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VOL. 1.
¿! i
Designing Methods
{
Reinforced Concrete Construction
Włoch
MAY, 1908
BUILDINGS
GENERAL PRINCIPLES OF DESIGN
This is the first volume of a series of bulletins
on designing methods for reinforced concrete
construction.
مسکر

THE JUNE BULLETIN WILL CONTAIN:
Each issue will be devoted to a particular type
of structure; the problems relating thereto will
be discussed and methods of analysis given,
illustrating same by a detailed design.
REPRINT, MARCH, 1909
Detailed design of typical building.
Complete analysis of the strength of
rectangular and T-shaped beams.
No. 1
PRICE, 25 CENTS
National Bank of Commerce Building
SAINT LOUIS
CORRUGATED BAR COMPANY
(FORMERLY EXPANDED METAL AND CORRUGATED BAR CO.)
DNC



COPYRIGHT, 1908,
BY
Corrugated BAR COMPANY
Ke Lib
hips
10-9-1920
REINFORCED CONCRETE BUILDINGS.
Reinforced concrete is the most efficient and logical construction
for buildings where permanence and fire-proof qualities are prime
requisites. There is only one perishable element, and it is so thor-
oughly imbedded in and surrounded by the concrete that, protected
from all corrosive influences, the reinforcement may be said to
have assumed imperishable properties itself. Concrete is recognized
by all as the best fire-resisting material used in building construction
and it is impervious to all atmospheric influences. There are but three
types of construction that may be considered for buildings of even
moderate size; mill or slow-burning construction, reinforced concrete
construction, and steel frame, fire-proofed, construction. It is proposed
to treat of these three types of construction very briefly and then bring
out some of the superior properties of reinforced concrete.
Mill
The only arguments that can be advanced in favor
of mill-constructed buildings is their low initial cost
and the ease and rapidity with which they may be
erected. The main objection is that the construction is not fire-proof
and that a fire of usual severity will destroy the interior structure and
necessitate complete renewal. There is, moreover, very little structural
stability to such a building, since the columns and beams are not rigidly
connected and dependence is placed upon the exterior walls for the
ability to resist shocks, wind pressure and such causes as act upon the
building as a whole. The structural elements are perishable and are
subject to periodic repair.
Construction.
A well-designed steel frame building properly fire-
proofed and with adequate protection for the steel
columns and beams is a "fire-proof" and permanent
structure. The question of the absolute protection of the structural
elements, however, is one that deserves consideration. Owing to the
large size of the beams and columns, planes of separation may be pro-
duced in the fire-proofing material, or the concrete may shrink away
from the sides of the members, thus affording opportunity for corro-
sion. The deflection under given loads and the vibration due to
moving loads or machinery are very much greater in a steel frame
construction than in one of reinforced concrete. The vital objection,
however, to a steel frame building is the excessive first cost without
any corresponding advantage over a reinforced concrete one.
Steel Frame
Construction.
Reinforced
In a reinforced concrete building the structural
elements themselves are of a permanent and fire-proof
Concrete
nature, the concrete improves with age and the steel
Construction.
is completely enveloped and protected. The entire
structure, floors, beams and columns are rigidly connected, and the
building as a whole is capable of resisting severe shocks and unusual
stresses. A reinforced concrete building is accordingly very rigid,
there being practically no vibration, even under heavy moving
machinery, and the construction admits of easy adaptability to various
requirements of span, loading, etc.
The cost of a reinforced concrete building in gen-
eral will be slightly in excess of that of a mill con-
structed one, although in some localities it may be
even less; in all cases, however, the cost will be less than that of a fire-
proofed steel frame construction. With the heavier floor loads such
as occur in factories and warehouses, the economy of a reinforced
concrete over a steel frame construction becomes more marked.
Costs in
General.
In order to get a proper idea of the relative value of the different
types of construction it is necessary to consider not only the first cost,
but also in connection therewith the rate of depreciation, maintenance
charges and insurance rates incidental to the particular construction.
The total investments, as represented by the first cost plus the capi-
talized charges are the proper bases of comparison, and viewed in this
light the economy of reinforced concrete construction is decidedly
apparent.
The advantages of a reinforced concrete building may then be
briefly stated as follows:
(1) Comparative low first cost.
(2) Absolute permanence.
(3) Superior fire-resisting properties.
(4) Low insurance rates.
(5) Rapidity of construction.
(6) Use of local material and labor.
GENERAL REMARKS ON THEORY OF DESIGN.
The primary conception of the action of a reinforced concrete beam
is very simple; a deficiency of tensile strength in the concrete is
supplied by the imbedded steel. It is only when we come to the consid-
eration of internal or web stresses, and endeavor to provide for shearing
and diagonal forces, taking into consideration the local stresses devel-
oped in the transference of stress from one material to another, that
the subject becomes complicated. Some engineers maintain that when
web-reinforcement, either as attached members or as "loose" stirrups
sp
{
-4
are used, a truss action results. It would seem, however, no matter
what form the web members take, that the action is very similar to
that of a plate girder; it is obvious that this condition must obtain
throughout the length of the beam until the concrete has been stretched
beyond its normal capacity. After cracks have developed, the same
action exists, the continuity, however, being destroyed.
The variation in the intensity of the stress in the reinforcement is
produced through the bond with the concrete, and the value of the
bond at any point may be determined by an analysis similar to that
used to determine the rivet spacing in a plate girder.
It is thus seen that although the web reinforcement plays a most
important part in providing against failure by shear or diagonal ten-
sion, and thus increases the load-carrying capacity of the beam, it is
necessary that the bond between the steel and the concrete be of
sufficient strength to develop the requisite stresses in the horizontal
reinforcement.
The adherence of concrete to metal is small, and is a negligible
quantity in practical work. The bond or adhesion (used in contrast
to adherence) of plain bars imbedded in concrete is mainly due to the
gripping of the bar by the concrete caused by contraction in setting.
supplemented by the microscopic irregularities of the surface, and the
variation in the size of the bar at different sections. As might be
expected, the values obtained for bond vary between wide limits in
the case of plain bars, and engineers, as a rule, prefer the use of a bar
having a mechanical bond with the concrete. This is, undoubtedly,
the best practice, as it eliminates all uncertainty in an element of vital
importance to the structure.
High-elastic-limit steel is now recognized as the most desirable
reinforcing material, as it is well known that the strength of a normally
reinforced beam is reached as soon as the stress in the steel passes the
elastic limit. With a given percentage of steel then, the strength of
the beam, assuming failure by tension in the steel, is directly propor-
tional to the yield point of the reinforcing bars. To get the full benefit
resulting from the use of a high-elastic-limit steel, it is necessary, in
most cases, to use a mechanical bond bar.
5
ON REINFORCED CONCRETE CONSTRUCTION
IN GENERAL.
Assuming a correct design and good materials, proper methods
of construction are essential to the success of a reinforced concrete
structure. The reinforcement is purchased under specifications which
insure the physical properties assumed in the design-and check tests
can easily be made.
With the concrete, however, the case is different, and although we
can assure ourselves of the quality of the sand, cement and "stone"
the strength of the resulting concrete depends upon the thoroughness
of the field work. Accurate proportioning, machine mixing and care-
ful placing will result in a concrete that will meet with the require-
ments of the design.
We see, therefore, that it is easily possible to secure proper
materials and the next element, and one which is vital to the success
of the work, is the accurate location of the reinforcing steel.
not possible to check the location of the steel after the concrete is
placed—this must be assured by proper methods of construction.
Should the reinforcing steel not occupy the exact position indicated
on the drawings the strength of the structure will be affected. This
statement applies to stirrups and "secondary members" as well as to
the main reinforcement. Exact location of the reinforcement must
be attained. To accomplish this (1) the bars and stirrups must be
bent or otherwise fabricated in exact accordance with the details.
(2) The bars and stirrups comprising the reinforcement for any
member should be assembled by adequate means so that the whole may
be inspected and placed in the forms as a unit. (3) The means
adopted must be such that the reinforcement cannot be displaced by
the operation of placing the concrete.
More accurate methods are desirable than those which are used
under the ordinary or "loose bar methods" of construction and con-
structors should give their best efforts toward the attainment of this
end.
6
REINFORCED CONCRETE BEAMS.
A complete mathematical analysis of the strength of reinforced
concrete beams of both rectangular and T-section will be given in our
June Bulletin.
The formulæ following are in simple and usable shape, and are
sufficiently accurate for all designing purposes. We will use the
following nomenclature:
Let the accompanying figure represent the cross section of a rein-
forced concrete beam, the shaded area above the neutral axis, a-a,
being the area of the compressive stresses.

b
a
E. Modulus of Elasticity of
the steel.
E-Initial Modulus of Elas-
ticity of the concrete in com-
pression.
F-Elastic limit of the steel.
fe Ultimate compressive
strength of the concrete.
q=Area of metal in width b.
p=Ratio of reinforcement in terms of bd=area of metal (g)
divided by the cross section of the beam above the plane of the
metal (bd).
M. Moment of resistance corresponding to unit stress, s, in
the steel.
M. Ultimate Moment of Resistance of the beam in inch pounds.
K, a constant for any given concrete.
All dimensions in inches and all stresses in pounds per square inch.
The critical percentage of reinforcement is that percentage which will
develop the compressive strength of the concrete when the steel is stressed to
its elastic limit-a smaller amount of metal would presuppose a tension failure,
a larger, failure in the compressive side.
7
Rectangular Beams
Ultimate Resisting Moment.
FORMULAE GIVING THE ULTIMATE STRENGTH OF BEAMS.
F Taken as 50,000 lbs.
Class No. 1.
Average
Rock Concrete.
Class No. 2,
Good
Rock Concrete.
Class No. 3,
Cinder
Concrete.
BEAM FORMULAE.
for
Rock Concrete.
This class is meant to include all concretes
having a compressive strength of 2000 lbs.
per square inch; fe then=2000 and taking
E=2,600,000, we get for the ultimate resisting
moment :
M。=370 bď² for 9=.0085 bd..... (1)
By using a 1:2:4 mix and good rock or
gravel we get a concrete of much greater
compressive strength, but with a higher
modulus of elasticity. For such a concrete
we may assume E. 2,800,000 and f=2700,
and we get:
M。=570 bď² for q=.013 bd... ... (2)
For a 1.2 5 mix of cinder concrete we
may assume E-750,000 and fe=750; we
then have :
Mo=207 bd2 for g=.0047 bd.....(3)
The ultimate strength of reinforced concrete
beams reinforced with any percentage of metal,
General Formula so long as the reinforcement is kept below
the critical percentage for the particular
concrete used may be expressed by a general
formula as follows.
M.=.86 F p bd2.
.. (4)
•
EXAMPLE-What size beam is required for an ultimate moment (live and
dead load moments times factor of safety) of 1,500,000 inch lbs, assuming aver-
age rock concrete?
M。=1,500,000=370 bd2.—(Formula No. 1).
Area of metal required
Assuming b=10" and solving for d we get d=20"
=q=.0085 bd=1.700"; say three 34" N. S. corrugated bars. The distance d is
from the top of the beam to the plane of the metal; there would be about 2″ of
concrete below the metal, making the outside dimensions of the beam 10"x22".
8
For convenience in designing. Table No. 1, page 23, has been
prepared, which gives the necessary depth and the amount of rein-
forcement required for beams 12 inches wide (of either average or
good rock concrete), reinforced with the percentage of steel noted,
corresponding to the ultimate resisting moments given.
FORMULAE BASED ON WORKING STRESSES.
By arbitrarily choosing the unit stresses in the concrete and steel,
and the ratio of the moduli, we at once fix the percentage of rein-
forcement.
The June Bulletin will contain curves for quickly finding p for any combina-
tion of stresses assuming various moduli of elasticity for the concrete. The
plates also give the resisting moment corresponding to the stresses noted.
Es
Let n=
Ec
and s unit stress in steel, c=unit stress in concrete.
Assuming n=12, s=16000, and c=800, we have, neglecting the
tensional value of the concrete in resisting flexure:
Mr=130 bď² for q=.0096 bd...
.. (5)
The strength of reinforced concrete beams may be assumed to be
directly proportional to the percentage of reinforcement as long as the
amount of steel is kept below that required to cause failure by com-
pression in the concrete. With the proportion of metal below the
critical percentage then, the stress in the steel need only be considered,
and a very simple formula results. The distance from the plane of
the metal to center of the compressive stresses may be taken as .86d
for all ordinary percentages of reinforcement. The moment of resist-
ance then becomes:
pow:
D =
M=sqX0.86 d=.86 sp bd'
. (6)
which is the general formula to be used in the design of beams.
•
Note. In applying this formula, that the unit stress in the steel,
s, should in no case be greater than 1-3 of the elastic limit of the
metal; and 9, using high elastic limit Corrugated Bars, must be less
than.0085bd the average rock concrete or.0132bd for good rock concrete.
If ordinary commercial steel is used, considerable more metal is
required to develop the compressive strength of the concrete and g
should then be less than .0125 bd for average rock concrete and not
greater than .02 bd for good rock concrete.
EXAMPLE-Given n=12, s=16,000, and c=800; what size of beam is
required to carry a load (including dead load) of 1,000 lbs. per ft. run, on a span
of 16'? Beam considered as simply supported at the ends.
The bending moment at the center of the beam then-Mr=%×1000×162×12
=384,000 in. lbs.
Assuming d=20", we get by substitution in formula No. 5, b=7.4″, and from
the equation, q=.0096 bd, we get the area of metal required 1.4 square inches.
9
T-BEAMS, APPROXIMATE FORMULAE.
In all T-beam designs, where the ends of the beams are restrained,
the strength of the section at the support should be investigated, and
its ability to resist the negative moment, there developed, determined.
A complete analysis of the strength of T-beams will be found in
the June Bulletin, and the reader is referred to the discussion there
given so that he may become acquainted with the limitations of the
approximate formulæ now given.
ó'
b

b
P
3° bl
F
x
Let-
=Length of span in feet.
"Maximum shearing
strength of the concrete in
pounds per sq. inch.
All other dimensions as shown, in inches.
The resisting moment of a T-section may be expressed with
approximate accuracy in terms of the stress in the steel by the general
formula recommended for the design of retangular beams :
(6)
M=.86 sq x d=.86 sp_bď² ...
This formula is not based on shearing values, and it will be neces-
sary to investigate the beam for possible failure in shear and diagonal
tension. In this investigation the width of the stem, b, must be used.
Experiments would seem to indicate that even when the area of the
reinforcement is as much as 3% to 32% of the rectangle enclosing.
the stem (bd), that the horizontal shearing strength of the concrete
will not be exceeded.
It is advisable, however, to compute the maximum permissible
amount of steel that may be used by the formula:
go
where q。=maximum amount of steel permissible.
For methods of investigating the vertical shear and diagonal ten-
sion see page II. •
The maximum width of floor-slab that may be considered as acting to form
the T-section may be computed by methods that will be given in the June Bulle-
tin, which will also contain a table for the rapid design of T-beams. This table
will be found convenient, and its use will result in a harmonious or balanced
design requiring no investigation of horizontal shearing stresses.
10
The recommendations of foreign authorities are given below:
AUTHORITY.
•
Prussian Regulations, 1904 .
French Rules, 1906...
PERMISSIBLE WIDTH.
Less than 3 length of beam.
Less than 3 length of beam, or
less than 3/4 spacing of beams.
Report Committee, Royal Institute British Architects, 1907.
Same as French rules.
SHEARING STRESSES IN BEAMS.
Reinforced concrete beams seldom fail in shear, yet an under-
standing of the distribution of the shearing stresses is necessary for
the proper consideration of the question of metallic web reinforcement.
In addition, the shearing stresses are indicative of the diagonal tensile
stresses developed, and it is through this diagonal tensile stress that
failure may be expected. It is desirable and proper to neglect the
tensile strength of the concrete in a consideration of the shearing
stresses, since we are mostly concerned with the conditions that exist
after the concrete has developed its full resistance; in other words,
the metallic web members must carry the entire stress, or its com-
ponent in their direction, after the concrete fails. Near the points of
maximum bending moment at the stage of loading considered, the
concrete has been stretched beyond its elastic limit, and cracks have
developed; and near the supports the diagonal tensile stresses are of
such magnitude that no reliance can be placed on the concrete.
Since, with horizontal reinforcement only, the concrete carries all
secondary or web stresses, it is evident that the total shearing stress
on a horizontal plane, directly above the reinforcement, must, for a
given length of beam, be equal to the variation in total stress in the
reinforcement between the limits of length selected. The variation
in the tensile stress between the given limits must, of course, equal
the variation in the compressive stresses in the same limits. We have,
therefore, since we have assumed that the concrete has no tensile
strength, a constant value for the horizontal (and vertical) shear
between the reinforcement and the neutral axis. Above the neutral
axis the value of the horizontal shear at any plane is equal to the
total compression existing above that plane.
11
The following diagram, then, represents the variation in the value
of the horizontal or vertical shear throughout the depth of the beam:
A



Id-z
B
b
la
ยะ
STRESS DIAGRAM.
SECTION OF BEAM.
SHEAR DIAGRAM.
Let v=value of the unit shearing stress below the neutral axis;
the value of v may be determined as follows:
V=v b (d-z) from which
V
b (d-z)
Let Vrepresent the total vertical external shear at any section A-B.
Considering the internal stresses and taking moments about a point O
in the plane of the reinforcement, and at unit distance from A-B
we have:
14.
b v =
v
BONDING STRESSES.
The total bonding stress developed between the reinforcement and
the concrete in one inch length of beam, is equal to the difference in
the total stress in the bars at the limiting points. This difference for
a length of one inch equals
V
(d-z)
From this total bonding stress, knowing the number of the rein-
forcing bars, and their perimeter, the intensity of bonding stress on
one square inch of bar surface is readily obtained.
SHEARING PROVISIONS.
There are two general methods of reinforcing against diagonal
and shearing stresses which may be used singly or in combination.
One of these consists in bending all, or part, of the longitudinal bars
up toward the supports at various points. The other method involves
the use of stirrups, either vertical or inclined, which may, or may not,
be attached to the main reinforcement. When loose stirrups are used
they should be placed vertically.
It is desirable, and, in fact, essential, in all beam work, that some
provision be made for web stresses; otherwise, if failure should occur,
it would be sudden and without warning. When the two systems of
web reinforcement are used in combination the analysis becomes uncer-
tain, and it is impossible to predicate the distribution of the stresses.
We will, accordingly, consider the two methods separately.
12
BENDING UP THE MAIN REINFORCING BARS.
When bars are bent up at intervals near the supports, the inclined
portions act as the diagonals of a truss, taking tension, and the stress
carried is a function of the inclination. Adjacent diagonals should
overlap sufficiently to insure "truss" action.
It is necessary that the inclined bars have sufficient length of im-
bedment above the neutral axis to develop the requisite stress. Since
the length of imbedment is necessarily limited, a bar with a strong
mechanical bond is especially useful for such purposes. In this con-
ception of the action occurring in the beam, all the bars are considered
to act as a unit, owing to their rigid connection through the concrete,
the bent-up bars acting as attached diagonals to the main member.
STIRRUPS.
Stirrups are generally used vertically, and we will consider only the
case of vertical stirrups carried under the longitudinal bars and extend-
ing to the top of the beam. Assuming no tension in the concrete, or that
this discussion applies only after the concrete is itself unable to resist
the diagonal tensile stresses developed, we may say that the stress in
any stirrup is equal to the variation in the total stress in the
longitudinal reinforcement in the distance tributary to that
stirrup. It is assumed that the stirrups carry only vertical stresses,
all horizontal stresses being transferred to the longitudinal bars
through bond with the concrete. Vertical stirrups should be investi-
gated for sufficiency of bond above the neutral axis of the beam, and
owing to the short length of imbedment available, it will be desirable
to use a mechanical bond bar, if no form of anchorage is provided.
Formulæ for Stirrup Spacing: The following formulæ afford a
ready means of determining the required spacing of vertical stirrups
at any part of the beam for any particular size of stirrup.
We will use the following symbols:
P=Total stress in one stirrup=total cross sectional area of the
vertical legs of the stirrup times the allowed unit stress.
V External vertical shear at any section.
v=Vertical unit shearing stress that the concrete is assumed to be
capable of taking.
V=Total vertical shearing stress assumed to be carried by the
concrete bd x vc.
}
y=Spacing, center to center of stirrups, required at any section.
All dimensions in inches and stresses in pounds per square inch.
As before v=
V
.86 bd
P—by v = b V y
.86 bd
V v
.86 d
13
P is fixed by the dimensions of the stirrup and the allowed working
stresses, and solving for y we have
.86 d P
V
·(7)
The above formula applies only when the stirrups are assumed to
carry all the vertical shearing stresses.
Should it be desired to make an allowance for the vertical shear
assumed to be carried by the concrete the formula becomes :
.86 d P
V-Ve
y =
ข
y =
WEB STRESSES IN T-BEAMS.
The preceding remarks are based on a consideration of rectangular
beams, but they apply also to T-shaped beams. As before, we have.
where b is the width of the stem.
The shearing stresses and the diagonal tensile stresses developed
in T-beams are much higher than the unit stresses developed in rectan-
gular beams, owing to the much greater load-carrying capacity for a
given width of stem, b. It is, therefore, especially necessary to inves-
tigate T-beam designs as to their ability to resist web stresses.
In all investigations for shearing stresses the value of (d-z) may
be taken as .86 d.
V
b(d-z)
•
The following table gives the allowable unit shearing stresses in
beams with horizontal reinforcement, recommended by the authorities
noted. It should be borne in mind, however, that the allowed stresses
in shear are intended to really limit the diagonal tensile stresses to
safe values, as it is through diagonal tension that failure would prob-
ably occur:
Committee German Architects
and Engineers, 1904....
Joint Committee, Royal Insti-
AUTIIORITY.
Prussian Regulations, 1904....64 pounds, or less than one-fifth
shearing strength.
65 pounds.
tute, British Architects, 1907. 60 pounds.
Commission du Ciment Armé,
appointed by the French Gov-
ernment, 1906...
*Safe Shearing Stresses =
V
*SAFE SIIEARING STRESSES
PER SQUARE INCH.
.028 compressive strength.
bd
NOTE.-Beams will fail in diagonal tension when the intensity of the stress
reaches the tensile strength of the concrete. Failure from this cause usually
occurs when the value of the vertical shearing stress as computed by the equation
V
reaches from 100 to 150 pounds, the higher value corresponding to
very rich mixes-(d-z) may be taken for the ordinary percentages of reinforce-
ment used, as 86 d.
V=
b(d-z)
14
>
COLUMN DESIGN.
There is a scarcity of comparable experimental data on the strength
of reinforced concrete columns from which reliable formulæ might be
derived. Experiments to determine the compressive strength of con-
crete, even when made on carefully-prepared test specimens, give widely
divergent results, and conservative values for the allowed stresses
should be used in designing. It is, of course, desirable in column de-
sign to use high-unit stresses, in order that the column may be com-
paratively small. There are three methods of increasing the allowable
compressive stress in a column:
(1) The use of a very rich mix for the columns only.
(2) The introduction of longitudinal reinforcement, which is usu-
ally tied together by bands.
(3) Hooping the column to prevent lateral expansion when under
load. The hooping may be accomplished by means of hoops or bands
at close intervals, or else by means of spirally-wound rods.
It is desirable and necessary that there be some longitudinal rein-
forcement in a concrete column, since, owing to the monolithic character
of reinforced concrete buildings, more or less bending moment will be
put into the columns; this applies particularly to the end or wall col-
umns. When the bending moment can be figured, as is the case when
the eccentricity of loading is known, or when the design contemplates
the resistance of wind stresses through knee-brace action, the stresses
due to both the direct load and the bending moment must be taken into
consideration.
In the discussion following, the columns will be considered as
axially loaded. We will take up the three methods outlined above.
A simple and reliable method of increasing the strength of a given
concrete is by the use of a greater amount of cement. The use of a
special mix for the columns only of a reinforced concrete structure, is
attended with considerable practical difficulty, and the practice is not
to be recommended unless the work is inspected with the utmost thor-
oughness. It is also essential that the column shaft be carried up
through the beam and girder levels, using the special mix, and this
presents many difficulties.
The following table gives the compressive strength of broken stone
concrete from Kimball's tests, made at Watertown Arsenal, in 1899,
on 12-inch cubes, and while the results may not be good averages for
the strength generally, yet they show the comparative strength of the
various mixtures at different ages:
15
MIXTURE.
I: I
: 3
I : 2
: 4
1:22:5
:6
I:3
1:32
3/27
I : 4 :8
I : 5
I :6
ооел со о сост
Crushing Strength in Pounds per Square Inch.

COLUMN
NUMBER.
9
13
15
: 10
: 12
18
7 DAYS.
1,600
1,525
1,300
1,230
I,100
1,000
800
600
SIZE AND
LENGTH.
The following table gives the crushing strength of plain concrete.
columns, and has been arranged from the results reported by Prof.
Talbot, in Bulletin No. 10, of the University of Illinois Experiment
Station. These columns were made of a 1:2:4 mix of limestone con-
crete, and should represent average conditions met with in practice. It
is the purpose of the table to show what strength may be expected of
such columns, and also to bring out the wide difference between the
compressive strength of the concrete, as determined from tests on cubes
and cylinders of the dimensions noted.
Column Tests-University of Illinois Experiment Station.
MAXIMUM LOADS CARRIED
TOTAL LBS. PER SQ. IN.
POUNDS. OF GROSS AREA.
12"x12"x12'0"
9″x 9″x12′0″
12"X12"x12'0"
12"X12"x12′0″
12"x12"x 6'0"
9″x 9"x 6′0″
X
Averages.
I MONTH.
•
2,750
2,460
2,225
2,060
1,875
1,700
1,350
1,000
250200
162000
236000
254000
176000
90300
3 MONTHS. 6 MONTHS.
3,360
2,944
2,670
2,440
2,210
1,980
1,520
1,060
*
1710
2004
1610
4,300
3,900
3,400
3,100
2,800
1709
1189
1079
1550
2,500
1,900
1,300
AGE IN
DAYS.
69
64
65
61
Above columns, limestone concrete 1:2:34 mix.
Average crushing strength of columns-1550, Age 65 days.
Average crushing strength of 12" cubes-2100, Age 64 days.
Average crushing strength of cylinders 8" in diameter and 16" long
-1490, Age 73 days.
63
65
65
4.
16
C
COLUMNS WITH LONGITUDINAL REINFORCEMENT.
As has been stated, longitudinal reinforcement is necessary in col-
umn construction in order to properly take care of such secondary
stresses as may exist. Necessarily, also, it carries direct compressive
stress in proportion to its deformation. It is well known that although
the modulus of elasticity of concrete may, without appreciable error,
be considered as constant up to stresses of about 600 pounds per square
inch, yet for higher stresses the error becomes increasingly larger, and
can not properly be neglected, owing to the greatly increasing deforma-
tion per unit increase of stress. Owing to this stress-strain relationship
for concrete the proportion of load carried by the longitudinal rein-
forcing bars in a column varies with the load, the proportion increas-
ing as the load on the column increases.
An investigation of column tests shows that computations for the
stresses in the steel and concrete, based on the assumption of a constant
ratio of the moduli, give results largely in error for the higher stresses.
This leads to the conclusion that while it may be justifiable to design
for working loads assuming a constant ratio, yet in a consideration of
ultimate strength to determine the factor of safety obtained, the vari-
able relation known to exist can not be ignored. The ratio of the
moduli will vary from about 12 at initial load to 25 at ultimate load.
In columns with longitudinal reinforcement (we are not now con-
sidering spirally wound or hooped columns), it is necessary that the
reinforcing bars be so banded or secured by ties that the unsupported
length of the rods between bands is not too great to permit of column
action.
Formula for Concrete Columns with Longitudinal Reinforcement.
In general in concrete column design the ratio of length to the least
side is such that the question of length need not enter into the formula
for strength.
It is good practice to assume that the outer one inch of concrete is
available for fireproofing purposes only, and should not be considered
as carrying load-the size of column given by the following formula
does not include the protective portion.
Let A₁=Net cross section of column=gross section minus area of
reinforcement, in sq. in. (A-g).
q=Sectional area of reinforcement in sq. inches.
Es
Ec
c=unit stress in concrete.
n c unit stress in reinforcement.
P total load on column.
Then:
n=
n is a variable as above explained.
P=A₁ c + ncq.
17
}
For normal working stresses " may be taken as 15, and we have:
FOR WORKING LOADS: P c (A,+15 q) or with sufficient
accuracy, P=c (A+15 q).
It is evident that when (as is usually the case) q is but a small percentage o
A that A may be substituted for A1, in the formula.
The following diagram gives the average unit stress in reinforced
concrete columns with varying percentages of longitudinal reinforce-
ment based on allowable unit stresses, c in plain columns of 500, 600
and 700 pounds per square inch, the ratio of the moduli being taken
as 15.

Average Permissible Stress per of Column.
1100
1000
*
900
#
800
#
700
600
500
EXPERIMENTOR.
Average
Tests.
TALBOT
Average of II
'Tests.
WATERTOWN
ARSENAL
}
of 4
C =
700
C = 600
The following values for the crushing strength of columns, with
longitudinal reinforcement from tests on full-sized specimens, should
be of value as bases of comparison:
1%
2%
Percentage of Reinforcement = q ÷ A
DIAGRAM PLOTTED FROM FORMULA
P = C (A + 158)
Average
Crushing Strength
P
per sq. in.
A
1746 pounds.
C =
2015 pounds.
500
Average
Percentage of
Reinforcement.
1.39
1.34
Mix of
Concrete.
I: 2:4
3%
I: 2:4
Age in
Days.
66
I04
کی موت رانی کی
18
COLUMNS WITH LONGITUDINAL BARS AND CIRCULAR
BANDS OR SPIRAL WINDING.
The theory of hooped columns may be broadly stated to rest on the
conception that if lateral movement is prevented by an encasing shell,
the pressure that may be borne by the contained concrete becomes a
function rather of the tensile strength of the shell than of the compres-
sive strength of the concrete. While experiments, particularly those
made by M. Considere, show enormously high values for the ultimate
strength of spirally wound columns, and experiments, both in the
United States and abroad, show that concrete can be subjected to great
distortion by cubical compression without apparently losing its struc-
tural integrity, yet we are not familiar with the various stages of
change which the concrete undergoes as the load is increased. When
the hooping bands or spirals are sufficiently close together so that the
concrete will not break out between them, a shell effect is produced.
It does not seem good practice to stress even hooped concrete beyond
its compressive strength, as determined on cubes, or short specimens.
Hooping undoubtedly prevents sudden failure (and a hooped col-
umn may carry very great loads with considerable deformation) by
preventing failure through shear on an oblique plane.
We may
accordingly assume as the maximum stress to which the concrete in
hooped columns can be properly subjected, the compressive strength
of the concrete as determined on short specimens. The arbitrary limit
of ultimate stress would not be approached in practice by the working
loads, but the factor of safety should be based upon it, and owing to
the action of the hoops this factor may be considerably less than that
which should be used for plain concrete columns, or those with longi-
tudinal reinforcement only.
All hooped columns should, of course, have sufficient longitudinal.
reinforcement to take care of bending stresses; the available cross-
section considered being that within the hoops.
In the design of hooped columns, therefore, the strength may be
figured as for columns with longitudinal reinforcement, but higher
working stresses may be used.
MONOLITHIC CONSTRUCTION.
In a reinforced concrete structure all the elements are rigidly con-
nected, and those members which resist flexure act under conditions
of end restraint. In end panels the beams framing into the wall col-
umns are subjected to the restraining action due to the stiffness of the
columns, and at the interior columns they are further fixed by the con-
tinuous action due to their extension through the columns into the
adjacent panels. It is not necessary here to go into the theory of con-
19
tinuous beams, as the discussion is based on the assumption of a
uniform load on all parts-a condition not likely to be met with in
an actual building. The whole question is indeterminate, but we
know the limits of the bending moments that may be developed.
The following formulæ have been adopted as representing the con-
ditions with sufficient accuracy, and tests seem to justify their use.
For interior beams, continuous over several supports, we may take:
Moment at Center=M.=_WI
Moment at End
Ι
12
I
12
=span center to center of supports.
=Me= WI
For beams in end panels take:
Mc
Me=
I
IO
I
IO
WI
WI
In floor slabs we have not only continuous action in the direction
of the main reinforcing bars, but also such arching and plate action as
exists in interior panels surrounded on all sides by beams. All of these
causes tend to reduce the moment at the center. These considerations
w12
have led to the adoption of the formula M.= for determining
16
the moment at the center of the span, and numerous tests show that
this is a conservative and safe practice. It is understood, of course,
that when the moment is taken as 1/16 w² that proper provisions
will be made (by means of top reinforcing bars over the supporting.
beams) to take care of the negative moment. The amount of metal
used should be the same as that in the center of the span. This top
reinforcement may consist of loose bars, extending on each side of the
floor beam at least to the quarter points, or alternate bars of the main
reinforcement may be bent up and carried over the support. It is not
necessary, under ordinary conditions, to make any provisions for shear
in floor slabs.
For floor slabs then we recommend the following formulæ for
determining the bending moment at the center of the span :
Interior Panels Me Me=1/16 wl²
End Panels M. Me 1/12 w/2
(w=Load per sq. foot.)
20
"
CONCRETE DATA.
Rock or gravel concrete weighs, approximately, 150 pounds per
cubic foot.
Cinder concrete weighs, approximately, 90 pounds per cubic foot.
The crushing strength of concretes vary largely; the
quality of the materials and methods of mixing affect
the strength very materially. The crushing strength
of stone or gravel concrete, when subjected to direct compression,
will, under average conditions, agree fairly well with the values given
in the following table:
Compressive
Strength.
MIX BY PArts.
1:3
I 1:3
I:3
1:22:5
I: 2
I : 2
:8
: 6
: 5
Tensile
Strength.
: 4
: 3
Shearing
Strength.
COMPRESSIVE STRENGTH. LBS. PER SQ. INCH.
AGE, 30 DAYS.
AGE, 6 MONTHS.
The values in the above table have been taken from "Concrete Plain and
Reinforced," by Taylor and Thompson, with the author's permission.
See Column Design for results of compression tests.
The tensile strength of concrete may be taken, roughly,
as 1/10 of the compressive strength, and the maximum
proportionate elongation to which concrete may be
subjected before cracking is, on the average, .00015.
Temperature
Stresses.
The shearing strength of concrete, as determined by
punching flat plates and shearing short beams, is, ap-
proximately, one-half of the compressive strength:
for general purposes assume shearing strength as one-third of com-
pressive strength.
1810 pounds.
1950 pounds.
2030 pounds.
2180 pounds.
2440 pounds.
2540 pounds.
The amount of steel required to reinforce against
temperature cracks may be taken as 0.4 per cent of
the cross section, if high elastic limit bars, with a
yield point of 50,000 pounds are used.
21
40000
1
2440 pounds.
2580 pounds.
2740 pounds.
2940 pounds.
3300 pounds.
3420 pounds.
}
ن (
f
The following table gives the quantities of materials required for
one yard of concrete. The results given have been taken from a similar
table in Concrete Plain and Reinforced by Taylor and Thompson, with
the author's permission to use this copyrighted matter:
Proportions
by Parts
Cement
Sand
QUANTITIES OF MATERIALS FOR ONE CUBIC YARD OF RAMMED
CONCRETE BASED ON A BARREL OF 3.8 CUBIC FEET.
13
13
1 3
1
1
1.5 1
2 1
1 1 2.5 1
1 1 3 1
1 1.5 2 1
1 1.5 2.5 1
11.5 3 1
11.5 3.5 1
1 1.5 4
1 1.5 4.5 1
1 1.5 5 1
1 2 3
1 2
1 2
1 2
1 2
1 4
Stone
Packed
Cement
Proportions
by Volumes
I
3.5 1
01207000
1
14
1
1 5
10
1 6 12
Loose
Sand
Percentages of Voids in Broken Stone or Gravel
50% Broken
45%
40%
Stone Screened Average Con- Gravel or
to Uniform Size
dition
Mixed
Loose
Stone
Cement
Sand
Stone
Cement
I
Sand
4
1
4.5 1
5 1
1 2 5.5 1
1 2 6 1
1 2.5 3 1
1 2.5 3.5 1
1 2.5 4 1
1 2.5 4.5 1
1 2.5 5 1
1 2.5 5.5 1
BblCu. Ft Ca Ft. Bbl. Ca.Yd ca Yd
Cu Bbl. Cu.Yd. Ca. Id. Bbl|Cu. Yd. Ca. Yd. Bbl. Ca. Yd.|Cu. Yd
3.8 5.73.190.45 0.67 3.080.43 0.65 2.970.42 0.63 2.78 0.39 0.59
3.8 7.62.85 0.40 0.802.73 0.38 0.772.620.370.74 2.430.34 0.68
3.8 9.52.570.36 0.902.45 0.340.862.34 0.33 0.82 2.15 0.30 0.76
3.811.42.340.330.992.220.310.94 2.120.30 0.90 1.93 0.27 0.82
5.7 7.62.490.530.70 2.40 0.510.682.31 0.490.652.160.46 0.61
5.7 9.52.270.480.802.180.460.772.090.44 0.741.94 0.410.68
5.711.42.09 0.44 0.882.00 0.42 0.841.910.400.81 1.76 0.370.74
5.7 13.31.94 0.410.961.84 0.39 0.911.760.370.87 1.610.340.79
5.7 15.21.800.381.01 1.71 0.36 0.961.630.34 0.92 1.480.31 0.83
5.7 17.11.69 0.361.07 1,600.34 1.01 1.510.32 0.96 1.370.29 0.87
5.719.01.590.341.121.50 0.32 1.06 1.420.30 1.00 1.280.27 0.90
7.6 11.41.890.530.80 1.81 0.51 0.761.740.490.74 1.610.450.68
7.613.31.76 0.49 0.87 1.68 0.47 0.83 1.610.450.79 1.480.420.73
7.6 15.21.650 460.931.57 0.44 0.88 1.500.42 0.84 1.380.39 0.78
7.6 17.11.55 0.44 0.98 1.48 0.42 0.94 1.41 0.40 0.89 1.28 0.36 0.81
7.6 19.01.47 0.41 1.031.390.390.98 1.320.37 0.93 1.200.340.81
7.620.91.39 0.39 1.081.31 0.37 1.01 1.250.35 0.97 1.13 0.32 0.87
7.622.81.320.371.111.25 0.35 1.06 1.180.331.00 1.06 0.30 0.89
9.5 11.41.720.610.731.660.580.70 1.60 0.56 0.68 1.490.520.63
9.5 13.31.62 0.57 0.80 1.55 0.55 0.761.49 0.52 0.73 1.380.49 0.68
9.5 15.21.520.540.861.460.51 0.82 1.400.49 0.79 1.290 45 0.73
9.5 17.11.440.51 0.911.37 0.48 0.87 1.310.46 0.83 1.200.42 0.76
9.5 19.01.37 0.480.96 1.30 0.46 0.92 1.240.440.87 1.130.40 0.80
9.520.91.30 0.46 1.01 1.23 0.43 0.95 1.170.41 0.911.07 0.380.83
9.522.81.24 0.441.05 1.17 0.41 0.99 1.110.390.94 1.01 0.360.85
9.524.71.180.421.08 1.12 0.39 1.02 1.06 0.37 0.97 0.96 0.34 0.88
9.526.61.13 0.40 1.11 1.07 0.38 1.05 1.01 0.36 0.99 0.910.320.90
11.4 15.21.42 0.60 0.801.36 0.57 0.771.300.550.73 1.21 0.51 0.68
11.4 17.11.340.57 0.851.28 0.54 0.81 1.230.52 0.781.130.480.72
l
11.4 19.01.280.54 0.901.220.520.861.17 0.49 0.82 1.07 0.450.75
5.5
.5 1 11.420.91.220.520.94 1.16 0.49 0.90 1.110.47 0.86 1.01 0.430.78
1
1 2.5 6 1
1
1 2.5 6.5 1
1 2.5 7 1
13 4
1 3
13
1 3
4.5 1
5
1
1 11.4 22.81.160.490.98 1.11 0.47 0.94 1.05 0.44 0.89 0.96 0.410.81
6.5 1 11.424.71.120.47 1.02 1.06 0.450.97 1.01 0.43 0.92 0.920.39 0.84
7 1 11.4 26.61.070.45 1.05 1.01 0.43 0.99 0.96 0.40 0.95 0.87 0.37 0.86
7.5 1 11.428.51.03 0.44 1.09 0.97 0.41 1.020.920.39 0.97 0.830.350.88
1 11.430.40.99,0.421.11 0.93 0.39 1.05 0.880.37 0.99 0.800.340 90
1 15.219.01.13 0.64 0.80 1.08 0.610.761.040.59 0.73 0.960.54 0.68
1 15.222.81.04 0.590.88 0.99 0.56 0.84 0.95 0.54 0.80 0.870.490.73
15.226.6,0 96 0.540.95 0.920.52 0.91 0.880.50 0.87 0.800.45 0.79
1 15.230.40.90 0.511.01 0.85 0.48 0.96 0.81 0.46 0.91 0.740.42 0.83
1 15.234.2 0.84 0.47 1.06 0.80 0.45 1.010.760.43 0.96 0.68 0.380.86
1 15.238.00.79 0.44 1.11 0.75 0.421.060.710.401.00 0.64 0.36 0.90
1 19.038.00.730.521.03 0.69 0.490.97 0.660.46 0.93 0.600.420.84
1 22.845.5 0.62 0.52 1.04 0.58 0.49 0.98 0.56 0.47 0.94 0.50 0.420.84
NOTE-Variations in the fineness of the sand and the compacting of the concrete may
affect the quantities by 10% in either direction.
Use 45% column for average conditions and for broken stone with dust screened out.
Use 50% column for broken stone screened to uniform size.
t
Stone
Cement
Use 40% column for gravel or mixed stone and gravel.
Use 30% column for scientifically graded mixtures.
Sand
Stone
I
C
Cement
I
30%
Graded
Mixtures
Sand
Stone
22
·
M。=370 bd².
M
50
75
100
• 150
200
250
300
350
400
450
500
550
600
650
700
750
800
850
900
950
1000
1200
1400
1600
1800
2000
2250
2500
2750
AVERAGE ROCK CONCRETE.
11
d
3 36
4.12
4.74
5.81
Table No. 1.
TABLE FOR USE IN DESIGNING REINFORCED CONCRETE BEAMS, 12" WIDE.
q=.0085 bd.
M。=570 bd².
6.71
7.50
8.22
q
0.343
.420
.484
.593
.685
.765
.840
.905
.968
1.025
1.080
1.135
8.88
9.50
10.06
10.60
11.12
11.62
1.185
12.09
1.232
12.55 1.281
12.98
1.322
13.42
1.370
13.82
1.410
14.23
1.452
14.62
1.491
15.00
1.530
16.43 1.675
17.75 1.810
18.98 1.937
20.12
2.050
21.21 2.165
22.50 2.295
23.72
2.418
24.88 2.540
M
3000
3250
3500
3750
4000
4250
4500
4750
5000
5500
6000
6500
7000
7500
8000
8500
9000
9500
10000
11000
12000
13000
14000
15000
16000
17000
18000
19000
20000
}
d
25.97
27.05
28.06
29.05
30.00
30.92
31.84
32.70
33.55
35.18
36.75
38.22
39.68
11.08
42.42
43.72
45.00
46.22
47.43
49.72
51.95
54.10
5.520
56.12 5.725
58.10
5.925
60.00
6.120
61.80
6.310
q
2.648
2.760
2.861
2.962
3.060
3.157
3.250
3.332
3.418
3.590
3.745
3.900
4.050
4.190
4.330
4.455
4 590
4.718
4.840
5.075
5.295
63.60 6.480
65.40
6.670
67.10
6.850
M
50
75
100
150
200
250
300
350
400
450
500
550
600
650
700
750
800
850
900
950
1000
1200
1400
1600
1800
2000
2250
2500
2750
!
GOOD ROCK CONCRETE.
M
3000
3250
3500
d
2.70
3.31
3.82
4.68
5.40
6.04
6.62
7.15
7.64
8.11
8.54
8.96
9.36
9.74
10.10
10.46
J
19.10
20.03
I
q
0.428
.524
.605
.742
.856
.956
1.050
1.134
1.210
1.285
1.352
1.420
1.484
1.544
1.601
1.658
10.81 1.714
11.15
1.764
11.47
1.817
11.78
1.866
12.08
1.915
13.23
2.100
14.30
2.266
15.28
2.421
16.21
2.570
17.09
2.704
18.13
2.875
3.025
3.180
3750
4000
4250
4500
4750
5000
5500
6000
6500
7000
7500
8000
8500
9000
9500
10000
11000
12000
13000
14000
15000
}
16000
17000
18000
19000
20000 |
q=.0132 bd.
d
20.95
21.78
22.62
23.40
3.705 .
3.828
24.17
24.92 3.950
25.65 4.063
26.32
4.170
4.282
1.190
1.691
4.875
5.070
3.245
34.20
35.422
35.25
5.580
36.28
5.741
37.26
5.901
38.22 6.050
10.10 1 6.355
11.85
6.628
43.60
6.908
45.22
7.165
7.418
27.03
28.35
29.60
30.80
32.00
33.10
1
T
3.320
3.443
3.582
I
16.80
48.35
49.83
51.30
52.70
54.10
To use table first
7.660
7.892
8.125
8.350
8.570
The moments given in the table are the ultimate moments of resistance of the sections in thousands of inch pounds.
apply desired factor of safety to actual moments.
M-Ultimate bending moment of external forces in thousands of inch pounds=Mo.
d=Distance from top of beam to plans of metal in inches.
q=Number of square inches of metal required in beam in width of 12 inches
23
Table No. 2.
TABLE SHOWING THE NECESSARY SPACING OF DIFFERENT SIZES OF CORRUGATED BARS
FOR A GIVEN AREA OF STEEL PER FOOT WIDTH OF SLAB.
OLD STYLE,
TYPE A.
C to C❘
9/4" % "
of /2" 3/4" 8" 1" 11"
Bars
יים
יים יים יים
יים וייס
O'
C to Cl
of
Bars
21.08 2.22 3.30 4.20 6.43
22" 0.86 1.78 2.65 3.36 5.14
3" 0.72 1.48 2.20 2.80 4.28|
32" 0.62 1.27 1.89 2.40 3.67
40.54 1.11 1.65 2.10 3.21
42" 0.48 0.99 1.47 1.86 2.85
50.43 0.89 1.32 1.68 2.57||
52" 0.39 0.81 1.20 1.52 2.34
6" 0.36 0.74 1.10 1.40 2.14
62" 0.33 0.68 1.02 1.29 1.97
7″0.31 0.63 0.94 1.20 1.83
72" 0.29 0.59 0.88 1.12 1.71
8" 0.27 0.55 0.82 1.05 1.60
8" 0.25 0.52 0.77 0.99 1.51
9"0.24 0.50 0.73 0.93 1.43
92" 0.23 0.47 0.69 0.88 1.35
0.66
10" 0.22 0.44 0.66 0.84 1.28
1170.20 0.40 0.60 0.76 1.17
12/0.18 0.37 0.55 0.701.07
CORRUGATED SQUARES,
TYPES B AND D.
ཉ
1/4" 1/2" 3/8" 1½″|
26" |
" | 1
5/8" 3/4" 8" 1" 11"
CORRUGATED FLATS,
UNIVERSAL TYPE.
יים ים ים ים ים ים ים ים ים
C to Cl
of No. 1 No. 2 No. 3 No. 4 No. 5 No. 6
Bars
CORRUGATED ROUNDS,
יים
יים
יים
יים
יים
יים
lo"lo"
Tog
f
I
|
[
ଘ
O
2" 0.36 0.66 0.84 1.50 2.34 3.36 4.626.00 9.37 2" 1.14 1.92 2.46 3.24 3 90 4.80 20.66 1.18 1.84 2.65 3.61 4.71 5.96 7.36
L
22" 0.29 0.53 0.67 1.20 1.87 2.69 3.70 4.80 7.50 22" 0.91 1.54 1.97 2.59 3.11 3.84 22" 0.53 0.94 1.47 2.12 2.89 3.774.77 5.89
3" 0.24 0.44 0.56 1.00 1.56 2.24 3.08 4.00 6.24 3" 0.76 1.28 1.64 2.162.60.3.20 3" 0.44 0.79 1.23 1.77 2.41 3.14 3.98 4.91
32" 0.21 0.38 0.48 0.86 1.34 1.92 2.64 3.43 5.36 32" 0.65 1.10 1.41 1 85.2.23 2.74 32" 0.38 0.67 1.05 1.51 2.06 2.69 3.41 4.21
4" 0.18 0.33 0.420.75 1.17 1.682.313.00 4.68 4" 0.57 0.96 1.23 1.621.95 2.40 4' 0.33 0.59 0.92 1.33 1.80 2.36 2.98 3.68
42" 0.16 0.29 0.37 0.67 1.04 1.49 2.05 2.67 4.16 42" 0.51 0.85 1.09 1.44 1.73 2.13 42" 0.29 0.52 0.82 1.18 1.60 2.09 2.65 3.27
I
5' 0.14 0.26 0.34 0.60 0.94 1.34 1.85 2.40 3.75 5″ 0.46 0.770 981.30 1.55 1.92 50.26 0.47 0.74 1.06,1.44 1.88 2.39 2.95
52" 0.13 0.24 0.31 0.55 0.85 1.22 1.68 2.18 3.41 52" 0.41 0.70 0.89 1.18 1.41 1.74 52" 0.24 0.43 0.67 0.96 1.31 1.71 2.17 2.68
6" 0.12 0.22 0.28 0.50 0.781.11 1.532.00 3.12 6" 0.38 0.64 0.82 1.08 1.30,1.60, 6/10
6'0 22 0.39 0.61 0.881.201.57 1.99 2.45
62" 0.11 0.20 0.26 0.46 0.72 1.03 1.42 1.85 2.88 62" 0.35 0.59 0.76 1.00 1.20,1.47 62" 0.20 0.36 0.57 0.82 1.11 1.45 1.84 2.27
70.10 0.19 0.24 0.43 0.67 0.96 1.32 1.72 2.68 7" 0.33 0.55 0.70 0.93 1.11 1.37 7" 0.19 0.34 0.53 0.761.03 1.35 1.70 2.10
|
72" 0.10 0.180.22 0.40 0.62 0.89 1.23 1.60 2.50 72" 0.30 0.51 0.66 0.86 1.04 1.28 72" 0.18 0.31 0.49 0.71 0.96 1.26 1.59 1.96
1
8" 0.09 0.17 0.21 0.380.59 0.84 1.15 1.50 2.34 8" 0.28 0.48 0.62 0.81 0.97 1.20 8" 0.17 0.29 0.46 0.66 0.901.18 1.49 1.84
82" 0.08 0.16 0.20 0.35 0.55 0.79 1.09 1.42 2.20 82" 0.27 0.45 0.58 0.76 0.92 1.13 82" 0.16 0.28 0.43 0.62 0.85 1.11 1.40 1.73
9"0.08 0.15 0.19 0.33 0.52 0.75 1.02 1.33 2.08
Į
9" 0.25 0.43 0.55 0.72 0.87 1.07 9" 0.15 0.26 0.41 0.59 0.80 1.05 1.33 1.64
92" 0.08 0.14 0.18 0.32 0.49 0.71 0.97 1.26 1.97 92" 0.24 0.40 0.52 0.68 0.82 1.01 92" 0.14 0.25 0.39 0.56 0.76 0.99 1.26 1.55
10" 0.07 0.13 0.17 0.30 0.47 0.67 0.921.20 1.87 10" 0.23 0.38 0.49 0.65 0.78 0.96 10" 0.13 0.24 0.37 0.53 0.72 0.94 1.19 1.47
L
11" 0.07 0.12 0.15 0.27 0.43 0.61 0.84 1.09 1.70 11" 0.21 0.35 0.45 0.59 0.71 0.87 11" 0.12 0.21 0.33 0.480.660.861.081 34
120.06 0.110.14 0.25 0.39 0.56 0.771.00 1.56 120.190.320.410.54 0.65 0.80 120.110.19 0.30 0.44 0.60 0.78 0.99 1.22
77|1.0
1
I
Net
Net
Net
Net
32/0
65|0.
10.190.
15
Sec- 0.18 0.37 0.55 0.70 1.07 Sec- 0.06 0.11 0.14 0.25 0.39 0.56 0,77 1.00 1.56 Sec- 0.19 0.32 0.41 0.54 0.65 0.80 Sec- 0.11 0.19 0.30 0.44 0.60 0.78 0.99 1.22
tion
tion
tion
tion
.550.
C to C
of 3%"
Bars
C.
9/6" | | | |
1/2" 5/8" 3/4" 8" 1" 11" 1"
יים יים


