\ inch larger than that of the nvet that
is to go into the hole, and then again the die is larger, m proportion to the
thickness of the plate, than the punch.
This results in the punched holes being larger at the lower surface of the
plate than at the upper, and the actual diameter of the holes is from 10 to 20
per cent larger than the nominal diameter of the rivets.
Strength of iron rivets in punched holes in iron plates.— li will be seen,
therefore, that the rivets filling punched holes will have a considerably larger
shearing area than the rivets shown in the drawings. , , ^ ,
If rivets of the figured sizes are strong enough to withstand the calculated
stress, then the actual rivets in the work will evidently be unnecessarily strong,
and their number might, so far as this point alone is concerned be reduced.
On the other hand, however, it must be borne in mmd that the stress
upon a riveted joint is not divided equally among all the rivets, as assumed
in the calculations, but that in consequence of bad workmanship, and for other
reasons (see p. 135), increased stress may come upon some of the rivets while
others do not take their share. n i . j
For this reason engineers sometimes add 10 to 20 per cent to the calculated
number of rivets to make up for uneven stress. . ^ . , ^
Thus on the one hand, though fewer rivets would be required m the work
than those calculated, because the rivets as clenched are actually of larger
sectional area, yet on the other hand more rivets would be required m order
to make up for defective workmanship and consequent inequality of stress.
Strength of actual riveted joint may be practically taken the same as figured.
—The result is that these two causes affecting the number of rivets actually
required as compared with those calculated may be considered to cancel one
another,3 and there is no practical error involved in calculating the rivets
according to their figured dimensions. This is the usual practice, and it is
followed in these Notes. .
Punched holes in iron tension joints.— mth regard to the deduction to be
made for rivet holes in tension joints the case is somewhat different, ihe
diameters of punched holes are, as mentioned above, from 10 to 15 per cent
larg-er than the nominal size of the rivet, and in addition to this, m roughly
punched work, the edges of the holes are somewhat torn, so that the effective
1 " In order to get good riveted joints the rivet must be properly put in and must
fill the hole when cold, but as a matter of fact this is really never the case m conse-
quence of the rivets contracting laterally when cooling. This is not so much the case
witb steel rivets, as they are or should be worked at a dull red heat. The most per-
fect joint would be one in which drilled holes were used and the nvets turned and
closed quite cold. "-Moberley, quoted in Stoney on the Theory of Stresses, p. 684.
2 Stoney. . , .
^ There are other minor circumstances affecting both sides, e.g. nvet iron is gener-
ally better than other iron, and might be subjected to a higher shearing stress. The
friction between the plates assists the rivets; on the other hand an iron rivet
sheaired on two surfaces has not twice the resistance of one surface, but only Ig times
that; resistance.
138 NOTES ON BUILDING CONSTRUCTION
area of the plate is reduced by a width greater than that of the nominal
diameter of the rivet.
Mr. Stoney says : " Probably the most accurate method for making an
allowance for the injurious effect of punching would be to add a certain per-
centage, say Jg- to i of its diameter, to each hole when calculating the effective
net area of a punched plate. Mr. White (Director of Naval Construction)
states that they were accustomed to allow 4 tons off the very best iron for
punching— 4 tons off 22 tons. Civil engineers, however, rarely make any
similar allowance in riveted girder work, probably through inadvertence, or
because the rivet holes are generally pitched farther apart in girders than in
ships." 1
It will be seen, therefore, that in the case of the holes as well as that of
the rivets it is the usual practice to base the calculations upon the figiured
dimensions of the rivets, it being generally considered that the working
stresses taken in practice are so low as to leave a sufficient margin to cover
all extra stresses caused by enlarged holes, rough pimching, injured plates,
and other defects in workmanship.
In some cases, however, engineers allow for the damage done to plates; by
punching, making a rough addition to the diameter deducted to cover this,
e.g. in the case of a f inch rivet they deduct an inch from the eftecttive
width for the plate. In all important tension joints this question should be
carefully considered and some allowance of the kind made if conside;red
necessary.
Bimering out.~li the hole is punched about 1 inch less than the requiired
diameter, and this margin is afterwards bored out to the required diamelter,
the damaged metal is removed and no harm has been done by the punchiing!
Drilled holes. — In the case of drilled holes their diameter is only abiout
^ inch larger than the nominal diameter of the rivet. The edges round the
hole are not damaged, so that there is no practical difference between the
actual sizes of the holes and those figured for the rivets.
Steel rivets fill the holes better than iron ones, because they are not , so
hot when clenched and contract less in cooling.
PIN JOINTS.
These joints are much used for connecting the various trie-
bars of roofs to the other members, and examples can be seen < on
reference to Parts I. and II. Such joints are also occasionallly
used for open girder work.
These joints are formed with liniks
somewhat of the shape shown in Ffig.
2 12. A little consideration will shcow
that this link is liable to fail Iby
^^g- tearing, as shown in the figure. Tfhe
continuity in the fibres of B is interrupted by the hole for tlhe
1 Stoney in Theory of Stresses, p. 650.
PIN JOINTS
139
pin P ; moreover I is further liable to be split from the inside by
P (if it is too small) bearing upon it, so that the intensity of
stress over the cross sections cc is by no means uniformly distri-
buted. However, the calculations that would be required on
Shaler Smith,
(hammered ) Howard.
Fig. 213. Fig- 216. Fig. 215.
Berkley.
Fig. 214. Fig. 217.
the: assumption that the stress is not uniformly distributed are
far too difficult for this book, and would certainly not be carried
out; in practice. We must therefore resort to experiments.
Such experiments were made by Sir C. Fox, C.E., and more
recently by Mr. David Kirkaldy, C.E., and by Mr. "C. Shaler Smith
I40 NOTES ON BUILDING CONSTRUCTION
in America. Figs. 213-220 give the shapes of links deri/ed
from these experiments, and also other shapes used in acmal
practice. It will be seen that there is a considerable diversit; of
opinion. In each of the cases shown, the various dimension; of
the heads are given in terms of the width of the body of the Ink
as the unit. On the whole, perhaps, the American experimmts
are the most reliable, and in these experiments the effect of the
method of manufacture and of the proportion the diameter of the
pin bears to the width of the bar was investigated.
Table IX. embodies the results of these experiments, and we
proceed to work out an example by its help.
Pin Joint for Roof-Truss.
Example 26. — Determine the dimensions required for the various pirts
of the joint shown in Figs. 219 md
220.
This is the joint at the poitt F
of the roof worked out in Example
42. From Table H, p. 213, we
find that FA is subject to a stress
of 7-83 tons, FG to 3-88 tons,
FC to 4-13 tons, FD to ]-69
tons.
The dimensions of the strut have
already been determined in Exan.ple
19, p. 119, that of the tie-rod FA in
Example 13, p. 108, and of the tie-
rod FG in Example 27, p. 147 ; and
it will be observed on referring to
these Examples that if" is the effect-
ive diameter of the rod FA and that
1"'0 is the effective diameter of the
rod FG.i
Pin holt. — It will be seen that
for the double eye of the tie-rod FA
the pin has four shearing surfaces,
but that there are only tioo shearing
surfaces for the single eye of the
tie -rod FG. Clearly, therefore, the
Fig. 220.
diameter of the pin will be regulated by the tension in the tie-rod with
the single eye, namely 3-88 tons. Let d be the diameter of the pin, then
since it is in double shear
2 X 0-78(?2x
3-
^ The actual diameter 1^ is reduced by the screw in coupling.
PIN JOINTS
141
taking r^ = '^ tons, we find
(^ = 0-79 = i|" nearly.
Practically a -I" jsin would be used, as some allowance must be made for the
pin not fitting tightly. Professor Unwin recommends as a full alloM'ance
that only |ths of the shearing resistance should be reckoned on. This Jull
allowance would require a pin diameter.
Tie-Rod FG. — The eye at the end of the smaller tie-rod is shown in
Fig. 2 22j and we have to determine three dimensions, viz. a, the thickness
of the link ; h, the width at the back of the eye ; and c, the width of the
sides.
The dimension a has a minimum value depending on the hearing resistance
of the eye which must be equal to 3*88 tons. Hence, taking the pin as |-"
diameter,
« X f X »'6 = 3-88.
Or, taking r^, = 5 tons, ^
3-88 X 8
7x5
■ 0-88 inch.
To allow for the pin not being quite tight it will be best to make a =
1 inch.
Eeferring now to Table IX, To use this Table we must first find the
ratio diameter of pin d
which is |--Mv^«=0-74.
width of the bar
The nearest number given in the Table is 0*75, and we find 0-67 as the
number in column 2 of the Table.
^ . 3-88 .
Hence, smce is
5
the effective area of the tie-rod.
ca X 5
3-88
0-67,
ha = ■
,\ c = 0*52 inch.
Next to find h ; for a hammered eye the cross section at the back of the
eye is equal to the efi'ective cross section of the tie-bar. Hence
3 -88
whence
6 = 0-78 inch = I" nearly =0-9".
Tie Rod FA. — "We next have to find the dimensions of the double eye
of th e larger tie-rod FA. Each half can be calculated as a single eye reckoning
on half the stress.
As before, to find a we have
7-83
«x| X 5 = ^-.
Or
a = 0-9.
This is rather wide, and will make the joint look clumsy. To diminish
a tke diameter of the pin must be, increased. Let, therefore, the pin be
1 " American engineers who have great experience in pin bearings . . . usually
limit the bearing pressure to from 10,000 to 12,000 lbs. (4-46 to 5-36 tons) per square
inch on the projected area of the pin, and this, no doubt, is a safe rule for pins." —
Stomey on Strains.
142
NOTES ON BUILDING CONSTRUCTION
made 1" in diameter ; this necessitates re- calculating for the pre;vions
eye.
Eye of FG. — It will be found that for this eye
3-88
«= = 0-78,
or say ^" to allow for a loose pin. We now have
diam. of pin d
■ fa f i = - = 1 ^ ItV ^
width 01 bar a
This number does not appear in Table IX., but by interpolation we find
that 0-7 is the corresponding number of column 2. Hence
ca X 5
and
3-88
0-7 X 3-88 X 8
5x7
= 0-62,
Whence
^iaitiiiiiiiii liiiiiiiiiiii
iflllillliiliJIIIIIIiillllllllllllllilllilli
Fig. 221.
Elevation of Eye of FG.
Fig. 222.
Flan of Eye of FG.
& = 0-9 inch.
These dimensions are given in Fig. 22 1.
E]ie o/FA. — Returning to the double eye we have for a pin 1 inch diameter
7'83
rt X 1 X 5 = ,
2
a = 0'78 inch.
3
IBll
Fig. 223.
Plan of Eye of FA.
Or, allowing for a loose pin, each half of the double eye can be made |"
PIN JOINTS
143
diam
thick. But the ratio ^^^r^ for each half of the double eye is 1 -^ If = 0-73,
so that
ca X 5
= 0-67,
Further
ha —
ix 7-83
.•. ct=0-6 inch,
(ix 7-83)
5 '
.-. 6 = 0-9 inch,
or very nearly the same dimensions as those found for the single eye.
Practically the double eye would be of the form shown in Fig. 221, and the
single eye also of the same shape.
Rod. FC. — There remains to be designed the double eye for the small
tension-rod FC having a stress of 4-13 tons. In this case we have
4-13
a X 1 X 5 = — — ,
or a = -4 13 inches.
Fig. 224.
Flan of Eye of FC.
But as this thickness would be barely sufficient to make the arms of the
jaw of equal strength to the body of the tie-rod, we will make a = i An
increase of strength in this part will be judicious.
If the diameter of the rod FC be taken as 1" as shown in Figs. 373, 374,
1 ,1 . , , „ , „ , diam. of pin
and the width of the arms of the iaw also 1 , then -t^-t — , = l -m = 1,
' width of tie-bar '
and the corresponding number in column 2, Table IX., is 1*50. Hence
c + c=lY c = |". The total width of the eye will therefore be nearly
the same as that of the jaw shown in Fig. 221, which will do very
well.
The cross section at the back of the eye is as before equal to that of the
tie-bar, or = "78 square inch — h therefore equals 0-78" or say |" ; but as it is
not desirable to make minute alterations in the patterns of these eyes, we
will take h = -9", and the shape of the eye will then be similar to that in Fig.
221. The complete jaw is shown in Fig. 224.
Fig. 220 shows the complete joint. It is wide in the direction of the
bolt, and therefore somewhat clumsy in appearance. It would be improved
by taking A = f " instead of |". The student is recommended to try this.
144
NOTES ON BUILDING CONSTRUCTION
Joints at Apex and Feet of Truss.
Example 26a.— The connections at the apex and feet of the roof-truss are
sometimes made by castings.
These, however, are clumsy and unreliable, and apt to be broken in transit.
As the principals are usually sent to their place of erection in parts (so
as to avoid injury, and for convenience of carriage), it is better in the case of
small roofs where there are not many principals to use bolts rather than
rivets, because there would be so few rivets that it would not be economical
to send a gang of workers witli their forge and tools, and because with bolts
under ordinary circumstances all the necessary work in erection can be done
by a carpenter. 1 i i i
For the roof, Tig. 373, p. 215. The rafter is a T iron 4i x 4^ x ^, and
the stress it sustains = 7-83 tons (Table H, p. 213).
7'83
Taking rj,= 5 tons, bearing area required for bolts = = 1"56 square
incli.
For such roofs it is better therefore to use bolts, and connections with
iVi'Bolts.
Fiff. 225.
Fig. 225a.
bolts are shown in Figs. 225, 225a. That at the apex will now be calcu-
lated.
Joi7it at apex. — Assuming l^" bolts.
Bearing area of each bolt = ^" x li" = -62 incb.
1-56
Number of bolts required on each side of the connection = = 2-5,
say 3.
As there will be a plate on each side of the T iron the bolts will be in double
shear, and the shearing strength will not require investigation, as it will be
very much in excess of what is required.
The united thickness of the plates should be at least equal to the thick-
ness of the stem of the T iron, i.e. |" ; but as it is better to make them a little
in excess, in this case two plates each |" thick will do.
Having calculated the joint for the T irons, the bars FC and CG (see Fig.
373, p. 215) must be considered. The diameter of the bolts should first be
calculated in proportion to the bars, as explained at p. 140, then the bearing
area of the bolt on the connection plates investigated. If the thickness of
plates calculated for the T irons is not sufficient, then the plates or the
diameter of the bolts connecting FC and CG must be increased ; in this case
the bars FC and CG are only 1" diameter, and it can be seen by inspection
that the plates are amply thick. Fig. 225a shows the connection.
SCREWS
145
Joint at foot. — The shoe of the principal can also be made of wrought
iron, as shown in Fig. 225. The stress and the bolts will be the same as
at the apex.
SCKEWS.
the
Screws are of constant use in building construction in
shape of nuts and bolts, screw connectors, etc.
The thread usually employed is that shown in Fig. 226; it
will be observed that it is cut into the bolt or rod, and is therefore
called a minus thread. The strength of the bolt or rod is thereby
Lliniis thread
Fig. 226.
Plus thread
Fig. 226a.
weakened, and to obviate this loss what is called a 'plus thread is
sometimes used, as shown in Fig. 226a. For such a thread the end
of the bolt or rod has to be thickened by upsetting, and it is a
matter for consideration, in each case, whether the saving in metal
is worth the additional cost of cutting a plus thread. As a
matter of fact their use is somewhat limited.
The thread on the bolt is called a " male " thread, and the
thread in the nut into which it fits is called a " female "
thread.
The resistance of the screw to being torn out of its nut
depends on the resistance
of the threads to being
sheared off. With the V
thread generally used in
building construction, the
area to be sheared is the
surface area of a cylinder
of height I and of diameter
D (Fig. 227). D is the
diameter of the bolt, taken
at the bottom of the
threads, and is called the
Fie. 227.
effective, diameter, and the ratio between D and d depends on the
B.C. IV. L
146
NOTES ON BUILDING CONSTRUCTION
pitch and the kind of thread employed. In the case of Whit-
worth's screws, which are in general use in this country, the
ratio is
D = 0'96Z-0-06 . . . (79),
which can also be written
^ = ^^^'^^ = 1-1 xD + 0-07 nearly . (80),
D and d being expressed in inches.
Now the area of the cylinder, that is the shearing area
required, is
Hence Es = 7rDL'?v
As regards the value to be given to i\, it is to be observed
that the metal is weakened in cutting the thread, and for this
reason Humber recommends a value of 2 tons to the square inch
(instead of 4 tons, the usual value). Substituting for o\ and D
we get
E, = 7r(0-9£?-0-06)Zx 2.
Practical formula. — This equation will be found to reduce
approximately to the following, which is a form suitable for prac-
tical use —
E^ = (5-6c^-0-4)Z . . . (81).
Theoretically the resistance to stripping of the thread should
be equal to the tensile resistance of the bolt. The latter is
^ 7rD2
So that
r, 4'
= -^(0-9«5-0-06).
4r,
find
Substituting 5 tons for and 2 tons for r, and reducing, we
Z = 0-5 GtZ- 0-04 nearly
(82).
It thus appears that I requires to be rather more than
The usual practice is to make I = d, which is a good, safe rule.
SCREWS
147
Screw of Tie-Eod.
Example 27. — The tie-rod of a roof is subject to a stress of 3-88 tons
(see Example 42, p. 209), and is provided with a screw coupling as shown
in Fig. 228. Find the diameter of the tie-rod and of the coupling.
I
I
i« .3 J." >^
Fig. 228.
It will be noticed that one end of the coupling is cut with a left-handed
thread and the other with a right-handed thread ; thus when the coupling
is turned in one direction the tie-rod is drawn together, but is slackened off
when the coupling is turned in the other direction.
We must first find the effective diameter of the tie-rod. Referring to
Equation 58, p. 106, it will be seen that
3-88 = 5 -{- ,
4
■n2 4 X 3-88
5 X 77
or D = 1 -0 inch.
Hence from Equation 80
d=\-\ X l-O-f-0-07,
= X-I7 = l3-V' nearly.
Therefore following the rule l = d, the female screw in the coupling
should be made inch long. The length of screw thread to be cut on each
half of the tie-rod depends on the amount of play desired ; under ordinary
circumstances about 3 or 4 inches of screw on each will do, and consequently
the least length of the coupling over all would be 6 or 8 inches.
To find the outside diameter of the coupling theoretically required for
strength, make the cross section of the coupling at A B equal to the effective
cross section of the tie-rod. If d^ is the outside diameter of the coupling, then
d^^ = T)'^ + d\
Or in this instance
(^^2 = (1-0)2 + (1.1 7)2.
Whence c?^ = 1 -5 4 = 1
Practically this diameter is quite insufficient, for a clearance must be
allowed beyond the outer diameter of the thread, or the female screw could
not be cut in the coupling ; the inner diameter of the coupling would there-
fore have to be 1-|. The thickness of metal would also be, say, ^ inch,
making the outer diameter of the coupling 2|-, or 2 x d — a very proper
proportion.
NOTES ON BUILDING CONSTRUCTION
A usual form of
COTTEK JOINTS.
these joints is shown in Fig. 229 (see also
Part I.) Such a joint tends
to fail as follows —
1. By shearing of the
gib G and cotter C (double
shear).
2. By the cotter cutting
into the tie-rod T.
3. By the gib G cutting
into the sleeve ss.
Fig. 229, Let h be the thickness of
the cotter or of the gib bearing against the rod or the sleeve, and
c the width ^ of the gib and cOtter ; t the thickness of the flattened
part of the rod, and the thickness of each side of the sleeve, then
Es = 2kc X r^.
For the rod Vv^^htx r^.
For the sleeve R'b = '^^k ^n-
Moreover, the effective section of the flat part of the tie and
of the sleeve must be proportioned to the resistance to tension of
the tie-rod, namely K^.
For a theoretical joint these resistances should each be equal
to E^, hence 2A;cr, = E^ . . . (83),
yt^rj = E^ . . . (84),
and clearly t=2t^ . . ■ (85).
If the value of k is assumed, the remaining dimensions for a
given value of E^ can be found.
Example 28. — The tie -rod of a roof is connected to the principal,
and to the shoe at the abutments, by means of a cotter joint, as shown in
Fig. 229. The tension in the tie-rod is 7-83 tons (see Example 42, p. 209,
and Table H, p. 213). Find suitable dimensions for the cotter joint.
Let k = ^", then from Equation 84, taking = 5 tons per square inch
^xix 5 = 7-83,
whence i = 3*1 inches. This is too large a value for t. It will therefore be
necessary to take a larger value for 7c, for instance 1 inch, then
7-83 X 1
Practically 1|" will do.
t =
0
Hence (85)
t
1-56.
And
2 X 1 X c X 4 = 7-83,
c = 1 inch nearly.
A taper of 1 in 24 to 1 in 48 is usually given to the cotter.
JOINTS IN WOODEN STRUCTURES
149
This is the minimum value of c. Practically, more room is required to
fit in the wedges and cotter ; not less than 2 or 2^ inches.
Lastly, to find the depth a of the sleeve. The effective section of the
7*83
sleeve is 2 x (a — = — -
5
2(a-l)f = ^,
a-\ = 1-04,
or a = 2'04 inches.
Some allowance ought to be made for possible errors in manufacture, so
that practically it would be well to make a = 2i",
The depth of the fiat part of the tie-rod can also be made 2^ inches.
The length of the sleeve should be made sufl&cient to prevent the cotter
shearing a piece out of each side of it, and the same is true of the flat part of
the tie-rod ; the calculations are, however, left to the student. The dimen-
sions given in the figure are slightly more than those theoretically necessary.
JOINTS m WOODEN STEUCTUEES.
Numerous kinds of joints are employed by the carpenter in
framing wooden structures, and a large number of these were
described in Part I. The proper proportions for such joints have
been determined by long experience, and the necessity to make
any calculations in connection with them is never likely to arise.
It is, however, thought that the consideration of the principles on
which the strength of such joints could be calculated will be a
useful exercise for the student, and a few of the joints given will
therefore be examined with this object in view.
"Fished" Joint for Wooden Beam.
Example 29. — For instance, let us inquire into the joint shown in Fig.
230, which is shown in most books on
Carpentry, and in Part I.
If the bolts fit tightly into the ^
wood it can be considered that part uuuu juuu
of the stress is transmitted by them,
and the remainder will therefore have
to be transmitted by the tabling.
Clearly only the resistance to bearing
Fig. 230.
of the bolts need be considered. If the fish-plate is made of the same wood
6 — 2i
as that to be joined, the thinner part t should be made equal to — - — , i being
the depth of the indent. Hence
2t=='b-2i . . . . . (86).
The resistance to tearing of the fish-plates will then be equal to the resistance
to tearing of the beams to be joined. It will also be seen that then the
resistance to bearing of each bolt of diameter d is
2i X d X rj).
ft ft f, rt.A f. ^ A_
'LI I I ]i 1 1 . y I I II I I'
I □
i
NOTES ON BUILDING CONSTRUCTION
So that if there are n bolts on each side of the joint the total resistance
Therefore the resistance to be demanded of the tabling is
— 2ntdr]).
But (remembering that we must deduct one bolt hole),
Therefore the resistance of the tabling must be equal to
2t{a - dy^ — 2ntdrij.
Now the indent must be of sufficient depth to prevent the fibres being
crushed ; this condition is fulfilled by
2t{a - d)rt - 2ntdrf, = 2iarc . . . (87).
Again, the length of the indent and the distance AB must each be
sufficient to prevent shearing off of the wood. The resistance to shearing of
either is
ears,
therefore t{a - d)rt — ntdr^) = eaVg . . ■ (88).
From these equations the various dimensions can be found.
As an example take the following data, which assume that the wood is
of fair quality Baltic fir.
a =12",
b - 8",
n = 4",
rf = 12 cwts. per square inch, Table I a, p. 328.
rg = 1 0 cwts. „ „
rs= 1-3 cwt. „ „
rft = 12 cwts. „ „
Substituting these values in Equations 86 and 87, we get
2t=8-2i,
or t = 4~ i.
And «(12-|)12-4i!x|x 12='ix 12x 10,
or 99i=120i.
Therefore 99(4 - i) = 1 20i,
219i=396,
4 = 1"8 inch,
and < = 2*2 inches.
Again, from Equation 88 we have
2-2 X (12 - f) X 12 - 4 X 2-2 X f X 12 = ex 12 X 1-3,
whence e = 1 4 inches,
or say llf inches.
It would probably be better in practice to neglect the resistance offered
by the bolts, and take them as only holding the joint together. To adapt
the above equations to this case it is only necessary to omit the term
2ntdri). The equations then become
2t = b- 2i,
t{a — d)rt = iavc,
t(a — d)ri = eavg.
Taking the same data as before,
t = 4-i,
^(12 -1)12=1 X 12 X 10,
JOINTS IN WOODEN STRUCTURES
whence
Hence
and from Equation 88
(4 -i)135
. 135x4
255
i = 4-2-l
= 120i,
= 2'1 incliea.
= 1-9,
1-9(12 -1)12 = ex 12 X 1-5,
1-9 X 11-25 X 12
or e= ,
12 X 1-5 '
= 14^ inches nearly.
It appears strange that the value obtained for e in this case should be
so little greater than in the first case, but the reason is that the beam is
much more weakened than in the first case owing to the greater depth of
the indent, as will be seen by the following.
In the first case the resistance to tension of the joint is
2<(a-d)rt = 2 X 2-2 X 11-25 X 12,
= 594 cwts.
In the second case it is
2i'(a-d!)re = 2 X 1-9 X 11-25 X 12,
= 513 cwts.
Further, the safe resistance to tension of the full section of the timber,
namely 8" x 1 2", is
8 X 12 X 12 = 1152 cwts.,
from which it is evident that this is a very uneconomical form of joint, more
than half the strength of the timber being lost.
Scarfed Joint for Wooden Beam.
Example 30. — Calculate the safe strength of the joint shown in Fig.
231, which is a bad form, as pointed out in Part I.
Fig. 231.
Preliminaries. — The various dimensions are given in the figure, taking
as before
= 1 2 cwts. per square inch,
»c=12 ,, „
Vg—l'S ,, „
'•6=12 „ „
This joint can fail in three ways : (1) by tearing across da or be ; (2) by
shearing along ca; (3) by the fibres at ab or a'b' being crushed.
For the first we have
E^ = 9 X 2f X 12 = 297 cwts.
For the second
R« = 24 X 9 X 1 -3 = 280 cwts.
152
NOTES ON BUILDING CONSTRUCTION
For the third
R, = 3i-x 9 X 10 = 315,
Fiff. 232.
It thus appears that this joint is most liable to fail by shearing the
fibres along ca, and that 280 cwts. is the safe stress that it could be
subjected to.
The strength of the full beam = 9 x 9 x 12 = 972 cwts., so that this
joint is still less economical than the last.
Joint at Foot of Principal Rafter (Wood).
Example 31. — To ascertain whether the joint shown in Fig. 232 is of
sufl&cient strength to resist a
compression of 61 cwts. in the
principal.
Preliminaries. — We will as-
sume that this joint is made with
inferior timber, and therefore
take
?•( = 10 cwts. per square inch.
— 7 5,
Tg — 1'3 „
^•6=7 „ „
The joint may fail in two
ways : (l) By crushing the
fibres at ad ; (2) by shearing the tie-beam along cd.
As regards (1), the bearing surface is ade, allowing nothing for the tenon,
which would not be fitted to bear ; de is inclined to the pressure, and only its
equivalent at right angles to the pressure must be taken, therefore the bearing
surface is the whole cross section of the principal, less the tenon, that is
4x2 = 8 square inches.
So that
R^=:8 X 7 = 56-5 cwts.,
which is not quite sufficient.
As regards (2), R^ = 6x4x1 '3 = 31-2 cwts.
But it will be observed that it is only the horizontal component of the stress
in the principal that tends to shear the tie-beam along cd. This horizontal
^ component can be very simply found by a graphic
^" " method, which will be more fully explained in the
sequel (see Chap. VIII.) Draw a triangle ABC (Fig.
233); AB being vertical, BC horizontal, and AC
parallel to the principal, then CB will represent the
horizontal component required, to the same scale
that AC represents the compression in the prin-
cipal. In Fig. 233 the scale chosen is 50 cwts.
to one inch, AC is plotted as 61 cwts., and the
horizontal component CB is found to scale 53 cwts.
It thus appears that the resistance to shearing of the tie-beam along cd (31 '2
cwts.) is quite insufficient.
The additional strength required can be obtained by means of a strap as
shown in Fig. 234. To find the stress in this strap we must resolve the por-
tion of the horizontal component it has to bear parallel to the strap. The
}< 53 cwts
ScaU 50(fi to 1 inch.
Fig. 233.
JOINTS IN WOODEN STRUCTURES
153
portion of the horizontal component which the strap has to bear is 53 — 31-2
= 21-8 cwts., and the resolved part of the stress can be found by means of a
triangle as before, and as shown in Fig. 235. Thus it is found that the stress
in the strap is 43-6 cwts. = 2-18 tons.
u- 21-8 cwts >1
I
Fig. 234. Fig. 235.
Supposing the strap is made of |-" iron, then the necessary breadth h can be
found from
26x;|x5 = 2-18,
whence 6 = 0"87inch.
A strap 1" wide would be used.
By a calculation similar to that given in Example 27, it will be
found that a -I" screw and nut are required.
A bearing plate is required to prevent the strap cutting into the upper
side of the principal at e ; its length will be 4", the breadth of the principal
rafter, and to find the least breadth we have
4 X 6' X 7 = 43-6 cwts.,
whence h' = 1^ inch nearly.
The bearing plate ^ at/ would be made of the same width.
The above calculation supposes that the horizontal component is partly
borne by the strap, and partly by the wood. Any shrinkage of the wood
would, however, throw the whole stress on the strap, and it is therefore safer
to consider that the whole of the horizontal component is resisted by the
strap. The stress in the strap would then be (referring to Tig. 235)
53
43-6 X —106 cwts. = 5-3 tons, and making the strap of |" iron,
21-8 ^
2& X f X 5 = 5-3 tons,
6=1-4. .
So that a strap 1|-" x f " would be used.
The bearing plates at e and / would be made about 4" wide, and the
screw bolt of |-" diameter.
1 Unless there is any objection to weakening the tie-beam at this point, /would
be better notched in to prevent slipping (see Part I. )
Chapter VIII.
PLATE GIRDERS.
4
THE subject of plate girders was discussed to a certain extent
in Part I., and it is now proposed to enter more into detail
and to show how the various calculations can be effected.
General form. — Fig. 236 shows in outline the
general form of the cross section of a plate girder
with a single web. The flanges consist of one or more
plates (the thickness and number depending on the
stress to be borne) riveted to a couple of angle irons ;
the web consists of a thin plate, stiffened, however,
against crumpling up, by means of vertical T or L
irons. On referring to the distribution of the stresses
in a rolled iron girder, p. 83, it will be seen that the
\ K web of a plate girder being very thin can bear but a
^^^^^ very small part of the direct stresses. Practically the
Fig. 236. whole of the direct stresses may be taken as borne
by the flanges, and the whole of the shearing stresses by the
web.
Practical points to he attended to. — The following practical points should
be borne in mind in designing a girder.
The depth of plate girders varies from ^ to —g- the span ; ~ is said to
be the most economical proportion. In floors of buildings much shallower
girders are often used to save height.
For calculations the " effective span" that is, the span between the centres
of the bearings of the girder upon the abutments, should be taken, and the
" effective depth," that is, the distance between the centres of gravity of the
flanges (see Plates A and B).
The width of the flange under compression should not be less than 3^ to
^ of the span, or it will be liable to buckle sideways.
No plates of less than ^ inch thickness should be used, or they will soon
be destroyed by corrosion.
There should be as few joints as possible, especially in the tension fla.nge
and web.
Plate girders should be constructed with a camber or rise in the centre
of about -j-g-g- to their span, or at any rate in excess of their calculated
PLATE GIRDERS
155
deflection, so that if they should droop beneath their loads they may not sag
below the horizontal line.
Care should be taken to use market sizes of plates and angle irons ^ (see
Part III.) as far as possible. Expense is incurred by using extra sizes. It
may, however, be cheaper in some cases to use extra lengths, in order to
reduce the number of joints.
All parts should be so arranged as to be got at easily for riveting.
The ends of girders should be bedded on lead or felt.
FLANGES.
As regards the direct stresses, we need therefore only consider
the part of the girder shown in Fig. 237, and from ^^a^^^
Equation 54, p. 87, it will be seen that the moment "n I —
of resistance of this section is
M = r,A,DGr rAT) • (89),
according to which is least, D being the " effective
depth " for the whole cross section.^
Algebraic method of finding lengths of plates, — By equating
the moments of flexure to the moments of resistance at differ-
ent points along the girder, the number of plates required at
those different points can be found. Such a method is, how-
ever, cumbersome and tedious, and the required results can
be obtained more easily and rapidly by the following
graphic method.
Graphic method of finding lengths for plates in tension flanges.
— It has been shown (p. 33 seq^ that the moments of flexure
at the various points along a beam can be represented graphically
by a curve ; for instance, by a parabola when the beam is uni-
formly loaded.
Thus, in Fig. 238, akc represents the moments of flexure at
every point of a uniformly loaded beam of span ac.
JL
Resistance of angle iron.
„ „ 1st Plate.
„ „ 2d Plate.
„ „ 3d Plate.
a a' a" <
Fig. 238.
Now considering the tension side of the girder, the moment of resistance offered
by the two angle irons will be the same at every point, since the depth of the girder does
not vary. Let ab represent this moment of resistance and draw M parallel to ac, thus
we obtain by the intersection with the parabola two points a' and c' ; and the portions
aa' and c'c require only the two angle irons so far as strength is concerned. Practi-
cally, however, the first plate is also generally continued from one end of the girder
to the other. Let now ef represent the moment of resistance of the first plate. As
before, we obtain points a" and c" where the second plate ought to commence, and
^ The weights of angle and tee irons can be found by means of Table X.
2 At is effective area of tension flange, Ac is effective area of compression flange.
156 NOTES ON BUILDING CONSTRUCTION
ill like manner the point a'" and c'", where the third plate ought to commence, can
be found, and so on for as many plates as may be required to afford the necessary
resistance.
The same process is clearly applicable to the compression
flange of the girder.
Flanges for Plate G-irder of 20 feet span.
Example 32. — The matter will be probably made clearer by an example.
Let the effective span be 20 feet and the load 50 tons, equally distributed
so that the moment of flexure at tbe centre is 1500 inch-tons ^ (see Fig.
240). Then we can draw tbe parabola akc as explained in Appendix III.^
Assume further that the depth of the girder to the backs of the L irons is
1' 7", and that 3" x 3" x ^" L irons and i" plates 10" wide are used.
Tension, or lower flange. — It will be seen from Plate A that the cross
section contains one rivet (|") hole in each angle iron, and two rivet holes in
each plate.^ It will be found that the " effective depth," i.e. distance between
centres of gravity for the L irons, is 17*1 inches.^ Therefore the moment
of resistance of both angle irons in the tension flange (taking r^=b tons) is
2(3 + 2|--|)xix 17-1 X 5,
= 395 inch-tons.
Hence we can draw Id by making a6 = 395 inch-tons to the scale chosen for
the figure, namely 3000 inch-tons to an inch, and we then find by measure-
ment that aa' and c'c= 1*4 foot.
Again, the effective depth of the first plate is 19"5", hence its moment of
resistance is
[10-(2 x|)]xlx 19-5 X 5,
= 402 inch-tons.
Of the second plate
[l0-(2 x|)]xlx 20-5 X 5,
= 423.
Of the third [10 - (2 x |)] x 1 x 21-5 x 5,
= 443.
We can therefore draw a series of parallel straight lines at distances
representing these several values in inch-tons, until we just pass beyond the
apex of the parabola. In this manner we obtain the points /, g, etc., and by
dropping perpendiculars the points a', a", a'", etc., so that the number of plates
and length of each are shown in Fig. 239, and can be found by measurement.^
Compression, or upper flange. — On the compression side the moment of
resistance of the two angle irons (taking = 4 tons) is
1 M,. = ^ (see p. 33)= g = 1500 inch-tons.
^ A circular arc can be substituted for the parabola if the rise of the arc does not
exceed |th of the span (see p. 30).
^ In many cases, when the plates are wider, they are riveted together also at
their edges, so that there would be four rivets to deduct.
* It is as easy and more accurate to take " effective depths " separately for the pair
of L irons and for each plate.
^ As the sum of the above moments is 1663 inch-tons, and by the data only 1500
are required, it is evident that the third plate need not be more than yV' thick. It
will be found that the moment of resistance of such a plate is 275 inch-tons.
PLATE GIRDERS
157
2(3 + 2-l-)xix 17-14x4,
= 377 inch-tons.i
And of the first plate
lOxlx 191 X 4,
= 390 inch.-tons.
Of the second =410.
Of the third = 430.
Eepeating the ahove process as shown in Fig. 239 we obtain the number
and lengths of plates given in Fig. 240.
Resistance in inch-tons,
3d Plate, 430.
2d Plate, 410.
1st Plate, 390.
Angle irons, 377
Total . 1607
Incli-tons.
1
'^\^
^1 is
11 a'
\a"
a'" c'"
c' c\
/
1 U-M-JS
Fig. 240.
Fig. 241.
Fiff. 239.
Angle irons, 395
1st Plate, 402
2d Plate, 423
3d Plate, 448
Total . 1603 inch-tons.
As already observed, the first plate is continued in both the tension and com-
pression flanges for the whole length of the girder, and is so shown in Figs.
239-241.
It is to be observed that in practice the plates (after the first) will be
somewhat longer than shown on the diagram, for they must include the next
row of rivets beyond the point indicated in theory.
WEB.
Calculations for the Web — Shearing stress. — As already
mentioned, the web can be considered as taking the whole of the
shearing stress, and no great error will be involved by further
assuming that the stress is uniformly distributed over the cross
section. The methods of finding the amount of the shearing stress
have already been explained in Chap. III., p. 55 et seq.
The thin webs of plate girders tend to fail, much in the same way as
long compression bars do, by bending or buckling, as shown in Fig. 242, and,
to understand this, we must consider for a moment the stresses acting on a
small portion of the web. Let Fig. 243 represent a small square marked on
the web. We know from p. 61 that vertical shearing stresses V act along
^ The rivet holes are not deducted, as the rivets are supposed to fill them, so that
the pressure is transmitted over the whole section. Some engineers would take
rc=5 tons over the section, deducting the rivet holes. M of the angle irons would
then be the same as for the tension flange 395 inch-tons.
158
NOTES ON BUILDING CONSTRUCTION
AB and CD, and that horizontal shearing stresses H act along CB and AD.
Now it is clear that these forces will distort the square and make it take the
shape of a lozenge, as shown in an exaggerated manner in Fig. 244. This
Fig. 243.
Fig. 244.
distortion clearly extends the material in the direction BD, and compresses it
in the direction AC; and the tensions and compressions thus excited in the
material will just balance the shearing stresses. This may perhaps be more
clearly understood by imagining the square AC formed of four rods jointed
B C VN^^-^"^H
Fig. 246.
at A, B, C, and D, and two springs introduced as shown in Fig. 245. When
the forces H, H, V, V commence to act, the square will be deformed as shown
in Fig. 246 ; the spring BD will be stretched and the spring AC compressed ;
and (supposing the spring AC is not allowed to bend) the extension and com-
pression of the springs will continue until they balance the applied forces.
Now Professor Rankine has shown that the intensity of tension and
compression excited as described above is equal to the intensity of the
shearing stresses which produce them. We may therefore replace the
vertical and horizontal shearing stresses, acting on the web of a plate
girder, by a tensile and compressive stress of equal intensity, and acting
in directions inclined at an
angle of 45° (see Fig. 247).
Now it is the compressive
stress CC which tends to cause
the buckling of the web, and
for purposes of calculation we
can consider the web as made
up of a series of strips, in-
clined downwards towards the
Fig. 247. Fig. 248. supports at an angle of 45°,
nr.
JET
PLATE GIRDERS
159
and each, acting as a long column. The length of these long columns is clearly^
\/2 X S,
where S is the unsupported depth of the web between the angle irons (see Fig.
248), and they can be considered as "fixed" at the ends. The thickness (ij
of the web is the least diameter of the long columns in question. We can
therefore apply Gordon's formula (see p. 113, or Tables V. and XI.) to calculate
the necessary thickness to be given to the web. It, however, frequently happens
that an undue thickness would have to be given to the web to make it stiff
enough. In such a case the additional resistance to buckling can be obtained
"by riveting vertical T or L iron stiffeners to the web as shown in Fig. 249,
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Fig. 249.
Fig. 250.
and if S is the clear space between the stiffeners, the length of the long
columns into which the web is supposed to be divided is
Sx s/\
and Gordon's formula or Table V. can be applied as before.
As stated below,^ the base and perpendicular of the triangle, of which
the column is the hypothenuse, are equal. Therefore, when the distance S
between the stiffeners is less than S will represent the base and perpen-
dicular of the triangle, and the length of the column will be shortened.
From this it will be seen that stiffeners are theoretically useless unless they
shorten the length of the assumed columns, and to be effective must be closer
together than the depth.
Example 33. — The depth between centres of rivets of a web is 21" and
the unsupported depth (S) is 18", the thickness j-^^, and the shearing stress
to be resisted is 9-3" tons. Find whether any stiffeners are necessary, and
if so, at what distance apart they ought to be placed. Find also what thickness
ought to be given to the web to obviate the necessity of stiffeners.^
^ Length of the long column = \/2 x 5.
sm 45
2 The length of the columns is represented by the length of a line drawn across
the web in the direction that the web will fail, and as this direction is at an angle of
45°, its length can be represented by the hypothenuse of a right-angled triangle whose
perpendicular is equal to the depth of the web. The formula V2 x 5 which expresses
the length of the hypothenuse in terms of the depth of the web is arrived at from
Euclid, Book I., Proposition 47.
2 The web being very thin the shearing stress may practically be considered as
imiformly distributed, not as in Fig. 97. The depth of the web to resist shearing
being taken as equal to the distance between the centres of the rivets connecting it
to the flanges, 21" in the present example.
1 60 NO TES ON B UILDING CONS TR UCTION
Substituting in Gordon's formula (see p. 113) as follows —
rc = 4 tons,
1
2500'
Z=18x ^/2,
A = -
we obtain —
Safe intensity of compression in tbe web =
1 +
1
(^2x18)2
,2500 " {-i-^f
= 1"1 tons per square inch.
Using Table V., n being 3-5, the ratio is 3'5
whence the safe intensity of compression is O'BB ton.
Now the intensity of shearing stress per square inch is
-^-~ = l-42 ton.
21Xt^
Evidently, therefore, the web is not quite stiff enough.
To find the distance apart of the stiifeners we have
1-42.
^2 X 18 X 16
, or 285,
1+:
(v/2.S)2'
2500 •
Whence we find by reduction
S2 = 221,
S= 14-9 inches.
The distance apart of the stiffeners can also be found by means of Table
The ratio corresponding to a stress of 1'42 ton is 215. Hence
V.
3-5-
S =
215, where 3-5 in Table V.
13 "5 inches.
To find the thickness of web required to obviate use of stiffeners we have
V'218
from Table V. as above
3-5-
= 215,
= 0-425 = —g inch.
We can also use Table XI. for this purpose. The stress per foot of depth of
9"3 X 12
the girder is — — = 5-3 tons, which for a net unsupported depth of 18"
corresponds to a thickness of web of |".
The shearing stress we have been dealing with is the maximum stress at
the ends. The intermediate stresses diminish towards the centre (see Fig.
90), and the thickness of web may accordingly be reduced ; but it is never
advisable, for practical reasons, to use less than Y' plates even at the centre
of the girder.
When the load on the girder is concentrated at one or more points it is
usual to add stiffeners to the web at such points, to transmit the load equally
to both flanges.
PLATE GIRDERS
i6i
o
o
O
Q
Q
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O
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Fig. 251.
Joints in the Web. — These joints are arranged as shown in Fig.
251. One or two covers can be used as shown in the sections,
but it is better work to use two
covers. The thickness of one cover
should be the same as that of the
web. When two covers are used
their combined thickness should be
at least equal to that of the web, so
long as neither of them is made less
than 1 inch thick.
As regards the number of rivets
required to make the joint, it is to
be observed that they must be suffi-
cient on each side of the joint to take up the" shearing stress
at the section of the girder where the joint is made.
Example 34. — For instance, taking the same data as in the two previotis
examples, namely — Shearing stress = 9-3 tons.
Thickness of web = |- inch,
Diameter of rivets = |- „
the bearing resistance of one rivet is 2-62 tons ; so that the number of rivets
required on each side of the joint is
= 3-5 nearly.
These rivets are arranged as shown in Fig. 251, and the pitch would be
18 . ,
— = 5 inches nearly,
o'o
which is too great, being 13 times the thickness of the plates riveted. The
pitch may be taken at 12 times that thickness,^ or practically at 4".
Width of cover plates. — As regards the width of the cover plates, follow-
ing the rule, p. 136, that the centres of rivets are not to be nearer the edge
of a plate than li times their diameter, it will be seen that the width must
not be less than Ax d = 6 d.
In the above example, therefore, the least width of cover plate is
6 X I = 5^ inches, say 6 inches,
and this would be a suitable width to take.
Connection of the Web with the Flanges.
The web is connected to the flanges by being riveted to the
angle irons, and the number of rivets should be sufficient to with-
stand the horizontal shearing stress. The easiest way is to calcu-
late the amount of the shearing stress per foot, and then to find
the number of rivets required for bearing and for shearing.
Example 35. — In the above example, for instance, the shearing stress
at each end per foot
B.C. — IV,
^ See Stoney's Rule, p. 136.
M
i62 NOTES ON BUILDING CONSTRUCTION
9-3x12
= — — — = 5*3 tons.
Now if 1^" rivets are used, tlie resistance of one rivet to bearing in web (taken
as thick) is = 2-62 tons.
{The resistance to shearing need not be considered, as it is much greater.)
Hence the number of rivets required per foot
5-3
= =2,
2-62 '
so that a 6-inch pitch would be sufficient as regards bearing, but this would
be too great in proportion to the thickness of the plates, and a pitch of not
more than 12xf = 4|" should be adopted. In practice I" rivets with 4"
pitch would probably be used in both this and Example 34.
We have now considered the main points in the calculations
required for plate girders. In the two following examples some
further matters of detail will be considered.
Example of Plate Girder -with Single Web.^
Example 36. — Design the girder shown in outline in Fig, 26, p. 21.
Load. — The load to be borne by the girder is 23 tons placed 10 feet
from the left support, and from Fig. 54 it appears that the greatest bending
moment occurs at this point. The weight of the girder itself must, however,
be taken into consideration, and this weight can be found by using either
of the formulae given in Appendix XV. These formulaj can only be expected
to give approximate results, but the error is, as a rule, too small to be of any
practical importance ; they are based on the supposition that the load is
uniformly distributed. To apply them in the present case we must therefore
first find the " equivalent " uniformly distributed load, that is the uniformly
distributed load that would produce the same bending moment at the centre
of the girder that the actual loading does. Now if W is this load, we know
by Case 7, p. 33, that the bending moment in the centre is
WZ W. 30 ,
= -r- = — - — foot-tons,
o y
And from Equation 2, p. 25, the bending moment due to the actual loading is
R^x 10'= 15-33 X 10 = 153-3 foot-tons,
W= — x 153-3 = 40-9 tons.
30
This does not include what would be required to carry the weight of the
girder itself, but the additional material would be small (about -16 ton), so
that we can take W = 41 tons.
A'p'proxi'mate, weight of girder. — Using Unwin's formula (see Appendix XV.)
and taking the data given there, and taking D = 30 as below, we obtain
41 X 302
Weight of girder =. ^^^^^^^ 30 _ 302 »
= 2-62 tons.
As a check we may use Anderson's formula, and thus we obtain
Weight of girder = X 41 X 30,
O D (J
= 2-2 tons^
1 See Plate A, p. 352.
PLA TE GIRDER WITH SINGLE WEB
163
The former result agrees witli tlie weight of the girder given at p. 21 (2-6
tons), and we will assume it to be the weight of the girder.
