m WORKS BY L. A. WATERBURY PUBLISHED BY JOHN WILEY & SONS, Inc. Stresses in Structural Steel Angles with Special Tables v + 77 pages. 51 x 8. Illus- trated. $1.25 net. Laboratory Manual For the Use of Students in Testing Materials of Construc- tion. 12mo, ix + 270 pages, 68 figures. Cloth, $1.50 net. Cement Laboratory Manual A Manual of Instructions for the Use of Students in Cement Laboratory Practice. 12mo, vii + 122 pages, 28 figures. Cloth, $1.00. A Vest-pocket Handbook of Mathematics for Engineers with Tables Second Edition, Revised and En- larged, xii + 213 pages, 80 fig- ures. Morocco, $1.50 net. STRESSES IN STRUCTURAL STEEL ANGLES WITH SPECIAL TABLES BY L. A. WATERBURY, C.E. M. AM. Soc. C.E. PROFESSOR OF CIVIL AND ARCHITECTURAL ENGINEERING, UNIVERSITY OF ARIZONA FIRST. EDITION NEW YORK JOHN WILEY & SONS, INC. LONDON : CHAPMAN & HALL, LIMITED 1917 Copyright, 1917 BY L. A. WATERBURY PRESS OF BRAUNWORTH & CO. BOOK MANUFACTURERS BROOKLYI N. V. PREFACE ALTHOUGH steel angles are extensively used in structural framing, there are important factors affecting the stresses in such members that are not commonly considered in analyses which are made for purposes of design. This volume has been prepared in the hope that it may serve to indicate the nature of some of these factors, and that it may furnish the means for their consideration in practical problems. On account of lack of symmetry of the section, the product of inertia is involved in the computation of bending stresses, but the ordinary structural handbooks do not include values of this element. The author has computed the values of the products of inertia for commercial angles 2 by 2 inches and larger, which values are included hi Table I and Table II at the back of the book. The section modulus polygon, which is a very useful device for studying and for determining the flexural stresses in unsymmetrical sections, is explained, and for commercial angles 2 by 2 inches and larger coordinates for the vertices of the polygons are given in Table III. The subject of end connections and their efficiency is considered, and in Table IV and Table V are given a consid- erable number of values of efficiency, of equivalent effective area, and of total tension allowable for a maximum unit stress of 16,000 pounds per square inch, computed in accord- ance with the method which is outlined in the text. L. A. W. TUCSON, ARIZONA, January i, 1917. 359875 TABLE OF CONTENTS THEORY AND DISCUSSION ARTICLE PAGE 1. Relation between Bending Moment and Flexural Stress i 2. Expressions for the Section Modulus 2 3. Product of Inertia 5 4. Section Modulus Polygons 6 5. Neutral Axis 9 6. Plane of Loading 12 7. Combined Stresses 13 8. Flexure for Angles in Pairs 15 9. Transfer of Stress by Shear to an Outstanding Leg 17 10. Efficiency of End Connections 19 TABLES TABLE I. Elements of Angles with Equal Legs 31 n. Elements of Angles with Unequal Legs 34 III. Coordinates of Section Modulus Polygons for Angles 44 IV. Efficiency and Allowable Tension for Angles Riveted through One Leg to a Rigid Connection Plate with Two Lines of Rivets 50 V. Efficiency and Allowable Tension for Angles Riveted through One Leg to a Rigid Connection Plate with One Line of Rivets 60 v STRESSES IN STRUCTURAL STEEL ANGLES THEORY AND DISCUSSION Art. 1. Relation between Bending Moment and Flexural Stress * FOR a s^Tnmetrical section, with the plane of loading coin- cident with one of the principal axes of the section, the relation between the bending moment and the bending stress at an extreme fiber is in which /=unit stress at the extreme fiber, due to bending; M = bending moment at the section under consid- eration ; / = moment of inertia of the section; y = distance from extreme fiber to a line through the center of gravity of the section, the line being taken coincident with or parallel to the neutral axis of the section. The value of - is a constant for any given section and is ordinarily called the section modulus or section factor. If * For an excellent treatment of this subject, see " An Analysis of General Flexure in a Straight Bar of Uniform Cross Section," by Professor L. J. John- son, Trans. Am. Soc. C. E., Vol. LVI, p. 169, 1906. 2 . STRESSES IN STRUCTURAL STEEL ANGLES the letter s is used to designate the section modulus, Eq. (i) becomes >=? (2) I . If - is retained as the value of the section modulus. Eq. y (2) will be applicable only for those conditions which obtained for Eq. (i), viz., for symmetrical sections, with the planes of loading coincident with principal axes of the sections con- sidered. However, a more general value of the section mod- ulus can be derived, which, if used in Eq. (2), as the value of s, will make this equation applicable to any section, whether symmetrical or unsymmetrical, and for a plane of loading extending in any direction from the center of gravity of the section. The term section modulus, unless otherwise spe- cifically stated, will be used to designate the general value, and for any case of pure flexure the extreme fiber stress will then be equal to the bending moment at the section divided by the section modulus. Art. 2. Expressions for the Section Modulus For any given section subject to bending stresses, let the rectangular axes of reference, X-X and Y-Y, be taken with their origin at 0, the center of gravity of the section, as indicated in Fig. i. Let OP be the plane of loading and nn the neutral axis. (It is frequently assumed that the neutral axis is perpen- dicular to the plane of loading, but this is not true for the general case.) Then the general expression for the section modulus is cZA (I v sin BJ cos 6)y+(I x cos 0J sin 6)x (3) THEORY AND DISCUSSION 3 in which s = section modulus of the section for the partic- ular plane of loading; . 7 X = moment of inertia of the section about X-X; I v = moment of inertia of the section about F-F; / = product of inertia of the section for axes X-X, 7-7; x, y = coordinates of the extreme fiber; = angle from OX to OP (see Fig. i). For certain special cases Eq. (3) can be simplified. Thus, for any section which is symmetrical about either axis, J be- comes zero, and the expression for the section modulus becomes I v sin 0-y+Ix cos 6-x For a symmetrical section for which the plane of loading coincides with F-F, 6 becomes 90 degrees and the expression for the section modulus becomes ' ' ; -7 ........ (s) For a symmetrical section for which the plane of loading coincides with XX t 6 becomes zero and the expression for the section modulus becomes (6) In order to derive the expression for the section modulus which is given in Eq. (3), let the section be indicated by Fig. i, a/, y' being the coordinates of any infinitesimal area (dA), and a being the angle which the neutral axis (nn) makes with X-X. M is the bending moment for which the plane of action is OP, and the other quantities are as defined under Eq. ( 3 ). For a combination of bending and direct stress the neutral 4 STRESSES IN STRUCTURAL STEEL ANGLES axis will not pass through the center of gravity of the section, but since the resultant stress can in this case be obtained by combining the stresses obtained for pure flexure and for axial loading, the derivation of the section modulus will be made for the case of pure flexure, for which the neutral axis will pass through the center of gravity. For any point for which the coordinates are x 1 ', y', the distance of the point from the neutral axis is (/ cos a x' sin a). Therefore, if the stress is proportional to the distance from the neutral axis, f y f COSCE x' sin a , ^ f y cos a x sin a ' in which /' = unit stress at the point for which the coordinates area/, /; /=unit stress at a point for which the coordinates are x, y, or since x, y, are the coordinates the extreme fiber, / is the extreme fiber stress. The bending moment may be resolved into two com- ponents, one of which acts in the plane of OF and the other in the plane of OX. Also, for equilibrium, the component of the moment which acts in the plane of OF must be equal to the sum of the moments of the stresses about the axis X-X, and the component of the moment in the plane of OX must be equal to the sum of the moments of the stresses about the axis F-F, or using for any stress the value of /' obtained from Eq. (7), Msin 8=Cfy'dA = '- : Cy'(y'cosa-x'sma)dA, (8) y cos a x smaj and Mcos 0= Cf'x'dA == 1 | yf(y f C os a - x' sin a)dA . . (o) y cos a x sin a J THEORY AND DISCUSSION 5 Eqs. (8) and (9) m ay be simplified by substituting for C(y') 2 dA its equivalent I x , for J (x') 2 dA, its equivalent I v , and for (x'y'dA its equivalent /. Then, by equating the value of /obtained from Eq. (8) to the value obtained from Eq. (9), sin _ cos e , . I x cos a J sin a J cos a I y sin a from which By substituting the value of tan a in Eq. (9), solving for - and by substituting for its equivalent s, the value obtained for the section modulus is that which is given in Eq. (3). Art. 3. Product of Inertia The product of inertia (/), which is involved in the gen- eral equation for the section modulus, may be defined by the mathematical expression, J=f x'y'dA, (12) in which x', y', are the coordinates of any infinitesimal area (dA). Also J=J cg +Akk, (13) in which / = product of inertia for any given rectangular axes; J cg = product of inertia for parallel axes through the center of gravity; A = area of the section ; k, h = coordinates of the center of gravity with ref- erence to the given axes for which / is desired. STRESSES IN STRUCTURAL STEEL ANGLES For any symmetrical section J cg is zero, or for such a section / becomes Akh. The product of inertia may be either a positive or a negative quantity. For an angle the value of / will be positive for axes taken as shown at a and d of Fig. 2, and the value will be negative - for axes, as shown at b and c of the same figure. The numerical values of J for structural steel angles are given in the tables at the back of the book, to which the proper sign must be applied for each particular case. Art. 4. Section Modulus Polygons If, for any given section, the values of the section modulus are computed for values of d from o to 360 degrees, and if ab FIG. 2. FIG. 3. Types of lection Modulus Polygons. these values are plotted as radii vectors, the outline of the resulting figure will be a polygon, having a side for each salient angle of the polygon which bounds the section. For some common sections the general forms of the corresponding section modulus polygons are indicated in Fig. 3. THEORY AND DISCUSSION 7 Having the section modulus polygon for a given section, the value of the section modulus may be obtained by scaling the radius vector corresponding to the plane of loading. For example, for the rectangular section indicated at a of Fig. 3, the vector OK is the graphical representation of the section modulus for the plane of loading OP, and the bending stress at the vertex B is obtained by dividing the bending moment by this value of the section modulus. It will be observed that by using the section modulus polygon, not only is the required section modulus readily obtained, but also, the lines of the polygon which are intersected by the line of direc- tion of OP indicate the points of the section at which the critical bending stresses will occur. In order to plot the section modulus polygon for a par- ticular section, the coordinates of each vertex of the polygon may be determined, and the polygon can then be drawn by connecting each two successive vertices with a straight line. If x aj y a are the coordinates of a and x 6 , y b , the coordinates of b, two successive vertices of the polygon which bounds the section, the coordinates x ab , y ab of the vertex of the section modulus polygon corresponding to the side ab of the section will be, _ (x a - x b )J - (y a - y b )I ( Xa - Xb )I x -(y a -y b )J - If ab is parallel to X-X, J (16) 8 STRESSES IN STRUCTURAL STEEL ANGLES If ab is parallel to F-F. x I, Xa = J_ Xa (18) (19) cd de FIG. 4. An inspection of Eqs. (14) to (19) will show that for many cases the vertices of the section modulus polygon can be easily located, since the values of - and - can be obtained y x from handbooks, for structural sections, and since / is zero for symmetrical sections, which includes I-beams, channels, THEORY AND DISCUSSION 9 and tees. For angles, Eqs. (14) and (15) may be used, but in order that the users of this book may avoid this labor, the coordinates have been computed and are tabulated at the back of the book for the commercial sizes of steel angles, and for variations in thickness of J inch. For thicknesses in odd numbers of sixteenths of an inch, the values of the coordinates may be obtained by interpolation. A typical section modulus polygon for an angle is shown in Fig. 4. If OP represents the direction of the plane of loading, the two points at which critical stresses will occur are b and d of the angle, but if the plane of loading is shifted to OP' the points of critical stresses will be a and d. The most advantageous plane of loading is that for which there is the greatest sec- tion modulus. For an angle used as a purlin, for a vertical load, a study of Fig. 4 will show that the purlin should be set as indi- cated at a of Fig. 5, not as shown r IG. O. at b. It will be observed that for an angle in tension or com- pression, the application of the load at the gauge line is not far from the most advantageous point of loading. Art. 5. Neutral Axis For pure flexure the neutral axis will pass through the center of gravity of the section, assuming some direction, as for example nn (Fig. i), which direction will depend upon the plane of loading and upon the section, but the neutral axis will not necessarily be perpendicular to the plane of loading. In Art. 2, the relation of a to (see Eq. n), was obtained in the derivation of the expression for the section modulus. 