24
Thickness of
Table No. 3.
FLOOR SLABS, ROCK CONCRETE, .85% REINFORCEMENT.
TABLE GIVING SAFE LOADS, IN POUNDS PER SQUARE FOOT, FOR FLOOR SLABS, CONTINUOUS OVER SUPPORTS
WITH .85 PER CENT OF CORRUGATED BAR REINFORCEMENT, ULTIMATE STRENGTH OF
THE CONCRETE TAKEN AS 2000 LBS. PER SQUARE INCH.
SPAN IN FEET.
Slab in inches.
10
Area of Steel in
12″ width.
3 0.220" 3" C.R. 6"
8
32 0.280" 3/8"
5"
4
0.320" 2"
42
0.370" ½"
5 0.430" "
2
Corresponding
Size and Spacing of
Corrugated Rounds.
Size. Spacing.
82
9
9%
52
5%½
0.480"/2½"
6 0.520" 2"
6½
7
7%
8
(เ
"
66
"
66
66
0.570" 5" 66
8
0.63" 5"
66
"
66
440 265 170 115 75
55
38
44
675
72" 970
420 270 185 135 95 65
600 400 280 150 105 80
385 280
55
50
6½" 1300
800 | 545
120
95
65
57
130
100
80 69
63
65
69
90
75
75
180 150
115
95
80
82
335
270
225 180
150
5½"
120 100 75
88
315
255 210 175
145 115 95 80
215 160
5½″ 1700 1070 715 505 375
375 285 215 170
5″ 2130 1340 905 645 480 365 280 220
175 135 110 90
42" 2600 1650 1120 800 600 455 355 230 220 180 140 115
6½"
1980 1360 970 | 730
730 555 435 345
345 275 225
2450 1600 1160 865 665 520 415
2750 1825 1320 990 760 595❘ 480 380
2130 1545 1160 900
900 710 565 455
2450 1780 1340 1040 | 815 655 535
2770 2020 1525 1185 930 750❘ 615 505
2280 1720 1345 1060 860 700 580
2550 1930 1500 1190 965
965❘ 790 655 545 465
.
465 390 330 280 240 200 125
For end panels or slabs free at one end, use % of above loads.
For single panels or slabs free at both ends, use ½ of above loads.
0.66" 5/8"
94
5½"
370
310 255 215
0.710" 5%"
5″
175 145 125 100 100
210 175 145 125 107
0.770" 34" "
375 305
7"
435
250
415 350 295
0.810" 34"
6%
6½"
245 210 175 150 113
340 | 290 240 210 180 119
480 400
0.860"¾4"
8/11 64
6"
0.910" 4 5½"
Safe loads given are in addition to weight of slab.
This table only applies when high elastic limit corrugated bars are used.
"
4
•
LO
5
CO
6
7
•
8
91 10 11 12 13
་
14 15 16 17 18
:
:
V:
•
⠀⠀
•
·
19 20
...
:
•
Weight of Slab,
in lbs. per
8q. ft.
25
Table No. 4.
FLOOR SLABS. ROCK CONCRETE. I PER CENT REINFORCEMENT.
ULTIMATE STRENGTH OF
TABLE GIVING SAFE LOADS IN POUNDS PER SQUARE FOOT, FOR FLOOR SLABS, CONTINUOUS OVER SUPPORTS, WITH
ONE PER CENT OF CORRUGATED BAR REINFORCEMENT.
THE CONCRETE TAKEN AS 2500 LBS. PER SQUARE INCH.
SPAN IN FEET.
Slab in Inches.
Thickness of
3
32
4
42
10
5
52
6
62
6½
7
7½
8
Area of Steel in
Width.
12″
92
10
~
Corresponding
Size and Spacing of
Corrugated Rounds.
Size. Spacing
0.27" 3" C.R. 5″
8
4″
6"
0.330" 3%" 46
8
1
0.390"2"
(6
""
"L
C
"
66
46
66
0.450" 1/2"
0.510" 2½"
0.570" ½"
0.630" 5%"
0.69" 5%"
0.750" 5%"
0.800" 5/8"
0.86" 34"
82 0.92" 3/4"
9
0.98" 34"
4%
1.040" 34"
1.100" 134″
Safe loads given are in addition to weight of slab.
This table only applies when high elastic limit corrugated bars are used.
"
"
66
66
4
42"
6"
LO
5½"
5½"
5″
5
CO
6
525
325 215 145 100
810 505 335 235 170
1160 720 485 340 250
5½"1550 970 660 470 345
4½" 2000 1260 850 615 455
4"
1580 1080 775 575
1950 1335 955 715
2370 1620 1165 870
6"
5½"
5″
7
8 9
10 11 12 13 14 15
:
:
16
·
17 18 19
70
125
90
190
145
110 85
260 200
155 120
95
345 270
270 210 165
130
105 80
440
345
270 220 175
140 115
90
550
430
180❘ 150
120 100
670
525
225 190
155
130 105
275 230
190
160 135 110
395
325 270
220
185 155 130 110 94
345 275 225
420 340 280
510 410 335
580 475
685 560 460
790 640 535
615 515
515 | 430
700 590 500
385 320
270
225 190 160 135 100
445 375
315
265 225 190 165 107
1920 1380 1040 805 635
2170 1580 1185 915 725
1840 1380 1070 850
2110 1590 1235 975
1810 1410 1115 905 740
2050 1600 1275 1030 850
2300 1800 1430 1165 960 795
For end panels or slabs free at one end, use % of above loads.
For single panels or slabs free at both ends, use ½ of above loads.
365
310 265 230 195 113
420 360 310 265 230 119
665 565 485410 355 305 260 125
:
20
:
Weight of Slab
in Lbs. per
Sq. Ft.
38
44
50
57
63
69
75
82
8888
26
→
~
STANDARD SIZES AND TYPES OF CORRUGATED BARS.
Corrugated Bars are Rolled to the Sizes and Weights Given Below.
Unless otherwise specified, we furnish corrugated bars rolled from a high
elastic limit steel, with a yield point of 50,000 pounds or over. We are prepared,
however, to roll any grade of steel desired, and can supply corrugated bars to
meet special requirements as to elastic limit, from open hearth or Bessmer stock.
OLD STYLE-TYPE A.

SIZE
NET
SECTION
WEIGHT
PER FOOT
SIZE
NET
SECTION
WEIGHT
PER FOOT
1/2"
.180"
.64 lb.
{
1/4"
3/4"
.370"
SIZE
NET SECTION
WEIGHT PER FOOT
1.35 lbs.
7"
}
.550"
1.95 lbs.
NEW STYLE-TYPE B.
1"
.700"
2.70 lbs.
3/4"
1/3 3/8" 1½" 5%"
I" 114"
1
.060".110".140" .250" .390" .560".770" 1.000" 1.560"
7/8"
1/4"
1.070"
4.00 lbs.

CORRUGATED SQUARES-TYPE D.
110 MITT
.24 lb. .38 lb. .48 lb. .85 lb. 1.33lbs. 1.91 lbs. 2.60 lbs. 3.40 lbs. 5.31 lbs.
CORRUGATED ROUNDS-TYPE C.

18"
14"
3/8" 1/2" 9/16" 5/8" 3/4″ 7/8" 1"
SIZE
NET SECTION .110″.190″.250" .300 .440" .600″ .780″ .990" 1.220″
WEIGHT PER FOOT .38 lb..66 lb..86 lb. 1.05lbs. 1.52lbs. 2.06 lbs. 2.69 lbs. 3.41 lbs. 4.21 lbs.

"
1/4" 3/8 72 5/8"
3/4" 7/8" 1″ 1%" 14"
SIZE
NET SECTION .060".140".250" .390".560".760" 1.000" 1.260" 1.550"
WEIGHT PER FOOT .22 lb. .49 lb. .86 lb. 1.35 lbs. 1.94 lbs. 2.64 lbs. 3.43 lbs. 4.34 lbs. 5.35lbs.
CORRUGATED FLATS-UNIVERSAL TYPE.

No. 6
No. 3 No. 4 No. 5
.410" .540" 650" .800"
1.35 lbs. 1.97 lbs. 2.27 lbs. 2.85 lbs.
A variation in weight of 5% either way is required.
No. 1 No. 2
.190" .320"
.73 lb. 1.18 lbs.
27
WEIGHTS, AREAS AND CIRCUMFERENCES OF ROUND BARS.
DIAMETER OF
BAR IN INCHES.
0
1
2
+100
moto
18
5
18
soko
17/8
HA
0/0 0 0 0 n/a No 10/0
198
1 1
16
12108
13
16
7
15
1 8
too no + ja
3
18
18
Coco
No to alo aia0 m od 00/000
7
10
9
16
1 1
呆
​}
1-60
15
18
WEIGHT OF
BAR 1 FT. LONG.
.042 lbs.
.094
. 167
.261
.375
.511
.667
.845
1.043
1.262
1.502
1.763
2.044
2.347
2.670
3.014
3.379
3.766
4.173
4.600
5.049
5.518
6.008
6.520
7.051
7 604
8.178
8.773
9.388
10.020
10.680
AREA OF BAR
IN SQ. INCHES.
0123
.0276
.0491
.0767
.1104
.1503
.1963
.2485
.3068
.3712
.4418
.5185
.6013
.6903
.7854
.8866
.9940
1.1075
1.2272
1.3530
1.4849
1.6230
1.7671
1.9175
2.0739
2.2365
2.4053
2.5802
2.7612
2.9483
3.1416
CIRCUMFERENCE OF
BAR IN INCHES.
.3927
.5890
.7854
.9817
1.1781
1.3744
1.5708
1.7671
1.9635
2.1598
2.3562
2.5525
2.7489
2.9452
3.1416
3.3379
3.5343
3.7306
3.9270
4.1233
4.3197
4.5160
4.7124
4.9087
5.1051
5.3014
5.4978
5.6941
5.8905
6.0868
6.2832
28
1
je vagy by s ma
SIDE
IN INCHES.
0
general
1
WEIGRTS. AREAS AND PERIMETERS OF SQUARE BARS.
2
+4
3
18
+ *** yo
13
142424
1988
f
÷1
#
180
+ 1000
18
* ∞ + do
Å
ap
16
1424
apo vabo por act as
1 1
6
1
book o
WEIGHT OF
BAR 1 FT. LONG.
.053 lbs.
.119
.212
.333
.478
.651
.850
1.076
1.328
1.608
1.913
2.245
2.603
2.989
3.400
3.838
4.303
4.795
5.312
5.857
6.428
7.026
7.650
8.301
8.978
9.682
10.410
11.170
11.950
12.760
13.600
AREA OF BAR
IN SQ. INCHES.
.0156
.0352
.0625
.0977
.1406
.1914
.2500
.3164
.3906
.4727
.5625
.6602
.7656
.8789
1.0000
1.1289
1.2656
1.4102
1.5625
1.7227
1.8906
2.0664
2.2500
2.4414
2.6406
2.8477
3.0625
3.2852
3.5156
3.7539
4.0000
PERIMETER OF
BAR IN INCHES.
.500
.750
1.000
1.250
1.500
1.750
2.000
2.250
2.500
2.750
3.000
3.250
3.500
3.750
4.000
4.250
4.500
4.750
5.000
5.250
5.500
5.750
6.000
6.250
6.500
6.750
7.000
7.250
7.500
7.750
8.000
29
11
WEIGHTS OF VARIOUS BUILDING MATERIALS.
Lbs. per cu. ft.
150
125
90-100
170
140
165
160-170
165
170
145-150
175
62%
I 20
90-105
100
150
90
140
450
480
100
Brick, Pressed
Brick, Common
Earth, Rammed.
Granite...
Granite Rubble Masonry.
Granite Masonry, Well Dressed.
•
Limestone.
Limestone Rubble Masonry
Marble...
Sandstone
Slate....
•
Rock Concrete.
Cinder Concrete.
Plaster
•
Water (1 cu. ft. 7.48 U. S. gals.).
Gravel
Sand, Dry.
Mortar.
Cast Iron.
•
•
•
•
•
Steel..
Paving Asphaltum..
•
•
•
•
Weight of Brick Walls, per Superficial Foot.
9-inch wall.
84 lbs.
22-inch wall.
13-inch wall
121 lbs.
26-inch wall.
18-inch wall.
168 lbs.
A bar of steel 1-inch square and I foot long weighs 3.40 lbs.
•
•
•
USEFUL INFORMATION.
Circumference of circle-diameter X 3.1416.
Diameter of circle Xo.8862=side of equal square.
Base of triangle X½ the altitude area.
0.7854 Xproduct of diameters=area of ellipse.
Surface of sphere=circumference X diameter.
Cubical contents of sphere surfaceX % diameter.
Area of trapezoid=altitude X½ sum of parallel sides.
Cubical contents of cone-area of base
I meter=39.34 inches.
1-inch=2.54 centimeters.
...205 lbs.
243 lbs.
perpendicular height.
30
SAFE LOADS, UNIFORMLY DISTRIBUTED FOR RECTANGULAR SPRUCE OR
WHITE PINE BEAMS ONE INCH THICK. FROM "CARNEGIE.”
Span
in
Feet.
6678 –›
5
R=224 99500 222** 99≈≈≈
12
13
14
15
16
18
19
20
21
23
24
27
6"
28
29
600
500
יי7
820❘ 1070
680
890
430
580
760
380 510
330 460
300
410
270 370
250
340
230
310
210 290
25 120
26
8"
อ
230
210
670
590
200
270 360
190 260 330
180
240
310
170
290
160
280
9"
1350
1120
160
110 160
110 150 200
110 140 190
110 140 180
530
490
440
560
410 520
380
670
610
960 1190
840❘ 1040
750 930
830
760
690
640
480 590
450
420
400
370
360
150 200
270 340
140
190
260
320
140
190
240
310
130
180 230
130
170
220
DEPTH OF BEAM.
290
280
210 270
210 260
10" 11" 12″ 13" 14" 15" 16"
1670
1390
2020
1680
2400 2820
2000 2350
1440 1710 2010
1260 1500 1760
1120 1330 1560
1010 1200 1410
920 1090 1280
840 1000 1180
780 930 1080
720 860
1010
560
520
490
460
440 530
420
390
380 460
360 440
350 420
510
480
330 410
320 390
250 310 370
240 300 360
230 290 350
1090
670 800 940
630 750 880 1020
710 830 960
560 670 780 910
590
630 740
860
8888888
3270
3750 4270
2730 3120 3560
2330 2680 3050
2040 2340 2670
1810❘ 2080 2370
600 710
570 670
540 640
520 610
500 590
1630 1880
1490 1710
1360
1260
1170
820
780
740
710
680
2130
1940
1560 1780
1440 1640
1340
1530
1250 1420
1330
1260
1180
1100
1040 1190
990 1130
940 1070
890 1020
850
970
810
920
780
890
480 560
750
660
460 540 630 720
440 520 610
430 500 580
410
860
820
690 700
670 760
490 560 640 740
For oak increase values in table by 3.
For yellow pine increase values in table by 3.
To obtain safe load for any thickness: Multiply values for 1 inch by
thickness of beam.
To obtain the required thickness for any load: Divide by safe load for
1 inch.
31
SAFE LOADS FOR RECTANGULAR WOODEN PILLARS (SEASONED)
FROM "CARNEGIE."'
Ratio of Length
to Least Side
l/d
12
14
16
18
2 2 2 2 28
20
22
24
26
30
32
34
36
38
40
1=length of pillar in inches.
d-width of smallest side in inches.
SAFE LOADS IN POUNDS PER SQUARE
INCH OF SECTION.
Yellow Pine
(Southern).
995
955
913
869
825
781
738
697
657
619
583
549
516
487
458
•
White Oak.
818
785
750
715
678
642
607
575
541
509
479
451
425
400
377
White Pine and
Spruce.
707
679
649
618
587
556
525
495
467
440
414
390
367
346
326
32

FORMULAE FOR BENDING MOMENTS.
W= total load; w-load perft. Mc⋅ moment at center; Me=moment at end; My moment at load
W↓
l
Max.Mom. Mε = wl.
គ
WI
K½l
1½l
Max. Mom. = Mc = 4 Wl.
W
Max. Mom. = Mw= Wab
wab.
JW
1/2 l
1/2 l
Max. Mom. =Mε = Mc = &wl.
++
b
Max.Mom. = Mε = ½½ w l²
kkk
l
Max. Mom.: Mc = 1 w l²
1
डोट
Max. Mom = Mw = 1/2 Wα.
l
Mε = 1/2wl? Mc=214 wl?
33
↓
•
APPROXIMATE BENDING MOMENT FORMULAE
MONOLITHIC CONSTRUCTION.
FLOOR SLABS.
For interior panels surrounded on all sides by beams, take
Mc=Me=16212.
For end panels, take Mc=Me (at inside beam)=wl.
For slabs supported at ends only take Mc=&wl²; Me=0.
A
BEAMS.
A
Uniform Loads.
For interior beams, continuous over a number of supports, take
Mε Me=1 zwl².
2
For end beams take Me Me (at inside support)=%w/.
Concentrated Loads.
For interior beams, continuous over a number of supports, use 2/3
of moment which would exist in a simple beam.
For central load take M.—Me=/ WI.
For end beams, centrally loaded, take Mc=M。 (at inside support)
= WI.
34
REQUIREMENTS OF VARIOUS
BUILDING CODES GOVERNING
REINFORCED CONCRETE
CONSTRUCTION.
Mix...
Compressive Strength, 28 days
Concrete in direct compression
allowed stress
COLUMNS.
Spiral winding or Hooping,stress
per "..
Maximum ratio height to least
diameter
Insulation..
**
..
SLABS.
Minimum thickness.
Insulation..
•
.
BEAMS.
Maximum fibre stress on con-
crete, compression...
Working stress in steel
Tensile stress in concrete.
Shearing stress in concrete..
...
Adhesion of steel to concrete....
Beams...
•
Ratio of moduli..
Shearing stress for steel..
Width of slab allowed for T-beam
action
Insulation.
··
BENDING MOMENTS.
·
Slabs, continuous
Square panels supported on four
sides, reinforced in both direc-
tions....
TESTS.
Load..
Deflection allowed....
Method of calculation ……..
BOSTON. ST. LOUIS
1-5
2000
350
120 times
least radius
gyration.
1½
3/4
500 to 600
16000
None
60
{
Beams, Slabs
60
15
Cols.-10
10000
½ Span.
12
W I
10
Contin.
W I
10
1:2:4
Rock or
Gravel,
2000
500
Special.
15
2
1
800
Medium Steel,
14000
High E. Limit,'
20000
None.
25
†Medium Steel,
50
High E. Limit,
30
1/4 Span.
12
πι
12
W l
8
}
SAN FRAN
CISCO.
1:6
2000
450
{
700
15
12
32
1
500
| *% Elastic
Limit.
None.
75
75
15
10000
5 times thick
ness of Slab.
1½ times
diam. bars.
πι
12
W I
20
WI
8
2 times Safe
Load.
1/700
Safe Load. Safe Load.] Safe Load.
PHILA.
DELPHIA.
1:2:4
Stone & Gravel,
500
Slag, 300
1000
15
2
1
Stone & Gravel,
600
Slag, 400
16000
None.
Stone & Gravel,
75
Slag, 50
Stone & Gravel.
50
Slag, 40
Stone & Gravel,
12
Slag, 15
20 times thick-
ness of Slab.
2, Bottom
132 Side
W I
10
W l
20
Figure as
T-Beams,
½ W l
2 times Safe
Load
*Steel with an elastic limit above 40000 lbs. shall have mechanical bond.
+Deformed bars. Value determined by test.
NEW
FRENCH
RULES
1906.
28% of
Strength at
90 Days
60% of
Strength at
90 Days.
18
3/4
(
3/4
50% and 40%
of E. Limit.
1/10 Allowed
Stress in
Direct Comp
1/10 Allowed
Stress in
Direct Comp.
8 to 15
Span, or s
Spacing of Ribs
3/4
Built in, End
1/12 W l.
Center,
1/24 W l.
Partially
Fixed,
1/10 Wl.
Safe Load for
24 hours.
Deflection
must cease
after 15 hrs
No Sign of Cracks.
Ult. Strength, Safe Load.
4× (LL+DL).
35
Brief Specifications for Materials and Labor Used in
Reinforced Concrete Construction.
CEMENT.
Only Portland cement conforming to the requirements of the speci-
fications adopted by the American Society for testing Materials, June
14, 1904, shall be used.
SAND.
Sand to be clean and coarse, and free from organic matter, a graded
sand, with coarse grains predominating, is to be preferred.
COARSE AGGREGATE.
Broken stone to be hard, durable limestone or its equivalent, free
from dust and foreign materials; maximum-sized particles to pass
through a one-inch ring; fines to be removed by passing over a one-
quarter inch screen. The fines may replace part of the sand. Under
special conditions making for uniformity crusher run may be used, ·
but this is not desirable.
Gravel shall be clean and of graded sizes; the sand carried to be
removed by screening as for broken stone.
PROPORTIONS OF MIX.
Concrete for columns, beams and slabs to be mixed in the propor-
tion of one part cement to six parts aggregate; proportions by volume
taking one bag containing not less than 94 pounds of cement, equal to
one cubic foot of cement.
The proportions of fine and coarse aggregrate used shall be chosen
so as to give a concrete of maximum density; in no case, however,
may the amount of fine aggregate be less than 50 per cent of the coarse.
REINFORCING STEEL.
Steel may be made by either the Bessemer or open-hearth process;
bars to be rolled from billet stock. Re-rolled material will be accepted
under conditions insuring rigid inspection.
The elastic limit and percentage of elongation shall be determined
by tests on accurately-machined specimens, and shall conform to the
following requirements:
Elastic limit to be from 50,000 to 60,000 pounds per sq. inch.
The percentage of elongation in 8" must not be less than given by
the formula:
Percentage of elongation=
I,400,000
Ult. Strength
-5.0.
BENDING TEST.
Bars as rolled shall bend cold, 90 degrees, to a radius=four times
the least diameter of the specimen, without sign of fracture.
36
MIXING.
All concrete to be machine mixed if possible. Concrete for rein-
'forced concrete work shall be mixed wet, sufficient water being used
to make a mass that will flow readily and be of such consistency that
the reinforcing steel will become covered with a protective coat of
fluid mortar. Excess water should be avoided as it tends to a separa-
tion of the parts.
FORMS.
Suitable forms must be provided so that the finished work will
accurately conform to the dimensions shown on plans and must be
such that the finished work will present a smooth and pleasing
appearance. Exposed corners should be beveled where possible.
PLACING.
CONCRETE-All concrete will be of a mushy consistency and care.
must be exercised in the placing so that the concrete is not too thin
or the wheel too long so as to permit settlement and consequent separ-
ation of the parts. It will not be permitted to drop the concrete any
considerable distance. All concrete to be thoroughly puddled, and
exposed faces worked so that a smooth finish will result.
In floor and beam work the concrete must be placed to the full
depth at one operation. Joints in the work to be made as directed
by the architect or engineer. It will not be permitted to wheel over
or otherwise disturb freshly laid concrete.
STEEL-All reinforcing steel must occupy the exact location
called for by the drawings, to accomplish this the following rules
must be observed:
I.
Bars and stirrups must be bent or otherwise fabricated in
exact accordance with the details.
2. The bars and stirrups comprising the reinforcement for any
member shall be assembled by adequate means so that the whole may
be inspected and placed in the form as a unit.
3. The means adopted must be such that the reinforcement can-
not be displaced by the operation of placing the concrete.
The brief specifications here given are intended to be suggestive-when our
materials are used we will furnish complete specifications covering the reinforced
concrete construction, if desired.
:
37
Adhesion (adherence)..
Advantages of Reinforced Concrete Con-
struction...
ALLOWABLE STRESSES-
Building Ordinances
For Columns..
Average Rock Concrete Beams, Formula.
BARS-
▸ ·
crete....
Safe Loads on Wooden...
Shearing Provisions.
Shearing Stresses.
... •
24
Corrugated, Squares, Flats and Rounds. 27
Spacing for Required Areas-Table...
Weights, Areas and Circumferences,
Rounds...
28
Weights, Areas and Perimeters, Squares 29
BEAMS-
Bending Moments Due to Various
Loadings..
Bonding Stresses.
Formulæ for Moment of Resistance.
General Formula for Reinforced Con-
Stirrups.
Strength of Reinforced Rectangular...
Strength of Reinforced T Sections...
Table for Design of Beams 12 inches
Wide ...
Webb Stresses in-See Shearing
Stresses.
Webb Stresses in T-Beams.
BENDING MOMENT FORMULE-
Advantages of Concrete...
Mill Construction.
Tests ...
Varies with Mix
··
•
· *
Rectangular Beams, Ultimate.
Rectangular Beams, Working Stresses.
T-Beams..
· ·
• ·
INDEX.
• • •
...
Page
6
··
·
• •
• +
..
•
•
• ..
•
8
9
10
Bending Moments, Due to Various Loads. 33
6
Bond, Importance of.
BUILDING CONSTRUCTION-
• •
·
5
•
355
Reinforced Concrete..
Steel Frame....
Building Laws, Various Cities.
Building Materials, Weights of.
Cinder Concrete, Beam Formula.
COLUMNS
Allowed Stresses, Building Ordinances 35
Design....
15 to 19
Diagram for Design..
18
Formula for Design.
Plain Concrete, Tests
Tests..
With Longitudinal Reinforcement...
With Spiral Winding or Hoops..
COMPRESSIVE STRENGTH-
15
•
က
•
33
12
8
∞
9
31
12
I I
13
7
10
23
14
10
5
4
4-5
17
16
16, 18
17
19
4
35
30
8
16
21
CONCRETE-
Compressive Strength
Data...
Quantities of Materials for.
Shearing Strength
Specifications for Materials
Tensile Strength, Maximum Extension.
Continuous Beams, See Monolithic Con-
struction.
Corrugated Bars, Standard Sizes, Sec-
tions and Weights...
Costs, Building Construction, Com-
parative ..
••
M
Critical Percentage Depends on..
High Yield Point is Desirable.
Of Corrugated Bars...
Specifications for.
Elongation, Specification for.
Critical Percentage of Reinforcement....
Crushing Strength of Concrete, Table.
Diagram for Column Design.
ELASTIC LIMIT.
•
··
Flats, Corrugated, Universal Type....
FLOOR SLABS-
Stirrup Spacing.
T-Beams
1
For Bonding Stresses.
General, Design of Beams.
Monolithic Construction
Rectangular Beams..
•••
•
•
•
Formulæ for Design.
Tables for Design of....
FORMULE-
Based on Working Stresses, Beams....
Bending Moments in Beams..
Columns with Longitudinal Reinforce-
ment
• •
··
·
• •
...
··
...
•
...
•
...
•
Page
Useful Information, Miscellaneous
General Remarks on Theory of Design...
Girders, See Beams.
High Elastic Limit Desirable...
Hooped Columns..
Importance of Bond
·
[
LOADS-
Safe Loads, Columns, Long, Reinforce-
ment
Safe Loads on Floor Slabs, Tables
Safe Loads, Wooden Beams
Safe Loads, Wooden Posts....
Modulus of Elasticity of Concrete Varies.
Moments of Resistance of Beams, 12 inches
Wide
Monolithic Construction
Notes on Design in General ...
Portland Cement, Brief Specifications
Posts, Safe Loads on Wooden
•
………♥
•
·
·
··
21
21
22
21
36
21
19
···
27
5
7
16
18
.20-34
25, 26
9
6
2 3 333
27
36
36
27
9
33
17
12
9
34
8
13
IO
18
25, 26
31
32
17
30
5
6
19
6
23
19-34
5
36
32
1
38
Page
Quantities of Materials Required per
Cubic Yard Concrete.
REINFORCED CONCRETE-
Advantages of this Construction.
Digest of Building Laws.
Theory of Design
Reinforcing Steel, Specifications for.
Rounds, Corrugated
Rounds, Plain...
Safe Loads, See Loads
Sand, Brief Specifications
SECTIONS-
Weights and Areas of Corrugated Bars.
Weights, Areas and Circumferences of
Round Bars
Weights and Areas and Perimeters of
Square Bars....
...
...
...
..
...
•
pop
··
· •
•
· ·
· • • •
•
22
•
5
35
5
♡♡♡ 2008
36
27
36
Shear, Allowed Values, Building Ordin-
ances
35
Shearing Provisions..
12
21
Shearing Strength of Concrete.
Shearing Stresses in Beams
Slabs, Floor Slabs, Tables for Safe Loads. 25, 26
II
27
28
29
Page
24
Spacing of Bars for Required Areas.
Specifications, Concrete Materials.
Squares, Corrugated....
36
27
36
Steel, Brief Specifications for.
Stirrups, Formula for Spacing.
Stresses, Allowed by Building Ordinances 35
13. 14
10
14
T-Beams, Formula..
T-Beams, Web Stresses
Tables, See Particular Subject.
Temperature Stresses
Tensile Strength of Concrete
Tests. See Particular Subject.
Theory of Design..
•
··
...
• •
•
··
•
• •
·
•
•
► • •
Universal Type, Corrugated Flats..
Useful Information...
Web Stresses in Beams, See Shearing
Stresses....
•
·
••
II
Webb Stresses in T-Beams
14
Weights of Various Building Materials... 30
Working Stresses, Allowed by Ordinances 35
Working Stresses, Formula for Beams
Yield Point, High Elastic Limit is Desir-
able
9
•••
21
21
5
27
30
6
39

Designing Methods
Reinforced Concrete Construction
VOL. 1
JUNE, 1908
BUILDINGS
DETAILED DESIGN OF TYPICAL BUILDING
COMPLETE ANALYSIS OF THE STRENGTH OF RECTANGU-
LAR AND "T" SHAPED BEAMS
The design of the typical building is based on the methods outlined
in our May bulletin. The analysis for the strength of rectangular
and T-shaped beams has been included in this bulletin, owing to
the large demand for our former discussion. The matter here given
is similar to that which appeared in our 1906 catalogue, but the
discussion has been revised and considerably extended.
THE JULY BULLETIN WILL TREAT OF:
No. 2
Highway bridges and culverts of the flat
slab and girder type.
REPRINT, MARCH, 1909
PRICE, 25 CENTS
CORRUGATED BAR COMPANY
(FORMERLY EXPANDED METAL AND CORRUGATED BAR CO.)
National Bank of Commerce Building
SAINT LOUIS
COPYRIGHT 1908, BY EXPANDED METAL AND CORRUGATED BAR CO.
DESIGN OF A REINFORCED CONCRETE BUILDING.
The following detailed design is given to illustrate methods of
computation and to bring out our practice in the application of the
formulæ given in the May bulletin.
The design will be based on the following data:
Lot plan; corner lot, 54'-0" x 100'-00".
Column and beam arrangement to be as shown on plan.
Building to be five stories and basement, as indicated on longitud-
inal section.
Exterior walls, 13" brick walls on spandrel beams.
Allowable soil pressure, 5,000 lbs. per □′.
Roof to be designed for safe live load of 40 lbs. per '.
Floors to be designed for safe live load of 150 lbs. per '.
Concrete to be a 1:2:4 mix, good rock or gravel.
All members subjected to flexure will be designed for their break-
ing or ultimate load, using a factor of 4 on the live load, and 2 on the
dead load.
In the design of beams, girders and slabs, allowance will be made
for "continuous action." The roof will be figured to carry full live.
load on all members-the floors to carry full live load on slabs and
beams, and 85% of full live load on girders.
In the design of columns the percentage of live load for which they
are to be figured is usually prescribed by ordinance. For a building
of this kind the average ordinance would permit the columns to be
figured to carry 85% of full live load.
NOTE-The loads specified agree with those used in light warehouse prac-
tice, and the design is much heavier than would be required for hotel or office-
building construction,
ROOF SYSTEM.
SLABS.
Span: 10'-0" c. to c. of beams.
L. L. 4x40 lbs.
160 lbs.
D. L. 2x45 lbs.
90 lbs.
Ultimate load per '.
Ultimate load per '.
Designing load=250 lbs.=Ultimate load per □'.
The bending moment at the center of span for roof and floor slabs
will be taken as 1 w2 for interior panels and as w² for end panels.
6
All slabs will be figured on a basis of .85% of reinforcement.
-
1
2
INTERIOR PANELS.
Ultimate moment=1X250X10²= 1,560 ft. lbs.
18,720 in. lbs.
For this percentage of reinforcement Mo=370 bd². (See formula 1,
page 8.)
42
b=12", therefore M. 4,440 d=18,720" lbs., from which d=2.06".
q=.0085 bd=.0085 X 12 X 2.06=0.21".
For practical reasons, it is not advisable to make the slab less than
3½" thick, in which case d may be taken equal to 2½". We will use
this thickness of slab and reduce the amount of reinforcement as found
2.06
above in the ratio ; using 0.173 sq. inches of metal.
2.5
Make slab 3½" thick, ½" corr. sq. N. S. 8″ cts.
Since the above design assumes reverse bending moment over the
beams, the same amount of reinforcement should be placed in the top
of slab at that point. This reinforcement will consist of " corr. sq.
N. S. 8" cts. extending to the quarter points of the panels.
It is desirable to have transverse reinforcement in slabs to prevent temper-
ature of shrinkage cracks, and also to act as distributing bars.
•
END PANELS.
Mo=12 wl²=24,800 inch lbs. (ult.)
The ultimate resisting moment of a 3½" slab, assuming d=2½",
reinforced with .85% reinforcement=M。=370_bd²=370X12 X 2.5²=
27,750 in. lbs. It is desirable to keep the thickness of slab uniform,
and to illustrate a different method we will determine the amount of
steel by the formula Mo-FqX.86d=24,800" lbs. 50,000XqX.86X
2½=24,800, from which g=0.230"-3" corr. squares, N. S. 6" cts.
BEAMS.
We will design a typical interior beam; B5.
Span of beams, c to c of supports,=18′-0″.
Distance between beams, c to c.10'-0".
Ultimate load on beam:
c,
Live Load =40X10X 18=7,200 lbs.; X+=28,800 lbs.
D. L. Slab =45X10X 18-8,100 lbs.; X2-16,200 lbs.
D. L. Beam=10X 10X 18=1,800 lbs.; X2= 3,600 lbs.
Actual, 17,100 lbs.; Ult., 48,600 lbs.
Mo=13 WI=1X48,600 X 18-72,900' lbs. 874,800" lbs. (ult.)
In beams and girders we will use 1.3% reinforcement, and design
by formula 2, page 8; Mo=570 bd. Assuming b=9", we have M。=
570X9d²=874,800, from which d=13.1″.
g=.013 bd=.013×9×13.1=1.54 sq. inches.
d-distance from top of beam to plane of metal.
Make beam 9"x15"; 4-%" corr. squares N. S.; bend up two
bars and extend to quarter points of beam in adjacent panel. This
design gives about 2" of fireproofing on the under side of the bars.
43
For complete discussion see page 12.
It will be assumed that the concrete will carry safely 50 pounds of
vertical shear per square inch of cross-section of beam.
We will then have Ve=bd v.=9X13.1 × 50=5,900 lbs. Stirrups in
roof beams will be 4" corrugated squares, bent in U-shape; allowed
working stress=16,000 lbs. per sq. in. Then P=2X0.06X16,000
lbs. 1,820 lbs. total allowable stress in one stirrup.
To determine stirrup spacing, use formula on page 14.
0.86 d P 0.86 X 13.1X1820 20500
V-Ve
y=
V-5900
V-5900
The spacing at any point may be determined by substituting the
value of the external shear, V, at that section in the above.
The following table illustrates the method:
Distance
from End of
Beam.
SHEARING PROVISIONS.
I
2'
4
Size of
Stirrups.
I
1/4"
Corrugated
Squares
New Style.
Vert. Ext.
Shear
=V.
85'50 lbs.
7600 lbs.
6650 lbs.
4750 lbs.
Vc
5900 lbs.
5900 lbs.
5900 lbs.
5900 lbs.
GIRDERS.
V-Vc
2650 lbs.
1700 lbs.
750 lbs.
Spacing="
0.86 d P
V-Vc
8"
12"
1
Beams in end panels to be figured in a similar manner, using formula
Mo=1% WI.
27"
Typical interior girder, G4.
The actual dead and live load concentrated at the center of the
girder is 17,100 pounds, corresponding to an ultimate load of 48,600
pounds, not including the weight of the girder itself.
Ult. moment at center M=% WI. (For a beam simply supported
this would be 4 WI. for center load.)
Mo=%X48,600 X 20=
162,000 ft. lbs. (ult.)
Assume girder weighs 250 lbs. per ft., then
M due to its weight=X250X20, times 2=
TOTAL ULTIMATE MOMENT
16,700 ft. lbs. (ult.)
178,700 ft. lbs. (ult.)
=2,144,400 in. lbs. (ult.)
Assume b=12". As before 570X12Xd=2,144,400 from which
d=17.7". q=.013 bd.=2.76 sq. in.
Make girder 12"x20"; 5-%" corrugated squares N. S.
The same amount of reinforcement should be used in the top of
the girder over the supports. This will be provided for by bending
up two bars in each girder, and using an additional 4-inch bar, placed
as shown on drawing.
44
い
​$
SHEARING PROVISIONS.
Neglecting the weight of the girders, the shear is constant from the
column to center of span.
V_17,100
8,550 lbs., neglecting weight of girder.
2
Ve=12X17.7X50=10,620 lbs.
Although at the allowed stress of 50 lbs. per sq. in. in the concrete,
no stirrups would be required, it is essential that at the columns and
where the beams frame into girders, that stirrups be provided to dis-
tribute the concentrated loadings at these points.
GIRDERS IN END PANELS.
Design girders G3 in a similar manner, using the formula M=} WI.
(The formula WI and WI are for one concentrated load at center of span.)
In the design of end beams and girders it is desirable to increase
the percentage of reinforcement, keeping same dimensions as for
interior beams, rather than make the percentage of reinforcement
constant and change the size of the beams.
FLOOR SYSTEM.
L.
L. 150 lbs. X4=600 lbs.=ult. load per '.
D. L. 60 lbs. X2=120 lbs. ult. load per '.
SLABS, INTERIOR PANELS.
Span=10′-0″ c to c beams.
ult. load per □'.
4,500 ft. lbs. (ult.)
54,000 in. lbs. (ult.)
As before Mo=370 bd=54,000, from which d=3.5 in.
9=.0085X12X3.5=0.360 sq. inches per foot width.
Make slab 42" thick; 2" corrugated rounds 6½" cts.
Design end panels by formula Mwl.
BEAMS.
720 lbs.
Mo=1'e wi²=16X720 X 10²=
Typical interior beam, B5.
Span, c to c of supports,=18′-0″.
Spacing of beams, c to c,=10′-0″.
L. L.,
D. L. (Slab),
D. L. (Beam),
TOTALS...
ACTUAL LOAD ON BEAM.
150X10X18=27.000 lbs. X
60X10X18=10,800 lbs.
15×10×18= 2,700 lbs.
.40,500 lbs.
•
X FACTOR
XXX
422
ULTIMATE LOAD ON
BEAM.
108,000 lbs.
21,600 lbs.
5,400 lbs.
135,000 lbs.
45
Mo=12 WI=12X135,000X18= 202,500 ft. lbs.
=2,430,000 in. lbs. (ult.)
Using 1.3 per cent of reinforcement, Mo=570 bd²=2,430,000.
Assuming b=12"; d=18.9" and 9=2.960".
Make beam 12"x21"; 4-" corrugated squares, N. S.
Bend up two bars in each beam.
Figure beams in end panels similarly, using formula MWI.
1
I
V at end-40,500
=20,250 lbs.
2
Ve=12X18.9X50=11,350 lbs.
Using "U"-shaped stirrups, 3%" corrugated rounds, we have:
SHEARING PROVISIONS.
P=2X0.11X16,000 lbs.=3,520 lbs.
.86 d P .86X18.9×3,520
20,250-11,350
y at end=
V-Ve
y, three feet from end=26″.
L. L. 0.85X27,000.
D. L. (Slab)
D. L. (Beam)
TOTALS.
Typical interior girder, G4.
All girders in floor system will be figured to carry 85% of the total
live load.
•
ACTUAL CONCENTRATED LOad on Girder. X FACTOR
GIRDERS.
C
=6½".
=22,900 X
10,800 X
2,700 X
• •
•
·
36,400
=
42 2
|| || ||
ULTIMATE LOAD.
91,600 lbs.
21,600 lbs.
5,400 lbs.
118,600 lbs.
Mo, concentrated load= WIX 118,600 X 20-.....395,000 ft. lbs.
The girder will weigh approximately 300 lbs. per ft.
Mo=2XM due to weight of girder=2X(X300X202)=20,000 ft. lbs.
TOTAL..
.415,000 ft. lbs.
M。=570 bd²=4,980,000 in. lbs. (ult.)
Assume b=14"; d=25″ and q=4.55 ″.
Make girder 14"x28"; 6-%" corrugated squares, N. S.
Bend up three bars in each girder.
•
}
•
46
B
•
V may be assumed constant and equal to
V=14X25X50=17,500 lbs.
V-V. 700 lbs., indicating that the concrete alone is capable of
carrying the shear at the allowed stress. As before stated, it is desir-
able to have stirrups to distribute concentrated loadings, and they will
be provided as indicated on drawings.
5th
Height=11′-0″
Story.
3d
Height 11'-0"
Contributory area=18' X 20' 360 sq. ft.
The unit stress in the concrete will be taken as 600 lbs. Referring
to diagram on page 18, the average permissible stress per sq. inch,
using 1% reinforcement, is found to be 690 lbs.
This figure will be used to determine the required area—one inch
of concrete will be added all around for fireproofing.
As stated, the live load on columns will be taken as 85% of full
live load.
4th
Height=11'-0" Col.
Basement
Height=10′-0″
•
Story
Loads.
1
L. L.....14400
D. L....28800
Col. 1100
2d
Height 11'-0" Col.
SHEARING PROVISIONS.
TYPICAL INTERIOR COLUMN.
...*
44300
L. L....45900|
D. L....39600
..... 2900
88400
L. L....45900
D. L……..39600
Col.
.... 4500
90000
IL, L....45900|
D. L...
.39600
·
... •
91500
|……. L………….45900
D. L....39600
....10500
Height 15'-0" Col.
1st
COLUMN DESIGN.
96000
L. L....45900)
D. L……….39600
Col. .. 9500
95000
Column
Load=P.
44300
132700
6000 314200
222700
410200
505200
Area Req'd
=P-690.
640"
193"
323"
36,400
2
4550"
0930"
7320"
47
COLUMN SECTION.
External Vertical
Dimensions. Reinforced.
12"x12"
16"x16"
=18,200.

20"x20"
24"x24"
27"x27"
29"x29″
4-2½" C. R.
4-3/4" C. R.
8-3/4" C. R.
8-%" C. R.
8-1" C. R.
8-1%" C. R.
Hooping.
¼4" C. R.
12" cts.
4" C. R.
12" ets.

