)1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The...

183
I I. A 110. f.y SIS .' J he Jnl?d I r/lre.e - ,mo.me;z1 , /.A:J uSed solve' -t-b )1-1 L · , Y .+ 1+1 5 lA E I . t'+1 ;//+, +- 1+1 . =-0 {l{J Wh evr a struc.fur,e ·fYOJ?1 'oJ?e anoTher'" will' <?J-f' ClPfcl de-foY h'1a "the5'e" e hCl HIe$' J(}/I/ ,tttrvt e&lu5'(? u-arr:::r- or' moments Jt{M" Cl/Xt'al forCe? S; df - . -rAe mefl-?.6er t I; w; If k e, d F, .(!} f .f2.&lcA me"" be y. d c< e to -r'h e fDYCe of S' .. $"t-t e-I. e::t (/'T:t I u. e wiLP e>t:!cq.y w6en 4 If//ed lead of f-here 'l'tOlTt:nS' are

Transcript of )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The...

Page 1: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

I

I. 7heDre1t~a/ A 110. f.y SIS .'

Jhe Jnl?d I f/~d r/lre.e - ,mo.me;z1,

/.A:J uSed ~ solve' -t-b /),1)~!.e.h-7.

)1-1

L · , Y.+ 1+1 5 lA

E I . t'+1 ;//+, +-1+1 .

=-0 {l{J

Wh evr a struc.fur,e faSSt?~ ·fYOJ?1 'oJ?e ekf~r,mq1;;'" eon~i,-<-raf-~

~fo anoTher'" rhe~e will' i~.e 'et~7~ <?J-f' Yeac~c'ns-' ClPfcl

de-foY h'1a t(~ns·.,.. "the5'e" e hCl HIe$' J(}/I/ ~ ,tttrvt ~ e&lu5'(? u-arr:::r­

1~S' or' moments Jt{M" Cl/Xt'al forCe? S; df - . ~hcl -rAe mefl-?.6er

'r()tq1i~~~? ,c.o~;elue~ t I; ~ ~ w; If k

e, ~n d F, .(!} f .f2.&lcA me"" be y. d c< e to -r'he

fDYCe

of S'..

rnst¢b,'/,'~

$"t-te-I. e::t (/'T:tIu. e

wiLP e>t:!cq.y w6en 4 If//ed lead j,q~ V~etc/'ed.

~f6c"t/' of f-here 'l'tOlTt:nS' are S"aT;~f,'e.c1

Page 2: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

pw

I

sQfoYe bl.l,ckl"Vll

VL=V~::= p!

HL = 10~ = - f-I ~If )

I L .LT C~) /vi/... + IiI S:LH~ + 'F2... = 0

F b b

c, -= (~I/ft)~ C(- til ~ eJ ~I f; )

C ~ (k~"-)'( 1- R.k..,.t k.L}

~ I (k~L .)'2. =- C~LP' ~ k.L - I

FL::= IN L (-'- I-~k~l _ -J- )H k2.L S'~k'~L ~

W~eYR

k. ~ J:~1<. =1 E~,

('3 )

Page 3: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

c h~a' I. C. f h. D

Te c hno {{)0 J I / 9 S- 7..

.e~ ~CL t ~ (:1. J. ~ kl2£ "t-he 1-6 YI?1

~ (1- "'. h"-" f U~) + k, h-k.. I<,;:L - ~L ( k~l T_ ~L - +-) ~ 0 - . <41

.. I / d I wL cIr-6Y tYIQ eqS-e IVI+h I'"oof ~4. W Oft';" 1-hS2rl P ~ 2. "a;-r Y.C4/

t ke5'" the f-4J'm

Ne.ct. , '/ ; rk e?u; 1/ brt 'Unt 0 f' mem /;(;0-5 w),en.,.k fro 171 eoeC?1S'r de r tJ

s-ttfe ~'.f" un·sfa/;/e ~U/!/b"Y/~#'l", -rk hi/owl? O-!CLQ.t~ s

1/ ' -, <lIf:.A YL = -:J.f T

II - 4RA V« = 2-f L

L1I1L == -AI-l h + f ..&.~

A 0 ~ == - AH h - fA/<

("I)

-f

= - 2.. -L

CJ v'- = 0 .;JH

= p

- -r

Page 4: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

TO {)6fa(~ e J

Qfter b~ckl(7 '"

w:-fl be llS"e d .

r~ re fer i/l7?t$")/{/' be7we € n /-/ eo n d p, I ~ nUI? c:I,~ Ie:; ~

-rhR "'fhy(~.R- mt:)I'HR.tf l2/cL_fc;"'" 707 71~?lrra.1 e!,q{~/,rbr':~

Fin'" n7e~ .6 e r s ,At3 &7 n cI B 1/ ~ 7 otR • r" I 6.ec-<!l n< e .so

("'i' J '

S:' '/''''''1 a, J ..,.~ .eIQat~ ftJr n?em~e;-s

;;b (S~ 4 Mt. + ~ s: HL + Gd Nt< -f.dG M~) +- E-t (CJ .dH/? -I- 4"G MI()

Can 6eeach

=-0+ AI( +.4 h.h

<I f(,. A 1-1 4 re Tk

et(f res-Sed II-? of tAer~

Af =:: 41f + :~ 4.1-/i

The lI'e:ly,'c:et/lS>1 c,r ~ -f"y h?ehlter A/3 IS c:/er;vecl as folllJw~

sl'm,'lar

/- l, ..d H

..d'~" - 5'4 / .d' H

.d (3 = -<::{ CI

(3)

Page 5: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

[h . L ( h I.2.H h I- c1 + -- c~-<;;;) + - C, -- - -=---] P Ll'R

Ere ET b ~ Ere L Ph

--. (/4)

5'1'V1 <:e L1 R ::f:. t> ~ftr?Y hu:cb/.·V11", ..,h(? ~fY{?SS(~'" In

Fe'r'o

11 . . L h c I .2.Hh I-- C, + -:-. (~-S'~) +-- L --- - CJEI I P hFTc FIb - c

F'rct')1 (3)

(C;L -Sz ~I

(L- ~~L~t~:a.L )- -

(~:a.LY2.

- - fl')

(

C, = dC', -AA" d f>, (- d ~I dr I

=(k. h)L

Si\'\ :l. (I<t l h) - ~ J;.

·r' T r I t.....I-..L I·~t-.O €.:r. (7) . . the'5t.~b,· f,'TJt· cond,'T/6->?t1eS'e<2[Q.~'crnS' :,,·ubJ'r{u.(~S' ~. () , ... (j

j.;q {'( ~Q IJ 0 6 fa (~ Q d (~ -t-hR +c y m ..

/tl,.th :l-

If, h~i Jr, h - ~~ (2 _ k,L eJ k~L )+ ~~ ("- -i<,hd Jr, h - , .... 'k,h )=D

-. - (If)

Page 6: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

Ii, 1

~ .:~ ."

:~ ..j'.

"

f

1 '

"

...

;.

"""

",-

2. ·tifp/a-<~ ~L. 7 !J~/e;

15o~r!./ (3,,A,

/v~~ ~-dr r ~ 'eJ~ 7

~·.~.I~ ~.

;T. crt ,A~~~ S~,r~~'

tJJ, /tp. I. 3 9 7 ((9~'9)

Page 7: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

STABILITY OF· "ELASTIC AND .PARTIALLY PLASTIC FRAMES

by

Le-Wu Lu

FRITZ EN~'NEERING

LABORATORY UBRARY;

A DISSERTATION

Presented to-the.Gradtia:te Faculty·of·- LEHiigh. Uriiversi ty

in candidac·y .. for-the ·-De gree of~ctor of Philosophy

Lehigh University1960

1

Page 8: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

ii

.Approved and recommendeq. for acceptance as a disser-

tation in partial fulfillment of the requirements for the

degree of Doctor of Philosophy.

(Date)

Accepted,

; i

George C.Driscoll, Jr •.~ofessor in Charge

(Date) ,

Special committee directing the doctoralwork of Mr. Le-Wu Lu

Professor William J. Ehey, Chairman

Professor Lynn S. Beedle

Profes,sor Voris V. Latshaw

Professor George Co Driscoll, Jr.

Page 9: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

iii

ACKNOWLEDGMENTS

The author is greatly indebted to Dr. George C.

Driscoll, Jr., professor in charge of the dissertation,

for his encouragement, advice, and valuable suggestion~

during the preparation of this dissertation. The guidance

of Professors William J. Ehey, LYnn S. Beedle, and Voris V.

Latshaw, chairman and members respectively of the Special

Committee directing the authorVs doctoral work, is grate­

fully acknowledged.

The work described in this dissertation is part of a

project on "Welded Continuous Frames and Their Components"

being carried out under the direction of Professor Lynn S.

Beedle. The project is sponsored jointly by the Welding

Research Council and the Uo So Navy Department under an

agreement with the Institute of Research of Lehigh Universityo

Funds are supplied by the American Ins.ti tute of Steel Con­

struction, Office of Naval Research, Bureau of Ships, and

the Bureau of Yards and Docks. The Column Research Council

of the Engineering Foundation acts in an advisory capacity.

The w.ork was done at Fritz Engineering Laboratory, of which

Professor William Jo Ehey is Directoro

The author wishes to express 'his appreciation to

Mr. T. R. Higgins, Director of Engineering and Research, .

American Institute of Steel Construction, for his oontinued

interest in the progress of the research work reported herein.

Page 10: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

,

iv

Sincere gra~itUd~ _is..~x.~ended ~~ all the author's

associates at Fritz LaboratorYJI especially to Mro Morris

Ojalvo for fruitful discussions and for the use of certain

beam-column nomographs which he developedo

The help of Messrs 0 '"Donald Ao "Recchio and Yu~Chin Yen

throughout the experimental investigation is gratefully. . " ...- .~ ..- - . -- - .

appreciated Jl and acknowl~dgment is also due to Mro Kenneth Ro

Harpel, foreman of the LaboratorYJI for preparing the test

set~ups and specimenso

The manuscript was typed with care by Mrso Lillian Morrowo

Her cooperation is appreciatedo

,;

Page 11: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

v

TABLE OF CONTENTS

ABSTRACT

10 INTRODUCTION

101 The Criteria of Structural Design)

102 The Phenomena of Frame Instability

103 ,Scope of Investigation

20 THEORETICAL METHODS CURRENTLY AVAILABLE

1

3

3

5

10

12

2;1 Structures'Carr~ingno Primary Bending Momento 12

20101 Methods of Analysis and Criteria ofSt~bilityo

2 0102 Buckling in the Ine'a.~stic .Range.

12

20

2 02 Structures Carrying Primary Bendi~g Moment 22

20201 Buckling in the Elastic Range

20202 Buckl~ng in the Inelastic Range

30 BUCKLING STRENGTH OF ELASTIC FRAMES

301 Theoretical Analysis

302 Experimental Investigation

3.3 The Effect of Partial Base Fixity

3.'4 Summary I .

22

29,

32

32

39

41

43

40 ELASTIC-PLASTIC ANALYSIS OF RIGID FRAMES WITHSIDESWAY PREVENTED

401 Assumptions and Limitations

4.2 Moment-~otation Curve of B~am

403 Moment-Rotation Curve of Column

444546 "

50

Page 12: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

4.4 The Method. of An~lysis

4.5 ·Illustrative ·Examples

4.6 Summary

vi

51

$2

56

50 BUCKLING· STRENGTH OF. PARTIALLY YIELDED STEEL FRAMES 58

5.1 Assumptions 58

5.2 · Stiffness of Members after Yielding 59

5.201 Determination of Stiffness and Carry-overFactors for the Beam 60

5.2.2 Determination of Column Stiffness 62

5.3 Method of Solution

504 Development of F~ame Buckling Curves·

505. Experimental Work

5.6 .. Buckling of Multi-Story Frames

50 7 Summary

6. APPLICATIONS TO DESIGN

6.1 Development of Design Charts

6.2 Illustrative Examples

63

66

73

75

76

78

78

85

603 Comparison of Theoretical Results with theAISC Design Rule. 87

70 SUMMARY AND CONCLUSIONS

8. NOMENCALTURE

9. TABLES AND FIGURES

100 APPENDICES

Appendix A

Appendix B

II. REFERENCES

12. VITA-

90

95

100

153

153

159

166

171

Page 13: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-1

ABSTRACT

The objective of this dissertation is to determine the

buckling' strength of pinned-base portal frames in the elastic

and partially plastic' rangeo The frames are acted upon by a

uniform load on the beam and t~o equal concentrated loads on

the columns 0 It is assumed that failure occurs by sidesway

instability in the plane of the applied loadso

It is first pointed out that frame design may be based

on one of the following three types of limiting strength~

the plastic strength, the inelastic buckling strength and

the elastic buckling strengtho Th~importance of 4evelop~ng

methods for predicting these strengths is indicatedo

An exact solution for elastic buckling of portal frameB .

is obtainedo ' Using this solution, the effect of 'primary bend­

ing moment present in each member at the instant of buckling

is clarifiedo The solution is verified by bucklin~ tests on

model steel frames o'

A precise method of elastic~plastic analysis for rigid

frames is developedo It takes into consideration such effects

as axial thrust, yielding, deformations, and residual stresses o

The method may be used. to deter~ine the stiffness of columns

at any stage of loading and also the strength of frames which

are prevented from sidesway movemento

Page 14: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

,"

utilizing the results from the elastic-plastic analysis,

the problem of sidesway bucklipg of partially yielded frames

is solvedo The method of solution is based on a modified

moment distribution procedure in which all the necessary.constants are modified for the combined effects of axial force

and inelastic actiono The method, is illustrated in connec-

tion with the development of a frame buckling curveo

A discussion on the application of the solutions 00-"

tained to tp.e design of building frames is includedo A, .

method whereby design charts for elastic and inelastic buck-

ling may be derived is also pI;esentedo

Page 15: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

1. ,INTRODUCTION

1.1 The Criteria of structural Design

According tO I the present=day concept of limit design of

engineering structuresjl three, cri teria for struct,ural damage

are generally considered to be important 0 One of these is

the ultimate or plastic strength,criterion. When this is

used in designing steel structures the procedure is usually

referred to as the "Plastic Method". Valuable 'advancement

has been made in the past twenty years in this field. For a

general introduction to the concepts involved·in such methodsjl

reference can be'made to (1) and (2).*

A second criterion is thebuc~ling strength. This has

led to the concept of design for,stability. The phenomenon

ofr:'buckling can be described as .the. process by which E/.ny" I

structure passes from onedeformat~on configuration into

another without a change in the'lo,.d. This phenomenon may

occur when the structure is stress~d either below or above

the elastic limit. Each case ,requ~res a different mathemati~

\cal treatment. Although elas'bic: ,bu¢kling has been studied

extensively in the past c~nturYjl few solutions exist in the

area of inelastic buckling. The,knowledge accumulated in this

* The numbers in parentheses refer to references given atthe end of the dissertation. They are listed in orderof their occurrence.

Page 16: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

r

,:,,4

field has been surmnarized by Bleich in Ref. (3) and by the

Column Research Council in Ref. (4).

In addition to the above,mentipned strength limits, the

change of shape at any stage of loading could also be con­

sidered as the limit of usefulnesspf a structur~. For any

structure the deformationundEJr.a given load must not exceed

a certain specified amount, or.. its function may be impaired.

This is often referred to as a; stiffness criterion. At pres­

ent n~erous methods are available for calculating the deforma=

tion of structures in the elastic as well as in the plastic

range.(S) The allowable deflection of a particular type of

structure is usually specified in design codes.

Ideally for any structural .des~gn, each member. should be

selected in· such a way that all the~e criteria are met, and

the structure as a whole should, also fulfill these require-

ments. When the plastic methoq is a~opted in design, it is a

common practice to proportion ,the· ·members on the basis of

their plastic strength, .then perform separate checks to see

if the stability and stiffness, cr.iteria are satisfied. Thus

the member sizes may have to be altered in order to fulfill

all these conditions.

Of course there are other criteria that can be used as

the design basis, such as design for brittle fracture,

fatigue, dynamic energy absorption, etc. However, these con=" '

ditions are not commonly used in designing civil engineering

. \

Page 17: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=5

structures. This dissertation will; only be concerned with the

proportioning of statically loaded structures to meet the

strength, stability and deformation criteria.

, ,

1.2 The Phenomena'of'Frame Instability

One of the important assumptions made in the plastic

theory of structural analysis .is that no instability of any

type should occur prior to th~ formation of 'the f~ilure

mechanism. In general three types of instability are en=

countered, these'are~

(1) The local instability of elements which make up

the cross section of ,the member.

(2) The instability of individual members under the

action of axial force and bending moment~

OJ The overall instability of the whole structure •

.. Detailed analytical and :~xperimental work on the first

type of instability has clarif~ed~e problem and led to

simple design recommendation~,~(6) ··.-In the see.ond type .of.

instability,:. problems inv?lved may be further classified into

two catagories~ (1) Instability. o~ beam=columns due .to ex=

cessive bending and (2) Lateral~torsional buckling•. Extensive

studies have been made on the problems in both catag!Jries in

the past 50 years 0 Some recent developments in this fi.eld

can.be found in Refso(?); (8) and (9).

The third type of ins·tabiIity is the subject. of this

Page 18: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

dissertation. The structure under,investiga~ionwill be a

single story rigid frameo The problem treated here is often

referred to as ""Frame Instability".

The phenomena of overall instability are illustrated in

Figo 1-1 for a portal frame which is not prevented from side­

sway. The following two types of failure may occur~

(I) When a symmetrical frame is loaded by symmetrical

vertical forcrs, it is possible that t~e frame may pass from

a symmetrical stable deformation configuration to an unsymmet­

rical, unstable configuration (Cas$s 1 and 2). At this ins-

tant the total resistance to any lateral force or lateral

movement becomes zero. The load-deflection curve is charac-

terized by a sudden shift from a situation where the applied

load. can increase as the deformation increases to one where

large deflection develops .wi thout an increase of load. The

behavior is analogous to that of a centrally-loaded column,

in which bifurcation of the equilibrium position is possible

at a certain critical loado*

(2) When a frame (symmet~ical or unsymmetrical) is sub-

jected to a combination of ~orizontal and vertical forces

(Case 3), or when an unsymmetrical frame carries beam loads,

the frame deforms la~~rally upon the application of the first

load. The resulting change in geometry may alter the carrying

* In this dissertation, no distinction will be made betweenbuckling load and critical loado

Page 19: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-7

capacity of the individual columns g since the column top is

no longer directly over the column 'base and hence additional

bending moments' are introduced by the vertical load. The

whole structure becomes unstaple ~n this deforme,d position

much like an eccentrically-loaded oolumn. A load-deflection

curve for this type of instability;is shown as a dot-dashed

line in Fig. I-I. At a certa~n .cri.tical conditionp the

structure continues to deform with ,a decrease of load. At the

present time only very limited information is available con-

cerning this type of failure.

This dissertation deals only with frame instability prob;:

lems of the first type. A close examination of eases (1) and'

(2) may reveal that the nature of these two cases is quite

different, even thOUgh they both may be considered as initi~l

motion problems. In case (l)p where the loads are applied\

directly to the columns, the f'rame',usually carries little or

no primary bending moment. Tperefore, only the column action

alone needs to be considered in the bu~kling analysis. If the

analysis is c~rried beyond the elastic limit p then the well­

known tangent modulus concept', for:.6entrally-loaded columns

may be used. CIO ) A brief survey ot~he presently available

methods in solving this type of stability problem is given in

Art. 2.10

However, most rigid frames are primarily designed to

support loads by bending action such as Case (2) of Fig. l-l~

Page 20: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=8

All the members of the frame are subjected to both axial force

and bending momento Therefore g not only the column action

should be taken into account 1~ determining the critical loads

but also the bending action. The problem becomes. more compli=

cated in that yielding Dlay, occur in part of the member.prior

to t~-occurrence of instability.~\ ,

To clarify·the problem furthe~g a discussion 0f the l0ad­

deformation characteristics of such frames is given here~­

Referring to Fig. 1=2g which shows a load=Q.eflection curve of

the frame under considenatlon g it 1s obvious that at any load=

ing stage before buckling the,_ struqture is deformed in a sym­

metrical manner 'and that very, little horizontal deflection at-.

theicolumn top may be_e~pected. If the fr!lDle Is properly

braced against lateral movement g tllen the +oad can increase

up to the ultimate load .calculatad,by the plastic th~o~yo For..

the case of an unbraced frame '1 there exists the possibility of

sidesway buckling. This type of buckling may occur at a load

level below the yield (shown as :pdint A in Figo l-2)g but- more

frequently would take placewhan the appli~d load has caused

yielding in some portion of the ·frame (shown as point B).

Since the formation of a plastic-mechanism can also qe con­

sidered as the limit of sta~ilitY-'of the whole frameg then it

is clear that for unbraced 'frames the inelastic buckling load,-

should have a value between the yield and the ultimate loads.

The similar~ty between the buckling of centrally=loaded

Page 21: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=9

columns and that of the frame~ inv~stigated in this disserta=

tioh is illustrated in Figo 1-30 It can,be seen from this

figure that for 'frames with slender columns sidesway buckling

would occur before the stress, at any point has reached the

yield level. This situation is analogous to the elastic buck­

ling of centrally=loadEl'd columns (EulerUs problem). For frames

with columns of 'intermediate slenderness ratios buckling takes

place when parts of the members have already been yielded.

