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1996
''
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,.-iPjq,r't:2.-3 Corpciration.
n-
I
are registe¡ed rrade ma¡ks
M$'FORTRAN is
U
i
a registered
of Lotus Devølopment
.
:
tade ma¡k of Microsoft Corporation.
(
"
DEDICATED TO MY MOTHER
Dßclaimei The computer
contained in this book have been prepared with great
efforts. r-, lh€ author andPTry*t Publisher makq no warranty of any kind, expressed or iirplied, with ,.egjJg.F.telrograms orthe documentation .ootainrd in this book. The author l'ublisher shall not be liable in any event for incidental or consequential damages ' 7 âIld in connection with, or arising out oî the fumishing, performance or use of these fprograms' The usER shoui
,program¡
\*: COPYRIGHT @ NEM CHAND & BROS, ROORKEE,
í .
pT
L
(, a, f
of.this publication may bq Rrinted or reproduced by any means, now known or J{o nereatter urvented, or_translated in any language, without the prior written permission of the Author and the publisher.
ISBN 8l-85240-60-4
,.;
(.-;
(, _Production Supervisor : Arbor Consultants, Roorkee
(.)
(, i
1996
Published by {em Chand & Bros, Civil Lines, Roorkee 247 667,lndia -and Photocomposed and printed at the
,loorkee Press, Mahavir Marg, Roorkee 247 667.
t a
U {") (iv)
t-,' Ç)
O {.f {-.'1
U
o i) ü ü t-i
PREFACE
of ùe two prograæ ue providcd so tlrat the r€ader may have a. direct cxpos¡re to the computcr spplicdioûs and develops confidence. Earlier was planned that the sourr¿ listing of thc nropÐgams srAP-3D and CABLE-3D will be included in the book, but later it v¡Es dccided to make it available on a floppy so that the ¡r¿der does not have to go through tte &udgery of feeding into a computer and thcn checking the same till fi¡ll c¡nfidenec is dcveloped. Those who have gone through this exercise wilt appreciate ttb &cisioa. The floppy is available with the publisben of this book at a nominal
it
price.
/&1:
CONTEI\ilTS
.!
Ahttottgh written mginly.for the undergraduatc students, the practicing engineers e4ually useñrl. The inticacies of structural analysis have Ueeñ explaineA
will find it wiú
the help of over one hundred and fifty solved examples. over one hr¡ndred ana nry problems are included at the end of thc chapters and answers of most of them are given at the end ofeach chúþter.
I
am graæñrl
to
my students who
we¡p taught from
Chapter
1.
the manuscript over the
proriding heþfiri comments, for including Orpir"i -examples and improving its contents. The critical comments offerpd uy *y colleagrres have been utilized in the ftrat text. Dr. p.w. co¿uole deserves ã speciat _¿, mention who gladly provided I copy of hii cable program for use in this book. rñ"n¡.s to shri l.P.S.vcrma for meticulously and neatly drawing the tacings. I am especially grateftl to my wife saríta and children payal and Gaurav foi oeir patience anã past fifteen years for inadveræntly
Page
i.:
(_' (,
encouragement in Writing another last book.
I,
1995
Ashok K.
Jain
. 2. MATRTX ALGEBRA
2.1 Intoduction 2.2 Definitions 2.3 Matrix Algebra 2.4 Application of a Work Sheet 2.5 Solution of Linea¡ Simulaneous
(,, (,_
)
t
t-,
Equations
(i
PART
a'
I 4 7,
l0 l0
ll
t2 l3
26 - ¡¡e 26 26 29-
36 38
t : TLEXIBILITI,METHODS
3. METHOD OF CONSISTENT DEFORMATIONS
41 -91
\-/
4l
t
3.1 ¡¡¡svggvf¡9¡¡ Introduction 3.2 Choice of Redundants 3.3.
(_
e
"*inate
1.4 Stiffiress Method . 1.5 Sysüem Approach vs. Element Approach 1.6 Choice of a Method 1.7 Degree of Static Indeterminacy 1.8 Degree of Kinematic Indercrminacy 1.9 IllustativeExamples
:;
December
I vs. rndeterminare
1.3, Flexibiliry Method
_;
ü
l*i"ojìiln Stn¡ctures
r
,
f,-ì
1 -25
BASIC CONCEPTS
':'r
r,
3.4 J.5 3.6 3.7
Beams with One Redundant Beams with Two or More Redundants Reactions due to yielding of Supports Frames Trusses
Problems
42 42 53
66 69 79 o.t
Preface
,j
l
Ë.,
The aim of this book is to present up-to-ãate methodologies-in the arylvs]s oj earlier boók statically indeterminate structurìs. Thisbookis,a companign of.my entitled Elementary Structural Analysis which deals with statically determinate wide spectrum of structural strucû¡res. Thus this set of two books completes the into the flexibility classified a¡e anaþsis. The methods of sÍuctur.al. analyses methods and classical the both present' book, methods and stifftress methods. In the entirely to is devoted attention The detail. in matrix based mçthods are discussed There is sfuctures' indeterminate of statically the behaviou¡ develop understanding of sfrong emphasis throughout the text on the use of computers'
with Each chapter begins with the introduction and develops the algorithms along are examples The method. suiøble sign convãntion which'is peculiar to eqch classical chosen anã solutions arranged so that the fure clearly brought out. They selve
to ampliff
points of structural
.analysis are
and supplement the theory'
Chapter I provides an overview of basic concepts fo¡ the analysis of statically indeteräinate ìttt"tut"t. Chapter 2 deals with the elements of matrix algebra'. Now onwards the book is divided in two parts : Flexibility methods and stiffrress methods.
I
of
consistent deformations, three moment equation, strain influence cogffrcient method , influence lines method, energy method, column analogy form part of the flexibility- methods but are not do chapters two last and ãiches. The
Part
covers the method
included here for the sake ofcompleteness.Part2coverstheslope deflectionmethod, moment distribution method, di¡ect stiffiress method : 2'D elements, anü 3-D elements , and salient features of STructural Analysis Program Sfnp-¡O. A key feature of this book is a comprehensive fieatment of non-linear analysis of structures covering theory of plastic análysis, material as well as geometric non-linea¡ problems. The concepts of nysieresis models, unbalanced load vector, updating of stiffrress matrix, ductility and incremental and iterative methods of solution are introduced. The sequence of formation
of, ptrastic hinges is explained throu.eh exarnples. The salient features of C*lil,n-¡O program are also discussed. Several listings of sample input data and ouÞl¡t
\")
i
l*
{) {)
(vi)
{J
4. THREE. MOMENT EQUATION
f
{"}
CONTENTS
¡
92 - 106
4.1 Infioduction 4.2 Derivation of Th¡ee ' Moment Equation 4.3 Beams to Yielding orsupports
{.)
ii
l.
srRA,ru ENERoY
5.2 5.3 5.4
L]/ û
5.6
r) (,,ì
J
t08
Strain EnergY EnergY Theorems
109
t_.'
(i (-r
6.7 6.8 6.9
(.';
6'10
(
:
i-2. wrueucl
t22
Problems
149 154
157 -192
SymmetrY
Problems
190
,,)
7.5
'7.4
t
193 - 231 193
t94
Force Diagrams
195
Graphical Method of Integration
197
IllustativeExamPles
198
Problems
229
a-' S.INFLUENCE LINES
l8l
2'¡2 - 242
Ll R7
Introduction M¡¡ller - Breslau Princiole
232 232
243 244 247
252 254 257
259 272
274 276
279 - 942 279
Introduction
DevelopmentofSlope' Deflection 279
Equatiòns
10.3 Equations of Equilibrium l0'4 ,. Beams 10.5 Frames : No Side SwaY ' 10.6 Frames : \Vith Síde SwaY 10.7 Frames with SloPing tegs 10.8 Flexibility and Stiffiress Matices 11. MOMENT DISTRIBUTION METHOD
J
ll.l DevelopmentoftheMethod ll.2 Distribution Factors I1.3 Sign Convention I1.4 Be'ams and Frames with no Side Sway I1.5
ll.6 ' I1.7
SwaY
331
343 - 397 343' 345 347 347 361
375
Comments on the Moment Disüibution
Method Problems
12. DIRECT SNFFNESS METHOD. z D ELEMENTS
l2.l
315
Beams with Uneven Support Settlement 3& Frames with Side Frames with Uneven Support Settlements 372
ll.8 SYmmeryandAnti-SYmmetrY I1.9
282 283 290 295
341
Problems
184
\l
lO.2
l7l
Pqrtal Frames with no SYmmeûry
coEFFløENT METHoD
l0.l
l6l
177
lii;|rff'swith
9.1 Intoduction 9.2 Two-HingedArch 9.3 IllustrativeExamPles 9.4 Fixed Arch 9.5 Symmetrical Fixed Arch 9.6 Elastic Centre 9.7 lllustr¿tive ExamPles 9.8 Influence Lines for a Hinged Arch g.g Influence Linês fora Fiied Arch
10. SLOPE. DEFLECqION METHOD
160
168
24t 24¡'F278
158
163
235
PART 2 : STIFFNESS METHODS
t57
Stiffiress and Carry Over Factors Beams with Va¡iable Cross-Sectíon Portal Frames with One Axis of
one Axis of
AROHES
Problems
157
Uniform Cross-Section
it7.3 $ä'#h""
(
J.
"
ll3 ll6
Frames - Illustrative Examples
6.1I /
ExamPles
6.1 Introduction 6.2 Stess in a Column 6'3 DeveloPment of the Method 6.4 Sigrr Convention 6.5 Analogous Column Sections 6.6 Fixed End Moments in Beams of
L
'
107
6. COLUMN A:NALOAY METHOD
{,,
r0l
Work and ComPlementary Work
5.5. Beams' Illustative
a
Problems
to7 - 156 Tl,t",on?o0.,.,,""
IllustativeExamPlês
96
105
Problems
8.3
92 92
t02
i:.1 ïffi::"'due
'(vii)
COì¡TENTS
Developmentof.stifhess Matrices Tn¡ss element
396 397
398 - ¿r78 398
LJ
* {.)
(viii)
,,J
t2.2
Properties of Stiftress Matrices
t2.3
Transformation of Coordinates Element Load Vector Assembly of Global Matices Stiffiress matrix Load vector Effect of node numbering
12.4
{,,)
t2.5
* t:)
12.6 IllushativeExamples 12.7 Boundary Conditions 12.8 Support Reactions . 12.9 Incllred Roller Support 12.10 Summary of Direct Stiffness Method
{:
t;
l2.ll
ü
12.12
t, i-')
CONTENTS
CONTENTS
13.l
4ll
l'5.9
4t2
15.
l5.ll 423
15.12
430 434 434
(,
13.8 14.
l-r
(ì
fllustrativeExamples Comparison of Flexibility and Stiffiress Methods Problems
476 477
15.16 Ductilþ 15.17 Illusûative
Element Element Element
Stiffiress Matrix - Truss Stiffiress Matrix - Beam Stiftress Matrix - Grid
479 - 501 479 4g0
Method
'
Element
STAP.SD COMPUTER PROORAM
l4.l Inûoduction 14.2 Salient Features
5. NON-LINEAR ANALYSß : MATERTAL NON;,LINEAR|Tí
l5.l
lS.2
Inhoduction SFess-Süain Curve of Steel
15.3 Theory of plastic Analysis 15.4 plastic Hinge and Mechanism 15.5 Moment - Curvatu¡e Relation 15.6 Plætic Analysis l5 7
Statical method Mechanism method llhrctmtiwe Þv¡-^l.o
(
5¡ß - 589 543 543
544 546 547
548
APPENQ'X
16.2
Geometric Stiffiress Maüix 2D Truss Element Non-Lineqrity of Cable Suspension
591
Systems
594
Non-Linea¡ Solution Algorithms Iterative method
595
A- aROORAMS FOR SOLUT//ON OF
SIMULTANEOUS ÊQUATIONS A.l Gauss Elimination Method 4.2 Gauss-Jordan Method 4.3 Cholesþ Method A'.4 Successive Over Rela¡
590 - 622 590
16.5 Convergence Criteria 16.6 CABLE-3D program 16.7 User'slnstruciions 16.8 IllustativeEiamples
SO2
Sl2
E-xamples
Incremental method Incremental cum Iterative method
so2.542 502 506 506
579 580
Introduction
16,4
4gg
576 576
16.l
16.3
Transformation Matrix - 3D Beam
Element
573
583 583 585
16. NON-LINEAR A,NALYSIS : OEOMETRTC NON-L|NEARìT1
495
569
579
Incremental Displacementand Load Vector Unbalanced Load Vector Step-by-Step Incremenral Analysis
15.14 15.15
14.3 Adding'New Elemenrs ûo the Library 14,4 User's Instuctions 14.5 Illustrative Examples 1
Incremental method Hysterèsis Loops Assumptions Member Stiffiress Matrix 2D Beam element 2D Truss element Modification of the Strucural Stiffiress
Matrix
15.13
13.2 13.3 4g3 13.4 Stiffiress Matix - Shea¡ WallElement 4g4 13.5 Stiftress Matrix - Beam with Rigid Ends 496 13.6 Stçped Members ' 488 13.7 Transformation Maûix - 3D Truss
(-
l0
Non-Linear Stiffness Marix Analysis Iterative method
435 437
13. DIRECT STIFFNESS METHOD. SD ELEMENTS
i..,
15.8
405 407
(ix)
-
SLOPES AND DEFLECTIONS
C. FIXED END ÙNOMENTS
5n 598
600
602 L//NEAR
623 - 634 623 625 627 631
635 - 636
6t7 - 6¡t8 639 - 6¡tO
CHA.PTER
one
BASIC CONCEPTS
{J
o ü (,; 1.I
INTRODUCTION
A
framed structure is a network of a number of units known as members or elements;
(_;
A floor may consist of beams and slabs. A building may consist of beams, columns, floors and ioundations. Sim ilarly a roof truss is made up of top chords, bottom chords; diagonat members, purlins. ties and cover shcets. A bridge may consist of longitudinal Ueartrs, transverse bàams, deck slabs and even cables. Such cornponents are.referred to
(:
A.beam or a çolumn element, and a truss or a cable elem.ent are most frequently
(*,
O
C (,j
(l
as nembers or elements.
employedlin framôd sfuctures. These are one dimensional (f -D) elements since their' cross-sectional dimensions are very small as compared to theil lengths. The scope of this. book is restricted to the study of various analytical rnethods employed for the analysis of various 2-D and 3-D framed structures consisling of these l.-D elements. There are three basic requirements for a.unique structural analysis :
l. stress-strain relationship of the materi¿ils in the structure, 2. equationi of static equiiibrium, and à
conditions of compatibility or liinematics.
ifhes,e were discussed in detail in chapter 2 of volume 1 of this book.
(-r
STATICALLY DETERMINATE VS. INDETERMINATE STRUCîURES
(*) L
(
(bearns'and columns) connected by rigid or
semi-rigid jolnts' 'lt isrcomplete-ly
BASIC CONCEPTS
STA1'ICALLY DETERMINATË
rioith¡hl.{ o,'only axiál members, that is, pin-connected bars. A structure wlrethera truss, a çutìlinuoufi beam or a rig-id frame is either . stable ór unstable. and either sratically dglgrminate or statically indetennínate depending upon the number and ariangement of ¡t'tçrnl)cls, internal joints and . extemal supports. A structure uray be extêrnally illrlr-'lr'nrtirli¡te. or internalþ indeternrinate or both. Idealization of internal joints and tflitii[D¡11 $upports, deteÛirinacy and stability of structures were discussed in volunie I of lhil br¡ok, Figure l.l shows typical statically determinate beams rvhile Fig. 1.2 shows
(q
VS,
INDF,]'ERMINATE STRUCTURES
) 3 SPAN C0NTINUOUS
BEAM
t),picn I statically índeterminate beams.
Sirniltrly, typical statically detenninate and statically indetermínate franres are slrown irr liigs, Li and 1.4 respectively; and rypícal statically deterrrrinate änd indeterminate Ilrss"r ,u'b shown ín Figs. l.5a and 1.56 respectively. The stafically determinate
(b)
stlucturcs can be fully analyzed by usìng the equations ofstatic equilibriurn:
ÐF*= 0, X-Fy:0.XM. =-g I lr,''0. Ð Fr:0, ÐF,=0. E M-=0,E Mr=0, ÐM,:0 'f
structure) (3-D structure) (2-D
Fig. 1.2 Statically indeterminate beams
( l. I a) (.1. I
ç0NTINU0US BEAM
b)
ltc statically indeterminate structures cannot be analyzed directly and need additional
rrclr¡ntions due
to
conditions of conrpatibílity.
I ':
'(o) SlMPLE
(o)
P0RTAL FRAME
(b) CANIILEVER
BEAM
Fig. 1.3 Statically determinate frames
(b) BEAM WITH
(c
)
OVERHANGS
CANf ITEVER BEAM
LI
I
(d)
BEAM WITH INTERNAL
lt HINGE
(o) Fig. L
l
Statically determinate beams
(b)
Fig. t.4 Statically indeterminate frames
FRAME
T.-j
FLEXIBILITY METHOD
*4
BASIC CONCEPTS Qnch,'point
*
of the application'of the
redundant
force. The solution of these linear
sirnulta¡reous,equations provide the magnitude and direction of each redundant necessary to maintain cornpatibility of the system. Finally; the member end forces, end displacernents and support reactions can be determined.
d]
o ï
'í-,j
Consider a threè span continuous beam shown ín Fig. 1.6a. The total number of reactions is six. Theref'
{_} ,{"-)
{i f; Ç (.._i
(--, (
-,,
(.) (_'.
(', t)
(b) truss (b) Statically indeterminate truss dcterniinate (a) Statically FiE. t.5 ACTUAL STRUCTURE AND L,OADS
better itrength, stiffness and economy. Therefore, it ís essentiai to study the development of various methods of analyzing such structures. One of the important conditions for the analysis of structures is that the principle of superposition is valid. lt means that the structure is linear and deflections are small. The stresses in the structure are below the
proportional limit. The first fourtqen chapters áre devoted to the analysis of linear statically indeterminate stn¡ctures. Practical considerations require that there is
(b)
RELEASED STRUCTURE
considerable strength in the nonlinear region of a stiucture which should be fully exploited. The non-linearitl, may be due to material or geometry.. Material nonlinear analysis is discussed in chapfer 15, while geometrical nonlinear analysis is discussed''in chapter 16.
I.3 FLEXIBILJTY METHOD
{
c)
The most powerful method for the analysis of staticälly.firdeterminate structures is the method ofionsisteht deformations. It is also called as the flexibititlt melhod or lhe force method or the compatibility method. The application of the flexibility method requires that fìr}t the structure be reduced to a stable, statically determinate system. The problem then reduces to establisþing a set of independent; simultaneous equations in-terms of unknown forces or.redundant actions so that they may be evaluated and anaþsis ofthe indeterminate sfiucture completed.
(
(q)
ln practice.staticätty inaeièrminate structures are frequently encountered because of-
When the statically determinate or the released stiucture is subjected to the applied it will uridergo deformations that are inconsistent with the behaviour of the original structure.. However, by applying forces equal to the released actions, the deformations of the released structure are made consistent with those of the original struiture. In analyzing the released Structure, the displacement at the point of application and along the line of action of each redundant must be êvaluated. Although the rnagniiude and direction of the redundant actions are unknown at this stage, the reledsed structure can be analyzed due to the application of4n assunled unit value ofeach. redundant, successively. Finally, considering displacement of released structure at the point of application of each redundant action, due to both the applied loads and the individual unit values of the redundants, a set of compàtibility equations can be written. These equations/ describe the actual unknown displacement ofthe origînal structure at
DEFORMATI0NS DUE
tn= Ø¡e
(d)
FORCE
'M4
t
1
OEFORMATIONS DUE TO UNIT FORCE RB
I R^= I
loads,
¡'ì
T0 UNIT
1*r=
(e
)
DEFORMATI0NS DUE
Fig. 1.6 Developmênt
T0 UNIT FORCE
of flexibility
method
R.C
t_i
{} {} {"}
{)
,r.-:'1,, [Iy renroving .r,,,
\";.::. lt.
',tJ ..'{-.,
.)..,:4.,
,{_:
.
the supports at B. C and D. it reduces to a deteiminate cantilever
hë$tn.
the support A, it reduces to an unstable continuous beam, hence it ls not a feasible solution. ßy rernoving the restraint at A, and the supports B and C, it reduces to a simply supportcd and stable beam. ,
'."å,
l.ct
0oo
rotation
'lheir magnitudes and Therc are three unknown actions or redundants, Mn, Rn arfd diroctions are not known. Let us assume that the moment Mo is anti-clockwise and the ¡S¿rçtions Ro and R. are acting upward. Let us apply unit redundants on the $tructure, one by one, and d€tennine the deformations at the points of application and direction of each redundant as shown in Figs. 1.6c, l.6d and 1.6e. The slopes and deflections can be de(erntined by using any one of the sevéral rnethods ,(viz. mornent-area method, conjugate beam me-thod, or unit load rnethod) as discussed in volume I of this book. The number of equations is equal to the.number of redundants. 0e = that is,
fe^o
I *looo fooo öou- oo.1 [vtn I Âss o'. *'
(-)J^Bol
.",
[AcoJ LÂ.o Âcn
(...,
'
'
tFl
I
(,)
Acc
jlJIRc Jf
{^rq} + [F]{R}
f
0
o
(t.3)
0
'
t
air,
tFl
{R}: FR:
{^,o
written
as:
}
À
(
l.6a)
(
r.6b)
This equation represents the defonnation - foròe relationship. For a single elemept. the term F represe nts deforntation per unit force and is called as flexibility. Knowing Å and F, ti can be evaluated which leads to the solution of the c-omplete structure. The ¡nethod of consistent defonnation is one of the earliest methods in vogue (Circa 1860). Theothermethodsthatcantàll inthiscategoryarêthreemotnentequation,strain energy rnêthod, column analogy method and influence coefficient method. These will be discussed in detail in the subsequent chapters.
I.4 STIFFNESS METHOD Another approach for the analysis of statically indeterminate*structuies is fhe stffie.ss tl'te tlisplacement method orthe equilihriuh method. Letvs reconsider the three span continuous bearn which was used to develop the flexibility mèthod. The application of the stiffness method requires that a staiically indeterminate structure be first reduced. to a kinetnaticalb) determinate system. A kinematically deterrrrinate iystem is the one whose end displacements are known. Kinematics relates the deformations and displacements of elements. The first step is to identify the dègree 'of kinemetic indeterrninacy and. there-fore. the unrestrained joint displacements. Now. the correspondi4g artificial restraint5 must be introduced so as to make it a kinematically determinate struclure, that is, a reslrained sffucture.
metlzod. tt is also called as
each restrai¡ring reaction is equal to the sum
=o
(t.4)
(a) (b)
= {^r} (
t.5)
deformation vector due tô the actual applied loads; the first subscript 'i' represents the location of the redundânt where the displacement is computed and the second subscript 'o' represents the applied loads. the elements 4 ¡ of this rnatrix represent the displacement Â, in the ditection and at the location of the redundant \ caused by a unit action redundant R, acting at the location j. This is called flexibility redundant foröe vector
total known displacernents at the location and direction of
the redundants in the actgal structure. Some of these displacements may not be zero as'in the case ofsettlenìent ofsupports.
of
the fixed end forces required to constrain the ends of the members that franìe
i¡rto thatjoint,
natrix.
lRl= t \ s
is zero. Eq. 1.5 can be
ot
[^cJ Â,.
}
This structure must now be analyzed for the aitual loading imposed on the original structure. The suoport reactions of the restrained stluctüre for any loading condition are sirnþly the action required to constrain the various joint displace¡nents. At each joint,
r^'Ì
=lo"f
-À.o* FR:
or
: where {4....ì \ ¡('l .::.i
-
or
ì..:
:.
('.
(.,
(1.2)
in rnatrix notatbn
t)
\..
Â.:0
+ ôaaMa + 0^reR, + OocR.:0¡: - Âso * Âa¡-M¡ * Ass Rs + Ânå Rc - Âs : .- Aco * Àco Mo o .Âcs Rs * Âcc Rc : Ac :
and
r\
\a.-,)
lr-
0, À, :0,and
Ooe
{.,j
{",:
(..
'l'he compatibility condition iequires that
'{),
(.
ih Fig. 1.6b. The simply supported beam at A, and deflections Âro and A.o at B and C, respectively..
rrs aclopt the third option as shown
Unclerg,ocs
{.)
ln case vector {4.
. By rcuroving
"ì,
tlt) t")
STIFF}{ESS METHOD
BASIC CONçEPTS
and
'
the action equal and opposite to the force acting directly at tllejoint.
The restrained structure also needs to be analyzed for displacements of lhe artificially joints. But magnitude and direction of dísplacements of the unrestrained joints are. unknown, It is, therefore. convenient to assume unit .value of each of the artificially restrained joint displacements and analyze the reStrained structure for'each displacement individually. Since. the artificial restraining support actions do not exist in tlre original structure. a set of independent linear sirnultaneous equations can be written irr terms of unknown joint displacernents. The number'of equations is equal to the number of constraints required to make the displacements zero. The solution of.this systetn of equatíons gives the mâgnitudd ahd direction of the joint displacements neçessar.y to maintain equilibrium of the systern. Finally. member end forces and
restrained
"$upport
reactions can be deterrnined.