Be^pth of girder. — The depth of girders is usually taken at
from -|- to of the span. It should be observed that the deeper
the beam the more liable is the web to buckle, and therefore a
thicker web or a greater number of stiffeners will be required.
On the other hand, a shallow girder requires more metal in the
flanges, and is less stiff. We will assume a depth of 30", that is
jlg- of the span. The distance between the centres of gravity of
the flanges may be taken as 30".
Deflection. — It remains to be seen whether the depth chosen (30") is
sufficient to ensure that the deflection does not exceed a certain amount, say
-^-q" per foot of span. The total admissible deflection would be
= J^x 30 = 0-75 inch.
Now, as the beam is to be subjected to certain limiting stress, we can
use Equation 46, p. 67, namely —
ED '
As regards the value to be given to n', it is to be observed that the
beam will be stifl'er than a beam of uniform strength, but less stiff than a
beam of uniform section, for the former w' = |-, and for the latter (witli
central load) »' = yV- will therefore be a fair approximation to take
w' = TV-
Farther, let = 4 tons,
Tt = 5 tons,
Z = 360 inches.
As regards the value of E it should be observed that the full value
(29,000,000 lbs.) for wrought iron cannot be reckoned on in built-up girders,
owing to the unavoidable imperfections of riveting. A value between
16,000,000 and 18,700,000 is generally assumed. We will take 17,920,000
lbs., or 8000 tons. Hence
^_ tVx 9 X 3602
~ 8000 X 30 ' ■
A = 0-49 inch.
So that the girder is amply stiff enough.
Width of flanges. — The width of the compression flanges should be
sufficient to prevent buckling sideways. It is found by experience that for
this purpose the compression flange should be made about to of the
span, that is in this case 12 to 9 inches. We will take 9 inches. The width
of the tension flange is not subject to the same condition, but for uniformity
we will assume it to be also 9 inches wide.
Thickness of flange plates. — The thinner the plates used, the more nearly
can the girder be made to approximate to a beam of uniform strength. On
the other hand, corrosion affects thin plates to a serious extent, and the
number of plates to be joined together is increased. It should also be
6bserved that very thin plates are liable to buckle along the edges when
placed in the compression flange. Professor Unwin says that the thickness
l64 NOTES ON BUILDING CONSTRUCTION
should not be less than of the span. In practice the thickness is limited
between ^' and |"; usually |-" or ^' is found to be suitable. We will in this
example take -I" plates.
Angle irons for connecting the flanges to the web. — The thickness will be
taken the same as that of the flange plates, namely |". A table 3 inches wide
gives sufficient room for riveting, but in the present case we will use angle
irons 3 1" x 3^" x f", as they will afford more room for riveting.
Calculation of flanges. — We are now in a position to proceed
with the calculation of the flanges in the manner explained at
p. 156.
From Fig. 54, p. 35, it will be seen that the diagram for the moment of
flexure due to the single load consists of a triangle, the height of which is
the maximum moment of flexure. This maximum moment of flexure due to
the load is in the present case equal to
X 10 X 12 = 15-33 X 10 X 12,
= 1840 inch-tons.
We thus get the triangle AcZB shown in Fig. 252.
The weight of the girder itself produces a different diagram of bending
moments, namely the arc of a circle (see p. 34) ; and the height of the arc
is the bending moment at the centre, that is
2-6 X 30 X 12
3 = 117 inch-tons.
We thus obtain the arc AcB shown in Fig 252.
Fig. 252.
The diagram for the total bending moments can of course be found by
adding the ordinates of the arc to those of the triangle, and thus, finally, we
get AeB as the diagram of bending moments.
Tension flange (see Plate A). — The rivet holes must be deducted
from the cross section to obtain the effective section. Deductinsr
one f ' rivet hole from each angle iron, and taking the depth to the
backs of the angle irons at 29", i.e. 1" less than the distance be-
tween the centres of gravity of the flanges, whence the "effective"
PLA TE GIRDER WITH SINGLE WEB
165
depth ^ of the angle irons is 2 7" we obtain the moment of resist-
ance of both angle irons
= 2(3^ + 31-1-1)1x27x5,
= 595 inch- tons.
Two rivet holes must be deducted from each plate, so that moment of
resistance of the first plate is
= (9-2 x|)|x 29|x 5,
= 413 inch-tons nearly.
In the same way it will be found that the moment of resistance of the
second plate is = 423 inch-tons, and of the third plate 434 inch-tons. The
total moment of resistance of the L irons and the three plates is therefore
1865 inch-tons, which is less than the maximum bending moment of 1944
inch-tons, but practically suflScient.
Following out the process explained at p. 155, we obtain the
completed diagram given in Fig. 253, which shows the number of
plates required and the length of each. On referring to the Tables
in Part III., giving the market sizes of iron, it will be seen that
plates of the full length can readily be obtained, and that therefore
no joints are needed ; but as an exercise for the student the joints
i Plate
- Ji \ex tended
; to Jorin
I cover
Fig. 253.
have been so arranged in Fig. 253 that no plate will exceed
13 feet in length, and, moreover, the joints have been so
placed that " grouped " joints can be formed. Each of these
joints requires to be designed and calculated on the principles
explained in Chap. VII., and in fact one of the joints has been
worked out in Example 24, p. 132.
Compression flange. — The diagram of bending moments just
^ i.e. distance between centres of gravity of angle irons.
166 NOTES ON BUILDING CONSTRUCTION
obtained for the tension flange holds good for the compression
flange, and is re-drawn in Fig. 254, but upside down, so as to
show the plates in their proper relative positions.
Fig. 254.
As already remarked at p. 136, it is not necessary in this flange to deduct
the rivet holes to find the effective section, but the safe stress is reduced to
4 tons per square inch.
The moment of resistance of both angle irons will therefore be
2(3| + 3i -Dfx 27x4,
= 537 inch-tons.
The moment of resistance of the first plate will be
9 xf X 29|x 4 = 396,
and so on.
The total moment of resistance of the L irons and of the three plates will
be found to be 1754 inch-tons, but the maximum bending moment is 1944
inch-tons. It will be therefore seen that three f" plates do not give sufficient
resistance at the point of maximum moments of flexure, but practically they
are quite safe. By making the outside plate Y thick the moment of resist-
ance will be increased to 1897 inch-tons, which is near enough.
Repeating the operation we can obtain the number and length of plates
required, and also ascertain in what position the joints ought to be placed if
full-length plates cannot be obtained.
The flanges can now be considered as designed, with the exception of the
joints, for which the student is referred to Example 24, p. 132.
Weh. — To design the web the first step is to find the shearing
stress in each portion of it.
The simplest way is to make use of the diagrams given in
Chap. III. As the girder is subjected to two loads, we must
combine the diagram for a uniform load (Fig. 90, p. 59) with that
for concentrated load placed at any point (Fig. 92).
PLATE GIRDER WITH SINGLE WEB 167
Thus, in Fig. 255, Ae/B represents tlie shearing stress due to the weight
of the girder, and AghklB the ^
shearing stress due to the load of
23 tons, so that AlmnoB repre- ^
sents the shearing stress to which e,
the web is subjected. j
It will be observed that owing !
to the weight of the girder being ^
small in comparison with the load
to be carried, the shearing stress is
almost constant from A to D and again from D to B. It will therefore be theo-
retically best to make the web of a certain uniform thickness from A to D, and
of another less thickness from D to B ; these thicknesses must now be found.
For the part AD we have a maximum shearing stress of 16 "6 tons to deal
with (seep. 22), and the depth of the web (S) between the angle irons may be
taken as 22" and the distance between the centres of rivets as 25 '1 (see Plate A).
Assuming the web to be ^ thick, the intensity of shearing stress is
16-6 ^ . '
— — =r= 1"32 ton per square men.
25-1 X 1^ r 1
Now referring to Table V., we find the value of - for the equivalent long
column to be, taking n = 3'5, _
3.5X^^=218,
2
and hence the safe intensity of stress (for a column fixed at both ends) to be
nearly 1-40 ton ; so that a i" web will do very well.
We can also find the thickness of the web by means of Table XI. The
net unsupported width of the plate is 22" and the shearing stress per foot is
^^25^1^^ ^ ''''^^ tons, and it will be seen on referring to the Table that ^"
is the thickness required.
It should be noticed that the f rivets, which have been selected, are a
little small for a ^" plate, but this difficulty can be overcome by drilling the
holes, or by using a f " plate for this part of the web, supported by stiffeners.
Pitch of rivets to connect the web to the flanges. — The stress per foot is
1^ X 12 = 7 "94 tons nearly.
Taking rg = 4 tons and = 8 tons the resistance to bearing of a f ' rivet
in a ^" plate is 3-0 tons (Table VIII.), and the resistance to double shear
is 1*8 X 2 = 3'6 tons ; so that the number of rivets per foot
7-9
= ¥-o = '-'-
Practically a pitch of 4" would do.
For the part DB of the web we have a maximum shearing stress of 9-0 tons
(see p. 22). Assuming the web to be f " thick, the intensity of shearing stress is
9-0
" =0-95 ton.
I . 25-1 X I
But the value of - is
1 See Appendix V., p. 298.
i68
NOTES ON BUILDING CONSTRUCTION
f
And from Table V. the safe intensity of stress is 0-86 ton, so that a |" plate
is somewhat thin. It should be observed, however, that the maximum shear-
ing stress occurs only at the abutment B, and it will also be found that -Z^"
is thicker than necessary according to the Table. There can, therefore, be
no doubt that f" is thick enough.
To use Table XI. we have : net unsupported width of plate =22", and
shearing stress per foot = 4-4 tons, which correspond to a thickness of |".
Fitcln, of rivets to connect the web to the flanges. — The stress per foot is
9-0
X 12 = 4-3 tons,
25-1
and the resistance to bearing of a
number of rivets per foot
rivet in a |" plate is 2-2 tons. Hence the
4-3
or a 6" pitch will do as regards bearing ; but in order to keep the plates
together the pitch should not exceed 12 x |" = 4|-", and may practically, for
the sake of uniformity, be made 4" as in Plate A.
Joints in the web. — Two joints will be required in the web, so as to keep
the plates within the ordinary market sizes. One of these joints will be best
placed immediately under the load, and the joint can be made by means of
two _L irons, which will also act as a stiffener to transmit the load to both
flanges of the girder. Keferring to p. 160 we see that the least width of the
table of the ± iron should be 8 x | = 6 inches. If the stem of the JL iron is
I" thick, there will be enough room for riveting, as shown in Fig. 256.
It will benoticed that this joint is placedat the point where there is noshear-
ing stress in the web, so that theoretically there is no need for a cover ; but prac-
U'l pitch
U; pitch
s/i rivet
Packing
c:7
rig. 256.
Q
@
© ''© 0
\
Q
0
Q
©
h"j'itch
1
Q
<«'-
Q
Q
©
Q
Q
Q Q
©
0
© © ©
I'ig. 257.
tically it would not do to dispense with it, and, moreover, in this before
said, it fulfils the important duty of assisting to transmit the load of 23 tons.
Theoretically, no rivets are required to connect the X ^^"^ to the web, but
practically they would be placed at about 4" pitch to put the girder together.
Fig. 257 represents the joint.
The second joint will be best placed midway between D and B. The
shearing stress at this point found from Fig. 2 5 5 is 8'4 tons.^ The cover plates
^ The student is recommended to draw Fig. 255 to a large scale to obtain this value.
PLATE GIRDER WITH SINGLE WEB
can be made 6" wide, and or -j-y is a suitable thickness. To
number of rivets, the bearing resistance of |" rivets in a |" plate is 2
therefore number of rivets
169
find the
'25 tons.
2-2
nearly,
or a 4" pitch will do.
End pillars. — The ends of the girder should be
stiffened to resist the shearing stress at those points.
The most economical way is to stop the longi-
tudinal angle irons at the ends of the girders,
and to rivet two vertical angle irons to the web,
using packing pieces as shown
in Fig. 258.
The ends of the girder
are generally bedded on
sheet lead. The rivet heads
should be countersunk to
allow of this, and four of
the rivets would be left out
at one end to enable the
girder to be bolted down to
the top of the column. oJ
Fig. 259 and Plate ^ A ^
show the girder as arranged to comply S
with the calculations made above. The
total length of the girder is 31' 0", 1' 0"
more than the distance between centres
of bearings, or 2' 0" more than the clear
span, thus allowing 12" at each end to
rest on the cap of the C.I. column and
on the wall. The bearing area is there-
9
fore X 1 = 0-75 square foot, so that the
9"0
pressure at the wall end is =12-0 tons
075
per square foot, and this is not more than
a hard stone can bear. If, however, the
stone is soft, a casting ought to be placed
under the end of the girder to distribute
the pressure.
<2>
[■^ !
•'J I
ll^ i
1-^ J
11^
L__i_
i
^ In Plate A the joggles in the angle iron are omitted for the sake of simplicity
and economy in construction.
NOTES ON BUILDING CONSTRUCTION
Example of Box Girder.
Example 37. — Calculate and design a box plate girder to act
as a bressummer, as shown in Tig. 260 (see Plate B, p. 353).
Load. — The first step is to find tlie load to be borne by the girder.
The portion of the wall supported by the girder is shown in dotted lines ^
in Fig. 260, and the cubic content of this portion is
Fig. 260,
^ It might be objected that the load due to the wall ought to be taken as the
weight of the wall contained between vertical lines drawn from the ends of the
BOX GIRDER 171
(20 + 14 164-14 ■) 14
f 14 + 10 16 + 14 ) 9
+ |-^x2 + 8x8 + -^xl j-
= 229 cubic feet,
A cubic foot of brickwork weighs 1 cwt. (see Table XVII.),-^ so that
weight of brickwork to be supported = 229 cwts.
The first floor is supported over a length of 1 8 feet, and since the floor is
20 feet wide, area supported = 18x10 = 1 80 square feet.
The second floor is supported for a length of 14 feet by the girder, hence
floor area supported = 14 x 10 = 1 40 feet.
The total for both floors is therefore 320 square feet.
Assuming that these are warehouse floors, and that the load is Z\ cwts.
per square foot, the total load produced by the two floors is
320 X 3-5 = 1120 cwts.
Lastly, the length of the roof supported is 14 feet, that is, a horizontal
area ofl4xl0 = 140 square feet is supported. Allowing 40 lbs. per square
foot (Tredgold), the load due to the roof is
140x40 _ .
= 50 cwts.
112
The total load on the girder (except its own weight) is therefore
229 + 1120 + 50 = 1399 cwts.,
= 70 tons very nearly.
A'pfroxi'maU weight of girder. — Clearly this load can be taken as uniformly
distributed.
Using Anderson's formula (Appendix XV.) to obtain the weight of the
girder we get the weight per foot run of girder
2240 x 70 . c,or^^■U
=4 X 70 lbs. = 280 lbs.
560
Hence total weight of girder ■ * "
= 280 lbs. X 20 = 5600 lbs.
= 2"5 tons.
Therefore the total distributed load the girder has to sustain is
70 + 2-5 = 72-5 tons.
Deflection. — As the load is considerable we will take the depth at of
the span, that is 24". It remains to be seen whether this depth will give
sufiicient stiffness. In the formula (Equation 46)
n'(rc + r()Z2
A = -
ED
We have in this case
rc = 4 tons,
rt = 5 tons,
1 = 240 inches,
E = 8000 tons (see p. 163),
D = 24 inches,
girder ; it is considered, however, that the bond of the brickwork causes it to take
part of the weight.
1 Under some circumstances it would be better to take 130 lbs, per cubic foot as
the weight of the brickwork, to allow for the bricks being saturated with water.
172 NOTES ON BUILDING CONSTRUCTION
and for n we must give some value between that for a beam of uniform
strength and depth (i), and for a beam of uniform cross section loaded
uniformly = ■^-^). We will take n' = i-
Hence ^_ ^x9x240^
8000 X 24 '
= 0-3 inch,
60 that the girder is amply stiff enough.
It should be noticed that the value thus found for A is only approxi-
mate, because it depends on the value of n' and of E, both of which can
only be approximately estimated ; but the error would not be sufficient
to affect the result practically, and does not, probably, amount to more
than 10 per cent.
Calculation of flanges. — As the girder has to support a 14" wall, the flanges
might also be given a width of 14" (see Plate B). This width, it will be
observed, is much more than is required to prevent side buckling ; but if, on
calculation, a less width be found to give a better arrangement of the plates
it might be adopted so long as the width is not less than 9", which is too
narrow for a box-section, and would require very large corbels for the brick-
work.
The size of the L irons used for connecting the webs to the flanges
must first, however, be determined ; there are two L irons to each flange.
The shearing stress is considerable, and therefore the thickness of the metal
must be such as to give sufficient bearing area to the rivets without having
to place them too close.
Referring to Fig. 263, p. 173, it will be seen that the vertical shearing
stress at the abutments is 36-25 tons, which is distributed over a length of 24
inches, and at a foot from the abutments it is reduced to
36-25 X 9
— = 32-6 tons.
Hence the average horizontal shearing stress for the first foot close to the
abutments is
12 36-25 -f- 32-6
^ X ■ ^ = 1 7-2 tons,
and since there are two webs the stress the rivets have to resist per foot is
8-6 tons.
If I" rivets are used, it will be seen on referring to Table VIII. that
if the resistance to bearing is taken at 8 tons per square inch, the resistance
of one rivet in a plate whose thickness is t will be
7-0 X t,
and if the pitch of the rivets is 4", that is 3 rivets per foot, the total bearing
resistance of the rivets per foot will be
7 X i X 3 = 21 X i nearly.
8*6 '
Hence = — = 0-41 inch,
a thickness of will therefore do, and L irons 3" x 3" x would be a
suitable size to use. It will be found that the centres of gravity of these L
irons are 22" apart. Hence their moment of resistance on the compression
side is
BOX GIRDER
173
2(3 + 3-T-V)/^x22x4,
= 428 inch-tons.
Bat the moment of flexure at the centre of the girder is 2175 inch-tons.
Hence the plates have to supply a resistance of
2175 - 428 = 1747 inch-tons.
We can assume as a sufficiently near approximation for the present
inquiry that 25 inches is the mean effective depth for the plates. If is
the total thickness of the plates at the centre and 6 the breadth, we have on
the compression side
Hence if
and for this, one y
If
6x e X 25 X 4 = 1747,
.-. 6<= 17-47.
6 = 14 inches,
i = 1*25 inch,
-V' and two -g-" plates would do.
6=12 inches,
i = 1 '45 inch,
and three y' plates would do.
If 6=10 inches,
t = 1*74 inch,
and one ^ and three -I" plates would be required.
Either of these three arrangements would require about the same weight
of iron, but the first two are preferable in having fewer plates, and it does not
appear to be a matter of much moment which of them is chosen. We will
adopt the first arrangement, so that a width of 14 inches will be chosen.
Fig. 261.
Compression boom
^ — ^ >k plate ^-'---^^
\ ^^x-^^ opiate ^^-^
L jrons
20' 0" - -
>i
[-<■
1
Scale UOOO inch-tons to one inch
L iro7is
\ 'Is plate ;
</2 plate
Tension boom
Fig 262.
The number of plates, length and position of joints, can be found
the previous example, by constructing
Figs. 261 and 262.
Calculation of toeh. — Eeferring to
Fig. 90, it will be seen that the dia-
gram for Shearing stress is simply an
inclined straight line,^ as shown in Fig.
263. In the present case the shear-
ing stress diminishes rapidly towards
Scale. 80 ions to 1 inch
263.
^ Fig. 263 differs from Fig. 90 in that, on the right-hand side, the diagram is
drawn above instead of below the centre line (aee Appendix Y. )
174 NOTES ON BUILDING CONSTRUCTION
the centre, and it will also be observed that it is only necessary to make
calculations for one half of the girder.
Commencing at the centre and working outwards, let us see up to what
point a Y plate can be used. The depth of the web between centres of rivets
is 19|-" and between the angle irons is about 17", so that, referring to Table
V. and p. 333
- = 3-5 = 336 nearly.
This value of ^ is beyond the range of the. Table, but from Fig. i66, p.
110, it will be seen that the available intensity of stress is about 0-5 ton.
The area of the web is 19*5 x ^ = 4-8 stjuare inches, so that the greatest
shearing stress it will bear is
4-8 X 0-5 = 2-4 tons,
and, as there are two webs, it follows that a ^" plate can be used up to
the point where the shearing stress is 2 x 2-4 = 4-8 tons, that is to the point
4-8 , „
^^7^x10=1 4 (nearly) from the centre. Strictly, however, the point
can be shifted half the width of the web (viz. 9") farther away from the
centre, because the shearing stress of 2-4 tons can be taken to occur at the
centre of the imaginary long column inclined at an angle of 45°. Hence a
Y plate 2(1' 4" + 9") = 4' 2" could be placed in the centre of the girder. This
is rather a small plate, and it will probably be better to use a -^-^ or |" plate.
Repeating the above calculations for a |" plate, it is found that such
a plate is available up to a point 5' 10" from the centre ; that is, plates 11' 8"
long and |" thick can be used at the centre of the girder for the webs.
Taking the next available thickness of plate, namely /g", we find, as
before, that such a plate can resist a shearing stress of nearly 15 tons, and
2x15
would therefore be available up to a point - x 10 = 8' 4" from the centre.
There remains 1' 8" of the web at the abutments, for which a plate is
insufficiently thick. It would be bad construction to put in a i" plate (which
would be strong enough) of so small a size, causing an additional joint, and
there are therefore two courses open : (1) to make the webs Y thick from the
point where the f " plate ceases ; (2) to strengthen the plates near the abut-
ments by adding stiffeners.
The first arrangement is the simpler, and as the amount of metal would
be about the same in either case, it would be the best arrangement to adopt.
The calculations for the second arrangement will, however, be given. It is
required to find the maximum distance between stiffeners. Let c be this dis-
tance, then is to be substituted in Gordon's formula for -. On the
d
V 8"
other hand, the intensity of stress at — - from the abutment is for each web
Ji
plate
18-12 X 9-11 1
X TTT^ Sr-= 1-9 tons.
10 19-5 X -/;
1^
BOX GIRDER
175
So that
l-90 = -
1 +
3000
16c
From this equation the value of c can be found, an exercise which is left to
the student.
The result can be obtained more rapidly by using Table V. It will be
I
found that a safe intensity of stress of 1*90 ton corresponds to a value of -
of 170.
whence
Hence taking n = 3*5 we have :
16c 170
c = 1 5 inches nearly.
The stiffeners can therefore be arranged as shown in Fig. 264, one stiffener
being placed immediately at the edge of the abutment, the other just where
the plate commences to be strong enough without assistance.
Joints in the web. — There will be two joints ; the calculations for these
are precisely similar to those already made, and are left to the student.
Joints in the flanges will not be necessary, as the longest plate is only 20'
long.
i« lu".
Q O O
© O
' 71 "
000)
PacTiing
J>iece
Section m.n.
Kg. 264.
Rivets to connect the web to the flanges. — The calculations for these are also
left to the student, being similar to those already explained.
Vic
■Web
^weh , I,
11 8--
Joint
riff. 265.
176 NOTES ON BUILDING CONSTRUCTION
End pillars. — The clear span is 20 feet, but th^
girder must be made longer at each end so as to bear
on the stone sills, say 9 inches ; and as the shearing
stress is considerable, it would be advisable to adopt the
arrangement shown in Fig. 264.
The completed girder is shown in Fig. 265, the
thickness of the plates being considerably exaggerated,
also in Plate B.
Intermediate diaphragms. — In a large girder, two or
three of these should be inserted. They are pieces of plate
iron riveted to both webs to prevent racking, that is, to
Fig. 266. prevent the girder bending over as shown in Fig. 266.
Saddle-backed or Hog-backed Girders have a curved upper flange such as
that of the Cast-hon Girder, Fig. 144. Both flanges can be made of nearly uniform
section throughout instead of as shown in Fig. 238.
Such girders are however very seldom used in connection with buildings, aud
need not be further described.
Chapter IX.
BRACED OR FRAMED STRUCTURES.
General Remarks. — Before proceeding to the consideration of
lattice girders and roofs, which are braced structures, we must
investigate a few general propositions in connection with such
structures, and we will also give two methods of finding the
stresses in them.
Let us consider in the first place the simplest case of a weight
W supported by two rods, as shown in Fig. 267, A and B being
Fig. 267. Fig. 268. Fig. 269. Fig. 270.
fixed points. The two rods are evidently in compression, and as
they are equally inclined to the direction of the weight they will
be subject to equal stresses. The same remarks apply to the
rods shown in Fig. 268, with the exception that they are in
tension instead of in compression.
Fig. 271. Fig. 272.
If the inclination of the rods to the vertical is diminished, as
in Figs. 269 and 270, the stress in them will also be diminished;
and on the other hand, if the inclination is increased, as in Figs.
271 and 272, the stress will be increased.
B.C. — IV. N
178 NOTES ON BUILDING CONSTRUCTION
In Figs. 273 and 274 the rod AD is shown at a greater inclina-
tion to the vertical than the other rod. Clearly the stress is less
Fig. 273.
in AD than in BD.
Fig. 274.
If, as in Figs. 275 and 276, BD is vertical,
D
w
Fig. 275. Fig. 276.
that is, the inclination to the vertical is nothing, the stress in BD
will be equal to "W, and there will be no stress in AD.
Having thus obtained an idea of the manner in which the
position of the rods relatively to the weight affects the stresses in
them, we proceed to show how the actual value of the stresses
(in terms of W) can be found.
Eeturning to Fig. 267, draw la (Fig. 277) parallel to W, Id
parallel to BD, and ad parallel to AD, then by a theorem called
the " triangle of forces," ^ if la is drawn to scale to represent W, dh
3 will represent C2 and ad will represent
to the same scale. Supposing, for instance,
that W= 1*5 ton, and that la is made 1"5
inch, then it will be found that
= 0-93 inch, a^^= 0*93 inch,
or
02 = 0-93 ton, Ci = 0-93 ton.
Further, la represents the direction of W, and ad will give
the direction of Ci and dl that of Cg, as indicated by the arrow-
heads in Fig. 277. Transferring these directions to Fig. 267 we
see that the rods AD and BD are in compression.
It is very important to observe that in Fig. 277 the arrows
follow each other, as it were, in the same direction of rotation.
Again, applying the theorem to Fig. 268, we obtain Fig. 278.
As before, if
W= 1'5 ton, and 1*5 inch,
dl = 0-93 inch, ad = 0-93 inch,
0-93 ton, Ti= 0-93 ton.
^ A proof of this theorem will be found in any work on Statics.
Fig. 277. Fig. 278.
we obtain
FRAMED STRUCTURES
179
In this case the arrows follow each other in the reverse
direction of rotation as compared with the previous case, and
transferring to Tig. 268 we find that the rods are in tension.
Applying the theorem in succession to Figs. 267 to 270
and Figs. 273 to 276, we obtain Figs. 279-286,^ and assuming
in each case that W=l-5 ton as before, we find the stresses
marked on the figures. The arrows showing the directions of the
forces are also marked. The student is recommended to compare
these results with the preliminary remarks.
The, theorem of the triangle of forces can be thus stated : —
If three forces acting on a point are in equilibrium, then, if a
triangle be drawn whose sides are parallel to the directions of the
three forces, the length of the sides will represent the magnitude
of the forces, and the directions of the forces will follow each
other round the triangle.
Thus, taking a general case, if F, G, H are three forces acting
on the point P as in Fig. 287, then Fig. 288 will be the triangle of
Fig. 287. Fig. 288. Fig. 289.
forces ; and if the magnitude of one of the forces be known, that
of the other two can be found.
Fig. 290. Fig. 291. Fig. 292.
Now supposing we replace the force H (Fig. 289) by two
forces K and M, which together have the same effect as H (in
other words, H is the resultant of K and M), then the triangle cda
(Fig. 290) would be the triangle of forces for K, M, and H
(strictly H reversed), so that we obtain the polygon abed to repre-
sent the forces Gr, F, M, and K. In the same way, if F were re-
placed by two forces IST and 0 (Fig. 291), we should obtain the
polygon aelcd (Fig. 292), and so on.
1 Figs. 279-286. See Plate I. at end of volume.
i8o NOTES ON BUILDING CONSTRUCTION
This result is called the " polygon of forces," and it can be
stated as follows : —
If any number of forces acting at a point are in equilibrium,
then, if a polygon be drawn having its sides parallel to the foices,
the sides of this polygon will represent or be proportional to the
magnitude of the forces ; and the directions of the forces will
follow each other in rotation round the. polygon.
In the case of the ^polygon of forces it is only possible to find
the magnitude of two of the forces when the magnitude of all
the others is known. Thus in Fig. 290, if only the magnitude of
G were known, all that could be done would be to draw cb and
the direction of cd, but the point d could not be fixed, as the
magnitude of K is not known ; but if the magnitude of K is
known, then d can be fixed, and drawing da and la we see
that the magnitudes of M and F are found.
The student is recommended to study the above very care-
fully, as these results will be found of the utmost importance in
the sequel.
Some different forms of framed structures. — A framed structure
is composed of a num-
ber of bars jointed to-
gether. It is usual to
assume that these joints
are hinged, and offer
therefore no resistance
to the movement of
the bars. Such an as-
sumption, unless pin
joints are used, is not
true, but it errs on the
side of safety.
The simplest frame
is a triangle, as in Figs.
293 and 294, and such
a frame is perfectly
braced; that is, it is
rigid or stiff, and can-
not change its form.^
This being the case, the reactions and can be found by the
1 A very minute change takes place in consequence of the elongation and shorten-
ing of the members produced by the stresses, but this may practically be disregarded.
FRAMED STRUCTURES
i8i
Fig. 295.
Fig. 296.
Fisr. 299.
rule given at p. 18 for beams, and thus, taking the numerical
data given in Fig. 294, we find
= 3:0 X 7 = 5 cwts.,
= T4 X = 2 cwts.
A frame composed of four bars, as in Fig. 295, is, however,
not perfectly braced, and
would collapse under
the action of the load
W. But if an addi-
tional bar DB be added
(Fig. 296), it will be-
come perfectly braced.
A moment's considera-
tion will show that this
additional bar, called a
"brace," is in tension.
The frame could also
have been braced by
adding a bar AC (Fig.
2 9 7), which would, how- n
ever, have been in com-
pression, and C3
would then be unneces-
sary.
If the bars AD and
CB are equally inclined,
and another weight W
be added at D (Fig.
298), the frame would
retain its shape, but the
least deviation would
cause it to collapse; it
is, in fact, in unstable
equilibrium, and practi-
cally a brace would have
to be added. If only
one brace is added, it must be capable of resisting both com-
pression and tension, that is, a simple rod would not do ; or two
rods capable of resisting tension only may be added, as in Fig. 299.
In this case the frame is over-hracecl, and it is possible, by drawing
up the rods by means of coupling screws, to introduce stresses in
l82
NOTES ON BUILDING CONSTRUCTION
the frame in addition to those produced by the weight. In such
a case the frame is said to be self -strained — a state of things
which is of course to be avoided as much as possible.
In Fig. 300 is shown
a frame composed of
five bars. If a brace
EC be added (Fig. 301),
the frame is still in-
B completely braced, and
another brace, either AC
or EB, must be intro-
duced. Figs. 302 and
303 show different ways
in which a five -sided
frame could be perfectly
braced, and Fig. 304
shows the frame over-
braced.
From the above it
will appear that to per-
fectly brace a frame it
must be divided up into
triangles.
Fig. 302.
Fig. 304.
Maxwell's Diagrams.
Eeturning to Fig. 294,
p. 180, which is redrawn
in outline in Fig. 305,^
and applying the triangle
of forces to the joint A,
we obtain the triangle
shown in Fig. 306, and
therefrom the stresses
given. The triangle also
shows that the bar CA
is in compression, as re-
presented by the arrow
pointing towards A, and that AB is in tension, as represented by
the arrow pointing away from A.
1 Figs. 305-311. See Plate II. at end of volume.
MAXWELL'S DIAGRAMS
For the joint B we obtain in the same manner the triangle
shown in Fig. 307.
And lastly, at C we obtain the triangle given in Fig. 308, and
we see that, as should be the case, the values obtained for
and Cg are the same as before.
Method of drawing a stress diagram} — The above three triangles
can be combined together in one diagram, Fig. 309, thus: Commence
by drawing the triangle for joint A. We now have C^, and on this
side of the triangle we can draw the triangle for the joint C, as in Fig.
308, taking care to reverse the direction of C^. Lastly, the triangle
for joint B can be drawn on C^, Fig. 309, and if the diagram has
been correctly drawn, it will finish off or " close " where it com-
menced. Sucli a diagram is called a stress force diagram,aTid is one
of the methods of finding the stresses in a framed structure. These
diagrams were originally proposed by Professor Clerk-Maxwell.
It will be noticed that in this example we could have com-
menced by drawing the triangle for the joint C, and then obtain
the other two triangles, thus finding the values of and with-
out previous calculation ; but it is not always possible to do
this, as will be seen by the following example : —
In Fig. 310, which is an outline copy of Fig. 297, but loaded
with two weights W and W^, if we were to commence at the joint
C we should have four forces to deal with, namely, the weight W
and the stresses in the three bars connected together at that joint.
Therefore, as already seen at p. 180, we cannot determine the
stresses. Finding, however, E^ and Eg as usual, we can draw the
triangle for joint B, as shown in Fig. 311. Now proceeding to
joint 0 we form a polygon and obtain the stresses C3 and C^.
This polygon is drawn by reversing the direction of C^, drawing
W vertically to scale, then drawing C3 to meet C^, drawn parallel
to their respective bars. Taking D as the next joint, we find
C^, and the polygon for the joint A can then be added.
This diagram could, however, have been drawn by commenc-
ing at the joint D, without determining E^ or E3.
The student is recommended to see what effect altering the
values of W and has on the shape of the diagram. He will
find, for instance, that when W = W^, then Cg = T and C^ = 0 ;
or, again, when = 0, then C3 = 0 and C^ = 0, which is, more-
over, evident on inspection.
^ In Appendix XYI. will be found more detailed rules for drawing a Maxwell's
diagram.
i84 NOTES ON BUILDING CONSTRUCTION
The above method will be applied in the sequel to the
determination of the stresses in roofs.
Bow's system of notation has not been used for the simple roof trusses here
dealt with, but it is admirable for complicated structures. It is explained in Ap-
pendix XVII. and used for the braced girders at p. 354.
METHOD OF SECTIONS.
We can now proceed to the consideration of the second method of de-
termining the stresses in a framed structure.
Imagine the frame shown in Fig. 3 1 2 cut in two parts along the line PP ;
then, considering the left portion, it is clear that this portion could he kept
in equilibrium by applying to each bar that is cut, a force of the same mag-
\ 1 A ^
Fig. 312.
Fig. 313.
nitude and direction as the stress existing in the bar before the section is made.
Thus the forces R^, W, C3, Cg, and T, acting on the portion of the frame shown
in Fig. 3 1 3, are in equilibrium. Hence the sum of the moments of these forces
about any point is zero (see App. II.) Now it is quite inmiaterial about which
point moments are taken. If, therefore, we wish to find Cg, we can select the
point A, which is the intersection of Cg and T (the other two unknown forces),
and so we obtain an equation containing Cg in terms of W, and the perpen-
dicular distances from A on to the directions of Cg and W, which are called the
lever arms (R^ does not appear in the equation, as it also passes through A).
If I is the lever arm for W, then IW is the moment of W ; and if
is the lever arm for Cg, Z^Cg is the moment of Cg — the sign being negative,
because Cg tends to turn the portion of the frame in what we have assumed
to be the negative direction (see p. 26).
Hence IW -\G^ = 0,
or 03 = ^ W.
I and l-^ could be found by calculation, when the dimensions of the frame
are known, but the process is rather complicated, and, moreover, they can be
obtained graphically very simply and rapidly, and with sufficient accuracy for
all practical purposes, by drawing perpendiculars from A (which can be called
the " turning point ") to the directions of "W and Cg, and measuring the
lengths of these perpendiculars to the same scale that the drawing of the
frame is made to. Carrying this out in the present example we find (Fig. 314)
Z = 6-70 feet, = 3 -93 feet,
METHOD OF SECTIONS
185
and consequently
6-70
T93'
W-1-7W.
To find T we must take moments about the point D, the intersection of
Co and Cg. Measuring the lever arms we find ^
-9-5T-8-3W + 15R^=0.
Hence
T = l-58R^-0-87W.
And lastly, to find Cg we must produce T and Cg to meet at 0 (Fig. 315);
and dropping perpendiculars from O on to the directions of W and R^ we find
-6-7C5+19-3W- 12-6R^ = 0.
Hence
C5= - (r88R^ - 2-9W).
E
A
<—-/-—»
Fia;. 314.
Fig. 315.
This method is called the Method of Sections, and is a modification of that
originally proposed by Professor Rankine. In the present form it was, it
is believed, originally suggested by Professor Ritter.^
Rule fok Method of Sections. — This method can be stated as
follows : — •
Sever the structure in two parts by means of an imaginary line (which need
not be a straight line), cutting if ^possible through three bars of the structure.
Select one of the parts, and apply to the severed ends of the bars forces equal
in magnitude and direction to the stresses in those bars. This part of the
structure will then be in equilibrium. To find the stress in any one of the
severed bars, take moments about the point of intersection of the other two
bars. An equation will thus be obtained containing the unknown stress it
is required to* find, any loads that may be acting on the part of structure
selected, any reaction that may be acting on it, and the lever arms of the
various forces involved.
The following terms are used in connection with this method, and are
tabulated as follows : —
Turning ]}oint is the point about which moments are taken, and is generally
the intersection of two of the unknown stresses.
Lever arm of a force is the perpendicular distance from the turning point
1 It will be observed that the forces applied to the ends of the several rods in
Figs. 313, 314, and 315 are all tensions. Hence clearly tensions are positive but
compressions are negative, if the forces are thus drawn.
2 See Iron, Bridges and Eoofs, by Professor Ritter, translated by Lieut. H. R.
Sankey, R.E.
1 86 NOTES ON BUILDING CONSTRUCTION
to the direction of the force. Practically these lever arms are obtained
graphically.
It was mentioned above that the section should if possible be taken
through three bars. This is not always possible, and the method must then
be slightly modified.
The above is only an outline of the method, which is a very powerful
one. The student is referred to Professor Eitter's work above mentioned,
where he will find the method very fully described.
We have now considered two methods of determining the
stresses in a braced structure, and it will be found that in some
cases the iirst, and in other cases the second method, gives the
result in the quickest and simplest manner ; and again in other
cases the methods are equal in this respect. Experience will
soon show which method to apply, and it does not appear to be
of any use laying down general rules on the subject.
It should be observed that both methods require a drawing
to sc^le of the structure, and to obtain accurate results this draw-
ing should be made to as large a scale as possible.
We now proceed to apply these methods to some simple cases
of framed structures.
Chapter X.
BRACED OR OPEN WEBBED GIRDERS.
"DEACED or open webbed girders are much used in engineering
works, but have only a limited application in building con-
struction. It will therefore only be necessary to consider one or
two simple cases. Moreover, the general theory of these girders
is difficult, and beyond the scope of this book.
Warren Girders. — A very usual form of braced girder is shown in Fig.
Fig. 316.
316, and names are given to the different parts as follows : —
AB is the compression boom or flange.
*io*i9 tension boom or flange.
a^ciii, «2^i2' ^3*^13' ^^°-5 called braces, and are tension braces
or compression braces according to the stress in them,
a^, a^, etc., are the apices.
A modification of this girder would be obtained by turning it upside
Fig. 317.
down, as in Fig. 317, when the nature of the stresses in the diagonals is
of course reversed. These forms of braced girders are known as "Warren
girders.
Fig. 318.
Lattice Girder. — Fig. 318 shows another form of girder, which is called
a lattice girder. Vertical braces are added at the ends, and are called end
pillars.
i88 NOTES ON BUILDING CONSTRUCTION
It is most usual to incline the braces of a Warren girder at 60° and
those of a lattice girder at 45°.
M" Girder.^ — Fig. 319 shows another form of braced girder, called the
N girder. In this the upright braces are called verticals.
Mr A
Fig. 319.
X
X
X
X
X
X
X
X
X
X
Fig. 320.
N (braced) girder. — Sometimes additional braces are added, as in Fig. 320 ;
but, as explained at p. 181, the girder is then over-braced and may be self-
strained, so that these braces should not be added unless the girder is subject
to a moving load, and then only to a certain number of bays next the centre,
depending on the proportion of the moving load to the weight of the girder.
Application of tlie Load; — The load is supposed to be concentrated at
the apices, either at the top or at the bottom ; occasionally, but unusually,
at both top and bottom. To do this, arrangements must be made to transmit
the load to the apices by means of secondary bearers. The weight of the
braces and booms can also be considered as applied at the apices, and practi-
cally the weight of the girder can be taken as uniformly distributed.
Proportion between span and depth of girder. — As in former
cases, the depth of the girder must be sufficient for stiffness, and
it is found that a ratio of about J^- gives the best results. The
exact depth must be arranged so as to get an exact number of
triangles, and will therefore depend on the angle of inclination
chosen for the braces.
Stresses in Warren Girder.
Example 38. — We will take as an example the girder shown in Fig.
321 supporting loads of W-^, Wg, Wg, etc., tons on the upper apices.
w, w, W3 w w, w, w, W3 w,
^ la, iflj ^a^ lil y la^
w
•11 "-12
<.'l6 "17 '^IS ^1!
a, 4 dy.
Fig. 321.
To find the number of triangles we have, d being the depth,
= tan 60 = ^3,
_ M
1" V^'
^ Also called WhiiJple-Murphy girder.
STJiESSES IN WARREN GIRDER
If n is number of triangles,
189
2c3!
n(Aa^) = span = 1 2d = n -y^,
n=10-4.
whence .. - - . -, , , x ^-u
Therefore ten triangles will do, and will give a depth rather greater than
-jL- span, as shown in the figure.
K- a.
Fig. 322.
To find the stresses, take a section and select the left portion of the
girder for calculation (Fig. 322). We will adopt the following notation :—
C4 5, the compression in the bar a^a^ of the top boom.
B^,;^^, the stress in the brace a^a-^^.
Ti3,;l4, the tension in the bar a-^^a^^ of the lower boom,
a, the distance between the apices.
And similarly for the otlier bars.
To find the str
obtain the equation
To find the stress €^,5 we must take moments about a^^, and we thus
Supposing that
= W, = W,
as is most usually the case, then
R^=4-5W,
and
Or
and
But
Therefore
+ 2cZxC4,5-a(W + 3W + 5W + 7W-9x4-5W) = 0.
+ 2dxC4,5 + 24-5Wxa = 0,
C, , = - 24-5W
4.5 2d
^ = cotan 60'
a
1
C
24-5
— — W.
The - sign shows that 0^,5 is in compression. _
Ao-ain, to find T^g^^^ we must take moments about a^, and we obtain
_dxTi3,i4-W3xa-W2X 2a-WiX3a + R,x4ct = 0.
As before, when the apices are equally loaded, we get
- d X T, o ,4 - a(W + 2W + 3W - 4 X 4-5 x W) = 0.
igo NOTES ON BUILDING CONSTRUCTION
a
^
5:
5 —
'•"lb
^
^ v«
5 —
^ —
5 —
a"
5
<
That is,
„ 12a ia
According to the rule, to find B^^^^ we ought
to take moments about the intersection of a a.
and a^^a^^. But these lines do not meet, or, in
other words, their intersection is at infinity. All
the lever arms would therefore be infinite. Although
the result can be obtained by following the rule, it
is simpler in this case to resolve vertically (see Apjoen-
dix II.) the forces acting on the portion of the girder
shown in Fig. 322. It will be observed that both
^4.5 ^'^^ '^\^,\'^ ^^"^^ resolved part vertically, since
they are both horizontal, and that the resolved part of
B
x^in 60°,
Hence
^4.14
Or, if the apices are equally loaded,
_ w
In the same manner the stresses in all the
other bars can be found. These have been worked
out, and are marked on Fig. 323.
In the above example the loads are supposed
to be applied to the upper apices. If they were
applied to the lower apices, the stresses would be
slightly modified. The student is recommended
to re-work the example with the loads applied to
the lower apices.
Stresses in Whipple-Murphy Girder.
Example 39. — Find the stresses in the girder
shown in Fig. 324, the lower apices being loaded
with 3-0 tons each.
To find the stresses in the bars a^a^^ a^^^g^
and ayy^Yii take an oblique section along the line
00, as shown in Fig. 325.
The stress in a^a^ can be found by taking
moments about ct^o, thus (the depth being 4 feet) —
+ 403^^-4 X 3-0-8 X 3-0 + 12 X 13-5 = 0.
WHIPPLE-MURPHY GIRDER
191
Whence
€3^^= — 31 "5 tons.
a^a^ is therefore in compression.
To find T^i,i2, take moments about dg,
so that o.Y)P^Y2. tension.
- 4Tii,i2-4 X 3-0 - 8 X 3-0 + 12 x 13-5 = 0,
'^11.12 ~ +31'5 tons,
cli as Oi as «6 ^
Ob 9t=13-5ions
Fig. 324.
Fig. 326.
Fiff. 327.
thus-
The stress in the vertical be found by resolving vertically,
'^3,12 - - ^'^ + 13-5 = 0,
19 = + 7'5 tons.
Another section must be taken to find the stress in the brace cu^f'ix,
shown in Fig. 326.
Kesolving vertically,
Bg^j^ X cos 45° + 13-5 - 3-0 - 3-0 = 0,
V-5 X v/2,
Bo
= - 10-6 tons.
To find the stresses in the bars a^a^ and (t^a-^^^, we can take a section
NOTES ON BUILDING CONSTRUCTION
cutting through only two bars (Fig. 32 7).< —
Resolving vertically,
Bo^;^ X COS 45° + 13-5 = 0,
or ■'^o,!" -13'5x a/2)
= - 19-1 tons.
To find Tq^jq Ave can take moments
about any point in aga^, for instance, about
P, vi^hose distance from is c, then
- Tq^^q X c sin 45° + 13-5 x c cos 45° = 0.
But sin 45° = cos 45°,
so that Tq,;^q= +13-5 tons.
The stresses Bq_^ and T^^^q can also be
found very simply, by diagram, as shown in
Fig. 328.
Fig. 328.
/
i
\
J 5-
'S ions.
Fig.
329.
It will be observed that a straight section
taken to obtain the stress in a^a-^^ will cut
through four bars, but by taking a circular sec-
tion, as shown in Fig. 329, we need only cut
through three bars, and by resolving vertically
we obtain at once
^5,14 = + 3 tons,
a result which is self-evidently correct.
The stresses in the remaining bars of the
girder can be obtained in precisely the same
manner to the above, and as an exercise the
student is recommended to carry out the cal-
culations for each bar and check the results
he obtains by those given in Fig. 330. He is
also advised to find the stresses when the
girder is turned upside down, as shown in
Fig. 331-
PRACTICAL FORMULA
193
General Eemarks. — It is to be noticed that in both the
above examples all the top horizontal bars are in compression,
and all the lower horizontal bars in tension. Further, all the
braces sloping downwards away from the centre are in com-
pression, and all the braces sloping downwards towards the centre
in the Warren girder and the verticals in the Whipple-Murphy
girder, supported as in Fig. 330, are in tension. The stress in
both top and lower booms is greatest at the centre, and gradually
diminishes towards the ends ; the stress in the braces is least at
the centre, and increases towards the ends.
The student is recommended to compare carefully these stresses
with those in a plate girder.
Dimensions of the various bars. — The stresses being obtained,
the dimensions of the various bars can be found as explained in
Chap. VI.
For the small braced girders used in building construction,
the booms are generally made of horizontal plates riveted to-
gether as in plate girders, and connected by L irons with a
vertical web plate, to which the braces are secured. The ties
are generally made of flat bar iron, and the struts of L, T, or
LJ iron.
Short practical Pormula for all open webbed Girders.
Tlie stresses in the flanges Bg may practically be taken as in
a plate girder.
Applying this to Example 38, p. 188, for a plate girder (uni-
formly loaded) we have from Equations 15 and 89
o
.:BsXd==~ and Bs= — ,
w= — in Example 38: d= ,
a r 7 2 '
W (10a)2 2
a 8 ^3
25aW
J3 '
aW liere = W in Example 38,
25W
V3
B.C. IV.