This is similar to the buckling of ,columns in the inelastic

range (problem of Engesser and Shanley) •

In obtaining the curves ~or frames the simple plastic

load w is taken as the maximum ca~rying capacity, correspond=uing to a b~am mechanism. Curve (al, of Figo l-3b shows the

actual load carrying capac!ty"of ",s'14ch frames if sidesway is

properly prevented. This curve is -determined also for the

beam me~hanism but with the plastlq hinge moment at the column

tops reduced for the beam=column.actiono ,The inelastic buck=

ling load of unbraced frames is plotted as curve (b). It may

be expected that· with the column, slenderness ratio less than

a certain limiting val~e the unbraced frames would carry

approximately the same load as -brace,d frames. Then: within

this limit the problem of frame -stability may be disregarded

and the plastic strength ~an be us~d as the design ,basis. If

this is not the case the inelastic buckling strength should

be considered as the proper criterion. For very tall frames

it ma~,be even necessary to design against the elastic

Page 22: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

~lO

buckling strength (see curve (c) of Figo l-3b)0

From the above discussion it is evident that in order to

design unbraced frames efficiently, methods by which the elas­

tic and the inelastic buckling str~ngth of .such fr8ll1es can be

precisely determined are necessaryo; Unfortunately after an

extensive survey of literature, it ,was found that only very

few cases of this type of buckling problem had been studied

and that these studies were limited to the elastic rl\lnge onlyo

No ·attempt has yet been made to develop methods for predicting

the buckling strength in the;i.nelaatic·rangeo Some empirical

formulas base.d on experimental results have recently.been,

suggestedo These will be discussed in Arto 20.20

It· is the purpose of this dissertation to investlgate

this type of instability problem ex~ensively and present the{-~ ,:. .",;" ,:::

results in a form that is suitable for designuseo

103 Scope of Investigation

This dissertation contains the results of anlnves~lga­

tion of the folloWing four p~asesg

f.. ; - ,I

,(1) Theoretical and experimental studies on the elastic

buckling strength of slngle~toriY-rigid frameso Figo 1-4

shows the 'dimensions of the frame and types of loa4ing

assumed in the investigation. The, parameter a whlchrelates

the magnitUde of the concentrated load P to the intensity of

the uniform load w, 'is kept,.constant for each qaseo This

Page 23: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

= 11

"implies that these two types ~f lo~d are assumed to increaseI

simultaneously with a fix~d ratio a between themo This par­

ticular loading condition would simulate a condition 'that may

be'expected in the lower stor~es of a tall buildingo, '

(2) ,A precise method of,indeterminate analysis appli-

cable to elastic as well as p~rtl~lly plastic structures o

(3) b,evelopment of a numerical method by which the

inelastic buckling strength of frames as shown in Figo 1=4

may be predictedo'

("4) c.onstruction of charts to facilitate design by

utilizing the results obtained in phases (I), (2) and (3)o

Throughout this dissertation, it is assumed that the

frames,are sUfficiently braced in the lateral direction so

that buckling can only occur in the plane of the applied

loads 0

Page 24: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-12

2. THEORETICAL METHODS CURRENTLY AVAI:J;.ABLE

In this chapter,~ brief descriptions of some general

methods of approach which have been.developed for the so­

lution of various frame instability problems are presented.

They merely serve to indicate' the ~esent knowledge i~ this

field; An extensive survey wiil be contained in a forth­

coming report prepared by the author.(ll)

2.1 Structures Carrying No Primary B.ending:Moment

This section will outline some' important methods for

.determining the critical load of structures which are

stressed principally by axial forces. All the methods

resernb Ie very much those ordinarily used in' a:naJyzipg:~~ta';'

tically indeterminate structures, except ,that some modi­

fication is made to include'the bending effect of the axial

thrus~. Asystematic survey of the existing theory, cover-

ing both exact and approximate methods, was given by

K · ...,.,( 12)av 9:11a 51-'" .,

2.1.1 Methods of Analysis and Criter-la of Stal:rility

Essentially three different approaches have been intro- ."P,

duced to solved stability problems considered here. Those

are (1) the analyt.ical·,·niethods (2) the convergence methods

and (3)' the energy method-., ,Each method is based on a par­

ticular type of stability criterion.,

Page 25: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-13

1) The Analytical Methods

The analytical method is often referred to as the de­

terminant method o It includes all the procedures of. analysis

which involve the solution of a system of linear simultaneous

equationse The unknowns 1n these equations are the addi­t

tional forces and moments induced by buckling, and the join~_

displacements and rotations of the memberse The coefficients

of these unknowns are usually functions of the dimensions of

the members and the axial forces o In general it is possible

to set up as many equations as there are unknowns e The

trivial solution of this system of equations represents the

state of stable equilibrium o Finite values of the unkpowns,

indicating the existance of additional forces and deforma­

tion, can occur 'only when the determinant 4. of the system

is equal to zero o . There are an infinite number of solutions

to the equation ~ =0, corresponding to an infinite number

of possible unstable equilibrium configurati ons e The con­

figuration·which gives the smallest value of load~g de­

termines the critical load of the structure e

Depending on the unknown quantities selected, there

are three different methods that can be used to set up these

simultaneous equations, namely:

a) Bleich's four-moment equation method

b) The slope-deflection method, and

c·) Mises method

Page 26: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-14

The first two methods are commonly used and will be dis­

cussed briefly in the succeeding paragraphs. The procedure

involved in using Mises Method can be found in Refs. (13)

and (14).

In the .four-moment equation -method, the unknowns 'se­

~~ctesi are the end moments, additional forces and the rota­

tion 'of members p. Referring to Fig. 2-1, which shows two

consecutive members Li and Li +l of a structure before and

after buckling, the four-moment equation can be shown to

have the following form:(3)

L p ..._i (s.M. 4-c.tv\.) +E1. L l L L

L

L~i" (c M~ +5 rv{) +(P. -f) = 0E I. ttl ii" _+1 It·1 \ Ltl i.,+, (2.1 )

where tv! = End moment

L = Length of members

f = Rotation of members

and

5 - _I(A_ l ].- (i!L)z ~inRL

~ =I :1(:;- = -'- ( I -12L Cot1J.]

(1tL/ .(2.2)

(2.3)

"p = Axial force in members

This equation represents a relation between the four end

moments of the two adjacent members and the bar rotationsp •

The values of s and c are tabulated in Refs. (3) and (15) as-

functions of kL ~ . Equatiops of this type form the first group

of the stability equations. To solve an actual.problem two

Page 27: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

additional groups of equations are needed •. They are obtained

by considering 1) the geometricat. relations. between rota­

tions and deformations, and 2) the equilibrium conditions- .- -

of the structural system when it is in the state of unstable

equilibrium. The existence of non~zero solutions of the

variables in these sets of equati ons defines the limit of

stability.

The slope-deflection method for stability analysis was

first adopted by Chwa1la' and Jokisch for analyzing multi­

story frames.(16) In this method, the angular rotation a of

the joints and the bar rotations f are used as the unknowns

and the terminal moments are expressed 'in terms of these

unknowns. Analogous to the slope-deflection equations in

indeterminate analysis, the pair of equations has the form

M AK ( 6 A ... Ce B -(ITC)j)

M B K ( c.e A + ee -(I-C))]

'where

Ke- El-- stiffness factorC 2 _S2 L

C = 5 =.carry-over factorc

Two equations of this type can be written for each-member,

expressing the moments at the two ends in terms a and f .The end moments are then eliminated by substituting these

expressions into the equilibrium conditions. In general two

types of equilibrium conditions are necessary,ione is ob­

tained by considering 2: M = 0 at each joint, and the other

Page 28: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

:'16

by considering. the equilibrium of the whole structure .. After

eliminations of the end moments, a system of equations linear

and. homogeneous in e and f is obtained. The stability

criterion ~ ~ 0 can then be applied to these equations for

the determination of- the critical load. Tabulations of the

constants K and C can be found in Refs. (3) and (15).

Except for very simple cases, the direct solution of

the equation L:::.. = Ois not possible because terms involved

in these simultaneous equations are transcendental in

character. The usual procedure is to solve for the values

of ~ corresponding to a number of different values of the

applied loads and to plot ~ vs. the loads. The point where

.the curve first crosses the load axis gives the lowest

buckling load.·

2) The Convergence Methods

Because of the difficulties involved in obtaining exact

solutions by the analytical approach, various types of nu­

merical methods have been developed. These include the

moment distribution procedure adopted by Lundquist(17, 18)

and the relaxation methods applied by BOley.(19)

The moment distribution method for stability analysis

is based on the same concept as that originated by Cross for

analyzing continuous beams and frames, but with all the con­

stants, such -as the stiffness factors and carry-over factors,

being modified for the .effect of axial loads. Values of

Page 29: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-17

these factors have been tabulated in Refs. (15), ((20) and (21)

as functions of kL for the cases when the far ends of the

members are pinned or fixed.

For structures without joint translation, Lundquist

has established a stability criterion, known as the Stiffness

Criterion, (he al~o developed the Series Criterion by further

reasoning) 0 The pbysical meaning of this, criterion can be

interpreted as the following~ At an~ joint of a structure

where several members are rigidly connected, a unit moment

is applied, causmg the joint to rotate through an angle Q,.

the magnitude of which ,depends, on the sum of the stiffnesses

L K of all these 'members. As the applied load increases,'

the value of [K tends to diminish gradually. The limit of

stability is reached when the total stiffness appro~ches

zero, that is [K = 0 0 ; This means that at the instant' of

buckling no external moment is required to rotate any joint

in the structure. Lundqulst 9 s criterion can be applied

quite successfully in analyzing structures such as bridge

trusses, roof trusses, and aircraft frames.(18, 21)

The convergence of the,moment distribution method and, (22)the uniqueness of the results were proven by Hoff • His

proof has led to the following important theory for the

determination of the stab iIity of frameworks: II If a frame­

work is subject to external moments and the Cross mf3thod as

modified for axial loads is used in the calculation of the

distribution of bending moments of the bars, and a fmite,

Page 30: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-18

unique set of values is obtained, then this result is a

necessary and suffic ient conCIi tion for stab iIity.n

A finite set of values means that the mqment distribu-I

tion method converges, ~nd uniqueness implies that the so-

lution must be independent of the order of balancing.

. - " .

To" analyzestruc"tures with joint translation, the

moment distribution procedure as described above cannot be

applied without modification. Several such modifications

have recently been proposed by Winter et al (15), Perri (23) ," - - (21)

and Livesley and Chandler. -In the following paragraphs

the method developed by Winter et al is described, since

this part icular me thod will be adopted in Chapter 5 to in­

vestigate the buckling ~trength of yielded frames.

For analyzing the s idesway buckling load of the" frame

shown in Fig. 2-2, an arbitrary lateral displacement d is

first introduced at each column top. Then from Eq.(2.4)

the fixed end moments are

dMFA = ~H!, = K 1 (I + C,) L";"

(2.6)

The joints-are then released and the unbalanced moments are

distributed, the resulting c?lumn moments will be Mi, M~

etc. The condition of equilibrium of each column requires

that the shear force be given.by~

I( I I )0 1 = """'C" fv'\ AT VI 13 - PI d _

Page 31: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-19

From Eq. (2.6), the displacement d carr be writterr as

d = (2.8)

Substitution of Eq. (2.8) into Eq. (2.7) results in the

following expressions for the shear forces:

Q1

I () I ~; L El,MF1 J= - .MA -+- Mp, -L1 K. (1+ C)

(2.9)

I ( II ~; L E12

Mn JO2 = ~ Me. + IV\ -D k 2 ll+C:..)

wherek is as defined in Eq. (2.3). As the applied load

increases the last terms in the parentheses of Eq. (2.9)

increase in absolute value relative to the first two terms.

Hence, the shear force Q decreases, "indicating a descreasing

resistance of the frame to the imposed displacement. The

critical condition i:sreachedwhen

= o (2.10 )

that is, when an imposed lateral displacement is no longer

counteracted by opposing shear forces.

The relaxation method is fundamentally identical with

the moment distribution method. The same stability cri-

terion is used in both cases. For a complete discussion of

the mathematical theory involved, reference can be to a

paper by Boley(19) and its discussion by Pullaczck

Geiringer(24) ~

Page 32: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-20

3) The Energy Method

This method states that when the second variation of

the total potential of the system is greater than ze~o the

equilibrium is stable, and when it is equal to zero the

equilibrium is neutral (limit of stability). Once the

total potential is properly formulated, the problem can

then be solved by using variation principles. The appli­

cation of this method to solved buckling problems has been

discussed by Mises(13), Ratzersdorfer(14), Kasarnowsky and

Zetterholm(25) and F. Bleich and H. Bleich.(26)

2.1.2 Buckling in the Inelastic Range

In the previous discussions, it is assumed that all

the.members of the structure are stressed within the elastic

limit. However, in most praotical Cases instability of the

structural system would occur when the stresses in some

members have already exceeded that limit. Therefore, the

methods outlined in the last article have to be modified

in order to take the inelastic effe·ct into consideration.

(Since alL.the members are SUbjected to axial force

only, uniform yielding may be assumed for every section

along each member. Then according to the tangent modulus

concept (no strain-reversal) the f'lexural rigidity EI should

be replaced by EtI, where Et

is the slope of the stress­

strain diagram of the particular material determtned from

a coupon test. But for material like steel which has

Page 33: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=21

discontinuously varying Et values 9 this type of modifica=

,tion cannot be applied •

. Re~ently investigations in Fritz Laboratory have''; .

shawn that the inelastic buckling strength of steel columns

is influenced considerab ly by residual stresses 9 and that

a total stress-strain diagram of the whole section obtained

from a stub column test shoulq be used to determine th7 Et

, (27)values. At the present time considerable i.nformation

has been obtained concerning the magnitude and distribution

of these residual stresses and their effect on column in=

stability.(28) By utilizing such available information 9 a

numerical procedure is suggested here for analyzing stability

problems in the inelastic range.

For a given loading system9 the axial load p and the

stresses a in all the members are first calculated. Then

from Fig. 2-3, the ratio of Ie can be determined for each, ' I

member, where Ie is the effective moment of inertia of the

section carrying the axial Ipad p (if no strain=reversal

is assumed in the analysis Ie is simply the mo~ent of ,

inertia of the unyielded parts of the section). To obtain

k values to be used in Eq. (2.2)9 the flexural rigidity

EI in Eq. (2.3) should be replaced.by EIe • The 'same sub=

stitution should also be made in'Eqs. (2.1) and (2.5).

The buckling analysis can then be carried out by any me'thods

described in Art. 2.1.1.

Page 34: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

2.2 Structures Carrying Primary Bending Moment

The discussion on structural stability contained in

Art. 2.1 was under the assumption that no bending moments

(or flexural deformation) are present in the individual

members at the instant of instability. But in most engi-

neering structures this condition is not satisfied. Gener­

ally, bending moments and deformations are introduced as

Boon as external load is applied. Therefore, it is neces-

sary to study buckling problems in the presence of such

effects. The frame stability problem investigated in this

dissertation represents an important example of such prob-

lems.

Unfortunately~ the solution of stability problems of

this type becomes very involved and so far only a few

attempts have been made to solve them. These attempts will

be briefly discussed in this section.

2.2.1 Buckling in the Elastic Range

1) .Chwallav s Investigations

In Ref. (29) Chwalla presented an exact solution to

the problem of sidesway buckling of a rigid frame under~

transverse loads as shown in Fig. 2-4. He showeq·analyti­

cally that for a symmetrical frame carrying symmetric loads

the type of instability is characterized by a bifurcation

of the equilibrium positions, with a sudden change of the

Page 35: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

':'23

load-deformation relationship •. By solving a system of

second order differential equations defining the equilibrium

at the buckled state, he obtained the following two condi­

tions which should be satisfied simultaneously at the limit

o:f s tab iIity:

Equilibrium Condition:

Il (1_ fl, hCot l,h) Cos ilJ +C'D.s ~.L - CM ilz (.1-_ -L) - ~,~ H),'n P.1 = 0p2 . 2 2 2 n .p 2

Buckling Condition:

(2.11a)

(2.11b)

and·

Using these two equations, the buckling loads o:f a :few

cases were computed by the author and the results are shown

in Table I. Two types of :frames were selected, the :first

type with L = h and the second with L = 3h. Moments o:f

inertia of the beal1l and the column were. assumed to be the

same lb = Ie = I. The number n in the table indicates the,

position o:f the applied loads P (see Fig. 2-4a). The loads

P act at the center of the beam when n equals 2 and at the

top of columns when n = 00. In the latter case, the problem

is reduced to the type discussed in Art. 2.1 anq its solu­

tion can be easily obtained. By comparing the values

listed in the first row o:f· T'able . r- it can be seen that,the critical load o:f such :frames is reduced when the loads

are applied away from the column tops. For the :frame with

Page 36: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

L = h the maximum reduction amounts to less than 3%, whereas

for the frame with L = 3h the reduction can be as much as

11%.

In the same paper, Chwalla suggested that the presence

of the axial force in the beam (horizontal reaction H at

the base) is responsible for part of the reductions, because

the stiffness of the beam is decreased due to this:'force.

To check this suggestion, the values p 9 shown in thecr.second row of Table I were calculated for the +oading

system shown in Fig. 2-5. T~e horizontal force H applied

to the beam was determined from Eq. (2.10). It is noted

that for the case L = 3h the buckling loads p 9 determinedcrfrom this simplified loading condition are very close to

the exact values Pcr It may be concluded from this study

that for practical purposes the actual loading system can

be substituted by the loading,system assumed here in de­

termining the critical loads.

Experiments were conducted on small model frames by

Chwalla and Kollbrunner to check the theoretical solution. , ( 30)

discussed in this article. ' Excellent correlation was

observed in, all the tests.

b) Puwein9 s Investigation

Following Chwalla9 s suggestion, Puwein obtained an

approximate solution to the sidesway buckling of portal

frames carrying uniformly distributed 10ad.(31 ) He

Page 37: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

used a simplified loading system similar to that of Figo

2-50 Therefore, the buckling load computed from his', solu~

tion would be slightly higher than the exact value~ In

Chapter '3 of this di~sertation an exact solution to the

problem will be presentedo

Puwein also extended his work to portal frames with

partial base fixity and to gable frames. A similar sub=

stitution of loa~ing system was used in all the cases.

c) Recent Developments by Masur, et al

Very recently, Masur, et aI, succeeded in extending

Bleich's method, the slope=deflection method, an~ the.m.oment

distribytion method, as desctiibed in Art. 20102, to the an=

alysis of the stability of frames carrying primary bending

moment(32 )0 'By using these extensions, the s~ability prob=

lems of the type und~r consideration can be solved in a

systematic manne~o

. 'In the first part of their paper, the original four=

moment equation was 'modified by includ~ng the effect of

the transverse loads to the following form.

L· ( P--' s.M.+E1· L l

l

r ) L· ( ~A Pc· . + ~ . IVI·eML fl. (L+-I t+1

, Ltl

+ L r. - f. ) + ( F.r

- F/ )\ .+1 l . .1'1

o (2 ~-12)

Page 38: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

h F ·1,- d Frwere i· an 1+1 represent respectively~ the loading

=26

terms at the left end of the member i, and at the right

end of the member i + 1. These loading terms depend on

the applied transverse load and the axial force in the

member. They have been derived explicitty in Ref. (3j')

for some simple loading conditions~

If only two members meet at the joint i, MI becomes

equal to MI+l and Eq. (2.12) simplifies to the three=

moment equation..

L s· WI! + -( --h c· + k c· ) Ml l r L. lEI' <+1EI ~ c • .4-1

+ --h 'S' tv'\ r + If. - F ) +( \=.t" - F~) = 0,,.1 i,.,., \. '+1 ~ ..., l .E1 ~+1

Equatioris (2.12) and (2.13) can also be used to solve

statically indetermipate structures~ including the

effect of axial forces.

When a structure passes from·one deformation configu=

ration into another~ there will be changes of react+ons

and deformations~ these changes will. in turn~ cause varia=

tions of. moments 6M~ axial forces 6P ~ and the member rota=

tions 6f. Consequently~ there will be ·~ariations of s~

c, and F ~ of each member due to the change of axial force.

~?~ev~I'~ all these variations should satisfyEq. (2.12) in..:.··

its variational formg

Page 39: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=27

L'+ ( ,1 l ~J1t f1~lIr M r)+_L_' C' ~M. +C>.C L'., IV 1;>- +S. IVI;~I+DS, 1'+1EI . ...... l1"1 T .... I 1+1 'T . v+-I. L4-1

r Q )+ (D,f'. - A f; ) + L11 'F. - DFe\ l+-t .• l+1

= 0

Eq. (2.14) is called the four-moment equation for neutral

equilibrium. The critical load of structures maybe obtained

by solving simultaneously equations of the types (2.12) and

(2.14). This method will be adopted in Chapter 3 to obtain

an elastic solution to the frame stability problem under

investigation.

Following a similar procedure~ Masur~ et al~ also

modified the slope~deflection equations (Eq. (2.4) to

include the effect of transverse load by introducing

terms which are commonly called the fixed=end moments.

The generalized slope=deflection equations~ ~hen~ have

the form~

K[ BA+ CBe "( 1+ C ) r] + M'A 1K [ ~eA +e.-(I+C) r] ± M,. )

with and

Page 40: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

I- .

=28

where FA and ··Fa are the loading terms f'or the ends k and B.

Equation (2.15) was also derived by Winter, et al,in Ref.

