THE UNIVERSIT Y OF MICHIGAN COLLEGE OF ENGINEERING Department of Civil Engineering Final Report STRENGTH OF STEEL BEAM-COLUMNS FOR STEEL YIELD POINTS BETWEEN 33 KSI AND 100 KSI, AND INCLUDING EFFECTS OF ENDRESTRAINT, BIPLANAR BENDING, AND RESIDUAL STRESS I. A. E1Darwish" B. G. Johnston, / —ORA Project:'05 154 under contract with: DEPARTMENT OF THE NAVY BUREAU OF YARDS AND DOCKS CONTRACT NO. NBy-45819 WASHINGTON, D.C. administered through: OFFICE OF RESEARCH ADMINISTRATION ANN ARBOR August, 1964

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TABLE OF CONTENTS Page LIST OF FIGURES v ABSTRACT vii 1. INTRODUCTION 1 1.1 General 1 1.2 Previous Studies 2 1.3 Summary of Present Investigation 3 2. ANALYSIS OF THE STRESSES, STRAINS, AND CURVATURES OF THE SIMPLIFIED FOUR POINT CROSS SECTION 5 2.1 The Simplified Cross Section 5 2.2 Stresses and Strains at the Load Causing Initial Yield in the Unrestrained Beam-Column Under Planar Bending 5 2.3 Stress-Strain Relationship Modified by Residual Stress 6 2.4 Regression of Stress 8 2.5 Stress-Strain Relations in Planar Bending for General Stress-Strain Relationship 8 2.6 Stresses and Strains Beyond the Load at Initial Yield of the Square Simplified Section in Biplanar Bending for Elasto-Plastic Stress-Strain Relationship 10 2.7 Stresses and Strains in Biplanar Bending for the Square Simplified Section for the General Stress-Strain Relationship 11 2.8 Effect of Initial Curvature 19 3. APPROXIMATE DESIGN PROCEDURE FOR BEAM-COLUMNS 23 3.1 Unrestrained Beam-Column Design Based on Load at Initial Yield 23 3.2 Load at Initial Yield for the Unrestrained Beam-Column (Simplified Four Point Cross Section) in Planar Bending Under Uniform Lateral Load (kP) 24 353 Beam-Column Design Based on the Load at Initial Yield for Other Cross Sections and Loading Conditions for Planar Bending 25 3.4 Unrestrained Beam-Column Design Based on the Load at Initial Yield for Biplanar Bending Case 28 3.5 Load at Initial Yield for the Restrained Beam-Column of Simplified Section in Planar Bending 32 iii

TABLE OF CONTENTS (Concluded) Page 4, ULTIMATE STRENGTH OF UNRESTRAINED BEAM-COLUMNS 33 4.1 Ultimate Strength of Unrestrained Beam-Column of the Four Point Section in Planar Bending, for the General StressStrain Relation 33 4~2 Ultimate Strength of the Unrestrained Beam-Column of the Four Point Section in Biplanar Bending for the Geneial Stress-Strain Relation 38 5. ULTIMATE STRENGTH OF RESTRAINED BEAM-COLUMNS IN PLANAR BENDING 45 5ol Modified Slope Deflection Equations for Compression Members With Transverse Loads and End Moments Within the Elastic Limit.45 5.2 Symmetrically Restrained Beam-Columns Above the Elastic Limit 46 6. SUMMARY 59 7. REFERENCES 61 APPENDIX Ao DESIGN TABLES 63 APPENDIX B, FLOW DIAGRAMS AND RELATED MAD PROGRAMS FOR COMPUTER ANALYSIS OF VARIOUS BEAM-COLUMN PROBLEMS 83 iv

LIST OF FIGURES Figure Page 1. The simplified cross section. 5 2. Simplified stress-strain diagram for steel. 3a. Assumed residual stress distribution. 7 3b. Compressive stress distribution after a central compressive load initiates yielding. 7 4. Average stress-strain relation resulting from residual stress distribution in Fig. 3. 9 5. Stresses and strains beyond the load at initial yield for biplanar bending and elasto-plastic stress-strain behavior. 10 6. Stresses and strains beyond the proportional limit stress for the biplanar bending case. 12 7. Interpolation procedure. 18 8. Permissible variation in straightness for wide flange shapes by ASTM Spec. A6. 20 9. Diagonal principal axes used for biplanar bending. 40 10. Deflection of an unrestrained beam-column in planar bending. 44 11. Beam-column with end moments. 45 12. Restrained beam-column under uniform lateral load. 46 13o Numerical evaluation of end slope. 49 14. Linear interpolation for satisfying the end condition of a restrained beam-column. 50 15. Yield condition of beam-column with equal end restraints. 51 16. Load deflection curve of a restrained beam-column in planar bending. 57 v

ABSTRACT The purpose of this investigation is to develop simple and accurate design procedures for beam-columns composed of four corner angles closed on four sides by lacing bars. Whenever applicable, these procedures are extended to cover other cross sections. The simplified section was adopted to avoid the complications of stress redistribution across the section. Emphasis is on analysis by use of the electronic digital computer, without which the calculations would have been too time consuming. Coverage herein is limited to the cases of unrestrained and restrained beam-columns in planar bending, and unrestrained beam-columns in biplanar bending. Tables that give the average compressive stress at the load causing initial yield for the unrestrained beam-column subjected to uniform lateral load W = kP, together with the axial load P, are provided. These tables are given for all currently used yield stress levels of structural steels. The applicability of these tables is extended by providing shape and load conversion factors to cover other cross sections under various conditions, both in planar and biplanar bending. Tables also are given that provide the actual ultimate load of the unrestrained beam-columns of the four point section under uniform lateral loads W = kP, together with the axial load P, in planar bending. The effect of residual stresses on the individual corner element is included in the analysis. The ultimate load values are obtained by an incremental increase in load, determining the column shape by a numerical integration procedure at each step up to the load beyond which this procedure does not converge. The ultimate strength of the unrestrained four point section in biplanar bending, under uniform lateral loads WX = kxP and Wy = kxP together with the axial load P was also investigated. A computer program is given to solve such a problem using the same procedure as that of the planar bending case, but no tables have been prepared. Restrained columns, both in planar and biplanar bending, were investigated by procedures essentially the same as for the unrestrained column. For a given P, the restraining end moments are determined so as to be compatible with the deformed geometry of the restraining members. A computer program is given for the case of restrained beam-columns in planar bending. vii

INTRODUCTION 1.1 GENERAL This final report for phase one of Contract NBy-45819 is in compliance with the contract requirements covering work done during the period June 4, 1962 to June 1, 1964, under The University of Michigan Research Contract 05154, on the subject of the ultimate strength of framed beam-columns of high strength steel. "Beam-columns" are members subjected simultaneously to axial loads and bending moments caused, either by transverse forces or by eccentricities of the axial force at one or both ends. The general analysis of a beam-column may involve end restraints to lateral displacements, twist, or rotations. The treatment of the problem to include all of these effects simultaneously is very complex. Column design may be based either on the load, at which the maximum stress reaches the yield point or the ultimate load capacity of a column in a given structure. This investigation takes up the effects on ultimate load of residual stress, end restraint, and biaxial bending. To eliminate other complications, such as the torsion of open sections, and to simplify the effects of cross-sectional shape, the column section considered in this investigation consists of four corner angles closed on the four connecting sides by lacing bars. It will be assumed that the lacing is continuously effective and that deflections due to shear have a negligible effect on column behavior. Although in some of the analyses the effect of residual stress in altering stress-strain characteristics will be included, each of the corners will be considered as a point concentration of area. The results pertaining to this simplified section may in some cases be used as a basis for estimating approximately the behavior of other cross sections. The accurate determination of the load-deflection history of a beamcolumn in the inelastic range is a time consuming process, requiring an evaluation of the deflected shape of the beam column at successive increments of load. The use of an electronic computer is essential for the efficient study of this problem within reasonable limits of time and cost. The IBM 7090 computer at The University of Michigan was used and instructions to the computer were written in the MAD (Michigan Algorithm Decoder) language. There are two approaches to the beam-column problem. If the material is steel, with no consideration given to residual stress, and with lateraltorsional buckling prevented, the design may be based conservatively on the column load at which the maximum stress due to combined axial load and bend1

ing moment reaches the yield point. When end restraint exists this procedure becomes still more conservative. If, however, a nonlinear stress-strain relationship is considered, either inherent in the material or effectively caused by the presence of residual stress, an incremental analysis involving the determination of the maximum load on a load-deflection curve is required. 1.2 PREVIOUS STUDIES The first rational approach to the determination of the maximum load at which an eccentrically loaded column, without end restraint, passes from a stable to an unstable equilibrium configuration is due to von Karman.1 For a given nonlinear stress-strain relationship, von Karman outlined the steps whereby the relationship between column curvature and applied moment could be established. For the eccentrically loaded column at any location, the applied moment, M = P(e+y). Thus for a given value of P and e, a relationship between curvature and deflection could be established, and by numeric and/or analytic procedures a particular column curve could be approximated. A series of curves obtained by numerical integration then served as the basis from which, finally, a relationship between average column stress (P/A) and maximum deflection could be obtained. Chwalla2 made tests confirming the correctness of von Karman's work and followed these by analytical studies regarding the effect of cross-section shape. A review of the work of von Karman and Chwalla, as well as that of later investigators who introduced various simplifying modifications as to shape of stress-strain curve or column deflection curve, is presented by Bleich.4 In recent years emphasis has been directed toward the effects of end restraint on beam-column strength. Most columns are part of a continuously framed structure. Although Chwalla made initial rigorous but laborious studies of this problem, it has been attacked more extensively in recent years through investigations at Cambridge and Lehigh in connection with plastic design developments, and at Cornell, where the problem was studied primarily within the elastic range. The work on columns (stanchions) at Cambridge (initiated in 1938) together with other investigations is reviewed with great descriptive detail in Chapters 13-15 of Ref. 5, by Baker, Horne and Heyman, covering both experiment and the ultimate strength elasto-plastic analysis of framed beamcolumns. The work at Cornell, initiated in 1948 under sponsorship of Column Research Council, included very extensive elastic analyses of the buckling and instability strength of framed columns and beam-columns, respectively, with approximate extension to elasto-plastic behavior, as summarized in a paper by Bijlaard, Fisher, and Winter.6 At Lehigh, in connection with many full scale tests, analyses in the 2

early 1940's were initially related to the ultimate strength of unrestrained beam-columns, including for the first time the effects of residual stress caused by hot rolling and cooling of structural steel shapes, as summarized. for a Column Research Council Symposium in 1955 by Ketter, Kaminsky, and Beedle.7 Later, in a series of papers growing out of the initial Lehigh work, by means of ingenious nomographic charts, Ojalvo adapted the method of von Karman and Chwalla to procedures for calculating the ultimate strength of eccentrically loaded restrained beam-columns in the inelastic range. Very complete ultimate strength tables for unrestrained beam-columns, with unequal end-eccentricities, giving critical combinations of length, endmoment and axial force have been prepared by Galambos and Prasad9 for wide flange beam-columns bent by end-moments about their major axis. The Ojalvo charts and Galambos tables have been prepared only for structural steel in the 33-36 ksi yield point range. Most investigations of the beam-column problem have been concerned with the case whereby bending moent is introduced only at the ends of the column, either by hypothetical brackets with a fixed eccentricity or by the angular rotation caused by the bending of restraining members framed to the columns In 1962 KetterO1 compared the effects of end-moment and lateral load on the maximum strength of the unrestrained beam-column. For an overall review of the beam-column problem, reference may be made to the Column Research Council "Guide to Design Criteria for Metal Compression Members."ll 1.3 SUMMARY OF PRESENT INVESTIGATION It has been the purpose of the present investigation to explore in a preliminary way, with the development of design aids, the laterally loaded beam-column problem, with limitations and scope outlined as follows: 1. Simplified 4 point area closed section. 2. No consideration of local buckling, torsion, or shear deformation. 3. Residual stress considered on a point area average basis. 4. Elastic stress analysis as well as ultimate load (instability) analysis. 5. Ends either unrestrained or restrained. 6. Uniform lateral load, with approximate load conversion factors to other lateral load distributions or to end-eccentricities. 3

7. Planar or biaxial bending. 8. Various steel strengths between 36 ksi and 100 ksi yield points. Most of the objectives of the investigation were realized. The use of a computer comparable to the IBM 7090 at The University of Michigan was essentialo However, the computer program for the final phase, biaxial bending with end restraint, was not entirely successful, with failure to achieve convergence near ultimate load. Considerable effort was expended on this portion of the program development but it is not reported upon and the work (under the authorized extension) was terminated prematurely by the early recall of the principal investigator, Dr. I. ElDarwish, to Egypt. The successful program for planar bending with end restraint, and biplanar bending without end restraint, are covered herein, and these also provide the bases for the second phase that is not reported and is now in progress (October, 1964). 4

2. ANALYSIS OF THE STRESSES, STRAINS, AND CURVATURES OF THE SIMPLIFIED FOUR POINT CROSS SECTION 2.1 THE SIMPLIFIED CROSS SECTION The cross section shown in Fig. la, four angles boxed by lacing bars, is commonly used for very long columns. The size of the angles is assumed to be comparatively small compared. to other dimensions of the member, in which case the areas of the angles may be considered as concentrated at points as shown in Fig. lb in which: 2c = depth of the assumed cross section (distance between angle centroids). 2b = width of the assumed cross section (distance between angle centroids ). 2c - ~ 2c L 2b X=b 2b a b Fig. 1. The simplified cross section. One advantage of such a simplified cross section is that its momentcurvature relationship in either the elastic or plastic range may be expressed analytically, obviating the need of curves or tables. 2.2 STRESSES AND STRAINS AT THE LOAD CAUSING INITIAL YIELD IN THE UNRESTRAINED BEAM-COLUMN UNDER PLANAR BENDING In an unrestrained beam-column, when the yield stress is reached at the top fibers of the section shown in Fig. Ib, the resistance of that section 5

to rotation will be at a maximum and the member cannot sustain additional column load. In the restrained beam-column, however, there may be some reserve strength left beyond the load at initial yield. (See Section 4.) 2.3 STRESS-STRAIN RELATIONSHIP MODIFIED BY RESIDUAL STRESS A stress-strain curve to represent the average behavior of a steel angle in compression may be computed by assuming a residual stress pattern as shown in Fig. 3a. It is assumed that the stress-strain diagram in the annealed state would be as shown in Fig. 2. Stress E Strain Fig. 2. Simplified stress-strain diagram for steel. If (ay-caR) < aa < ay, the average stress on the section is aa = ay - aR (Xe/c)2 (1) The average strain is a E + -a 2(aR/ay)(xe/c (2) a = + y) Eliminating (Xe/c) from Eqs. (1) and (2), Ea = E + aR -2 R(ya) ) or 6

a = y R ) EEa (4a) - CR Alternatively, for p = ay-CR E2(ea-Ep)2 aa = -(v+bEa) a ) 4(~ IIyIy- 2 O'Xe,. Y R C b Fig. 3. a. Assumed residual stress distribution. b. Compressive stress distribution after a central compressive load initiates yielding. From Eq. (4-a), the effective tangent modulus may be determined Et = d = [ay+aR-Ea]. () d- a 2 2% Alternatively, Et could have been determined directly from Eq. (2) by means of the relationship that Et = e E = e E. (6) A c 7 e diAlthough test information afis limitedr a central comptheressive load indications that thyielding. ratio mR/Y decreases as) the effective tangepoint moduuincreases. may be deterassuming edR = 3 y

for steel of yield stress of 50 ksi, and. aR = 0.2 ay for steel of yield stress of 100 ksi, the following equation representing a linear interpolation will be assumed for UR/ay R =.40 --. (7) G y 500 When the stress-strain relationship is assumed to be modified by residual stress, it will be referred to as the "general stress-strain relationship" in contradistinction to the "elasto-plastic stress-strain relationship." 2.4 REGRESSION OF STRESS If the compressive strain at any point in the cross section decreases during a positive load increment, the stress will decrease locally at the constant elastic rate "E" as shown in Fig. 4, between points r and s. If ar and Er are the stress and strain at point r after which regression of stress exists and as and es are the regressed stress and regressed strain at point s then Es = r - (ar-cs)/E (8) The possibility of strain regression must be considered in the accurate analysis of a beam-column to obtain its ultimate loads. 2.5 STRESS-STRAIN RELATIONS IN PLANAR BENDING FOR GENERAL STRESS-STRAIN RELATIONSHIP If the section shown in Fig. lb is subjected to axial load P, bending moment M, and stresses at upper and lower corners, respectively, of aU and aB: A P (aU+*B) (9) M = c(cU-agB) 8

SR/E (R/E Stress I E..1 ()( ) Point of Tangency / //^^s) / s(oYoR L a =y -R [( + =) + _ E I~L? Fil6: CT: CT~R i' a/ EL[Y+RYR2 y^ ] J Strain (a ) Fig. 4. Average stress-strain relation resulting from residual stress distribution in Fig. 3. Solving P + M au = - U A S (10) P M A S in which A = total area of the section S = section modulus of the section = Ac2/c = Ac. 9

The strain at the upper and lower fibers will be obtained by substituting in Eq. (3). Denoting these strains by EU and eB, the curvature at this particular location, assuming a linear distribution of strain, is CU-CB UB (11) 2c Thus if the moment and axial load are given at a particular section, the curvature at the same section could be computed. 2.6 STRESSES AND STRAINS BEYOND THE LOAD AT INITIAL YIELD OF THE SQUARE SIMPLIFIED SECTION IN BIPLANAR BENDING FOR ELASTO-PLASTIC STRESSSTRAIN RELATIONSHIP Mx is taken as greater than My. For convenience in the computation, the diagonal axes (alternate principal axes for the square section), will be taken as reference x and y axes, as shown in Fig. 5. ^ y CYcTI2C 2 2 2 /3 2\cD= Strain Stress D= 2c 14 D D Fig. 5. Stresses and strains beyond the load at initial yield for biplanar bending and elasto-plastic stressstrain behavior. At any point along the beam, if the initial yield strain level is exceeded at one corner, the stresses and strains will be as shown in Fig. 5. From geometry and statics, letting D = 2 c 10

A P = T (ay+92+3+4+) MX = D( y-C4) (12) 4 A My = AD(o2-o3) solving 4Mx 4 = aY AD 2P 2(Mx+My) 52 = - + 2+ A(13) Y A AD 2P 2(Mx-My) 3 = - aY + A AD The curvatures will be: the curvature in the vertical plane 1 (s2+C3)-a4 a2+a3-2 ED ~2ED 1 the curvature in the horizontal plane =,_2-o. = My 2ED A D x 2ED EI 2.7 STRESSES AND STRAINS IN BIPLANAR BENDING FOR THE SQUARE SIMPLIFIED SECTION FOR THE GENERAL STRESS-STRAIN RELATIONSHIP For convenience in the computations the diagonal principal axes are again considered as reference axes. The stresses and strains for the case of biplanar bending and with a stress-strain relation similar to that shown in Fig. 4, at any section of the beam-column, will be as shown in Fig. 6. From statics we have 11

P = C (l+a2+a3+J4) Mx = D(a-) (1) My = D( 2-a3) solving in terms of a, 4Mx (74 = 1- - AD 2p 2(Mx +My) a2 = - + 2P + (16) A AD = - 1 2P + (MxM)' A AD in which D = 2 c. D2 Y 3 iD D D Fig. 6. Stresses and strains beyond the proportional limit stress for the biplanar bending case. Equation (16) is not enough to determine the four stresses a1,a2,a3, and a4. The additional condition results from the linearity of the strain distribution. From Fig. 6, 12

1 + 4 = 2 + 3 ~ (17) If the stresses and then the strains are known, the curvature will be: x = (ci-4)/2D (18) Oy (E2-E3)/2D. If both a2 and a3 are less than ap then Oy =My/EIin which ox, y are the curvatures in the vertical and horizontal planes respectively. If the stresses a2, a3, and a4 are within the proportional limit stress, then from Eqs. (17) and (4) c2 + 3 - =4 1 or E E 2ay - ap) - 2 (ay-al)(ay-ap)) By adding the components of Eq. (15) together and solving for al one gets = (- ( 9 ) aR-2F) + /(- aR-2F)2-4(F) 2 - t aR)) (19) 9 9y in which 1 4P 8Mx F = A + A (2ay-ap) After assuming a2, a3, and a4 less than ap, if it turns out that the stress a2 exceeds the proportional limit, then the relation between a2 and E2 is not linear and should be obtained from Eq. (4). In this case the equation for ao will take the following form: a1 = AI+ 5My +. (y 2P 2(M+ ) )-aR * (20)

Equation (20) may be solved by trial and error. If the assumption that only al and 02 exceed the proportional limit is incorrect, then Eqs. (16) and (17) may be solved directly by trial and error. If ol = cyy there is still some reserve strength, and the stresses will be obtained by putting oa = ay in Eq. (16). The resulting values of c2, a3, and o4 will be those given in Eq. (13). The strains c2, ~3, and E4 will be obtained from Eq. (9) and the strain e~ is = ~2 + C3 - E4 If oa and a2 both reach the yield point there will be no reserve strength left of the section. The ultimate for unrestrained beam-columns in such a case is that load at whicho 2 just reaches the yield point Example: For a simplified section of the following properties A = 40 sq in.; c = 15 in.; E = 29,000 ksi cy = 50 ksi; up = 35 ksi and subject to an axial force of 1,200 kips and bending moments of Mx = 8,485 K-inch and My = 4,243 K-inch, find the stresses, strains and curvatures. Solution: D = 15 = 21.213 in., I = 2 x 10 x (21.213) = 9,000 in. By the ordinary flexural formula P MXD = A+ - = 30 + 20 = 50.00 ksi A Ix thus Oi>ap. From Eq. (19) =R = 15 ksi p = 50-15 = 35 ksi lke

1o + 8 x 8.485 10o+35 = 45.0 F, 1 120^ +I 8OQ45 10 + 3 45 0 3 40 x 21.213 J ai - ( 15 90) + - 905 9) 4(452 4 50 x 15 48.3 ksi. By Eq. (16) 2 = - 48.3 + 60 + 2 x 12.728 483 +.90 = 41.7 ksi 40 x 21.213 In this case c2 > aPL, and Eq. (20) is applied: ~J 30 +r ^^x8,485-4.243 + o0-6 -5 0+a-600 -2 15 0= + x21.213 + - 40x21.213 J then = 55 + 3.873 t ~ -f i -J -1 0 Solution by trial Assumed ol* Computed cL 48.30 48.89 48.60 48.22 48.41 48.65 48.53 48.38 48.46. 48.55 48.51 48.43 48.47 48.52 48.50 48.46 48.48 48.50 Thus 14 = 48.49 ksi. *First assumed value of al is that obtained by assuming a2 < apL. Other assumed values of ol = (last assumed al+last computed 01)/2. 15

If Eqs. (15) and (16) are used to compute the other stresses, then 2 = 41.51 ksi, a = 21 51 ksi, and a4 = 8.49 ksi From Eq. (3) = 1 ((50 + 15) - 2 (50-48.49)13 } = 1.910 x 10-3 29000 E2 = (5(0 + 15)- 2 (50-41.54)15 = 1.464 x 103 29000 and 3 21. - 0.743 x 10-3; c = 29000 29000 As a check ~2+E3-c4 = 10-3(1.464+.743+.292) = 1.915 x 10-3 = E From Eq. (18) the curvatures are = (.P4l) = (1.91-.292)103 /42.426 = 3.8137 x 10-5 rad/in. = (D) = (1.464-.743)103/42.426 = 1.6994 x 10 5 rad/in. The same example will now be solved b trial: From Eq. (16) the stresses U2 (73, and U4 in terms of oI are 4Mx 4 x 8,485,4 = 1 =: - 40= I AD 40 x 21.213 2P 2(M x+My) 2 x 1200 2 x 12,728 _ 2 = - 1 +A + AD 40 40 x 21.213 2P 2(Mx-My) 2 x 1200 2 x 4,243 + ~ s= -AD + ~ + -- + AD _7 "3 = ^^^40 + ~ 40 x 21.21 - 16

First. Cycle: MxD Assumed l = + = D 0 ksi Assumed Ei = 65 A I5 Thus Ca = 40 ksi Ee2 = 40.5 U3 = 20 ksi Ee3 = 20 o4 = 10 ksi Ee4 = 10 Computed Eei = E(E2+E3-E4) = 50.5 Second Cycle: Assumed Ee1 = (Assumed Ee1 in the first cycle + computed EE1 in the first cycle)/2 65+5o. 50 G1 = 49.12 ksi Assumed EE1 = 26 5 = 57 C2 = 40.88 ksi EE2 = 41.61 U3 = 20.88 ksi Ee3 = 20.88 4 = 9.12 ksi EE4 = 9.12 Computed Eel = E(E2+E3-E4) = 53 37. At this stage, an interpolation between two results is very helpful. Consider Fig. 7. If T1c = assumed strain ci = computed strain The linear interpolation between the two results (TEl,GO & Te, Ek) is Te~-E~(TcE-Te~)/(c[-E~) l-(T= -Te) /(. -e) 17

Assumed Strain o o / Interpolated Value \45~I Computed Strain eI Fig. 7. Interpolation procedure. Points of less difference between Te1 and. e1 should be used in the interpolation. Thus, Third Cycle: (T )E -65-50.5(57.75-65)/(55.7-50.5) 5461 1-( 57-.75-65) /(53.37-50 5) a1 = 48.20 ksi Eel = 54.61 G2 = 41.80 ksi Ee2 = 42.82 s3 = 21.80 ksi EE3 = 21.80 c4 = 8.20 ksi EE = 8.20 Computed Eel = E(e2+C3-c4) = 56.42. Fourth Cycle: (T1 )E 54.61-56.42(57.75-54.61) /(.57-56.42) = 553 1-(57.75-54.61) /(53.7-56.42) 18

C = 48.50 ksi EEI = 55.53 c2 = 41.50 ksi EE2 = 42.42 a3 = 21.50 ksi Ee3 = 21.50 4 = 8.50 ksi Ee4 = 8.50 Computed Eei = E(c2+E3-c4) = 55.42. Fifth Cycle: (Te)E 54.61-56.42(55.53-54.61)/( 5542-56'42) 55.48 1-(55 53-54.61) /(55.42-56.42) 71 = 48.49 ksi Ee1 = 55.48 C2 = 41.51 ksi EE2 = 42.43 a3 = 21.51 ksi Ee3 = 21.51 a4 = 8.49 ksi Ee4 = 8.49 Computed EE1 = E(E2+E3-E4) = 55.45 Good agreement is obtained between the assumed and computed e1. 2.8 EFFECT OF INITIAL CURVATURE The effect of initial curvature is of comparable importance to residual stress, especially so in the case of high strength steels. In the case of rolled sections, the specified tolerances (ASTM Specifications A6) for out-ofstraightness may be used as a basis for estimating the initial curvature, as shown in Fig. 8 by the full line. On the basis of the dashed line approximation for out-of-straightness tolerance, the permissible variation bi is bi(initial curvature) =.000833L (21) in which 5i and L are in inches. If the additional bending moment due to initial curvature is assumed to take the same form as that due to a uniform lateral load on the beam, then an 19