¼4" C. R.
12 cts.
4″ C. R.
12" cts.
4" C. R.
12" cts.
¼" C. R.
12" cts.
Story.
Story
Loads.
L. L.... 7200
D L....14400
3500
25100
L. L....22950)
D. L....19800
Wall...19000
5th
Height=11'-0" Col.
4th
Height=11'-0" Col.
• •
TYPICAL WALL COLUMN.

3500
65250
L. L....22950Į
D. L....19800
30
Wall...19000
Height=11'-0" Col.
....
3500
65250i
IL. L....22950
D. L....19800
2d
Wall...19000
Height 11'-0" Col. 4100
....
65850!
IL. L……..22950
D. L... 19800
1st
Wall...19000
Height=15'-0" Col. .. 7000
...
68750
|L. L……..22950|
D. L....19800
Basement Wall...27000
Height=10'-0" Col..... 6000
75750
Column
Load=P.
25100
90300
155600
Required area=
221400
290200
366000
Area Req'd
=P÷690.
360"
1310"
2250"
320"
4220"
5300"
COLUMN SECTION.
External Vertical
Dimensions. Reinforced.
* 13"x24"
13"x24"
13"x24"
15"x24"
19"x24"
24"x24"
Load at base of column..
505,200 lbs.
Approximate weight of footing= 25,000 lbs.
530,200 lbs.
4-34" C. R.
530,200
5,000
106 sq. ft.
Make footing 10'-4"x10'-4".
4-3/4" C, R.
6-34" C. R.
6-%" C. R.
* Wall columns are made of uniform width throughout to simplify brick work.
6-1" C. R.
FOOTINGS.
TYPICAL INTERIOR FOOTING.
Hooping.
1/4" C. R.
12" cts.
4" C. R.
12" cts.
4" C. R.
12" cts.
1/4" C. R.
12" cts.
6-1%" C. R. 4" C. R.
12" cts.
4" C. R.
12" cts.
In the design of the footing we will not allow the vertical shearing
stress to exceed 125 lbs. per sq. in.
Required depth of footing then=-505,200
=35".
4X29 X 125
As the shear decreases toward the edge of the footing, and this
depth of footing is not required to resist the bending moment devel-
oped, we will step off the footing, as shown.
++
48
C
X
→
CG
Eat
10-4"
2'-5"
3'-6"
COLUMN BASE
FOOTING
1
У
•
D
"1
D
• •
10
5,000
"
is resisted by a section in the plane x-y, the width considered avail-
able, may be taken equal to the side of the column plus 1½ times the
depth of the footing=29"+(1½ X23")=5'-3½".
The column base is used to distribute the load over the footing and
its dimensions are determined from shearing considerations only. The
footing and column base must be built monolithic.
The moment per foot width of section=11,970.000÷5.29=2,260,000
inch lbs. (ult.)
Area required=991,200
26
For footings it is necessary to have a comparatively large amount
of concrete to resist shear. We will therefore use .85% reinforcement.
M。=370 bď²=2,260,000 for b=12″, whence d=22.5″.
The upward pressure on the tri-
angle C-O-D and the downward
pressure on the triangle A-O-B are
each equal to one-fourth of the
column load, neglecting the weight
of the footing.
The maximum bending moment
is in the plane x-y, and may be
obtained by taking moments about
this line. Let a distance from x-y
to center of gravity of triangle.
C-O-D and b=distance to center of
gravity of triangle A-O-B. Then
___505,200 (a-b)=
505,200
M
'(41.3″—
4
4
9.7'')=3,990,000 in. lbs.

q=.0085 X 12 X22½=2.29 sq. inches per foot width.
//
Make d=23″ and use i corr. sq. new style, 5" cts. both ways in
footing.
Using a factor of safety of 3.
the ultimate moment equals M。=
11,970,000 inch lbs. This moment
COMBINED FOOTINGS.
Since the footings are not to extend beyond the inside lot line, it will be
necessary to use combined footings on this side of the building. The combined
footing acts as a beam resisting the upward reaction of the soil, and supported at the
ends by the columns. By making the center of gravity of the footing plan corres-
pond with the resultant of the column loads, a uniform soil-pressure is produced.
Load on wall column.
Load on interior column
TOTAL..
Assuming footing to be 4'-0" thick, its weight would
be about....
TOTAL ON SOIL..
=198 sq. feet.
..366,000 lbs.
.505,200 lbs.
871,200 lbs.
I20,000 lbs.
991,200 lbs.
49
Since the column loads are unequal, the footing will have a trape-
zoidal shape, and by fixing its length, since we have the area, we can
determine the length of the two parallel sides for conditions of uniform
load.
Q = 17 88
a+b
2
Katalogla
É-COLUMN
•
***
ola
1 = 20'-6"
COLUMNS
G = 1085'
w
RESULTANT, COLUMN
LOADS AND CENTER
GRAVITY OF
-FOOTING
Jos
El
17'-0"
COLUMN
f
36
44
N
2011
side

Let G distance
from small end to
center of gravity of
footing and to the
resultant of column
loads.
If A area of
trapezoid, we can
determine the ends
"a" and "b" from
"C
A
must equal
9.65 feet, or a=19.3-b.
198
ι 20.5
Substituting in the formula we find b=11.42 feet, a=7.88 feet.
The load on soil equals 5,000 lbs. per sq. foot; the weight of
footing is 600 lbs. per sq. foot; upward reaction equals 4,400 lbs. per
sq. foot.
04
To design footing, we will find the moment on a strip 12" wide.
Using a factor of 3 on the bending moment, we have
Mo=3 (% wl²)=3 (4400 X 17² X12)=5,720,000 in. lbs. (ult.)
By means of table No. 1, page 23, average rock concrete, we find
that the section corresponding to this moment requires d=36" and
9=3.700"
f
Since we have fixed upon the total depth of the footing as 48",
will make d=44″ and use X3.70″=3.03″ of metal per foot width.
we
the following for-
mula:
A3
· b=
6_24(3 G—1)
I
The average width
Make footing 48″ deep, d=44″, and use 1" corr. squares,
N. S., 4" cts. in top of footing.
The distance, c to c given, is on the line m-m through the center
of gravity of the footing.
#
At the ends of the footing the portions under the columns act as
transverse beams. We will figure the reinforcement required for this
cantilever action at the larger end, under interior column. Ultimate
505200
2
II.42
moment Mo=
X12X3=26,000,000 in. lbs. This will be
X
assumed to be distributed over a width of five feet.
4
50
L
1
"
M。 per foot width equals 5,200,000 in. lbs. (ult.); this is practically
the same as the reverse moment in the longitudinal direction and the
same reinforcement will be used.
SHEARING STRESSES.
Limiting the vertical shear to 125 lbs. per sq. inch, the minimum
depth of footing under interior column will be found to be 35 in.; as
the footing is 48 in. deep the punching shear at the end need not be
considered.
The wall column will require investigation, since only three sides
may be considered in determining the intensity of the shear. To bring
this within safe limits, and also to avoid high stresses at the edge of
the footing, the column will be arranged as shown on drawing.
In order to avoid any possibility of the concrete shearing along
diagonal planes at the ends, horizontal and vertical tie bars should be
' introduced, as shown on detail.
Note-No provisions have been made in this building for resisting wind
stresses, as, owing to its low height in comparison to its width, such stresses
may be neglected.
51
18-0
0-81
18
54-0″
LOT LINE
2
COL./
COL.2
LINE 7
BUILDING
COL.3
COL.4
20-0
COL 15
COL. 6
COL.7
COL 8
COL. 9
BUILDING
20-0′
COL /O
COL./1
COL.12
LINE
COL 13
20-0
100'- 0'
"
COL.14
COL.15
COL16
COL
20-0
p
COL.18
|COL.19|
COL.20
COL. 2/
20-0
"
COL.22
COL.25
COL24
FOUNDATION PLAN
TYPICAL REINFORCED CONCRETE BUILDING

52
1214
Z
ម
រ
7}
!
MANA DAW
0-11
,0,11
0.11
15 0
100
LONGITUDINAL SECTION
REINFORCED CONGRETE BUILDING

53
54'-0"
18-0
18 0
0.81
LOT LINE
BUILDING LINE
COL.I.
COL.2
84
COL. 3
ზა
COL 4
10:0
G,
G,
A m
10:0
COL.5
COL 6
COL. 7
COL. 8
10-0
G₂
COL. 9
10-0
ELEVATOR
COL 10
COL.//
COL. 12
Gr
10-0°
STAIR
WELL
G5
G
·BUILDING LINE
100-0
10-0
COL.13
29
COL.14
COL.15.
||| COL. 16
10-0
Gr
G
G₂
10-0
$9
COL.17
COL.18
COL.19
COL. 20
10.0
C
GI
COL.2/❘
COL.22
COL.23
COL.241
10-0
TYPICAL FRAMING PLAN.
TYPICAL REINFORCED CONCRETE BUILDING.

54



CORR SQS
·
Bs · 9″ 15½"BEAM
G₂- 12"x 201 GIRDER.
SLAB - 31.
'CORR SQS.- B'CTS.
CTS.
TRANSVERSE REINFORCEMENT & CO
10.0
zas zudsaSAS “A
"
G
caune
199
PLAN
CORR. SQS.
CORR SQS ·2·0˚CTS.
NOTE - ARROW
INDICATES DIRECTION
NFORCEMEN
OF MAIN REINFORCEMENY
10.0
12"x 20½ GIRDER
6
LNESSMITUSAMMEN
stada ka plat
STIRRUPS MADE OF CORR SQS
G3
NOTE -
MAOVUSTU
15½
AND SPACED AS SHOWN
II
CORR SQS. - 2 BENT UP AS SHOWN
12
DETAILS OF ROOF CONSTRUCTION
TYPICAL REINFORCED CONCRETE BUILDING
ALL CORRUGATED SQUARES
ARE NEW STYLE
32 SLAB
CORR JQS.- 16°CTS
·TRANSVERSE BARS - ✯ CORR SQU
2-0˝CTS AS SHOWN
CORR. SQS
12' 20½ GIRDER
SECTION SHOWING REINFORCEMENT IN ROOF GIRDERS
·STIRRUPS MADE OF CORR, SQS. - SPACED AS SHOWN.
HOPEDAG.
B5
-1- CORR SQ - STRAIGHT· 10-0 LONG
9.15 BEAM
SECTION SHOWING REINFORCEMENT IN ROOF SLAB
Z
C& CORR SQS - 8
BI
5 § CORR SQS - 2 BENT UP AS
SECTION SHOWING REINFORCEMENT IN ROOF BEAMS
9"x 15½" BEAM
SHOWN
G4
8"CTS.
4 CORR SQ5,- 2 BENT UP AS SHOWN.
ANBUTANIAVĒDUS
2 BENT UP AS SHOWN.

55

4 · É CORR. JUS
AM
20
64
14:28 GIRDER
SLAB - 43
É CORR ROUNDS
6 ½ CTS
TRANSVERSE
REINFORCEMENT
CORR ROUNDS
4 PER PANEL
10:0"
"
PLAN
6-7 CORR SQ5
NOTE
ARROW
INDICATES DIRECTION
OF MAIN REINFORCEMENT
IGL
-
10-0
85
✓
B,
G3
IBAR
7-} CORR SQ5. - 3 BENT UP AS SHOWN
2 BARS
5.1 CORR SQS. - 2 BENT UP AS SHOWN
NOTE
14
ALL CORRUGATED SQUARES ARE NEW STYLE,
12
4½ SLAB
({"CORR ROUNDS - 6½" CTS.
TRANSVERSE BARS & CORR
STIRRUPS MADE OF CORR. SECTION SHOWING REINFORCEMENT IN GIRDERS
ROUNDS AND SPACED AS SHOWN
HE
N
DETAILS OF FLOOR CONSTRUCTION
TYPICAL REINFORCED CONCRETE BUILDING
The
SECTION SHOWING REINFORCEMENT IN FLOOR SLAB
STIRRUPS MADE OF CORR ROUNDS SPACED AS SHOWN
TILL
(14"x 28" GIRDER
~12 x 21" BEAM.
SECTION SHOWING REINFORCEMENT IN BEAMS.
24
ROUNDS 4 PER PANEL
CORR.ROUND BARS- 6½ CTS.
6.70
G4
2 BARS
IBAR.
CORR SQS - 3 BENT UP AS SHOWN
W
B5
SQS
4. & CORR. 5Q5 - 2 BENT UP
KZ
AS SHOWN
JUIX
PURA
56

ROOF
12412
5TH FLOOR
16x16*
4TH FLOOR
20.20
3RD FLOOR
24x24
2ND FLOOR
27127
1ST FLOOR
29.29"
BASEMENT
4-Ź CORR ROUNDS
+
4 CORR ROUNDS
CORR ROUNDS THROUGHOUT COLUMN 12"APART
8 - CORR ROUNDS
8- CORR ROUNDS
8-1´CORR ROUNDS
HOOPING
8-1½"CORR ROUNDS
HEIGHT VARIES WITH
SLOPE OF ROOF
"1
O
"/
0-11
0.11
0-,91
DETAIL OF TYPICAL
INTERIOR COLUMN
13-24
0.01
13:24"
13x241
15:24
19:24
24.24
4- CORR ROUNDS
4- 2 CORR ROUNDS
HOOPING CORR ROUND3 - 12″ APART THROUGHOUT COLUMN
6 CORR ROUNDS
11 42
6 & CORR ROUNDS
CORR ROUNDS
9
6 1½ CORR ROUNDS
YN
"
COLUMN AND FOOTING DETAILS
TYPICAL REINFORCED CONCRETE BUILDING
TYPICAL WALL COLUMN
NOTE
CORR SQS ·
ALL CORRUGATED SQUARES ARE NEW STYLE
20
6
EVERY 3RD BAR IN TOP BENT
AS SHOWN
Kedskoen
SECTION COMBINED FOOTING
10-4
·EVERY 3RD BAR BENT
PLAN- COMBINED FOOTING
SHOWING REINFORCEMENT IN TOP OF FOOTING
-
•½ COPP SQS
+ +
PLAN COMBINED FOOTING
SHOWING REINFORCEMENT IN BOTTOM OF FOOTING
/"CORR SQS AS SHOWN
1
"CORR SQS BOTH
DIRECTIONS AS SHOWN
10-4
"CORR SQS AS
I
CIL
DETAIL OF TYPICAL INTERIOR FOOTING
}
788
1226
SHOWN
57
THE STRENGTH OF RECTANGULAR BEAMS.
In the following analysis for determining the resisting moment of a reinforced
concrete beam, it will be assumed that the bond between the concrete and steel
is sufficient to bring them into the necessary intimate stress relationship, and
that proper and adequate provision for all web stresses has been made. While
it is evident that simple formulæ are desirable, and that there is no objection
to the use of strictly empirical formulæ, yet it is essential that the designer be
familiar with the complete analysis, so that when using a simplified formula, he
will be familiar with its limitations.
Tension has been considered as existing in the concrete up to that fiber in
which the extension equals that to which plain concrete can be subjected without
cracking. While the influence of the tension in the concrete, on the resisting
moment, is large for very low unit stresses, yet its effect is negligible when the
stresses due to ultimate, or even working, loads are considered. It was thought
advisable to include it in the analysis for the sake of making the discussion com-
plete mathematically.
By first assuming a definite law of variation between stress and strain, for
the concrete, it is possible to readily evaluate the resisting moment of the section
for any given percentage of reinforcement by further assuming the stress in the
steel. The principle of invariability of plane sections, although it does not
strictly obtain in the case of reinforced concrete beams, in conjunction with the
requirement that the total tension must equal the total compression is sufficient
to determine the position of the neutral axis. The resisting moment may then
be obtained by taking moments about either the neutral axis or the centroid of
compression or tension. The form of curve representing the stress-strain rela-
tion in the concrete was chosen after a thorough investigation of experimental
data on deformation, and a study of the actual carrying capacity of beams, and,
we believe, represents the actual conditions with sufficient exactness.
In the following discussion it is assumed that a section plane, before bending,
is plane after bending. It is further assumed that the modulus of elasticity of
concrete varies, its value decreasing as the stress increases, and that its instan-
taneous value may be represented by the tangent to parabola.
To obtain an equation for a parabola that would represent the variations of
the modulus, an inspection of a number of stress-strain diagrams was made,
which led to the conclusion that if the compressive strength be taken as two-
thirds of the stress corresponding to the observed deformation at failure, using
the initial modulus of elasticity noted, that the parabola so obtained would
represent closely the actual stress-strain diagram. The tensile stresses in the
concrete, between the neutral axis and that plane at which the unit elongation
has the limiting value At, are considered in the discussion.
We have, then, for Rectangular Beams, the following discussion:
h
12
Figure 1.
d
λε
λε’
LASH
Ac"
FASH
5
y₁
Y2
e
Figure 2.
Z
0
ft
c
£c
ان
-214x35
It
λο
Figure 3.
Yr
Y2
X
A

58
•
Fig. 1 is a cross section of a reinforced concrete beam.
Fig. 2 represents the strain or deformation diagram at any instantaneous
load.
Fig. 3 is the stress diagram corresponding to the above strain diagram.
Let Es Modulus of elasticity of steel in pounds per square inch.
Ec Initial modulus of elasticity of concrete in compression in pounds
T
F
fc
f = Compressive stress on extreme fiber in pounds per square inch.
f may have any value less than fc.
c':
S
ft
20
20"
λό
Unit deformation of extreme compression fiber at ultimate stress fc.
Unit deformation corresponding to stress c'.
Note that λe' and c' deal with conditions after the ultimate strength of the
concrete is passed, and have no value except in determining the curve, etc.
2t
Unit elongation of concrete corresponding to stress ft.
25
Unit elongation of steel corresponding to stress s.
b= Width of beam in inches.
Mo
-
****====******
Ma
per square inch.
Elastic limit of steel in pounds per square inch.
Compressive strength of concrete in pounds per square inch.
**
Abscissa to stress diagram at vertex of parabola.
Any assumed unit stress in steel, pounds per square inch.
Modulus of rupture of concrete in cross bending in pounds per
square inch.
Unit deformation of extreme compression fiber corresponding to a
stress f.
Y/1 Distance from neutral axis to extreme fiber in compression in inches.
Distance from neutral axis to plane of reinforcement in inches.
Y2
e = Distance in inches from plane of metal to extreme fiber on tension side.
d=y₁+y2=Effective depth of beam.
p= Ratio of reinforcement in terms of bd=q÷bd.
α
Ratio of reinforcement in terms of bh=q÷bh.
M = Bending moment of external forces in inch_pounds=resisting
moment of beam.
***
Distance from top fiber to center of gravity of compression area in
inches.
St
Po
Pc = Total compressive stress in concrete in width b.
Pt
Total tensile stress in concrete in width b.
q= Area of metal in width b, in square inches.
Ultimate moment of resistance of cross section in inch pounds.
Total stress in metal in width b.
Referring to Fig. 3, the shaded area above the neutral axis represents the
compressive stress diagram of the concrete, O Y being the axis of proportionate
elongation, and O X the axis of stress per square inch.
Before getting the area of the compression diagram, it will be necessary to
get the equation of the parabola referred to the axes O X and O Y. We have
Ee which is represented by the tangent to the parabola at the origin O, and
have also imposed the condition that the compressive stress at rupture is two-
thirds Ece". The equation for the parabola then becomes:
From which λe"=
And
f = Ec λc
fc = 3 Ee λc"
3 fc
2 Ec
9 fc
20=
4 Ec
Ec 22
2λ'c
►
·
(8)
(9)
59
·
Substituting in the general equation and solving for λe, we get:
-):
..(10)
λc = 76 (1
76 ( 1 − √1
Or
We can now get a value for y₁ :
From the strain diagram,
Therefore, y=
Pc
b
Pt
b
2
S
Yi
(11)
The expression for the arca of the compressive stresses may be written in
the form,
K
8f
370)
ft Xt
2 2c
9fc
4 J'
I2
λ c \ E c λ c
2
Yi
S
:(=-
y₁=Dy₁.
For the area of the tensile stresses we may, without appreciable error, con-
sider the parabolis area as a triangle (since the allowed stress is very small, the
tangent and parabola practically coincide), and can express the area by the
equation,
Therefore, p= 311 X
d
Es λc
s + Es λc
Y1
d — Yı
λed
28 +λc
Ec λt
2λc
xć r
A
D-G
S
Y1, where r
4r
d………
Since the sum of the compressive and tensile stresses must equal zero, we
can write Ps Pc — Pt
λε
M = pbds(d-z)+ by₁ G(y₁ + 3 Y¹ ø
λε
ag
but s
yi Gy₁...
M=Ps (d-z)+ Pt
We have, taking moments about the center of gravity of the compressive
stresses, the following expression for the moment of resistance of the section:
λε
20
S
Es
(2 ht
Yi + yı
λε
M。 = pbd F'(d-2)+ by₁ G(yı + 3 yı
•
- z)……….
•
- =)
•
(12)
(13)
•
(14)
. (15)
It is to be noted that the above discussion is perfectly general, and we may,
by assuming any fiber stress, f, and any stress in the steel, s, find the percentage
of reinforcement required, and the corresponding resisting moment of the section.
We are, however, mainly interested in the ultimate strength of the beam,
reinforced with the critical percentage of metal (it being taken for granted that
the designer will apply his factor of safety to actual moments, designing the
section for the ultimate moment so obtained), which condition obtains when the
percentage of steel is so chosen that the beam is equally strong in tension and
compression; or, differently expressed, that the stress in the steel reaches the
elastic limit at the same time that the compressive stress on the extreme fiber
becomes the ultimate strength of the concrete.
Putting these values in the general equation, No. 15, we get the following:
(15a)
(16)
60
The size of beam needed to develop a required moment of resistance can be
obtained from the above equations, when the constants dependent upon the par-
ticular materials used are known.
If it is desired to neglect the tension in the concrete, the last term in equa-
tion “15" and "15a" becomes zero, and we have:
For no tension in the concrete:
M = pbds (d-z).
M。=pbd F(d-z)
r
Note If the extension of the steel corresponding to the assumed unit stress
should happen to be less than the value for λ chosen, the formula would not
apply. The value of 2t will, in all cases, be taken as .00015; the unit stress in
the steel corresponding to this extension being 4,350 pounds. If a lower stress
were assumed in the steel t would then have to be taken equal to the exten-
sion of the steel, 2,
To illustrate the application of the formulæ, the following example has been
worked out in detail:
9
From equation (9)′ = ½
fc
4 Ec
} λc'
Also 2"
Example-Determine the critical ratio of reinforcement and the ultimate
moment of resistance of a beam, taking the compressive strength of the con-
crete as 2,000 pounds per square inch, and assuming the elastic limit of the
steel to be 50,000 pounds per square inch.
For average rock concrete when fc=2,000 pounds, Ec may be taken as
2,600,000 and 2t .00015. F is given as 50,000 pounds and Es may be taken
as 29,000,000, and, in this particular case, where ffc, e becomes equal to
".
= X
.0017308
λε
2.c
21/0
4 - r
•
•
Y₁ = 18 yı
Yi
Ec λc
D= (1-3210)
2
and from equation 13,
(.b & — 1 ) =
G
p=
12
47
To obtain y₁, we substitute in equation II,
Es 2c
33463
d=
s+Es 2c 83463
We will now obtain values for D and G, these quantities being respectively
proportional to the amount of compression and tension in the concrete, only.
From equation 12:
Ee de
2
Ec λi
220
2000
2,600,000
.0011539
→
d=.401d.
(17)
. (17a).
= 3 x
·
(Eq. 16)
(Eq. 16)
I143.3
50000
2,600,000 X .0011539
2
25.4
The percentage of metal necessary may now be readily obtained by means
of equation 14. This will be the critical ratio of reinforcement, since we are
using fc and F as the stresses in the materials.
1166.7
31 12 X
= .401 X
0.00915
Yı D-G
F
The ultimate moment of resistance may now be obtained by substituting the
proper values in equation 15a.
•
λε
M。 = pbd F(d-e)+ by₁G(yı + Y₁ ——≈)= 392 bd² + 3 bd² = 395 bd²
61
In a similar manner we obtain the following values for a good rock concrete
beam, when fc may be taken as 2,700 pounds, and Ec= 2,800,000, F= 50,000.
λ=.0021696
2c"=.0014464
p=.01418
Mo=596.6 bd
Note-The average F for corrugated bars will be found to be about 55,000
pounds. With this value the critical ratio for average rock concrete becomes
.00785, and for good rock concrete .01224.
It is desirable to have a ready means of determining the resisting moment
of a section reinforced with a particular ratio of reinforcement, and the plates,
Nos. 1 to 7, inclusive, give this information for various conditions.
Plates 1 and 3 represent graphically the results obtained by substituting in
formula 15, the constants for the two grades into which we have arbitrarily
divided all rock concretes. For average rock concrete, Plate 1, fc=2,000 and
Ec=2,600,000. For good rock concrete, Plate 3, fc=2,700 and Ec=2,800,000.
In Plate 2 and 4 the effect of the tension in the concrete has been omitted. A
comparison of Plate I with Plate 2, and of Plate 3 with Plate 4, shows that for
high stresses in the steel the effect of the tension is negligible. For example,
assuming a ratio of reinforcement of three-fourths of 1 per cent, and a stress in
the steel of 50,000 per ", we have:
From Plate 1; M=328 bd2 and f=1840 pounds.
From Plate 2; M=326 bd² and f=1825 pounds.
Plates 5, 6 and 7 are based on the values for the constants shown, and in
the computations the effect of the tension in the concrete has been omitted.
EXPLANATION OF PLATES.
Each plate consists of two series of curves independent of each other, but
with common abscissæ representing the ratio of reinforcement. The curved
lines show the relation existing between p and s for the various values of f noted.
For example, it is desired to develop 1,800 pounds extreme fiber-stress in average
rock concrete when the stress in the steel is 50,000 pounds: what is the corres-
ponding ratio, p, assuming tension to exist in the concrete?
Referring to Plate 1, follow the ordinate 50,000 until it intersects with the
1,800 curve; the ratio of reinforcement at this point is .007.
The straighter lines diverging from the origin give the relation between p
and M for the various values of s given. Interpolate for values not given by
curves. To get the moment of resistance of a section given p and s, we proceed
as follows: Assume average rock concrete and include tension in the concrete.
From Plate 1, follow up the line .007
up the line .007 = p to its intersection with the 50,000
curve. The ordinate of this point is 310 = K. Therefore, M = 310 bd².
Conversely-If M and p are known, the stresses corresponding may be deter-
mined as follows: Let M÷bd²
Let M÷bd² = 300, and take p = .0075; average rock con-
crete, assuming no tension. Referring to Plate 2: The point of intersection of
the ordinate 300 with the abscissa .0075 is seen to be near the 50,000 curve. By
interpolation its value may be estimated as 47,000 pounds. The stress in the con-
crete may now be found, since we have s and p. The intersection of the ordi-
nate 47,000 and the abscissa .0075 is near the curve for f=1,800 pounds, and by
interpolation is seen to be about 1,750 pounds.
62

S
*
STEEL
IN
STRESS
{
55000 550
50000 500
40000 400
Kbd².
1000원
​30009 0300
10000
=
O
20000 200
COEFFICIENT;
K =
100
O
O
£100
019
200
1600
1800
2000
.005
RATIO OF REINFORCEMENT
01
ASSUMED DATA
TENSION IN CONCRETE CONSIDERED
TI
015
AREA OF STEEL : pbd.
-
Ec- 2600,000
fc 2,000
30000
COMPRESSION
25008
02
MAT
20000
ऊए
VO
8000
.025
4006
12000
10000
PLATE. NO. 1
.03
63

5
*
STEEL
IN
STRESS
55000 550
50000
40000 400
20000
Kbd²
10000
O
JA
Mo
30000 300
;
500
COEFFICIENT
H
x
200
100
O
O
Р
*
20
200
жаба
P
1000
2000
.01
ASSUMED DATA
Ec= 2,600,000
fc. = 2.000.
TENSION IN CONCRETE NEGLECTED
००००
.005
015
.02
RATIO OF REINFORCEMENT ; AREA OF STEEL = bdp.
T
$5
doo
025
120000
8000
dog
-14000
12000
loose
PLATE NO.2
.03
*9

.50 000
n
(1
STEEL
IN
55000
STRESS
30000
W 20000
K bd
10000
C
40000 400
COEFFICIENT
750
700
K-
600
500
500
200
100
O
0
BOO
P
=
X200
100
1000
55000
.005
RATIO OF
ASSUMED DATA.
Ec - 2800,000
fc
2700
.01
REINFORCEMENT
=
TENSION IN CONCRETE CONSIDERED.
.015
AREA OF STEEL
1700
=
02
pbd
COMPRESSION
50000
25000
XXX
IN
.025
NG
Vong
AGIN
B
ENOVKY
5000
4000
.03
PLATE NO. 3.
65

ហ
=
STEEL
IN
STRESS
55000
50000
40000
30000
20000
10 000
O
Kbd2
1
-
Mo
COEFFICIENT
8
=
750
700
600
500
400
300
200
100
O
O
P
**
4 = 1000
HH||
ASSUMED DATA
AGO
Ec
fc = 2700
=
2 800,000
2000
50000
005
이
​RATIO OF REINFORCEMENT
KA
2800
500
TENSION IN CONCRETE NEGLECTED
015
AREA OF STEEL
00000
=
130000
02
pbd
2
SSION
85008
120000
5-18000
16000
1000
025
PLATE NO.4
03
99.

►
"S"
་
STEEL
IN
STRESS
55.000
50000 500
40000
K bd².
20000
ព
550
30000 Σ 300
COEFFICIENT
=
400
200
10000 100
O
0
Jo
4600
K
800
0000
ASSUMED DATA
Ec-2,000,000
fc. 2000
Pa
.01
S
TENSION IN CONCRETE NEGLECTED
-
200
.005
.015
RATIO OF REINFORCEMENT. AREA OF STEEL
T
SION
=
p bd
30000
.02
25009
.025
30000
In
n
8000
16001
2007
17000
.03
PLATE NO 5,
67

S
STEEL
IN
STRESS
55000
30000
K b d 2
20000
N
50000 500
10000
-
40000|Σ 400
750
700
COEFFICIENT
600
K =
300
200
100
O
=
800
"
ASSUMED DATA
Ec. 2,500,000
fc = 2400
50
1800
1600
2000 A
COMPRESSION LIMIT
-
Do
2400
TENSION IN CONCRETE NEGLECTED
Ol
005
.015
RATIO OF REINFORCEMENT AREA OF STEEL
=
pbd
02
30000.
155
200
18000
16000
4000
025
PLATE NO 6
Deg
03
68

S
u
STEEL
IN
STRESS
55000
50000
30000
20000
10000
Kbd2.
O
Σ
40000
Mo
750
700
COEFFICIENT:
си
600
500
400
300
200
100
O
=
RATIO
600
1400
ASSUMED DATA
1800)
2000
Ec 3,000,000
fc
--
= 3000
MV
005
OF REINFORCEMENT
Dod
이
​Rich
COMPRE
00005
hoo
TENSION IN CONCRETE NEGLECTED
015
AREA OF STEEL
3000
Es
10
*
pbd
.02
80000
k
20000
15
IS:
SE
400
10090
025
PLATE NO 7
N
Po
oa
.03
69
*
T-BEAMS.
1
In all T-Beam designs, when the ends of the beams are restrained, the
strength of the section at the support should be investigated, and its ability to
resist the negative moment there developed determined.
The resisting moment of a "T" section may be determined by simple approxi-
mate formulæ, and with sufficient accuracy for most purposes, when it is assured
that the proportions of the beam are such that the integrity of the entire section
will be preserved. Manifestly no more stress can be put into the flange than can
be transmitted into it through the shearing strength of a horizontal section of the
stem of a length equal to half the length of the beam, or into the wings than can
be transmitted through the vertical planes of junction with the stem. An analysis
based on these shearing values will then afford a means of determining the max-
imum amount of stress that can be developed in either tension or compression,
and thus gives a means of determining the maximum amount of reinforcement
that may be used.
The amount of compression being determined, the width of slab that should
be considered tributary to the beam can be obtained, assuming some law of de-
crease in stress from the center of the beam to the edges of the wings.
It is for the purpose of bringing out these values and showing their relations
more clearly that the following analysis is given :
B'
4
Pc"
Pc'
Sh
b
P
LOCATION OF NEUTRAL AXIS.
The location of the neutral axis is dependent on the simultaneous deforma-
tions in the concrete and steel, and may be found, as in the case of rectangular
beams, by means of equation II.
mad.com
Yı
Y,
92
Should the neutral axis be found to be in the flange, the analysis for rec-
tangular beams applies strictly, b' replacing b in the equations.
Es λc
d
s + Es 2 c
Investigation for failure by shearing or diagonal tension should be made in
this case based on the width, b, of the stem.
In the analysis no
account has been
taken of any hori-
zontal tensile stresses
that may exist in the
concrete.
Let Sv=Total shear in pounds along the two vertical planes of attachment
between the wings and the beam from one end of the beam to the
center of span.
Total shear in pounds along the horizontal plane of attachment
of the flange and stem from one end of the beam to the center
of span.
•

70
。= Maximum shearing strength of concrete in pounds per square inch.
Y3
K
=
/= Length of span in feet.
Pc Compression existing at point of maximum bending moment, at ulti-
mate load, in that portion of the section between the neutral axis and the
underside of the flange.
P'c=
Compression existing at point of maximum bending moment at ulti-
mate load in the flange.
bo Limiting value of b'.
V = Total vertical shear at any section in pounds.
v=Intensity of vertical shear at any point in pounds per square inch.
ď Distance from plane of reinforcement to centroid of compressive
stresses, in inches.
All other functions as shown on the cut and in inches.
3/1
The sub-figure o denotes limiting or ultimate values.
We will now determine the ultimate resisting moment, Mo, of a T-section.
There are three methods of failure above the neutral axis:
1. By compression in the flange.
2.
By deficiency in Sh owing to smallness of b.
3. By deficiency in Sy owing to smallness of t.
Condition I need not be considered, as the analysis which follows is based on
a maximum stress of fe in the extreme fiber of the concrete.
We will limit the discussion to a consideration of Sh, assuming the section
to be so proportioned that Sy is greater than Sħ.
The unit vertical shearing stress is a maximum at the end of a uniformly
loaded beam, and its value decreases uniformly to zero at the center, and at any
point the horizontal stress is equal in intensity to the vertical shearing stress.
In a reinforced concrete beam, assuming the concrete to have no tensional value,
the intensity of the shearing stress at any section is constant up to the neutral
axis. Considering then, a uniformly loaded beam, the total shearing stress on
the horizontal plane of junction of the stem from one end of the beam to the
center Sh.
Sh = 30 bl⇒ P¿"…….
(18)
Sv 6 o t l...
. (19)
We can now get the necessary relation between b and t, such that we need
consider failure through Sh only. For equal shearing strength, t=2b. But
the stress transmitted into the wings through Sy is less than that transmitted
through Sʼn by the amount of compression in that part of the flange directly over
the stem. Also the floor slab reinforcement, running at right angles to the
length of the beam, increases the shearing strength on the vertical planes.
It may accordingly be safely assumed that if t is not less than of b, failure in
Sv need not be apprehended. We will assume at once that t will not have a
value less than b.
Similarly,
The intensity of the vertical (and also horizontal) shear at any point below
the neutral axis is given by the equation
V
(20)
bď
It is desirable to get an expression for Sh in terms of fe. The theoretical
relation between σ and fe is:
心
​σ
fc
2 tan 9,
where is the angle made by the
0
71
plane of rupture on a compression specimen, with a plane at right angles to
the direction of the applied stress. (See "Johnson's Materials of Construction,'
page 19.) For concrete
A is about 60°, hence o
Although the above relation between compressive strength and shearing
strength holds for homogeneous materials, we think it desirable to use a lower
value for shearing strength in concrete than is given by this equation. The
shearing strength is affected to a much greater extent by variations in uniformity
of structure, than is the compressive strength. We will accordingly take twice the
factor in shear that we do in direct compression, and derive the following value
for Pe":
Pe" = 3 0 b l = x
Pefe yı K² (1 — ¹3 K) b .
fc
1
Mo
P"
3 fc b l
3.5
If we replace the stress-strain diagram by a parabola, with its vertex at the
top of the beam, and coinciding with the stress-strain diagram at this point and
at the neutral axis, the area included by this parabola will closely approximate
the actual stress-strain area. By using this parabolic area, we simplify the mathe-
matics, and get results sufficiently accurate for T-beam analysis. We then have
as a value for the compression in the stem :
and P. Ps = Pc'+ Pc" = F q.
=
Taking moments around the plane of the metal we have :
N
fc
3.5
·43 b l fc. ....
•
bo
Po' (Y2 + 2½ Y3) + Pc” (d — 1)
2
Malaga Mada
..(23)
This Mo is the maximum moment of resistance of the section as determined
from a consideration of the strength of the connection between flange and stem.
The maximum amount of reinforcement that may be used in a T-beam is,
accordingly :
qo
Pe+ Pe"
F
To determine the width of flange that should be considered tributary to the
beam so that the stress will be zero at the edges of the wings, it will be neces-
sary to get Pe" in terms of bo. We have :
Pe"
04
fc Y₁ (2 + K³ — 3 K²) 2
•
6
2 y₁(2+K³ — 3 K²)
fc
bfe (y₁ K² (1 — 1/2 K) + .431) (24)
2
स
……..
•
I
Pc" (2 + K³ — 3 K²) fc bo Y1
3
bo
on the assumption that the stress is the same in the outer portion of the wings
as in the middle; this would not be the case, the stress varying as the ordinates
to a parabola from zero at a distance from the center line to a maxi-
mum over the stem. We must, therefore, take two-thirds of the above value for
the stress in the wings, and doing so, obtain the following :
2
fe b ¼¹ (2 + K³ — 3 K²) + 3 (fc yı (2 + K³ - 3K²) (↳。 — b)
Po"
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
fc i
3
—
fc_Y¹ (2 + K³ — 3 K²) (b+bo) ·
–
..(25)
9
And we have for bo
bo
........
(21)
•
(22)
•
.43 blfc, and substituting this value in equation 26 we have:
3.871
-1).
.(26)
• •
(27)
72
Problem-Required the size of T-beam necessary to carry a total ultimate
load of 600 pounds per square foot on a span of 32', ribs to be 9 feet apart.
8,300,000-inch pounds.
For this spacing of beams the floor-slab should be 4″ thick, and we will
assume d 20". Using good rock or gravel concrete, we have from Eq. (11),
assuming F 55,000 pounds.
=
Yi
Y2
K
K2
K3
bo
11
Mo
8.66
11.34
.538
.289
.155
12 X 9 X 600 X 1024
8
Mo
Pe' = fo y₁ K² (1 — K.).
3
Pe"
43 blfc = .43 X 32 X 2700
Pc = Ps = Fq= Po + Pe"
M。 = Pé' (y2 + # 1/3) + Pe" (a– ½)
2
= 5500 X 1755b +37100 X 186
b
==
q
Pe"
& fc Y₁ (2 + K³ — 3 K²)
8300000
764300
42600 X 10.8
55000
3 σ bl
• •
• 43 b l fc
10.8″
b i te (a –
( a − 1 )
2
go
FORMULAE OMITTING P
The neutral axis will, in the majority of T-beams, be found to be below the
flange. It will also be seen by reference to the previous example that Pe is
very small, and that its effect on the resulting resisting moment may be considered
negligible.
We may then write, with sufficient accuracy for most purposes:
Pc"
Sh
t
M₁ = P." (d)=3ebi (d)
a 30b1
Mo
-
(a -
2
2
The value of y₁ is obtained as before, from the equation
y
b
2
b=2700 X 8.66X.289 X .82b = 5500b
8.4 square inches.
37100 X 10.8
X 2700 X 8.66 X 1.29
Esc
s+ Esλc
That value for σ, which applies to the particular concrete used, should be
substituted in Eq. (29). The value chosen, however, should not be more than
one-seventh of the compressive strength of the concrete.
If we expresso in terms fe as before we have
96500b+
667800ს
764300ს
Pc"
The maximum amount of metal that may be used qo F
8,300,000
667,800
37,100 X 12.5
55,000
d
10.8
2
The value of bo is obtained as before by means of Equation 26 or 27.
Applying these formulæ to the preceding example, we have:
b
: 12."5
•
8.4 square inches.
(28)
(29)
= 37100
42600
(30)
54.6"
73
REMARKS ON PRECEDING ANALYSIS.
In the analysis it is assumed that a sufficient width of slab (bo) is available
so that the compressive stress is zero at the edges. From the assumptions made
bo as obtained from the equations is 50 per cent greater than the width required
on the assumption of a uniform distribution of stress over the flange. It is evi-
dent that bo may be reduced quite considerably without appreciably affecting the
strength of the beam, since the outer portions of the flange carry very little stress.
When the width of flange is limited, it may, however, be necessary to make a
corresponding reduction in Pec".
EXPLANATION OF T-BEAM TABLE.
The following table will be found useful for the rapid design of T-sections.
To illustrate the use of the table the preceding problem will be solved by its use.
The table is based on ultimate values; the compression in the concrete being
limited to 2,700 pounds per square inch. The stress in the steel has been taken
as 50,000 pounds per square inch:
Example:
1
蠱
​M。=8,300,000 inch lbs; t=4″, and l=32′-0″.
Assuming d=20" as before, we have from table for 4″ slab,
M。=b (90,300+20,9007)=8,300 000 inch lbs.
=b (90,300+20,900X32)=759,100 b.
8,300,000
759,100
10.9 in.
Steel required
55,000
whence b
=b (0.1264+.02321)=bX0.8688=9.45 sq. in.
bo=172 bl-
b
2
Note—In the solution of the problem by the formula F was taken 55,000
pounds. In this table a more general value: 50,000 pounds, has been adopted.
If F had been assumed at the higher value, the amount of steel required would
have been 50,000X9.45-8.5″. This agrees with the amount before obtained.
55½".
.1
GOOD ROCK CONCRETE.
t
3½"
4
42"
5"
TABLE FOR THE DESIGN OF TEE BEAMS.
S=50000
fc=2700
52"
d
10
11
12
13
14
15
16
17
18
19
20
10
11
12
13
14
15
16
17
18
19
20
22
24
26
10
11
12
13
14
15
16
17
18
19
2222228
24
26
30
231
14
15
16
17
18
19
20
22
24
26
1*2800*
30
32
34
GROFAR22*2*O***
14
16
17
18
19
20
24
26
28
30
32
34
36
AREA OF STEEL
b( 0122+.0232 l)
b( 0222+ 0232 l)
b( 0337+.0232 1)
b(.0464+0232 )
b(.0600+0232 )
b (.0736+ 0232 )
b( (882+ 0232 1)
b(.1030+.0232 l)
b(.1182+ 0232 l)
b(.1336+.0232 ()
b(.1486+.0232 1)
b( 0035+.0232 1)
b(.0103+.0232 )
b( 0195+0:32 l)
b(.0302+.(232 1)
b(.0424+0232 )
b(.0518+.0232 )
b(0681+ 0232 )
b(0824+.0232 )
b( 0964+0232 1)
b(1112+.0232 l)
b( 1264+ 0232 )
b(.1561+.0232 ()
b(.1876+.0232 l)
b( 2160+0232 )
b(.00004+.0232 1)
b .0027+.0232 ()
b(.0087+.0232 1)
b(.0171+.0232 l)
b(.0271+0232 1)
b(0382+.0232 l)
b(.0507+.0232 )
b(.0633+.0232 )
b.0770+.0232 )
b (.0909+.0232 1)
b( 1050+.0232 1)
b(.1342+0232 )
b(1650+.0232 )
b(.1960+0222 1)
b(.2268+.0232 1)
b(.2580+.0232 1)
b(.0021+.0232 ()
b(.0073+0232 )
b(.0136+.0232 ()
b(.0246+.0232 ()
b(.0350+.0232 7)
b(.0465+ 0232 ()
b( 0590+ 0232)
b(.0718+.0232 1)
b( 0854+.0232 ()
b(.1130+.0232 ()
b(.1426+ 0232 )
b(.1726-.0232 ()
b.2036+.0232 ()
b(.2344+0232 /)
b(.2660+0232 l)
b( 2980+.0232 7)
b( 0062+.0232 ()
b( 0132+.0232 ()
b(.0220+0232
b(.0318+.0232 ()
b(.0429+.0232 ()
b( 0550+0232 /)
b(.0673+ 0232 ()
b(.0936− .0224 7)
b(.1216+.0232 7)
b(.1510+.0232 ()
b(.1916+.02.32 ()
b(2120+.0232 ()
b(.2420 +.0232 ()
b( 2736+.0 321)
b(.3050+.0232 )
| ULTIMATE MOMENT |
b( 3750+ 9580 l)
b( 7810+10750 l)
b( 13200+11900 )
b( 20160+13080 )
b( 28600+14250 )
b( 38200+15400 l)
b(49500+16550 )
b( 622 0+17700 )
b( 76500+18850)
b( 92000+20050 7)
b(108700+21200 l)
b( 1035+ 9290 )
b( 3430+10450 l)
b( 7320+11610 )
b( 12600+12780 l)
b(19420-13920 )
b( 27500+15100 )
b( 37250+16250 l)
b_48460+17400 7)
b( 60700+18600 l)
b( 74700+19750 l)
b( 90300+20900 l)
b(125000+23200 )
b(166000+25500 l)
b(209500+27900 7)
11+ 9000 Z)
880+10160 )
b( 3120+11320 )
b( 6880+12500 l)
b( 12020+13650 l)
b( 18560+14800 l)
b( 26800+16000 )
b( 36200+17120 )
b( 47200+18300 l)
b( 59000+19460 )
bi 73200+20600 l)
b(105000+22920 1)
b(143000+25250 l)
b(187000+27600 )
b(235400+29900 )
b(290υ00+32200 1)
8 (
bl
bl
b
725+11050 2)
2830+12200 ()
b( 6430+13350 ()
b(11550+14500 ()
b( 17900+15700 )
b( 25750+16850 ()
b( 35200+18000 Z)
b( 45800+ 192007)
b( 58200+20300 ()
b( 86.00+22650 ()
b(121500+25000 ()
b(161700+27300 ()
b(208000+29600 l)
b(259000+31900 ()
b(316200+34250 l)
b(380000+36600 7)
b( 2560+13050 )
bi 6000+14200 7)
b 10900+154007)
b(17100+16550 )
b(24800+17700 ()
6( 31200+18900 )
b( 44600+20000 )
b( 70000+22350 ()
b(101500 +24700 ()
b(138800+27000 /)
b(182200+29300 ()
b(230800+31600 ()
b (284000+34000 ()
b(344000+36300 ()
b(410000+38600 7)
Pe"=1161 bl.
.230bl-b/2
.220bl-b/2
.214bl-b/2
.210bl-b/2
20561-6/2
.202b6l-b/2
.200bl-b/2
.1986l-b/2
bi
.196bl-b/2
.196bl-b/2
.194bl-b/2
.217bl-b/2
.205bl-b/2
.196bl-b/2
.190bl-b/2
.186bl-b/2
.183bl-b/2
.179bl-b/2
.177bl-b/2
.175bl-b/2
.173bl-b/2
.172bl-b/2
.170bl-b/2
.168bl-b/2
.168bl-b/2
.212bl -b/2
.195bl-b/2
.185bl-b/2
.177bl-b/2
.172bl-b/2
.168bl-b/2
.164bl-b/2
.161bl-b/2
.159bl-b/2
.157bl-b/2
.156bl-b/2
.154bl-b/2
.152bl-b/2
.1516-6/2
.150bl-b/2
.149bl-b/2
.178bl-b/2
.169bl-b/2
.162bl-b/2
.157bl-b/2
.153bl-b/2
.150bl-b/2
.147bl—b/2
.145b-b/2
.143bl—b/2
.14167-6/2
139bl-b/2
.137bl-b/2
•
.136bl-b/2
.135bl-b/2
.13467-6/2
.134bl-b/2
.156bi-b/2
.15067-6/2
.145bl-b/2
.14rbl-b/2
.138bl-b/2
136bl-b/2
.1336/-b/2
.130b/-b/2
•
.128hl-b/2
.127bl-b/2
.125bl-b/2
.124bl-b/2
.123bl-b/2
.122bl-b/2
.1226l-b/2
75
A
TABLE FOR THE DESIGN OF TEE BEAMS-Continued.
GOOD ROCK CONCRETE.
S=50000
P." 1161bl.
t
6"
7"
8"
ď
27228222*-****
17
18
19
24
26
30
32
36
38
16
17
18
19
20
22
24
26
28
30
32
********
34
36
38
40
*****O**
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
AREA OF STEEL
-
b(.0118+.0232 )
b(.0198+.0232 l)
b(.0292+ 0232 ()
b( 0398+.0232
b(.0510+.0232 /)
b(.0756+.0232 l)
b(.1028+.0232 l)
b(.1304+.0232 ()
b(.1592+.0232 1)
b(.1892+.0232 l)
b(.2190+.0232 )
b(.2506+.0232 ()
b(.2820+.0232 l)
b(.3130 +.0232 7)
b(.0007+.0232 l)
b(.0038+,0232 l)
b(.0091+.0232 l)
b(.0161+.0232 l)
b(.0244+.0232 l)
b(.0444+.0232 l)
b(.0674+.0232 l)
b(.0930+.0232 1)
b(.1196+.0232 /)
b(.1470+.0232 l)
b(.1760+.0232 l)
b (.2064+.0232 /)
b(,2360+.0232 l)
b(.2660+.0232 l)
b(.2970+.0232 1)
b(.0071+.0232 ()
b(.0206+.0232 l)
b(.0390+.0232 l)
(.0604+.0232 1)
b(.0844+.0232 ()
b(.1096+.0232 l)
b( 1368+.0232 7)
b(.1646+.0232 )
b(.1928+ 0232 /)
b(.2224+.0232 )
b(.2526+.0232 )
b( 2820+.0232 l)
b(.3130+.0232 ()
b(.3430+.0232 l)
b(3760+.0232 ()
Page
BEAMS-
58
Complete Analysis, Rectangular.
Complete Analysis, T-shaped
70
Plates for Design of Rectangular. 63-69
Table for Design of T Sections......... 75
BUILDING DESIGN-
•
•
•
• • • •
INDEX, JUNE NUMBER.
See Design.
Buildings-General Principle of Design
37
•
fc=2700
ULTIMATE MOMENT
b( 5630+15100 l)
b( 10300+16250 ()
b( 16400+17420 Z)
b(24100+18600 )
b( 33050+19750 ()
b( 55400+22050 )
b( 84000+24400 )
b(117500+26700 l)
b(157000+29050 ()
b(202500+31400 )
b(253000+33700 )
b(311000+36000 l)
b(374000+38300 l)
b(441500+40700 l)
16
291+14520 )
(b_1860+15700 l)
(b 4820+16830 Z)
(b 9200+18000 )
(b_15000+19150 ()
(b 31100+21500 )
(b 52800+23800 /)
(b 80800+26100 /)
(b114000+28450 l)
b152500+30750 ()
(b198000+33100 )
6249500+35400 l)
6305000+37700 l)
(6368000+40100 Z)
(6434000+42400 )
b( 4140+18600 l)
b( 13720+20900 l)
b( 29300+23200 )
b( 50600+25500 Z)
b( 77600+27900 l)
b(110200+30200
b(149200+32500 l)
b(193300+34800 l)
b(243000 +37200 l)
b(299000+39500 l)
b(361000+418007)
b(426500+-44200√)
b(500000+46500 l)
b(578000+48750 ()
b(664000+51100 ()
..
•
···
bi
.139bl-b/2
.134bl-b/2
.131bl-b/2
.128bl-b/2
.125bl-b/2
.122bl-b/2
.120bl-b/2
.118bl-b/2
.116bl-b/2
.115bl-b/2
.114bl-b/2
.113bl-b/2
.113bl-b/2
.112bl-b/2
•
133bl-b/2
.127bl-b/2
.122bl-b/2
.118bl-b/2
.115bl-b/2
.110bl-b/2
.106bl-b/2
.104bl-b/2
.102bl-b/2
DESIGN OF TYPICAL BUILDING-
Roof System.
Floor System..
Column Design
Footings...
Drawings.
FORMULAE—
.101bl-b/2
.100bbb/2
.099bl-b/2
.098b/-b/2
.097b1-6/2
.097bl-b/2
.109bl-b/2
.102bl-b/2
.09861-b/2
.095bl-b/2
.093bl-b/2
.091bl-b/2
.090bl-b/2
.088hl-b/2
.087bl-b/2
.086bl-b/2
.086bl-b/2
.085bl-b/2
.085bl-b/2
.084b/-b/2
.084bl-b/2
•
•
Page
•
•
47
48
.52-57
42
45
Rectangular Beams, Rigid Analysis.... 58
T-Beams, Rigid Analysis..
T-Beams, Rigid Analysis..
70
70
(May Number)
See page 38 (May number) for Index on General Principles of
Design, Building Construction.
..
76

Designing Methods
Reinforced Concrete Construction
VOL. 1
JULY, 1908
BRIDGES AND CULVERTS
FOR HIGHWAY TRAFFIC
FLAT SLAB AND GIRDER TYPE
Discussion of the special problems met with in the design
of highway bridges and culverts.
No. 3
Detailed design of flat slab bridge, box culvert and
girder bridge, illustrating the application of the methods
suggested.