( 15) •

The tec'hnique of' using variations of deformations, re=

act.ions, moments, and axial f'orces, occurring' immediate ly

af'ter buckling, for solving stability problems can be ap=

plied equally well in the slope=def"lection method. The

variational form of Eq. (2.15) may be shown to be ~

ilMA

A similar expression can be written for flIVB. By separate

applications of' Eq. (2.15) and Eq. (2.16) two groups of'

slope=deflection equations are obtained. Equations of'the

f'irst group lead to the relationships between the·reactions

and the applied loads when the structure is'in stable equi=

librium.· From equations of' the second group, similar rela=

tionships may be obtained for the unstable state. The

critical load of' the structure can be determined by solving

, these two types of relat'ions 's·lmultan.eous ly.

__ .. ,The moment distribution procedure' developed by Winter,

et aI, as desc·r.ibed in Art. 2.1.1, was extended further to

the investigation of stability problems considered in this

section. The extension is principally based on the varia=

t ional f'orm of. the s-lope=deflect ion equat ions as;'given ab ove.

Page 41: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=29

The criterionI.Q = 0 at the limit of stability can also be .

used in the present case. Since the method involves success=

ive numerical approximations, the buckling load can generally

be determined without solving simultaneous equations.

It should be pointed out that the methods described by

these authors can be successfully applied only to very simp~

structures. For more complicated cases, the evaluation of

those variational terms contained in Eqs. (2.14) and (2.16)

becomes ve~y cumbersome.

2.2 .2 Buckling in the Inelastic Range

In recent years, because of the rapid development in

the plastic method of design, the problem of frame stability

in the inelastic range has received special attention. Num=

erous test programs were carried out at several institutions

in England to study the reduction of plastic strength of

single and multiple story rigid frames due to instability

(34, 35, 36, 37). Based -on the' results obtained from these

t~sts, several types of empirical formulas have been proposed

for estimating the inelast~'buq'kling loads( 38). Neverthe­

less, due to the complexity of the problem, very little

theoretical work has been done in this field. The analytic=

al studies contained in Chapter 5 of this dissertation may

be considered as the first attempt to solve this type of

stab llity prob lem. '

Page 42: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=30

In the following paragraphs, brief discussions on some

empirical methods that are presently available will be pre=

sented:

a) Generalized Ranking Formula Proposed by Mercharit(3.8 )

Merchant has recently suggested that the actual load

carrying capacity WF (or the inelastic buckling strength)

of unbraced frames may be assumed to depend linearly on two

parameters: 1) the s·tmple plastic" strength Wu , and 2) the·

elastic buckling strength wcr , and that a simple relation=

ship of the followin~ f-orm~-tre us-~rdror es-timating wi

This equation is called the "Generalized Rankine Form=

ula". Tests on model frames of rectangular sections con=

duct~'d at Manchester University and at Cambridge University,

have shown that, in general, the buckling load given by Eq.

(2.17) is very much on the safe side(34," 35, 37). So far;

no test has been performed on full scale frames of WF shapes

to check the validity of this formula.

b) The AISC Formul~

To safeguard"·against frame in!3tability, the following

formula is recommended in the AISC Plastic Design Manual(39)

for pr.oportioning the columns in unbraced frames~

2~P~

I h+---10 r ~ 1.0 (2.18 )

Page 43: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

~31

in which P is the axial force in the column when the frame

carries its maximum load~ Py , the axial yield load of theh

column, and r , the slenderness ratio. The justification

of this rule can be found in Ref. (2). It has also been

partially confirmed by tests(40 ).

In the preceding sections, currently available methods

for the solution of two types of instability problems are

discussed. For the problem of the first type, where the

members are primarily stressed by axial forces, methods are

readily available for determining the buckling loads either

in the elastic or in the inelastic range. Whereas for the

second type of the stability problem, where bending moment

as well as axial force effects have to be considered, only

very limited theoretical investigations have been made.

These investigations are limited to elastic buckling. At

present no method is in existence for solving inelastic

buckling problems of this type. Therefore, the main object

of the present research is to develop such methods. In the

next chapter the elastic buckling problem will be studied

in detaiL These studies will constitute the first step

toward the solution of the problem in the inelastic range.

Page 44: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

, \,

-32

,3. BUCKLING STRENGTH OF ELASTIC 'FRAMES

Theoretical and experimental studies on the buckling of

,rigid frames with the loading condition shown in Fig. 1-4

will be presented in this :,cha.p!ier~ Also included is a dis­

cussion concerning the effect of partial bas~ fixity on the

buckling strength of such frames.

3.1 Theoretical Analysis

The modified three-moment equation method as d~scrib~d

in Art. ,202.1 is used to solve the present problem. , Two"

.governing equations will be derived by separ~te applications

ofEq,s. (2013) ,and (2.14). They both define relatio!lships

between the applied load and the horizontal reactio~ a~ ~he

base, .one for the stable state and the other for theun~:

stable state. Instability will occur when the applied load

has reached such a value that both of these equations are

satisfied.

,Figure 3-1 shows the moments and forces acting on a

frame before and 'immediately after ~uckling. Considering

the equilibrium of the frilme 'with symmetric deformation, the

rollowing relations are obtained~

v = PR

~A = - Hh'YIR

}

Page 45: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

On appltcation of Eqo (2013) to members AB and BD, the three=

moment equation for equilibrium reads~, "

where the sUbscripts 1 and 2 refer to nlelnbers ~ ,and BD,.,;

respectivelyo The coefficients c, sand F are defined as ­

follows~

<: I = (fll'h)Z ( 1- 'k,h tot ll,h )

C Z ~Lz (1- -klL Lot tl<LL)\:R,z )

Sz I (~2-L I)l't; lY <)in ~z l

F =2wL ( IH ~zL

1- Cos izL'S>irl ~ <L L -+)

with

'"Tije expression forF2 is taken from Refo (33)0 On sub=

stitution of these expressions and the expression for ML

and Eqo

the frame

and MR from Eqo (301) into Eqo (302) and rearranging terms,

the'equilibrium condition becomes~

~ (1- R,h Cot u',h) +~2h tan t;L - ~l (1llzL tan t;L -+)= 0, = wL

For the case with roof load w only, then P =~ ,(304) takes the form

'2H('Lh tL1 ) .t.h' ~lL wL'(:'t,t.l I')~ 1- 11.1 CvI(lIt1 + /(l Tan "2 - ~ thl "flz"-Z = 0

Next consider the equilibrium of members when

is iIi the state of unstable' equilibrium as shown in Figo 3=lb,

the following equations can be r~~dily obtalned~

-' - ~Rvl +tNl =p.,-zp T

V + tN, = -p -1-2 -p ~11. 11. L

"M L +t..M L = -(H +AH)h + VL tiR

M R+ II M11 = - ( H+ AH ) h - VI!. AR }

Page 46: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-34

Combining Eqo . (306) withEqo (301) leads to~

or

l::!.VL =t:,.YR =

~ML=

l\MR =

- t.R- 2PT

2P oARL

-6~ h+ pl1R

-61-1 h -PoOR

('NL P ;) VL .-2- --=0

aR L aH"

.~ 2P .2.Y..E... = 0=JR. L JH

.:lM L - J.ML = -h= pdR dH

. aMI/. = - dMg = -hodD..

- P JH~

To obtained the relationship between Hand P immediately

after buckling, the three-moment. equation for neutral

equilibrium will be usedo For members AB and BD Eqo (2014)

becomes~

()010)

Similarly, the equation" for' members BD a.nd DE IB~

---l (51 6.M L+ AS2 ML+C1AM~+oOCz MI/.) ... ~ ((3 .o~R+ .0(1 Mil ). EI H~

~+ <1.t + t!J. F,. = 0

It can be seen from Eqo (308) i;;h~:t after buckling the

quantities 6.R and ~H may be regarded as the independent

variables of the systemo' So the variation of axial force

p in each member can be expressed in terms of these variables 9

that is

Page 47: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=35

The variation of cl for member AB is derived as follows:

dCl. -- -- - .-

,where cl =- The last step is obtained by us ing Eqs.

dPl•• - ••• _..•_ ••• _. _. "_I ._.

(J.l) and (J.8). In a similar manner, the following ex-

pressions for L::::.c2, 6.S2 and .6C3 are obtained:I

A(z = (2 AH

D. 52 = S/ llH

~C3"'" ~~~(i

}.Substituting Eqs •. ()"012) and (J.13)·and the expressions for

~L and.6MR from Eq. (3.7) into Eqs. (3.9) and (3.10) the

equations of neutral equilibrium become:

and

-[Jl. (I + -l-(fz-S:z.)+ _h_ c/ 21·H, - ~h] rhR~I(.. Elb El, L P

-(...h...c'-T·--'=-l(~;-'SJ+~l c.c:+s;)]h ..6H+~F= 0EL. U" I: 1"

Subtracting Eqo (3015) from (3014) leads to~

= 0

Page 48: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

Since6R should be finite after buckling, the expression iri

the bracket has to be zero~

I---ph =0

From Eqo (Jo 3) the following relation may be deduced~

( . _ _1_ (2 _Jo Lr '.11h' )Cz - 5z ) - (11. LY 1('1 \..OJ?

-l J

Upon substitu~-1ng the. expression for cl from Eqo (30)) andI .

the above'expressions for (C2 = s2) and cl intoEqo (3017),

the stability condition is finally obtained in. the form~

This equation may be seen to be identical withEq~ (20l1b)

obtained by Chwalla for the loading condition shown in Figo

2~4~ This implies that the actual condition of load appli­

cation is i~terial in analyzing Eq~ ()019) as long as the-

.prope~ value for P is used in every case (P will always be/.

equal tG half, of the total loadfl)0

Both Eqso ()04)·· and C3ol9) express impliei~iy thehori=J! -..' .

zontal reactionH as a function of the applied load Po lIt

would be impossible to obtain simultaneous solution of these. ~. "

two equations for Pcr and Hcr in explicit forms 0 Hence,

separate computations have to made for individual cases with

the aid of some numerical technlq~eso The method of solving

Page 49: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=37

these equations is illustrated here for the case of a square

frame (L = h) with I c = Ib = I sUbjected to a uniformly dis=

tributed load w on the beam (that is a = 0)0 Let L = h 1n

~9so (305) and (3019), and introduce a non=dimensional para='

meter~ which is defined as~

p'=~and note the relationships

11, ~ I ~~~ and 1,~ I EHI ~ 11, ~The following two equations are obtained~

2These equations are then solved for the value of ~ for each

assumed value of klh by a trial=and=error prpcedureo For

instance from Eqo' (3020.) the follo,wing values of ~2 are ob=

tained 000980 g 0009766~ 000953 corresponding to klh = 005g

100g 10350 Jilor'klh = Og the value of ~2 can be determined by

or~inary structural theorYll because in this case the axial

force is assumed to have ho effect, on the reaction Ho Using!

these values of klh and ~2 the applied load wand the re=

action H can be determined from the equations 0

wh =Lth)~

}2 ' hZ and

= pZ w; ~ 2 n (3022)H = l-R1 ~) - hI:

Page 50: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

~ I

The results of the above computation are plotted as curve (I)

in Figo 3-20 It represents the relationship betweeti wand H

as derived from the non-linear theoryo The'same relation-

ship based on ordinary structural theory is shown as a dotted

line in the figure 0 A small deviation between these two

curves can be noticed, indicating the effect of axial force

in indeterminate analysis 0'

Following the above procedure, Eqo (3021) is also solved

and the final results are plotted as curve (II), in :Figo 3-20

This curve gives the relation between wand H when the frame

is in the state of unstable equilibrium~ The point of inter­

action of curves (I) and (II) determines the exa'ct buckling;

load of the structure 0 As shown in Figo 3-2, the critical

values are~

wh) EI(-l-e.t= 1.1817 and I--l -= 0 li05 B-

~r . h2

It can also be noted that the .point where curve (II) int,er=

sects with the abscissa corresponds to the' .critical load of

the frame when all t~e loads are assumed to act along the

centerline of· the columnso

Similar computations were carried out for the case a. = 2

and for the frame with L = 3ho The results are listed in

columns (1) and (2) of Table II for L = h and in columns (6)

(7) -arid for. L = 3ho The values of Per for the case when. all

the'loads are assumed to act at the column tops'are also

given in the table and compared with that of the actual load­

ing conditions (a. = 0 and a. = 2)0 For the case L = h the

Page 51: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-39

difference is less than 2% when a = 0 and less than 1% when

a = 2, and for L = 3h the corresponding differences are 607%and 204%. The reason that t he difference becomes smaller

for larger values of a can be explained by considering the

relative magnitude of the axial forces in the beam at the

instant of bucklingo For higher values of a, the portion of

the total load that applies to the beam is reduced, conse­

quently the horizontal reaction H at tpe base (which is

equiyalent to the axial force in the beam) is also reducedo

The stiffness of the beam is therefore increased slightlyo

This results in an increase in buckling loado

From this study, it is suggested that for larger column loads

with a = 2 or greB. tell' the influence, of axial force H may be

ignored in computing the stiffness of the beam and that in

practical cases the critical load of such frames can be com-

puted by assuming all the loads acting at the top of columnso

Such simplifications will be adopted in subsequent chapters '.

of this dissertationo

3 0 2 Experimental Investigation

To check the theoretical work presented in Arto 301

experiments on model steel frames were conductedo The cross-

sectional shape and the loading arrangement of the test framesi

are as shown in Figo 3=30 The uniform load w assumed in the

analysis was replaced by two concentrated loads PI each

applied at a distance 003 L from the center line of the

Page 52: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

~".

-40

columns. Figure 3-4 sho~s the test setup and the fixtures

used for transmitting the loads to their points of applica­

tion. A dead weight and lever system was used to produce

the downward force in the sling. The loading system as a

whole could sway freely with the frame at all stages of .the

test. Details of the test procedure and the experimental

techniques employed will be described in a forthcoming

report. (40).

All the· important information pertaining to these model

tests, inclUding the'frame dimensions, load ratio a, theo­

retical predictions and the test results, is s~arized in

Table III. The "test load" reported in the table is not

the buckling load, but the maximum load observed. in each

test'. Because of the unavoidable imperf'ections of;.. the test

specimens and inaccl~acy in load application, it was 'im~

possible to detect exactly when the test frame started to

buckle. However, in general very little increase of load

can be expected after the initiation of sidesway moveme~t,

so the ultimate load' observed from test should be very

close to the actual buckling load.

Figure 3-5 shows the load-deflection curves of the

test frame P~4. The curve shown in the right is very

similar to the load-deflection curve usually obtained from

a centrally-loaded column test. Figure 3-6 shows the s~me

frame after unloading, typical. sidesway buckling can be sean.

It can be seen from the comparisons given in Table III

that satisfactory correlation between the theory and the tests

Page 53: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-41

has been obtained. For both .frames P-3 and P-4, the test loads·

are a few percent lower than the prediction. These discpre­

pancies were mainly due to local yielding at the welded joints,

where several yield lines were observed.

3. J .~p.e Effect of Partial Base Fixi.ty

In the above described theoretical and experiment

studies, the base of the frame was assumed to be perfectly

pinned. However, in actual structures this condition usually

does not exist. In most of the so-called "pinned" column

bases, the rotational restraint at these bS:ses may be rather

appreciable. The actual· amount of base restraint depends on

the details used in construction and on the foundation soil.

Recently Galambos has shown that the buckling strength

of portal frames with small amount of foundation restraint

can be considerably higher than that of pinned-base frames. (41)

This has also been observed experimentally in some recent

tests. (40)

In this section, it is intended to indicate how the·

theoretical solution discussed in this chapter can be used

to determined the buckling load of frames with partial pase

fixity. As was done in Ref. (41), the base restraint can be

simulated by inserting a restraining beam between the two

column bases as shown in Fig. )-7. This beam restrains the

column ends in the same way as would be done by an actual

base consisting of base plates, the fo·oting and the soil.

Page 54: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-42

With this simplification, the problem can then be solved

approximately according to the following steps~

10 Assume a distance hI from ~he column top to the

inflection point in the column, then h2 = h - hl o

As can be seen from the moment diagram given in

Figo 3-7b, the given frame may be considered as two

separate pinned-base frames with their respective

column heights equal to hI and h2~

20'. -v .

Determine the buckling load Per for the upper frame

using the theory just presentedo

30 Compute the buckling load PHr for the lower (in­

verted) frame using the methods discussed in

Art 0 20 10'10

4 ~v -" .o Compare rcr with Pcr , if they are not equal, a new

value of hI should be assumed and the process re-

peatedo . The correct buckling load is obtained when

the assumed hI value gives identical critical loads

for bothfrain~s'.,that is P~r = P~~o

.'Calculations using the procedure here described have

also shown that the effective length of colUmns in a frame

can be reduced appreciably when a small amount of rotational

restraint at the base was taken into accounto Future re-,

search in this field should include. the development of methods

by which the restraints offered by different types of foot-

ing can be evaluatedo

Page 55: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

304 Summary

The investigation pres~nted in this chapter may be

summarized as the followingg

(1) An exact solution.to elastic buckling of pinned­

base frames was described in detail and some typical results

were shown in Table 110

(2) The effect of axial force in the beam on the

buckling strength of the whole frame was examinedo It was

found that for practical frames this effect could be ignored

for computing "the sidesway critical loado

(3) Results of model frame tests were presented in

~able 1110 Satisfactory agreement between the theoretical

and phe experimentalelasticb~cklingloads was found 0

~43

Page 56: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-44

ELASTIC-PLASTIC ANALYSIS OF RIGID FRAMESWITH SIDESWAYPREVENTED

Since the portal frame uPder investigation is stati­

cally indeterminate, a me,thod of analysis fo~ par~ial~y ~~astic

redundant structures has to be developed before any attempt

can be made to solve the buckling problem in. the inelas~ic"

range .. This chapter will present-a method by means of which

the distribution of moments in a structure can be precisely

determined at any stage of loading. In order to obtain re-

suIts accurate enough for buck~ing analysis, it is necessaI"_y

in the present analysis to take into account not only the

inelastic action, but also the effects of axial force and

the deformations due to bending. The method can also be

used to obtain the following information:

(a) Load carrying capacity of braced frames (curve (a)

of Fig. 1-3b)

(2 ) D~flections of frame at any loading stage."

0) Rotation of the first plastic hinge.*

In the following sections the method will be described

specifically for analyzing the type of frame shown In Fig..

1-4, because 'the results of the analysis will be used in the

* This is often "called the hinge angle which is defined as"t~e angle through which a plastic hinge must rotate inorder to redistribute the moment and form a plasticmechanism..

..

Page 57: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-45

next chapter to invest~gate the stability of frames of this

,particular typeo The appro~ch is equally applicable to the

solution of many other types of frames 0(42)

401 Assumptions and Limitations

,

The following assumptions are made in developing the

present method:

(1) The material possesses the idealized elastic-fully

plastic stress-strain relationship in tension and

in compression as indicated,in Figo 4-lbo

(2) Plane cross sections 'remain plane after loadingo

(3) The shear torce present at any section of the fr~me

is small and its effect on yielding may be neglectedo

(4) Throughout the loading history there is no strain

reversal of material stressed beyond the elastic

...

(5) Only d,eformations (elastic or plastic) due to bend-

ing are consideredo

(6) The axial force in beam members 'is small as com~

pared to the thrust in the columns and itserfect

may be ignoredo

(1) All the members are prismatico

(2) No lateral (sidesway) displacements at column ends

are allowedo

Page 58: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

.(402)

-46

(3) No transverse load is applied to the columns except

at the endso

(4) No instability of any type as discussed in Arto 102

should occur prior to the ~oading stage at which

the analysis is performed 0

The first step in the deveaoped of the method is to ob~

tain separately the moment=end rot~tion (M-e) curves of beams

and columns o The procedure employed to construct these curves

will be described in Art0402 and 4030

The first step in the development of the method is to

obtain separately the moment-end rotation (M-e) curves of

beams and columns o The .procedure employed to const~ct thes.e

curves will be described in Arto 402 and 4030

402 Moment-Rotation Curve of Beam'. ':

Referring to the beam in Figo 4-2, ~y ignoring the

effect of H the bending moment at distance x from the center

is given byg2

M = M - wx. (401)c 2

in which Me. is the center momento A positive bending momen~

is one that causes compressive stresses·on the outside of the

frame 0 For convenience of numerical computation, Eqo (401)

may be non~d1mensionalizedby dividing both side~ by My, ~he

nominal yield moment of the section (My = cry 8)0 This re­

sults in the expression-l{L _~ _ W:(l

M~ - tv1~ 2. M1

Page 59: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

~47

z zM = ~_ wn rb (40)

M~ M~ 2M~

.From ~he definitions of ~ and Myp the above equation can be""1.

put in the form

-.M...- = ~ _ nZ~ dM

1M j ' 4 ~ A

in which d is the depth of the. section and A is the area o

By letting w== cryA and Ji. = >-.- the following. equation may bea. W

obtained from Eq~> (404)

.l:1.- _~ _n2~tv1~ _. M

14-

The parameter A. is cal·led the "loading ratio" 0 For' practical

frames its value is in the range from 00001 to 00010 With a

fixed value of I\. ; Eqo (405) determines the bending moment at

a distance nrb from the center of the spanp where a value of

Me/My is assignedo By knowing the moment at all the sec.tions·p

the deflect,ed shape of the beam may be established by a

numerical integration procedu.re similar to that developed

"" ,by Karman for investigating the strength of eccentrically'

loaded columns 0 (43) The proce4u.re is based on the moment­

curvature (M~.¢l relationship' of the beam section. In this

dissertation the M-¢ curve for a typical beam section p the

27WF94, is used for all the computationso It was determined

on the basis of assumptions (l)p {2)p (3) and (4) of Arto 401by the method described in Hefo (44) and is plotted in a

Page 60: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

, ,

-48

non-dimensional form in Fig. 4~3. A symmetrical residual

stress pattern as shown in Fig. 4=4 was included in determining

this curve. It has been shown in earlier ~ork that the M=¢

curves for most of the WF sections that are commonly used as

'beams are approximately the same. (45) Therefore the, results

obtained for that particular section, after being properly non-I

dimensionalized, can also be applied to other sections as well.