1.21 t) 3i2 z.8......... __,,__/_ (co I-6 ^ //, fon wid ^lnges / ASTM Spec. A6 ~z.4~~ / %"Approximate (8, L) Relation 0..___________________________________ 0 20 40 60 80 100 L IN FEET Fig. 8. Permissible variation in straightness for wide flange shapes by ASTM Spec. A6. 20

equivalent value (Wi) of this lateral load may be taken as WiL 8 = Pi = P x.000833L If the lateral load Wi is given as a ratio of the axial load P, then L kiP = P x.000833L or ki =.00667 ~ (22) In which ki is the ratio of an equivalent uniform lateral load, introduced to allow for initial curvature, to the axial load. If the ultimate load of a beam-column is obtained without the effect of initial curvature, a value of uniform lateral load equal to kiP should be added to the original lateral loads in both principal directions. 21

3. APPROXIMATE DESIGN PROCEDURE FOR BEAM-COLUMNS 3.1 UNRESTRAINED BEAM-COLUMN DESIGN BASED ON LOAD AT INITIAL YIELD A lower bound to the ultimate load capacity of a beam-column is the load at initial yield. Such an estimate is restricted to those cases in which failure is due primarily to excessive bending. It should be noted that if the initial yield procedure is to be used to estimate the strength of columns with small eccentricities or small lateral loads, it is necessary to assume that some unintentional eccentricity exists, in order that the initial yield strength curve approaches a proper limit i.e., the strength curve for a concentrically loaded column, as the lateral load or end eccentricity diminishes to zero. The effect of all imperfections such as initial crookedness, nonhomogenity, residual stresses, etc., are lumped into the equivalent initial eccentricity A good approximation for estimating the maximum bending moment in columns with lateral loads is given by M = Mo + (23) - PPe in which Mo and 60 are the maximum moment and deflection respectively without regard to the added moment caused by deflection. Equation (23) can be written as P__ 1 + P e M = Mo ~ e (24) where 5EI a structural handbook. Some values of t for different cases of loading are given in Table II. The maximum combined stress is therefore am = A+ (26) 23

in which a = P/Pe' The load at initial yield (Pp) can be obtained. simply by putting am =Oy in Eq. (26), then solving for Pp. Equation (26) gives estimates that are in error by less than 1% for all usual cases of load distributions at all levels of P. As mentioned in Section 2.2, the unrestrained four point cross section in planar bending reserves no resistance to rotation beyond the load at initial yield. Thus, the ultimate load for unrestrained beam-columns of the simplified cross section in planar bending is the load at initial yield. 3.2 LOAD AT INITIAL YIELD FOR THE UNRESTRAINED BEAM-COLUMN (SIMPLIFIED FOUR POINT CROSS SECTION) IN PLANAR BENDING UNDER UNIFORM LATERAL LOAD (kP) For the case of the unrestrained beam-columns in planar bending under uniform load kP Eq. (26) may be written: kPL P 3- rl+.0280 8 =m A+ [ Ar (27) m A ^ ^ J Ar2 Putting cm = ay and solving.for P/A, kLc + kLc N)2 O.028kLc + +1 ( + 1+ -4 (- O2 A e A = ae.r..e. - (28) L0^ " 2(1-.028 kLc/8r2) in which oe = Euler stress equal to 72E/(L/r) r = radius of gyration. For the simplified section in which I = Ac2 or r = c Eq. (28) reduces to 24

2 CT.028kL P = { + (e 8 l 1 + 8 ) kL ) (1 _ k028kL + }4 P Ge 8rj e 8r G e 8r A =....~ (29) 2(1-.028 kL/8r) Equation (29) is a function of L/r, E, oy, and k. The modulus of elasticity of steel is usually taken as 29,000,000 psi. Column slenderness ratios L/r vary between 0 and 200. The various yield stresses, that cover the current constructional steels, are ay = 33, 36, 42, 46, 50, 60, 70, and 100 ksi. In Table I, Eq. (29) has been solved to cover this wide range of parameters, together with variation of the load ratio k between.01 to 0.3, in increments of.01. 3.3 BEAM-COLUMN DESIGN BASED ON THE LOAD AT INITIAL YIELD FOR OTHER CROSS SECTIONS AND LOADING CONDITIONS FOR PLANAR BENDING The solution, as tabulated in Table I is for c/r = 1, applicable to a column with the area concentrated at the full depth, as in the case of four corner angles or tubes connected on all four sides by lacing bars. Modification to other values of c/r may be made by simple multiplication. In using Table I, a value of k = 0.01 should be added to take care of column imperfections. This allows for an initial curvature of 20% more than allowed by usual mill tolerances for camber or sweep and is based on an assumed maximum deviation from straightness of L/800. It is 50% more than the value of 0.0067 arrived at by ASTM tolerance specifications. The additional amount will roughly compensate for other imperfections, such as residual stress, that are not explicitly included. Thus, the equivalent lateral load moment is kPL PL = - - and k = 0.01 8 800 In addition, for nonuniform lateral loads, an equivalent uniform load may be established by multiplying k by (1) A load distribution factor, and (2) A deflection factor. The load distribution factor is the same as that tabulated in the beam tables of the AISC Manual and is IF = 8Mo/WL. (30) 25

The deflection factor is p DF = 1 (31) 1+0.0281 P e but for design purposes may be approximated simply by DF = 1+ ( (32) e Some values of both the load and deflection factors are tabulated in Table II. For end eccentricities el and e2, where el < e2, the load factor is kecc = Ls1] + 0.23 (33) L PeJ but is not to be less than kec=3.2e2 F +0.25 kecc = + 023 Pe (34) Finally, the combined equivalent load factor must be multiplied by the ratio of c/r appropriate to the cross section being used. Equation (33) in different form corresponds to CRC and AISC recommendations. In summary, then for any beam-column situation, 1. Calculate kw = W/P. 2. Add kI = 0.01 for column imperfection. 3. Add kecc by Eq. (33) if there are applied end eccentricities. 4. Estimate approimmate P/Pe (or ra/%e). 5. Determine load distribution factor (Eq. (30) or Table II), deflection factor (Eq. (32) or Table II), and c/r for section type to be used. 26

6. Determine equivalent uniform load ratio Qor use in Table I, as follows: ke [(kw)(LF)(DF) + kcc + (0.01)]. (35) 7. Enter Table I with the chosen ay, the computed ke, and the estimated L/r. 8. Determine P/A at yield and divide by the desired safety factor. 9. Select column and redesign if necessary as may be indicated. For uniform lateral load with no eccentricity of column load,'the procedure is simplified because the load and deflection factors are unity, and kecc = 0. In numerous other cases the load factor is close to unity and may be taken as such. (See Table II.) Example 1. Design by Use of Table I a Beam-Column for Load Conditions as Illustrated Below: 16K K 300300- L- 60o j, L8" 9/16" 24" ry 50Ksi After preliminary trial design the above cross-section is arrived at for check analysis. A = 42.41 in.2 I = 3054 in.4 27

Sxx = 3054/12.28 = 248.7 in.3 The average stress P/A (load factor = 175) =300x1.75/42.41 = 12.38 ksi. L/r = 84.8. Pe = (2900) 39.80 ksi. (84.8)2 Moment Factors for Use in Table I 1. Load distribution factor = 2. 2. c/r = = 1.414. 3. Deflection factor = 1-.018 x 1.8 = 944. 39.80o 4.8x8 4. Eccentricity kecc = (1+0.23(11.67/39.80)) = 0.057. 720 The dead weight of the tube is 7.68/37.7 x 42.41 = 8.64 K for which kw = 8.64/300 = 0.029; crookedness allowance k = 0.01. ke = (.029 x 1 x 1 +.02 x 2 x.944 +.057 +.01)1.414 = 0.190 Referring to Table I, and interpolating between L/r of 80 and 90, for oy = 50 ksi and k = 0.190, the maximum stress reaches ay = 50 ksi at aa = 12.59. This being greater than the trial Ga = 12.38 ksi, the design is confirmed. 3.4 UNRESTRAINED BEAM-COLUMN DESIGN BASED ON THE LOAD AT INITIAL YIELD FOR BIPLANAR BENDING CASE Table I may be adapted to the case of biplanar bending. Under the assumption that the maximum moments in each of the principal planes of bending occur at the same location, Eq. (27) may be adapted to biplanar bending (modified for the simplified section by putting c = r): p f+.028ax P L +.028 p L 1 a A h kxe 8A r ) k ye 8A r (36) in which kxe and kye are the equivalent uniform load ratios, for bending about the principal axis x and y respectively, ac and ey are the load ratios P/Pex and P/Pey, and rx and ry are the radii of gyration about the principal axes x and y respectively. Where Pex and Pey are EulerTs loads about the principal axis x and y respectively. If the following two factors are introduced: 28

1e The (slenderness-ratio) ratio factor SFR = L-/L = (3) ry rx ry 20 The amplification ratio factor AFR = 1+.o28y / l+o0280x (38) l-oy I-aX which can be approximated by: AFR 1- x Then, Eqo (36) may be rewritten as +l.028s "' P L k =a A + l- kxe 8A r }1 + kxe (SFR)(AFR)[ (39) If the biplanar bending equivalent uniform load factor kre is introduced as k^e k= xe + (SFR)(AFR) k(.) (40) Then Table I, can be used to find the load at initial yield for unrestrained biplanar bending. The procedure will be as follows: 1. Assume P. 2. Find the equivalent uniform load ratios for bending about both principal axis from Eq. (35). 3. Find (the slenderness-ratio) factor ratio (SFR), from Eq. (37). 4. Find the amplification factor ratio (AFR), from Eq. (38). 5. Find the biplanar equivalent uniform load ratio from Eq. (40). 6. Enter Table I with the chosen ayy the computed kBe, and the estimated (L/rx). 29

7. Determine P/A at yield and divide by the desired safety factor. 8. Select column and redesign if necessary. Example 2. The beam shown is subjected to a uniform vertical load of 20 Kips, together with a hrizontal concentrated load of 4 Kips acting at mid section. Calculate the maximum load P if this yield point is 33 ksi. Plate 18 x.88", Uniform vertical load = 20K 1.8U58 " 35= 0" 1 5.68Horizontal concentrated load 18" = 8 Plate 18 x y 4 A = 2 x 16.98 + 2 x 18 x 3/4 = 60.96 in. Weight of beam = 7.29 Kips Ix = 2 x 670.7+2 x 18 x.75 x (9-375) = 1541.4+2373.05 = 3714.5 in.4 Iy = 2 x 18.5+2 x 16.98 x (5.68)2+2 x 3/4 x 183/12 = 37+1095.6+729 = 1861.6 in.4 rx = 43714.5/60.96 = 7.81 in. Cx = 9.75 in. ry = 41861.6/60.96 = 5.52 in. Cy = 9.0 in. xL/r = 55.78 Sx = 380.97 in.3 L/ry = 76.10 Sx = 206.84 in.3 cx/rx = 1.25 cy/ry = 1.63 3o0

aex = T'cE/(L/rx)2 = 98.96 3ksi ey = - 2E/(L/ry)2 = 49.42 ksi SFR = rx/ry = 1.415. Assume P = 600 Kips, with assumed factor of safety = 1.65 ca = 600x1.65/60.96 = 16.24 ksi Ox = 16.24/98.96 =.164 y = 16.24/49.52 =.329 Bending About xx Bending About,y Load factor 1 2 c/r 1.25 1.63 Deflection factor 1 1-.18x329 =.9407 27.29 04554 kw =.0455.0o67 ke (,0455xlxl.+01) (.0067x2x.9407+.01) 1.25 =.0694 1.63 =.0368 AFR 1-.164 = 8 = 1246 1-.329.671 kBe = 0694(1+1.415x1.246 x ) =.1343.0694 By interpolation from Table I for L/r = 53.78 and k =.1343, gives ~a = 15.9 ksi. Assume P = 580 Kips with assumed factor of safety = 1.65 aa = 580x1.65/60.96 = 15.70 ksi Cx = 15.70/98.96 =.159 y = 15.70/49.42 = 3.18 31

Bending About xx Bending About yy Deflection factor 1 1-.18x.138 =.9428 27.29 4 kw 2" =.o0471 =.00oo69 580 580 ke (. 0471xlxl+ 01) (.0069x2x.9428+. 01) 1.25 =.0714 1.63 =.0375 AF = 1-.159/1-.318 =.841/.682 = 1.233 kBe =.0714(1+1.415xl.233 x.0375/.0714) =.1368 By interpolation from Table I for L/r = 53.78 and k =.1368, gives oa = 15.77 ksi > 15.70. 3.5 LOAD AT INITIAL YIELD FOR THE RESTRAINED BEAM-COLUMN OF SIMPLIFIED SECTION IN PLANAR BENDING Usually there is no direct solution for the load at initial yield for a restrained column in planar bending under uniform lateral loads. The solution may be obtained indirectly, using the computer, by increasing the loads on the beam-column until the maximum stress, at either the end or middle section, equals the yield stress. In Table III, Appendix A, the average stress at the load at initial yield for the simplified section for the symmetrical case under uniform lateral load is given only for a yield stress of 50 ksi and E = 29000 ksi.* This average stress is given as a function of the following dimensionless parameters: 1. Slenderness ratio L/r, 2. Ratio of lateral load to axial load k, 35 r = 3L/1OEI, in which P is the rotational stiffness summed for all the members, except the beam-column itself, rigidly connected at one end of the beam-column. L/r ratio ranges between 10 and 200 with increments of 10. k ranges between.02 and.3 with increment of.02 and i = 0.2, 0.4, 0.6, 0.8, and 1.0. The case q = 0 is that of the unrestrained column and is given in Table I. Reference should be made to Section 5.1 for equations used in this solution. *This particular program was developed near the end of the contract period, in 1964, and funds were not available to tabulate results for all yield points. This work will be completed under the Phase 2 continuation of the project. 32

4. ULTIMATE STRENGTH OF UNRESTRAINED BEAM-COLUMNS 4.1 ULTIMATE STRENGTH OF UNRESTRAINED BEAM-COLUMN OF THE FOUR POINT SECTION IN PLANAR BENDING, FOR THE GENERAL STRESS-STRAIN RELATION Incremental procedure will be used to determine the ultimate strength of the unrestrained beam-column of the simplified section in planar bending. The procedure is as follows: 1. Find the load at the proportional limit stress from Eq. (28), by putting am = ap and solving for P. For a beam-column of the simplified cross section subject to a uniform lateral load W = kP, P 2 x (l- L28 ) J ---- + 10 + -b-28 - + -4 - 028% r a 8 r' r 2. Increase the load P by an arbitrary increment AP, say Py/200, and use Newmark's numerical procedurel2 to find the beam deflection. A good assumption for the initially assumed deflection curve is the deflection curve due to the lateral loads multiplied by the amplification factor 1/(1-P/Pe). 3. Repeat Step 2, until the numerical procedure method does not converge, in which case the nonconvergent load is greater than the ultimate load. Convergence in this report is considered to occur when the ratio between the assumed deflection and the calculated deflection curve at any nodal point is less than or equal to 1 ~.005. Divergence will be considered when this condition is not satisfied within 200 cycles. If the maximum stress at any stage equals ay, the ultimate load will have been reached. Table IV in the Appendix gives the ultimate average stress (P/A) for unrestrained beam-columns of the four point cross section subject to uniform lateral load W = kP. A comparison between the ultimate strength as given in Table IV (with the effect of residual stress included) and the strength based on the load at initial yield (ultimate without the effect of residual stress in case of the simplified section), shows that the ratio between the two strength's is much closer to unity than the ratio between proportional limit determined by the residual stress and the yield stress. This average stress at ultimate is given as a function of the slenderness ratio, which ranges from 10 to 120 with increments of 10, and the k ratio, which ranges from 0.02 to 0.30, with increments of.02. Each table corresponds to a certain yield stress. The yield stresses include 33, 36, 42, 46, 50 60, 70, and 100 ksi which cover the current constructional steelso E is assumed to be 29,000 ksi. The effect of residual stress is included and is assumed to be present as discussed 33

in Section 2.3 and in the proportion given by Eq. (7). Illustrative Example 3 is given to show how the ultimate strength may be computed. The numerical calculations are made by using the IBM 7090 digital computer. In Appendix B, Flow Diagram No. 1, is given for finding the ultimate strength of the unrestrained beam-column under uniform lateral load W = kP, for the simplified section with the effect of residual stresses. This flow diagram was the basis for a computer program that requires as data the following, with computer names shown in parentheses: L (L) = length of beam in inches c (C) = half depth of section in inches k (K) = the (lateral load)/(axial load) ratio A (A) = area of section in square inches E (E) = modulus of elasticity of steel in ksi ap (SIGPL) = proportional limit stress (ay-aR) in ksi ay (SIGY) = yield stress level in ksi N (N) = half number of beam divisions. This program (No. 1) is given also in the Appendix. Example X. For an Unrestrained Beam-Column of the Simplified Section, of the Following Properties: L = 900 in. c = 15 in. A = 40 sq in. E = 29,000 ksi cp = 35 ksi Oy = 50 ksi 34

and subjected to axial loads and uniform lateral load of ratio.04 to that of the axial load, find the utlimate load. 1. With L/r = L/c = 60, I = 9000 in.4 and S = 600 in.3, compute P by means of Eq. (41). ae -=:2E/(L/r)2 = 79.505 ksi. By substituting in Eq. (41) Pp = 974.57 K. 2. Take AP Ary/200 = 40x50/200 = 10 Kips. For an increment of P of 10 Kips one cycle of the integration to determine deflection will be illustrated. The beam is divided into 8 divisions. For each load, the deflection at nodal points are computed, the assumed deflection having been arrived at as the iterative result of many earlier applications of the same procedure, as follows: W =.04 P P 1124.6?-=112.5" —- --- 112.5"< >^ r 112.5" ~ < ~ 112.-5"1~> Common Factor Moment due to lateral load, Mo(K-in.) 2,214 3,796 4,745 5,061 Assumed deflection, in. 1.0069 1.855 2.419 2.617 Pxy(K-in.) 1,132 2,086 2,720 2,943 M = Mo+Py 3,06 5,882 7,465 8,004 M/S (ksi) 5.58 9.80 12.44 13.34 aU = P/A + M/S (ksi) 33.70 37.92 40.56 41.46 OB = P/A - U/S (ksi) 22.54 18.32 15.68 14.78 EU 1.162 1.315 1.422 1.465 10-3 eB.777.632.5407 -5097 10-3 eU-EB 385.683.875 -955 10-3 Curvature $.128.227.292.318 10-4 Concentrated angle change 1.413 2.52 3.248 3.528 10Average slope 8.945 7-532 5.012 1.764 10-3 Deflection 8.945 16.477 21.489 23.253 \x10-3 Deflection (in.) 1.0063 1.854 2.417 2.616 Assumed deflection/ computed deflection 1.0006 1.0005 1.0008 1.0004 35

The above Newmark numeri.cal calculation of deflection and moments for the case of an unrestrained beam-column under uniform lateral load in planar bending is shown only for the final cycle. Whenever a < ap, then e = a/E, otherwise Eq. (3) may be used to calculate the strain. The-average stress, ca, at the foregoing load of 1124.6K is 28.11 ksi, the foregoing being a manual computation of one stage of the computer print out as marked by arrows in the following tabulation: P/A DEFLECTION IN INCHES AT NODAL PCINTS 24.61.8C32 1.4767 1.9217 2.0769 24.86.8151 1.4986 1. 503 2.1C79 25.11.8272 1.5210 1.9794 2.1394 25.36.8396 1.5438 2.CC92 2.1717 25.61.8522 1.5671 2.0397 2.2C48 25.86.8651 1. 5908 2. 709 2.2386 26.11..8782 1.6151 2.1C28 2.2732 26.36.8928 1.6423 2.1387 2.3123 26.61.9C70 1.6687 2.1135 2.35CC 26.86.9218 1.6961 2.209r5 2.3892 27.11.9370 1.7245 2.2469 2.4298 27.36.9528 1.7539 2.2857 2.4720 27.61.9711 1.7880 2.3307 2.5208 27.86.988b 1.8201 2.3739 2.5678 -_28.11 1.0069 1.8549 2.4191 2.616928.36 1.0261 1.8907 2.4664 2.6685 28.61 1.0462 1.9283 2.5162 2.7226 28.86 1.0696 1.9722 2.5744 2.7859 29.11 1.0924 2.0150 2.6311 2.8477 29.36 1.1166 2.0605 2.6915 2.9135 29.61 1.1425 2.1092 2.7562 2. 840 29.86 1.1703 2.1615 2.8257 3.C599 30.11 1.2 2C37 2.2244 2.9C95 3.1513 30.36 1.2374 2.2880 2.9943 3.2440 30.61 1.274b 2.3586 3. C887 3.3472 30.86 1.3214 2.4466 3.2066 3.4764 31.11 1.3163 2.5506 3, 3.465 3.6301 31.36 1.4473 2.6858 3.5295 3.8321 3. For the next increment of load P beyond 1254.4 Kips (31.36 ksi), at which the maximum stress is P kPL PYmax 1254.4.04x1254.4x900 1254.4x3.821 + + 6max -A 8S S 40 8x600 600 = 31.3 + 9.408 + 8.01 = 48.778 ks i the maximum stress was found to be equal to oy = 50 ksi. Thus the ultimate load is closely estimated as 1255 Kips.

The use of Table IV will now be illustrated by Example 4. Example 4. By use of Table III, find the required area of cross section for proportion as indicated based on a required ultimate load capacity as shown. W 43.33Kips Kr 2 I~,~^~^~A~ ~ ~ ~ ~ P=1000 c —o ~ ^ ~ ^ ~ ^ ~ ^ ~ ^ ~ ^ ~ ^ ~^ ~,~ - ~ 2c= / o - ~1100" L oy=50 Ksi k = 43.33/1000 +. 6 =.050 Initial Curvature Allowance L/r = L/c = 1100/20 = 55. By interpolation from Table III (Appendix) P/A = 31.57 ksi or A = 31.68 sq in. 4L 3-1/2 x 2-1/2 x 7/16 of A = 8.3 x 4 = 33.2 sq in. will be chosen. If the load P is not known, find the ultimate load this member can carry, for an area A = 40 sq in., and c = 20 in. Assumed P/A k Computed P/A From Table III 30.0361+.00667 =.0428 32.76 33.0328+.00667 =.0395 33-36 33.4.0324+.00667 =.0391 33.43 (thus Pult = 1336K) If P = 100K, W = 43.33K, A = 40.00 sq in., find the smallest c k =.05 P/A = 25 ksi. From Table IV by interpolation between L/r = 70 and 80 for k =.05, L/r = 73.6, (c=r) or c = 14.95 in. _ 15 in. 37

4.2 ULTIMATE STRENGTH OF THE UNRESTRAINED BEAM-COLUMN OF THE FOUR POINT SECTION IN BIPLANAR BENDING FOR THE GENERAL STRESS-STRAIN RELATION The procedure for finding the ultimate strength of the unrestrained beam-column in biplanar bending is essentially the same as that for planar bending. The procedure may be summarized as follows: 1o Find the load at the proportional limit stress from Eq. (28), by putting am = ay and c = D (see Fig. 5). This value will be dependent on the largest lateral load in one of the diagonal directions compared to the other direction. For a beam-column subject to uniform lateral loads: (42) Wx = kXP In which ky and kx are the ratios of the lateral loads to the axial load in the diagonal directions and ky > kx. The load at the proportional limit stress is ( + 1 + kyLD/8r2) - aP + 1 + kLD/8r2) -. (.28ki) PP = ACre 2 A Pp = Aae e "e y e 8r8r C~~i ~2(1-.028kyLD/8r) (43), 2. Increase the load P by an arbitrary increment AP, say Py/200, and use the numerical procedure to find the deflections in both directions. This can be accomplished by assuming deflection curves in both directions equal the deflection curve corresponding to the lateral loads in the same direction multiplied by the amplification factor l/(1-P/Pe) or 1/1l-p. After the deflection curves in both directions have been assumed, the bending moments Mx and My will be tentatively evaluated, and thus the curvatures at any nodal point may be obtained as outlined in Section 2.7. The new deflection line will be computed in both directions. The procedure will be repeated until a desired degree of correspondance is achieved between the assumed and computed deflection curves in both directions. 3. Repeat Step 2, till the numerical procedure method does not converge, this means that this load is greater than the ultimate load. Convergence in this report is assumed when the ratio between the assumed deflection and, the calculated deflection is less or equal to 1 +.005. Divergence will be assumed when this condition is not satisfied within 200 cycles. If it happens that both the stresses ao and a2 at any location equal the yield point, then no reserve strength of the beam-column will be left. In such a case the ultimate load is that load at which al and a2 equal at any location the yield point. 38

In Appendix B, Flow Diagram No. 2 is given for finding the ultimate strength of the unrestrained beam-column under uniform lateral load Wy = kyP, and Wx = kxP, for the simplified section with the effect of residual stresses. This flow diagram was the basis for a computer program that requires the following data: L = length of beam in inches c = half depth of section in inches ky = ratio: (vertical uniform lateral load)/(axial load) Q = ratio: (horizontal lateral load)/(vertical lateralload)i.e.,kx/ky A = area of section in sq ino E = modulus of elasticity of steel in ksi Up = proportional limit stress (.y-ER) in ksi N = half number of beam divisions. It may be noted that both ky and kx/ky should be related to the x and y axis not to the diagonals. The program converts these ratios as follows: k* = ky(l+Q)/f (44) k = ((l-Q)/(1+Q))k* in which Q = kx/ky and k-y and kx are related to the diagonal axis x* and y* as shown in Fig. 9. 39

y y W =K P \/45 Wx=KP=QWy X Fig. 9. Diagonal principal axes used for biplanar bending. Example 5. An unrestrained beam-column has the following properties: L = 900 in., c = 15 in., A = 40 sq. in., E = 29,000 ksi, CR = 15 ksi, aPL = 35 ksi, y = 50 ksi. It is subjected to vertical lateral load of Wy = 3 F/100 P, and to horizontal lateral load Wx = 2/200 P. Find the ultimate load. /=15" 1. Loads in the -y* direction = (3 1/100 + 2 /100) x P/ =.04P. Loads in the -x* direction = (3 /100 - /100) x P/ =.02P. 2. Load at proportional limit from Eq. (42). I = 9,000 in., r = 15 in., D = 21.213 in., S = 424.26 in.3, k3 = 0.4, oe = 2E/(L/r)2 = 79.5 ksi, PPL = 879.74 Kips. 40

3- Increase the load P by increments AP = Aay/200 = 10 Kips and perform numerical procedures similar to that illustrated in the next page. Stop computation when the procedure does not converge, or when al = 2 = oy. Typical Newrnark12 deflection calculations at nodal points by the Newmark procedure shown only for the last cycle, meeting the convergence criterion, at one of the intermediate load stages, are as follows: Wy =.04P = 40.39K, Wx =.02P - 20.19K P= 1009.74 l P/A = 25.2435 ksi ~- l2~ -- L = 900 in. Moments and Deflections in the x Direction Common Factor Mox (Kips:x in.) 1,988 3,408 4,260 4,544 Assumed y(in.).846 1.558 2.030 2.196 Py (Kips x in.).854 1,573 2,050 2,217 Mx =Mo+Py 2,842 4,981 6,310 6,761 P/A + Mx/S (ksi) 31.942 36.984 40.116 41.179 xEI 2,842 5,004 6,419 6,934 Concentrated x 33,424 59,301 76,128 2x41,089 X/12EI Average slope 209,942 176,518 117,217 41,089 %/12EI Deflection 209,942 386,460 503,677 544,766 x2/12EI y (in.).848 1.562 2.035 2.201 Yassumed/Ycomputed 9976.9974.9975.9977 Moments and Deflections in the y Direction Moy (Kips x in.).994 1,704 2,130 2,272 Assumed x (in.).417.766.997 1.077 Px (Kips x in.).421.773 1,007 1,087 My = Moy+Px 1,415 2,477 3,137 3,359 ~EI 1,415 2,477 3,137 3,359 Concentrated EI 16,627 29.322 37,206 2x19,932 X/12EI Average slope 103,087 86,460 57,206 19,932 -/12EI Deflection 103,087 189,547 246,685 266,617 X2/12EI x (in.).417.766 -997 1.077 Xassumed/Xcomputed hii

Details of computations for Ox EI and 0y in the foregoing table are as follows: At 1/8 L F = — + AD- (2y -aPL) 3 100974 + - 65 27.65 3 A AD 1340x21.213 = - (9 R-2F) + 9 yR-2F)-4(F 9 y<R) = (9 - 55 )+ 348.6) -4(764.62-333.) = 36.96 ksi C = r - 4Mx/AD = 36.96 - 4x,98 = 13.48 ksi 40x21.213 va2 = - - + 2P/A + AD =-36.96+50.49 + 31.11 ksi 40x21.213 2(Mx-My) 2(45981-2,47.3 = - -. + 2P/A + AD -36.96 + 50.49 + 77) 19.43 ksi AD 40x21.213 E1E = ((2yPpL)- 2 (( i)R 3 = (65-7.746 50-36.96 ) = 3707 /xEI = ((E1-e4)/2D)EI = 9000 E(E1-e4) 5004.1 2x21.215 $yEI = My If a2 and 03 < YPL, yE I = My.