10 STRESSES IN STRUCTURAL STEEL ANGLES This equation and two of its equivalent forms are: I X J tan 6 f x tana = - -, (20) J-Iy tan i x cot e-j f N tana = - , (21) /cot e-iy tan = y I x cos 6J sin /cos BIy sin 0* For a combination of flexure and direct loading the neutral axis will not pass through the center of gravity of the section, but it will be parallel to the position which it would have for pure flexure, and the distance between the two par- allel lines will be, _F (y cos a x sin a) , ^ A / in which v = distance from the neutral axis to the center of gravity; F = total stress acting normal to the section; A =area of the section; /"=unit stress at the extreme fiber due to flexure; x, y = coordinates of the extreme fiber; a = angle which the neutral axis makes with X-X. By the use of Eqs. (20) to (23) the position and the direc- tion of the neutral axis can be determined, but by the use of the section modulus polygon these computations can usually be avoided for practical problems. For example, consider .an I-beam, as indicated in Fig. 6. For a loading in the plane OP, the section modulus is represented by the line from to the vertex ab. Had the plane of loading intersected the sides a and c of the polygon, the critical stresses would have been at A and C of the beam, and had the plane of loading inter- THEORY AND DISCUSSION 11 ab sected the sides b and d of the polygon, the points of critical stress would have been B and Z>, but since the plane of loading passes through the vertex ab the bending stresses at A and B are equal, or the neutral axis must be parallel to the line joining A and B. Also, if the plane of loading, for pure flexure, is parallel to a side of the sec- tion modulus polygon, the neu- tral axis will intersect the corre- sponding vertex of the section. As another example, let the plane of loading be taken as OP' (Fig. 6). The lineOtf now da( represents the section modulus for the point B, and by pro- longing the side c to c', Oc f is obtained which is the section modulus for the point C of the beam. The bending stress at B is to the bending stress at C as Oc' is to Ob'. Since the stress is proportional to the distance from the neutral axis, the distance from C along the prolongation of the line join- ing B and C to the point N at which the neutral axis for pure flexure is intersected, will be, CN = BC (24) ov I To observe the conditions which obtain for the angle, consider the section and polygon shown in Fig. 4. It will be seen that when the plane of loading coincides with OG the section modulus for the side DE of the angle is represented by the line from to the vertex de of the polygon. Since the 12 STRESSES IN STRUCTURAL STEEL ANGLES plane of loading (OG) intersects the vertex de, the neutral axis must be parallel to DE of the section, as has been explained for the case of the I-beam. However, while for the I-beam the neutral axis happened to be perpendicular to the plane of loading, in the case of the angle such an assumption is not even approximately correct. A study of the angle and its polygon will show that if the plane of loading coincides with OF, the stresses at D and at E, for pure flexure, will be of opposite sign, since the lines d and e of the polygon cut the plane of loading on opposite sides of O; or, for pure flexure with OF as the plane of loading the neutral axis will inter- sect the side DE of the angle within the limits of the section. As the plane of loading shifts from OG to OF the change of sign for the stress at E will occur when the neutral axis passes through E, and the neutral axis will intersect E when the plane of loading is parallel to the side e of the polygon. Art. 6. Plane of Loading For those cases in which the member is restrained from bending in some direction, the determination of the equiva- lent plane of loading may require some study. If the char- acter of the restraint determines the direction in which bending must occur, then the position of the equivalent plane of loading can be determined by noting the direction of the neutral axis. Since the deformation is proportional to the distance from the neutral axis, the neutral axis must be parallel to the line joining any two points whose deformations are equal in amount and of the same sign. In such a case tan a becomes a known quantity, and the position of the equivalent plane of loading may be computed by the use of Eqs. (20) to (22). From Eq. (20) the corresponding equation for 6 becomes Ix /tana / N tane = (2S) THEORY AND DISCUSSION 13 For practical purposes the section modulus polygon may be used to determine the equivalent plane of loading, avoiding the necessity for using Eq. (25). For illustration, consider again the angle and its section modulus polygon, shown in Fig. 4. When the neutral axis is parallel to DE of the angle, the section modulus is represented by the line from to the vertex de, or if the angle is riveted to a connection plate along the face DE, the angle being restrained from rotation parallel to the plate, and if the point of application be taken at the face of the connection plate, then the equivalent point of application of the load is G. If the gauge line, and, there- fore, the actual point of attachment, is at H, the restraining couple which is called into action will have a moment arm equivalent to GH. It is sometimes thought that complete restraint parallel to the connection plate implies the exist- ence of a couple having a moment arm equivalent to MH, but this is not the case; complete restraint parallel to the connection plate will move the equivalent point of applica- tion from either M or H to G, or, rather, the equivalent plane of loading to the plane OG. When the neutral axis is parallel to DE it will also be parallel to BC, and hence, the plane of loading which includes O and the vertex de must also pass through the vertex be. Art. 7. Combined Stresses In determining the critical stresses for a combination of direct tension or compression with flexure, the usual methods may be used, substituting the correct value of the section modulus for the more commonly used special value. For a member which is short enough to be considered as a prism, >- '- 14 STRESSES IN STRUCTURAL