The complete procedure of integration is illustrated in

Appendix A for the follo~ing two cases~

(1) ?\ = 0.002 ' Me = 0.5RY(2) J.... = 0.002

Me= 1.0

My

In these computations it was necessary to consider a specific

value of Young's modulus ofE = 30 x 106 psi and a value of

yield stress cry = 33 x I(P psL These values, are typical for, .

ordinary structural ,steel (ASTM A7).

'The beam=deflection curve determined py the integration

process gives the slope e and the deflection y of points at

distance:a n~ from the center of the beam.* Figure 4=5 shows

the deflection curve in non~dlmensional form for the two cases

listed above. The slope e at all the points along the curve--

is plotted against the mo~~nt "M/My at the corresponding points

in Fig. 4=6 and against the position of the points in Fig. 4=7.

* The condition that the slope @ and the deflection yare bothequal to zero at the center section is assumed as the initialcondition in the integration.

Page 61: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-49

The integration,process terminates at a section where the

bending moment is equal to MP or where the slope of the de­

flected shape become's zero 9 ,whichever is reached firsto The

slope at the point where the moment is equal to ~ Cor

M/MP = 1014) becomes discontinuous 9 indicating the formation

of a plastic.hingeo

Similar. computations can be made for other values of

MC/My'o Thus for a specific value of 1\ (that is at a given

load W)9 a.set of beam~deflection curves may be obtainedo

The set of curves for A ~ 000Q2 are plotted in Figo 4-8 9 the

corresponding sets of moment-slope and coordinate-slope curves

are shown in Figo 4~9o The desired moment-rotation (M-e)

~urve of the beam shown in.Figo 4=2 may be determined graphlc~

ally in the following mannerg For a given beam member of

length L the ratio L/2rb is first determinedo A horizontal

line (a) is then drawn on the lower diagram of Figo 4-9»

starting from the particular L!2rb valueo The interse,ction of

the line with each of the curves giv~s the value of G for

which the equilibrium condition is satisfiedo A vertical line

is carried up from each interse-ction to the corresponding

moment-slope curve 0 This gives the moment at the end of the'

beam in equilibrium with ,the assumed center moment Meo By

joining a number of poin~s in the upper diagram 9 curve (b)

is ob.tainedo This curve jd.efin~s the relationship between the

moment and rotation at the end, when the beam is carrying the

assumed load w = I\wo

Page 62: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-50

Com~utationswere carried out for A values ranging from' .

OoOOI'to 0001 to cover mast of the practical caseso Results

obtained for X = 00002, 00004, 00006, 00008 are presented in

Figso 4-8 to 4-15 inclusive 0

403 Moment=Rotation Curve of ColUmn

The moment-rotation, curve of the column required in the

present method of analysis may be determined by the procedure

developed by Ojalvo in a' recent dissert~tiono(46) The loading

condition investigated is that shown in Figo 4-2 'for the,

column of 'a pinned-base portal frameo The procedure involves

the construction of a set' of column defIee,tion curves for

each assumed value of axial loado' A n~erical integration

process which is based on, the thrust-moment-curvature rela-, "

tionship of the column s~ctlon is n~ce,ssary to construct these

curves 0 In this section the thrust-moment-curvature~relation-

ships for the,8WF31 section as shown in Figo 4=16 are usedo

They weJ;"e taken from Refo (44lo A residual stress pattern

the same as that shown in Figo 4-4 was assumed in computing

these curves 0 'Stric:tly speaking the curves obtained can only

be applied to this partioular sectiono However it may be

shown that the results ,are eq~ally applicable to those sec­

tions which have their radius of gyration to depth ratio rid

close to that of the 8WF31 sectiono Fortunately this is the

case for most of the coltimn sectionso(46)

By using the method'outlined in Refo (46) the moment~

Page 63: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=51

rotation curves shown in Figso, 4=18 through 4=21 were deter=

mined for columns with slenderness. ratios hire ra~gipg from

30 to 130 subjected to axial loads equal to 0012~ 002 p 003 and

004 Pya The M=@ curves for the columns with no axial thrust

(Fig. 4=11) were obtaiQedby a, method similar to that des=,

cribed in Art. 4.2. The values of YoungUs modulus and yield

stress adopted in const~cting these curves were the same as,

those·used for the beam=q@f::t(;tction curves.

Afte,r all the M=@ curves are made available, the f'rame

shown in Fig. 4=2 may be analyzed by the graphical method

presented .in Art. 4040 The completed procedure will be illus­

trated with examples in Arto 4.5.

404 The Method of Analysis.... !

.To analyze a given fI;"tilJ.1ejFig. 1-4) subjected to certain

applied loads p the following in-forma'tion is usually knowng

span length Lp column height h~ member sizes of the beam and

the columns (their.I and,MP values)~ material properties (E

and cry) and the loadso From this inf'ormatlonthe ratios L/2rb

and A of the beam and h/rcand p/py of the column can be

determined. Using the results presented In Art. 402 and 4.3

two moment-rotation curves·,may. be obtained p one for the beam.

and one for the column. When ,they are plotted together as In

Figo 4=22 the point of' interse,ction gives the moment and rota­

tion at the knee for the equilibrium conf'iguration of' the .

structure 0 That is p of all the combinations of'momentand

Page 64: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-52'. .. . '\'1

rotation at B (or D) that are possible in the column.and·of

all the combinations that are possible for the beam, the cor­

rect one is one that is simultaneously cQmpatible.w~th the

beam and the column. Onc;e the moment M and rotation e at B

are determined, it is thus possible to determine th~ forces

in the rest of the frame by staticso. The deflection at the

center can be obtained immediately from the beam-deflection

curve corresponding to a particular value of Me which is in

equilibrium with the applied load .w and the corner moment MB. 'The procedure is indicated in Fig. 4-5.

By successively increasing the applied loads, a complete

load-deflection curve of, the s,tructure can be ·obtained. with

this methodo The load .carrying capacity of such frames'can

then be estimated precisely from this curve.

4.5 IllustratlveExamples

Example No.1 As a ~irst example consider the portal

frame shown'in Fig. 4-23~From a design handbook the follow­

ing properties for the sections 24WFl20 and 24WFIOO are taken

(assuming cry = 33 ksi) :

24 WF 120 A = 35029.' 2

d = 24·31 in. :";10.15 inoin. rb

Py = 1165 kips My= 822.ft"\"kips· ~= 926 ft~kips.,

24 WF 100 A 29.43 2 d 24.00 in. 100 0.8: in.= in. = r c '=

Py = 971 kips My= 684 ft-kips ~= 765 ft-kips

~

Page 65: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-53

The.parameters needed in the present analysis may be deter­

mined as follows~

For beam BD

L211>

. W=(J ~ = 33 35.29 = 47.9 kips =y 24.31 i~.

=50 x 12 = 29.62 x 10.15

For column AB (or DE)

JL =.42 x 12 =50r c 10.08

574.9 kips·ft.

The values of w that are to be considered are w = 2.30 i 3.45

and 4.60 kips per ft. The corresponding values of A are

0.004i 0.006 and 0.008 and the average compressive stresses

in the columns are pIA = 6.6i 9.9 and 13.2 ksi. From Figs.

4-11i 4-13 and 4=15 the moment=rotation curves(~D=@) for

the beam having L/2rb = 29.6 are obtained i they are plotted

as the dot-dashed li~es in Fig., 4=24. Shown as dashed lines

are the MBA = @ curves for the columns i they are taken from

Figs. 4=19 i 4=20 and 4=21 for hire = 50. The intersection of. . "

each pair of the curves gives the moment M and rota7:~on Q for

each value of w. The solid .. line in' the figure is ".the .complete

moment-rotation curve s.t:B (or. D). By knowing the value of

~i the moment at· the center Me can be easily determined by

equilibrium considerations. The deflection at the center ()

can then be obtained, directly from Figs. 4=lO'i 4-l2 i and 4-14.

The complete load-deflection curve of the frame is shown in

Page 66: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-54

Fig. 4-25. The maximum load t;h.at can be carried is seen to be

783 kips. The ultimate load based on simple plastic theory is

920 kips. There is a 15% difference between simple plastic

theory and the exact theory.

It may be noticed from Fig. 4-24 that the pair of M-9·

curves with A= 0.008 and ,P/P1

= 0.4 becomes tangent when the

applied load reaches the.maximum. Any further increase of ~

would produce no intersection (that is, no equilibrium con­

figuration can exist at this increased load), thus indicating"the capacity of the frame has been exhausted.

The example selected here has the special feature that

the formation' of the plastic hinge at the center and the

attainment of the maximum moment capacity of the columns

occur at the same timeo ,Therefore no rotation capacity is

required for either of the members 0 Th.is will b.ecome. evident. ,

after comparing Fig. 4-24 with Fig. 4-27 of the next example •.

Example No. 2 The,f'rame to be considered in the second

example is that shown in.Fig. 4-26. This example is chosen:.

mainly to show how the deflection and hinge angle for such

frames can be obtained by using the method developedo

Following the same procedure as' that, .used in Example Noo 1

moment-rotation (Men -.9)- curv~'s of the beam forA = 0.00315,

0000526 and 0000783 are determined by interpol~tion from

curves shown in Figso 4-11, 4-13 and 4-15 and other curves of

a similar nature which are omitted from this dissertation for

Page 67: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

The ~oment-rotation (MBA - Q) curves

-55

the sake of brevityo The axial forces in the columns corres­

ponding to these selected values of A are equal to 0012, 002

and 003 P respectively.y

of the column for these ~alues of axial force are then obtained

from Figs. 4-18, 4-19 and 4 ...20. Figure 4-27 shows the two

sets of curves and the. complete moment-rotation relationship

of the columns (solid line).

When the applied load reaches 3.34 kips per.ft.

(7\= 0.00730), the first plastic hinge f.orms at the' center of

the beamo At this instant ,the rotat'ionatthe column top is

designated as G' (0.0205 radian) and the moment as ME (00705

MY). The corresponding vertical deflection at the 'hinge is

01 . (6.53 ino) as shown in Figs. 4-28 and 4-29a. The beam is

still b~nt in a continuous· curve at this load. . Upon increas­

ing the load further, th~, ~o;J..U1UIl ends would rotate in the

manner indicated in Figo ~4-27 by the portion of the solid line

beyond the point Ho Correspon~ing to a small increment ot

rotation ~Q of the col~ the, increase pf center deflection

would be ·~o =t::"Q"L!2 and the ~ngle change at the plastic hinge

.would be 2 l:::.Qo

I~ maybe noted from Fig. 4-27 that the load reaches its.

maximum value when the ,moment carrying capacity of the columns..

is exhausted (on the flat portion of the M-Q'curve). To

. attain th~s maximum moment, the top of the columns has. to

rotate through an angle equal to 9"-9 1 (000145 radian). Thus

the hinge angle at the center is equal to 2(Q"-QI) and the

Page 68: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-56

deflection at the. ultimate load is llu = 6' + (@"~@') L/2 =

12~62 inch& The complete load~def1ection curve and the load~

hinge rotation curve of this example are plotted in Figo 4~29o

The total load that can be carried by t~e frame is 752 kipso,.

The maximum load based on simple plastic theory is 879 'kips~

•indicating a reduction of 1404% df the load carrying capacity

due to beam~co1umn actiono

406 Summary

This chapter presents a,precise method of analysis for

indeterminate frames loaded into the plastic rangeo The,

method has been illustrated by reference to pinned-base

symmetrical frameso The following summarizes this chapter~

(1) A method for obtaining moment-rotation curves of

beams with uniformly distributed load was developedo For the

convenience of actual computation a set of non-dimensional

nomographs was prepared by w~ich the M=@ relationship may be

determined "gr~phically fqr given values of A and L/2rb

(Figs 0, 4=9 ~. 4~11~ 4~13 and 4~15) 0

"c'2) Following the procedure described 'in Ref 0 (4.6) ~

moment~rotation curves for columns having slenderness ratios

between 30 to 130 and subjected to axial forces of 0012~ Oo2~

003 and 004Py were obtai"nedo (Figso 4~18 through 4-210')

(3) By utilizing the moment~rotation relationships for

beams and columns thus obtained~ a method of elastic-plastic

Page 69: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-57

analysis was establishedo The. chief importance of the method

is .that it provides information with which the inelastic frame

stability problem may be investigatedo This will be discussed

in the next chapter.-,.

(4) ~wo examples of portal frames were solved to indicate

the use of the nomographs and the method developed.

(5) As illustrated in these examples, this type of analy=

sis can be used 'to determine precisely 1) the load carrying

capacity of 'frames which are restrained against sidesway

buckling, 2) the complete load=deflection relationship, and

3) the required rotation capacity at the first plastic hingeo

Page 70: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

"".

-58

5. BUCKLING STRENGTH OF PARTIALLYYIELDED STEEL FRAMES

In Chapter 4 a method of analysis for partially yielded

steel frames .was described and its use was illustrated by

reference to portal frames which were assumed to be braced

against sidesway movement. The method can be applied to

determine exactly the distribution of moments and the varia­

tion of yielding along the members at any stage of loading.

By incorporating such a technique with the existing method of

bucklIng analysis described in Chapter 2, a procedure for

determining the inelastic buckling strength of frames is

developed in this chapter. Illustrations will be given to·

show how the procedure can be used to construct a complete

frame buckling curve. (See Fig. 1-3b)

5.1 Assumptions

Because the results obtained from the elastic-plastic

analysis of the previous chap~er--are used'd~rectly in the

deve16pment, the present procedure is also subject to the

assumptions ..and limitations stated iIi Art~; 4.1. In addition

to those, the following two assumptions are made~

(1) The axial force in the beam is small and its effect

on the bending stiffness may be ignored in the

buckling analysis. The justification of this

assumption was discussed in Chapter 3 in connection

with .the elastic buckling problem.

Page 71: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-59

(2) The frame deforms in a perfectly symmetrical shape

(no lateral movement should occur at the column tops)

up to the instant of bucklingo This implies that the

method of analysis presented in Chapter 4 can be used

to determine the yield conf~guratio~at any load level

below.that which causes sidesway instabilityo . (see

limitation 2 of Arto 401)

Analogous to the method commonly used in determining the

inelastic buckling strength of centrally-loaded columns~ the

procedure here developed also requires proper evaluation of

the reduction of bending. -stiffness (buckling constant) of all

the members due to yieldingo By using these reduced stiff­

nesses in the analysis~ the problem of inelastic buckling may

be treated ina manner similar to that for the elastic case o

Figure 5-1 shows schemat'1cally·the yield configuration of the

beam and the columns at a certain lo~do The methods of de­

terminingthe inelastic· stiffness of these members will be

outlined in the. next sectiono;

502 stiffness of Members·After Yielding

Since the modified moment distribution procedure de­

scribed in Ai-to 20101 is adopted in the next section to solve

the present problem~ it is necessary to obtain the following

buckling constants:

10 For beams - stiffness factor. Kb (assuming far end

fixed) and carry-over factor Cb

Page 72: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

,For columns - stiffness factor KQ

end hinged}

(assuming far

-60

50201 Determination of Stiffness and CarryoverFactors for the Beam

For a given set of loads wand P the 'bending moment a t­

B (or D) is first determined by the elastic-plastic analysis

of Chapter 40 Knowing the two end moments, the moment dia­

gram of the beam may be easil~ constructed by staticso

Figure 5-2a shows a typical example of such a diagramo

According to the elementary theory of strength of materials,

.the flexural behavior (in the elastic and plastic range) of

any section of the beam is governed cpmpletely by the moment­

curvature relationship of the,member. In the elastic range

the slope of the moment';'curvature diagram is constant.and

equal to the flexural rigidity of the section -Elb. When the

applied moment exceeds the elastic limit, the slope ,(or

rigidity) starts to decrease and approaches zero when the

moment is near the value Mpo The effective flexural rigidity

(EIb}eff of the section can thus be determined as the in­

stantaneous slope on the M-¢ diagram corresponding to the

applied moment. In this chapter the moment-curvature rela-

tionship of a typical beam section, )27WF94 as shown in Figo

4-3 is adoptedo The variation.ofEIb of this shape with the

applied moment is shown in Fig. 5=3. It may be seen from

Fig. 4-3 that the actual yield moment of this section is

only 70 percent of the nominal value of My. Therefore,

Page 73: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

~6l

yielding will occur at sections where the moment is greater

that 007 Myo This is indicated in Figo 5-2b for the beam

under considerationo

If no strain reversal is assumed to take place at the

moment of buckling, the stiffness of the beam can be deter­

mined by considering a beam with variable ~o For the

elastic part the flexural rigidity of the beam IsEIb while

for the plastic part the effective flexural rigidity is re-

duced as if the yielded portions of the beam were removedo

The effective flexural rigidity (Elb)eff is plotted as a'

'function of the applied moment tn Figo 5-30 Figo 5-4a shows

a symbolic representation of abeam corresponding to the

yield configurat.1on indicated·. in Figo 5-2bo. It is called

the ureduced beam"::tnthis dissertation. The stiffness of

this beam can be evaluated by: the method of column analogy

which is commonly used in indeterminate analysis 0(47) The

analogous column of the reduced beam is shown in Fig• .5-4bo

The width of the column at each section is inversely pro­

portional to the flexural rigidity.of that sectiono To

determine the bending moment at B induced by a unit rotation

at B, a unit load of one radian is ,then applied to the

analogous column at the end Bo The stiffness of the beam at

B is equivalent to the stress on the analogous column at

that point, that is

·1-'- f-

A

L L).-.­2 2

I

Page 74: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-62

in which A ,is the area of the analogous c'olumn and I is the

moment of inertial.about the centroidal axis ~Go Similarly

the moment at D is equal to~

IM =--D A

L L1'-'­2. 2

I-

The carry~over factor is simply the ratio of the moment atD

to that at B.5l or

If it is known that both ends of the beam would rotate'

through the same angle and in the same direction,at the

instant of sidesway.5l the stiffness ,may be computed'by using

the analogous colmnn shown in, Figo .5~5bo' The centroidal axis

G-G is now at the right end of the column and the area is

assumed to be infinityo The stiffness Kb" of the beam at the

left end is

Me.I

L L,._.-Z 1.

I

(505)

(506)

where I is the moment of inertia about axis ~Go ,The carry~

over factor is not needed in this caseo

5 0 2 0 2 Determination of Colmnn stiffness

. vThe stiffness factor Kc of a column with hinged ends

can be determined as the slope of, the moment-rotation curve

of a beam=column as shown in Fig. 5~60 Within the elastic

range the stiffness is given by

J{' = _' EIc.c:: C h

whe~e C = _1-ll-12.hUlt~h) andl1l..hf

Page 75: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-63

As the applied moment increases beyond the elastic limit, the

stiffness of the colunm decreases and approaches zero when

the moment is near the maximum valueo At this instant the

colunm has lost completely its resistance to any further in-

crease of bending momento However, if a moment of opposite

sense is applied, the colunm will behave elastically again

and. its stiffness is equal to that given by Eqo (505)0 This

is indicated as "unloading" in Figo 5-60

vThe variation of the stiffness K~ with the applied end

moment for colunms having slendern~ss ratios ranging from 30

to 130 sUbjected to axial forces of 0012, 002, 003, and 004

Py are shown in Figso 5-7 and 5-80 They are obtained by

measuring the slope on the moment-rotation curves presented

in Figso 4-18, 4-~9, 4-20, and 4-210 Knowing the axial force

-P and the end moment M of the colunm, as determined from the

graphic analysis of Arto 404, the column stiffness can be

obtained by interpolation from these curves 0

503 Method of Solution

From the results of the investigation presented in

Chapters 3 and 4 Jnd the concept of modified stiffness d~s­

cussed in Arto 502, a method for determin~g the inelastic

buckling strength of frames is developedo It is based on

the modified moment distribution procedure due to Winter,

et al (15), in which the stiffnesse.s are modified for the

effect of axial force present in the members at a given

loado In the present method the modification is made not

only for the effect of axial force.but also for yieldingo

Page 76: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-64

When the dimensions and loading condition of a frame are

specified, its buckling strength can be 'determin~ in the

following manner~

(1) Carry out a complete. elastic-plastic analysis of

the frame by assuming that no sidesway instability

occurs at all the stages 'of loading.

(2) Select a suitable ~oad level wI and determine the

moment at the column topso The values of Kb.' Cb,

and K~ can then be obtained by the procedure

des.cribed in Art. 5,02.

(3) Introduce an arbitrary lateral (sidesway) dis-'

placement of d (see,Flg. 2-2) and perform a moment

distribution computation for.the frame. As indi­

cated in Eqo (2.6) the applied fixed end moment canvbe taken to be proportional to the stiffness Kc of

each columno Using the end. moment values resulting

from the distribution process; the horizont~l shear

Q of each column may be determinedo The sum of

these shears'LQshould be positive if the selected

load wI is belo~ th~ critical valueo In other

words, a lateral force is required to displace the

frame laterally. In the momentd;stribution pro­

cedure, it is required .that assumption (1) of Arto

5.1 be va.lid, ioeo,. the effect of small axial load

on the bending stiffness of the beam is negligibleo

Therefore, thEE,stability of "tHe" frame. may. be' examined

by simply consideripg the loading system shown in

Page 77: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-65

Fig. 5-9c. As explained in Chapter 3~ the buckling

load thus determined will be very close to the

exact value. Although it is possible to obtain

more precise results~wlth the same procedure~ by

taking the horizontal force into account (Fig. 5-9b)

the work involved is formidable.