ä 3
BASIC CONCEPTS
SYSTEM APPROACH VS. ELEMENT APPROACH
Ë :l;
ionsider the three span continuous beam shown in f:ig-
t.lu.
The unrestrained
is shown clisplacernents are shown in Fig. l .Tb and the corresponding restraiúed structure end actions fix-ed and in Èig. t.2". The unrestrained displacements are 0u,0. These are l.7d' in Fig. are shown structure due ¡o the applied loading on the original shown in are reactions support net joint and the displacements, required to constfain the
eo llt
Fig.
1.7e.
Now remove the applied loading and-impose unit joint displacement,_successively, and evaluate the support reactions as shown in Figs. l3f'.l.7gand l:1h'. [, , represents moinent at joint i due to a unit displacement imposed at jointi. Thus, Ki, is force per unit displacement and is called stifftess or stifness coeficient' Similarly' R' j rep-resents vértical reaction at joint i due to unit displacement'imposed at joint j. The equilibrium equation can be written at each joint, that is.
Z'Ms = or,
Mso
and
Mco Moo
+
*
+
0, tMc:0
KsB eB + KBc ec Kcs, eB + Kcc ec Kps 0É + Koc 0c
unq EMD:
0
+ Kro 0o:
0
+K.o0o:0 *Koo0o=0
(1.7)
l'
I
i.: LI {.' ,:,
Ivr*J
{P} + [K] {^'}
or, where,
() (,.,
It (
{ (
LKo"
or
or,
(1.8)
:
K". r"o'l Ie'l Kcc *." llo. l=9
{o)
ACTUAU STRUCTURE ANO LOADS
.11
t' ti1
rli
roe
-
.
Kp.
: tKl {^} if : - Â' for conveniencè ^ p=KÂ (l.lOb) {P} : equivaient nodal load vector due to the actual applied loads; the f,rrst'subscript represents location ofthejoint and second subscript 'o' represents the applied loads. elements of this matrix k, represent the force.M, at joint i in ' due to a unit displaôement the direction of the conitraint joint matrix. called stffiess This is applied at unknswn disPlacement vector
. {^} : P : KÂ represents
RESTRAINED STRUCTURË
j.
the force-deformation relationship. For a single The pquation represents element. the term K .þrce per unit deform.ation and is. called as. stffiess. Knowi¡rg P and K, À can be evaluated which leads to solution of the complète structure. The.slope-defleition method and rnoment distribution method äre in vogue since l9l 5
and 1930, respectively. The other method that falls in this category is the direct stiffness method. A relatively recent method, it has revolutionized the concept of
ffi&,
^ÐL-<ø e, e41)=.--)-4,
1 1-
I
11
ACTIONS DUE TO APPLIED LOADS
I Í-L t frrín.+, I"'ry+ry Mc9 Mao
, xae {¡'Rle (fI
eo eo M
l-1¡ 'R
l^
'Roo
øe=
W Kea
t-1.,
rþxac 'Rac
-1.'
f0
Koa
'1, 'ROe
UNIT RO¡ATIoN SB
lw \+, ^cc 'Rcc
*î, ^r, Rec
REACTI0NS DUE TO
Kca
'RCg
'ReB
REACTIoNS DUE
(9)
r.f¡ Moo
{-,'Rco
REACTI0NS DUE TO APPLIED LOADS
.tLítl
(l.l0a)
{0}
too
UNRESTRÂINEO DISPLACEMENIS
(c)
(e)
KooJ leoj
\oc .
(b)
(l.e)
{P}
tK] :
Ê
'$r 'Rro
IM"o] lr"' 1M.o l*l K.u
1..:
iã
id),FIXED
in matrix notation
i
ç
.UN
IT
ROIATION
q
Kco CD
ec
r-.oD=
Jrrro 'R¡D (h)
'J-r
xeo
'ReD
1
xoo {-,1.n {,'Roo ' Rco
REACTTONS DUE TO UNIT ROTATION 9o
Fig. 1.7 Development of stiffness ¡iethod
€"-J
t0
r.)
powèrful eléctronic etfUçtufttl nnalysis due to the simultaneous development of chapters' the subsequent in in detail be discussed w,ill methods içif,pUi*,,*, I'ltesc
{:)
* {-)
\) {¿,
iJ
t) {) {)
t:
BASIC C()NCEPTS
i¡# SYSTEM APPROACH VS.'ELEII{ENT APPROACH weie developed and perfected over a number Vnrious methods of structural analysis 'century. There were no. computers at that timg, ol. ycnrs during the mid-nineteenth physicál reasoning.of the structure' lh(ìreforc, methods of analysis had to be basLd on the well known, but it could not atfract the very was algebra matrix of nfit Ough the concept as the consistent deformation such Methods reasons. obvious for analysts the oite,rtioi of method and moment distribution method slope-deflection method, energy strain tngth6d, considered the structure as a whole. matrix w¡th the advent of digital computers around the middle of twentieth century' were methods and flexibility approach became attractive and the stiffness methods the through developed were structure the entire for alvefop"O. The F and K matrices as be known to came approach This constitutive.elements. for the ,.rp."iiu. rnatrices a digital computer' by analysis an automatic for appropriate is tnore lt nfu)uunt ctpproach. Having understood the concept of flexibility and stiffness methods. the analysts took discover that the another look at the conventional methods. tt was a pleasant surprise to method' Since earlicr methods could be classified either as a flexibility or as a stiffness approach' as system was called they considered the structure as a whole, this approach and quite tenn new relatively is a approach eletnent or ]-hus tt¡e classification of system
(
convett ient.
(
The flexibility method is also known as the force method because the forces are is'imposed to treated as unknowns. Since the condition of compatibility of displacements the Simi{arly. method. the compatibiliry as known is also it equation,,, final the generare
(,, (,) /.
-,1
(., (. ,ì
\.) t,, (,' (-
nrethod is also known as the displacement method because the displacements are imposed to generate Ireared as unknowns. Since the condition of equilibrium of forces is is necessary to method. the equilibrium as thc fi¡ral equation. it is also knôwn
ititfn.$
lt
understand that both these approaches make use of equilibrium as well conditions.
as compatibility
It iS convenient to write a general purpose "oniput", programme using the stiffness very efficient nretlrod for static or dynamic analysis of any framed structure- lt is also method. for íhe non-linear analysis ofsuch stiuctures. lt is referred to asthe usetfriendly selecting of possibilities l"lre flexibility nlethod cannot.be generalised due to several piOper reduniants and the associated difficulties in writing simple and straight forward aigåritlrms. l{ence, it is not much in use. Nevertheless.-the classical flexibility methods prõvide n deep insight into the physical understanding of the structural behaviour and are ncccssary to study.
I.6 CHOICE OF A METHOD analyzed using the force approach or the displacement approach' number The question is how to decide whjc method is bctter. As discussed earlier, the
A structure can be
il
DEGRËE OF S'I'A'I'IC INDETERMINACY
*
of equations in the force method is equal to the nr,¡mber of redundants, whereas, the
numher of equations in the displacetnent method is equal to the number of constraints required to make the displacernents zero. ln other words. size of the flexibility matrix is equal to the degree of statical indeierminacy (qs), whereas, size of the stiffness matrix is equal to the degree of kinematic índeterminacy (a*). Obviously, a method which leads to a fewer number of equations. is preferable. Thus. first step is to determine the degree of statical indeterminacy as well as the degree of kinematic indeterminacy and the method of analysis can be selected accordingly keeping in víew the convenience involved. It is also possible to adopt a mixed approacå in certain types of problems. The mixed approach is beyond the scope ofthe present text book.
I.7 DEGREE OF STATIC INDETERMINACY lf all unknown reactions in a structure can be uniquely
determined with the help
of
equations of static equilibrium, the reactions of the structure are referred to as statically
deternlinate. Otherwise. the reactions of the structúre are referred to as statically indeterrninate. The degree of indeterminacy is equal to the number by which the unknowns .exceed the available equations of statics. . The total .degree of static indeterminacy of a structure is considered as sum of the following two types of indeterm.inacies:
(?) degree ofexternal indeteiminacy and (b) degree ofinternal indeterminacy 'The external indeterminacy
is_ related to the number and type of supports.,, The intÞ¡nal indeterminacy is concemed with the determination of all member forces knowing'the support reactions. In a pin-jointed truss. if the number bf members rneeting at a joint is just suffìcient to preserve its geometry. the truss is internally determinate, otherwise the truss is internally indeterminate. Similarly, a rigid- jointed frame is internally determinate if its members forrn an open configuration. that is.there are no closed cells, otherwise the rigid-jointedtframe is internally indeterminate. Alternatively,,if more member actions are presen{ than can be solved for from statics . the frame is internally
indeterminarc
a 2- Dpin-jointedtruss. a, = (m +r) -2j For a 3-D pin-jointedtruss, o, = (m+r) -3j For a !-D rigid-jointedframe, as: (3m+r) -3j For a 3 - D rigid-jointed frame, qs: (6m+r)-6i ,
(l'l2b)
is said simply to be statically
indeterminate without stating whether it is
For
(l.l la) , (t.l lb) (t.t2a)
The total static indeterminacy is given by the above relations. The degree of external indeterminacy is easy to calculate. The total number of external reactions minus the Irumber of available equation of static equilibrium gives the degree of: external indeterminacy. Thus, the degree of internal indeterminacy çan be computed. Often a
$tructure
indeterminate internally, externally or in both manners.
-
6'.--)
¿- ,\
e)
* ()
t) Ç
t)
BASIC COI.ICEPI'S
t2
ILLUSTRATIVE EXAMPLES
I.8 DEGREE OF KINEMATIC INDETERMINACY when determining deformations, there are no
I.9
equations which are analogous to the
{) (,;
and the applications of stiffness.and
Example
.ì-
Soiution
Pin--jointedtruss,
qx:2i - S' -St For a3 - D pin-jointedtruss, cfk : 3j - Sr'- Sz For a 2 -D
(-,;
'
(_
or.: 3j - S' -St
(l.l4a)
dr.: 6j-
(r.r4b)
For a 3-Drigid-jointedframe,
St -S?
great extent.
:
Total nurnber ofexternal reactions r 2x =6 Total number of.members m = Total number ofjoints j 6
+ 3 +.2 =
g
6
Q'
. wherp, -
\
(,.j
dku:2i -
(l
lf
r"\ a
tlre number
S¡
=
(2Sr+ 2Sh+Sr+m)
Total members
Eq.
(
cko
2t+3'2r12:=
0
'
, ..
.
'.;.....
6)
.
I
t.2
of
I
(1. l
J
3,
18, reac-tions r.= 4 :. 4l + 4 - 2 x lB: .9 ' Degreeofexternal i¡determinacy = r:3:2+, I+ l.-3= Degreeof intemal indeterminacy = 9- I:8
pl3nètifid-joiite.l frame
':j"l T;.
(m+ r)-2j= determinate.
Total statical indeterminacy g.
the total degree of kinematic indeterminacy
6
.
Plane Íruss (Fig:
number of fixed supports
a*t
m:21, joints j =12, ¡svç¡itrs r = :
I.llagives,. d, :
1,5b) Total memberr r'= 41, joints j:
l. r5)
: numbenof hinged suPPorts S, : number of roller suPPorts m': number of members is
6
=
The truss is statically
S,,
ofjoint rotations
5
Plane truss (Fig. l.5a)
/.
.
3:
{rr:3x6+6-3x6= Degreeofexternal indeterminacy : r-3= 2x3- 3: Degreeof internal indeterminacy = o. -3 :6-3 = 3
The number of independent joint translation in a plane frame is glven by tlre'equation:
'\.
('
3
frante (Fig. l.ab)
Hence,
possible Under certain conditions, in rigid-jointed plane as well as space frames, it is dgformations' axial the by ignoring indeterminacy kinematic of degree the to reduce irt"t, ¿"*p"tatiãns in the classical methods of structural analysis wére simplified to a
l
r :
numberofmembers m =
d, : (3m + r) - 3j :3x6+8-3x7=5 Degreeofexternalíndeterminacy : r-3:3+3 +2- 3: Degreeof internal indeterminacy *- crs -5 :5-5:0 Plane
(,,,
(..
j - 7
using Eq. l.l2a,
(r.r3b)
rigid-jointed frame,
ln most practical situations. the supports are unyielding and; hence' St
,,''
frame (Fig. l.4a)
nurnberofjoints
(l.l3a)
i_;
('
Plctne
Total number of independent external reaction'i
Total statical indeterminacy
For a 2-D
{_)
(--;
l.l
Determine the degree of static indeterminacy of plane frames shown in Fig. 1.4, and the pir-jointed trùsses shown in.Fig. 1.5.
Jirpluc"rn"nt, is 3 or 6 depending upon whether it is a plane frame or a'rspacc or the number of independent joint displacements is also called as the degrce'olfreedom joint or of the node. If the number of fully restraint support displacements is S' and' ino*n support dispiacements is. S, the degree of kinematic indeterminacy can be computed by the following equations:
{.:")
f';
ILLUSTRATIVE EXAMPLES
The following examples illustrate the concept of static and kinematic indeterminacy, fleribility methods in simpie cases. .
exceptions' are equations of static equilibrium. Therefore, all struòtures. with certain a nrenrber are known' then of ends the Of deformations If indeterminate. kinematically A fixed-end the deformation at any other point in the member can be easily deterrnined' is kinematically beam supported simple a beam is kinematically determinate and equal to the indeterminate..The degree of kinematic indeterminacy of a structurc is ¡nember of the tl'uss, pin-jointed a ln components. displacement number of independent is a plane it whether upo¡ joint 3 depending 2 or is' each at translations independentjoint independent of 'ln a rigid-jointed frame, the number trúss or a space truss. frame' The
L,{
t3
'Þete¡ntin"
the dcgrees of itatic and kinematic indeterminacies for the multistoreyed e shown in Fig. 1.8a. Assume that axial effects, that is, changes in member lengths bo ignored.
IILI"JSTRATIVË IiXAMPtES
BASIC CONCEP'I'S
l.Ël
K i nena!
. .'.
I5
ic i ndetcnn i nacy Total number ofjoints =33 Degree of freedom perjoint - 3 Known zerosuppondisplacementsSl = 3 x3 =9 Degree of kinematic indeterrninacy cru: 3 x 33 - 9 = 90
If axia! deformation is ignored,
.'.
floor : I : l0xl:10 Vertical displacement at each joint = 0 : I Rotation at eachjo¡nt Total rotations in the frame 30 x I :30 Total unknown displacements in the frame : l0 + 36 = 49 Degree of kinematic indeterminacy an : 40 Lateral displacement at each
Total lateral displacements
I Thus, the degree-of kinematic indeterminacy ' axial deformations in the members.
cr* is reduccd by 50 by neglecting the
Alternatively, Kinematic indeterminacy can atso be determined using Eq.l.l6.
j: i3, Sr:3, S¡,:0, S, = 0. m :50 dku : 2x 33- (2x3+0+0+50).: l0
(q) Fig. 1.8
s
í.!r'. l:i: t::.' ii.:-¡1
¡:r;. !:r:,r
r,
ii¡.: ¡iì:r,
í
.!irj.r
.
.:i:l ll!::'
Solution ,{ilat
0te
=30
clk
= dku+
Okg:
t0+30:40
o. K.
ic i nde te r nt i nacY
'..',
Totalnumberofexternalreactions = 3x 3=9 Degreeofexternal staticindeterminacy = r- 3:9'3:6 applied to the left
.ltl a structure is cut at any point, three forces or actions must' be ,"i6oÈitt" cut and three equal and opposite member forces must be applied at .
.¿pîUti¡Ar¿nA o{the cut to maintain *hoâr.lbrçe and a bending momenl
continuity. These three, forces are
:
the
an axial force. a
Simply is a convenient method to determine the degree of static indeterminacy. but determinate to be,statically a slructure to reduce intr,Oduqe as many cuts as necessary is stable' ensure,that each cut portion
ïhis,
'Ihe sturcture is reduced to three ln the present frame, introduce a cut in each beam. 3 unknowns per cut. Thus' total and 20 cuts, are There ¡n¿euerr¿eit vertical cantilevers. j degree of internal static Therefore"the x 60' 3 is 20 ;ü;Ëtt". ;*tic indeterminacy indctciminacY, Ís 60
- 6: 54.
'l'he totål decree of indeterminacy can also be found using Eq' I ' l2a'
Total
: 50, Totaljolnts j : 33î Total reactions .r 5 cls:(3m+r)_3j
ñiembers m
.,:i.. , i
;
.
iÈ:;t.:r r:,
.
= 3x50+9_-3x33:60
9
o. K.
Po as in
ILLUSTRATIVE EXAMPLES
BASIC CONCEPTS
l7
tf*o bruÄ!-+åe-v P
A
@
displacement
'forePring
t'
^Po :, =K,
or if
t'o
= ¡g kN' Kr =
" '--tor snrinsrfr,
100
n. ='lo -r 100
and
(4,
Po: Kz(^r -
-
150 kN/m
¡)'
(ii)
A¡)
The equilibrium
TlJ
ofjoint
a, =
or
ar = jl' *6'¡ î50
::ils;-l-r:
rç{ìr,
= o'167m
Pn
lK-+K:ìÂ:
'
.t,, .i,
same
p-
o,
Á=
Po
K,+K, r:i.i jÉiriì ¡f Po : lOkN, Kr : l00kN/m, K, : l50kN/m,
DÞ-P^ =.ü.Ë, uno. ^, = rî
P-
(iD
Po
Lii.
: let ùs rearrange Eqs' (i) and (ii); To obtain the equivalent rpring tii-ftn"ss, Lì.
Fz:
..:,,. lt may be seen that spring I elongates by Â, whereas spring 2 shortens by the ,,i,¡,. amount. Substituting the values of F, and F, in Eq. (ii) gives,
f,+a'
(i)
B
Fr +
or
+A
K2 F2 F? ---:+ o---_-4Å,¡V\Ày'rl\ÁÄfu-< + ia @,¡, tc' Fig. l.l0 and F2: K2 á!s' F¡ = K¡ À ^ gives,
= o.lm
net elongation in spring 2 is
,.,:tl'
Kr:
(b)
(o)
Fl
+e---{¡\¡\,\¡\d¡\'t\,t\y'e--:io+
' kN/m
Kt
Fl
(i)
,;'
,.ii $;¡i:,'
Po
'
^=
lo roo+ l5o
=0.Ò4m
:1,:..
KrK, or, ^"0 = KlK,
r' '': l'hus, ,pringt Eq.(iii), which
ExtmPlc .;t
I
*itt
(iiÐ
whose stiffness'is given by and2 can be replaced with a s;ingie spring'
gin" the same displacement'
load Po' Determine system shown in Fig. l.lOa is subjected !9 T Tiul B' ofjoint ,fto'inttttit"äi.aã i" the two springs and displacement
.,;;*'*urings
$oruui;
:
springs aro $hown springs: ;ñ;
ü"
ir nËr.ì. iõu""¿
-
f1e1'bodVdiagrams of the can be wrinen for relations J. rrr"ç,rotæiefoniation
and F'' ¡,ett'forccs iróduceA in the two springs-bef'
''
. . .: ',
1.4
:
Fl = 100 x 0.04 : 4 kN tension ;; i'". and F2 : 150 x (-0.04): -6kN compression. coml 'i.ì, .-' ,,. Example l.$
lhe
.j,
t
'A
Fig:l.lla is subjected to axial loads P, and pr. d,¡, and spring forces, if they are .connectéd through
four springs systern shown in
Determíne the displacements a weightless rigid body Q.
À,
and,
,o,u'on
ra
:,lrìfi Since the springs are intercqnnected through a rigid body .Q, the compatibility ,,-: c.ondition requires^that ^springs.l, 4 and 3 must elongate by Â, each. The free body ,;ji-Ëdiugtutt of
ii;liã'.,
each of the four springs are shown in Figs.
Fnr¡c-rlcfnmafinn ralqfinnc give oir¡p Force-deformation relations
l.l lb to d.
BASIC CONCEPTS
l8
ffrr¿¡arJe¿rl
P2'aZ
I
,ct
lixanple
K,
*,
l\ '.
.51¡
kN/rn. K, = 75 kN/m,
50
kN/m'
Ko = 60 kN/m, P, = 20
kN and
fzoì
L-uo roojla,J: iroJ f
0r
¡,
ì = [o.rt tl
iorl
(i) eive
lo.:ozJ*
{l}j {ri,ïi-
1;1.:
:'r :r
:
Irrs -60lf^,ì
Ëq. (iv) gives,
Eos.
K.,
40 kN.
[Fo
.
13.34
J
(.,.ì
&rgmplc l.6
(_;
. ' Weightless bar ABCD is supported on two spiings as shown in Fig.l.l2a. The *$ritrg flexibilities are f, and fr. Delermine the spring foices and rotati
(_ì
F1e P
(_,;
F1
l,
(e)
D
Frr+
(--) ì
(:,
Fig'
(.,'
(.'
(,
I
Fr
: Kr Lz,
Fr= Kz
Ap
l'll F, = K, Â'
'
F¿
= K¿
(^2
' ^l)
,L/3,L/3,t/3, f-T-î.----------i . (o)
(i)
Joint equilibrium equations givei 'l
and
:- Fr-Fo :Pt Fr+Fo+F,:P,
There are six relaiions and six unknowns F1 , give' values of Fr to F4 in Eqs. (ii) and (iii)
and
(Fig'l'lle) F2
, Fr,
Fo
(ii) (iii)
, A, and A"' Substituting the
KzÂr-K4(42-À')=Pr KlOl+Ko(42 - Ar)+KiAt:Pt
(b) Fig.1.12
$olution '1i.. ,,1
¡,,i¡:.:
,[f ,bal AD rotates by 0, (Fig.
^:
-JJ
l.l2b),
Lo. ¡, = !o
the compatibility,condition requires that
and
L,,
=Lg a
(Ð
2t
ILLUSTRATIVE EXAMPLES
PL
=
Rr
l."rT
'o-Lr-2Lz 3\'3f2
(iv)
; :l. t' Substituting the values of À, and Â, from Eqs' (i) in Eq' (iv)' R2r A2
-o=!9*L9 9fi_ 9 rz
o "
""' " l'
Let
,' ';''
11
, ,
R4, A4
fll =(4fier'E + fr) \ L/
(v)
,1,
"tl.
.1,'
P=20kN, L=6*'fr=0'005m/kN,ft:0'010m/kN' e=0.05radian':2.86" À, :0'l0nì' az :0'2oin' Rr :20kNarid Rz:20kN
rhl
,:i;¡-.,f1',.r
,,.., fhe momerti equilibrium equation gives, r Mo: o :a1.,
.
'.i
'.,l*.
=:0, t,
Âr=ae and Âo=Lo = =fe, ã
PL:Rr*!+R2xf**r"**RoxL-
or
iîffiou'*uting
Ar
Eqs' (i) and
Ëîi.
,=[*,(*)'.,.,(;)'.*,(;)'.*.]*
(D
A
Rj:Kra, ,.t.,1'.tta,, Rl=Kl^l , R2:KzLz, whcro, Kt
, K2, K3, K4 :
*:,'*r, \,
R¿
(ii i)
(ii) in Eq' (iii)'
by A, and Ân' Spimgs I and 3 compress by Â, and A, whereas, Springs 2 and 4 elongate Tlíe.force-dcformation relations give, ,:
I
Pl
=
spring stiffnesses sPring forces
and
Ro:K¿a¿
(iv)
(ii) Eianple
Let
K, :30kN1m, Kr=
40
kNim, Kr:50kN/m, Ko = 60kNim,
ZJ
ILLUSTRATIVE EXAMPLES BASIC CONCEPTS
1l:
: 20 kN and l-= 6m e : 0.033 radian : l.9lo, R, = - l.485kN comPression, R, = -' 7 .425 kN comPression,
Solution
P
[ìxnntplc
Lt:
0.198 m Rz= 3.96 kN tension R¿= I 1.88 kN tension
The joint o occupies position o' under the application of vertical load W as showil in Fig. l.l4b. Displacements are small. o'o" = u, oo" = v Considering the equilibrium ofjoint o'
I-et
EF*= ÐFy:
1.8
Eqs.
Apin.jointedthreebartrussisshowninFig.l.l4a.Determinethememberforcesand
0, T, cosa = T¡ cosP 0. T, sino *Tz* TtsinP : Y
(i) and (ii) can be arranged in the matrix form
(i)
(ii)
:
loittt displacements.
foì l-coscr
tùl=l .or. A3,E
A2rE2 .12
A1,El L1
30"
L3
450
.
:1?1.