STEEL ANGLES in which /=unit stress due to both flexure and direct load; F = total load or force normal to the section; A =area of the section; M = bending moment at the section; 5 = section modulus. If the bending moment (M) in the last equation is due to an eccentricity (e) of the force (F), then, (27) For a slender member, subject to both direct loading and bending, the following formula is applicable, and will be found convenient on account of its wide range of application: <-> KEz in which M = bending moment neglecting the increment of the moment which is due to the deflection of the member itself; L = length of the member (in inches if A is in square inches) ; E = coefficient of elasticity of the material; z= distance from the extreme fiber to a line through the center of gravity of the section, parallel to or coinciding with the neutral axis; K a constant, the value of which is 9.6 for a simple beam uniformly loaded and 12 for a simple beam with a load at the center, and for which the value 10 is suggested for general purposes of design, Other quantities in the last equation are as denned under Eq. (26). THEORY AND DISCUSSION 15 In Eq. (28), the negative sign is used in the denominator for compression and the positive sign for tension.. The stress computed by Eq. (28) will be the stress at the point of maximum deflection, which for a member in com- pression will evidently be the critical stress, but fo"r a member in tension the deflection reduces the eccentricity at the mid- length, and for this case the critical stress will usually be at an end section. For the stress at the end section, the com- putation may, in general, be made by use of Eq. (26), but for angles the increase in stress on account of the outstanding legs may need special consideration, Art. 8. Flexure for Angles in Pairs The case of two angles of the same section, riveted together at frequent intervals, and subjected to bending in a plane 00 0| FIG. 7. parallel to the connected legs, is indicated in Fig. 7. For such a condition the restraint which each angle exerts upon the other may be assumed to produce a deflection in the plane of bending and the neutral axis may be assumed to be per- pendicular to the connected legs. The loading for each angle will then be equivalent to some loading in a plane passing through 0, the center of area of the section, and through the vertices be and de of the section modulus polygon, which is indicated for each angle in Fig. 8. Let m denote the bending moment for one angle,. which if acting in the plane of 0, be, and de would produce the same deformations as those which result from the actual loading for the pair of angles. The 16 STRESSES IN STRUCTURAL STEEL ANGLES actual moment must be equal to the resultant of the two moments m, or M = 2m sin 6, . . . . (29) in which M = actual bending moment for the pair of angles; m = equivalent bending moment for one angle ; 6 = angle which the plane for m makes with the .neutral axis for the pair of angles, de FIG. 8. The bending stress for each angle is equal to m divided by the section modulus in its plane, or M 2S sin 0' (30) in which / = unit stress at the extreme fiber; s = section modulus for the plane of m. In Eq. (30), the value of s sin 6 may be considered as the equivalent section modulus for one angle of the pair for half of the total vertical load. For the case illustrated, this value is equal to the ^-coordinate of be, for the edge BC of the angle, or to the ^-coordinate of de for the edge DE of the angle. THEORY AND DISCUSSION 17 Thus, for angles in pairs, the equivalent section modulus for one angle can be obtained directly from a table of coordinates of section modulus polygons. However, even this is unnec- essary if a table of properties of sections is at hand, for the values of the coordinates in question are each obtained by dividing the moment of inertia by the distance from the neutral axis to the extreme fiber. In other words, the stress for the pair of angles, for bending in the plane of symmetry, may be computed as for a single symmetrical section, provided the two angles are firmly attached to each other. For the case shown in Fig. 7, independent action upon the part of each angle would tend to crowd the angles toward each other at the midspan. i Art. 9. Transfer of Stress by Shear to an Outstanding Leg In a member consisting of one or more angles, the critical stress is likely to occur at a section near the connection of the member to one of its connection planes, par- ticularly in the case of tension members. At such a section there may be considerable variation in the distribution of the stress, due to a difference in the deformations of the connected and the out- standing legs of each angle. In order to make an approximate determination of the stress dis- tribution, due to the transfer of stress to the outstanding leg by shear, consider the case shown in Fig. 9, in which there are two angles connected to opposite sides of a connection plate. In this case the stress at a section 6' r'' i 6 ' a ' C ) c ) c } d' d L 1 1 ilium FIG. 9. 18 STRESSES IN STRUCTURAL STEEL ANGLES taken at the middle of the length of the member will be assumed * to be uniformly distributed, but at the end of the member the point a is subject to slightly less deformation than the point b. If it be assumed that the stress increases at a uniform rate from zero at point a to the average unit stress at the center of the length of the member, then at any section distant x from the point a the unit stress at the extreme fiber of the outstanding leg will be , 2X F in which /' = unit stress at the extreme fiber of the outstand- ing leg; x = distance from the end of the angle to the sec- tion at which f is taken; L = length of the angle; F = total stress or force acting lengthwise of the angle; A = area of cross-section of the angle. Also, let / be the unit stress in the connected leg of the angle at the section at which/ 7 is taken. Then, for equilibrium at the section x distant from the end of the angle, ..... (32) in which A i = area of the outstanding leg; A 2 = area of the connected leg. Or, substituting the value of/ from Eq. (31) in Eq. (32), and solving for/, /=2 x I ~ A (33) * The reasonableness of the assumptions of this article will be better under- stood by a study of the results of tests by Prof. Cyril Batho, reported in an article on " The Effect of End Connections on the Distribution of Stress in Certain Tension Members," Journal of the Franklin Institute, August, 1915. THEORY AND DISCUSSION 19 Art. 10. Efficiency of End Connections The efficiency of a member may be taken as the relation of the average to the maximum stress, or if, in Eq. (33), x