194 NOTES ON BUILDING CONSTRUCTION
wHch is practically near enough to the result in Example 38, viz.
24-5W
V3
The stress on any har is equal to the shearing stress at the
point multiplied by the cosecant of the angle of inclination of the
bar to the horizontal, and divided by the number of triangulations.
Thus in Example 38 the shearing stress at 'B^,ji= — W^ — Wg
_W2-Wi + E^= -4W + 4-5W=r-5W.
By above rule the stress in B4,i4 = '5W cosecant 60°
2 W
which agrees with the result by the other method (see p. 190).
In vertical bars (the cosecant of 90° being 1) the stress is equal
to the shearing stress at the point (see Vgjig, p. 191).
Graphic Method.
Figs. 1 and 2, Plate C, show Example 38 worked out by a Max-
well's diagram, and Figs. 3 and 4 show the same for Example 39.
In Example 38, 2 tons per upper apex has been assumed
as the load. The notation is by Bow's method, which is explained
in Appendix XVII.
By measuring the different lines of the diagrams on the scales
given it will be seen that the stresses found graphically agree
with those found by calculation.
The thick lines denote compressions, the thin lines tensions.
Lattice girders may be calculated by the method given for
Warren girders at p. 188, or by the graphic method.
A separate calculation or diagram should be made for each
triangulation.
Bowstring Girders have a curved upper flange or low, wliich for an uniformly-
distributed load should theoretically be a parabola, but is practically made an arc
of a circle (see p. 34) ; the lower flange or string is horizontal.
For an uniformly distributed dead load only vertical bars are required, suspending
the string (on which is the load) from the bow.
Where Z = span, w = weight of bow and load per unit of length, (i = depth of
girder at centre, = distance between the vertical bars.
wP
The horizontal thrust (T) throughout bow = tension in string =g-^
The compression along the bow at any point distant x horizontally from the
centre = ^w^x^ + T".
Thrust on bow at spriuging=T x secant of angle of tangent of bow at this point
with horizontal.
Tension on each vertical bar = mw.
Advantages. — The stress on the string and verticals is uniform, that on the bow
varies only from 8 to 12 per cent, so that the bars and plates may be of uniform
dimensions. As the girder can be made of a depth equal to 4 its span, the sections
of the flanges will be light and economical in comparison with those of shallower girders.
Bowstring girders are very seldom used in connection with buildings and need
not be further described.
Chaptee XI.
TRUSSED BEAMS
O EVEEAL kinds of trussed beams were described in Part I., and
^ it will be noticed that the construction of these beams is
simple ; the calculations to obtain the stresses accurately are, how-
ever, difficult, and beyond the scope of these Notes. It is clear that
the beam itself bears a certain proportion of the load, and the
trussing makes up the difference. The part borne by the beam
depends on the comparative stiffness of the beam and of the
trussing, and this varies with the temperature, which affects the
length of the tie-rods. Moreover, the beam is not in a simple
state of bending stress, but is also subject to direct compression
arising from the trussing, and liable at any time to alteration by
tightening or slackening the nuts of the tie-rods. These con-
siderations will show that the accurate calculations are compli-
cated and practically useless, as they would have to be based on
assumptions which might at any time cease to be correct.
It is therefore usual in practice to assume that the beam does
not bear any of the load ^
as a beam, and this as-
sumption amounts to the
same thing as to suppose
that the beam is jointed
at the point or points i^ —
where the struts are con- *
FLfr. 332.
nected to it. Thus the
trussed beam shown in
Eig. 332 is assumed to be jointed as in Fig. 333. This assump-
tion clearly errs on the safe side, and makes a rough allowance
for the compression produced in the beam by the tension in the
tie-rod.
With this assumption there is no difficulty in finding the
196 NOTES ON BUILDING CONSTRUCTION
stresses by either of the methods given in the preceding chapter.
For instance, employing Maxwell's diagrams it will be seen that
Fig. 334 is the diagram of the stresses in the trussed beam shown
in Fig. 332.
Having thus obtained the
stresses, the dimensions of the
'« various parts can be found as ex-
plained in Chaps. III. and VI.
■^^S" Trussed beams of the form
shown in Fig. 335 are, as explained at p. 181, incompletely braced,
and are therefore unsuitable if the load is not uniformly distri-
njPlui
thread
Fig. 335.
buted. Fig. 336 shows the change of form which would take
place if the beam were jointed and one point only were loaded •
Fig. 336.
this figure exhibits, therefore, the tendency when the beam is con-
tinuous, although the stiffness of the beam controls the deforma-
tion. For uniform loads, however, such beams are suitable, as the
stiffness of the beam would be quite sufficient to counteract any
small unevenness in the load that might occur.
Example 40. — Design a trussed beam 1 8 feet span to carry a load of 8
cwts. per foot run.
The first step is to find how many struts to employ. If we place one
ptrut at the centre, the beam will be divided into two portions of 9 feet each,
and there will be a load of 9x8 = 72 cwts. on each of these portions.
Considering each, portion as a beam loaded uniformly and supported at both
ends, it will be found on referring to Tables II. and III. that for strength
6" X 9" deep, and for stiff'ness 3|" x 1 1" deep, is required. If, therefore, timber
3f X 11, or its equivalent in section, is available, one central strut will be suffi-
cient, and the trussed beam would be of the shape shown in Fig. 332. We
will, however, suppose that timbers of this section are not obtainable ; it will
therefore be necessary to use more than one strut. If two struts are used, as
shown in Fig. 335, the total span will be divided into three portions of 6
feet each, and if each of these portions is looked upon as a beam uniformly
TRUSSED BEAMS
197
loaded and supported at both ends, it will be found tbat for strength and for
stiffness a beam 4" x 7" deep is needed. We will decide on this section.
The depth of the trussing must next be fixed. If it be made deep, the
stresses will be much reduced, but in many cases a deep trussing is incon-
venient. A rule applicable in general cases is to make the depth of the
trussing from centre of beam to tie-rod about of the span. In the present
Ra=AS cwts
IS avis.
Fig. 337.
case, therefore, a depth of l' 6" would do very well, so that the struts will
be 1' 2" long. Fig. 337 shows the trussed beam in outline.
Distribution of the Load. — We may consider that half the load on AC and
half the load on CD is concentrated at C ; that is, the load at C is
1(6 X 8 -1- 6 X 8) = 48 cwts.
The remaining half of the load on AC is transmitted direct to the point
of support, and it therefore does not affect the stresses in the trussing. At
D there will clearly be the same load as at C, hence the reactions E.^ and
Eg are each equal to 48 cwts.
We wiU now find the stresses both by the method of sections and by
Clerk-Maxwell diagram.
Stresses by Method of Sections. — To find the stresses in AE and CE
take the section shown in Fig. 338, and for the
stress in AE take moments about C.
- l-44T^-i-6 X 48 = 0,
6 X 48
= 200 cwts..
1-44
= +10 tons,
the stress in
CE take moments
To find
about A.
+ 6 X 48 -f- 6 X Cj^ = 0,
C-^ = — 48 cwts.,
= — 2'4 tons.
The stress in EF can be found by taking the section shown in Fig. 339,
and taking moments about C, then
- 1-5 X Tg-f-e X 48 = 0, Yf
T„ =
6 X 48
cwts.
Fiff. 339.
1-5
= 9-6 tons.
The stress in FB will clearly be equal
to that in AE, and the stress in DF to that in CE. We do not require
the direct compression in AC, CD, or DB for the approximate method we are
employing.
198 NOTES ON BUILDING CONSTRUCTION
Stkesses by Clerk-Maxwell Diagram. — Commencing at the abutment
A we obtain the triangle fgh (Fig. 340), from which we find T-^ = 10 tons as
before. Proceeding to E we obtain the triangle 1dm (Fig. 341), from which
we obtain
Tg = 9-6 tons,
Cj^ = 2*4 tons.
h
Fig. 340.
Scale 100 cwt. — 1 inch.
Having obtained the stresses, the diameter of the tie-rod can be found as
Fig. 341.
Scale 100 cwt. = 1 inch
explained in Example 13, and is given in Fig. 335. The stress in the struts
is so small that no calculation of dimensions is needed.
Chapter XII.
ROOFS.
MANY different kinds of roofs were described in Parts I. and
II., and Tables of the dimensions used in practice were given.
Cases may, however, occur in which the conditions are different
from those assumed in these Tables, and it then becomes necessary
to find the stresses in the various members of the roof, and there-
from deduce the dimensions.
The general method of finding the stresses in roofs will
therefore be described in an elementary manner, and wiU be
illustrated by examples.
Loads to be borne by Roofs.
Roofs are subject to several kinds of loads, but these loads can
be classed under two heads —
1. Permanent Loads. — These consist of the weight of the
roof truss itself, of the purlins, rafters, and roof covering;
frequently also the weight of a ceiling has to be borne by a roof,
and sometimes the weight of a lantern.
The weight of the truss itself must be estimated much in the
same way as the weight of a girder, but there does not appear to
be any formula available corresponding to that of Unwin's for
girders. Table XII., however, gives the information required for
several different kinds of roofs, and even in a special case the
weight required can be deduced therefrom with sufficient accuracy
for practical purposes, care being taken in large structures to
check the assumed weight by that deduced from the completed
design, modifying the dimensions if found necessary.
The weights of purlins, rafters, roof covering, etc., are also
given in Tables XII. and XIII.
The use of these Tables will be better shown when working
out the examples.
200 NO TES ON B UILDING CONSTR UCTIO N
2. Occasional Loads.— The occasional loads consist of the
weight of snow and of the wind pressure.
Bnow.~T\i% weight of snow a roof is Hkely to have to bear
depends on the locahty in which it is to be erected, and at best
only an approximate estimate can be given. It is usual to assume
m England that a roof of very flat pitch may have a depth of 6
inches of snow on it, and the depth will of course diminish as the
pitch increases. It can be further assumed that the weight of a
cubic foot of snow varies from 6^ to 11 lbs., according to the
amount of consolidation, so that an allowance of 5 lbs. per square
foot of horizontal surface covered by the roof is ample.
Wind pressure.~The wind pressure is more difficult to allow
for, and the experimental data on the subject are at present very
deficient. It is, in fact, only since the Tay Bridge disaster that
Enghsh engineers have inquired at all closely into the subject, and
the estimated allowance to be made for wind pressure has vlried
very considerably. In Unwin's Iron Bridges and Hoofs, however
although published long before the Tay Bridge disaster, the subject
IS treated at some length, and the following is principaUy derived
from that work.
Assums that the wind blows horizontally. When it meets the inclined
surface of a roof it will produce a pressure, normal (i.e. perpendicular) to the
surface of the roof, and there will also be a force exerted along the surface of
the roof. Now as air is almost a perfect fluid, this force along the surface of
the root will be extremely minute and can be neglected. Therefore we can
state that the wind pressure acts perpendicularly to the surface of the roof
Let P be the horizontal wind pressure per square foot, that is P is the
force exerted by the wind on a vertical surface one square foot in'area and
let P, be the pressure normal to the surface of the roof per square 'foot
then accordmg to Hutton's experiments '
Pj, = P(sin i)i-84cosi-i . . C90)
where i is the angle of pitch of the roof. ' ' ^ ^'
_ Amount of wind pressure to he calculated for.— As regards the value to be
given to P, the data at our disposal are not so numerous or so trustworthv
as might be wished ; the general practice, however, is to give P some value
between 40 and 60 lbs. per square foot.i For ordinary roofs, unless in positions
of exceptional exposure, 50 lbs. per square foot will be sufficient to take.
Table XIV. is derived from Equation 90, which it must be
remembered is only empirical, on the assumption that P= 50 lbs.
per square foot.
' A wind pressureof 41 lbs. per square foot has been registered at Greenwich.
Tlie gale which destroyed the Tay Bridge was calculated to have a force of 42 lbs
per square foot. The Board of Trade require 56 lbs. per square foot to be taken for
railway bridges which are often much exposed (Wray, Seddon).
LOADS ON ROOFS 201
Stresses in roofs of special shape.— It may occur in large roofs of special shape that
the wind pressure produces (say) compression in some member of the roof, in which
the permanent load and the snow produce tension, or vice versa. In such a case the
wei4t of the snow actually assists the member in question, but as the maximum
wind pressure may occur when there is no snow on the roof, this assistance from
the snow should not be reckoned on.^ Hence the stresses due (1) to the permanent
load (2) to the snow, (3) to the wind pressure, should be determined separately, and
that' combination of (1), (2), and (3) which produces the greatest stress in any member
must be adopted in designing it. However, in the comparatively small roofs used
in ordinary building construction, such as those illustrated m Parts I. and II., the
wind pressure does not reverse the stresses in any of the members.
Stresses in ordinary roofs. — In these it is only necessary to
obtain the stresses produced —
(1) By the permanent load and the snow combined, (2) by
the wind pressure ; and from a consideration of these stresses the
maximum stress to be borne by each member can be found, as will
be shown in the sequel.
Tredgold assumed that the wind pressure acted vertically and
uniformly over the whole surface of the roof. Such an assump-
tion is clearly erroneous, nevertheless the scantlings given on his
authority in the Tables, Parts I. and II., are dependent on this
assumption. The lighter scantlings for wooden roofs given in
Parts I. and II., and in Table XV., were obtained on the assump-
tion that the wind pressure acts perpendicularly to the surface of
the roof, and only on one side, and are recommended for use.
Distribution of the Loads.
The weight of the roof covering of the snow and of the wind
pressure is borne in the first instance by the common rafters, and
by them transmitted to the purlins. Taking a simple case, in
which there is only one purlin on each side of the roof, each
common rafter will be supported in the centre by the purlin, at
the top by the ridge piece and at the bottom by the wall plate.
If the rafter is continuous, and these three points are accurately in
a straight line, it can be shown (see Appendix XL) that f ths of the
load on the rafter will be supported by the purlin, y^gths by the
ridge piece, and j-^^ths by the wall plate. If, however, the three
points of support are not in a straight line, this distribution
would be changed. Supposing the rafters were not continuous,
but (as it were) jointed just over the purlin, then clearly the dis-
1 " It is suggested that, except perhaps in very cold climates, the snow need not
be considered at° all. Since long before such a wind force as 50 lbs. per foot square
could take effect upon a roof all the snow would have been blown offit."— Wray,
Instruction in Construction (Seddon).
202 NOTES ON BUILDING CONSTRUCTION
tribution would be as follows : — half the load on the purlin,
quarter on the ridge piece, and quarter on the wall plate. This
latter distribution causes greater stresses in the roof, and it will
therefore be better to accept it. Now let a, I, c (Fig. 342) be three
purlins (and we will suppose, as is
usually the case, that ab = he and that
w is the load ^ on the portion db of
the rafter and also on the portion he),
then, if the rafter is considered to be
jointed over each purlin, half the
load on the portion ab ^or ^ will be
transmitted to the purlin b, and half the load on the portion he
[ov I) will be transmitted to the same purlin. On the whole,
therefore, the purlin h will have a load = w transmitted to it from
the rafter in question. Applying this to the case where there are
Tig. 343.
•w
s
two purlins on each side of a roof, as shown in Fig. 343, the two
purlins will each support a load w and the ridge piece and wall
plate — .
The total load on a purlin will be that due to the sum of the
loads transmitted by all the common rafters supported by that
purlin. Eeferring back to Fig. 342 it will be seen that purlin
1 The loads are shown acting vertically, but the argument holds equally for the
wind pressure.
LOADS ON ROOFS
203
6) supports the load on a strip of the roof — ^ — ^^'^
toending the whole length of the purlin, that is, from one truss
s^upporting the purlin to the next. This is indicated in Fig. 344.
The purlins transmit the load to the principal truss, and,
Hooking upon the purlin as a beam supported at each end, half
tbhe load on the purlin will be borne by each truss supporting it.
Mg. 344.
Fig. 345.
TThus, referring to Fig. 345, at each point where a purlin is fixed
tto the principal rafter of the truss there will be a load
= \ load on purlin AB + -J- load on purlin BC,
tthat is the load due to the hatched portion of the roof. Fig. 345.
Usually the trusses are equidistant. In such a case the
lload at every point of the principal rafter where a purlin is fixed
iis equal to the load on a strip of the roof equal to
distance apart of purlins X distance apart of trusses.
So far we have tacitly assumed that the purlins are placed
i immediately over the joints of
tthe truss as shown in Fig. 346.
^Sometimes, however, smaller pur-
]lins are used, and they are dis-
Itributed along the principal rafter
fas shown in Fig. 347. In such
fa case we can consider ab as a
Ibeam supported at I, and Ic as
f another beam supported also at
II, so that half the load on each
( of these beams will be transmitted
Ito the joint at &. The load on
Ithe joint at I will therefore be
1the same as if there were only
( one purlin at I. The above
Fig. 347
refers strictly only to the loads due to roof covering, rafters and
204 NOTES ON BUILDING CONSTRUCTION
purlins, wind and snow. The weight of the truss itself should
strictly be considered as applied at all the joints, but it is
sufficiently accurate to consider this part of the load as applied
to the same joints {i.e. the upper joints) as the other loads.
In every case, therefore, the load acting on a joint of the
principal rafter (and in almost every case these are the only joints
that have to be considered as loaded) is the load on a strip of
roof whose width extends half-way from the joint in question to
the joints on either side of it, and whose length is the distance
between the trusses.
Example 41. — To take a numerical example, let it be required to find the
loads on the joints of the iron roof shown in Fig. 348, the trusses beino-
supposed to be 10 feet apart. °
Permanent Load and Snoio. — First as regards the permanent load and the
snow, we want to know the load per square foot of roof surface.
Referring to Table XIII. we find
Per square foot of roof surface.
Eoof covering (countess slates)
Slate boarding (1" thick)
From Table XII.
Common rafters and purlins
Principal
Snow ....
8 lbs.
3-5 „
11-5 lbs.
Per square foot of covered area.
. 2-0 lbs.
. 3-5 „
. 5-0
10-5 lbs.
Fig. 348. Fig. 349.
This last part of the load must be reduced to the load per square foot of
roof surface. Referring to Fig. 349 it will be seen that it is the load on
1x1-15 square foot of roof surface, so that this part of the load is, per
square foot of roof surface,
10-5
r— -r=9-l lbs.
I'lo
Therefore the total load per square foot due to permanent load and suow
= 11-5 + 9-1 = 20-6 lbs.
The load at joint A is due to a strip of roof 5 feet wide as indicated by the
arrows in Fig. 348, and 10 feet long, but as this load is supported directly
by the abutment, it does not cause any stresses in the roof, and need not
therefore be considered.
LOADS ON ROOFS
205
The load on joint D is due to a strip of roof 5 + 5 = 10 feet wide by 10
feet long. Hence load at D
= 10 X 10 X 20-6 = 2060 lbs.,
= 0-92 ton.
The load at C is due to two strips, each 5 feet x 10 feet, so that the load at
C is the same as at D.
Clearly the load at E is also the same.
Load due to Wind. — Next, to find the load due to the wind. Supposing
that the wind is blowing from left to right, then the left side only of the
roof will have any wind pressure to bear. As already mentioned, the
direction of the wind pressure is taken normal to the surface of the roof,
and from Table XIV. we see that for a pitch of 30° the wind pressure to be
allowed for is 33-0 lbs. per square foot.
The load at joint A is due, as before, to a strip of roof 5 x 10, so that
the load equals
5 x 10x 33 = 1650 lbs.,
= 0-73 ton.
The load at joint D is
10 X 10 X 33 = 3300 lbs.,
= 1-46 ton.
And the load at joint C is
5 X 10 X 33 = 1650 lbs.,
= 0-73 ton.
The load on any roof can be found in a similar manner.
Reactions at the Abutments.
As regards the permanent load and the snow, the reactions at
the abutments can be found as explained at p. 23 (Example 3).
The reactions due to the wind pressure cannot, however, be
found in so simple a manner.
Let Fig. 350 represent a roof
(the arrangementof the bracing, ^y,
being immaterial, is omitted), il*
then, regarding the roof as a I
whole, the total wind pressure
Pj, can be considered as concentrated at D, the centre point of
the side of the roof.
Further, P^, can be replaced by two forces — one a vertical
force, P^, and the other a horizontal force, Pg. The reaction at
each abutment can also be replaced by two forces, one vertical
and the other horizontal, as shown in the figure.
and Vg can be found at once, as being the reactions due
to a vertical load P^, and thus (Eule I., p. 18)
V = 4^ X P = ^P
2o6 NOTES ON BUILDING CONSTRUCTION
.'A
IS
V — 4 V P — IP
B — s~ ~
As regards and H^, however, all we can say is, that
H, + H3 = P„ . . . (91),
but we cannot, from the conditions of equilibrium, determine the
value of or H^.
This will be easily understood by considering that, if the
support B were quite smooth, and therefore unable to offer any
horizontal resistance, the support A would have to furnish all
the horizontal reaction. "We must therefore endeavour to find
some other condition to determine
the relative values of and H3,
or else make some assumption. ISTow
in the case of wooden roofs a fair
assumption to make would be, that
^ T.. ^ each abutment will iust afford suffi-
Jbig. 351. • . 1 • 1 .
cient horizontal reaction to make
the total reaction parallel to the normal wind pressure, as is
shown in Fig. 351. In this case
EB(m + = P^ X m,
Eb = -^P. . . (92),
m + n
and
K = -^'P. . • (93).
And m and n can be found by measurement.
In iron roofs, however, the case is rather different. To pro-
vide for the expansion and contraction due to variations of
temperature, one end of the roof is fixed and the other end left
free to move ; and so to enable this end to move freely it is
usual in large roofs to place steel roUers under the shoe.
In small roofs, where the shoe at the free end simply rests
on a stone template, the horizontal resistance of the free end
cannot exceed the frictional resistance. Now the coefficient of
friction for iron upon stone can be taken at about 0--i5, and to
find the resistance to friction we have only to multiply the
vertical pressure by this number, so that, supposing B to be the
free end, we have, if Vg is the vertical reaction due to the wind
pressure, roof covering, roof truss, and snow,
H3 = 0-45V3. . . . (94),
REACTIONS DUE TO WIND PRESSURE ON ROOFS 207
and this is the greatest value H3 can have, and = - O^SV^,
so that, knowing H^, we can easily find H^. The assumption
made in the case of wooden roofs is therefore only tenahle in the
case of small iron roofs, so long as it does not require a greater
value of than the above.
Now, for ordinary pitches, the assumption that the reactions are
parallel to the wind pressure holds good, and equations 92 and 93
are applicable, but for high pitches it does not hold good.
Let us inqiTire at what pitch the change tales place. Fig. 352 is the
outline of a roof, the reac-
tions at the abutments being
parallel to P,,. It will be
seen that the inclination of
I^B to Vb is a, the angle of
pitch. Hence
Bb cos a = Vb,
E, sin a = H„.
A fixed end
Fis. 352.
Free end
Fig. 353.
also
Eesults which are also evident from the triangle of forces, Fig. 353.
But by taking moments about A,
|P^ = 2cos2axEB.
Hence
H =■
sin a
and V =
4 cos a
4 cos ^a'
Now if L denote the permanent load,
_ P, L
Y = — - — h-
^ 4 cos a 2
Therefore the maximum value of Hb, from equation 94,
= 0-45 —^ + -
V4 cos a 2
and the value of a required can be obtained from the equation
P^sina ^ ^ / P L
-r — ^ = 0-45 +-Z
4 cos \4 cos a 2
It will be seen that a depends on L, and it will be found on trial that
for very light roof coverings (such as zinc) a = 36°, and for heavy roof cover-
ings, such as in Example 41, a = 45°.
When an iron roof expands by an increase of temperature, the
abutments will resist the expansion by exerting horizontal forces
and h^, which will be equal , ^
and opposite to each other ; but ^>v!l
as soon as these forces exceed
the frictional resistance of the -p. y- -^/^
free end, that end will begin to ^dedend ^'ig- ^54. r^eeend
move. Now supposing that a strong wind is blowing from right
to left, whilst the roof is expanding, the wind causes a horizontal
reaction TL'^, and the increase of temperature a horizontal reaction
2o8 NOTES ON BUILDING CONSTRUCTION
in the opposite direction, as shown in Fig. 354. Supposing, as
may well happen, that
then clearly there will be no horizontal reaction at the abutment
B, or, in other words, abutment A will have to supply the whole
of the horizontal reaction, that is
h; + a, = p,.
A similar case may occur when the wind is blowing from left
to right, and the roof is at
P»^JJ^ ^ the same time contracting ;
f then, as in Fig. 355, if
H, + A, = P^,
and the horizontal reaction at the free end will be zero.
Cases to be Considered.
Eeviewing what has been said with regard to small iron roofs,
supported on substantial abutments and with the free end not
resting on rollers, it will be seen that there are four cases to be
considered, as follows : —
The free end being on the right.
■ Eeactions parallel to normal wind press-
ure for pitches from 36° to 45°, ac-
cording to the roof covering, or hori-
Case 2. Wind on right j zontal reaction at free end equal to
frictional resistance for pitches above
45°.
Case 3. Wind on left 1 tt • ^
Case 4. Wind on right | horizontal reaction at free end zero.
It will be necessary to find the stresses in each case, and then
pick out the maximum stresses to which the various bars may be
subjected.
On the other hand, in large iron roofs, where the free end of
the trusses are supported on rollers, or, again, in small iron roofs
having one end supported by a column, which can offer little or
no horizontal resistance, the horizontal reaction at the free end
can be taken as zero. In such roofs, therefore, only two cases
need be considered, namely —
Case 1. Wind on left 1 tt • ^ i
o TTtr- J • 1 i. r -Horizontal reaction at iree end zero.
Case 2. Wind on right j
Case 1. Wind on left
EXAMPLE OF IRON ROOF 209
We have now shown how to find the loads and reactions
acting on roof trusses, and the next step is to show how the
stresses can be obtained. This we proceed to do by means of a
numerical example.
Stresses in an Iron Boof Truss.
Methods of finding the Stresses.— The stresses can be found either by
the method of sections or by the graphic method (Clerk-Maxwell diagrams)
If the stresses m all the bars are required, then the latter method will be the
better one to use ; but if it is only required to find the stresses in one or
two of the bars, then the method of sections will give the result, as a rule
more quickly. As an illustration, however, the stresses will be found by
both methods, by the graphic method below, and by the method of sections
m Appendix XVIII.
Example 42.— Find the stresses in the iron roof truss shown in Fig
373, P- 215, and 356,1 Plate III, assuming the following data :—
Span 40 feet.
Distance apart of trusses 8 feet.
Purlins placed over joints of truss.
Eoof covering — Countess slates on |" boarding.
Pitch 21|° or i. span.
Rise of tie-rod -^^ of span.
Stresses due to Permanent Load and Snow.
Preliminaries.— To find the permanent load and snow on the roof we refer
to Table XIII., whence
Per square foot of roof surface.
Eoof covering and boarding . .10-5 lbs.
Per square foot of covered area.
Common rafters and purlins . .2-0 lbs.
Principal . . . . _ .3-5
Snow
5-0
10-5 lbs.
1
1^'^ ^ r08 ^^^^^y P^^ square foot of roof surface.
Hence total load per square foot of roof surface
= 10-5 + 9-7 = 20-2 lbs.
Now the length of principal rafter is
V202-}-(i/)2
= 21-6 feet.
Hence the loads are (see p. 203) at joint D
21-6
-—X 8x20-2 = l745 lbs.,
= 0-78 ton.
And joints C and E will be loaded to the same amount.
The reactions at A and B will be each equal to
I X 0-78 = 1-17 ton.
1 Figs. 356-359. See Plate III. at end of volume.
B.C. IV.
2IO NOTES ON BUILDING CONSTRUCTION
The external forces due to the permanent load and the snow are therefore
as drawn in black in Fig. 356, Plate III.
Maxwell's Diagrams. — Commencing at joint A (Fig. 356), Plate III., by
drawing ah parallel to AC, he to AF, etc., we get the triangle ahc (Fig. 357), which,
measured on the scale of forces, gives the stresses in AD and AF, and shows
that AD is in compression and AF in tension. We next proceed to joint D :
there three bars meet, and there is also one external force, hence we obtain the
quadrilateral ahdea, and we thus find the stresses in DC and DF. Next pro-
ceeding to joint F we get the quadrilateral chdfc ; as this quadrilateral is of
peculiar shape it is shown separately in Fig. 3 $8, and from it we find that both
FC and FG are in tension. The student is strongly recommended thus to draw
separately the diagram for each joint until he has made himself completely
master of the subject. We have thus practically found the stresses in the left
half of the truss, and the stresses in the right half can be deduced at once by
symmetry ; for clearly the stress in CE will be the same as that in DC, and
so on. The diagram has, however, been completed for the whole truss.
At joint C we get the five-sided figure eghfd (Fig. 359).
At joint E the quadrilateral ghlk (Fig. 357).
At joint G the quadrilateral fhlc. To draw this quadrilateral all we
have to do is to draw from Z a line parallel to GB, the other sides being already
determined. This line ought, if the diagram has been correctly drawn, to
pass through c, or, as it is expressed, the diagram ought to " close." We
thus have a very useful and simple check on the accuracy of the work, for
if any mistake has been made the diagram will not close.
Finally the triangle ckl gives the stresses at joint B, and the numerical
values of the stresses can be obtained by measuring the lines representing
them on the scale given (| inch = 1 ton), the values being then filled in
to column 2, Table H., p. 213.
Stresses due to Wind Pressure.
1 We now proceed to find the stresses
''^K \Pc due to the wind. On referring to
Table XIV. we find that, for a roof of
\ p ^^sT^"'^ -^ 2lf° pitch, the wind pressure normal
G^^^=~^**jB to the roof should be taken at 25*2
-11-5 -X- , -^^^^ g^y. 25 ii3g_ per square foot.
Fig. 360. Following the method explained a,t p.
205 we then obtain the load caused by the wind on the joints D and C, as
shown in Fig. 360, namely—
P _!l:^x 8x 25 lbs.,
= 0-96 ton.
P, = iP, = 0-48 ton.
P^=0-48 ton,
and Pj,, the total wind pressure, is
P^=2P„=l-92 ton.
We will suppose that A is the fixed end of the truss, and that the end
B is free to move, but not placed on rollers.
Case I (see p. 208). Wind on left. Reactions parallel to wind pressure.'^ —
1 Since the pitch is less than 45° (see p. 207).
EXAMPLE OF IRON ROOF 211
Eeferring to Equations 92 and 93, p. 206, and to Figs. 360 and 361 we see
tliat 1 '
and
^ 40 ^
28-4
X 1-92 = 1-36 ton,
Rb = -^x 1-92 = 0-56 ton.
The external forces due to the wind from the left are therefore as shown in
Fig. 361.2 ^
Maxwell's Diagrams. — Commencing at joint A we find two external forces,
and ; as, however, these two forces are acting in exactly opposite
directions, we need only plot their difference, namely —
1-36 - 0-48 = 0-88 ton.
"We thus obtain the triangle ahc, which gives us and S^^ (Fig. 362).
Proceeding now to joint D we get the quadrilateral cade, in which ad
represents P^ and is drawn to scale equal to 0-96 ton.
For joint F we obtain the quadrilateral hcef. This quadrilateral is also
shown separately in Fig. 363.
Next taking joint C we get the five-sided ^guTe fedgh (see also Fig. 364).
Then proceeding to joint E we get the straight line gh as the diagram.
This shows that S^^ = and = 0.
For joint G we obtain the triangle fhh, and not a quadrilateral, since
= 0- This triangle will be a check on the accuracy of the drawing, for
the diagram ought to close at h.
Lastly for joint B we get the triangle hhg, and we have thus obtained the
stresses in all the bars. They can be measured to scale on the diagram and
are then inserted in Table H., p. 213.
Case 2, Wind on right. Reactions 'parallel to normal wind pressure. — A
moment's considerati6n of Fig. 365 will show that this case is exactly the
Fig. 365
opposite of Case 1, and we can immediately write down the values of the
various external forces, as has been done in the figure, and it is further
manifestly unnecessary to go through the calculations for the stresses, as we
can deduce them from those obtained under Case 1. Thus the stress in CE
in the present case will be equal to the stress in DC in Case 1, and so on.
Column 4 has thus been filled in in Table H.
Case 3 (p. 208). Wind on left. Reaction at free end vertical. — In this case
_28-4
m + 7i ~ 40 ■
2 Yigs. 361-364. See Plate IV. at end of volume.
212
NOTES ON BUILDING CONSTRUCTION
(Fig. 366) it is a little more difficult to find the values of and R^, because
the external forces acting on the roof are not parallel to each other, and we
ft^O-52
Fig. 366.
know neither the direction nor the magnitude of R^. Now when three forces
acting on a body are in equilibrium, their directions meet in a point,^ so that,
as we know the direction of and E3, we can find the direction of by
joining A with the point of intersection of Pj, and R^. But with flat roofs,
as in the roof under consideration, this intersection is a long way below the
roof, and the intersection is acute, and therefore cannot be obtained with
great accuracy.
In such a case, therefore, a better way is to find the value of R^, which
can be done by taking moments about A. We obtain
+ 10-8xP^-40xR3=0,
10-8 X 1-92
^3 = - ^ = 0-52 ton.
Now draw ah (Fig. 367) parallel to P^ and measure along
it 1'92 ton to scale ; next be parallel to R^, and measure
along it 0-52 ton a to scale ; then ac will be the direction of
R, and we also find by measuring ac that
R^=l-45 ton.
We can therefore now proceed to find the stresses.
Maxwell's Diagrams. — Commencing at joint A (Fig. 368)2 -we find two exter-
^,\>.K nal forces, P^^^ and R^. Let ad parallel
to P^ and ac parallel to R^ represent
these forces (Fig. 369), then adec is the
complete diagram for joint A.
It will be found that the remainder
of the diagram for the roof can be
drawn as in Case 2 ; the description is
not therefore repeated, but the diagram
is given in Fig. 369.
Case 4. Wind on right. Reaction at
free end vertical. — In this case we can
obtain the reactions very simply by dia-
gram. Thus produce Pj, and R3 to
meet at K (Fig. 370) and join AK; then
AK is the direction of R^ (see Fig. 367).
Now, plotting KA to scale to represent
Fig. 370.
^ This can be easily proved by taking moments about the intersection of two of
the forces. The moments of these two forces will be nothing, but for equilibrium
the sum of the moments of three forces is nothing, therefore cleai-ly the moments of
the third force about this point must be nothing ; that is, its direction must pass
through the point, or, in other words, the three forces meet in a point.
2 Figs. 368, 369. See Plate V. at end of volume.
EXAMPLE OF IRON ROOF
213
(1-92 ton), and drawing ah parallel to E^, we obtain the triangle of forces
for the three forces P„, E^, and E.„ and by measurement we find
E^ =0-87 ton,
E3 = 1-26 ton.
As a check, taking moments about A, we have
-4OXE3 +26-3 X 1-92 = 0,
E3 = 1-26 ton.
Maxwell's Diagrams. — Fig. 37 ji shows the loads for this case, and the corre-
sponding diagram is given in Fig. 372, and the stresses obtained therefrom
are entered in Table H.
Tabulation of Stresses.
We have thus obtained the stresses in the various members of the roof
under various conditions. These stresses must now be tabulated, so that we
may be able to pick out the maximum stress to which each member of the
roof truss may be subjected. This has been done in Table H below, and
the stresses^ that are to be added together, to obtain the maximum stress, are
printed in italics. The seventh column gives the maximum stresses, and it
will be seen that the maximum stress to be borne by corresponding members
on either side of the roof is not quite the same. This is due to the fact that
one end of the roof truss is fixed and the other end is free to move. But as
these differences are very small, and it would be practically unwise, at any
rate in so small a roof truss, to make corresponding members on either side
of different scantlings, these differences will be ignored and the scantlings
will be calculated to meet the stresses given in column 8.
It should be noticed also that the stress in the principal rafter is greater
in the lower part than in the upper part ; but, in a small roof truss like the
one under consideration, no difference would practically be made in the
scantling of the principal from A to C. In column 8, therefore, the stress in the
upper part of the principal rafter has been put equal to that in the lower part.
TABLE H.
Compressions — Tensions + ?
Stresses.
2.
Perma-
nent load
and snow.
Wind.
7.
Maximum
stresses.
8.
Stresses
to be used
in calcu-
lating
scant-
lings.
Reactions parallel
Reaction free end
vertical
3.
On left.
4.
On right.
5.
On left.
6.
On right.
-3-27
-1-89
-3-38
-1-62
-7-83
-7 -83
+ 3-38
+ 1-57
+ 3-67
+ 0-83
+ 7-83
-f7-83
-0-73
-0-96
0-00
-0-96
0-00
-1-69
-1-69
-If- 16
-3-28
-1-89
- 3-33
-1-62
-7-49
-7-83
+ 1-93
+ 2-17
+ 0-32
+ 2-20
+ 0-16
+ 4-13
+ 4-13
+ 2-34
+ 1-30
+ 1-31
+ 1-54
+ 0-69
+ 3-88
-1-3 -88
+ 1-93
+ 0-32
^2-17
+ 0-38
+ 2-02
+ 4-10
-f 4-13
-4-16
-1-89
-3-28
-1'98
-3-0
-7-44
-7-83
-0-73
0-00
-0-96
0-00
-0'96
-1-69
-1-69
-V4-16
+ 1-57
+ 3-38
+ 1-88
+ 2-70
+ 7-54
+ 7-83
-4-45
-1-90
-3-27
-1-98
-3-00
-7-72
-7-83
^ Figs. 371, 372. See Plate V. at end of volume.
2 See p. 185. Many writers take compressions as -|- and tensions as -
NOTES ON BUILDING CONSTRUCTION
Calculation of the Scantlings.
Having obtained the stresses in the various members of the roof truss the
next step is to decide on the form, and to calculate the scantlings of the vari-
ous members, so that they may be able to safely resist these stresses.
Suitable forms, such as are used in practice, have been described in
Parts I. and II., and the student is therefore referred back for information
on that part of the subject.
As regards the calculations of the different members and joints, several
examples have already been worked out in Chap. VI. of this Part, and the
student is specially referred to Examples 13, 19, 20, 26, 26a, 27, and 28.
The roof truss can now be considered as completely designed, and the
result is given in Fig. 373. The same truss is shown in Fig. 374, when the
purlins are distributed along the length of the principal rafter, in which
case the principal rafter is increased in section but no alteration is made in
the other members of the truss.
The calculations to find the size of the principal rafter in the
case where a number of small purlins are distributed along its
LENGTH would be as follows : —
We must first find the load on the principal. The permanent load and
snow consists of
Per square foot of roof surface.
Koof covering and boarding , . .10*5
Per square foot covered.
Purlins 1"5
Snow 5-0
Or
6"5
— 6 lbs. per square foot of roof surface.
Total=10-5 + 6 = 16-5 lbs.
This load is vertical and must be resolved perpendicularly to the principal
by multiplying by cos 2lf° thus —
16-5 X cos 2lf° = 15 lbs.
The normal wind pressure is 25 lbs. per square foot (see Table XIV.), so
that the total load per square foot is
15 + 25 = 40 lbs.
Hence total load on one half of the principal rafter is
Distance
Length, apart of
trusses.
40 X 10-8 X 8 = 3456 lbs.,
= 1-54 ton.
This load produces a bending stress in the principal, the compressive
element of which must be added to the direct compression. The moment of
flexure is wV^ ^ 1-54 x 10-8 x 12
~8"~"8"~ 8 '
= 25 inch-tons,
whether we consider the principal as jointed at the strut, in which case
the lower half of it is a beam supported at both ends, or whether we suppose
it to be continuous over the strut, when the lower half is a beam supported
at one end and fixed at the other.
2 1 6 NO TES ON B UILDING CONS TR UCTION
We must now assume some section of T iron, and see if it is strong
enough. Assume, for instance, a section 4" x 5|-" x y, the table being 4"
wide.
According to the rule given in Appendix IV. the distance of tbe centre of
gravity from the edge of the table is
- 4x1 + 5 X 3 , „ . ,
X = f — =1-8 inch.
4 + 5
And according to Appendix XIV. the moment of inertia about an axis passing
through the centre of gravity is
I - H 4 (1 -83 - 1 -33) + 1 (1 -33 + 3-73)},
= 13-65.
Treating 'principal as jointed. — Hence the greatest compression due to
bending (at the edge of the table), on the supposition that the lower half of
the principal is supported at both ends.
-r. ^Vo 25x1-8 . ,
Irom Equation 53 r„ = — = i^.q^ = tons per square inch.
The direct compression produces a stress per square inch (see note below)
7-83 7-83
= ^=(ST5)l=i'^^°^^^^^^^- .
Hence total maximum compression
= 3'3 + 1*7 = 5 tons per square inch.
Practically this section would do.
Treating principal as continued. — If, however, the principal is considered
as continuous over the strut, and is supposed to be supported at one end and
fixed at the other, the maximum compression due to bending at the edge of
the stem becomes
25 X (5-5 -1-8)
So that on this supposition the section is too small.
Next, try a section 3" x 5^", the table being i" thick and the stem
It will be found that the distance of the CG from the edge of the stem is
3""92, and that 1= 15-9. Hence maximum compression (at edge of stem)
25 x 3-29
/ = =5'l7tons.
15-9
And the total compression is
5-17 + 1-49 = 6-66 tons.
This section is therefore apparently too small, on the supposition that
the principal is continuous over the strut. This supposition, however, re-
quires that the three
points A, D, and C
D^'^^^^^^^"^^ ""^^^t"*^ shall remain in a straight
line, but the maximum
compression occurs in
Fig. 375. t]je principal AC when
the wind is blowing from the left, and this wind pressure bends the roof as
shown in Fig. 375 in a very exaggerated manner, and it is evident that this
bending of the principal will considerably relieve the compressive stress on
EXAMPLE OF IRON ROOF
217
the stem of the T iron, so that it can be taken that this section is quite
strong enough.
iyofe.— It has been assumed that the direct compression is uniformly distributed over the cross
section of the T iron. The detiection due to the distributed load tends, however, to increase the
intensity of the stress towards the edge of the table. The increase is, however small If for
instance, the deflection were i", the moment of flexure due to the direct compression would be
3-2x^8 ^ ^^^^'^'^ ^^'^^'^ "^"^"^ "^'^^ ^ ''^^^^ 4"x5|"xi" a maximum compression of
ton per square inch, to be added to the I'T ton already found.
Diagrams for various Roof Trusses.
In Figs. 376-395 ^ diagrams for different kinds of roof
trusses are given, with the hope that they may be of assist-
ance to the student. It is strongly recommended that, as an
exercise, some span be as-
sumed, and that the stresses
be obtained by following
these diagrams ; the results
obtained being occasionally
checked by the method of
moments.
In these diagrams only
one case for the wind
pressure has been shown,
since the other cases pro-
duce very similar diagrams.
It shouki be observed that
the diagram for the roof truss
shown in Pig. 392 cannot be
drawn by the graphic method
alone, because either at the
points E or M the stresses in
three of the bars are unknown.
But the stress in MN can easily
be found by taking a section as
shown in Fig. 396 for permanent
load and snow, and in Fig. 397
for wind. The value thus found for S
diagram completed.
WOODEN EOOFS.
It is not worth while to give examples of the calculations for
timber roofs. These can very seldom he necessary, and, if required,
ivould he similar to those for iron roofs.
Tables of Scantlings for Wooden and Iron Roofs are given in
Tahle XV. p. 340, and in Parts L and II.
1 See Plates VI., VII., VIII., IX., and X., at end of volume.
2 For Figs. 398-430, see Appendix XVIII.
Fig. 397.2
can then be inserted, and the
Chapter XIII.
STABILITY OF BRICKWORK AND MASONRY
STRUCTURES.
TT is proposed in the following chapter to show, in an elementary
manner, how the dimensions for various kinds of brickwork
and masonry structures can be calculated.
The mortar with which the stones or bricks of a wall are
bedded assists the wall by its resistance to tension ; but, on the
other hand, the resistance to compression of mortar being less than
that of stone or brick, it reduces the resistance to crushing of the
wall. On the whole, the effect of the mortar is to increase the
strength of the wall as regards overturning.
In important structures, however, the tenacity of the mortar is
not taken into account, because settlements are liable to occur
which may dislocate the joints, and such walls are treated as if
built up of uncemented blocks.
We will first consider the following problem.
Single Block. — A rectangular block is supported by a flat surface, and
is pressed against it by an inclined force F (Fig. 431). Find in wbat manner
tbe pressure between the block and the
flat surface is distributed.
It will be observed on looking at the
plan that, for simplicity, the force F is
assumed to act in the central vertical plane
of the block, and is supposed to include the
weight of the block.
The point C, where the direction of the
force intersects the supporting surface, is
Fig. 431. called the centre of pressure.
Now F may be resolved into two components — one perpendicular to the
supporting surface or joint, and the other parallel to it. Y and X are these
components. The component X tends to make the block slide along the
surface and is resisted by the friction ; but at present we are not concerned
with this part of the question.
A K
R
G
c.
H
" F
E
D
Y
A
STABILITY OF BRICKWORK AND MASONRY
The normal component Y is the one that causes the pressures between
the block and the surface.
Since C is not in the centre of GH it is clear that these pressures will
not be uniformly distributed, but will be greater towards AE than towards BD.
In the first place, since GH is midway between AB and ED, and C lies
on GH, the pressure at all points of any line KL drawn parallel to AE will
be the same, and therefore we can take the pressures along GH to represent
the pressures over the whole surface of the joint.
Secondly, on the supposition that the joint is a perfect one (and in a
masonry structure the principal function
of the mortar is to make the joint prac-
tically perfect), we can say that the
pressure along GH diminishes xmiformly
from G towards H. The pressures along
GH may therefore be represented by a
triangle GMO as shown in Fig. 432, so
that the pressure p at any point V is
given by the ordinate VN, and this or-
dinate will also give the pressure at any
point on KVL (Fig. 431).
First. Let 0 be situated between G and H ; this means that there is no
pressure on the joint from 0 to H.
Now the system of parallel forces acting over the whole joint and repre-
sented by the triangle GMO have a resultant K acting through the centre of
gravity of the triangle, and for equilibrium R must be equal and opposite to
Y, that is
R = Y.
Again, R is represented by the area of the triangle GMO, and is equal to
the area x AE. That is
^xGOxAE = Y.
2
Fiff. 432.
R:
Therefore
MG =
2Y
GO X AE
But since Y passes through the centre of gravity of the triangle GMO we have
GO = 3GC,
and for MG we can write X'{mas. >
2Y
Hence
(95),
SGCxAE •
and since Y, GO, and AE are known, f^-oisx.) can easily be found.
Example 43. — As a numerical example let
Y = 4 tons,
GO = 0-5 foot,
AE = 1 foot,
2 4
then =
= 5*3 tons per square foot.
Second. Let O and H coincide, as in Fig. 433. In this case we have
GC = i-GH,
and this is the condition which must exist in order that the pressure may
just vanish at H, or more strictly along DB.
NOTES ON BUILDING CONSTRUCTION
Let GH = 3 feet, and the remaining data be as before, then
Fig- 433.
Now if 0 were situated at the middle point of GH, the pressure would clearly
be uniformly distributed over the joint, and its intensity would be
Y ,
^ GH X AE
from which we see that when the centre of pressure is so situated that the
pressure vanishes along one edge of the joint, then the pressure along the
opposite edge of the joint will be double what it woiild have been had the
centre of pressure been at the middle point. This is an important result, and
should be carefully remembered. ^
Thus far we have not considered the horizontal component X of the force
F. This component is balanced by the resistance to friction of the joint so
long as slipping does not occur. Now the resistance to friction is
where fx is called the coefficient of friction, and is simply a number (always
less than 1) depending on the nature of the surfaces. Thus for stone against
stone
If, however, in a masonry or brickwork joint the strength of the mortar
is taken into account, the resistance opposed to X will be the resistance to
shearing of the joint. In practice, however, it is not usual to take account of
the strength of the mortar. It will be observed that the horizontal compon-
ent X and its effect on the joint are independent of the position of the centre
of pressure.