,:
'
'
[,
forces
equilibrium matrix
^l:âo'='ae-eo' Àr .vsina-ucoscl
Or
Âz= oo" =
(iv)
(v)
v
Member 3 elongates by Â,
co': cd + do' ^3: À¡:vsinP.+ucosB
or or .
|.¡'] [ -cosa {Â"i=.1 "[ 0
Io,i
(''':,
.cosB
sinc¿l, ,
,J u] tTl
Â:RU
or,
iiC
(vi)
Eqs. (iv), (v) and (vi) can be arranged in the matrix form:
¡'--t:,'
(vii)
Here R is the kinematic matrix relating the member elongations vector A to the joint displacement vector U. The transpose of the equilibriummatrix is the kinematic matrix. This can be proved using tlre principle of.virtual :
li
,iL)
work. ,
,--.,
\,,'
Force
-
deformation relations give
I
ic
\
.\
I
i(.'
(iiD
Member 2 elongates bY A.
'
t\
|
fl ì ji;lf
.",81 ,,''p
{Tl , Tz, T.,} is'the vector of internal {0, W} is the vector ofjoint loads
"' or
r"ì
'i,
I
P: RrT
where T : P : Rr :
(o)
î,:!,'
0
There are thlee unknown foices but only two equilibrium equations, hence compatibility do¡ditions are required to obtain a solution. Considering the compatibility conditions, Member I elongates bY Â,
lì
a'\:.';:
3
sinc'
-
Fig.
l.l4
Pin-jointed three bar truss
^,'
=
T,L,
-:'!:L ArEr
ILLUSTRATTVE ÊXAMPLES
BASIC CONCEPTS
AtEt^,
T,,LI -
'
T..L, =
Similarly,
T" = 'L3
and
(viii)
AzEz
AsE¡
...
o.
o
oI
L.o
o
n3.r5]
k: | 0
(ix)
o,
fr:r.se
400 0 |
K=Rrk*-frt-zoo t_3.420
(x)
from Eq. (xi)
-3.4201 s60.s22J
K is square and symmetric,matrix
Those in matrix notation are
P= KU
4rEr [1] {Gl=
AzEz
o
[t,J
0
{^,}
K
(xi)
A¡E¡ [^;J
0_
P =
where
Io']
0
L2
;ï,; KU
(xii) (xiii)
fromEq.(vii)
(xiv) (xv) (xvi)
"
RrkR
$nowing u and v from Eq. (xv), member forces can be obtained from Eq. (xi) or (xiii)'. {[l-is metho¿ is referred to as the displacement or stifftress method since displacements are the üiiknowns. "
Numertcal Example
.'.r"',,, L,"t. Ar : Az : r., Er :E¡ :
As: 6
cmz
, L"-: 300 cm, ø : 60o Q: Ez:2 x 104 kN/cm2,
45"
8 x 103 kN/cm2,
j'.'ì.w=l00kN ,
,
",44,
i;,1::.1;tri:r:;[-ij,ii,.
= r¡a.se kN i :,
;:
t+ cm, .' L2
=400 kN /cm,
4+ L3
Eq. (vii) gives,
-o't
*=ll
o
L0.707
o.soo'l
rl
0.707 J
Qf,
or'
L3
k^T T : kRU B1 nremurtinrying both,"î;ï:' or,
or'
;
o
Ll
¡'
=
[-o's 10.866
o 0.7071 t 0.707 )
= r r3.r5 kN /cm
and Eq. (x,iii) gives
I o I lgt.zoo -3.a2ol t'ul
t'o,i=l-t,oro
rr;;j1;i
f"ì= [ o.oooz I
{"/ {-o.rz*ol "*
tl
I
[r,
J Ir+.roJ
[zo.ss]¡
1t,l= {zr.sof m
25
¿l 27
DEFINITIONS i:,
A is the symbol selected to represent the m x n aftay of elements where element A, , is the element in the i th row and j th column. Note that element A,u may be distinguiihed from matrix Ar* n by the times ( x ) sign between the subscripts. Where ambiguity may occur a comma is used to separate ntimerical i, j values. Thus a,r, o indicates the element of A in the l2th row and 4th column, while A,r *o denotes a
ëi..IAFTHR
means that
.two
IT{ATRIX ALGEBRA
matrix of ,'l2 rows and 4 columns containing
l2 x4
= 48 elements.
The subscripts m. n indicate that matrix A contains m columns and n rows and A is said to be of order m by n or m x n. The subscripts are often omitted and the matrix symbol may be inclucled in square brackets. Thus A, lAl, lAl, * n,.and A, * n are equivalent ways of representing the matrix given by Eq. 2.1
(b) (c) (d)
A square matrix i5 a rectangular matrix for Úhich m A row vector is defined by m : l. .'
Xr *3= {x,
:
n.
x2 x-.},*,
A column vector is defined by n : l.
z.t INTRoDucitoN ,t':. l
i:..ìl
{:+,il :.r.i::,,
ìïf
l,jÙ
They are u.ser friendly as Matrices have a special importance in structural analysis' is.usually limited and computer a of capacity well as computer friéndly. The memory problems'cannot-be large curtailed, is space data and unless progåm space is Ãinimired, making the op^erations arithmetic many too to leads normally algebra solved. fio muctr numerous algebra, of calculations too unwieldy. To avoid getting lost in a labyrinth This is individually' than groups rather in vâriables entefing a probíem should be treated by making use of matrices' , possible
fP' ì p,,,: {p,l 't-l
IprJr*, ln this text-book a set ofcurly brackets {} is reserved for row or column u..tort. (e) .A diagonal matrix.is a square matrix with all off-diagonal terms equal to 0.
problems, the. matrix approaph From the point of view of generalizing the solution of matrices need not be offers many advantages. Normally an algorithm written using written to. analyze program A is increased. changed iust becauseìhe size ofthe problem to 50' is-increased of number the when even a 5 -"stoiey frame can be easily used -storieq language' c õr of FORTRAN variables subscripted and loops ln combination with Do to large approach hèlps in the production of efficient and general solutions
Thus
dtr o 0 0d2 0 D: 00 d¡¡
o
ihe matrix
.size
0 o doo
problems in structural engineering is a diagonal matrif of order four by four. (Ð A ùnit or identity matrix is a diagonal matrix with all diagonal terms equal to and is written with the symbol I. Thus
'The basic principles of matrix algebra required to understand the structural analysis proceclures described in this book are now explained'
2.2 DEFINITIONS
'
0
(a) Alnatrix
is a rectangular alray or fable
[ltt o o'l r=10 I 0l loorJ
of numêrical quantities or mathematical
symbols represented by a single symbol' Thus
Iu,, I
A,n*u
I itr r
]
t' I
Lt,n
(2.r) 1,
.
is an identity matrix of order three by three. A null matrix contains all zeros and is written with the symbol 0 or [0]. G) (h) A symmetric matrix is one in which each A¡¡ : A¡ i . (i) A skew symmetric (or anti symmetric ) matrìx occurs if each A¡j: A, ¡ ald cach 4,, : 0 0) A skew matrix occuis if each 4,, : - Aj, and nôt all A,, = 0.
I
{r-}
Ç)
MATRIX ALGEBRA
MATRIX ALGEBRA
str
simultaneous equations expresses
(lg, l,lncar sítnult'neous Equatio¡¿s A set of linear È.rðlationbetweenanafiayofdependentvariablesandanarrayofindependent variablei-that is ,
ü. i'l',:'
,:.ï{,,:,-,
xt
[0,
lb. lu.
X7
xl *ât2 Xz a âl¡ x¡ * .'..:.'.......* atr X.: bl ar'r*, *urrx, + arrx¡ +""'*""""1ft2,n Xr: b2
1:.t11,:.
ât I
iiiilÈ
1r:'
x: {x},n*r :
(2-2)
x-3
, b
={b}"*r
:
i--¿¿
::¿jfÍ.::,j:::
29
t
nl
Xl * ân2 X2 + âu] X¡ * "".'"""' *
anil
xt =
xm
bn
variables are 'the m independent variables aie xr ' *i, *r '""""ix'-and n deqendllt These coefficients' constant the b, , br , b. , .......... b,,. The terms a' , ãt2 ""'represent as notation may be written in matrix iiir*oi.lgËutu¡. "qu"i'iont
f ia,r
r
lilr,
t-'
t:
Lan'
¿t27 âtl
ar*l [*,1 ui*l
3¡2
u,,.1,,",,,
èrz
ât¡
ân3
l.:l
fu'l
:lo:l
[*ì'J--,
(2.3)
[d",l"-'
Addition Addition of two matrices is performed by adding corresponding terms in each matrix.
Thus A=B+C implies that u, j : 4¡ + ",, for all i and j. Obviously, A, B, and C must be of the same order. Example 2.1
variables
i,i¿-r.*"g"far block oi elements is called a matrix. It may be a rectangular iii. t* * ,i o, a square matrix of size (n x n)- The úertical array of elements
Find the sum B + C.
l) or a row vector of 1
is called a x m). The
pf
Bquation 2.3 can be'written in an abbreviated form as
:
r',.,r
:
tA],,
: *,n {x}, * ¡
. ,or
{b}n *
B:l- 4J L3 ^l
matrix of
size (l the number represents letter the second and rows il*i- t.Ur, t.piesents the number of columns.
T*,ìì. Ii *uv u" å column vecìor of size (.
,. .' :.
A=B+C: l(r *Ð Subtraction
Q.4a)
¡
A x:b
(z:4b)
luu A:
[A]n *,n = | ":'
atz ":'
1"", u",
ait 3.2t
u*'l '"azn
'l
I
u",
unrl
c=l;
-41
Lt
0J
ta =lto
_21
Solution
l(:+z)
(z-t)l (a+o)J
4l
Subtraction of two matrices .is performed by subtracting corresponding elements matrix. Thus
each
implies that
where,
Iu"
2.3 MATRIX ALGEBRA
brackets'_ The ind-ependent The coefficients 'a's are enclosed in a pair of square a pair ofcurly brackets' within enclosed açe b's variables x's and the-dependent
.
l. t't.
D = E- F dij = for all i and-j. "'¡ 4:
Commutative Law The commutative law for addition or subtraction states that
and
.
Associative Law
B+C: C+B - F = -F + E
E
The.associative law for addition or subtraction states that
€.J
* Ç) r"_) {,_:t
{) {,J
,{)
(B + c) = (A + B)+ c
Ai 'IransPosition 'fhc transPose
of A.
n'
r tho transpose and each a¡ i "
is ATn
:
*,
",;'äd
(read. "A'transpose") where superscript
i:
1,2......"m,
j = l' 2"""'n
,... .!
t-r 2
rhe associätive raw in
4l 31
therefore,
L4 5 6lr*, I
"
=lj 5l
C**p = Ar*n.x
a)
(2.5b)
=),a¡rxbr¡
i : l' 2'."'""þ'
of m x n x p multiplications
ExamPle 2.3
\ (
Find the Product A x B'
the
6) = -- I i
I
Partitioning is performed by dividing a rnatrix into two or more sub matríces. Example 2.4 Indicate one possible partition of A.
must be
-Thus-a Ttal (n) in A must equal to the number of columns
of
;;; freaicieA without sarisfied. The or.der "f ï;î";it ooerations. Thus scalar x matrix =matiix matrix x column = column' row x matrix =row row x column :scalar = matrix column x row " not possible' niuttipti"utioíof a matrix by a column ar€ otheÊmulriplications uo"
of A times
A xB * B xA
except for sofne special cases sûch as A or B =
denote the location "f postmultiplied by B' premultiplied by A while A is said to'be are as long as the multiplicæion.conditioni , Matrices can be multiplied by vectors the out càrrying actuallv
(.;
( 3 x 5 ) + (-- 5 x
in row
A x (B+C):A x B + A xC
c. - o '-i;;-;l¡mber j on B) rows(n)inBfor.lanostou."onro*ableformultiplication.ThessubscriptsonA t"u*'¡pï' Ihe outei subscripts (i onBAisand and B in Eq, 2.5bu." tnã i"*t to be said Eq'2'5a' éì;;;t;;, *'the c. matrix' tn
(,,
II
The commutative law generally does not hold. Thus
(2.5a)
Bn*p.
ll
made in computing
each eiement
The distributive law in rnultipiication gives
where each C,, term is given bY c¡.¡
of
Matrix Partitioning
The product C of A times B is
1, 2.......m,
. r _st[t 4 zf _ [-r -n 53 -t3l 22J-lz[r 4 ã jL" ¿
L28
*",,lJi¿*,jltJ"1l. (A xB) xC=A x (B xC)
6 Jr*,
MultiPlication
i=
1363J
Ðlement c,., is computed as the product associated elements in column 2 of B. Thus
= A if A is sYmmetric' matrix equals the matrix' Thus Ar The transpose of a symmefic
and
or,
of A'
Solution
Â=l
B=12
Solution
C: AxB
h'ind the transPose
.{'7
lí't,
T indicates
3l
fr 4s 2l0l
-:1, ^:l) i
ExnmPle 2.2
:'* :..::i :aÃ
MATRIX ALGEBRA
MATRIX ALGEBRA
30
î:: ; ît: lîr; : o
î:il
; A;; o;¿ lo;, ^)., Aot Aqz : A¿¡ Aoo I LA' As: , 4r, AroJ
g
I
I
Partitioning of A is performed by drawing rines as shown above to produce.the desired si¡b matrices. The partitioned matrix may be written as
ie,t i a,.l
A=1...'l-.:.-. li. lar, ; arrJ
I
ir"
^"'
=
[l]: i',:]
^'',:ll]l il]
,..il
¡i. ,å.
E
Ð ,J
rã
MATRT*AL.EBRA
, \-' *
A,r,
.,'-i,, . .,+ :"J '', d_ ,
=
[o' nrrl lRo, eqzl LAI ArrJ
MA TRIXALGEBRA
$
*
:å.
|-a' Ai, : lAo, o*
¡.
:Fr
LAs
I
lr-
The transpose of C
I l-t lz .....t=l vzz-) L_32 30
.
^tlt
A scalar quantity associated with Determinant of a secsnd order matrix
a
square marrix
: 16 tOl
' -i; ïsl is called its
deterfiinant.
ol A=fu dJ ,r.lenoted bv lu bl uno it given bv det A : ad - bc
.
Find the product A x
r 1.. I vlt
Determinant
ßxamnle 2.5
{
.\-1
¡v2trl lzq'sz : .55 207
r
is ct = I :.1:.
AroJ
l¡artitioned matrices may be'added, subtracted, multiplied, and transposed keeping in mind tltat each "element" of a partitioned matrix is in turn a sub matrix which must sâtisry the rules of matrix operations.
4'
.,
Arol
JJ
B making use of partitioning of matrices.
lc
I'u
lc
al
(2.6)
The determinairt of a 3 x 3 matrix
n=
i¡ì',,r.;:f.. ¡.1-al..,.',,:fi;; .?-il:,:nr.:'1.1,,
t .
l+ 2 : ll : B,rll3 0 : 0l lB,, B=l .' : '.. 1=1.. .. , ....1 .ïl nr2l li ïI :; 7l lBr, LBrt : Brrllz 150:_3J
1,
1 t t
tt'
| ' :
solution
u, az O, b2 | Lç, f
ur
b,
l
is denored
I ., "¡-.|
by
la,
a,
arl
lb¡ b2 b3l and is given as .rl. 1.,
.,
lb, b.l -l-a.tllb, ' b.l "l+a.ll¡, ' b"l'l atl-. -lcz c¡ -lc, czl I lcr .¡ I Determinant of a higher order matrix is obtained by repeatedly using the formula
det A=arArr -atzl.iz+a,r4,r.......+(-l)'*1a,,A,.+....(-l)n*la,nA,n e.7) where, 1,
¡
u.? elements of first row (or any convenient row) and A, are of (n - l)th order obtained by suppressing fìrst rori, and .the jth
determinants
-
column. Bycontinuousieductionoftheorderofthè¿eterminant.tofinalsecond order determinants, det A can be calculated as in the following example.
Example 2.6 Find the determinant of A , r.
(,
'., '''
"
Solution '
dl l
C t-. \- ..i ( (
The {inal C rnatrix is
: -32f lzq -l :.30 n
ln
':l;l l0 :: 155 16 L20
a:
|'
-t7l l8J
lr 2 3l ¿etl¿
l,
s
ol=1.15
8 el
18
:l-,1: Í1.,'l; ;l
= _3_2(_6) + 3(_3) =
0
Though this method oi calculating determinants is convenient for third or fourth order
matrices,
it is very
inconvenient
for
higher order matrices. usually, the
35
MATRIX ALGEBRA d¡ttçrmin¡rtt is obtained as a by-þroduct while solving a set of simultaneous equations whlelt h¡rs tho same coefficient matrix as the matrix of the determinant.
2l 'z l*1, 4J
Inversltln
.L-
Mnffix division is undefined. Thus
if
AxX:C
tlic ftrmiliar arithmetic operation
X
:
z*' x7"î
(2.8)
îb[i
C/A
ì'l[å
îJ"ih,[',Í'
=' '%]"[å.?]
tllÉ Inyersq of lt$hoi order matrices is Inuch more involved te, lt(iw€vêr, thnt invense of any diagonal matrix Z is another gonnl ale¡nent$ are reciprocals of the Z, , terms.
mfly not be performed except for the trivial case where x, c, and A are of order one by ¡''Önc, Instead, both sides of Eq.2.8 must be pre multiplied by the inverse of A. The ' 'lnvorse of A is denoted by 4-l where A-l (read "A inverse"). is the matrix which rvhei F nrultiplied by A gives the identity matrix I. Thus
AA-r=41 4=¡ Pre
,: i'
multiplying both sides of Eq. 2.8 by A-r gives
A.IAX IX X
of or
Inverr" ofA
.,
A.I .C A-IC
ltoool lo s o ol '=l; ; ;
adioint A
Q9)
lAl A square matrix is invertible if and only if A is non-singular, that is I Al +
(_;
A is defined to be the transpose of the matrix A, , and is denoted by adj.A, where 4,, is cofactor of the element a,, of matrix A.
Ç-¡
Let A be the 2
C;
(';
\ (, { (,., .
. I
The.adjoint of
: ;
rhe-n :
2 matrix given by
0.
'.: '
'Find the inverse
of
'Att
D:ZBC = (z B C)r = çr D-t: (zBC)-t : c-r
:i412-l
0 0
t/5
+
0,
A
4l
l-o 2
: (:t-
Since lAl
Îr¡åns,pose and Inverse of a product
0l 1l
0)-l(4 -0)+4(s -6)
2 is invertible
:2
Now, BT ZT
B-t Z-l
4,, = Cofactor of
Examplel.T Azt
Find the inverse of Z.
0
A:l 4 -r ól l_-o 2 t)
|
¡Al
-Arrl A-r = -----1-f A¡422-AuAtzL-Az, A,,J
Dr
0 0 t/6 0
$olution
O,,I LAz' Az')
Let theh and
0 t/9 0 0
lz r ¿l
.
A: IA,,
,f
B-'|=
t/7 0 0 0
,Exnmple 2.9
lr'
x
;l 10005J
A-IC
A-' = --:TJ
is defìned as
o.K.
t-t 0t = I ^ _i -1, 12 tl
: +7, Azz: Arz :+
A¡l = 4,
A,z
26, 16,
=
lq 0l lq -rl -l llt: -4. e', : l-6 l_o ,l=, A''t : -lo A¡¡ : -6
q-) $6
{)
MATRIXALGEBRA
:'
t.l Y"J.:.
,.,
-
.
f-')..1.
,,.,
adj A
:1..:
I
.
7
l-r 26 t6l =l-4 L2 -ro _61
'ij.:.tr.-;..,
*.
:,:,-.1 i;i
A-,
ì
tl:
,¡-' '.t
:
ugjl lAl'
1
':':l '-;:,.6'-',
ehpck
AA-I:
I
APPLICATION OF A WORK SHEET
"ï
41
Lct us enter the matrices . A and B in the
(¡) ADDITION
rStep
In cell A6, enter the captìon MATRTX ADDTTTON rAl + tBl:[c] which r¡,ill extend upto cell C6 In celi 88, er¡rer the formula:
I
:., 1-1
-4"1+lx:4+0 ;l'l6xl-2x4+lx2 ':..
'.
Step 2
+1x26-4x10 2x4+lx16-4x6f i4x7:1x26+0 4x4=lxlb+0 I -6x7 +2 x 26- I x l0 -6x4+2x16-tx6] +2'x7
:
U (-,
Copy this formula in cells Bg to DlO using the copy command.
copy frbm cell : 88 copy to cells : 88 . DlO
-5 'ol f* 160 6s -el
2.4 APPLICATION OF A WORK SHEET .
itr' iv,l (._;
+82 +F2 Step 3
o. K.
LOOU
l':'"J. :.
'
/C
frool l0 I 0l
ì,.-:.. .
A work sheet or a spread.sheet is an electronic sheet consisting of rows and columns. A cell is formed where each column and row intersects. Each ceir can hold one piece of ínformation - a number, a forrgula, u øu"rr" r¡. .o¡urn, by retters
to zz), and rows are referenced
*
*"-."Ëä by numbers.. lorus L2-3 is;; ;ñ;;"st Ëu¡¡cnoÑf-most
(A
Matrix [C] looks
::ffiilr:tr*cruraranatysis
*"ll
u..r"J;rñ;
It is desirable rhat
various capabilities
The following example illustrates the applications of
opcrations.
LoruS
"f ;;;;.rî:"î
as
(iÐ SUBTRACTION
popurar softwares. with the help of @ mathematical operations can be conveniently carried out. Dãø arrarysis ,""iriiqu., provide m4trix commands for multipliôation and inversion. The detaiís of the software can be seen in the user,s manuar supplied by the LoruS Developme;i c"éàrtrl"n, New york.
i,
worksheet cells as indicated:
IAI in cells B2 to D4; [B] in cells F2 toH4
7 ll ,l-t 26 : ;l-o t6l : L2 -10 -61
ll-2"1-tx4+4x2
Lorus
Step
I
67
J
In cell Al2. enterthecaption
Step 2 Step 3
a
Irzs -zs
MArRrx SUBTRACTTON IAI _ tBl: tc1 which will extend upto cell DI2 'In cell B 14, enter the formula: + BZ _ F2 Copy this formula in cells B14 toDl6 using the oopy command.
/C copy from cell : B14 copy ro cells: B14 . Di6
spreadsheet for ¡natríx
E*sniilciz,to
(-j ,"0:".--ffi;iriårtff,::y*t* (..
manix operations on a
Lorus
worksheer, dnd prinr ail input
l
(,,
(ì (-
lro
15
lol
fy
_zo
o1
[n]=f:s 20 of,[n1=fzs 4s.el 156 -20 4sJ Lu., _s 22] (Ð A+B=c (,ii) A-B:c (iii) AB=C
(iv) AB+B=C
Màrrix [C] looks
(iii)
l-to
35 lol
L_l1
231
as Itt IO -zS _ll
g
I
MULTIPLICATION Step
I
Step 2
In cell AlB, enterrhe caption MATRIX MULTIPLICATTON IAI x tBl: tcl Use DATA MATRTX MUr.TIparcAilóN àoä*an¿ a,
follows:
/D M M
37
:
MATRIX ALGEBRA
range of first matrix : B2 . D4 range of second matrix : F2 . H4 range.of output matrix :B'20 . D22
Mstrix [C] looks
rOs rr'l 200 -t801 I lt+w -242s I tTol I
as
raor
t6eo
,,... ffV/ MIJLTIPLICATION AND ADDITIOU ,',,. . . .Matrix [C] obtained in part (iii) can be added to matrix [B] as explained in part , (i). the resulting matrix [D] can be srored in cells P;24 to D2.6. ,:,,.::,
.2ß
soLUTIoN oF LINEAR sIMULTANEous peuATIoNS
t. Gauss elimination ryrethod (without or with partial pivoting)
)
J.
Gauss-Jordan method
Cholesky method for symmetric and positive definite matrices. The matrix A is decomposed
where,
into LDLr,
L=
lowertriangular'fnatrix
D=
diagonal matrix
Succçssive Over Relaxation method
PART
1
FLEXIBILITY }'ÎETHDDS (-i' (-;
, ,,.,,.;,,,
'-'t "
(.i
() (-;
(r (,
t-,!:,ii ,,';.;:;:',
:i., iÊtl: :
LJ
* ,ü, .'i
CHAPTER
j
three
1t
"!r,
tr-l'ì
,,';',,:11t'¡..
iüt-, .N
METHOD OF' CONSISTENT DEFORMATIONS
i,v'-.;t:L.-; ,i ..
#' i
a:
1!
rf
g
,
iäÍi¡:!'
:l
Ð,
'rÍ '¿
iÐ
'¿i;È
3.T INTRODUCTTON
,i .:;
f'ti
.