is taken so that / will be the maximum stress, the equation for the efficiency for the transfer of stress by shear to the outstanding leg will be, _, iA+A 2 in which EI is the efficiency, and the other quantities are as denned under Eqs. (31) and (32). In place of the areas A\ and A 2 the widths of the legs of the angle may be used. Using il\ for AI and tl 2 for A 2 > the equations corresponding to Eqs. (32), (33), and (34) become . (35) SI 2 or . . (36) ^1 L H T^2 J and For use in designing, the critical section will usually be taken through the connection rivet or rivets most distant from the end of the angle, in which case the net areas of the legs will be used for A\ and A 2 , or in Eqs. (35), (36) and (37), the values of the widths of legs must be corrected to their net values. For such a case the equations become n 2 d), . (38) 20 STRESSES IN STRUCTURAL STEEL ANGLES or and Ei=- in which /= maximum unit stress at the critical section; EI = efficiency of net section relative to gross section ; /i = width of the outstanding leg;* /2 = width of the connected leg;* ni= number of rivet holes in the outstanding leg which are cut by the critical section; H2 = number of rivet holes in the connected leg which are cut by the critical section; d = diameter of each rivet hole ; x = distance from the end of the angle to the crit- ical section; L = total length of angle. For a connection with lug angles, the length of the lug may be added to both x and , in the equations just given, but 2 this must be considered as a rather crude approximation. Since the equations of articles 9 and 10 are developed upon the assumption that the angle is restrained from bending, they are applicable in particular to pairs of angles riveted to opposite sides of a connection plate, but, as will be shown later, they may also be used in the design of single angles. Structural specifications are inclined to be both indefinite and unsatisfactory with reference to the requirements for connections. If any clause relating to this matter is included, * For exact work reduce the outside dimension of the leg by half the thick- ness of the angle. THEORY AND DISCUSSION 21 it usually stipulates that each angle must be connected by both legs or that only one leg shall be considered as effective area. The results of tests indicate that the allowance of only one leg as effective area is too rigid a requirement, even though but one leg of the angle is connected; while on the other hand the use of a lug angle to connect the outstanding leg results in too liberal an allowance for the capacity of the angle. Eq. (40) furnishes a ready means for estimating the efficiency of angles used in pairs, and as already mentioned may be used for connections with or without lug angles. For an angle without lugs, n\ will become zero. For this case the author has computed the efficiency of those angles which are most used, the values for which are given in Tables IV and V, at the back of the book. In the same tables, there is given for each angle, the area which if entirely effective at the max- imum unit stress would be equal in strength to the connected angle for the computed efficiency. Also, the total allowable stress for a maximum unit stress of 16,000 pounds per square inch at the critical section is tabulated. In the case of a single angle attached to a connection plate, there may be nearly complete restraint parallel to the plate with but little restraint perpendicular to the plate. In this case bending and shear each have an effect, but within the limits of working stresses the effect of the longitudinal shear may usually be neglected. In the development of Eqs. (31) to (40), inclusive, it was assumed that the stress at the center of the member was uniformly distributed. While this con- dition will be nearly fulfilled for angles in pairs, a single angle, with a small load, will be subjected to bending throughout its length, in which case the stress at the extreme edge of the outstanding leg will be less than the average stress, or may be of opposite sign. For the last-named case the stresses at the midsection may be as indicated in Fig. 10, in which j\ represents the unit stress at the face of the connected leg and [2 the unit stress at the extreme fibers of the outstanding leg. 22 STRESSES IN STRUCTURAL STEEL ANGLES At some distance b from /2, the horizontal shear will be zero, and the stress which must be transferred by horizontal shear from the connected to the outstanding leg will be equal to the sum of the stresses between connected leg and the section Ja of zero horizontal shear. Furthermore, if it be assumed that the angle is restrained from bending until the horizontal shear necessary to transfer stress to the outstanding leg has occurred, then at the critical section the moment arm tending to produce bending will FIG. 10. ke the distance from the centroid of the stresses at the section to the line of action of the applied force in the connection plate. But, on account of the variation of the stresses across the section, due to the longitudinal shear, the centroid of the stresses will lie between the center of area of the section and line of action of the force in the connection plate. In other words the neglect of the horizontal shear for this condition will be partially counterbalanced by the reduction in the moment arm due to the shear, and designs may be made on the basis of the bending and direct stresses. In this case for one angle in tension, the stress may be computed by Eq. (27), or the efficiency for direct tension and flexure will be in which Ez = efficiency of a single angle for eccentric ten- sion; s = section modulus of the critical section for the given loading; A =area of section; e = eccentricity of the applied load at the critical section. THEORY AND DISCUSSION 23 In using Eq. (41) the critical section will be at some distance from the extreme end of the angle. Hence, the value of e to be used will be less than that at the end section, due to the deflection of the angle. Since this reduction in e may result in an appreciable increase in the value of 2, its determination may be desired. In Fig. n, e c represents the resulting eccentricity at the center, e is the eccentricity FIG. 11. at the critical section, and e\ is the eccentricity at the end. The maximum stress at the midsection in terms of e c is or in terms of e\ it is , F. Fe + , . (43) KEz in accordance with Eq. (28), to which the reader is referred for nomenclature. Equating the values of /in Eqs. (42) and (43), ^ KEzs in which e\ is eccentricity of the applied load at the end sec- tion, and the other values are as stated for Eq. (28). At the instant that the angle