Uncemented Block Structures. — If we consider any masonry
or brickwork structure, such as a retaining wall for earth or water,
a boundary wall or an arch, we see that such a structure can be
dissected as it were into a series of blocks separated by joints,
such as the one we have just been considering. There will be a
resultant external force acting at each joint through the centre of
pressure of the joint, and this force will be resisted by the com-
pressions (and tensions in some cases) acting over the surface of
the joint.
^ See Appendix XIX.
STABILITY OF BRICKWORK AND MASONRY
221
Bef. — The line joining the centres of pressure is called the
" line of pressures," or " Moseley's line of resistance."
Thus in Fig. 434, representing a portion of an arch, the broken
line dbcdef is Moseley's line
of resistance; a, h, c, d, e, and
/ being the centres of press-
ure at the respective joints.
Conditions of Stability.
I. For important struc-
tures, or for structures the
failure of which might lead
to serious consequences,
such as loss of life :
{a) No portion of any
joint should be brought into
tension. Eef erring to Fig. 433, p. 220, we see that, to conform to
this condition, the centre of pressure must not be nearer the edge
of any joint than one-third the width of the joint (GH); or, in other
words, Moseley's line of resistance must lie within the middle third
of the structure.-^
(6) The intensity of pressure should nowhere exceed what
the material is capable of safely resisting.
That is,p(^ax.) must not exceed the safe resistance to crushing
or cracking of the material and mortar (see Table Ia).
(c) The resistance to sliding must be greater than the
horizontal component of the external force at each joint.^
II. For unimportant structures (such as boundary walls) :
{d) The intensity of pressure should nowhere exceed what
the material or mortar is capable of safely resisting.
(e) The intensity of tension should nowhere exceed what the
mortar can resist.
(/) The safe resistance to shearing of any joint must be
greater than the horizontal component of the resultant external
force acting at the joint.
These conditions will now be applied to some numerical
examples.
The middle half is sometimes taken for unimportant structures, as in Example
51, p. 254. The middle third should be taken for all important arches, as in the
Example, Appendix XX.
2 See p. 248.
222 NOTES ON BUILDING CONSTRUCTION
Brick Pier.
Example 44. — Find whether a brick pier 14 inches square and 8
feet high can resist a thrust of 3 cwts. inclined at an angle of 55° and
applied at the top of the pier, as shown in Fig. 435. If the pier is not
p strong enough, find what additional thickness of brickwork
sfnvts. is required, and how high it ought to be carried up. The
\ tenacity of the mortar not to be taken into account.
Preliminaries. — We will assume that one cubic foot of
brickwork weighs 1 cwt. The total weight of the pier
will therefore be
„ 14x14
11 cwts. nearly,
and acts vertically through the centre of gravity of the
pier, which is easily found.
The direction of the weight W intersects that of the
force P in the point A, and these are the only two external
forces acting on the pier. Their resultant can be found
in the usual way by means of the triangle A5c.
To find the centre of pressure, produce A6 to intersect
the plane of the base of the pier. It will be observed
that this point falls outside the actual base of the pier,
and, since the tenacity of the mortar is not to be taken
into account, it follows that the pier is not strong enough
to withstand the pressure of 3 cwts.^
It will therefore be necessary to increase the thickness
of the pier. Suppose the thickness is increased to 18
inches, which would appear to be about what is required, as the centre of
pressure will be brought nearer to the wall owing to the additional weight.
Fig. 436.
Fiff. 437.
^ The student will also observe what a very small thrust is sufficient to overturn
a pier of the given dimensions.
STABILITY OF BRICKWORK AND MASONRY
223
then the first question is how far up is it necessary to make the pier this
extra thickness.
Joint KL. — As a trial, suppose the increased thickness commences 3 feet
from the top of the pier, then it must be ascertained whether the joint KL
at the point of change is safe (Fig, 436). The external forces acting on the
upper three feet are the thrust of 3 cwts., and the weight of the portion of the
pier EL acting through its centre of gravity. Determining the resultant of
these two forces in the usual way, it will be seen that its direction passes
very near to K, so that the portion EL of the pier would be on the point of
overturning.
Next try if increasing the thickness of the pier 2' 6" from the top will
do. Carrying out the same construction as before, we find that the centre
of pressure now lies 0-9 inch within the edge K^.
Crushing. — We must now find the maximum intensity of compression, that
is,P(max.) The vertical component (Y) of the total pressure on the joint K^L^
can be found by dropping a perpendicular M on to ScZ the direction of W
(Fig. 437), and it is thus found by measurement that
Y = 5-85 cwts.
Hence applying Equation 95 we have
5-85
P(max.) = 2 X 3^0-9^ ^4 = cwt. per square inch.
And since i5,max.) can safely be as much as 0-5 cwt. per square inch (Table Ia.)
there is no fear of crushing, and condition (&) is therefore fulfilled.
Sliding. — As regards condition (c), the force tending to cause sliding is
the horizontal component of the thrust of 3 cwts. By measurement as above,
it is found that
X= 1-7 cwt.
The resistance to sliding is found by multiplying the vertical component
(Y) of the pressure on the joint by the coeflicient of resistance ^ to sliding,
which can be taken at 0-47. Hence the resistance to sliding is
0-47 X 5-85 = 2-7 cwts.,
jcrictuts.
Fig. 438. Fig. 439, Fig. 440.
^ 0-47 is the co-efficient of friction for masonry and brickwork with wet mortar
(see Table XVa.)
NOTES ON BUILDING CONSTRUCTION
so that tlie joint is safe against sliding.^ Finally, therefore, it is safe to
commence thickening the pier at 2' 6" from the top.
Joint GH. — It remains to be seen whether the conditions (b) and (c) are
fulfilled at joint G^H, Fig. 438. We have now to consider the portion L^G-^
of the pier. The external forces acting on this portion of the pier are the
resultant pressure acting on joint K^L-^, the value of which is found by-
measurement from Fig. 437 to be 6"1 cwts., and the weight of the portion
L^G-^ of the pier, namely 9*6 cwts., acting through the centre of gravity of that
portion, as shown in Fig. '438.
The centre of pressure of joint G-^H is found to be quite close to G^^, so
that the pier is still not thick enough at the bottom. We have therefore
to find how far below the joint K-^L^ the thickness of 18 inches can be
carried. Assume that it is carried down 4' 6" below K^L^ to joint MN (Fig.
439), then, proceeding as before, the centre of pressure of joint MN is found
to be 0'3 inches within the edge M, and from Equation 95, p. 219,
13-7
i'(max.) = 2X 3 ^ 0.3 ^ 14.
= 2'18 cwt. per square inch.
Orushing. — The maximum intensity of compression is therefore more than
the safe intensity, and consequently MN must be taken nearer to K-^L^, say at
3' 6" below K^L^i (Fig. 440). We then find
11-9
^'dnax.l-S X 3 ^ Q.g ^
= 0-63 cwts. per square inch.
M,N,,
'U-U cwts.
Fig. 441.
Fig. 442.
1 According to Kanldne, there is no danger of sliding if the angle S (Fig. 437)
between the normal to the bed joint and the direction of the resultant pressure is less
than from 25° to 36° the angle of repose of fresh masonry. In this case, the angle at
S is only 16°.
CHIMNEY
225
Below K^L^, therefore, the pier must again be thickened to the next brick
dimensions, that is to 23 inches.
Lowest portion of pier. — We now have to consider the portion of the pier
shown in Fig. 441. The external forces acting on this portion are its weight
(4-4 cwts.) and the resultant pressure of 12-1 cwts. acting at Cg as found by
measurement from Fig. 441.
Crushing and sliding. — It is then found that the centre of pressure is Co,
which is 3*5 inches from the edge, and the vertical component of the resultant"
pressure on the joint is also found to be 16-4 cwts., so that
16-4
P(m^^.) = 2 X 3 X 3-5 X 14'
= 0-22 cwt.
Condition (b) is therefore fulfilled. As regards condition (c), the force
tending to cause sliding is found by measurement to be 1-7 cwt. ; and since the
vertical component of the pressure is 16-4 cwts. the resistance to sliding is
0-7 X 16-4 = 11-5 cwts.
The pier has now been designed, and is shown in Fig. 442.
once
most
Chimney.
Example 45. — Find what wind pressure the chimney shown in Fig. 443
can safely resist.
It will be seen at
that AB is the joint
likely to fail.
Now the fall of a chimney
may lead to serious conse-
quences, and it is therefore
well to lay down the rule that
the centre of pressure shall be
kept so much inside of one edge
that no tension will be excited
at the other edge. Refer-
ring to Appendix XIX. we see
that for a hollow square chim-
ney, AC must not be less than
^ AB in order that the above
condition maybe fulfilled ; and
this value can be applied with
sufficient accuracy to the case
iinder consideration. But AB
is 22'5", so that least value of
AC = 3-75".
It will be found that there
are 8 7 cubic feet of brickwork
above the joint AB. There-
fore (allowing 1 cwt. per cubic
foot for weight of brickwork)
the weight of the chimney
above the joint ABis 87 cwts.
B.C. IV.
Fiff. 443.
226
NOTES ON BUILDING CONSTRUCTION
Let P be the normal wind pressure ; the total wind pressure on the side
of the chimney will therefore be
ll'3"x6xP = 67-5P cwts.,
and this force is applied at the centre of gravity of the side of the chimney.
The part of the chimney above the joint AB is therefore under the action
of three external forces — namely, 87 cwts. acting vertically through C.G.,
67-5P cwts. acting horizontally at E, and R the reaction to the resultant
pressure acting upwards at C.
Taking moments about C we have
/22-5
V 2
- 3-75 187
11-25
X 12 X 67-5P = 0,
whence
Pr= 0-1428 cwt.= 16 lbs.
It therefore appears that this chimney does not fulfil the condition laid
down, and although the tenacity of the mortar would materially increase the
powers of resistance, this chimney cannot be considered as safe.
Crushing. — The maximum intensity of compression with 16 lbs. wind press-
ure should also be found. In this case the vertical component of the resultant
pressure on the joint AB is equal to the weight of the chimney, namely, 87
cwts., and this pressure is distributed, but not uniformly, over an area of
9-9 square feet, or 1425 square inches. Following the rule given at p. 220
we have
, = 2 X
= 0-122 cwt. per square inch nearly,
f'dnax.) — " '- -J 425
so that there is no fear of crushing Avith a wind pressure of 16 lbs. per
square foot.
Sliding— As regards sliding, the force tending to cause sliding is the wind
pressure equal to 16 x 67-5
= 9-6 cwts..
112
: 61 cwts.
and the maximum resistance of the joint AB to sliding is 0-7 x 87 =
so that there is no fear of sliding.
Stay. — This chimney can, however, be
made perfectly safe by fixing an iron stay as
shown in Fig. 444. Let it be ascertained what
the cross section of this bar ought to be.
We have seen that the chimney itself
can safely resist 16 lbs. wind pressure.
Taking 50 lbs. as the maximum wind press-
ure, it follows that the iron will have to
resist 50-16 = 34 lbs. per square foot. So
that the wind pressure the stay has to resist is
67-5 X 34 ^ ^
= 20-5 cwts.
112
take
To find S, the stress in the stay,
moments about A (Fig. 443)
7 x 8- 5-62 x 20-5 = 0,
S= 16-5 cwts. = 0-82 ton.
When the wind is blowing in the direction shown in Fig. 443, the stay
has to act as a strut, and it must therefore be calculated as a long column
ENCLOSURE WALLS
227
with ends fixed. By the help of Table V. it will be found that an L iron
2" X 2" X f " can safely bear 0-9 ton (ends rounded), and will therefore be
strong enough either as strut or tie.
■■'/i/m////mm/"'
Enclosure Wall.
Example 46. — Find the wind pressure per square foot a long enclosure
wall of the section and plan shown in Fig. 445 can safely resist.
Ignoring tenacity of the mortar. — If the tenacity of the mortar is neglected,
it can be shown that, with a wind pressure of 16 '2 lbs. per square foot, the
centre of pressure would be at A, so that, to say nothing of crushing the
bricks, the wall would be on the point of overturning. This investigation
is, however, left as an exercise for the student.
Long enclosure walls should therefore be built in good tenacious
mortar, and it is well to build the lower courses in cement, especially if the
situation is exposed.
By a long enclosure
wall is meant one
whose unsupported
length is more than
about ten times its
height. Shorter walls
than these receive
considerable support
from each end.
Considering tena-
city of the mortar. —
We will now suppose
that this wall is built
in a good hydraulic
lime mortar, and that
the adhesion of this
mortar to the bricks
can be safely taken at 0-06 cwt. per square inch (see Table Ia). The tenacity of
the mortar itself is greater, but of course we can only reckon on the adhesion,
as it is less.
Now, looking at the plan, we see that we can take a portion of the wall
from E to F as a unit. The weight of this portion of the wall (allowing, as
before, 1 cwt. per cubic foot of brickwork) is found to be 59 cvvts. This
weight is distributed uniformly over the area of the joint AB in section and
portion EF on plan, which is found to be 1215 square inches. The pressure
due to the weight of the wall is therefore
59
j^j-g = 0*05 cwt. per square inch nearly.
Let P be the wind pressure per square foot, then the total wind pressure
acting at the centre of gravity of the unit of the wall is
10' 6"x 6' 7-|"x P = 69-5P.
The full height of the wall is not taken on account of the inclination of the
coping.
The wind pressure will clearly produce tension on the windward side,
that is at B, and compression at A, and the intensities of these stresses can be
—0 0 ><r-i (;->
Fig. 445.
228
NOTES ON BUILDING CONSTRUCTION
found by considering the portion of the wall as a beam fixed at one end,
namely, at the joint AB. Referring to p. 50 (footnote) we see that we must
first find the moment of inertia of the section, and this we can do by adding
together the moments of inertia of two rectangles thus —
I = _i_ X 1 8" X (13-53) + 3-^9 X 1 2) X 9^,
= 10251.
The distance of the extreme fibre from the "mean" fibre is sensibly
13-5 . ,
= — — 6 ■ 2 5 inches.
So that moment of resistance
10251
inch-cwts.
But moment of flexure
= 69-5P X
Hence P =
6-25
(6' 7i") X 12
2
10251 X 2
6-25 X 69-5 X 79-5 "
To find P we must therefore decide what value to give to r^. Now we
have seen that the mfe adhesion is O'l cwt. per square inch ; but can be
given a greater value than this, because the compression produced by the
weight of wall (viz. 0'05 cwt. per square inch) relieves the tension. Hence
we can take
= 0-1 + 0-05,
= 0-15 cwt. per square inch.
Hence
P = -089 cwt. = 10 lbs.
Therefore even if the wall is built in good hydraulic lime it cannot mfehj
resist more than 10 lbs. per square foot of wind pressure. It would of course
take a greater wind pressure than this to cause the wall to fail. Thus, taking
the ultimate adhesion of best hydraulic lime to hard stock bricks as 36 lbs.
per square inch or 0"32 cwt., we can put
r(, = 0-32 + 0-05,
= 0-37 cwts.
And
10251 X 2 X 0-37
"'6-25 X 69-5 X 79-5'
= 0-22 cwt.,
= 24"5 lbs. per square foot.
In an exposed situation, therefore, this wall would undoubtedly in course
of time be blown down. In an unexposed situation, however, the wall miglit
be safe enough, especially as the friction of the wind against the ground is
known to lessen considerably its force near the surface.
The wall can be materially strengthened by building the first ten or twelve
courses in cement. In this case the safe adhesion can be taken at about 0'6
cwt. per square inch, and hence
10251 X 2 X 0-6
~6-25 X 69-5 X 79-5'
= 0-35 cwt.,
39'5 lbs. per square foot,
so that the wall is now strong enough.
ENCLOSURE WALLS
229
It is clearly unnecessary in this case to find the maximum intensity of
compression, neither need any calculations be made as regards the resistance
to sliding, as the resistance we can now reckon on is not merely the friction
of brick against brick, or stone against stone, but the shearing strength of the
mortar, and this is much more than enough.
Chapter XIV.
RETAINING WALLS.^
A EETAHSTING- wall is a wall built for the purpose of " retain-
ing " or holding up earth or water. In engineering practice
such walls attain frequently enormous proportions, being used in
the construction of railways, docks, water-works, etc. In building
construction, however, their dimensions are never very large, and
the difficulties attending their design and construction, which in
engineering works are often very great, are therefore much reduced.
Usual crqss Sections for Eetaining Walls. — The form of
cross section given to such walls varies considerably according to
circumstances, and often according to the fancy of the designer.
A few of these forms, such as might be useful in building con-
struction, are given in Figs. 446-456.
Fig. 446 shows a simple rectangular wall. Such a section is
suitable only for very low walls (say not exceeding 5 feet), as it
is wasteful of material.
Figs. 447 and 448 show a better section, in which the back
of the wall is sloped. Except in the case of a wall for retaining
water, the alterations in the thickness of the wall would be carried
out by means of offsets, as in Fig. 448.
1 For Practical Formula see pp. 235, 239, 244, and App. XXL
RETAINING WALLS
231
rig. 449 shows a wall with vertical back and with a " battered"
face. In point of stability a wall with a battered face may or
may not be, according to circumstances, more stable than a wall
with a straight face and sloped back. It has the appearance of
being more stable, and there is the advantage that a small altera-
tion in the batter, caused by rotation or inclining forward, is not
noticeable, whereas the eye is offended if a wall with a vertical
face rotates ever so little. To prevent lodgment of water, how-
ever, the batter should not exceed 1
In Figs. 450 and 451 walls with battered face and sloping
Fig. 450. Fig. 451.
backs are shown. It will also be noticed that the footings are
inclined, thus giving greater resistance against sliding forward.
The section shown in Fig. 450 is, however, clearly not suitable for
retaining water.
In the case of walls for retaining earth, counterforts are some-
times added, as in Figs. 452 and 453, and Professor Eankine has
shown that a slight saving in masonry is thereby effected.
In the preceding cases it has been supposed that the top-
surface of the earth to be retained is horizontal, and also level with
the top of the wall. Frequently, however, a bank has to be sup-
232 NOTES ON BUILDING CONSTRUCTION
ported, as sliown in Fig. 454 or 455. Such walls are called
" surcharged " walls and clearly require greater strength.
Fig. 452. Fig. 453.
Another case in which greater strength is required is when a
building is placed close to the wall as shown in Fig. 456, because
Fig. 454.
Fig. 456.
Fig. 455.
the weight of the building tends to overturn the wall by increasing
the natural thrust of the earth.
RETAINING WALLS FOE WATER.
The calculations for finding the dimensions of walls for retain-
ing water, so long as the depth of water to be retained is not
great, are comparatively simple.
Such a wall is acted on by two external forces, namely, the
pressure of the water and the weight of the wall.
As regards the pressure of the water, it is known from hydro-
statics that it is :
(1) Directly proportional to the depth below the surface.
(2) Normal to the surface exposed to the pressure.
Applying this to the case of a wall with a vertical back
(Fig. 457) ; at the lowest point B the pressure will be proportional
to the depth of water AB, and will be horizontal, since it is
normal to the surface pressed. The pressure of the water at B
RETAINING WALLS FOR WATER
233
can therefore be represented by BD, which is drawn horizontally
equal to the depth of water at B. At A the pressure is clearly
zero ; hence ordinates from AB to AD represent the pressures at
all points along AB.
Fig. 457.
The area of the triangle ABD therefore represents the total
water pressure on the wall (see p. 219), and the resultant
pressure will pass through the centre of gravity of this triangle.
The centre of pressure E is therefore two-thirds of the way down,
that is AE = f AB . . . (96).
And if we consider 1 foot of length of the wall, the resultant
pressure (P) on this portion of the wall will be equal to the
weight of a prism of water whose cross section is the triangle
ABD and length 1 foot (Eig. 458).
Fig. 458. Fig. 459.
The volume of the prism is
ABxBD , AB2
where H is the height of water above the point B. And, since a
cubic foot of water weighs 0*557 cwt., we have
NOTES ON BUILDING CONSTRUCTION
P= 0-557 x-—cwts. . . (97).
A
We therefore know the magnitude, point of application, and
direction of P, which is all we require to know.
In the case of a wall with sloping back some modifica-
tions must be introduced. Thus in Fig. 459 the pressure at B
will be represented by BD drawn normally to AB and equal to
H. AD will then represent the pressure along AB, so that the
triangle ABD will represent the total pressure, and
AE = f AB.
Also
w AB X H
P = 0-557 X — 2 — • •
P is therefore completely determined.
As regards the second external force, namely, the weight of
the wall, if we assume a trial section the weight can easily be
obtained, and we can then ascertain whether this trial section is
suitable by applying the rules given at p. 221.
In the case of a wall for retaining water it is of the utmost
importance that no cracks should be formed, and for this it is
necessary that no tension shoiild be excited. Hence the centre
of pressure must not approach the outer edge nearer than one-
third the width of the joint.
The various points that have been considered will now be
illustrated by means of an example.
Reservoir Wall.
Example 47. — A wall to retain water to a depth of 6 feet is required ;
the top of the wall is to be 1 foot above the surface of the water and the
batter of the face i?-- The wall to be built in brickwork and cement.
Preliminaries. — So far as stability is concerned, the top of the wall might
be built to a feather edge, but if the wall were thus built the top would soon
be in want of repair ; moreover, the saving in brickwork would be exceedingly
small, and would not compensate for the additional difficulty in building.
A width of 9" at the top, although somewhat narrow, will be assumed.
Since the batter of the face of the wall is given, it remains only to find
the batter of the back of the wall, or, what comes to the same thing, to find
the width of the wall at the footings.
Now it can be shown that if the wall were of triangular section, i.e. with
a feather edge at the top, that all joints would be of equal strength, and the
effect of giving a certain width to the top of the wall is to make the lowest
joint the weakest ; we need therefore only consider this joint.
The courses will be built at right angles to the face of the wall, so that
the joints will have a slope of -^q.
RETAINING WALLS FOR WATER
235
As a first trial, assume the thickness of the wall at the footings to be 4
feet. The weight of one foot of length of the wall will be
0-75 + 4
^ — X 6-6 X 1 cwt. = 15-6 cwts.,
and this force will be applied at the centre of gravity marked C.G. in Fig.
460, which can be found graphically by a method explained in Molesworth's
Pochet-Book}
The water pressure is equal to 6 4 6
0-557 X ■ — ^ — ,
= 10-7 cwts.
It is perpendicular to AB, and is applied at the point E.
ii-S cwts.
Fig. 460. Fig. 461.
The resultant of these two forces intersects the joint DB in the point C,
DB
and as DC is 1-05 foot and is less than — the wall must be made thicker.
Next try 5 feet ; the weight will be 19 cwts. and the water pressure
11-2 cwts. It will be found that DC = 1-7 foot, which is within the middle
one-third of the width of the joint. This thickness is therefore sufficient,
and the section of the wall will be as shown in Fig. 461.
Strictly it should be ascertained that con-
ditions (5) and (c), p. 221, are fulfiUed. This
is left as an exercise for the student.
As a further exercise, the student can find
the proper dimensions to give to this wall
supposing the back to be vertical.
Practical Kule for thickness of walls to
retain water. — The following is an old rule
for retaining walls for water —
"Width at bottom = height x 0*7,
Width at middle — height x 0-5,
Width at top = height x 0-3,
1 Page 386, 22d edition.
236 NOTES ON BUILDING CONSTRUCTION
and a wall designed accordingly would be as shown in Fig. 462. It will be
seen that this section has a smaller base but somewhat greater area than the
one just designed. Its weight per foot run is 19-6 cwts. The resultant in
this case also falls just at the one-third of the width of the base within the
toe. On comparing Figs. 461 and 462 it will be seen that the width at the
top has a considerable effect on the cross section of the wall.
EETAINING WALLS FOE EAETH.
The section of a wall for retaining earth could be at once ascer-
tained in the manner just described for walls for retaining water
if we knew the direction and magnitude of the pressure exerted by
the earth on the back of the wall. We saw in the last example
that the water pressure on the back of a wall can be easily found,
but it is not so with regard to earth pressure, which varies with
every different kind of soil, and also with the conditions in which
the soil is ; for instance, whether loose or compressed, wet or dry.
Various theories on the subject have been propounded by Professor
Eankine, Boussinesq, and others, but they do not appear to be very
satisfactory, and in any case are far too complicated for these Notes.
Under these circumstances it will only be possible to give a mere
outline of the subject.
If a steep bank of earth is left to itself it will, under the
action of the weather, gradually
crumble down until it has taken
up a certain slope as shown in Fig.
463. The angle of inclination of
the slope at which crumbling ceases
is called the " angle of repose," and
this angle varies with each different ^
kind of earth. This angle is gener- -^'S- ^63.
ally denoted by In Table XVI. will be found the angle of
repose for several of the usual kinds of earths, but this angle can
be found experimentally in any particular instance by shovelling
the earth up into a
heap and measuring
the angle of the slope
so formed.
Now suppose that
the earth behind a re-
; taining wall is loose,
^ and that it would im-
Fig. 464. mediately fall down to
RETAINING WALLS FOR EARTH
237
the angle of repose were the wall removed, then it will be seen
from Fig. 464 that the wall, aided by the friction of the earth,
supports the wedge of earth represented in the figure by the
triangle ABF, which is tending to slide down BF.
The resultant pressure of this wedge of earth will be some
force such as E, and it is to find this force E that the various
theories already alluded to have been proposed. Once found, the
dimensions of the wall can be ascertained precisely as in the case
of the reservoir wall (Example 47). If, however, the earth is
at all consolidated, so that it would not fall to the natural slope
immediately on removal of the wall, the pressure on the back of
the wall is much less than that due to the wedge ABC. The
earth may even be sufficiently consolidated to stand vertically on
removal of the wall, in which case there is practically no pressure
on the back of the wall, and if it can be ensured that the earth will
always remain in this condition, the wall need only be built thick
enough to resist its own weight. It then acts simply as a covering
to the earth to prevent its being degraded by the action of the
weather. Such walls are called " breast walls." On the other hand,
however, infiltration of water will increase the pressure of the earth
against the back of the wall, beyond what it is when the earth is
quite loose. The student will at once see that to frame a theory to
meet all these different conditions must be a difficult matter.
"The presence of moisture in earth to an extent just sufficient to expel
the air from its crevices, seems to increase its coefficient of friction slightly ;
but any additional moisture acts like an unguent in diminishing friction, and
tends to reduce the earth to a semi-fluid condition, or to the state of mud.
In this state, although it has some cohesion, or viscidity, which resists rapid
alterations of form, it has no frictional stability; and its coefficient of friction,
and angle of rejDose, are each of them null.
" Hence it is obvious that the frictional stability of earth depends, to a
great extent, on the ease with which the water that it occasionally absorbs
can be drained away. The safest materials for earthwork are shivers of
rock, shingle, gravel, and clean sharp sand, whether consisting wholly of
small hard crystals, or containing a mixture of fragments of shells ; for those
materials allow water to pass through without retaining more than is bene-
ficial. The cleanest sand, however, may be made completely unstable, and
reduced to the state of quicksand, if it is contained in a basin of water-holding
materials, so that the water mixed among its particles cannot be drained off.
" The property of retaining water, and forming a paste with it, belongs
specially to clay, and to earths of which clay is an ingredient. Such earths,
how hard and firm soever they may be, when first excavated, are gradually
softened, and have both their frictional stability and their adhesion diminished
by exposure to the air. In this respect, mixtures of sand and clay are the worst ;
for the sand favours the access of water, and the clay prevents its escape.
238
NOTES ON BUILDING CONSTRUCTION
" The properties of eartli with respect to adhesion and friction are so vari-
able that the engineer should never trust to tables or to information obtained
from books to guide him in designing earthworks when he has it in his
power to obtain the necessary data either by observation of existing earth-
works in the same stratum or by experiment." ^
Mathematical Formulse. — Eankine's Theoey — Earth, liori-
zontal top. — It lias been shown by Eankine that in the case of a
wall retaining loose granular earth (i.e. not consolidated), the top
surface of the earth being horizontal and the back of the wall
vertical, the pressure P of the earth acts horizontally, at a point
^d of the height from the base, and that its magnitude is
P =
ivIL^ 1 — sin ^
(99),
2 1 + sin </) '
in which cf) = the angle of repose, w = weight of earth per unit of A^olume, H = AB.
Graphic Method. — This result can be expressed graphically
as shown in Pig. 465 by laying off the angle ABC = cf), and mak-
ing CP = CA. Then ^w(BP)^ lbs. is the magnitude of the earth
thrust, for it can be shown ^ that
1 + sm 0
so that Pz=i^BP)nbs. . . (100),
and hence the earth thrust is represented by cd.
If the hack of the wall slopes, AB can be taken as a vertical
line just clear of the wall, and the earth resting on the back of
the wall adds to its stability.
^ A Manual of Civil Engineering, by Prtf Ranldne, F.R.S., etc.
2 BP- (BC - AO- (-A^ - AB tan Oii^^^H^l^.
^ \cos ^ / cos-<^
RETAINING WALLS FOR EARTH
239
WTien the surface of the earth is at a slope instead of being
horizontal a similar construction wiU give the earth thrust, by
drawing its direction, according to Eankine, parallel to the slope
of the earth. These constructions are shown in Figs. 466 and
467. It will be observed that Fig. 467 shows the general case,
and that Figs. 465 and 466 can be deduced from it. According
to Dr. Scheffler's theory the direction of the thrust is always in-
clined at an angle cf) to the
horizontal, but the magni-
tude is greater than given
by Eankine, except in the
case shown in Fig. 466. /
The overturning power d'^'^'
of the thrust as found by \^
Scheffler's hypothesis is less
than that of the thrust as
found byEankine's method,
but it has been proved by experiments made by G. H. Darwin,
as well as by the results of actual practice, that Scheffler's
Fis. 467.
Fig. 468. Case 1. Fig. 469. Case 2, Fig. 470. Case 3. Fig. 471. Case 3a.
theory can be safely depended upon. This hypothesis, in fact,
makes some allowance for the cohesion of the soil, whereas, as
already stated, Eankine's method is based on the supposition that
the earth is granular and loose.
Practical Formulae. — A formula is given in Hurst's Pocket-
Book,'^ based on Eankine's investigations, which is shown below,
with certain modifications.^
1 Page 123, 14th edition.
2 The following process is recommended for designing retaining walls ; first find
the section by means of the formula, and then apply the graphic method as a
check.
240 NOTES ON BUILDING CONSTRUCTION
Let
T = thickness of wall with vertical sides.
Tj = mean thickness of wall with either face or back sloping
{i.e. thickness half-way up the wall).
H = height of the wall.
w = weight of a cubic foot of the earth at the back of the wall.
W = weight of a cubic foot of the wall.
/90-(f>\
Ki=0"7 tan ( — z — ), where (j) is the angle of repose (see
Table XVI.)
Then
Case (1) Wall with vertical sides and hacldng horizontal at the
top (Fig. 468). _
T = . • (101).
(2) Sloping walls with vertical bach (Fig. 469).
Batter of face
Tf T^ = 0-86T.
8 1\ = 0-80T.
fi T^ = 0-74T.
Case (3) Wall with vertical face and sloping hack (Fig. 470),
or (Case 3a) with offsets (Fig. 471).
Ti=0-85T.
Fig. 472. Case 4.
Fig. 473. Case 5.
RETAINING WALLS FOR EARTH
241
Fig. 474. Case 6.
Case (4) Suecharged Eevetments. — Substitute for H in the
formula the vertical height H' measured to the point G, found by
setting off the distance H along the earth slope, as shown in Eig.
472.
Case (5) When the lanh is of less height above the wall than the
distance IT would give, take the actual height of the bank, as
shown in Fig. 473.
Case (6) When the earth slopes from the front edge of the loall
the point G- should be found, as shown in Fig. 474.
"Values for w and W will be found in Table XVII.
Retaining Wall for Light Vegetable Earth.
Example 48. — Design a wall to retain light vegetable earth, consolidated
and dry, to a depth of 8 feet, the backing being
horizontal at the top and flush with the top of
the wall. The back of the wall to be vertical
and the face to battel- A. The wall to be built
in ordinary brickwork.
Preliminaries. — From Table XVII. we find
^y = 90 lbs. per cubic foot for vegetable earth.
W = 1 1 2 „ „ for brickwork in
mortar.
Also from Table XVI.
K = 0-26 for moist vegetable earth.
B.C. IV.
NOTES ON BUILDING CONSTRUCTION
Calculation by practical formula. — Hence Equation 101, p. 240,
= 1-86 foot ;
and = 0-80 X 1-86,
= 1-49,
or say 1' 6" ; and as the batter is y the thickness at the top will be 1' 0",
and at the bottom 2' 0".
The foundations can, witli advantage, be inclined so as to resist the
pressure which, tends to make the wall slip forward. We thus get a wall as
shown in Fig. 475.
Graphic method. — To compare the formula given above with Kankine's and
Scheffler's methods, draw a cross section of the wall, and apply the graphic
construction as shown in Fig. 476, taking the angle of repose as 49°. It
will be found that BP is 2-96 feet, so that
90
lw(BP)2= —(2-96)2 = 394 lbs.
Also the weight of one foot length of the wall= 1345 lbs.
Taking Rankine's method first, we find R as the resultant of the weight
and the thrust, and it will be seen that R intersects the plane of the base
sufficiently inside the wall. the resultant found according to Scheffler's
method, intersects the base still more inside the wall. It is left as an exercise
for the student to find the maximum intensity of compression at the edge.
RETAINING WALLS FOR EARTH
243
The student should observe that an extreme case has been chosen for this
example ; the earth is very light and the angle of repose large. If the earth
were not consolidated, the angle of repose would be 29°, and it will be found
that the mean thickness (T^) would then be 2-4 feet.
Eetaining Wall for London Clay.
Example 49. — Same wall as in the previous example, but designed to
retain London clay recently excavated and saturated with water (see Fig. 477).
In this case we find, on referring to Table XVII.,
i«= 120 lbs. per cubic foot, and
K = 0-53.
Hence Equation 101, p. 240, T = 0-53 x 8 \/If| = 4-4 feet,
Tj = 0-80 X 4-4 = 3-5 feet.
Therefore a wall of the section shown in Fig. 477 is required.
^London clay
aturated
'with water
Tis. 478.
Surcharged Retaining Wall (Loamy Earth).
Example 50. — Find the dimensions of the surcharged retaining wall
shown in Fig. 478. The earth is supposed to be of a loamy nature ; the
iace of the wall is to be vertical, and the back to have a batter of about |,
the wall to be built of rubble masonry.
Preliminaries. — From Table XVII. we find
w= 80 lbs. per cubic foot,
W = 140 „
and
K = 0-33.
On measuring H along the slope we find
= 9-3 feet,
and therefore
T = 0-33 X 9-3 ^^ = 2 -3 feet.
Hence from (3), p. 240,
Tj = 2-3 X 0-85,
= 2 feet nearly.
The dimensions in Fig. 478 give a mean thickness of 2 feet and a batter
of f at the back.
244
NOTES ON BUILDING CONSTRUCTION
Baker's practical Rules for Thickness of Retaining Walls.
Proportion of thickness to height. — "As a result of his owu
experience " Sir Benjamin Baker " makes the thickness of retain-
ing walls in ground of an average character equal to of the
height from the top of the footings."
He also says that "A wall quarter of the height in thickness, and
battering 1 inch or 2 inches per foot on the face, possesses sufficient
stability when the backing and foundation are both favourable."
Also that "under no ordinary conditions of surcharge or heavy back-
ing is it necessary to make a retaining wall on a solid foundation
more than double the above, or ^ of the height in thickness."
Eqiiivalent fluid pressure. — He says further that " Experiment
has shown the actual lateral thrust of good filling to be equivalent to
that of a fluid weighing about 10 lbs. per cubic foot; and allow-
ing, for variations in the ground, vibration, and contingencies, a factor
of safety of 2, the wall should be able to sustain at least 20 lbs. fluid
pressure, which will be the case if ^ of the height in thickness."
These rules may be usefully applied to check the calculations
made by other methods.
Foundations.
Ordinary firm earth will safely bear a pressure of about 1 to
1-^ ton per square foot, while moderately hard rock will bear as
much as 9 tons.
It is therefore not worth while to make any calculations for
the foundations of ordinary walls because the pressures they
cause are so small.
The foundations of retaining walls are subject, however, to
considerable pressures, which moreover are not uniformly distri-
buted. They should therefore be of such a width that the
maximum intensity of pressure is not greater than the soil can
safely bear, and the centre of pressure should not be nearer the
outside edge than ^ the width of the foundation.
Example 50a. — Pressure. — As an example suppose Nc, Fig. 478a. (the
normal constituent of the resultant of the weight of the wall for 1 foot in
length and the pressure of the earth upon it) to be equal to 2 tons.
Assume foundations such as those shown in Fig. 477, the centre of
pressure c being at J the width ef of the foundation from the outer edge,
then the maximum intensity of pressure will be at e (see p. 219), and it will
be equal to 2 x -I = f tons per square foot.
It is evident, therefore, that on firm earth, clay, or harder soils the maxi-
mum intensity of pressure would not be too great.
If the soil were very loose and unfit to bear weight then the concrete
RETAINING WALLS FOR EARTH
245
foundation would have to be extended so that the maximum pressure upon it
should not exceed what the soil could bear, and in extreme cases the founda-
tion would have to be prepared by driving piles.
If it is very important
for any reason that the
pressure on the earth should
be uniform, the toe of the
concrete must be extended so
that c should be the centre
of the earth pressed upon.
Concrete, base. —
Care must be taken
that the concrete is
made so thick that
there is no danger of
its breaking across.
In Fig. 478a ahde is
in the condition of a can-
tilever nearly imiformly loaded and tending to break off at Id.
Thus see Appendix VII.
6 '
taking /o= 100 lbs. per square inch,^ '
wl-
2 X 12(x 18)2 X 100
6 X 24
= 5400 lbs.
= 2|^ tons nearly.
Therefore the cantilever will bear 2|- tons distributed or ij ton per foot
superficial without breaking.
Whereas the most intense pressure at the end is only ^ ton per foot
2^
(giving a factor of safety of = nearly 4), and the mean pressure distri-
buted only ^^n.
The concrete therefore is amply thick enough. If made 12" thick it
would bear about i ton per foot-superficial.
Sliding on concrete. — To prevent danger of the wall sliding
forward on its concrete base, the tangent of the angle MQE (between
the normal NQ to the base and the resultant pressure EQ) must
not be greater than -J x the coefficient of friction of brickwork
with damp mortar, Table XVcc, i.e. not greater than ^ x '74 = -59.
The angle EQN measures 21°, the tangent of which is '38,
so that the condition is amply fulfilled and the wall quite safe
against sliding.
Sliding on clay. — In the same way by using the proper co-
efficients from Table XVa it will be found that the wall would
slide on wet clay but would be safe against sliding on dry clay.
I Baker.
Chapter XV.
ARCHES.
IN Parts I. and II. some preliminary remarks were made in
connection with arches ; the names of the different parts were
given, and also some practical details as to the construction of
brick arches.
It is proposed to show in this chapter how to ascertain the
thickness of the arch ring in order that the arch may safely resist
the load it has to bear.
The calculations in connection with the stability of the arch
ring are amongst the most difficult that occur in engineering, but
there is a simple graphic method by means of which the desired
result may be obtained by a series of approximations, and it is
this method which will be described.
It should, however, be observed that in the large majority
of cases the arches used in building construction have but a small
span, and are, moreover, only repetitions of what have been built
before, so that there is ample experience to fall back upon. There
is therefore no necessity, in such cases, to make any calculations,
and the dimensions can be obtained by referring to a table such
as that given in Molesworth's Pocket-Booh^ or Table XVIIa.
Graphic Method of determining the Stability of an Arch.
Let I, J, K, L be any voussoir^ of an arch. The load on
this voussoir consists of its own weight and of the portion of the
weight carried by the arch included between the dotted lines
(Fig. 479). Let w be the resultant of both these weights acting
through their common C.G. The voussoir is also subjected to
the pressure P of the voussoir above it, and to the reaction E of
the voussoir below it. These are the only three forces acting on
the voussoir, and if we know w and P we can find E and also its
1 Page 108, 22(i edition.
2 The student is referred to Parts I. and II. for the meaning of the various
technical terms employed.
ARCHES
247
point of application C, which is the centre of pressure for the
joint I, J. Now we know (see p. 221) that for
stability the point C must lie inside I, J, and
that, moreover, to prevent crushing the material
of the voussoir, C must not approach I or J
nearer than a certain distance, depending on
the value of K, and on the resistance to crush-
ing of the material. On this point Dr. Scheffler
has laid down a general rule that the point C
should lie within the middle, hcdf of the joint, ^'ig-
but other authorities recommend that C should lie within the
inner third of the joint,-^ in order that no tension be excited (see
p. 221). Dr. Scheffler's assumption is amply safe for all ordinary
cases, and will therefore be accepted for these Notes.
Now, for purposes of calculation, the arch can be considered
as divided up into a number of imaginary voussoirs, as shown in
Fig. 480, and we can suppose that the centre of pressure at each
joint has been found, and that these centres of pressure are then
joined together by a broken line. If the arch had been divided up
into a greater number of imaginary voussoirs, as in Tig. 481, the
line joining the centres of pressure would be almost a continuous
curve, and if a still greater number of voussoirs had been assumed,
the line would hardly be distinguishable from a curve. This
line is called the line of resistance, and from it we can obtain the
point of application (centre of pressure) and direction of the
pressure at any section of the arch.
Applying, therefore, the results arrived at in connection with
one voussoir, we see that for safety the line of resistance must lie
within the middle half of the arch ring, as shown in Fig. 485.
For mere stability, if the material of the arch ring were un-
crushable, the line of resistance might just touch the outside of
the arch ring.
The line of resistance also gives the direction of the resultant
^ For important arches the line of resistance should always be kept within the
middle third of the arch ring, as in the Example, Appendix XX.
248 NOTES ON BUILDING CONSTRUCTION
pressure on any joint of the arch ring, this direction being a tangent
to the curve ; and should this direction be too much inclined to
the joint the arch would fail by sliding.^ As a matter of fact, how-
ever, the inclination seldom, probably never, is sufficient to produce
failure in this manner, except in straight arches (see Fig. 1 4, Part I.)
To find the line of resistance. — The problem therefore is to find the line of
resistance, and herein lies all the difficulty.
On looking at Tig. 481 it will be seen that the line of resistance is hori-
zontal at the point D ; at this point therefore
the pressure is horizontal.
\^ I n Suppose the portion DB of the arch is re-
^ ot moved (Fig. 482), then we have the left-hand
// j \ portion of the arch kept in equilibrium by —
// Y (1) The portion of the load it has to sup-
ll-an \ P«^t (Wj).
L^jr__n_.:^ y. ^2) The horizontal pressure H at D.
/ ^ (3) The pressure at the abutment A.
/ pjg_ 482. Now, a force is determined when we know
(1) its direction, (2) its point of application,
and (3) its magnitude. is, or can be, so determined. Of H we only know
the direction ; and as to we know nothing except that it is equal and op-
posite to the resultant of "Wj^ and H. We therefore cannot solve the problem
until we know the point of application and the magnitude of H. To obtain
these we must resort to what is known as Moseley's Principle of Least Resistance.
Professor Eankine states this principle as follows : —
" If the forces, which balance each other in or upon a given body or
structure, be distinguished into two systems, called respectively active and
passive, which stand to each other in the relation of cause and effect, then will
the passive forces be the least which are capable of balancing the active forces,
consistently with the physical condition of the body or structure."
In the case under consideration W^^ represents the active forces and H and
the passive forces.
It therefore follows that, in the actual arch, the magnitude of H will be the
least possible consistently with the physical condition of the arch. And it will
also be seen that to each value of H there is a corresponding line of resistance.
Combining this with the condition relative to the line of resistance
(p. 247) we arrive at the following result.
The value of H will be such that the corresponding line of resistance
is just included ivithin the inner half of the arch ring.
This line of resistance is called the line of least resistance.
Theory of grafhic metJiod. — To determine the true line of least
resistance mathematically is most difficult, even in simple cases ;
but by the graphic method of trial and error already referred to,
the line of least resistance can be found very expeditiously — only
^ " To insure stability of friction the normal to each joint must not make an angle
greater than the angle of repose with a tangent to the line of pressures drawn through
the centre of resistance of that joint" (Rankine, Civil Engineering).
ARCHES
249
approximately, it is true, but with ample accuracy for practical
purposes. This method is as follows : — ■
Referring to Fig. 482, let be the lever arm of H from the centre of pressure
of the joint at A (viz. the point of application of Pj^), and a the lever arm of
from the same point. Then taking moments about this point we have
■WjX(i-Hx/!- = 0,
or H = rWi .... (102).
Hence H vs^ill be least when a is least and It is greatest, and that is the case
when the point of application of H is as high up as possible (that is at E, Fig.
483), and the point of application of as low down as possible (that is at K).
Fig. 483.1 / " ' Pig. 484.1 Fig. 485.1
On the other hand, the greatest admissible value of H will be when a is great-
est and In, least, that is when H is applied at F, and P^^ at G, as in Fig. 484.
The true value of H lies somewhere between the least and greatest
values thus found, and is the least value of H that allows of the line of
resistance being just included within the inner half of the arch ring. Thus,
supposing that it has been found that a certain value gives the line of
resistance shown in Fig. 485, then any less value of H would cause the line
Fig. 487.
1 In Figs. 483, 484, 485 and others, the curved dotted lines represent the centre
half of the arch.
250
NOTES ON BUILDING CONSTRUCTION
of resistance to pass below K, is therefore the least admissible value of
H, and the corresponding line of resistance is the line of least resistance.
Clearly, therefore, the next step is to establish a method of finding the line
of resistance corresponding to any assumed value of H.
Suppose that the portion AD of the arch ring is divided up into a number
of imaginary voussoirs, as shown in Fig. 486, and let the load on each of these
voussoirs be found, namely w^, w^, w^, etc.
Now the first voussoir is kept in equilibrium by three forces, as shown in
Fig. 487. We know H and w^, but P^^ has to be found. To do this, draw ah
(Fig. 488) to represent H, and aa-^ to represent ; then (by the triangle of
forces, p. 179) will represent P^. We therefore have only to draw P^
parallel to through the intersection of H and to find Cy
Proceeding to the second voussoir we have forces as shown in Fig. 489 ; and
Fig. 490 gives the triangle of forces. Cg therefore can be found as before. The
same process can be repeated for each succeeding voussoir, and by this means
the various centres of pressure can be found, and by joining them the line of
resistance can at once be obtained (Fig. 491).
It will be observed that Figs. 488 and 490 can be combined to form one
H
Fig. 492.
ARCHES— LINE OF LEAST RESISTANCE 251
figure, as shown in Fig. 492, and similarly the triangles of forces for all the
succeeding joints can be combined into one figure (Fig. 493), thus saving a
considerable amount of drawing.
It will also be observed that the broken line formed by the forces H, P^,
Pg, etc. (Fig. 491), approximates to the line of resistance, and if double the
number of voussoirs had been taken, the approximation would be still closer,
as shown in Figs. 494 and 495. With a very large number of voussoirs, the
line of resistance and the broken line formed by the direction of the pressures
at the joints practically coincide.
Practical Graphic Method of finding the Line of Resistance
in an Arch with a Symmetrical Load. — The line of resistance
corresponding to any particular value of H can therefore be found
by the following very simple graphic method.
Trial line. — Find the loads w^, w^, w^, etc., acting on the arch
252
NOTES ON BUILDING CONSTRUCTION
(Fig. 496), and draw the diagram of forces, Tig. 497, thus obtaining
the forces V^, Pg, P3, P^, etc. Then find the intersection of H and
w^, then the intersection of P^ and w^, and so on. Join these
points of intersection, and the broken line so obtained will be
approximately the line of resistance required.
It will be observed that this line of resistance (Fig. 496) is
not included within the inner half of the arch ring. The reason
is that the point of application of H has been chosen too high.
Fig. 498. j,y Fig. 499.
True line. — Lowering the point of application, which will
also increase H to (Fig. 498), we obtain a new diagram of
forces (Fig, 499) and a new line of resistance, which is now found
to just fit in within the inner half of the arch ring, and is there-
fore the line of least resistance required.