(4) Repeat' steps (2) and (J) for several values ofw_

that are in ,the range between the yield load and

the ultimate load. By plotting the total shear I:.Q

against the load w for each case~ a curve such as

that shown in Fig. 5-10 is obtained. The inter­

section of,,'this curve with the load axis giv~s the

critical load of the frame which will cause it to

sway without the application of lateral load.

In determining the stiffness of the,members~ the follow­

ing rules are adopted with re~ard to unloading of the yielded

portion:

(1) No strain reversal is assumed to take place for the

plaetic portion of the beam at the instant of side=

sway buckling.

(2) For the case when the first plastic hinge forms at

the center of the beam~ no unloading of the columns

is a'8sumed. This is the situation that usually

occurs for tall frames or frames with slender

columns.

Page 78: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=66

Both (1) and (2) are in agreement with the

generally = accepted concept of inelastic buckling

due to Engesser ard Shanleyo (10)

. (3) When no plastic hinge forms in the beam~ one of the

columns may be assumed to unloado This assumption

is adopted from Refo (36) and has been checked with

experiments 0

The procedure outlined above will be illustrated by two

complete examples in the following section in connection with

the development of a frame buckling curveo The dimensions of

the frames in these examples are such that the reduction of

the load carrying capacity due to sidesway buckling is

appreciable 0

504 Development of Frame Buckling Curves..

As shown in Figo 1=3b the carrying capacity of frames

having the same span length ~nd acted upon by the same load=

ing .system can be related to the slenderness ratio of the

columns by a frame buckling curveln a manner similar to that

for the case of centra1ly=10aded columnso The curve generally

consists of twosegments~ onecsegment defines the load=

slenderness ratio relatio:pship ..for buckling in the elastic

range~ the other segment for the inelastic rangeo This

section is concerned with theidetermination of such curves

by the method presented ip Chapter 3 and in Arto 5030 To

illustrate the complete procedure ~" a buckling curve for the

series of frames shown in Figo 5011 will beconstructedo

Page 79: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

The span length L of the frames is assumed to be 80r

(88 02 fto), where r is the radius of gyration of the 33WF130

shape 0 A value of 2 00 is assigned for the loading parameter'.ao

The cross sectional properties and the material constants

adopted are as follows:

,A = 38026 in0 2 d = 33010 ino I = 669900 in0 4 .S = 40408 in0 3

r = 13023 ino E = 30xl03 ksi cry = '33 ksi

Py = 1263 kips My = 1113 ft - kips Mp = 1282 ft:- kips

~ = 40 W= 45707 ~~:s wL = 400369 kips

The ultimate:load determined by the simple plastic theory

(correspondIng to a beam mechanism) is Pu = 349 kips (wL =

232 kips) 0

In the elastic range, the buckling load of the frames can

be determined approximately by the expression:

_ nlEA

Pet - l { +y (50 7)

where k is the effective length factor of columns o The value

of Ie can be taken from Figo 5~120 More accurate results of

the buckling lo~d may be determined by the solution presented

in Chapter 30 By assuming different values of column'height,

a set of }lcr can be obtained o It is plotted non=dimensionally

as the dot=dashed line (line C) in Figo 5=190 This curve is

valid only for frames with slenderness ratios greater than

that corresponding to point Y shown on the curve 0 At that

point the elastic buckling load is equal to the load which

causes initial yielding at the most highly stressed sectiono

Page 80: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=68

For frames having column slend~rness ratios less than that

value~ inelastic buckling will generally 'govern their load­

carrying capacity.

To obtain the buckling curve applicable in the inelastic

range ("curve (b) in Fig. 5-19) ~ .it is necessary to determine

the strength of several frames with various slenderness

ratios. In this example two frames having h = 60 and 80 are, r

chosen for illustration. The method of analysis will be ex=

plained in detail for these two cases.

First Case = Frame having h =60 r

The procedure presented in Art. 5.3 is applied here to

compute the inelastic buckling strength of this frame. The

complete analysis consists of the following steps~

step 1 Perform an elastic=plastic analysis of the frame~

using the method described in Chapter 4~ to determine~

I. The exact load-carrying capacity of the frame if

aidesway buckling is prevented. The ultimate load

thus obtained gives one point on curve (a) of Fig.

2. The bending moment'at the column tops for any

applied loads.

The resulting moment=rotation curve from this analysis

is shown in Fig. 5=13.

step 2 Select a trial load Al :,0.0048 (wL = 194 kips).

The intersection of the corresponding moment=rotation curve

Page 81: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=69

of the beam with that of the column is indicated as 01 in

Figo 5~~30 The moment MB at the column top for this load is

equal to< 00830 ~yo The axial force in the columns is .

p = 2 i\l wL = 29007 kips2

and therefore

By interpolation from Figs 0 5=7 and .5=.$ the stiffness of

columns subjected to this combination of bending moment and

axial force are determiued to be.

K~ = 2400 My (loading)

K~= 48.91 My (unloading)

The computations involved in the evaluation of the

stiffness facto~ Kb and carry=over Cb of the beam are con=

tained in Appendix Bo The values of these factors thus

obtained areg·

Kb = 46063 My

Cb =,007125

Since there is no plastic hinge forming at the center

section of the beam for this trial load (see Appendix B)~

then according to rule (3) of Arto 503 one of the columns

may be as~umed to unload in the buckling analysis~ If the

frame is assumed to sway to the right~ the left coiumn will

be the unloading column.

Page 82: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

..",=70

step 3 A rixed end moment due to a lateral displacement at

~he column tops or MFL :::; 100 rt=kips is arbitrarily assigned

to act on the unloading column. Then according to Eqo (20b)

the rixed end moment or the loading column is equal to

MFR =~ x 100 = 4900 rt-kipso These moments are dls=48.91 i

tributed and balanced as shown on Figo 5=14.

The resulting shear rorce may be determined by Eqo (2'09)!

as rollowsg

Ql = +(M~ - ~~ MFL·h]

= _I (53 10- ZgO.1 ·\oo·~o· \.1025 ,J= -hi (11.191h . 48.91' 11\3 J

t (M~ - ~~ MF~' h]_I (43.l5 ~ 290.1 . 49·60 '\'1025 J = l-h [Tg9)h t 24.0' II , '3

~Q = t (25.78) '> 0

Consequently, by the criterion or Eqo (2 0 10), the rrame is

stable at this selected load levelo

step 4 Select a sec.ond trail load with A.2 = 000050 (wL =202 kips) and repeat steps 2 and 30 The total resulting

snear rorce ror this trail load is LQ = ~ ( 0061J 0 This

indicates that the selected load is very close to the actual

buckling load. In Fig. 5c d5 the total shear L Q is plotted

against the loading ratio A ror these two trials, the

critical· load is determined as tne intersection or this

Page 83: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-71

curve with the ~'-axis, that is Acr = 0.00501. The tot~l

load corresponding to this value of A.. is Per = '~ 0 A cr wL =

303 kips, therefore, the ratio Per =1Qi = 0.8690 ThisPu 349

furnishes one point on the inelastic buckling curve.

Second Case - Frame having h = 80 r

Step 1 Carry out a complete analysis the same as that for,

the first case. The resulting moment-rotation of the column

is shown in Fig. 5-160

Step 2 As a first try, a load of Al = 000048 (wL = 194 kips)

is selected. The point of intersection of the moment-rotation

curve of the beam for this value of ~ with that of the column

is marked as 01 in the figure 0 The moment MB corresponding

to point 01 is 0.770 My. The axial force in the column 'is

p = ~ Al -wL = 290.7 kips

and the ratio p= 290.7 = 0.230

Py 1263

For this combination of axial force and end moment, the

column stiffness is

20096 My (loading)

It may be seen from Figo 5-16 that the bending mo~ent

at the center is equal to Mp for this trial load. Then

according to rule (2) of Art. 5.3, no unloading should be

assumed for either of the columns. Therefore, the beam will

, be bent in an anti-symmetrical form at the instant of buck-

ling. The stiffness factor of the beam may be determined by

Page 84: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-72

the simplified procedure shown in Fig. 5-5. Numerical com­

putations involved are similar to that of the first case o,

They are also included in Appendix Bo The value thus

determined for stiffness with anti-symmetrical bending is

Kb" = 70.87 My.

step 3 Introduce a fixed end moment due to lateral dis-

placement' of MF = 100 ft-kips for each column. These moments

can,be distributed and balance-d in one cycle as indicated on

Figo 5-17.. The resulting shear force is:

~ Q = +(Wl'l - ~~ M~Lo h]...L [11018- 290.1 '\00. 80'101025] ='--'[£'538)

h 20.9(,'1113 h . ~

T4is indicates that the trial ~oad Al is higher than the

critical load and that a smaller A value should be assumed

for the next try.

step4 Use A2 = 0.0045 as the second trial load and repeat

steps 2 and 3. The resulting shear force at the column tops

1 Jis [Q = - h (4. 90 , indicating that' the selected A is slightly

higher than the correct value. Using these results, the

inelastic buckling,load of the frame can be determined

graphically as shown in Fig. 5-18. The value of A cr is

equal to 0.00447_and the total load P = ~ ". A cr .' wL = 271 '

kips, the ratio Pcr = 271 = 00776. This gives another pointPu 349

on the inelastic buckling curve.

Page 85: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-73

Similar analyses may be perfonmed for frames with

'different values of h. These analyses will result a seriesr

of points, each of which gives the.buckling load ofa.

particular frame. .By passing a curve from Y through~hese

points, an inelastic buckling curve is obtained (labelied as

curve (b) in Fig. 5~19). At point T this curve becomes

tangent to curve (a) which defines the strength of frames if

they are restrained 'against sidesway buckling. For any

frame with a slenderness ratio less than that corresponding

to point T, its load carrying capacity will not be reduced

significantly by lateral instability. Therefore within this

region the problem of frame stability may be safely ignored

and the design can be based on the plastic strength.

5.5 Experimental Work

Two series of model steel frames having column slender­

·ness ratios in the range between 40 and 100 were tested to

determine their buckling strength in the inelastic range.

The cross sectional shape and the dimensions of these frames

are the same as those shown in.Fig. 3-3. .The following values

of column height were adopted: h = 15, 20, 25, 30, and 35

inches. The corresponding slenderness ratios were ~, = 4107,

55.6, 6904, 8303, and 9702.

The frames were loaded in a manner similar to that shown

in Figso 3-3 and 3-4. In the first series of tests (Series A)

Page 86: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=74

a loading ratio a = 7.5 was selected. For this selected value

of a, the total axial thrust in the columns at the computed

ultimate load (simple plastic theory) is approximately equal

to 0.4 Pyo The a value for the frames of the second series

(Series B) was kept at 3.35. The corresponding axial force

in the columns is Oo2Py o In this section, the results ob­

tained from Series B tests are briefly discussed o The com­

plete iest ~esults will be presented in a forthcoming report 0 (40)

The experimentally observed loads of frames of Series B

are plotted nondimensionally against the column slenderness

ratios in Fig. 5-20. The curves in this figure can be com­

pared with those shown in Fig. 5-19 which were obtained

theoretically for frames of WF section and with an a value

equal to 2.0. OWing to the difference in cross sectional

shape of members (different moment-curvature relationships)

and loading conditions, the comparison between these two

figures can be made' only on quali tative basis 0 It may be

seen' from Fig. 5-20 that the" general trend of the experi­

mental inelastic. frame buckling curve (curve (b) ) is similar

to that of the theoretical buckling curve and that the re­

duction of load-carrying capacity due to sidesway buckling

is about the same order of magnitude as that obtained by

theoretical computations 0

In order to check more exactly the inelastic buckling

theory developed in this chapter, experiments on larger

Page 87: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

~75

scale frames of WF s~ction are being planned.

5.6 Buckling of -,Multi':"stor'Y Frames

The theoretical solution contained in this chapter can

be used to check approximately the inelastic stability of

multi-story frames. The procedure of applying this theory

may be ill~strated with-reference to the single-bay three­

story frame -shown in Fig. 5-2la. The frame is loaded by a

uniform load w on each floor. Due to the variation of axial

thrust in each story, the columns may be made of different

member sizes.

In gen~ral~ threepo.ssible modes of sidesway buckling

may be expected" for this frame, these are il1ustr~ted in

Fig. 5-21b. Corresponding to each buckling mode a fictitious

single-story portal frame may be obtained. Figur~ 5-21c

shows three such frames arid their loading conditions for the

assumed failure modes. It may be seen in this figure that­

in obtaining these frames the following simplifications have

been made:

1) All the bases are assumed to be pinned.

2) The effect of the bending moments transmitted

from the upper and the lower frames is neglected.

The above simplifications may be shown to yield results on

the safe side. The buckling loads of the frame.s can then

be computed by the method developed in this chapter. The

lowest value among the computed buckling loads of these

Page 88: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

frames determines approximately the buckling strength of .. the

multi-story frameo

r

At the present time there is no experimental ev~dence to

justify this approximate methodo ~ch more extensiv~ in-

vestigations on this problem are needed before any conclusive

answer can be obtainedo

507 Summary

The investigations presented in this chapter can be

summarized as the following~

(1) A method for solving the problem of sidesway buck~

ling of portal frames in the inelastic range was developedo

It is a modification of the moment distribution procedure

commonly used, for analyzing the stability ofela,stic frame,so

In the present method~ all the required stiffness and ca.rry­

over constants are modified for the combined effects of axial

force and yieldingoThese effects can be precisely evaluated

with the aid of the elastic- plastic analysis described in

Chapter 40

(2) To illustrate the procedure presented~ two numerical

examples were solved in detail and the results were shown in

the form of a frame buckling curve o (Figo 5-19)

(3) A qualitative comparison between the frame buckling

curve computed by the theoretical procedure and tp.at obtained.

from model frame tests was describedo Although the cross

sectional shape and the loading condition of the test frames

Page 89: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

\

-77

diff.er from those assumed in the theoretical computatfons j)

the general trend of those buckling curves was found to be

similaro (Figo 5-20)

(4) Using the results of theoretical solutionsj) ,n ap­

proximate method for checking the stability of multi-story

frames was developedo (Figo 5-21)

Page 90: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

". _."

6 0 APPLICATIONS TO DESIGN

It may be noticed from the buckling curve obtained in

Arto 504 that for most practical frames overall instability

generally ta~es place in the plastic range and that the re~

! .;""

duction of load-carrying capacity due to this type of failure·

is quite significant for frames with columns of intermediate

slenderness ratioso Thepefore, for certain frames, it may be

necessary to base the de~1gn ~n their inelastic buckling

str~ngth. However, the determination of this strength by.

the procedure proposed in this dissertation requires lengthy

computations and tedious.graphic technique 0 It would be im­

possible to use such procedure in actual practiceo There

exists a need to develop design charts for practical elppli- .

cations. In this chapter a method fo~ constructing such

charts is describedo Some typical samples will be .presented

and their use will be illustrated by solving two design

problems 0

601 Development of' Design Charts

In designing any portal frame of the type considered

in this investigation the following information is usually

given:

10' Loads - wand P (also P and the ~altie of a)

20 Dimensions - Span length L and height h ..

It is required that the designer selects a member which

Page 91: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-79

will be sufficient to carry the given loadsoc For single

sto~y building frames, it is usually economical,touse the~...

same membE;tr size for both beam ,and column 0 This is ass,umed

to be the case in developing the design charts presented in

this'articleo

In the following discussion, It is assumed tha,t the de­

sign will be based on the.limiti'ng strength (plastic strength... ' '.~" ','

or buckling strength) and that the frames are allowed to sway.- .

only in the plane of the applied loads 0,

To develop charts for design use, all the above-mentioned

factors are to be taken into considerationo After investiga=

ting several possible ways of grouping these fa'ctors, it was,

found convenient to fix the following two for each chart~

10 Loading parameter a (Leo, fixing theratl0 between

the uniform load wand the concentrated load P)" ,,/

20 Span length L

Therefore the variables involved in each chart were reduced

to~

10 Column height h (or the ratio,~)

20 Member size

30 Load-carrying capacity - (or the ~)Pcr ratiol>u

Then corresponding to each specified member size, a fraDie'

~_cr hbuckling curve (_ vSoP L-u

similar to that shown in

Page 92: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-80

Figo 5~19 may be computedo It will be shown, i'n the sUbsequent

paragraphs that the shape of the non-dimensional buckling.. ,

curves depends mainly on the radius of gYration r of the

members 0 Therefore for.each,chart it is only necessary to

compute several of these curves for a few WFshapes having

various r valueso

At firs·t,it will be shown that for frames which have

(1) same span length L and height h,

(2) same loading condition (constant a), and

(3) uniform member size for both beam anq column,

, ~the ra~io pcr for buckling in the elastic range is dependent

uonly on the radius of gyration of the memberso According to

Eqo (501) the elastic buckling load is given by

21( E I

Pet =. l i h)2 (601)

It may be seen from Figo'$=12 that for.all the frames which

satisfy the above conditions the value ofk is independent of

'the member size usedo ,On. the basis of simple plastic theory

Pu can be computed by the equation

- ~MP = (ITo!)~u L

pTherefore the ratio -.£!: is

s'"

Pu"

-

~r

Page 93: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

~81

l IT\ L ] I d.= (ih)z ~(I"'o() l~ f

[lT~ELI :)., I ' ( 6•3)

= (-khy I~ll+.() ~.. f.£ . r

where l' is the shape factor Qf the. member. The quantity in

the bracket is a constant for all these frames. The values

of l' • ~ for all the connnonly used. beam sections are in the

range from 0.460 to 0.474 and those for all the column sec~

tions are between 0.470 and 0.491.. These variations may be

considered to

computations.

on r alone.

be small (amount to 3 to 4%) in practical

PerHence, the ratio can be assumed to dependPti

Next, the same ratio wl~l be examined for buckling in

the inelastic range. As illustrated in Art.• 5.4, the f'ollow~

ing two cases are usually encountered in determining the

inelastic buckling strength~

(1) Buckling takes place when the first plastic hinge

has formed at the. center of the beam.

(2) Buckling takes place bef'ore the formation of the

plastic hinge in the beam.

For the first case, the buckling load can be expressed by

the equation

Page 94: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-82

- wGrLPc.r ",;" II +cJ..) -2-

= (lr"d 4( ~p+ M~)L

in which Mo is the moment applied to the column tops at the

instant of sidesway buckling" This moment can be expressed

as a·: fraction of ..~ by the relation Mo = )J Mp ' where.? is

alway-s less than unitYo Sub&titut1on of this relation into

Eqo (6.4) results in the following equation for the buckling

load-p. _ II +J.) 4.( It)J ) tv1 p

c.r \..:L

The ratio Pcr can be obtained by dividing.Eq. (605) byPu

Eqo (602), thus

(605)

..

The value of p will be known'!p,the moment Mo is determinedo

The determination of this moment requires tGt elastic-plastic

analysis described in ChapteF4 and the inelastic buckling

analysis presented in Chapter 5. The parameters involved in

the first type of analysis are the following~

For .the beam

For the column

L and Acrwcr--

2r ~

h andPcr

r: Py

Page 95: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

For frames having the same dimensions, the values of .L and2r

.~ are inversely proportional to r and the values of Acr and

p_cr can be shown to be related to rasfollows:Pu

'.

By definition, w= cry ~ and wcr

is then

A<r;TI ~ (1+ fI ) . f . r Z

LZ

Pcrand the ratio isPy

= 8(1 +;U)

L2,. the ratio /\cr

-~r

P~(J; . A

.~

4 (I+.,() (I.,. fA) ( of· -£- ). r (6.8)

As explained before the product f x ~ may~be considered as

constant for most WF shapes. After 'performing numerous com-·

putations of this type it is found that the values ofAcr-. Pcrand --- do not change signi~icantlywithp and f. From the

Py

.above discussions it may be'concluded that the res·ults of the

elastic=plastic analysis depend mainly on the value of ro

Since these results are used directly to determine the stiff-

ness of members, the buckling strength determined by~sing

these stiffnesses is also dependent mainly on r. Therefore,

pthe ratio ~ may be considered approximately asa function

Pu

Page 96: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=84

of r onlyo

By extending the above explanation further a .similar

conclusion may be arrived at.for tpe case that buckling occurs

before the fo~ation of a plastic hinge in the beam. But in

Per 1this case the ratio does not vary appreclablywith memberP'u

sizes 0 (See· .the curves shown on Figs 0 6-1 and 6=2 in the

range of h'= 004 to Oo8o)L

From all the foregoing discussions, it is possible to

conclude that the shape of the frame buckling curves (for

fixed values of L and a) is dependent ma~n~y on the radius..

of gyrationr of the members and that design.charts may be

developed by computing several frame buckling curves for some. ,

typical WF shapes with various r valueso

In Figso 6-1 and.6~2 two sample design charts are pre=

sentedo Thes·e charts are computed for the following two caseS8

( I)

(2)

L = 70 l'to

L = 90 l'to

The structural shapes selected for calculating the

buckling curves are 36 WF 26Q.(r=. 15000 ino}, 33 WF 130

(r = 13 0 23 ino), 27 WF 102 (r = 10096 in.), 21 WF 73. .