begins to deflect, the moment at every section is Fei and the elastic curve, for this value of the moment, would be a parabola. As the load increases, the deflection at the midsection approaches zero 24 STRESSES IN STRUCTURAL STEEL ANGLES as a limit, and if for this condition e be assumed propor- /L \ n tional to ( x) , the resulting elastic curve will give e pro- XL \ n+2 portional to ( x) , or the limit toward which the curve (L \ tends is that for which the value of e is proportional to ( x ) . The significance of this may be appreciated by reference to Fig. 12. The bending would all occur at the end of the unsup- ported length, and the centroid of the stresses throughout this un- F IG , 12. supported length would coincide with the line of action of the applied forces. In reality this limit can not be reached, for some bending occurs throughout the unsupported length, and at any section this resisting moment must balance the moment which is equal to the resultant force F multiplied FIG. 13. by the distance from the centroid of the stresses to the line of action of the applied force. Evidently, the curve of the neutral surface for the unsupported length of the angle will approximate the form of the parabola, and at the end of the unsupported length the maximum bending stresses will be developed. For an angle attached to a plate the section of the sharp bend will be transferred to the connection plate and will occur close to the end of the angle, as indicated in Fig. 13. Therefore, the curve throughout the length may ordinarily be taken as a parabola, for which the value of e in Fig. ii becomes 2 .... (45) THEORY AND DISCUSSION 25 in which e = eccentricity at the critical section; e c = eccentricity at the midsection; e\ eccentricity at the end section; x = distance from the end of the angle to the critical section ; L = length of angle. It is evident that for large stresses and for practically all cases of ultimate loads the condition indicated in Fig. 12 will be nearly fulfilled and that the stresses at the critical section will tend to become identical with those which were determined in Art. 9. In this case the efficiency becomes identical with the value derived for pairs of angles, as given in Eqs.-(34) and (40). Of the two efficiencies which have been derived, one for the transfer of stress to the web and the other for the bending stress, the lower of the two should control in the design. As the load becomes large the value of e at the critical section becomes small, and the efficiency for transfer of stress becomes dominant. That value may, therefore, reasonably be used as a basis for design, for tension members. It may be of interest to compare some values obtained by experiment with the values for efficiency computed by Eq. (40). Table A contains a summary of the average values for tests made by Prof. F. P. McKibben and reported in Engi- neering News, Vol. 56, page 14, and Vol. 58, page 190. In studying the results for Table A, it should be remem- bered that Eq. (40) is developed for a straight line relation of stresses at the critical section, whereas at the ultimate load the stress is not proportional to the deformation and that in this case it will cause the computed efficiency to be less than that observed for the tests. On the other hand, influences which produce bending parallel to the outstanding leg of the connection tend to make the efficiency obtained by experi- ment lower than the efficiency computed by Eq. (40), which neglects this bending. On the whole the results obtained 26 STRESSES IN STRUCTURAL STEEL ANGLES TABLE A COMPARISON OF COMPUTED AND OBSERVED EFFICIENCIES OF CONNECTIONS FOR SPECIMENS TESTED BY PROF. .McKiBBEN MAIN ANGLES.* LUG ANGLES. Holes EFFICIENCY IN PER CENT 1 o m 3 ^ ' ^JH fe (M d O Remarks. fl s J Jj h h L'gth | Length. II X "3 0" H S O'-iU" 7.25 1.52 1 1 61.5 77.4 B 5 1 3 4 5'-4" 1 J ;! 0'-11|" 7.25 1.52 1 1 61.5 79.0 B 6 1 3 4 5'-4" 1 3 3 O'-Hi" 7.25 1.52 1 1 61.5 70.3 B 7 1 3 4 5 / -4 // 1 3 3 O'-Hi" 7.21 1.50 1 1 61.5 79.1 B s 1 3 4 5^ /7 1 3 3 O'-ll" 7.19 1.48 1 1 61.5 70.9 C A 1 4 6 5'-4' 5.33 1.51 1 79.7 71.6 C 4 1 4 6 5'_4" 5.33 1.51 2 68.7 70.5 C 5 1 4 6 5'-4" 1 3 4 '-!" 7.49 2.09 2 68.7 69.4 A 2 3 3 SM" 5.33 1.37 1 67.7 68.9 A 2 3 3 3'-4" 2 3 3 '-0" 7.33 1.83 1 1 55.2 67.4 * Thickness of metal for A, A, A, and Z> 4 was specimens f inch, inch, and for all other THEORY AND DISCUSSION 27 by computation appear to be sufficiently conservative to warrant the presentation of the method for practical use, until a better method shall be proposed. For a single angle in compression, the critical section for bending will be at the mid-length, the maximum stress may be computed by Eq. (28), and the efficiency for direct compres- sion and flexure will be KEz in which E$ = efficiency of a single angle for eccentric com- pression; s = section modulus of the critical section for the given loading; A = area of section ; e= moment arm at the end section of the couple producing flexure; F = total applied force; / = length of member; K = a constant, for which the value 10 may be used; E= modulus of elasticity of steel, usually taken as either 29,0x30,000 or 30,000,000; 2 = distance from the extreme fiber to a line through the center of area of the section and parallel to the neutral axis. The efficiency of the connection may be taken identical with the values given for tension, and if this is lower than the value obtained from Eq. (42), it should govern the design. TABLES TABLE I ELEMENTS OF ANGLES WITH EQUAL LEGS * .i F 8 ! / = moment of inertia jL /= product of inertia ^L~_^~_~jx 2 r- radius of gyration Weight Area Axis 1- 1 and A xis 2-2. Axis 3-3. Size. Foot. of Section. I r I l-x z j r min. Inches. Pounds. In.2 In.* In. In.s In. In.* In. 8X8XU 56.9 16.73 98.0 2.42 17.5 2.41 57.3 1.55 8X8X1;^ 54.0 15.87 93.5 2.43 16.7 2.39 54.7 1.56 8X8X1 51.0 15.00 89.0 2.44 15.8 2.37 52.1 1.56 8X8XM 48.1 14.12 84.3 2.44 14.9 2.34 49.6 .56 8X8X1 45.0 13.23 79.6 2.45 14.0 2.32 47.1 .56 8X8XH 42.0 12.34 74.7 2.46 13.1 2.30 44.2 .57 8X8X1 38.9 11.44 69.7 2.47 12.2 2.28 41.2 .57 8X8XH 35.8 10.53 64.6 2.48 11.2 2.25 38.3 .58 8X8X1 32.7 9.61 59.4 2.49 10.3 2.23 35.4 .58 8X8XA 29.6 8.68 54.1 2.50 9.3 2.21 32.1 .58 8X8X 26.4 7.75 48.6 2.51 8.4 2.19 28.9 .58 6X6X1 37.4 11.00 35.5 1.80 8.6 1.86 20.6 .16 6X6XH 35.3 10.37 33.7 1.80 8.1 1.84 19.6 .16 6X6X1 33.1 9.73 31.9 1.81 7.6 1.82 18.6 .17 6X6XH 31.0 9.09 3Q. 1 1.82 7.2 1.80 17.5 .17 6X6X1 28.7 8.44 28.2 1.83 6.7 1.78 16.4 .17 6X6XH 26.5 7.78 26.2 1.83 6.2 1.75 15.4 .17 6X6X1 24.2 7.11 24.2 1.84 5.7 1.73 14.3 .17 6X6X& 21.9 6.43 22.1 1.85 5.1 1.71 13.1 .18 6X6X 19.6 5.75 19.9 1.86 4.6 1.68 11.9 .18 6X6X3^ 17.2 5.06 17.7 1.87 4.1 1.66 10.6 .19 6X6X1 14.9 4.36 15.4 1.88 3.5 1.64 9.2 .19 * The values of J were computed by the author. The remaining values in this table are from the Carnegie Pocket Companion, and are used with permission of the Carnegie Steel Company. 