THE AUGUST BULLETIN WILL CONTAIN:
Standard Designs for Flat Slab Bridges,
Box Culverts and Girder Bridges, for the
three classifications by loadings, adopted.
REPRINT, MARCH, 1909
PRICE, 25 CENTS.

CORRUGATED BAR COMPANY
(FORMERLY EXPANDED METAL AND CORRUGATED BAR CO.)
→
National Bank of Commerce Building
SAINT LOUIS
COPYRIGHT 1908, BY EXPANDED METAL AND CORRUGATED BAR CO.
REINFORCED CONCRETE BRIDGES.
Reinforced concrete is now largely used for the construction of
bridges and culverts, and its popularity is well deserved. No other
structural material possesses, to the same degree, those qualities of
permanence and freedom from maintenance charges so desirable for
public works. Of all engineering structures bridges are probably the
most exposed to the action of the elements and receive the severest
load treatment. It is essential, therefore, that bridges be designed
with unusual care and that the materials used in the construction be
carefully selected. Concrete gains strength with age and is not
affected by atmospheric influences—a concrete bridge is stronger
at the end of ten years than it was at the age of three months.
Wooden, as well as steel bridges, are subject to annual repairs and
periodic renewals, while for concrete bridges the initial appropriation
represents the entire outlay, present and future. And the cost of a
reinforced concrete bridge is not prohibitive; in fact, the cost is but
slightly in excess of that timber construction and usually considerably
less than that of steel. Sand and gravel (or crushed stone) may
usually be obtained near the site and the construction may be done
with local labor.
A reinforced concrete bridge may be designed with the same exact-
ness as a steel or wooden structure.
Reinforced concrete is more reliable than plain concrete or masonry,
the construction is lighter, requiring less extensive foundations.
Reinforced concrete is not affected by atmospheric or corrosive
influences.
A reinforced concrete bridge is practically free from vibration.
In viaduct work, where steam or electric cars are to be carried,
reinforced concrete should be used, as it practically eliminates all noise.
Local labor and materials may be used in the construction of a
reinforced concrete bridge.
FLAT SLAB AND GIRDER BRIDGES.
A flat slab design will in general be found more desirable and
economical for spans up to twenty feet; for longer spans a girder type
bridge should be used. By a "girder bridge” is meant a comparatively
thin reinforced concrete decking carried by girders extending from
abutment to abutment; these girders should preferably be entirely below
the decking. In some cases, however, the side girders may be carried up
78
/
above the slab to form the side rail. Girder bridges are economical
under the usual conditions for spans of from eighteen to thirty-five feet;
for longer spans an arch bridge will probably be more desirable. Gir-
der bridges have been built for spans as great as sixty or seventy feet;
these larger structures, however, should be specially designed, and we
have made no attempt to include such unusual structures in the standard
tables given in these bulletins.
CLASSIFICATION BY LOADINGS.
Highway bridges must be designed to safely carry the heaviest load
likely to come upon them, and as this maximum load varies with the
locality, we have arbitrarily adopted three standard classifications by
loadings, which should cover all usual conditions.
In short span bridges, such as we are now considering, the con-
centrated loads are the determining factors in the design-the uni-
formly distributed loads usually specified (100 to 150 pounds per square
foot) causing smaller stresses.
Class No. 1-Light highway specification answering the purposes
of ordinary county traffic where the heaviest load may be taken as a
twelve-ton road roller. Uniformly distributed load, 100 pounds per
square foot.
Class No. 2-Heavy highway specification, designed for localities.
where heavy road rollers, up to twenty tons, and electric cars of a
maximum weight of forty tons must be provided for. Uniformly dis-
tributed load, 125 pounds per square foot.
Class No. 3-City highway specification, designed for heavy con-
centrated loads and large interurban cars. This classification should
be adopted for all city work; the weight of the maximum car has been
taken as sixty tons. Uniformly distributed load, 150 pounds per
square foot.
LOAD DIAGRAMS.
The following diagrams represent the loadings adopted in the
above classifications and used in the design of the culverts and
bridges shown in this bulletin:
47-0


Q
D₁Q
8:6-
24-0"
5:03
+5·0
Figure 1. Standard Car, Class No. 2-40 tons on eight wheels.
79


56-0"
99
Figure 2. Standard Car-Class No. 3-60 tons on eight wheels.
MIETTE
30:0″-
The concentrations due to a steam roller will be taken as indicated
by Figure 3; two-thirds of the total load being assumed on the rear
wheels.
عراس
11-0"
KIKI
1
2W
3
8:6-

7-6″.
Figure 3. Road Roller Loading Diagram, Class 2.
NOTE-Reinforced concrete slab bridges are very stiff and that
part of the slab directly under the concentrated load is materially
assisted by the adjoining sections. To assist this lateral distribution
of load transverse reinforcement should be used in all slab bridges.
Earth Fill-When there is an earth fill over the culvert or bridge
it acts to distribute concentrated loads and our methods of dealing
with such conditions will be briefly outlined.
DEAD LOAD.
The dead load includes the weight of the bridge or culvert top,
and the weight of that part of the fill supported by the bridge includ-
ing the paving or track system. In all our designs the culvert will be
considered as supporting a volume of fill as indicated in Figure 4.
80
Oku
*
std.
Roadway
Kusm

S
Df= 100d
Figure 4.
The trapezoid abcd represents the amount of fill considered as sup-
ported by the culvert; assuming the weight of the fill to be 100 pounds
per cubic foot we may write
(2 std)
2
пола
50d (2 s + d)
where D-load on 12" length of culvert due to fill.
sidered as a uniformly distributed load across the culvert.
Dr will be con-
To obtain the total dead load add the weight of a 12" strip of cul-
vert top, also the weight of paving or track system.
LIVE LOADS.
A uniformly distributed load shall be considered as causing the
specified pressure per square foot on the bridge regardless of depth of
fill.
A minimum fill of twelve inches is required on all bridges.
Wheel or road roller concentrations shall be considered as acting
on a line whose length equals the out to out tread of the wheels.
Loads on car tracks shall be considered as uniformly distributed
over a width of roadway equal to the length of the ties and in the
direction of the track for a distance of two feet on both sides of single
wheels and for a distance of the wheel base plus two feet for trucks.
The above distribution of load is at the level of the roadway. The
following methods of finding the loads on the bridge itself are sug-
gested:
81
Wheel Loads on Roadway-Assume distribution of load by fill
to be only in the direction of the roadway and to be carried down
on a slope of ½ to 1. The following diagram, Figure 5, showing the
distribution of road roller concentrations, illustrates our method.
11-0"

4:03
을​을
​dddd-
Figure 5. Showing Distribution of
Loads due to Road Roller.
Wheel Loads on Tracks-See distribution by track system, page
81. These loads will be considered as distributed in a manner
similar to that adopted for wheel loads on the roadway, excepting that
the distribution will be assumed to be in both directions. It should,
however, be borne in mind that on double track slab bridges the width
of slab considered as supporting one track can not be taken as greater
than the distance c. to c. of tracks.
Impact-When the fill is less than five feet add 25% for impact
for rapidly moving loads.
The following diagram (Fig. 6) shows the assumed distribution of
standard truck load, forty-ton car.
akuminan
-5′0″
With this arbitrary distribution
of loading it will be noted that for
a strip, the width of the front
wheel, the loaded areas overlap
when the depth of fill is greater
than the distance between axles.
In this case, consider the load as
uniformly distributed over an area
of slab 7'6" wide by (d+11'0")
long.


70"-
d+70″-
24-0"-
Tie Length

d+Tie Length
Figure 6. Load Distribution, 40-ton Car.
Tuu
82
Treatment of Loads for Girder Bridges-The distribution of
loads through the fill will be as above outlined; in this type of bridge,
however, the girders must be so located as to properly take care of
the track loads. The girders under the tracks being assumed to carry
the full load.
Abutments and Side Walls-For the design of abutments and
side walls take the horizontal component of the earth pressure as one-
third of the vertical pressure at that depth, assuming the resultant to
act at a distance one-third the height above the base. The intensity
of the horizontal pressure due to live load may also be taken equal to
one-third of the vertical intensity at any depth; assuming that the
planes of zero pressure, bounding the supporting prism of earth to have
a slope of one-half to one.
Weights and Dimensions of Electric Cars-The weights
assumed for the electric cars in the preceding classification may seem
rather large, but it should be remembered that the stresses in the
bridge depend not only on the weight of the car, but also on the wheel.
base, distance between trucks, etc. The dimensions vary with the
locality and the weights and dimensions chosen are, in our opinion,
justified.
If it is desired to make a special design the following data on electric
cars may be of use. The values given must be taken as approximate
averages. The weights given are for the loaded car and include the
weight of the trucks.
Small cars, such as are used in small towns, four wheels on two
axles, seating twenty-eight persons. Car body, 20'0"x8'3": over all
length, 29'0"; distance c. to c. axles, 8'o"; weight, 11 tons.
City car for heavy service, seating fifty-two persons. Car body,
34′0″x8′6″; over all length, 47′0″; wheel base, 4′0″ to 6′0″; c. to c.
trucks, 24′0″; weight, 15 tons.
Large interurban cars, seating 72 persons. Car body, 50'0"x8′6″;
over all length, 56′0″; wheel base, 6′3″; c. to c. trucks, 30′0″; weight,
42 tons.
METHODS OF DESIGN.
Having found the loads on the bridge by the methods above outlined,
the bending moments and shears are obtained by the ordinary principles
of mechanics. All bridges of single span will be considered as beams.
freely supported at the ends and the bending moments must be figured
on this assumption.
83
Box culverts, however, are reinforced against reverse moment at
the ends and the designing moment may be taken as eight-tenths of
that which would be developed in a simple beam.
All bridges and culverts will be designed on the basis of ultimate
strength with a factor of safety of two on the dead and four on the live
load, the minimum designing moment, however, to be taken as three
times dead load plus live load moments. The elastic limit of the steel
has been taken as 50,000 pounds.
The above method of designing gives a maximum working stress in
the steel of about 16,500 pounds per square inch.
Since it is desirable to use local materials, and a dense hard aggre-
gate may be difficult to obtain, the ultimate compressive strength of the
concrete will be taken as 2,000 pounds per square inch. This com-
pressive strength should be easily obtained with average aggregates
using a 1:2:4 mix. The critical amount of reinforcement, correspond-
ing, using steel with an elastic limit of 50,000 pounds would be .0085 bd.
Girder Bridges-For single span bridges figure the inside
girders as T-beams; in this case the cross section of the reinforcement
should not exceed 22% of the area of the rectangle enclosing the stem
(bd). For bridges with two or more spans the girders should be fig-
ured as continuous rectangular beams and provision made for reverse
moment over the supports.
Percentage of Reinforcement-Use .85% of reinforcement in
slabs and rectangular beams for concrete of 2,000 pounds compressive
strength. In T-beam designs limit the amount of reinforcement to
22% of bd. These figures are based on the use of mechanical bond
bars with an elastic limit of 50,000 pounds per square inch.
a
b
Figure 7. Rectangular Beam.
a
¿'
b

b
Figure 8. T-Beam.
P
2
พง
84
FORMULAE TO BE USED IN DESIGNING.
Let M=moment of resistance of section in inch lbs.
Mo-ultimate moment of resistance of section in inch lbs.
q=area of reinforcement.
p percentage of steel-g÷bd.
Felastic limit of steel-50,000 lbs.
All dimensions as shown, in inches, and stresses in lbs. per sq. in.
Then we have for rectangular beams:
and for T-beams:
Mo=370 bd for g=.0085 bd.
M。=.86 Fp bd²=.86×50,000×p bď³.
=43.000 p bď, using high elastic limit corrugated bars.
(For complete discussion of these formulæ, see May and June bulletins.)
WORKING STRESSES.
If it is desired to design for working stresses in the steel, use the
formula:
M=sg×.86 d, where s=unit stress in the steel.
Shearing Provisions-In flat slab bridges it will not be necessary
to make other provision for shearing stresses than by bending up part
of the main reinforcing bars as required.
In girder bridges, however, special provision must be made against
failure by diagonal tensile stresses, as the girders carry all the shear.
The concrete may be assumed to carry safely fifty pounds of vertical
shear per square inch of cross section (bd). In girders some of the
reinforcing bars should be bent up near the ends of the beam and
stirrups put in as required.
In all T-beam designs it is especially necessary to have a number
of stirrups for the purpose of bonding the flange and stem.
85
BRIEF SPECIFICATIONS FOR MATERIALS
AND LABOR.
Cement-Only Portland cement conforming to the requirements
of the specifications adopted by the American Society for Testing
Materials, June 14, 1904, shall be used.
Sand-Sand to be clean and coarse, and free from organic matter;
a graded sand, with coarse grains predominating, is to be preferred.
Coarse Aggregate-Broken stone to be hard, durable limestone,
or its equivalent, free from dust and foreign materials; maximum-sized
particles to pass through a one-inch ring; fines to be removed by
passing over a one-quarter inch screen. The fines may replace part of
the sand. Under special conditions making for uniformity crusher
run may be used, but this is not desirable.
Gravel shall be clean and of graded sizes; the sand carried to be
removed by screening as for broken stone.
PROPORTIONS OF MIX.
Concrete for columns, beams and slabs to be mixed in the propor-
tion of one part cement to six parts aggregate; proportions by volume
taking one bag containing not less than 94 pounds of cement, equal to
one cubic foot of cement.
The proportions of fine and coarse aggregate used shall be chosen
so as to give a concrete of maximum density; in no case, however,
may the amount of fine aggregate be less than 50 per cent of the coarse.
REINFORCING STEEL.
All reinforcing steel used in bridge and culvert construction shall
be rolled to such form that it has a positive mechanical bond with the
concrete. Adhesive bond will not be considered sufficiently reliable
for this class of structures.
Steel may be made by either the Bessemer or open-hearth process;
bars to be rolled from billet stock. Re-rolled material will be accepted
under conditions insuring rigid inspection.
The elastic limit and percentage of elongation shall be determined
by tests on accurately-machined specimens, and shall conform to the
following requirements:
Elastic limit to be from 50,000 to 60,000 pounds per sq. inch,
ultimate strength not less than 1 Xelastic limit.
86
водить и умер и
The percentage of elongation in 8″ must not be less than given by
the formula :
Percentage of elongation=
1,400,000
Ult. Strength
5.0.
Bending Test-Bars as rolled shall bend cold, 90 degrees, to a
radius=three times the least diameter of the specimen, without sign
of fracture.
Forms
Forms must be smooth and true to dimensions, with
close joints to prevent leakage, and must be of sufficient strength to
carry the load without appreciable deflection.
Removal of Forms-The time of removal of forms should be
left to the discretion of the engineer, as the time required for the con-
crete to gain sufficient strength is dependent upon temperature and
weather conditions and on the particular cement used. Forms for
slabs should not be removed in less than two weeks under the most
favorable conditions; girder forms should not be removed in less
than three weeks.
Mixing Machine mixing is to be preferred in all cases-hand
mixing to be allowed at the discretion of the engineer. Concrete for
reinforced work shall be mixed wet, sufficient water being used to
make a mass that will flow readily and be of such consistency that
the reinforcing steel will become coated with a protective coat of
fluid mortar. Excess of water should be avoided, as it tends to a
separation of the parts.
Placing-In girder bridges and in all T-beam designs the con-
crete must be placed the full depth (to the top of the slab) at one
operation. If possible the work should be carried on continuously
to completion.
87
!
DETAILED DESIGN OF A FLAT SLAB BRIDGE
The following example illustrates the application of the methods
above outlined:
Problem-Design a flat slab bridge, resting on abutments, clear
span 16'0", with an earth fill 2'0" deep. Roadway to be 16'0" wide in
the clear, Class 2 loading. See Figure 9.
WININ
A
S
EARTH FILL 20" THICK.
S=16-0"
Figure 9.
In the design we will consider only a strip of bridge 12" wide as
this simplifies matters. The section will be made constant across the
width.
DEAD LOAD.
Total....
Weight of fill=50 d(2 s+d)=100(34)=3,400 lbs.
Weight of slab (assuming thickness=16")=
13X150X16=3,200
Bending moment - WI-X 6,600 X 16
|
KIK

LIVE LOADS.
Actual dead load moment=158,400 inch lbs.
(6
..6,600 lbs.
13,200 ft. lbs.
For this span maximum stresses will be caused by the concentrated
loads, the uniform load will not be considered. We will determine
the bending moments due to the twenty-ton roller and to the forty-
ton car, using the larger in the design.
88
W
Về
1
Road Roller-Maximum moment occurs with rear wheels at
center of span. Load on rear wheels equals two-thirds of 40,000
pounds=26,700 pounds. This load as previously explained (see page
82) acts on a line 7'6" long; the distribution on the slab is shown by
the following diagram (Figure 10); the broken lines indicate the
area of slab over which the load is distributed.

TRI
1
Figure 10.
7-6″-
8-0
Spry

3560
نف
16 0'
The load per square foot on area 2′0″X7′6″_27,600
Figure 11.
نش
2X7.5
=1,780 lbs.
On a strip of bridge 12″ wide, the load would be as shown by
Figure II.
Maximum moment at center of span, on strip 12″ wide,
=M=(1,780X8)—(1,780X2)=13,400 ft lbs.
=161,000 inch lbs.
89
Electric Car-The maximum moment occurs with one truck
on center of span. Distribution of load on slab is as shown by diagram
(Figure 12), assuming ties to be eight feet long.

,0-01
80
£
TRACK
7-0'-
d+7·0=90'
Figure 12.
ار
The load on a strip 12" wide would be as shown by the following
diagram, Figure 13:
The full line shows area over
t which truck load is distributed by
track system; the broken lines
indicate loaded area of slab.
Load per square foot of loaded
area=40,000÷90=445 lbs.

18:0"
D
9-0″.
4000#
16-0″
Moment at center=M=(2,000X8) — (2,000X24)= 11,500 ft. lbs.
=138,000 in. lbs.
Figure 13.
Adding 25% for impact, moment=172,000 in. pounds.
This moment is larger than that due to the road roller and we will
use it in the design.
(
Using a factor of safety of two on the dead load and four on the
live load we have
Ultimate moment, dead load=2X158,400
((
((
=4X 172,000=
live
Designing moment=Mo=1,004,800 in. lbs.
316,800 in. lbs.
688,000
((
90
We can determine the depth of slab and the amount of reinforce-
ment required by the formula:
Mo=370 bd, for q=.0085 bd.
Mo
1,004,800=370X12Xď.
from which d=15"
9=.0085X12X15=1.53 sq. in.
d=distance from top of slab to the center of the reinforcing bars,
we will add 1½" of concrete, giving 1' on underside of bars.
I
I
Make slab 16½" thick; 1" corrugated rounds spaced 6" centers,
Bend up every third bar at the sixth point, say 2′6″ from the abutments.
Tranverse Reinforcement-To properly distribute concentrated
loads and to tie the bridge in the transverse direction ½" corrugated
rounds will be placed (over the main reinforcing bars) crosswise of
the bridge, and 12″ on centers.
Shearing Investigation-The dead load shear on a strip 12″ wide
is 3,400 pounds.
The maximum live load shear occurs when the rear wheels of the
road roller are 12″ inside the abutment, and is equal to
3,560 X 15
16
Total shear=3,400+ 3,340=6,740 lbs.
F3.340 lbs.
At the allowed stress of fifty pounds the concrete alone is capable
of carrying 12X15X50 9,000 pounds of vertical shear. This would
indicate that no provision for shear need be made; every third bar will
be bent up, however, as stated.
Side walls for retaining fill. It will not be necessary to figure
these. They will be made 12″ thick and reinforced as shown.
Waterproofing-Some form of waterproofing should be used and
the top surface of the slab arranged for drainage. The top surface of
the slab will be as shown on the drawings.
Bearing on Abutments-All concrete bridges resting on abut-
ments shall have at least 12" bearing; a maximum pressure of fifty
pounds per square inch will be allowed for slab bridges.
91

147"
6:1
ESI
/ // | / \ | || | / \ || / // // ///
DRAIN PIPE
ܫ ܫ
SIDE ELEVATION
•#•
I
I CORR ROUNDS 6″ CTS.
15=
18-
d
LONGITUDINAL SECTION.
-
#ST
GM QU
EVERY THIRD BAR BENT AS SHOWN.
1
三​三百
​8'0'
SUMM
Wa
"
ཌ་
TRANSVERSE BARS CORR ROUNDS 12" CTS
Ź CORR. ROUNDS 12 CTS
8'-11"
SECTION AT CENTER
OF
NOTE: ALL BARS ARE CORRUGATED ROUNDS.
DETAIL PLANS
LUGHA TI
CLEAR SPAN 16-0"
LOADING-CLASS 2.
2:0*
FLAT SLAB HIGHWAY BRIDGE.


92
BOX CULVERTS.
Box culverts will be built monolithic and reinforced against reverse
moment at the corners. The methods suggested for designing this
type of bridge will now be illustrated by a detailed design.
Problem-Conditions of span and loading as before, side walls to
be eight feet high.
Design of Top Slab-The live and dead load moments will
be figured as in the preceding example. For the designing moment,
however, use eight-tenths of the moment that would be developed if the
ends were free.
Mo, then-X 1,004,800 in. lbs.=803,840 in. lbs. (ult.)
Solving for d and q as before, we have
d=13.4″ and 9=1.36 sq. inches.
Make slab 15" deep and use 1" corrugated rounds 7″ on centers.
Outside corner reinforcement use one-half the amount of steel re-
quired in slab and fillet the corners as indicated on drawings.
Bottom Slab-Make same as top slab.
As before transverse reinforcement will be required and bars will
be introduced as indicated on the drawings.
SIDE WALLS.
In addition to supporting the top slab the side walls must be capable.
of resisting the horizontal pressure of the earth. This moment is of
opposite sign to that produced in the side walls owing to the rigid con-
nection with the top and bottom slabs. As before we will take the de-
signing moment as eight-tenths of that which would be produced in a
beam free at the ends.
The horizontal intensity of earth pressure will be taken as one-third
of the vertical intensity. The horizontal pressure due to the live load.
will be based on a uniform load of 450 pounds per square foot on the
roadway. This will give a horizontal pressure of 150 pounds per
square foot, regardless of depth of fill, and should be ample to cover all
contingencies.
93
al
1
L_b
KUKTIKTIKTIETISTIKTI
MINTINIK NIINI UNUNUN
SÜRÜN
Figure 14.
Luise 200

The inclined broken line represents
the intensity of the horizontal earth
pressure at any point.
a=
b=
100X3.3-110 lbs.
3
100X11.3–375 lbs.
3
Average=242 lbs.
It will be sufficiently accurate to assume this average pressure as
uniform over the side wall.
Average load per ′ then=242 lbs. +150 lbs. 392 lbs.
M=}{} wl²=§×392X8²=3,130 ft. lbs.
Using a factor of 3 on dead and live load, we have an ultimate
moment of
3X3,130X12=112,500 inch lbs.
Designing moment-Mo=1X112,500-90,000 in. lbs.
In all box culverts the side walls will be made the same thickness
as the slab; d, then, would equal 13.4".
The amount of reinforcement required would be
_90,000×1.36=.15 square inch.
808,000
Use ½" corrugated rounds, 14" on centers.
The spacing of the reinforcing bars in the side walls should be a multiple of
the spacing of the main reinforcing bars in the top slab, as this facilitates the
placing of the steel.
NOTE-In the above design that part of the culvert directly below
the roadway and for a distance of one-half d on either side, has been
considered. Beyond these limits the stresses due to dead load decrease,
the side slope usually being one and one-half to one. If desired the
amount of reinforcement and thickness of concrete may be reduced
toward the ends of the culvert.
94
Longitudinal reinforcement should be used in all culvert construc-
tion. On account of unevenness in the supporting power of the earth
and similar causes, box culverts often must act as beams spanning
these weak places. Wing walls with large footings and slabs to pre-
vent scour, when built monolithic with the culvert sometimes produce
a bending moment in the whole structure, owing to the uneven soil
pressure occurring under these conditions.
Aprons Box culverts should be provided with aprons to prevent
scour, and the wing walls should be figured as retaining walls.
95

,9-,01
E=2
Lauren
16
Ź CORR ROUNDS 14'0C
15
E
=
뚝
​==
===
I" CORR ROUNDS - 7″ac.
BEND EVERY THIRD BAR AS SHOWN.
_______
OUTSIDE CORNER BARS I CORR ROUNDS 14"QC. LAP AS SHOWN.
BARS
NOTE: ALL LONGITUDINAL BARS ARE Ź CORR ROUNDS.
F
#
||
= it
L
#
1
}
11
m
PRE
}}
!
18-6″
CROSS SECTION
I
{"
16:0"
+
HALF PLAN
SHOWING OUTSIDE CORNER BARS. CORNER
DIAGONALS
DIAGONALS AND OUTSIDE LONGITUDINALS-
1
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==
1
→X
ព
J
C
Ź CORR ROUNDS 14.0.C
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11
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11
we make se
F
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11
==
11
=== p
11
===
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==== ==F
16"
1
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HALF PLAN
SHOWING MAIN SLAB REINFORCEMENT
AND INNER LONGITUDINALS.
1
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==
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makan bila dah sem
SHOWING LAP OF OUTSIDE CORNER BARS ·
11
NOTE
==
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pa pa
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===
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DETAIL PLANS
=
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Car S
VERTICAL LONGITUDINAL SECTION.
OUTSIDE CORNER BARS NOT SHOWN IN ELEVATION.
med 143
.SI
OF
TYPICAL BOX CULVERT
16-0″x 8-0″ OPENING.
..0-8
CULVERT IS SYMETRICAL ABOUT BOTH CENTER LINES.
TOP AND BOTTOM SLABS ARE SIMILARLY REINFORCED.
ALL BARS ARE HIGH ELASTIC LIMIT CORRUGATED ROUNDS.
15'


96
GIRDER BRIDGES.
The following detailed design will illustrate the application of the
methods advocated to the design of a girder bridge:
Problem-Design a girder bridge, resting on abutments; clear span
32'0"; earth fill 15" deep. Bridge to be 24'0" wide in the clear, with
two 4′0″ sidewalks and car track on center line. Class 2 loading.
The cross section to be as shown by "Proposed Section," Fig. 15.
The detailed drawings are based on this section but the computations.
apply to the alternate section as well.

F
G1.
5-0″
ISIA
G2
5:0″
SUPIENFIFIREUENEUENIEMS VETEMEN
ALTERNATE SECTION
G3.
G3
Dead load per square foot:
510"
5
12
1
5:0°
DESIGN OF SLAB.
15
2
Gz
Figure 15.
Floor Slab The minimum thickness of floor slabs will be taken
as 5″.
This thickness of slab should take care of extraordinary
concentrated loads such as might be caused should a car be derailed
on the bridge.
PROPOSED SECTION.
To provide for such contingencies all slabs for girder bridges will
be designed for a live load of 500 pounds per square foot, in addition
to weight of slab and fill, using a factor of two on the dead and four
on the live load.
Moments will be figured by the formula M-2 wl, since the slabs
are continuous over three or more supports; l=distance c to c of
beams.
Slab, X150= 62 lbs.
Fill, X100=125 lbs.
Total, 187 lbs.
G1
5:0"
97
Dead load moment
wX187X25=390 ft. lbs.
Live load moment
X 500X25=1,040 ft. lbs.
Designing moment=M=(2X 12X390)+(4X12X1,040)=59,350
in. lbs.
Taking a strip of slab 12" wide, we can find the thickness of slab and
the amount of reinforcement required from the formula Mo=370 bd²,
in which q=.0085 bd.
Inserting the values for M. and b in this formula, we find that
d=3.7 inches, and q=.38 square inches, where q is the section of rein-
forcing steel required in a 12-inch width of slab.
A
Since we have made the thickness of the slab 5" d will be 4", which
is greater than required by the formula. The amount of steel required
may accordingly be decreased, and is equal to
3.7.38 sq. in.=.35 sq. in.
4
•
Slab will be 5" thick, reinforced with ½" corrugated rounds placed
7" on centers.
In the design we have considered the slab as partially fixed on the
beams and to provide for the reverse bending moment developed,
reinforcing bars will be placed in the top of the slab over the beams;
the amount used will be one-half that required in the bottom of the
slab and we will use ½" corrugated rounds, 3'0" long, spaced 14″ on
centers.
NOTE―That part of the slab under the sidewalks will be the same
as that under the roadway.
Girders-In all girder bridge designs the length center to center
of bearings will be taken equal to the clear span plus one foot. This
length, c to c of bearings, will be used in computing the stresses
developed. It is desirable to have brackets at the ends of the girders
when conditions permit, so as to reduce the unit vertical shearing
stresses and gradually unload the reaction at the abutment into the
girder. In all girder designs special provisions for taking care of
shearing and diagonal tensile stresses should be made. Some of the
main reinforcing bars should be bent up near the ends of the girder
and stirrups used throughout the length.
98
Girder G1-This girder will be figured for the dead load and a
live load of 125 pounds per square foot on the walk.
Dead load on girder:
Sidewalk, X150X2½ X32= 4,000 lbs.
Fill, 12X100X22X32= 7,350 lbs.
Slab, 1X150X22X32= 5,000 lbs.
Girder (assumed 12"X36")=14,400 lbs.
Total...
=30,750 lbs.
4
1 2
1
8
Dead load moment- WI-X30,750 X 33=127,000 ft. lbs.
Live load, 125 lbs. per square foot.
Live load on girder=125X2½ X 33=10,000 lbs.
Live load moment= WI-X 10,000 X 33=41,200 ft. lbs.
To get the designing moment, use a factor of 2 on dead load and
4 on live load, minimum to be, however, 3 (d. 1.+l. 1.)
Mo=3(127,000+41,200) X12=6,050,000 in. lbs.
Applying the formula M。=370 bď, and taking b=12″, we find that
d=37″; q=.0085 bd=3.76 square inches.
We will make girder 12" wide and 40″ deep, using five 1" corrugated
rounds and bending up two bars as shown, at a point 4'0" from each
abutment.
Shearing Provisions-The maximum external vertical shear at
the end of the girder, due to full live and dead loads equals 20,375
pounds.
In all girder designs the concrete will be assumed as capable of
carrying 50 pounds of vertical shear over the cross section bd. Accord-
ingly, if V total shearing value of the concrete, we have:
Ve=12× 37×50=22,200 lbs.
This would indicate that no special shearing provisions are neces-
sary. It is advisable, however, in all cases to make some shearing
provisions, and we will use U-shaped stirrups of ½" corrugated
rounds, spaced 18" on centers throughout the length of the girder.
Girder G2-This will be designed for the average of the stresses
in girders GI and G3, so we will accordingly figure girder G3 first.
99
Girder G3-Class 2 loading requires that the design be based on
the maximum stresses produced by either a twenty-ton road roller or
a forty-ton electric car. (The alternative live load of 125 pounds per
square foot causes much smaller stresses than the concentrated loads.)
The two girders G3 will be designed to carry the total car load.
Each girder may, however, carry two-thirds of the road roller
concentrations; the full load on the front wheel and one-half of the
load on the two rear wheels.
All interior girders on single span bridges should be figured as
T-beams.
Dead load on girder:
Fill,
X100X5X32=20,000 lbs.
5
Slab, X150X5X32=10,000 lbs.
1 2
Girder (assume 450 lbs. per ft.)=14,400 lbs.
Total....
=44,400 lbs.
Dead load bending moment:
R₂H-.
ASIDA
Live Loads-Maximum moments due to road roller.
We will assume that only one road roller will be on the bridge at
any one time. The maximum load on one girder then may be repre-
sented by two concentrated loads of 13,300 pounds each, 11'0" On
centers. The maximum mo-
ment will occur with one of the
loads 2'9" off center of span, as
shown by Figure 16.
31M
M=} WI=1}{X44,400 X 33=183,000 ft. lbs.
8
1
8
110+
R₁=
po
15
1 2

32-0
33-0
294
वर
R,
+
Figure 16.
13,300X8.25, 13.300X19.25
33
33
Since the fill is but 15" deep,
the effect of the fill in distribut-
ing the loads will be neglected
in determining the moment on
the girder.
=11,100 lbs.
M=13.75X11,100=153,000 ft. lbs.
J
+
Maximum moment due to electric car.
For assumed distribution of load by track system, see Figure 6,
page 82.
2
100
T
The maximum moment will occur with one truck at the middle of
the span, the other truck being off the bridge. (Two cars following
each other will, for this span, produce practically the same moment as
one car. See sketch of standard forty-ton car on page 79.)
16-6
KAJAKAUGELATIN
7,0
20000*
32,0
33 0
i
Figure 17.
To this static moment add 25% for impact for rapidly moving loads,
giving a moment of 184,000 foot-pounds.
The loading for maximum
moment will be as shown by
Figure 17; where the load
given is that on one girder
M=(10,000X16%)-(10,000
X1.75)=147,500 ft. lbs.

The maximum moment then due to the specified live load is 184,000
foot-pounds.
Designing moment :
Mo=((2×183,000)+(4×184,000))12=13,224,000 in. lbs.
For the design of T-beams we will use the formula
M。=.86 Fp bď²=43,000 p bď², using high elastic limit corrugated
bars, See page 85.
Assume d=32″ and b=14″, we then have
Mo=13,224,000=43,000XI4X32 XP
from which p=.0215
9= 0215×14×32=9.65 square inches.
We will make the girder 36" deep over all and use eight 14" cor-
rugated rounds.
P
For this length of beam there is no danger of failure by horizontal
shear along the horizontal or vertical planes of attachment of the stem
to the flange. The distance between beams is 5'0", and the amount of
reinforcement used=0.215 bd, where b=14'; corresponding to an
average percentage of reinforcement for the full width of slab of one-
half of 1%. This indicates that there is ample width of slab between
beams for T-beam action.
101
Shearing Provisions-The vertical external shear at the end of
the beam, due to dead load is 22,200 pounds, the load per foot of
girder being 1,380 pounds.
The shear at the end of the girder due to the car would be practically
a maximum when the center of one truck is 3'6" from the abutment ;
this total vertical shear may be taken equal to 20,000 pounds.
The total maximum shear at end of girder=42,200 pounds.
In providing for vertical shear we will assume that the concrete.
carries fifty pounds per square inch on the section bd, and put in steel
to carry the excess.
Steel for reinforcing against diagonal tensile and shearing stresses.
will consist of bent up main reinforcing bars and loose stirrups.
In the design we will neglect the effect of the bent up bars.
(If bent up bars are figured to carry the diagonal component of the vertical
shear in the "panel" in which they occur, limit the direct tensile stress to 12,000
lbs. per sq. inch.)
Loose vertical stirrups will be figured by the formula
.86 d P
V-Ve
.86 d P
V-50X bd
y=
Where y-spacing of stirrups required at any section.
P-total stress in one stirrup=total cross sectional area of
the vertical legs of the stirrup times the allowed unit
stress (16,000 lbs.)
V=external vertical shear at any section.
Ve=total vertical shearing stress that the concrete is assumed
to be capable of taking=v.×bd.
(For the derivation of this formula see page 13, May issue.)
If the stirrups are to be figured to carry all the vertical shear with-
out assistance from the concrete, use the formula
y=
.86 d P
V
102
•
Should it be desired to include that part of the vertical shear
assumed to be carried by the bent up bars the formula becomes
.86 d P
y= v(Ve+Vs)'
V₁ amount of vertical shearing stress carried by bent up bars.
The following table gives the data necessary to determine the
required stirrup spacing, neglecting the effect of the bent up bars:
Stirrups U-shaped ½" Corrugated Rounds, P=6,080.
Distance from
Abutment.
02+0∞ ~
O
4
6
8
12
}
Vert. Ext.
Shear,
V.
42,200
38,400
33,200
28,400
24,200
16,100
Ve
in which
22,400
22,400
22,400
22,400
22,400
22,400
Į
V - Vc
19,800
16,000
10,800
6,000
1,800
Required
Spacing,
y.
"
8.4
10.4
15.5
27.9
"
We will make spacing nine inches for a distance of six feet from
the abutment, increasing the spacing to eighteen inches beyond this
point.
Mo, then=½ (6,050,000+13,224,000)=9,637,000 in. lbs.
9,637,000=50,000 9X.86X 32
from which q=7.0 square inches.
"!
Bent up Bars-Bend up two reinforcing bars at a point 6'6" from
abutment, and two additional bars 3'3" from abutment.
Girder G2-In designing this girder we will take the average of
the moments in girders GI and G3.
This girder will be made the same size as G3; the amount of
reinforcing steel required may be determined by the formula
M。=s_qX.86×d
103
Make girder 36" X 14" as before, using seven 18" corrugated rounds.
Bend up one bar 6'6" from end and two bars 3'3" from abutment.
Stirrups: use ½" corrugated rounds same spacing as in G3.
Bearing of Bridge on Abutment-In order to properly distri-
bute the load and provide for sufficient bearing area the bridge will
be made solid for the full depth of the girders, where it rests on the
abutment. This construction is desirable on all girder bridges, owing
to the rigidity and general stiffness given by the solid end.
104
12+
61
4:0″
5-0°.
Laz
Ź CR TCTS
G2
SECTION AT CENTER OF SPAN.
MAMMURTEBETE SATSUKUKUEUEN
5-0′
14.
..
16-01
16
63
"
5.0
SIDE ELEVATION
63
32-0'
RIWAY
02.
5-0'
14-
NOTE: SLAB 5" THICK - Ź "CORR. ROUNDS 7′CTS.
TOP BARS & CORR ROUNDS - 3′ LONG, 14″ CTS.
1
4-0
'CR 3'OLONO
G1
5.0"
CL
L
NOTE: ALL BARS ARE HGH ELASTIC CORR. ROUNDS
DESIGN BASED ON CLASS 2 LOADING
DETAIL PLANS
LO
TYPICAL GIRDER BRIDGE.

SHEET NO 1.

105


Ź´ CORA ROUNDS.
IN
15
4
15
+
4-0.
REINFORCEMENT $-1" CORR. ROUNDS --TWO BARS BENT AS SHOWN; STIRRUPS ½" CORR. ROUNDS SPACED AS SHOWN.
32
主
​8 SPACES @ 9¨*= 6·0°
F
HTT
=
//
REINFORCEMENT ¹7-12
==
•
GIRDER 61
11
15
5 SPACES 18″-7-6
"CORR ROUNDS THREE BENT AS SHOWN; - STIRRUPS 3 CORR. ROUNDS
32-0″
GIRDER 62.
N
…………
8 SPACES @ 9^-6:0″ "1 15
5 SPACES @ 18″=7-6″
REINFORCEMENT 8'-11" CORR. ROUNDS FOUR BARS BENT AS SHOWN;
32-0
GIRDER 63
U
U
[ ] ] ]
NOTE: ALL BARS ARE HIGH ELASTIC LIMIT CORRUGATED ROUNDS.
3-3°
SPACED AS SHOWN.
3:3
WESHa
"
3-3″
Dans
STIRRUPS {"CORR. ROUNDS SPACED AS SHOWN.
3-3′
15"
CORRR.
1
12"
DETAIL PLANS
..9:2
3.0~
3:0
OF
TYPICAL GIRDER BRIDGE.
SHEET 2



106
Bridges-Advantages of Reinforced Concrete Construction
Classification by Loadings.
Formulæ to Use in Designing..
Formulæ Based on Working Stresses
INDEX, JULY NUMBER.
Load Diagrams Adopted ...
LOADS-
..
·
Assumed Load Due to Fill.
Live Loads-Distribution.
Wheel Loads on Roadway-Distribution
Wheel Loads on Tracks-Distribution.
Dead Load.
ELECTRIC CARS-
•
•
•
Weights and Dimensions of Standards Adopted in Classification
Weights and Dimensions of Ordinary Cars
Methods of Design.
Specifications for Materials and Labor..
DESIGNS-
Detailed Design of Flat Slab Bridge.
Detailed Design of Box Culvert .
•
Detailed Design of Girder Bridge
Flat Slab Bridge-Detailed Design ..
Box Culvert-Detailed Design
•
Girder Bridge-Detailed Design
•
•
·
·
•
•
Page.
78
79
85
85
79
80
81
82
82
80
79, 80
83
•
83
85
87
93
97
87
93
97
See page 38 (Bulletin No. 1) for Index on General Principles of Design, Building
Construction.
See page 76 (Bulletin No. 2) for Index on Detailed Design of Typical Building
and Complete Analysis of the Strength of Rectangular and T-Shaped Beams.
107

}
Designing Methods
Reinforced Concrete Construction
VOL. 1
AUGUST, 1908
STANDARD DESIGNS
-FOR-
BRIDGES AND CULVERTS
FOR HIGHWAY TRAFFIC
FLAT SLAB AND GIRDER TYPE
The standard designs given in this Bulletin are based on the
methods outlined in our July number. Attention is here called to
the fact that these plans contemplate the use of High Elastic Limit
Mechanical Bond Bars-we can not be responsible for their accuracy
when inferior forms of reinforcing bars may be used.
BULLETIN No. 5, VOL. 1, WILL TREAT OF
Retaining Walls of the cantilever and
buttress types.
THIS NUMBER WILL BE PUBLISHED IN OCTOBER.
No. 4
REPRINT, MARCH, 1909
PRICE, 25 CENTS
CORRUGATED BAR COMPANY
(FORMERLY EXPANDED METAL AND CORRUGATED BAR CO.)
National Bank of Commerce Building
SAINT LOUIS

COPYRIGHT, 1908,
BY
CORRUGATED BAR COMPANY
P
STANDARD DESIGNS.
The standard designs given in this bulletin were made in accordance
with the methods outlined and the specifications contained in our July
Bulletin.
Each type of bridge has been designed for the three classes of
loadings adopted. The designs for Flat Slab Bridges and Box Culverts
include spans of from four to twenty feet. Girder Bridges have been
designed for spans of twenty to forty feet.
It should be remembered when using these standard plans that the
designs are based on the use of reinforcing bars which are rolled to
such shape that they have a positive mechanical bond with the con-
crete—the steel to have an elastic limit of 50,000 pounds per square
inch.
We wish to emphasize the fact that the proper factor of safety may
not be attained unless the specifications given in the July Bulletin are
strictly adhered to. On account of the importance of the subject we
are here repeating the specifications for the reinforcing steel and the
concrete.
REINFORCING STEEL.
All reinforcing steel used in bridge and culvert construction shall
be rolled to such form that it has a positive mechanical bond with the
concrete. Adhesive bond will not be considered sufficiently reliable
for this class of structures.
Steel may be made by either the Bessemer or open-hearth process;
bars to be rolled from billet stock. Re-rolled material will be accepted
under conditions insuring rigid inspection.
The elastic limit and percentage of elongation shall be determined
by tests on accurately-machined specimens, and shall conform to the
following requirements:
Elastic limit to be from 50,000 to 60,000 pounds per square inch,
ultimate strength not less than 13 Xelastic limit.
111
The percentage of elongation in 8" must not be less than given by
the formula:
1,400,000
Percentage of elongation Ult. Strength
5.
Bending Test-Bars as rolled shall bend cold, 90 degrees, to a
radius equal to three times the least diameter of the specimen, without.
sign of fracture.
CONCRETE.
Cement-Only Portland cement conforming to the requirements
of the specifications adopted by the American Society for Testing
Materials, June 14, 1904, shall be used.
Sand-Sand to be clean and coarse, and free from organic matter;
a graded sand, with coarse grains predominating, is to be preferred.
Coarse Aggregate-Broken stone to be hard, durable limestone,
or its equivalent, free from dust and foreign materials; maximum-
sized particles to pass through a one-inch ring; fines to be removed by
passing over a one-quarter inch screen. The fines may replace part of
the sand. Under special conditions making for uniformity crusher run
may be used, but this is not desirable.
Gravel shall be clean and of graded sizes; the sand carried to be
removed by screening as for broken stone.
K
Proportions of Mix-Concrete for columns, beams and slabs to
be mixed in the proportion of one part cement to six parts aggregate;
proportions by volume taking one bag containing not less than 94
pounds of cement, equal to one cubic foot of cement.
The proportions of fine and coarse aggregate used shall be chosen
so as to give a concrete of maximum density; in no case, however,
may the amount of fine aggregate be less than 50 per cent of the coarse.
CLASSIFICATION BY LOADINGS.
CLASS 1 BRIDGES.
Designed for Class No. 1 Loading-Light highway specifica-
tion, answering the purpose of ordinary county traffic where the
heaviest load may be taken as a twelve-ton road roller. Uniformly
distributed load, 100 pounds per square foot.
112
CLASS 2 BRIDGES.
Designed for Class No. 2 Loading-Heavy highway specifica-
tion, designed for localities where heavy road rollers, up to twenty
tons and electric cars of a maximum weight of forty tons must be
provided for. Uniformly distributed load, 125 pounds per square foot.
CLASS 3 BRIDGES.
Designed for Class No. 3 Loading-City highway specification,
designed for heavy concentrated loads and large interurban cars. This
classification should be adopted for all city work; the weight of the
maximum car has been taken as sixty tons. Uniformly distributed
load, 150 pounds per square foot.
113
FLAT SLAB BRIDGES.
Reference Drawing: Plate 1.
Reinforcing Steel: Mechanical Bond Bars; Elastic Limit,
50,000 lbs.
The following tables, in conjunction with Plate 1, give detailed
designs of Flat Slab Bridges for Spans of from six to twenty feet, with
fills up to eight feet. A minimum depth of fill of twelve inches is
required on all bridges.
The tables not only give the size and spacing of the reinforcing
bars but also the total number and lengths of each size required for
the complete structure.
The standard detail is based on a width of roadway of 16'0" in the
clear. This width was adopted as the most usual one for this type of
bridge, and it was necessary to select a definite width in order to give
a complete bill of material of the reinforcement required for each
design. We have given the spacing of all bars so that the design may
be adapted to any desired width of bridge.
114

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ww
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SIDE ELEVATION
BARS-D-CORR ROUNDS 12'CTS.
wy
TRANSVERSE BARS-B
Ź CORR ROUNDS 12 CTS
---
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==
ܕ
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G S
at o
[
fi
BARS-E
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CLEAR SPAN´S.
P
Jung (genting volunt
LONGITUDINAL SECTION
pa Spa Maga
Hom
P
BARS-COCCUR ONLY WITH FILLS OVER 20:
£*CORR ROUNDS -12″CTS.
10
C
·MAIN REINFORCING BARS-A EVERY THIRD BAR BENT AS SHOWN
-
Sto
8-0
BARS-E-1"CR 9°CTS STAGGERED
BATTER IIN 12"
UK MAMMASIKIN
BARS-D-1ˆCR 12CTS
BARS-C
BARS-B-ŹˆCR 12″CTS. - 17-0” LONG.
BARS-A.
||
SECTION AT CENTER
K
VINY
DETAILS
IN
W
NOTE ALL REINFORCING BARS ARE HIGH
ELASTIC LIMIT CORR ROUNDS.
6₁
STANDARD FLAT SLAB
HIGHWAY BRIDGES.
PLATE 1.