,i,Ð',
:li
li
equations. The supports play a very important roie. The number of reactions existine at a support depends upon the type ofsuppot. A roller.support restrains only one transiátion normal to its surface. A hinged.support permits oniy a rotation and restrains against any trans-lation. Thus, there are two mutually perpendicular reactions at a hinged support. Ã
t1
iÇ:,
iÐ
It has been demonstrated in Chapter I that a st¿tically indeterminate strucrure cannot be analyzed by the laws of statics alone. In these situations the conditions of compatibility are required to be enforced to generate the additional required number of
i.t "*rt,,
1i
,ti
':lil
'li . ]l*
.(l
fixed support.does not permit any movement at all and there exisitwo mutuallv
perpendicular reaçtions and a moment.
;-1
.é
The steps for the analyiis of any statically indeterminate structure whether it be a beam' a rigid frame or a truss were enumeratecl in, Chapter A statically indeterminate structure is made determinate while maintaining its staUitity by removing all redundant
Ê
,O
l.
$
ì
r:r
ü
f:}
ü
reactioi'rs.
,
Îi
',
.:1l ..
C,I
iË'
o
fi,
Ift,
(-ì
$ti 'år'i
the
:åt.j
i$
..
il
(-,
:
'.'' -:
'gir'1,
,.
.:
É t'ì'. ,€ ,",.' $, ';',.
(,]
.É
'
.t.,,
,3',ì.'.',. liï
c
ons is
tent defor m at ion met hod.
The unknown redundant forces together with the forces of the applied loads must to deform such that the deformations are consistenì.wrth the support conditions. Enforcing rhe consisrenr deformations or rhe comparibiljt ;;;;;r;iiÄ;; the structure leads to the governing equations in terms of all ihe unknown redundants. This means that the actidn of one tãdunãunt will affect trre Aisplacemenìs *¡m the compatibility equation of another redundant. This is known as ^io.iut"¿ coupling of cause the sifucture
#.,
(,.:
'
.
'
lii ,:.i
support is removed and the reaction is replaced by an unknown
reacting force, the compatibility condition requires that the defleciion theré must be zero. a fixed support is changed into a hinged support, and the moment reaction is replaced ff by an unknown reacting moment, the compatibility condition is that the slopê there must be zefo- If a fixed support is completely removed and the reactions are replaced by two
unknorvn mutually perpendicular reacting forces and a reaeting moment, the compatibility condition requires that both the deflectipns and the slopl there must be zero. The unknown redundants can thus be computed. This method of analysis is c¿lled
ììì:
(_)
(
'
If a roller
-
':::
i.'l
4"_)
Ç)
-4å
METHOD OF CONSISTENT DEFORMATIONS
,-,_.:'
i
.:,iri':.
BEAMS WITH ONE,REDUNDANT
43
::,r:, r
: deglec.
of ficedom. that is . the compatibility
equations are coupled. Such equations need to be solved simultaneously. all equations but independently solved CnnnOt bc
:.ì,i.:
;ttiì;: aa:a
.,,ã.tr
a
:"f,.
!r.:'l
4
CI{OICE OF REDUNDA.NTS
.',,;,,¡Qncg[he degree of static indeterminacy is determined, a.released structure must be ,.eliOssri by ídenti$ring suitable redundants. A released structure which is also known as ¡¡ig,'primae strptuie rs statically dëterminate and stable. The reactions and forces in r¡xcess of those required to make a structure determinate and stable are the redundants. Redundants may be eittier support reactions or internal actions such as moment. shear or lra*ial f'orce. The question is, how do we identify the redundants or unknown forces of the .., $tntcture? Generally a structure can be made statically determinate and stable in more ' than one way. Some indeterminate structur€s and the corresponding released stluôtures and redundants are shown in Fig.3.l. The choice ofredundants is usually based upon ..r,êonvenience. Some released structures lead to more cumbersome computations than others. By a judicious selection of the redundants, however, numerical computatíons can : be nrinimized. The following"points may be kept in mind while selectíng redundants: .
l,
m
, GIVEN
GIVEN STRUCTURE
4--
STRUCTURE
aa ,,Ç'---TG-TR. .
RELEASEO STR. I
RELEASEO 51R.
RELEASEO STR. I
ffiaç RELEASEO STR,2
2
Iol
ffia6
RELEASED
ffi .
A released structure shor¡ld be such that the effect ofthe various loading conditions
is localized as much as possible. 5 Take advantage of symmetry of the structure, if possible The positive sense of a redundant may be chosen arbitrarily. 4. A statically detçrminate reaction component must never be chosen as redundant as it rnay lead-to instability.
"Ð
-
SIR.
3
aMBr .Mc. RELEASEO STR.4
(b)
6IVEN STR.
Sign Convention
Rox
Shear force at any traísverse cross-section is the algebraic sum of all forces acting transverse lo the member on.either side-of the section. A shear force is said to be positive scction if the right hand portion of the beam tends to slide upward with reqpect to the ,ilqftr ïtgf¡ tiand pqrtign. ln other words. it is positive if the resultant of forces on the right hand side is upw4rd
r
REL.srR.
lRoY
GIVEN SlRUCTURE
:,,'
Morl.
M¡
Rox
REL. STR,2
düÇr
(i f-;
(,
tt
,t
..:FEAMS WÍTT{ ONE REDUNDANT
REL. STR.
Ëxnrrrpl:3i1,
"
(cl
Solution (
3
RELEASEO STRUCTURE
2
(d)
Al'pfül¡ped':caRtilever-beam is shown in Fig. 3.2a. Draw shear foice and bending
momen! diagra¡ns.
(_
;, Fig. 3.1 Typical released structures
.
Thedegreeïfi':ítfdetérrhinacy is
-;:
l.
Choose R" as redundant, remove the support C
and obtain a statically determinate beam AC as shown in Fig. 3.2b: The value of R" make the deflection of the beam at C equal to zero.
' besuch that will
will
BEAMS WITH ONE RbPUWPATT
METIIOD OF CONSISTENT DEFORMATIONS
.'.
"sH
þ Ll'z ,þ Ll\
à
(o) REAL BEAM
!ëEI
and
ol¡anau
DIJE T0 Rc
ftI,.,
5P r6
(f)
æ_
redundant, introduce mornent reiease at
A
and obtain a simply
SHEAR
FORCE
DIAcRAM DUE To p
,,/ T
á (d)
Alternatively supported beam AC as shown in Fig. 3.3a. This beam is also known as a released beam. The value of Mo will be such as to make net slope at A equal to zero.
RETEASED BEAI,I
S
The resulting shear and moment diagrams are shown in Figs 3.2f and g. A saggìng bending moment is taken as positive, whereas a hogging. benãing moment is taken aò
choose Me
#lÌÞ,-tct
M, : t6vxj,=*vt (sasei"Ð Mn : -LL-* ,1.r"=-LrpL (hogging) 2 t6 t6--
negative.
IP
(b)
'5Iq
45
(9}
f'3
(o)
BENDING MOMENT
RELËASËD BEAM PL
l¿El
l^'
BEAM UNDER REOUNDANT Rc
(b)
DTAGRAM DuE To p
HH
Fig.3.2 Propped cantilever beam niornent of the MÆI loading dia.gram due to rhe applied load between A and B about C (Fig. 3.2c) (moment - area theore.m)
Ycr
å
tct
I PL, L (L zI\ 2 2Etx-xl-+-z \2 32)
$
DTAGRAM DUE To.MA
I
moment of the M/EI loading d{agram due to the redundant
(Fis.3.2e)
,.,,'!cn For compatibility, oft
or,
+
Fig.3.3
R. 0o,'
I Rf-Lx -Lx-2L
PL, 5 8EI 6
-
R.=:P "
Ycz
eA2 R.L1 g 3EI
A due ro_ MryI loading diagram due to rhe applied loading befween A and C in the coqiugate bèam (Fig. 3.3b) shear at
: lPr-,t ;,ffixtx1
2ET3
Ycr :
=
-
rh..*
at
A due to M/EI loading diagram due to the redundant (or
unknown)
,, : lMo-2 v¡2
Mo
t"Eî*"*t
16
(= Reaction) (Fig.3.3c)
(:
Reaction)
:F
L_]
'
:g.
{)
lfr
{.)
Ft¡t'
;.# iiÜ '.'¡1 l:.
Me
llxtrnplc
r^,, .,:.\-i
iiv. itÐ
3.2
,..,.,-.,
iliì",,-'' l-:l;Y'
ysl :
i,;i!!.''
,i*'':' j';¡'':,,, .
'ct
i
'
=
. j__4.¡ s R,,ì
''r3
Cìorrrpatibility condition
requires.
Ysr
:
æ
tj ("')
_
sts4
(_, (--)
',,
=
'laking nìornent about A, 0
:
#+-#)]
= 20xl3-R,-ec=
25.4-skN
r¿
poìritive, maximúm negative and zero bending moments.
"
M*,*rm'Negative Bending Moment
ï
s" ¡zl t208.34fr- '---'L' 2xze )
Ro
64'65 kN
l. -: .:
..'.''^
Lct us draw detailed bending moment diagram, and lobate the points of maximum
'l'he rnaximum negative bending moment accurs at the intermediate support, that is, at
At x=5mfromA, B.M. due to
U.D.L,
M, = ('o!"1,.s-zo"i z \ 2 )
6967
..Ê.,,,.;,, .!!4 ax i tn u m P o s i t iv
momênt of the M/EI loading due to the redundalit R" between A and B about point B in the conjugate beam (Fig. 3.4c)'.
- *[;x5xRs.T[', -1"').i"8xR,
xT(3"')]]"
-*t+x5xRp'T,.;]
e
:
ooo
n"*
{vR
alone, Mz : 169.9 x l3 r;ìiì;'l, .'. Net bending moment M M, - M, : B.M. due to R"
EI yp2
-
,', ,.,. :::''
13m
(-.j
i,)'
ysz
6967 = 4l ptKt El RB : 169.9 kN
Rc
t-,
ü
=
t3
2
l-,,,r,.Ì ,
2x5)x[t
ErJ
2o*l3x11- 169-9x5-Rcx13=
center of gravity of the M//EI loading diagram between A and B
r, : }{; *20,#,s-zo,l(3x13-
5x8 7::-!rlx:__"_
+21.33||x
n
lleact,ir¡ns
from A
'-i
(,
i-L R,
IEt'
0r,
.z - a(qu-za\ ^\ X=:-l ¡= !-(3t-¿al. __------_-1, a:)m, l2\3L-za)' 12' where A = area of M/EI loading diagram between A and B
l::1¡:,/'l
(-ì
:
17
EI "
moment of the M/El loading due to the applied loads between A and B about point B in the conjugate beam (Fig.3.ab)
ír''
(,,
{t-lr¡ro
or,
Choose R, as redundant, remove the support B and obtäin a statically determinate beam AC. The ialue of R" will be such that will make the deflection of the beam at B equal to zero. The deflection at B can be computed using the conjugate beam method.
,
'( j
:
16
qs:4-3=l
;,Ër, {'
t,
hogging
The degree of static indeterminacy is equal to
iiï¡,'','
;::,:'d-l
:1.'i\-.-;:
3PL
i
Solution
,:,:r',.i''.''
1.,.
iÈ
oot =
BEAMS WITH ONE REDUNDANT
$ .',t'
Anatyze a two span continuous beam using the method of consistent deformations as shown in Fig. 3.4a and draw shear force and bending moment diagrams.
t
.',
cornpatibility. 0o'
-,
',
(), K.
t::;:
':ir'r
4
METHOD OF CONSISTENT DEFORMATIONS
of,
..:.;.
'..::. .l¡,
,.a:
.-t,,
B e nd i ng
M o me n t
=
-122.77 kNm
,.1. "l'he maximum positive bending moments occur at the point of zero shear force. The '' | ¡hear force diagram can be drawn as shown in Fig. 3.4d knowing the su¡iport reactions. Þ
thc span AB, zero shear force occurs at a distãnce of 1.27nt from A, while in span BC, ,-;;. :'r . ,{n ;^..-r- ---- r-,-shear f-orce- occurs at a distance of 4.77m from C. z$t'(t ;i;ij:' 'Ál x = l.2i nt front A.
:_-i _:: r:i
bending moment M
:
25.45
x 1.27 -20
*
l'2f
' :
16.19 kNm
METHOD OF CONSISTENT DEFORMATIONS
4S
20
BEAMS WITH ONE REDUNDANT
kN/m
:'r'r.
::
Let us assume that point of inflection occurs at x
, 8t
.t
(o) 2 SPAN CONTINUOUS
ta,
a,
l3 2ox2 . -^,g * Ax 2 ^- 2:roy. 13
x = 2.54m
of,
( SpanB-C
BEAM
Let us assume that point of inflection occurs at x
8Et DIAGRAM DUE TO UNIFORM LOAD
*
5x8
(D
B
13*- 2Oi
Firsg the bending moment due to the redundant R" is plotted. It is hogging in nature and , hence, negative. It is triangular in shape. Next, the bending moment due to uniform load in the released stucture IS plotted. It is sagging in nature and, hence, positive. It is parabolic in shape. It ís drawn so as to superimpose on the hogging moment diagram. The area common to both the hogging and sagging moments consists of znro bending moment, hence ignored
(ii)
DIAGRAM DUE
S
C.
or. x:6.46m The bending moment diagram is plotted as follows:
RB
E¡
tcr
from
20 x
B. M. at.x,
(b)
from A.
20 x
B,M, at x.
h r,5m +
49
(iiD
(iv) The locations of þoints of inflection,. and points of maximum positive
and,
negat¡ve bending moments and their values are indicated on the diagram. The shear force and bending moment diagrams are shown in Figs. 3.4 d and e.
(d)
Example 3.3
SHEAR FORCEI KN
Analyze the beam shown in Fig. 3.5a using the method of consistcnt deformations and draw shear and moment diagrams. Treat $o as redundant and make use of the law of reciprocal deflections.
l-1122'77
Solution
6.16m
.2'5¿
í-----------
(r)
The released beam is shown in Fig. 3.5b. The condition of compatibility requires that the deflection at the free end A due to the combined action' of the applieiloads and the redundant Ro is zero, that is,
-r
BENDiNG MOMENI kNm
-i
Yor : -
'
Fig.3.4
(
r
At x= 4.77mfromC, . bending
(
moment'M:
64.65 x 4.77
¿zl2 -20 "
=-
:
80,85 kNm
(Figs. 3.5b and d ) RnY¡z Moment of the MÆr loading due ro the appried roads about point A in the conjugate beam of Fig. 3.5c Taking momenr of the loading ro the right of hinge B about B, (Fig. 3.5c)
yn¡ : MsR =
0,.
+"
#x
r.15x
|U-i"
(z.s+ r.rs).
:0 fr (t,,t,;,:tt + r.r5)+ R¡ x5
,U
t)
(ö
*
^.
44.98
EI
{:} {-} {^:
F
t's (ql
l-Þ
i
¡l 2's
5l
BËAMS WITH ONE REDUNDANT
METHOD OF çONSÍSTENT DEFORMATIONS
: or'1 ¡l':2;l:m '{
.,
*o
+
R.
R.:^EI
2:SPAN'CONTINUOUS BEAM
,trL
),
+
1
136142 x 3.85 : ;- Ov'2.65 -;" O
0
1.83
'Iaking moment t f the loading to the Ieft of hinge B about B, (Fþ. 3.5c) -l
['
11.83 lvl^ | 36x ,...1.5 :_ u^ l.fxEI^zEI3
RELEASED BEAM
-xJ_
'r',
I
M!:'*:vo, ^EI
moment of the M/EI loading due to the unit redundant Ro about point the conjugate beam ofFig. 3.5e. {c
I
$
ot
ncnau:::, ON
î:':5:.'o^o'
=0,
L1 EEAM A CONJUGATE
'r¿¡,.oÉÉlecrED sHAPE UNDER RA=
.
-. Þ.
1'r or, Ro+R. =t",,-t
:0,
A
t2 =-/ EI
1"¿x5x'l=R^x5
2Et
1
3
c
'LEI R^
:
2'5
3
E¡
{et
R. : ^EI
$ otronau oue
9'5
^ x3-+,.+*s"]-vo 2 EI 3
Ro
=0
:24 : ¿^¿ v^^ I EI yÀl : t Rn yaz (here Ro: ieaction at A in the original M^
1
( f"}.SHEAR FORCE KN
bpam of F,ig.3.5a)
))
"":R^*"* EI^EI
11.37
{
s-)
R^ =
BENOING Mo.MENÌ kNm
Ileactions
Fig.3.5
I
Mc = o, (Fig.
3.5a)
0.917
)L kN = 0.92 kN
in
-U {} *
:
52
Ir l' i
Ti:i !i.
tri,l-) ll: r,.
qE xz.S + 24 x6'5-Ro-x Rs = 53'73 kN
or,
I {--' f'.-, I
BEAMS WITH TWO OR MORE REDI.'NDANTS
METHOD OF CONSISTENT DEFORMATIONS
Ës, =-0,
Rr*Rs"h
I-
=Îi;li '-
ofr
(i)
Rn x 5 = 0 Maxwell's law of reciprocal deflections gives,
(ii) ..
¡.rti
- ^- x2'5=-33'25 ^o kNm Momentat B " == {il: x5-4E kNm xt.5=-33'25 ,á-zi 3'5 f and g' in Figs' shown are diagrams The resulting shear and to*ât
o'
K' = YeE
a{ yon=frxt5-24', -1"!:xl.s*!l--lo'30 EI 3 ff=y-
AlternativelY Law
of
ReciPr ac a I Deflec t ions''
TheMaxwellslawofreciprocaltheoremmaybestatedasfollows(see9.l4'volumel): o load P øt pciint 2 is equal to the The deflection ò[ point I on,a structure !u: ,'o p at point l. ofcourse, the deflections referred to deflection ofpoint 2 due';; tie load
ïíin llith
rn"
¿^e
relerence
directions øs the applied loads'
to
JJ
yap due to 24 kN toad at D
-
E
=
yo,
deflections gives' Figs' 3'6 a snd b' Maxwell's law of reciprocal
due to 48 kN load at
24 Ll-0'30
EI 48 x
J =toJ:' EI
4.69
EI
225.12 + =_ EI |
Ye¡
olt TvAA
t_.-
(o) RELEASED BEAII
UNDER UNIT LOAD
AI
Example 3.4
TYao (b) RETEASED EEAM UNDER UNIT LOAO AT
D
T¡ (c
I
1.:3;4 BEAMS WITIf T1VO OR MORE REDUNDANTS
A
. Calculate the end reactions in the fixed ended beam of Fig. 3.7a. Draw shear force ', ' and bending moment diagrams. I' "solut¡on : Th" axial deformation in the beam is ignored. The degree of indeærmin acy is 2. ' choose Mo and MB as two redundants and a simply supported beam AB as shown in Fig. b, is obøined. The values of Mo and M" will be such so as to make the net slopes 0, . 3:7 :, and 0" equal to zero.
RELEASED BEAM UNDER UNIT LOAD AT E
.
,
:.,
0n, :
2a ' .:,'l
3'0
(dI GONJUGAÎE BEAII WITH M/EI Fig.3.ó
=
'
onz
LOAOS
slope at A due to externat toading. shear at
A in the conjugate beam ofFig. 3.7c.
l2 wL2 = -x-x-x|, 23881 : slope at A due to redundant l*Mo = 2Et3 *Lr?
Mo
in the conjugate beam of Fig.3.7d
\^-.'^'""
"t
o *
BIïAMS WITH TWO ORMORE REDUNDANTS DEFORMATIONS METHOD OF CONSTSTENT
5'\
w[,j
MaL_11¿=o
uEt
3Et
M^ Reaction
(o)
:
Y:
6El
Ms
{hogging)
R^=+:Rs
FIXED BEAM
wú
M¡
24
(sagging)
are shown in Figs. 3.7
Rr (b) RELEASED BEAM
(c) lÉ- DIAGRAM DUE EI
5_5
f
and
g. Bending
wl-2 8E1
TO
UNIFORM LOAO
M¡
ter
DUE
T0 MB
$DIAGRAM
.'üxample 3.5
:.',,i:i¡d!!alyze the beam shown in Fig. 3.8a using the consistent deformation method. : ..: l åolution ..'' , l: The beam is indetermin#e to a degree equal to 3. If, its axial deformation is ignored, .
l!! 2
SIIEAR
of indeterminacy reduces to 2. Choose Ro and M".as redundants and remove support C. The compatibility cònditions require that the net vertical deflection at C .'.¡¡,iffi slope at C under the combined action of the applied load and redundants must be :'.,ißro, that is, (Figs. 3.8 b, d and f): ' :ftSrdegree
.:,1::i.tllÞ
(g)
BENDING M0¡'IENT
i..t.: Fig.3.7
0n¡ :
¡iil, iii:ill
I Mo ,.. = _x-_ex-LX__3
ll( j,,i: ii
i:i
("
slope at A due to redundant 1
Mt
in the conjugate beam of Fig'3'7e
' t'i
,' 0ç : ,,
zBl
f)ue to sYmmetrY' Mn For comPatibilifY'
-
=
0o, - 0o, - 0n¡ =0
(i)
0c* 0'c* e"" = 0
(ii)
to external load on the beam [using the moment-area method, Fig. 3.8b]
s.lope at C due
Ma EI
Ms
Yc :
+ Y'6'+
Y"":9
.,Yc
clockwrse
deflection at C due to extemal load
5li
DEFORM4'TIONS METHOD OF CONSISTENT
.-t!*,,',.' 'l , ''t .;,:;L:.1:;..,,'.t ,iÍ -.:!ì: ..i:..r .ri
/
=
,'ìtl.it..",,- - 0a
=
rl',h
slope at C due to unit upward load at C (Figs. 3.8 d and e) L2
\ 2Et anti clockwise
deflection at C {ue to unit upward load at C
r] 2, 2EI 3
RELEASED BEAM
f-T-ls I I
"{o-;J - Ma(a+2b), EI zEI zBt
y'c :
L3 3EI
^
slope at C due to unit clockwise rotation at C (Figs. :.'S f anO g¡
IE¡
R;
SHAPE DUE T0
r (d) OEFLECTED
lr-t-l->ierÆ |
(iiÐ
,", Soraonau
DUE
ro R.'
1
_Mc
Tr (f)
DEFLECIED SHAPE DUE T0 Mc
1
111
EI t
srf,
:ì/
o\
I LxL
FIXED BEAM
ffit (b)
,
''.r..:ì:Ì.:1t t
*l
(q)
BEAMS WITH TWO OR MORE REDUNDANTS
DIAGRAM
ï ouE To M6 =
.Ë
""!1r.3.-$*.n ú v.=o
;$$9*""of 1,,
(iv)
Eqs.liii¡ åa1iu¡ giurr,
;r,.;i::$: .
Mc : fftZi-ù
"r :_ 6Maba _ri^
D
1
^^
-
-l
MA M4
: ffQu-u¡
'lVo possible.b"lgttg moment diagrams
(h) BENOING
M tre shown in Figs. 3.8 h and i.
depending upon the relative values of a, b and
..
MOMENT '.rj'
Exampte
M¡
, '
(¡)
::: BENDING MOMENT
'Fig;3.8
(-
-.
:.
'Analyzc the 6am shown in Fig 3.9a using the consistent defoimation method.
':, ,Sólution ..,,i
i-,
.''''. ,
j
', l
S.O
l'¡r'. . t-:
$ti
ugrgoo
oF coNSISTeÑr oeroRMATIoNS
.:'
:-
.REAMS WITH TWO ORMORE
r:,':.':Flope and deflection at '-
D underthe cornbined action of the applied toad and redundants
- 0'oRo+o"oMo:0 and yo - y'o Ro + y"p Ml¡ = 0 eD : slope at D due ro extemal load on the beam (Fig. 3.9b and c)
Ær. 'ÈlREt-EASED BEAM
wb3
(iù
wfuo*É*{ì=-
6Er 2t
å[u' 6EI t
2
+ 3ab(a +
2)Et
b)l 'J
clockwise
deflection at D due'to external load on the beam
J,D
wbZ
2El
f 3bl w ., *u)i.*o* a(iu++a)] f"*7)* ruuo|.u t 6(a+bJ
wb3
6Er
(c) M/EI
(i)
(using the moment - area rnethod)
b+c=e o+b'dr (o) Flx ED SEAM
(b)
ie
lÌtust be 4ero, (Figs. 3.9 b, d and e):
0D
ffi '
i.,,ì
REDUNDANTS
ttD
J
slope at D due to unit upward load at D (Fig. 3.9 d)
LOAOING DUE
t2
- ,"
-r'1 3EI
JD o"o
å (d) DEFLEcIED.sHAPE aHo LOADING DUE fO
Rgr
$
=
0"o :
1
slope at D'due to unit clockwise rhoment at D (Fig. 3.9e)
I
clockwise
" y"o : ..
Û
2EI
ing the values of
J
anti clockwise
, r
0',
and y'5 in the compatibility equations;:
EI
(e) OEFLECIED SHAPE AND å L0AD|N6 OUE T0 MP = 1
(ii i)
Fig' 3'9 Ro and Mo as 'fhe beam is statically indeterminate to a degree equal to 2' Choose redundantsandremovettr"supportD.Thecompatibilityconditionsrequirethatthenet
(iv
tl
,r:.¡i,l'äi,ì.
_.,\_.7_,-.*-,-...-.,'-.,.,.