At./8 L F = 100.974 + 8x6,1 - 65 = 31.82 3 40x21.21_3 ac = [ (- - 63.64) + (56.97)2-4(1012.51-333.33) ] = 39.99 ksi 2 9 2 = - 39.99 + 50.49 + 2(6,310+3,137) 32.77 ksi 21.213x40 3 = - 39099 + 50.49 + 2(6310-3,137) = 17.98 ksi 40x21.213 0 = 39.99 -4x6 3 = 10.24 ksi 40x21.213 1E = (65 - 7.746 50-39.99 = 40.50 ~XEI = 9,000 E(El-e4) = 6,419 2x21.213 oiI = My At L/2 F = (100.974 + 8x6761 65) = 3357 3 40x21.213 C1 = (2 -(+ - - 67.14) + ( 60 47)24( 1126.9433333 ) 3 = 41.21 ksi 2 9 C = - 41.21 + 50.49 + 2(6,761+33559/40x21.213) = 33.13 ksi C3 = - 41.21 + 50.49 + 2(6,761-3,359/40x21l213) = 173.0 ksi 4 = 41.21 - 4x6,761/40x21.213 = 9.34 ksi iE = (65 - 7. 746 50-41.21 ) = 42.03 xEIl = 9,000/2x21.213 E( 1-E4) = 6,934 yEI = My The load deflection curves in the x and. y directions, plotted from the computer output for this particular problem are shown in Fig. 10 43

1200 1190 11801160 1140c,) / Deflection in Direction x a. 1120 z 1100 - a. 06 1080- Deflection in Direction y 0 10601040- / 1020 1000.8 1.0 1.2 1.4 1.6 1.8 2. 2.2 2.4 2.6 2.8 3. 3.2 3.4 3.6 3.8 DEFLECTION IN INCHES Fig. 10. Deflection of an unrestrained beam-column in planar bending.

5. ULTIMATE STRENGTH OF RESTRAINED BEAM-COLUMN IN PLANAR BENDING 5.1 MODIFIED SLOPE DEFLECTION EQUATIONS FOR COMPRESSION MEMBERS WITH TRANSVERSE LOADS AND END MOMENTS WITHIN THE ELASTIC LIMIT The following is limited to cases where lateral relative movement of the ends of the beam-column is prevented. (Fig. 11) Mab WI W2 Mbo P P L Fig. 11. Beam-column with end moments. Within the elastic range of behavior, the usual form of the slope deflection equation (modified for axial load) are applicable to case of compression members with transverse loads.4 13,l4 The slope-deflection equations (with no lateral movement of ends) are: EI Mab = L (Cla+C2b) + MFab (45) EI Mba E- L (C2a+C.9b) + MFba where GaGb = end slopes at a and b respectively, considered positive when clockwise. MFabMFba = fixed end moments at a and b respectively, considered positve when acting on a member in a clockwise direction. C = (1-2u cos 2u) (46) (tan u 1) u

C - =2u csc 2u-1 C2 - (47) (tan u - 1) r = carry over factor = S2/S1 =2u cose 2u (48) 1-2u cot 2u where pL2 2u = ES' Tables (Ref. 4, 13) and graphs (Ref. 14) give values of C1 and C2 and factor r, for different values of 2u. For a uniform load w pounds per unit length, the fixed end moments are: MFab = uMFba - 1 Lt wL-. (49) Reference 14, Chapter 5, gives a chart for fixed end moments in beam-columns for uniform and other load distributions. For the symmetrical case of loading u and end restraint, Eq. (45) reduces to Mab = E (C1-C2)a + MFab ~ (50) 5.2 SYMMETRICALLY RESTRAINED BEAM-COLUMNS ABOVE THE ELASTIC LIMIT If the maximum stress in the beam shown in Fig. 11 exceeds the proportional limit, Eq. (45) is not applicable and the solution may be obtained by a numerical integration procedure. If the beam shown in the same figure is framed continuously to other members at its ends, the end moments Mab are not known, adding to the complexity of the problem. The analysis of the beamcolumn shown in Fig. 12 will be divided into two parts. Initially the load ~~~~P P Fig. 12. Restrained beam-column under uniform lateral load. 46

at which the maximum stress almost equals the proportional limit stress will be determined. Assume a load P let "M" equal the summed stiffnesses of all members at joint a or b except ab.- = rotation at end a. Then the slope deflection equations at a will be: EI (C1-C2) L + MFab + = 0 (51) and Nab = - 5. (52) The deflection and, bending moments along the beam will be y = cos + B' sin -p Mo + wL2/4u ) (53) Mx = A' cos 2u + B' sin 2xu wL2/4u2 (54) L L where x = distance from the end of beam A' = Mab + wL2/4ua B' = -(A' cos 2u + Mba - wL2/4u2)/sin 2u The maximum deflection and maximum positive moment at the center of the beam are obtained by substituting x/L = 0.5 in Eqs. (53) and (54), respectively. Then, by trial, the load is increased until the maximum stress P+ m ax is almost equal to the proportional limit stress. Secondly, the deflection curve of the restrained column is determined. after the stresses exceed the proportional limit. The analysis would be essentially the same as for the case of the unrestrained column, using the numerical integration procedure, if the end moments Mab = -"~ were known. Since these cannot be known in advance, they also must be assumed and then modified by trial until the condition Mab = -A9 is satisfied. To obtain the ultimate strength of the restrained, beam in planar bending, under uniform lateral load W = kP, the procedure for computer analysis was as follows: 47

1. Divide the beam into a reasonable number of divisions (2N). 2. Assume an initial value P and corresponding w. A good start for P is the load at the proportional limit stress assuming no end restraint, as given by Eq. (41). 3. For this P find 2u, C1, C2 from Eqs. (46) and (47) and the fixed end moments MFab as given in Eq. (49). 4. Find the end slope 9 from Eq. (51), then Mab follows from Eq. (52). 5. Find the maximum positive moment from Eq. (54), by putting x/L = 0.5. 6. Calculate the maximum stresses from the equation P Mmax Gmax -A Mmax is the larger of end moments Mab and the maximum positive moment obtained in Step 5. 7. If the maximum stresses do not exceed the proportional limit stress, increase P by any arbitrary amount say AP = Py/200 and repeat from Step 3 to Step 7, otherwise proceed to Step 8. 8. With the last value of Mab as given in Step 4 and the deflection corresponding to the last value of P and w as given by Eq. (53) calculate at all nodal points the bending moment by the equation M = M + Py- Mab () 9. For the end moment Mab and the assumed deflection curve, find by the numerical procedure the deflection at all nodal points. 10. Calculate the end slope 9, from the equation: (See Fig. 13). 9 = (96Y-72y2+52y3-6y4)/24X. (56) 11. If the ratio between the assumed Mab and 39 is within a reasonable limit (a reasonable limit in this report is assumed when this ratio is within limits 1 ~.005), then the boundary conditions are satisfied, and one proceeds to Step 12. If not, assume a new value of 48

w Mab Y2 y 3 Fig. 13. Numerical evaluation of end slope. Mab and repeat from Step 8 to Step 11. As a second assumption of the end moment it may be taken as Mab = (first assumed Mab + P9/2). (57) A linear interpolation in the other trials is efficient in treating this problem, referring to Fig. 14, the linear interpolation between two values is as follows: [0G0-Mb (P' G -POGO)/(M'b-Mb)] ( Mab = (58) [1-(1 G -0o0o()/(Mab-Mab)] Points of the least difference between Mab and. A9 should be used in the interpolation. 12. Increase the load. P by AP and assume new deflection curves and end moments as follows: PN+ = PN + AP w = k x PN+ (59) Mab(N+l) = MabPN+l/PN YN+1 = YN x PN+1/PN 135. Repeat from Step 8 to 12 with the new value of P. 49

B8 i39 Mab, o Interpolated value of Mab Mab, 450 Fig. 14. Linear interpolation for satisfying the end condition of a restrained beam-column. 14. Ultimate Strength: Upon increasing loads P on the beam by increments of say, AP = Py/200, the ultimate load is finally reached when either of the following two conditions is satisfied: A. The numerical calculation procedure does not converge (convergence is assumed when the ratio between the trial deflection and the computed deflection is equal to 1 +.005. Divergence is assumed when this condition is not satisfied within 200 cycles.) B. When both the end section and middle section of the beam yield. After one section yields, there is still reserve strength left in the restrained beam-column whereas in the case of the unrestrained beam-column, there is none left. The strain which corresponds to the yield stress at the section which yields first is indefinite. Since yielding exists only in a very small portion of the beam, the strain at this location initially may be assumed as that strain corresponding to Oy as given by the equation: 5o

cy = (2ay - apL)/E (60) If the end section yields first, which will happen when the beam is subject to large restrains, the end moments will be independent of the restraint 3G and will be given by: (See Fig. 15) Mab = (P - P)c (61) in which Mab is the end moment. PPY 8 or Me 3 or Me t' c P P 22 equal end restraints. 0 P 2 Bottom fibers yield Top fibers yield Fig. 15. Yield condition of beam-column with equal end restraints. If the ultimate load is reached after the end section yields, then this ultimate load will be the same for all larger end restraints. If the middle section yields first, which will occur for small end restraints, the corresponding bending moment at the middle section will be known. For a given end moment Mab the deflection at the centerline 5 will be: 5 = ((Py-P)c + Mab - Mo)/P. (62) If both the middle and end section yield simultaneously, the moment at both locations will be given by Eq. (61). The maximum deflection in such a case will be 6 = 2c( - p~. (63) 51

From Eq. (63), it may be seen that as P decreases increases. In such a case the ultimate load is that load at which both end and middle section yields. Illustrative Example 6 is given to show how the ultimate strength has been computed. In Appendix B, Flow Diagram No. 3 is given for determining the ultimate strength of the restrained beam-column under uniform load W = kP, for the case of the simplified section'in planar bending. This flow diagram should. be translated to a computer program, and that program requires as data the following: L = length of beam in inches c = half depth of section in inches N = half number of beam divisions k = ratio of the lateral load/axial load A = area of section in sqaure inches E = modulus of elasticity in ksi = stiffness of all members at end joint except the beam in K in./rad ap = proportional limit stress in ksi a = yield stress level in ksi Example 6. A restrained column of the simplified section has the following properties: L = 300 in., c = 15 in., A = 40 sq. in., E = 29,000 ksi, cp = ksi, cy = 50 ksi. If subjected to a vertical load W =.15 P, and, restrained by members having D = 4,350,000 Kips in./rad, find its ultimate strength. 1. The load at the proportional limit stress, without effect of end restraint is about 1000 Kips. 2. AP = Py/200 = 5Kips. 3. At the load P = 1167.85, the stresses computed by the usual elastic theory as outlined in Section 3 were found to be greater than op at the end section. 52

4. For increment of P of 5 Kips, the numerical integration procedure will be illustrated. The beam is divided into 8 divisions. For each load, the deflection at nodal points are computed. 5..Beyond the load P= 1,672.85 K, no convergence can be obtained, thus this load is the ultimate load. = 4,350,000, K in./rad, c= 15 in., A = 40 sq in., E = 29,000 ksi, ap = 35 ksi, Oy = 50 ksi, L = 500 in. W =.15 P = 249.43 P = 1662.85 K P/A= 41.57 ksi 37.5" 375" =37.5" 37.5" Mab = 5057.27 L=300" Common factor Mo (Kips x in.) 4.092 7,015 8,769 9,354 Assumed y(in.).0577.1236.1741.1933 Py (Kips x in.).096.206.290.321 M = Mo+Py -5,057 -.869 2,164 4,002 4,618 M/S = ksi -8.43 -1.45 3.61 6.67 7.70 ca = P/A + M/S 33.14 40.12 45.18 48.24 49.27 aB = P/A - M/S 50. 43.02 37.96 34.90 33.87 eU 1.143 1.402 1.655 1.887 2.013 10-3 EB 2.241 1-535 1.316 1.203 1.168 10-3 Curvature 3 -3.66 -.443 1.130 2.28 2.817 10-5 Concentrated - -6.96 13.137 26.747 2x16.365 105 12 Average slope 49.289 56.249 43.112 16.365 12 -5 Deflection 49.289 105.538 148.650 165.015 - 10-5 Deflection (in.).0578.1237.1742.1934 Yassumed/Y calculated.9983.9992.9994 ~9995 53

9 = (96x.0578-72x.1237+32x.l742-6x.1934)/(24x37.5) = 1.1738 10-3 39= 5.106, M/59 =.9904. The foregoing illustrates for a specific case, just short of maximum load, the numerical calculation of deflection and moments for the case of a restrained column under uniform load in planar bending. The following tabulation shows the computer printout for loads above those causing initial yielding for the foregoing example, together with a graph of the results in Fig. 16. It is of interest also to compare these results with the load at initial yield of the same column, assuming no residual stress, as discussed in Section 3.5 with solution tabulated in Table III. L L 4,550,000x500 0.50 10EI 10x29000x9000 L/r = 20. By interpolation, from Table III, 9a = 41.52 ksi, or P = 40x41.52 = 1661 Kips, which is to be compared with the 1673 Kips ultimate strength with residual stress included. It should not be assumed that these two different procedures will always agree so well.

N = 4, k =.15, = 4,,350,000 K in./rad, L = 3000.0 in. c = 15.0 in., A = 40.0 in.2, E = 29,000 ksi aPL = 35.0 ksi, cy = 50.0 ksi End Moments P(KipS) w(Kips) Deflection in Inches at Nodal Points (K in.) 1167. 85 175.18.0332.0682.0937.1030 -3180.D6 1172.85 175.93.0333.0685.0941.1034 -3193.68 1177.85 116.68.0335.0 88.0945.1039 -3207.29 1182.85 _.177.43. 0336..0691. 0950. 1044 -3220.91 1187.85 178.18.0338.0694.0954.1048 -3234.52 1192.85 __ 178.93.0339.0697.0958.1C53 -3248.14 1197.85 179.68.0341.0700.0963.1058 -3261.75 12u2.85 180.43.0342.0704.09I67.1063 -3275.37 1207.835 181.18.0344.0707.0971.1068 -3288.98 1212.85 181.93.0346.010.0976.1073 -3302.60 1217.85 182.68.0347.0713.0980.1078 -3316.21 1222.85 183.43.0349.0717.0985.1083 -3329.83 1227.85 184.18.0349.0717.0986.1084 -3352.79 1232.85.84.93.0350.0720.0991.1089 -3366.44 1237.85 "1..68.0352.0724.0995.1094 -3380.10 1242.85 186.43.0354.0727.1000.1100 -3393.75 1247.85 187.18.0356.0731.1005.1105 -3407.40 1252.85 187.93.0357.0735. 1010.1111 -3421.06 1257.85 188.68.0359.0738.1015.1116 -3434.71 1262.85 189.43.0361.0742. 1020-.1122 -3448.36 1267.85 190. 18.0363.0746.1026. 1128 -3462.02 1272.85 190.93.0363.0747.1028.1131 -3481.60 1277.85 191.68.0365.0751.1034.1137 -3495.27 1282.85 192.43.0367.0755.1039.1143 -3508.95 1287.85 193.18.0369.059.1-045.1149 -3522.63 1292.85 193.93.0370.0761. 1048 1152 -3542.87 1297.85 194.68.0372.07b5. 1053.1158 -3556.58 1302.85 195.43.0374.0769.1059.1165 -3570.28 1307.85 196. 18.0374.0770.1061.1167 -3594.74 1312.85 196.93.0376.0774.1067.1173 -3608.48 1317.85 197.68.0378.0779.1073.1180 -3622.22 1322.85 198.43.0380.0783.1079.1187 -3635.97 1327.85 199.18.0383.0788.1085.1194 -3649.71 1332.85 199.93.0384.0790_.1089.1198 -3670.30 1337.85 200.68.0386.0795.1095.1205 -3684.07 1342.8 5 201.43.0386.0796.1098.1208 -3707.76 1347.85 202.18.0389.0801.1105.1215 -3721.56 1352.85 202.93. 0391.0806.1112.1223 -3735.37 1357.85 203.68.0394.01 1.1119.1231 -3149.17 1362.85 204.43.0395.0814.1122.1235 -3770.53 1367.85 205.18.0397.0819.1130.1243 -3784.36 1372.85 205.93.0399.0822.1134.1248 -3804.91 1377.85 206.68.0401.0828.1142.1257 -3818.76 1382.85 207.43.0402.0831.1146.1262 -3839.73 1387.85 208.18.0405.0836.1154.1270 -3853.61 1392.85 208.93.0406.0839.1159.1275 -3874.88 1397.85 209.68.0409.0845.1167.1284 -3888.79 1402.85 210.43.0411.0848.1171.1290 -3910.36 1407.85 211.18.0414.0854. 1180.1299 -3924.30 1412.85 211.93.0415.0857.1185.1304 -3946.19 1417.85 212.68.0418.0864. 1193.1314 -3960.15 1422.85 213.43.0419.0867.1198.1319 -3982.38 1427.85 214.18.0422.0873.1207.1329 -3996.37 55

1432. 5 214.93.0424.08 7.1212.1335 -4018.97 1437.65 215.68.0427.0883.1221.1345 -4032.99 1442.e5 216.43.0428.0887.1226.1351 -4056.09 1447.85 217.18.0432.0894.1236.1362 -4070.15 1452.85 217.93.0433.0897.1241.1367 -4093.18 1457.85 218.68.u436.0904.1251.1379 -4107.87 1462.85 219.43.0438.09<8.1256.1384 -4132.07 1467.85 220.18.0441.0915.1267.1396 -4146.20 1472.85 220.93.0443.0919.1272.1402 -4171.21 1477.b5 221.63.0446.0926.1283.1414 -4185.37 1482.85 222.43.0448.0930.1289.1420 -4210.78 1487.85 223.18.0450.0936.1297.1429 -4230.8k 1492.85 223.93.0453.042.1305.1439 -4251.00 1497.85 224.68.0455.0948.1314.1448 -4271.38 1502.85 225.43 U.458.0954.1322.1458 -4291.94 1507.85 226.18.0461.0960.1331.1468 -4312.68 1512.85 226.93.0463.0966.1340.1478 -4333.61 1517.85 227.68.0466.0972.1349.1488 -4354. 74 1522.85 228.43.0469.0979.1359.1499 -4376.05 1527.85 229.18.0472.0985.1368.1509 -4397.57 1532.85 229.93.0475.0992.1378.1520 -4419.30 1537.85 230.68.0478.0999.1388.1531 -4441.24 1542.85 231.43.0481. 106.1398.1542 -4463.39 1547.85 232.18.0484.1C 13.1408.1554 -4485.77 1552.85 232.93.0481.1020.1418.1565 -4508.37 1557.85 233.68.04 90.1 027.1429.1578 -4531.19 1562.85 234.43.0494.1035.1440.1590 -4554.25 1567.85 235.18.0497.142.1451.1602 -4577.55 1572.85 235.93.0500.1 50.1462.1615 -460.1..10 1577.85 236.68.0504.1 58.1474.1628 -4624.90 1582.85 237.43.0507.1; 66.1485.1641 -4648.96 1587.85 238.18.0510.1174.1497.1655 -4673.29 1592.85 238.93.0514.1i82.1510.1668 -4697.91 1597.85 239.68.0518.1091.1522.1683 -4722.82 1602,85 240.43.0521.199.153 5.1697 -4748.02 1607.85 241.18.0525.1108.1548.1712 -4773.54 1612.85 241.93.0529.1117.1562.1728 -4799.37 1617.85 242.68.0533.1127.1576.1743 -4825.53 1622.b5 243.43.0537. 1136.1590. 1760 -4852.02 1627.85 244.18.0541.1146.1605.1777 -4878.84 1632.85 244.93.0545.1157.1621.1794 -4905.99 1637.85 245.68.0549.1167.1637.1613 -4933.44 1642.85 246.43.0554.1178.1653.1832 -4961.15 1647.85 247.18.0559.1190.1671.1852 -4989.00 1652.85 247.93.0564.1202.1690.1874 -5016.72 1657.85 248.68.0569.1216.1710.1897 -5043.44 1662_.65 249.43.0577.1236.1741.1933 -5057.271667.85 250.18.0626.1327.1864.2'069 -4982.21 1672.85.250.93. 0090.1449__. 2 33.2262 -49 7.2 7 NO CONVERGENCE 33.64 50.00.11601E-ij2.22411E-02 40.70 42.95.14266E-02.15332E-02 45.81 37.83.16945E-02.13103E-02 48.91 34.73.19626E-02.11977E-02 49.95 33.69.21842E-22.11617E-02 56

1678 _{Eq. 1678 ___ __ __' 1668 - 1658 1648 1638 cO X. 1628z 1618 / -34,350,000'KIP INCH/RAD, C=15 INCH, O 1608- / A=40SQ.INCH, E=29,000 KSI, Oy=50KSI, < / L= 300 INCHES 3 1598 1588 1578 1568 l.16.17.18.19.20.21.22.23 MIDDLE SECTION DEFLECTION 8 IN INCHES Fig. 16. Load deflection curve of a restrained beam-column in planar bending.