(r = 8064 ino) and 18 t 5407 (r = 7007 ino)o It is expected

that curves···for members having other radii of gyration may be

fitted in by interpolation between these curves 0 The use of

these charts will be i~tustrated by examples in Arto602:0

Page 97: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=85

Example Noo 1 As a first example consider a portal

frame loaded as shown in Figo 6=30 The following is given~

Span length

Column height

Working loads ~

L _. 70 fto

h = 56 fto

'h = 008-L"w = 1030 kips/fto

p = 91 kips,

a = 200, :

. According to current design :r,eccmmendations p (2 p 39) a load

factor of 1085 is adoptedo The design ultimate loads are

then equal to

Wu = 20405 kips/fto and

Pu = 168035 kips'... , " .._,,':

On the basis of simple plastic theory ~he required pl~stic

. moment is

M ==p

=

Wy L2

16 .. 2

2.405 ~ 1016

136.5

From the plastic moment table of,.R~fo (39) the most economical

section supplying this value is the 27 WF 94 having r =: 10087

incho To check this member size for inelastic frame buckling,

the design chart of.Figo 6=1 'can be usedo Using the curve

Page 98: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=86

Per iseorresponqing to r = 10096 in the .chart g the r~t~o

Pu

determined to be 008550 . The inelastic buckling strength of

the frame then equals

P = 00855 Pucr

.,/

be done in the' followingP'

of =cr may be applied toPu

manner~ Assume that the same value

This indicates that the se1eoted'member 27 WF 94 is not

sufficient to carry the design ultimate loado Ther~foreg the

design has to be revisedo This can

the frame of the new member size and let Pe~" be equal to the'

design ultimate loado Then the required Mp' value is

M Wu L2• I

P = If, 0.855

861 4· tt :':"K.ips

The most economicai section is the 30 WF 108 .supplying 950

ft=kipso This shape should be adopted for the final designo

Example Noo 2 The frame of the second example is 'shown

in Figo 6=40 The following information is given~

Span length :L = ,·90 fto

h = 36 fto'hL = 004

Page 99: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

Using the same load factor as that in Example No. 1 the

design ultimate loads are

w =u

p =·u

2.22 kips/ft.

199 0 8 kips

The required plastic moment value is

2.22" 902

Mp = 1123 f+-Kip5If,

The 33 WF 130 may be selefted.with M = 1282 ft-kipsp

and r = 1).23 in. For the chosen member size g the ratio-Pcr is determined to be 0.92 'from the curve r = 13.23 onPuFig. 6-2. Therefore

W cr = o. 92 gx 128 2 .90 2

= 2.~3 Kir/n > 2.22 \('p'/ft.

This show~ that the 33 WF 130 s&ct1on is sufficient.

6.3 Comparison of Theoretical Results with the AlSCDesign Rule

Since a great de~l of computations were carried out for

obtaining the design charts pre'sented in Figs. 6-1 and 6-2 g

it. is now possible to check the AlSC design rule dis'cussed in

Art. 2.2.2 with the th~oretical results. The AlSC design

Page 100: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-88

rule for frame stability requires that the axial load and the: .,..;~

slenderness ratio of columns ishould be so.propo:rtloned that

P I h2-- +.- - ~ 1.0p) lO r,-

Within this limit the reducti.on.of load carrying capacity due

to sideswa,-y buckling would be ·negl~gibly smallo In" Figo 6-5

results obtained from theoretical solutions are plotted and

compared with the a~ove ruleo.The.computed inelastic buckling

loads are 'expressed as percent ()f the ultimate load 'determined- j

by the exact analysis pr~~~nted in. Chapter 4, in wh~ch t~e

reduction of strength due to col~ instability alone iscon­

sidered (that is the frames are assumed to be preven,ted from

sidesway)0 The P values used are the axial forces in the coi..." .:J

umns at this computed ultimate loado

It may be sQ.~om Fig. '6-5 that, .within the range of-~ and ~ that has been cover.~d by, the theoretical computa-y c

tions, the design rule is som,what', conservative and that the1:

buckling strength of some frames having the combination of

P and!L considerably outside of. the safe regipn definedP r .y c· .by the rule can be more than ,95% of the plastic strength. On

the basis of these observations. the following rule is tenta- .

tively proposed~

85

\

1.:'0 'J ' .. (6010)

It is plotted as the dotted line on Figo 6-5. The validity

Page 101: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

...... j

of this rule for frames sUbjected to high axial forces and

for frames having high column slenderness ratios should be

examined by more extensive computations 0

=89

Page 102: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=90

70 SUMMARY AND CONCLUSIONS

This di~sertation deals with t~e stability of steel

portal frames which are acted upon by a uniform load on the

beam and concentrated load on the columns 0 The frames are

assumed to fail by sidesway b~ckling in the plane of the

applied loadso The following problems are included in the

investigationg.

(1) Buckling of frames in the elastic rangeo

(2) Method of analysis for structures with non=linear

behavior (due to axial forces g deformations and

yielding) 0

(3) Buckling of frames in the inelastic rangeo

The natu~e of these problems has been examined in detail

in Chapter 10 A survey of the current approaches was pre=

sented in Chapter 20

The contributions contained in· this dissertation may be

summarized as followsg

(1) An exact solution for determining the elastic buck=

ling strength of pinned=baseframes is obtainedo Using this

solution g the effect of bending moment present in each

member at the instant' of buckling is clarifiedo

(2) Results of the elastic solution are compared with

those obs~rved from buckling testso Satisfactory correlation

Page 103: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-91

I

between theory and experiments has been obtained. (Table III)

(3) A procedure by which the elastic buckling strength

of frames with partial base fixity may be determined is pre­

sented. Preli~nary calculations show that the bucklin~

strength can be increased appreciably due to base restraint.

(4) A precise method of analyzing statically indeter­

minate frames in the plastic range is established. It takes

into consideration such effects as axial thrusts~ residual

stresses~ yielding and deformations due to bending. The

method involves the determination of a moment and rotation

at a joint by the intersection of moment-end rotation curves

for beams and columns.

(5) A numerical integration procedure for obtaining the

moment versus end rotation cUrves of beams is developed. The

results of computation are presented in the form of a nomo-

graph which allows the rapid determination of these curves.

(6) Using the above method of analysis~ it is possible

to determine .exactly a) the load-carryiing capacity of frames

which are restrained from sidesway, b) the complete load­

deformation relationship and c) the required rotation·capacity

at the first plastic hinge.

(7) The problem of sidesway buckling of partially

yielded frames is solved. The method of solution consists of

the following steps~

Page 104: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-92

10 Perform a precise analysis. of the frame under' in­

vestigation for all stages of loading~ as~uming

that no sidesway instability would take pla~e.

2•. Select a suitable t~ial load and determine the

stiffness of members by the method outlined in

Chapter 5.

3•. Introduce an arbitrary lateral displacement at the

column tops and perform a moment distribution

computation' for the frame. Using the end moment

resulting from the distribution process, the hori-'

zontal shear of each column may be determined.

If the sum of the shears is positive .then the trial

load is less than the buckling load.

4. Select another trial loa~ which is higher than the

first one an~ repeat steps 2 and 3. The critical, , ,

condition will be reached when the sum of the re­

sulting shears becomes zero. The load at which

this occurs determines the inelastic buckling

strengt~ of the frame.

(8), To facilitate the solution of actual pr.oblems, a

method whereby design charts may be derived is presented.

Sample charts are obtained and their use is illustrated.

The f9110wing conclusions can be drawn from the results

of the investigation presented in the preceeding. chap,ters~

Page 105: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-93

(1) . For unbraced portal frames with slender columns,

the elastic buckling strength should be used as the basis of

design. Methods fo~ predicting such strength have been de­

veloped in this dissertation.

(2) In the elastic ·range~ the reduction of buckling

strength due to the presence of primary bending moment is

small and ~an be neglected in practical computations.

(3) The ultimate(plastic)strength of frames which are

restrained a~ainst sldesway can be precisely determined.

In the determination of this strength the moment-rotation

relationship of the columns has to be considered. (Figo 4-24)

(4) For most practical frames s'idesway buckling takes

pla?e in the inelastic range o This will usually cause a re­

duction of the load-carrying capacity of the frames 0

(5) The prediction of this type of failure can be made

by the method .presented in this investigationo Due to the

complexity of the method, it is.recommended in practice to

proportion the members such t~at the possibility of overall

may be eliminated. This can be achieved by selecting member

sizes to mee~ the proposed design rule.

The theoretical and experimental studies presented in

this dissertation constitute the first ~teps in obtaining

solutions to the buckling of multi-story frames o Further

problems that need be, studied are:

Page 106: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=94

(I) Experimental veri£ication o£ the inelastic buckling

theory developed in this dissertation.

(2) Inelastic buckling o£ portal £rames with partial

base £ixity.

(3) U).·timatestrength o£ £rames subjected to combined

vertical and horizontal loads.

(4) Inelastic instability o£ multi-story building £rames o

Page 107: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-95

80 NOMENCLATURE

A ,=, Cross sectional area

C = Carry-over factor

E = Young's modulus of elasticity

Est = Strain-hardening modulus

Et = Tangent modulus

EI = Flexural rigidity

(EI)eff = Effective flexural rigidity

F = Loading term

H = Horizontal reactIon ,at the base

- Axial force in .;beam

I = Moment of inertia

= Moment of inertia ,of beam section about its strongaxis

= Moment of inertia of oolumn'section about itsstrong axis

= Effective moment of inertia

= Moment of inertia of ficticious base restraint beam

Ie

Is

K = Stiffness factor"

Kb = Stiffness factor of beam with far end fixedn

Kb = Stiffness factor of beam having equal end rotations,

Kc = Stiffness factor of column with far end hinged

L = Span length

L/2~

= Length of member

= A non-dimensional pa~ameter used in determiningmoment-rotation curves of beams

M = Moment

Page 108: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-96

M' = Column moment resulting from moment distributionprocedure

MB = Moment at column top

MC = Moment at the center of beam

MF - Fixed end moment

MQ = Moment at column top at the instant of buckling

Mp = Full plastic moment = OyZ

My = Nominal yield moment = 0yS

P = Concentrated load applied at column top

= Sidesway deflection

= Half of the total ultimate loads based on simpleplastic theory

= Half of the total .applied loads corresponding toinitial.yielding

= Axial yield load = OyA·

= Shear force at column top

=."Applied beam load (See Figo· 2-4 )

= Elastic buckling load of frames shown in Fig~ 2-4

= Elastic buckling load of frames shown in Fig. 2-5

= Elastic bucklir;lg load of upper frame (See Figo 3=5b~

= Elastic buckling load of lower frame (See Figo 3=5b)

= Half of the total applied loads = (l+d)wL/2

= Half of the total buckling loads (elastic or inelastic)

= Half of· the total buckling loads observed fromexperiments

Fr.cr,\ Per

nPer

)-= : p=Per=Pexp

=Pu

=Py

Py

Q

R

= Radius of curvature

S = Section modulus

v = Vertical reaction

WF = Inelastic buckling load of frame - (SeeEqo (201n )

Page 109: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

It,:,.

=97

Wu "

Z

b

c

d

f

h

n

= Ultimate load of frame by simple plastic theory(SeeEq. (2.17) )

= Elastic buckling load of frame (SeeEq. (2.17) )

-Plastic modulus

= Flange width

= Non-dimensional coefficient in four=moment equationdefined by Eq. ,,(2.2)

= Derivative of c with respect to p

= Depth of section

= Horizontal displacement of column top

= 'Shape factor = z/s

= Height of frame

= Height of ficticious frames (See Fig. 3-7)

= Slenderness ratio of column

= /P/EI= Inte'gers defining the position of points OIl a beam

deflection curve.

= Int~gers defining the position of the beam loads(See Fig. 2-4) r ., ,.~

p = Axial force in member

r = Radius of gyration

= Radius of gyration of the beam section about itsstrong axis

= Radius of gyration of the column section aQout itsstrong axis

s = Non=dimensional coefficient in four=moment equationdefined by Eq. (2.2)

Sl = Derivative of s with respect to p

t = Thickness of flange

w = Intensity of uniformly distributed load

= Thickness of web

Page 110: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

Wcr

=w

x

y

=98

= Intensity of uniformly distributed load at the limit- of stability

= OyA/d

= Coordinate parallel to beam

= Coordinate perpendicular to beam

= Deflection of 'beam-'

..

a = Loading parameter which relates the concentratedload P to the uniformly distributed load w

=fJA = Increment of a "quantity

= Determinant

a =, Deflection at a point

a' = Deflection at the c~n~~r section of beam when thefirst plastic hinge forms at this section

= Deflection at ultimate load

e = Strain

est = Strain at the onset of strain~hardening

ey = Strain corresponding to initial yield point stress

e = Slope of beam deflection curve

= End rotation of member

= Rotation of column top at the instant when the·first~lastic.hinge forms at the center sectionof beam -

= Rotation of Qolumn top when the ultimate load isreached

fo

= Loading ratio = w/w

= -He.tio of, colunm end moment at the ins,tant of buck=ling to plastic-moment

= Rotation of member

= Stress

Page 111: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

CJRC = Maximum residual compressive stress

CJRT = Maximum residual tensile stress

CJy = Yield stress

~ = Curvatur~

~y = 2CJy = Curvature corresponding to initial yieldEd under pure moment

=99

Page 112: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

90 TAB LES AND FIGURES

=100

Page 113: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

Table I . ELASTIC BUCKLING LOAD OF FRAMES SHOWN IN FIG. 2-4

L = h L = 3h

Loading Position Disc repanc ie s Loading Position Discrepancies(1) (2) ( 3) ( 4) ( 5) '(6 ) .( 7) (8 ) ( 9) (10)

n =2 n =--3 n'=oo' (3)~(1) (3)-(-2) n =2 n = 3 n =00 (8)-(6) (8)-(7)Center Third Column (~) ( ~) Center Third Column (8 ) (8 )Point Point Tops Point Point Tops %' %

EI1.775 1.821 2.8 2.5 1.058 8.8PCr/ h2 1.770 1.032 1.160 11.0-

P~r~; 1.815 1.817 1.82i 0.3 0.2 1.046 1.071 1.160 9.8 7.7

P~r-Pcr "

'P % 206 2.4 0 ". 1,.4 1.2 0~-~ ~-- -~ ~-~

cr

P == Actual buckling loadcr•Per = Buckling· load determined on the basis of the loading condition

as shown in Fig. 2-5.

0'I' •• ; ••••

Page 114: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

Table II

, "

ELASTIC BUCKLING LOAD OF FRAMES SHowN IN FIGo 1~4--

.. '

.L = h . .- - - ~ - . - -- -- - - . - L = 3hLoading Position Discrepancies Loading Position Dis crepanc ie s( 1). d 2) ( 3) (4) ,. (5) (6 ). (7 ) (8) , ( 9) (10 )

,

0(= 0 0(= 2 (3)=(1) (3)-(2) r)..= 0 .<X = 2·(8l~(6)·· .-{8}":'~7}Roof :.. Column ( 3) ( 3) Roof Column (8 )Load Load - Load 'Load . (8

Only Only .% % Only Only % %

- ~I 1.787 1.811 1.821 ... 0.5. 1:82 1 0160 607 2.4Pcr h2 1.9 1.132~ '. I',: .

• ' i._

Table 111- ELASTIC BUCKLING TEST RESULTS

Span Column Slenderness Yield Predicted Test LoadTest P ~ 0< Bucklmg PexpLength Height Ratio:. -- Load - Load 2Pexp kipNo o PI 2Py kiP.,

~.L in o . h in. h/r 2Pcr kip

P=3 50 30 83.4 8.5 16 014 8.50 8015 0.96

p=4 50 35 97.2 7.7 14012 6.61 6046 0.98

nI----loI\)

Page 115: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

o Point of Instability

Load

.l, J

F~-T~,I~ Point of Bifurcation

I--·----~HII MI - --- II I II \ (2) ,';.-_-o--_~............ I ()) ,'.

I / ~ "I /': /'/'

,

Horizontal Deflection of Column Top

Figo 1-1 TYPES OF FRAME INSTABILITY

Ultimate

~103

A

------..... , Inelastic Buckling

'"w

Elastic Buckling-----......,

'\

Horizontal Deflection-o\. ,I j"[

Figo 1=2 LOAD-DEFLECTIONlRELATIONSHIP"OF ASYMMETRICAL FRAME LOADED SYMMETRICALLY

Page 116: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-104

1.0Yield Load------ - --- -- --- - - -,

\\

\

",.," Euler,(Hyperbola

ppY

P

h

@P

oSlenderness Ratio

Design forElastic BucklingStrength

n--- Beam-Column: I InstabilityI I ., .,

(b) :--

L

n-----. I !I I I

II

Design for Inelastic :Buckling Strength :

rT"MrTT"I"":;""TTTT""TT"rn III

II

Design forPlastic Strength

oSlenderness Ratio of Column h-r

Fig y 1-3 LOAD-CARRYING CAPACITY OF COLUMNS AND FRAMES

..

Page 117: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

p .p

wL

BCIb

h

HA E

H-t· 'w t

VL VR

wL Lp ='a (~). ~ =: {l+a)i=.

,.

Figo 1-4 FRAME .DIMENSIONS AND LOADS

~l05

Page 118: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-- - .

I"1\ tE/,-.--.~E~-,~l-~~---'---';::::""""~

Mt /J?int'/ L-I

//

//

/r/.----_':~

L-I -__ A r---\:!.i..

Moments and Member Rotations f as shown are assumedto be positive

Figo 2=1 NOMENCLATURE USED IN FOt1R~MOMENT ~UATION

=106·

Page 119: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

\=107

L II L 12 L I)

I 1_ A

C

Figo 2-2 SIDESWAY BUCKLlNG OF A RIGID BENT

Proportional. Limit

(j~

cry Weak Axis<)

o

I

Flgo 2=3 MOMENT OF INERTIA REDUCTION FACTORFOR WF MEMBERS ACCORDING TO TANGENTMODULUS THEORY (Taken from Refo 28)

Page 120: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=108

R PCI' PCI' 6R

---i----- FlIIIII

lei II I

, I II I

, I " I--

H t'-~__L=- ~ t H H* t~' ~'~*

L L- P Pn n---.--l=::d- Ib__

IIIII IeIII

h

S~able ~qu111br1um Unstable Equl1ibrlum

(a) (b)

F1go 2-4 FRAME STABILITY PROBLEM STUDIED INREFS 0 (29) and (32)o '

H

p p

~i----__---:l...!-n '

. Per

Her l .,----III

RPer

___ ! Her-

h

I~·------=L=---,-1

F1go 2=5 THE EFFECT OF AXIAL FORCE H IN THE BEAMON THE BUCKLING STRENGTH OF FRAMES

.;-'.'

Page 121: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=109

w

h Ie Ca) Ie I(b)

II

H A _~~I

fL

t .tVL VR VL+.6VL

P = a(~)

- ( )wLP = 1+«" r

//

Figo 3-1 EQUILIBRIUM OF MOMENTS AND FORCES

W

00 I./

I I ' . hH

EI- h~2

001 (II) FromEqo (3=21)

o """-------L-_-----'----------J------..-.:.-'-----G>------'--_

o 004 008 102 1 0 6 . 200

Wh/ !!,.2 h 2

Fige 3=2 DETERMINATION OF BUCKLING LOAD BY GRkPHIC MFrHOD

Page 122: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=110

1"

1 1/2"

~~ Bending Axis

Cross Section PPI PI !D~ !

1 t;1i I C I l'~w P =: a.PlI -P =: {l+a)Pl

P

lr.=!========~===;i

LoadingArrangement

--~

I'50"

_ ~E

..I

Figo 3-3 MODEL FRAME TESTS

Figo 3~4 ELASTIC BUCKLING TEST

Page 123: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-Ill

Figo 3-5 LOAD-DEFLECTION CURVES OF FRAME p-4

Figo 3-6 FRAME p-4 AFTER TESTING

Page 124: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=112

p p

Ib

hIh Ie Ie,

Is h2

\L f t\

(a) (b)

Figo 3~7 THE EFFECT OF PARTIAL BASE FIXITY

. .~ ~

, ,/i,

.I f, J

I {i ;1

.1.("I, ;1

·11\

Page 125: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

stress(J

tan E

strain e

II

.1IIrIIIIII e·'

=113

./

/

stress(J

Ca) Typical Stress=strain Curve ofMild Steel

tan.E

strain e(b) Idealized Stress~Strain Curve

Figo 4~1 ST~ESS=STRAINDIAGRAMS OF STEEL

; ,

Page 126: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=114

w

---...'C wL-2

p

H~~II

I .I

Ie IIIII

~-----·tp

PMB'

-H

p

h

H

Figo 4=2 MOMENTS AND FORCES ON BEAMAND COLUMNS

27WF94

I

~ ----------MY =: 10144 .

08 ProportionalLimit

M 06My

04

02 .

0a ··~·5 1 .. 0

;~

~Figo'4=3 MOMENT=CURVATURE RELATIONSHIPj)

INCLUDING INFLUENCE OF RESIDUALSTRESSES

Page 127: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-115

aRC = 003 ay

ORT =[bt+W(~:2t)J aRC

,~ "aRT __'y. ~:>~~RC

b =Flange Wldthd = Depth of sectlont :: Thlckness of flangew = ~lckness of Web

Flgo.4-4 ASSUMED COOLING RESIDUAL STRESSPATTERN

M.::.Q.=l 0My0

Center __1-----/

De.t:lectlon

ZeroSlope

20 30 40 50 ·6.0 70

Flgo 4-5

n or L210

BEAM DEFLECTION CURVES OBTAINED'BY NUMERICAL INTEGRATION (A = 00002)

\\

\.l

Page 128: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

(~116

M ',.', I

- = 10J4My

M 0MY 02

04(+) 06,

08

100

1 0 2 1------'----_100

08(~) 06

04

Figo 4-6 MOMENT~SLOPE RELATIONSHIPS(A.= 0.002)

10Plastic Hin e

Me_ = 1.1My

·60

50

n 40or Me

~= 0.5

-L 30 My

2rb

20

10

00 5 10 15 20 25

..