To find where the line of resistance is horizontal. — On refer-
ring to p. 248 it will be seen that the section D was taken through
the point where the line of resistance is horizontal, and we must
now show how this point can be obtained.
7V
Fig. 500.
Fig. 501.
Taking the simplest possible case, when the load is uniformly
distributed, as in Fig. 500, it is clear from symmetry that D will
be situated at the centre of the arch ring.
ARCHES— LINE OF LEAST RESISTANCE
253
Or, again, if the load is symmetrically distributed D will still
be at the centre of the arch ring (Fig. 501).
In both these cases the line of least resistance will be
symmetrical about the point D.
Now let us investigate the effect a concentrated load,^ placed
at any point on the arch ring, has on the position of the point D,
as shown in Fig. 502. Let W be this concentrated load, and let
it be applied at the point P. If we take any section of the arch,
as in Fig. 503, the thrust at the section can be represented by
the horizontal thrust H and a vertical force V. What we want
to find is the section at which V is nothing. If we regard V in
the light of a shearing stress we see on referring to p. 5 6 that
since, as a consequence of Eule 2, the shearing stress is nothing
under the resultant of all the loads, therefore the section of the arch
we require will be under the resultant of W and w^, . . . w^, w^.
In any case, therefore, the point D can be found by determining
the resultant load on the arch, and this can be done very simply,
by finding the centre of gravity of all the loads, as explained in
Appendix VI.
Thickness of the arch ring. — The thickness of the arch ring
should be sufficient to allow of the intensity of pressure being not
greater than what the material is able to bear safely. Moles worth
gives a rule, based on Eankine's, for finding the thickness of the
arch at the crown, or, in other words, the depth of the keystone.
It is as follows : —
Depth of keystone = 71 ^radius at crown . . (103),
where n= OS for blbckstone,
= 0"4 for brickwork,
= 0'45 for rubble stonework.
The depth thus found is expressed in feet.
1 A different method of treating an arch with an nnsymmetrical load will be
found in Appendix XX.
254 NOTES ON BUILDING CONSTRUCTION
The actual intensity of pressure at any section of the arch
ring can, however, be found, as soon as the line of least resistance
is known, by applying Equation 95.
Semicircular Brick Arch.
Example 51. — Find the line of least resistance in the symmetrically
loaded semicircular brick arch shown in Fig. 504 when loaded with a uni-
form load of 1 ton per foot run, including its own weight.
Fig. 504.
Preliminaries. — It will be found, by applying Equation 103, that the
thickness of the arch ring at the crown is
= 0-4 v^5 = 0-9 foot nearly.
14" is therefore the nearest brick dimension, and we will call this 1'2 foot.
The arch being uniformly loaded, we need only consider one half of the
archj and the thrust at the crown will be horizontal, as shown in Fig. 505.
Fig. 505.
The total load on the half arch is 6*2 tons, and its direction is at a distance
ARCHES— LINE OF LEAST RESISTANCE
255
of r9 foot from tlie point A. Now, as seen at p. 249, H will be a minimum
wlieu applied at the point E, and its lever arm will then be
5 + (fx 1-2) = 5-9 feet.
The lever arm of the load is
l-9 + (|x 1-2) = 2-2 feet.
Hence
2-2
- 2-3 tons.
Again, H will be a maximum when applied at F, and when moments are
taken about G. In this case the lever arm of H is
5 + (^x 1-2) = 5-3 feet,
and of the load
l-9 + (fx 1-2) = 2-8 feet.
Hence
_2-8
= 3-26 tons.
The true value of H, giving the line of least resistance, must lie between
these two values, provided of course the arch ring is able to withstand the
load put upon it.
As a first trial draw a line of resistance through E. Following the pro-
cess explained at p. 251 we obtain the line of resistance given in Fig. 506,
"Iton
that is, draw H = 2"3 tons horizontally through E to intersect the first load
of 1 ton, from this intersection draw a line parallel to P-^ (Fig. 507) to meet
the second load of 1 ton, then from this intersection a line parallel to Pg to
meet the third load, and so on until a line parallel to P^ is drawn to meet the
last load. From this point of intersection a line parallel to Pg is drawn and
gives the direction of the pressure at the springing ; Pg in Fig. 507 being
NOTES ON BUILDING CONSTRUCTION
the magnitude. This line of resistance passes below the soffit of the arch,
and therefore the value of H must be increased. It should be observed that
the direction of Pg passes through K, the point about which moments were
taken, and this forms a valuable check on the accuracy of the drawing.
As a second trial increase H to H(max.), namely, 3-26 tons. The point of
application of H will in this case be F, and the line of resistance shown in
Fig. 508 will be obtained, which also cuts through the arch ring at the soffit.
As a last trial apply H at E and take moments about G, so that the
line of resistance will pass through G. We find
TT 2-8
H = —x 6-2 = 2-95 tons,
and the corresponding line of resistance shown in Fig. 510 is obtained. This
line of resistance is within the arch ring at all points, but at the haunches it
is outside the inner half of the arch ring, and therefore does not comply
with Dr. Scheffler's rule.
To raise the line of resistance at the haunches so that it will be within
the inner half of the arch ring, we can either increase the load above the
haunches or increase the thickness of the arch ring at the springing. The
first alternative is left as an exercise for the student.
Thickening arch. — Supposing that the thickness of the arch ring is in-
creased to 1' 6" at the springing, as shown in Fig. 511, the point G is now
at a distance of f x 1-5 = I'lS from the inner point of the springing, and the
lever arm of the load becomes l-9xl-13 = 3-0 nearly. Hence, if H is applied
at E,
3-0 , ,
H = X 6-2 = 3-15 tons, and the
5'9
True line of resistance shown in Fig. 5 1 1 is obtained. This line of resistance
is practically everywhere included within the inner half of the arch ring, and
is therefore the line of least resistance required. Practically, however, a brick
J
ARCHES— EXAMPLE
257
arch would not be built of varying depth owing to the expense of cutting
the bricks, and the arch ring would therefore be made 1' 6" deep throughout.
This expedient could, however, be adopted if the arch were built in masonry.
Thickness to resist crushing. — The thickness of the arch ring must be suffi-
cient to prevent the intensity of pressure being greater than the pressure the
material can safely bear. Now on inspection it would appear that the
greatest intensity of pressure occurs either at the point M (Fig. 513) or at
Fis. 510
Fig. 510a.
the abutment, and to find the required thickness it will be necessary to ascertain
at which of these points the intensity of pressure is greatest. At the point
M the line of least resistance is practically perpendicular to the joint between
the voussoirs, and on referring to Fig. 5 1 2 it will be found that the pressure
at the point M is 5-1 tons. We must now employ Equation 95, p. 219,
and we can write
Y = 5'1 tons,
GC = J^ inches,
AE = b (the width of the arch ring).
_ 2x 5-1x4 0-97
Hence
or
3 X 14 X 6
0-97
Now proceeding to the abutment we find from Fig. 5 1 2 that the pressure
is 6-8 tons ; but this pressure is inclined to the joint, and we must resolve
perpendicularly to the joint. "We thus find (Fig. 514)
Y = 6 tons,
also GC = i^ inches.
2x6x4 0-89
Hence
3x18x6
B.C.
-IV.
S
2s8
NOTES ON BUILDING CONSTRUCTION
0-89
K = .
E
F
Fig. 512.
Fig. 511.
Therefore I is greater than \, or, in other words, the maximum intensity of
pressure is greater at M than at the abutment.
-.S' 0"-
Fig. 514. Fig. 515.
If the arch is built in ordinary brickwork in mortar, p can be taken at
about 0"5 cwt. per square inch. Hence
, 0-97 x 20 . ,
6 = — ^-^ — = 38 inches.
ARCHES— EXAMPLE 259
But if the arch were built in cement, f can be taken at 0"8 cwts. per sq^uare
inch, and then
^ 0-97x20
0 = — — — =24 inches.
U'o
Arch with unsymmetrical Load. (See Appendix XX.)
Abutment for Arch.
Example 51a. — As an example it will be assumed that the
arch is situated at one end of the wall of a building, as shown in
Fig. 515. The forces acting on the abutment are the thrust of the
arch, namely, 6*8 tons as found above, and the weight of the brick-
work in the wall immediately over the wall, including the weight of
the abutment itself. It will be assumed that this weight is 10 tons,
acting as shown in Fig. 515. These two forces intersect at A, and
completing the triangle of forces we obtain B as the centre of pressure
at the joint DC. In order that no tension be excited at C we must
make DB = -g-BC. It must also be ascertained that the intensity of
pressure is not too great. The vertical 'pressure, on the joint DC is 1 6
tons. Hence taking the width of the arch as 2 3" we have from (95)
_ 16x20
6'6"x23"
= 0"36 cwt. per square inch nearly.
It will be seen from Table I A that O'S cwt. is safe.
The maximum pressure at the joint at A will, however, be
somewhat greater; the vertical pressure will be reduced by
about 1 ton (viz. the weight of the abutment below A) ; and as
A is (according to the dimensions taken) at the width of the
joint from the outer edge
^ ^ 15 x 20
= 0*48 cwt. per square inch nearly.
Resistance to sliding can be found as before.
The concrete foundations to the abutment can be designed on the
same principles as above, taking care that the maximum intensity
of pressure is not greater than the soil can withstand with safety.
Table of thickness of arches.^ — The arches used in building
construction are not generally of very great span, and they do
not as a rule require to be calculated. Their thickness is often
governed by appearance, or it may be found in Tables founded
upon experience. One of these is given at p. 343, and others in
Molesworth's and Hurst's Pocket-Books.
^ See Appendix XXI.
Chapter XYI.
HYDRAULICS'
(As applied to Building Construction).
HYDEAULICS is a subject which is related to building con-
struction only to a limited extent, and in fact only two
subdivisions of the subject need be considered, namely —
1. The motion of liquids through pipes in connection with
water supply and disposal of sewage.
2. The delivery of water from a jet in connection with stand-
pipes.
We must first consider and define some of the terms in use.
Pressure. — If we imagine a small cube immersed in a liquid as
shown in Fig. 516, the pressure exerted by the
liquid on each of the six faces of the cube will
be normal to that face, and if the cube is sup-
posed to be very small indeed, all these pressures
will be almost equal. If the cube be now sup-
posed to diminish without limit it will become a
point ; the pressures will become exactly equal,
and will then be the pressure at a point, as at P, Tig. 516.
for instance, in Fig. 517. It is shown in works on Hydrostatics
that the pressure at a point in a liquid is propor-
tional to the depth below the surface of the
liquid, and in fact that the intensity pressure,
that is, the pressure per unit of area (a square
inch, for instance), is equal to the weight of a
column of liquid of a height equal to the depth
below the surface and one unit of area in cross
Fig. 517. section. Thus if D is 50 feet and the liquid
is water, a cubic foot of which weighs almost exactly 6 2 "4 lbs.
' For Practical Formulce, see pp. 267, 281, and App. XXI.
HYDRA ULICS— DEFINITIONS
261
at a temperature of 52°"3 Fahr., the pressure on a square inch at
the point P would be
50 X -^x 62-4 = 21-67 lbs.
144
The intensity of pressure at the point P is therefore 21*67 lbs.,
or as it is usually expressed, 21*67 lbs. per square inch.
Head of pressure. — The depth of the point P (Fig. 5 17) is also
called the head of pressure at the point P, or simply the " head,"
and is generally expressed in feet.
Head of elevation. — The height of the point P (Fig. 517) above
some datum level is called the head of elevation of the point P.
Loss of head. — "When a liquid is in motion, each particle is
constantly moving from a place of greater head to a place of lesser
head, and the difference between the two heads is called the loss of
head. This loss of head may be entirely a loss of head of pressure, or
entirely a loss of head of elevation, or, again, partly a loss of head
of pressure and the remainder a loss of head of elevation.
The following examples will assist to illustrate the above —
AB (Fig. 5 1 8) is a pipe connected to a tank T ; at A there is
a tap. When the tap A is closed, the head at the point P will be:
Head of pressure, EP ;
Head of elevation, FP.
At A the head of pressure is G-A, and there is no head of
elevation with reference to the datum chosen.
Dahitn level.
F ' "
Fig. 519.
When the tap is opened, however, the pressure at A is reduced
262 NOTES ON BUILDING CONSTRUCTION
to zero, and if the resistance of the pipe to the flow of the water is
uniform, the pressure at each point of the pipe can be represented
by the straight line AL, as shown in Fig. 519. Therefore the
head at P will now be :
Head of pressure, KP ;
Head of elevation, FP.
It will be observed that the head of elevation is not altered,
but that the head of pressure is reduced by the amount KE.
With reference to a second point P', the head of pressure is K'P',
and the head of elevation Y'^' ; but the " loss of head " between
the two points is K'O. This loss of head is made up of K'M, the
loss of the head of pressure (KM is drawn parallel to PP0> ^-nd of
MO, equal to P^N, which is the loss of the head of elevation.
If the pipe is horizontal, as in Fig. 520, it is clear that the
^ Fig. 520.
loss of head between the two points P and P' is entirely loss of head
of pressure, and there is no loss of head of elevation.
On the other hand, if water is flowing in an open channel,
as in Fig. 521, the loss of head is entirely loss of head of eleva-
tion ; in fact, the water is not flowing under pressure.
Another way of looking at the matter is as follows : — The water, when flowing
in the pipe or in the channel, has to overcome the resistance of the pipe or
of the channel, which means that the water has to do a certain amount of tvork,
and this work is proportional to the loss of head. Supposing, for instance,
that 1 lb. of water, in flowing through a pipe, loses 1 0 feet of head, then the
work done by that pound of water is 10 foot-lbs.
Wetted perimeter. — In an open channel, or in the case of a
pipe not flowing full, the portion of the cross section of the
channel, or of the pipe, wetted by the liquid (from E to F, Figs.
522 and 523) is called the wetted perimeter. It is also called
the " harder."
Hydraulic mean depth. — The quotient of the area of the
HYDRA ULICS— DEFINITIONS
263
cross section of the liquid divided by the wetted perimeter is called
the hydraulic mean depth, and will be denoted by E.
Fig. 522. Fig. 523. '
MOTION OF LIQUIDS IN PIPES.
The motion of a liquid in a pipe is due to gravity, and if the
pipe were perfectly smooth, and offered no resistance, the velocity
would increase without limit. But, as already mentioned, pipes
do offer a resistance depending on their diameter and on the
state of their inner surfaces, so that the velocity, instead of in-
creasing continually, is limited. Moreover, the velocity of the *
liquid is not uniform throughout the cross section of the pipe,
but is greatest in the centre, diminishing gradually at first, and
then rapidly towards the sides. Close to the sides the liquid
does not flow freely, but is disturbed by numerous small eddies.
It was at one time thought that neither the diameter nor the
state of the surfaces had any influence on the resistance, and to
account for this it was supposed that these eddies formed, as it
were, a liquid lining to the pipe.
Darcy has, however, shown by means of an elaborate series of experi-
ments that this view, due to De Prony, is erroneous, and has also shown that
the mistake arose from combining experiments on pipes of large diameter
but with rough surfaces, and experiments on smooth pipes but of small
diameter, and that the resistance offered by the rough surface in the one set
happened to exactly equal the resistance due to the small diameter in the
other set ; so that it appeared that neither the state of the surface nor the
diameter had any influence on the resistance.
Although this supposition is now known to be erroneous, yet
in many of the formulae in general use for obtaining the discharge
of pipes, the effect of the diameter on the resistance is not taken
into account, nor is any notice taken of the degree of roughness
of the surface ; practically this latter is of little moment, for
whatever the surface may have been originally, it will speedily
be covered with sediment. These formulae are empirical, and are
264 NOTES ON BUILDING CONSTRUCTION
based on numerous experiments that have been made on the
flow of liquids in pipes. It should, however, be observed that
these experiments, made at various times by different observers,
are neither numerous enough nor were they sufficiently system-
atically carried out to obtain results of scientific accuracy,
although they may be good enough for practical purposes.
We have to consider two cases of the flow of liquids through
pipes, namely —
1. When the pipe is flowing full and the liquid is therefore
impelled to move by the pressure of the head of liquid.
2. When the pipe is flowing partially full. In this case the
liquid is not under pressure, and simply flows down, owing to
the slope of the pipe, as it would in an open channel.
We will consider each of these cases separately.
Discharge feom a Pipe flowing full under Pressure.
As already mentioned, numerous formulae have been proposed
to find the discharge from a pipe, but we will only consider four
and then select for future use the one best adapted for our purpose.
One of these, known as Eytelwein's formula, is
where V= 108^^ X H_ oas . . (104),
V is the nYiean velocity of the water in feet per second ;
H the effective, head of water in feet, that is, the actual head
reduced by the various losses of head due to bends, etc. (see
p. 268);
L is the total length of the pipe in feet ;
D is the diameter of the pipe in feet.
Neville's formula is said to be more accurate than Eytelwein's, but it is
more complicated and is as follows : —
Both in Hurst's and Molesworth's Pocket-Booh will be found tables calcu-
lated from this formula, which are of great assistance in practical calculations.
Thrupp's formula. — The latest, and probably the most accurate formula, is
that recently elaborated by Mr. Edgar C. Thrupp, which is as follows : —
„ , /z-R
V = -^ ^ ^ x(S)» . . . (106),
where K is the hydraulic mean depth (which is equal to ^ for pipes flowing
full, see p. 280), S is the slope of the pipe fi.e. and x, y, z, and c are con-
PIPES FLOWING FULL UNDER PRESSURE 265
stants depending on the material of which the pipe is made and on the sur-
face of the pipe. For further information on this formula, and for values ot
the constants, reference is made to the Proceedings of the Society of Engineers
(Dec. 188Y). The following cases may, however, be useful :—
For wrought iron pipes over 3 J" diameter,
■nO-65 1
0-004787
For wrought iron pipes l" diameter.
For new cast iron pipes,
V =
0-004787
X (S)i-8
x(S)^
(107) .
(108) .
(109) .
"0-006752
Darcy's formula.— Th.Q fourth formula we will mention is
Darcy's, namely —
v.oy5,H . . . (no).
where C is a coefacient to be found from the following table.^
TABLE J.
Diameter of pipe in inches
= 12xD.
1
2
3
4
5
6
7
8
Value of C.
65
80
93
99
102
103
105
106
107
Maximum value of C for very large pipes, 113-3.
To compare these formula we will work out some examples.
Example 52.— Find the velocity in a pipe 1 inch in diameter, 100
feet long, the effective head being 1 foot.
We have
By Equation 104 (Eytelwein' s formula),
V = 108 sJl X X - 0-13,
= 1-43 feet per second.
By Equation 105 (Ne ville's formula ),
V = 140 ^/i X jJ^^ X - 11 v/i >^ T2 X T^TT'
= 2-02 - 0-65,
= 1*37 feet per second.
By Equation 110 (Darcy's form ula), taking C = 80 for this size of pipe,
V = 80x/ix'^io<^,
= 1-15 feet per second.
By Thrupp's formula (for wrought iron 1" diameter). Equation 108,
1 From Molesworth's Pocket-Boole of Engineering Formula.
266 NOTES ON BUILDING CONSTRUCTION
1
y — V? ^ 12^ X lT(ny/)
0-004787 '
0-0786
"0-004787 X 1-29'
= 1-27 feet per second.
Example 53.— Find the velocity in a pipe 1 inch in diameter, 100
feet long, the effective head being 10 feet.
By Equation 104,
^= 108 ^f^^A^^- 0-13,
= 4-8 feet per second.
By Equation 105,
140^f^^^^- 1 1 ^i^T^^,
= 4-99 feet per second.
By Equation 110,
V = 80 a/1 V 1 V 1 0
V V 4 X 3-2- X Yo^,
= 3-65 feet per second.
By Thrupp's formula (Equation 108), i
0-004787 '
^ 0-0786
0-004787 X 3-594'
= 4-59 feet per second.
Example 54.— Find the velocity in a pipe 4 inches in diameter, 100
leet long, the effective head being 1 foot.
By Equation 104,
V= 108 v/ix 3^x^^0-0-13,
= 2-99 feet per second.
By Equation 105,
= 6 01 teet per second.
By Equation 110, taking C= 102,
= 2-94 feet per second.
By Thrupp's formula (for cast iron. Equation 109),
V = k_^T2) , 1 a
0-006752 ^^^^ '
0-209
0-006752 X 10'
= 3-09 feet per second.
Example 55.— Find the velocity in a pipe 4 inches in diameter, 100
feet long, the available head being 10 feet.
By Equation 104,
■^=108x/|xAxtV^-0-13,
= 9-73 feet per second.
PIPES FLOWING FULL UNDER PRESSURE
267
By Equation 105,
V = 1 40 ^/i X T-*^ X T-i^ - 11 -^i X t\ X
= 10"53 feet per second.
By Equation 110,
V = 102v/ix i*2XtV%
==9-31 feet per second.
By Thrupp's formula (for cast iron, equation 109),
^ ~ 0-006752 ^^100'' '
0-209
0-006752 x 10^
= 9-79 feet per second.
Practical Formula (Darcy's).
It will be observed that in all cases Darcy's formula gives
the lowest velocity, and practically it is better to under-estimate
the velocity than to over-estimate it. It is also a simple formula
to use, so that on the whole it is to be preferred to Thrupp's
in cases where great accuracy is not essential, and it will therefore
be used to work out any further examples in this book.
The discharge of the pipe can of course easily be found as
soon as the mean velocity is known ; thus
Discharge in cubic feet per second = V x
r = 0-785D2V . (111).
ttD^
Discharge in gallons per minute = V x x 6-25 x 60,
G = 294D2V . . . (112).
In practice, however, the question is generally to find what
diameter of pipe is required for a given discharge, the effective
head and length of pipe being known. Combining equations
110 and 112 together we get
/D H
G = 294D2xC^^x
whence, by squaring and simplifying,
==7:^(s-IT • • •
D being expressed in feet.
Or d=l'e3[^.-^,) . . • (114),
when d is expressed in inches.
or
or
268
NOTES ON BUILDING CONSTRUCTION
To obtain the fifth root of the expression in brackets,
either logarithms can be used, or else a table of fifth roots,'
such as published in Molesworth's Pocket-Booh
Equation 114 can, however, be written
H«' = C<T^) •• ■ (115).
/ d \^
C^f——J depends only on d, or, in other words, for
every value of d there is a corresponding value of the expression
H
Table XVIII. gives the value of — . for various sizes of
H
pipes, and also shows the diameter of pipe that should be practi-
cally adopted to allow for the incrustation which takes place in
water-supply pipes.
^ Table XIX. is a similar table for larger pipes, the discharge
being reckoned in cubic feet per second, and is obtained from the
formula
(116),
L 7r2 / d\^
where F is the discharge in cubic feet per second.
These Tables simplify the calculations very considerably, as
will be shown farther on.
Loss of Head.
On referring to p. 264 it will be seen that H is defined as
being the effective head. In Fig. 524 A'G is the total head, and the
Fig. 524,
effective head AD is somewhat less, the loss being due to certain
resistances which will be mentioned later on. It will also be seen
that the effective head is equal to the loss of head between A
and B, due to the resistance of the pipe ; in fact, the flow of water
in the pipe is such as to produce this equality.
LOSSES OF HEAD
269
Further, if we know the loss of head between two points P, P'
in the pipe and the length of pipe between these two points we
can find the velocity and the discharge.
The resistance of the pipe causes by far the greatest loss of
head, but there are some minor losses produced as follows : —
Loss of head, due to orifice of entry. — The orifice of entry
obstructs to a certain extent the flow of water into the pipe,
causing therefore a loss of head. This loss depends on the form
Fig. 525. Fig. 526. Fig. 527.
of the orifice, as will readily be seen from Figs. 525, 526, and
527, and it can be found from the formula
Ho = Y2xC . . . (117),
where is the loss of head due to orifice.
C = 0'007849 for round orifices such as the end of the pipe,
Fig. 525-
= 0*000444 ditto when splayed or bell-mouthed. Fig. 526.
= 0*0 14846 when the pipe projects into the cistern (diameter
uniform). Fig. 527.
Example 56. — Find the loss of head due to a round orifice projecting
into a cistern when the mean velocity in the pipe is 2 feet per second.
From the above we have
Ho = 22 X 0-014846 feet,
= 0-06 feet.
Loss of head due to velocity. — The water in the reservoir or
cistern is at rest, and a certain amount of head is lost in causing
the water to take up the velocity in the pipe, or, as it would be
more scientifically expressed, a certain amount of energy of position
(which is measured by the head) has to be converted into energy
of motion. The energy of motion is measured by
Y2
64-4'
Hence if be the loss of head due to velocity
Y2
H^ = —— = V2x 0-0155
64-4
(118).
270 NOTES ON BUILDING CONSTRUCTION
Example 57. — Find the loss of head due to a velocity of 2 feet per
second.
From the above = 22 x 0-01 55 feet,
= 0-062 feet.
Loss of head due to bends and elbows. — Every bend in a pipe
causes a resistance and a consequent loss of head, and elbows do
so to a still greater degree.
These losses can be found from the following formulae, some-
what modified from those given in Molesworth's Pocket-Book and
in Hurst's Pocket-Book.
Calling the loss of head due to bends
Hg we have
For bends = SAV^ . (119),
where & is a coefficient depending on
R
the ratio, of the radius of the bend
d
to the internal diameter of the pipe,
and A is the change in direction caused
y by the bend measured in degrees (see
Fig. 528. Fig. 528). Table XXL will be found to
give values of b for various values of
d
Example 58. — Findtheloss of headdueto a bend of 30° ina pipe 2"diameter,
the radius of the bend being 3-1 inches, the velocity being 2 feet per second.
Ttr I, R 3-5
We have — = =1-7 5.
d 2
In Table XXI. we find for
and for
6 = 0-000015,
6 = 0-000013.
We may therefore take 6 = 0-000014
Hence = 0-000014 x 30 x 2^,
= 0-0017 foot.
For elbows = e .V^ (120),
where e is a coefficient depending
on the angle A of the elbow (see
Fig. 529), and Table XXII. gives
values of e for certain values of A.
Example 59. — Find the loss of head
due to an elbow of 30° when the velocity
is 2 feet per second.
From Table XXII. we find e = 0-0011. Fig. 529.
LOSSES OF HEAD
271
Hence
: 0-0011 X 22
= 0-0044 foot.
The above constitute all the minor losses of head, and, as will
be seen by the following example, these losses, unless the bends
and elbows are very numerous, are small in comparison with the loss
due to the resistance of the pipe, and can practically be neglected.
Example 60. — Water is flowing at the rate of 2 feet per second in a
pipe 2" diameter ; find the loss of head due to the resistance of the pipe in
100 feet length.
From Equation 110,
and substituting,
squaring, etc.,
V=93
/D H
V 4-L'
2-93^ ixy^^x -
00
H:
4 X 4 X 12 X 100
932 X 2 '
:1-11 feet.
It is thus clearly shown that the losses due to entry, velocity,
etc., are in ordinary cases insignificant in comparison with the loss
due to the resistance of the pipe. Of course if there are a great
number of bends and elbows it might be necessary to allow for
the loss of head due to them.
PEACTICAL EXAMPLES.
We will now proceed to work out some practical examples.
Water Supply to a House (Intermittent Service).
Example 61. — A house is supplied with water on the intermittent
system : 400 gallons is the quantity of water used per diem, and the water is
turned on for four hours every day, and flows into a supply cistern. The ball
cock in the cistern is 30 feet above the main in the road, and the service pipe
is 120 feet long. The head in the main, at the junction with the service
pipe, is 40 feet. Find the diameter of the service pipe.
At the tap the available head, ■i.e. the head of pressure less the head of
elevation, is
40-30 = 10 feet,
and the required discharge into the cistern to fill it in four hours,
= 1-67 gallons per minute.
4 X 60 ^ ^
DiaTneter of pipe by Darcy's formula. — Hence substituting in Equation
113, and taking C = 65 since the pipe will be a small one,
272
NOTES ON BUILDING CONSTRUCTION
1 / 120 l-672\i
65^ ) '
— (^\^
7-37 \126-8/ '
7-37 X 2-63'
= 0-0517 foot,
or (Z = 0-62 inch.
Strictly the value C = 65 is for a |" pipe, and for a pipe 0-62 inch
diameter C = 70 would be a more correct value. It will be found by re-
peating the above calculation with the new value of C that the diameter of
the pipe is 0*052 foot, so that there is practically no difference. Adding ^
to allow for incrustation,
fi = 0-72 inch,
so that a ^" pipe would practically do.
Diameter of pipe by Table.- — This result may be obtained more readily by
the use of Table XVIII. We have
H ^ 10 X i ,
= 33-3.
And on referring to the Table it will be seen that |" is the nearest market
size of pipe, allowing for incrustation.
Tap. — It has tacitly been considered in the above that the tap in the
cistern has the same bore as the pipe — a small reduction in the bore does not
affect the flow very much, but if the reduction is at all considerable the flow
is much impeded, as will be seen by the following.
The data being the same, find the diameter required for the delivery pipe :
1. When the bore of the tap is
2. When the bore of the tap is
3. When the bore of the tap is
As in each case the discharge must be 1*67 gallon per minute in order
that the cistern may be filled in four hours, it follows that the smaller the
diameter of the tap, the greater must be the velocity of the issuing stream.
If V is this velocity expressed in feet per second, the volume of water issuing
in one second is
V X — . TT cubic feet,
or V X — . 77 X 6-24 gallons.
And since G is the discharge in gallons per minute we have
D2
G = Vx-— XTTX 6-24 X 60,
whence by reduction
where D is in feet, or
where d is expressed in inches.
V = 0-0034 x^ .... (121),
V=0-49x|^ (122),
WATER SUPPLY TO A HOUSE
273
Applying Equation 122 to the first case under consideration we have
1-67
V = 0"49 X -tjvq- = 3'3 feet per second.
There must therefore he sufficient head at the tap to produce a velocity of
3 '3 feet per second. The required head can be found from Equation 118,
thus
TT 3-32
xi„ = — — = 0*17 foot nearly.
64-4
The available head to force the water through the supply pipe instead of
being 10 feet, as before, is reduced to
10 -0-17 = 9-83 feet,
but clearly this reduction in the head is not sufficient to make any practical
diff'erence in the size of the pipe required.
In the second case, when the diameter of the tap is only \ inch, we have
V-0-49X —
V-0 49 x^^^,,
= 13"1 feet per second.
Hence
1 2
H, = —- =2-7 feet.
64-4
Let us see whether this greater reduction will make any diff'erence in the
size of the pipe required. We have, using Table XVIII. ,
L 120
— .G2=— — — X 1-672,
H 10-2-7
= 46,
so that a f " pipe is still sufficient, allowing for incrustation.
In the third case we have
V = 0-49 X 'r^^ = 23-3 feet per second.
TT 23-32
^^==6?4=^"^^^^'-
The available head is therefore
10 - 8-4= 1-6 feet.
Hence
^. G2=^x 1-672 = 209.
xi Vb
This number corresponds in Table XVIII. to a 1" pipe.
Bends. — Find the effect of 10 bends of 90° each and 4 inches radius.
On reference to Equation 1 1 9 it will be seen that the smaller the diameter
of the pipe, the greater is the resistance of a bend in it. Therefore to be on
the safe side we ought to reckon the resistance of bends when the diameter of
the pipe has been diminished by incrustation, or, in other words, resistance of
the bends should be reckoned on the diameter found to be required before
incrustation has been allowed for. From Table XVIII. and also from p.
271 it will be seen that in the present instance this diameter is 0-6 inch
nearly.
Eeferring to Equation 119 we see that we must first find the value of h.
B.C. IV. T
274
NOTES ON BUILDING CONSTRUCTION
R
Now ^ ^ 0-6 ^
Hence from Table XXI.
6 = 0-000011.
We next require V; from Equation 122
V = 0-49xi:^^„
(0-6)2'
= 2-27 feet per second.
Hence from Equation 119
H3 = 0-000011 X 90 X 2-272,
= 0-005 foot.
And for ten bends the loss of head will be only O'OS foot. This loss of head
is much too small to have any practical effect on the size of the pipe, and
thus the opinion expressed at p. 271 is confirmed.
Water Supply to a House (Constant Service).
Example 62. — The water supply to a house is represented in Fig. 530.
Eind the diameter of the pipes in order that the tap at A may discharge, three
A-<— V
+30
B
+10'ff-
+10
Fig. 530.
— >D
gallons per minute, and the tap at B two gallons per minute, when both taps
are open, allowing for incrustation. Find also the discharge from each tap
when the other is closed. The available head in the main is 90 feet. The
levels are shown thus +10 with reference to the main.
To FIXD SIZE OF BC AND AC. — The best way to proceed is to assume the size
of one of the branches. For instance, supposing BC is taken as ^" diameter.
L
Then, from Table XVIIL,
H
G2:
Hence if we take L = 30 feet and G =
from C to B, thus
H:
2 gallons, H will be the loss of head
30x 22
= 24 feet.
Now since B and C are at the same level the available head at C ought
to be 24 feet.
In the pipe AC the water has to rise 19 feet, hence the available head is
24-19 = 5 feet.
WATER SUPPLY TO A HOUSE
275
Hence ^-02 = ^x32 = 36.
n. 5
So that (Table XVIII.) a pipe is too small and a f" is too big. "We liave,
however, made no allowance for the various minor losses of head. It will
be well to see whether in this case they are of sufficient magnitude to take
into account.
Minor losses of head. — In the first place, if the pipe AC is connected to
BD by means of a T piece, as shown in Fig. 531, the general flow of the
water will be along DB, as shown by the arrows. Consequently in entering
the pipe CA there will be a loss of head due to the change in the direction of
the flow, a part of which may be taken as the loss of head required to impart
the velocity in the pipe AC, but, in addition, eddies are formed (called gurgli-
tation) which cause another loss of head,
generally taken as Uvice that due to velo-
city. On the whole it is usual to reckon
the loss as three times that due to velocity.^
We must therefore first find the velo-
city :— Supposing a |-" pipe is provision-
ally decided upon, then we must reckon
the velocity Avhen the diameter of the
pipe has been diminished by incrustation.
From Table XVIII. it will be seen that
the diminished diameter is 0-64 inch.
Hence from Equation 122
V = 0-49 X
(0-64)2'
= 3*6 feet per second,
3-g2
and = = 0-2 foot.
^ 64-4
So that the loss of head is
3 X 0-2 = 0-6 foot.
Bends. — We will neglect the loss due to bends ; it must be small owing
to the low velocity.
Tap. — The tap, if smaller than the pipe, will caiise a slight loss of
head. Supposing that the bore of the tap is |", or 0"48 inch, making a slight
allowance for incrustation, then to discharge three gallons per minvite the
issuing velocity must be
= 6*4 feet per second
TT 6-42
or H= ——= 0-63 foot.
64-4
But the velocity in the pipe is 3*6 feet per minute, which represents a head
of 0'2 foot as already seen. Hence the additional head required to produce
the issuing velocity, or, in other words, the loss of head due to issue^ is
0-63 -0-2 = 0-43 foot.
Total minor Loss of Head. — Altogether, therefore, we must reckon on
a loss of head of 0-6 + 0-43,
See Hurst's Architectural and Surveyor's Handbook.
276 NOTES ON BUILDING CONSTRUCTION
= r03 foot,
or say 1 foot.
The available head is therefore reduced to 4 feet
we thus have
20
— X 32:
4
45.
It thus appears (Table XVIII.) that a f " pipe is the proper size.
To FIND Size of CD. — We now have to find the diameter of the pipe
CD. We found that the available head at C ought to be 24 feet, and there
is a loss of head in the pipe CD of 10 feet, due to difference of level.
Hence the available head is
90 - 10 - 24 = 56 feet.
Therefore - . G2 = — x (2 + 3)2 = 45,
H 56
which corresponds to a |" pipe.
Since the pipes AC and CD are both of the same diameter, the joint at C
would be made as shown in Fig. 532, instead of as in Fig. 531, and this would
prevent the loss of head of 0'6 foot, and
the loss would occur in the ^" pipe.
Referring back to p. 274 it will be seen
that the diameter of the pipe BC was
assumed as Had a larger diameter
been assumed the loss of head in BC
would have been less, so that the available
head at C producing the discharge in AC
would also have been less, and conse-
quently this pipe would have to be made
bigger. On the other hand, the available
head in CD would have been greater,
hence this pipe would have been smaller ; that is, the service pipe would be
smaller than the branches, which is not advisable. Thus the arrangement
worked out is practically the best.
To find the discharge from one tap when the otlier is closed.- — First, when tap
A is open, it is clear that the available head is 90-29 = 61 feet, or, allowing
for the various minor losses of head, say 60 feet.
Hence from Table XVIII.
100 + 20
Fig. 532.
-.G2:
H
60
X G2 = 48.
Thus
or G = 4-9 gallons per minute.
When tap B is open and tap A closed, we have to deal with two sizes of
pipe. Let H, be the head at the point C, then 90 — 10
available head in the portion CD of the pipe.
L
H
and similarly for the pipe CD
L 100
H
Hj^ will be the
Hence for the pipe BC
30
G2 = S^.G2 =
■^1
(a),
G2:
80 -H,
G2 = 48
(6).
Now the discharge G is the same in both pipes,
(a) by Equation (6),
Hence, dividing Equation
WATER SUPPLY TO A GROUP OF BOUSES
30 5H,
277
whence
100 48(80 -Hj)'
3x 48(80-Hi) = 50Hi,
^ 3 X 48 X 80
ill =
194
= 59-5 feet.
Inserting this value in Equation {a),
5 X 59-5
G2 =
30
9-9,
or G = 3'15 gallons per minute.
Of course we ought to get the same result from Equation (b\ namely —
^,^48(80- 59^^
100
Water Supply to Eight Houses in a Street.
Example 63. — Eight houses, as shown on plan in Fig. 533, are to be
supplied with water from a branch main connected to the main at the point
Fig. 533.
A,
i 2U0
—2U0-
■r -SAO *h—r20—>
A
H,
,1
0 Main,
Hr5
Vertical scale 10 times horizontal scale
Fig. 534.
A. The levels of the branch main are shown in Fig. 534, and each house is
fitted up as in Example 62. Find the diameter of the branch main
capable of supplying each house approximately with 5 gallons of water per
minute.
On referring to Example 62 it will be seen that the highest tap was
29 feet above the main, that it was connected to the main by 120 feet of f"
pipe, and that the discharge from it was 4-9 gallons per minute when the
head in the main was 90 feet. It will also be seen that the discharge is
almost the same when both taps are running. W^e will therefore ignore the
lower tap in the present example.
Size of Main — Half the taps opened simultaneously. — It is very un-
likely that all the houses will require water at the same time, and a very
278 NOTES ON BUILDING CONSTRUCTION
safe assumption to make is that half the number of houses require the full
amount of water at the same time. Supposing, therefore, that the upper
taps in houses 1, 2, 1', 2' are opened simultaneously and that an average
of 4-9 gallons is running out of each, then, from Example 62, the head
at A2 must be 61 feet more than the height of the tap, that is
H2 = 61 +31 = 92 feet.
Hence the available head in the portion AgA of the branch main is
92-5 -92 = 0-5 foot.
Again, the discharge at Ag must be *
4 X 4'9 = 19*6 gallons per minute.
L 2 •
Hence the value of the expression g . (jt is
600 ,
— - X 19-62=460,000 nearly.
0-5 '
And referring to Table XVIII. it will be seen that a 4" pipe is the nearest
market size, allowing for incrustation.
Theoretically the branch main ought to be made to diminish after the
junction to each house, but the practice is to maintain the same diameter, for
some distance at any rate.
All tipper taps opened simultaneously. — As an exercise let us inquire what
the discharge would be supposing all the upper taps were open. Although
it may appear paradoxical, the discharge from each tap will not be very much
diminished. The direct solution of the question, however, leads to several
cumbersome quadratic equations, and the best way is to work by approximation.
Thus as a first approximation let us suppose that the average discharge
per tap is 4-0 gallons per minute when all the taps are open, then the dis-
charge at A^ must be
8x4 = 32 gallons per minute.
Now from Table XVIII. we have for a 4" pipe
^. G- = 420,000.
Hence
120 X 32^
II:
420,000 '
= 0-29 feet,
which is the loss of head between A and A^. Hence H^, the head at A^,
92-5 - 0-29 = 92-21 feet
since there is no difference in level.
The discharge from the tap in house No. 4 will therefore be
2 48 X (92-21 - 29)
^4 = 120 '
or . = 5 gallons per minute.
It follows that the discharge at Ag will be
32 -2x5 = 22 gallons.
Hence
240 x 22^
420,000 =^'^^'
or Hg = 92-21 + 5 -0-28,
= 96-93.
WATER SUPPLY TO A GROUP OF HOUSES 279
Therefore discliarge in house No. 3 is
48(96-93 - 24)
or Gg = 5-4 gallons per minute.
The discliarge at will therefore be
22 - 2 X 5-4 = 11-2.
Hence
„ 240 x 11-2^ ^
H= 420,000
or H2 = 96-93- 7 -0-07,
" = 89-86.
Hence discharge in house No. 2 will be
48(89-86 -31)
^2 - 120
= 4-8 nearly.
We still have the discharge from house No. 1, so that clearly the original
assumption of 4 gallons per minute is considerably too small.
As a second trial assume 4-9 gallons as the average discharge of each tap
per minute, then repeating the above calculations we will find
H4 = 92-5 -0-44 = 92-06,
= 5 gallons.
H3 = 92-06 + 5 - 0-48 = 96-78,
G3 = 5 -4 gallons.
H2 = 96-78 - 7 - 0-19 = 89-59,
G2 = 4-8 gallons.
= 89-59 - 0-04 = 89-55,
G^ = 4-8 gallons.
Practically, therefore, we get the same values as we did before, when only
half the number of taps were running, and the reason is, that a small alteration
in the head produces a considerable alteration in the discharge of a 4"
pipe, but no practical difference in the discharge of a f " pipe.
As a further exercise let us find the discharge in the pipe connecting
house No. 3, Avhen only the corresponding tap is opened. We have for the
f'P^P' ^ 2_48(H,-24)
'3
120
and for the 4" pipe ^ ^ 420,000(92-5 + 5 - H,
^ 240 + 120
Therefore 420,000(97-5 - H„)
48(H3 - 24) = \
„ / „ 420,000\ 420,000 x 97-5
H3U8 + '—J = '—^ + 48 x 24,
whence
H3=97-5 very nearly,
that is to say, that the flow of water from A to H3 is so little, that there is no
appreciable loss of head. Therefore
28o NOTES ON BUILDING CONSTRUCTION
, 48(97 5-24)
^3 - 120 '
= 5 "4 2 gallons per minute.
When the pipes are new, the discharge will of course be greater, for
instance in the last case, since the corresponding value from Table XVIII.
for a pipe 0-75 inch diameter is 117.
^, ^ __ 117(97-5 -24)
3 120
G'3 = 8-4 gallons per minute.
DiSCHAEGE FROM PiPES FLOWING PAETIALLY FULL.
It has been shown by experiment that the mean velocity of
flow of a liquid in a pipe is roughly proportionate to
Vis,
where E is the mean radius or hydraulic mean depth (see p. 262),
and S is the slope of the pipe (see p. 264).
An approximate formula would therefore be
V = CVlS . . . (123),
the units of measurement being the foot and the second, and C being
a constant. Different values have been ascribed to C by various
experimenters as follows : —
Beardmore . . . 95,
Downing . , . . 100.
But Darcy makes C vary according to the diameter of the pipe,
as in the Table given at p. 265.
Neville's formula — namely,
V=140^RS-11^IS . . . (124)
— is considered to be more accurate, but as in practice the choice is limited
to certain sizes of pipes, the additional accuracy is not worth the considerable
complication introduced into the calculations.
The student will observe that these formulae are similar in form to those
given for pipes running full under pressure. Now for a pipe running full
7rD2 1 _D
and ^ =
Li
So that
D H
from which it will be seen that the formulae for pipes running full are a
special case of those now given. A distinction must, however, be made
PIPES FLOWING PARTIALLY FULL
281
H
between j- and S ; the former is
the head of water divided by the
length of the pipe, whereas S is
the slope of the pipe, and may
vary at different points. Thus in
^'ig- 535 if the pipe were running
full we would have to substitute
Fig. 535.
H H
^ — ^Y- for ^ in Equation 110, and the velocity of the water would be the
same throughout the pipe ; but if the pipe were only partially full, then for
the portion AB
and for the portion BC
and the velocity would be greater in BC than
in AB.
Darcy's formula. — For the same
reasons as those given at p. 267, Darcy's
formula is preferred, and will be used
in working out examples.
To find the discharge we must multiply
the velocity by the area of discharge. Let
A be this area, then
F = A.C^ES . (125).
To find A in circular pipes we must sub-
tract the area of the triangle NKM (Fig. 536) from the area of the sector
NKM, thus— ^ j)2 D2
^ ^^^^--T sin 4>
^ = 360 ^'^4
whence
R = .
J / 7r<^ sin
~V360 ¥~
D2
■7r(f>
360
(126),
• (127).
D
These expressions are given to show how the discharge could
be calculated if necessary, but in practice what we require to
know in connection with drain pipes is whether, when conveying
the average quantity of sewage, the velocity is sufficient to keep
the pipe clean, without flushing.^ Of course the size of the drain
pipe depends on the maximum quantity of liquid to be conveyed.
Velocity. — The question therefore is, knowing the diameter of
the pipe and the discharge, to find the velocity. We have frqm
(123)
1 See p. 283.
282 NOTES ON BUILDING CONSTRUCTION
A
But R = p'
■where P is the wetted perimeter expressed in feet. Further,
F
^ = V
Hence = . ^^^^ . S,
, c^r . s
or Y^=—^ . . . (128).
jSTow we do not know P, and we must therefore assume a trial
value P^, and thus we obtain an approximation (V^) to V. But
F = A V
where is the corresponding value to P^ obtained from Table
XXIII.
If F^ is greater than F, then is also too great or P^ is too
large, and vice versa. Two or three trials ought to give a suf-
ficiently accurate value of V.
Example 64. — A 15-inch drain pipe, laid at a slope of -g-^o, is discharg-
ing 0'2 cubic foot per second. Find the velocity.
Assume = 0-6 x D,
= 0-75 foot.
„ 1102x0-2
Then from Equation 128 v^"' = -— — — —
or Vj^ = 2'5 feet per second.
P
But from Table XXIIl. the corresponding value of A^^ when ^ = 0'6 is
0-034 X D2,
= 0-053 square foot.
Hence 1^ = 0-053 x 2-5,
= 0-132 cubic foot per second,
so that clearly P^ is too small.
Assume as a second trial
P2 = 0-7 X D,
= 0-87 foot.
. o 1102x0-2
Then V 3
^2 -0-87 x 200'
or V2= 2-4 feet per second.
But A2 = 0-052 xD2,
= 0-0813 square foot.
Hence == 0-195 cubic foot.
Vo is therefore a sufficiently near approximation. We will, however, try
the next value of P from Table XXIIL, namely —
PIPES FLOWING PARTIALLY FULL
283
P3 = 0-8D,
= 1-0 foot.
, 1102x0-2
Then
3 1-0x200
= 2-3 feet per second.
But
A3 = 0-075 X D2,
= 0-117 square foot.
Hence
F3 = 2-3 X 0-117,
= 0-27 cubic foot.
Drain Pipes.
A drain pipe must be large enough, to convey the maximum
discharge, due to heavy rains, without running quite full, so that
there may be no danger of the pipe being under pressure, and
the velocity of discharge should be ascertained to see whether a
heavy rainfall will flush the drain. The next step is to find the
velocity of discharge when the pipe is only conveying the average
quantity of sewage ; it will then be seen whether arrangements
must be made for periodical flushing during dry weather.
In the case of small drains (6 to 9 inches diameter) the
velocity ought to be not less than 3 feet per second, and for larger
pipes, in which the sewage is generally well diluted with water, the
velocity ought to be not less than 2 feet per second, in order
to dispense with periodical flushing.^ The method of calculation
will be best illustrated by an example.
Simple System of Drains.
Example 65. — Fig. 537 shows two branch drains BC and BD con-
nected to a main drain AB. The lengths of the pipes and the levels of the
bottom of the pipes are also shown on the figure. The pipes are laid at an
even slope between these levels. The discharges to be reckoned upon are as
follows : —
Cubic feet per second.