115
STANDARD DESIGNS-FLAT SLAB BRIDGES.
Class No. 1
Loading.
d=depth of
All.
t=thickness
of con-
crete.
î-bearing
on abut-
ment.
d.
|
t.
2′ 8″ 12"
4' 8" 12/
6' 9" 12"
8' 10" 12"
2' 10"
4' 10"
12"
12"
6' 11"
12"
8′ 12″ | 12”
2' 11" 12"
4' 12" 12"
6' 13"
12"
8' 15" 12"
2′ 12″ 12"
4′14″ 12"
6' 15" 12"
8′17″ 12″
2' 14'' 12''
4' 15" 12"
6' 17'' 15"
8' 19" 15"
}
BARS A
Number
Number
f. Size. Spac. No. Length.
Size. Spac. No. | Length. Required. Required. No. Length. No. Length.
No. Length. ||No. Length.
CLEAR SPAN, 6′-0″.
2′ 16″ | 16″
4′19″ 15"
6' 21" 18"
23" 20"
8' 23"
15"
18"
2' 18"
4′ 21″ 15″
6' 23'
18"
8' 26" 20"
Main Reinforcement
Corr. Rounds
Every third bar bent up
as shown
5
@ 10 10 11
Co\C
Jand
2′ 15″ 12”
4' 17''
12"
1
6′ 19″ 15" 1
8′21″ 15″
1
1
"1
8
8
770
8
7/"
ངས་
<< < <
1
7/8"
%"
"/
%'
%"
1
1
"/
6
52"
572"
7
1
1%
11/8"
1666]
7
LIGHT HIGHWAY SPECIFICATION.
5½
5
78.
5½" 38
7/" 6½" 32
8"
52"
38
62" 32
1
6
5'
6%"
6
==
//
"/
42"
1%" 6%
6½"
6
5½"
4"
CO LO LO LO
6
5
5
188888888
35
38
62" 32
52"
6
38
30
こここ
​52
30
35
233882
35
5%" 38
42
2025585
35
38
32
35
20588080
5
5' 38
38
38
88230000
35
46
7'-6"
7'-6"
7'-6"
7'-6"
35
42
38
42
9'-6"
9'-6"
9'-6"
9'-6"
11'-6"
11'-6"
11'-6"
11'-6"
CLEAR SPAN, 8′-0″.
13'-6"
13'-6"
13'-6"
13'-6"
Trans-
verse
Reinforce-
ment.
"Corr.
Rounds.
17'-0" Long
BARS B.
15'-6"
15'-6"
16'-0"
16′-0″
9
9
9
9
CLEAR SPAN, 10′-0″.
17'-6"
17'-6"
18'-0"
18'-0"
42 20'-0"
20'-0"
20'-6"
20′-9″
11
11
11
11
13
13
13
13
CLEAR SPAN, 12'-0".
22'-0"
22′-0″
22'-6"
22′-9″
15
15
15
15
17
17
17
17
Trans-
verse Re-
inforce-
ment in
top of slab.
"Corr.
Rounds.
17'-0" Long
BARS C.
CLEAR SPAN, 14′-0″.
•
· 19
19
19
19
None.
9
9
9
2222
None.
11
11
11
CLEAR SPAN, 16'-0".
21
21
None.
13
23
23
24
24
13
13
CLEAR SPAN, 18'-0".
None.
15
15
15
None.、
17
17
17
None.
19
19
19
CLEAR SPAN, 20'-0".
None,
21
22
22
None.
23
24
24
Verticals in
Side Walls
"Corr.
Rounds.
Number re-
quired for
both walls.
BARS D.
18 2′-9″
18 4′-9″
6'-9"
8'-9"
00000000
18
18
2222
22226
20333333
8888888888888
12-tou
road roller.
38
34
3'.3"
34 5'-3"
34
34
2′-9″
4′-9″
7'-0"
9'-0"
3'-3"
38 5'-6"
3'-0"
5'-0"
7'-0"
9'-3"
coco00000
3'-0"
5'-3"
7′-3″
9'-6"
38 7′-6″
46
38 9′-9″
48
7'-6"
9'-6"
48
42 3'-3"
5'-6"
42
7'-9"
10' 0"
44
44
3'-6"
5′-9″
46
8'-0"
10'-3″
Horizontals
in Side
Walls.
Y" Corr.
Rounds.
Number re-
quired for
both walls.
BARS E.


10
14
20
26
10
14
20
26
10
14
20
26
24223
10
14
20
26
7'-8"
7'-6"
7'-6"
7′-6″

10
14
20
26
9'-6"
9'-6"
9'-6"
9'-6"
10
14
20 13'-6"
26
13'-6"
10
14
20
26
11'-6"
11'-6"
11'-6"
11'-6"
13'-6"
13'-6"
10 17'-6"
14 17'-6"
20 18'-0"
26 18'-0"
15'-6"
15'-6"
16'-0"
16'-0"
20'-0
20'-0"
20'-6"
20′-9″

22'-0"
22'-0"
22'-6"
22′-9″
116
d.
Class No. 2.
Loading.
d-depth of
t-thickness
of con-
crete.
f-bearing
ou abut-
ment.
STANDARD DESIGNS-FLAT SLAB BRIDGES.
20-ton roller or
40-ton car.
t.
f.
2' 9" 12"
4' 9" 12"
6' 10" 12"
8' 10" 12"
2' 11" 12"
4' 11" 12"
6'12" 12"
8' 13" 12"
2′ 12″ | 12″
4' 13" 12"
6′ 14″ 12″
8′ 15″ 12″
2' 16" 124
12"
4′ 17″
6'18"
8' 20"
2' 19"
4′ 21″
15"
15"
2′ 14″ | 12″
4' 15" 12"
6' 16"
12"
8′ 17″ 12" 1
15"
15"
6'22"
18"
8' 24" 20"
|| ε
15"
2' 21"
4' 22" 15"
6' 24"
8' 27"
18"
20"
5/8"
5/"
8
3711
83
14
Every third bar bent up
as shown.
Size. Spac. No. Length.
$\&\3\7\
Main Reinforcement
Corr. Kounds.
106
==
5576
3711
∞LALAA
= = = =
४
1
1
1
311
7/8"
2' 18" 12" 1
4′18″
12"
1
6' 20" 15"
1
8' 22" 15"
1/8"
7/8",
"/
"
"
12"
1%"
HEAVY HIGHWAY SPECIFICATION.
"
%85
72"
BARS A.
5%"
532"
52
7
à visto
6"
5"
6"
<< < <
5
62"
6
5½"
6"
5
6
"/ 42
42
28
==
5
"
5½"
512"
5
"/
6
5½"
"
"
"/
"
22235
1
5
1% 6
1
12"
12"
5
43½"
??
888888
42
32
35
38
18132159335
42
5" 38
4% 46
6
35
42
38
42
332333
35
38
42
35
2524
35
16
7'-6"
7'-6"
7'-6"
7'-6"
15'-6"
42
15'-6"
16′-0″
16′-0″
Number
Required.
CLEAR SPAN, 6′-0″.
9'-6"
9'-6"
9'-6/
9'-6"
Trans-
verse
Reinforce-l
ment.
"Corr.
Rounds.
17'-0" Long
BARS B.
11'-6"
11'-6"
11'-6"
11'-6"
CLEAR SPAN, 8'-0".
9
9
9
9
17'-6"
17′-6″
18'-0"
18'-0"
11
11
11
11
20'-6"
20′-9″
CLEAR SPAN, 10′-0″.
13
2232
13
13
13
GGGG
17
17
17
17
Trans-
verse Ke-
inforce-
ment in
top of alab.
"Corr.
Rounds.
17′-0″ Long
BARS C.
CLEAR SPAN, 12 0".
15
15
13'-6"
13'-6"
13'-6"
13'-6"
15
15
CLEAR SPAN, 14'-0".
Number
Required.
19
19
19
19
2222
None. 18 2′-9″
9
4'-9"
9
6′-9″
8′-9″
9
21
21
None.
11
11
11
CLEAR SPAN, 16′-0″.
None.
13
13
13
None.
15
15
15
None.
17
CLEAR SPAN, 18′-0″.
20′-0″
20'-0"
17
17
None.
19
19
19
None.
21
22
22
CLEAR SPAN, 20′-0″.
23
23
22'-0"
22' 0"
22′-6″
22′-9″
24
24
Verticals in
Side Walls.
1½" Corr.
Rounds.
Number re-
quired for
both walls.
Horizontals
in Side
Walls.
1½" Corr.
Rounds.
Number re-
quired for
both walls
BARS E.
BARS D.
No. Length. ||No. Length.


None.
23
24
24
00000000
18
18
18
2222
26
26
8888
30
30
9:0
30
****
8888888888888
34
34
3'-3"
5'-6"
7'-6"
34
9′-9″
34
42
42
44
44
46
38 3'-6"
00000000
3'-0"
5′-0″
7'-0"
9′-0″
146
3'-0"
5'-0"
7'-3"
9'-3"
48
7'3″
9'-6"
3'-3" 10
5'-3"
14
20
26
5'-6"
7′-9″
9′-9″
3'-6"
5'-9"
7'-9"
10'-0"
3'-9"
5'-9"
8′-0″
48 10'-3″
10
14
20
26
#
10
14
20
26
10
14
20
26
10
14
20
26
10
14
20
26
10
14
20
26
7'-6"
7'-6"
7'-6"
7'-6"

9'-6"
9'-6"
9'-6"
9'-6'

11'-6"
11'-6"
11'-6"
11'-6"
13'-6"
13'-6"
13'-6"
13'-6"
15'-6"
15'-6"
16′-0″
16′-0″

17'-6"
17'-6"
18'-0"
18'-0"
20'-0"
20'-0"
20,-6"
20'-9"
}
10 22'-0"
14 22'-0"
20
22'-6"
26 22′-9″
117
!
d.
STANDARD DESIGNS-FLAT SLAB BRIDGES.
Class No. 3.
Loading.
d-depth of
t-thickness
of con-
crete.
f-bearing
on abut-
ment.
t.
f.
2' 11" 12"
4' 11" 12"
6' 11" 12"
8' 11" 12"
1
2′12″
12″
4′ 12″ | 12″
6'12" 12"
8′ 13″ 12″
2' 14" 12"
4' 14" 12"
6' 14" 12"
8' 15" 12"
2′ 16″ 12″
4' 16" 12"
6' 16" 12″
8′ 17″ 12"
2' 19" 12"
4' 19" 12"
Every third bar bent up
as shown.
2' 24' 15"
4' 24" 15"
6'24' 18"
8′ 27″ | 20″
"
BARS A.
Size. Spac. No. Length.
coco
Main Reinforcement
Corr. Rounds.
3711
32" 6%
6
8
Immo
2′ 21″ 12" 1
4′ 21″ 12" 1
6' 21" 15" 1
8' 22" 15"
====
"/
"
1/2"
7/"
78
7"
18
****
\R\R\R
"
720
* * * *
1
6' 19" 15" 1"
8' 20" 15"
1"
"
"
"/
1%'
2' 22" 15" 1%
4' 22" 15" 11
6'22" 18"
8′24″ 20″
11/2"
11
"/
1/8
1%
1
"
62'
61/2"
LO LO LO CO
5
5
5
6/2"
is is is io
67/2,1
6
5½"
5"
5"
6"
CITY HIGHWAY SPECIFICATION.
Trans-
Trans-
verse Re-
verse
inforce-
Reinforce-
ment in
ment.
top of slab.
" Corr.
½" Corr.
Rounds.
Rounds
17'-0" Long
17′-0″' Long
BARS B. | BARS C.
Number Number
Required. Required.
5"
is is is co
6½" 32
5"
シュシュ​シュシュ
​6"
6"
1
6"
5″
2223
35
2222
42
42
42
5½" 38
52" 38
5%"
5
42
22538
32
22235
42
42
42
2223
42
42
42
35
35
35
35
42
5½" 38
5½" 38
5
4½" 46
I
I
CLEAR SPAN, 6'-0".
7'-6"
7'-6"
7'-6"
7'-6"
9'-6"
9'-6"
9'-6"
9'-6"
CLEAR SPAN, 8′-0″.
13'-6".
13'-6"
13'-6"
13'-6"
9
9
9
17'-6"
17'-6"
18'-0"
18'-0"
11
11
11
11
CLEAR SPAN, 10′-0″.
20'-0"
20′-0″
20'-6"
20′-9″
1
11'-6"
11'-6"
11'-6"
11'-6"
CLEAR SPAN, 12′-0″.
13
13
13
13.
22'-0"
22'-0"
42 22'-6"
22'-9"
555
15
15
15
15
17
17
17
17
CLEAR SPAN, 14′-0″.
15'-6"
15'-6"
16'-0"
16′-0″
19
19
"
19
19
None.
9
CLEAR SPAN, 16′-0″.
I'
7522
9
9
21
21
None.
11
11
11
23
23
24
24
None.
13
13
13
CLEAR SPAN, 18′-0″.
None.
15
15
15
None.
17
17
17
None.
19
19
19
CLEAR SPAN, 20′-0″.
None
21
22
22
None
23
24
24
Verticals in
Side Walls.
" Corr.
Rounds.
Number re-
quired for
both walls.
BARS D.
00000000
No. Length. No. Length.
|
18 3′-0″
18
5'-0"
7'-0"
9′-0″
18
18
2222
33333
26 3'-3"
26 5'-3"
26 1
7'-3"
26
9'-3"
30
30
30
30
34
34
34
888888888888
38
38
2214
20-ton roller of
60-ton car.
44
Lo 60 000 000
1 46
46
3'-0"
5'-0"
7'-0"
9'-0"
48
3'-3"
5'-3"
44 7′-9″
10'-0"
7'.3"
9'-6"
3'-6"
5'-6"
7'-6"
9′-9″
3′-9″
5′-9″
7'-9"
9′-9″
"
3′-9″
5'-9"
4'-0"
6'-0"
8'-0"
10'-3"
Horizontals
in Side
Walls,
½" Corr.
Rounds.
Number re-
quired for
both walls.
BARS E.
v
10
14
20
26
10
14
20
26
10
14
20
26
10
14
20
26
1
10
14
20
26
10
14
20
26
7'-6"
7'-6"
7'-6"
7'-6"
9'-6"
9'-6"
9'-6"
9'-6"
11'-6"
11'-6"
11'-6"
11'-6"
13'-6"-
13-'6"
13'-6"
13'-6"
15'-6"
15'-6"
16'-0"
16′-0″
17'-6"
17'-6"
18'-0"
18'-0"
10
14
20′-0″
20'-0"
20 20'-6"
26 20'-9"
22'-0"
22′-0″
10
14
20 22'-6"
26 22'-9"
118
BOX CULVERTS.
Reinforcing Steel: Mechanical Bond Bars; Elastic Limit,
50,000 lbs.
Reference Drawing: Plate 2.
The following tables, in conjunction with Plate 2, give detailed
designs of Box Culverts for spans of from four to twenty feet; with
fills up to twelve feet. A minimum depth of fill of twelve inches is
required on all culverts.
Owing to the varying conditions met with in practice no standard
lengths have been adopted. The tables give, however, the size and
spacing of the bars and also the number required, where this is possi-
ble. With the information given the complete bill of material of the
reinforcement required for any design is easily obtained.
4
The style of culvert to be used at a particular location, whether of
the box or open type, will depend upon the conditions. For a soft
ground, or one of uncertain character, the box type should be used,
but when a rock, or other substantial foundation, can be obtained,
with no danger of scour, the open culvert may be used.
OPEN CULVERTS.
The tables for Box Culverts may be used for the design of Open
Culverts of a construction similar to that shown by the accompanying
sketch.

UK
CLEAR SPAN-S
BAFFLE WALL
KINIKUNIKNUNINNIKUNUNIA
RIN
16757
119
TWIN CULVERTS.
Reference Drawing: Plate 3.
The tables for the design of Box Culverts, when used in conjunc-
tion with Plate 3, apply also to Twin Culverts. It should be noted
that, of the main reinforcing bars, "A," every third bar is bent up
and carried over the dividing wall into the adjacent slab.
NOTE-All box culverts should be provided with aprons at each
end, which extend down into the ground for a distance sufficient to
prevent scour. Wing walls should be designed as retaining walls and
may or may not be built monolithic with the box.
*
Open culverts should be provided with baffle walls at intervals of
about four feet; the baffle walls not only prevent scour, but also act
as struts between the side walls.
Johannes, F
120
Y
ť
UNHA KMSM
| SCENE KRAJE Rods 199jji p
BARS D
TEMENT
ROADWAY
BARS-E
MAIN REINFORCING BARS-A-
BARS-C PNE "CORR ROUNDS IN EACH SIDE WALL
FOR ALL VALUES OF h
BARS B
ਭਾਦ
CORNER DIAGONALS BARS E É CORR ROUNDS
SAME SPACING AS BARS F LENGTH=/41/2t+1-6)
BARS-A
EVERY THIRD BAR BENT AS SHOWN
BARS F
CLEAR SPAN-S.
La
£
1
HALF PLAN
SHOWING OUTSIDE CORNER BARS, CORNER
DIAGONALS AND OUTSIDE LONGITUDINALS
CROSS SECTION.
katy jako hindi paran
LONGITUDINAL BARS B ₤ CORR
ROUNDS - FB´CTS STAGGERED
→ Joma in
·OUTSIDE CORNER BARS-F
-
P
SIDE WALL REINFORCEMENT
BARS D
my jea
BARS
·BARS-
pema kwa
ale, porta
#
*******
jok
Jan
HALF PLAN
SHOWING MAIN SLAB REINFORCEMENT AND
INNER LONGITUDINALS
DIAMETERS
CORNER BARS,
SHOWING LAP OF OUTSIDE
NOTE
BARS
BARS!!!!
pak za
|pdates, para
-
-E
PEARL-0
+6+
#1
--.
pada p
·BARS C
Sa pamama
pada pe
samang agam
VERTICAL LONGITUDINAL SECTION A A.
OUTSIDE CORNER BARS NOT SHOWN IN ELEVATION.
CULVERT IS SYNNETRICAL ABOUT BOTH CENTER LINES.
TOP AND BOTTOM SLABS ARE SIMILARLY REINFORCED.
ALL BARS ARE HIGH ELASTIC LIMET CORRUGATED ROUNDS.
DETAILS
BOX CULVERTS.
STANDARD BOX CULVERTS.
PLATE 2.
Į



121
Class No. 1
Loading.
d=depth of
All.
h-height of
culvert.
t-thick
ness of
concrete.
d.
2/
4'
6'
8'
10'
12/
4'
6'
8'
10'
12'
10′
12'
2'
6'
10'
h.
2'
294667
JOGACON
4'
5'
6'
4'
66L6OA
bòìbbà
t.
8'
6"
6"
6"
ייף
C00127
7"
8"
9"
9"
10"
11"
12"
9' 13"
8"
9"
10"
11"
10"
6' 11"
12"
711
5/"
7" 57"
58,
"
8.
13"
14
12′ | 10′ | 16"
1
STANDARD DESIGNS-BOX CULVERTS.
11"||
12"
14"
8' 15"
Main.
Reinforcement,
Every third bar
bent up as shown.
Corr. Rounds.
BARS A.
!!!
"
"
8
3/11
8
Size. Spac Lgth. No.
58
"
8
87"
8
9711
10'
17" 1
12′ · 10′ 18″|| 1
1
Lall
11
1811
3/4
97"
7
881∞∞∞
72'
CO CO
6
5½"
5
62"
5%"
87" 15
6½"
6
LIGHT HIGHWAY SPECIFICATION.
ミミミ
​52'
578
765
4'-9" 20
4′-9″ 20
5½" 5'-0" 20
= = =
"/
6'-9"
6'-9"
7'-0"
7'-0"
7'-2"
7'-4"
"/
9'-0"
9'-0"
9'-2"
9'-4"
Longitudi-
nal Bars.
"Corr.
Rounds.
Number re-
quired for
both Slabs.
4'-9" 20
4′-9″ 20
4'-9"
9'-6"
9'-8"
BARS BARS
B.
C.
6 11'-8"
5½" 11-10"
52" 12′-2"
52" 13'-4"
5 '" 13'-6"
6
" 13'-10"
52
14'-0"
14'-4"
62
6
14'-6"
:|:
222222
20
CLEAR SPAN, 4′-0″.
222222
a co co co co co
30
30
30
30
30
30
11'-2'
32
11'-4" 32
11'-6"
32
32
32
32
1
1
CLEAR SPAN, 6'-0".
088888888888888
10
10
10
888888
38
10
10
10
10
10
10
10
10
10
CLEAR SPAN, 8'-0".
No. Size. Spac. Lgth. Spac. Lgth Size. Spac. Lgth
222222
10
10
10
10
10
10
10
10
10
10
Verticals
in
10
10
Side Walls.
Corr. Rounds.
888888
A
10
10
10
10
10
10
BARS D.
1½"
17/1
72
22″
"}
"
"
EV100
"
"
a ja on or av or
1/11
ገዛ
Vi
1%
52,
5211
8
16" 2'-8"
3'-8"
16"
16″
15"
4′- 8″
5'-10".
12" 6'-10"
11" 8'-0"
CLEAR SPAN, 10′-0″.
"
12″ | 2′-10″
12" 3'-10″
11" 5'- 0"
10" 6' 2"
14″ 7'- 4″
12″ 8'-6"
====
11
10
5'- 2"
6'2"
13" 7'- 4"
"8'-6"
5
9'- 8"
62" 10'-8"
"10′-
"
11
CLEAR SPAN, 12′-0″.
8"||
"/
"/
14" 6' 4"!
12″ 7'-6"
10" 8'-8"
13" 9'-8"
11" 10-10"
11" 12- 2"
Corner
Diagonals.
2" Corru
gated
Rounds.
11" 6' 6'
10″ 7′ 8″
12″ | 8′-10";'
11" 10'- 0"
13" 11'- 4",
12" 12′ 6″
BARS E.
16" 1'-2"
16" 1'-2″
16″ 1-2″
15" 1'-5"
12" 1'-5"
11" 18"
12"
12"
11"
10"
14″
12"
I
1'- 8"
1'-8"
1-10"
11"
2'- 4"
2'- 4"
10"
13"
11"
2- 7"
2'-10"
10" 3'- 1″
13″ 3′- 4″
1
2-10"
3'- 0"
3'- 3"
14"
12"
10"
13" 3'-6"
11" 3' 9"
4'- 3"
11" 4'-
•
12-ton
Road Roller.
Outside
Corner
Bars.
Corr. Rounds.
11″ 3″ 3″
10″ 3′ 6″
12″ 4'- 0"
11" 43"
13″
13"
12'
4′-9″
5'- 0"
2'- 1"
2′ 4″|
27" $20
4
16"
5/8"
5/8 16"
5911
16"
(8,11
5/811
15"
12"
11"
7/8
/8
BARS F.

5/8"
37"
B
Bu
8
620
8
4
23
18
4%
It
18,
Mar 10
"/
12"
12"
11"
10"
14"
12"
4′-0″
4'-6″
5'-0"
5'-6"
6'-0"
6'-6"
10"
13"
4'.9"
5'-3"
6'-0"
6'-6"
7/3"
7′-9″
11" 7'-0"
10"
13"
7'-6"
8'-3"
11"
8"-9"
9'-3″
9'-9"

14"
12"
8'-3"
9'-0"
10" 10'-0"
13" 10'-6"
11" 11'-0"
11" 11'-9"

11"
9'-6"
10" 10'-0"
12" 10'-9"
11" 11'-6"
13″ 12′-0″
12″ 12′-9″
122
Class No. 1
Loading.
d=depth of
h-height of
culvert.
t-thick
ness of
concrete.
d. | b. 1
h.
NOORAN
NOORAN
8'
STANDARD DESIGNS-BOX CULVERTS.
5' 12"
6 14"
7' 16"
8' 17" 1
9' 19" 1
12′ | 10′ | 21″|| 1
NOORAN
Main
Reinforcement,
Every third bar
bent up as shown.
Corr. Rounds.
3/4" 5
6
2/1
t. Size. Spac. Lgth. No.
| ||
6' 14" 1/8"
16"
"
8' 17" 1
9′ 19″
9'
10' 10' 21"
12' 11' 23" 1"
LIGHT HIGHWAY SPECIFICATION.
BARS A.
"
15" "
17" 1
19" 1
21" 1
10' 10' 23" 18"
12' 11' 25" 18"
9'
"
6
52"
6
"
5"
4/2
1
"""
6
6′ 16″*
16 %
7' 19' 1
8' 21" 1 "
8' 9' 23' 1'
11"
10' 10' 25" 112"
1%
12' 11' 28" 14"
//
"/
5½"
6
5½
5
Longitudi-
nal Bars.
" Corr-
Rounds.
Number re-
quired for
both Slabs.
BARS BARS
B. C.
No.
15'-6"
-6″
44
15′-10″ 44
16′-2″ 44
16'-4" 44
16'-8" 44 10
17'-0" 44 10
CLEAR SPAN, 14′-0″.
999228
10
10
10
10
17'-10" 46
18′-2″ 46
18'-4" 46
18'-8" 46
19'-0" 46 10
19'-4"
46
10
888888
60
888888
10
10
10
10
52" 20'-0" 54 10
6 20'-4″ 54 10
52" 20'-8" 54 10
42" 21'-0" 54
5½" 21'-4″ 54
21'-8"
10
10
54
10
CLEAR SPAN, 16'-0."
5 22''2"
52" 22′-8" 60
42" 23′-0″
5½ 23'-4″
5
60
60
23'-8" 60
10
52" 24′-2″ 60 10
10
10
Verticals
in
Side Walls.
Corr. Rounds.
888888
10
10
BARS D.
CLEAR SPAN, 18′-0″.
1
1½"
1/2"
5711
8
Size. Spac. Lgth.
Size. Spac. Lgth. Spac. Lgth. Size. Spac. Lgth.
"
CLEAR SPAN, 20˚-0″.
1/2"
1/2"
ミニ
​1/"
10" 6'-8"
12" 7'-10"
1/2"
1/2'
"
11" 9'- 2″
12" 10'-4″
11" 11'- 8"
10" 13'- 0",
5/8
13″ | 7′-10″
11″9'- 2″
12" 10′- 4″
11" 11- 8"
9" 13'- 0"
12" 14'- 4"
11" 8'-0"
12" 9'- 4″
11" 10'- 8"
9" 12'- 0"
11" 13'- 4″
10" 14'- 8",
10" 8'- 2″
11" 9'- 8"
9' 11'- 0"
17/1 11" 12'- 4"
7)
Corner
Diagonals.
"Corru-
gated
Rounds.
BARS E
10" 13'- 8"
11" 15'- 2"
10" 3'-
3'- 9'
12"
9"
+/- 3/7/11
11"
12"
11"
10″
13" 4'-5"
11″ 4'-11"
12″ | 5′- 2″
11" 5' 8"
9″ 6′- 2″
12" 6'- 8"
11"
12″
11
4′-9″
5'- 0
5'-6"
6'- 0"
9"
11″
10"
12-ton
Road Roller.
Outside
Corner
Bars.
Corr. Rounds.
-
10" 10'-6"
7/2 12" 11'-0"
11" 11'-9"
12″ 12′-6″
1
Jauk þenk Jak
1
1
1
1
1%"
BARS F.

=
5'
7/8"
5'-5" 1
"1
6' 6 1
7'- 0"|| 1½"
'7'-6" 11/8
//
1
1
11" 13'-3"
10" 14'-0"

13"
11"
12"
11"
9″ 15′-0″
12"
15'-9″
12'-0"
12'-6"
13'-3″
14'-3"

11" 13'-5"
12" 14'-3"
15′-0″
15'-6"
11" 16′-0″
10" 16′-9″
11"
9"
10"
11"
5'- 6′
6'-1'
9" 6' 7"
11" 17'- 1"
9"
1" 11"
10" - 6" 11/2"
10"
11" 8' 2'' 14'
11"
" 10" 14'-0"
11"
14′-9″
15'-9"
16'-6"
17'-3"
18'-3″
123
Class No. 2
Loading.
d=depth of
fill.
h-height of
t-thick-
ness of
concrete.
d.
270002
10'
12'
NOXRAN
2'
10'
12'
NOORAN
6′
NORGAN
Main
Reinforcement.
Every third bar
culvert. bent up as shown.
Corr. Rounds.
h.
2'
2QCTICN
ཆེ བེཤྩ s io če t-
i
t.
7"
"y"!
7"
8"
9"
9"||
9"
9"
9"
10"
7' 11"
11
Size. Spac. Lgth. No.
7' 12"
8' 12"
9′ 13″
5' 11'
6′ 12″
13"
14″
8′
9'
10'
15"
12' 10' 16'
STANDARD DESIGNS-BOX CULVERTS.
5'13"
6' 13"
4' 10"
5' 10" 3
B
11"
8' 16"
5/11
6911
9'
12′ 10′ 18″
8
SPIL
"
"
68
-
33,
17"
17" 1
5/8" 5/2 7'-0"
522" 7'-0"
5/2" 7'-0"
"/
7'-0"
7′-2″
7'-4"
$3
BARS A.
AI
HEAVY HIGHWAY SPECIFICATION.
I
6
6
8
4'-9"
4'-9"
4'-9"
4′-9″
5'-0"
5½" 5′-0″
6
6
5
6
"52"
"
%"
72"
2816
7711
15"
52" 11-4"
52" 11'-6"
7
11 -8"
Longitudi-
nal Bars.
½" Corr.
Rounds.
Number re-
quired for
both slabs.
BARS BARS
B. C.
6" 13'-8″
"13'-8"
"14'-0"
6
5" 14'-2"
6
14'4″
52" 14'-6"
222222
222222
CLEAR SPAN, 4′-0″.
6" 11-10"
32
51/2" 12'-0"
222222
7"
8
グン​!! 52" 12′-2″ 32
32
32
32
30
62" 9'-2"
61/2" 9'-2″
30
9'-4"
30
9'-6"
30 10
9'-6" 30 10
9'-8"
30
10
32
0808080808080
888888
38
10
10
CLEAR SPAN, 6′-0″.
38
10.
38
10
10
10
38
10
10
10
888888
No. Size. Spac. Lgth. Spac. Lgth. Size. Spac. Lgth.
10
10
10
CLEAR SPAN, 8′-0″.
888888
10
10
10
888888
Verticals
in
10
10
10
Side Walls.
Corr. Rounds.
CLEAR SPAN, 10′-0″.
388888
1/2"
1½"
10
10
1½"
5/11
57"
10
10
10
10
16"
177
Leu
Y"
10
10 5/8"
"/
BARS D.
€21
·67"
12" 2'-10"
12″ | 3'-10"
12" 4"-10"
5′-10″
CLEAR SPAN, 12′-0″.
12″
12"
11"
7/2/11
12,
Kul
"
7'- 0"
8'- 2"
13"
13"
12"
12"
8'-8"
10"
9'-8"
12" 10'-8"
Corner
Diagonals.
"Corru-
gated
Rounds.
BARS E.
11" 3'-2" 11" 2'-1'
11" 4'-2" 11" 2'-1"
11" 5'-2" 11" 2'1'
10" 6'-2″ 10" 2'-1"
12" 7'-4' 12" 2'-4"
11" 8'-6" 11" 2'-7″
12"
12"
12"
12"
5'-4"
6'-4"
13"
13″
7'-6" 12"
20-ton roller or
40-ton car.

12"
11"
12"
10"
12"
1'-5"
1'- 5"
1'-5"
1'-5"
1'- 8″
1-11"
<
2' 7'
2′- 7″|
2'-10"
3'- 1"
3'- 1"
3'- 4″
11"
6'-6" 11" 3'-0"
11" 7'-8" 11" 3'-3"
14" 8'-8" 14" 3'-6"
13" 9-10" 13" 3′-9″
11" 11'-0" 11" 4'-0"
11" 12'-2" 11" 4'-3"
13"
6'-8" 13" 3'-9"
12" 7'-8" 12" 3'-9"
Outside
Corner
Bars.
Corr. Rounds.
12" 9'-0" 12"
4'-3"
11" 10'-2" 11" 4'-6"
12" 11'-4" 12" 4'-9"
11" 12'-6" 11" 5'-0"
"/
5/0
8
"1
5211
5
8
5% 12"
12"
11"
1
"/
4
$35.
BARS F.


8
12"
12"
12"
""
11"
11"
11"
10"
12"
11"
13"
13"
12"
12"
10"
12"
//"!! 13"
72"
18,
4201
18
%
4'-0"
4'-6"
5'-0"
5'-6"
1
6'-0"
6'-6'
4'-9"
5'-3"
6'-0"
6'-6"
7'-3"
7′-9″

7'-0"
7'-6"
8'-3"
8'-9"
9'-3"
9'-9"


11"
11"
14"
13"
11"
11'-0"
11" 11'-9"
8'-3"
9'-0"
10'-0"
10'-6"

9'-6"
12" 10'-0"
12" 10'-9"
11"
11'-6"
12" 12'-0"
11"
12′-9″
124
Class No. 2
Loading.
d-depth of
All.
h-height of
t=thick
d.
NOBRAN
6'
ness of
concrete.
8'
NORDAN
246002
Longitudi-
Main
nal Bars.
Reinforcement." Corr.
Rounds.
Every third bar
culvert. bent up as shown. Number re-
Corr. Rounds. quired for
both slabs.
10'
12′ | 10′ | 21″
4'
6'
·
h. t.
8'
5' 15"
6' 15"
7′ 16″
8' 18"
9'
19"
2'
4'
6'
8'
STANDARD DESIGNS-BOX CULVERTS.
6' 16"
7' 16"
8' 18" 1
8' 9' 20" 1
10' 10' 22"
12′ 11′ 23″
BARS BAKS
B. C.
Size. Spac. Lgth. No. No.
1
Jak jedi jend
2' 6' 17" 1
1
7' 18" 1
8' 20"
20" 1
9'
"
8'
==
1"
1″
"
8
6' 19" 1
7' 20" 1
***NA
"
"
22"
22" 1%
BARS A.
HEAVY HIGHWAY SPECIFICATION.
"}
12′ | 11′ | 26″|| 1%" | 5
8' 22"
22"|| 1%"
9' 24" 1"
10' 10' 26" 18"
12' 11' 28" 14"
44
//
44
52" 16'-0"
52" 16'-0" 44
16′-2″
5
5½" 16'-6"
52" 16'-8"
17'-0"
44
44
//
5
44
5
5
10' 10' 24"|| 1%½" | 5½″ | 21′-6″
"21'-10"
18'-2″ 46
18'-2"
46
5½" 18'-6" 46
5 "18"-10" 46
6 19'-2″
5½"″ 19′-4″
//
"1
"/
6 "20'-4" 54
5" 20'-6" 54
5 "20'-10"
54
21'-2″ 54
6
54
54
"}
CLEAR SPAN, 14'-0".
".
46
46
10
10
52" 22'-8"
|
6
60
5 ""22'-10" 60
23'-2" 60
52" 23'-6" 60
5 23'-10" 60
52" 24'-2" 60
888888
10
10
10
10
Verticals
in
Side Walls.
Corr. Rounds.
CLEAR SPAN, 16′-0″.
Size. Spac. Lgth.
Spac Lgth.
10
10
10
10
1/2"
"
72
1/11
17"
2
B211
10 1/2"
10
10
10
10
10
10
10
5/8
CLEAR SPAN, 18′-0″.
BARS D.
//
72
23,1
62,.
77
11" 7'-0"
8'-0"
11"
10" 9'-2"
11" 10'-6"
11" 11'-8"
10" 13′-0″
10" 8'-2″
10" 9'-2"
-11" 10'-6"
10" | 11'-10"
12″ | 13′-2″
11" 14'-4"
12" 8'-4"
11″ 9'-6"
10" 10'-10"
12" 12'-2"
CLEAR SPAN, 20′-0″.
Corner
Diagonals.
"Corru-
gated
Rounds.
BARS E.
11" 4'-6"
11" 4'-6"
10" 4′-9″
11" 5'-3"
11" 5'-6"
10"
6'-0"
Size. Spac. Size. Spac. Lg th
Size. Spac
6'- 5″
10 //½" 11" 8'-8" 11"
10 1/2"
10" 9'-10' 10"
10 Y!!
12'' 11'-2″
12"
10 Y"
11"' 12'-6"
11"
10 5/8"
10" 13'-10" 10"
11″ | 15′-2″ 11"
"
10
5/8"
20-ton roller or
40-ton car.

8" ||
6′ 8″
-
- pat gand
Outside
Corner
Bars.
Corr. Rounds.
8 11"
11"
8
"
7%
10"
11"
11"
10"
ייל
1
1
=
10" 4-11" 7" 10"
10" 4'-11"
11" 5'-.5" 1
10"
5'-11" 1
12'-0"
12'-6"
10"
11"
13'-3″
12"
10" 14'-3"
12" 15'-0"
11"
11″
15'-9"
BARS F.

1″
1%"
"
12″ 5'-5" .1
11" 5'-7″ 1
10" 6'-1″ 1
12" 6'-6"
11" 7'-0" 13" 11"
12" 13'-6"
11"
14'-3"
15'-0"
15'-6"
11" 13'-6"
16′-0″
10" 14'-10" 10" 7'-6" 18" 10" 16′-9″
6' 1'
1
6'- 4'' 1
6'-10" 1%
7'- 3" 1"
7- 9" 1%"
8-2 1"
10'-6"
11'-0"
11'-9"
12'-6"
13'-3"
14′-0″

1
"
"1
10"
12"
11" 14'-0"
10" 14′-9″
12" 15'-9"
11" 16'-6"
10″ 17'-3″
11" 18'-3"
125
d=depth of
All.
hheight.
of cul-
vert.
t-thick
ness of
concrete.
1.
2'
4'
6'
10'
12
NOORAN
10'
12'
8' 5'
2'
8'
10'
12'
Class No. 3
Loading.
FOODAN
NÀY CÒN
2' 8"
∞ ∞ ∞ ∞ o
6'
h. t. Size. Spac. Lgth. No.
8"
8"
8"
8"
9"
2' 10"
3' 10"
4' 10"
5' 10"
6' 10"
7' 11"
11
5' 11"
6' 11"
12″
8' 13"
9′ 14″
2'
4'
6'
7' 13"
8'
14"
9'15"
10'
12' 10′ 16″
5′ 12″
6'12"
STANDARD DESIGNS-BOX CULVERTS.
5'15"
6' 15"
7' 15"
8' 16"
10' 9′ 17″
12' 10' 18"
Main
Reinforcement.
Every third bar
bent up as shown.
Corr. Rounds.
5
'8
8
"/
8
H
1
BARS A.
8
33,
"/
3311
7/8
7/7/7
4311
7/"
78
770
Za
====
\&\&
8
CITY HIGHWAY SPECIFICATION.
!
6
6
******
//
//
"
6
52"
5
5
72"
71%'
7
6
5½"
Longitudi-
nal Bars
½" Corr.
Rounds.
Number re-
quired for
both slabs.
52" 9'- 4"
52" 9' 4"
52" 19′ 4″
51/2" 9'-6"
7
9'- 8"
6" 9'-10"
BAKS BARS
B. C.
No.
5'-0"
5'-0"
5'-0"
5'-0"
5'-0"
20
5'-0" 20
7'-2" 22
7'-2″ 22
7'-2'
72"
22
7'-2" 22
7'-2"
22
7'-4" 22
"/
11-6
" 11'- 6'
"11'-8"
11'-10"
62'
5½" 12'-0"
12'-2"
5
6
14'-4"
5%" 14'-6"
222222
CLEAR SPAN, 4′-0″.
30
30
30
30
30
30
32
32
32
32
32
32
5/4" 14'-0" 38
5" 14'-0"
52" 14'-0"
38
38
5" 14'-2"
"/
888888
10
8888888888888888
10
38
10
38
38
10
10
10
888888
CLEAR SPAN, 6′-0″.
10
10
10
10
10
10
10
10
10
10
10
Verticals
in
Side Walls
Corr Rounds.
10
10
10
10
10
10
BARS D.
CLEAR SPAN, 8′-0″.
10
10
1/2 12" 3'-0"
12" 4'-0"
1/2
1/2 12"
11"
10"
Size. Spac. Lgth. Spac. Lgth. Size. Spac. Lgth.
Size. Spac |Lgth ||Spac. 1
Spac Leth
"/
"/
1½"
1/2
1/2
CLEAR SPAN, 10′-0″.
1/2"
1/2"
13,,
12"
"/
5711
8
1/2
13,
5%
(8,1
"
CLEAR SPAN, 12′-0″.
B
12" 5'-0"
6'-0"
7'-0"
8'-2"
Tok at et at
1
13,,
8
11" 5'-6"
11" 6'-6"
11" 7'-6"
11″ 8'-8"
14" 9′-8″
13" 10'-10"
15" 3'-4" 15"
15" 4'-4' 15"
15"
5'-4" 15"
6'-4"
14"
14"
12"
7' 4"
8'-6"
11"
12"
11"
10
1/2"
10
2,
10
10
Corner
Diagonals.
½" Corr.
Rounds.
BARS E.
1/2" 11" 7'-0" 11"
11"
11"
11"
11" 10'-2" 11"
8'-0"
11" 9'-0"
12" 11'-4"
11" 12'-6'
12"
11"
20-ton roller or
60-ton car.
12" 1'-8"
}
12" 1'- 8″
1'- 8"
12"
12″
1'- 8″
.1'- 8″
1'-11"
11″
10"
6'-8" 10"
10"
10" 7'-8" 10"
14"
14" 8'-8"
9'-10"
13"
13"
11" 11'-0"
11"
10" 12′- 2"
10"
11" 2'-10"
11" 2'-10"
11" 2'-10"
3'- 1"
11"
14"
13"
3'- 4″
3'- 7″
2'-4"
2'-4"
2'-4"
2'-4″
2'-4"
2'-7"
Outside
Corner
Bars.
Corr. Rounds.
I
1 3
3'3"
3'-3"
3'-6"
3'-9"
4'-0"
4'-3"
8
5%"
8
5/11
8
8
57"
"
8
BARS F.
4
4'-3'
4'-3"
4'-3"
4'-6"
1
4′-9″
5'-0" 1
3/4"
====
7
7/8"
770
%"
7
"/
8.
"1
18
12"
12"
12"
12"
11"
10"
15"
15″
15"
14″
12"
11"
11"
11"
14"
13"
10"
10"
4'-0"
4'-6"
5'-0"
11"
7'-0"
11" 7'-6"
8'-3"
8′-9″
9'-3"
9′-9″
*
5'-6"
6'-0"
6'-6"
4′-9″
5'-3"
6'-0"
6'-6"
7'-3"
7′-9″•
8'-3"
8′-9″
14"
10'-0"
13" 10'-6"
11" 11'-0"
10" | 11′-9″
11" 9'-9"
11" 10'-3″
11" 10'-9″
11'-6"
11"
12" | 12′-0″
11" | 12′-9″
I

126
-depth of
All.
h-height
of cul
vert.
thick.|
ness of
concrete.
t
5' 17"
6′ 17″
6'
7' 17"
8'
8' 18"
9'19"
10'
12′ | 10′ 21″
21802
Class No. 3
Loading.
4'
6'
BARS BARS
B. C.
d.b. t. Size. Spac. Lgth.|| No. No.
A
6' 19"
7' 19"
8′ 19″ 1
9' 20"
1
10' 10′ 22″ 1
12' 11′ 23″
STANDARD DESIGNS-BOX CULVERTS.
8'
10' 10' 24"
12' 11' 26"
སོཝཡྻཱཨོསུ
2' 6' 20" 1
7′ 20″ 1
8' 20" 1
1
Main
Reinforcement.
Every third bar
bent up as shown.
Corr Rounds.
1
10′ | 10′ 26″
11′ 28″
12
gol gol gad pat
シシミ
​1"
1'
1"
17/8"
9′ 22″ 1%"
1%"
1%"
1
-
18
BARS A.
11
78
1%"
1
78
14"
CITY HIGHWAY SPECIFICATION.
6
∞ ∞ COŁO LO LO
"16'-4"
44
16′-4″ 14
16'-4" 44
5½" 16'-6" 44
44
44
6
Longitudi-
nal Bars
½" Corr.
Rounds.
Number re-
quired for
both slabs.
5½" 16'-8″
"
5 17′-0″
CLEAR SPAN, 14′-0″.
கககககக
888888
"20'-10" 54
"20'-10" 54
"' 20'-10" 54
"21'-2"
54
5½″ 21'- 6″
5
"21'-10"
54
54
10
10
10
52" 18'-8" 46
5" 18'-8" 46
5½" 18'-8" 46 10
"18"-10"
46
"19'-2"
46
5%" 19'-4" 46
5
10
10
10
10
6′ 22″ 1%"
4' 7' 22" 11/11
1% 6
8' 122"
9' 24"
6 "' 23'-2'' 60 10
"23'-2'
"23'-2"
6
60
5½" 23'-6"
523'-10'')
52" 24'-2″ 60
60
60
10
10
888888
60
888888
CLEAR SPAN, 16′-0″.
10
10
10
10
10
10
{
888888
Verticals
in
Side Walls.
Corr. Rounds
10
10
10
10
10
BARS D
20 as it is not to t
CLEAR SPAN, 18′-0″
1/2"
1/2'
Size. Spac Lgth Spac Lgth Size. Spac. Lgth
1½" 11" 8'-8"
1½" 11" 9'-8"
1½"
11" 10'-8"
"1
8
B"
CLEAR SPAN, 20'-0".
12,,
1211
72
12" 7'-4″
12"
8'-4"
9'-4"
12"
11" 10'-6"
11" 11'-8"
10" 13′-0″
1/11
Y"
1/2"
5711
8
BIN
8
Corner
Diagonals.
½" Corr.
Rounds
BARS E
10" 11'-10" 10"
12″ 13'-2″ 12"
11" 14'-4″
11"
12″ 9′-2′′
12" 10'-2"
12″ 11'-2″
11" 12'-6"
10' 13'-10"
11" 15'-2"
20-ton roller or
60-ton car.