6. SUMMARY By means of a simplified "four point area closed section" a conservative estimate of column strength has been provided. The problems of stress redistribution as well as local buckling and torsion have not been considered. Computer programs have been developed both to determine the beam-column load at which the maximum stress reaches the yield point and, alternatively, the maximum column load that the beam-column can carry as it reaches a condition of instability. Residual stresses are considered on a point area basis. Complete design tables for each of these approaches are furnished for the unrestrained case in planar bending for various yield points of steel from 33 to 100 ksi and various ratios of total uniform lateral beam load. to column load up to 0.30. A partial table for design use (for yield point of 50 ksi) is provided to give the load at which the maximum stress reaches yield point for the end-restrained symmetrically loaded beam-column. It is shown that the unrestrained beam-column in biplanar bending can be designed by means of the. tables prepared for planar bending. A program is developed for the determination of the ultimate strength in biplanar unrestrained bending including effects of residual stress. The latter part of the report is somewhat exploratory in nature and, does not provide complete design tables as does the initial part of the report but it should point the way to further research in this important area of work. Computer facilities of The University of Michigan were used in obtaining the numerical results. The work was initiated through the interest of Mr. A. Amirikian, Special Design Consultant, Bureau of Yards and Docks, U.S. Navy. The results on Phase I as here reported include some preliminary partial results for Phase II. A wide range of design information is provided, including a rather detailed comparison of the two entirely different procedures of analysis, one based on initial yield and the other based on ultimate strength. For the simplified section that was considered herein, the results in most cases have shown remarkably close agreement between the two procedures. 59

7. REFERENCES 1. Karman, Th. v., "Untersuchungen uber Knickfestigkeit," Mitteilungen uiber Forschungsarbeiten auf dem Gebiete des Ingenieurwesens, No. 81, Berlin (1910). 2. Chwalla, E., "Uber die experimentelle Untersuchung des Tragverhaltens gedruckter Stabe aus Baustahl," Der Stahlbau, v. 7, p. 17, 1934. 3. Chwalla, E., "Der Einfluss der Querschnittsform auf das Tragvermogen aussermittig gedruckter Baustahlstabe," Der Stahlbau, v. 8, p. 137, 1935. 4. Bleich, F., "Buckling Strength of Metal Structures," McGraw-Hill Book Co., Eng. Monograph Series, 1952. 5. Baker, J. F., Home, M. R., and Heyman, J., "The Steel Skeleton: II, Plastic Behaviour and Design," Cambridge University Press (1956). 6. Bijlaard, P. P., Fisher, G. P., and Winter, G., "Eccentrically Loaded, End-Restrained Columns," Trans. ASCE, v. 120 (1955). 7. Ketter, R. L., Kaminsky, E. L., and Beedle, L. S., "Plastic Deformation of Wide-Flange Beam-Columns," Trans. ASCE, v. 120 (1955). 8. Ojalvo, M. and Fukumoto, Y., "Nomographs for the Solution of BeamColumn Problems," Welding Research Council Bulletin No. 78, June, 1962. 9. Galambos, T. and Prasad, J., "Ultimate Strength Tables for BeamColumns," Welding Research Council Bulletin No. 78, June, 1962. 10. Ketter, R. L., "Further Studies of the Strength of Beam-Columns," Trans. ASCE, v. 127 (1962). 11. Column Research Council, "Guide to Design Criteria for Metal Compression Members," First Ed., 1960 (Second Edition in Preparation). 12. Newmark, N. M., "Numerical Procedure for Computing Deflections, Moments, and Buckling Loads," Trans. ASCE, v. 108 (1943), p. 1161. 13. Lundquist, E. E., and Kroll, W. D., "Tables of Stiffness and Carry-Over Factor for Structural Members under Axial Load," NACA TN No. 652, 198. 14. Gere, J. M., "Moment Distribution," Van Nostrand, 1963. 61

I

APPENDIX A DESIGN TABLES 1. Average column loads for which combined stress in unrestrained beamcolumns with uniform lateral load just reaches yield point. 2. Load and deflection coefficients to give equivalent lateral loads for use in Table I. 3o Average column load for which combined stress in restrained beamcolumns with uniform lateral load just reaches the yield point. (This table limited to yield point of 50 ksi.) 4. Ultimate strength of unrestrained beam-columns under uniform lateral load, in planar bending, including the effect of residual stress. 63

TABLE I AVERAGE COLUM{N LOADS FOR WHICH COMBINED STRESS IN UNRESTRANED BEAM-COLUMNS wITH UNIFORM LATEA LOAD JUST RECHES YIELD POIT~ L/R VALUES OF P/A IN KS][ FOR k=.01.02.03.04.05.06.07.08.0_9.... 0........1]__...._2.... 13.14 ~ 15__ 2 032.16 31.36 30.60 29.88 29.19 28.54 27.91 27.31 26.74 26.20 25.67 25.17 24.68 24.22 23.77 30 31.68 30.47 29.35 28.32 27.37 26.48 25.66 24.88 24.16 23.48 22.83 22.23 21.65 21.11 20.59 19.08 18.51 17.97 50 30.39 28.27 26.49 24.97 23.64 22.46 21.41 20.47 19.62 18.84 18.13 17.47 16.86 16.30 15.78 60 29.45 26.86 24.81 23.13 21.71 20.49 19.42 18.47 17.62 16.86 16.16 153!49!44 392 70 28.17 25.17 22.96 21.22 19.79 18.57 17.53 16.62 15.81 15.08 14.43 13.84 13.30 12.80 12.34 90 24.20 21.07 18.97 17.39 16.12 15.06 14.17 13.39 12.71 12.10 11.56 11.06 10.62 10.21 9.83 100 21.65 18.88 17.01 15.60 14.46 13.52 12.72 12.02 11.41 10..87 10.38 9.'94 9.54 9,17 8.83 110 19.09 16.79 15.19 13.96 12.97 12.14 11.43 1.0.81 10.27 9.79 9.35 8.96 8.60 8.27 7.97 120 16.74 14.88 13.54 12.49 11.63 10.91 10.29 9.74 9.26 8.84 8.45 8.10 7.78 7.49 7.22 130 14.68 13.19 12.08 11.19 10.45.9.82 9.28 8.81 8.38 8.01 7.66 7.35 7.07 6.81 6.57 140 12.92 11.72 10.80 10.04 9.41 8.87 8.40 7.98 7.61 7.27 6.97 6.70 6.44 6.21 6.00 150 11.43 10.46 9.68 9.05 8.50 8.03 7.62 7.26 6.93 6.63 6.36 6.12 5.89 5.68 5.49 160 10.16 9.37 8.72 8.17 7.71 7.30 6.94 6.62 6.33 6.07 5.83 5.61 5.41 5.22 5.05 170 9.08 8.43 7.88 7.41 7.01 6.65 6.34 6.06 5.80 5,57 5.35 5.16 4.98 4.81 4.65 180 8.16 7.61 7.15 6.75 6.40 6.09 5.81 5.56 5.33 5.12 4.93 4...76 4....59 4...44__ 4.3 0 0O\ 190 7.37 6.91 6;51 6.16 5.86 5.58 5.34 5.12 4.91 4.73 4.56 4.40 4.-2 5 4.12 3.99 200 6.69 6.29 5.95 5.65 5.38 5.14 4.92 4.72 4.54 4.37 4.22 4.08 3.95 3.82 3.71 L/R VALUES OF P/A IN KS! FOR k=.16.17.18.19.20.21.22.23.24.25.26.27.28.29.30 10 27..45 27.17 26.89 26.62 26.35 26.09 25.83 25.58 25.33 25..09 24.85 24.62 24.39 24.1i6 23.94 20 23.34 22.93 22.53 22.14 21.77 21.41 21.06 20.72 20.39 20,08 19,77 1....]_948 19.19 18.91 18.64 30 20.10 19.64 19.19 18.77 18.36 17.97 17.60 17.25 16.91 16.58 16.26 15.96 15.67 15.39 15912 40 17047 17-00 16.55 16,12 15.72 15.34 14,98 14.63.14.30 13.99 13.69 13.40 13.13 12.86 12.61 50 15.29 14.83 14.40 14.00 13.62 13.26 12.92 12.59 12.29 12.00 11.72 11.46 11. 2.0 10.96 10.73 60 1311~413-303-]!2 It63 1._2 —w2 6.. 1][,09!L -11,58- -1 1,26 I0. 1069.104!Q1....993 9.71 9.49 9.28 70 11.92 11.52 11.16 10.82 10.50 10.19 9.91 9.64 9.39 9.15 8.93 8.71 8.51 8.31 8.13 80 10.61 10.25 9.92 9.61 9.32 9.05 8.79 8.55 8.32 8..11 7.91 7.71 7.53 7.35 7.19 90 9.49 9.16 8.87 8.59 8.33 8.08 7.85 7.64 7.43 7.24 7.06 6.88 6.72 6.56 6.41 100 8.52 8.23 7,97 7.72 7.48 7,26 7,06 6.86 6,68 6.51 6.34 6.19 6.04 5.90 5.77 110 7.69 7.43 7.19 6.97 6.76 6.56 6.38 6,20 6.04 5.88 5.74 5.60 5.46 5.34 5.22 120 6.97 6.74 6.53 6.32 6.14 5.96 5.79 5.'64 5.49 5.35 5.022 5.09 4.97 4.86 4.75 130 6.34 6.14 5.94 5.76 5.59 5.44 5.29 5.14 5.01 4.88 4.76 4.65 4.54 4.44 4.34 {.40- -.5...80.5.61 5.44...2.7.5_2........12 4.98 4.84 4. 7.2_...___ 4....__48 4...._37 4.2 4. 17 4.08 3_.99 150 5.31 5.15 4.99 4.84 4.71 4.58 4.45 4.34 4023 4.12 4.C3 3.93 3.84 3.76 3.68 160 4.89 4.74 4.59 4.46 4.34 4.22 4.11 4.01 3.91 3.81 3.72 3.64 3.55 3.48 3.40 170 4.51 4.37 4.24 4.12 4.01 3.91 3.80 3.71 3.62 3.53 3.45 3.37 3.30 3.23 3.16 180 4.17 4.05 3.93 3.82 3.72 3.62 3.53 3.45 3.36 3.28 3.21 3.14 3.07 3.00 2.94 190 3.87 3.76 3.65 3.55 3.46 3.37 3.29 3.21 3.13 3.06 2.99 2.93 2.86 2.81 2.75 200 3.60 3.50 3.40 3.31 3.23 3.15 3.07 3.00 2.93 2.86 2.80 2.74 2.68 2.63 2.57

TABLE I (Continued) (YIED STRESS = 36 KSI) L/R VALUES OF P/A IN KSI FOR =.01 _Q2.03.04.05.06.07.08.09.10.11.12.13.14.15 10 35.55 35.11 34.68 34.27 33.86 33.46 33.07 32.69 32.32 31.96 31.60 31.26 30.92 30.59 30.26 20 35.08 34.20 33.37 32.58_31.83_31.11 30.43 29.78 29.15 28.56 27.98 27.43 26.90 26.40 25.91 30 34.54 33.21 31.99 30.86 29.82 28.84 27.94 27.09 26.30 25.56 24.85 24.19 23.57 22.97 22.41 4.. 3_390 32.07 30.47_290D4_7!75_!-k659-_25._53_ 24.57 23.68 22.85 22.09 21.38 20.72 20.10 19.52 50 33.07 30.71 28.75 27.08 25.62 24.34 23.20 22.17 21.25 20.40 19.63 18.92 18.26 17.66 17.09 60 31.952906 26.81 24.98 23,44 22,11 20.95 19.93 19.02 1819 17.45 16.77 16.14 15.57 15,,03 70 30.38 27.06 24.66 22.78 21.24 19.95 18.83 17.85 16.99 16.21 15.51 14.88 14.30 13.77 13.29 80 28.22 24.75 22,38 20.,57 19.11 17_9Q16..8 15,96 15.16 14.45 13.82 13.24 12.72 12.24 11.79 90 25.49 22.25 20.07 18.42 17.10 16.00 15.07 14.25 13.54 12.90 12.33 11.81 11.34 10.91 10.52 100 22.52 19.75 17.86 16.42 15.26 _14.29 13.46 12.74 A12..111.54 11.C4 10.58 10.16 9.77 9.42 110 19.65 17.43 15.84 14.61 13.61 12.77 12.04 11.41 10.85 10.36 9.91 9.50 9.13 8.79 8.48 120 17.11 15.35 14.04 13.01 12.15 11.43 10.80 10.25 9.76 9.32 8.93 8.57 8.24 7.94 7.66 130 14.94 13.54 12.47 11.60 10.87 10.25 9.71 9.23 8.80 8.42 8.07 7.75 7.46 7.20 6.95 140 13.10 11.99 11.11 10.38 9.76 9.23 8.76 8.34 7.97 7.63 7.32 7.04 6.79 6.55 6.33 150 11.56 10.67 9.93 9.32 8.79 8.33 7.93 7.56 7.24 6.94 6.67 6.42 6.19 5.98 5.78 160 10.27 9.53 8.92 8.40 7.95 7.55 7.20 6.88 6.59 6.33 6.09 5.87 5.67 5.48 5.31 170 9.17 8.56 8.05 7.60 7.21 6.87 6.56 6.28 6.03 5.80 5.59 5.39 5.21 5.04 4.88 180 8.23 7.72 7_29 6.91 6.57 6.27 6.00...5.75 5.53 5.33 5.14 4.96 4.80 4.65 4.51 190 7.43 7.00 6.62 6.30 6.00 5.74 5.50 5.29 5.09 4.91 4.74 4.58 4.44 4.30 4.17 200 6.73 6.37 6.05 5.76 5.51 5.27 5.06 4.87 4.70 4.53 4.38 4.24 4.11 3.99 3.87 L/R VALUES OF P/A IN KSI FOR k = > ~.16.17.18.19.20.21.22.23.24.25.26.27.28.29.30 10 29.95 29.63 29.33 29.03 28.74 28.45 28.17 27.90 27.63 27.36 27.10 26.85 26.60 26.36 26.11 20 25.44 24.99 24.55 24.13 23.72 _23.33 22.95 22.58 22.23 21.88 21.55 21.22 20.91 20.61 20.31 30 21.88 21.37 20.88 20.42 19.98 19.56 19.16 18.77 18.40 18.04 17.70 17.37 17.05 16.75 16.45 40 18.97 18.46 17.97 17.51 17.07 16.66 16.27 15.89 15.54 15.19 14.87 14.56 14.26 13.97 13.70 50 16.56 16.07 15.60 15.16 14.75 14.36 14.00 13.65 13.32 13.00 12.70 12.42 12.15 11.89 11.64 60 14.54 14.08 13.65 13.25 12.87 12.51 12.18 11.86 11.56 11.27 11.00 10.75 10.50 10.27 10.04 70 12.83 12.41 12.02 11.66 11.31 10.99 10.69 10.40 10.13 9.88 9.63 9.40 9.18 8.98 8.78 80 11.39 11.01 10.65 10.33 10.02 9.73 9.46 9.20 8.96 8.73 8.51 8.31 8.11 7.93 7.75 90 10.15 9.81 9.50 9.20 8.93 8.67 8.43 8.20 7.98 7.78 7.58 7.40 7.23 7.06 6.90 100 9.10 8.79 8.51 8.25 8.CO 7.77 7.56 7.35 7.16 6.98 6.80 6.64 6.48 6.33 6.19 110 8.19 7.92 7.67 7.43 7.21 7.01 6.81 6.63 6.46 6.29 6.14 5.99 5.85 5.72 5.59 120 7.40 7.16 6.94 6.73 6.53 6.35 6.18 6.01 5.86 5.71 5.57 5.44 5.31 5.19 5.08 130 6.72 6.51 6.31 6.12 5.94 5.78 5.62 5.48 5.34 5.21 5.08 4.96 4.85 4.74 4.64 140 6.12 5.93 5.75 5.59 5.43 5.28 5.14 5.01 4.88 4.77 4.65 4.55 4.44 4.35 4.25 150 5.60 5.43 5.27 5.12 4.98 4.85 4.72 4.60 4.49 4.38 4.28 4.18 4.09 4.00 3.91 160 5.14 4.99 4.84 4.71 4.58 4.46 4.35 4.24 4.14 4.04 3.95 3.86 3.78 3.69 3.62 170 4.74 4.60 4.47 4.35 4.23 4.12 4.02 3.92 3.83 3.74 3.66 3.58 3.50 3.42 3.35 180 4.37 4.25 4.13 4.02 3.92 3.82 3.73 3.64 3.55 3.47 3.39 3.32 3.25 3.18 3.12 190 4.05 3.94 3.83 3.73 3.64 3.55 3.46 3.38 3.31 3.23 3.16 3.09 3.03 2.97 2.91 200 3.76 3.66 3.57 3.47 3.39 3.31 3.23 3.15 3.08 3.02 2.95 2.89 2.83 2.78 2.72

TABLE I (Continued) (YIELD STRESS = 42 KSI) L/R VALUES OF P/A IN KSI FOR k =.01.02.03.04.05.06.07.08.09.10.11.12.13.14.15 10 41.47 40.96 40.46 39.97 39.50 39.03 38.58 38.13 37.70 37.28 36.86 36.46 36.06 35.68 35.33 20 40.91 39.88 38.91 37.98 37.10 36.26 35.46 34.70 33.97 33.27 32.60 31.95 31.34 30.74 30.17 30 40.27 38.69 37.24 35.91 34.68 33.54 32.48 31.49 30.56 29.69 28.87 28.10 27.37 26.68 26.02 40 39.45 37.27 35.35 33.66 32.15 30.79 29.55 28.42 27.39 26.43 25.54 24.72 23.95 23.24 22.56 50 38.36 35.51 33.17 31.20 29.49 28.00 26.68 25.50 24.43 23.46 22.57 21.75 21.00 20.30 19.65 60 36.80 33.30 30.65 28.52 26.74 25.23 23.90 22.74 21.70 20.77 19.92 19.15 18.44 17.79 17.19 70 34.50 30.59 27.85 25.72 23.99 22.53 21.28 20.19 19.22 18.36 17.58 16.88 16.23 15.64 15.09 80 31.33 27.49 24.89 22.92 21.33 20.01 18.87 17.89 17.02 16.24 15.54 14.91 14.33 13.80 13.32 90 27.57 24.26 21.99 20.27 18.87 17.71 16.71 15.84 15.07 14.39 13.77 13.21 12.70 12.24 11.80 100 23.82 21.19 19.31 17.85 16.66 15.66 14.80 14.04 13.38 12.78 12.24 11.75 11.30 10.89 10.51 110 20.48 18.45 16.93 15.72 14.72 13.87 13.13 12.48 11.91 11.39 10.92 10.49 10.10 9.74 9.41 120 17.66 16.09 14.87 13.88 13.04 12.32 11.69 11.14 10.64 10.19 9.78 9.41 9.07 8.75 8.46 130 15.32 14.09 13.11 12.29 11.59 10.98 10.45 9.97 9.54 9.15 8.8C 8.47 8.17 7.90 7.64 140 13.38 12.41 11.61 10.93 10.34 9.83 9.37 8.96 8.59 8.25 7.94 7.66 7.40 7.16 6.93 150 11.77 10.99 10.33 9.77 9.27 8.83 8.44 8.09 7.77 7.47 7.20 6.95 6.72 6.51 6.31 160 10.42 9.79 9.24 8.77 8.35 7.97 7.64 7.33 7.05 6.79 6.56 6.34 6.13 5.94 5.77 170 9.29 8.77 8.31 7.91 7.55 7.23 6.93 6.67 6.42 6.20 5.99 5.80 5.61 5.45 5.29 180 8.33 7.89 7.51 7.16 6.85 6.57 6.32 6.09 5.87 5.67 5.49 5.32 5.16 5.01 4.87 190 7.51 7.14 6.81 6.51 6.25 6.00 5.78 5.58 5.39 5.21 5.05 4.90 4.75 4.62 4.49 200 6.80 6.49 6.20 5.95 5.72 5.50 5.31 5.13 4.96 4.80 4.66 4.52 4.39 4.27 4.16 L/R VALUES OF P/A IN KSI FORk = ON.16.17.18.19.20.21.22.23.24.25.26.27.28.29 30 10 34.93 34.56 34.21 33.86 33.52 33.18 32.86 32.54 32.22 31.91 31.61 31.31 31.02 30.73 30.45 20 29.62 29.10 28.59 28.09 27.62 27.16 26.72 26.29 25.88 25.47 25.09 24.71 24.34 23.99 23.64 30 25.40 24.81 24.25 23.71 23.20 22.71 22.24 21.79 21.36 20.94 20.55 20.16 19.80 19.44 19.10 40 21.93 21.34 20.78 20.24 19.74 19.26 18.81 18.38 17.97 17.57 17.20 16.84 16.50 16.17 15.85 50 19.05 18.48 17.95 17.45 16.98 16.53 16.11 15.71 15.34 14.98 14.63 14.31 14.00 13.70 13.41 60 16.63 16.10 15.62 15.16 14.73 14.33 13.95 13.59 13.25 12.93 12.62 12.33 12.05 11.78 11.53 70 14.59 14.12 13.68 13.27 12.89 12.53 12.19 11.86 11.56 11.27 11.00 10.74 10.50 10.26 10.04 80 12.87 12.45 12.06 11.69 11.35 11.03 10.73 10.45 10.18 9.92 9.68 9.45 9.23 9.02 8.83 90 11.41 11.04 10.69 10.37 10.07 9.78 9.52 9.27 9.03 8.80 8.59 8.38 8.19 8. 1 7.83 100 10.16 9.84 9.53 9.25 8.98 8.73 8.50 8.27 8.06 7.86 7.67 7.49 7.32 7.16 7.00 110 9.10 8.81 8.55 8.30 8.06 7.84 7.63 7.43 7.24 7.07 6.90 6.74 6.59 6.44 6.30 120 8.19 7.94 7.70 7.48 7.27 7.07 6.89 6.71 6.55 6.39 6.24 6.09 5.96 5.83 5.73 130 7.40 7.18 6.97 6.77 6.59 6.41 6.25 6.09 5.94 5.80 5.67 5.54 5.42 5.30 5.19 140 6.72 6.52 6.33 6.16 5.99 5.84 5.69 5.55 5.42 5.29 5.17 5.06 4.95 4.85 4.75 150 6.12 5.95 5.78 5.63 5.48 5.34 5.21 5.08 4.96 4.85 4.74 4.64 4.54 4.45 4.36 160 5.60 5.44 5.30 5.16 5.03 4.90 4.78 4.67 4.56 4.46 4.36 4.27 4.18 4.10 4.02 170 5.14 5.30 4.87 4.74 4.63 4.51 4.41 4.31 4.21 4.12 4.03 3.95 3.87 3.79 3.71 180 4.73 4.61 4.49 4.38 4.27 4.17 4.07 3.98 3.90 3.81 3.73 3.66 3.58 3.51 3.45 190 4.37 4.26 4.15 4.05 3.96 3.87 3.78 3.70 3.62 3.54 3.47 3.40 3.33 3.27 3.21 200 4.05 3.95 3.85 3.76 3.68 3.59 3.51 3.44 3.37 3.30 3.23 3.17 3.11 3.C5 2.99

TABLE I (Continued) (YIELD STRESS = 46 KSI) L/R VALUES OF P/A IN KSI FOR k =.01.02.03.04.05.06.07.08.09.10.11.12.13.14.15 10 45.42 44.86 44.31 43.78 43.25 42.74 42.25 41.76 41.29 40.82 40.37 39.92 39.49 39.07 38.65 20 44.80 43.67 42.60 41.58 40.61 39.69 38.81 37.97 37.17 36.40 35.66 34.96 34.28 33.63 33.01 30 44.07 42.32 40.73 39.26 37.90 36.65 35.48 34.40 33.38 32.42 31.52 30.68 29.88 29.12 28.41 40 43.14 40.70 38.58 36.71 35.04 33.55 32.19 30.95 29.82 28.78 27.81 26.91 26.08 25.30 24.56 50 41.84 38.64 36.05 33.87 32.01 30.38 28.94 27.65 26.49 25.44 24.47 23.59 22.77 22.02 21.32 60 39.91 36.00 33.09 30.77 28.85 27.21 25.78 24.53 23.41 22.41 21.50 20.67 19.91 19.21 18.57 70 37.01 32.75 29.80 27.53 25.69 24.14 22.82 21.66 20.63 19.71 18.89 18.14 17.45 16.82 16.24 80 33.06 29.08 26.38 24.33 22.67 21.29 20.11 19.07 18.16 17.35 16.62 15.95 15.34 14.79 14.27 90 28.63 25.37 23.09 21.34 19.92 18.72 17.69 16.80 16.00 15.29 14.65 14.07 13.54 13.05 12.60 100 24.46 21.95 20.12 18.67 17.48 16.46 15.59 14.82 14.14 13.52 12.97 12.47 12.00 11.58 11.19 110 20.88 18.98 17.52 16.34 15.36 14.51 13.77 13.12 12.53 12.00 11.53 11.09 10.69 1C.32 9.98 120 17.92 16.47 15.32 14.36 13.54 12.83 12.21 11.65 11.15 10.70 10.29 9.91 9.56 9.24 8.94 130 15.50 14.37 13.45 12.67 11.99 11.40 10.87 10.40 9.97 9.58 9.23 8.90 8.60 8.32 8.06 140 13.51 12.62 11.87 11.23 10.67 10.17 9.72 9.32 8.95 8.62 8.31 8.02 7.76 7.52 7.29 150 11.87 11.16 10.54 10.01 9.54 9.11 8.73 8.39 8.07 7.78 7.51 7.27 7.04 6.82 6.62 160 10.50 9.92 9.41 8.97 8.57 8.21 7.88- 7.58 7.31 7.06 6.82 6.61 6.40 6.21 6.04 170 9.35 8.87 8.45 8.07 7.73 7.42 7.14 6.88 6.65 6.42 6.22 6.03 5.85 5.68 5.53 180 8.38 7.98 7.62 7.30 7.01 6.74 6.50 6.27 6.06 5.87 5.69 5.52 5.36 5.22 5.08 190 7.55 7.21 6.91 6.63 6.38 6.15 5.93 5.74 5.56 5.38 5.23 5.08 4.94 4.80 4.68 200 6.83 6.55 6.29 6.05 5.83 5.63 5.44 5.27 5.11 4.95 4.81 4.68 4.55 4.44 4.32 L/R VALUES OF P/A IN KSI FOR k = o~.16.17.18.19.20.21.22.23.24.25.26.27.28.29.30 10 38.24 37.85 37.46 37.08 36.70 36.34 35.98 35.63 35.28 34.94 34.61 34.28 33.97 33.65 33.34 20 32.41 31.83 31.27 30.73 30.21 29.71 29.22 28.75 28.30 27.86 27.44 27.02 26.62 26.24 25.86 30 27.73 27.08 26.47 25.88 25.32 24.79 24.27 23.78 23.31 22.86 22.43 22.01 21.61 21.22 20.85 40 23.88 23.23 22.62 22.04 21.49 20.97 20.48 20.01 19.56 19.14 18.73 18.34 17.96 17.61 17.26 50 20.66 20.05 19.47 18.93 18.43 17.95 17.49 17.06 16.65 16.26 15.89 15.54 15.20 14.88 14.57 60 17.97 17.41 16.89 16.40 15.94 15.50 15.09 14.71 14.34 14.00 13.67 13.35 13.05 12.77 12.50 70 15.70 15.20 14.74 14.30 13.89 13.51 13.14 12.80 12.48 12.17 11.88 11.60 11.34 11.09 10.85 80 13.80 13.36 12.94 12.56 12.20 11.86 11.54 11.24 10.95 10.68 10.42 10.18 9.95 9.73 9.52 90 12.19 11.80 11.44 11.10 10.78 10.48 10.20 9.94 9.69 9.45 9.22 9.01 8.8J 8.61 8.42 100 10.82 10.48 10.17 9.87 9.59 9.33 9.08 8.85 8.63 8.42 8.22 8.03 7.85 7.68 7.51 110 9.66 9.36 9.09 8.83 8.58 8.35 8.13 7.93 7.73 7.55 7.37 7.21 7.05 6.89 6.75 120 8.67 8.41 8.16 7.94 7.72 7.52 7.32 7.14 6.97 6.81 6.65 6.50 6.36 6.23 6.10 130 7.81 7.58 7.37 7.17 6.98 6.80 6.63 6.47 6.32 6.17 6.03 5.90 5.77 5.65 5.54 140 7.07 6.87 6.68 6.51 6.34 6.18 6.03 5.88 5.75 5.62 5.49 5.38 5.26 5.16 5.05 150 6.43 6.25 6.09 5.93 5.78 5.64 5.50 5.38 5.25 5.14 5.03 4.92 4.82 4.72 4.63 160 5.87 5.71 5.56 5.42 5.29 5.16 5.04 4.93 4.82 4.72 4.62 4.52 4.43 4.34 4.26 170 5.38 5.24 5.11 4.98 4.86 4.75 4.64 4.54 4.44 4.35 4.26 4.17 4.09 4.01 3.93 180 4.94 4.82 4.70 4.59 4.48 4.38 4.28 4.19 4.10 4.02 3.94 3.86 3.78 3.71 3.64 190 4.56 4.45 4.34 4.24 4.14 4.05 3.97 3.88 3.80 3.73 3.65 3.58 3.51 3.45 3.39 200 4.22 4.12 4.02 3.93 3.84 3.76 3.68 3.61 3.53 3.46 3.43 3.33 3.27 3.21 3.15