30

Slope Q Ra~ x 10-3

~. ,, I. ,

( Fig. 4~1 n VSo SLOPE CURVES (A..= 00002)----- -

Page 129: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

7030L/2IQ

2010o

2.0

1.81\.= 0.002

MC1.6 R=y

I.

1.2

1.0YIQ

.8

.6

Fig. 4=8 BEAM DEFLECTION CURVES FOR A = 0.002 .

Page 130: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-0.

-Q2M- 0My

0.2

0.4

0.6

0.8

1.0

70

50

40

30

20

10

Flgo 4-9

~=0.002

10 15 20

e Rod. x 10-3

NOMOGRAPH FOR OBTAINING MOMENT=ROTATIONCURVE OF BEAM (A = 00002)

=118

Page 131: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

\

1.~'-----------------------------,

LI"b

.8

.4

/\. = 0.004'

MC 'MY = 1.1

.6

0 I

0 8 16 24 32 40 48L/2rb

4=10 A~ 0 0 0048

Fig. BEAM DEFLECTION CURVES FOR .....(--'...0

Page 132: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

~=O.004

-0.4

-0.2

.M- 0 eMy

Q2

0.4

0.6

Q8

1.0

30

20

10

2 4 6 8 10 12 14 16 18 20 22 24 26

e Rod. x 10-3

Figo 4~11 NOMOGRAPH FOR OBTAINING MOMENT-ROTATIONCURVE OF BEAM (A= 00004)

~120

Page 133: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

05

MCM"= 1.1y}\. == o. 006

00- 4 8 12 16 20 24 28 32 0

L ',-.,I r-rb

Fig o 4=12 : BEAM DEFLECTION CURVES FOR A.= 00006 G/o.JI'\)F'

02

01

04

05

O.6r--------------------,.-------------

Page 134: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

M-My

~=0.006

o 10------1..-+--+-......--+--......---+---_e

50

10

2 4 6 8 10 12 14 16 18 20 22 24

e Rod. x 10-3

Fig 4~13 NOMOGRATH FOR OBTAINING MOMENT~ROTATIOrfCURVE OF BEAM (A. = 00006)

~122

Page 135: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

005,--------.----------------------------------.

04

.3

01

MC _Ie - 101

Y

.6

Fig e 4~14 BEAM DEFLECTION CURVES FOR A.:= 00008 o......I\)w

Page 136: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=124

~=0.008

0 ....---+-......-+...-04...--.....-.-.---t---9

2 4 6 8 10 12 14 16 18 20

9 Rod. X 10-3

Fig. 4=15 NOMOGRAPH FOR OBTAINING MOMENT=ROTATIQNCURVE OF BEAM (A = 00008)

Page 137: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

oo

-125

L = 0Py_------...L...-

_--------0~12

_-------0 0 2_-------003

~-------004

8WF31

~JE-

Figo 4=16 THRUST-MOMENT-CURVATURE RELATION­SHIP INCLuDING INFLUENCE OF RESIDUALSTRESSES

Page 138: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

" 1.1

"1.0

" .9

.8

" .7

M- .6My

.5

.4

.3

.2

.1

o .02 .04 .06 .08 .10e "RADIAN"

.12

~126

Figo 4=17 MOMENT-ROTATION CURVES OF COLUMNS (P ; 0)

Page 139: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

1.1

1.0

.9

.8

.7

14 .6

My.5

.4

.3

.2

.1

00 .02 .06 ,08

: =0.12y

«TRC :: 0.3 «Tytry =33 ksi

.10 .12 .14 .16 .18

e Radian

h7=70e

.22 .24 .26 .28 .30 .32

h7=30c

Fig o 4=18 MOMENT=ROTATION CURVES OF COLUMNS (p =

Page 140: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

h1.0 -=30re

.9

.8

.7

.6

M MrhMy.4

P~:O.20

crRe =0.3 crytTy =33 ksi

tp.1

Fig8 4-19 MOMENT-ROTATION CURVES OF COLUMNS (P = 082Py ) UI-'f\)ex>

Page 141: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

11=30.9 rc

.8

.7

~6

M .5

MY M

.4 -t=O.30\ .

Y.3 CTRC =0.3cry

cry =35 ksi

.2

is.1

.01 .02 .03 .04 .05 .07 .08

e Radian

Fig o 4-20 MOMENT-ROTATION CURVES OF COLUMNS (P == 0 0 3Py)

Page 142: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

r

08

07

06

M_ 05

My.4

03

.2

.1

t:; 0.4y

CTflC =O.3 CTyCTy=33 ksi

.01 .02.' .03 ...04 ..05 .06 .07 .08 009 .10 .11 .12

e Radian

Fig.' 4-21 MOMENT-ROTATION CURVES OF COLUMNS (P = a .. 4Py )

If-IWo

Page 143: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

MColumn

~__ Beam.

=131

Flgo 4-22 DETERMINATION OF MOMENT ATPOINT B

p P

C

24WF120

A

oorl

~C\J

I"

P =: 204(~)2

_. 41.wL"p =: 3~ \'21

L = 50 v

.­,; Flgo 4=23 ILLUSTRATIVE EXAMPLE NOol

\'.

Page 144: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-132

P

M\\\

\ \\ \ >-.= 00008

'\P. I "'--,p -0 .. 2 /- .. : ,,__\ y /1-=0 0 ~=.:.;:,.;(-J-----

, "'1/ P\ '" / /'7, 1/"/-\ 1/ I L = 004, I/, / Py

/

X/\

/ \ ..

\ / .'\'= 0 0006I .

\ I \

l\ \\ \ P

.....--,---r--.--r-r-r-r""T""T""1r::!.-

0L...- -'-- --L.. ----''----__-----''

300

100

600

500

400

'200

M. (ft~kips)

o 001 002

Q (Radians)

MBD-G curves for beam

MBA- Q curves for column

Figo 4-24 MOMENT-ROTATION CURVE(EXAMPLE NOo 1)

/I

,f

Page 145: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-133

432·1OL...-.._...L..-_-L-_--I-._----J._-----'-'

°

200

600

:.. '"

i.J

2P(kips) 400

Center De~lectionin inches

Figo 4-25 LOAD-DEFLECTION CURVEOF EXAMPLE NO'. 1

pp

BC

33WF13<f··

WL)P = 2 (­2 .~ \

'. "~

.A L = 70'rf--------~I

E

.Il"ig. 4-26' ILLUSTRATIVEEXAMPLE

NO. 2

I:'

Page 146: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-134

\\\ "A= 0~00783:

\, " ,1 (Hinge Angle)\ " '2

A. =0. 00'3l'i, ' "-, 1'--;---------1

\ P ':0 ;,1-7/ -_. -~----~---I",-p. 41 / /_~

'; ,{~ / ///, .

\ MB'. /' g'l~My , I / ~ -_0_, -_. -'--

\ -p;. =0.2/I. '/ First Plastic:\ 'Py IIJ 'I 'Hinge at ,Center

. I g'. '\ l /

I. / P =,'0·3/ - ' ..\ I. / Py. \ I. /

002001

III//1 '\

/i'l ,/' ' \ 'A..= 0.0,0526

\ I, I ".,

"'­\

\ A.. = 0.00315 ,I

\\

\

oo

G (Radians)

Figo 4-27 MOMENT-ROTATION CURVE (EXA~~ NO~2)

Page 147: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-135

Angle .Change = 2 t:::. 9

L

Figo 4-28 DEFLECTION AND HINGEROTATION DUE T0.69

l-

v---" Hinge AngleI- = 0029 Hado

l-

I-

I-

"I- ,

I-

800 2 Pulto = 752 kip-------

700

600

2 P 500

(kips) 400

300

200

100

0L...----JL.----J_----lL.-----l

oDeflection in inches

(a)

o 001 002 003 004Hinge Rotation in Radians"

(b)

Figo 4-29 LOAD-DEFLECTION AND LOAD-HINGE ROTATIONCURVES (EXAMPLE NOo 2)

Page 148: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

p p-136

A

Ie , -Ie

E

h

Figo 5-1 YIELD CONFIGURATION OF THEFRAME

MaI~_ Me 4rO:7My

O.o~~11TTTTIllllJ:Y".. L . ,j-"- -~ ,

:. I (a) I,i ,I ,I I 'II .1 I I'I 1 I I

I . I I II I I II I I I

! I : It: == J

(b)

Fig~ 5-2 MOMENT DIAGRAM AND VARIATIONOF YIELDING ON THE BEAM

Page 149: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

"iI

-137

l~O,,80604020'1-__----L. ...L-__----1 -L... L.-_--l..--I

o

1 0 0 r-----..:...---------...;;:--------~

MMy

Figo 5-3 THE EFFECT OF YIELDING ON THE 'FLEXURALRIGIDITY OF THE ~7WF94 SECTION

(EI~)~££. (E1b)e££. (E1b)e££.-~ H.. E1b + ~ rt----E--=---Ib_.....,•• 1

B C ~:::=::::: ) D (a) Red¥eed,. G Beam

! 1 tad.

rhl L

M= 1° '2", (e) Loads on

tQe Analog'ous, ColuDm

I _

I

DETERMINATION OF 'Kb and Cb BYCOLUMN ANALOGY

A ,

Page 150: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

L2

L"2

-138

Ca) I" t-~1~--;>I

,

'f. ,

(b) ~JJ///2T~

IG

.Analogous Column

A = c::X:l

Ioo is finite

.' .

MomentM

(e)I 1 Rado

Fig. 5..5

Load on Column

" ..DEl'ERMINAT.ION OF Xb FORMEMBER WITH EQUAL ENDROTATIONS

Loading

Unloading..p

M

h

t-p

Fig. 5-6

Rotation Q

STIFFN]SS FACTOR OF A BEAM­COLUMN WITH CONSTANT AXIALFORCE

Page 151: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

, 04 06 08 100 102M!My ,'

~fRIATIONS 'OF COLUMN STIFFNESS FACTORKc WITH THE;A PPLIED MOMENT

. _t.~.

• 60Kc '

MY 50

40

30

20

10

0

0 02

Flg~ 5-7

~139

110 -100 P = 0012!L.-:: 30 Pyr c

90

80

70

6050

'~My i

500

4090

30 110

20 1 0

10."

0. 110 -100 .

.!L,= 1.:. = 00230 Py

90.rei

\\

\

Page 152: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

, ·-140

110 -100

P = 003h PY= 30 (1r O90

80

70

60 50K'. .,:"

0--- 50My ,

40 ° /30 0

20 110

1 010

0110

100h = 30

P = 004- Py99 r o

80

70

60 50

K~50M:y

40 70

30 90

20

10'0

00 02 04 06 08 100 102

M/Myi~igo 5-8 VARIATIONS' OF COLUMN STIFFNESS FACTOR

K'..

WITH THE APPLIED MOMENT'0 . -;1'

"'1'"

" ~:.\'.. " ...v

r

Page 153: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

p PwL

I I I I I I I I I I I

) L(P =(l+a)J:L. 2

p

~

p

~

-141

(a)

Loading Conditionfor DeterminingBuckling Constants

(b) (c)

Loading Conditionfota Anal,-.irigBuckling Load

Figo 5-9 SIMPLIFICATION OF LOADING CONDITrONS

, .

[Q

-or Per

Flgo .5~10 D~EijMINATION OF BUCKLING LOAD

Page 154: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

P p

BC

33WF130

4

A

Figo 5...11

h (Variable)

L =.80 r

P =uJwL). 2

; ~ (l+a)~ a = 2'0'02

EXA'MPLE FOR ILLUSTRATINGTHE .CONSTRUCTION OF FRAMEBUCKLING pURVE

3

k 2

I t

MI1 Ie Ie h

I ~ ·1

oo 1 2 3 4 5

...Figo 5~12 EFFECTIVE LENGTH FACTOR k OF

PINNED-BASE FRAMES (TAKEN FROMREF 0 4)

I.

Page 155: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

~143

h=60

P

L=80r

P

I

"- '- ...... __A 1= .000<:>48------

00830~"--

, .

:0 L...- :.....J....... --:-L ---=--L--__~

o 001 002

Q (Radians)

ob4

F;1go 5~13 .' ELASTIC=PLASTIC ANALYSIS OF FRAMEWITH h = 60 r

Page 156: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=144

MFL = t~OO~OO ~~~. -51~20

+11~80.-8037+1093-1037+0 .. 31

8ML =+53010

Cb = 007125

0~660 +49~00 = MFHo ...16:~66rt"\ +li~82

o~. +0032 l'

+43 02? := MR

<.: -

Figo 5-14 MOMENT DISTRIBUTION COMPUTATIONFOR THE CASE h = 60 r . "

40

30[Qxh

20

10

0

00040 00044

A = 000501cr

Figo 5-15 DETERMINATION OF CRITICAL LOAD(h = 60 r) . -

Page 157: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-145

he80

-Ir,=80r

\\

\ \\ \, \, '\, '\, '\

\ "-',', Hinge at

'\ '" ,Center,'\. ':. \'\. ~, 8"'- 01..... __A I = 0.:004 .

,O

2...... ,

'-_ A..z.= 000045~-----

02 .'

Ma 06-My

p. P /,.

04

o 001 002

.Q (Radians)

Figo 5=16 ELASTIC-PLASTIC ANALYSIS OF. FRAME WITH h = 80 r

Page 158: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

o 18MFL = ~lOOoOO ~ .

. "22082C\J==::::;1 C\J

Mr, +77018 a

~7I~i8

-146

.I,I/

I

Flgo 5-17 MOMENT DISTRIBUTION COMPUTATIONFOR THE CASE h = 80 r ...

+10

0 . /\.LQxh

-10

-20

-30

-40

-50

-60

-70

Figo 5-18 DErERMINATION OF CRITICAL .LOAD (h = 80 r)

Page 159: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

---------

LaSimple Plastic Load

---~----------------------'\-

\VElastic. Frame

. Buckling

~.{c)--~

't

1/ / (b) ~Inelastic.,Frame .-Buckling

P Pw

P= rx (~L)., .33WF1300

C"'"'\ wLr-l h P = (1"+ ()( )

~2

C"'"'\ ()(.= 2.0.2 L = 80r

.8

.6

40 60 8.0 100 120

Slenderness Ratio of' hColumn r

Fig. 5=19 ,yiILLUSTRATION OF FRAME BUCKLING CURVE

20OL..----...:....--'!- --l ...l- :-----l... L.- -.J

a

Page 160: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

1 0 -- ---- - - -- --- -- -~ ----,o ~----- ~aU-.:~'_

-pPu

P PI PPI

(b)

\ ", ,". (e),

'\."

"

'"

L }1--1,---1

P =aPl

P = (1+tt) PI

a = 3035 f :::: 1~27

40 60 80 100 120

h/r

INELASTIC BUCKLING TESTS ON MODELSTEEL FRAMES .(SERIES B)

Fig" 5-20

oL--__....L..-__---l..-__--,-L_.:,....-----'L.....-__....Io.--__............__

020

w P P wL(1)

ifP = a-

2

a = 200

P P

if a = 100(2)

w

~nI I a = 0L

·1 ','(3)

(a).\/:.;

(e)

(b).Figo 5-21 SIDESWAY BUCKLING OF MULTI=STORY FRAMES

Page 161: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

100.----------------------------------.

08

06

04

02

p

I· L = 70 9

p

h

ElasticLimit ---

r = 7007 ft

.r- 8064ft

r =10 0 96 ft

r-= 1)02)ft..P = D( (~L)2_

P = (1+()() ¥.IX. == 2 0 0

Ol-----~----l...;_------L_:__---~------l-----L---:-------Jo 0.20040.6 008 1 00 1 02 1.4

h/L

Fig. 6-1 SAMPLE DESIGN CHART FOR L :;:: 70 FEET AND 0<. == 2.0

Page 162: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

r,

1.0r------------------------------------.

.S

.6

PPu

P p

r =: 8.64"

.4 r == 10.96"

r - 13.23"15.00"

.,,~:..h

.. .r =

p. = ex (~L).2

1- L 90 9 P (1'+()() wL= ·1 = 2.

0( = 2.0

o '=""0----:0:-1-.--=-2---0-::.1·-..4----oL• 6'----'---0....L.--:-

S----1L.O-·---1...1...:-2---1 -.1.

4hlL

Fig. 6~2 SAMPLE DESIGN CHART FOR L :;; 90 FEET AND (){ = 2.0o

!-'\no

Page 163: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

P P

=151

h = 56'

L' =:,70' -IWorking Loads2

w = 1.30 kips/ft.P = 91 kips

Fig. 6-3 ,DESIGN EXAMPLE NO. 1

Pw

• ~l P

I· -\Working Loads~

w = 1.20 kips/ft.P = 108 kips

Fig. '6=4 ' DESIGN EXAMPLE NO. 2

Page 164: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

0.6 r----------------------'------------------------.,

c08208

8506

D

8200

c7704

o92 0 50-

9309

o9107

0 0 95099603

2 P+ 1 h-_ 1 0Py 70 r c =- 0

AISC Rule

c L = 70~

o L =: 90 1l

~~ Proposed Design Rule:

~-

~ P 1 h'" 2- + 'C'i' - = LO~ Py u;> r e

""'-"--"-- Approx 0

'a ",,100% 0

"" 9608~pproxoo 0 '-.,.100%

""o ApP186%'-.,.

'-.,.'-.....

"--"" '-.,.

"" '-.,.'-.....

'-..... -'-.....

0L-____L.. ..L.--__~ __I__--_____J1--____L..__~>L_____1._..::::.....__....L._______l

05

03

04

01

o 10 20---- -30 40 -" ---~ 50

hire60 70 80 90 100

FIg o 6=5 ,- COMPARISON OF THEORETICAL RESULTS WITH THE AISC DESIGN RULE BI-J

\nI\)

Page 165: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

,"

=153

100 APPENDICES

Appendix A

Typical Numerical Integration bomputationsrorObtaining Beam Deflection Curves

This appendix will illustrate a numerical integration

procedure for determining the slope and deflection or any

ppint in the length CD or the beam pictured in Figo 4020

In all the computations~ the moment=curvature relation=

,s,hip of a typical beam section~ the 27 WF 94» is usedo The

section properties and the material constants adopted for

-"th( computations are the followingg

I = 32660,7' in0 4

E = 30xl06 psi

~y = 668 ft=kips

S = 24208 in0 3

cry == 33xlO 3 ps i

¢y = 8l.76xlO=6

r ::: 10.87 ino

rado/ino

The n~erica1 procedure used to obtain the beam deflection

curve is illustrated in tabular form on pages 156 and 157 ror

the following two casesg

( 1) /\. = 0.002

(2) A = 0.002

MC/My = 0.5

Me/My ::: 1.0

The procedure is basically the same ror both c~seso

In computing slope and derlection by numerical integra=

tlon~ the beam is broken into a number of short segments of

length "a~". (a constant times the radius or gyration of

the beamo) Average values or bending moment and curvature

\

Page 166: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

(A.l )

'.

-154

are determined for each segment and are used to obtain values

for the change in slop~ and deflection occurring within that

segmento By summing the changes in slope and deflection

from an initial starting point~ the total slope and deflec­

tion at any desired point may be determinedo The initial'

point used here is the center of the beam (point C) at which'

the slope Q and the deflection y.are assumed to be zero.

A~ average value of bending moment for any segment is

first calculated by statics from Eqo (405)

M'l "M,. 2, 'A.~ .= .:.-.loI£- - h -n~ . n~ 4

aveWhere n l is a number such that lnlIQ" is the distance from

the center of the beam to the center of any segment. The

average curvature ¢n.J corresponding to MnJ is picked¢y ave ,My ave

from an appropriate M~, ¢ curve' such as Figo 4030

The change in slope is the average curvature times ·the

length of segment (column 6 of Tables Al and A2)

8. en = CArb <fnave 1" (A. 2 )The change in deflection i'rom':the properties of the

tangent offsets of a small circular arc is (column 12)

(A. 3)

Figure A=l shows these changes occurring within the segment

"a~"~ The slope at one end of the (n+l)th segment is

(column 7)

Page 167: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

~155

and the corresponding deflection is (column 8)

As may be seen from Tabl'es Al and A2, the integration

process for the first case (,\,= 0 0002 and Me =005) termi-My

nates at the point where the slope of the beam deflection

curve becomes zero, whereas that for the second case termi-

nates when the moment reaches the Mp value.