12" 5'-0"
12" 5'-0"
12" 5'-0"
11"
11"
10"
10" 8'-10" 10"
10" 9'-10" 10"
10" 10'-10''| 10"
12″ 12′-2″ 12"
11" 13'-6" 11"
10" 14'-10" 10"
11"
5'- 8"
11" 5' 8"
11" 5'- 8"
1
5'-3" 1
5'-6"
5'
5′-9″
1
Josh park but
Outside
Corner
Bars.
Corr. Rounds.
1
1
1"
1"
"
"
5' 11" 1
-
6' 5" 1%
6′- 8" 13 8
"
6'
6'
6'.
1
6'-6'
6'-6" 1
7'-0" 13
7'-6" 11"
BARS F
"
12" 6'-10" 1"
12″ | 6′-10"|| 1%"
12" 6'-10", 1"
11″ | 7′- 3''|| 1%
10 7'- 9" 1"
11" 8- 2". 134″
//
[
12"
10'-y"
11'-3″
12″
12″ | 12′-0″
11"
12'-6"
13'-3″
14'-0"
11"
10"
11"
11"
11"
12'-6"
13'-0"
13'-6"
10" | 14'-3″
12″ 15′-0″
11"
15'-9"
14'-0"
10''
10′′ ¦ 14′-6''
10" 15'-0"
12" 15'-6"
11'' 16'-0"
10" 16′-9″
12" 14'-6'
12″ | 15′-3″
12″ | 15′-9″
11" 16'-6"
10″ | 17′-3″
11" 18'-3"
127

ji
ANNIKSANAISHISHKINUKI
T
NOTE.
BARS-F
BARS-D
DAVIS HRSZAWAN
BARS C
BAAS-E
Jay
BARS A
BARS-A
CLEAR SPAN-S
-BARS B
что
COLAR SKA
1/2 + 1/1/
PART CROSS SECTION AND ELEVATION TWIN BOX CULVERT
K
REINFORCEMENT, SAME AS FOR CORRESPONDING STANDARD BOX CULVERTS, EXCEPT-
ING BARS A
BARS-A: BEND UP EVERY THIRD BAR AT THE Ê POINTS AS INDICATED, EXTENDING
THEM / UP OVER MIDDLE SUPPORT) INTO ADJOINING SLAB AS SHOWN.
NOTE
HOLY
CULVERT IS SYMMETRICAL ABOUT THE CENTER LINE.
TOP AND BOTTOM SLABS ARE SIMILARLY REINFORCED,
ALL BARS ARE HIGH ELASTIC LIMIT CORRUGATED ROUNDS
DETAILS
STANDARD TWIN CULVERTS.
TASHISI
PLATE 3.
128
}
1
GIRDER BRIDGES.
Reference Drawings: Plates 4, 5 and 6. (See also Detail Sheets for
Girders G1, G2 and G3.)
Reinforcing Steel: Mechanical Bond Bars;
Elastic Limit, 50,000 lbs.
The following tables, in conjunction with Plates 4, 5 and 6, and the
three sheets of details, showing slab and girder construction, give the
complete design of Girder Bridges for spans of twenty, twenty-five,
thirty, thirty-five and forty feet.
The standard bridges have been figured for the three classes of
loadings, but with only one depth of fill-eighteen inches. A mini-
mum depth of fill of twelve inches is required on all girder bridges.
The slab has in all cases been made five inches thick.
The two girders under the car tracks have been figured to carry
the full car load. Girders GI in Class I and Class 2 Bridges, and
Girders G2 in Class 3 Bridges have been designed for that proportion
of the roller load which may come upon them. For the sake of
uniformity Girders G2 in Class 3 Bridges have been made the same
depth as Girders G3.
The standard designs for Class 3 Bridges are based on the sections
shown on Plate 6, page 134. The tables, however, apply just as well
to the
Alternate Section," which may be preferred by some
engineers.
("
120
18
b
a
GI
·
5-8
INFIFIYAKATAIRIEIETI
x
16-0
G2
5-8
"
"
G2
CLEAR SPAN - S
SIDE ELEVATION
SIP
SECTION AT CENTER.
5-8
ENE
GI
461
NOTE
SLAB - 5 THICK - Ź CORR ROUNDS,
6"CTS IN BOTTOM OF SLAB
TOP BARS -Ź CORR ROUNDS- 3-0 LONG
12"CTS OVER G2 TOP BARS -
TOP BARS - CORR.
ROUNDS 20ʻLONG
12´CTS AT GI
2- ½ CORR ROUNDS LENGTHWISE OF BRIDGE IN EACH
PANEL
STANDARD DETAILS,
CLASS I BRIDGES
PLATE 4
•


130
Clear Span h.
20' 0" 32"
25'-0"
30'-0"
35'-0"
20′-0″
25′-0″
38"
10'-0" 53"
30'-0"
14"
50″
25"
29"
34"
35'-0" 39"
40'-0" 47″
|
b.
12"
12"
12"
14"
15"
12″
14"
14"
14"
14"
{
CLASS 1-BRIDGES.
GIRDERS G1.
See Detail Sheet, Page 136.
f.
15" 6-4" Corr. Rounds.
15" 6-7% Corr. Rounds.
3-%"
3-1"
Corr. Rounds.
Corr. Rounds.
15"
18" 6-1"
21" 8-1"
Reinforcement.
15'
GIRDERS G2.
See Detail Sheet, Page 137.
Corr Rounds.
15" 6-1 Corr. Rounds.
8-1"
Corr. Rounds.)
18"
Corr. Rounds.
15" 8-1" Corr. Rounds.
4-1%" Corr.
4-14" Corr.
Rounds.
Rounds.
21" 8-14" Corr. Rounds.
Bent Bars.
Point
6 Bars
Bend up 1 Bar at the Point
Bend up 2 Bars at the
In Beams with
Point
Point
6 Bars
2 Bars at the
Berd up 1 Bar at the
In Beams with
Bend up
Point
In Beams with 8 Bars
Bend up 2 Bars at the & Point
Bend up 2 Bars at the
Point
Point
Bend up 2 Bars at the
Bend up 2 Bars at the
In Beams with 8 Bars
SLAB.
5" Thick; " Corr. Rounds, 6" Cts. in Bottom of Slab.
Top Bars," Corr. Rounds 3'-0" long, 12" Cts. over G2.
Top Bars, ½" Corr. Rounds 2′-0″ long, 12" Cts. at G1.
2-2" Corr. Rounds lengthwise of Bridge in Each Panel.
Stirrups.
Shown on Detail Drawing
½" Corr. Rounds Bent as
Spacing: 12″ to the Point,
18" Beyond.
½" Corr. Rounds Bent as
Shown on Detail Drawing
Spacing: 9" to the Point,
18" Beyond.
131
18
Оп
의
​AISE
GI
-461
10
ПОППОППЛЛЛЛЛЛЛЛЛЛЛОЛ
5-8"
ó
SEISUSURIENZENNEMSIENSISININ SIENTAIBIAN, VIGNJILSINSIMILATENENSISTIN
→
16-0
G2
5-8
"
CLEAR SPAN = S
SIDE ELEVATION
G2
SECTION AT CENTER
5-8
GI
+61-
4
NOTE -
10
-
SLAB - 5" THICK - Ź CORR.ROUNDS.
6 CTS. IN BOTTOM OF SLAB.
TOP BARS - Ź CORR ROUNDS-3-0 LONG 12 ČTS OVER G₂
•
TOP BARS CORR ROUNDS-20 LONG 12CTS AT GI
"1
2-20
CORR ROUNDS, LENGTHWISE OF BRIDGE IN EACH
PANEL
STANDARD DETAILS,
CLASS 2 BRIDGES.
PLATE-5.

132
Clear Span
20'-0" 36"
25'-0"
30'-0"
35'-0"
40'-0"
20'-0"
30'-0"
h.
35'-0"
40'-0"
43"
47"
52"
25'-0" 34"
58"
31″
39"
43"
47"
b.
12"
12"
14"
14"
16"
14"
14"
14"
14"
17"
CLASS 2-BRIDGES.
GIRDERS G1.
See Detail Sheet, Page 136.

f..
15"
15"
15"
21"
Reinforcement.
3-34"
3-1″
3-1"
Corr. Rounds.
3-1" Corr. Rounds.
18" 6-1%" Corr. Rounds.
Corr. Rounds.
Corr. Rounds.
3-%" Corr. Rounds.
3-1″ Corr. Rounds.
15"
4-1" Corr. Rounds.
4-1%" Corr. Rounds.
GIRDERS G2.
See Detail Sheet, Page 137.
15" 8-1" Corr. Rounds.
15' 8-1%" Corr. Rounds.
4-1%" Corr.
4-14" Corr.
Rounds.
Rounds.
18" 8-14" Corr. Rounds.
21" 10-14"Corr. Rounds.
Bent Bars.
In Beams with 6 Bars
Bend up 1 Bar at the Point
Bend up 2 Bars at the Point
Point
Point
8 Bars
In Beams with
Bend up 2 Bars at the
Bend up 2 Bars at the
Point
Point
Beams with 8 Bars
Bend up 2 Bars at the
Bend up 2 Bars at the
Point
Point
Bend up 2 Bars at the
Bend up 3 Bars at the
In Beams with 10 Bars
SLAB.
5″ Thick," Corr. Rounds, 6″ Cts. in Bottom of Slab.
Top Bars, " Corr. Rounds 3′-0″ long, 12″ Cts. over G2.
Top Bars," Corr. Rounds 2′-0″ long, 12″ Cts. at G1.
2-2" Corr. Rounds lengthwise of Bridge in Each Panel.
Stirrups.
Shown on Detail Drawing
" Corr. Rounds Bent as
Spacing: 12" to the Point,
18" Beyond.
Point,
Shown on Detail Drawing
" Corr. Rounds Bent as
Spacing: 9" to the
18" Beyond.

133
GI
Gr
P
|| ||||
Job
ALTERNATE SECTION.
b
VELGIA
24'-0"
TENSISISI, UKUSUSUEMINENSE
G3
G3
5=0"
CLEAR. SPAN 5.
MEMEMEMEMEM
π
SIDE ELEVATION
G2
5=0"
SECTION AT CENTER
·
4
"
GI
5=0°
BE
NOTE-
"
SLAB - 5 "THICK & CORR ROUNDS-6CTS IN
BOTTOM OF SLAB. TOP BARS - CORR ROUNDS
3-0"LONG-12 "CTS OVER G2 AND G3.,
A
TOP BARS - ½ CORR ROUNDS - 2-0 LONG-12CTS. AT GI
·
2-₤ CORR ROUNDS -LENGTHWISE OF BRIDGE IN EACH
PANEL.
STANDARD DETAILS,
CLASS 3 BRIDGES.
-
PLATE-6


134
Clear Span h.
20'-0"
25'-0"
30'-0"
35'-0"
40'-0"
20′-0″
25'-0"
30′ 0″
35'-0"
40'-0"
20'-0"
30'-0"
30"
35′-0″
38"
45"
50"
56"
34″
39"
44"
47"
54"
25'-0" 39"
34"
44″
47"
40'-0" 54"
CLASS 3-BRIDGES.
GIRDERS G1. See Detail Sheet, Page 136.
b.
12"
12"
12"
14"
12"
14"
14"
GIRDERS G2.
14"
15″ 21″ 8-1″ Corr. Rounds.
14″
f.
14'
14"
15" 6-4" Corr. Rounds.
17"
15" 6-%" Corr. Rounds.
15"
17"
18" 6-1' Corr. Rounds.
15"
GIRDERS G3:
Reinforcement.
15"
3-%" Corr.
3-1' Corr.
4-1%" Corr. Rounds
4-14" Corr. Rounds
14" 21" 8-14" Corr. Rounds
18'
Rounds.
Rounds.
15" 8-1" Corr. Rounds.
15"
1
6-1" Corr. Rounds.
4-1″ Corr.
4-1%" Corr.
See Detail Sheet, Page 137.
Rounds.
Rounds
15" 8-1/4" Corr. Rounds
4-1%" Corr. Rounds
4-14" Corr. Rounds
15" 8-14" Corr. Rounds
Bent Bars.
18" 10-14" Corr. Rounds
Point
Point
21" 10-14" Corr. Rounds
In Beams with 6 Bars
Bend up 1 Bar at the
Bend up 2 Bars at the
Bend up 2 Bars at the
Bend up 2 Bars at the
In Beams with 8 Bars
Point
Point
See Detail Sheet, Page 138.
--|-
In Beams with 6 Bars
Bend up 1 Bar at the
Bend up 2 Bars at the
Point
Point
Point
In Beams with 8 Bars
Bend up 2 Bars at the & Point
Bend up 2 Bars at the
Point
In Beams with 8 Bars
Bend up 2 Bars at the Point
Bend up 2 Bars at the
Point
Point
3 Bars at the
In Beams with 10 Bars
Bend up 2 Bars at the
Bend up
Stirrups.
Shown on Detail Drawing
"Corr. Rounds Bent as
18" Beyond.
Spacing: 12" to the Point,

Shown on Detail Drawing
½" Corr. Rounds Bent as
Spacing: 9" to the Point,
18" Beyond.


" Corr. Rounds Bent as
Shown on Detail Drawing
Spacing: 9″ to the Point,
18" Beyond.


SLAB.
5″ Thick, " Corr. Rounds, 6" Cts. in Bottom of Slab.
Top Bars," Corr. Rounds 3'-0" long, 12" Cts. over G2 and G3.
Top Bars, ½" Corr. Rounds 2'-0" long, 12" Cts. at G1.
2-2" Corr. Rounds lengthwise of Bridge in Each Panel,
135
•
---
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CORR.ROUNDS-
17-6"LONG-6 CTS IŹ
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TYPICAL DETAIL GIRDERS GI FOR CLASS-1-2 AND 3 BRIDGES.
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TYPICAL DETAILS - SLAB AND GIRDERS GI, CLASS I BRIDGES.
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2- CORR ROUNDS.
2-4 CORR.ROUNDS.
ALL REINFORCING BARS ARE HIGH
ELASTIC LIMIT CORRUGATED ROUNDS.
STIRRUPS - Ź CORRUGATED ROUNDS.
BENT AS SHOWN.
DETAIL SHEET,
GIRDERS -GI.


136
SININ
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18
TYPICAL DETAIL, GIRDERS G2, FOR CLASS. 1-2 AND 3 BRIDGES.
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18
CORR ROUNDS
17'6 LONG-6^CTS.
SPAN
CLEAR SPAN - 5.
CORR ROUNDS - 2·0" LONG-12 CTS,
5-8
2
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2-Ž CORR ROUNDS IN EACH
PANEL,
G2
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ing
ROUND'S
TYPICAL DETAILS - SLAB AND GIRDERS G2 -CLASS 2-BRIDGES.
NOTE
-
5/5
5/10
CORR ROUNDS.
ALL REINFORCING BARS ARE HIGH
ELASTIC LIMIT CORRUGATED ROUNDS.
STIRRUPS / CORR ROUNDS - BENT AS SHOWN.
·
DETAIL SHEET,
GIRDERS - G 2.


137
GI
2
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11
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GIRDERS G3, FOR CLASS 3 BRIDGES.
CŹ CORR. ROUNDS · 3-OʻLONG 12°CTS.
5-0
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TYPICAL DETAILS - SLAB AND GIRDERS G3, CLASS 3 BRIDGES.
A
NOTE
D
3
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T
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5/5
5/10
3
4
CORR ROUNDS.
ALL REINFORCING BARS ARE HIGH ELASTIC
LIMIT CORRUGATED ROUNDS.
n
STIRRUPS - CORRUGATED ROUNDS BENT AS
SHOWN
DETAIL SHEET,
GIRDERS-G 3.


138
INDEX, AUGUST NUMBER.
Box Culverts-Standard Designs
Classification by Loadings
Concrete, Specification for
Flat-Slab Bridges-Standard Designs
Girder Bridges-Standard Designs
Reinforcing Steel, Specifications for.
•
• •
Page
Irg
II2
•
I 12
II4
129
III
See page 38 (May number) for Index on General Principles of Design, Building
Construction.
See page 76 (June number) for Index on Detailed Design of Typical Building and
Complete Analysis of the Strength of Rectangular and T-Shaped Beams.
See page 107 (July number) for Index on Methods of Design and Standard Load-
ings for Highway Bridges.
139.

Designing Methods
Reinforced Concrete Construction
VOL. 1
OCTOBER, 1908
RETAINING WALLS
OF THE
CANTILEVER AND BUTTRESS TYPES
Discussion of the special problems met with in the design
of retaining walls, with a brief discussion of the subject
of earth pressures.
No. 5
Detailed designs of cantilever wall to retain earth fill for
both horizontal surface and with surcharge. Detailed
design of buttress type of wall, surface of fill horizontal.
BULLETIN No. 6, VOL. 1, WILL CONTAIN
Detailed design of rectangular
reservoir.
THIS NUMBER WILL BE PUBLISHED IN DECEMBER.
REPRINT, MARCH, 1909
PRICE, 25 CENTS
CORRUGATED BAR COMPANY
(FORMERLY EXPANDED METAL AND CORRUGATED BAR CO.)
National Bank of Commerce Building
SAINT LOUIS

COPYRIGHT, 1908,
BY
CORRUGATED BAR COMPANY
RETAINING WALLS FOR EARTH.
A reinforced concrete retaining wall can be built at less cost than
a gravity section of equal stability, under practically all conditions.
Moreover the amount of saving that may be effected by the adoption
of a reinforced concrete design increases with the height of the wall;
comparative estimates on actual designs would show a saving of 20
to 25% on walls 30 to 40 feet high.
The dimensions of a gravity section or "heavy masonry" retaining
wall are determined, not from a consideration of the strength of the
material used in resisting tensile or compressive stresses, but by mak-
ing it of such size that the weight alone enables it to resist the over-
turning moment of the backing. The usual condition imposed is that
the resultant of the forces acting on the section shall pass through
the middle third of the base, thus avoiding any possibility of tension
on the inner face, but in most cases causing very heavy soil pressure
under the toe. This pressure, which in the case of a gravity section,
may exceed the safe bearing power of the soil-necessitating the use
of piling-can very often be brought within the allowable amount by
the adoption of a reinforced concrete design. When piling must be
used in any event, a large saving in the the cost of the foundation
results from the use of a reinforced concrete wall.
A reinforced concrete retaining wall may be described as a section
which derives its stability (against overturning) mainly from that part
of the fill which rests upon the base; its structural integrity depending
upon its ability to develop tensile as well as compressive and shear-
ing stresses. The dimensions are not determined from a consideration
of the weight of the wall as in gravity sections, but from considerations
of strength as a structural material.
A reinforced concrete retaining wall will then be in almost all cases
of less cost than a gravity type of wall; the total pressure on the
foundation will be less, and the maximum soil pressure at the toe may
be kept within required limits, owing to the possibility of extending
the base. Differently expressed this means that with a reinforced
concrete wall a smaller and more uniform soil pressure is produced—
a most desirable condition. Owing to the small sections used in a
reinforced concrete design, it is practicable to provide against tem-
perature and shrinkage cracks by means of longitudinal reinforcement
without resorting to expansion joints.
143
EARTH PRESSURE AGAINST RETAINING WALLS.
It is not proposed here to discuss the various theories of earth
pressure against retaining walls. The active pressure exerted by the
fill varies with the material, its compactness, the percentage of con-
tained moisture, etc. If no provision for drainage is made, and there
is an opportunity for water to collect behind the wall the pressure.
may have any value up to hydrostatic pressure.
Experience has shown that masonry walls of the gravity type will
be stable under all conditions usually met with when the width of the
base is made of the height. Reinforced concrete retaining walls,
therefore, should be so designed that the stability against overturning
will be at least as great as that of a solid masonry wall the base of
which is of the height. It is therefore recommended that for
ordinary fills (horizontal surface), where the material is not likely to
become saturated, that the design be based on an active horizontal
pressure, the intensity of which at any point is 1/3 of the vertical
intensity at that point. A wall designed on this basis will have a
slightly greater factor of stability than the gravity section referred to.
10
For convenient reference Rankine's formulæ for earth pressures.
are given, as they are in general use. It is to be noted that for a
horizontal surface and 30°, the intensity of the horizontal pressure,
becomes the vertical.

NATURAL SLOPE
KUSTIK
e
H-Pcos
1/3
4
Let Figure 1 represent a retain-
ing wall, the surface of the fill
making an angle with the hori-
zontal. The symbols may be defined
as follows:
Figure 1.
w=weight of fill per cubic foot.
P the resultant pressure corresponding to a depth of fill h.
P=H, when @ is equal to zero.
144
0-angle with the horizontal made by the surface of fill.
angle of repose, or natural slope of the material constituting
the fill.
p-intensity of the lateral pressure at any depth, in pounds per
sq. foot.
Then:
P=
P=
wh²
2
P
When 6, which is the usual value in surcharge design we have:
wh²
..(2)
2
In all cases
COS
cos e X
If the surface is horizontal, equals zero, and we have:
wh2 I sin &
X
2 1+ sin &
.. (3)
H=P cos 0……….
cos e — √ cos² 0
COSA + √ Cos² 0
2
P=H=
2
cos² Φ
cos² &
=H……….
wh
and p=
3
...( (1)
·(4)
It may be noted that the formulæ take no account of such cohesion
as may exist between the particles of the fill and that equation 3 will
give the pressure due to a liquid if is made equal to zero, the equa-
Ф
tion reducing to the form P=½ wh².
Since the intensity of the horizontal pressure is assumed to increase
uniformly with the depth the resultant, P, will act as a distance h
from the base of the wall.
For ordinary fills, where the material is not likely to become sat-
urated, ☀ may be assumed to have a value of 30°; substituting this
value in equation 3 we have for a horizontal surface
2
wh²
6
….(5)
(6)
Equations 5 and 6 should be used for earth fills when the surface
of the fill is horizontal; these values it will be noted agree with those
previously recommended.
Superimposed Loads-A uniformly distributed load, covering
the entire surface of the fill will be assumed to exert a uniform hori-
zontal pressure, the intensity of which is 3 of the vertical, regardless
of the depth.
145
The following table gives the intensity of the horizontal pressure
at any depth, h, the total pressure H, above the section considered
and the overturning moment, M, in inch lbs., at the section A-B:
h
1 2 3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
2272
19
20
feet. pounds. pounds.
21
HORIZONTAL SURFACE.
w=100 lbs.
23
24
25
26
27
28
29
30
p=3 wh
TABLE I.
33
67
100
133
167
200
233
267
300
333
367
400
433
467
500
533
567
600
633
667
700
733
767
800
833
867
900
933
H
967
1000
214
}
H=P
=% wh2
17
67
150
267
417
600
817
1067
1350
1667
2017
2400
2817
3267
3750
4267
4817
5400
6017
6667
7350
8067
8817
9600
10417
11267
12150
13067
14017
15000
Overturning
Moment
M= wh3X12
inch pounds.
67
533
1800
4267
8333
14400
22867
34133
48600
66667
88733
115200
146467
182933
225000
273067
327533
388800
457267
533333
617400
709867
811133
921600
1041667
1171733
1312200
1463467
1625933
1800000
h
feet.
1 2 3 4 10
6
7
8
∞ ∞
9
10
11
12
13
14
15
16
17
18
19
20
21
2****5*28
23
24
26
27
29
30
}
Surcharge, 0=30°.
w=100 lbs.


p cos A—
3/4 wh
pounds.
75
150
225
TABLE II.
300
375
450
525
600
675
750
825
900
975
1050
1125
1200
1275
1350
1425
1500
1575
1650
1725
1800
1875
1950
2025
2100
0.30:
2175
2250
H=P cos A
38 wh2
pounds.
38
150
338
600
938
1350
1838
2400
3038
3750
4538
5400
6338
7350
8438
9600
10838
12150
13538
15000
16538
18150
19838
21600
23438
25350
27338
29400
31538
33750
B
Overturning
Moment
M=% wh³×12
inch pounds.
150
1200
4050
9600
18750
32400
51450
76800
109350
150000
199650
259200
329550
411600
506250
614400
736950
874800
1028850
1200000
1389150
1597200
1825050
2073600
2343750
2636400
2952450
3292800
3658350
4050000
}
146
"
NOTES ON DESIGN.
In the design of retaining walls it is necessary to investigate the
stability of the structure, not only against overturning but also
against sliding upon its base. To prevent sliding the total pressure
upon the foundation multiplied by the co-efficient of friction of con-
crete upon the soil, must be greater than the horizontal thrust of the
fill. It is good practice to provide a projection or toe extending into
the foundation to assist in preventing such movement.
The maximum soil pressure at the toe of the retaining wall should
be computed in all cases, as it is very often the determining factor in
the design of the base.
In the design and construction of retaining walls, provision should
be made for adequate drainage. This is an important feature as the
development of an hydraulic head back of the wall might result in
its failure.
Provision should be made in all retaining wall work against the
development of shrinkage or temperature cracks. On account of
the small sections required in a reinforced concrete design, it is prac-
ticable to prevent the development of such cracks by the use of
longitudinal reinforcement. Some longitudinal reinforcement should
be used even if expansion joints are provided.
It is usually considered that about 10% of reinforcement will be
sufficient to prevent cracks, using high elastic limit steel.
Retaining walls of the cantilever type will generally be found
economical up to heights of 18 feet; for higher walls the buttress
type should be used.
147
The following diagrams, Figures 2 and 3, show the forces acting
upon a retaining wall which we will consider in the design:
We weight of section of wall one foot long; the line of action
passing through the center of gravity of the cross section.
We weight of earth, per lineal foot, resting on base of wall; the
line of action passing through center of gravity of the
figure C, D, E, F.
p2 soil pressure at toe, lbs. per sq. foot.
p soil pressure at the heel, in lbs. per sq. foot.
R=resultant of the earth pressure and (We+We).
Other symbols as previously defined.
H
A
162
8
G

Wc
ENETISIENST
ом том
We
H = P
Figure 2.
VETEMENKVIENE
p
3/3
Diagram showing forces acting upon a retaining wall, surface of
fill horizontal.
148

Н
A
ре
B
6
ом
ам том
We
E
P
ZIF
А
Ө
$14
ろ
​Figure 3.
Diagram showing forces acting upon a retaining wall, surface of
fill not horizontal.
149
SPECIFICATIONS FOR MATERIALS.
All formulæ for the strength of reinforced concrete beams are
necessarily based on the physical properties of the concrete and steel.
The formulæ proposed in this bulletin are based upon a concrete with
a compressive strength of 2,000 pounds per square inch and upon the
use of mechanical bond reinforcing bars which have an elastic limit
of 50,000 pounds per square inch; the critical amount of reinforce-
ment, under these conditions being .0085 bd.
Reinforcing Steel-All reinforcing steel used in retaining wall
construction shall be rolled to such form that it has a positive.
mechanical bond with the concrete. Adhesive hond will not be con-
sidered sufficiently reliable for this class of structures.
Steel may be made by either the Bessemer or open-hearth process;
bars to be rolled from billet stock. Re-rolled material will be accepted
under conditions insuring rigid inspection.
The elastic limit and percentage of elongation shall be determined.
by tests on accurately-machined specimens, and shall conform to the
following requirements:
Elastic limit to be from 50,000 to 60,000 pounds per square inch,
ultimate strength not less than 13 Xelastic limit.
The percentage of elongation in 8″ must not be less than given
by the formula:
Percentage of elongation=
I,400,000
Ult. Strength
5.
Bending Test--Bars as rolled shall bend cold, 90 degrees, to a
radius equal to three times the least diameter of the specimen, without
sign of fracture,
150
Cement-Only Portland cement conforming to the requirements
of the specifications adopted by the American Society for Testing
Materials, June 14, 1904, shall be used.
Sand-Sand to be clean and coarse, and free from organic matter;
a graded sand, with coarse grains predominating, is to be preferred.
Coarse Aggregate-Broken stone to be hard, durable limestone,
or its equivalent, free from dust and foreign materials; maximum-sized
particles to pass through a one-inch ring; fines to be removed by
passing over a one-quarter inch screen. The fines may replace part
of the sand. Under special conditions making for uniformity crusher
run may be used, but this is not desirable.
Gravel shall be clean and of graded sizes; the sand carried to be
removed by screening as for broken stone.
Proportions of Mix-Concrete for wall, base, and buttresses to be
mixed in the proportion of one part cement to six parts aggregate;
proportions by volume taking one bag containing not less than 94
pounds of cement, equal to one cubic foot of cement.
The proportions of fine and coarse aggregate used shall be chosen
so as to give a concrete of maximum density; in no case, however,
may the amount of fine aggregate be less than 50 per cent of the
coarse.
b
FORMULAE FOR USE IN DESIGNING.

P
Let M-moment of resistance of section in inch lbs.
Mo-ultimate moment of resistance of section
in inch lbs.
q=area of reinforcement.
p=percentage of steel=q÷bd.
F-elastic limit of steel-50,000 lbs.
All dimensions as shown, in inches, and stresses in lbs. per sq. in.
Then we have for rectangular beams:
Mo=370 bd for g=.0085 bd...
and for various percentages:
.. (7)
Mo=.86 Fp bd²=.86X50,000xp bd²....
..(8)
=43,000 p bď², using high elastic limit corrugated bars.
(For complete discussion of these formulæ, see May and June bulletins.)
151
WORKING STRESSES.
If it is desired to design for working stresses in the steel, use the
formula :
M=sqX.86 d. where s=unit stress in the steel.. (9)
REMARKS ON THE APPLICATION OF THE
FORMULAE.
In retaining wall design it will often be found that the section
required to resist the bending moment, as determined by formula 7,
is smaller than would be desirable to use in heavy work. In this case
the value of "d" may be assumed, and the amount of reinforcement
required determined by means of equation 8.
Equation 8 should also be used in the design of buttresses, as
there is more or less T-beam action. The amount of reinforcement
used, however, should not exceed 2% of the cross section of the
buttress.
It is to be noted that equations 7 and 8 give the ultimate or
breaking strength of the structure, and in applying them the actual
moments must first be multiplied by the factor of safety.
152
CANTILEVER WALLS.
The two examples following illustrate the methods of design
advocated:
Problem 1-Design a cantilever wall, 16'-0" high, to retain an earth
fill weighing 100 lbs. per cubic foot; surface of fill horizontal, the
value of to be taken as 30°.
Ф
Maximum allowable soil pressure 3,000 lbs. per square foot.
We will assume the section shown herewith and investigate its
stability and strength:
4267
*
9-
p2°2872
G
5.56
4.64'
3.17'
12250
B
2.66
2:0"
FF

・Wc = 4455*
11490
16
A
=
Wc+ We
We = 7035*
P-H-420
+420
6:0"
"1
9:01
"0-,91
1267
„9-1
*
To determine the position of the center of gravity of the concrete
section and of the earth prism take moments about the point “A.”
153
Center of Gravity of Wall-
SECTION
CONSIDERED.
ADIH..
K B C L.
BCFG..
·
·
Totals.
SECTION
CONSIDERED.
CFF'..
CF' ED...
Totals..
Dag my blog
AREA IN
SQ. FT.
12.0
2.5
15.2
29.7
Distance from "A" to c g=
AREA IN
SQ. FT.
=3.17 ft.
Weight of wall per lineal foot=29.7 X 150-4445 lbs. W。.
Static moment about "A" =4445×3.17=14110 ft. lbs.
Center of Gravity of Earth Prism-
5.I
65.25
70.35
P-H=
MOMENT.
ARM.
391.8
70.35
4.00
2.75
2.58.
94.I
29.7
MOMENT
ARM.
wh² 110X16²
6
6
3.25
5.75
=5.56 ft.
AREA
MOMENT.
48.0
6.9
39.2
94. I
Distance from "A" to c g=
Weight of earth prism per lineal foot=70.35X100=7035=We.
Static moment about "A"=7035X5.56=39180 ft. lbs.
AREA
MOMENT.
To obtain the position of the resultant divide the sum of the static
moments by the total weight:
16.6
375.2
391.8
14110+39180_53290
4445+703511480-4.64 ft.
The horizontal thrust H, on the wall may be determined by means
of formula 5.
R=√(42672+11490²)=12250 lbs.
=4267 lbs.
(The pressure might have been taken from the table on page 146).
This thrust is assumed to be concentrated at a point 5′-4″ above
the base. Combining this force with the force We+ We we obtain the
final resultant R.
154
•
The tangent of the angle which the resultant makes with the
4267
vertical is
.372, and R would cut the base at a distance,
I 1490
5.′33X.372=1.98 ft. from the line We+We, or 2.66 ft. from the
edge A-H.
Since the resultant R passes through the third point the soil
pressure at the heel is zero, and that at the toe twice the average.
Average soil pressure=
=1436 lbs. per sq. ft.
I 1490
8
Maximum soil pressure=p2=2X1436=2872 lbs. per sq. ft.
Stability Against Sliding-The total horizontal thrust=4267 lbs.
Assuming that the foundation is good clay the coefficient of friction.
between the concrete and the soil may be taken as .50. The frictional
resistance to horizontal displacement would then equal 11490X.5=
5745 lbs. It is desirable, however, to have a projection on the base
as shown in the drawing as the coefficient of friction may be as small
as .33 should the footing become wet.
The earth fill in front of the wall assists to a slight extent in resist-
ing sliding, but its action is not positive, and should not be included
in the design.
REINFORCING STEEL.
Having assured ourselves of the stability of the section, when
acted upon by the assumed forces, it will be necessary to examine its
structural integrity and provide reinforcing steel where required.
The bending moment, per foot of length, in the face wall at the
plane A-D is equal to
2
3
100X14.5×14.5— 100X14.5=16900 ft. lbs.
6
3 18
=202800 inch lbs.
Neglecting the small fillets the effective depth of the beam, d, may
be taken as 16".
We will base the design on the ultimate strength of the section
using a factor of safety of 3, and determine the amount of reinforce-
ment required by means of formula 8.
Mo=3X202800-608400-43000 pbď².
155
b=12" and d=16", substituting these values in the equation we
have p=.0046.
q=16X12X.0046=.88 sq. in. per foot of wall.
Use 3/4" corrugated rounds, spaced 6″ on centers.
It will not be necessary to carry this amount of reinforcement to
the top of the wall, as the bending moments decrease rapidly toward
the surface of the fill. The bending moment at a point 10'o" below
the surface is 66,667 inch pounds.
Mo=3X66,667=200,000 in. lbs. (ult.)
The thickness of the wall at this point is 15.6″, d=13.6″. The
amount of metal required, 9,=.34 sq. in. per foot of wall, which would
indicate that one-half of the reinforcing bars may be stopped off at
this point.
It is necessary in all cantilever designs to investigate the anchorage
of the vertical bars in the base. If the length of imbedment is not
sufficient the bars may pull out, due to the shearing of the concrete
around them. Using round corrugated bars a length of imbedment
of 25 diameters will insure sufficient shearing strength. The vertical
bars will accordingly be extended into the projection on the base as
shown on detailed drawing, page 158.
Design of Base-That part of the base back of the wall will be
made sufficiently strong to lift the fill resting upon it in addition to
its own weight. We will determine the bending moment on a vertical
plane through the point C.
Taking moments about C, we have:
Moment due to fill =7035X2.06=14492 ft. lbs.
Moment due to base-1012X2.25= 2277 ft. lbs.
16769 ft. lbs.
=201228 in. lbs.
The effective depth d, is 16", and the amount of reinforcement
required is .88 sq. in. per lineal foot of wall.
Use 3/4" corrugated rounds, 6'-6" long, spaced 6" on centers in the
top of the slab.
In the design of the toe, since the overhang is small, we will
assume a uniform upward pressure of 2872 lbs. per sq. foot. The
footing weighs 225 lbs. per sq. foot, and the effective upward pressure
is 2872-225-2647 lbs.
>
156
+
Taking moments about B:
M=(2X2647)XI=5294 ft. lbs.=63530 in. lbs.
M。= 3×63530=190590 in. lbs., and applying equation 8.
q=
=.276 sq. in.
190590
43000X16
Use 5%" corrugated rounds 4′-0″ long, spaced 12″ on centers.
It is desirable to fillet the corners, where the face wall joins the
base, providing sufficient reinforcement in the rear fillet to carry the
bending moment. Owing to the depth available for beam action 3/4"
corrugated rounds, spaced 12" on centers will be sufficient.
Longitudinal Reinforcement-As before stated the amount of
longitudinal reinforcement required in the face wall is dependent upon
the particular conditions-the position of the wall, its exposure and
the probable range of temperature. Some longitudinal reinforcement
is necessary in any event in both wall and base. We will use ½"
corrugated rounds spaced 24" cts. in each side of the wall. If it is
desired to provide fully for temperature stresses the amount of
longitudinal reinforcement above the grade line should be increased
to about .004 of the cross section.
157
9-67
9N07, 9-GI-SL7, 9:
5/8 C.R.-12 XTS
4-0"LONG
VERTICALS - É CA
2-0
VERTICAL BARS•
3/4 CORR. ROUNDS 12 CTS.
16.4 "LONG.
HORIZONTAL BARS ·
1/2" C.R-2·0"CTS. IN EACH
SIDE STAGGERED.
·VERTICAL BARS. & G.R.
12"CTS.-6'6" LONG
18"
C.R.-12 CTS-4-0 LONG
"C.R.•6" CTS-6-6" LONG
LONGITUDINALS -
12. C.R. AS SHOWN.
8′-0″
CROSS SECTION
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IN D
WIZJI.
HORIZONTAL SECTION
DETAILS.
-
==
"
ALL REINFORCING BARS ARE HIGH
ELASTIC LIMIT CORRUGATED ROUNDS.
CANTILEVER TYPE.
++++++++++*
Poocom
RETAINING WALL
16-0 HIGH.
PROBLEM I.


158
h
Problem 2-Design a cantilever wall, 16'-o" high, to retain an
earth fill, the surface of which slopes upward at an angle 0=4=angle
of repose of the material.
Maximum allowable soil pressure, 5000 lbs. per sq. ft.

2:0"
A
H
3=0"
K
G
B
0099*
R-27200
-
ору
*...*.
2:0
15038***
C
-
Wc + We
10-0″
0-0-30°
We-8438
566°
396
265
6.52'
E'.
| P= 16400
I
4.21
„0-01
6.49
2:0"
,90·61 =,4
„0.91
We will determine the position of the center of gravity of the
concrete section and of the earth prism by taking moments about the
point I.
159
Center of Gravity of Wall-
SECTION
CONSIDERED.
G F C B ....
BCLK.
ADIH
•
Totals
•
FCF ..
EFE..
FCLDE.
SECTION
CONSIDERED.
Totals ..
¿
AREA IN
SQ. FT.
... . . .
18.75
5.25
20.00
44.00
Center of Gravity of Earth Prism-
P
248.5
44.0
Distance from I to c g
=5.'66.
Weight of wall per lineal foot=44X150=6600 lbs.=Wc.
Static moment about I=6600X5.66=37356 ft. lbs.
AREA IN
SQ. FT.
6.25
10.38
67.75
84.38
MOMENT
ARM.
223.4-2.65.
84.38
6.22
6.00
5.00
·COSO
2
37356+22350_59706
6600+8438
MOMENT
ARM.
Distance from I to c g
Weight of earth prism per lineal foot=84.38X100-8438 lbs We.
Static moment about Z=8438X2.'65=22350 ft. lbs.
Distance to resultant, We+ We,
5.33
2.00
2.50
AREA
MOMENT.
15038-3.96.
The thrust P, acting parallel to the surface of the fill, may be
determined by means of formula 2.
wh2
2
100X 19.462
117.0
31.5
100.0
248.5
X.866=16400 lbs.
AREA
MOMENT.
33.3
20.7
169.4
223.4
160
The force P should be considered as applied on the plane I E, and
at a point h' above the base. Combining P and (We+We) we
obtain the resultant R=27200 lbs., which cuts the base at a point
6.'52 distant from I.
Soil Pressure-The vertical component of R-23238 lbs. acting
at a point 6.'52 from I. The average soil pressure is 2324 lbs.
The maximum pressure at H is obtained as follows:
The moment of the vertical component of R about the middle
point of the base 23238X1.52=35400 ft. lbs.
The moment of resistance of a rectangular section=M=&fbh²,
where f=stress on extreme fiber, b=width and h=depth of section.
In this case :
b=1'-0" and h=10'0".
M=35400-ƒX 100, from which f=2124 lbs.
The maximum soil pressure = 2324 + 2124
2324 + 2124 = 4448 lbs.
The minimum soil pressure 2324-2124 =
= 200 lbs.
―
Stability Against Sliding-With a coefficient of friction of .5 the
resistance to sliding would be .5X23238=11619 lbs. The total hori-
zontal force tending to move wall is P cos 0=H=14200 lbs. It will
be necessary therefore to provide a projection extending into the soil
as shown.
Reinforcement Required-The bending moment in the wall (per
lineal foot) in the plane B-C may be obtained from Table 2, on page
146, which gives M-614400 in. lbs. for an h of 16′-0″.
At this point d=22" and the amount of steel required as deter-
mined by formula 9, for s=16500, is 1.98 sq. in.
I
Use 1" corrugated squares 6" on centers. Table 2 gives the bend-
ing moments at various depths from which we can compute the
amount of reinforcement required at these points. This operation
leads to the conclusion that of the vertical bars should extend to
the top of the wall; 3 should extend to within 3′-0″ of the top, and
the remaining be stopped off at a point 7'-0" below the top of the
wall.
The vertical bars will have sufficient imbedment if extended to
the underside of base.
161
Design of Base-The bending moment on the base in the plane.
CF' is obtained by taking moments about the point C.
Diagonal reinforcement will be placed in the rear fillet in sufficient
amount to carry the bending moment at that point; the area required
is approximately ½ sq. in. per lineal foot of wall-use 3/4" corrugated
squares 12″ on cts..
a
In the design of the toe we will determine the bending moment
in the vertical plane through B.
9000
3:0°
11433
8
Moment due to fill =8438X28″=236000 in. lbs.
Moment due to base=1500X30″= 45000 in. lbs.
281000 in. lbs.
158
d=22" and q=.90 sq. in.
Use 1" corrugated squares 12" cts.
3174
7:0°
200
To determine the average
pressure on the toe draw the
diagram showing soil pressures.
The average pressure on the
overhang will be ½ (4448+3174)
=3811 lbs. per □'.

The resultant of this reaction
will pass through the center of
gravity of the trapezoid a b c d,
which is 1.58 ft. from bc.
Mo=1.58X(3X3811)X12=217000 in. lbs. (ult.)
To resist this moment there will be required .72 sq. in. of steel
assuming_d=22 inches, and we will use %" corr. squares 12" cts.
Longitudinal Reinforcement-We will not provide fully for
temperature stresses, but will arbitrarily use ½" corr. squares spaced
24″ cts. in each face.
162
#
07/
VERTICALS
1/2" C.S: 2·6075
3-0
"!
··
.....
78 CORR. SQ. 12" CTS-
4:6 LONG
://
12"
(VERTICAL BARS-I "C.5.-18 CTS.
16-6"LONG
7
VERTICAL BARS - 1 "C.S.-18 "CTS.
10 LONG
HORIZONTALS - 2 6.5. ·2·0 CTS.
IN EACH FACE STAGGERED
1
2-0
K
VERTICAL BARS - 1 C.5. 18 CTS.
9:6 "LONG
"
2:0"
¾/14 0.5. 12 (TS. 5'O'LONG
ICORR. SQ. 12 CTS-70
LONG,
10=0"
CROSS SECTION
N
===
P
====+=+==+=+==+F
U B U U U
REAR ELEVATION
NOTE.
퓨
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sit
HORIZONTAL SECTION
DETAILS
Patt
==
CANTILEVER TYPE.
meta
ALL REINFORCING BARS ARE
HIGH ELASTIC LIMIT CORRUGATED SQUARES.
=
RETAINING WALL
16-0" HIGH.
PROBLEM 2.



163
The following example illustrates the methods of design suggested
for this type of retaining wall.
Problem 3-Design a buttress wall 25′-0″ high to retain an earth
fill weighing 100 pounds per cubic foot; surface of fill horizontal; the
value of to be taken as 30°. Maximum allowable soil pressure
4000 pounds per square foot.
A
H
G
BUTTRESS WALLS.