TABLE I (Continued) (YIELD STRESS = 50 KSI) L/R VALUES OF P/A IN KSI FOR k.01.02.03.04.05.06.07.08.09.10.11.12.13.14.15 10 49.37 48.76 48.16 47.58 47.01 46.46 45.92 45.39 44.87 44.36 43.87 43.39 42.92 42.46 42.00 20 48.69 47.45 46.28 45.17 44.11 43.11 42.15 41.23 40.36 39.52 38.72 37.96 37.22 36.51 35.83 30 47.88 45.95 44.20 42.59 41.11 39.74 38.47 37.29 36.18 35.14 34.16 33.24 32.38 31.56 30.78 40 46.81 44.11 41.77 39.72 37.90 36.27 34.80 33.45 32.22 31.09 3C.05 29.07 28.17 27.33 26.54 50 45.27 41.71 38.86 36.49 34.46 32.70 31.14 29.75 28.50 27.37 26.33 25.38 24.51 23.70 22.95 60 42.90 38.60 35.44 32.94 30.87 29.12 27.60 26.26 25.07 24.00 23.03 22.15 21.34 2.60 19.91 70 39.30 34.74 31.62 29.23 27.29 25.67 24.27 23.05 21.97 21.01 20.14 19.35 18.62 17.96 17.35 80 34.53 30.48 27.73 25.62 23.91 22.48 21.26 20.19 19.24 18.39 17.63 16.94 16.30 15.72 15.18 90 29.49 26.32 24.06 22.30 20.87 19.65 18.60 17.69 16.87 16.14 15.48 14.88 14.33 13.83 13.36 100 24.97 22.59 20.81 19.39 18.20 17.19 16.31 15.53 14.84 14.21 13.65 13.13 12.66 12.22 11.82 110 21.20 19.43 18.03 16.89 15.92 15.08 14.34 13.69 13.10 12.57 12.09 11.64 11.24 10.86 10.51 120 18.14 16.79 15.69 14.77 13.98 13.28 12.67 12.12 11.62 11.17 10.75 10.37 10.C2 9.70 9.39 130 15.65 14.61 13.74 12.99 12.34 11.j6 11.24 10.78 10.36 9.97 9.61 9.29 8.98 8.70 8.44 140 13.62 12.80 12.10 11.49 10.95 10.46 10.03 9.63 9.27 8.94 8.64 8.35 8.09 7.84 7.62 150 11.95 11.29 10.72 10.22 9.76 9.36 8.99 8.65 8.34 8.06 7.79 7.55 7.32 7.10 6.90 160 10.57 10.03 9.56 9.14 8.75 8.41 8.10 7.81 7.54 7.29 7.C6 6.85 6.65 6.46 6.28 170 9.40 8.96 8.57 8.21 7.89 7.59 7.32 7.07 6.84 6.63 6.43 6.24 6.06 5.90 5.74 180 8.42 8.05 7.72 7.42 7.14 6.89 6.65 6.44 6.23 6.05 5.87 5.71 5.55 5.40 5.27 190 7.58 7.27 6.99 6.73 6.49 6.27 6.07 5.88 5.70 5.54 5.38 5.24 5.10 4.97 4.85 200 6.86 6.60 6.35 6.13 5.93 5.73 5.56 5.39 5.23 5.09 4.95 4.82 4.70 4.58 4.47 L/R VALUES OF P/A IN KSI FOR k =.16.17.18.19.20.21.22.23.24.25.26.27.28.29.30 $^l\ 10 41.56 41.13 40.71 40.29 39.88 39.49 39.10 38.71 38.34 37.97 37.61 37.26 36.91 36.57 36.23 20 35.18 34.55 33.94 33.36 32.79 32.25 31.72 31.21 30.72 30.24 29.78 29.33 28.90 28.48 28.07 30 30.04 29.34 28.67 28.04 27.43 26.85 26.30 25.76 25.25 24.77 24.30 23.85 23.41 22.99 22.59 40 25.79 25.09 24.43 23.81 23.22 22.66 22.13 21.62 21.14 20.68 20.24 19.82 19.42 19.03 18.66 50 22.24 21.59 20.97 20.39 19.85 19.33 18.84 18.38 17.94 17.53 17.13 16.75 16.39 16.05 15.71 60 19.27 18.67 18.12 17.60 17.11 16.65 16.21,15.80 15.41 15.04 14.69 14.35 14.C3 13.73 13.44 70 16.78 16.25 15.76 15.30 14.86 14.46 14.07 13.71 13.37 13.04 12.73 12.44 12.16 11.89 11.64 80 14.69 14.22 13.79 13.39 13.01 12.65 12.32 12.00 11.70 11.41 11.14 10.89 10.64 10.41 10.18 90 12.93 12.52 12.15 11.80 iY.47 1i.fr5 10.86 10.58 10.32 10.07 9.83 9.61 9.39 9.19 8.99 100 11.44 11.09 10.77 10.46 10.17 9.90 9.64 0 9.6 9 9.17 8.95 8.74 8.54 8.36 8.18 8.30 110 1J.18 9.88 9.59 9.33 9.07 8.84 8.61 8.40 8.20 8.01 7.82 7.65 7.48 7.32 7.17 120 9.11 8.85 8.60 8.36 8.14 7.93 7.74 7.55 7.37 7.20 7.04 6.89 6.74 6.60 6.47 130 8.19 7.96 7.74 7.54 7.34 7.16 6.99 6.82 6.67.6.52 6,37 6.24 6.11 5.98 5.86 140 7.40 7.20 7.01 6.83 6.66 6.49 6.34 6.19 6.05 5.92 5.80 5.67 5.56 5.45 5.34 150 6.71 6.54 6.37 6.21 6.06 5.91 5.78 5.65 5.52 5.41 5.29 5.18 5.08 4.98 4.89 160 6.12 5.96 5.81 5.67 5.54 5.41 5.29 5.17 5.06 4.95 4.85 4.76 4.66 4.57 4.49 170 5.59 5.45 5.32 5.20 5.08 4.96 4.86 4.75 4.65 4.56 4.47 4.38 4.30 4.22 4.14 180 5.13 5.01 4.89 4.78 4.67 4.57 4.47 4.38 4.29 4.21 4.13 4.05 3.97 3.90 3.83 190 4.73 4.62 4.51 4.41 4.32 4.22 4.14 4.05 3.97 3.90 3.82 3.75 3.68 3.62 3.55 200 4.37 4.27 4.17 4.08 4.00 3.92 3.84 3.76 3.69 3.62 3.55 3.49 3.42 3.36 3.31

TABLE I (Continued)................ ~_Y!ELDSTI:ESS =60 KSI..... L/R VALUES OF P/A IN KSI FOR k=.01.02.03.04.05.06.07.08.09.10',1.12.13.1.1'lO 59.24 58.51...57.79 57.09 56.40 55.73 55.08 5-4".'45..'5-3-83 53.22 52.62...52.04 51.48 50.92 50.38 2 0 58.41. 56.90 55.48 54.13 52.85 51.64 50.48 49.37.48.32 47.31 46.35 45.42 44.54 43.69 42.87 30 57.36 54.99 52.84 50.87 49.07 47.41 45.87 44.44 43.11 41.86 40.69 39.58 38.54 37.56 36.63 40 55.90' 52,51.. 49,62 47,11 44.91 42.94 41.16 39.56...38-09 36,74 35.50 34.35 33.28 32.2.8 31.35 50 53.61 49.12 45.62 42.76 40.34 38.25 36~42 34.79 33.33 32.01 30.81 29.70 28.68 27.74 26.87.....60 49.82..44.56 40.84 37.9-4 35.57 33.56 31.83 30.30 28.95 2'7.74 26.64 25.64 24.72 23.87 23.09 70 44.03 39.03 35.62 33.01 30.89 29.11 27.58 26.24 25.05 23.98 23.02 22.15 21.35 20.61 19.93 80 37.24 33.32 30.54 28.38 26.61 25.1!1 23.81 22.68...21.67..20.76 19.94 19.19 1'8.50 17.87 17.28 90 30.99 28.15 26.02 24.30 22.87 21.64 20.58 19.63 18.79 18.02 17.33,16.70 16.12 15.58 15.08 100 25.85 23.81 22-19 20,85 19.70 18.71 17.84 17,06 16./36 1..5.72!15.14 14.61 14.12 13.66 13.24 11O 21.76 20.26 19.02 17.97 17.06 16.26 ls.55 14.90 14.32 13.79 13.30 12.85 12.44 12.05 11.69 120 18.51 17.38 16.43 15.60 14.86 14.21 13.63 13.09 12.61 12.16 11.75 11.37 11.02 10.69,10.38 130 1..5.92 i5.05 1..i4.29 13.63 13,03 12.50 12.02 11.57 11.17 10,79 1...i0.44....0.12 9.82 9.54 9.27 140 13.82 13.13 12.53 11.99 11.50 11.06 10.66 10.29 9.95 9.63 9.33 9.06 8.80 8.56 B.33 150 12.10'11.55 11.06 10,.62 10(.22 9...85 9.51 9'20 8.91' 8.63 8.38 8.14 7.92 7.71 7.51 160 10.68 10.24 9.83 9.47 9.13 8.82 8.53 8.26 8.02 7.78 7.56 7.36 7.17 6.98 6.81 170 9.50 9...13..8.79 8.48 8.20 7.94 7-6-9;.7,46...7.25 7......705 6.86 6'68 6,51 6.35 6.20 ~180 8,50 8:19...7.91 7.64 7.40 7.18 6.97 6.77 6.58 6.41' 6.24 6.09 5.94 5.80 5.67 190 7.65 7.39 7.15 6.92 6.71 6.52 6.34 6.17 6.00 5.85 5.71 5.57 5.44 5.32 5.20 200 6.92 6.69,6049.6.30 6_-1_2 _5_95__5,79 5064 5.50 5.36 5.24 5.11 5.00 4o89 4.79 \,0 L/R VALUES OF P/A IN KSI FOR k=.16.17.18.19.20.21.22.23.24.25.26.27.28.29.30 10 49.85 49,33..48,82 48J32 47_83 47.35 46.8-9..46.43 45.98 4,5.53 45.10 44.68 44.26 43,85 43.45 20 42?08 4...,33 40.60 39.90 39.22 38.57 37.93 37.32 36.73 36.16 35.61 35.07 34.55 34.05 33.56 30 35.75 34.92 34.12 33.36 32.64 31.95 31.29 30.66 30.05 29.47 28,91 28-.38'-27.86 27.36 26'88 40 30.47 29.65 2.8.87 28.14 27.44 26.78..26-16 25.56.25.00 24.45 23.94 23.44 22.97 22.52 22.08 50 26.05'25.29 24.58 23.91 23.27 22.68 22.11 21.58 21.07 20.59 20.13 19.69 19.27 18.87 18.4% 60 22.36 21.69 21.05 20,46 19.90 1..!9-.38 18-88..18,41 17,96 17.754 17.14 16.76 16.39 16.04 15.71 70 19.29 18.70 18.15 17.64 17.15 16.70 16.27 15.86 15.47 15.11 14.76 14.43 14.1il 13.81 13. 52 80 16.74.1.6.23...15,76 15.32 14.90 14.51 14.13 13.78 13.45 13.13 12.83 12.55 12.27 12.01 11.76 90 14.62 14.18 13.78 13.40 13.04 12.70 12.3.8 12.0'8 11'79 11.52..1'1-.26 11.0i 19.77 10.55 10. 33 100 12.84 12.47 12,.12 11.80 11.49 11.20 10.92 10.66 10.41 10.17 9.95 9.73 9.53 9.33 9.15 110 11.35 11.03 10.73 10.45 10'19...9.94 9.70 9.47 9.26 9.05 8.86 8.67 8.49 8.32 8.16i 120 10.09 9.82 9.56' 9.32 9.09 8.87 8.67 8.47 8.28 8.10 7.93 7.77 7.61 7.46 7.32 130 9,'02 8,79....8.57 8....-36 —8 -1- 6 J- 7 - -7...-7....7.62 7."45..7.30 7,15....7.00 6.87 6.73.~ 6.61 140 8.11 7.91 7.71 7.53 7.36 7.19 7.03 6......,88 6.74 6.60 6,47 6....,34 6.22 6.11 5.99 150 7.33 7.15 6.98.6.82 6.67 6.52i 6.38 6.25 6.12 6.00 5.89 5.77 5.67' 5.56 5.46 160 6.65 6.49 6.34 6.2'0 6.07 5.94 5.82 5.70 5.59 5.48 5.38 5.28 5.18 5.09 5.09 170 6.06 5.92 5.79 5.67 5.55....5.43 5.33 5.22 5.12 5.03 4.93 4.84 4.76 4.68 4.6: 1SO 5.54 5:42...5.31 5....20 5...-09 4-99 4.....489 4,80 4.71 4.62 4.54 4.46 4.38 4.31 4.24 190 5.09 4.98 4.88 4.78 4.69 4.60 4.51 4.43 4.35 4.27 4.19 4.12 4.05 3.99 3.92 200 4.69 4.59 4.50 4.41 4.33 4.25 4.17 4.09 4.02 3.95~ 3.89 3.82 3.76 3.70 3. 6/

T-ABIE I (Continued) L/R VALUES OF P/A IN KS[ FORl=.ol.02.03.04.os.06.07.08.09. zo. ll.12.13.14. is 20 68.11 66.33 64.66' 63.07 61.56 60.13 58.77 57.47 56.24 55.06 53.93 52.85 51.81 50.82 49.86 30 66.81 63.96 61.39 59.06 56.93 54.9'7 53.16 51.49 49.92 48.47 47.10 45.82 44.61 43.47 42.39 50 61.55 56.07 51.94 48.62 45.84 43.46 41.37 39.52 37.87 36.38 35.02 33.77 32.63 31.57 30.58 60 55.7 4 49.73 45.58 42.37 39.76 37.55 35.65 33.98 3'2.49 31.15 29.95 28.84 27.83 26.90 26.04 70 47.4.3 42.39 38.89 36.17 33.95 32.08 30.46 29.03 27.77 26.63 25.60 24.66 23.80 23.00 22.27 80 38.97 35.38 32.71 30.58 28.80 27.29_25.96 24.79 23.75 22,80 2 1.95 21.16 20.44 19.77 19.15 90 31.9'2 29.43 27.47 25.84 24.46 23.25 22.19 21.24 20.39 19.62 18.91 18.26 17.66 17.10 16.58 100 26.40 24.64 23.19 21.95 20.87 19.92 19.07 18.30 17.61 16.98 16.39 15.85 15.36 14.89 14.46 110 22.11 20.83 19.73 18.78 17.93 17.18 16.50 15.88 15.31 14.79 14. 31 13.86 13.44 13.05 12.69 120.18.76 17.79 16.95 16.20 15.53 14.93 14.37 i3.87 13.40 12.97 12.57 12.20 11.85 11.52 11.21 130 16.09 15.35 14.69 14.10 13.56 13.06 12.61 12.20 11.81 11.45 11.12 10..80 10.51 10.23 9.97 140 13.95 13.37 12.84 12,36 1l.92 11.52 11.14 10.80 10.47 10.17~ 9.89 9.62 9.37 9.14 8.91 ----—...........................-. —-------------- 150 12.20 11.74 11.31 10.92 10.55 10.22 9.91 9:62 9.35 9.09 8.85 8.62 8.41 8.21 8.01 160 10.76 10.38 10.03 9.71 9.40 9.12 8.86 8.62 8.38 8.17 7.96 7.77 7.58 7.41 7.24................................................................. 170 9.56 9.25 8.95 8.68 8.43 8.19 7.97 7.76 7.56 7.37 7.20 7.03 6.87 6.72 6.57 180 8.55 8.29 8.04 7.81 7.60 7.39 7.20 7.02 6.85 6.69 6.54 6.39 6.25 6.12 5.99 190 7.69 7.47 7.26 7..06 6.88 6.70 6.54 6.38 6.23 6.09 5.96 5.83 5.71 5.59 5.48 200 6.95 6.76 6.58 6.42 6.26 6.111 5.96 5.83 5.70 5.57 5.46 5.34 5.24 5.13 5.03 L/R VALUES OF P/A IN KSI FORk o ~~~~~~~~~.16.17.18.19.20 -.21.22.23.24.25 026.27.28.29.30 i() 58:3 57o52 56.92 56.34 55.77 55.21 54.67 54.13 53.60 53.09 52.58 52(09 51.60 51.12 50065 30 41.37 40.40 39.47 38.60 37.76 36.96 36.20 35.46 34.76 34.09 33.45 32.83 32.23 31.66 31011 40 35.00 34.06 33.17 32.33 31.53 30.78 30.06 29.38 28.73 28.12 27.5'3 26.96 26.42 25.90 25.41 60 25.24 24,49..23,79 23-!3 22-5l1 21.93 21.38 20.86 20.37 19.90 19.45 19:02...8.62 18.23 17.86 70 21.58 20.94 20.35 19.79 19.26 18.76 X8.29 17.85 17.42 17.02 16.64 16028 15.93 15.60 15.28 80 18.57 18.04 17.53 17.06 16.61 16.19 15.79 15.41 15.05 14.71 14.38 14.07 13.78 13.49 13022 90 16.10 15.64 15.22 14.82 14.44 14.08 13.74 13.42 13.1.1 12.82 12.54 12028 12.03 11.78 11.55 1oo 14.05 13.67 13.31 12.97 12.65 12.34 12.05 11.78 11.52 11.27 1l.03 10.80 10.59 10.38 lo.18 110' 12.35 12.02 11.72 11.043...6 1.Z —i —~90-' — 1.... 10'.4'2 10.79 —9.98 9.....:77 9... —58 9.39 9.21 9.04 120 10.92 10.65 10.39 10.14 9.91___9.69 9.47 9.27 9.08 8.89 8.72 8.55 8038 8.23 8.08 130 9o72 9.49...9.26 9.05 8.85 8.66 8.48 8.30 8.13 -7.97 7.82 7.67 7.53..7.39...7.26 140 8.70 8.50 8.31 8.13 7.95. 7.79 7.63 7.48 7.33 7.19 7.05 6.93 6.80 6.68 6.57 150 7.83 7.66 7.49 7.33 7.18 7.04 6.90 6.76 6.64 6.51 6.40 6.28 6.17 6.07 5.97 160 7.08 6.93 6.79 6.65 6.52 6.39 6.27 6.15 6.04 5.93 5.83 5.73 5.63 5.54 5.44 170 6.43 6.30 6.17 6.05' 5.94 5.83 5.72 5.62 5.52 5.42 5.33 5.24 5.15 5.07 4.99 180 5.87 5.75 5.64 5.53 5.43 5.33 5.24 5.15 5.06 4.97 4.89 4.81 4.74 4.66 4.59 190 5.37 5...27 5.17 5.08 4.99 4.90 4.82 4.73 4;66 4o58....4.51...4.44 4.37 4.30 4,24 200 4.94 4.85 4.76 4.68 4.60 4.52 4.44 4.37 4.30 4.23 4.16 4.10 4.04 3.98 3.92

TABLE I (Concluded) (YIELD STRESS = 100 KSI) L/R VALUES OF P/A IN KSI FOR k.01.02.03.04.05.06.07.08.09.10.11.12.13.14.15 10 98.72 97.47 96.26 95.08 93.92 92.80 91.70 90.63 89.59 88.57 87.57 86.59 85.64 84.71 83.80 20 97.18 94.53 92.05 89.71 87.50 85.40 83.42 81.54 79.74 78.04 76.41 74.85 73.36 71.93 70.56 30 94.89 90.45 86.53 83.02 79.86 76.99 74.36 71.93 69.69 67.61 65.67 63.85 62.14 60.54 59.02 40 90.68 83.97 78.67 74.29 70.55 67.30 64.41 61.83 59.50 57.38 55.44 53.64 51.98 50.44 49.00 50 81.76 73.69 68.04 63.63 60.01 56.93 54.27 51.91 49.81 47.91 46.18 44.60 43.15 41.80 40.55 60 67.07 60.69 56.15 52.57 49.62 47.11 44.92 42.99 41.26 39.70 38.28 36.98 35.78 34.67 33.64 70 52.60 48.54 45.39 42.80 40.60 38.70 37.01 35.51 34.16 32.93 31.80 30.76 29.80 28.91 28.08 80 41.47 38.89 36.75 34.91 33.32 31.90 30.63 29.48 28.44 27.48 26.60 25.78 25.02 24.31 23.64 90 33.30 31.57 30.08 28.77 27.59 26.54 25.58 24.70 23.89 23.14 22.45 21.80 21.23 20.63 20.10 100 27.25 26.04 24.96 24.00 23.12 22.32 21.58 20.90 20.27 19.68 19.13 18.62 18.14 17.68 17.25 110 22.67 21.79 21.00 20.27 19.60 18.98 18.41 17.87 17.37 16.90 16.46 16.05 15.66 15.29 14.94 120 19.15 18.49 17.88 17.32 16.80_16.31 15.86_15.43 15.03 14.65 14.30 13.96 13.64 13.33 13.04 130 16.38 15.87 15.40 14.95 14.54 14.15 13.79 13.44 13.12 12.81 12.52 12.24 11.97 11.72 11.48 140 14.17 13.77 13.39 13.03 12.70 12.38 12.09 11.80 11.54 11.28 11.04 10.81 10.59 10.38 10.18 150 12.37 12.05 11.74 11.46 11.18 10.92 10.68 10.44 10.22 10.01 9.81 9.61 9.43 9.25 9.08 160 10.90 10.63 10.38 10.14 9.92 9.70 9.50 9.30 912 8.94 8.77 8.60 8.44 8.29 8.15 170 9.67 9.45 9.24 9.05 8.86 8.67 8.50 8.34 8.18 8.03 7.88 7.74 7.63 7.47 7.35 180 8.64 8.46 8.28 8.11 7.95 7.80 7.65 7.51 7.38 7.25 7.12 7.00 6.88 6.77 6.66 190 7.77 7.61 7.46 7.32 7.18 7.05 6.92 6.80 6.69 6.57 6.46 6.36 6.26 6.16 6.07 200 7.02 6.88 6.76 6.63 6.52 6.40 6.29 6.19 6.09 5.99 5.89 5.80 5.71 5.63 5.55 L/R VALUES OF P/A IN KSI FOR k =.16.17.18.19.20.21.22.23.24.25.26.27.28.29.30 t -'P r10 82.91 82.04 81.19 80.35 79.53 78.73 77.95 77.18 76.43 75.69 74.96 74.25 73.56 72.87 72.20 20 69.25 67.99 66.77 65.60 64.48 63.39 62.35 61.33 60.36 59.41 58.50 57.61 56.75 55.92 55.12 30 57.59 56.23 54.95 53.72 52.56 51.45 50.39 49.37 48.40 47.47 46.58 45.72 44.89 44.10 43.34 40 47.65 46.38 45.19 44.07 43.00 42.00 41.04 40.13 39.26 38.44 37.65 36.89 36.17 35.48 34.81 50 39.38 38.29 37.27 36.31 35.40 34.54 33.72 32.95 32.22 31.52 30.85 30.22 29.61 29.32 28.47 60 32.67 31.77 30.93 30.13 29.38 28.67 28.00 27.36 26.75 26.17 25.62 25.09 24.59 24.11 23.65 70 27.30 26.57 25.88 25.24 24.62 24.04 23.49 22.97 22.47 22.00 21.54 21.11 20.70 20.30 19.92 80 23.02 22.43 21.87 21.35 20.85 20.38 19.93 19.50 19.09 18.70 18.33 17.97 17.63 17.30 16.98 90 19.59 19.12 18.67 18.24 17.83 17.45 17.08 16.73 16.39 16.07 15.76 15.46 15.18 14.90 14.64 100 16.84 16.45 16.08 15.73 15.40 15.08 14.78 14.49 14.21 13.94 13.68 13.43 13.2) 12.97 12.75 110 14.60 14.28 13.98 13.69 13.42 13.15 12.90 12.65 12.42 12.20 11.98 11.77 11.57 11.38 11.19 120 12.77 12.50 12.25 12.01 11.78 11.56 11.35 11.14 10.95 10.76 10.57 10.40 10.23 10.06 9.91 130 11.25 11.03 10.82 10.62 10.42 10.23 10.05 9.88 9.71 9.55 9.40 9.25 9.10 8.96 8.83 140 9.98 9.80 9.62 9.45 9.28 9.12 8.97 8.82 8.68 8.54 8.41 8.28 8.15 8.03 7.91 150 8.91 8.76 8.60 8.46 8.32 8.18 8.05 7.92 7.80 7.68 7.56 7.45 7.34 7.24 7.14 160 8.00 7.87 7.74 7.61 7.49 7.37 7.26 7.15 7.04 6.94 6.84 6.14 6.65 6.56 6.47 170 7.23 7.11 7.00 6.89 6.78 6.68 6.58 6.49 6.39 6.30 6.21 6.13 6.05 5.97 5.89 180 6.56 6.45 6.36 6.26 6.17 6.08 5.99 5.91 5.83 5.75 5.67 5.60 5.52 5.45 5.38 190 5.97 5.89 5.80 5.72 5.63 5.56 5.48 5.41 5.33 5.26 5.20 5.13 5.07 5.00 4.94 200 5.47 5.39 5.31 5.24 5.17 5.10 5.03 4.96 4.90 4.84 4.78 4.72 4.66 4.61 4.55

TABLE II LOAD AND DEFLECTION COEFFICIENTS TO GIVE EQUIVALENT LATERAL LOAD FOR USE IN TABLE I CASE LODING CONDITION EQUIVALENT / 1IN DEFLECTNO. L LOAD FACTOR ION FACTOR W i I ___________ I I I.028 2 ~,__T1^ 1.026 0.003 w W 4 2 - 0.178 L/2 L/2, a____________, b 2 78 a I o^ b L-' 3L2 ) L6 I L 1.333 + 0.051 L/3 L/3 L/3 3 3 3?7 ^ ^ ^,1.333 -0.023 -L/4,L/4, L/4 L/4 * At load point * At Center Note: For design purposes it would be satisfactory to assume tjO, with resulting deflection factor of 1.0, for cases I,2,3,6, and 7. 72