ì BEAMS WITH TWO ORMORE REDUNDANTS
.l.iiii' DEFORMATIONS METHOD OF CONSISTENT
(t(l
,;::'i¡,¡,, .1]':':''''
É,q,
(iii) gives'
*" = {å[03
+ 3ab(a + b)1.
å"']?
(iv) gives' value of Ro in Eq' Substituting the
*,
=
c) + a(¿"]'O]t:lttt3+rau(a 3b)+6au[e(a+ uXu+
dþþ¡3(4c+
Substitutingc
=
L- d'
= " " îJþ[or
b=d
+
u)l]
{o)
-
(a¡ - :d) - a'(+r' -
:ul
ctockwise and
hossing
(bI
beam' equilibrium of the By considering the
Mn Example
w f-¡/4L-3"¡-"'(+l-æl \
I
/T-
È
B
3'7
Determine
RELEASED BEAM
2
anicloctw*eandhogging
rl7["
=
FIXED 9EAM
a andsimpliflingleadsto:
fcr
Fig. 3'l0a ¡ . the non-prismâtic -^ -^ñ-ñ?isñ'atic beam of D in -L-
:*-:10^"Tffi:åî
deror using the consistent
$
E
r-oAolxo DUE To w
å'",åflt
t"trl,îî"* is staticanv irrdeterminate
l":
o:f'-ï"îIll"rî"Xt'"åi*'as
redundants
I,tl'"*¡;Í**fm;;"ltllii;'¡å:; ît11:ïffi ,,Um*åfr must D( load and redundants iiäïï;
(i)
"ooiied 0e - 0'n -0"4 = 0
and
0o
-0'o -0"P
OEFLECIED SHAPE AND M/EI LoADtNc DUE ro MA
(iÐ
=0
',:ri:î!ll'Yz'í3ål''å3:L 0"nand-',!i,:,:l:,,:"å"i:ii:iåËÍîiil'åîffi uot"¿ï"JU.
lnot shown). Thp slop€s
ìntn*"try,
evaluäted.
onlY the sloPe
3'r0 Moment at B in Fig.
M/Er ordinate
t
b=
(e)
BENDING MOMENT
+ "i- ""[il "l=$*e
l.-2=++
-(Fig'3'l0c)
Fig.3.l0 M/Elordinate e -3=
*4 8EI
rlat€ areas of various M/EI loading segments and their center of gravity.
wElordinare
B-l=åH
A
- B-
I (Fig.3.l0 c):
ól
A,
-
Area of segme ntB
:
*f :ì[3L-2x"1 ti1"
2-
r t-a .r\
,rr "'
3
BEAMS WITH TWO OR MORE REDTINDANTS
DEFORMATIONS
-*"":::î"''*:ìr
{,;r
5
= rur
c.g.of segrnentC- D-6from
wL3
u
c.g.of segment
E (Fig' 3'10 c)
Areaofsegment
(,J lrìl r _ ll wr.3 - ão)1"= 384 EI
o,n
=
usingtheconjugatebeammethod
64 '[t-l*='þ=' 2 )Et Er
,
c.g.of segmentA- B -2- lfrom
A
'' L :4 3L+1.!l s]',zl= 42 3 4l J*t L s'z I
Rn =
- t- *':
o,o
- Rn :
compatibility condition
5L
6
6
l*+ :#.*i" N.W-il 90 MAI-
384
EÌ
lvl¿* ll * Mol*ll I+"ry*+* Er 42 4Et z+' onu" al L64 : Substituting values of rhe slopes at A in Eq.(i) gives
lr )wr-3_ eo tvr¿_ 54 MAL : I s * 384) Et 384 Er
I
37,
t-- !
_ 54 MAL _ A,, *A 384 EI
[¡ ...- tl
c.g.of segmentA-B- 2-lfromD
D -ófromO =
-
: A- B-2- I (Fig'3'l0d) ,[
O: ?r.! =! 34 6
Llsing the conjugate beam method.
A' : ål'{i i'ft"' 0n = Ar +Az '
C-
(ìr
\768
=al {2"
or. MA :
384 EI
**r,
n Y
hogging
positive bending moment occurs at the midspan where shear force is zero. ¡l..¡i,r "Mnximum
Areaofsegment
b- c- 5-
3(Fig'3'10d):
rH*=*ï
Momenr
at L/2
- ,y!2 * !-
M.
-*l!ìt \z)
1 2
wt] 3wt] wl]
Ir,"-¡l
c.g.of segment
B- C-5-3fromD
L*!* Llq-!^ ¿,l=9t
+-t^zl I l'o
: t-#t =#L c.g.of segmentB- c-5 . 3frornA Areaofsegment
I L MÂ M¡,L C-D - 6(Fig'3'l0d) : t" 4t8EI=64E1
:
.
the continuous beam shown in Fig. 3.1la using the consistent deformation fnllyze. \ ûl€thod and draw shear and mornent diagrams.
to a de.qree 2. Selecr R. and Ro acring upward lf.ï::"ttj:i,:ly.indeterminare a1 rodundanrs; By removing the supports at È and o, simpty'sufforted "" beam Ae is ,1-.
,
ëlloscn as the released beam as shown in
¡¡ì*t',,
Fig.3.llb. The cornpatibility
conditions
U{''"\
6"t
*
il
BC þ3.5.f2.s+ ¿ * (c)
BHAMS WITH TWO OR MORE
ot
,t
CONTINUOUS BEAM
r.:)
# at¡tl Ynn = where. Y¡ ¡ =
Y'ocRc
l' ,
about B in the conjugate beam
Yei¡
,.1
:r*ì
'*
1*'=t ,
' '
(c)
^it
Li i..
DEFLECTED SI{APE AND LOADING OUE
/7Ð7
f
no=r
ofFig.3.ll
s5.56
c
r0 ls+lslo I xox_x_ . 2
6
(d)
l".o = y'oc using
DEFLECIED SHAPE AND M/E¡ LOADING DUE
f0+rs'\z.s r x¿.)x^- r. 2.5 Et 3 sEr \ 3 )ts 2
l .- 18 - _xl)x-^l2
Y"oo
T0 R9=l
:
the reciprocal theorem (Figs. 3.1
I
SHEAR FoRCE, kN
--t6'3E
(' (f
)
\ 3 )ts 2
Using the Maxwell's.reciprocal theorem: -
39.1
BENDING MOMENT hNm
Fig.3.ll
Ycs
:
100
xy'r.: #
Y'oc
:
100
xY""o:
3046 EI
Using the compatibility Eqs, (i) and (ii) 3300 55.56 Rc + 54.67 RD
:
.
3046
=
EI
c and d)
r - l8 6 | -- 18 fs+rsì0 l---xox-x-=-
(.)
30.46
moment of the M/EI loading diagram about D in the conjugate beam ofFig. 3.1 I d
= -xl)x-xl 2 5Er
t;
(
l0 5
EI moment of the M/EI loading diagram (due to the uriit load at D) about B in the conjugate beam ofFig. 3.tl d
JBD-
(,,
{
c
23EI\3)ts2_EI3
iO n.=t
¡
t_,
ofFig.3-l I
r0 /to+lsl 5 I
| .__xl)x
M/EI
(,j
(
(ii)
54.67
{_;
{..,
+ Y"opRo
deflection at i due to certain load applied atj.
beam
í'' ;
(i)
I
.
i
65
Y"cP Ro
y'^._: -xl)x-xl lx---x)x-x-=-JLL 3El 3 EI 3Et \ 3 / 15 2 2 y'oc : moment of the M/EI loading diagram about D in the conjugate
'{"''¡
{L
*
R.
Y'cc
l*15,. lo xf lo+l5Yäl-1 ,2.s, 5 *a=L 3EI 3 EI 3EI \ 3 )\ts) 2 2
(b.) RELEASED EEAM UNOER APPLIEO LOAO
eùr'
REDúINDANTS
require that the dcflections at both C and D of the released beam under the combined action of the applied loads and redundants R. and Ro must be zero, that is : ( FlÊ, 3,I I b' c and d)
r00kN
I
"¿ {:)
.
METHOD OF CONSISTENT DEFORMATìO¡¡S
54.67 Rc
+
64.80 RD
¡Er 3
64.8
Er
Q._)
6r)
6ô
ü
Mstrix solution,
i) {:)
{)
METHOD OF CONSISTENT DEFORMATIONS
s4.67'lfRcl_{r:oo}
or.
i:
.
IIIÍAC'rIONS DUE TO YIELDING OF SUPPORTS
77.28\
f*.1= [ lRoJ [-18.20J ls+.at 64.80J[RDJ 13046J tJsing the equations of static equilibrium, RA '= 39.l0kN and RE : 1.82 kN fss.so
The shear and moment diagrams are shown in Figs. 3.1 I e and f.
3.5 RbACTIONS DUE TO YIELDING OF SUPPORTS it: d-''t -i;. *'
il.,, t,i
.'1,
,,,
t;ì
{"\
:.ìr-.'*'
,iirr , \i
"-¿
,ìt
j li
ti:"Ç)
,.:i:.
Example 3.9
Determine reactions in a fixed ended ,beam due to a vertical settlêment of A at support B as shown in Fig. 3.12
{--,
:
':\¡
1,.:.
,, _.'
i[) 'Ç-t
Tþe degree of indeternrinacy is 2. Select R" and M" as redundants. By removing the support B, a cantilevêr beam AB is obtained as the released beam. The compatibility conditions require that under the combined action of the applied load and re
Yu+Y's-Y"a:^ 0s+0's-0"e:0
i{; '('
fCI M/ET
wL3 3L
wL4
(
wL3 6EI
EI
e', = JB_
Ail
,(-
(
MB
2
Y
R'L' 2Et
SHAPE AN{D
M/ET T¡ADING
OUE
'I
i;RoB (Fig-3'l2d) DEFLECTED SHAPE AND M/E..LOAO|NG DUE T0 Rg
trig 3.r2e) å# 3EI
unti clockwise
tZEIA\ / wL-TT)
T0 Mg I
'å L'g
clockwise
=
I'B
6l-
I o;MJ
*?r :r*:1 2EI 3
't',t
(d) DEFLECÎED (Fig.3.l2 c)
clockwise
Msr..L : Y's :
i( ,tl
LOADING DUE T0 w
,
!
'6EI 4 8EI eB : aÍeaof the M/EI diagram
(r'
(,,
RELEASED BEAM UNDER w
is applied upward while Mu is applied clockwise. 'Ihe slopes and deflections can be obtained using the moment area theorems. ye : moment of the MÆl loading diagram about B in the beam of Fig. 3.12 c
. R,
,(,r ..::
shear force and bending moment diagrams.
Solution
.,:tt-) ,ril ii1
a. Draw
Fig.3.12
(f ) ,SHEAR
\2
FORCE
H:d_!î (ç1
BENo|NG MoMENT
67
r.'
METHOD OF CONSISTENT DEFORMATIONS
6rT
RnnerroNs DUE To yrELDrNc or supponrs
69
Subrtituting the values in Eqs. (i) and (ii) give,
M"L2 R"L' _ 8EI zEI --=a 3ET ^ wL3 , MrL R"Lt _^ 6EI EI 2EI
wL¡ .
and
(iii¡ (iv)
(o)
CONTINUOUS BEAM 2
Solution of Eqs.(iii) and (iv) gives,
wL lz9l Ite: 2-Lro
.
(b)
8.43
r3.96 kN sHEAR F0RCET '!
7.33
vÉ-go M^: 'l2Lt
and
IF,:0gives,n^=f*Sa I Me: 0gives,
Me
l2Et.lwt] 6Et. --;";'^- wt] "+l(wL lL--:::-+ \2 L' ^ ) t2 L''. 2 -0
(c)
- wt] *9o ^12L,
BENDING MOMENT, kNm
Fig.3.13
M^
er
Thus, yielding of support B results in. an increase in moment at A and decrease in moment at B. The shear and moment diagrams are shown in Figs. 3.12 f wrd g.
Rc: 42.39 kN I Ro: - 35.76 kN .l.
Example 3.10
,:ì: ' lfirlhe
Ãnalyze the 6eam shown in Fig.3.l3 a due to settlement of support C by l0 mm using the method of consistent deformation. Take El : constant.
.,.i1:f,Í3 kN J. respectively.
Solution The degree of indetèrmin acy is 2. seleci R" and Ro as redundants. By removing the supports C and D, the released structure is obtained. The compatibility conditions require (Refei Figs. 3.1I c and d of Example 3.8) : y'cc Rc + y"co RD - 0.01 : 0 (i)
I'pc Rc + y"po Ro : 0
and The values
of
y'cc , yloc , y"çp and
(ir)
y"p¡
cân be obtained from Example 3.g. It should be.noted that in Example 3.8,.R. and Ro are assumed to be acting upwarcl.
55.56 Rc
+
54.67
54.67
or.
fss.se
RD
-
0.01 EI =
Rc + 64.80 Ro
:
0 0
54.67]fRcl_/o.orer]_/+oo]
Ls4.67 64.80J[RDJ
[ 0 J, [0J
pr€sent problem, since support C sinks, the corrqct direction of
R.
is downward
.$n!,thât of Ro is upward. Therefore, the reacrions Ro and R, are l¡.ÞO tN
-rïiieshear force and bending moment diagrams are shown in Figs.
uå],f;
3.
I
an¿
l3 b and c.
n*urs
'lËin¡iiplÈ
'
3.rl-
., filgure3.l4 a shows a portal frame with hinged supports A and F. nttd,hending moment diagrams and élastic curve ofthe frame.
Draw
shear force
.!loi¡rtlon statical indeterm-inacy of the frame is l. choose R as redundant. Replace the * , -'l .l"he Itittgs support by a roller supprort at F to obtain the releasêd structure as sbdwn in Fig, .1.14 b' The compatibility condition requires that the net horizontal deflection at F under tl¡ç eoi'lnlrined action of the âpplied loads and redundant Ro* should be zero. .The elcUections can be calculated using the unit load method as discuiied in sec 9. I I and 10.3 çf volume I of this book. The free body diagram due to applied loads on the frame is lhown in Fig. 3.14c, and that due to unit redundant R.* : I in Fig. 3.14f.
{"":)
7l
FRAMES
{.}
METHOD OF CONSISTENT DEFORMATIONS
?(!..,
I
tr1
{J
iÍÈ¡ -ll
T
'
,t.':,
3m
rî; ll^ (o)
7s
-r.' tltì',
,
l,i' (9! f o.ss
(d}
2-I{INGED FRAME
(h) SHEAR
FIN¡ REAcTloNs FINAL ..Ìi;frl1,.' ;fii,. , :t:rTJ:,' ze i'?.20,=< !
DEFLECIEO SHAPE
-J'-',
",:i';¡',,: t-.r.Ì'i-.,' t:1,..i.i, .
.'.'j : i,
FORCE KN
K
.''.,
,.,;,
(b)
RELEASED FRAME
(c)
:;,,:,..
DEFLECÍED sHAPE
(¡l
75
A
f
(c)
rz.es
l¡ rz.rsf
FREE BODY DIAGRAM DUE IO APPLIED LOAD
Fig.3.l4 Portal frame with hinged suppoits
llorizontal Deflection at F due to Applied Loads Segment
FE
ED
DC
CB
Origin Lirnits I M (Fig. 3. l4 c)
F'
E
D
.c
0-4 0-3.5 I
0
FREE EODY DIAGRAM DUE TO
RFro
t
Fig. 3:14 Portal frame with hinged supports (continue)
."
3.t4 Ð
2t 67.85x
X
-3x 4+I
q-3
0-3.5m 2t
I
67.85(x+3.5)-50x
300
-- (17.85x + 237
m (Fig.
(f)
DEFLECTED
"
,,,." .''.:
(¡}
BENDING MOMENT kNm
JX
B
0 -4m I
(300
-
75x)
.48)
++](x+:.s)
=).)+-
BA
(7-
- (x+3) :4 _ x
7
x)
U {}
FRAMES
METHOD OF CONSISTENT DEFORMATIONS
t2
{r} t-
{-}
I
ar,*
244.t6 Mmdx
EI
J EI ar* + a'r*
L} 3.5
{")
= o*
{")
/
^\
3.5
ioz.ss-lo-*l**"i(tz.ss* \ 7)2Er {{ 1
{ri
,.
J[¡oo(z -
+237.48)
" lt.t.*l*.* 7)2Et \
{)
-
ü
1
J.
*)9 [(¡oo- zs*X¿ - *)31 'EI 'EI * J'
o
10541.86/ 244.16
: (-) 43.13 kN
Rn^ is acting towards left. Final reactions are shown in Fig. 3.14 g. The .reaction ing shear force and bending.moment diag4ams are shown in Figs. n and i. The shape of the frame is shown in Fig.3.l4 j. tnple 3.12
0
-Figure 3.15 a shows a porral frame with fixed supports A and F. Draw shear force
Lt3.5
) ^ -l gllnr.,rrL
tt'gzslo*z
73lo* ttl Er l2 .1{l'"
{")
-:
¡.li
{"_i
I
Rr, : (-)
or,
Rr.*
2
bending moment diagrams.
+t.sst+
r306.r4x +
-t
4
2lo Erllr.-41' 2lo*_]_l,rooxErl 6004*zs4lo 2 ¡l¿
ror.zs4l +
{,:
^:-Fx
3
2'
EI
Horizontal Deflection
,l'
T
10541.86 att
F due to a Unit Load at
o '¡zs lso t c\
lso
F
{m
mmdx ^, _ - Jf EI ^Fx
I
.0
:i#.f(..+)'#.Í,,.ï)' #.lt',-*)'#*Jto-.1'fr ,3-".4
'
r
I
t, (o} FIXED
FRAME
+
150
(c)
0.0
{l- *--Ll,ux+¡.¿:{+0. 3 lo 2Etl 2
r,
--
I
.L
E
FREE BODY DIAGRAM
13.5
rs¡or-l
:MF
3lo
t lru.rr* +q.ttt+als¡d{l '*alon*-,04*dl' *zBtl 2 3[ Erl 2 '3[
Rpri
\-..+Rrx
't
(b) RELEASED
FRAME
(dI RELEASED FRAME UNDER REDUNDANTS
r
^
-Á
I llo*-8*'**'l * Erl 2 3lo
Fig. 3.15 Portal franre with fixed supports (continue)
t,)
*
METHOD OF CONSISTENT DEFORMATTONS
14
t)
c
E
'rþr
î
L)
ü '*
tg)
r. r;--c{
fr
li-ftt
I I'
lr' I
{_;
*
:A
\
{_;
(l lj.
(T) FREE
MF=t
Segment
F
Origin I
FREE BODY DIAGRAM
M(Fis.3.1.5 c)
rf zs
ii-'o.l
36.19
86.12 '
2
(hI
:
{_;
t., (., ,il
:
(r ( ,(' (
I
t
4-
4
l(x
+ 3.5
-t
x
4-
0-4m li5 -75 x (x+
3): I -x
1
1
:l
-t zs*)(r
-
*)S
| )
¡ 2¡iì5
l2lo
¡3 I
)
¡14
rzsll- - ¿xi + l-rzs** rool- + zsI-l 2 3lo
12 lo |
F¡NAL REACTTONS
_ _225
Fig. 3'15 Portal frame with fixed suppoits (continue)
3.53,4
Solution
M. The frame is staticaliy indeterrninate to a degree 3' Choose Rr* '. Rr, and èantilever is a strucfure redundants und ,"rou. the fixed supFûrt F. The released
{;
B
2l
EI
{, ,,
(-l
BA
c
-.)# * j{-,rrff * jt-,2Ð{o 00
* * looll-l +
lil:
iii'
CB
s.7s1
f zr. zs
BODY DIAGRAM
DC D
0-3,5m 0-3 -50x - t75
Lirnits m (Fig. 3.15 e) m (F'ig. 3.15 Ð rn (Fig. 3. ls g)
RFY=I l1'
fr
I
zþr
E
lí
ri9,
I
I
r")
l,-t
H tl
{} * {}
shown in Fig. 3. 15
b.
Free bbdy
,iJiagrarn due
as as
ôi'*ôP*z*ôt*t+ôt*o=0
(Ð
(b)Thesumofverticaldisplacementsat_Fduetotheappliedloadsin Fig. 3.15 b and redundants in Fig. 3.15 d must be zero'
+ôrrz+ôrrilôrro=o
(c)ThesumofclockwiserotationsatFduetotheappliedloadsin Fig. 3.1'5 b and redundants in Fig' 3'15 d must be zero'
2loErl3 r.r4l"
- ?114 *
The sum of horizontal displacements at F due to the applied loads in Fig. 3.I5 b and redundants in Fig' 3'I5 d must be zpro'
ôlr1
r.Ð# * J(-rzsXz)fr . Jt-rzs- zs-)zS
to applied loads is shown in Fig: 3. I 5c.
The compatibiliry conditions may be u'ritten as follows:
(a)
J{-so*X.. 000
_
'|
t3668.23 EI
3.5 (ii)
-lrrrr**szstl 2lo I
1225t ¡u xtL
J
t-so.)f-r)
000
ff
3 * (-r zsX-r) J
4-
$
. t-r zs - zs,.)(J
r)
$
',LJ ',{,"}
,il ,l* I
:
rzslxll* ':' l,rr^ |
+l+1". 't-to
,. = upt
.rr*l' 2lo
:l:
s
Displacements at F Due io Unit Horizontal Load at Segment
FE
Origin
F
0-
timits
,l* 'i*'
EC
CA
c 0-7m 0-7m E
4m I
2l
I
X
4
4-x
0
,x
7
I
m (Fig.3.l5 e) m @ig.3.15 f) m lFie. 3.15 e)
;^., *i
-l
-l
-l
iii' n' =-# j':'. o,, ^ = i¿* i¿* l¿*
i¡,
::ìlì;.liiÈ
ii"-
.
^ òt'*2
EI
ili
li
t. ,.
77
: o+l*'l'* -,Tlo "'l^ to lzt*-tt-l'
öryz = -00
o.JT#.Jto-ozfr
,,
^ ÒFrr:
73.5
U
,î : j-.#.i,-0,#.i,-0.,.i* = I-+1,- zt^tt.l-*.*11,,
iic, li
,,
t07.67
:
ili' l,lr-,,
[lit {i,, $¡''
?{
llr.
,^ = ,vr¡
a,
ti
i,'
ii
t
ii
(
i. li i.
^
:
5
=ôr*o F
73.5
-
Substituting the values of õr* ,
ôr,
and 0, in the compatibility Eqs. (i), (ii) and
. 73.5^ 400.17_ 6t.2s Er .. ,, *r* *_;, *0, __Èl_Me:o
t3668.23 ::',.
i'...
). ôrv¡:r.j#.i#=
l6h
.
,1.,
+
l1ly,
lRearranging the above equâtions in the matrix form:
'.
lSymmetric
7
o.ldl'+
,
ïi # ** - uå?t *r,
ßs -zs.s l[n..] [ zzs ì I rot.n 4oo.t7 -6t.2s ll*r, I = ltzae*zsl I _uso j l+.s j[v.J
EI
7
14.5
J--I
¡,
Displacements at F Due to Unit Vertical Load at
ò"r"
fn-{reI.{,
:ÒF*3
E,
orz =-"
l-"^
,.
':,.1
:i-:..:
li:'
-, EI
F
,[ \._,,
i,i
itii,
400.17
Er
.i?'
t,
.
^rYr= '. .Ò¡ry-t
77
1980
jr)
:l
FRAMES
METHOD OF.CONSISTENT DEFORMATIONS
=
1
ii,
ê
76'
t¿ex ¡
[
=e
(iii)
&} j
i,{'z
TRUSSES
METHOD OF CONSISTENT DEFORMATIONS
78
{'2
:,JRUSSES
O
t: {)
or,
[noì
[-ze.s
[*.,|
[-zs.zo
{*., i= |
diagram can be drawn in three Parß:
(i)
L,
(ii)
ti
(iii)
alone is drawn as shown First end moment diagram due to the redundants
in Fig.3.l5i. to the applied loads assuming Next, free span moment diagram is drawn-due in Fig' 3' l5j' as shown each member to be simply supported are superimposed to get the Now" the bending moment diagrams so obtained The location of the Fig''3'l5k' in net bending to.ãnt diagram as shown moment and tleir bending negative and positive foint of iriflection, -*i^'n' each .magnitudes der.rmined Uy considering the free body diagram of
.an'bË
."*b"t
l$: truss may be externally indeterminate or internally indeterminate or both. If a truss ternally indeterminate, the redundants may include çxcess reactions over three so the truss remains stablo. However, it is possible to obtain a released structure by ing only member forces as redundants. when a reaction is chosên as a redundant,
zt.ts
in Fig. 3.15h. The bending moment The final reactions in the frame are shown
{:)
I
79
(beam and column) separately'
TheshearforcediagramisshowninFig.3.l5/.Itisdrawnwiththehelpoffreebody diagram due to the applied loads and final reactions
ll.. e,"on.touint in the direÕtion of the'reaction is removed and its action is replaced by ,â,qnknown force R as done in the case of a statically indetenninate beam or a frame. The ' c(inrpatibílity condition is that the deflection in the direction of the constraint must be
'
t"era.