Page 168: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

~156

Table A1

NUMERICAL INTEGRATION COMPUTATIONS FOR THE CASE A. = 0.002 AND

MC/My = 0.5

(1) (2) (3) (4) (5) (6) (7)

Xn 9nx10-3 Mn~]ave GnJ 0 x10- 3 ar 0 xlO- 3

(1)+(6)~ rJ y ave

nave b nave

0 0 0.5000 0.4995 0.4995 0.04084 0.8878 0.8878

2 r b 0.8878 0.4980 0.4955 0.4955 0.04051 0.8807 1. 7685

4 rb 1.7685 0.4920 0.4875 0.4875 0.03985 0.8665 2.6350

6 rb

2.6350 -=

54. 4rb 0.2448 0.9797 0.9851 1.088 0.08895 0.1934 0.0514

54.6rb 0.0514* 5.9906

(cont. )

(8) (9) (10) (11) (12) .(13)2

Xn Yn Yn/rb arb9n(arb) AYn

2 ~nave (10)+(11) (8)+(12)

0 0 0 0 0.00965 0.00965 q.00965

2 rb 0.00965 0.00089 0.01930 0.00957 0.02887 0.03852,

4 r b 0.03852 0.00354 0.03845 0.00942 0.04787 0.08639

6 r b 0.08639

3.61844

3.61876

0.33288

0.33291

0.00053 0.00021 0.00032 3.61876

* This is considered to be very close to zero

Page 169: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=157

Table A2

NUMERICAL INTEGRATION COMPUTATION FOR THE CASE A.. = 0.002 AND

Me/I). = 1.0

(1) (2) (3) (4) (5) (6) (7)

~ 8nxlO- 3 ~ ~] ~n] ~ xl0- 3arb~ xlO- 3

(1)+(6)My My ave 0yave

nave . nave

0 0 1.0000 0.9995 1.128 0.09222 2.004~ 2.0049

2 rb

2.0049 0.9980 0.9955 1.115 0.09116 1.981.8 3.9867

4 rb

3.9867 0.9920 0.9875 1.095 0.08953 1.9463 .5.9330

6 rb

5.9330

6~.2rb 17.3236 1.1255 1.1320 2.100

65.4rb

16.9504 1.1386*

0.17169 0.3732 16.9504

(cant. )

(8) (9) (10) (11) (12) (13)

Yn yn/rb arb8n (arb)2 b.Yn~2 llave (10)+(11) (8)+(12)

0 0 0 0 0.02179 0.02179 0.02179

2 r b 0.02179 '0.00200 0.04358 0.02154 0.06512 . 0,08691

4 rb

0.08691 0,00799 0.08667 0.02116 0.10783 0.19474

6 rb 0.19474

65.2rb 13.97349

65.4rb 14.01074

1.28551 0.03766

1. 28874

0.00041 0,03725 14.01074!

* This value of moment is almost equal to the plastic moment

Page 170: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=158

y

R

OL.-~~::::::::::'_----*--=--------I----_x

= arb.¢nave

(arb) 2

= arb9n ... 2

INCREMENT· OF· SLOPE AND DEFLECTIONIN THE IN~VAL a~

Page 171: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=159

Appendix B

Determination of Stiffness ,of PartIally Plastic Beams b,Ithe Column Analogy

This appendix contains the numerical computations in=

volved in determining the stiffnesses and carry=over factors

for the beams of the two frames considered in Arto 504~ The

dimensions of these frames and the applied loads (selected

trial loads) are as followse

(See Fig o 5011)

Member size

Radius of gyration» r

Span length» L

Column height» h

Distributed load» w"Loading ratio, /\.

First Case

33WF130

1302~ i:qo

80 r

60 r!

2020 kips/fto

000048

Second Case

33WF130

13023 ino

80 r

80 r

2 020 kips/fto

000048

For each ease the moments at the ends of the beam are

first determined graphically by the elastie=plastic analysis

of Cha.pter 40 The moment at al~~, the sections of the beam

can then be computed by staticso From the computed moment

values the effective flexural rigidity of all the sections

can be determined from the flexural rigidity=moment relation=

ship of the 33WF130 section (similar to that shown in Figo 5=3)0

Thus, in effect» a beam of variable Err is obtainedo In the

buckling analysis» it is required to evaluate the s~iffness

factor and carry=over factor of this beamo This cap be done

Page 172: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

or

,.,.',

=160..

conveniently by the method of column analogy. The procedure

. of applying this method was discussed in Arto 5 02 010

Detai~ed computations for the beams of the frames considered

here will be explained belowo

First Case = Frame having h = 60r

Referring to Figo 5=13 g the moment at the column top

for AI= 000048 is equal to 0083 My = 924 ft=kipsg thei

moment at the center can then be determined by statics~

w L Lf1c = -z- 4"" - nB = 2 I .3 6 - :3 2 4- =- 12. I 2. ft - Ki rs

~:L 09n =j.,d.Ou~'

This indicates that the center section of the beam is not

fully plastified 'at the trial loado Figure B=l sho~s the

distribution of bending moment of the beam and the Corres=, I

poridirig yield configuration. To compute the stiffQess of

this partially yie~ded beam by the method of column analogyg

it is convenient (and also accurate enough) to diviqe the

yielded portion into segments g each having a length of one

foot. Within each segment the flexural rigidity mar be

assumed to be cons tanto The area and. the moment of inertiar;.

of the an~logous column can then be calculated nume~ically

as tabulated in Table BI.

The moment at the center section of each segment ~s

first computed from the known values of MB and ~. (listed in

column (3) of Table Bl) 0 The effective flexural rigidity

(E[)eff of these sections is then determined from theEI

Page 173: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-161

flexural rigidity - moment curve constructed for the 33WF1300

This gives the values shown in column (4)~ Since the width

of the analogous column at any- section is inversely p:ro-

. portional to the flexural rigidity of that section, the

reciprocal of the values of column (4) gives di:rectly the

width of the analogous cOrUmn at the center sectionso For

example the width of the first segment is Ooi34

~ =

. 7E¥6, where 1rr. is the flexural rigidity of the section in the

el~stic rangeo The area of each segment of the analogous

column can be computed by multiplying its width by its

length (listed in column (6». The moment of inertia of

each segment with respect to axis G~G may be computed by

using the parallel-axis theoremo The values obtained for all

the segments.are listed in column (7)0 As shown in Table Bl,

the total area and the total moment of inertia thus obtained

are 18~04 and 61.0~±o08 respectivelyo The stiffness of

the beam or the moment at end B induced by an imposed unit

+

rotation at B is given by Eq·o

IKp =Ma= -I&~1;"".o"-t"'--

EI

= O. 00J 7/:9 EI

(50 1) ~

I· (44.1) ·(41-. I)61,o61.QB

EI

where the unit rotation applied at end B is represented by a

unit load applied to the'analogous column at point B. The

stiffness facto~ may be expressed in terms of My by sub­

stituting My/¢y forEI, that is,

K - O. 00 ~ 7 19 M 46 6 -:l Mb - cf~ ~ = ..J 'j

Page 174: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

-162

The moment at D is

= 00002650 =00003719

I' (44.1) '(44: I)61,061.08

EI

I1'1 0 = '&7.0"t

E1= - 0.002.650 EI

The carry-over factor is therefore equal to Cb

Second Case - Frame having h = 80 r

It is required to determine the stiffness of the beam

when it is bent into an anti-symmetrical configurationo The

bending moment diagram of 'the beam and the yield configura­

tion are shown in Figo B-2. In this case only half of the

analogous column needs to be considered and its area ~ay ~e

assumed to be infinity. Table B2 contains all the co~puta­

tions involved in determining the moment of inertia of the

half column about the axis G-Go The stiffness of the beam,

according to Eqo (504), is

(·(44.1)·(4Lt-.1) =O,05653E.l' ..)4J 405 .l.2.

when expressed in terms of My, the stiffness if Kb" = 70 0 87 My.

Page 175: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

Table B1

DETERMINATION OF STIFFNESS BY COLUMN ANALOGY (h = 60 r)

( 1) (2) (3) . (4) (5) (6) (7)

Distance Moment ofSection from M (EI) eff 1 Area Inertia About

Axis G-G My EI (4) Axis G-G

0 - 1 0.5 1.088 0.13~ 7.46· 7.46 .2.49

1 2 1.5 1.087 0.134 7.46 7.46 17.412 - 3 2.5 1.083 0.140 7.14 7.14 45.22·

3 - 4 3.5 1.'077 0.160 6.25 . 6.25 77.08

4 - 5 4.5 1.069 0.172 5.81 5.81 118.13

5 - 6 5.5 1.059 0.190 5.26 5.26 159.56

6 - 7 6.5 1.047 0.216 4.63 4.63 196,01

7 - 8 7.5 1.033 0.250 4.00 4.00 225.33

8 - 9 8.5 1.018 0.290 3.45 3.45 249.55

9 - 10 9.5 1.000 0.338 2.96 2.96 267.39

10 - 11 10.5 0.980 0.400 2.50 2.50 275.84

11 12 11.5 0.958 0.470 2.13 2.13 281.87

12 - 13 12.5 0.935 0.550 1.82 1.82 284.53

13 - 14 13.5 0.909 0.640 1.56 1. 56 284.44

14 - 15 14.5 0.882 0.720 1. 39 1.39 292.37

15 - 16 15.5 0.852 0.800 1.25 1.25 300.41

16 - 17 16.5 0.820 0.872 1.15 1.15 313.18

17 - 18 17 .5 0.787 0.930 1.08 1.08 330.84

18 - 19 18.5 0.751 0.972 1.03 1.03 352.61

19 19.9 19.45 0.716 0.996 1.00 0.90 340.53

19.9 - 42.6 31.25 1.000 1.00 22.70 23,142.69

42.6 - 43.1 42.75 0.723 0.990 1.01 0.51 992.93

43.1 - 44.1 43.6 0.787 0.930 1.08 1.08 2053.13

L= 93.52 30,533.54

Page 176: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

Table B2

DETERMINATION OF STIFFNESS BY COL~ ANALOGY (h = 80 r)

=164

(1) (2) (3) (4) (5) (6)

Distance M .(EI) eff 1 Moment of. Section from My EI (4) Inert~a About

Axis G-G Axis G-G

0 - 1 0.5 1.150 0.010 100.00 33.331 - 2 1.5 1.150 0.010 100.00 233.33

2 - 3 2.5 1.145 0.015 66.67 422.25

3 - 4 3.5 1.140 0.020 50.00 616.67

4 - 5 4.5 1.132 0.040 25.00 508.33

5 - 6 5.5 1.122· 0.060 16.67 505.66

6 - 7 6.5 1.110 0.085 11. 76 497.84

7 - 8 7.5 1.096 0.120 8.33 . 469.25

8 - 9 8.5 1.080 0.148 6.76 488.97

9 - 10 9.5 1.063 ,0.180 5."56 502.25

10 - 11 10.5 1.043 0.225 4.44 489.88

11 - 12 11.5 1.021 0.280 3.57 472.43

12 - 13 12.5 0.997 0.350 2.86 447.12

13 - 14 13.5 0.972 0.427 2.34 426.66

14 - 15 14.5 0.944 0.515 , 1.94 408.43

15 - 16 15.5 0.915 0.620 1.61 386.93

16 - 17 16.5 0.883 0.720 1.39 378.54

17 - 18 17.5 0.849 0.806 1.24 379 ;-85

18 - 19 18.5 0.814 0.884 1.13 386.83·,

19 - 20 19.5 0.776 0.945 1.06 403.;16

20 - 21 20.5 0.737 0.983 1.02 428.74

21 - 21.4 21.2 0.708 0.988 1.01 181. 57

21.4 - 43~3 32.35 1.000 1.00 23,794.13

43.3 - 44.1 43.7 0.735 0.986 1.01 1,543.07

I= 34,405.22

Page 177: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

G

Figo B-1 . DETERMINATION OF BEAM STIFFNESSFOR THE CASE h = 60r .

B 1:::!:::!:::::!:!Jv:;=;=;=:::;=Y=j==j=j=::;=;=iCf=r=;:=;=:;==r=;=;=r=;=~!::::!::!::::l D E0 770My

L15My = Mp

G

I.

Fig o B=2 ,., DETERMINATION OF BEAM STIFFNESSFOR THE CASE h = 80r

=1(,5

Page 178: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

~166

11. REFERENCES

1. Beedle, L. S.PLASTIC DESIGN OF STEEL FRAMES, John Wiley & Sons,Inc., New York, 1958

2. WRC - ASCE Joint CommitteeCOMMENTARY ON PLASTIC DESIGN IN STEEL, Journal of~~e Engineering Mechanics Division, ASCE, Vol o 85~Nos 0 EM3, EM4 (1959), VoL 86, Nos. EMl, EM2 (1960 )

30 Bleich, F oBUCKLING STRENGTH OF METAL STRUCTURES, McGraw-HillBook Coo, New York, 1952

40 CoRoCoGUIDE TO DESIGN CRITERIA FOR METAL COMPRESSIONMEMBERS, Column Research Council, 1960 .

50 Driscoll, Go C., Jr.ROTATION CAPACITY REQUIREMENTS FOR BEAMS AND FRAMESOF STRUCTURAL STEEL, Ph.D. Dissertation, LehighUniversity, 1958

60' Haaijer; Go and ThHr limann , BoON INELASTIC BUCKLING IN STEEL, Journal of theEngineering Mechanics Division, ASCE,yolo 84,No 0 EM2, (1958)

70 Galambos, To V. and Ket.ter, R. L.COLUMNS UNDER COMBINED BENDING AND THRUST, Journalof the Engineering Mechanics Division, ASCE, Vol. 85,No. EM2, (1959)

8. Ojalvo, MoRESTRAINED COLUMNS, Fritz Laboratory Report 278 02,Lehigh"University, 1959

9. Lee, G. C.A SURVEY OF LITERATURE ON THE LATERAL INSTABILITY OFBEAMS, Welding Research Council Bulletin Series No.63, 1960

10. Shanley, Fo R.INELASTIC COLUMN THEORY, Journal of AeronauticalSciences, Vol o 15, No.5, po 261, (1947)

Page 179: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

~167 .

110 Lu, Lo VoLITERA~URE SURVEY ON STABILITY OF STRUCTURALFRAMEWORKS, Fritz Laboratory Report 27602, LehighUniversity, (in preparation)

Kavanagh, To CoINSTABILITY OF PLANE TRUSS FRAMEWORKS, DoctoralThesis, New York University, 1948

von Mises, R 0 •

. tmER DIE STABILITXTSPROBLEME DER EIASTIZITKTSTHEORIE,Zeitschrift fllr Angewandte Mathematik und Me chanik ,Volo 3, No 0 6, po 406, (1923)

Ratzersdorfer, J oDIE KNICKFESTIGKEIT VON STXBEN UND STABWERKEN,Julius Springer Verlag, Vienna, 1936

Winter, Go, Hsu, Po To,Koo, B o, and Loh p Mo HoBUCKLING OF TRUSSES AND RIGID FRAMES, Cornell Uni­versity, Engineering Experiment Station BulletinNoo 36, 1948

16.

180

Chwalla, Eo' ~nd Jokisch,.F otJBER DAS EBENE KNICKPROB LEM DES STOCKWERKRAHMENS,DerStahlbau, 'Vol o 14jl po 33jl (1941)

Lundquistjl Eo EoSTABILITY OF MEMBERS UNDER AXIAL LOADjl NationalAdvisory Committee~'for Aeronautics jl Technical NoteNoo 617, (1937) .

Lundquistjl Eo EoPRINCIPLE OF MOMENT DISTRmUTION METHOD APPLIED TO .STABILITY OF STRUCTURAL MEMBERSjl Proceeding, FifthInternational Congress of Applied Mechanicsjl po 145jl(1938 )

Boley jl BoA 0

NUMERICAL METHODS FOR THE CALCULATION OF ELASTICINSTABILITYjl Journa~ of Aeronautical SciencesjlVol 0 14, p 0 337 jl' (:1947 )

200 Lundquistjl Eo Eo and Knolljl Wo DoTABLES OF STIFFNESS AND CARRY=OVER FACTORS FORSTRUCTURAL MEMB·ERS UNDER AXIAL LOAD, NationalAdvisory Committee for Aeronautics, Technica~ NoteNo. 652, 1938

21. LivesleYjl Ro Ko and Chandler, Do B oSTABILITY FUNCTIONS FOR STRUCTURAL FRAMEWORK&,Manchester University Press, 1956

Page 180: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

=168

Hoff, N. J.STABLE AND UNSTABLE EQUILIBRIUM OF PlANE FRAMEWORKS,Journal of Aeronautical Sciences, Vol. 8, No.3,p. 115, (1941)

Perri, J. G.'MODIFIED METHOD OF MOMENT DISTRIBUTION FOR RIGID

. FRAMES WITH SIDESWAY AND FOR ANALYZING COMPRESSIONMEMBERS AS PARTS OF RIGID FRAMES, Doctoral Thesis,New York University, 1948

Pollaczek Geiringer, H.DISCUSSION ON REFERENCE 19, Journal of AeronauticalSciences, Vol. 14, p. 850~ (1947)

.Kasarnowsky, S. and Zetterholm, D.ZUR THEORIE DER SEITENSTEIFIGKEIT OFFENER FACHWERK=BRttCKEN, Der Bauingenieur, VoL 8, p. 760, (1927)

26. Bleich, F. and Bleich, H. .BEITRAG ZUR STABILITKTSUNTERSUCHUNG DES PUNKWEISEElASTISCH GESTUTZTEN STASES, Der Stahlbau, Vol. 10,p. 17, (1937) .

27. Huber, A. W. and Beedle, L. S.RESIDUAL STRESS AND THE COMPRESSIVE STRENGTH OFSTEEL~ Welding Journal, Vol. 33~ No. 12, p. 589=s,( 1954J

28. Beedle, L. S. and Tall, L.BASIC COLUMN STRENGTH Journal of the StructuralDivision, ASCE, Vo~. 86, No. ST7, (1960)

29 •. Chwalla, E. .DIE STABILITKT LOTRECHTBELASTETER RECHTECKRAHMEN,Der Bauingenieur, Vol. 19, p. 69, (1938)

Chwalla, E. and Kollbrunner, C~ F.BEITRXGE ZUM KNICKPROBLEM DES BOGENTRKGERS UND DESRAHMENS, Der Stahlbau, Vol. 11, p. 94, (1938)

31. . Puwein," M. G.DIE KNICKFESTIGKEIT DES RECHTECKRAHMENS,Die Bautechnik, -VoL 18, p. 32, (1940)

32. Masur, E. F., Donnell, L. H., and Chang, I. C.3TABILITY OF FRAMES IN THE PRESENCE OF PRIMARYBENDING MOMENTS, Paper presented at the Conferenceon Structural Mechanics, Purdue University, (1960)

33. Chang, I. C. --- -.--ON THE BUCKLING STRENGTH OF FRAMES, Ph.D. Disserta=tion, .Illinois Institute of Technology, .1957

Page 181: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

34.

35.

36 •.

37.

38.

39.

40.

41.

42.

43.

44.

45.

~169

Boltong A.STRUCTURAL FRAMEWORK» PhoD o Dissertationg ManchesterUniversitYg 1957

Salemg ,A 0

STRUCTURAL FRAMEWORKg,PhoD o Dissertation, Manche.sterUniversitYg 1958

Gurney, To RoFRAME INSTAB ILITY OF PARTIALLY PLASTIC STRUCTURES,British Welding Research Association, ReportFE 1/56/68, 1957' .

Low g Mo W0

SOME MODEL TESTS ON MULTI ...STORY RIGID STEEL FRAMES,Proceedings, Institution of Civil Engineers gVol o 13, p. 287, (1959)

Merchant, W., Rashid, Co Ao, Bolton,A o, and Salem, AoTHE BEHAVIOR OF UNCLAD FRAMES, Paper presented atthe Fiftieth A~iversary Conference of theInstitution of Structural Engineers, (1958)

PLASTIC DESIGN IN STEEL g American Institute of SteelConstruction g Inc 0, New York, 1959

Lu, L. Vo and Driscoll g Go Co, Jr'oBUCKLING TESTS ON MODEL FRAMES g Fritz LaboratoryReport 276.3 g Le.high UniversitYg (in preparation)

Galambos, To V0

INFLUENCE OF PARTIAL BASE FIXITY ON FRAME STAB ILITY,Journal of the Structural Division, ASCE, Vol o 86,No o ST.5,' (1960)

Ojalvo, Mo and LU, Lo W.PRECISE ANALYSIS OF FRAMES LOADED INTO THE PLASTICRANGE, Fritz Laboratory' Report 27606, LehighUniversity, (in preparation)

" .-

von Karman, To'UNTERSUCHUNGEN nsER KNICKFESTIGKEIT, Mitteilungenllber Forschungsarbeiten, herausgegeben vom VereinDeutscher Ingenieure, No o 81, (1910)

Ketter, R. Lo, Kaminsky, Eo Lo and Beedle, Lo SoPLASTIC DEFORMATION OF WIDE=FLANGE BEAM=COLUMNS,Transactions of the ASeE, yolo 120, po 1028, (19.55)

Galambos, To V0

Private Communication

Page 182: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

460 Oja1vo, Mo

RESTRAINED COLUMNS, Ph.D. Dissertation, LehighUniversity, 1960

470 Wang:, Co K.STATICALLY INDETERMINATE STRUCTURES, McGraw=Hi11Book Co., New York, 1953

=170

Page 183: )1-1 · 2012-08-01 · the problem of sidesway bucklipg of partially yielded frames is solvedo The method of solution is based on a modified moment distribution procedure in which

..

-171

12. VITA

The author was born as the second child of Wei-Yen and

Choong-Bei Lu on JUne 5, 1933 in Shanghai, China.

He attended National Taiwan University, Taipei, Taiwan,

China "from September 1950 to" July 1954. There he received

the degree of Bachelor of Science in Engineering.

After graduation he served for 11 months in the Chinese

Army and was commissioned as a second 'lieutenant in the

Corps of Engineers in August 1955.

He came to the United States in September 1955 and

started doing his graduate work in Civil Engineering at Iowa

State University of Science and Technology.- He was awarded

the degree of Master of Science with a major in Soil Engi­

neering in December 1956. From September 1955 to January

1957 he was employed by the Iowa Engineering Experiment

Station as a graduate assistant and engaged in Soil Stabili-

zation research.

The author joined the staff at Lehigh University in

February 1957 as a research assistant. In July 1959 he was

appointed research associate. He has:been associated with" '

the research project on Welded Continuous Frames and Their

Components.