TABLE III AVERAGE COLUMN LOAD FOR WHICH COMBINED STRESS IN RESTRAINED BEAM-COLUMNS WITH UNIFORM LATERAL LOAD JUST REACHES YIELD POINT cy = 50.000000 )= *200000 L/R VALUES OF P/A IN KSI FOR k =.02.04.06.08.10.12.14.16.18.20.22.24.26.28.30 10 49.13 48.33 47.58 46.82 46.11 45.39 44.71 44.06 43.39 42.76 42.16 41.59 40.98 40.47 39.92 20 48.27 46.73 45.23 43.86 42.59 41.39 40.20 39.12 38.13 37.17 36.22 35.34 34.53 33.71 32.94 30 47.33 45.03 42.93 41.03 39.33 37.74 36.24 34.91 33.67 32.55 31.48 30.50 29.54 28.66 27.84 40 46.29 43.16 40.52 38.20 36.15 34.32 32.70 31.23 29.87 28.65 27.56 26.51 25.55 24.66 23.84 50 44.96 41.11 37.94 35.31 33.U1 31.13 29.44 27.93 26.59 25.34 24.28 23.25 22.31 21.45 20.71 60 43.40 38.80 35.30 32.44 30.12 28.14 26.47 25.02 23.68 22.54 21.46 20.53 19.62 18.84 18.12 70 41.36 36.16 32.47 29.67 27.31 Z5.4Q 23.77 22.40 21.13 20.05 19.07 18.18 17.42 16.66 16.01 8U 38.19 33.36 29.66 26.9 24.70 22.93 21.40 20.06 18.91 17.94 17.00 16.26 15.51 14.83 14.24 90 35.75 30.42 26.89 24.37 22.32 20.63 19.26 18.05 17.02 16.09 15.29 14.57 13.89 13.27 12.74 10O 32.34 27.44 24.31 21.96 20.08 18.63 17.34 16.25 15.32 14.48 13.76 13.10 12.49 11.98 11.50 110 28.92 24.69 21.88 19.81 18.13 16.76 15.67 14.68 13.84 13.07 12.42 11.82 11.32 10.86 10.36 w"< ~ 120 25.66 22.08 19.65 17.80 16.35 15.18 14.19 13.29 12.53 11.89 11.30 10.74 10.29 9.86 9.46 130 22.79 19.80 17.69 16.09 14.84 13.78 12.88 12.06 11.43 10.84 10.30 9.79 9.37 8.98 8.61 140 20.24 17.73 15.96 14.57 13.44 12.47 11.72 11.02 10.44 9.90 9.40 8.99 8.61 8.24 7.90 150 17.98 15.96 14.42 13.21 12.24 11.42 10.66 10.09 9.55 9.05 8.65 8.21 7.92 7.58 7.26 160 16.09 14.38 13.03 11.99 11.16 10.41 9.77 9.24 8.75 8.35 7.91 7.56 7.29 6.97 6.74 170 14.46 13.02 11.84 10.95 10.19 9.55 9.02 8.53 8.07 7.70 7.35 7.03 6.72 6.48 6.20 180 13.05 11.79 10.82 10.06 9.36 8.77 8.28 7.82 7.45 7.11 6.79 6.48 6.25 5.97 5.76 190 11.83 10.73 9.90 9.19 8.60 8.11 7.65 7.23 6.89 6.56 6.32 6.03 5.82 5.56 5.36 2C0 10.72 9.82 9.11 8.45 7.96 7.51 7.08 6.74 6.42 6.12 5.83 5.62 5.42 5.17 4.99

TABLE III (Continued) Oy = 50.OCOOOO =.400000 L/R VALUES OF P/A IN KSI FOR-k =.02.04.06.08.10.12.14.16.18.20.22.24.26.28.30 10 49.26 48.58 47.96 47.32 46.68 46.08 45.52 44.94 44.39 43.82 43.28 42.78 42.30 41.78 41.30 20 48.58 47.23 45.98 44.80 43.71 42.64 41.64 40.68 39.75 38.86 38.03 37.22 36.47 35.71 35.00 30 47.83 45.84 44.06 42.41 40.83 39.43 38.12 36.91 35.73 34.68 33.67 32.69 31.79 30.97 30.15 40 46.98 44.35 42.02 39.95 38.09 36.38 34.89 33.48 32.24 31.03 29.94 28.95 27.99 27.10 26.28 50 45.96 42.67 39.88 37.50 35.36 33.51 31.88 30.43 29.09 27.84 26.78 25.75 24.81 23.89 23.09 60 44.78 40.80 37.61 34.94 32.74 30.77 29.09 27.58 26.24 25.04 23.96 22.97 22.06 21.22 20.50 70 43.30 38.72 35.22 32.42 30.13 28.21 26.52 25.02 23.75 22.61 21.57 20.61 19.19 18.97 18.26 80 41.41 36.36 32.73 29.93 27.64 25.74 24.15 22.74 21.54 20.44 19.44 18.57 17.76 17.08 16.37 90 39.18 33.86 30.21 27.49 25.32 23.50 21.95 20.67 19.52 18.52 17.61 16.75 16.02 15.39 14.74 100 36.46 31.19 27.74 25.15 23.08 21.44 20.03 18.81 17.70 16.79 15.95 15.23 14.55 13.92 13.38 110 33.48 28.57 25.32 22.93 21.06 19.51 18.23 17.12 16.15 15.32 14.55 13.88 13.26 12.67 12.17 120 30.41 25.95 23.09 20.92 19.23 17.81 16.63 15.61 14.72 13.95 13.30 12.68 12.10 11.61 11.15 130 27.42 23.55 21.01 19.02 17.53 16.28 15.20 14.31 13.49 12.78 12.17 11.60 11.06 10.60 10.17 140 24.61 21.36 19.08 17.38 16.00 14.85 13.90 13.08 12.38 11.71 11.15 10.68 10.17 9.74 9.40 <-~ ~ 150 22.10 19.34 17.35 15.84 14.62 13.61 12.79 12.02 11.36 10.80 10.28 9.83 9.42 9.02 8.63 160 19.90 17.57 15.84 14.49 13.41 12.53 11.17 11.11 10.50 9.97 9.47 9.06 8.66 8.35 8.05 170 17.96 15.96 14.47 13.32 12.31 11.55 10.83 10.22 9.70 9.20 8.79 8.40 8.09 7.73 7.45 180 16.24 14.54 13.26 12.25 11.36 10.64 10.03 9.45 9.02 8.55 8.16 7.79 7.50 7.22 6.95 190 14.77 13.29 12.15 11.25 10.47 9.86 9.28 8.79 8.32 7.94 7.57 7.28 7.01 6.74 6.49 200 13.47 12.19 11.23 10.39 9.71 9.13 8.64 8.18 7.80 7.43 7.08 6.81 6.55 6.30 6.05

TABLE III (Continued) ay 50.000000 =.600000 L/R VALUES OF P/A IN KSI FOR k =.02.04.06.08.10.12.14.16.18.'20.22.24.26.28.30 10 49.32 48.77 48.14 47.57 46.99 46.45 45.89 45.37 44.89 44.38 43.91 43.40 42.92 42.47 42.05 20 48.70 47.54 46.36 45.30 44.27 43.27 42.33 41.43 40.57 39.79 38.97 38.22 37.47 36.77 36.13 30 48.02 46.28 44.62 43.10 41.64 40.30 39.12 )7.91 36.86 35.80 34.86 33.94 33.04 32.22 31.46 40 47.29 44.91 42.77 40.83 39.09 37.51 36.07 34.73 33.49 32.34 31.25 30.26 29.36 28.48 27.66 50 46.40 43.42 40.82 38.62 36.61 34.82 33.26 31.80 30.47 29.28 28.15 27.13 26.19 25.32 24.52 60 45.40 41.80 38.80 36.31 34.12 32.21 30.59 29.08 27.74 26.54 25.46 24.41 23.50 22.66 21.87 70 44.17 39.97 36.66 33.92 31.69 29.78 28.08 26.59 25.25 24.11 23.01 22.05 21.17 20.34 19.63 80 42.66 37.86 34.35 31.56 29.33 27.37 25.72 24.31 23.04 21.88 20.88 20.01 19.14 18.39 17.68 90 40.75 35.61 32.02 29.24 27.01 25.19 23.57 22.24 21.02 19.96 18.98 18.13 17.39 16.64 15.99 100 38.52 33.19 29.68 26.96 24.83 23.06 21.59 20.31 19.20 18.23 17.33 16.54 15.80 15.17 14.56 110 35.92 30.69 27.32 24.81 22.81 21.14 19.79 18.62 17.59 16.63 15.86 15.13 14.45 13.86 13.30 <-i( 120 33.04 28.20 25.09 22.74 20.91 19.43 18.13 17.04 16.09 15.26 14.55 13.87 13.23 12.68 12.21 130 30.17 25.80 23.01 20.84 19.21 17.85 16.63 15.69 14.80 14.03 13.36 12.72 12.18 11.67 11.24 140 27.36 23.55 21.02 19.13 17.63 16.35 15.34 14.40 13.63 12.90 12.27 11.74 11.23 10.74 10.34 150 24.73 21.46 19.23 17.52 16.18 15.04 14.10 13.27 12.55 11.93 11.34 10.83 10.35 9.95 9.57 160 22.40 19.57 17.59 16.12 14.91 13.91 13.02 12.24 11.62 11.03 10.53 10.06 9.60 9.22 8.86 170 20.27 17.90 16.15 14.82 13.75 12.80 12.02 11.34 10.76 10.20 9.73 9.34 8.90 8.61 8.26 180 18.43 16.35 14.82 13.62 12.67 11.83 11.15 10.51 10.02 9.48 9.10 8.67 8.31 7.97 7.70 190 16.77 14.98 13.65 12.63 11.72 10.98 10.34 9.79 9.32 8.88 8.45 8.10 7.76 7.49 7.18 200 15.28 13.75 12.61 11.64 10.90 10.19 9.64 9.12 8.67 8.25 7.90 7.56 7.24 6.98 6.74

TABLE III (Continued) ay = 50.OCOCOO T). =.800000 L/R VALUES OF P/A IN KSI FOR k =.02.04.06.08.10.12.14.16.18.20.22.24.26.28.30 10 49.32 48.64 48.02 47.45 46.80 46.26 45.71 45.12 44.58 44.07 43.53 43.03 42.55 42.10 41.61 20 48.64 47.36 46.17 45.05 43.96 42.96 41.95 41.05 40.13 39.29 38.47 37.66 36.90 36.21 35.50 30 47.95 46.09 44.37 42.78 41.33 39.93 38.68 37.48 36.36 35.30 34.29 33.38 32.48 31.66 30.84 40 47.23 44.78 42.58 40.58 38.84 37.20 35.70 34.29 33.06 31.90 30.81 29.16 28.80 27.91 27.09 50 46.40 43.36 40.69 38.44 36.36 34.57 32.94 31.49 30.15 28.90 27.78 26.75 25.81 24.39 24.09 60 45.46 41.80 38.80 36.25 34.06 32.14 30.41 28.89 27.55 26.29 25.14 24.16 23.18 22.34 21.56 70 44.36 40.16 36.85 34.11 31.81 29.78 28.08 26.52 25.19 23.98 22.88 21.93 20.98 20.16 19.38 80 43.10 38.36 34.79 31.99 29.64 27.62 25.90 24.43 23.10 21.94 20.88 19.94 19.08 18.26 17.56 90 41.56 36.42 32.71 29.87 27.51 25.56 23.88 22.49 21.21 20.09 19.11 18.19 17.3) 16.64 15.99 100 39.65 34.32 30.62 27.77 25.52 23.63 22.03 20.69 19.51 18.48 17.51 16.67 15.93 15.23 14.63 110 37.30 32.07 28.51 25.74 23.63 21.83 20.35 19.05 17.97 17.01 16.11 15.32 14.63 13.98 13.42 120 34.66 29.70 26.40 23.86 21.85 20.18 18.82 17.61 16.59 15.64 14.80 14.12 13.48 12.86 12.34 130 31.92 27.30 24.32 22.02 20.21 18.66 17.38 16.25 15.30 14.47 13.73 13.04 12.43 11.92 11.42 140 29.17 25.05 22.33 20.32 18.63 17.22 16.09 15.08 14.19 13.40 12.11 12.11 11.54 11.06 10.59 150 26.54 22.96 20.54 18.71 17.24 15.98 14.91 13.96 13.18 12.43 11.84 11.21 10.73 13.27 9.82 160 24.09 21.01 18.84 17.18 15.91 14.78 13.83 12.99 12.25 11.60 10.97 10.43 9.98 9.54 9.17 170 21.90 19.21 17.34 15.88 14.69 13.67 12.83 12.03 11.38 10.76 10.23 1.78 9.34 8.92 8.57 180 19.93 17.60 15.95 14.62 13.61 12.70 11.96 11.26 10.64 10.05 9.54 9.10 8.69 8.34 8.01 190 18.15 16.17 14.71 13.56 12.60 11.80 11.09 10.48 9.89 9.38 8.95 8.53 8.20 7.81 7.49 200 16.60 14.88 13.61 12.58 11.71 10.94 10.33 9.80 9.30 8.81 8.40 8.00 7.67 7.36 7.05

TABLE III (Concluded) Oy = 50.000000 ll = 1.000000 L/R VALUES OF P/a, IN KSI FOR k =.02.04.Oh.08. 10. 12. 14.16. 18.~0.22.24.26.28.30 10 49.26 48.58 47.96 47.32 46.68 46.08 45.52 44.94 44.39 43.82 43.35 42.78 42.30 41.78 41.30 20 48.58 47.29 46.04 44.86 43.77 42.71 41.70 40.74 39.82 38.98 38.10 31.34 36.53 35.83 35.13 3D 47.89 45.97 44.18 42.6041.08 39.68 38.37 37.10 35.98 34.93 33.92 32.94 32.04 31.22 30.40 40 47.10 44.60 42.3340.33 38.53 36.82 35.32 33.91 32.68 31.47 30.37 29.32 28.42 27.48 26.66 50 46..27 43.17 40.51 38.19 36..11 34.26 32.57 31.11 29.78 28.53 27.40 26.31 25.38 24.45 23.65 60 45.40 4,1.68 38.61 36.00 33.74 31.83 30.09 28.58 27.18 25.92 24.83 23.78 22.81 21.97 21.12 70 44.30'40.04 36.66 33.92 31.56 29.53 27.77 26.27 24.88 23.67 22.57 21.55 20.67 19.84 19.07 80 43.10 38.30 34.73 31.81 29.39 27.43 25.65 24.18 22.85 21.63 20.63 19.63 18.76 18.0{ 17.24 90 41.62 36.42 32.71 29.74 27.39 25.38 23.70 22.24 21.02 19.84 18.86 17.94 17.14 16.39 15.74 100 39.90 34.44 30.62 27.77 25.46 23.56 21.91 20.56 19.32 ~8.29 1'7.33 16.48 15.74 15.04 14.38 110 37.92 32.32 28.63 25.81 23.63 21.83 20.29 18.9g 17.84 16.82 15.98 15.19 14.45 13.79 t3.23 120 35.60 30.20 26.65 23.99 21.91 20.18 18.'75 17.54 16.4~' 15.58:14.73 13.99 13.35 12.74 12.21 130 33.04 28.05 24.69 22.21 20.28 18.72 17.38 16,25 15.30 14.40 1'3.67 12.97 12.37 11.79 [1.'30 140 30.42 25.92 22.83 20.63 18.81 17.35 16.15 15.08 14,,19 13.40 12.65 12.05 11.48 10.93 10.46 -,1 150 27.79 23.90 21.10 19.09 17.43 16.11 14.98 14.02 13.18 12.43 11.78 11.21 10.67 10.209.76 — 1 160 25.28 21.95 19.53 17.68 16.16 14.97 13.96 13.05 12.25 11.60 10.97 10.43 9.98 9.54 g.l/ 170 23.02 20.21 18.03 16o38 15.Oh 13.92 12.96 12.15 11.45 10.83 10.23 9.78 9.28 8.92 8.51 180 20.99 18.54 16.70 15.18 13..98 L2.95 12.09 11.32 10o70 10.11 9.60 9.10 8.69 8.34 7.95 190 19.21 17.04 15.46 14.13 13.04 1.2.1[ 11.2810.60 10.01 9.50 9.01 8.60 8.20 7.81 7.49 2C0 1'7.60 15.69 14.29 1.3.14 12.15 11.26 10.58 9.93 9.36 8.87 8.4'6 8.06 7.67 7.36 7.05 TI~E ESTIMATE EXCEEDED.. AT LOC 20076

TABLE IV ULTIMATE STRENGTH OF THE UNRESTRAINED BEAM COLUMNS OF THE SIMPLIFIED SECTION IN PLANAR BENDING UNDER UNIFORM LATERAL LOAD (W = kP) Yield Stress = 33 ksi, Residual Stress = 11.02 ksi Values of k: L/r.02.04.06.08.10.12.14.16.18.2.22.24.26.28.30 10 32.00 31.32 3C.50 29.87 29.26 28.50 27.93 27.37 26.83 26.31 25.79 25.30 24.81 24.34 23.ee 20 31.13 29.66 28.28 27.14 26.07 25.06 24.10 23.19 22.31 21.64 20.84 20.23 19.65 19.10 18.57 3C 29.91 27.86 26.16 24.61 23.18 22.C2 2C.94 19.94 19.00 18.11 17.44 16.81 16.05 15.48 14.94 40 28.66 26.C6 23.93 22.21 2C.66 19.43 18.30 17.27 16.32 15.59 14.76 14.13 13.55 13.00 12.49 50 27.18 24.23 21.92 19.92 18.49 17.C6 15.93 15.07 14.13 13.42 12.78 12.18 11.62 11.10 10.61 60 25.46 22.22 19.73 17.90 16.46 15.20 14.09 13.26 12.35 11.67 11.06 10.49 9.96 9.47 9.17 70 23.65 20.19 17.86 16.14 14.72 13.50 12.42 11.63 10.92 10.28 9,69 9.16 8.66 8.36 7.93 8C 21.58 18.32 16.15 14.31 13.09 11.92 11.06 10.30 9.63 9.03 8.64 8.14 7.67 7.40 6.99 90 19.57 16.45 14.34 12.89 11.74 10.78 9.97 9.26 8.63 8.6.....7.71 7,23 6.79 6.54 6.32. 100 17.48 14.78 12.92 11.55 10.47 9.57 8.81 8.31 7.72 7.18 6.86 6.41 6.16 5.93 5.56 11C. 15.68 _13,_14 L.518__C.3C 9.45 8.61 7.90 7.44. 88 6.55 6.25 5.83 5.59 5.22 5.03 120 13.89 11.73 1C.30 9.28 8.51 7.73 7.22 6.64 6.28 5.97 5.53 5.30 5.08 4.73 4.56 Yield Stress = 36 ksi, Residual Stress = 11.81 ksi Values of k: L/r.02.04.06.08.10.12.14.16.18.2.22.24.26.28.30 1C 34.94 34.19 33.29 32.6C 31.75 31.1C 30.47 29.86 29.26 28.69 28.13 27.58 27.05 26.53 26.02 2C 33.98 32.37 3C.85 29.60 28.24 27.13 26.26 25.26 24.30 23.56 22.86 22.02 21.38 20.77 20.19 30 32.64 30.38 28.51 26.8C 25.23 23.96 2.2.78 21.68 20.65 19.86 18.94 18.24 17.59 16.97 16.20 4C 31.25 28.38 26.04 24.15 22.46 21.10 19.87 18.74 17.69 16.90 15.98 15.30 14.66 14.C6 13.50 5C 29.43 26.17 23.62 21.62 2C.04 18.48 17.24 16.30 15.27 14.50 13.80 13.14 12.53 11.96 11.43 60 27.50 23.92 21.43 19.37 17.79 16.42 15.21 14.30 13.31 12.57 11.90 11.28 10.71 10.35 9.85 7C 25.28 21.65 19.10 17.23 15.68 14.53 13.36 12.50 11.73 11.03 1040 10.00 9,45 8.95 8.66 8C 23.1C 19.54 17.19 15.37 14.C5 12.78 11.85 11.03 10.31 9.83 9.24 8.69 8.36 7.89 7.62 9C 2C.63 17.44 15.34 13.C6 12.37 11.34 10.64 9.88 9.20 8.59 8.21 7.69 7.39 6.94 6.70 10C 18.43 15.55 13.58 12.12 11.14 10.19 9.37 8.83 8.19 7.80 7.27 6.97 6.52 6.27 6.C5 11C 16.38 13.72 12.08 10.95 1 C.2 9.13 8.54 7.87 7.46 6.92 6.60 6.32 5.9C 5.67 5.47 120 14.35 12.33 1C.85 9.78 8.97 8.15 7.62 7.17 6.61 6.28 6.00 5.56 5.34 5.13 4.95

TABLE IV (Continued) Yield Stress = 42 ksi, Residual Stress = 13.27 ksi Values of k: L/r.02.04.06.08.10.12.14.16.18.2.22.24.26.28.30 IC 4C.83 39.74 38.89 38.C6 37.06 36.29 35.55 34.83 34.12 33.44 32.78 32.14 31.51 30.90 30.30 20 39.49 37.57 35.99 34.30 32.91 31.59 30.56 29.37 28.46 27.37 26.55 25.76 24.80 24.08 23.39 30 37.895..20 32.98 3.30.96 29.32 27.81 26.42 25.12 23.90 22.97 22.10 21.07 20.30 19.57 18.88 40 36.00 32.79 3C.02 27.78 25.99 24.38 22.93 21.60 20.38 19.44 18.58 17.78 17.03 16.33 15.66 50 33.99 30..1C 27.09 24.93 22.87 21.24 19.80 18.69 17.70 16.59 15.76 15.00 14.29 13.83 13.20-. 60 31.41 27.35 24.41 22.00 2C.15 1855 17.35 16.29 15,35 1449 13.71 12.99 12.32 11.91 11,32 _.128...63 24.54..21.57 19.40 17.81 16.28 15.14 14.14 13.25 12.65 11..92.11.25 10.83 10.25 9.91 8C 25.65 21.72 19.22 17.14 15.64 14.39 13.32 12.39 11.76 11.01 10.54 9.91 9.53 8.98 8.68 9__ _2215.2_-1,12...16.97.9..15,22_13....82 12.66 11,.87 11.00 10...44... 9,..31....8.,72__..3_8 78. 7.859_6... -7 100 19.93 16.8C 14.83 13.43 12.35 11.27 10.56 9.76 9.24 8.59 8.20 7.86 7.34 7.06 6.82 110 I..7^3i.4 149__1.3..1.7.1-.C....10...,.01.. 9..37.. 8*83..816..7 76.. 74 _Q..6..-.0.6 635 6 13. 120 15.31 13.1C 11.75 10.63 9.76 9.06 8.48 7.78 7.37 7.00 6.68 6.40 5.94 5.71 5.51 Yield Stress = 46 ksi, Residual Stress = 14.17 ksi \o Values of k: L/r.02 -04.06.08.10.12.14.16.18.2.22.24.26.28.30 10 44.62 43.64 42.47 41.56 4C.68 39.83 38.78 37.98 37.21 36.46 35.73 35.02 34.55 33.88 33.22 20 43.36 41.24 39.26 37.62 36.08 34.63 33.26 32.18 30.94 29.97 29.06 28.19 27.13 26.34 25.58 3C 41.59 38.6C 36.14 33.91 31.86 30.20 28.66 27.46 26.12 25.09 23.90 22.99 22.14 21.34 20.58 40 39.24 35.67 32.83 30.35 28.15 26.38 24.78 23.55 22.20 21.17 20.22 19.34 18.52 17.74 17.01 50 36.75 32.65 29.54 26.93 24.89 23.11 21.52 20.31 18.99 18.00 17.09 16.25 15.70 14.97 14.29 60 33.81 29.30 26.28 23.64 21.85 20.1C 18.79 17.63 16.60 15.67 14.81 14.02 13.29 1284 12.20 70 30.61 26.11 23.08 20.95 19.00 17.56 16.32 15.23 14.26 13.62.12.82 12.0911.64 11.00 10.63 80 27.18 23.12 2C.43 18.41 16.79 15.45 14.29 13.28 12.61 11.80 11.29 10.60 10.19 9.60 9.27 9C 23.79 20.17 17.90 16.03 14.76 13.52 12.68 11.75 11.14 10.38 9.92 9.28 8.92 8.59 8.29 100 20.57 17.78 15.73 14.26 12.89 11.98 11.22 10.36 9.82 9.34 8.70 8.33 8.00 7.70 7.21 110 17,88 15.52 13.7112.63 1,39 190.57 9.9C 9.33 8.84 8.19 7e.81 7.48 7.18 6.69 6.45 120 15.59 13.64 12.30 11.14 1C.24 9.52 8.92 8.41 7.74 7.35 7.02 6.72 6.45 6.21 5.76

TABLE IV (Continued) Yield Stress = 50 ksi, Residual Stress = 15 ksi Values of k: L/r.02.04.06.08.10.12.14.16.18.2 22.24.26.28.30 10 48.64 47.31 46.28 45.28 44.07 43.14 42.24 41.37 40.52 39.69 38.89 38.11 37.35 36.61 36.14 20 47.0 44.68 42.76 4C.96 39.27 37.68 36.17 34.99 33.62 32.57 31.57 30.36 29.46 28.59 27.75 30 45.05 41.76 39.06 36.61 34.61 32.79 31.10 29.79 28.32 27.19 25.89 24.90 23.97 23.09 22.26 4C 42.45 38.51 35.38 32.66 3C.50 28.57 26.82 25.47 24.00 22.87 21.84 20.87 19.98 1913 18.34 50 39.65 35.12 31.71 28.e6 26.63 24.69 23.20 21.89 20.45 19.37 18.39 17.73 16.88 16.C8 15.34 60 36.33 31.36 28.07 25.44 23.25 21.60 20.18 18.68 17.81 16.80 15.87 15.02 14.48 13.74 13.04 70 32.43 27.76 24.49 22.19 20.34 18.78 17.45 16.28 15.23 14.54 13.68 13.14 12.40 11.96 11.31 80 28.49 24.3S 21.52 19.36 17.64 16.45 15.21 14.13 13.41 12.53 11.99 11.50 10.81 10.42 9.81 90 24.61 21.C8 18.69 16.97 15.64 14.31 13.43 12.43 11.79 11.23 10.49 10.05 9.66 9.C6 8.74 100 21.47 18.4C 16.29 14.75 13.57 12.61 11.82 11.15 10.33 9.83 9.39 8.75 8.40 8.09 7.80 11C 18.47 16.15 14.31 12.96 11.92 11.07 10.37 9.77 9.26 8.81 8.42 7.82 7.51 7.22 6.97 120 15.95 14.C9 12.75 11.58 1C.66 9.91 9 29 8.76 8.30 7.90 7.55 6.98 6.7g0. 6. 22 00 Yield Stress = 60 ksi, Residual Stress = 16.8 ksi Values of k: L/r.02.04.06.08.10.12.14.16.18.2.22.24.26.28.30 1C 58.33 56.71 55.45 54.22 53.04 51.9C 5C.79 49.72 48.68 47.67 46.69 45.74 44.81 43.9C 43.32 2C 56.32 53.46 51.09 48.88 46.81 44.85 43.31 41.86 40.19 38.90 37.67 36.50 35.39 34.33 33.31 3C 53.58 49.81 46.47 43.76 41.31 39.07 37.C1 35.40 33.61 32.23 30.95 29.74 28.61 27.54 26.53 4C 50.30 45,6E 41.82 38.78 35.83 33.78 31 65 30.02 28.53 26.87 25.61 24.75 23.66 22.64 21.67 5C 46.35 41.CC 37.12 33.64 31.25 28.9C 27.12 25.53 24.11 22.81 21.63 20.53 19.81 18.86 18.26 6C 41.62 36. 1 32.12 29.28 26.96 25.C1 23.32 21.83 20.50 19.29 18.49 17.48 16.83 15.95 15.42 70 36.65 31 37 27.82 25.13 23.27 21.46 19.9C 18.82 17.60 16.79 15.77 15.14 14.27 13.75 13.28 8C 31.28 26.95 24.0C 21.84 19.87 18.52 17.4C 16.15 15.32 14.61 13.67 13.11 12.60 11.85 11.43 9C 26.70 23.CC 2C.71 18.83 17.36 16.17 1.5.18 14.04 13.32 12.68 12.12 11.33 10.88 10.48 10.11 10C 22.62 19.9C 18.00 16.38 15.1C 14.07 13.20 12.46 11.82 10.97 10.47 10.03 9.64 9.28 8.96 11C 19.51 17.11 15.57 14.18 13.08 12.47 11.72 10.77 10.21 9.72 9.28 8.89 8.54 8.22 7.94 120 16.81 14.92 13.68 12.49 11.54 10.76 10.1C 9.53 9.04 8.60 8.22 7.87 7.56 7.28 7.C2