,i.;.| ,' lf a truss is internally indeterminate or externally as well as internally indeterminate, ,:.Ël.nernber forces may be chosen as redundants. The redundant members are removed from ,,,fi ihc given truss such that the truss remains stable. A pair of unknown member forces are
pulling on the joints. This change will not aiter forces in other members of the which has .now become perfect and statically determinate. The compatibility ì,:.iì'.llusrt ::r.,,,¡tpplieci
, eondition requires that the relative movement of the joints away from each other is equal ,..:j. r", rl.^ ^l^--^¿:^^a.L^ removed from thejoints. -^-L^ ¡,i$ tttu elongation of the member ':ë:. a truss shown in Fig. 3.16a. Remove membgr 3 - 5 and apply a pair of ^^--t)^;. consider :.,,¡,iti¡known member forces. Mathematically it may be written as:
t_' {,_ì
(l
(bl
86.4 2
(¡)
END MOMENTS
FREE SPAN MOMENTS
kNin
kNm
I' ij
(¡t
;
::.
Fig. 3.16 Statically indeterminate truss
$JôinÏs 3 - 5 (Fig.3.l6 b)
r,
...,4
il"l
. '
.
'n I ltLI
: ^-Tô
!1
'l
Fy...tr¡iglqq¿ method,
:5'F'uL
(-:
L
06.42
a-':
(k)
BENDING
MOMENÍ
TII
\.
('
Q.2a)
SHEAR FORCE,KN
*-*fn
,t'
¡.p.
:
Fig. 3.15 Portal framc with fixed supports
:tu2L Lt
AE
(3.2b)
METHOD OF CONSISTENT DEFORMATIONS
80
: u : Tl : = ^ . ô =
where ,
F''
where n
member forces due to the applied loads member foices due to a unit load pulling on the joints redundant force
lf
axial displacements due to. the applied loads joints
The unit ldad method was discussed in section
9'll of volume I
:
total number of members in the given truss
there are k redundant members, there
ban be sirnultaneously solved
where
will
be k equations similar to Eq.
3.5.
These
to get the redundants. If there are more than one
redundants, it is more convenient to use Eq. 3.5 rather than Eq. 3.1.
redundant members have been removed. axial displacements due to a pair of unit loads pulling on the
8t
TRUSSES
[,xample 3. l3
of this book.
Find forces in various members of statically indeterminate truss shown in Fig. 3.17 a.
Altematively, the compatibility eondition may be expressed as: Joints 3 and 5 wili come closer by an amount =
But rhember, -
,
,",U elongate by an amount
y2
For compatibility,
FLu
AE
f H
2 (2O)
T
: +!t A¡Er
*
(20)
_1ì!r_ ArE,
:
g
oÍ, f;:,
-t,
í'
(3.3)
3 @ /.00cM
(g) INDETERMINATE
#þXIX 25(cl VALUES 0F Fr
TRUSS
s0
-0.707
Q'4)
T, in Eq. 3.3 which can be easily evaluated.
5-F'L,f þ AE L AE + tu2LI
K. I
400
p'+ uT¡ The total force in any member p = There is only one unknown quantiry can be expanded as:
-s0 tl7s
lTskN
3t
ÆlxlÈx
Eq. 3.3
- 0.707
(bI
IL, :0 * AIEI
I*- ffi
RELEASED TRUSS :D TRUSS
=0
(d) VALUES 0F u¡
Fig. 3.17 Statically indeterminate truss
Solution The last term may be rewritten =
1t+, AIEI
since the force in this member is unity
(u:l).
'h",,
äî#. rË*=o
The truss is statically internally indeterminate by degree l. Let us remove the member 3-5 and the released truss is shown in Fig.3.17b. Analyze the truss due to the applied .Ì'. loads as shorvn in Fig. 3.17c. Now apply a pair of unit loads at joints 3 and 5. Due to l;,,.this unit force, there will be no support reactions and member forces are shown in Fig. t;..:.. .lZ¿. The physical properties of the mernbers and stresses due to external loads and
',
It can now be included in the sècond term.
(3.s)
Force in mernber 3
$ r'ru or?
or, t, : (-)i
su_-L
2
a'e
- 5,
F¡
Total force in a member : (3.6)
_, F
'
: - 1262ë : -n.6BkN lll.4 +
ux
(-
compression
17.68), that is; as shown in last coltimn of
:'-ir',1
l
,',--.1' t:; ;::
METHOD OF CONSISTENT DEFORMATTONS
Table 3.2 : Computations for Example 3.14
Table 3.1 : Computations for example 3.13
Length
Member
L cm,
2-3 2-6 2-5 5-6 3-5 3-6
ru
Area
A cm2
400.0 565.6
20
400.0 400.0 56s.6 400.0
25 20
Force F' KN
25
KN
A
0.707 1.000
35.35 0 25 0
l5
F'uL
-
-50
l5
Force u
-
-
-25
0.707 0.707
-
Total
A-
force kN
10.0 JI.I
-
37.50 17.67
8.0
t2.s0
10.0
37.50
c
-17.68
0.707
282.80
37.7 8.0
t
-12.50
I .4
-ve forces : compression Example 3.14 Ãnalyze the statically externally indeterminate truss shown in Fig' 3.18 chord
mlmbers:
100 cm2, and area of web members
:60
a.
Area of
cm2'
Solution roller support at joint
8. The truss can be made statically determinate by removing the The truss remains stable. The released truss is shown in Fig. 3.18b. Let us determine unit load in the members due to the external loads, and forces u, due to forces also in Table and and d; in Figs. 3.18c joint are shown forces member The 8. applied af 3.2.
a
F',
¡- F'uL LAE '
Lcm
Member
353.s0
969
r-2 2- 3 3- 4 4- 5 6- 7 Bottom 7-8 unoro 8- 9 9- l0 l0-lr n-t2 Web l-6 Member I - 7 t- 8 .2-
33-
E
100
'600
100
600
100
600 600 600 600 600 600
r\l
l-
(
r00 100
r00 60 60 60 60 60 60 60 60 60 60 60
t000
8
800
8
1000
9
800
3
- l0 4- l0
1000
-5-10
1000
5-12
1000
800 800
Force u
Force F'
F'uL A
KN
KN
-76.67 -76.67 - 93.33 - 93.33
0.500 0.500
1.500 1.500
-
280.00 280.00
35.00 35.00
- 0.750 - 292.50 - 0.750 -292.50 - 0.2s0 - 32.50 - 0.250 - 32.s0
- 63.89
0000
0.837
2 -3 3 -4 4 -5 6 -7 7 -8 8 -9
F'l! T 4 A^E R:y-:-L \L¡n (-) !ls-\'qz that is, R : (- , ' 60.51 = + 68.6 kN
9-t0 l0- ll upward
The net force in each member can be determined by superposition, that is, (F', + u' R). The final member forces are shown in Table 3.3.
tt-t2
Force kN
-
7.866 7.866
-
58.933 58.933 46.067 46.067
-
t3.399 13.399
4.466 4.466
1.500
3.370 3.370 0.37s 0.375
63.89 - 0.837 - 891.27 11 .670
0000 2.920 0000 - 36.10 0.416 -25030 2.920 0000 36.10 - 0.4t6 -250.30 2.920 0000 - 36.t0 0.416 -2s0.30 2.920 - 63.89 '- 0.416 443.00
L-
4150.42
must be zero,
| -2
1.500
- 891.27 11.670
Table 3.3 Final member forces
I
A6.000 6.000
- 11.67 - 0.500 - n.67 - 0.500 65.00 65.00 21.66 21.66
u2L
1.000 -460.00 1.000 -460.00
Total
2 AE
i ;--.
100
800
Member
l
100
t00
1000
s'u'L 60.51 u AE,, E
yF'ur-*Tl&n=o AAE
re '(.
100
600 600
Top Chord'
5-ll
4150.42
For compatibility, net deflection at support
Area A crn2
Length
u2L
t.000 Tota
+ve forceS : tension
707.00 1332.93 0
83
TRUSSES
.
i': ,tå
Member
6-l 7 -1 I -8 8 -2 8 -3, 9 -3 0- 3 0- 4 0- 5
l-
Force kN
- õ.))) 0 6.555 0;
-92.5s6 0
-
7.444
'0
5
5.- l2
7.444 0
-
7.444
TRUSSES
METHOD OF CONSISTENT DEFORMATIONS
84
,Ito
*n
¡ 50kN
zo (o)
lr
uo
(ro)
l,
T
Js0rH 20 ? -20
tol r
I
6@6m-3'¿ñ
þ
85
------{
I I
INDEiERMINAIE IRUSS
100 I I I
30 (b) RELEASEO IRUSS
-r
3
t ,--76.6? Ifcorn r -91.31 -91 31 4+
I I I
sokN
30
ç o
400 I
ai .¡l
i
50rN
t
I
I
300cm
,
,
(b) F, vALUE's
(o.)
-
5r
0.60
tlo.st(¿) vALUES oF u¡ I
\ Fig.3.r8 Example 3.15 member A truss shown in Fig- 3.19a represents an industrial frame' Determine the forces.
Solution
Thetrussisstaticallyindeterminatetodegreetwo.Itcanbemadestatically
example determinate in several ways selecting any two redundants, for
:
(a) selectmembers I - 4and3-6 (b)' selecìmembers 2- 3and4-5 (c) select reaction Rr* and member I - 4 (d) select reaction Rr* and member 3 ' 6 ìi" bst choice is b¿d ás there is no member at joint 6 to balanee the horizontal force Ru*. Let us select the first oPtion
(c). U1 vALUES
(d) U2
Fig.3.19
VALUES
TRUSSES
METHOD OF CONSISTENT DEFORMATIONS
86
PROBLEIúiS
The compatibilitY conditions are
$r'u,r--o 4tv
.
(D
""d ËFi+=o 7^E
where" I- u,
u^
: :
4 iorces in members due to unit load iir location l6 forces in members due to unit load in location 3-
F'uzL *T u,ur! ^-r aano ' L AE ', 't-4 *t+3F¡_a L AE - L-AE n_,
A ¡n the span AB.
=o
(iii)
=o
(iv)
,rl'W (b) (c)
Eqs. (iii) and (iv) become,
and
Fig.
(a)
ThevaluesofF',u'andu,areshowninFigs:3.l9b,cand'dandalsoinTable3.4.
'
in
structures and indicate the redundants.
æT
give' Substituting for F in the compatibility conditionS ur2L ç,L F'url, +T n-, *I!*F¡-o u AE AE L¡B-'-"
itrt momcnt nt
Rcduce the following structures
(ii)
F: F'+ urFr¿ + urFr-u
but
momen(s ere shown as pos¡t¡ve while anticlockwise moments are shown as :j,l¡t.'illl problcms. M6" represents moment at B in the span BA. Similarly, M"6
4953.7.2+147.2Fr-¡+10.8F1-o:0 + 10.8 F, - 4+ 136'4 F¡-o : 0
7733.88
lo'8.l|.E-41 f-+ess'tzl = or' lut.e r:o.+llrr-u/ ro.s l-zzr:.stJ I
.r
{l-:}={-ii Íl}
are shown The net force in each memÈer can be computed by Eq.(ii), and the values in Table 3.4 Table 3.4 : Computafions for example 3.15
Fig.
P
3.1
P3.l a-g into all possible
releaseil
í,.J {"-}
Ð {J Ç).
{,)
{) {)
o
METHOD OF CONSISTENT DEFORMATIONS
t8
Determine reactions in the beains shown in Figs. P3.2 - P3.7 using the method deformations and draw shear force and bending moment diagrams. Also draw deflected shape ofthe beams. Take EI : constant.
{:)
a {i (r'
J.J
of consiStent
zl R¡, -- 2s kttl, ^ Mo : - 47.55 kNm, (p3.3) R..., : S9.Sr kl,rT, M., : 54.14 kNm, (p3 4) R:, :" s7.5 kt\rl, nu, = 173.5 kM|, Ru, : tg khrl, Mun : t05 kNm,^ Mn, : -105 kNm (p3 5) Rt,, = 65.83 kl\rl, R", = I14.17 khrl, Ru, : I0 kl\r|, M"n : 85.0 kNm (p3 6) Ror, : 29.t5 kt\rl, Rn, : 95.74 kNî, M¡,, : 43.27 kNm, : M", 43.67 kNm. R;, = 30.05 k^fT, (p3.7) Rn, : 84.36 rr1, n", : 33.90 kìrî, M¡u : 79.4 kNm, M,¡ : 33.3 kNm,
Determine all reactions due to a vertical settlement of 7.5 mm at support B of the beam in Fig. p3.4. Determine sropes at sections A, B, and D oithe finar elastic curve and draw shear and bending moment ¿¡ugrurr.'iut" B : ;õ;0;
kN/m2 and I: 150 x 10-6 ma. (P3.4)Re.t:63.7s klû, Rhv= 159.75 klll, R,,,:26.5 kìr|,
(pi
¿t 'r-)
PROBLEMS
3.4
M*:
3.5(a)
'o
- d into a,
possibre rereaseJ
P 3.2
,ã -lt
i
n'--" Þ_3r l.t.s * .
50k
N
P 3.3
r.s
20 kN
/m
T
..
P3.4
T
2m
2m
+'
+
I
3m
I
3m
{',, _:
58.04 kNm,
Reduce the portar frames shown in Figs. p3.g a structures and indicate the redundants.
25kN/rn
{.
E
I r Qg¡51.
1
E
I
= C0NST.
10kN/m
(b)
:
i'{-,
A
P 3.5 B
þl-+zt*¡+2r. 3 O3.sm",
:{; 1.
-,
':,(
i
(' i,:
l(' I
{.
67.5 kNm
lSkN/m
)
{...
Mo":
Determine all reactions due to a vertical settleient of g mm at support B of the beam in Fig. P3.7. Determine slopes at A. B, c and D of the final elastic curve and draw shear and bending moment diagrams. Take E = 200 x 106 kN/m2 and I :200 x l0-5 ma. (P3 7) !,r: 6,r,t t lvl. R",= 52.45 kM|, Mh,,= 40.46 kñn,
T
P 3.6
5m
þ-4 L---a-z ,f. 3 .*¡ (c)
i
Þ_ 6.
75kN
CD -Ð-Ð,
Fig.P3.2-3.7
P 3.7
Figs. P 3.8
.rr.s,
U ü {) {J
3,5,(b)Éindthereactionsinportalframesshow¡ìinFigs.P3.8.a-dusingthemost '*"-' and bending moment appropriate released rì.u"tu.". Draw thrust, shear force diagrams.
: M¿" = (P3.sb) R;; = (P3.8c) R,, : Muo :
(P3.8a) R,^,
{."}
'{) Ç)
{:)
.ü
{i
3.6
{)
3.7
\.,:
{_,
(.)
=
-
r21
27.84 kNm'
8.65 kN-+.
R;
30'22 kNm'
R*
kNm' 8.14 kN<-' 41.86 kN<-,
143.58 kNm
T
=
'=
t¿;,
13.05 kNm,
8.65 kN<-,
t8'40 kl{T'
kNm 5.30 kN+, 45.30 kNT,
=
= R", = M¡o :M¿" :-
3t.60 k^rl. Mot,
M¿"': '27'46
R.,,,.:
I
6m I
1
5
32'53 kNm
:
122-14 kNm
Determine all reactions due to a vertícal settlement of l0 mm at support A and portal frame also a rotational slip of 0.003 radian clockwise of support A of the mÓment bending and force shear fiee-body, Draw P3.8c, in Fig. shown : diagrams. Take E = 200 GPa and I 300 x l0{ ma. : 23.61 kN-+, Ro, = 24-32 khrl, Moh : 03 0! kyf' ' ¡pi.8c1 ' R,,, 14.95 kNm, R;; : ß.61 kN<-, Ro, : 2s.68 kt'rÎ, úi;,, : kNm kNm' Mro : 47'60 M",! = -
rf
?
2ookNl
v a@6m=24m
84.44 kNm
support Determine all reactions due to a l5mm vertical settlement at the hinged arid bending force shear E of the frame shown in Fig. P 3.8 a. Draw free-body, x l0-ó ma' I 300 and = moment diagrams. Take E:200 GPa 25'57 kN+-, R"' = 2't'43 kN+-, M"o -27'84 kNm' (P3.8a) R;
M¿":
r,)
(-,
M-^.: -
:
{j
M"b
Analyze the truss shown in Fig. P3.10. Take area of top and bottom chords: :70 cmz. Treat R. as redundant. (p3.t0) Rrr: 82.3 krl, nor:285.5 kN1, Rs,,: 132.2 k^rÎ. F1.z:35.51 kN, 100 cm2 and area of ^web members
122'14 kN'm
6t.o7ktû,
M*t = -/5'80
(P3.8d) R,, : &¿. :
24.43 kN<-,
25'57 kN<-'
9t
PROBLEMS
METI{OD OF CONSIS'TENT DEFORMATIONS
froorN
Fig. P3.10 Analyze the truss shown in Fig. P3.l areas in cm2.
l;
Numbers on the members indicate the
..
(a) (b)
Treatmembers 5:2 and 2-7 as redundants. Treatmernbers I -6and6-3 asredundants. (P3.1t) Rrv : 256,30 kNî, Rr, : 265.70 k^f|,
Fi_n: r26.30kN, Fi_r:
r2t.4okN,
! tsol 2 (so)
Fs-z :-23.95kN, F2-- :-ll'l9kN,
3
TI lm
.-70'43
3'8
Determine force in member 2 - 4 olthe truss shown in Fig. P 3.9. Area of each
member: l5
I I
cm2.
(P3.9) Ft-t:
8-08 kN,
Fz-t = 4.62
kN,
{,:
5
Fs-t = - 2'31 kN
þ-31!*+--!e--¡
llsokN
I
f75
2oo I
Fig. P3.l t Analyze the truss shown in Fig. P 3.12. Treat member 2- 3 as intemal reïundant and Rr, as extemal redundant.
(P3.12)
(i
¿
4
Rr, :
-59.43
R
42.38
, :
kN<-, R,"
kN1,
R-;
= :
62.34
kN1, R¡* :
45.28 kN1,. F;_4
Fz-¡:-59.94kN,
r
59.43 kN+, 88.15 kN,
:
(60)'.
2
T 6m
I {
{
ito
i
r, H
Fig. P 3.9
Fig. F3.12
tl,,ro*"
þ-_em+sm,
LJ
{J
DERIVATION OF THREE-MOMENT
ÇI"IAPTËR
four
û
EQUATION
93
illhe total loading on any sþan can be sprit in two parts assuming it to be simpry rported and loaded with applied loads on the span, and support moments Mn and M".
L)
{) * {)
THREE - MOMENT EQUATION //
/" fiT
(_)
I
hc
rl
/")
{) {,) {_)
{,;
t. {.: :
{:
L (l
í:
(o) 2-SPAN
. 4.t INTRODUCTTON
Ë
-
continuous beams are commonly used in buildings and bridges. The spans may be different and cross sections of the beams in different ,pun, ãluy also be djfferent. Moreo'er, there may be unequar settrement of supports. such båms can arways be analyzed by the method of conSistent deformations, However. this will involve computations of a large numbet of deflections or slopes. The thrþe moment equation is a more convenient approach. tt is based on the coniinuity of the elastic curve of a beam over any intermediate support. The unknown moments at the supports are treated as redundants' The compatibility condition requires that slopes of the liastic curve at e¡ther side of an intermediate support rnust be equal. In this manner each span may be treated individually acted upon by the roads on it anÇ the nioments at both ends, if any. The computation of slopes at either. side of a ruppoi requires a knowredge of loading on the two adjacent spans as well as the bending moment at three successivJsupports includíng the one before and one after the support under consideration. Sincå this approach requires three bending moments at supports, it is cailed as three - moment also clapeyron equatior who proposed it in rg57. ft was rater modified "nrr;;;;.-k-;; by Mohr -called in I 860 to account for the effect of settlement of suppofts.
4.2 DERMTION OF THREE-MOMENT EQUÀTION Let us consider trvo adjacent spans AB and BC of a iontinuous beam as shown in settlements, the supports A,and c ,are ut tigh", elevations relative to the support B by amounts h.n and tr.. ttre erastic curve p^r"rî.ough A, B , and
Fig..4.la. Due to uneve¡
lt {
C0NTTNUOUS BEAþt
c'shown
in dotted rines. The free span beñding moment diagrams due to the appried loads on these spans AB and B.c a1e shown in Fig. 4rb. A free span moment diagram is drawn assuming a span to be simpry supported. if th" rupport moments ur. tøo, M" and en-d momenr diagrams are shown in Fig. 4.Ic. T-his aspect may be examined in Yc,.jh"
detail as follows.
(b) FREE SPAN
,.
Ll (c) END
BENDING MOMENT
at
* _
_.{
MOMENT DIAGRAM
Fig.4.l .
The moment diagram can arso be sprit acóordingry. The net mornent is the sum of free
u, shown in Figs. 4-2b and 4.2c. ev ,up",po,itån, lii*i::yjgïl"T:iïI l:t mom€nr diagram is obtained as shown in rig. l.za. Th" fr.;; momenrs
;á"
ithr
are end mäments are'hogging. rherefore due carå shourd be raken
[TL!,i]tr"::gl1g,:lr]:]n" superposing. the momenrs.
rhe quggi;!:';;a';;;rö ä;#iH'iä:ii l;ipositive and negative moments, respectively'anã, l::lÍne as shown shgw¡
*h:.ur
ll"ijgiil
are drawn o-n opposite sides of a base be drawn on the same side as shown
f.io â.7 a. â A-rternatìvely, Ât+ôh^+:,,^r". ¿L^-in Fig-4.2 they may
bi¡_r
11 ,desirable way is
jr ¡s difricutr to g.t nåi oøÍnur" in rhis manner. rnererorel ,h""ïo; to plot the two diagra., u, ,to*n ,-pË'-i:å.-il;Jï#ï;
I
t.,to-tr'. . " ì,if ðd. , I , dlngf¡rm
'
ÍIfi.EJVÁ.YÍ g-,êiF, l'tLR[iß"MOMEN'Ë EQUATtoN
TI"IREE-MOMENTEQUATION
portion' The superposition of loads or inoment diagrams
is shown by the shaded
'
.,,'. ': ,
l::::i:a'::'
':. .l4i
is permissible within the elastic range only'
area of the free span moment diagrams on spans ln Fig.4. lb, Al and of the end moment diagram on span AB while area represent Ao unå a, ¡\B and õC.
A, represent
{} {) {}
I
"o".
tF
I
M¡
{:)
{,.
uRJMB (b)
{_i
(o) NET MOMENT
FREE SPAN
.(c)
END M0MENI'
Substituting Eqs.4.2
Lr
of the end momerit diagram on span BC. The free span M¡, Mn and M¿ are to be determined.
A, and Au represent
Lr
In Fig. 4.Ia, Iines A, B and BC, are tangents to A'B and BC', at B respectively. Since the elastic curve A'BC' is continugus at B, the two tangents at B must be in the same
i,,-¡
straight line.
area
whère,
-**or,'-Jr,t,']
ttr
+Lo,u, -år"r,'
=çE
AA, : h¡ .- A'4, , and CCI = C'Ct - hc A'4, : úertical dqflection of A with respect to the tangent at B C'C, : vertical deflection of C with respect to the tangent at B.
(4.3)
-
Both A'4, and C'C, can be obtained with the help of moment area theorem. Let a, and a, be the distances of the.center of gravity of the M/EI loading on spans AB and BC from A and C, respectivelY.
(4.4)
t..
, ,---.r1 -åMeLt'-iM"L'-l lo,u,
-å
"ur,'
-å
r.t,']-
f
(t \
(l-,,trì ., lbl zHn . *'[tJ J l\ rr / '"[t ,; )-
rønl
- -6.Ala ,llt¡ - lzLz *.6Eho Lr
(4.1) (4.2\
-þ."t:l
, 4.3 and4.4 inEq.4.l,
6{za,
AAr
h _^^+l
Multiplying every rerm in the above expression by 6E and upon simprifiing,
môments are known while the support moments
. "LrLz
[o'^'
_A.,
x
+lo'u'-a'!-^'?]
: ,,!
Ç
(
,I
EIrL¡L'u'
Fig.4.2
(l
=
no- ,[o
\i
a.'
#[o,",
=
{_.
(.')
"=
:
L;
{';
;Jr*i " ni;TMerx
c'c, =#Ï'*. , JiYjs EI,
t";
i:_',
"
L)
Mg
{)
i'
Ms
ffi=ffi"È__1D 4è=á\"*^l:11
::)
1:':.1.
95
+
6Ehc
(4.5)
L2
This is the three - nioment equation: ho and h. are taken as positive if the supports A and c are higherthan support B. A sagging mõment is consiäered as positive whire a hogging moment is considered as negative. If solution of three - *orn"nt gives a positive value of MA , MB or Mç , it means hogging, while a riegative "quations varue means
sagging.