Maximum.
Average.
Discharge in BC
Discharge in BD
4
3
0-2
0-03
^ "These velocities should be increased for small pipes, say for house drains
under 6 inches in diameter, and for mains under 12 inches in diameter " (Hurst).
284
NOTES ON BUILDING CONSTRUCTION
Find what size of pipes should be used, and whether any arrangement
for periodical flushing ought to be provided.
We will work out this example by means of Tables XIX. and XXIII.
Pipe BG. — For the maximum discharge we have
F = 4 cubic feet per second.
H = 4-7 - 1-6 = 3-1 feet.
L = 320 feet.
Hence
H-^ ~ 3-1
= 1650.
On referring to Table XIX. it will be seen that this number corresponds
to a 12-inch pipe, and that this pipe will run very nearly full when dis-
charging the maximum quantity.
To find the velocity of discharge we may consider that the jjipe is
running full. The discharge in that case can be found from
L „
jj-F2= 1850,
1850 X 3-1
~ 320 '
or F = 4'23 cubic feet per second.
Hence V. =4-23,
4
or, since D = 1 foot,
4-23 X 4
TT X 1^
= 5"4 feet per second.
The velocity is therefore well above the limit for flushing.
Next, to find the velocity when the average quantity of sewage is being
discharged. Proceeding as in Example 64, assume
P = 0-8D,
then, since (Table I.) C= 109*5 for a 12" pipe,
f T? Trq 109-52 X 0-2 X 3-1
from Equation 128 V,^ = ,
^ 1 0-8 x 1 x 320 '
= 3-07 feet per second.
Hence, since (Table XXIII. j = O-OTSD^,
Fj = 0-075 X l^x 3-07,
= 0-23 cubic foot.
This is a sufficiently near approximation, and since the velocity is a
little over 3 feet per second, there is no necessity to make any arrangements
for flushing.
Pipe BD. — For the maximum discharge we have
F = 3 cubic feet per second.
H= 13-8- 1-6 = 12-2 feet.
L = 416 feet.
Hence
L o 416 ,
On referring to Table XIX. it will be seen that a 9" pipe will do.
SIMPLE S YSTEM OF DRAINS
285
Let US inquire how full this pipe will run when discharging the maxi-
mum quantity. Assume
then (Equation 128)
2-2 X D = 2-2 X 0-75,
3_ 108^ X 3 X 12-2
1 ~2-2 X 0-75 X 416'
== 8"5 feet per second.
Hence, using Table XXIII.,
Fj = 0-699 X 0-75^ X 8-5,
= 3*2 cubic feet per second.
For Pg = 2-0 X D we find
V9 = 8"8 feet per second,
= 2*95 cubic feet per second.
It can therefore be taken that the wetted perimeter is 2-1 x D, or 18-9
inches. The un-wetted perimeter is therefore
28-3 - 18-9 = 9-4 inches,
and the angle subtended at the centre by an arc of 9-4 inches is
9*4
— — X 360 = 120° nearly.
28-3 ^
Hence, when discharging 3 cubic feet per second, the
pipe will be filled to the depth shown in Fig. 538.
Check ly Neville's forimila. — As a further exercise let us
test the above result by means of Neville's formula (p. 264).
We have, using Table XXIII.,
• 0-633 X 0-75
Fig. 538.
2-1
- = 0-225 foot.
V = 140
7'
0-225x^-11
4I0
7
0-225
12-2
416'
11-36-2-06
= 9-3 feet per second,
so that there is a fair agreement between the results.
To find the velocity when the average discharge is flowing.
0-6x0-75,
„„ 108^x0-03x12-2
then
Assume
P
0-6 x 0-75 x 416
= 1-32 foot per second,
Fi= 1-32 X 0-034 X 0-752 = 0-025.
Since the average discharge is 0-03 cubic foot per
minute, the velocity just found is a sufficiently near
approximation, and as it is less than 2 feet a second, it is
clear that this drain would require periodical flushing
during dry weather. The depth of liquid in the pipe
for the average discharge is shown in Fig. 539. The
calculations to find this depth are left as an exercise for
the student.
Pi'pe AB. — For the maximum discharge we have
F = 4 + 3 = 7 cubic feet per second.
H = r6.
L=1283.
Fig. 539.
286 NOTES ON BUILDING CONSTRUCTION
1283
Hence — — x 7^ = 39,300 nearly.
1-6
And from Table XIX, it appears that a 24-incli pipe is the nearest market size
that will do.
Assume P^^ = 2-0 x D,
111-52 X 7 X 1-6
then
2-0 X 2 X 1283 '
V, = 3-0,
F^ = 3-0x 22 X 0-595,
= 7-1 cubic feet nearly.
The average discharge is
0-2 + 0-03 = 0-23
Assume P^ = 0*6 x D,
then Vj'^
lll-52x 0-23 X 1-6
0-6 x 2x 1283 '
= 1 -44 feet per second,
r^= 1-44 X 22 X 0-034,
= 0-196.
As a second trial, if we assume = 0-65 x D we will find
1-40 feet per second,
= 1-40 x 22 X 0-042 = 0-236,
/ n\ which is near enough.
/ / \ \ This drain will therefore require periodical flush-
/ L 21^". X \ ing, more especially because the velocity is low even
I I ,A, I I when discharging the maximum quantity. When
\ \ ^/~..'\ j I discharging the average quantity the drain will be
\\ / \/ / filled to the extent shown in Fig. 540, and it will be
seen that the wetted perimeter is large in comparison
with the amount of liquid, and this is one cause of
Fig. 540. ^i^e lo^^ velocity.
Egg-shaped Sewer.
The wetted perimeter can be reduced by using an egg-shaped
sewer as shown in Fig. 542, and the velocity of discharge will
therefore be somewhat increased. The j)rincipal advantage, how-
ever, is that the depth of the liquid is increased, which is im-
portant in the case of sewage. Such sewers are therefore suitable
when the slope is small, the maximum discharge large, and the
average discharge small.
Example 66. — Design an egg-shaped sewer suitable for the pipe AB
in the previous example.
Egg-shaped sewers can be, and are made of various forms, but the most
usual form is as follows : — ^
Let (Fig. 541)
N = diameter of bottom of sewer,
M = diameter of top of sewer,
O = radius of sides,
K = depth of sewer,
1 See also Part II.
EGG-SHAPED SEWER
287
then
N = |, M = ^, 0 = K.
Fis. 5-11.
Fig. 542.
In the present case we must, in the first place, ensure that the sewer is
large enough to carry the maximum discharge without running full. We
have already seen that a circiilar pipe of 24 inches is required, and it will be
near enough for practical purposes if we make the area of the egg-shaped
sewer equal to that of the circular sewer. Xow the area of an egg-shaped
sewer of the form given above can be found from the formula
Area = 0-525 x . . . . (129).
And the area of a circle 2 feet in diameter is
3-14 square feet.
Hence
0-525 X K2 = 3-14,
K = 2-45 feet.
Therefore the depth can be taken as 2' 6". Hence N= 10", M= 1' 8", and
applying these dimensions we obtain the sewer shown in Fig. 542.
Let us now see to what extent the velocity is greater than with the 2-foot
drain pipe when the average discharge is flowing. The quantity flowing
being small, it will probably not rise above the invert, so that we may regard
it as flowing in a 10" pipe. Supposing that
10
= 1-2 X
then
and
12'
lOO^x 0-23 X 1-6 X 12
1-2 X 10 X 1283
10^
= 1-5 X 0-216 X
= 0-225,
which is a very close approximation, and thus we see that the velocity is only
very slightly increased.
Check hy Nevilles formula.— To check this result by using Neville's formula
we have
. 102
288 NOTES ON BUILDING CONSTRUCTION
Hence
V = 140^ 0-15 x^^l-li;/ 0-15 xA|,
= 1-92 -0-63,
= 1 "3 feet per second nearly.
The results therefore agree fairly well.
It appears therefore that, at any rate in this case, an egg-
shaped sewer does not increase the velocity to any material
extent.
JETS.
It is sometimes necessary to know, in connection with fount-
ains and hydrants, the height to which a jet of water will rise.
When a stone is thrown vertically into the air it rises to a
height depending on the velocity with which it was thrown. If
the stone is not thrown vertically, but at an angle, it will describe
a curve depending on the velocity and on the angle of elevation,
i.e. the angle at which the stone is thrown. In the same way a
vertical jet of water will rise to a height depending on the velocity
with which it issues from the orifice (or nozzle), and an inclined
jet will describe a curve depending on the velocity and the angle
of elevation.
Now it is shown in books on dynamics that in the case of a stone thrown
vertically, and neglecting the resistance of the air, the height to which it will
rise is given by ya
" = 6—4
If the stone is thrown at an angle, then the cnrve described by it is repre-
sented by the equation ^2
1/ = xtan B - 16'1 5-7),
cos*^ vr
where V is the initial velocity in feet per second, and Q is the angle of eleva-
tion, as shown in Fig. 543.
Fig. 543.
To jini the range, or the distance from 0 where the stone strikes the
horizontal plane, that is OA = R, put y = 0, then
JETS
289
tan (9= 16-1
R
E = TTTT • sin ^ cos
1d"1
(131).
R
Further, to find the greatest height the stone rises to, or BD, put V = '^^ tlien
BD = /(-(max.) = ta^l ^ sin 6 COS
16-1
V2cos2^" 32-22
V2 sin2 ^
(max.) —
64-4
sin- ^cos^^.
(132).
These formulae are applicable to jets projected in vacuo, and
to make allowance for the resistance of the air the simplest way
is to multiply the initial velocity by a reducing factor, or, in other
words, to calculate the path for a less initial velocity than the
actual.
The result is not quite accurate (see Fig. 544), but is
near enough for practical purposes. The reducing factor dimin-
I Full velocity
Z Reduced velocity
S Actual path
Fig. 544.
ishes as the ratio of the diameter of the jet to the head increases,
and can be found from Table XXIV., which is based on experiment.
Example 67. — A jet of water is issuing from a |^-inch nozzle with a
velocity of 60 feet per second. Find :
(a) Tlie height to which the jet will rise when it is thrown vertically.
(h) The distance to which the jet will reach, and the greatest height it will
attain, when thrown at the angles of 15° and 55° respectively.
A velocity of 60 feet per second corresponds to a head of
60^
64^
56 X 12
H =
56 feet.
Hence d being diam. of nozzle
•= 1350.
We can therefore take (see Table XXIV.)
J = 0-91.
Consequently for case (a), from Equation 130,
(0-91 X 60)
64-4
46 feet.
B.C. IV.
290
NOTES ON BUILDING CONSTRUCTION
— 1- It should be observed that if the jet is perfectly ver-
tical, and there is no wind, the water falls back on the jet,
and thus reduces the height as shown in Fig. 545.
Again for case (b), when the angle of elevation is 15°,
we have from Equation 131
R =
(0-91 X 60)^
sin 15° X cos 15'
16-1
= 46 feet ;
and from Equation 132
(0-91 X 60f sin^ 15°
^(max.) —
64-4
Fig. 545.
= 3-1 feet.
Lastly, when the angle inclination is 55°, we have from
Equation 131
(0-91 X 60f
R =
16-1
87 feet ;
sin 55° X cos 55°,
and from Equation 132
^(max.)
(0-91 X 60)
64^4
= 31 feet.
X sin^ 55°
Issuing velocity of a Jet. — We must now see how to obtain
the issuing velocity of a jet. If the nozzle offered no obstruction
to the issuing stream of water, it is clear that the velocity would
be that due to the head at the nozzle ; but the nozzle does offer
an obstruction, or, in other words, causes a loss of head, the
amount of which depends upon the shape of the nozzle.
The reduction in velocity caused by the nozzle has been
ascertained by experiment as follows : —
A well-shaped nozzle, such as the one shown in Fig. 546,
reduces the velocity to
0-97 X velocity due to the head.
Fig. 548.
JETS
291
1 to
\
4 "
\_
13 >'
2 5 "
1
3
1_
12
JL_
2 4
1
A converging nozzle, as in Fig. 547, reduces the velocity to
0-9 4 X velocity due to head.
And a cylindrical nozzle, as in Fig. 548, causes the following
reductions : —
When the diameter is
length . . . . 0-81 \
0-77 f Multiplied by
n.^TO / velocity due
^ 16 i to head.
0-68 j
The reduced velocity obtained by applying the above is of
course the initial velocity of the jet to be used in finding the
height or the range.
Example 68. — A 1" nozzle for a vertical jet is connected to the end of
a 1" pipe, 300 feet long. The head at the farther extremity of the one-inch
pipe is 100 feet, and the nozzle is 10 feet higher than this point. Find the
height of the jet and the quantity of water that will be discharged.
Let "V feet per second be the mean velocity of the issuing jet of water ;
then if G is the discharge in gallons per minute, we have from Equation 122
•49'
= 2-04x jij-x V,
= 0-126 X V.
Now if the nozzle is of a good shape (see Fig. 546) the velocity due to the
head at the nozzle will only be greater than V. Hence if H is the
head at the nozzle / y
Hence
or
0-97; =^^-^H'
V2 = 60-7H.
{~^\ =60-7H,
VO-126/ '
G2=0-96H nearly.
Again, we have from Table XVIII. for a 1" pipe, allowing for incrusta-
tion and remembering that the nozzle is 10 feet above the farther extremity
of the 1" pipe,
100-10-H
G2 = 250 x
300
Hence
0-96H = ^^90-2^.H,
300 300
or H = 41 -7 feet.
The ratio of H to the diameter of the nozzle is therefore
41-7x12
= 2000.
Hence by interpolation from Table XXIV. the factor J is 0-82.
from the above equation
And since
NOTES ON BUILDING CONSTRUCTION
we have
V2=60-7 X 41-7,
_ (0-82)2 X 60-7 ^ 41.7
64-4
= ^6-4 feet.
Lastly for the discharge we have
G' = 0-96 X 41-7,
or G = 6'3 gallons per minute.
Water Supply for House and Jet.
Example 69. — A jet, to rise to a height of 30 feet above the nozzle,
is required for an ornamental piece of water. The water is to be obtained
from a reservoir which is also to supply the house, the arrangements for
which are the same as in Example 62. The various lengths and levels
are given in Fig. 549. Find the diameters of the pipes that will be required,
and also how much higher the jet will rise when no water is being supplied
to the house.
We must first find the head required to produce a jet 30 feet high. To
do this we require to know the issuing velocity.
Datum level.
Vertical scale exaggerated S Unus.
Fig. 549.
40x12x8 T
neaiiv,
3
The head required at the nozzle will be about 40 feet, hence the ratio of
the head to the f " diameter of the nozzle is
H
'd '
= 1280.
So that, referring to Table XXIV., it will be seen that the factor J can be
taken at 0*90. Hence if V is the issuing velocity by Equation 130
3^^(0;90^
64-4
or V = 49 feet per second nearly.
If the nozzle is well shaped, the velocity due to the head will be
49
= 50'5 feet per second.
0-97 ^
Hence the head required to produce a jet of 30 feet with a good nozzle is
50-5'^
6^4
It is to be observed that the head assumed (namely, 40 feet) for the
purpose of calculating the factor J is very nearly the calculated head. Had
39-6 feet.
JE TS— EXAMPLE
293
there been any considerable difference the calculated head would have been
looked upon as a first approximation, from which a more accurate value
of J could have been found.
The next step is to find the discharge from the jet, which will be, from
Equation 122,
2"04 X 49 X 3^
G = — . — = 14 gallons per minute nearly.
On referring to Example 62 it will be seen that a head of 90 feet
is required at the point B to obtain the necessary water supply to the house.
Hence the available head from B to A is
90 + 12-39-6,
= 62-4 feet.
Applying, therefore. Table XVIII. we have
62-4 '
and to this number corresponds a 1;^" pipe.
To find the diameter of the portion BC of the pipe we have the available
head : 112-90-12 = 10 feet, and the discharge is : 14 + 5 = 19 gallons per
minute. Hence
'-1^^^=206,000 nearly,
a number which corresponds in Table XVIII. to a 3|^ inch pipe.
The jet will not be exactly 30 feet high, because the available diameters
of pipes are not quite those required to produce a jet of this height under the
conditions given. Working backwards we find —
Loss of head from C to B,
'^:^^^^ = 9-9feet,
208,000
and the loss of head from B to A will be
2^^^.60-5 feet.
810
Hence the head at A will be
112 -9-9 -60-5 = 41-6.
The velocity due to this head is
\''41 -6 X 64'4 = 51-8 feet per second,
hence the issuing velocity will be
51-8 X 0-97 = 50-3.
Therefore height of jet = Q'^Q' x 50-3^ _ ^^.^
64-4
When no water is being supplied to the house, the jet will be a little
higher. In this case the loss of head from C to B
5700 X 142
208,000
Hence the head at A will be
= 5-4.
112 - 5-4 - 60-5 = 46-l.
The ratio of the head to the diameter of the nozzle is
294
NOTES ON BUILDING CONSTRUCTION
Hence from Table XXIV. the value of J is about 0-89.
Therefore the height of the jet will be
46-1 X (0-97 X 0-89)2 = 34-3 feet.
Strictly a slight correction ought to be made, because the issuing velocity
of the jet is slightly increased. Hence the discharge will be greater — con-
sequently the loss of head somewhat greater than calculated above ; but as the
formulae used are only approximate it is not worth while going into such
niceties.
APPENDICES.
APPENDIX I.
Factors of Safety.
The following Table shows the Factors of Safety recommended by various
eminent engineers for application in several cases that arise in practice.
Authority.
Nature of Structure.
Nature of Load.
Factor
of
Safety.
Remarks.
Cfl-st Iron,
U
Dead
4
U
Live or varyin*^
(3
Stress of one kind only
u
"
10
Equal alternate stresses
TT
U
1.
Varying with shocks
15
of different kinds
"
Dead
3 to 4
6 to 8
Dead
3
i>
Live
6
s
» •
Live
6
s
Dead
s
Pillars
6
8
]) subject to vibr3,tiou . .
Live
8
S
y transverse shock
10
"Wrouglit Iron.
u
General .....
Dead
3
u
Live or varying
5
Stress of one kind only
u
8
Equal alternate stresses
u
Varying with shocks
12
of different kinds
R
Dead
3
R
vjirciers ......
4 to 6
s
Dead
3
8
6
B
Bridges
Mixed
4
Riveted joints ....
Dead
4
U
Roofs
4
8
Tension and compression bars .
Live with shocks
6
88
4i to 7
According to proportion
8
Compression bars
Dead
4
of length to least dia-
meter
Steel.
U
Dead
3
U
Live and varying
5
Stress of one kind only
u
8
Equal alternate stresses
u
Varying and shocks
12
of different kinds
R
Dead
3
R
Live
4 to 6
C
Mixed
4
Timber.
u
Dead
u
Live and varying
10
Stress of one kind only
u
15
Equal alternate stresses
u
Varying and shocks
20
of different kinds
R
10
8
Exposed to weather
Dead
10
8
8
8
For temporary purposes
4
Brickwork and Masonry.
U
Dead
20
U
Live
30
R
Dead
4 to 10
S
20
B. Board of Trade. C. Commissioners. R. Rankiae. 8. Stoney. 88. Shaler Smith. U. Unwin.
296
NOTES ON BUILDING CONSTRUCTION
The Factor of Safety determined upon for any particular case will depend
upon what is accurately known of the str,ength and quality of the particular
material to be used, upon the importance of the structure, etc.
APPENDIX II.
Wote on Equilibrium.
When the forces acting on a body lie in a plane, there are three conditions of
equilibrium, namely —
1. The algebraic sum of the resolved parts of the forces along some straight Hne
is equal to zero.
2. The algebraic sum of the resolved parts of the forces along another straight
line perpendicular to the former is also equal to zero.
3. The algebraic sum of the moments of the forces about any point is equal to
zero.
By the resolved part of a force along a straight line, is meant the effect the force
produces along that straight line. Thus if a force is parallel to a straight line, the
resolved part is equal to the force ; but if the force is perpendicular to the straight
line, the resolved part is zero. When the force is inclined, the resolved part is equal
to the force multiplied by the cosine of the angle included between the direction of
the force and the straight line. Thus the resolved part of X along AB (the direction
A to B being taken as positive). Fig. 29, is + X cos 6 and the resolved part of Y wdl
be - Y cos <f>.
By the moment of a force about a point, is meant the magnitude of the force
multiplied by the length of the perpendicular drawn from the point to the direction
of the force. The moment measures the power the force has of turning the body it
is acting upon about the point ; and if one direction of rotation (say like the hands of a
watch) be taken as positive, then the reverse direction will be negative. Thus in
Fig. 29 the moments of the forces X and Y about the point P will be - Xa and + Yb.
■< a >-
Fig. 29. Fig. 30.
As an example, let the conditions of equilibrium be applied to the case of a ladder
resting against a wall shown in Fig. 30.
Resolving horizontally, Rg - Rj = 0,
,, vertically, Ro-W = 0,
Moments about A, Rj x 0 - R3 x 0 + W x a - Rj x 6 = 0.
APPENDICES 297
From which it appears, by solving these three equations, that
R2 = W,
^3 = ^ . W.
Sometimes a solution is arrived at sooner hy taking moments about more than
one point, instead of resolving.
APPENDIX III.
To draw a Parabola.
Suppose the object to be to find the curve of bending moments for a
Span
A C B
1^
ii
D
Fig. 550.
c
I
B.
(15
1S\
Fig. 551,
298
NOTES ON BUILDING CONSTRUCTION
uniformly loaded beam. The data given are the two points of support AB
(or, in other words, the span) and the maximum bending moment at the centre
CD ; that is, the three points A, B, and D (Fig. 550) are given, and it is
required to draw a parabola through these three points.
Produce CD to F, making DF = CD ; join AF and BF. Divide AF and
BF into an even number of parts by repeated bisections, and number the
points thus obtained as shown in Fig. 551. Then join 1, 1 ; 2, 2 ; 3, 3 ;
etc. etc. These lines are tangents to the required parabola, which can then
be drawn in. (The parabola has not been drawn in the figure, as it would
confuse the construction, and the tangents sufficiently indicate the shape.)
APPENDIX IV.
To show that the Neutral Axis in a Beam passes through the
Centre of Gravity of the Cross-section (see p. 44).
For a rectangular cross-section.
Let To be the stress in the outside layers of fibres (o'g' or^'j-', Fig. 62).
the depth of the beam.
X, the distance from the neutral axis of any layer of fibres situated above
the neutral layer, such as c'd', Fig. 62. Ax the thickness of the layer.
6Aa;, the area of this layer of fibres (h being the width of the cross-section),
x' and 6A«' the same for a layer such as e'f.
Then the total tension over the cross-section of layer c'd' is, by assump-
tions 2 and 3, p. 44,
X
X rJ)Llx.
Hence the sum of the compressions above the neutral axis
= — ^ZjXIXx,
a '
and similarly the sum of the tensions below the neutral axis
= — f-2a;'Aa;'. •
a
But for equilibrium the tensions must be equal to the compressions.
Hence
2xAcc — Sx'Acc' = 0.
If now X be the distance of the centre of gravity from the neutral axis,
we have, by taking moments about this axis,
«(62Ax + b2Ax') = &2ccAx - h^x'Ax',
= 0.
that is, a; = 0) or the neutral axis passes through the centre of gravity.
The same reasoning can be applied to a cross-section of any shape, with
the difference that the breadth of each layer, instead of being constant, varies.
APPENDIX V.
Diagrams of Shearing Stress.
These, for beams supported at the ends, are often differently arranged
from those given at pages 58 to 60.
APPENDICES
299
By many writers the shearing stresses acting in one direction, say tlius
are called positive, and those acting in the other direction mgative, or
vice versa.
The diagram of positive shearing stress is shown on one side of the line
representing the beam, and that for the negative shearing stress on the other.
Thus in Fig. 263, p. 173, which represents the shearing stress caused by an
uniformly distributed load, the diagram for the shearing stress of this direction
is below the line, that for the shearing stress ■^y^- is above the line representing
the beam, instead of the whole diagram of shearing stress being below the line as in
Pig. 90, p. 59.
Fig. 255, p. 167, is similarly arranged for a distributed and a concentrated load,
both diagrams from A to D being below the line, and both from D to B above the
line, instead of being as in Fig. 91, p. 59.
Similarly in Fig. 87, p. 58, half the rectangle denoting the shearing stress from
one abutment to C might be below the line and the other half above the line.
In Fig. 4, Appendix VI., the diagi-am of shearing stress with direction may be
called positive, and is above the line from A up to the point where the direction
changes to from this to B it may be called negative, and is below the line.
The area of the shearing stress diagram between any point and either abutment
is equal to the bending moment at the point.
This is evident in Fig. 82, p. 57, where (P being the point), the area of the
shearing stress diagram at P (being the rectangle under PB)=:PB x W = ^Yx and the
bending moment at PMp = "\Va;.
In Fig. 83, P being the point, the area of shearing stress diagram at P (being the
triangle under PB) = xx=
and Mp = irax-=— •
When the portions of shearing stress diagram between the point and the
abutment chosen are of different signs or directions, the difference of these
areas must be taken.
Thus in Fig. 90, if the area be taken of the diagram under PB, we have to take
the area of the triangle under BC minus the area of the triangle under PC
_ivl I ^
_ wlx lOX^
~~2 2"
and
wl X
Mp= — xx-wxx-
_wlx wr?
~~2 2"'
By working out other cases the student will find that the rule given above
always holds good, and it is sometimes very useful.
APPENDIX VI.
Graphic method of finding the Bending Moments and
Shearing Stresses in a loaded Beam.
In Fig. I AF is a beam carrying the loads W3, W^, at the points
BCDE, through which draw vertical lines downwards as dotted.
Draw another vertical line YZ (Fig. 4) at a convenient distance. Along
300 NOTES ON BUILDING CONSTRUCTION
YZ lay off W^, Wg, W^, on any convenient scale of loads. Take any
point 0, which is called the "j;o/e," and draw the lines «, 6, c, d, e, joining to
it the extremities of the lines indicating the loads to the pole.
Bending Moments. — Starting from any convenient point X (Fig. 2) verti-
cally under A, draw in succession a, h, c, d, e, parallel to a, b, c, d, e of Fig. 4,
respectively— intercepted between the verticals from A to F of Fig. i, and
terminating at V on the vertical from F. Join XV, Fig. 2. The polygon
thus formed is called the "Funicular pohjgon." In Fig. 4 draw the
horizontal line oH, and the dotted line oN — the latter parallel to XV.
Then YN will represent and NZ will represent Ep measured on the scale
of loads.
Moreover in Fig. 2 the ordinate UT of the "funicular polygon" vertically
under any point P measured on a proper scale represents the bending
moment at P.
Take a foot as the unit of the linear scale, and a lb. as the unit of the
scale of loads.
Then if Ho is equal to 1 foot on the linear scale, UT on the scale of loads
will represent the number of foot pounds of the bending moment at this point.
If, however, Ho is more than one foot of the linear scale in length, then
the length UT on the scale of loads must be multiplied by Ho to find the
number of foot pounds of the bending moment. Of course any other units
may be substituted for the foot and pound — for example, the inch and ton.
Shearing Stresses. — These are shown in Fig. 3, the construction of which
is obvious.
The dotted horizontal lines are drawn through the extremities of the loads
set off along YZ, and are intercepted between the verticals from Fig. i as shown.
A dotted horizontal line is drawn also through the lower extremity of
Ra on Fig. 3, and at this point the shearing stress changes from to
or + to - .
The shearing stress at any point of AF is equal to the portion of the
vertical from that point intercepted between the upper and lower lines of
the shearing stress diagram (Fig. 3) . Thus SQ is the shearing stress at the
j)oint P.
Example.— In Fig. 1 take AF = 20 feet.
AB = 4 feet. 80 lbs.
BC = 2 „ W2 = 120
CD = 6 20 „
DE = 4 „ 60
EF = 4
Take the linear scale to be 10 feet = l inch, and the scale of loads 120 lbs. =1 inch.
Lay off AB, BC, etc., on Fig. i on the linear scale, and Wi, Wg, Ws, "W4, Fig. 4 on
the scale of loads. Complete the figures as described above.
Measure oH. It equals 10 feet on the linear scale.
Measure YN on the scale of loads. It measures 168 lbs., which should be equal
to Ra. NZ measures 114 lbs., which should equal Er.
To check this by calculation (see page 60) we have
^ 80x16 120x14 20x8 60x4
Ra = 1 h - 1
20 20 20 20
3360
Similarly Rf = 114 lbs.
=168 lbs., which agrees.
APPENDICES
301
100 50
Scales.
10
Lifiear Scale 10 ft. = i inch.
) 100
Scale of moments 1200 ft. lbs,= \ inch.
20 feet
Scale of loads 120 ids. = 1 inch.
ICKDO 500 O 1000
200 pounds
2000 foot pounds
302 NOTES ON BUILDING CONSTRUCTION
To find the scale of bending moments we have
Ho in feet on the linear scale x number of lbs. 2}er inch on load scale = the
number of foot pounds in one inch on the scale of moments.
10 ft. X 120 = 1200 foot pounds
.•. on scale of moments 1200 ft. lbs. = 1 inch.
Draw the scale.
Then to find the bending moment at any point P, draw PUT a vertical inter-
cepted by the upper and lower sides of the polygon. UT will be the bending moment.
Measuring UT on the scale of moment we find it = 500 foot pounds. ^
Shearing 5'<rcss. —Prolonging PT through the shearing stress diagram we have
the shearing stress at P=SQ = 168 lbs. By calculation we know that it equals
Ra = 168 lbs., which agrees.
In the same way the bending moment and shearing stress at any other point can
be found, and a similar process will enable the bending moments and shearing
stresses for any other distribution of load to be found. When a load is uniformly
distributed it may be divided into equal portions, and each of these treated as a
load ; but in this case the bending moments may be obtained more easily by drawing
the parabola described at p. 34.
Centre of gravity or Eesultant of system of parallel forces. — If the lines
a and e be prolonged so as to meet at G, then a vertical drawn upwards
through G will cut the beam at the centre of gravity of the loads — that
is, the point through which their resultant Avould pass.
APPENDIX VII.
Bending moments, Breaking weights, and Shearing stresses for
Beams and Cantilevers with various distributions of load.
The Table opposite shows for Cantilevers and supported Beams with different distributions of load
the Bending moments and the Breaking weights for the cases pp. 28 to 40, and the Shearing stresses
for the cases pp. 57 to 60.
Comparison of Concentrated and Distributed Loads.
From the table opposite it will be seen that if the total distributed weight
wl be taken as W, then,
Case 1. Cantilever. Load at free end W — _/^^'^''
61
Case 2. Cantilever. Uniformly loaded wl=\Y = ^^°^'^
61
Case 6. Supported Bea7n. Load in centre W = ^^'^^^'
61
Case 7. Supported Beam. Load distributed wl=^ =
^ , . 6^
J^rom this we see that
W in Case 2 is twice as great as in Case 1.
Hence fhe load uniformly distributed over its length that a cantilever can hear is
twice the load that it can hear at the free end.
W in Case 7 is twice as great as in Case 6.
Hence the load uniformly distrihuted over its length that a heam supported at
the. ends can hear is twice the load that it can hear at the centre.
^ If we had not made a scale of moments we should measure UT on scale of
loads— X = 50 lbs.— and multiply by Ho = 10 feet. 50 lbs. x 10 feet = 500 foot pounds.
APPENDICES
303
Again,
If a cantilever can bear W concentrated at its free end
It can bear 2W uniformly distributed over its length.
A beam of the same length supported at both ends and loaded in the
centre can bear 4W.
A beam of the same length supported at both ends and loaded uniformly
throughout its length can bear SW.
In all these cases the weight of the beam itself is omitted, being assumed
to be so small compared with the loads as to be practically insignificant.
APPENDIX VIII
Comparison of the Strength, Stiffness, and Shearing Strength,
of Beams of Uniform Cross Section throughout their Length
Symmetrical above and below the neutral axis and sup-
ported at both ends, with those fixed at one or both ends,
showing the effects produced by fixing one or both ends.
Arrangement
of Beam.
Arrange-
ment of
Load.
Bending
Moments.
Effect of
fixing on
Strength.
Deflection
value of n
in formula
^ = ^
see p. 66.
Effect of
fixing on
Stiffness.
Shearing
Stress.
Effect of
fixing on
Shearing
Stress.
Supported
both ends
Fixed both
ends
Centre
Centre
WZ
— at centre
4
m
-g- at fixed ends
-g- at centre
Increased
as 2 to 1
h
\
Increased
as 4 to 1
W
— at ends
W
— at ends
0 in centre
No effect
Supported
one end, fixed
at otiier
Centre
3W
^rj- at fixed end
lb
5WZ
at centre
-g- at centre
Y^- at fixed ends
2^ at centre
Increased
as 4 to 3
Increased
as 2-3 to 1
}JW at fixed
end
,=bW at sup-
ported end
Increased
as If to 1
Reduced
Supported
botli ends
Fixed both
ends
Uniformly
distri-
buted
Do.
Increased
as 3 to 2
3 It
Increased
as 5 to 1
'"^ i J
at ends
^ do.
No effect
Supported
one end, fixed
at other
Do.
-g- at fixed end
M'^" at centre of
16 beam
Q ,9 at centre of
— supported
128 portion
No effect
Increased
as 5 to 2-08
^wl at fixed
end
\vil at sup-
ported end
Increased
as 5 to 4
Reduced
From the above table it will be seen that
The. effect of fixing both ends of a beam as compared with leaving both ends
supported is
With Load at Centre. Strength doubled. Stiffness increased as 4 to 1.
Shearing stress not altered.
With Load uniformly distributed. Strength increased 50 per cent.
Stiffness increased as 5 to 1. Shearing stress not altered.
The effect of fixing one end of a beam as compared loith leaving both ends
supported is
304
NOTES ON BUILDING CONSTRUCTION
With Load, at Centre. Strength increased 33 per cent. Stiffness increased
as 2-3 to 1. Shearing stress increased at fixed end, reduced at supported end.
"With Load uniformly distributed. Strength not altered. Stiffness in-
creased as 5 to 2. Shearing stress increased at fixed end, reduced at
supported end.
APPENDIX IX.
Deflection of Rectangular Beams under different
Conditions of Load.
The formula (47), p. 68,
A =
i2^w;3
md^
(47)
is for rectangular beams. It may, by the substitution of the different values
of n from Table C, p. 67, be stated in a convenient form for different con-
ditions of load.
Take the case of a beam fixed at one end, free at the other, with a load
at the free end. Substituting in (47) from Table C for », we have
^ - ~ Wd^
Substituting similarly values of n for other cases we have the following
Table.
Arrangement of Beam.
Load.
Formula.
at free end
A
. 4W^^
Beam fixed one end, free other .
= EM^
A
3W/3
uniform
A
Beam supported both ends
centre
A
5> '» • •
uniform
~ 32md^
Beam fixed both ends
centre
A
^ T^'^Ebd^
A
)) )) ...
uniform
= s\Ebd^
A
Beam supported one end, fixed other .
centre
A
)) )) "
uniform
APPENDIX X.
Relative Strength of Beams of different Sections.
The following are given as they may be of interest to the student, but they are of
no practical value.
APPENDICES
Square Beam and Cylindrical Pole of Diameter equal to ^
SIDE OF Beam.
_
M of sqimre learn =fol _ 12,
yo~ d
2
6
In a square beam i=c?.-. M = -:^ ...
6
M of cylindrical pole='^= — —
4
4
8x4
_ /oX3-14xc;g
32
= . . . n.
from I. and II.
M square beam : M circular pole : : \fod^ : i-^fod^
° O
: : Wod^ ■■ W
: : 10 to 6.
The Strength of a Square Beam side Vertical is to that of a
Square Beam with Diagonal Vertical
:7 :5
A Triangular Beam of the same area and depth as a Rectangular
Beam is only one-half the strength of the latter.
The strength of a cylindrical pole to that of a square beam of the same area is as
1 to 1-18.
The strength of a cylindrical beam to that of the strongest beam that can be cut out
of it is as 1'53 to 1.
APPENDIX XI.
Continuous Beams of Uniform Section loaded uniformly with
Points of Support equidistant from one another and in the
same Straight Line.
On pp. 74, 75 are shown the forms assumed by a uniformly loaded girder
of uniform strength extending over two spans.
B.C. IV. X
3o6 NOTES ON BUILDING CONSTRUCTION
In practice, however, small continuous beams are generally of uniforik
section.
When such beams are fixed at the ends and continuous over one or
more supports, each portion may be treated as a beam fixed at both ends
(see p. 72, Case 2), and the moments of flexure, etc., will be found accord-
ingly.
When, however, a beam is continuous over one or more supports, and
Total load = 2iwZ.
C
Fig. 1.
3wl
Ra =
Ma = 0
, /
lOwl
Mc =
wZ2
Mb = 0.
supported only at the ends — which is sometimes the case in practice — the
Fig. 2.
Total loa.i=3wl.
C D
.„,./.„-± /—J
Ma = 0 Mc = t^' ^iD = i~ Mb = 0.
resulting stresses are different, and are very difficult to find entirely by
calculation.
Total load=4wZ.
Fig. 3.
— -/-
RA=iiwl Rc=
Ma = 0
Mc=?-^ Me = ^— Md = |— Mb = 0.
Reactions and Bending Moments. — Figs. 1 to 3 show the reaction and
the bending moment at each point of support for beams continuous over from
2 to 4 spans.
Each span = I
Load on each span = wZ
Deflection. — This will be a maximum in the two end spans, and may be
obtained for those from Case 3, p. 73, and Appendix VIII., though if the
two end spans are not reduced their inner ends will not be perfectly fixed
and the actual deflection will be in excess of that calculated.
Points of Contrary Flexure. — These, if required to be known, may for
each, of the intermediate spans be found as in the case of a beam fixed at
both ends (p. 72), or for each of the end spans, as in the case of a beam fixed
at one end and supported at the other.
To find the Bending Moment from the Reactions. — When the reactions are
known, the bending moment at any point may be found by the ordinary
method described in Chapter III.
APPENDICES
307
Thus in Fig. i
The bending moment at a section infinitely close to C will be
M. = R X AC - X -
C A 2
= %wl X I - —
« 2
= - \wl'^.
Again in Fig. 3
Mo = \\wl
Beams continuous over several Supports.
The following table gives the reaction on each support caused by uniformly
loaded beams of uniform section continuous over several supports.
Uniformly Loaded Beams continuous over several Supports.
Table of Distribution of Loads.
No. of
Span.
Number of each Support, and Load borne by it in terms of W. or of load on each span.
1st
2d
3d
4th
5th
6th
rth
8th
9th
10th
1
2
3
4
5
6
7
8
9
1
3
?
4
\ts
1 1
1 5
4 1
TlTi
6 6
15 2
2 0 9
1
T
1 0
1 1
nr
3 2
4 3
lis
TT5T
16 1
TTT
4 4 0
6 0 1
"5 "SIT
3
1 1
TiT
2 6
■iTS-
■Si
10s
TiTJ
13 7
TTT
374
TTS-
S 1 1
4
TTT
3 2
IS
37
10 6
TTTT
14 3
TTT
3 9 2
Tg¥
6 3 5
TTiT
1 1
Tg-
4 3
TS-
10 8
TtTf
1 4 3
TTT
3 8 6
■j-rs-
5 2 9
1 5
118
TUT
13 7
TTT
3 9 2
Trs
5 2 9
4 1
TOT
161
TTT
3 7 4
TT?
5 3 5
5 6
TTT
44 0
5 11
15 2
6 0 1
200
Illustration. — The principal rafter of the roof dealt with in Chapter XII.
is treated in j)age 214, first as if it were in two portions each supported
at both ends, and next as a beam continuous over two spans. In both
cases it is uniformly loaded.
In this example the maximum bending moment is the same under
either supposition, being in the case of each supported portion -— - in the
8
8"
support.
In the case, however, where a rafter extends over several points of
support (see p. 202, Fig. 343) the maximum bending moment to which it is
subjected will be very different if it is a continuous beam, compared with what
it would be in a jointed and supported beam.
centre, and m the continuous beam at the intermediate point of
3o8 NOTES ON BUILDING CONSTRUCTION
Thus in Fig. 4, if it be assumed that there are joints at the points of
Fig. 4.
support, then AC, CD, DB are supported beams (see Case 7, p. 33), and the
maximum bending moment is — at the centre of each.
If AB is assumed to be continuous, then (see Fig. 2 above) the maximum
bending moment is at C or D, i.e. -| x - being less than — , and therefore if
the rafter is really continuous, all the points being accurately in a straight line
a less scantling will do for it than if it were jointed and supported.^
APPENDIX XII.
Method of finding the Centre of Gravity of an unsymmetrical
Cross Section of a Girder by Calculation.
Divide up the cross section into rectangles all parallel to each other and
A to the bottom edge of the section.
Rule. — Multiply the area of each rectangle by
the distance of its centre of gravity from the bottom
edge, and add these products together. Divide by
the whole area of the section ; the quotient is the
distance of the centre of gravity from the bottom
edge.
Example. — Thus the cross-section of the cast
iron girder shown in the figure can be divided up
into three rectangles, and the products are
4x 1 X 19-5 + 18 X 1 X 10+ 16 X 1 x 0-5 = 266.
The whole area is 4x1 + 18x1 + 16x1= 38.
Therefore distance of CG from bottom edge
= — = 7 mches.
Since the section is symmetrical vertically, the
centre of gravity is determined as shown in the
figure. Should, however, the figure be unsym-
CG.
^ In reality the case lies between the two, because C and D are not rigid
supports.
APPENDICES
309
metrical in all directions, the above process must be repeated with reference
to an axis perpendicular to the first one. Thus in the present case, to find
the distance of the centre of gravity from the axis AB the products are
4x1x8 + 18x1x8 + 16x1 x 8 = (4 + 18 + 16)8.
Hence distance of centre of gravity
(4 + 18 + 16)8 ^. ^
= 77, — T77~ = 8 inches.
4+18 + 16
APPENDIX XIII.
Graphic Method of finding Centre of Gravity of unsymmetrieal
Cross Section of Girder.
Draw the polygon OPQR having its sides parallel to o-pcp respectively, and
intercepted between verticals drawn through the centres of gravity of ABC.
From the intersection of R and 0 at Y draw a vertical. The point at which
this cuts the centre line of the girder will be its centre of gravity.
The figures show that this agrees with the calculations above (App. XII.)
NOTES ON BUILDING CONSTRUCTION
APPENDIX XIV.
Moment of Inertia (I) of Various Sections.^
Section.
(1) Rectangle (breadth h, deptli d) . . . . .
(2) Isosceles triangle (base h, heiglit d) ....
(3) Circle (radius r) .......
(4) Annidus (E, outer radius, r inner radius)
(5) Semicircle either thus \^ or thus ^'^"^
Value of I.
hd:^
T2
36
^(E4-r^)
0-1 Ir*
(6) T
(see Fig. i).
(7) I girder divided into rectangles as in Fig. 2 —
neutral
Fig. 1.
Fig. 2.
GENERAL RULE FOR ANY SECTION MADE UP OF A NUMBER OF SIMPLE FIGURES.
a. Find the moment of inertia of each of the simple figures about an axis,
traversing its centre of gravity parallel to the neutral axis of the complex figure.
b. Multiply the area of each simple figure by the square of the distance be-
ttveen its centre of gravity and centre of gravity of the whole figure.
Add the results so found (by a and b above) for the moment of inertia of the
whole figure. (Rankine.)
I for Balled Beam. — Applying No. 7 formula to find I for the section of
the rolled iron girder, p. 83, we have
T-V{|x93 + 4(103-93)},
= tV{^ X 729 + 4(1000 - 729)},
= 120-7.
^ In each case the moment of inertia is taken about an axis parallel to the neutral
axis passing through the centre of gravity of the section.
TT is the ratio between the circumference and diameter of a circle, and its
numerical value is 3'1416.
APPENDICES
I for Cast Iron Girder. — Applying Rankine's rule to the section of the
„ ■< — 4 — >
cast iron girder, Fig. 3, we have
a. I of upper flange
I of web
I of lower flange
4 X 1^
4
~ 12
^ 12'
1 X 183
5832
12
16 X 13
16
~ 12
~ 12'
Total
5852
12
487-6.
h. Area x square of distance of centre of
gravity from the neutral axis
Of upper flange = 4x1 2-5'^ = 625,
Web = 18 x 32 = 162,
Lower flange = 16 x 6"52 = 676,
Total of a and & = 1950 = I.
so
■16
Fig. 3.
It should be observed that the thickness assumed for the lower flange,
namely, 1 inch, is rather little for the width, but it simplifies the example.
APPENDIX XV.
rormulsB for finding approximately the Weight of Girders.
The weight of a girder can be found with a fair degree of accuracy by
either of the following formulae, the first being, however, the more accurate.
Weight of girder in tons
W X
= ^^fz — (Prof. Unwin's formula),
W X L
~~560~ (Anderson's formula),
where
W = distributed, or equivalent distributed, load to be borne by the girder,
in tons.
L = span in feet.
C = a factor whose value is from 1400 to 1500 for small girders, and
from 1500 to 1800 for large bridges.
rc = stress in compression boom (generally about 4 tons per square inch).
D = eft'ective depth of girder in feet.
APPENDIX XVI.
Bules for drawing Maxwell's Diagrams {see p. 183).
The diagram of the truss itself is called the frame diagram. The
diagram of the forces is called the stress diagram or "diagram offerees."
312
NOTES ON BUILDING CONSTRUCTION
In the stress diagram lines are drawn parallel to the several lines, of the
frame diagram, thus
1. Draw each hiown force in succession in the direction in which you
know it acts.
2. ^ Then from the two " open ends " draw the two unknown forces parallel
to their directions on the frame. The first unknown force working in this
direction -> is at the end of the last known force, that is, at the end to which
the arrow mentioned in next paragraph points.
3. Place arrows on the stress diagram beginning with the known forces,
and make them all follow each other pointing in the same direction.
4. By transferring these arrows in imagination back to the frame, and
considering them with reference to the point in q^uestion, you know whether
each member is in compression or tension.
The above rules are applied to each joint of the frame in succession.
Example.— Thus in the example illustrated in Figs. 305 to 308, Plate II.
Fig. 1. Frame Diagram.
Fig. 2. Stress Diagram.
we know (pp. 180, 181) that \V = 7 cwts. Ra = 5 cwts. Rb=2 cwts.
• Commencing at joint A in the frame (Fig. i) draw for the stress diagram the
known Ra = 5 cwts., and mark upon it the arrow head showing its direction, i.e. upwards.
Then working in the direction draw from its end q, to which the arrow points, the
APPENDICES
313
unknown T parallel to AB of the frame, and from the other end 0 draw Ci parallel
to AC. These intersect at ^, and by measurement
22^ = 3-94 cwts. =T
0|i = 6-4 cwts. =Ci.
ISTow place arroM's on the stress diagram (Fig. 2). We have marked the arrow on
Ra pointing upwards — following round in the same direction the arrow on T points
towards p and that on Ci towards O. Transferring these to the frame we see that Ci
is in compression pushing towards A, and T in tension pulling from A.
Proceeding to the next joint 0 on the frame, we have Ci and W known and C2
unknown.
On the stress diagram we have Ci already drawn. Draw W vertically = 7 cwts. ,
and draw a line from the end s of W parallel to C2 ; this will be the line sO.
We know the direction of the arrow for W is downwards. Following on in the
same direction the arrow for will point toward O and that for Ci toward p. The
latter is in a different direction from what it was before, because Ci is now in com-
pression toward C.
Transferring these arrows in imagination to the frame, we find Cj C2 both in
compression.
We come now to the last joint, B, and have Rb Co T all known.
Draw Rb = 2 cwts. and a line T = 3 "94 cwts. from s parallel to AB of frame. This
will be the line sr.
Again the arrows are placed — on Rb pointing upwards, on C.2 toward s, and on
T towards r. Transferring these to the frame we find Cj in compression toward B,
and T in tension from B.
If a line drawn through s parallel to AB and equal in length to 3 "94 cwts. did not
come exactly to r, it would be certain that some mistake had been made in the
drawing.