TABLE IV (Concluded) Yield Stress = 70 ksi, Residual Stress 18.2 ksi Values of k: L/r.02.04.06.08.10.12.14.lb.18.2.22.24.26.28.30 IC 68.01 66.44 64.58 63.12 61.71 60.35 59.03 57.75 56.52 55.67 54.50 53.36 52.26 51.18 50.48 20 65.61 62.18 59.34 56.70 54.58 52.25 50.41 48.68 46.69 45.15 43.69 42.30 40.98 39.72 38.86 30 62.28 57.74 54.08 5C.48 47.55 45.24 42.79 40.88 39.11 37.12 35.60 34.17 33.18 31.92 30.72 40 58.23 52.62 47.98 44.36 41.22 38.78 36.28 34.34 32.59 30.99 29.51 28.14 27.21 26.01 24.88 50 52.87 46.74 42.12 38.38 35.57 32.83 30.74 28.89 27.58 26.07 24.69 23.41 22.58 21.46 20.77 60 46.72 40.52 36.24 32.96 3C.29 28.04 26.10 24.74 23.21 21.82 20.90 19.73 18.99 18.32 17.35 70 39.78 34.50 3C0.87 27.85 25.76 23.72 22.30 21.09 19.70 18.79 17.98 16.90 16.25 15.67 14.78 80 33.50 29.31 26.14 23.81 21.99 20.51 19.27 17.87 16.96 16.16 15.45 14.47 13.90 13.39 12.92 90 28.09 24.80C 22,.4_20.51 18.95 17.67 16.61 15.69 14.55 13o16_ 13.24 12.69 12.20 11.75 11.34 100 23.89 21.08 19.23 17.57 16.25 15.16 14.24 13.46 12.77 12.17 11.63 11.15 10.72 10.32 9.97 110 19.95 18. _5.36__15. 32 14.20 13.27 13.27 48 1 1.80 11..21 10_l__._68__10.22 9.80 9.42 9.07 8.76 120 17.26 15.64 14.17 13.36 12.41 11.61 10.94 10.35 9.83 9.37 8.96 8.59 8.26 7.95 7.68 Yield Stress = 100 ksi, Residual Stress = 20 ksi co Values of k: L/r.02.04.06.08.10.12.14.16.18. 2.22.24.26.28.30 IC 96.99 94.59 92.25 90.C5 87.9C 86.33 84.33 82.40 81.02 79.,20Q_77,44 76.22 74.55 73.43 71.85 20 93.25 88.47 84.59 80.53 77.26 74.22 71.40 68.76 66.28 63.94 61.73 59.63 58.14 56.23 54.42 3C0 87.86 81.15 75.42 7C.S.1 66.46 62_.93 59.7 _56.8_7 54.,22 51._78 50.01 47.89 45.90 44.52 42.75 40 80.21 71.7C 65.58 6C.62 55.95 52.35 49.69 46.86 44.31 42.48 40.34 38.85 37.C0 35.76 34.13 50 70C.01 61.55 55.2 5 C. 44 4697..43.60..40.67 3S.58 36.75. 3.4,6J.133 19 3140. 30.22 29.16 28.18 60 58.09 50.75 45.61 41.78 38.76 36.28 33.69 31.89 30.33 28.95 27.22 26.12 25.12 24.21 23.39 70 46.9 41.29 37.61 34 47 31,97 29,.... 28.16 2673Z5.352L491%_12i4._ 20.89 20.12 19.42 80 37.88 33.83 31.09 28.59 26.57 24.89 23.45 22.21 21.12 20.15 19.29 18.51 17.8C 17.16 16.57 90C 30.85 8.3 25.55 2 4. C9 225.24...4721.9Z_18,90 1. 14. 67 4_4.16 100 25.24 23.25 21.88 2C.32 19.02 17.91 16.95 16.10 15.34 14.67 14.06 13.50 13.0C 12.54 12.11 110 21.24 19.89 18.40 17.16 16_.1 15.70 14.91 14.20 13.57 13.00 12.49 12.0211.59 11.19 10.82 120 18.09 16.66 15.98 14.99 14.13 13.38 12.72 12.12 11.59 11.11 10.67 10.27 10.40 10.06 9.75

APPENDIX B FLOW DIAGRAMS AND RELATED MAD PROGRAMS FOR COMPUTER ANALYSIS OF VARIOUS BEAM-COLUMN PROBLEMS 1. Ultimate strength of unrestrained beam-columns of simplified section under uniform lateral load. (Used for the printout of Table IV, Appendix A.) 2. Ultimate strength of unrestrained beam-columns of simplified section in biplanar bending under uniform lateral load. 3. Ultimate strength of the restrained beam-column of the simplified section in planar bending under uniform lateral load. 83

FLOW D.IAGRAM NO.1- ULTIMATE STRENGTH OF UNRESTRAINED BEAM COLUMNS OF THE SIMPLIFIED SECTION N PLANAR BENDING UN DER UNIFORM LATERAL LOAD W-KP T =Trme....~, ~F = rase TAR) T~ | ~ A R)l READ L,C,N i Ir11 K, A,E,Q, L ~ --,P... F:r r >^- -T E7Bj =_p^~ =~_ -:3-I)'X BJ: L.. i.2L AF %L/r,~~~~182N4 > L/EA LC t N p^^ ITr T F 1 PIRIT N QA'K(L$2iX * ~~~~~Mx:'( (STARTI) 81.//~~~~~~~ LE ~ -....:AR

FLOW DIAGRAM NO. 1_ CONT. ALPHA.,L =. -J> N^\ N-~ AS C. NtIJI-J 2......., TT T "j PRINT PRINT ^I' ~8 = PINC I AI W.KP Ti I-T YJ /Yi~ l >\,__ ________ i S JJ /< RJ\ REPEA5 Rj Rj-l F

MAD PROGRAM NO. 1 $COMPILE MAD,EXECUTE,DUMP RPROGRAM FOR FINDING THE ULTIMATE STRENGTH OF UNRESTRAINED RBEAM COLUMNS IN PLANAR BENDING UNDER UNIFORM LATERAL LOAD, RFOR THE 4 POINT SECTION RTHIS PROGRAM REQUIRES AS A DATARL=LENGTH OF BEAM IN INCHES RC=HALF DEPTH OF SECTION IN INCHES RN-HALF NUMBER OF THE BEAM DIVISIONS RK-RATIO OF THE LATERAL LOAD TO THE AXIAL LOAD RA=AREA OF SECTION IN SQ.INCHES RE=MODULUS OF ELASTICITY IN KSI RSIGPL=PROPOTIONAL LIMIT SRESS IN KSI RSIGY=YIELD STRESSS IN KSI START READ DATA C,NA,ESIGY,L,SIGPLK PRINT COMMENT$1$ PRINT RESULTS L,CNK,A,EIGPLSIGPLGY Lll LAM=L/(2.*N) SR=L/C SIGE=286218.5276/(SR*SR) PE=A*SIGE H=SIGPL/SIGE G=-(H+1.+K*SR/8.) F=(1.-.03*SR*K/8.) R(-G-SQRT*(G*G-4**H*F))/(2**F) SIGA=R*SIGE PPL=A*SIGA I=A*C*C WPL=(5.*K*PPL*L.P.3)/(384.*E*I)*(1./(l.-R)) PRINT RESULTS SR,SIGEPPLgWPL INC=A*SIGY/200. P=PPL+INC W=K*P PRINT FORMAT TYTLE VECTOR VALUES TYTLE=$44H P/A DEFLECTION IN INCHES AT NODAL 1POINTS*$ REPEAT AFAC=1./(1*-P/PE) THROUGH INIT,FOR J=l,1,J.G.(N+1) X=(J-1)*LAM INIT Y(J)=(W*X*(LeP.3-2.*L*X*X+X.P.3)/(24*E*I*L))*AFAC PLAST C THROUGH ULTFOR J=2,1,.JeG(N+I) MX(:J)=W*L*(le-(N+1-J)*(N+I-J)*1I/(N*N*l.))/8e+P*Y(J) SIGU=P/A+MX(J)*C/I SIGB=P/A-MX(J)*C/I WHENEVER *ABS*SIGU.GE.SIGY.OR*.ABSeSIGB~GE.SIGY PRINT RESULTS PW PRINT FORMAT EXCEED VECTOR VALUES EXCEED=$11H SIGM:SIGY *$ TRANSFER TO START END OF CONDITIONAL WHENEVER *ABS.SIGU.G.SIGPL AA=SIGU/.ABSoSIGU SIGU=*ABS*SIGU EPSU=(1.*-SQRT(lI.-(SIGU-SIGPL)/(SIGY-SIGPL)))/(E/(2'*(SIGY1SIGPL)))+SIGPL/E EPSU=EPSU*AA OTHERWISE EPSU=SIGU/E END OF CONDITIONAL 86

WHENEVER *ABS.SIGB.G*SIGPL BB=SIGB/.ABS.SIGB SIGB=.ABS.SIGB EPSB=(1.-SQRT.(1.-(SIGB-SIGPL)/(SIGY-SIGPL)))/(E/(2.*(SIGY1SIGPL)) )+SIGPL/E EPSB=EPSB*BB OTHERWISE EPSB=SIGB/E END OF CONDITIONAL WHENEVER *ABS.(LSIGU(J) )G.SIGUEPSU=LEPSU(J)-(LSIGU(J)-SIGU) 1/E WHENEVER.ABS (LSIGB(J) ) *GSIGBEPSB=LEPSB(J)-(LSIGB(J)-SIGB) 1/E CSIGU(J )=SIGU CSIGB(J)=SIGB CEPSU(J)=EPSU CEPSB(J)=EPSB'JLT PHI (J)=(EPSU-EPSB)/(2.*C) PHI ( )=O.O PHI(N+2) PHI(N) THROUGH LOAPFOR J=21,J.G.*(N+1) LOAP CPHI(J)=(PHI(J-1)+lO**PHI(J)+PHI(J+1))*LAM/12 ASL(N+l)=CPHI(N+l)/2. THROUGH LOOP1,FOR J=N -1J.L. 2.OOP1 ASL(J)=ASL(J+1)+CPHI(J) Y(1)=0.O R( 1 ) =OO THROUGH LOOP2,FOR J=2,1,J.G.(N+1 ) TRY(J)=Y(J) Y(J)=Y(J-1)+ASL(J)*LAM R(J)=.ABS.(1.-(TRY(J)/Y(J))) WHENEVER R(J).L.R(J-l),R(J)=R(J-l) LOOP2 R=R(J) Ll=Ll+l WHENEVER R.G.(.005).AND.L1.L.200,TRANSFER TO PLAST C WHENEVER RoG.(e005).AND.Ll.G.200 PRINT RESULTS PW PRINT FORMAT NOCON VECTOR VALUES NOCON=$39H NO CONVERGENCE BEYOND THE LAST P GIV 1EN*$ END OF CONDITIONAL WHENEVER R.LE.(.005).ANDLlILE-200 SIG=P/A PRINT FORMAT RESULTSIG,Y(2)...Y(N+l) VECTOR VALUES RESULT=$F8.2,52,4F8.4*$ THROUGH REVERS,FOR J=2t1J.G.*(N+l) LSIGU(J)=CSIGU(J) LSIGB(J )CSIGB(J) LEPSU(J)=CEPSU(J) REVERS LEPSB(J)=CEPSB(J) P=P+INC W=K*P TRANSFER TO REPEAT END OF CONDITIONAL DIMENSION Y(50) MX(50),PHI(50) CPHI (50) ASL(50)TRY 1(50),R(50)1 DIMENSI.ON LSIGU(50),LSIGB(50),LEPSU(50),LEPSB(50) DIMENSION CSIGU(50),CSIGB(50),CEPSU(50)'CEPSB(50) INTEGER JN,NLI 87

TRANSFER TO START END OF PROGRAM $DATA L=900*C =15*,N=4,K=:04,A=40 E=29000. SIGPL=35SIGY=50* 88

FLOW D LARAM No.2? ULTIMATE STRENGTH OF UNRESTRAINED BEAM COLUMN5 OF TH. S. MPLI.FED ECTION IN NIBLANAR BENDINQ UNDER UNIFORM LATERAL LOADS Wy K P, _ _ P. ^,NtA L ASTC T =TRUE LU, j L L(TJ Lcr3J- L C LrJ'= ALFA LCiJ = LE2 L = L E4. O F =FALSE READ& PRIIT 7T J-J+ \J=2 | LC, K,Q A > N+i ~ rF T I:F IR IA R'- IMJ' WXL (I) (r I I7 IR-+ L/C P- t Ie= L 2J! J~ 1,,1 IINC +QA'c) TESTiL2,3 EST= K= PRR V, /+L = I SR = ~ I If ) Z1 =Z-^ ^ / \ fF IT _ T / F U T l FSp^ R) C(R)J 11t t13 (e) t (a P.! x= N='(Jl) C]PPINC'. x\ I 1x, =. P Pu=P-INC -- ______ =_L _ J.TEST 24EI/F PAS STAT r-IDOLL) i: (ALFA 89 >Ntl >/~~~~~~~~~~~ 8T

FLOW DIAGRAM NO.2. CONT T =Tr'ue False TEIJ - EJ CJE2J + EJJ 4 +Y., X X^Vis-dy, 0. |T ACuy - | 0x(NF')=.' T fl^,X G I T X.. X^,., CI I - IINC J^ 4....... I AS +S +CJ \('Ai?i \i E~ i TEVTI 4i~ ^~{/ T +I) A04 d

FLOW DIAGRAM NO.2_ CONT. (Cv^l~) T T=RUE.F =FALSE - X>N\ Li L TYJ Y,TXJ-XJ /\ =II-T/X' fI( - Ij t /R>,00\ R-A / PRINT PWW 1W3 1r IJ R = NT P, Iy,'j J R-R JN+I. + AT P =P+INTc RYJ<RW= RRTJY j L3J(J1: Li.T)j R.... T IJ1F Tr^ |LJ =lJLCJP= P 2JLC =IJ,, L RJrJLE, C..4...9 9:L

$COMPILE MAD,EXECUIE 002844 10/22/63 2 41 33.3 PM $PUNCH LI iRARY, PUNCH OBJCCTFULL DUMP MAD (21 GCT 1963 VERSICN) PROGRAM LISTING PROGRAM F(R FINDING THE ULTIMATE STRENGTH OF UNRESIRAINED BLAM COLUMN IN BIPLANAR BENDING UNDER UNIFORM LOAD,FOR THE 4 PCINT SECTION THIS PROoRAM REQUIRES AS A DATA L=LE GTH OF BEAM IN INCHES C= HALF UEPTH OF StCTION IN INCHLS N=HALF NUMBER OF BEAM DIVISIONS KY=RATIO OF VERTICAL LATERAL LOAD /AXIAL LOAD q=RATIC UF VERTICAL LATERAL LOADS/HORIZONTAL LATLRAL LOADS A=AR~A CF SECTION IN SQ.INCHES E=MODUJLUS OF ELASTICITY IN KSI SIGPL=PR(JPCTIONAL LIMIT STRESS IN KSI SIGY=YIELD STRESS IN KSI BEG1N READ DATA L,C,N,KY,QA,E,SIGY,SIGPL *001 PRINT COMMENT$1$ *002 PRINT RESULTS L,C,NKY,Q,A,E,SIGY,SIGPL *003 TEST=4 *004 THROUGH REVER1,FOR J=2,1,J.G.N+1 *005 CSIG1(J)=O.0 *006 CSIG2(J)=0.0 *001 CSIG3(J)=0.0 *008 CSIG4( J)=0.0 *009.0 CEPSI J)=0.0 *010 CEPS2 J )=0.0 *011 CEPS3(J)=O.0 *012 CEPS4(J)=0.0 *013 LSIG1(J)=O.O *014 LSIG2(J)=O.G C015 LSIG3(J)=0.0 *016 LSIG4(J)=0.0 *017 LEPSI(J )=0.0 *018 LEPS2(J)=0.O 019 LEPS3(J) =0. *020 REVERI LEPS4(J)=0 (. *021 INTEGER L3 *022 RR=( 1.-w) / (1.+Q) *023 KY=KY*(1.+Q)/ (S(SRT.(2.)) *024 KX= R.t* KY *025 PRiNI' RESULTS RR,KY,KX *026 D=C*SwRT.(2.) *027 LAM=L/(. *N) *028 SR=L*S(KT. (2.)/0 *029 I =A*I-C/2. *030 Z=1/0 *031 PE=A*SIOL *033 ALL=(SIuPoPL+-,IGE+PE*KY*L/(8.*Z)-SORT. ((SIGPL+SIGE+PE*KY*L/(8.* *034 1 Z)).''.2-4.*fSlhESIGPL*(1.-.U3*KY*L*A/(b.*Z))))/(2.*SIGE*(I.- *034 2.03*KY*L*A /(8.*Z))) *034 PPL=ALL*PL *035'k 1I RLSULTS -,LAM,SR, I,PE,PPL U36 I.C= oA*SIY/2G. *037

P=PPL+ l C *038 WY=KY*P *039 WX=KX*P *040 L=l1 *041 REPEAT AFAC=l./(l.-P/Pc) *042 THROUGH INITFOR J=1,l,J.G.(N+1) *043 X=(J-1)*LAM *044 X(J)=(WX*X*(L.P.3-2.*L*X*X+X.P.3)/(24.*E*I*L))*AFAC *045 INIT Y(J)=(WY*X*(L.P.3-2.*L*X*X+X.P.3)/(24.*L*I*L))*AFAC *046 ~PL'W`^S~TC''"" THROUGH ULTFOR J=2,1,J.G.(N+1) *047 - 11=1 *048 INTEGER II *049 TEST=l *050 L2=1 *051 L3=1 *052 MY(J)=WX*L*(l.-(N+l-J)*(N+1-J)*1./(N*N*1.))/8.+P*X(J) *053 MX(J)=WY*L*(1.-(N+I —J)*(N+ 1-J) *./( N*N*1.))/8.+P*Y(J) *054 PHIX(J)=MY(J)/(E1*) *055 PHIY(J)=MX(J)/(E*I) *056 SIGi=P/A+MX(J)/Z *057 WHENEVLR SIGI.LE.SIGPL *058 STG4=P/A-MX(J)/Z *059 SIG2=P/A+MY(J)/Z *060 SIG3=P/A-MY(J)/Z *061 EPS1=SIGI/L *062 EPS2=SIG2/E *063 EPS3=SIG3/E *064 P"S'"" 4=SIG4/E *065 TRANSFLR TO DOLL *066 END OF CONDITIONAL *067 -WHENEVER S.IGl.GE.SIGY *068 SIGI=LSIGI(J)*P/(P-INC) *069 %l WHENE VLR SIGI.G.SIGY,SIG1=SIGY *070 L2=1 *071 TEST=1 *072 TRANSFER TO REP *073 END OF CONDITIONAL *074 WHENEVER SIGI.G.SIGPL *075 O=(4.*P/A+8.*MX(J)/(A*D)-2.*SIGY+SIGPL)/3. *076 SIGI=(-4.*(SIGY-SIGPL)/9.+2.*O +SQRT. ((-4.*(SIGY-SIGPL)/9.+2. *077 I *0).P.2-4.*(O04-4.*SIGY*(S IGY-SIGPL)/9.)))/2. *077 SIG2=-SIG1+2.*P/A+({X(J)+MY(J))/(.5*A*D) *078 WHENEVER SJG2.GE.SIGY *079 PRINT FORMAT END *080 VLCTOR VALULS END=$20H SIGM = SIG2 = SIGY *$ *081'"THROUGH KAP,FOR J=2,1,J.G.N+1 *082 KAP PRINT FORMAT ABCD,LSIGL(J),LSIG2(J),LSIG3(J),LSIG4(J),LEPS1 *083 I (J),LE PS2(J),LEPS3(J),LEPS4(J) *083 TRANSFER TO START *084 END OF CONDITIONAL *085 WHENEVER SIG'2..G.SIGPL *086 S=P/A+(3.*MX(J)-MY(J))/(A*D) *087 TOP TSIGI=SIGI *088 1 (SIGY-TSIGI))*SQRT.(SIGY-SIGPL) *089 WHENIVER SIGl.GE.SIGY *090 SIGI=LSIGI(J)*P/P~-INC) *091 WH[NLVLR SIG1.G.SIGY,SIGI=SIGY *092 L2=1 *093 TEST= *094

__________TRANSFER TO REP *095 END OF CONDITIONAL.096 C_________ JL3=L3+1 *097 ij`WHENEVER'.ABS.(SIGL-TSIGi).G..005.AND.L3.G.200 098 TRANSFER TO DONE ___ 099 END OF CONDITIONAL "''100 ____________WHENEVER.ABS.{SIGl-TSIGI).G.0.005.AND.L3.LE.200'101 SI G1=(SIGI+TSIG ) /2 - *102 WHENEVER SIGI.GE.SIGY *103 ~ STIGI=LS -IG IJ ) *P / ( P-INC~) 104' WHENEVER SIGI.G.SIGYSIG1=SIGY *105 L2=1 *106 TEST=l *107 TRANSFER TO REP ---- - *108 END OF CONDITIONAL *109 TRANSFER TO TOP *110 END OF CONDITIONAL *111I END OF CONDITIONAL -112 REP SIG4=SIGI-4.*MX(J)/(A~D).113 SIG2=-SIG1+2.*P/A+(MX(J)+MVY(J))/(0.5*A*D) 114 WHENEVER SIG2.GE.SIGY *115 PRINT FORiAT-EN ~"-"* 116 ~____________VECTOR VALUES END=$20H SIGM = SIG2 = SIGY $ *117 THROUGH KBP,FOR J=2,1,J.G.N+1 *118 KBP PRINT FORMAT ABCD,LSIGI(J),LSIG2(J),LSIG3(J)tLSIGJ) J),LEPSI *119 I (J), LE P S2(J),LS3 tLE 3(J),EPS4J) - - *119 __________ TRANSFER TO START *120'f ND7fTDT —"JONDIfX0VNAV - *121 SIG3=-SIGI+2.*P/A+(MX(J)-MY(J))/(0.5*A*D) *122 EPS2=F (SIGY,SIGPLSIG2,ELEPS2(j),LSIG2(J)) *123 EPS3=F.(SIGY,SIGPLSIG3,E,LEPS3(J),LSIG3(J)) *124 EPS4=F.(SIGY,SIGPL,SIG4,E,LEPS4(J),LSIG4(J)) *125 _______ EPSI=F.(SIGYSSIGPL,SIGIE,LPSI(J),LSIG1(J)) *126 *127 __________ EPS 1=EPS2+EPS3-EPS4 *128 TRANSFER TO DOLL.129 ________ END OF CONDITIONAL *130 ~TE P~S1=E.PS - - *.. *131 _________ EPSI1=EPS2+EPS3-EPS4 *132 L-4z~r +.133 _ ___ _WHENEVER.ABS.((LPSI-TEPS1)*IOOU).G..OOO.AND.L2.L.800 *134 WHENEVER II.E.2,TRANSFER TOLOOP13 *135 ___________ OEPSL=EPSI *136 OTEPS1=TEPSI --- - -- 137 _____ EPSI=(OEPSI+OTEPSI )/2. *138 ~ STGI=E4EPS1-((E*EPS1-SIGPL)Y.P.2)/(4.*'(SIGY-SIGPL)) * 139 ____ ___ WHENEVER SIG1.L.LSIGI(J) *140 SIGI=LSIG1 (J)-E*(L-EPSI(J)-EPSI) *141 END OF CONDITIONAL *142 WHENEVER SIGI.GE.SIGY * 143 ____SIG1=SIGY *144 "TrEST=2 *145 _____ __ TRANSFER TO REP *146 END OF CONDITIONAL - 147 1=2 *148 TRANSFER TO REP *149 LOOP13 EPSII=EPSI. *150 EPS1=(OTEPSI-OEPSL*(EPTEP&1-OTEPS1I)/(EPSI-OEPSI))/(1.-(TEPSl- *151 I OTEPSI)/(EPSI-OEPSI)) *151 SIG1=E*EPS1-((E*EPSI-SIGPL).P.2)/(4.*(SIGY-SIGPL)) *152

WHLNLVER SI'G.L.LSIGI(J) *153 SI G1=LSIGI( J)-E*( LEPS( J) -EPS 1) 154 END OF CONDITIONAL * 155 WHENEVER SIG1.GE.SIGY *156 SIG1=SIGY *157 TEST=2 *158 TRANSFER rC REP *159 END t3F CONDITICNAL *160 WHENEVER.ABS.(EPS11-TEPS1).L..ABS.( EPSI-CjTEPS1 ) *16 OEPS1=EPSI' L *162 OTEPSI=TEPSI *163 END OF CONDITICNAL *164 TRANSFER TC REP *165 END UF CONDITIONAL *166 WHENEVER.ABS.((EPS1-TEPS1)*L000O).G..OO10.AND.L2.GE.800 *167 PRINT RLSULTS P,WX,WYJ,L2 *168 PRINT FURMAT ABC *169 VECTOR VALUES ABC=$-7H NO EQUILIBRU1M OF STRESSES POSSIBLE *$ *170 THROUGH KCPFOR J=2,1IJ.G.N+l *171 KCP PRINT FORMAT ABCI,LSIGi (J),LSIG2(J),LSIG3(J),LSIG4(J),LcPS1 *172 1 (J),LEPS2(J),LEPS J),PJ LEPS4(J) 172 TRANSFER TO START *173 END OF CONDITIONAL *174 DOLL CSIGI(J)=SIG1 *175 CSIG2(J)=SIG2 *176 CSIG3(J)=SIG3 *177 CSIG4(J)=SIG4 *178 CEPSI(J)=EPS1 *179 CEPS2(J)=EPS2 *180 \) CEPS-3(J)=EPS3 *181'1 CCCEPS4(J)=EPS4 *182 PHI.X(J)=(LPS2-EPS3)/(2.*D) *183 PHIY(J)=(EPSI-EPS4) /(2.*D) *184 ULT END OF CONDI ITIONAL * 185 PHIX( )=.O0 *186 PHIY(1)=O.0 *187 PHIXI(N+2)=PHIX('4) *188 PHI Y Y(N+2 ) =PHI Y ( ) *189 THROUGH LOAP,FOR J=2,1,J.oG.(N+1) *190 CPHIX(J)=(PHIX(J-1)+l PHI X()+ IJ )+PHIX(J+) )*LAM/12. *191 LOAP CPHIY(J)=(PHIY(J-l)+1O.*PHIY(J)+PHIY(J+1) )LAM/12. *192 ASLX(N+1)=CPHI. X(N+1)/2. *193 ASLY(\+1 )=CPHIY (N+1)/2. *194 THROUGH LOOP1,FOR J=N,-1,J.L.2 1 95 ASL J SL =ASLX(J+ 1 )+CPHIX( J) *196 LOOPl ASLY(J)=ASLY(J+ )+CPHIY(J) *197 Y( 1)=.O *198 X( 1)=0.0 *199 RX(1)=O.0 *200 RY(1)=O.0 *201 THROUGH LOOP2,FUR J=2,1,J.G.( N+1) *202 TRYY(J) =Y(J) *203 TRYX(J)=X(J) *204 Y(J) =Y(J-1)++SLY(J)LAM *205 X(J)=X(J-l)+ASLX( IJ)*LAN, *206 RX(J)=.AI3S. (.-(TRYX (J)/X(J))) *Z07 RYJ) =.ABS.( 1.-(TRYY(J) /Y (JJ ) *208 tHEiVL-E RX(J).L.RAX(J-1), RX(J)=RX(J-1) *209 WtIEiJLVER RY(J). L. RX( J,<Y( J ) =RX ( 21