The three - moÍnent equations can be written for all intermediate supports of a continuous beam. Hence number of three - moment equations id the bame as the ,.rJ
t"-)
û
o t) {) {)
efl
BEAMS
THREE.MOMENT EQUATION
Mnn=-50kNm,
ì
Mo, =-20 xl.5:-30kNm
Msc : -
50
in Fig. 4.6a irsing the three - moment equation. Draw
..iiúom shown
Due to moment-joint equilibrium
dingram.
kNm, and Moc =
-
30 kNm
There will be only one three.- moment equation for one unknown, that is, consider the free span moment in span CD as shown in Fig.-4.5b.
M".
Let us
tlreåFpliu{tlourJs,areshowninFig.4.6b. ByinspectionMD:25x3=T5kNmhogging. A is fìxed, an imaginary span A'A of length Lo and with I : æ is added.
-gincá rupport
(")
50kN
{J
li't: .,:,.
20kN
:;:' :
r
:,
:i,ii:: . - i;.'::...''.
*
A,
,Ð--
..:. ,"¡:;
{:)
(q)
,,,
(O} GIVEN BEAM
{; (b)
{-:t
{"t
50
50 122 5
d;
(ö)
FREE SPAN
MOMENTS
{,'' 35.
{-,
(c)
(_;
BEN0ING M0MENt kNm.
d';
B
Fig.4.5 71.28
Area A:l/2x5x60=150 and a:2.6?mfromD
d) {.-)
(r
Three-rnoment equat¡on at support
(¿'\
(.,)
.
o
MOMENTS ON SPAN tsC
becomes,.
,, l¿.sl ., lsl j l= - 6xo - Msl l-2M"1I l- pr^l "t(' l-: +',a)-'"'o[.tj "\ ;; r.s/ 4 x 1.5
(-.r
(,'
C
(c)
4
or,.
-50x _-10.34Mc-30x2.5=_240.3 1.5
or,
M. :
3.09 kNm
s'l
6x150x2.67
5x2
st
:8!
::BJ
îþi .Ë;j
ìh ..aa i:.i:t
(d)
BENDING MOMENT kNm
'
The resulting bending moment diagram is shown in Fíg. 4.5c.
Fig.4.6
Ë
\J
,,
C'
tüå
fi
Snp¡xtrt C
ü
o {.)
FRAMES
THREE-MOMENT EQUATION
-
u'(T o') *o -,".(+.f)""(i) ""(f)=
of,
- 4M" -
20Mc
: -0.015EI : -
i
600
,::::t::
::.
-o'1f""Ì _ {roso} f-'r L4 -2oJLMCJ [-6ooj
t) *
f
or,
r"l
F.-5
(oI A,
It gives M"
as sagging and M" as hogging moments. These are the same values as obtained by the method of cohsistent deformation in.Example 3.10.
{_;
4.5 FRAMES
{,)
Example 4.5
{.; L,]
(-; (-.,
PORTAL FRAME
= {-er:ro}
[M.J [ 43.e8J
{'t
t_)
Aü KN/m '::ì t:. .:u.: ì
Matrix Solution
o
Ç_,
-,çr- Ll
þ
A
40kN/m
l5kNm
+(b)3.+2+
8m
.Solution Since joints B and C are rigid, the portal frame may be considered as a continuous beam having three spans. For the sake of convenience, the vertical members may be opened on to form a straight ccintinuous beam as shown in Fig. 4.8b. Since support A is fixed, an imaginary span A'A of length L, with I æ is added. Similarly, an imaginary
(c)'FREE SPAN MOMENI kNm
:
span
DD'is
added.
Free span moments are shown in Fig. 4.8c. Let us first calculate areas and centers span moments.
SpanAB c.g. from
{,)
(-
B: 2+l :3m,
c.g. from B
(,' (
I AreaAE : -;x3x9:
AreaEB
(':
(
L2 2, 3 3' {. .z +2+
Analyzethe portal frame shown in Fig. 4.8a using the three - moment equation.
)
t.,
25kN m
EOUIVALENT CONIINUOUS BEAM
of
gavity of the free {..
r03
:
c.g.from
-l3.5hoggingmoment 2m
A=
I
=;x2x6=6 c.g. from
1.33 m,
A= 3+ ? = 3.67 m 3
Span BC
Area c.g.
=
n
;x320x8: :4m
(d)
BENDING M0MENT kNm
1706.67
Fig.4.8
D,
¡l
t-)
,
t04
THREE.MOMENT EQUATION
FRAMES
r05
.:l
{")
ü {) {)
c.g. from D = 3+
J
Area FD
{")
c.g. from D
O
:
$f
-ro -5 o
-l 2
:2m,
x3
x 15 = -22.5 hogging
c.g.from
-5
moment
o
C=2+l-3m
{")
-
¡x3 +6x,
l
f _ J_r+oz.r0l
-s.34 --20.6i -5 ll vr. |
|
_:+o:.ls
-5 -roj[v,j I os.o+
(v)
I
J
Jt"l=Jr4s'6ol
(vi)
"(*)-r"^(*.i)-""[i)=-e(-r¡ oI,
The resulting bending moment diagram is shown in Fig. 4.gd.
(i)
-
t.;
Note: ctDckwise moments are shown anricrockwise momenrs ere shown âs ":i3r,TLtHÌe in all problcms. IVt6".represents moment at'B in thc sprn BA. simirarry, M"6
negative
Support B
)
{_;
represenfs moment at A ín thc span AB.
4.1
"^(;)-r""(;.å)- ".(å)=.f
{-,,
r,. 2 + 6x3
-
67)
or, - 5 Mo -
?0.67 MB
-
5.34
{,_,:
Mc
ofthe elastic curve at all supports. Mh,, = 160 kNn, M"u : I0 kNm (P4.2) M¡u : 34 92 kNm, M"; : 24.70 kNm, M"" : 40 kNm (P4.3) Mtu: 117,0 k-Nm, M,t,: 134.30kNm, U*=_øZ.tS *Wm
I)
= _ 3407.36 "fu
-
{"_;
tl t'[/r\*J-zvt.[å.
5l t"[i.,|= /s\ i,J-
6
4.2
5.34
MB-
-,rfu
Qtoø.et,n)
¡(lo"
20.67 Mc
4.3 3.67
_22.5x2)
- 5 Mo : _ 3403.3g
".ff)-""(ï.*)- "r(*)= f,{,',.
U*:
tt.t
kNm
Analyze the continu-ous, begm of Fig. p4.2 for a 5 mm settrement of support B without the applied loads. Draw shear force and bending moment diagrams. Take E:200 GPa and I I50 x l0-ó ma.
Analyze the continuous beam of Fig. p4.4 for a 5 mm setillment of support B. Draw shear force and bending moment diagrams. Take E = 200 x 106 kñ7m2,
l= 300x I0-óm4. (P4.4) M,,t, = - I8g.8t kNm,
(iii)
4.4
Support D
-
37.60'kNm,
:
6
-
:
(P4.2) Rh = 12.3t kNJ, R,:9.6kN1,Mh,,:_32.t5kNn, M,n :-14.87 klnm
,
{r or,
(P4.4) Mo¡ :-101.2 kNm, Mto
(iÐ
Support C
{_;
Analyze the continuous beams shown in Fig. p4.l - p4.4 using the three moment equation. Draw shear force and bending moment diagrams. compute
slopes (P4.
$toe.at *+)
(
l|.r^] I t,
|."^l fa.to] lr. | 1t46.301
or.
33)
-loMA-5M":+39
U
i'
(iv)
IMoJ l_7e.66]
t"":
{';
o
-20.67 -5.34 o lJ y,
o o
ïf #
Support A
=65.04
MaÍrix Solulion
l.urm,.c.g. from C=1.33m
-"^[+)-,'"[+.+)-".[f)=
{)
{.j (.
-5Mc-l0MD
or,
: ! x2x,l0 =10
Let us write the three - moment equations for' each interior support.
{-}
(r,
AreaCF
Span CD
:
-SS.2S kNm,
M", :77.g7
kNm
Analyze the portar frame shown in Fig. p4.5 using the three-moment equation a1d draw shear force and bending moment diagrãins. Arso draw its deflected snape.
t33-22 5x3)
Mn,,
(P4.5)
M,,o M,tn
= 13.05 kNm, Mnu : = -27'46kNm
30.22 kNm. Mou
=-
I5.g
kNn
{.-l
,,
t) "Lr
'i,,,
r.
::..
THREE-MOMENT EQUATION
l0ß
CHAPTER
{t
five
[) {)
20 kN/m P
{.r
STRAIN ENERGY METHOD
{..}
tti {)
10k N /m P 4.2
D
30kN /m P 4.3
5.I
t)
A structure undergoes elastic deformations under the application of extemal forces or other causes. The applied loads suffer displacement and work is done. This work is Stored as strain energy in the structure. The strain energy is proportional to the strains produced in the structure. The work done by external forces or actions on an elastic system is equal to the strain energy stored internally on the system. The external forces are assumed to act gradually on the system from zero to their final values so that there is no dynamic effect. The external and intemal actions or forces are in static equilibrium at all times.
C l'
4.4
{,)
f---f4
d' i:_,
Ll
T
P 4.5
5m
{',
I
(, þ--l--{**-"5f!--..*i {.,,'
{l (_)
(ì {-I (
INTRODUCTION
Fig.P4.l-P4¡5
The slopes and deflections produced in a structure depend upon the strains developed as a result of extemal actions. The strains may be axial, shear, flexural or torsional. Apparently, there is a relationship between strain energy andrd-eformations in a structure. This relationship can be used to determine the slopes and driflections in a structure both statically determinate as well as statically indeterminate. Thus, strain energy method is one of the most powerful tools for the analysis of any structure. The energy principles discussed in this chapter are applicable to linear elastic or nonfor the computation of deflections in statically determinate structures. Nevertheless it is desirable to understand scope of their application to the analysis of statically indeterminate structures. Strain energy and complementary energy methods are used for the analysis of structures. Both these methods may be considered as special cases of the principle of virtual work. The principle of virtual work may be associated with the principle of virtual forces. Thus, energy theorems may be grouped as follows:
linear elastic deformable systems. These principles are applieé generally
l.
Strain energy theorems: Castigliano's first
.(a)
theorem
(b) Principle of virtual displacement or unit displacement (c) Principle ofstationary potential energy.
method
STRATN ENERCY
STRÄIN ENERGY METHOD
* '.
*
"ì
{--)
{)
-.
1'!.
r09
work done may also be obtained by considering the horizontal strip shown in
:r,,;å Conrplementary energytheorems: ...., .,, (a) Castigliano's second theorem
liig.
i:
The term W* represents com¡:lomentary rvorþ The load is increased by a amount dp. 't'hc total work don$ a$ the loûd increasos from .zero to its final value po is obtained by summing all the inffcm€ntlr of lvofk represcnted by the similar strips. This is equai to the ¿lfefl {liove the lbrce-defbnnqtion curve.
-5.1a
:
(b) Principle of virtual forces or unit load method (c) Principle of stationary complementary energy.
,r Tlie theorêms in the first group are used for the stiffness method or displacement mcthod of, analysis, while the theorems in the latter group are used for the flexibility rììethod or l'orce method of analysis.
*
A
(s.2a)
l.r
5.2
\J tjr,., r-,,r t\L L-.r
f w* = lÂdr : g* 'o
A force is defined
as an action that tends to change the state of motion of a body to which it is applied. Force is equal to mass times acceleration according to the Newton's laws of motion. In structural mechanics, it is defined asthe stress times area. When the point of application of a force moves. then a work is said to be done. The rvork is equal to force times the displacernent in the direction of the applied force. Consider a structure acted upon Èy a load Po producing a displacement A typical load- displacement ^o. characteristics is shown in Fig.5.la for a linear structure when the load is applied shaded strip .
dw=-Pd4
(5.
la)
(
. {\--l t ,
P,t
..
WORK AND COMPLEMENTARY WORK
The work done by Po is represented by the area OAB which is equal to the strain energy stored in the structure. The total work is found by summing all the increments of rÌork represented by'the similar strips as the displacement increases gradually from zero to the final value Âo. This is equal to the area under the force-deformation curve :
.'. w: lPd^:u
(5.
I
{".,¡
COMPLE MENTARY
STRAIN ENERGY
(,:
For a linear elastic system as shown in Fig. 5. I a,
W:W* U:U*
and
and it is therefore not necessary to distinguish between work and complementary work, or strain energy and complementary strain energy whenever the Hook's law holds. Then, the integrals representing the areas under load-deformation curves are reduced to area of the triangles. Because of,the relative simplicity in the evaluation of these expressions, Iinear elasticity is generally assumed in structural analysis if the material characteristic can be represented conveniently as linear.
For a nonlinear elastic system as shown in Fig. 5. lb, the work and complementary v0ork are not equal. Therefore, it is. necessary to clearly distinguish them. The area oAC
represents
the
comprementary
work or
complementary strain
energy.
lb)
structure using En_eesser's theorem.
5.3 STRAIN ENERGY Let us first derive the expressions for various types ofstrain
Energy
.
energy.
Axial Strain
i-l (;
sectionofthememberbeAatadistancexfromoneendasshowninËig's'z.
consider a member of length L subjected to an axial load p. Let the area of cross,
STRAIN ENERGY
dA
{,,
AAo
(_;
(')
(,
:
'
{)
t'
The
complementary strain energy is useful in computing defleciions in a nonlinear elastic
uo
,,
(5.2b)
D15
PLACEME NT
(o) 'Fig.
5,1 Load
r x ,, F-+|-F-
AÂo
Fig. 5.2 Axial deformation
DISP LACEMENT
ib)
displacement curves in elastic (a) linear structures and (b) non-linear structures
dX
Stress in nlember
o'
P
A
TJ
{}
STRAIN ENERGY METHOD
'.'
ú {--}
,' ,l
iir:,
::.
{:} Ç)
in $trnin j
member
: oP -=EAE
q
Ì
the force P was increased gradually from zero to P, average force causing the elongation ÁL
If
{; Tliereforc, work done
U:
Force x
- 0+P22
P
displacement z
Ç,¡
{'.' t"_; {. _)
?l u : tJ AE=Jl ¡s 00
or,
p'¿*
(5.3)
U :P,L 2AE
(s.4)
å
Consider a smáll element of a structure dx, dy and dz subjected to a shear force V as shown in Fig.5.3. It causes a.shear strain equal to 0 and a lateral displacement equal to 'a' atthe top fiber. Ifthe shear force is applied gradually,
$
{-,)
average shear
,
or'
(:
shear strain
{".'
(l ('
:
stress
I
0
or
,
(5.6)
trr> '
å' t
:
o=
1
G
'.1.',
r[ì='llt
o'lìl
t'L-ei-:
I ', * '.
modulus of rigidity
:G
u : ,o24"G ïl:10.
Ð,
F¡¡
or,
:;
fT,
u=""0t=1"., 'shear
du=l.vtd* zA.c
f::::r:,::: ,.
: qI = + 22
(5.5)
Eq. 5 .l becomes
$.r::,.'r.
Shear Strain Energy
(;
o*o, o,
€
Ir
force
.',
i
The strain may be tensile or compressive.
(_,,
)(
area ofcross-section in shear
iI 'iL
{, {.':
JJJ
.:+
i
If ,area of cross-section of the member is uniform,
Total strain energy
t:
enersy
:.r.
€
g f å' å,
{,,¡
p ?p¿*
î'otarstrain
å. .Ëv
(5.4)
!{a*aro,
å"ttnnu*ltivelv, wlrere, Ao o effective
-
d*
, -r
,, ,
å,
0
c
Vt .., V dU*Txnæ*oo-
I
^,=iH
a
nl
U'o¡ 'fbree x dfÉploúem6it
J
tot"l elongation in the member will be
L)
{:)
Work done '
AE
{;}
i_i
u=
Pdx
Elongation in a length dx
ïìe
STRAIN ENERCY
þ-q' r
T"
(b)qNLARGED ELEMENT
.
$'-tli-.: ßrl''; i, r,.. .. .1t.," $
.
-.
Fig. 5.3 Sheardeformation
1
:t/
it
STRAIN ENERGY METHOD
il? lÍlcxural Strain Energy
bending moment M
uniform of length ds subjected to a consider a small elemenr 5'4' lf moment is applied Fig' rrìoinln ui ¿ un ungr"îö"ii'ìii, Ï"ng,t
T
Work done
dU:--
graduallY'
but
du:-'
Tdx ._*
/de'r
t\
(j*
shcarmoclulus=
where
GJ
f*
torsional inortia
du:-
Hence
Work done
{,
{.¡
.v
1,,
(.)
Totål strain energy
i"-r
Torsional Strain EnergY
(;
uo = Mdt uv EI
du:
oroãä;; *git "r*iu
d0
(', {
M2ds
of length. 41,19i;"t"0
,( --\
Fig' 5'5'
/' --7À
_1F:r þ d* J\
Fig. 5.5 Torsionai'deformation (
2GJ
(5.8)
¡I P2dx I V2dxr fI M2dx J zAE J zA"G J zET -f
u
to a torsional moment
:
ll'torsion is also include!
-ñ
as shown in
(."",
(-,
by moment - area theorem
r I M2ds u:J2zll
Consider a small element
T2dx
these f''orces can be written as
M.^ dU: -- ou
Ç''
the member
In a general plane beam element, there òan be three force components: axial force p, $hear force V and bending moment M. Total strain çnergy under the cornbined effect of
'¡2
but
of
Combined Strain Energy
'_ 0+M _M
,'
,8,
2(l + v)
f I T2dx J2 G]
T'otal strai.¡r ener$y
Fig. 5.4 Flexural deformation
xdO
2
which produc.,
average bending moment
il3
ENERGY THEOREMS
T
:
(s.9a) !
-
f P2dx*Jf v2dx*JI Mzdx*Jf T2dx J 2AE 2Añ zu zct
(5.eb)
(5.7)
,:¡1',lfi:general, the'etïect of axial deformation and shear deformation is very small (< 5%), therefore, only flexural strain energy given by 8q.5.7 is used extensively.
which
',,'$.ENERGY ..'-,¡¡ A, Castigliano published two theorems in'I879 to determine
THEoREMS
iËJ., ,
deflections in structures
tf a linearly elastic structure is subjected to a set of loads, of the total strain energt with respect to
partial derivative
the the
cletlection at alry point is equal to the load applied at that point. AU _=p. aô, -
^i
(s. r 0)
L} ;-\ E
H *
ENERGY THEOREMS
STRAIN ENERGY METHOD
ll4
1i
v
Second Theorem
a set of loads' the partial If a..linearly elastic strucnre is subiected to to a load applied at respect with derivative o¡ tn"-ro'it strain enerfÐ)
ir
) I
d*j
{-}
point is equal to the deflection at that point' or torsion. The corresponding disilacement The load may be a force, moment, shear theorem states that: slope, súear deformation or angle of twist.'This any
will
es
ü {) {) {)
This means that the strain energy is either maximum or minimum. The maximum the value of strain energy corresponds to unstable equilibrium. For a stable equilibrium, comes work least of theorem of the välidity The minimum. be should energy . $train clirectly toiitre Castigliano's second theorem. Both these theorems are applicable to staticaily indeterminate beams, trusses as well as frames. By differentiating the total
strain energy of the structure with respect to each unknown redundant separately and equating to zÊro, there will be as many equations as the number of unknown reactions. The vaiues of these redundants can be determined by the solution of linear simultaneous equations. The following examples illustrate the application of these theorems in the solution of statically indeterminate structures'
be a deflection,
.i
U
where,
Pl
=
Mr :
;j , e,' : '
=6,, #=r,
(5. r
l)
Principle
total strain energy
AU
f;
If a body is in equilibrium under a virtual force syìtem and remains in'equilibrium while it is subjected to a small deformation, the virtual
âU
AM
¡ Mdx ôM
This is very useful in computing deflections in any structure using:
aP
¡ : JEIAP [99Y¿.
":
cl2)
or,
{'..,;
t,r
il. (-,t,
tj (,.,
(,,
(: (
TheintegralisevaluatedasitstandstogivedeflectionunderaloadP.Thevalueof of the load P' If no
the bending moment rr¡
"i ""1,
other sectioi is determined in. terms
can be divided into several hence the limits of the ini.g.ul, the beam deflecrion (or slope) is required at a'point or often, uíi"ã. Quite required, and the ,.rort, ìo load applied. In such cases, an imaginary load P is which
;;;; ;
introducedintherequireddirection,theintegralobtainedintermsofPandthen evaluated sçtting P equal to zero' use since it requires expressing the strain The first theorem is not very convenient to is very effective in rhe analysis of theorem i"i.r*ì"i¿"flections. The second the second theorem leads to the.application'of "n"äî In fact, statically indeterminate ,ouout.t. of virtual work' the same exprdssion as obtained by the method
laul
(s. 14)
l*J,=.
^:J# : M m
:
(5.15)
moment due to external load moment due to unit load
This is the same expression as derived by the unit load method.
t;.,'i , fne
main difference between the virtual work method and the Castigli4no's.method is method, ihe partial ¡r'Ys¡vet In the virtual work and integration. ¡rrLw6rs! of ulllç¡gl¡ll4l¡vrr differentiation srru tllE uluçl order u¡ 'furs È.i.,: ,:,rll lll rhe then summed up. In the and placed inside the integral is (M EI) m ds/ rerm .':: . differential 'i'i',... astigtiano's method, the strain energy is integrated first and then differentiated. :.',.+t:|,,
' Beui's Reciprocal Theorem :t.. This theorem is very useful and is applicable to any fype of structure. It was proved . by Betti in lï7Z-and may be stated as follows: ¡:trl
:'
""irirî^lrri*J"T;
^1,
,r^rr"uy
indeterminate structure and mav be stated as rôllows: redundønts should be
Foi' any statically indeterminate structure' jhe such so as to make the total strain energt withîit
a structure
a
minimum. {.
i
where
generalexpressionforMintermsofPcanbederivedtocovertheentirebeam,and segments as rnay be in a direction in
Virtual Work
work done by the external forces is equal to the virtual work done by the internal stresses due to those forces-
ap=aMaP=l Er
{r)
of
It may be stated as follows:
loads on the structure displacements in the direction of the loads
If u:J#
{:)
rhus,
fr=o
ì
l15
(s. r 3)
lf a structure is acted upon by two þrce systems Pn and Pu, in equilibrium separately, the external virtual work done by a system of forces P o during the deþrmations caused by another {ystems of forces Pn is equal to the external virtual work done by the Pn system during " the deþrmations caused by the P, system.
rtd
r,
BEAMS - ILLUSTRATÍVE EXAMPLES
srRÀIN ENERGY METHOD
HeÌti,s law is a glneralization
folloty'o: The deflection of point
!
of the lvfaxwell,s law. The tatter can be stated
Strainenergy
å
.o
2 on a structure due to a load P at point
0
ilriing Castigliano's second theorem
'"¡"""a
apPlied loads'
statically indeterminate structuies as well This law is very useful in the analysis of structur€s' in,constructing influence lines for 9 of volume These theorems were derived in chapter Engesser;s Complimentary Strain Energy
o^ '
as
I of tliis book'
Theorem
.i" Ifalinearlyonnonlinearþelasticstructureßsubjectedtoasetof wifh strain-ener¿gt loads, the partiat derivativi of iþe complèmentary õ' of the point displacement the to equål is respect to ony iooa P, action' of line its of ciirection the in that uppti"otion àf force
, ¡lM'dx U=J;-f
is
at poinÍ l' iqrAío tn" aelteit:ton of point 2 due to the load P àcting as the direction same tu are in'the Of coui¡se, the a"lntio*i
.
t-.
as
,'.
1
: uu - JÍM ôt¿* ôRc EI âRc
.
Iror U to be
minimum,EU :0 âRc
[iet us evaluate the integrals as follows:
of
Segment
CB
BA
Origin
C
B
Limits
#:ai. second theorem which is applicable This theorem is very similar to the castigliano's to a linear elastic structure' in the following examples' The The applications of various theorems is illustrated method discuss,ed in method is v"ry similar to the consistent deformation *ruin "nËrgy 3. chäpter
s.i nnnprs : ILLusTRATIvE'ExAMPLES' Example
M
Rcx
AM
x
I
:
Ll2+x
,L/
Erl3lo Erl -l_l 24Et-
redundant and remove the support C'
3\2 /
EtLj--
-:=g- = 0 for minimum âRc
(' t
t6
a
u- u l
=!'I;]_5PL3_^ 3EL 48Ef - ^c
(
(t
LlZ .'
Il Fig. 5.6
LlZ
.l
Px
'L/ R.l*'l;'2+_l-l I lR. / _+x L l'I ____l Px2L P*31"
Thebeamisstaticallyindetermiaatetodegreel.LetusconsiderreactionRcas
('
-
B.Lll _ - RcL3 I l&É_ PL'_g|_ -
Solution
(-]
R" (L/2+ x)
#; : +ii..'a* * fr i[..(:-.)-*](i..)'.
Fig' 5'6 Using Íhe strain Analyze the propped cantiliver þeam shown in :nergy : constant. ' EI method. Take
(..,
0.L/2
-L/2
aR.