In such a case the diagram is said not to "close" properly, and this property
of not closing when there is anything wrong in the diagram is very valuable,
because the error thus shown to exist can be investigated and corrected, whereas,
in using formuloe there is often nothing to show the existence of an error, which
remains undetected and makes the result incorrect. The non-closing may be due
only to inaccurate drawing. If this is suspected, the stress upon one or more bars
may be checked by Ritter's method (see p. 185).
APPENDIX XVIL
Bow's System of lettering Maxwell's Diagrams {seep. 184).
The following words quoted are from Mr. Bow's own description of liis
system.^
" This plan of lettering consists in assigning a particular letter to each
enclosed area or space m, and also to each space (enclosed or not) around or
bounding the truss, and attaching the same letter to the angle or point of
concourse of lines which represents the area in the diagram of forces. Any
linear part of the truss or any line of action of an external force applied to
it is to be named from the two letters belonging to the two spaces it
separates : and the corresponding line in the reciprocal diagram of forces
which represents the force acting in that part or line, will have its
extremities defined by the same two letters."
Example."^ — Figs, i and 2 give an example of a truss and a diagram of forces so
treated.
^ From Bow's Economics of Constructio^t.
^ Ibid. , but condensed.
314 NOTES ON BUILDING CONSTRUCTION
The spaces in the truss are lettered ABC, those around the truss DEFGH.
These letters are attached to the corresponding points of concourse in Fig.
Fig. 1. Truss.
Fig. 2. Diagram of Forces.
Each line in the truss is lettered according to the letters of the spaces it
separates. Thus the line separating spaces A and F is AF and is shown as AF on
the force diagram below. Similarly the line separating spaces B and C is BC.
If any triangle in the truss be taken, such as A, the stresses on its sides as shown
on the force diagram will be found to radiate from A. Thus, in Fig. 2, AB, AF, AD
radiate from A.
Examples of the application of Bow's method to girders will be found at p. 194.
APPENDIX XVIII.
Calculation of the Stresses upon the Iron Roof (Example 42,
p. 209) by the Method of Sections.
Referring to Example 42, p. 209, we now proceed to find the stresses by means of
the method of sections (see p. 185).
Stresses DUE TO Permanent Loads. — Commencing as before at joint A (Fig.
3565 Plate III.), we take the section shown in Fig. 398, which cuts through only
two bars ; therefere to find the stress in AF we can take
moments about any point in AD, round m for instance.
Ra = 1'17 tons (see p. 209). We find by measurement
that the lever arm of Ra is 8 '53 feet, and of the stress
in AF, 2-39 feet.
Hence
8-53 X 1-17- 2-39 xS^F = 0,
or
Saf —
8 -53x1 -17
2-39
= +4-16 tons.
Similarly to find Sad we can take moments about any point in AF.
for instance,
lever arm of Ra =10 "So feet,
Sat>= 2-72 feet.
We find.
APPENDICES
315
Hence
10-35 xll7 + 2-72xS^=0,
or
Sad= ~ 4 "45 tons,
so that AD is in compression.
To find Sdf we can take the section shown in Fig. 399, and the lever arms are
marked on the figure. Taking moments about A we have
10 X 078 + 10-78 xScF = 0,
SDf= -0-72 ton.
H=l-17
Fig. 399.
Fig. 400.
The same section will enable us to find Sdo ; the lever arms are, as shown in Fig.
400, for
Ra, 11-07 feet,
Wi, 1-04 „
Snc, 2-90 „
and therefore
whence
11-07 X 1 -17 - 1-04 X 0 -78 + 2-90 x 8^,^,= 0,
S„„= -4-20 tons.
Lastly, to find Sfg and Sfc we take the section shown in Fig. 401, and obtain the
lever arms as marked, whence
20 X 1-17 - 10 X 0-78 - 6-67 X 8^0 = 0, K- --20'0"
Sf„= +2-34 tons.
For Syo we must take moments
about the intersection of FG and DC,
thus
3-25 X 1-17 + 6-75 x 0-78 - 4-69 x S^^,
Spo= +1-93 ton.
On examination (see Table H) it
will be found that the two methods do
not in all cases give exactly the same
values for the stress. The differences
are due to slight errors in the drawing of the diagram and in measuring the lever arms.
Stresses due to "Wind Pressure. — It does not appear necessary to describe the
working of this method again in detail, and therefore the various sections to be taken
are simply marked on the figures, together \vith the turning points and corresponding
lever arms ; and the calculations are given below for each stress.
Case 1. Wind on left — Reactions ■parallel to loind pressure. — S.^} : Section 1, 1,
Fig. 402. Turning point a.
2-69 x 8^0 + 10-0x0-88 = 0,
Sad= -3-27 tons.
Fig. 401.
2-69
n'O-'
Fig. 402.
Fig. 403
31 6 NOTES ON BUILDING CONSTRUCTION
Saf: Section 1, 1, Fig. 403. Turning point I.
-2-61xSaf + 1O-0x0-88 = 0,
Saf = +3-38 tons.
Sdf : Section 2, 2, Fig. 404. Turning point A.
AD xSpF + ADx 0-96 = 0,
ScF= '0-96 ton.
That is, the compression in DF is equal to the wind pressure acting at D.
Fig. 404.
Fig. 405.
Sdc : Section 2, 2, Fig. 405. Turning point F.
2-9 xSdc + IO-Sx 0-88 = 0,
Si,c= -3-28 tons.
Thus we see (allowing for errors of measurement) that
Sad — Sec-
Fig. 406.
Sfo : Section 3, 3, Fig. 406. Turning point c.
- 4-65 xSfo + 3-62 X 0-88 + 7-18 X 0-96^0,
Sfc= +2-17 tons.
Fig. 407.
APPENDICES
317
Sfo : Section 3, 3, Fig. 407. Turning point C.
- 6 -67 X Sro - 10 -8 X 0-96 + 21 '6 X 0 -88 = 0,
Sf.o= +1'30 ton.
Fig. 408.
Section 4, 4, Fig. 408. Turning point G.
- 2-9 xScE- 9-81 X 0-56 = 0,
Scj.= -1-89 ton.
Fig. 409.
ScG : Section 4, 4, Fig. 409. Turning point d.
4-65 xSco- 2-62 X 0-56 = 0,
SeQ= +0-32 ton.
0-56
Fig. 410. Fig. 411.
SoE : Section 5, 5, Fig. 410. Turning point B.
-10-8xS<,E-0x0-56 = 0,
ScE-0.
Sob : Section 5, 5, Fig. 410. Turning point E.
+ 2-78 X Sob- 7-8 X 0-56 = 0,
S(,3=+l-57 ton.
Seb: Section 6, 6, Fig. 411. Turning point G.
- 2-9 xSeb- 9-81 X 0-56 = 0,
Seb = - 1 -9 ton.
0-56
31 8 NOTES ON BUILDING CONSTRUCTION
Again we find slight discrepancies between the vahies found by the two methods,
and, as before, these discrepancies are due to errors in measurement.
Case 2. Wind on right — Beactions parallel to normal wind pressure. — The stresses
can be deduced from Case 1.
Case 3. Wind on left — Reaction at free end vertical. — It should be noted that Pa and
Ra being no longer exactly opposite in direction (as they were in Case 1), we cannot
now use their difference only, but must introduce them separately into the equation.
1
Fig. 412. Fig. 413.
Sad '• Section 1, 1, Fig. 412. Turning point a.
+ 2-69 X Sab -10-0 X 0-48 + 9-58 X 1-45 = 0,
Sad= -3-38 tons.
Saf : Section 1, 1, Fig. 413. Turning point h.
- 2 -ei X Saf - 10 -0 X 0-48 + 9 -92 x 1 -45 = 0,
Saf= +3 "67 tons.
2
Fig. 414.
Sdf: Section 2, 2, Fig. 414. Turning point A.
AD xSdf + ADx 0-96 = 0,
Sdf= — 0 96 ton,
which is the same value as already obtained under Case 1. In fact, since this stress
does not depend on the reaction at the abutment, but only on the load at the joint
(as is evident from the equation of moments), it follows that it is unnecessary to
recalculate it each time.
Sdc : as in Case 2.
Scd = Sai,= -3-33 tons.
s
APPENDICES
Syc : Section 3, 3, Fig. 415. Turning point c.
- 4 -65 X S,c+ 7-18x0 -96 + 3-51 x 1 -45 - 3 -62 x 0 -48 = 0,
Spo= +2-2 tons.
Sfg : Section 3, 3, Fig. 416. Turning point C,
- 6 -67 x Sfo - 10 -8 X 0 -96 - 21 -6 x 0 -48 + 21 -4 X 1 -45 = 0,
S,o=+l-54.
Fig. 417.
Fig. 418.
ScB : Section 4, 4, Fig. 417. Turning point G.
- 2-9 xScE- 11-04 X 0-52=^0,
ScK = -1-98 ton.
Sco : Section 4, 4, Fig. 418. Turning point d.
+ 4-65 X Sco - 3-40 X 0-52 1:^0,
Sca = +0-38 ton,
Sro could also be found very easily from this section.
Fig. 419.
Fig. 420.
Sgk : Section 5, 5, Fig. 419. Turning point B.
- 10-8 xSqe- Ox 0-52 = 0,
Sge=0.
Sgb : Section 5, 5, Fig. 420. Turning point E.
+ 2-75 xSgb- 10-0x0-52 = 0,
S<,B= +1-88 ton.
Fig. 421.
S^E : Section 6, 6, Fig. 421. Turning point G.
-2-9xSbb-11-04xO-52 = 0,
Seb= - 1 -98 ton.
320 NOTES ON BUILDING CONSTRUCTION
Case 4. W ind on right — Reaction at free end vertical.
Sad : Section 1, 1, Fig. 422. Turning point F.
2-9 xS^„ + 5-40 X 0-87 = 0,
S^„=: -1-62 ton.
Fig. 423.
Section 1, 1, Fig. 423. Turning point D.
- 2-75 X Saf + 2-65 X 0-87 = 0,
Sai.= +0-84 ton.
Sdj. ; the stress in DF is clearly 0.
Sot '• Similarly, as in Case 3,
Sbc = Sai,= -1-62 ton.
Fig. 425.
Section 3, 3, Fig. 424. Turning point c.
-4-65 xSi,o + 0-88x 0-87 = 0,
Si,c= +0-16 ton.
Section 3, 3, Fig. 425. Turning point C.
-6-67x8^0 + 5-26x0-87 = 0,
S™= +0-69 ton.
Fig. 426.
Fig. 427.
ScE : Section 4, 4, Fig. 426. Turning point G.
- 2 -9 X ScK - 0 X 0 -96 - 11 -04 X 1 -26 + 10-8x0 -48 = 0,
ScE= - 3-0 tons.
Sco : Section 4, 4, Fig. 427. Turning point d.
+ 4-65 xSco- 7 -18 x0-96 + 3-62x0-48 -3-37 x 1-26 = 0,
ScG= +2-02 tons.
APPENDICES
321
Fig. 428. • Fig. 429.
: Section 5, 5, Fig. 428. Turning point B.
- 10-8 xSoE- 10-8 X 0-96 = 0,
Sge= ^ 0'96 ton.
Sqb : Section 5, 5, Fig. 429. Turning point E.
+ 2 75 X Sob + 10 -8 X 0 -48 - 10 -0 X 1 -26 = 0,
Sgb= +2-7 tons.
I
Fig. 430.
Sbb : Section 6, 6, Fig. 430. Turning point G.
- 2 -9 X - 11 -04 X 1 -26 + 10-8x0 -48 = 0,
Seb= - 3'0 tons.
APPENDIX XIX.
Centre of Pressure in Masonry Joints [see p. 219).
It can be shown by an investigation, similar to, but more complicated than
that given at p. 219, that the position of the centre of pressure for a joint of
any section -svhatever, when the pressure vanishes along one edge, depends on
the moment of inertia of the section. Rankine ^ gives the result as follows :—
Let A denote the area of the joint.
y, the distance from the centre of gravity of the figure of the joint to
the edge farthest from the centre of pressure.
h, the total breadth of tlie joint in the same direction.
I, the moment of inertia of that figure, computed as for the cross section
of a beam relatively to a neutral axis traversing the centre of gravity at right
angles to the direction of the deviation to be found.
S, the deviation to be found.
Then S = — .
Ay
1 A Manual of Civil Engineering, by W. T. M. Rankine, F.R.S., etc v 378
10th Edition. '
B.C. — IV. V
322
NOTES ON BUILDING CONSTRUCTION
If the joint is rectangular, as shown in Fig. 43 1, p. 218, and the centre of
pressure is situated in GH, then :
A = BD.GH,
GH
y = ^,
h = BD,
I = — .BD.GH3,
12 '
8 = ^ -GO.
2
GH 2 . BD . GH3
Hence
or
GC =
2 12.BD.GH2'
GC = IGH, as before.
The following values of GC for a few sections will be found useful in
practice : —
Descri2:)tion of Section.
Value of GC.
Eectangular .....
Circular .....
Hollow rectangular ) , .
1 > chimneys
,, circular j
IGH
|GH
IGH ) . ^
*GH j ^PP^o^^™^te
APPENDIX XX.
Arch with Unsymmetrioal Load.
The following memorandum by Mr. Henry Fidler describes how the
line of least resistance in an arch with an unsymmetrical load may be
drawn, by the method devised by Professor Fuller, and modified by Professor
Perry.
Let ahcd be a segmental arch divided into any convenient number of
imaginary voussoirs (in this case 9), and loaded unequally.
The positions and intensities of the loads are indicated by the vertical
arrows 1, 2, 3, 4, etc., and the direction of each vertical force is supposed to
pass through the centre of gravity of each voussoir.
The centre third of the depth of the arch ring is indicated by the dotted
arcs, and it is required to draw the polygon of forces corresponding to the
load and contained within the boundaries of the central third.
With any pole e, draw the force diagram. Fig. 2, shown below the arch,
plotting the vertical loads 1, 2, 3, etc. to any convenient scale. Draw
the polygon of forces JKLMNORSTUV in the usual way, terminating
with the closing line JV, to which ef is parallel, giving the vertical
components of abutment reactions fg, fh, for the right and left hand
abutments respectively.
For the sake of convenience repeat the polygon of forces jhlmnorstuv
above the arch to an increased vertical scale if necessary, plotting the
vertical ordinates from a horizontal base AB. On AB take any two
convenient points 2^, q, and join pr, qr.
324
NOTES ON BUILDING CONSTRUCTION
From the points I, m, n, o, etc. of tlie polygon draw horizontal lines
intersecting the straight lines pr, qr. From the points of intersection
let fall the perpendiculars I, II, III, IV, V, etc.
From the points of intersection of the perpendiculars 1, 2, 3, 4, 5, etc.
with the upper and lower boundaries of the central third draw horizontal
lines intersecting the corresponding perpendiculars I, II, III, IV, V, etc.
Join the points thus obtained and complete the irregular figure WXYZ.
Within the area W, X, Y, Z, select the pair of straight lines, Xx ; Zx ; meet-
ing at X (on the vertical through r) containing the smallest possible angle at x,
and contained wholly within the boundaries of the area W, X, Y, Z.
The straight lines Xx and Zx bear the same relations to the polygon
required to be drawn within the central third of the arch ring as the straight
lines pr, qr, do to the polygon jklmno, etc.
From the points of intersection of the lines Xx and Zx with the
verticals I, II, III, etc. draw horizontal lines intersecting the verticals 1, 2, 3,
etc. By joining the latter points thus obtained the required polygon lying
within the central third of the arch ring is drawn.
A second force diagram, Fig. 3, may now be drawn from the polygon last
obtained giving a polar distance T equal to the minimum horizontal thrust
of the arch required to meet all the conditions.
If it be not found possible to draw any pair of straight lines within the
boundaries of the area W, X, Y, Z, and meeting at x, the depth of the arch ring
must be increased until these conditions are met.
The above method is of course equally applicable to an arch with a
symmetrical load : and it avoids the necessity of drawing repeated trial lines
of least resistance, which is very tedious.
APPENDIX XXI
Short Formulae to be remembered.^
It is dangerous to trust to the memory for formulae that are not in
habitual use, and the invaluable Pocket-books of Molesworth and Hurst are
not always at hand. A student who has grasped the principles iipon which
the different calculations are founded can generally build up the formulas he
requires, but in everyday out-of-door life there is not always time for tliis,
and it is well to have a few simple formulfe in one's memory to aid in rapidly
forming an opinion when reference to books of any kind is impossible.
The following are good formulte to remember. In those for supported
beams the safe distributed load is given, in cwts. for timber, in tons for iron.
From this the safe weight for other distributions of the load can easily be
ascertained (see Appendix VII.),
^ = depth*^^}°^ beam or girder in inches.
S = span of beam or girder in feet.
^ These short formulae are derived from those given in this volume at the pages
mentioned in connection with each. They have been made shorter by substituting
in some cases feet for inches, cwts. for tons.
APPENDICES 325
Timber Beams, supported at ends and uniformly loaded.
Strength. — For Fir.
The safe distributed dead load in cwts. = J^" inches.
S m feet. * ' ^ '■'
(See Case 7, p. 59.) This assumes the modulus of rupture/,, =6500 lbs., or about
60 cwts., and a factor of safety of 5.
For English Oak and Pitch Pine multiply the result of (I.) byl^.
For Teak multiply by 2, and for Greenheart by 3.
Stiffness.— When a beam supported at the ends is loaded with the dis-
tributed safe weight found by (I.) then (see Equation 45, p. 67)
n . , S^ in feet.
JJenection at centre = — — — ; — • n\ \
•^^ dm. inches. " ' ' • \ •/
For _PVr, Oa\ Teak, Greenheart.
The distributed dead load in cwts. ) • • i
to cause a deflection of per \ = ^^^hes
foot of span. J in feet.
(See Equation 47, p. 68.) For dead load in centre to cause the same deflection,
multiply the result of III. by i.
In II. and III. the modulus of elasticity is taken at 1,210,000 for large scantlings.
Cast-iron Girders supported at ends and uniformly loaded.
The safe distributed dead load in tons = ^ . . . . (IV.)
S
a being the effective area of tension flange in inches— the depth of girder in
inches — S the span in feet.
(See Equation 57, p. 95.) This assumes c = %\ tons and factor of safety = 4.
Wrought iron, rolled, or Plate Girders, supported at ends and uni-
formly loaded.
The safe distributed dead load in tons = , . , (V.)
- (See Equation 54, p. 87.) a being the effective area of compression flange in
inches. The ultimate resistance to compression is taken at 18 tons per square inch,
the factor of safety as 4. The limiting stress per square inch is therefore 4 4 tons.
Open Webbed Girders.
Load is as in (V.) above.
Stress on any bar in web
_ shearing stress at the point x length of bar
depth of girder. ' (^^'^
(See p. 194.)
Retaining Walls.
For water (see p. 235) mean thickness =| of height . (VII.)
For average earth (see p. 244) „ =1 . . (VIII.)
Arches, thickness of, from 10' to 25' span.
Masonry — Block, 1 inch for every foot of span . . . (IX.)
Eubble, 1^ inch „ „ . . . . (X.)
Brick, ^ brick for every 5 feet of span . . (XI.)
^ H. E. Aide Memoire.
2 Seddon.
326
NOTES ON BUILDING CONSTRUCTION.
Water Supply. — Pipes flowing full (see p. 267).
G = Gallons supplied per minute.
H = Head in feet.
L = Length of pipe in feet.
d = diameter of pipe in inches.
For i" pipes take f, for 1" pipes take |, and for pipes of over 12" diameter take {h, of the result
found by XII.
For i pipes take d J larger than as found by XIII.
(XIII.)
(XII.)
a
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TABLES.
327
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328 NO TES ON B UILDING CONS TR UCTION
TABLE Ia.
Safe Eesistanee of Materials.
This Table only includes the materials most generally in use, and the values given
can be depended upon as quite safe for the ordinary quality of materials. ^
Cast iron
Wrought iron
built-up girders
"Wrought iron
rolled joists
Steel, mild .
Cast iron
Wrought iron
built-up girders
Wrought iron
rolled joists
Steel, mild .
Oak, English
Fir, such as Kiga
or Dantzic
Pine, American
yellow
Cast iron
Wrought iron
Steel, mild .
Cast iron
Wrought iron
Cast iron
Wrought iron
1"5 ton per
sq. inch
5 tons do.
6 do., do.
8 do., do.
Safe Resistance to Tension.
Oak, English
Fir, Baltic .
Pine, American
yellow
Pine, pitch .
Teak , Moulmein .
16cwts. per
sq. inch
12 do., do.
3 do., do.
10 do., do.
10 do., do.
Portland cement,
neat, after 9 men.
B'kwork in cement^
Do. in mortar (hy-
draulic lime) 2
Concrete, Portland
cement, 5 to 1
Safe Resistance to Compression.
8 tons per
sq. inch
4 do., do.
5 do., do.
6-5 do., do.
13 cwts., do.
10 do., do.
6 do., do.
Pine, pitch .
Teak, Moulmein
Portland cement,
neat, after 9
months
Mortar, common .
Concrete, lime
,, Portland
cement, 5 to 1
Concrete.Portland
cement, 10 to 1
lOcwts. per
sq. inch
12 do., do.
9 do., do.
O'o cwt. do.
O'o do., do.
2 cwts. do.
1 cwt. do.
Bricks, ordinary
stock
Brickwork in mor-
tar (good) 3
Brickwork in ce-
ment 3
Granite, Aberdeen
Limestone, granu-
lar
Sandstone, ordin-
ary
Masonry, rubble
Safe Resistance to Shearing.
2*4 tons per
sq. inch
5 do., do.
5 '5 do., do.
Fir, such as Riga
or Dantzic, along
the grain
1 '3 cwt. per
sq. inch
Oak
English,
across grain
II 11 along
the grain
I'O cwt., per
sq. inch
0'5 cwt., do.
0-1 do., do.
0-3 do., do.
0"S cwt. per
sq. inch
0-5 do., do.
0-8 do., do,
10 cwts. do.
9 do., da
5 do., do
0-4 cwt. do.
5 cwts. per
sq. inch
2 do., do.
Safe Resistance to Bearing
steel, mild .
10 tons per
sq. inch
5 do., do. I Oak, English
8 tons per il Fir, such as '
sq. inch or Dantzic
25 cwts. do.
3 "5 tons per
sq. inch
4 '5 do'., do.
Safe Modulus of Rupture.
Oak, English
Fir, such as Riga
or Dantzic
16 cwts. per
sq. inch
11 do., do.
Pine, American
yellow
Teak, Moulmein .
1 12 cwts. per
sq. inch
10 cwts. per
sq. inch
20 do., do.
1 For more detailed information see Part III.
2 These figures represent the adhesion of fresh mortar to bricks
The best hydraulic mortar cannot be safely taken as adhering to the best stock bricks after six
months with a greater ultimate force than 36 lbs. per square inch. The adhesion to soft place bricks
IS only about 18 lbs. per square inch.
Cement mortars after a short time exert a tensile and adhesive resistance nearly equal to the
tensile strength of good brick, i.e. about 280 to 300 lbs. per square inch.
3 The resistance of brickwork to cracking or crushing is much less than that of the bricks alone.
Alter allowing three to six months (according to the mortar) for setting, good stock brickwork will
begin cracking at a pressure of 200, 400, or 700 lbs. per square inch, according as it is laid in gray
Chalk lime, Lias lime, or Portland cement mortar. For ultimate crushing from one and a half time
to twice these pressures would be required. {Wray and Seddon.)
TABLES
329
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330
NOTES ON BUILDING CONSTRUCTION
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TABLES
331
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332
NOTES ON BUILDING CONSTRUCTION
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TABLES
I
333
For practical purposes the value of - cau be found with sufficient
accuracy from
I I
- = nr,
K 0
where n and b have the following values for various sections in general use. —
Section.
m
Holloiu
Metal i/w'-b
Solid
ITolloiij
Oi
Metal '/w'^b
3-5
2-5
3-0
4-0
3-1
Section.
Ti
1
ylrea of IVeb = l/i area of
flanges. ^ ^
I
Area o/lVei^'/s "f
flanzes.
6-0
4-9
4-2
3-9
4-2
Section.
I — I
+1
4-9
3-9
3-8
4-3
4-9
Example. — Find the safe stress for an L wrought iron strut, 3" x 3" x i",
10' 0" long, the ends being considered rounded. We have 6 = 3, » = 4-9,
Z=120; hence
I 120
- = 4-9 —-=196.
K 3
From the Table the safe stress per square inch is 0"70 ton nearly. Hence
Safe stress = 0*70 x (3 x + 2l x |) = 1-9 ton nearly.
334 NOTES ON BUILDING CONSTRUCTION
TABLE VI.
Strength of Wooden Struts.
Rounded Ends.
Wood supposed to be average quality fir, such as is used in roof work.
R = ratio of length least dimension of cross section,
rc = safe stress per square inch in ciots.
If both ends are fixed take value of corresponding to ^.
If one end is fixed and the other rounded take mean of values of r
R
corresponding to — and R.
R
R
R
R
5
9-55
19
3-85
33
1-56
54
0-58
6
9-25
20
3-60
34
1-43
66
0-55
7
8-85
21
3-88
36
1-35
58
0-52
8
8-30
22
3-18
36
1-28
60
0-49
9
7-75
23
2-92
37
1-22
62
0-46
10
7-20
24
2-75
38
1-16
64
0-43
11
6-70
25
2-60
39
1-11
66
0-41
12
6-25
26
2-41
40
1-06
68
0-39
13
5-80
27
2-25
42
0-96
70
0-37
14
5-40
28
2-11
44
0-87
72
0-35
15
5-05
29
1-96
46
0-79
74
0-33
16
4-75
30
1-82
48
0-72
76
0-31
17
4-45
31
1-71
50
0-66
78
0-29
18
4-15
32
1-60
52
0-62
80
0-27
Example.— Find, the safe stress that can be applied to a wooden strut
3" X 2", and 10 feet long.
10x12
- 2 — ' '•c = 0-49 ; safe stress = 3 x 2 x 0-49 = 2-9 cwts.
TABLE VIL
Strength of Timber Columns.
The following Table has been deduced by Mr. Stoney, C.E., from a series
of experiments made by Mr. Brereton, C.E., on square columns of American
yellow pine {Pinus strohus).
Ends adjusted as in ordinary practice.
Ratio of length to side of column
10
15
20
25
30
35
40
45
50
Breaking weight in tons per sq.
foot of section
120
118
115
100
90
84
80
77
75
TABLES
335
TABLE VIII.
Strength of Rivets (Iron Rivets in Iron Plates).
Safe resistance to shearing, 4 tons per square inch.
Safe resistance to bearing, 8 tons per square inch.
Diam.
of
rivet.
Resistance to
single, shear —
tons.
Resistance to bearing of one rivet in plates of various
thicknesses — tons.
A"
i"
1"
1"
1"
r
1"
i"
0-78
0-25
1-00
1-50
2-00
2-50
3-00
3-50
4-00
"1
1-23
0-31
1-25
1-87
2-50
3-12
3-75
4-37
5-00
3''
I
1-77
0-37
1-50
2-25
3-00
3-75
4-50
5-25
6-00
111
8
2-40
0-44
1-75
2-62
3-50
4-37
5-25
6-12
7-00
15"
1 «
2-76
0-47
1-87
2-81
3-75
4-69
5-62
6-56
7-50
1"
3-14
0-50
2-00
3-00
4-00
5-00
6-00
7-00
8-00
U"
3-98
0-56
2-25
3-37
4-50
5-62
6-75
7-87
9-00
li"
4-91
0-62
2-50
3-75
5-00
6-25
7-50
8-75
10-00
TABLE IX.1
Dimensions of Eyes of Wrought Iron Tension Bars.
Ratio of area of metal at
side of eye to
Ratio of diameter of
area of bar.
Ratio of maximum
pin to width of bar,
or diameter of bar
thickness of bar to
Hammered
eye.
Hydraulic forged
width. (Pin in
(if round)
(Metal section at
eye.
(Same section
single shear.)
(not to be < 0-67).
back of eye =
= tliat
at back as at sides
of bar.)
of eye.)
0-67
0-66
0-74
0-21
0-75
0-67
0-75
0-25
1-00
0-75
0-75
0-38
1-25
0-76
0-80
0-54
1-33
0-79
0-85
0-61
1-50
0-83
0-93
0-70
1-75
0-84
1-00
0-88
2-00
1
0-88
1-13
1-08
^ This Table is taken from Instruction in Construction (revised edition), and is
tbunded on that given by Mr. Shaler Smith.
336
NOTES ON BUILDING CONSTRUCTION
TABLE X.1
Weight of Angle Iron and Tee Iron.
1 foot ill length.
Sum of the 'Width and Depth in Inches.
Thick-
ness.
2
2i
03
2i
21
2|
Cliches.
lbs.
lbs.
lbs.
lbs.
lbs.
lbs.
lbs.
lbs.
lbs.
lbs.
lbs.
1
•57
•62
•68
•73
•78
•83
•88
-94
-99
1^04
1-09
■i
T<5"
i
•81
-89
•97
1-05
1-13
1^21
1-29
1-37
1-45
1^52
1-60
1-04
1-15
1^25
1-36
1-46
1^56
1-67
1-77
1-88
1^98
2-08
5
TS"
1-24
1-37
1-50
1-63
1-76
1^89
2^02
2-15
2-28
2^41
2-54
3
3*
3i
31
3*
3i
33
3^
4
4i
1
8
1-14
1-20
1-25
1-30
1-35
1^41
1^46
1-51
1-56
1-62
1-72
3
i
r68
1-76
1-84
1-91
1-99
2-07
2^15
2-23
2-30
2-38
2-54
2^19
2-29
2-40
2-50
2-60
2^71
2^81
2-92
3-02
3-13
3-33
5
TS-
2^67
2-80
2-93
3-06
3-19
3-32
3^45
3-58
3-71
3-84
4-10
3-13
3-28
3-44
3-59
3-75
3-91
4-06
4-22
4-38
4-53
4^84
7
TB'
3^57
3-75
3-93
4-11
4-29
4-48
4-66
4'84
5-02
5^20
5-56
4i
4|
5
H
5i
5|
6
6i
6|
7
i
2^70
2-85
3-01
3-16
3-32
3-48
3-63
3-79
3-95
4-10
4-26
3-54
3-75
3-96
4-17
4-38
4-58
4^79
5-00
5-21
5-42
5-63
6
4^36
4-62
4-88
5-14
5-40
5-66
5-92
6-18
6-45
6-71
6-97
5^16
5-47
5-78
6^09
6-41
6-72
7-03
7-34
7-66
7-97
8-28
7
5^92
6-29
6-65
7-02
7^38
7-75
8-11
8-48
8-84
9-21
9-57
i
6-67
7-08
7-50
7-92
8^33
8-75
9^17
9-58
10-00
10-42
10-83
9
TB'
7^38
7-85
8-32
8-79
9^26
9-73
10-20
10-66
11-13
11-60
12^07
n
71
8
8i
8i
8|
9
9i
9i
9|
i
5^83
6-04
6-25
6-46
6^67
6-88
7-08
7^29
7-50
7-71
7-92
5
TB-
7^23
7-49
7-75
8-01
8-27
8^53
8-79
9^05
9-31
9-57
9-83
3
5
8-59
8^91
9-22
9-53
9-84
10-16
10-47
10-78
11-09
11-41
11-72
7
TB'
9^93
10^30 10-66
11-03
11-39
11-76
12-12
12^49
12-85
13-22
13-58
1
2
11^25
11-671 12-08
12-50
12-92
13-33
13-75
14^17
14-58
15-00
15-42
9
12-54
13-01
13-48
13-94
14-41
14-88
15-35
15-82
16-29
16-76
17-23
13-80
14-32
14-84
15-36
15-89
16-41
16-93
17-45
17^97
18-49
19-01
10
11
Hi
12
12i
13
13^
14
14J
15
§
12-03
12-66
13-28
13-91
14-53
7
TB'
13-95
14-67
15-40
16-13
16^86
17-59
18-31
19^04
19^77
20-50
21-22
4
15-83
16-67
17-50
18-33
19^17
20-00
20-84
21^67
22^50
23-34
24-17
9
T¥
17-70
18-63
19-57
20 51
21^44
22-38
23-31
24^25
25^19
26-12
27-06
19-53
20-57 21-61
22-66
23^70
24-74
25-78
26-83
27^87
28-91
29-95
3
4
23-13
24-38
25-63
26-88
28^13
29-37
30-63
31^88
33^13
34-38
35-63
12
124
13
13i
14
15
16
17
18
19
20
23-70
24-74
25-78
26-83
27-87
29-95
32-03
34-12
36-20
38^28
40-36
3
4
28-1329-37
30-63
31-88
33^13
35-63
38-13
40-63
4M3
43-63
46-13
7
8
32-45,33-91
35-36
36-82
38-28
41-19
44-12
47-02
49^95
52-87
55-78
1
36-67,38-33
40-00
41-67
43^33
46-67
50-00
53-33
56^67
60-00
63-33
Note. — When the base or the •web tapers in section, the mean thickness is to be
measured.
^ From D. K. Clark's Rules and Tables.
TABLES
337
TABLE XI.
Web Plates— thickness of.
TahU 1 /or determining the thickness of loeb -plate, suitable for a given
shearing stress per foot depth of girder, at varying widths of plate between
supports (whether L irons of flanges or vertical stiffeners). The Table is
founded on the formula 48<
- - (Unwin's /rort Bridges and Roofs),
where T is the stress in tons ; t the thickness of web plate ; h the unsupported
width of plate in feet.
To use the Tahle.—Find the least unsupported width of plate in the top
row of figures, and underneath it find the number nearest in excess of the
given shearing stress in tons per foot depth of girder. The corresponding
thickness of plate will be that required.
For example, a girder 3 feet deep has a maximum shearing stress of 18
tons = 6 tons per foot of depth : the distance between the L irons is 30 inches
Under the number 30 we find 6-3 tons, and the corresponding thickness is
f , which is to be the thickness at abutments.
Tliick-
ness of
web in
inches.
Net unsupported width of plate in Inches, whether between flange L irons
or vertical stiffeners.
12
15
3-1
5-5
8-0
10-9
14-2
17-4
207
24-1
27-4
18
21
24
27
1- 2
2- 2
3- 5
5-3
7-4
9-8
12-3
15-0
17-9
30
33 36
1-0
1-8
3- 0
4- 5
6-3
8-0
10-8
13-4
16-1
39
42
45
•45
•9
1- 5
2- 3
3- 3
4- 6
6- 0
7- 6
9-5
48
1- 3
2- 0
3- 0
4- 2
5- 4
6- 8
8-6
51
36
7
■2
8
7
8
9
■3
Obviously the Table can be used for determining the thickness of web
plate at any part of the girder, as well as at the abutments, when the shearing
stress is known ; and it is easy to find thicknesses for unsupported widths not
given in the Table by estimation between the widths above and below. The
safe compression shearing stress used in the formula is 4 tons per square inch,
and as the assumed diagonal pillar, 12 inches wide, is not independent but
continuous with the adjacent parts, the stresses given in the Table are well on
the safe side. If, therefore, the given stress is but little over a tabular number
it will be safe to use the corresponding thickness instead of the next greater.
If, for example, the stress is 1 1 tons per foot of depth, and the unsupported
width 30 inches, it will be quite safe to use a web plate |" thick.
TABLE XIL2
"Weight of Roof rraming, Ceilings, etc.
Ceilings.
Lath and plaster ceiling . . . • 8 ) lbs. per square foot hori-
g!j:l^ jo^'^ts 3 I zontal s urface covered.
1 Kindly given to the wiiter by Mr. C. Light.
2 The weights of framing of iron roofs are taken from Unwin's Bridges and Hoofs ;
those for wooden roofs from Instruction in Construction, by Col. Wray, R.E.
B.C. IV. 2
338
NOTES ON BUILDING CONSTRUCTION
Framing, etc., Wooden Roofs.
Common rafters 3 ) 1^^- ^'l^^^^ ^^^^ ^""^
Collar beams • • ^ ■' suriace.
Framing' for wooden roofs, including " purlins and ridge boards, but exclusive
of tie beams —
20 feet span, king post, rise \ span
30 ,, )5 "
40 „ queen post „
50 5, )5 "
60 ,, )) "
Tie beams —
20 feet span, king post
30
40
50
60
queen post
lbs. per square foot of roof
surface.
2
5) "
2
J> "
3
)5 "
4
iy »
11 lbs.
per foot run of tie.
20
j> "
18
)) "
20
)) "
30
J) »
Framing of Iron Roofs.
Trussed Roofs.
Pent .
Common truss
Bowstring Eoofs.
Manchester .
Lime Street
Birmingham
Arched Roofs.
Small Corr. Iron.
)> "
Strasburg Rail
Paris Ex. large
Dublin
Derby
Sydenham .
))
St. Pancras
Clear
Span.
Distance
apart of
Prin-
cipals.
Feet.
15
37
40
54
55
72
84
50
100
130
140
50
154
211
40
60
97
153
41
8lh
120
72
240
Of Pur-
lins, etc.
Feet.
5
12
14
20
9
10
14
26
12
11
26
24
13
26
16
24
Weight per square foot of covered
area in lbs.
Of Prin-
cipal.
11
2-0
6-5
4-6
4-2
2-6
0-8
291
9-5
3-4
10-8
7- 9
8- 4
7-4
3-5
3-5
3-0
7-0
2-8
5-9
5-6
4-5
4-9
7-3
5- 5
7-3
6- 0
3-9
2-9
17-1
Total
Iron-
work.
3- 5
4- 6
5- 5
9-5
11-6
7- 0
8- 5
3-0
7-0
6- 4
9-6
11-0
12-0
15- 0
10- 7
16- 8
11- 8
11-3
24-5
Total
with
covering.
6-9
5-2
9-0
8-0
2- 5
3- 5
TABLES
339
TABLE XIII.i
Weight of Roof Coverings.
Lead covering, including laps, but not boarding or rolls
Zinc covering „ 14 to 16 zinc gauge
Corrugated iron, galvanised, 1 6 W.G. .
18W.G. .
20 W.G. .
Sheet iron, 16 W.G.
„ „ 20 W.G. ...
Slating laid with a 3-inch lap including nails, but not
battens or iron laths —
Slates, Doubles, 13 inches x 9 inches, at 18 cwts. per
1200 . .
„ Ladies, 16 inches x 8 inches, at 31-5 cwts, per
1200
„ Countesses, 20 inches x 10 inches, at 50 cwts. per
1200
„ Duchesses, 24 inches x 12 inches, at 77 cwts. per
1200 .......
Tiles, plain, 1 1 inches x 7 inches, laid with a 3-inch lap
and pointed with mortar, including laths and
absorbed rain ......
„ pan, 13|- inches x 9 J inches, laid with a 3-inch lap
and pointed with mortar, including laths and
absorbed rain ......
„ Italian (ridge and furrow), not including the board-
ing, but including mortar and absorbed rain .
Slate battens, Z\ inches x 1 inch —
For Doubles
For Countesses
Boarding, | inch thick .
)5 1 ))
Wrought Iron Laths, angle irons-
For Duchess Slates
For Countess „
Cast Iron plates, | inch thick
Thatch, including battens
Per square foot.
h\ lbs. to 8| lbs.
to If „
3i
2
2i
8l
°2
12
14
2
44
6J
TABLE XIV.
Wind Pressure.
Wind pressure normal to a roof- surface from Hutton's formula, viz.
PN-P(sin i)i-84cosi-i^ taking P=50 lbs. per square foot.
Pitch of Roof (i)
l6°
15°
20°
21° for
\ span
25°
26°Jor
\ span
30°
33° J or
J span
35°
40°
45°
50°
Normal wind pres-
sure in lbs. per
8q. foot (Pn)
12-1
18
22-6
25-2
28 -8
30-2
33-0
36-6
37-8
41 '6
43
47-6
^ This table is taken from Instruction in Construction, by Col. Wray, R.E.
340
NOTES ON BUILDING CONSTRUCTION
TABLE XV.
Scantlings for "Wooden Roofs.
The roofs are supposed to be of Baltic fir covered with Countess slates laid on
inch boards ; the maximum horizontal wind force is taken at 45 lbs. per foot super
acting only on one side of the roof at a time, equivalent to a normal wind pressure of
30 lbs. per square foot for a pitch of 30°, and 40 lbs. per square foot for a pitch of 45".
The common rafters to be 1 ft. from centre to centre, but in sheltered positions
they may be placed 1 ft. apart in the clear.
A EooFS WITHOUT CEILINGS. —Pitch up to 30°.
Nature of
Roof.
Span in
feet.
Common Rafters.
Collar.
In tixing
the collar
to the
rafter the
latter
should
not be cut
into.
Remarks.
"Walls cap-
able of
resisting
thrust.*
Collar
placed
J- way up.
Walls not
capable of
resisting
thrust, t
Collar
placed \-
way up.
* The following solid walls are strong enough, when
built in Lias lime mortar 1 to 2, to resist the thrust ol
roof; allowance must be made for door and wimlow
openings. Height to be taken from level of the tlooi
below roof.
Stone
Walls.
Brick Walls.
Couple.
8
10
12
Bdth.Depth.
2" X 3"
2"X3J"
2"x4"
16" thick,
not over
15 ft. high.
o e -irx ( 9" thick, not over 7' high
Span of roof 10 -j ' °
Span of roof 18' 1^^^' " " jq' "
t When the walls are not capable of resisting the tlrnis',
of the roof, place the collar low down ; but if th(
collar is required half-way up, the scantlings musi
be increased as follows :—
Rafters, add one-fourth to both breadth and depth
Collar, add J" to depth ; but it would be better to us(
the scantlings for walls capable of taking the thrust
and make some arrangement to prevent the walls fi on
spreading, such as tying the wall plates together ai
intervals.
Collar
Beam.
8
10
12
14
16
18
irx2i"
irx2r
li"x2i"
l|"x3"
2" X 3J"
2"X3|"
2"x3J"
2"x4"
2"x4j"
2J"x5"
2i"x5i"
2i"x6"
2"x2J"
2"x2A"
2"x2|"
2"x3"
2"x3i"
2" X 4"
King
Post.
Trusses
10' centre
to centre.
Tie Beam.{
Depth in-
cludes 3"
for joints.
Principal
Rafters.
King
Post.
Struts.
Straining
beam. .
Purlins.f
10 ft. bear-
ing.
Common
Rafters, j
20
22
24
26
28
30
3"x4i"
3"x4|"
3i"x4r
3i" X 4|"
4"X4}"
4"X4J"
3"x5"
3" X 5i"
3J"x5i"
3i" x 5f "
4"x5^"
4"X6"
3"x2i"
3"x2f"
3i"x2|"
3f' x2|"
4" X 2f "
4"x2}"
3" X 3"
3" X SJ"
SV'xSi"
i3|"x4"
4"x4" '
4"x4i"
5"x7i"
5"x7|"
5" x 8"
5" X 8i"
5"x8J"
6" X 8|"
2"x3i"
2"x3^"
2"x4"
2"x45"
2"xU"
2"x4f'
Queen
Post.
Trusses
10' centre
to centre.
32
34
36
38
40
42
44
46
4 J" X 4r
4i"x4|"
4f" X 4|"
4|"x4i"
4|"X6"
5" X 5"
5"x5i"
5i"x5i"
41"X4|"
4!"X5"
4|"x5"
4J"x5i"
4|" X bV
5" X bl"
5"x5|"
5J"X5J"
Queen
Posts.
4*"X2|"
4|"x2i"
4J"x2i"
4|"x2i"
4|"x2f"
5"x2|"
5"x2r
5i"x2r
4l"x2h"
4l"x2i"
4|" X 3"
4J"x3i"
4f"x3J"
5" X 3J"
5"X3J"
5i"x3r
4J"x5i"
4i"x6"
4|"X6J"
4J"x6|"
4|"x7i"
5"x7i"
5"x8"
5i"x8i"
5"x7r
5"x7f'
5"x8"
5"x8"
5"X8J"
5"x8J"
5"x8i"
5"xS|"
2"x?A"
2"x3-l"
2" X 4"
2" X 4"
2"x4i''
2"x4L"
2"x4.';''
2" X 5"
TABLES
341
TABLE ILY.— Continued.
B Roofs with Ceilings.— Pitch up to 30°.
Nature of
Roof.
Span
in feet.
Common Rafters.
Collar.*
* In fixing the collar to the rafter, the latter should
not be cut into. As regards walls capable or not cap-
able of resisting the thrust of roof see remarks under A.
Collar
Beam.
8 to IS.
Add J" to depths given
in A.
Add f" to
depths
given in
A.
King
Post.
Trusses
10' centre
to centre.
Tie-beam. J
Principal
Rafters.
King
Post.
Struts.
Straining
beam.
Purlins.§
10 ft. bearing.
Common
Rafters.
20
22
24
26
28
30
4" X 7"
4"x7r
4i"x8"
4J" X 8i"
4^X9"
4J"x9r
4" X 4"
4"x4r
4i" X 4J"
il" X 5"
4J" X 5J"
4|" X 5|"
4" X 3 '
4" X 3"
4J" X 3"
4J" X 3"
4J" X 3"
4J" X 3"
4" X 2i"
4"x3"
4i" X 3"
4^" X 3"
4i"x3r
4|"X3|"
5" X 7*"
5" X 7|"
5"x8"
5" X 8J"
5" X 8|"
5" X 8|"
2" X 3J"
2" X 3i"
2" X 4"
2"X4J"
2" X 44"
2" X 41"
Queen
Post.
Trusses
10' centre
to centre.
32
34
36
38
40
42
44
46
4rx7i''
4|" X 7|"
4|"x8i"
5"x8J"
5"x9"
5i"x9"
5rx9J"
5i" X 10"
4|" X 51"
4J"x5|"
4rx6i"
6"x6"
5" x 6J"
s^xer
6i"x6r
6rx7i"
Queen
Posts.
4|"x3"
4i" X 3"
4J"x3"
5" X 3"
5"x3r
5rx3r
5i"x3r
5i"x4"
4|"x2r
4f" X 2|"
4|" X 3"
5"x3"
5"X3J"
5J" X 3"
5i"x3i"
5i"x3|"
4|"x6r
4|"x7i"
4|"x8i"
5"x8i"
5"X9"
5J"X9"
5i"x9J"
5J"xlO"
5"x7r
5" X 7J"
5" X 8"
5" X 8"
5" X 8i"
5" X 8*"
5" X 8|"
5" X 8}"
2"x3i"
2" X 3|"
2" X 4"
2" X 4"
2" X 4i"
2"x4i"
2"x4r
2"x5"
For Roofs of 45° Pitch.— Add 1" to the depth of common rafters, purlins, and
struts, and I" to the depth of the principal rafters, as given in A and B.
t The joint of the tie beam with the principal rafter should be placed immediately
over the supporting wall. If this cannot be conveniently done, the depth of the
tie beam should be increased one or two inches.
§ If the purlins, instead of being placed immediately over the joints, are placed
at intervals along the principal rafter, increase the depth of the latter, given in the
Tables, as follows :
^4. f r '>^ii'^out ceiling 2", without ceiling U",
King post roof I ^ .^^ Queen post roof | ^ .^^ ^ |;
The purlins if placed 2 feet apart and with 10 feet bearing may be made 3"x 6".
The scantlings of the principal rafters, struts, and straining beam can be slightly
modiiied by means of the following rough rule : " For every J" deducted from the lesser
dimension of the scantling, add 4" to the other dimension, and vice versd." For the
tie beam, purlins, and common rafters, so long as the depth is about double the
breadth, I" deducted from the breadth requires J" to be added to the depth.
This Table is derived from the War Office practice.
TABLE XVa.
CoeflS-cients of Friction.
4> = Angle of Repose. /= tan (p = coefficient of friction.
9
/
/
Masonry and brickwork (dry)
,, with wet mortar
,, with slightly damp
mortar
,, on dry clay
„ on moist clay
31° to 35°
25^°
27°
isi°
•6 to -7
•47
•74
•51
■33
Wood on stone
Iron OD stone
Wood on wood (dry)
Metals on metals (dry)
Smoothest and best greased
surfaces
22°
35° to 163°
14° to 26J°
8^° to lli°
If to 2°
•4
•7 to -3
•25 to -5
•15 to -2
•03 to -036
^ Rankine's values.