LGOP2 R=RY(J) *212 LI=L1+1 *213 WHENEVER R.G.(.005).AND.Ll.L.200,TRANSFER TO PLASTC *214 WHENEVER R.G.(.005).AND.L1.GL.200,TRANSFER TO DONE *215 PRINT FORMAT RESULT,P,X(2)...X(N+1),Y(2)...Y(N+1) *216 VECTOR VALUES RESULT=$1F8.2,8F10.4*$ *217 P=P+INC *218 WX=KX*P *219 WY=KY*P *220 THROUGH REVERS,FOR J=2,1,J.G.N+l *221 LSIGI(J)=CSIGI(J) *222 LSIG2(J)=CSIG2(J) *223 LSIG3(J)=CSIG3(J) *224 LSIG4(J)=CSIG4(J) *225 LEPSI(J)=CEPS1(J) *226 LEPS2(J)=CEPS2(J) *227 LEPS3(J)=CEPS3(J) *228 REVERS LEPS4(J)=CEPS4(J) *229 Ll=l *230 TRANSFtR TO REPLAT *231 DONE PRINT RESULTS P,WX,WY *232 - PRINTT-R-FMAT CON *233 VECTOR VALUES CON=$16H NO CONVERGENCE *$ *234 THROUGH KDP,FOR J=2,1,J.G.N+l *235 KDP PRINT FORMAT ABCD,LSI-GI(J),LSIG2(J),LSIG3(J),LSIG4(J),LEPS1 *236 1 (J),LEPS2(J),LEPS3{J),LEPS4(J) *236 START TEST=4 *237 DTrMNSION X(100),MY(100),PHIX(100),CPHIX(100),ASLX(100),TRYX *238 1_ 100),RX(100),Y(100),MX({100),PHIYIO),CPHIY(100),ASLY(100), *238 \O 2 TRYY(100),RY(100) *238 0 ___________ VECTOR VALUES AbCD=$4Fl0.2,4E12.5*$ *239 DIMENSION LLPSl(50),LEPS2(50),LLPS3(50),LEPS4(50),LSIG1(50), *240 1 LSIG2(50),LSIG3(50),LSIG4(50) *240 DIMENSION CEPSI(50),CEPS2(50),CEPS3(50),CEPS4(50)tCSIGI(50), *241 1 CSIG2(50),CSIG3(50),CSIG4(50) *241 INTEGER Ll,L2,J,N *242 _ INTEGER TEST *243 TRANSFER TO BEGIN *244 END OF PROGRAM *245

$COMPILE MAD,PUNCH OBJECT 002844 10/22/63 2 41 47.9 PM MAD (21 OCT 1963 VERSION) PROGRAMLISTfNF^......... EXTERNAL FUNCTION (SIGYSIGPL,SIGE,LEPSLSIG) *001 ENTRY TO F. *002 WHENEVER SiG.L.$ST P~ — T EPS=SIG/E *004 OR WHENEVER.ABS.LSIG.G..ABS.SIG *005 AA=SIG/.ABS.SIG *006 EPS=(.ABS.LEPS-(.ABS.LSIG-.ABS.SIG)/E)*AA *007 OTHERWISE *008 AA=SIG/.AB-S"T'~..... -..wa..SIG=.ABS.SIG *010 EPS= 1.-SQR T (1.-(SIG-SIGPLJ/ (IGY-SIGPL})) )/{E / 2. * (SIGY- *01 1 SIGPL)))+SIGPL/E *011 EPS=EPA.. _. _._. _................................._.. * [ -'2 -- EPS=EPS*AA Z END OF CONDITICNAL *013 FUNCTIoTTNRETUR N S:" - 4-......END OF FUNCTION _ __ 015 -^)

L = 900.000000, C = 15.0C00000, N = 4, KY_ _.042426 Q =.333333, A = 4.0.OCCCOO, E = 2.900000E 04, SIGY = 50.000000 SIGPL = 35.000000 RR =.500000, KY =.040000, KX =.020000 0 = 21.213203, LAM = 112.500000, SR = 60.000000, I = 8999.999878 PE = 3180.205841, PPL = 879.743546 889.74.3480.6397.8323.8994.6960 1.2793 1.6646 1.7990 899.74.3534.6497.8453.9136.7069 1.2995 1.6909 1.8274 909 T.74.3'5'9.6598.8585.9278.7180 1.3200 1.7176 1.8563 919.74.3645.6700.8718.9421.7294 1.3408 1.7449 1.8859 929.74.3700.6803.8852.9566.7409 1.3621 1.7727 1.9161 939.74.3757.6906.8987.9712.7527 1.3839 1.8012 1.9470 949.74.3814-.7011.9123.9860.7646 1.4060 1.8303 1.9785 959.74.3871.7117.9261 1.0009.7778 1.4305 1.8625 2.0136 969.17 4'.3929.7223.9400 1.0159.7906 1.4543 1.8938 2.0476 979.74.3987.7331.9540 1.0311.8039 1.4788 1.9262 2.0827 989.74.4046.7439.9682 1.0464.8175 1.5042 1.9596 2.1190 999.74.4106.7549.9824 1.0618.8316 1.5304 1.9942 2.1566 1009.74.4166.7659.9968 1.0774.8462 1.5576 2.0300 2.1955 1019.74.4226.7771 1.0114 1.0931.8630 1.5888 2.0712 2.2403 1029. 74.4287.7884 1.0260 1.1090.8790 1.6187 2.1107 2.2832 Iu39.74.4349.7997 1.0409 1.1250.8956 1.6498 2.1518 2.3279 IU09.74 4411.8112 1.0558 1.1412.9130 1.6822 2.1947 2.3746 1059.74.4474.8228 1.0710 1.1576.9311 1.7161 2.2395 2.4233 1069.74.4539.8347 1.0864 1.1743.9500 1.7516 2.2865 2.4746 \0 1079.74.4605.8469 1.1024 1.1917.9720 1.7928 2.3412 2.5340 Co 1089.74.4674.8596 1.1190 1.2097.9934 1.8331 2.3946 2.5922 1099.74.4745.8728 1.1364 1.2285 1.0160 1.8755 2.4510 2.6537 1109.74.4819.8865 1.1544 1.2481 1.0397 1.9202 2.5104 2.7185 1119.74.4896.9008 1.1733 1.2687 1.0654 1.9687 2.5749 2.7889 1129.74.4978.9161 1.1936 1.2907 1.0958 2.0259 2.6512 2.8723 1139.74.5065.9323 1.2150 1.3141 1.1267 2.0843 2.7291 2.9572 1149.74.5158.9497 1.2381 1.3392 1.1606 2.1483 2.8146 3.0509 1159.74.5259.9685 1.2631 1.3665 1.1980 2.2190 2.9094 3.1546 1169.74.5372.9898 1.2914 1.3974 1.2451 2.3083 3.0296 3.2866 1179.74.5500 1.0139 1.3237 1.4329 1.3013 2.4151 3.1737 3.4452 1189.74.5658 1.0437 1.3640 1.4773 1.3774 2.5602 3.3710 3.6636 P= 1199.743546, WX = 23.994652, WY = 47.989268, J = 5 L2 = 800 NO EQUILIBRUIM OF STRESSES POSSIBLE 39.03 34.17 25.48 20.30.13569E-02.11782E-02.87851E-03.69984E-03 45.61 38.15 e 2.83 12.38.16622E-02.13218E-02.78734E-03.42692E-03 49.08 41.39 21.91 6.59.19858E-02.14575E-02.755636-03.22725E-03 49.87 42.92 22.02 4.17.21460E-02.153046-02.75928E-03.14372E-03

FLOW DIAGRAM NO. _ ULTIMATE STREt~TH OF TH E RESTRAINED BEPA COLUMN OF THE SIMPLIFIED SECTION IN PLANAR BENDINq UNDER UNIFORM LATERAL LOAD W= KP, TRUE F=FALSE STA I /=Jt I jTg READ% F T FJl f PRINT T 1 ~ L, C, N,' K A,) MABA T.L I IT.X1 P.MMAXC ( y | AK~J"l~ M I A I /J>N+i ITA MAX LE,, jL NC LI = I~P+I \^ I EAST ) __________Lt f L3, y t... PL'AST F,..9 "AL/,N \ / M\^.=L ( (wU) UT) 1:AC F= LJ _ C.. AP _l I^ 8 E' I I.l'J /NiF (S R) T C' /A4 M>. /Z,e~~F~~~ ^, J =(Mxj+[t=!~ ~,= ____ 99

FLOW DIAGRAM NO.3. CONT. T' TRUE _ F=FALSE i! c A: C m. A I T T E: T f t I | (F) Jnt4J 4 ~Bj=~J, N P/P-INc /B B 6! A C || I -P P +1 INC I IAS/\ 10J T JM A.~ =LC uJ' — J r.^- "P ~I~ I9 T J - < -= J TYR J FINT 6 /J=4 J. J+ AJ TzyP/P-iFNC TAR - " — " F TF-'/>LOB\ ~Ii=== 1 =60 EA ALEB E T~ f Fr,, R=9,T > J \ I 4|evJU.TXJ LA C1|aj0 (L b IafE I I IJEe K 100

$COMPILE MAD,EXECUTE,FuLL DUMP,PRINT OBJECT 004986 11/19/63 8 51 26.6 PM $PUNCH LIBRARY,PUNCH OBJECTFULL DUMP MAD (23 OCT 1963 VERSION) PRUGRAM LISTING. PROGRAM FOR FINDING THE ULTIMATE STRENGTH OF RESTRAINED BEAM COLUMN UF THE FOUR POINT SECTION IN PLANAR BENDING. THIS PROGRAM REQUIRES AS A DATA - L=LENGTH OF BEAM IN INCHES C=HALF DEPTH OF SECTION IN INCHES N=HALF NUMBER OF BEAM DIVISIONS K=RATIO OF THE LATERAL LOAD /AXIAL LOAD A=AREA OF SECTION IN SQ.iNCHES E=MODULUS OF ELASTICITY IN KSI BETA=STIFFNESS OF ALL MEMBERS AT END JOINT EXCEPT THE BEAM IN KIPS INCH /RADIAN SIGPL=PROPORTIONAL LIMIT STRESS IN KSI SIGY=YIELD STRESS IN KSI BEGIN READ DATA L,C,NK,AtEBETASIGPL,SIGY *001 PRINT COMMENT $1$ *002 PRINT RESULIS L,C,N,K,A,E,BETASIGPLSIGY *003 TEST=1 *004 I=A*C*C *005 THROUGH REVLR1>FOR J=1,1 J.G.(N+1) *006 CS IGU(J) T=U. -*007 CSIGB(J)=O. ______*008'H' CEPSU(J)=O.0 *009 0 CEPSB(J)=O.UO _ _ -*010 LSIGU(J)=O.U *'0 1LSIGB(J)=0.0 *012 LEPSU(J)=O.0 *013 REVERI LEPSB(J)=O.U *014 II=1 *015 Ll=1 *016 L3=1 *017 INTEGER L3 *018 LAM=L/(2.*N) * 019 SR=L/C *020 SIGE=286218.5276/(SR*SR) *021 H=SIGPL/SIGE *022 G=-(H+1.+K*SR/8.) *023 F=( 1.-.03*Sk*K/8.) *024 R= (-G-SQRT."'~" G-4. *H*F ))/(2. *'F). 025 PPL=A*R*SIGE *026 INC=A*SIGY/200. -027 P=PPL+INC ____ __028 W=K*P *029 ELAST C ALFA=SQRT. (*L*L/(E*I)) *030 F=ALFA/2. -- *-3 i F2=ALFA *032 MFAB=-WL*(I.-2.* S IN. (F)/(COS.(F)*ALFA))/(2.*ALFA*SIN.(F)/ *033 1 COS.(F)) *033 S 1=( 1.-ALFA*COS. (F2 )'/SIN. ( F2))/ (2.* SI N. (F)/.(COS.(F)*ALFA)-I.)'034 S2=SI*(ALFA/SIN.(F2)-1.)/(1.-ALFA*COS.(F2)/SIN.(F2)) *035 T = E * I / L............................................... CITA=MFAB/((S1-S2)*T+BETA) _ *037 MAB=-CITA*BLTA *038

M=MAB+W*L/(ALFA*ALFA) *039 B=(-M*COS..(F2)+MAB+-L/(ALFA*ALFALFA))/SIN.(F2) *040 MS=W*L/8.b+ MAB *041 0=F 2/ 2 *042 MMAX=MCOS.(O)+B*SIN. (O)-W*L/(ALFA*ALFA) *043 WHENEVER A6S.MAB.G. MMAX,MMAX=-MAB *044 SIGX=MMAX*C/ I+P/A *045 WHENEVER SuIX.L.SIGPL *046 P=P+INC *047 W=K*P *048 TRANSFER TO ELAST C *049 OTHERWISE *050 TRANSFER TO PLAST.051 END OF CONDI IONAL *052' PLAST THROUGH MOMLNT,FOR J=l,1,J.G.(N+I) *053 MOMNENT "" MOX(JJ)=WL*J(' (N+1-J)*(N+1-J)*1./(N*N*I.))/8. *054 THROUGH LOOP,FOR J=1,1,J.G.(N+1) *055 MS=MOX(J )+MAB *056 O=(J-I)*F2/(2.*N) *057 LOOP Y(J)=(M*COS.(0)+B*SIN.(O')-MS-W*L/(ALFA*ALFA))/P - *058 MAB=-MAB *059 PLAST C THROUGH ULTFOR=iI-',;J.G. (N+ *060 _____ MX'(J)=W*L*(i.-(N+I-J)*N+-J)*1./(N*N*1.) )/8.+P*Y(J)-MAB _ *061 SIGU=P/A+MX(J)*C/I *062 SIGB=P/A-MX(J)*C/I *063 WHENEVER.A SSI GUGE.SIGY. AND. (P/A+MA*C/i)'GES I GY "... *064 PRINT RESULTS P,W *065 PRINT FORMAI EXCEED *066 VECTOR VALUES EXCEED =$18H SIGY =SIGM *$ *067 HJ THROUGH KAP,FOR J=I,l,J.G.N+ l *068 0 KAP PRINT FORMAT ABCD,LSIGU(J),LSIGB(J),LEPSU(J),LEPSB(J) *069 VECTOR VALUES ABCD=$2F10 2,2E12.5*$ *070 TRANSFER TO START *071 END OF CONDITIONAL *072 WHENEVER.AbS.SIGU.G.SIGY *073 SIGU=SIGY *074 SIGB=2.*P/A-SIGY *075 MX(J)=(A*SIGY-P)*C *076 Y(J)=(MX(J)+MAB-W*L*( I.-(N+I-J-J)*(N+ -J)*1./(N*N*I.))/8.)/P *077 END OF CONDITbONAL *078 WHENEVER.AbS.SIGB.G.SIGY *079 SIGB=SIGY *080 S________ IGU=2.*P/A-SIGB *081 END OF CONDiTIONAL *082 WHENEVER.AbS.SIGU.G.SIGPL *083 AA=S IGU/.AB. SIGU *084 SIGU=.ABS.SIGU *085 EPSU=( -.-SQKT. ( 1.-(SiGU-SIGPL)/(SIGY-SIGPL )/ (E/(2'* (SIGY- *086 1 SIGPL)))+SI'PL/E *086 EPSU=EPSU*AAF'A *087 OTHERWISE *088........... -"co"-.-...... * o 089 END OF CONDiTIONAL *090 WHENEVER.AbS.SIGB.G.SIGPL " *091 BB=SIGR/.ABe. SIGB *092 SIG3= ABS.'>IGB *093 EPSB=(.-SQT. (.-SIGB-SIGPL ) / ( SIGY-SIGPL) ) )/(E/(2.*(SIGY- *094' I'GPL'))) SIUPL' " *094 EPSG=EPSB*80 *095 OTHERWISE. *096

EPSB-=SIGB/E *097 END OF CONDITIONAL *098 WHENEVER.ABS. (LSIGU(J)).G.SIGU,EPSU=LEPSU(J)-(LSIGU(J)-SIGU) *099 I /E *099 WHENEVER.AtS. (LSIGB(J) ).G. SIGB,EPSB=LEPSB(J)-(LSIGB(J)-SIGB) *100 1 /E *100 CSIGU(J)=SIGU _.*101 CSIGR(J)=SIGB *102 CEPSU(J)=EP.U - 103 CEPSB J) =EP.B -104 ULT PHI (J)=(EPSU-EPSB)/(2.*C) _*105 U L T _ _ P H.= ___'_ _ _ _................_.............. -...0 PHI (N+2)=PHi (N) *106 THROUGH LOAP,FOR J=2,1,J.G.(N+1) _ _107 LOAP CPHI (J)=(PHI(J-1)+10.*HPHI(J+PHI (J+1) )*LAM/12. 108 ASL(N+1)=CPH I (N+1)/2. *_109 THROUGH LOOP1,FOR J=N,-1,J.L.2 *110 LOOPI ASL(J)=ASL (J+1 ) +CPHI(J)........... *1 Y( )=0.0 *112 R(1)=0.0 *113 THROUGH LOOP2,FOR J=2,1,J.G. (N+1 ) 114 TRY(J)=Y(J) *115 Y( J)=Y(J-1) +ASL(J) *LAM *.116 R(J)=.ABS.( 1.-.TRY(J)/Y(J) ) * 117 WHENEVER R(J).L.R(J-1),R(J)=R(J-1) * 118 LOOP2 R=R(J).*119 LI=L 1+1 *120 WHENEVER R._.(.005).AND.LI.L.200,TRANSFER TO PLAST C *121 WHENEVER RG.(.I005).AND.Li.GE.200 tTRANSFER TO LOOP5 *122 WHENEVER TEsT.E.2,TRANSFER TO ENDY *123 CITA=(96.*Y(2)-72.*Y(3)+32.*Y(4)-6.*Y(5)) /(24.*LAM).*124 - 1L3L3+1 *125 0 WHENEVER. AS. (.-MA8/(CITA*BETA)).G.005.AND.L3.G.200.... *126 W>^~~ __ _ PRINT RESULIS PW *127'PRINT FORMAT ENDMT *128 VECTOR VALUES ENDMT=$36H END CONDITIONS CANNOT BE SATISFIED * *129 1 $ *129 THROUGH KUP,FOR J=,1,1J.G.N+1 *130 KUP PRINT FORMAT ABCD,LSIGU(J),LSIGB(J),LEPSU(J),LEPSB(J) *131 TRANSFER TO START *132 END OF CONDiTIONAL *133 WHENEVER.AuS.(1.-MAB/(CITA*BETA)).LE..005 *134 ENDY PRINT FORMAT TWO,P,W,Y(2)...YIN+1),MX(1) *135 VECTOR VALUES TWO=$2Fl0.2, 4F8.4,FlO.2*$ *136 P=P+INC *137 MAB=MAB*P/(P-INC) *138 WHENEVER (P/A+MAB*C/I).GE.SIGY *139 TEST=2 __140 MAB= A*SIGY-P)*C *141 END OF CONDITIONAL *142 THROUGH LOOP4,FOR J=2,1,J.G.(N+1) *143 LOOP4 Y(J)=Y(J)*P/(P-INC) *144 I I=1 *145 Ll=l *146 W=K*P *147 1HROUGH RCVi-RS,FOR J=1,1,J.G.(N+1) *'148 LSIGU(J )CSiGU(J) *149 LSIGB(J)=CSiGB(J) *150 L-EPSU'( J )=CE PSU(J ) *151 REVEKS LEPSB(J)=CE'SB(J) *152 1 RANSFER TO PLAST C *153

END OF CONDITIONAL *154 IHENEVER' II.E.2,TRANSFER TO LOOP3 *155 WHENEVER.ALS. (1.-MAB/(CITA*SETA)).G..005 *156 OMAB-MA=Ab *157 OBC=BETA *CirA *158 MAB= (MA[:+BObC)/2. *159._.I..-11=2....._.! *160 TRANSFER TO PLAST C *161 -END OF CONDITIONAL *162 LOOP3 M BI=MAB *163 MAB=(OBC-OMAB*(CITA*BETA-OBC)/(MAB-OMAB) )/1.-{CITA*BETA-OBC) *164 I /(MAB-OMAB)) *164 wHENEVER.AB. (MAB1-BEAITITA).L..ABS. (OMAB-OBC) *165 OMAB=MABI *166 OBC=BETA*C IIA *167 "END OF CONDiTIONAL *168 TRANSFER TO PLAST C *169 LOOP5 PRINT RESULIS P,W *170 PRINT FORMAI CON *171 VECTOR VALULS CON=$16H NO CONVERGENCE *$ *172 THROUGH KOP,FOR J=l,I,J.G.N+1 *173 KOP' PRINT FO'ARMA' ABCD,LSIGU(J),LSIGB(J),LEPSU(J),LEPSB(J) *174 START TEST=1 *175 DIMENSION MOX(50),Y(50),MX(50),PHI(50),CPHI(50),ASL(50),TRY *176 1 (50),(50) *176 DIMENSION L-IGU(50),LSIGR(50),LEPSU(50),LEPSB(50) *177 DIMENSION C.IGU(50),CSIGB(50),CEPSU(50),CEPSB(50) *178 INTEGER J,NJ,Ll,II,TEST *179 _ TRANSFER TO iEGIN *180 END OF PROGKAM *181 0 4r

L = 900.00U000, C = 15000000, = 4 K = 040000 A = 40.00u000, E = 2.900000E 04, BETA = 4.350000E 06, SIGPL = 35..0000 SIGY = 50.00000 1174.57 46.98.1526.3338.5124.5718 -3444.36 1184.57 47.38.1536.3564.5164.5762 -3476.35 1194.57 47.178.1552.3600.5216.5820 -3509.30 1204.57 4R.18.1567.3635.5267.5877 -3541.36 1214.57 48.58.1581.3668.5316.5932 -3574.62 1224.57 48.98.1597.3705.5368.5990 -3606.66 1234.57 49.38.1o11.3738.5417.6045 -3640.08 1244.57 49.78.1626.3774.5469.6103 -3672.98 1254.57 50.18.1641.3809.5521.6161 -3706.10 1264.57 50.58.1656.3845.5574.6221 -3739.55 1274.57 50.98.1672.3882.5627.6281 -3773.40 1284.57 51.38.1687.3918.5682.6343 -3807.65 1294.57 51.78.1703.3956.5738.6406 -3842.33 1304.57 52.13.1719.3995.5795.6471 -3877.25 -T1314.57 52.58.1735.4033.5853.6536 -3912.77 1324.57 52.98.1735.4080.5922.6614 -3948.74 1334.57 53.38.1771.4120.5983.6682 -3985.31 1344.57 53./8.1789.4163.6048.6756 -4022.05 1354.57 54.18.18U7.4208.6115.6832 -4059.18 1364.57 54.58.1826.4255.6185.6912 -4096.70 1374. 57 54.98''184'2.4296.6249.6984 -4135.57 1384.57 55.38.1863.4348.6327.7012 -4173.97 1394.57 55.78.1834.4400.6405.7161 -4212.88 1404.57 56.18.1910 4463.6499.7266 -4252.25 1414.57 56.58.1933.4520.6534.7362 -4292.23 1424.57 56.98.1956.4580.6674.7464 -4332.62 0 1.-4'34.57 57'.38'~ 1981.4642.6768.7570 -43713.56 1F4444.57 57.78.2007.4708.6867.7682 -4414.92 1454.57 58.18.2034.4775.6968.7796 -4457.00 1464.57 58.58.2067.4856.7087.7930 -4499.75 1474.57 58.98.2095.4928.7197.8054 -4543.44 1484.57 59.38.2125.5007.7315.8187 -4587.61 1494'.57 ""59.78"...2157.5087.7436.8325 -4632.26 1504.57 60.18.2189.5170.7562.8466 -4677.63 1514.57 60.58.2223.5259.7696.8618 -4725.01 1524.57 60.98.2258.5349.7833.8774 -4772.25 1534.57 61.38-.2293.5442.7974.8934 -4820.30 1544.57 61.78.2333.5545.8129.9110 -4870.45 2 54 61.546.8285.9287 -4920.65 1564.57 62.58.2411.5752.8447.9470 -4971.82 1574.57T 62.98.2461.5878.8637.9686 -5024.06 1584.57 63.38.2503.5992.8814.9888 -5078.55 1594.57 63.78.2553.6124.9016 1.0117 -5133.66 1604.57 64.18.2599.6252.9215 1.0345 -5190.58 1614.57 64.58.2656.6402.9446 1.0608 -5248.27 1624.5'7 64.98.2709.6352.9679 1.0875 -5307.25 1634.5' " 65.38.2732.6747.9978 1.1216 -5366.01 1644.57 65.78.3963.9048 1.3082 1.4630 -5331.47 P = 1654.56b802, W = 66.182751 NO CONVERGENCE 32.23 50.00.L1113E-02.22411E-02 38.71 43.52.134400-02.15644E-02

43.95 38.28.15844E-02.13294E-02 47.37 34.86.18078E-02.12024E-02 48.56 33.67.19208E-02.11610E-02 0 0'\

UNIVERSITY OF MICHIGAN 3 901111 1 1115 02841 205III I 3 9015 02841 2057

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Production Notes Form Univ. of Michigan Preservation/MOA 4 Project MOA 4 ID#: U/___ C&GS Shipment#: i __ Call #: Date Collated:./ l Collated by:_ Total# of Pages: ____ Illustrations: K Yes No - Foldouts/Maps: Yes X No Bookplates/Endsheets: Yes I No Missing Pages: Yes X No Irregular Pagination: Yes X No Other Production Notes: _Yes X No

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