31 32 TABLES TABLE I Continued ELEMENTS OF ANGLES WITH EQUAL LEGS / = moment of inertia J = product of inertia r radius of gyration F Ib Weight Area Axis 1- 1 AND A xi s 2-2. Axis 3-3. Size. Foot. of Section. 7 r / l-x X j r min. Inches. Pounds. In.* In." In. In. 3 In. In." In. 5X5X1 30.6 9.00 19.6 .48 5.8 1.61 11.1 0.96 5X5XH 28.9 8.50 18.7 .48 5.5 1.59 10.6 0.96 5X5X1 27.2 7.98 17.8 .49 5.2 1.57 10.1 0.96 5X5XH 25.4 7.47 16.8 .50 4.9 1.55 9.6 0.97 5X5X1 23.6 6.94 15.7 .50 4.5 1.52 9.2 0.97 5X5XH 21.8 6.40 14.7 .51 4.2 1.50 8.6 0.97 5X5X1 20.0 5.86 13.6 1.52 3.9 1.48 8.0 0.97 5X5XA 18.1 5.31 12.4 1.53 3.5 1.46 7.4 0.98 5X5X1 16.2 4.75 11.3 1.54 3.2 1.43 6.7 0.98 5X5X& 14.3 4.18 10.0 1.55 2.8 1.41 6.1 0.98 5X5X1 12.3 3.61 8.7 1.56 2.4 1.39 5.4 0.99 4X4XM 19.9 5.84 8.1 1.18 3.0 1.29 4.6 0.77 4X4X1 18.5 5.44 7.7 1.19 2.8 1.27 4.4 0.77 4X4XH 17.1 5.03 7.2 1.19 2.6 1.25 4.1 0.77 4X4X1 15.7 4.61 6.7 1.20 2.4 1.23 3.8 0.77 4X4X^ 14.3 4.18 6.1 1.21 2.2 1.21 3.5 0.78 4X4X1 12.8 3.75 5.6 1.22 2.0 1.18 3.2 0.78 4X4Xtfe 11.3 3.31 5.0 1.23 1.8 1.16 2.9 0.78 4X4X1 9.8 2.86 4.4 1.23 1.5 1.14 2.6 0.79 4X4X& 8.2 2.40 3.7 1.24 1.3 1.12 2.2 0.79 4X4Xi 6.6 1.94 3.0 1.25 1.0 1.09 1.8 0.79 3X3XH 17.1 5.03 5.3 1.02 2.3 1.17 2.9 0.67 3*X3*Xf 16.0 4.69 5.0 .03 2.1 1.15 2.8 0.67 3X3^XH 14.8 4.34 4.7 .04 2.0 1.12 2.6 0.67 3^X3^X1 13.6 3.98 4.3 .04 1.8 1.10 2.5 0.68 3X3Xfk 12.4 3.62 4.0 .05 1.6 1.08 2.3 0.68 3|X3X* 11.1 3.25 3.6 .06 1.5 1.06 2.1 0.68 TABLES 33 TABLE I Concluded ELEMENTS OF ANGLES WITH EQUAL LEGS 7 = moment of inertia J = product of inertia r= radius of gyration F L, Size. Weight Foot. Area of Section. AXIS 1-1 AND AXIS 2-2. Axis 3-3. / r J l-x X J r min. Inches. Lbs. In.2 In.< In. In.' In. In.* In. 3|X3Xf* 9.8 2.87 3.3 1.07 1.3 1.04 1.9 0.68 3iX3|X| 8.5 2.48 2.9 1.07 1.2 1.01 1.7 0.69 3fX3^XA 7.2 2.09 2.5 1.08 0.98 0.99 1.5 0.69 3^X3Xi 5.8 1.69 2.0 1.09 0.79 0.97 1.2 0.69 3X3X1 11.5 3.36 2.6 0.88 1.3 0.98 1.5 0.57 3X3X& 10.4 3.06 2.4 0.89 1.2 0.95 1.4 0.58 3X3X1 9.4 2.75 2.2 0.90 1.1 0.93 1.3 0.58 3X3X& 8.3 2.43 2.0 0.91 0.95 0.91 1.1 0.58 3X3Xf 7.2 2.11 1.8 0.91 0.83 0.89 1.0 0.58 3X3XA 6.1 1.78 1.5 0.92 0.71 0.87 0.87 0.59 3X3X 4.9 1.44 1.2 0.93 0.58 0.84 0.75 0.59 2^X2^X1 7.7 2.25 1,2 0.74 0.73 0.81 0.68 0.47 2|X2^X& 6.8 2.00 1.1 0.75 0.65 0.78 0.62 0.48 2^X2^X1 5.9 1.73 0.98 0.75 0.57 0.76 0.57 0.48 2|X2XA 5.0 1.47 0.85 0.76 0.48 0.74 0.49 0.49 2^X21X1 4.1 1.19 0.70 0.77 0.39 0.72 0.41 0.49 2^X2XA 3.07 0.90 0.55 0.78 0.30 0.69 0.32 0.49 2^X2|Xi 2.08 0.61 0.38 0.79 0.20 0.67 0.23 0.50 2X2X;& 5.3 1.56 0.54 0.59 0.40 0.66 0.30 0.39 2X2X1 4.7 1.36 0.48 9.59 0.35 0.64 0.27 0.39 2X2XA 3.92 1.15 0.42 0.60 0.30 0.61 0.24 0.39 2X2X1 3.19 0.94 0.35 0.61 0.25 0.59 0.21 0.39 2X2X& 2.44 0.71 0.28 0.62 0.19 0.57 0.16 0.40 2X2Xi 1.65 0.48 0.19 0.63 0.13 0.55 0.11 0.40 34 TABLES B g e J 1 1 1 fl S r I O M TO 1 ft "g n h T; / OOOO OCOOl>CO 0000005^001 i ^H -iM i-loo H|H w| H|H inloo ^IS rn|N H* xxxxxxx><:xx CO^OcDCOcOcOCDCOcD^O xxxxxxxxxx GOOOOOOOOOOOOOOOOOOO TABLES 35 oooooooooo ooooooooooo g I oooooooooo ooooooooooo 1 I 2 COt^l>OOGOClOO'-HC<) C5C5O' ii fC^OOOO-^fiOCO oo oo oe cc oc oo p p os o oooopppoppppp oooooooooo ooooooooooo ,1 t^* CO CO CO CO CO *O iO ^O lO b a o J3 ll 15 IOGO lO O5 ^ GO CO t> i ( .3 "3 eoioi>occ500r-fooo go > 's xxxxxxxxxx xxxxxxxxxxx CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO XXXXXXXXXX XXXXXXXXXXX OCOOOOOOCCOOOOOOOOOO t>-t^-tl>t>-t>.l>^t>-l>-t- 36 TABLES F HH M ^ H o J I | I fli *H -* a ^ o 2 3 3 S ^3 .2 O O T-J R2 -i>i>oo g OOOOOOOOOOOOOOOOOOOOOO ddodddddddd S ^ oiososwour^r^co ,_<* ,_; _H' ^ ^ ,-J rn' TH' o O O OOCO^C^OOOtOCOi-HOOCO CO CO CO CO CO (N (N O5OiO5 xxxxxxxxxxx TH^^TttrtH"^T^TiHTtio^cO'-HOoocOT^oio oiOTjHcofNi-Idci i iOOOt > -t > -I>-I>-l>-l>l>I>- dddddoOOO OOOOOOOOO oddddddod i-Hi- co co co co < COT^C S QOI>-*OCOOOO OOOiOCOi-HCiCO^fC^ i-HOJt^^fC^O ddddddddo oddddddod t-i 1-1 ,-H i- ,-H -< Ci IO CO i-H O OC I> !>O>OTfiCOi iOGOI> COi-H O O i-i i-i i-i i-i i-i -J i-< O C> ci(N O O i i iC i i i (C^COCOTtHiOCOb- OOOOOOOOGOOOOOOOOO ooooooaqoqoooooooo pppppcjcop ddddddddd ddddddddd ^'^^^,-1^^^^ QOQOGOGOl>t^I>t^O cOOOtOOiC>OTtiTti OOrOCO>- co co co IM IM OiO 1C lO O 1C O O oddddddddo OOoddddd OOOOOO 00 l2 CO O iOCOOOOCOTtHi I b*. *O CO ^H 00 CO pppppoooGooocb- i>i^i>t^pcococo i> t^. i> i> co co dodddddoOO C5OOOOOOO OOOOOO 00*OiO OCSOOt^COtOLOr^ 00 t^ O >O ^ * ^H'^J^^H^^HOOOO OOOOOO o'o OOOOOO iOiOOI>l>OOO5OO'-i OOOOi-il^t>-l> t>t^|>t>.t>.t>. o'odo'ooo'ooo dooddood ddddoo ^ 00 O ^ COi lOOOiOCOi tOOOCO l>O*O-*C<|iiC>t^ -^COC^OO5t>- cococo(N(NCiocooao- O5N-CO 1 ^cO'-HC5t> C^OOiOOCOO OJ(NT-lT-l,-l,-l,-!i-lOO ^Hi-ti-H^-ti-Hi-HOO i-Hi-HOOOO -^TfOCDt^l>GOOiO^H Ot^-OCC5O5Oi-((N i-tr-ifNCO^iO Pppppppp^H^H ppppp^Hi-lT-l OO5O3O5OO5 ^H^H^H^H_^H^^H^H^H ^H ^ _i ^H' _' _<' _ ,H" 000000 Ot^i lOO^Oi i|>COO5 -iOOCOC "ic H ^i Mk H COCOCOCOCOC'SCOCO xxxxxx COCOCCCOMCO 42 TABLES r f s I PQ < $ J is a " S CO 3 ^ 111 2w ^^ d '1 d 2 2 2 2 2 SJ SJ S S S5 2 2 TT "^ ^r '^ "^ "^ ^r ^r ^r ^r ^r ^r d d d o d doddddd 21 *^ S d Tt< l> i i O OS CO CO O -"f 00 C^l CO CO *O O TfH CO -* TjH rtl CO (N t^ < ij e d voiocot^t^ <>Ir^OOOOO5Oi-i io>o>oioio io^oio^o>ococo ooodd ooddddd ^ d t>. i i Tt< !> O5 rfHGOi-HOI^O5O OCOtOTfriCO OiOiO-^CO(Nt>i> I-H T-H T-H i i O OOOOOOO v H ^ L d O500CDTF O(MtO1^.00O5O OOOl>COO 1>COOTJHCOC-OOOOO5O ooooo ooooooo d i I O5 1C i 1 IO os i> >o co T i T-nooii>coioco S <: d o si d lOOCOt^Oi O 00 IO T i CO i i "3 (MOt^^r-H Ot^OCOOCOO OQ 1 fe"8 -3 a (N CO 1>OOO5O'-H CO i I CO "3 CO 1> 00 1 ft f^ 1 t^.COOiOT^ COcOiO^COC^i-* g fi Inches. H| HS l=o HS HM H| N HS | HS HH- HS H| xxxxx xxxxxxx (N(N(N(N(N CO CO ooo co co o i> ^ I-H (^ <^j CNJ ,_| ,_,,_, ooo'ooo o o o O I-H ^ T^ TF Tt^ o o o o co i I i I rH o o o 000 OS CO M odd os oo oo odd O CO i I OO GO 00 o d d C5 t>- lO l> l> l> o d d TjH O 00 ^ CO . t- l> d o d i-H OS CO l> 1C T^ odd 10 00 i-H t>- CO O O O O CO Tt< Tt< 1C TjH CO O O O i-H O O i-lrH^H^H c Xcd - Xde yde Xea yea 5X4X1 2.56 1.84 2.47 3.30 9.59 4.04 5.70 7.60 4.99 2.10 5X4X1 2.27 1.58 2.22 2.88 8.80 3.80 5.43 7.07 4.37 1.89 5X4Xf 2.00 1.35 1.91 2.46 7.78 3.40 4.91 6.34 3.72 1.63 5X4X 1.65 1.12 1.60 2.05 6.68 2.99 4.39 5.60 3.06 1.37 5X4X1 1.33 0.91 1.21 1.58 5.30 2.35 3.50 4.56 2.33 1.04 5X31X1 2.44 1.48 2.19 2.52 8.77 3.02 5.19 5.96 4.89 1.68 5X3|X| 2.21 1.31 1.96 2.24 7.94 2.80 4.90 5.60 4.27 1.51 5X3^X1 1.89 1.03 1.72 1.88 7.06 2.59 4.63 5.06 3.63 1.33 5X3^X1 1.63 0.84 1.43 1.54 6.02 2.23 4.06 4.40 3.00 1.11 5X3^X1 1.31 0.69 1.10 1.21 4.84 1.80 3.37 3.72 