4:0"
126
2'0",
= 104925
Wc
с
BUTTRESSES- 100" ON CENTERS
AND 18" THICK
7.63
Wc+ We = 296425*
We=191500
5.51'
14-0"
4.37'
8:6"
*********
8.44'
E
18" I
160; &
23-0"
2:0
25'-0"
H-104170
4-8
NOTE - FORCES GIVEN
FOR 10 FT. LENGTH
We will assume the section shown herewith, and investigate its
stability and strength.
To determine the position of the center of gravity of the wall, we
will consider a section 10'-0" long.
164
Center of Gravity of Wall, moments taken about I-
SECTION
CONSIDERED.
Coping..
Face Wall..
Fillet...
Base...
Buttress..
Totals..
SECTION
CONSIDERED.
CDEF....
FED……….
VOLUME
CU. FT.
Totals....
25.0
217.5
7.5
7.5
280.0
15.0
147.0
699.5
VOLUME
CU. FT.
1760.0
155.0
1915.0
8385
1915
|
5332.2
Distance from I to c g
699.5
Weight of ten foot length of wall=699.5X150=104925 lbs.
Static moment about I, for section 10'-0" long=
104925X7.63=800000 ft. lbs.
Center of Gravity of Earth Prism, moments taken about I-
MOMENT
ARM.
9.33
9.50
10.25
10.83
=7.'63.
=4.37.
7.00
0.75
6.16
MOMENT
ARM.
4.50
3.00
VOLUME
MOMENT.
Distance from I, to the resultant, We+W.=
800000+836000_____1636000
233.0
2065.0
76.7
81.3
1960.0
II.2
905.0
5332.2
Distance from I to c g=
Weight of earth per ten feet of wall=1915X100=191500 lbs.
Static moment about I, for section 10'-0" long=
191500X4.37=836000 ft. lbs.
VOLUME
MOMENT.
7920.0
465.0
8385.0
=5.'51.
104925+191500 296425
The horizontal thrust, H, on a wall 25'-0" high is 10417 lbs. per
lineal foot, see Table 1, page 146. The thrust on a ten foot length
would be 104170 lbs. Combining this thrust with We+ We, we obtain
the final resultant R, which has a value of 314000 lbs., and cuts the
base at a distance of 2.'93 from the line We+ We, or 8.'44 from the
point I.
165
Soil Pressure-The weight of the wall and earth prism per lineal
foot is 29642 lbs. The average soil pressure equals
29642
=2120 lbs.
14
By the methods illustrated in Problem 2, the maximum soil pressure
is found to be 3430 lbs. per sq. ft., and the minimum 810 lbs. per sq. ft.
The maximum and minimum soil pressure may be readily computed by
means of the following formula:
p=max. or min. soil pressure per sq. ft.
W=total load per lineal foot on soil, in this case Wo+We.
e eccentricity of R-distance from center of base to point where resultant
cuts under side of base.
l-width of base.
Stability Against Sliding-With a coefficient of friction of .5 the
resistance to sliding, per lineal foot of wall, would be .5X29642=
14821 lbs. The horizontal thrust per foot of wall is 10417 lbs. A
projection will, however, be provided at the heel, which will serve the
additional purpose of giving sufficient anchorage to the buttress
reinforcement
Reinforcement Required-Face wall. The face wall will be
designed as a continuous horizontal beam, and reinforcing bars will
be placed near the rear face at the abutments and extending to the
quarter points of the span. The thickness of the wall is 12" and d=10″.
I
Bending moments will be figured by the formula M———wľ².
12
Depth.
h.
W
бе
p= (1+6e).
仕
​ι
The following table gives the bending moment and the amount of
reinforcement required at various depths:
2'
6'
10'
12'
14'
16'
18'
20'
22'
Unit
Pressure.
67 lbs.
200 lbs.
333 lbs.
in which
400 lbs.
467 lbs.
533 lbs.
600 lbs.
667 lbs.
733 lbs.
•
Bend. Mom.
100P.
M= X12
12
6700 in. lbs.
20000 in. lbs.
33300 in. lbs.
40000 in. lbs.
46700 in. lbs.
53300 in. lbs.
60000 in. lbs.
66700 in. lbs.
73300 in. lbs.
q
M
.86 sd.
.05 sq. in.
.14 sq. in.
.23 sq. in.
.28 sq. in.
.33 sq. in.
.38 sq. in.
.42 sq. in.
.47 sq. in.
.52 sq. in.
Reinforcement.
Size and Spacing.
½" Corr. Rounds, 12″
½" Corr. Rounds, 12"
½" Corr. Rounds, 10"
"Corr. Rounds, 8"
½" Corr. Rounds, 7″
½" Corr. Rounds, 6"
" Corr. Rounds, 5½"
2" Corr. Rounds, 5"
"Corr. Rounds, 4½"
Vertical bars in front face; ½" corr. rounds, 2′-0″ on centers.
•

166
Base-That part of the base back of the face wall will be designed
as a continuous horizontal beam, and made strong enough to lift the
earth resting upon it as well as its own weight. It must also be
designed to resist the upward reaction of the earth when this exceeds
the weight of the materials above the section considered.
Longitudinal reinforcement in bottom of slab, in rear of face wall.
Owing to the continuous action moments will be figured on the basis
I wl2, and top reinforcement will be provided extending to the
12
For a section 12″
quarter points. The load per sq. foot is 2600 lbs.
wide,
3430
M=
4-0"
12220
K
12
d=21" and using a working stress of 16500 lbs. per sq. in.
q=.88 sq. in.
Use %" corrugated rounds 8″ on centers, top and bottom.
2,08
I
12/1214
I2
2680
w?²====X2600X10X12=260000 in. Ibs.
2
10=0"
DESIGN OF TOE.
810
is:
The bending moment in the base
at the vertical plane through B,
12220X2.08X12=305000 in. lbs.
If we neglect the fillet and take
d=21", the amount of reinforcement
required is 1.02 sq. in. per foot.

Use 1" corrugated rounds spaced
9″ on centers; make 2/3 of the bars
7'-0" long, and the remainder 13′-0″
long.
Buttresses-The buttresses act as vertical beams attached to the
base by the inclined reinforcing bars. The total stress in the steel
necessary to anchor the buttress is obtained by dividing the overturn-
ing moment (due to the thrust of the earth above the plane A-D on
ten lineal feet of wall) by the lever arm of the steel, which may be
taken as .86 Dr.
167
н
K
B
Figure 12.
9
From Table 1, page 146, the bend-
ing moment due to a fill of 23′-0″ is
811133-inch lbs. per foot of wall; for
one buttress the moment would be
10X811133=8111330 inch lbs. The
area of metal required at an allowed
stress of 16500 lbs. per square inch,
would be:

8111330
.86 X 96 X 16500
5.07 sq. in.
This amount of reinforcement will not be required for the full
height of the buttress; for example, the bending moment at a point
15'-0" below the top is 2250000 in. lbs., the q required is 2.03 sq. in.
Use 6-1%" corrugated rounds, two of which extend to the top of
the buttress, two to a point 5'-0" below the top, and two to a point
15'-0" below the top.
It is necessary to have some vertical reinforcement to transfer the
load from the base plate to the buttress; this will be as shown on the
detailed drawing.
The face wall should also be tied to the buttress by means of
horizontal bars as shown. These bars should have sufficient area to
transfer the horizontal thrust to the buttress without assistance from
the concrete, and should be hooked over the horizontal bars in the
face wall.
168
//
23
"
5 SP. @.12
1/2 CORR ROUNDS,
HORIZONTAL BARS -
SP.@ 10
@ 81
SP. 5 SP. 4 SP. 43P. 3SP. 35P.
@ 7
@5/2 @ 6
@4/2
FO
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•
VERTICALS
HORIZONTAL BARS REAR FACE
TTTT
=======
==
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BUTTRESS TYPE
==
Life Dama venta ef þ
D—— A ŞOU ŞUN CON
DO MERES CHICA CORA Čo — aus olla pass wa
+++
mama uma caušļa servo cacaan doa mo sa da se fa
La casa pa glass vases made van — « A
==
HORIZONTAL SECTION
1 - - soft cam c — A D A QU
**
CO COM QUÉ
jams pando C
Take a man
=* K Ca Scat sa ak
gaman sa
----more
SA A De fa co co je za d
P
NOTE-
ALL REINFORCING BARS ARE HIGH
ELASTIC LIMIT CORRUGATED ROUNDS.
DETAILS
RETAINING WALL
25-0" HIGH
2 cm my saken grant parts get man pat bus gerai pak sa DA MORAS SEEN BEDRA YOON METER MOUNT
PROBLEM 3.


109
USEFUL INFORMATION.
COEFFICIENTS OF FRICTION.
The following table gives average values of the coefficients of
friction. The values are subject to wide variation in individual cases:
Masonry on dry clay.
Masonry on wet clay.
Masonry on masonry
•
.50
· .33
.60
Brick work on brick work. .5 to .7
Timber on timber
Timber on stone
.2 to .5
.4
ANGLE OF REPOSE, 9.
The natural slope of most materials used for filling purposes is
roughly 1½ to I, corresponding to an angle of 34° with the horizontal.
With wet sand may become as small as 15°.
i
WEIGHTS OF MATERIALS.
Pounds per Cubic Foot.
Sand, dry...
Sand, wet.
Earth, loose..
Earth, rammed.
Gravel ..
Rock concrete..
Granite or lime stone masonry, well dressed
•
•
•
90-110 lbs.
110-120 lbs.
75-90 lbs.
100 lbs.
120-135 lbs.
150 lbs.
165 lbs.
•
#
170
INDEX, OCTOBER NUMBER.
Advantages of Reinforced Concrete Construction..
Buttress Walls-Detailed Design....
Cantilever Walls, Detailed Designs-
Problem I..
Problem 2....
Diagram of Forces Acting on a Retaining Wall-
Surface of Fill Horizontal...
Surface of Fill not Horizontal.
Earth Pressure-Discussion
Earth Pressures, etc., Tables 1 and 2
•
Formulæ for Use in Designing.
Notes on Design
Specifications for Materials.
Useful Information
•
•
•
• •
•
•
•
Page
143
164
•
• •
153
159
148
149
144
146
151
147
150
170
See page 38 (Bulletin No. 1) for Index on General Principles of Design, Building
Construction.
See page 76 (Bulletin No. 2) for Index on Detailed Design of Typical Building
and Complete Analysis of the Strength of Rectangular and T-shaped Beams.
See page 107 (Bulletin No. 3) for Index on Methods of Design and Standard
Loadings for Highway Bridges.
See page 139 (Bulletin No. 4) for Index on Standard Designs for Bridges and
Culverts for Highway Traffic.
171

Designing Methods
Reinforced Concrete Construction
VOL. 1
DECEMBER, 1908
RESERVOIRS
Discussion of the special problems met with in the de-
sign of reservoirs, with some remarks on this type of
structure.
Detailed design of rectangular reservoir, with vertical
walls backed with earth fill; reservoir divided into two
compartments by a partition wall.
BULLETIN No. 7, VOL. 1, WILL TREAT OF
CONDUITS AND SEWERS.
THIS NUMBER WILL BE PUBLISHED IN MARCH, 1909.
No. 6
CORRUGATED BAR COMPANY
(FORMERLY EXPANDED METAL AND CORRUGATED BAR CO.)
National Bank of Commerce Building
SAINT LOUIS

COPYRIGHT, 1908,
BY
Corrugated BAR COMPANY
4
REINFORCED CONCRETE RESERVOIRS.
The exterior walls of reservoirs must be designed as dams to resist
the lateral pressure of the water and also as retaining walls, capable
of holding the earth backing, should the reservoir be built in
excavation or the side walls be banked with earth. Reservoir walls
are then practically retaining walls, and a reinforced concrete design
will be superior to a heavy masonry or plain concrete structure.
In fact, the features which make reinforced concrete so desirable for
the construction of retaining walls make it even more desirable for the
construction of reservoir walls. Briefly summarized, the advantages
of a reinforced concrete design over these others are as follows:
/
I.
Greater stability at less cost.
2.
Lower and more uniform soil pressure.
3. The practicability of reinforcing against
temperature stresses.
4. The possibility of using such sections as
could not be constructed of the other ma-
terials generally used.
For the construction of the roof of covered reservoirs, especially
when an earth fill is to be placed upon it, reinforced concrete, owing
to its permanence, is superior to all other usual materials, as the ele-
ments of decay and corrosion are most active under such conditions.
Reservoirs of rectangular shape may be
Types of Reinforced
roughly divided into two classes; first, those
Concrete Reservoirs.
with vertical side walls; second, those with
sloping exterior walls. We are not considering those structures in
which the reinforced concrete construction merely acts as a lining for
the earth embankment.
TYPE I.
VERTICAL WALLS.-This type of reservoir is usually
constructed partly in excavation—the excavated material is generally
used as a backing for the walls. The backing must not, however, be
relied upon to assist in maintaining the stability of the wall. It is
possible that the fill may exert no active pressure on the wall; in fact,
when the fill is mainly clay it may even shrink away from the back of
the wall. (Such materials as sand or other granular substances with-
out cohesion between the particles, will, of course, exert an active
pressure.) The walls then should be designed as retaining walls to
hold back the earth when the reservoir is empty, and must also be
figured for the water pressure, neglecting such horizontal earth pres-
sure as may exist.
With this type of wall it is necessary to prevent the development
of an hydraulic head under the base. Adequate provision should also
be made against failure by sliding on the foundation and a projection
extending into the soil should be provided in all cases.
175
TYPE 2. SLOPING WALLS.-This type of wall, which is illustrated
in Figure 1, is desirable and economical under some conditions.
The deck B-C is carried on the
buttresses or walls B-C-D, which
in turn may be carried on a con-
tinuous footing or rest on isolated
footings. When a rock foundation
is available the buttresses are built
directly upon same and the apron
A-B let into the rock a suitable
depth. With independent footings
A
Figure 1.
a small leakage would not endanger the stability of the wall and
could easily be taken care of. This type of structure has been
extensively used for dams and has proven entirely successful.

B
D
There are various methods of making concrete
water-tight, using the term in a general way, but it
should be remembered that the thoroughness of the
mixing and the method of placing the concrete are most important
factors. The use of graded mixtures, the introduction of special
waterproofing materials into the mix, and surface treatment are three
well-defined methods.
Water-Tight
Concrete.
In regard to the use of graded mixtures, it may be stated that the
densest concrete that can be made with the given materials will be the
most impervious to water. The richer the mix the more nearly water-
proof will the concrete be. For practical reasons it is evident that an
excess of fine aggregate is required in concrete for this class of work.
Surface treatment may consist in plastering the wall, immediately
after the removal of the forms if possible, with a coat of cement and
sand mortar. This coating is usually from one-half to three-quarters of
an inch thick, and lime paste is sometimes added for smoothness in
working. It is essential that the surface of the coating be well
trowelled, as the hard skin developed by such treatment is practically
waterproof. A cement wash is sometimes used, applied with a brush.
On horizontal or inclined surfaces the surface of the concrete should be
worked and trowelled, giving the effect of a
a cement coating.
Another method of surface treatment is the use of asphalt of tar-
product mixtures used alone or in combination with felt or similar
waterproofing fabrics. In addition, there are various proprietary
waterproofing compounds of more or less merit. A standard way of
waterproofing is the use of an inch layer of asphalt poured hot.
between the face of the main wall and a thin protecting wall of brick
or concrete built in front of it and carried up as the work progresses.
176
The concrete walls may be made water-
tight by any of the methods outlined in the
preceding paragraphs. Very frequently no
special provisions are made except that of special care in the mixing
and placing of the concrete. The degree of water-tightness to be
attained governs the method to be adopted.
The design of the floor is a most important item in reservoir
construction. For small reservoirs it is desirable to put in the floor
monolithic with the walls and reinforce it in both directions for tem-
perature and shrinkage cracks. In larger work it is customary to
place the floor upon a layer of clay puddle--the concrete being sepa-
rated into blocks for its full depth and the joints run with hot asphalt
filler.
The Waterproofing
of Reservoirs.
SPECIFICATIONS FOR MATERIALS.
Reinforcing Steel-All reinforcing steel used in retaining wall
construction shall be rolled to such form that it has a positive
mechanical bond with the concrete. Adhesive bond will not be con-
sidered sufficiently reliable for this class of structures.
Steel may be made by either the Bessemer or open-hearth process;
bars to be rolled from billet stock. Re-rolled material will be accepted
under conditions insuring rigid inspection.
The elastic limit and percentage of elongation shall be determined.
by tests on accurately-machined specimens, and shall conform to the
following requirements:
Elastic limit to be from 50,000 to 60,000 pounds per square inch,
ultimate strength not less than 1Xelastic limit.
The percentage of elongation in 8" must not be less than given
by the formula :
Percentage of elongation=
1,400,000
Ult. Strength
5.
Bending Test-Bars as rolled shall bend cold, 90 degrees, to a
radius equal to three times the least diameter of the specimen, without
sign of fracture.
Cement-Only Portland cement conforming to the requirements.
of the specifications adopted by the American Society for Testing
Materials, June 14, 1904, shall be used.
Sand-Sand to be clean and coarse, and free from organic matter;
a graded sand, with coarse grains predominating, is to be preferred.
Coarse Aggregate-Broken stone to be hard, durable limestone,
or its equivalent, free from dust and foreign materials; maximum-sized
particles to pass through a one-inch ring; fines to be removed by
177
passing over a one-quarter inch screen. The fines may replace part
of the sand. Under special conditions making for uniformity crusher
run may be used, but this is not desirable.
Gravel shall be clean and of graded sizes; the sand carried to be
removed by screening as for broken stone.
Proportions of Mix-Concrete for wall, base, and buttresses to be
mixed in the proportion of one part cement to six parts aggrègate;
proportions by volume taking one bag containing not less than 94
pounds of cement, equal to one cubic foot of cement.
The proportions of fine and coarse aggregate used shall be chosen
so as to give a concrete of maximum density; in no case, however,
may the amount of fine aggregate be less than 50 per cent of the
coarse.
FORMULAE FOR USE IN DESIGNING.
All formulæ for the strength of reinforced concrete beams are
necessarily based on the physical properties of the concrete and steel.
The formulæ proposed in this bulletin are based upon a concrete with
a compressive strength of 2,000 pounds per square inch and upon the
use of mechanical bond reinforcing bars which have an elastic limit
of 50,000 pounds per square inch; the critical amount of reinforce-
ment, under these conditions, being .0085 bd.
ULTIMATE
STRENGTH
Let M=moment of resistance of section in inch lbs.
M。 ultimate moment of resistance of section
in inch lbs.

р
q=area of reinforcement.
p percentage of steel-g÷bd.
Felastic limit of steel=50,000 lbs.
b
All dimensions as shown, in inches, and stresses in lbs. per sq. in.
Then we have for rectangular beams:
Mo 370 bd2 for g=.0085 bd
(1)
Mo=.86 Fp bi =.86X50,000 Xp ba.
.(2)
=43,000 p bď, using high elastic limit corrugated bars.
(For complete discussion of these formulæ, see May and June bulletins)
and for various percentages;
•
WORKING STRESSES.
If it is desired to design for working stresses in the steel, use the
formula:
M=sqX.86 d, where s=unit stress in the steel..(3)
178
NOTES AND SUGGESTIONS ON DESIGN.
In the detailed design which follows we have endeavored to bring
out the important points, but the following disconnected notes may
have a suggestive value.
•
Reservoir walls when banked with earth must be designed to resist
the maximum thrust that the fill may exert, assuming the reservoir
empty. It is possible, however, that at times the earth may not
exert any active pressure on the wall and may even shrink away from
the back face. The wall must therefore have sufficient stability to
resist the water pressure, assuming the fill to exert no lateral pressure.
Reservoir walls and retaining walls have failed by moving bodily
forward while preserving their perpendicular position, and all designs
should be investigated for frictional stability against sliding. The
coefficient of friction of concrete on wet soil is low and a projection
or toe extending into the soil should be used in all cases.
The reservoir site should be drained, not only to preserve the
bearing power of the soil but also to prevent the development of an
hydraulic head outside of the reservoir which might destroy the floor
when the reservoir is emptied.
Longitudinal reinforcement should be provided in all walls, even
if expansion joints are introduced. It is usually considered that four-
tenths of one per cent of reinforcement, using high elastic limit steel,
will be sufficient to prevent cracks.
The maximum soil pressure should be computed in all cases, as it
is often the controlling factor in the design of the base.
179
The following table gives the intensity of the horizontal pressure,
p, at any depth, h, the total pressure H, above the section considered,
and the overturning moment, M, in inch lbs., at the section A-B:
.
1
- 1
h
، ܐ ܕ ܐ ܐ ܕ
feet.
1 2 3 ◄ LO
1
4
5
ON
6
78
9
10
11
12
13
14
15
16
17
18
20
21
22
EARTII, HORIZONTAL SURface.
w=100 lbs.
*****
23
24
25
26
27
28
29
30
ان
}
[
H=
% wh2
=3
pounds. pounds.
p=3 wh
33 29
67 SA
100 35
133/4
167/3
200/
233:
267 22
300: 7
3338
367
400-5
433
467 400
TABLE I. Lwin
yard
500+2
5335
56748
600
633 13
667, *.
700 le
733617
1
- 767
800
833 775
1000€
86774+
900772
933 8 01/
967 83b
-67 5
150/2
17/4
267 2
417-367
замен
P
H
600
817 7
1067 9/5
1350/
166754
514
A
/
}
案
​2017
2400
2817
3267** (>
3750
42675
4817
5400
6017
66672
7350'
8067 €
88177565
9600 7
×
10417 X
112677
12150 Pas
13067
14017
15000 (287
2-
1
13
2
wo
1
Overturning
Moment
M='s wh³×12
inch pounds.
6754
533
1800/54
42673647
3333 7/40
14400
22867
34133 24
48600- //
6666757
887337612
115200
146467
182933
225000/
273067
*
P
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हू
====
811133
921600
1041667
1171733
1312200
1463467
1625933 $
1800000, 51-
}
1
11
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feet.
h
}
32753317
388800 335%
18
L
457267 3 19
5333337 20
617400 21
7098677
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22
23
..
>
1
2
CT A CON
3
4
5
6
7
8
9
1
10
11
12
13
14
15
16
- 24
< 25
· 26
***27
28
29
1 30
2465/tal
FLUID PREssure.
w=62.5 lbs.


A
TABLE II.
p=wh H=2ph
pounds. pounds.
31
125
281
500
781
1125
1531
2000
2531
3125
3781
4500
5281
6125
7031
8000
9031
62.5
125.0
187.5
250.0
312.5
375.0
437.5
500.0
562.5
625.0
687.5
750.0
812.5
875.0
937.5
1000.0
1062.5
1125.0
1187.5
1250.0
1312.5
1375.0
1437.5
1500.0
1562.5
1625.0 -21125
1687.5
22781
1750,0
24500
1812.5
26281
1875.0
28125
H
10125
11281
12500
13781
15125
16531
18000
19531
h/3
Overturning
Moment
M=4Hh
inch pounds.
124
1000
3372
8000
15620
27000
42868
64000
91116
125000
166364
216000
274612
343000
421860
512000
614108
729000
857356
1000000
1157604
1331000
1520852
1728000
1953100
2197000
2460348
2744000
3048596
3375000
•

180
K
DETAILED DESIGN OF RECTANGULAR RESERVOIR
Problem.-Design a rectangular reservoir, with vertical sides, 200
feet long and 100 feet wide, and divided into two compartments by a
partition wall. Distance from floor of reservoir to top of walls to
´be 20' 0"; maximum depth of water, 18′ o"; distance from floor of
reservoir to average grade of ground to be 10 o". Earth to be
banked against walls as shown. Working stress in the steel to be
16000 lbs. per ".
/
4-0"

12:0
0-8/
CLAY PUDDLE
DRAIN
6-0"
1-6″
7-0"
Jos de
17:0
R=
*
F
Wc+ We
2:0
6.45'
GRADE LINE
H= 94075
11:0"
8.51'
3
18
Forces shown are for a section of wall 12′ 6″ long.
The resultant R does not include water pressure.
2.7
7.08'
P
20:0″
12 1-9
•
We will assume the section shown herewith and investigate its
stability and strength. Since the buttresses are 12′ 6″ on centers this
length of wall will be considered in the design.
181
Center of Gravity of Concrete Section, moments taken about P.
Table refers to a section of wall 12′ 6″ long.
SECTION
CONSIDERED.
Coping.
Face Wall..
(Above B. C.)
•
Buttress.
Fillet.
Base...
Projection.
•
•
Totals....
SECTION
CONSIDERED.
VOLUME
CU. FT.
HDEFG..
HEFG..
18.7
212.5
(Over Buttress.)
Totals...
138.0
I 19.0
371.8
18.7
878.7
7838.6
878.7
VOLUME
CU. FT.
MOMENT
ARM.
. Distance from P to cg=
Weight of section 12′ 6″ long=878.7X150=131800 lbs.
Static moment about P=131800X8.92=1,175,000 ft. lbs.
2135
118
10'.25
10.30
2253
6'.84
II.25
8'.50
o'.75
We will now determine the weight and the position of the center
of gravity of the earth prism, HCDEFG, resting upon the base of
the wall, as we will consider the vertical pressure due to the earth in
the design.
-8.'92.
MOMENT
ARM.
VOLUME
MOMENT.
5'. II
3.37
192.2
2188.2
943.9
1340.0
3160.3
14.0
7838.6

=5.02 ft.
VOLUME
MOMENT.
10910
398
11308
11308
Distance from P to cg=
2253
Weight of earth prism, 12′ 6″ long=2253X100=225300 lbs.
Static moment about P=225300X5.02=1,130,800 ft. lbs.
The distance from P to the center of gravity of the combined
earth and concrete sections
1175000+1130800___2305800
131800+225300 357100
-6.'45.
Having determined the weight of the section and the position of
the center of gravity, we will proceed to investigate its stability under
the pressure of the earth, assuming the reservoir to be empty, and also
when subjected to the pressure of the water.
182
I
•
I. STABILITY AGAINST EARTH PRESSURE.
We will assume that the intensity of the horizontal pressure
exerted by the earth against the wall at any point is equal to one-
third of the vertical pressure at that point; also that the resultant
horizontal pressure acts at a point, h, above the base.
(For a discussion of earth pressure against retaining walls, see
Bulletin No. 5).
In order to simplify the calculation, we will assume that the sur-
face of the fill is horizontal and of indefinite extent, at an elevation
of six inches below the top of the wall. Distance from under side of
base to top of fill is 21'-3″.
The horizontal thrust, H, for a fill of 21'-3", taking the weight of
the fill as 100 lbs. per cu. ft.=}{} wh²=
The thrust on a section 12′-6″ long would be
7526X12.5=94075 lbs. acting at a distance of 7.08 ft. above
the under side of base.
Combining this thrust with the weight of the combined earth and
concrete section (W.+ W.), we obtain the final resultant R, which has
a value of 369000 lbs. and cuts the under side of the base at a distance
of 8.31 ft. from the point P.
The soil pressure under these conditions would be practically
uniform, as the resultant passes very near to the center of the base.
Average soil pressure per sq. foot then equals:
357100
12.5 X 17
=1680 lbs.
The section will evidently be stable under the assumed forces, and
we will determine the amount of reinforcement required in the various
parts.
Face Wall-The face wall will be designed as a continuous hor-
izontal beam; bending moments will be computed by the formula
M= === wr².
I
12
The following table gives the bending moments and the amount of
reinforcement required in the front face of the wall at various depths:
1
X100X21.25²=7526 lbs. per lineal foot.

Depth.
h.
5'
10'
12'
14'
16'
Unit
Pressure.
167 lbs.
333 lbs.
400 lbs.
467 lbs.
533 lbs.
Bending
Moment.
26100 in. lbs.
52000 in. lbs.
62500 in. lbs.
73000 in. lbs.
83500 in. lbs.
g=
M
.86sd.
.22 sq. in.
.37 sq. in.
.41 sq. in.
.45 sq. in.
.50 sq. in.
Reinforcement.
Size and Spacing.
5%" Corr. Rounds, 12″ cts.
5%" Corr. Rounds, 10" cts.
5%" Corr. Rounds, 9" cts.
5%" Corr. Rounds, 8" cts.
5" Corr. Rounds, 7" cts.
183
i
The vertical bars in the front face, and the back face, carry the
shear due to the weight of the water on the toe into the face wall, and
will be 5/8" corr. rounds spaced 24″ cts. in each face.
DESIGN OF BASE.
Rear Part-That part of the base back of the face wall will be
designed as a continuous horizontal beam between buttresses, and
will be made strong enough to lift the earth resting upon it in addi-
tion to its own weight. It must also be designed to resist the upward
reaction of the earth should this exceed the weight of the material
above the section considered.
A
The load per square foot =2220 lbs.
For a section 12" wide
A
Toe-The toe acts as a cantilever beam carrying the reaction of
the earth beyond the face wall.
I
M= = wr² = 1 X 2220X12.52×12=347000 in. lbs.
I
2012
12
12
=1.4 sq. in.
Use 1" corr. rounds=7" cts. in bottom of slab.
I
P
B
3:01
6:0"
347000
d=18" and 9.86X18X16000
B
11:0"
Soil pressure is 1680 lbs., and
taking the average weight of the con-
crete toe as 300 lbs. per sq. ft. we
have, taking moments, about B,
M=(1680—300)6X3X12=298000

Figure 4.
=1.14 sq. in.
Use 1" corr. ronnds, spaced 12" cts. Make one-third of the
bars 16'-0" long and the remainder 12'-0" long.
Figure 5.
in. lbs.
Taking d=19" we have
298000
9¯¯¯.86×19×16000
DESIGN OF BUTTRESSES.

Buttresses-The buttresses act as
vertical beams attached to the base by
the inclined reinforcing bars. The
total stress in the steel necessary to
anchor the buttress is obtained by
dividing the overturning moment by
the lever arm of the steel, which may
be taken as .86 or.
184
}
From Table I., page 180, the bending moment due to a fill of 19'-6″
will be found by interpolation to be 495300 in. lbs. per foot of wall;
for one buttress the moment would be 12.5X495300=6190000 in. lbs.
6190000
9.86X100 X16000
4.5 sq. in.
Use six I" corr. rounds in rear of buttress. Extend two bars the
full height, two to a point five feet below the top, and the remaining
two to a point twelve feet below the top.
It is necessary to have some vertical reinforcement to transfer the
load from the base plate to the buttress; this will be as shown on the
detailed drawing.
The face wall should also be tied to the buttress by means of
horizontal bars as shown. These bars should have sufficient area to
transfer the horizontal thrust to the buttress without assistance from
the concrete, and should be hooked over the horizontal bars in the
face wall.
185
II. STABILITY AGAINST WATER PRESSURE.
This investigation will be based on the following assumptions:
I, That the back fill is in place before the reservoir is filled. 2. That
the fill exerts a vertical pressure on the base but may not exert an
active horizontal pressure on the face wall. 3. That the weight of
the water resting on the toe assists in maintaining equilibrium.
4. That adequate drainage will be provided so that an hydraulic head
cannot develop under the base.

12-0"
H=128125*
K
Ww-81800 #
6-0"
H
*0068€/
186
#
Wc+ We + Ww =
Wc+We = 357100
(62.5×12.5×6) (16.0+18.8
2
R• 457000#
11-02
7:85
6:45
14=0″
11-0"
~5.60
*
4140
The resultant R does not include any lateral pressure due to fill.
The weight of water resting on the toe, considering a section of
wall 12'-6" long, will be:
P
=81800 lbs.=Ww.
6;/
1
We will consider that the resultant water pressure acts at a distance
of 14'-0" from P.
The position of (We+ We+Ww) may be found as follows:
357100X 6.45-2304000
81800X14.00=1145200
438900
The horizontal pressure per foot of wall, due to a depth of water
of 18′-0″, is 10125. See Table 2, page 180.
For a 12'-6" length of wall, H=12.5×10125=128125 lbs.
Combining H and (We+ We+ Ww), we find R-457000 lbs., which
cuts the under side of the base at a distance of 5.60 ft. from P.
Since the resultant practically cuts the third point of the base, the
soil pressure at the heel will be twice the average, while that at the toe
will be zero.
1.
Depth
(from top
of wall.)
5'
10'
12'
14'
16'
3449200
438900
438900
17 X 12.5
Solid pressure at heel-4140 lbs. per sq. ft.
NOTE-Any horizontal pressure that may be exerted by the earth backing
will add to the stability of the wall. Whether or not an active pressure is exerted
against the wall depends on the nature of the fill and other conditions-in any
event, it should not be considered in the design.
pear
фило
wita tu
Average soil pressure=
Unit
Pressure.
*
sa?
{
}
3449200
=7.85 ft.
Bending
Moment.
187.5 lbs.
500.0 lbs.
29200 in. lbs.
78200 in. lbs.
625.0 lbs. 97800 in. lbs.
750.0.lbs. 117500 in. lbs.
875.0 lbs. 137000 in. lbs.
Ľ
M
Q .86sd.
C
pr
b
P
++
A
Fi-Fal
Face Wall-The face wall will be designed as a continuous hori-
zontal beam; bending moments will be computed by the formula,
M=2w12.
ا :
ܝܝ ܬ
=2070 lbs.
0.26 sq. in.
.58 sq. in.
.68 sq. in.
.71 sq. in.
.85 sq. in.
L
way
بشر
$
meri
ALU
J4 - L
Reinforcement.
1.
R
f
น
The following table gives the bending moments on a twelve-inch
strip of wall at the depths shown; also the amount of horizontal
reinforcement required in the rear face.
**
anf
ܐ
Ch
t
1 Lad
6-24
Size and Spacing.
+= +6 Fo
;
C
"/
the bene
the b
2 L
centi
ت
5%" Corr. Rounds, 12 cts.
5%" Corr. Rounds, 6" cts.
5%" Corr. Rounds, 5%" cts.
5/8
5%" Corr. Rounds, 5 cts.
5%" Corr. Rounds, 4½" cts.
"/
A
pom
187
DESIGN OF BASE.
Rear Part-That part of the base back of the face wall should be
designed as a continuous horizontal beam between buttresses to resist
the upward reaction of the soil minus the weight of the material
directly above it. The weight of the soil and base is (19.5×100+
1.75X150)=2112 lbs. The maximum upward reaction is at the heel
and is equal to 4140-2112=2028 lbs. This loading would require
the same amount of reinforcement as was used in the under side of
the base when designed to lift the fill resting upon it. In any event
since moments have been figured at w/ the amount of reinforcement
in the top of the slab should not be less than in the bottom.
" corr. rounds-7" cts. in top of slab.
We will use
I
Toe-The toe will be made suffi-
ciently strong to lift the water resting
upon it.
Taking moments about B, we have
M=6500X3X12=234000 in. lbs.
Assuming an effective d of 37″
234000
.86X 37X16000
.46 sq. in.
Use 34" corr. rounds-12" cts.
placed as shown.
Longitudinal reinforcement in toe will consist of 1" corrugated
rounds as shown.
A
6500lbs

3′0″
6:0
B
ཤ་
H=128125 lbs
A
11-0"
6=0"
B C
|
DESIGN OF BUTTRESS.
D
11-0"
q=
q=
E
9224000
103 X 16000
The buttress must resist the
overturning moment due to
horizontal pressure of the water
on a 12'-6" length of wall.

Taking moments about r in
the plane of D—E, we have
M=128125X6X12=9224000
inch lbs.
The reinforcing steel would
be placed vertically in the front
face of the buttress and the lever
arm may be taken as .86 of the
distance or=
.86X120"=103"
5.6 sq. inches.
L
"
188
Use seven I" corr. rounds.
corr. rounds. Extend three of these bars to the top of
the wall; two to within five feet of the top and stop off the remaining
bars at a point ten feet below the top.
In order to provide sufficient anchorage for these bars and to make
the fillet available for imbedment purposes, seven special U-shaped
3/4" corr. round stirrups will be placed in the base at each buttress,
as shown on the drawings.
STABILITY AGAINST SLIDING ON THE BASE.
Considering no active horizontal pressure of the earth, the force
tending to move the wall laterally is 10125 lbs. per foot of wall.
The vertical load per foot of wall==
=35100 lbs.
With a coefficient of friction of .33, for concrete on wet clay, the
frictional resistance to sliding is 35100X.33=11700 lbs., which is
greater than the lateral force tending to move the wall. The projection
at the heel gives additional security against sliding and also affords
sufficient depth for the imbedment of the inclined reinforcing bars at
the back of the buttress.
438900
12.5
TEMPERATURE STRESSES.
3
10
With thin walls it is usually considered that % to 1% of 1% of
reinforcement will prevent cracks. The face walls are reinforced with
longitudinal bars in both faces with an excess of metal in the upper
part of the wall. The longitudinal reinforcement should be sufficient
to take care of temperature stresses.
189
DIVISION WALL.
The top of the division wall will be 12" below the top of the side
walls and the under side of the bases of both walls will be at the same
elevation. This wall will be designed to withstand water pressure on
either side and will be symmetrical about a vertical center line.
We will assume the section shown herewith and investigate its
stability and strength.

H=10125*
M
b
A
18
*
Ww = 7380
1:69
4:60
n
We+Ww=20080
S
Blo
E
N
K
13700
=
Wc
R=22500
2=0"
16~0~
5.72
Forces shown are for one lineal foot of wall.
···
•
H
The weight of a section of wall one foot long will be determined.
Weight of DEFC=150 X 16X3.75,
2
4500 lbs.
Weight of BCFG=150X3.75X9.5=5350 lbs.
Weight of ABGH=150X1X6
Weight of KLIJ=150X2X1.5
=2400 lbs.
= 450 lbs.
Total ..
12700 lbs.
3-9
190
Center of Gravity and Weight of Prism of Water Resting
on Base-We will assume, for the purpose of computation, that the
water occupies the area MBCN.
The following table gives the information desired:
SECTION
CONSIDERED.
MnCb...
n NC...
Bb C.
Totals.
MOMENTS TAKEN ABOUT M.
VOLUME
CU. FT.
97.5
8.4
12.2
118.1
!
Distance from M to cg-
MOMENT
ARM.
7380X4.6
7380+12700
3.25'
6.87'
2.17
401.0
118.1
Weight of water per lineal foot=118.1X62.5=7380 lbs.=Ww.
F3.40 ft.
1.69 ft.
VOLUME
MOMENT.
To get the position of (We+ Ww), take moments about the center
line of the wall.
316.8
57.8
26.4
401.0
The horizontal pressure H, for one foot of wall for a depth of
water of 18'-0" is from Table 2, page 180, 10125 lbs., and is applied at
a point 6'-0" above the floor level, or 7.75 ft. above the footing.
SOIL PRESSURE.
Combining the horizontal thrust, H, with (We+Ww), we find the
resultant R to be 22500 lbs. and that it cuts the base at a distance of
3'.9 from the line, (We+W), or 5'.79 from the edge of the base;
5½ inches inside the third point.
Methods for finding the maximum and minimum soil pressures for
any position of the resultant are given in Bulletin No. 5. We will
compute these pressures, repeating the formula there given for con-
venient reference, although the position of the resultant shows that
the soil pressure at A is small and that at H about twice the average.
I L
The weight of the wall and the water on one side of the base is
20080 lbs. per lineal foot. The average soil pressure then is
20080+16=1255 lbs. per sq. ft.
191
Let:
W total vertical load per lineal foot, in this case We+Ww.
e eccentricity of R-distance from center of base to point where
resultant cuts under side of base.
l-width of base.
þ=maximum or minimum soil pressure per sq. foot.
Then
бе
p = W/ (1 + 60)
1
1
6X2.21
20080
16
16
Max. pressure=2295 lbs.; min. pressure=215 lbs.
It will be noted that the maximum soil pressure is much less than
that allowed. The resultant should, however, be kept within the middle
third in this type of structure, so as to insure pressure over the entire
footing.
Depths
Below
Top of
Wall.
It
For the case when both compartments are filled the resultant would
be vertical and pass through the center of the base. The average soil
pressure would be (7380+7380+12700)÷16=1720 lbs.
STABILITY AGAINST SLIDING.
The horizontal thrust per foot of wall is 10125 lbs. Using a co-
efficient of friction of 3, corresponding to concrete on wet clay, the
frictional resistance to sliding would be 20080X3=6693 lbs., leaving
an unbalanced thrust of 3432 lbs. The projection at the center of the
base is accordingly necessary.
REINFORCEMENT REQUIRED.
Wall-The wall proper will be designed as a vertical beam fixed to
the base; the maximum bending moment will occur at the plane C-F.
The following table gives the bending moments and the amount of
vertical reinforcement required at the depths given.
Bending
Moment.
=1255+1040
5'-0"
8000 in. lbs.
8'-0"
42868 in. lbs.
12'-0" 166364 in. lbs.
16'-0" 421860 in. lbs.
Thickness
of Wall.
17.5"
22.5"
29.3"
36.0"
M
q 13760d
Vertical Reinforcement.
.04 sq. in.
.15 sq. in.
.44 sq. in.
.92 sq. in.
Size and Spacing.
5%" Corr. Rounds, 16" cts.
5/8" Corr. Rounds, 16" cts.
5%" Corr. Rounds, 8" cts.
5%" Corr. Rounds, 4" cts.
Bending moments taken from Table 2, page 180.
~
;))
192

•
375
peop
215
A
6's
*Wr⋅ 7380*
4:60
BARS Ą
BARS C.
3.100
22500#
16-01
1450
q=
· BARS-B
6′50
H
2295
Moment due to weight of base=
360000
.86×16000X 52
Reinforcement in Upper
Faces-The base will be made
strong enough to lift the water
resting upon it in addition to its
own weight.
Total moment=360000 in. lbs.
BASE.
We will first determine the
bending moment in the plane
C-K, on a section twelve inches
long.
1830X2.17X12=47600 in. lbs.
975X3.25X12=38000 in. lbs,
85600 in. lbs.
Moment due to water=
7380X3.1X12=275000 in. lbs.
135200
.86X16000X36
F.50 sq. in.
We will also determine the bending moment and the amount of
reinforcement required at a plane 4'-0" from the center line of the wall.
Moment due to water (17.5X4X62.5)X24"-105000 in. lbs.
Moment due to concrete=( 2.1X4X150 ) X24":
30200 in. lbs.
135200 in. lbs.
.27 sq. inch.
The reinforcement in the upper side of the base will consist of 5%"
corr. rounds spaced 8″ on centers. Every other bar (Bars A) will
be continuous from B to G, and be bent at the center as shown. The
remaining bars (Bars B) will be straight and crossed at the center as
indicated.
REINFORCEMENT IN BOTTOM.
The transverse reinforcement (Bars C) in the bottom must be
proportioned to resist the moment due to the reaction of the soil.
The maximum moment will be in the vertical plane through F.
The diagram gives the soil pressure. The intensity of pressure
directly under the point F being 1450 lbs.
193
Upward moment due to reaction=
(1450X6.5)X3.25×12+(845×3.25)X4.33X12
=367000+143000 510000 in. lbs.
510000
=.68 sq. in.
.86X 54 X 16000
9
Use 34" corr. rounds spaced 8" on centers. Bars to be 13′-6″
long and placed with the ends of alternate bars near the same side
of the base.
LONGITUDINAL REINFORCEMENT.
Longitudinal bars in the wall section proper should be propor-
tioned for temperature stresses. The thicker the wall, however, the
less will be the average change in temperature. An amount of rein-
forcement equal to %% of the cross section should be sufficient for
the upper part of the wall. We will use 5/8" corr. rounds spaced 12″
cts. in each face.
The longitudinal reinforcement in the base will consist of 3/4" corr.
rounds spaced approximately 2'-0" apart, as shown.
FLOOR.
The floor of the reservoir will consist of a concrete slab of a
minimum thickness of seven inches and separated into blocks about
8'-3'x8'-3", by a joint extending the full depth. Joints to be ½"
wide and are to be filled with an asphalt filler.
The details of the junction of the floor construction with the side.
walls and division wall to be as shown on the drawings.
WATERPROOFING.
It will be desirable to back the reservoir walls with twelve to
fifteen inches of clay puddle. The division wall with the heavy rein-
forcement, both vertically and horizontally, should not develop cracks.
but may be given a surface treatment if desirable.
DRAINAGE.
The site should be drained and trenches containing drain tile
and filled with large broken stone well compacted, should be pro-
vided where indicated on the drawings.
}}}
194
||
Greife
o of
U
-0-8/
100
-
U
||
ㄱ
​PLAN OF RESERVOIR.
16.0
+
LONGITUDINAL SECTION_
100
-0
Mark Ang pang
1001
: ==|
=2]
11
|}
U
||
Q
n
n
X
SPAMMAL
F
100'- 0
Foot-our
TRANSVERSE SECTION.
GENERAL PLANS
TYPICAL
RESERVOIR



195
//
20-9
"
6 SP. @ 12
FRONT FACE-5/8" CORR ROUNDS
35A 7 33P. 8 * 3SP_ 9″ 65P @ 10
31/428. 12'675
SHO
ICR. AS SHOWNES
6 SP. @ 12'
VERT. BARS-7/8" CR-24 CTX BOTH FACES.
REAR FACE-51g"CORR ROUNDS.
55p. 52 9.50.@6Ž
1.6
9.9
Let me shot of th
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ས
№2-1°C R. -24-6 LG.
Berna
JÓN
ܡܕ
HORIZONTAL BARS-EACH
SIDE 1/2 GR. SPACED AS
SHOWN HOOKED OVER
BARS IN FACE WALL.
(2.1CR. 18-626.
P
4VERT, BARS
17:0″
CROSS SECTION
CR IN EACH
SIDE AS SHOWN
X2-16R.
11:6226
1½LR-X2 CTS EVERY 3RD
LONGITUDINAL BARS-
BAR 18-0 LG REST 120G)/"CR 7"CTS. TOP & BOTTON
6-0
8
24 TS
7-3/4 CR. STIRRUPS AT EACH BUTTRESS 18
"
11-0
ALP
121
__
G S
——
{{
S
HORIZ BARS HOR.BARS IN
IN FRONT
REAR FACE
FACE
12:6"
REAR ELEVATION
U
20
✔
=====
——||
Je me s
ܫ
ܐ
ܚ
ܫ
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mmg a
==
ܡܫ
C G
C 1 - -
7-1CR BARS AT EACH BUTTRESS
AS SHOWN •3@ 21-6"LONG 2@16:0 LG.
& 2 @ 11:0 LONG.
HORIZONTAL SECTION.

NOTE.
ALL REINFORCING BARS ARE
HIGH ELASTIC LIMIT CORRUGATED ROUNDS.
DETAILS
EXTERIOR WALL
TYPICAL
RESERVOIR
ܒ.