NOTES ON BUILDING CONSTRUCTION
TABLE XVI.
Retaining Walls.^
Angle of repose (<^) of various earths, and value of K = 0-7 tan
Description of Earth.
<P
K.
Description of Earth.
<P
K.
Fine dry sand .
Sand, wet .
,, very wet .
Vegetable earth, dry .
„ „ moist
,, ,, very wet
„ „ consoli-
dated and dry
1 —
37 to 31
26
32
29
45 to 49
17
49
0-35 to 0-40
0-44
0-39
0-41
0-29 to 0-26
0-52
0-26
Loamy earth, consoli-
dated and dry
Clay, dry .
, , damp, well
drained .
,, wet .
Gravel, clean
,, with sand
Loose shingle
40°
29
45
16
48
26
39
0-33
0-41
0-29
0-53
0-27
0-44
0-33
TABLE XVII.
Weights of Earths, Stone, etc.^
Pounds
avoir-
dupois
per
cubic
foot.
Pounds
avoir-
dupois
per
cubic
foot.
Basalt ....
180
Concrete, lime .
118
Bathstone
123
Earth, vegetable
90
Brick, common stock
115
„ loamy .
80-100
„ red facing
130
„ semi-fluid
110
„ fire . . .
150
Granite, Aberdeen .
164
Brickwork, in mortar
110
Gravel, Thames
112
,, in cement
112
Limestone, lias
156
Cement, Portland
87
Lime, ordinary quick, stone
53
112
Masonry, rubble
140
Chalk, solid . . -j
to
,, ashlar, Portland .
150
175
,, ,, Granite .
160
Clay, with gravel
130
Mortar, new .
110
„ ordinary
120
Portland stone .
145
137
Sandstone, Craigleith
145
Concrete, Portland cement |
to
Slate, Welsh .
181
142
Sand ....
119
^ Prof. Unwin's " Railway Construction " as regards the value of (p.
From R. E. Aide Memoire.
TABLES
343
TABLE XVIIa.
Table of Thickness* required for Arches, Semicircular
Arches, and Arches of 120^
Ordinary vaul
Semicircular.
s, bridges, etc.
Arches of 120°.
Remarks.
Block
Stone.
Brick.
Rubble
Stone.
Block
Stone.
Brick.
Rubble
Stone.
Span of 5 feet and less
/ //
0 8
1
0 10
0 9
1
0 lOi
6
0 9
1
0 11
0 10
1
1 o"
8
0 10
1
1 0
0 11
14
1 14
The ai'ches are
10
0 11
14
1 14
1 0
14
1 3
1 0
li
1 3
1 1
14
1 44
assumed to be
14 „
1 1
14
1 4
1 2
14
1 6
built in well-
16
1 2
14
1 5
1 3
2
1 7
made hydi'aulic
18
1 3
2
1 6
1 4
2
1 8
lime mortar.
20
1 4
2
1 7
1 5
2
1 9
„ 22
1 5
2
1 8
1 6
2
1 104
24
1 6
2
1 9
1 7
24
2 0
^ War Department practice ( Wray).
TABLE XVIIL
HYDRAULICS.
Flow of Water in Pipes running full (gallons per min.)
Calculated from Darcy's formula.
Value of the
expression
Correspond-
ing diameter
of pipe in
inches.
Internal
diameter of
pipe allowing
for incrusta-
tion, t
Value of the expression
Correspond-
ing diameter
of pipe in
inches.
Internal
diameter of
pipe allowing
for incrusta-
tion, t
1
0-32
3
8
34,600
60,000
2-14
24
5
0-48
1
2
2-36
18
0-54
Is
92,000
2-57
3
48
0-64
3
4
140,000
279
117
0-75
'Is
208,000
3-00
250
0-86
1
296,000
3-21
370
0-96
ru
420,000
3-43
4
810
1-07
H
760,000
3-86
4^.
1,410
1-18
1,340,000
4-29
5
2,260
1-29
u
2,150,000
4-71
5V.
3,400
5,020
1-39
ris
3,400,000
7,500,000
5-14
6
1-50
If
6-00
7
7,400
1-61
iVs
16,500,000
7-00
8
10,200
1-71
2
32,800,000
8-00
9
20,800
1-93
60,200,000
9-00
10
* L = length of pipe ; H = available head ; G = discharge in gallons per minute,
t The allowance made for incrustation is :
Jth of diameter for pipes under 6 inches in diameter,
1" for pipes over 6 inches diameter.
The pipes that are not market sizes are printed in italics.
344 NOTES ON BUILDING CONSTRUCTION
Example. — The length of a pipe is 500 feet, the available head is 30 feet,
and the discharge is to be 600 gallons per minute. What ought the
diameter to be, allowing for incrustation ?
We have -.02 = ^.6002=6,000,000.
H 30
A 7" pipe would therefore be required.
The discharge from this pipe when new will be
16,500,000 X 30
-^^^ = 995 gallons per minute.
And when incrusted
7,500,000 X 30
= 670 gallons per minute.
oOO
TABLE XIX.
HYDRAULICS.
riow of Water in Pipes running full (cubic feet per min.)
Calculated from Darcy's formula.
Value of the
expression
H
Corresponding
diameter of pipe
in inches.
Value of the
expression
L p2 *
H
Corresponding
diameter of pipe
in inches.
6'6
4
14,380
18
17-6
5
18,800
19
53
6
24,900
20
115
7
31,500
21
227
8
39,600
22
427
9
49,400
23
730
10
61,500
24
1,190
11
75,700
25
1,850
12
91,700
26
2,740
13
110,600
27
4,020
133,400
28
5,740
15
158,300
29
7,880
16
188,000
30
10,650
17
* L = length of pipe ; H = available head ; F = discharge in cubic feet per second.
No allowance is made for incrustation.
The pipes that are not market sizes are printed in italics.
This Table is used in the same way as Table XVIII.
TABLES
345
TABLE XX.i
Cast Iron Pipes— Thickness, Weight, and Strength.
Inside
Diameter.
Thickness
forrutila.,
Nearest
thickness
in six-
teenths of
Net weight
per foot
run for
thickness
in Column
Length of
pipe
equal in
weight to
the
Weight of a 9-
feet length of
pipe.
Burstin*^
pressure
per sq.
inch
reckoned
jc acTjor 01
safety for
Normal pres-
sure of 300
ft. of water
an inch.
3.
socket.
on Column
or 183 lbs.
3.
per sq. inch.
Whole
Inches.
Inches.
Six-
teenths.
lbs.
feet.
cwts.
lbs.
Times.
2
•31
7^09
•60
(6 feet) ^418
4900
36
■33
i
10-6
•61
(6 feet) -625
3920
30
3
■35
S
12^4
•62
1^06
3920
30
4
■375
i
161
•62
1^38
2940
22
5
•41
23-4
•63
2^01
2744
21
6
■45
27^7
•63
2^38
2290
17
7
■47
36^8
•64
3^17
2240
17
8
•50
\
41^7
•64
3^59
1960
15
9
•53
52^8
•65
4^55
1960
15
10
•56
58^3
•66
5-03
1764
13
11
•59
9
63^9
•66
5^51
1604
12
12
■62
§
77-5
•67
6^69
1633
12
13
•65
83^6
•67
7^22
1508
11
14
•70
11
T¥
99^1
•68
8 ■56
1540
12
15
•71
11
TS-
105^9
cwts.
•68
9^15
1440
11
16
•75
3
1^10
• .69
10^66
1470
11
18
•81
1 3
llT
1^43
•70
13-87
1415
10-6
20
•87
7
8
1^60
•71
15^54
1372
10-3
21
■90
7
8
1-68
•72
16-33
1307
10
24
•99
1
1^91
•73
18-58
1307
10
27
1^09
2-61
•75
25-45
1234
9-3
30
1^18
lA
3-24
•77
31-65
1241
9-3
33
1-27
U
3^75
•78
36-67
1190
8-9
36
1^36
li
4^50
•80
44-10
1198
8-9
39
1^45
5^11
•82
50 -IS
1156
8-7
42
1-55
5^98
•83
58-78
1167
8-8
45
r65
If
6'67
•85
65-70
1133
8-5
48
1-74
If
7^63
•87
75-31
1143
8-6
Note, to Table — Flanges. — The additional weight for a pair of flanges is reckoned
as equivalent to that of a lineal foot of pipe ; equal to 11 per cent extra for 9-feet
lengths.
1 From D. K. Clark's Rides and Tables.
^ The thicknesses in this column are obtained from the following formula deduced
by Mr. Clark from Mr. Bateman's practice —
^~ "^ + 9600'
where t = thickness of the pipe in inches,
H = the head of pressure in feet of water,
£Z=the inside diameter of the pipe in inches.
346
NOTES ON BUILDING CONSTRUCTION
TABLE XXI.
HYDEAULICS.
Loss of Head due to Bends.
When mean velocity of flow is 1 foot per second. For other velocities,
multiply by the square of the velocity.
Radius of Bend
Diameter of Pipe'
Loss of Head for each Degree of
Change of Direction.
1
0-000025
1-25
0-000018
1-5
0-000015
2-0
0-000013
3
0-000011
4
0-000011
5
0-000011
Example. — Find the loss of head caused by a bend of 20°, of 2 inches
radius in a 1" pipe ; velocity of flow, 4 feet per second.
Radius of bend 2
: = T- Hence loss of head = 0-000,013 x 20 x 41
Diameter oi pipe 1
TABLE XXII.i
HYDRAULICS.
Loss of Head due to Elbows.
= eV^, where e can be found from the following Table.
A
10°
20°
30°
40°
50°
60°
70°
80°
90°
e
0-0001
0-0005
0-0011
0-0022
0-0036
0-0056
0-0083
-0115
•0152
^ From Hurst's Pocket Book.
TABLES
347
TABLE XXIII.
HYDRAULICS.
Pipes flowing partially full.
Relation betvreen wetted perimeter, P=^.D, and cross section of flow.
"^"^(mo ~ ^~^r) ' ^^^^^ ^ angle subtended at the centre by the wetted
perimeter, and D is the diameter of the pipe (see p. 282).
p
A
P
A
P
A
P
A
P
A
D
U
Da
D
Da
D
Da
D
D2
0-1
0-00025
0-7
0-052
1-3
0-260
1-9
0-551
2-6
0-745
0-2
0-00137
0-8
0-075
1-4
0-307
2-0
0-594
2-6
0-760
0-3
0-0064
0-9
0-103
1-5
0-357
2-1
0-633
2-7
0-771
0-4
0-0105
1-0
0-136
1-6
0-406
2-2
0-669
2-8
0-779
0-5
0-020
1-1
0-174
1-7
0-456
2-3
0-699
2-9
0-782
0-6
0-034
1-2
0-203
1-8
0-505
2-4
0-725
3-0
0-785
TABLE XXIV.
HYDRAULICS.
Jets— Factors to find initial velocity.
H*
Factor
d
(J).
300
0-98
600
0-95
1000
0-92
1500
0-89
1800
0-84
2800
0-77
3500
0-71
4500
0-50
* H = head at nozzle, diameter of nozzle.
348 NOTES ON BUILDING CONSTRUCTION
NOTATION AND WORKING STRESSES.
TABLE XXV.
NOTATION.
The following is a list of the notation employed in this volume : —
General.
r-ft, safe resistance to bearing in lbs., cwts., or tons per square inch,
rc, „ „ compression „ „ „
rg, „ „ shearing „ „ „
Tt, „ „ tension „ „ „
E, raodiilus of elasticity.
Rj or Rj, total safe resistance to tension.
Ec, „ „ compression.
Es, „ „ shearing.
Rb, » " bearing.
Beams.
breadth in inches.
depth „
/o, modulus of rupture.
I, moment of inertia.
L, span in feet.
„ inches.
M, bending moment generally.
Mc, „ „ at centre of beam.
Mp, „ „ at any point P.
M) moment of resistance generally.
moment of resistance at point P.
R^, reaction at abutment A.
ro, limiting stress in lbs., cwts., or tons per square inch on the outside
fibres of a beam.
Sp, shearing stress at point P.
W, concentrated weight or load on a beam, also total distributed load,
distributed load per inch or per foot run.
distance of neutral axis from extreme fibre on compression side,
i/o, distance of neutral axis from extreme fibre on compression side, sub-
ject to stress Tq.
lit, distance of neutral axis from extreme fibre on tension side.
A, maximum deflection in inches.
Tension and compression bars.
A, effective cross-sectional area.
d, " least diameter " of a long column.
Riveted joints.
b, breadth of the bars or plates to be jointed.
d, diameter of rivet holes.
yfc, number of rivets in the first row.
TABLES
349
N, number of plates on. each side to be connected.
n^, number of rivets in each end group.
ft2> ), ii between the joints.
ftg, number of rivet holes to be deducted from the cross section of cover
plates.
Tft, total safe resistance to bearing.
Ts, ,, ,, shearing.
<, thickness of plates.
fg, ,, „ cover plates.
a, number of riTOts required for bearing.
„ „ shearing.
Plate girders.
A(,, effective area of compression flange.
Aj, „ tension „
D, „ depth, i.e. distance between centres of gravity of flanges.
tyj, thickness of web.
Braced or framed girders.
0^,5, compression in the bar a^,a^ of the top boom.
d, depth, i.e. distance between centres of gravity of flanges.
^4,14, stress (tension or compression) in brace a^,a.^^.
n, number of triangles.
T^3_^^, tension in the bar a^2,a^^ of the lower boom.
Roofs.
H^, horizontal reaction at abutment A due to wind.
P.
P.
K
A
B
B
horizontal wind pressure,
normal „
vertical „
reaction at abutment A.
stress (tension or compression) in member AD.
vertical reaction at abutment A due to wind.
B
temperature,
wind.
temperature.
Masonry structures.
p, intensity of pressure at any point of a bed joint.
'P(max)i greatest intensity of compression on a bed joint.
Tq see Beams.
Retaining walls.
H, height of wall.
K, constant depending on <^ (angle of repose).
T, thickness of wall with vertical sides.
Tj, mean thickness.
W, weight of a cubic foot of the material of the wall.
^, angle of repose of the earth to be retained.
w, weight of a cubic foot of the earth to be retained.
350
NOTES ON BUILDING CONSTRUCTION
Arches.
H, horizontal thrust.
P^jPg, etc., total pressure on joints between voussoirs. N
V, vertical component of pressures.
w-^^w^, etc., loads on each voussoir.
W, total load on an arch.
Hydraulics.
A, area of cross section of flow.
D, diameter of a pipe in feet.
d, „ „ inches.
F, discharge in cubic feet per second.
G, discharge in gallons per minute.
H, effective head of water in feed, i.e. the actual head reduced by the
various losses.
Hb, loss of head due to bends.
H^, „ „ elbows.
H„, „ „ orifice of entry.
H^, „ _ „ velocity.
J, factor for jets.
L, length of a pipe in feet.
R, hydraulic mean depth.
S, slope of a pipe.
V, mean velocity of flow in feet per second.
TABLE XXVI.
WORKING STEESSES.
The undermentioned are the working stresses that have been used in this
volume : —
Cast Iron.
Girders
Wrought Iron.
Rolled beams.
Built-up girders
Roofs .
Tension tars and rods
Riveted joints
Tension l| tons per square inch (see p. 94).
Compression 8
Shearing 2-4
(
Tension 5 tons per square inch (see p. 82).
Compression 4
Shearing 4
Tension 5 „ „ (see p.
Compression 4 „ » ( »
Tension 5 „ ,, (
Compression (depends on ratio of ( ,,
length of struts to least
diameter)
Tension 5 tons per square inch ( „
Shearing 4 „ „ ( „
Bearing 8 „ » ( »
).
).
156).
156).
108).
119).
106).
123).
» )•
Pin joints
Screws .
Timber.
Fir (superior)
Fir (inferior)
Oak
Brickwork.
In Mortar
In Cement
TABLES 351
Shearing 4 tons per square inch (see p. 140).
Bearing 5 „ „ ( „ 141).
Shearing 2 „ „ ( „ 146).
Tension 12 cwts. per square inch
(see p.
150).
Compression 10
53
35
,3 )■
Shearing I'S
J>
53
55 ).
Bearing 1 2
»
53
35
,5 ).
Tension 10
3>
5J
55
152).
Compression 7
)>
33
33
„ ).
Shearing 1-3
5>
)3
33
5, ).
Bearing 7
5!
33
33
,5 ).
Tension 1 6
55
33
33
328).
Compression 13
))
53
35
55 )•
Shearing 5
J)
35
35
33 )•
Bearing 2 5
S)
53
33
,5 )•
Compression 0*5 cwts. per square inch (see p. 223).
Adhesion 0-06
?>
35
(
55
227).
Compression 0"8
33
(
33
259).
Adhesion 0'5
55
35
(
55
5, ).
352
NOTES ON BUILDING CONSTRUCTION
NOTES ON BUILDING CONSTRUCTION
INDEX
A
Abutment for arch, 259.
Adams, reference to, 100.
Anderson's formula for the weight
of girders, App. XV.
Angle irons, resistance of. Table I.,
weight of. Table X.
Angle of repose of earths. Table
XVI.
Annulus, moment of inertia, App.
XIV.
Appendices : {see List of, p. xv.)
Area, effective, of tension bar, 106.
Area of resistance, equivalent, 47 ;
safe, 48 ; mechanical method of
finding, for timber beam, 48 ; for
rolled iron beam, 83 ; for cast iron
beam, 97.
Arches, 246 ; abutment for, 259 ;
example of brick arch, 254 ; graphic
method of determining stability
of, 248, 251 ; do. of drawing line
of least resistance for symmetrical
load, 251 ; do. for unsymmetrical
load, App. XX. ; line of least
resistance, 248.
Arch ring, thickness of, 253, Table
XVIIa. ; table of thickness, 259.
Axis, neutral, 42 ; note on, App.
IV.
B
Baker's practical rales for retaining
walls, 244.
Bars {see Compression bars, Tension
bars) ; wrought iron, resistance of.
Table I.
Basalt, weight of, Table XVII.
Bath-stone, weight of. Table XVII.
Beams, 24 ; bending moment, 25 ;
cantilevers, 28 ; cast iron, 94 ;
comparison of strength, stiffness,
and of fixed and supported, App.
VIII. ; continuous, 74, App. XI. ;
deflection, 66; fixed, 69; moment
of resistance, 40 ; reactions at
supports, 1 8 ; strength of rectangu-
lar beams, 54; rolled iron, 81;
rolled steel, 92 ; shearing stress,
55 ; strength and stiffness, com-
parison between, 66, App. VIII. ;
supported, 32 ; timber beams,
79 ; trussed, 195 ; uniform
strength, 62. {See above headings
in Index.)
Bearing stress, 7 ; resistance of rivets
to, 123, Table VIII. ; resistance
of pins to, 141.
Beech, ultimate resistance, moduli, and
weight, Table I.
Bending moment, 25 ; graphic method
of finding, App. VI. ; various cases
for cantilevers and supported beams,
28-40 ; for continuous beams,
App. XI. ; and for fixed beams,
App. VIII.
Bends, loss of head due to, 270, Table
XXI.
Block structures, uncemented, 220.
Booms of girders, 154 {see Flanges).
Box girder, 170 {see Plate girder) ; ap-
proximate weight, 171 ; deflection,
171 ; diaphragm, 176 ; effective
span, 170; end pillars, 176; flanges,
172 ; section of, 175, Plate B.
Bow's notation, 184, 194, App. XVII.
Bowstring girder, 194.
Braced frameSjperfect, 182 ; overbraced,
181.
356
INDEX
Braced girders, 187 ; application of
load, 188 ; different forms
of, 187 ; dimensions of bars,
193 ; graphic method for,
194 ; lattice 187, 194 ; N,
188; N (braced), 188; prac-
tical formula for, 193; pro-
portion between span and
depth, 188; Whipple -
Murphy, 188 ; stresses in,
188, 190.
„ structures, 177 ; different forms
of, 180.
Breaking load, 5.
„ stress, 8, 49 (see Ultimate
stress).
Brick chimney, calculations for, 225.
„ pier, calcidations for, 222 ; arch,
254.
Bricks, brickwork, coefficient of friction.
Table XVa. ; safe resistance of,
Tablela. ; weightof, Table XVII. ;
Brickwork structures (see Masonry).
Built-up X beams, 93.
C
Cantilever, cast iron, 101 ; timber,
76-79 ; wrought iron, 22.
Cast iron, resistance of, 94 ; moduli,
weight, iiltimate resistance,
Table I. ; safe resistance,
94, Table la., XXVI.
,, cantilevers, example, 101.
„ columns, examples of, IIG,
117.
,, girders, 94, 103 ; approximate
formulse for, 95 ; camber
for, 100 ; casting, 100 ;
centre of gravity of, App.
XII., XIII. ; depth, 100 ;
example of, 103 ; flanges,
99 ; graphic method for
finding section, 96 ; mathe-
matical method, 99 ; mo-
ment of inertia for Apja.
XIV. ; practical points,
99 ; resistance of, 94 ;
stress areas, 97 ; uniform
strength, 100 ; web, 100.
„ pipes, thickness of, Table XX.
Cement, Portland, safe resistance of.
Table la.; weight of. Table XVII.
Centre of gravity ; calculation for
cross section of girder, Apji.
XII. ; graphic method of
finding, for parallel forces,
App. VI. ; graphic method
for cross section, App.
XIII. ; mechanical method
of determining, 86.
„ pressure, 218, App. XIX.
Chimney, centre of pressure, App. XIX. ;
example of, 225.
Circle, moment of inertia for, App.
XIV.
Circular arc, substituted for parabola, 3 4 .
Classification of stresses, 8.
Clay, angle of repose. Table XVI. ;
weight. Table XVII. ; coefficient
of friction. Table XVa.
Clerk-Maxwell (see Maxwell).
Coefficients of friction. Table XVa.
Columns, cast iron, exanqjles of, 116,
117 ; timber, strength of. Table
VII.
Comi^arative effect of dead and live
loads, 5.
Compression, 6 ; resistance to, of cast
iron,94 ;wrought iron,
82 ; steel, 92 ; other
materials. Table I., la.
bars, 105, 109 ; ex-
amples, 115, 116-
121; Euler's formula,
110 ; Fidler's views,
110 ; Gordon's for-
mula, 113 ; long and
short. 111 ; Eitter's
and Rouleaux's form-
ula, 114 ; securing
ends of, 112 ; strengtli
of. Tables V-, VI. ;
Table for wooden, 115.
„ riveted joints, 136.
Concentrated load, 3.
Concrete, safe resistance of. Table la. ;
weight of. Table XVII.
Conditions of equilibrium, 14.
Connections, 122.
Continuous beams or girders, 74,
App. XI. ; example, App. XI.
INDEX
357
Contra-flexure, points of, 70-75, App.
XL
Cotterjointjformulseand example, 148.
Couple, 43.
Cover,plates, 124, 125, 128, 131, 132.
Covering (see Eoof).
D
Darcy's formula, pipes flowing full,
267 ; pipes partially full, 281.
Dead load, 4.
,, and live loads, comparative effect
of, 5.
Deflection, 12, 66 ; amount allowed,
69 ; of beams and cantilevers, 66 ;
of beams of fir, Table III. ; of
continuous beams, Ai)p. XI. ; of
fixed beams, Ajsp. VIII. ; of rect-
angular beams, A]3p. IX. ; of
rolled beams. Table IV. ; formulae,
66, 68 ; of supported beams, 66,
App. VIII.
Depth of girders, plate, 154; braced,
188 ; bowstring, 194 ; cast-iron,
100 ; rolled, 91.
Diagrams, Bow's system of lettering,
App. XVII.; Clerk - Maxwell's,
182 ; method of drawing, 183 ;
riiles for drawing, App. XVI. ;
for various roof trusses, 217; for
Warren and N girders, 194 ; for
trussed beams, 196, 198.
Dimensions for riveted joints, 136.
Distributed load, 4 ; compared with
concentrated load, App. VII.
Double-cover grouped joint, 131 ;
for plate girders, 132.
Drain pipes, 283 ; egg-shaped sewer,
286.
Drains, example, 283.
E
Earths, angle of repose. Table XVI. ;
weight of. Table XVII.
Effective area of tension bars, 106 ;
effective depth and span of girder,
154.
Egg-shaped sewer, example, 287.
Elastic limit, 10 ; of iron and steel.
Table I.
Elasticity, 10; modulus of, 12;
modulus for various materials,
Table I.
Elbows, loss of head due to, 270,
Table XXII.
Elm, ultimate resistance, moduli, and
weight. Table I.
Elongation, 11.
Enclosure wall, calculation for, 227.
Equilibrium, 13 ; conditions of, 14 ;
note on, App. II.
Equivalent area of resistance, 47.
Euler's theory, for compression bars,
110.
Examples (.see List, p. xvii.)
External forces acting on a structure,
15.
Eyes of wrought -iron tension bars,
139, Table IX.
Eytelwein's formula, flow of liquids in
pipes flowing full, 264.
F
Factor of safety, 9, App. I.
Fidler's rules for struts, 109 ; views
on compression bars, 110; Table
for strength of struts. Table V. ;
drawing of line of least resistance
for arch with unsymmetrical load,
App. XX.
Fir, ultimate resistance, moduli, and
weight, Table I. ; safe resistance.
Table la.
Fished joint, 149.
Fixed beams, 69 ; bending moment,
71-74; comparison of strength,
etc., with supported beams, App.
VIII.; objections to, 74 ; stresses
on, 69 ; various cases, 71-74.
Flanges of plate girders, 155, examples,
156, 164; width of, 163, 172;
of cast iron girders, 99.
Flexure, moment of 25 (see Bending
moment) ; points of contra-flexure,
70-75, App. XL
Flow of liquids in pipes, 263 ; pipes
running full, 264 ; pipes running
partially full, 280.
Forces, polygon of, 180 ; triangle of,
179.
358
INDEX
Formulse, arches, thickness of, 253,
App. XXI. ; beams, 54, App.
VII., XXI. ; beams, different sec-
tions, App. X. ; beams, continuous,
App. XL ; beams, fixed, 71, 72 ;
rolled, 87 ; bending moments for
different distributions of load, 28-
40, App. VII. ; App. VIII. ; cast
iron girders, 95, 99 ; centre of
gravity, App. XII. ; compression
bars, 113-115 ; continuous girders,
App. XI. ; cotter joint, 148 ; de-
flection, 66-68, App. VIII., IX.,
XXI. ; egg-shaped sewer, area of,
287 ; equilibrium, App. II.; fished
(wood) joint, 149, 150; fixed
beams, 71,72, App. VIII. ; girder,
approximate weight of, App. XV. ;
head, loss of, 268-271 ; jets, 288 ;
masonry joints, 219, App. XIX. ;
moment of inertia, App. XIV. ;
open webbed girders, 193 ; pipes
flowing full, 264-268 ; pipes
partially full, 280-282 ; plate
girders, 155 ; retaining walls, 234,
238,240,244; rivets, 136; riveted
joints, 123-131 ; rolled iron beams,
87, 88 ; roof, reaction at supports,
206; screw, 146; shearing stress,
56-60, App. VII.; short, for mem-
ory, App. XXI. ; struts, 113-115.
Foundations of retaining walls, 244.
Fracture, 9.
Framed structures, 177 ; different
forms of, 180.
Frames, 180 ; overbraced, 181 ; per-
fectly braced, 180, 182.
Friction, coefficients of. Table XVa.
Fuller, Prof., reference to, App. XX.
Funicular polygon, App. VI.
G
Gib, 148.
Girders, bowstring, 194; box, 170;
braced, 187; built-up, with rolled
iron beams, 93 ; cast iron, 94,
103 ; continuous, 74, App. XI. ;
hog-backed, 176 ; lattice, 187 ; N
188 ; N (braced), 188 ; open-
webbed, 187; plate, 153 ; uniform
strength, 64, 101 ; Warren, 187 ;
weight, formula for, App. XV. ;
Whipple-Murphy, 188. {See above
headings in Index.)
Gordon's formula for compression
bars, 113.
Granite, safe resistance of. Table la. ;
weight of. Table XVII.
Graphic method of finding moment of
resistanceinbeams,46; inwrought
iron rolled beams, 83 ; in cast iron
girders, 96 ; bending moments
and shearing stresses^ App. VI. ;
centre of gravity of cross section
of girder, App. XIII. ; resultant
and centre of gravity of parallel
truss, App. VI.
Gravel, angle of repose, Table XVI. ;
weight of. Table XVII.
Gravity, centre of {see Centre of gravity).
Greenheart, moduli, weight, ultimate
resistance. Table I.
Grouped joint, double-cover riveted,
formulse for, 129 ; examples, 131,
132.
H
Head (hydraulic), loss of, 261, 268 ;
due to bends, 270, Table XXI. ;
to elbows, 270, Table XXII. ; to
orifice of entry, 269 ; to velocity,
269.
Hodgkinson's experiments on cast iron
girders, 95.
Horizontal shearing stress, 55, 60.
House chimney, calculation for, 225.
Hydraulics, 260-294; definitions:
head of elevation, 261 ; head of
pressure, 261 ; hydraulic mean
depth, 262 ; intensity of pressure
at a point, 261 ; loss of head
261 ; pressure, 260 ; wetted peri-
meter, 262 {see Head, Pipes, etc.)
Hurst, references to, 121, 239, 259,
264, 270, 275, 283.
I
X section, moment of inertia for, App.
XIV. {see EoUed beams).
Inertia, moment of, 50, 86, 104 ; of
various sections, App. XIV.
INDEX
359
Intensity of stress, 8.
Iron, angle and tee, weight of, Table
X.
„ rolled, 81 ; girders (see Girders).
„ cast, resistance of, 94 ; moduli,
weight, ultimate resistance,
etc.. Table I. ; safe resistance,
94, Table la., XXVI.
„ columns (see Columns).
„ pipes, strength, thickness, weight.
Table XX.
„ roofs, example of, 209 ; principal
for, 120; struts for, 119.
„ struts (see Struts).
wrought, resistance of 82 ;
moduli, weight, ultimate re-
sistance, Table I. ; safe resist-
ance, 82, Tables la., XXVI.
J
Jets, 288; example, 289, 291, 292 ;
factors to find velocity, Table
XXIV. ; height, 289 ; issuing
velocity, 290 ; nozzles, 290 ; path
described by, 289 ; range, 288.
Joints, 122 ; at apex and feet of truss,
144 ; cotter, 148 ; in webs
of plate girders, 161, 168;
pin, 138, 140 ; riveted, 122
(see Riveted joints) ; screw,
145, 147.
„ in wooden structures, fished,
149 ; scarfed, 151 ; ut foot
of principal rafter, 152.
„ (see Masonry).
Joist, fir, 79 ; rolled iron, 89 ; steel,
92.
L
Lattice girders, 194.
Layer, neutral, 41.
Least resistance, Moseley's principle of,
248.
Lever arm, 185.
Light, C, reference to. Table XI.
Lime, quick, weight of, Table XVII.
Limestone, safe resistance of. Table
la. ; weight of, Table, XVII.
Limiting stress, 8.
Line of resistance, 221.
Links, various forms, 139.
Live load, 4.
„ and dead loads, comparative
effect of, 5.
Load, 2 ; breaking, 5 ; concentrated,
3 ; concentrated and distributed
loads compared, App. VII. ; dead,
4 ; distributed, 4 ; live, 4 ; safe,
5 ; safe for fir beams. Table II. ;
on supports, 18 ; working, 5.
Long compression bars (see Compres-
sion bars).
Loss of head (hydraulic), 268 (see
Head).
M
Masonry, safe resistance of. Table la. ;
coefficient of friction.
Table XVa. ; weight.
Table XVII.
„ joints, centre of pressure,
218, App. XIX.
„ structures, important, 221,
unimportant, 221 ; con-
ditions of stability, 221;
line of resistance, 221;
resistance to sliding, 221,
224; resistance to crush-
ing, 221, Table la.; re-
sistance to shearing, 221 ;
single block, 218 ; un-
cemented 220.
Materials, factors of safety for, 9, App.
I. ; ultimate resistance, moduli,
weight in, Table I. ; safe resistance.
Table la. ; weight, Table XVII.
Maxwell's diagrams, 182 ; for braced
girders, 194 ; for roof trusses,
217; for trussed beams, 196, 198;
Bow's system of lettering, App.
XVII. ; method of drawing, 183 ;
rules for drawing, App. XVI.
Mean fibre, 135.
Method of sections (see Sections) ;
graphic (see Graphic).
Modulus of elasticity, 1 2 ; of rupture,
52 ; of rupture for different sec-
tions, 53 ; of rupture and elas-
ticity for different materials,
Table I.
Molesworth, references to, 120, 247,
253, 259, 264, 265, 268, 270.
36o
INDEX
Moment, bending, 25 ; of flexure, 25 ;
of inertia, 50, App. XIV. ; of re-
sistance, 40 (see Bending Mo-
ment, Resistance, etc.)
Mortar, safe resistance of. Table la. ;
weight. Table XVII.
Moseley's line of resistance, 221 ;
principle of least resistance, 248.
N
N Girder, 188 ; examples, stresses
on, 190; Maxwell's diagrams for,
194 (see Braced girders).
„ (braced) girder, 188.
Neutral axis, 42 ; note on, Ap]3. IV,
„ layer, 41.
Neville's formula, flow of liquids in
pipes, 264, 280.
Newton's third law, 15.
Notation, Bow's, 184, App. XVII. ; no-
tation in this volume. Table XXV.
Nozzles for jets, 290.
0
Oak, ultimate resistance, moduli,
weight. Table I. ; safe resistance,
Table la.
Open-webbed girders, 187 (see Braced
girder).
Orifice of entry, loss of head due to,
269.
Overbraced frames, 181.
P
Parabola, to draw a, App. III., to
substitute circular arc for, 30, 34,
156.
Perfectly braced frames, 180, 182.
Perimeter, wetted, 262.
Perry, Prof., reference to, App. XX.
Pier, brick, calculation for, 222.
Pillars, cast iron, examples of, 116,
117 ; at end of girders, 169.
Pin joints, 138 ; example of, for roof
truss, 140, 144.
Pine, American yellow, and Pitch,
ultimate resistance, moduli, and
weight. Table I. ; safe resistance,
Table la.
Pipes, flow of liquids in, when run-
ning full, 264, Tables
XVIII., XIX. : formula;—
Darcy's,265,267 ; Neville's,
Eytelwein's, Thrupp's, 264 ;
when running partially full,
280, Table XXIII.: formulae
— Beardmore's, Downing's,
Neville's, 280 ; Darcy's, 280,
281 ; thickness, weight, and
strength of, Table XX.
„ drain, 283.
Pitch of rivets, 136.
Plate girders, 154 ; booms or flanges,
155, 163, 166; connection of
web, 161, 168 ; cover plates, 133;
depth of, 163 ; flange plates, thick-
ness of, 156, 163 ; flanges, width
of, 163; hog-backed, 176 ; joints
in flanges, 132, 165 ; joints in
web, 161 ; proportion between
depth and span, 154 ; web, 157,
166; web plates, thickness of.
Table XI. (see Box girder).
Plates, wrought iron, resistance of.
Table I.
Pole of diagram, App. VI. ; cylindri-
cal, strength of, App. X.
Polygon of forces, 180 ; funicular,
App. VI.
Portland cement, safe resistance, Table
la. ; weight. Table XVII. ; stone,
weight. Table XVII.
Pressure, distribution of, on masonry
joint, 219, App. XIX. ; hydraulic,
261 ; wind, 200, Table XIV.
Principal rafter, iron, calculations for,
examples, 120, 144 ; as jointed
and as continuous, 216 ; loads on,
203 ; joint at head and foot, ex-
ample, 144 ; wood joint at foot,
example, 152.
Punched holes in iron plates, 137 ;
in tension joints, 137.
R
Rafter, common, wood, loads on, 201,
continuous, App. XI.
„ principal (see Principal rafter).
Range of jet, 288.
Rankine, reference to, 5, 9, 44, 50,
INDEX
361
53, 66, 86, 158, 185, 224, 238,
242, 248, 353, App. I., XIV.,
XIX., Table XV«.
Eeaction at supports of loaded beams,
15-20 ; of a cantilever, 22 ; of
continuous beams, App. XI. ; of a
girder, 21 ; of a roof, 23 ; rules for
finding, 18, 20.
Rectangle, moment of inertia, App.
XIV.
Eectangular beams, strength of, 54 ;
deflection of, App. IX.
Repose, angle of, 236, Table XVI.
Resistance, 1 4 ; of masonry to crush-
ing, 221; equivalent area
of, 47 ; safe area of, 48 ;
of masonry to sliding,
221; least (Moseley's prin-
ciple), 248 ; ultimate, 49 ;
Avorking, 49 {see Working
stresses).
„ to tension, compression, and
shearing, ultimate and
safe — of cast iron, 94 ;
wrought iron, 82 ; of vari-
ous materials. Tables I., la
(see Working stresses).
,, to find moment of — beam,
by experiment, 51 ; by
graphic method, 46 ; by
reasoning, 40; rolled beam,
by graphic method, 83 ;
by mathematical method,
86 ; cast iron beam, by
graphic method, 96 ; by
mathematical method,
99.
Retaining walls, sections, 230 ; for
earth, 236, 239 ; examples of,
241 - 243 ; foundations, 244 ;
formula5, 238, 239, 244 ; graphic
method, 238 ; for water, 232 ;
example of, 234.
Revetments, surcharged, 241.
Ritter's formula for compression bars,
115 ; method of sections, 184,
App. XVIII.
Rivet holes, diameter and pitch of,
136 ; punched in iron plates,
137 ; rimered out, 138.
Riveted joints, 122 ; butt, with single
cover plate, 124; butt, with
double cover plates, 125, 128 ;
dimensions of, 136 ; double cover
grouped, 131 ; examples of, 128,
131-134 ; formulae for grouped,
129 ; general formulae for, 127 ;
in compression, 135 ; ia tension,
122 ; lap, 123 ; oblique, 134.
Rivets, iron, 136 ; in drilled holes,
138 ; in punched holes, in
iron plates, 137 ; strength
of, Table VIIL
„ steel, 138.
Rolled wrought iron beams or joists,
81 ; advantages of, 92 ; de-
flection, 91, Table IV. ;
depth, 91; examples of, 88-
9 1 ; example, load borne by
given beam, 88, 89 ; girders
built up with, 93 ; market
sections of, 92 ; moment of
inertia, App. XIV. ; practical
formulae for, 87, 88, 90 ; re-
marks on, 91 ; section re-
quired for given load, 89 ;
stress diagrams, 83, 85.
„ steel joints or beams, 92.
Roofs, iron, 196; calculation by
method of sections, App. XVIII. ;
calculation of principal, 120, 214 ;
diagrams for various trusses, 217 ;
distribution of loads, 201 ; ex-
ample, 209 ; joint at apex and
sj)ringings, 144 ; loads, 199 ; oc-
casional loads, 200 ; permanent
loads, 199 ; pin joint, 140 ; reac-
tions at supports, 22, 23, 205 ;
snow, 200 ; stresses on, 213 ;
strut, 119 ; weight of framing,
Table XII. ; weight of roof cover-
ing, Table XIII. ; wind pressure,
200, 208, 210, Table XIV.
Rouleaux, formulae for wooden struts,
115.
Rules for drawing Maxwell's dia-
grams, App. XVI. ; finding reac-
tions, 18, 20 ; shearing stress, 55
(.see Formulae).
Rupture, modulus of, 52 ; for dif-
ferent sections, 53 ; for various
materials. Table I.
362
INDEX
S
Safe load, 5.
„ stress 8 (see Working stresses).
Safety, factor of, 9, App. I.
Sand, angle of repose, Table XVI.;
weight, Table XVII.
Sandstone, safe resistance to compres-
sion, Table la. ; weight, Table
XVII.
Scantlings for wooden roofs. Table XV.
Scarf joint, in wooden beam, 151.
Scheffler's theory, retaining wall, 239;
arch, 247.
Screw of tie-rod of roof, 147.
Screws, 145.
Sections, method of, 184 ; applied to
iron roof, App. XVIII. ;
to Warren girder, 188 ;
to Whipple-Murphy girder,
190 ; lever arm, 185 ;
turning point, 185.
„ of box girder, 175, Plate B ;
cast-iron girder, 97 ; plate
girder, 154, Plate A ; rolled
beams, 83 ; market, of
rolled beams, 92.
Seddon, reference to, 53, 100, 200,
201 ; Tables I, la., IX., XII.,
XIII., XV. (see Wray).
Semicircle, moment of inertia for,
App. XIV.
Sewer, egg-shaped, 287.
Shaler Smith, references to, 139, Table
IX.
Shearing, resistance to, of cast iron, 95 ;
wrought iron, 82 ; of iron
and steel. Table I. ; of rivets,
123, Table Villi ; safe
resistance to, of iron, steel,
and timber. Table la.
„ stress, 7 ; beams (various
cases), 57-60 ; diagrams,
55, 60, 157, 166, App.
V. ; distribution of, 60 ;
graphic method of finding,
App. VI.; horizontal, 55,
60 ; resistance to, in rolled
beam, 9 1 ; rule for finding,
in beams, 55 ; vertical, 55 ;
in web of plate girders,
157, 166, 173, Table XI.
Shingle, angle of repose. Table XVI.
Short compression bars, 112 (see Com-
pression bars).
Shortening, 11.
Slate, weight of. Table XVII.
Sleeve, 148.
Sliding, resistance to, 221, 224, 245.
Stability of brickwork and masonry
structures, 218.
Steel, ultimate resistance, moduli,
weight. Table I. ; safe resist-
ance, Table la.
,, rolled joists, market sections,
92 ; resistance of, to compres-
sion and tension, 92.
Stiffeners, 159.
Stififest rectangular beam cut out of a
round log, 80.
Stiffness, 12 ; of beams compared,
App. VIII.
Stoney, references to, 136, 137, 138,
141, 161, Table VII.
Strain, 5, Table of, 9.
Strength of materials. Table I., la. ;
of beams of fir. Table II. ;
of rectangular beams com-
pared, App. VIII. ; of
beams of different sections,
App. X. ; of rivets, Table
VIII. ; of struts. Table V. ;
of wooden struts. Table
VI.
„ uniform, 64.
Stresses, 6 ; in beams, 41 ; bearing,
7 ; breaking, 8 ; classification of,
8 ; compressile, 6 ; in fixed
beams, 69 ; limiting, 8 ; intensity
of, 8 ; repeated, 1 1 ; safe work-
ing, 8, Table XXVI.; shearing, 7 ;
Table of, 9 ; tensile, 6 ; torsional,
7 ; transverse, 6 ; variation of, 1 1
(see Stress diagrams, Working
stress, etc.)
Strongest rectangular beam that can
be cut out of a round log, 80.
Structures, braced or framed, 177.
Struts (see Compression bars), 109 ;
examples — cast - iron columns,
116, 117 ; timber, 115, 118 ;
wrought-iron, 119, 120; Fidler's
rules for, 109 ; formula? for, 113,
INDEX
363
114, 115 ; Tables for, Tables V.,
VI., VII.
Supports, loads on, 18 ; I'eactions at
Rule I., 18 ; Rule II., 20.
Surcharged revetments, 241 ; example,
243,
T
T iron, moment of inertia for, App.
XIV. ; weight of. Table X.
Tables, various (see List, p. xvi.)
Taps, flow of water from, 272, 278.
Teak, ultimate resistance, moduli, and
weight. Table I. ; safe resistance,
Table la.
Tension, 6 ; resistance to, of cast iron,
94 ; of various materials.
Tables I., la. ; of wrought
iron, 82 ; riveted joints
in, 122.
„ bars, 105 ; effective area of,
106 ; examples of, 108 ;
eyes of wrought iron, 139,
Table IX.
Terms in use, 2
Thickness of arch I'ing, 253, Table
XVIIa.
„ of pipes. Table XX.
Third law, Newton's, 15.
Thrupp's formula for the flow of
liquids in pipes, 264.
Tie-rod, screw of, 147.
Timber beams, 24 ; examples, of, 78,
7 9 ; modulus of rupture,
52; of different sections
comj)ared, App. X. ; for-
mula? for, 54, App. XXI. ;
stiffest beam from log, 80 ;
strength of, 54 ; strength
of beams of fir, Table II. ;
strongest beam from log,
80 ; struts, 115, 118.
„ ultimate resistance, moduli,
and weight. Table I. ; safe
resistance. Table la.
„ cantilever, 76-79.
„ roofs, note, 217 ; scantlings
for. Table XV.
„ structures, joints in, 149.
Torsion, 7.
Transverse stress, 7.
Triangle of forces, 179.
Trussed beams, 195 ; example, 196.
Turning point, 185.
U
Ultimate resistance, 49 ; of cast iron,
94 ; of wrought iron, 82 ; of vari-
ous materials. Table I.
Uniform strength, beams of, 62 ;
constant breadth, 64 ; constant
depth, 65 ; cast-iron girders, 100.
Un win's formulas for weight of girders,
App. XV. ; references to, 1 25, 1 36,
140, 163, 200, Apps. I., XV.,
Table XVII.
V
Variation of stress, 11.
Velocity, loss of head due to, 269.
Vertical shearing stress, cases of, 55-
60 ; rules for finding, 55.
W
Walls, enclosure, 227 ; retaining,
230; retaining for earth, 236;
for water, 232 (see Retaining
walls).
Warren girder, 187 ; Maxwell's dia-
grams for, 194 (see Braced girder);
stresses on, 188.
Water supply, examples of, for a house,
constant, 274 ; intermittent, 271 ;
for a street, 277 (see Hydraulics,
Pipes, etc.)
Web of plate girders, 157, 166, 173 ;
connection to flanges, 161, 167 ;
joints in, 161, 168, 175 ; thick-
ness of, 160, 168, 174, Table XI.
Weight of earths, stones, etc., Table
XVII. ; of girders (Anderson's and
Unwin's formulae), App. XV. ; of
metals and timber. Table I. ; of
roof covering. Table XIII. ; of
roof framing. Table XII. ; of T
and angle iron. Table X.
Whipple-Murphy girder, 187 ; ex-
ample, 190 ; Maxwell's diagram
for, 194 (see Braced girder).
364
INDEX
Wind pressure, on chimney, 225 ; on
roof, 200, Table XIV. ; on wall,
227.
Wohler's experiments, 11.
Wooden roofs, scantlings for, Table
XV., note, 217.
,, structures, joints in, 149 {see
Joints).
„ beams {see Timber beams).
Working load, 5.
„ resistance, 49.
„ stresses or resistance, general.
Tables la., XXVI.; Brick-
work, Table la. ; cast iron,
94 ; concrete, Table la. ;
pin joints, 141 ; rolled
beams, 82 ; riveted joints
123 ; screws, 146 ; tim-
ber, Table la.
Wray (Seddon), references to, 100,
200, 201, Table la., IX., XIL,
XIII., XV.
Wrought iron, resistance of, 82 ; mod-
uli, weight, ultimate
resistance. Table I. ;
safe resistance, 82,
Table la, XXVI.
„ cantilever, 22 ; strut
for roof, 119; rolled
beams, 81; principal
for roof, calculations,
120, 214, 216.
END OF PART IV.
Printed by R. & R. Clark, Edinburgh.
PLATE I.
Compressions Red
Tensions Blue
Scale of Forces 1 Inch- = 1 Ton.
O'lS , , , . ^ 1 2 aTans
D
PIG 283,
PIG 284.
FIG 285. PIG 236.
PLATE 2.
Compressions Red
Tei\sions Blue
ScaLe of forces 1 Inch -3 Owt
J ^ 0 7 2 3 4 S 6 .7 B 9 OvtS
C
FIG 811
T
PLATE 4.
Compressions Red
Tensions Blue
Scale of Poroes % - 1 Ton.
FIG 361.
FIG 362.
FIG 363
PLATE 5.
Ccst^jressions Red
Tensions Blue
■ Scale cf Forces -J Ind\-1 Ton.
OS o 1 J 5 ficms
PLATE 7.
Compressions Red
Tensions , .Blue
PLATE 8.
Compressions Red
Tensions Blue