2.30 0.86 5X3iX^ 1.11 0.56 0.94 1.02 4.14 1.57 2.98 3.22 1.93 0.73 5X3Xif 2.14 1.04 1.82 1.73 7.52 2.10 4.53 4.30 4.46 1.24 5X3X1 2.08 0.97 1.71 1.62 7.18 2.01 4.40 4.16 4.17 1.17 5X3X1 1.85 0.84 1.50 1.41 6.34 1.83 4.12 3.88 3.56 1.03 5X3X1 1.59 0.67 1.24 1.16 5.42 1.60 3.73 3.47 2.92 0.86 5X3X1 1.23 0.45 1.00 0.87 4.35 1.36 3.29 2.86 2.24 0.70 5X3X& 1.11 0.40 0.82 0.78 3.75 1.13 2.80 2.64 1.90 0.57 4i X 3 X ii 1.75 1.02 1.57 1.72 6.24 2.00 3.67 4.00 3.62 1.16 4|X3Xf 1.71 0.97 1.46 1.60 5.95 1.90 3.53 3.87 3.38 1.08 4|X3Xf 1.42 0.81 1.29 1.38 5.31 1.77 3.38 3.62 2.88 0.96 4^X3X1 1.27 0.64 1.09 1.13 4.54 1.56 3.04 3.17 2.37 0.81 4|X3Xf 1.03 0.51 0.84 0.88 3.69 1.27 2.57 2.70 1.83 0.63 4JX3XA 0.92 0.44 0.70 0.75 3.20 1.09 2.22 2.36 1.54 0.53 TABLES 47 TABLE III Continued COORDINATES OF SECTION MODULUS POLYGONS FOR ANGLES COORDINATES. Size of Angle. Xab + yab Xbc ybc Xcd ycd + Xde + Vde Xea + yea 4X4XM 1.57 1.57 1.70 2.99 6.28 3.57 3.57 6.28 2.99 1.70 4X4X1 1.49 1.49 1.61 2.82 6.06 3.47 3.47 6.06 2.82 1.61 4X4X1 1.34 1.34 1.37 2.42 5.44 3.09 3.09 5.44 2.42 1.37 4X4Xi 1.12 1.12 1.13 1.99 4.74 2.71 2.71 4.74 1.99 1.13 4X4X1 0.86 0.86 0.91 1.54 3.86 2.28 2.28 3.86 1.54 0.91 4X4Xi 0.58 0.58 0.62 1.03 2.76 1.66 1.66 2.76 1.03 0.62 4X3IXM 1.54 1.28 1.51 2.30 5.74 2.65 3.24 4.96 2.96 1.36 4X3^Xf 1.50 1.22 1.41 2.16 5.45 2.54 3.12 4.77 2.74 1.28 4X3|Xf 1.24 0.98 1.26 1.83 4.96 2.40 2.98 4.33 2.36 1.14 4X3^X 1.05 0.85 1.04 .52 4.24 2.08 2.60 3.80 1.93 0.95 4X3^X| 0.83 0.65 0.83 .18 3.47 .73 2.19 3.12 1.50 0.75 4X3IX& 0.72 0.57 0.70 .01 3.05 .53 1.93 2.80 '1.28 0.64 4X3XH 1.42 1.01 1.31 .70 5.07 .87 2.87 3.72 2.85 1.06 4X3X1 1.35 0.93 1.25 .59 4.86 .83 2.83 3.59 2.57 1.01 4X3X1 1.13 0.75 1.13 .36 4.38 .75 2.76 3.33 3.28 0.91 4X3X1 1.00 0.62 0.92 .10 3.76 1.50 2.41 2.89 1.87 0.75 4X3Xf 0.83 0.47 0.72 0.86 3.12 1.25 2.05 2.43 1.47 0.59 4X3Xi 0.57 0.37 0.49 0.62 2.26 0.89 1.49 1.89 1.01 0.40 3X3XH 1.22 1.22 1.24 2.27 4.53 2.48 2.48 4.53 2.27 1.24 3|X3|Xf 1.13 1.13 1.19 2.13 4.35 2.43 2.43 4.35 2.13 1.19 3^X3|Xf 0.93 0.93 1.04 1.79 3.91 2.27 2.27 3.91 1.79 1.04 3^X3|X 0.80 0.80 0.86 1.48 3.40 1.98 1.98 3.40 1.48 0.86 3X3|Xf 0.65 0.65 0.68 1.16 2.87 1.68 1.68 2.87 1.16 0.68 3X3^X* 0.44 0.44 0.47 0.79 2.06 1.24 1.24 2.06 0.79 0.47 48 TABLES TABLE III Continued COORDINATES OF SECTION MODULUS POLYGONS FOB ANGLES Y COORDINATES. Size of Angle. Xab + Vab + Xbc Vbc Xcd ycd Xde + yde Xea + yea 3X3XH 1.11 0.90 1.09 1.63 4.06 1.79 2.24 3.37 2.20 0.97 3^X3Xf 1.05 0.83 1.03 1.52 3.89 1.74 2.19 3.23 2.06 0.92 3^X3Xf 0.91 0.75 0.91 1.34 3.51 1.62 2.07 3.04 1.76 0.82 3X3X 0.82 0.61 0.75 1.08 3.10 1.41 .82 2.61 1.48 0.68 3^X3X1 0.61 0.45 0.60 0.83 2.50 1.20 .57 2.17 1.11 0.54 3|X3X| 0.45 0.34 0.42 0.59 1.83 0.88 .16 1.64 0.77 0.37 3|X2|XH 0.85 0.59 0.85 0.98 3.23 1.18 .90 2.21 1.84 0.65 3X2^Xf 0.80 0.51 0.80 0.91 3.04 1.12 .87 2.13 1.69 0.62 3X2|X 0.70 0.45 0.67 0.78 2.67 1.00 .72 2.00 1.39 0.52 3X2X! 0.61 0.34 0.53 0.60 2.24 0.84 1.47 1.67 1.11 0.41 3|X2Xi 0.41 0.22 0.38 0.41 1.62 0.64 1.16 1.28 0.75 0.29 3X3X1 0.66 0.66 0.74 1.29 2.65 1.53 1.53 2.65 1.29 0.74 3X3X 0.55 0.55 0.63 1.06 2.36 1.40 1.40 2.36 1.06 0.63 3X3X1 0.50 0.50 0.47 0.85 2.02 1.12 1.12 2.02 0.85 0.47 3X3X1 0.29 0.29 0.35 0.55 1.43 0.89 0.89 1.43 0.55 0.35 3X2iX& 0.59 0.43 0.59 0.81 2.26 1.00 1.32 1.82 1.16 0.51 3X2|X 0.55 0.41 0.54 0.74 2.10 0.94 1.25 1.73 1.05 0.47 3X2^Xf 0.48 0.30 0.42 0.56 1.77 0.78 1.06 1.41 0.83 0.37 3X2Xi 0.33 0.21 0.30 0.40 1.32 0.60 0.83 1.-12 0.57 0.26 3X2Xi 0.47 0.27 0.45 0.47 1.76 0.59 1.10 1.16 0.99 0.33 3X2Xf 0.40 0.21 0.35 0.37 1.44 0.49 0.94 1.00 0.77 0.26 3X2Xi 0.30 0.14 0.26 0.26 1.11 0.39 0.80 0.80 0.55 0.19 TABLES 49 TABLE III Conduded COORDINATES OF SECTION MODULUS POLYGONS FOB ANGLES Y COORDINATES. Size of Angle. Xab + yab + Xbc ybc + Xcd yea + Xde + yae Xea + yea 2|X2iX 0.38 0.38 0.40 0.71 1.48 0.84 0.84 1.48 0.71 0.40 2iX2|Xf 0.30 0.30 0.33 0.56 1.29 0.75 0.75 1.29 0.56 0.33 2X2Xi 0.22 0.22 0.23 0.39 0.97 0.57 0.57 0.97 0.39 0.23 2|X2Xi 0.11 0.11 0.13 0.21 0.57 0.34 0.34 0.57 0.21 0.13 2|X2X 0.34 0.27 0.34 0.47 1.25 0.52 0.73 1.01 0.68 0.28 2X2Xf 0.28 0.19 0.28 0.36 1.10 0.48 0.69 0.88 0.54 0.24 2|X2Xl 0.22 0.15 0.19 0.25 0.82 0.35 0.52 0.68 0.38 0.16 2X2Xi 0.11 0.07 0.11 0.13 0.47 0.22 0.33 0.41 0.20 0.09 2XHXA 0.24 0.10 0.18 0.17 0.79 0.22 0.50 0.47 0.44 0.12 2^XUXi 0.20 0.08 0.15 0.14 0.67 0.19 0.45 0.42 0.36 0.10 2ixi*x& 0.16 0.06 0.12 0.11 0.54 0.16 0.40 0.37 0.28 0.08 2iXlXf 0.26 0.16 0.22 0.25 0.87 0.26 0.48 0.54 0.54 0.16 2JXUXI 0.21 0.12 0.19 '0.20 0.75 0.25 0.45 0.48 0.42 0.14 2JXUXJ 0.16 0.09 0.14 0.14 0.57 0.19 0.38 0.41 0.30 0.10 2|XHX& 0.12 0.06 0.11 0.11 0.45 0.16 0.32 0.32 0.23 0.08 2X2X^ 0.21 0.21 0.22 0.40 0.82 0.45 0.45 0.82 0.40 0.22 2X2X1 0.19 0.19 0.20 0.35 0.75 0.42 0.42 0.75 0.35 0.20 2X2Xi 0.13 0.13 0.15 0.25 0.59 0.36 0.36 0.59 0.25 0.15 2X2X1 0.08 0.08 0.08 0.13 0.35 0.20 0.20 0.35 0.13 0.08 50 TABLES e - a -* 2 3 Q CO CO (M 00 CNI t^ rH CO Oi Oi t>* CO CO i lO Oi rH O S! ^ fS t"* i-H tO oi OS 00 o co CO C CO Oi CO t^ rH CO OS 00 CO CO 1C 1C (N I 1> TH Oi GO OO 00 C^ CD O iO Oi t>^ CO id Oi Oi 00 OO OO OO 00 00 00 00 00 00 00 TJH CO CO 00 00 00 lOiOiOiOiOiOiOiOiC O O O O TH rH Oi Oi CO CO CO CO O5 CO Oi Oi CO CO o o CO CO 1C 1C 1C CO co co 00 00 Oi Oi O O co co CD CD CO CO CD CO CD H5 SfcH-SS-^^HS, oo TABLES 51 CO 00 00 Oi OO I s * CO ^4 ^O ^H 00 T*< CO OS rH rH rH O CO tO Tt* CO rH ^ 00 O> OSOGOCSb-COOT^GOC^ CO to to CO b* GO OS OS OS ^^ tOTfCOC<|rHOO3GObb- OS p OS GO GO CO M* CO CO CO rH O OS GO b- cd O Tt< CO CO b- rH W 00 ^ b^ cd cd CO (M t>. tO OS 00 00 b- 1C > $ 8 ~i o 5 3 3 t 1 1 4-i 02 S a -H i* < I "S i "rt '3 o" w ~ rS ^ CC5 rH t>. 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CM CM CM 00 l> CO CO tO O5 CO CO 00 rH os oo oo CO CM O 1> OS CO CO 00 O CM CO CO OS CM Tfl Tj< CO CO CM rH CO CM OS l> CM to CO CO t>- 00 O 0^28 CM rH OS CO cO cO to **^ i> o co to oo O O OS 00 I s *- CM "* (M i-i CO b- rH os oo oo 00 CM 00 CM rH t^ 00 l> OS O OS 00 00 t> CMt>-CMl>OrHrHrHOOOCO rHCOOSCOtO 5 CM ^^ CO OO OS CO CO CO '^ OS CO :> OS 00 b- CO to to 00 00 l> CO CO tO CO CM COOcOrHCO O CM to CO CM rH N. rH tO OS to to ^ eo b- t>. 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OS rHC^TCOt^OO COCO g 1010*0 CO CO CO O Oi Ci CO CO CO (N -^ CO GO rH rH rH rH CO CO CD CO CO co CO O CO tO CO rH T^ t CO CO CO CO CO CO CO OS O i I CO (MCOtOCO (MtOCOrH OSrHCQtO tOGOrHT^ rHtOOS rH rH rH CD O CO CO cD CO CO CO CD GO q co co OSrHCO OOOSrH OOO oo o co to co IO to to (N CO (M CO CO CO 00 C5 LO to tQ tQ tQ to ss HS THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO 5O CENTS ON THE FOURTH DAY AND TO $I.OO ON THE SEVENTH DAY OVERDUE. 29 1936 LD 21-100m-8,'34 UNIVERSITY OF CALIFORNIA LIBRARY ^S?.v.^' ".";_