196

A
5/8 CR 8"CTS."
ALTERNATE BARS
BENT & LAPPED
AS SHOWN
74 CR-8"CTS.
13-6.LONG.
STAGGERED.
191
19к
2-0
16-0
A
HORIZONTAL BARS
5/8" C.R. 12" CTS. BOTH
FACES.
VERTICAL BARS• Folk.
PATION
AS SHOWN ON ELEVATION
❤
LONGITUDINAL BARS.
14C.R. AS SHOWN
.....
CROSS SECTION.
-.02
1-6712
+
TT
+-+
==
SIDE ELEVATION
HO
7/8 CR: 8075 5/8" C.R.-16 CTS.
VERT, BARS.- BCR. BOTH FACES.
alp
$18°C.R- 4″C75.
+
NOTE-
+
atman, ma
W
HORIZONTAL SECTION A-A
ALL REINFORCING BARS ARE
HIGH ELASTIC LIMIT CORRUGATED
ROUNDS.
Kata Bran
DETAILS
DIVISION WALL
TYPICAL
RESERVOIR
197
INDEX, DECEMBER NUMBER.
Advantages of Reinforced Concrete Construction.
Detailed Design of Rectangular Reservoir..
Earth Pressure-Table of
Formulæ for Use in Designing.
Notes and Suggestions on Design
Specifications for Materials......
Types of Reinforced Concrete Reservoirs
Water Pressure-Table of
Water-Tight Concrete...
•
• ·
•
•
•
Page
175
181
180
178
179
177
175
180
176
►
•
•
See page 37 (Bulletin No. 1) for Index on General Principles of Design,
Building Construction.
See page 76 (Bulletin No. 2) for Index on Detailed Design of Typical Building
and Complete Analysis of the Strength of Rectangular and T-shaped
Beams.
See page 107 (Bulletin No. 3) for Index on Methods of Design and Standard
Loadings for Highway Bridges.
See page 139 (Bulletin No. 4) for Index on Standard Designs for Bridges and
Culverts for Highway Traffic.
See page 171 (Bulletin No. 5). for Index on Retaining Walls of the Cantilever
and Buttress Types.
199
ش

Designing Methods
Reinforced Concrete Construction
VOL. I
MARCH, 1909
CONDUITS AND SEWERS
Discussion of the special problems met with in the
design of sewers and conduits, with some remarks on
this class of structures.
Detailed designs of an aqueduct, sewer section and
pressure pipe.
BULLETIN No. 8, VOL. 1, WILL TREAT OF
SEMI-CIRCULAR ARCHES
THIS NUMBER WILL BE PUBLISHED IN JUNE, 1909
FIRST ISSUE
No. 7
PRICE, 25 CENTS
CORRUGATED BAR COMPANY
(FORMERLY EXPANDED METAL AND CORRUGATED BAR CO.)
National Bank of Commerce Building
SAINT LOUIS
COPYRIGHT 1909, BY CORRUGATED BAR COMPANY
CONCRETE CONDUITS AND SEWERS.
Concrete has been extensively used in the construction of conduits
and sewers and its use is becoming more general. No one questions
its suitability for the construction of aqueducts, conduits, and storm
sewers, but doubt is sometimes expressed as to the ability of concrete
sewers to withstand the possible action of acid sewage. This is a
question that has been investigated by sewage experts and the adoption
of concrete for the construction of sewers in many large cities, in some
cases after exhaustive tests to determine its value as compared to brick
construction, shows the confidence that the engineers place in this
material.
"Cement Sewer Pipe,"
The following extract from an article on
by Rudolph Hering, may be of interest in this connection.
"It should be added that the acid question must be viewed in a
reasonable light. When the dilution by sewage is sufficient, the
discharge of a small amount of even strong acid will not cause
objectionable effects, as evidenced by European cities by the use of
concrete sewers almost exclusively in some cities, as Paris and Vienna.
In England concrete sewers are also very common.
""
Many American cities have adopted concrete or reinforced concrete.
for the construction of large and important sewers, and the following
list refers only to the more important works: New York, Jersey City,
Harrisburg, Wilmington, Cleveland, Duluth, New Orleans, Louisville,
and St. Louis.
DURABILITY OF CONCRETE INVERTS.*
"Concrete inverts have proved in practice to be equal, if not
superior, in durability to the best hard-burned brick.
"C
The hardness and smoothness of surface obtainable with concrete
reduce the friction to a minimum and render it less liable to erosion
than are other materials. Concrete sewers built at Duluth, Minnesota,
furnish a practical example of the ability of Portland cement mortar
to resist erosion. After twenty years of wear, they show no appreciable
deterioration or enlargement in diameter, while brick sewers laid at
the same time required rebuilding after six or seven years. A section
of the Duluth drains, about 2,000 feet long and 4 feet in diameter,
was built on a 13 per cent grade where the velocity of the water was
42 feet per second, with an invert of flat granite flags laid with I: I
Portland cement joints. The flow of water during heavy storms was
tremendous, carrying down with it quantities of sand and boulders,
but after two years of wear the invert showed ridges of mortar between
the granite flags, indicating that the Portland cement mortar was
more durable than the granite."
*This quotation is taken from “ Concrete, Plain and Reinforced," by Taylor
and Thompson, with the authors' permission.
202
In connection with the subject of acid sewage and the durability of
concrete inverts, it should be borne in mind that strong acids may
affect the brick work as well as the mortar in the joints. The smooth
surface and true lines that are easily possible with concrete con-
struction cannot be attained in brick work. As a result, the concrete
surface offers less obstructions to the flow and is less liable to erosion.
One striking advantage of reinforced concrete construction is the
possibility of successfully constructing sewers in poor ground without
the heavy foundation that would be required for a brick or masonry
sewer. By the use of longitudinal reinforcement the sewer section
itself may be transformed into a stiff beam capable of spanning soft
spots or even openings in the supporting soil.
Some of the advantages of reinforced concrete construction will
be given:
Concrete is not affected by ordinary sewage. A concrete sewer is
durable-the surface is smooth and strong, with minimum opportunity
for erosion.
A concrete sewer has a smaller friction coefficient than one of brick
or masonry, resulting in larger capacity for a given section.
Concrete can be shaped to exact curves and dimensions. This is
impossible with brick work.
A reinforced concrete sewer can be built, as a rule, more cheaply
than a brick or masonry one and the construction is more rapid.
A reinforced concrete conduit or sewer can be designed to with-
stand bursting pressure due to an hydraulic head.
Concrete sewers, even of large size, can be moulded in advance
and laid in sections when this method of construction is desirable
or economical.
Concrete sewers are usually built in place; pressure pipes and small
sewers, say those under three feet in diameter, are often laid in sections
which have been moulded in advance. This latter method of con-
struction is economical on account of the thin sections of concrete that
may be used, accurate forms and "shop methods" permitting the
adoption of thinner sections than could be constructed in place.
SPECIFICATIONS FOR MATERIALS.
Reinforcing Steel-All reinforcing steel shall be rolled to such
form that it has a positive mechanical bond with the concrete.
Adhesive bond will not be considered sufficiently reliable for this class
of structures.
Steel may be made by either the Bessemer or open-hearth process;
bars to be rolled from billet stock. Re-rolled material will be accepted
under conditions insuring rigid inspection.
+
203
The elastic limit and percentage of elongation shall be determined
by tests on accurately-machined specimens, and shall conform to the
following requirements:
Elastic limit to be from 50,000 to 60,000 pounds per square inch,
ultimate strength not less than 1% Xelastic limit.
The percentage of elongation in 8" must not be less than given
by the formula:
Percentage of elongation=
I,400,000
Ult. Strength 5.
Bending Test-Bars as rolled shall bend cold, 90 degrees, to a
radius equal to three times the least diameter of the specimen, without
sign of fracture.
Cement-Only Portland cement conforming to the requirements
of the specifications adopted by the American Society for Testing
Materials, June 14, 1904, shall be used.
Sand-Sand to be clean and coarse, and free from organic matter;
a graded sand, with coarse grains predominating, is to be preferred.
Coarse Aggregate-Broken stone to be hard, durable limestone,
or its equivalent, free from dust and foreign materials; maximum-sized
particles to pass through a one-inch ring; fines to be removed by
passing over a one-quarter inch screen. The fines may replace part
of the sand. Under special conditions making for uniformity crusher
run may be used, but this is not desirable.
Gravel shall be clean and of graded sizes; the sand carried to be
removed by screening as for broken stone.
FORMULAE FOR USE IN DESIGNING.
All formulæ for the strength of reinforced concrete beams are
necessarily based on the physical properties of the concrete and steel.
The formulæ proposed in this bulletin are based upon a concrete with
a compressive strength of 2,000 pounds per square inch and upon the
use of mechanical bond reinforcing bars which have an elastic limit
of 50,000 pounds per square inch; the critical amount of reinforce-
ment, under these conditions, being .0085 bd.
204
р
ULTIMATE STRENGTH
Let M=moment of resistance of section in inch lbs.
Mo-ultimate moment of resistance of section
in inch lbs.
g=area of reinforcement.
p-percentage of steel-g÷bd.
F-elastic limit of steel-50,000 lbs.

b
All dimensions as shown, in inches, and stresses in lbs. per sq. in.
Then we have for rectangular beams :
Mo=370 bd2 for g=.0085 bd..
and for various percentages:
M.=.86 Fp bd²=.86X50,000xp bd²
….(2)
=43,000 p bď, using high elastic limit corrugated bars.
(For complete discussion of these formulæ, see bulletins Nos. 1 and 2.)
I
… … … … (1)
•
WORKING STRESSES.
If it is desired to design for working stresses in the steel, use the
formula:
M=sqX.86 d, where s=unit stress in the steel.. (3)
NOTES ON DESIGN.
The design of sewer sections is attended with some difficulty, the
stresses developed are to a certain extent indeterminate, as the
pressures that may be exerted by the surrounding earth cannot be
accurately predicated either in direction or amount.
The above statement refers more particularly to circular or curved
sections where the stresses developed by the vertical loads are so
largely determined by the amount of the lateral pressure.
In the case of rectangular sections these conditions do not exist
and the section may be designed with reasonable accuracy.
Longitudinal reinforcement should be used in all cases, even if
expansion joints are provided.
The design must provide for such stresses as might be developed
from oblique or unequal pressures during construction,
The restraining action of the fill on the sides has a marked effect
on the strength of a curved section.
The back fill should be carefully tamped to place.
A uniform soil pressure is desirable with circular sections; the
stresses are decreased by insuring that the main support is not along
the bottom element of the section.
$
205
The following table, which was given in bulletin No. 6, is here
repeated for convenient reference. In Table II, p is the pressure per
square foot, corresponding to an hydraulic head, h:
EARTH, HORIZONTAL SURface.
w=100 lbs.
h p=% wh
feet.
1 2 3 4 LO
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
*****2.
A
27
28
29
30
33
67
100
133
167
200
233
267
300
333
367
pounds. pounds.
400
433
467
500
533
567
600
633
667
700
733
767
800
833
867
900
933
967
1000
TABLE I.
H=
% wh2
17
67
150
267
417
600
817 -
1067
1350
1667
2017
75/4
2400
2817
3267
3750
4267
4817
5400
6017
6667
7350
8067
8817
9600
10417
11267
12150
13067
14017
15000
B
Overturning
Moment
M= wh3X12
inch pounds.
67
533
1800
4267
8333
14400
22867
34133
48600
66667
88733
115200
146467
182933
225000
273067
327533
388800
457267
533333
617400
709867
811133
921600
1041667
1171733
1312200
1463467
1625933
1800000
h
feet.
1 2 3 4 5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
*********
26
27
28
29
30
FLUID PRESsure.
w=62.5 lbs.


A А
TABLE II.
p=wh H=2ph
pounds. pounds.
62.5
125.0
187.5
250.0
312.5
375.0
437.5
500.0
562.5
625.0
687.5
750.0
812.5
875.0
937.5
1000.0
1062.5
1125.0
1187.5
1250.0
1312.5
1375.0
1437.5
1500.0
1562.5
1625.0
1687.5
1750.0
1812.5
1875.0
31
125
281
500
781
1125
1531
2000
2531
3125
3781
4500
5281
6125
7031
8000
9031
10125
11281
12500
13781
15125
16531
18000
19531
21125
22781
24500
26281
28125
514
Overturning
Moment
M=4Hh
inch pounds.
124
1000
3372
8000
15620
27000
42868
64000
91116
125000
166364
216000
274612
343000
421860
512000
614108
729000
857356
1000000
1157604
1331000
1520852
1728000
1953100
2197000
2460348
2744000
3048596
3375000
p the intensity of the horizontal pressure at any depth h.
H=total horizontal pressure above A-B.
M the overturning moment in inch lbs. at seçtion A-B.
>
206
1
RECTANGULAR CONDUITS.
Rectangular Conduits of the open type, built on the hydraulic
gradient, either in cut or resting on the surface of the ground, may,
so far as the design of the section is concerned, be considered as
reservoirs; their design offers no problems that are not considered
in Bulletin No. 6, which covers this subject. The design of an
aqueduct or flume supported at intervals offers some interesting
features, and a detailed design will be given.
Problem 1-Design a rectangular aqueduct, 4' deep and 8′ wide;
supported on bents at an elevation of twelve feet in the clear above
grade. Depth of water 3'-6".
We will make the section of the aqueduct as shown in Figure 1.
The distance center to center of the supports will be 18'-0". Allowable
soil pressure will be taken as 3000 lbs. per □'.
8:0~
L
Bottom Slab The bottom
slab acts as a transverse beam
between the side girders. The
bending moment at the center
will be less than that for a free
beam, owing to the partial fixity
of the ends. The minimum
reverse moment at the ends is
equal to the overturning moment
due to the horizontal pressure of
the water on the sides and for a
strip 12" long=M..
447 ft. lbs.
The maximum moment at the center of the span would be Mc=%
w2-Me. Assuming the weight of the slab to be 50 lbs. w=50+62.5
X3.5=269 lbs. and M.-1705 ft. lbs.=20460 inch lbs.
COLUMNS -18:0″
CENTER TO CENTER
8:6″
Me
5-6′
3.5X62.5
2
12:0
The stress in the reinforcing
steel in the sides and bottom of
the aqueduct will be limited to
14000 lbs. per "-in other parts
of the structure to 18000 lbs.
per "

3.5X
3.5
3
207
Using the formula M=sqX.86d, we determine q as follows: Assume
d=4″
20460
9.86X4X14000
F.42 ".
We will make slab 5" thick and reinforce with ½" corrugated
rounds spaced 5½" on centers.
The reverse moment at the corners, Me=447X12=5364 in. lbs.
q=
5364
.86X4X14000
The moment at these points may be greater than Me owing to the
stiffness of the sides-we will therefore increase the amount of rein-
forcement as figured, and use 3%" corrugated rounds 3'-0" long spaced
5½" cts. in the top of slab.
Water...
Bottom Slab..
Side.....
Sides-The sides act as vertical cantilevers fixed to the base, and
also as girders between columns. The minimum thickness allowed
will be 5"-the section to have a flange at the top to give stiffness.
The overturning moment at the base equals Mo=5364 in. lbs. The
vertical reinforcement in the inside face will be 3%" corrugated rounds.
spaced 5½" on centers. The bent-up ends of the bars in the bottom
of the slab carry the moment around the corner.
•
Considered as Girder. The load per foot run on each side is:
3.5X62.5X4= 875 lbs.
50X4 200 lbs.
150(.6X4.5)= 405 lbs.
=1480 lbs.
q=
IIO".
•
•
Total...
The girders will be designed as continuous beams with the same
amount of reinforcement over the supports as at the center of span.
M=12 wl²=1480 X 324-480000 in. lbs.
480000
.86X48X14000
=.83 0″.
Use 4=½" corrugated squares; bend two, as shown on drawings.
*Shearing Provisions-The vertical shear at the end of the girder=
1480X9=13320 lbs. The average unit shear would be approximately
13320
5X8 55 lbs. per ".
The concrete itself should be capable of carrying 50 lbs. of vertical
shear per sq. in., and on this assumption no special provisions to carry
shearing stresses need be made. The vertical bars in the sides will
act as shear members, and in order to have some vertical steel in the
outside face every third bar will be put in as a U-shaped stirrup.
*For a discussion of Shearing Stresses, see Bulletin 1, page 11.
208
3
Temperature Stresses-It is usually considered that .35 to .40 of
one per cent of longitudinal reinforcement will take care of temperature
and shrinkage stresses. We will use .35 of 1%. The amount required
in the bottom slab per foot width=.0035X12X5=.21 0″.
We will use %" corrugated rounds 12" centers in top and bottom
of the slab. The sides will be similarly reinforced.
Expansion Joints-Every tenth bent will be a double one, and an
expansion joint consisting of a bent sheet lead plate made in the flume.
COLUMN DESIGN.
The columns will be designed to resist the overturning moment
due to a wind pressure of 30 lbs. per sq. ft. on the sides of the aque-
duct in addition to the vertical load.
The vertical load on each column=18X1480=26640 lbs. The
horizontal shear in each bent due to the wind is 18X4.7 X 30=2540
lbs.=H.
In the analysis for stresses due to wind action, the two columns
comprising a bent will be considered as fixed in direction (by the
brace beam) at the top and practically free at the bottom. (Such fixity
as is given by the bases of the columns will reduce the stresses
computed on a free-end assumption).

H=2540
12
8:0°
V4320
H/1270
at
P
ཆ
a
V. 4320
.0.01
The forces acting on a bent, due
to the wind load only, are shown in
Figure 2.
M2=1270
The horizontal shear in each col-
umn being taken=½ H.
The value of V is found by taking
moments about the foot of one col-
umn.
The maximum moment in the
column will be in the plane a-a at
the foot of the knee brace.
Figure 2.
M=1270X13.25X12=202000 in. lbs.
The critical section in the brace beam is at the plane b-b; taking
moments about the center line of the beam at this plane we find the
moment to be
2540 X 14.5
V=
8.5
4320 lbs.
M=1270X14.5-4320X1.25= 13015 ft. lbs.
=156180 in. lbs.
209
The moment in the brace beam at the center line of the bent is
zero, which may be shown by taking moments about the point o.
M=1270X 14.5
2540X14.5×8.5—0
8.5
We will make the columns 12"X12" and the brace beam 10'X12'
in section and determine the amount of reinforcement required to
resist the moments developed. The knee brace will of course have to
be similarly reinforced.
Column Reinforcement
·
202000
9.86X10X18000
1.30 □".
This amount of steel will be required in both the outside and
inside faces.
Column section, 12"X12" with 4-%" corrugated squares, one in
each corner-1/4" bands, 12" cts.
Ф
Brace Beam Reinforcement
156180
.86X10X18000
2
q=
1.01□".
Beam section 10"X12" with 4-3/4" corrugated squares.
The two bars in upper face to be bent down into the column as
shown in the drawings.
The reinforcement in the knee brace will be two 34" corrugated
squares.
NOTE-If the columns were fixed in direction at the base as well as at the
top a point of contra-flexure would have been developed at mid-height. The
bending moments would have been reduced by one-half, the effect of fixing the
base being equivalent to shortening the column by the distance from the base to
the point of contra-flexure.
Stresses in the Column
Weight of Structure and Water per column=26640 lbs.
Weight of Column=14X150
Compression Due to Wind
= 2100 lbs.
4320 lbs.
=33060 lbs.
Total Maximum
33060
12 X 12
Average Pressure=
The maximum unit compression that can occur in the column will
be on the plane a-a, where we have a combination of direct compres-
sion with the maximum compressive stress due to the bending moment.
With this low unit stress, it will not be necessary to investigate the
maximum stress on the extreme fibre.
=230 lbs. per □”.
Footings-The load on the footing is 33060 lbs., requiring II square
feet of bearing area at the allowed soil pressure.
Make footings 3′-4"X3'-4"X15" deep, and reinforce with 4'-½"
corrugated rounds each way.
210
EVERY THIRD BAR BENT 10 U SHAPE
FORMING VERTICAL REINFORCEMENT.
IN OUTSIDE TAGE
[61] +=====SUPER -
8:0
LONGITUDINAL CARS-7/8 CORR ROUNDS
AS SHOWNI
46 CORE POUNDS.
5/14°CTS.
34°
2
•He CORP ROUNDS $$'(75
EVERY THIRD BAB IN BOTTOM BENT
AS SHOWN
1
3/4 CORP SQ5
--
-
T
3/4" CORR 303 TOP & BOTTO
24/4 CORR 303
74 CORP 303 DENT
AS SHOWN
2-4 CORR.S03
M
DETAIL SHOWING
BRACE BEAM REINFORCEME
4 & CORR ROUNDS.
IN BOTH DIRECTIONS.
AS SHOWN.
TRANSVERSE SECTION
4 ¾4 CORR SQS, IN COLUMN TIED EVERY 12 WITH & ROUND
CA
34
THA MOŽN
15
M
+
-3/8 CORR ROUNDS.
3.4°
ਮਨ
++
1/2"CORR SQ5 2 BENT AS SHOWN
SHOWING VERTICAL AND HORIZONTAL
REINFORCEMENT IN INNER FACE OF
SIDE WALLS.
PART LONGITUDINAL SECTION
3/8 CORR BOUNDS
SHOWING VERTICAL AND HORIZONTAL
REINFORCEMENT IN OUTER FACE OF
SIDE WALLS.
18 0"
PART SIDE ELEVATION.
GENERAL NOTE
ALL REINFORCING BARS TO BE
HIGH ELASTIC-LIMIT CORRUGATED
ROUNDS AND SQUARES.
DETAILS
OF
AQUEDUCT
PROBLEM NO.I

211
Closed rectangular conduits under fill built on the hydraulic
gradient are culverts, and their design is fully discussed in Bulletin.
No. 3 which treats of Highway Bridges and Culverts. The reader is
referred to this bulletin in which will be found a detailed design of a
box culvert. Unless joints are provided this type of conduit must be
considered monolithic and reinforcement must be introduced to take
care of the reverse moments that may be developed.
Closed rectangular conduits working under head must of course be
designed for the internal pressure in addition to the external loads.
In the design no reliance should be placed on any possible restraining.
action due to the lateral pressure of the surrounding earth. For con-
ducting water under pressure the circular type of conduit will be found
preferable, excepting in special cases.
CIRCULAR SEWERS.
The stresses in circular sections due to the surrounding fill cannot
be accurately determined. Not only does the active pressure exerted
by the fill vary widely for different materials but the pressures due to
any given material change both in direction and amount with its
condition. The method of back filling, the thoroughness of the
tamping and the moisture content in the fill affect the pressures to a
marked degree. This is true not only of the lateral pressure but also
of the vertical.
The lateral restraining action of the fill due to its resistance to
displacement is another influence that largely affects the stresses that
may be developed under certain load conditions. The compactness of
the soil in the bottom of the trench is another factor. In hard soil or
rock the section might rest almost entirely on the bottom longitudinal
element, a condition that gives maximum stresses.
In masonry designs the minimum thickness for a given section is
usually determined by the judgment of the engineer. In reinforced
concrete sections the ratio of the thickness of the ring to the diameter
is usually such that the analysis, for the stresses developed in thin
rings subjected to external loads, may be applied in the design;
the section being reinforced so that it can resist the tensile stresses
due to the bending moments.
To assist the designer in determining the probable limits of the
stresses the following discussions are given. In general it may be
212
1
fly
stated that the stresses in a circular sewer are a maximum at the ends
of the vertical and horizontal diameters and decrease toward the 45°
points. Oblique pressures and other causes may change or even
reverse these conditions.
In the following equations the plus sign (+) will be used to indi-
cate positive moments, i. e., tension on the inside fibre, and the
negative sign (-) to indicate moments causing tension on the
outside fibre.
Q
ELASTIC RINGS UNDER EXTERNAL LOADS.
Case 1-Concentrated loads at opposite points, Figure 3.
W
by

る
​W
Figure 3.
Wr
2
Ma
Mb=
THE
Mp=
W r
2
☺
Wr
12
KAD PERS
A=50°-28′, cos=.6366.
c=Unit compressive stress due to total thrust, C, which is considered
uniformly distributed over the section.
t-thickness of the ring.
bar
୫
W r
Q
(cos &—cos`q)....
W-Concentrated load on unit
length of ring.
Ma-Bending moment at a.
Mь=Bending moment at b.
Mp=Bending moment at any
point subtended by the
angle ?.
C-Point of zero
moment or
contraflexure the radius
to which makes an angle
o with a-a.
· :—cos p)—(1—cos ()
]
2
(:6366-1)=.1817 Wr..
··(4)
(S) · · · ·
(.6366~0)=+.3183 Wr....... ….(6)
213

COEFFICIENTS OF Wr
(CONCENTRATED LOADS)
*
0.4
0.3
0.2
0.1
O
0.1
0.2
W
16
G
0.5+
0.4
90° 80° 70° 60° 50°40′30°··0°
VALUES OF $
Moment diagram for concentrated loads. To
determine the moment at any point multiply
Wr by the coefficient corresponding to the
value of given.
a 'load, w'.
W
C= cos from which
2
W
cos ?.
$....
21
In order to illustrate the
change in bending moment,
the following diagram
has been prepared. The
full line gives the coeffi-
cients of Wr, which deter-
mine moments, at any point
p, due to the vertical load
W the dotted line, the
moments in the same quad-
rant due to the horizontal
The moment
corresponding to a com-
bination of the two loads
Sp
may readily be obtained
by the aid of this diagram.
In figure 4, page 215, the thrust at a due to the concentrated load
W
Wis evidently and acts in a vertical direction, the thrust at b is
2
zero. At any point p, the vertical component of the direct thrust is
W
and
2
Compressive stresses
in the ring due to direct
thrust.
K
It is necessary to consider
these direct compressive
stresses in conjunction with
the stresses due to the bend-
ing moment when design-
ing the section. The total
thrust at any point should
be considered as uniformly
distributed over the section.
….(7)
· NOTE-This analysis follows that given in Structural Mechanics, by Charles
E. Greene.
214
*
Case 2-Uniform loads in opposite directions, Figure 4. The
intensity of the load is assumed uniform over the projection of the ring
normal to its direction.

a
Ᏹ
Figure 4.
Ө
0
Ma
1/4 w r²
Mb = + ¼/4 w r²
2
from which
a
че 22
το
2
w=Load per unit width of
projection for unit length of ring.
Other symbols as for case I.
The general equations for the
moment at any point, p, which
are obtained by taking moments
about are as follows:
Mp=
-Mb
W r
че
2
os ² =½ or 0 =
2
cos 20 1/4 20 g 2
2
sin²q—Ma . . . . . . . .(8)
του ν
We know from other considerations that Ma is numerically equal
to My but of opposite sign. We therefore obtain the following values:
45°
2
2
os 24...
To find the point of zero moment let Mp = zero in equation (9).
Mp = 0 = Mo
COS
•
(10)
(11)
w r
2
2
(9)
COS
20
NOTE-For full discussion of this subject, see Bulletin No. 22 of the Univer-
sity of Illinois Experiment Station.-Tests of Cast Iron and Reinforced Concrete
Culvert Pipe, by A. N. Talbot.
215
In the following diagram the full line gives the bending moment
at any point (p) corresponding to the value of (4) noted, due to the
vertical loads. The moments due to similar horizontal loads are shown
by the dashed line. An inspection of the diagram will show that for
equal values of (w) the bending moment is zero throughout the ring.
In this case there is a uniform compression in the ring, the value of
which is (wr)

10.3
0.2
COEFFICIENTS OF WI 2
LOADS)
(UNIFORM
0.1
O
-0.1
0.2
'0.3
18
90° 80° 70° 60° 50° 40° 30°--0
VALUES OF $
a
Moment Diagram for uniformly distributed loads. To determine the moment
at any point multiply (wr²) by the coefficient corresponding to the value of (Ø) given.
In the usual case the intensity of the horizontal pressure will be
much less than that of the vertical. Let (wy) represent the intensity of
the vertical forces and (wn) the intensity of the horizontal forces then the
bending moment at any point may be obtained by means of the equation.
че
2
My = (1-10) (M. - 2012 cos')
(12)
Wh
W
Mp=
I
Mb
2
216
It is to be noted in equation (12) that when wh-wv, Mp=0.
Direct compression in the ring due to thrust.
As before stated, it is necessary to consider both the bending
moment and direct thrust when determining the stresses at any section.
In figure 4, page 215, the direct thrust at a is evidently wr, and that
at bis zero. At any point p, the vertical component of the direct.
thrust would be, wr cos & and the tangential thrust would be
C=wr cos² &
ων
cos². &
t
…….(13)
When the ring is subjected to both horizontal and vertical loads,
uniformly distributed over the projections of the ring, formula (13)
may be used to determine the thrust. When both these pressures are
of equal amount there is no bending moment in the ring and the
thrust is uniform throughout; its value may be obtained from the
following formula, C=wr:
APPLICATION OF FORMULAE.
The analysis for thin rings applies with sufficient accuracy to
concrete sections where the thickness of the ring is usually from to
of the diameter.
They also apply to sections similar to those shown on page 220
where the base is flattened to give more uniform soil pressure.
O
Figure C.
Sections as shown by the accompanying diagram
must be individually analyzed. The lower half not
being so flexible as the upper half, exerts a restrain-
ing influence on it-the invert acting as
an arch
between abutments to the extent of this horizontal
restraint. The treatment properly comes under a
study of semi-circular arches, and will not be given
in this bulletin.

Problem 2-Design a circular sewer section, internal diameter
Io'-0" with a depth of fill of ten feet over the crown. Soil stiff dry clay.
Case I, no horizontal pressure. Case 2, with horizontal pressure.
Limit compressive stress in the concrete to 650 lbs. per ". Maximum ten-
sile stress in the steel to be 15,000 lbs. per sq. in.
217

}
MEMEME
| 10:0~
1150°
FIRI
0:01
Case 1-It will be necessary to
assume a thickness for the ring and
determine the stresses and reinforce-
ment required. We will make the
thickness of the ring % of the diame-
ter; the diameter of the center line of
the section will be 11'-o," and the
radius corresponding to r in the
formulæ, is 5'-6". The average depth
of fill will be taken as 12′-0″, and
the vertical pressure taken as uniform
and equal to 12X100=1200 lbs. per
', over a width of 11′-0″.
We will determine the stresses at the critical points, a and b, on
the assumption that the vertical pressures on the upper and lower half
are equal and that no side pressure is exerted.
Crown, Point b-We will consider a section one foot long.
With vertical loads only, there would be no direct thrust at the
points b, and we need only consider the bending moment.
Substituting in formula II, we have
I 200
X5.52
4
Mb =¼/ wr² =
9075 ft. lbs.
=108900 inch lbs.
108900
.85 0".
q, at inside face,:
Use 3/4" corrugated rounds spaced 6" on centers.
For vertical loads only the maximum positive moments occur at
the points b. Since the direction of the pressure on the sewer is
uncertain, the reinforcement will be made continuous.
.86X10X15000
Points a—The moment at a is equal to that at b, but of opposite
sign, requiring reinforcement in the outer face.
Use 3/4" corrugated rounds spaced 6″ on centers in the outer face.
Since we have the maximum direct compression at these points it
will be necessary to determine the extreme fiber stress, due to direct
compression and cross bending; which is developed, to see if it comes.
below the prescribed maximum.
The average pressure due to the fill=
1200X55
12 X 12
46 lbs. per sq. in.
218
The compressive stress on the extreme fibre due to flexure may be
determined as follows: The neutral axis in a reinforced concrete beam
will, under usual conditions, bed from the compression side. The
total compressive stress, C, in a strip 12" long, must equal the total
tensile stress T, and is eqnal to the bending moment divided by the
lever arm of the forces, .86 d.
.86d
T
C=T=
~12" |
M
.86d
108900
.86X 10
I
Figure 2.
The total stress on the extreme fibre=530X46=576 lbs.
NOTE-This method for finding the extreme fibre stress is sufficiently accurate,
but involves an error, since it does not take into account the displacement of the
neutral axis due to the direct compressive stress.
The accompanying diagram rep-
sents the conditions. If frepresents
the compressive stress per sq. in. on
the extreme fibre we may write
C 2
f=
сх
12650X2
12X4
=530 lbs.
.4d
The section assumed appears to meet the requirements of the design.
In order to obtain a uniform bearing on at least part of the bottom,
and avoid concentration of pressure along the bottom element, the
section will be as shown in the drawings. The increased thickness at
the base will not affect the stresses to any large extent.
533
Longitudinal Reinforcement-Use ½" corrugated rounds spaced
about 24″ centers in each face.
Substituting in formula 12, we have
:) (-
Ma
F12650 lbs.

Case 2-We will assume that the intensity of the active horizontal
pressure of the fill is % of the vertical at any point. The average
horizontal pressure would be at the level of the center of the sewer
and would be 3X16X100=533 pounds per square foot. Owing to the
lateral restraint, we will make the section 10" thick; r=5′-5."
I 200
I2
I 200
4
4900 ft. lbs.
=58800 inch lbs.
As before M-Ma, and the direct compression at a would be
50 lbs. per sq. inch, and that at b,
533X5.46
II X 12
=22 lbs. per sq. inch.
2
X 5.42
58800
.86X8X15000
·57 0".
Use 5%" corrugated rounds 6½" on centers.
Longitudinal Reinforcement-Same as in Case 1.
219

12
"
苏
​314 CORR ROUNDS 6
6 CAS.
100
3/4 CORR ROUNDS 6 CTS.
Dies
LONGITUDINAL BARS
1/2" CORR ROUNDS AS SHOWN
CASE I
▸
—
سلام
3
←
45
NOTE
12
:117
3011
„0:01
15"
10
누수
​D
5/8 CORR ROUNDS 61/2 CTS 518 CORR ROUNDS 6/2CTS
10:0
LONGITUDINALS BARS
1/2 CORR ROUNDS-AS SHOWN.
CASE 2
===
ALL BARS ARE HIGH ELASTIC LIMIT
CORRUGATED ROUNDS LAP HOOP BARS
30 DIAMETERS LAP LONGITUDINALS 25 DIAMETERS.
45
DETAILS
10
"/
10°
"
0:01
13
SEWER SECTIONS
PROBLEM NO.2

220
อ
PRESSURE PIPES.
Conduits for the purpose of carrying water under pressure are
usually built in the circular form. In this case the steel reinforcing
bands are figured to carry all the tensile stresses due to the pressure
of the water, the concrete being considered as a filling-in material.
This type of conduit has been used in the construction of a large
number of inverted siphons and pressure pipes and have been found
entirely satisfactory. In connection with the Salt River Project, the U.S.
Reclamation Service constructed about six thousand feet of reinforced
concrete pressure pipes. The maximum head to which these pipes
are subjected is about 75 feet, the internal diameter being 5'-3″.
The question of leakage is an important one, and pressure pipes
are usually given a finish coat of mortar to increase the water-tight-
ness. It has been found in practice that such small cracks as may
develop in pressure pipes will soon be closed up if silt-bearing water
is conveyed. In pressure pipe work, as in other reinforced concrete
construction, the water-proofness depends to a large extent upon the
workmanship. Low-unit stresses should be used in determining the
amount of hooping required and a sufficient amount of longitudinal
reinforcement should be used to take care of temperature and
shrinkage stresses.
}
In the construction there is a tendency for the concrete at the top
of the pipe to dry out. Special means must be provided to insure
that the concrete is wet throughout the circumference of the pipe, as
any unequal drying will have a tendency to cause longitudinal cracks.
It is desirable to keep the pipe moist until put in use. Mechanical
bond bars should be used in all pressure pipe work.
Problem 3-Design a pressure pipe 4'-0" inside diameter,
maximum head 40 feet to center line of pipe. Pipe to be laid in
a wet sandy soil, with a 3'-0" fill over the top.
The reinforcement will be placed in two layers-one near the outer
face and one near the inner face of the pipe, and will be proportioned
to resist the stresses caused by the water pressure and the external
loads. The direct tensile stress in the reinforcing steel is not constant,
owing to the fact that the water pressure is less at the top of the pipe
than at the bottom. In the design we will neglect this variation and
221
assume an average normal pressure equal to that at the center line of
the pipe.
The stress in the steel will be limited to 12,000 lbs. per
square inch, and the design based on the static pressure assuming that
suitable provisions will be made to prevent excessive stresses due to
water-hammer, etc.
We will assume a thickness of 4½" for the concrete shell and
determine the reinforcement required.
The average water pressure
For a section 12" long, the direct tensile stress in the shell should
be 2,500 X 2 = 5,000 lbs.
5,000
12,000
The unit tensile stress in a ring subjected to a uniform fluid pressure is equal
to unit pressure times the radius. Since the pipe under consideration is horizon-
tal, the pressure is not constant, and bending moments will be developed in addition
to the direct tension.
9
Stresses Due to Fill-We will consider that the vertical pressure
due to the fill is uniform over the width and corresponds to a depth
of 4'-0"
Mb
40 X 625 2,500 lbs. per sq. foot.
Wv = 100X4400 lbs. per sq. ft.
With a soil such as is described, the intensity of the horizontal
pressure may be taken as % of the vertical at any point.
Ma
Taking the depth to the center line as five feet, the average horizon-
tal pressure would be
1/3 X 100 X 5 = 167 lbs. per sq. ft.
Substituting in formula 12, we have
Wh
W₂ ) 1/4
τον
"
.42 □″ per foot of pipe
9
του
22
w
I
M
400
3264
.86x32X12,000
167
100X2.212)
.09"
=272 ft. lbs.
=3264 inch lbs.
This amount of steel is required in both the inner and outer faces.
The total amount of steel per foot of pipe would be
.42 + (2 x .09) = .60 "
222
อ
We will use 3%" corrugated squares spaced 6" centers in each face.
This reinforcement may be put in as spirally wound bars or individual
hoops. The spiral method is to be preferred. If individual hoops
are used, lap bars at least 30 diameters and stagger laps around the
circumference of the pipe.
Longitudinal Reinforcement-This will be shown on the draw-
ing, corresponding to approximately 4% of reinforcement.
ม
4호
​TAMAN MEMAN
GRADE
Ni
PLASTER FINISH
SAND CEMENT MORTAR
12
31/8 cope SQUARES & "CTS.
PIB CORR SQUARES
AS SHOWN
PROPOSED SECTION ALTERNATE SECTION
3:0″
RAIRIEN
NOTE
ALL REINFORCING BARS
ARE HIGH ELASTIC LIMIT CORRUGATED
SQUARES.
-
USE ALTERNATE SECTION
WHERE IT IS NOT POSSIBLE TO
FORM THE PIPE TO ACCURATE
DIMENSIONS.
DETAILS
PRESSURE PIPE
PROBLEM NO 3
P

223
OLD STYLE,
TYPE A.
C to Ci
of
Bars!
TABLE SHOWING THE NECESSARY SPACING OF DIFFERENT SIZES OF CORRUGATED BARS
FOR A GIVEN AREA OF STEEL PER FOOT WIDTH OF SLAB.
½"
3/4"
8" 1"
11"
יים |יים |יים ייסויים
21.082.223.304.206.43
C to Cl
of
Bars
9″ 0.24 0.50 0.73 0.93 1.43
92" 0.23 0.47 0.69 0.88 1.35
10" 0.22 0.44 0.66 0.84 1.28
11/0.200.40 0.60 0.76 1.17
12/0.18 0.370.55 0.701.07
CORRUGATED SQUARES,
TYPES B AND D.
3/4" V½" 1" 11"
3/8" 1/2" 5/8" 34" 8" 1"
1"
\" \ "\/2″ | 1/2" |
ロッ
​"0" 0" 0" 0" 0" 0"0"
O"0"
CORRUGATED FLATS,
UNIVERSAL TYPE.
JC to Cl
of No. 1 No. 2 No. 3 No. 4 No. 5 No 6
Bars
ם
יים
יים ייםייםיים יים
CORRUGATED ROUNDS,
TYPE C.
!
C to C
1/2" | 5%
}
%"
of 3/8" 12" 5%"″ 34″ %″ 1″
Bars
}
T
1
{
i
1
}
1
}
{
1
I
1
T
1
20 36 0.66 0.84 1.50 2.34 3.36 4.62 6.00 9.37) 2" 1.14 1.92 2.46 3.24 3 90 4.80 20.66 1.18 1.84 2.65 3.614.715.967.36
Į
2½"0.861.782.65 3.36 5.14 22" 0.29 0.530.67 1.20 1.87 2.69 3.70 4.80 7.50 22" 0.911.54 1.97 2.59 3.11 3.84 22" 0.53 0.94 1.47,2.12,2 89 3.774.77 5.89
3" 0.76 1.28 1.64 2.162.60 3 20 30 44 0.79 1.23 1.772.413.143.984.91
30.72 1.48 2.20 2.80 4.28 3" 0.24 0.44 0.56 1.00 1.562.243.084.00 6.24
32" 0.62 1.27 1.89 2.40 3.67 32" 0.21 0.38 0.48 0.86 1.34 1.92 2.64 3.43 5.3632" 0.65 1.10 1.411 85 2.23 2.74 32" 0.380 67,1.05 1.512.06 2.69 3.41 4.21
4″ 0.541.111.652.103.21 4" 0.18 0.33 0.42 0.75 1.17 1.68 2.313.00 4.68 4' 0.57 0.96 1.23 1.621.95 2.40 40 33 0.59'0.92 1.33 1.80 2 36 2.983.68
Į ľ
42" 0.480.99 1.47 1.86 2.85 42" 0.16 0.29 0.37 0.67 1.04 1.49 2.05 2.67 4.16 42" 0.51 0.85 1.09 1.44 1.73 2.18 42" 0.29 0.52 0.82 1.18 1.602.09 2.65 3.27
I
I
50.43 0.891.321.682.57 50.140.26 0.34 0.60 0.94 1.34 1.852.403.75 5″ 0.46 0.770 981 301.55 1 92 50.26 0.47 0.741 06 1.44 1.88 2.392 95
52" 0.390 811.201.52 2.34 52" 0.13 0.24 0.31 0.55 0.85 1.221.68 2.18 3.41 52" 0.41 0.70 0.89 1.18 1.41 1.7452" 0.24 0.430 67 0.96 1.31 1.712 17 2.68
6" 0.36 0.74 1.10 1.40 2.14 6" 0.12 0.220.28 0.50 0.78 1.11 1.53 2.00 3.12 6" 0.38 0.64 0.821.081.30 1.60 6'0 22 0.39 0.61 0.881.201.57 1.99 2 45
62" 0.33 0.68 1.02 1.29 1.97 62 0.11 0.20 0.26 0.46 0.72 1.03 1.42 1.85 2.88 62" 0.35 0.59 0.761:00 1.20 1.47620 200 36,0.57 0.82 1.11 1.45 1.84 2 27
7″0.31 0.63 0.941.20 1.83 70.100.190.24 0.43 0.67 0.96 1.32 1.72 2.68 7″ 0.33 0.55 0.70 0.93 1.11 1.37 7" 0.19 0.34 0.53,0.76 1.03 1.35 1.70 2.10
Į
72" 0.29 0.59 0.88 1.121.71 72" 0.100.180.220.40 0.62 0.89 1.23 1.60 2.50 72" 0.30 0.51 0.66 0.86 1.04 1.2872" 0.18 0.31 0.490 710.96 1.26 1.591.96
I
8" 0.27 0.55 0.82 1.05 1.60 8" 0.09 0.17 0.21 0.38 0.59 0.84 1.15 1.50 2.34 8″ 0 280.48 0.62 0.81 0.97 1.20 80.17 0.29 0.16 0.
8"0.17 0.29 0.46 0.660 901.18 1.49 1.84
8" 0.25 0.52 0.77 0.99 1.5182" 0.08 0.160 20 0.35 0.55 0.79 1.09 1.42 2.20 82" 0.27 0.45 0.58 0.76 0.92 1.1382" 0.16 0.28 0.43 0.62 0.85 1.11 1.401.73
T
90.08 0.15 0.190.33 0.52 0.75 1.02 1.33 2.08 9" 0.25 0.43 0:55 0.72 0.87 1.07 9'0.15 0.26 0.41 0.59 0.80 1.05 1.331.64
92" 0.08 0.14 0.18 0.320.49 0.71 0.97 1.26 1.97 92" 0.24 0.40 0.52 0.68 0.821.0192" 0.14 0.25 0.39 0.560 76 0.99 1.26 1.55
! Į
10" 0.07 0.13 0.17 0.30 0.47 0.67 0.92 1.20 1.87 10" 0.23 0.38 0.49 0.65 0.78,0 96 10"0.13 0.24 0.37 0.53 0.72 0.94 1.19 1.47
|
11"0.07 0.120.150.27 0.43 0.61 0.84 1.09 1.70 11" 0.21 0.35 0.45 0.59 0.71 0.87 11 0.12 0.21 0.33 0.48 0.660.861.08 1.34
12" 0.06 0.11 0.14 0.25 0.39 0.56 0.771.00 1.56 120.190.320.41 0.54 0.65 0.80 120.110.19 0.30 0.44 0.600 780.991.22
40.60 0.78 0.99
}
}
1
I
1
1
Net
Net
Net
Net
06/0.11/0
560,71.00
19/0.820.410.540.650
180.
Sec- 0.18 0.37 0.55 0.70 1.07 Sec- 0.06 0.11 0.14 0.25 0.39 0.56 0,771.00 1.56 Sec- 0.19 0.32 0.41 0.54 0.65 0.80 Sec- 0.11 0.19 0.30 0.44 0.60 0.780.99 1.22
tion
tion
tion
tion

1/13/
יים ים ים ים ים ים
T
I
f
11" 11"
יים ייםיים
224
ร
STANDARD SIZES AND TYPES OF CORRUGATED BARS.
Corrugated Bars are Rolled to the Sizes and Weights Given Below.
Unless otherwise specified, we furnish corrugated bars rolled from a high
elastic limit steel, with a yield point of 50,000 pounds or over. We are prepared,
however, to roll any grade of steel desired, and can supply corrugated bars to
meet special requirements as to elastic limit, from open hearth or Bessemer stock.
OLD STYLE-TYPE A.

SIZE
NET
SECTION
WEIGHT
PER FOOT
SIZE
NET
SECTION
WEIGHT
PER FOOT
1/2"
.180"
.64 lb.
!
3/4"
.370"
1.35 lbs.
SIZE
NET SECTION
WEIGHT PER FOOT
7/8"
.550"
1.95 lbs.
NEW STYLE-TYPE B.
1"
.700"
2.70 lbs.
一
​3/4"
1/4" 1/3" 3/8" 1/2" 5%"
7/8"
11/4"
.060".110" .140" .250" .390" .560" .770" 1.000" 1.560"
14"
1.070"
1
4.00 lbs.

.24lb. .38 lb. .48 h..85 lb. 1.33 lbs. 1.91 lbs. 2.60 lbs. 3.40 lbs. 5.31 lbs.
CORRUGATED ROUNDS-TYPE C.
1″

SIZE
8
3/8" 1/2″ 9/16" 5/8" 3/4" %" 1″ 1%"
18
14'
NET SECTION .110".190" .250" .300" .440" .600" .780″.990" 1.220"
WEIGHT PER FOOT .38b..66 lb. .86 lb. 1.05lbs. 1.52 lbs. 2.06 lbs 2.69 lbs. 3 41 lbs. 4.21lbs.
CORRUGATED SQUARES-TYPE D.

TO
SIZE
3/8" 1/2" 5/8" 3/4" 7/8" 1″ 1%' 14"
1/4"
NET SECTION .060".140" .250″ .390″ .560″ .760″ 1.000" 1.260" 1.550"
WEIGHT PER FOOT .22 lb. .49 lb. .86 lb. 1.35 lbs. 1.94 lbs. 2.64 lbs: 3.43 lbs. 4.34 lbs. 5.35 lbs.
CORRUGATED FLATS-UNIVERSAL TYPE.

No. 6
No. 1 No. 2 No. 3 No. 4 No. 5
.190" .320"
.410" .540" .650" .800"
.73 lb. 1.18 lbs. 1.35 lbs. 1.97 lbs. 2.27 lbs. 2.85 lbs.
A variation in weight of 5% either way is required.
225
No attempt has been made in this bulletin to touch upon the
design of special sections or those larger sewers which must be de-
signed after the methods employed in arch analysis. The informa-
tion given is intended to cover the design of such sections as are
ordinarily constructed.
T
226
Advantages of Reinforced Concrete Construction....
Aqueduct Detailed Design-Problem 1.
ANALYSIS-
A
M
INDEX, MARCH NUMBER.
Elastic Ring Under Concentrated Loads.
Elastic Ring Under Uniform Loads..
Bar Spacing Table...
Circular Sewers..
• •
Detailed Design, Problem 2.
Corrugated Bars, Standard Sizes..
Durability of Concrete Inverts..
Earth Pressures-Table of.
Formulæ for Use in Designing.
Notes on Design..
Pressure Pipes....
•
Detailed Design, Problem 3.
Rectangular Conduits..
Specifications for Materials.
Water Pressures, Table of..
•
•
• •
•
•
Page
203
207
•
•
→
213
215
224
212
217
225
202
206
204
205
221
221
207
203
206
See page 37 (Bulletin No. 1) for Index on General Principles of Design,
Building Construction.
See page 76 (Bulletin No. 2) for Index on Detailed Design of Typical Building
and Complete Analysis of the Strength of Rectangular and T-shaped
Beams.
See page 107 (Bulletin No. 3) for Index on Methods of Design and Standard
Loadings for Highway Bridges.
\
See page 139 (Bulletin No. 4) for Index on Standard Designs for Bridges and
Culverts for Highway Traffic.
See page 171 (Bulletin No. 5) for Index on Retaining Walls of the Cantilever
and Buttress Types.
See page 199 (Bulletin No. 6) for Index on Reservoirs.
227

1
|
1
{
1
J
THE UNIVERSITY OF MICHIGAN
GRADUATE LIBRARY
MAY
MAV 71
DATE DUE

PRO
AN
UNIVERSITY OF MICHIGAN
\
3 9015 00889 4266
DO NOT REMOVE
OR
MUTILATE CARD
OINTED IN U. S. A.
Cat. No. 23 520