,
5.l
0
: '16
R^
5P
potential energy
3lo
tt7
l19
BEAMS - ILLUSTRATIVE EXAMPLES
.STMIN
lt8
ENERGY METHOD
_ = _yL Me: 162t6 PL
5PL
Mclment at A
Net moment at
B M" =
lll
au 5 wLa*- PL3 ôP 384 EI 48EI hogging
^b
,urr¡n,
=[#).=,
5
wl.a
384
ET
Deflection Due to Shear Deþrmations
ExamPle 5.2
ôvl aP2
Determinethecontributionsindeflectionduetobendingandsheardeformationsatthe of beam AB shown in Fig' 5'7a'
'"'d;;;
'¡
^.
(O) SIMPLE BEAM IP
^r' ,r*> ""'--ttZ-u--L/2 (b) BEAM
J
Liz
Solution
is the desìied deflection'
Pll dx
trL2
Fig 5'7
o-ïlligi".
-
lwL : ^f L I t
UNOER IMAGINARY
vertical load
[ål),=,
AU ¡.,âV dx AP J AP A"G
B
imaginary midspan of the simplg Otli li,'-ln order to determine'deflection at atP : minimized energy p applied uìôu. shown in Fig. 5.7b. The strain
=
zlz+ec
.wL2
PL
SAeG
4AeG
^ I lwL wx'
:
AeGl2
-r-X
2
tLl2
Pxl
2lo
*ti ^ _fauì -8A"c ^u-[apjr=o
Segment AC
Deflection due,to Bending Deþrmations
AM="
P -. wL *xz +VX, : _:=y.-: -
oo
a-'
(.,
.,ttf
t-"{t
(*ux 2
wx2 2
Defl eötion due to bending
aP2
: I aul =I'#åi [* J,=o
au
Deflectjon due to shear
Av. ab
'PxYxlax +-il-t-
2
)\2)Er
:
/c\(
t
9.61v ll -=i-
_ -
wt]
/
/84"G
1;FZ/384Et
\ |
\c /\ rA" /
is valid for simply supported beam ofany cross-section subjected to a
*- E '' 2(l + v)
if v:0.25.
G=0.4E
20
l"ttr a steel beam of I section
AJ
=
may be considered to be constant' A'"u of web whe¡e shear stress
241 _ 211_19ì Â*,=E+=;þz(r/
(ii)
- lolt = '[;J
1:
âRc
:
aM, fM t--¡
J EI âRc
The integrals can be evaluated as follows:
(iiÐ
241
used in beams' C varies as follows: For rolled steel sections generally
c
.{
1 tvt'¿x u : J2 [ EI
ôu
,: ul
For ISMB600,
8-
lJting Castigliano's second theorem ' .-
P = flePth of l-section ; = tták;"ss orweb of I section
C
.
Fig.5.8 $train enersy
For ISMB 100,
kN/m
f., 5- +
= Dto,
"
t2l
BEAMS - ILLUSTRATIVE EXAI\,TPLES
STRAIN ENERGY METHOD
t20
Segmenl
AB
Origin
A
Limits.
24x257'5.- :15.45 (-tO * O.a),. tO2
M due to UDL
24x91813 - a(
M due to
0-5m
-
l30x
RB
6;frj;æ
CB
-
0-8m 'y2
204
t30x
2
- å*,*
-
20$-
- å*,*
For a rectangular sectton bD3
A, (ii) gives Eq.-ab -
DIL .
If. Deflection
ratio
-å
"" t, 2lgl' 'L2xbD = \L/ t ,
0'8
0'5
0'2
0'l
#l','+-,0+
(%)
shear D/l- lies around 0'l' hence contribution of In the majority of practical cases' pier caps and well caps in bridge
Jtep'btamt such as deformation is very 'tuii'tin is quite appreciable' deformations shear of ,h. effeci ;;;;,
t-îlii"t"
Fig. 5.8 using the strain two span continuous..be4m shown in
: !i¡ro* . I3EI
energv
diagram' method ãnd draw bending moment
Solution Thebeamisindeterminatetoadegreel.LetustreatRBasredundairtandremovethe support B.
*,.X-å.)*.
=0
=
3.36
-
j(,,0-
- å-"+1, .
-,ox2 -
a
R".X-å.i#
#l,,oo'-,0+ - å.,+1,
20s.12R, + 5e733.35
-
32s.2 RBJ
for minimum strain energy
169.81 kN
is the same value as obtained in Ex.
3.2
using the method
of
consistent
o. K.
.:::.:
STRAIN ENERGY METHOD
122
EXAMPLES 5,6 FRAMES - ILLUSTRATIVE
'fotal strainenergy
0xamPle 5.4
shear due to bending' axial and in the total vertical deflection Constant' EI = Determine contributions Take 5'9a' Fig' in ;i;f ;;;l frame shown deformations at the ftee AE = 50 EI. A"G =25 El
Solution
moment, shear and thrust frame. Let us draw bending determinate a statically This is In order to determine in rig'' 19 b' c anã to due diasrams load w is aþplied at F loads are ¡emãu"¿,'u vertical u"Jical deflection at F, the extemal Figs. 5'9 f, gand h' The iir^r.iÃ.rrrrs are plotteã ," rho*n in ancr moment, shear and integrals can now be evaluated'
d'
appliedìJ; lo*n
uã.iou,
li
l_
Ittz *utz,,
(o)
(b)
MOMENT
FRAMES - ILLUSTRATIVE EXAMPLES
tlM2 f I V2 f lTzdx U. - J2EI !: - dx+l;-dx+l:JTA.G J2 AE
:rìÈ:i.: '¡
1:i'.Çastigliano's second theorem gives
:ii .' ,,
AU
âw
.
l+úember
EF
'.
..'llcnding moment M
'.'
:0,
:,. :::l r:.::rl
i¡:r;:
(d)
TH.RUST
{e} FRAME UNDER
,
ø--l
.,-Bh*r
h'r
"',',' :i,.:,
"
:
- v âva*=o IJA"Gâw
(2)
,''ThntstT:W
âT:l aw
(C) SHEAR
M
=0-
rwl
L!2
âT.
:lIT JAEôW --OX
(f)
(l)
J##o.=o
i-i;;'::,';
PORTAL FRAME
Fr]-
:
I l{dJAE 0
=
ôw
MOMENT
LOAO W
Lt2
rMAM. tox
(-w*)l-*)!r 'EI JI 0
=W,
v
(,.,
(
av.
A"G AW
(91
SHEAR
(h)
(3)
2AE
A\Á
Wx.
JEIAW
('
t23
-ox THRUST
diagrams due to different loadings Fie. 5.9 Fortal frame and force
t
.tUZ lwrl' | 3Er lo
wL' 24Et
(4)
ôv=_, aw
: 'li-*x-1¡-!l_ 'A"G i'
T:
O
WL .2A-.G
(5)
124:
STRAIN ENERGY METHOD
rT I
AT
J AE âIW
FRAMES - ILLUSTRATIVE EXAMPLES
dx:O
V=0
(6)
d* -
Jluôu AW A"G
Memher CD
M=-Px ôM
aw
t25
- wf:-.1 \2 )
(i l)
o
âT = T=-P-w. 'aw -r
(t- \ _ l_+X \2 )I
J'##. :''it *')#=(e*w)j,
.9
(t2)
0
JÍv9!49r AWEI
:
i
t,._
ln
r
*(;..)],_,(i..)#
It4ember AB :
M:_ A=p*-wL,
ïi++Px2+'(;.'.)']#
7)/
J[v9!4¿* AWEI
w(t \'lt" = ipL*t* Px3 , *T[.t**Jlo
|.
u#:-t
y+
ft 2 {I 0\
* wL3 - 83 PL3 IEI EI 2EI ,t-
plx+ wr., lg1
(13)
=0
5 PL3 7 wL3 48 EI 24 El
a) (14)
V= -P-W,
AV AW
-
-t
gr-41
-wx-r)-$'A"G= illl*rv){'1' JIv ôw A"c=''í,-, t {'
(e1tÐr-
.^o"o
dx = o J¡tâT ôWAE
(8)
?A"G
(15) (e)
Member BC
M: -
derivative terms obtained through Eqs. I to 15 can now be al deflection at F.
T-'',
ertical deflection at F due to flexura[ deþrmation alone, let us
rg$ 13. that is,
J¡"
ôM dx AWEI
PL3 wL3 4ET zEI
(10)
u. : laul
[.;çjw=o
bending îo
FRAMES - ILLUSTRATIVE EXAMPLES
STRAIN ENERGY METT{OD
126
:
^.
dx : Jf.,ôM '' âw Er
5 PL3 PL3 s pll ¿S
El
*
¿U
*
S El
=
T
3m
35 PL3 EI
+
43
¡
alone, let us evaluate the net vertical deflection at F due to shear deþrnations
combine Eqs. 2, 5, 8, I
I and
14,
4m
I
that is,
I
t,.av dx Âr: fauì JvôwAs [u*j*=r: shear
t27
PL PL = zep = 5oEI ^r
FE
To evaluate the ner vertical deflection at F ciue to axial deformatiow alone, let
âM M:0 r Rr*x:fr
us
=-
combine Eqs. 3, 6, 9, 12 and 15, that is,
laul t-ôT dx o':lã#j."=.:J'u*Ë thrust Total deflçction,
^.
:
if L:
J'j##
:rtt*r** *
PL * PL PL= PL zeg z¡t= AE 5oEI
'
3m
a, : å {#,.t,.*.*}
,
P- ': z7.l L (26.98+0.00+0.06): EI The proportion deflectiqn are
:
of
deflections due
bending gg.560/0,
l*''.,.1
to various
causes
sheet 0.22o/o, and
i''1
67.85
4.
:{*,*.'# =f l*"-'^ *'lo = Er[ 3lo
r . (o.T)*,.,
AM
âRr*
:
21.34
uRo
(l)
¿+!I 1
.-'
¡" -ðM dx : J ôRr* EI .
with respect to the total
ï f.,,,..(..ï)*,.](..+)#
a
llr,
thrust 0'22Yo
'-
Example 5.5 Analyzethe portal frame shown in Fig. 5.10 for horizontal thrust at F using the strain
* + 2e.o'xz. (o. ï)' *"
l#
: *lr',l.rn *f;.å(..+)'-..(1¡[
energy_ method.
= ( ro3e+3e.81*.-i*
Solution
l.
Introduce a roller support at F and The frame is statically indeterminate to a degree treat horizontal reactign at F, R * as a redundant. The bending moment diagrams in the released structure due to the applied loads and redundant R.* are shown in Figs. 3.13 c and f. The various integrals can be evaluated as follows'
(2)
DC
M=
17.85
x+237.48+( ?v\
lt't*;.,¡*'-,
AM
âRr*
: ).J+-3x 7
FRA M ES'
STRAIN ENERGY METHOD
lzlt
: +zr +t+(s'.ï)-.-](".+)jå J*, -r+* i ['"'. T
[r.*.'
+
=
r306.r4.(r.r.ï)'*.-]#
tee.e6x+
I
306.
)3
f, (3)
!tnrr-'s+6B-6eR*l
i
(7
aM -d] Jlrr¡
=
ARF* EI
-
x)
AM
R,,
â Rn*
j¡roo*{, - x)n..](z^-
:l,o* |
:
-)# : i¡rt*-
300x +(z
-.)'*.-]#
(300
-
aM d1 J[rr¡ âRr* EI
-of. 2¡l4.lt&- = -
.-..;;;,At!4y2" -', "1,.-.. ' ,,r{hhitor
- roo
*'
2-
(4gso+lf n..)
(?
=
æ
ç
i0541:79
+ilosz;ts + 6E.69.Rr* + 4950 + + 93'Rr* + 1600 + 21.34,Rn* I : .
i
r::
0
.
Rr*: -43.17kN
-
*)'l'
3
'
ill ... ::. tmn *e isstatically
-
loEr (4)
f,
the portal frame shown in Fig 5. I I a using the strain energy method.
indete¡minate to degree l. Let us replace the hinge support at D q:*oller support. Thê released structure is shown in Fig 5.1lb. The free body diagram qnd that due to the redundant R* is , ,;@rto úre applied loads is shown in Fig:S.Itc I ld. The integrals in Fig. 5. various cait tie evaluated as follows : ,,:.thown
r...1.-i.
.
Member B A
:
., i- '
(5)
: (7-x)
.
M
,
ôU
tzr.u n.* + 1039 + 39.tt
ìii.'ijlr.'..
Mernber C B
M:300+
+ 2t.34*r.l*
;îf.,,,
1 lr., * ,,.,ì Z *.- l" + tee.t'6t-* 2 3\ 3 ^^lo
l4x
+ '^[er
,3
13
(t600
129
(o;*)'*.-l
lrr* - 600+2 + r2ü)x -
ã,ir.. F*tinimumstrainenergY,
: -L l, u, d + zBrl 3 :
=
- I LLU-STRAT.IVE TEXAMPLES
'..:r.þ¡¿
tl ' ,,,îtlwther CcDD "',[iø¡her ' ttJil {i = 10.t25 x -
(using the orthogonal system ofaxes) Ro* * tan ó0o (Using of axes)
'.ïË*=-xtanóoo:-J3* 75x'¡ +
(4- x)Rr",
AM
ôRr*
:(4-x)
î'
l(+-*)9r . Jt¡OO-75x+(4-x)Rp*,. Er 0
rr
4
J 11200-600x 0
"'4-*)tn"-l9r +75x'+( .^r Er
,-, ,iiì!.,
= .'1.'.r,".
f,j1-rmr 'l
-
sxzJl+
r ,.o;.,tulnno
:x2R*þ r -
.
=
*l-,o.rrrtt*t.:.*l;
.ISTRAINüNEREY
t30 Member B C
M=
METHOD,
'-
''- .':, :": lo.t25x -' 3,1îB*,
32.475 +
aM dx . ó,1Ì ¡" = J ARDX
_
lpz.n?s + ¡o.s2s
EI
," t.:
.
:.
FRÁMËS - :ILLUSTRATIVE EXAMPLES
i
aM
';Ç
3JtR*
=
-
3'13
J(*ø)ff (bI
1 *J-,,ur =
¡
EII
-
ro
*-sç.4**er**n[
9?.125
l'I )"
l0l2.3t + O¡ p*¡
Member A B ti
.
l,l.= -25x
tan60o
f.. aùt¿x J
"i-* Ë
- Io.E2jx -
ur, =
-
.
RD,
'',
xtan60o
añn-
_ =
f\l
I ro.62s
'
__-,_ E¡l 3 +þr+-ruazsJi{*rn*{l' j , ,rlo Ef
32.175 c
-xß ,:
-:.-,- a:.rì :
'=l
'
'..,t,
AM
{ÞtA.-roE25x-**.JtJ(-,.Jr)# =
RELEASED,FRAME
FRAME
t- 56'3Í t z?,t*l
,''
,
)
,Ro, o {6Rot LtC d:' 'r ¡J3no, -tt
RDx
,
;
,..
For minimum strain energy,
' r
an - 0, combiningEqs. l,2and3 ;Rv r\Fx
or, or,
I
EI
.,
f-
t6E.75
R* = l?1¡o
+
2? RDx
_
l0t2.3E + St RD* _ Sø.
'.r::
ZO
* 27R* ¡ =ç
rN :
Net moment at B
-
97.425
_ 3.ß x 12.50 = 32.50kNm
The net bending momenr diagram is shown in Fig.
,.,
,r.
',
r.l' BENDtNG.MOMÉNf ' rXá S.tt Por{it'frrná rtfli ¡ncti¡ãd'tiiúgcd frg, (
F¡á.
i
t"'l
l*
,,
n32
'.STRAIN ENERG
,l
{,)
Y..
FRAMES
MET}iOD
Example 5.7
{")
ôU
Analyze the portal frame shown in Fig 5. l2a.
âRe*
ILLUSiIRA,T¡VE EXAMPLES
T
ôu :oand ôu -. i
:0.'
wilt lead to three linear
* E
I
{) r-
AM ¿Ror
Em
;tuI ÈEueIisED
(O) 6IVEN ERAME
{:
FFAME .t.
âM
dx aRD* EI
5 RDt
atü âMp
=0,
5
=t
:.
:{,'"-n*x)(-*)S
{.:
= ï-
r"t
rôM
50kN
{*'
(c)
{.,
MOMENT
(d)
a RDY
M.OMEI{,T: D.UE,:rO
dx
12.5
(e)
Fot
Mo
AM
tot I
M0MEN.T DUE TO ROy
dx
M = 0-JRo"
(f ) MoMENT DUE TO MD
Fig. 5.t2
.:.
:::j:
',
-q!!âRn*
:
Solution
;
The frame is staticaty indeterminate to a dégree 3 Let us remove the support D and the released frame is shown in r,g. i.iä.iä,#,*"-*, rhe shear and axiat deformarions and draw bending moment diagrim, d"; ,Jã;;;fied
Mo
as
shown in Fiss5.r2
"l¿, "
l-v"f
.rl5
***ii],*
L
(r)
EI
:0
Q)
sl
: J{v'-Rp*x)g 0"r"'
alr I = lvro*-n* x'l z,[ ET I
(3)
EI
Member B C
-:.
follows.
)
Mo + 4f .67 RD" I
= [5Mo-12.5RD.]
'
|
=
aRDy EI
aMD El
l:'
in terms of three unknown
-Ro*x+Mo
5m
f
o
{)
simultaneous equations
CD
T
z
{J
=o "'
The solution of these linear equations will give the values of the redundants.
Br
o
aMD
âRo"
133
un¿
i
"i¡ltuu.¡oustoads, ;;;*oï".
,
R*
and
integrars can be evaiuated as
f":oÍe" "rthre-e--¡edundunî, Ro* ,-Ro, and Mo . rhererore, ,JT:":[îî;iJii!,if iri1iq1i-29{ wil.h respçclto,ea"r, g"il"t-r-#lnlrh"r',3,
:
-5.
+ RDyx+Mn
' ,. "aM aM --:. =X.l
âRo,
grì'l: '-î I r
.
âMp
-:l
âM dx : ¡, J ARD* EI J#[**o*
+
R*x
+
vo](-s)ox
0
l'^ ¡,1r. :... r.':,,,.ltj'
' ¡l *2 ' l25xR^- -t:-Rov = 2Btl 2 I.:
:i
ti
r1 11
It
-5xMol
ro
:.
= EItl00RD* -80Rry -20MD l
(4)
ì'; li
t
rl
36i
STRAiN.ENERGY METTIOD
FRA
:"e"-)
ì'.
jo J
MES* ¡ILLI,STRA-TÍVE ÊXAMPLES
t37
,:.
tÞ
M
f
tÀ
ü {)
- Mo - å*sino,
=,,
ff
AUi(P
#;
{.)
=
i["..t*s"e)fr
=
l"re-åu*"|'
=
0
ds
= Rdo
rorminimumsrrainenergy
{)
t) (}
(o)
{)
'l' CTVEN R|NG
0.3t8
PR
(
b
)
;:
RIN6
ÏPlz
.'.
l
-jbuo
{) {-'
3e.s"
l,lo,
t": (c)
or,
{:
sinO
.
+ - += insine
,
o.tt2pR hogsing =
.
o
: +:t - + 0.636, 0r, i 5.
0.= ilg,Je,,,,. l3c.
Deflection of the Loaded Secrion
Solution
"!i: - "'.;.';¡,,'
Due to symmetry arong the diagonar AB, the ftee body diagram is shown in Fþ. The harf ring crqi"r_a n"a""q""iil'ìä *¿mon¡ent Mo ar A. The fra¡ne is indererminate to a degree 3. Due ry*Ã""y, le snear at A is zero, Let thrust at A. us ignore the
5'l3b'
A= + ôt
"
strainenergy
u
=
.tt,;
=
fl.Yià¡ El
t2
lq=:*"ãtuË , aU i_
At
,.:
-o.glepRsagging
The bcnding moment diagram is sttown in Fig.
Fig, 5.13 CircUh.l ¡in.g freme
i:- ,.
= - PR o 3Rr¡ng tt 2 --- -
?*
.:1.,. t.
".=-*
^
BENDING
..aSeine,, ::
;"''
.-. M. = Ooccurs"r
vJBa'
$
BendingmomentatC, Mc
at 0= e0o, Mc = -
{-)
{__;
Mo, = ,=
at,0=0o,
Vt'
{"-;
{,';
of,,,,,
sY_,M,ùtE TR|C
HAIF
o
arr¿
ds
any pointC,.t-rq mqnien¡ isgjv.qg
iy, . .
."r:-. ':,:.;-'^
,: ;., .;11':,,;-*.
ff,
'
'
abng rhe direcfion ofrhe
loadp
'
r1
{!it:.a;V.,-.,i,,.,, 2f
M:=-ds, :, oy
EIJ
where
t:.;;.¡'-'
::I.,;,:' :''l fr .i-. ^:
+i ,
M=
.:.i.
-
Þ'R
:
*
Pn'òitie'
:.,...
EIJ ^=. +fl-s**.¡nell-l*ns-¡nel*a, Z J -oL 1t ¿ Jt l
'i
' l
)]
IlT
pR3
STRA.TN.ENERGYMETHOD;..]
(^ sin20) l
I|
lu:
.--
.
FRAMES : II-LUSTWITIVE EXAMPLES
=
I'I
ftf^**.[=-f'.oul' t\/lo ^
': :
pR3r --- lzr'-81 " -r = 4nEI t
=
decrease in
#l-"(å-
ôU
1
':
(au |.âw
Deflection of the Horizontal Diameter
Let us appþ two equal and opposite horizont¡t loáds lV across the horizonal
or,
a
E^
diameter as shown in Fig. 5.13 d.
- Yt*-Rcos 2, , ,,
â, aw= _ltn_ 2' n"o, ér. i.
Strain energy stored
in part AEB,
å
Errnph '::
'
#rp-'f
5.9
diagrams.
Sol¡tbE
'
The frame and loads are symmetical about a ccntre line.passing-tlirough B as shown in Fig 5.14b. Ler us cr.rr rh: inro two equat.halves. rru aã" bodi aiagrams frir ryr!5,.!!c. various mçmbers are shown.il The shear is"zero ar B äùe t" ryrí,*"t. rr,"." fig arc two ur*noivns at B, axial'thn¡st Tand bending mofüênt M^ "Thesi fwo redundants ^ can be determined using the strain
ørergy:merhod.
Forminimum strain energy,
#=0, ff=o
Member'B C
M=
åi'["--]'"e-]{n-n"o,e)] f-l--n*¡Ð]nae i' [""
(1
-.o",!) . E,,(.'ne=
ir
_ Anatyze a box frame fo¡ a.culvert shown in Fig.S.l4a usingthe srain en€rgy method. Draw bcnding momenl and sheat force ,, ' :- ,
',¡.:
= Z"í'r'r9I¿" Er í aw'
=5
F
.'
.au asr=#I:
:
on zubstituting ror Mo
Ho¡izonAl deflection of E relative to
Horizontat movement of E relative to A,
#
#t+ +]
PR3 ¡," = frlr-rl
o)
As, =
' .,' t
= 2'f M'¿s 2Et
=
-,
¡r7-1-
tJ
ì.i{-Ð. +(;-,. i)-i(-'.i)]
ãF
lenglh'ôfthe Veiiic¿l diamefen
0<0<90o, M: Mo ,- lnsino Z
139
$) -,,+tt -
r
c*e *.o,2
eþ
-M-n'
AM âMo
-
_
50x (Taking :sE¡ging'moment
_ .aM l' -:=O AT
=
2J
as
' 'Y
positive)
-..',.
zl2'5 '-. = I {(t"+sox)ff lr*a"**so}L
#
-
\_)
*
STRAfN ENËR6Y.METHOD:
t40
*
Member C D
û {)
,aM r' fT
50 kN
i. ,i...
{) {,)
ü Ð
{i o
I
I
:r
E
2¡
4
) GrvEN FRAME lso lso T l.Mo ,l I
(a
Y(Mo+
C,.
C
,
r'
m
tj¡¡--q
l50 tl.s t -
{c)
;IiË:rr$lÎ,,
ru
I /T1'ru, z.s r ,uo*
¡
(4ì
er
'ôM' 'aT :
: - t.(Mg
.t"
:
ôM dx J¡, ôMo El
2.sT
=
'f[t,
+ t25-5ox+'rox'z
-z.sr]rS
0
FRE'E.BODY DIAGRAM
=
|
srx + Mox + 1251 ,-so4*,041"
z
f-z
:
( l.25 Mo ! lt4.:*,tl'-:.r2S r)
3
(eI
SHEAR FORCE KI\
J*#s
BENDING MOMENT kNm
' :;- ,
Fig.5.l4 Box frame
:
(1.25 Mo
=o Jf n¡9{$ ATEI
+ 78.125)
I
-
(t) (2\
AU :0 ^âMo
='í¡r.t
-1ruo
gJ2El
= [7.StT -
combining Eqs. 1,3 and
5Mo
-
=
ì1.
(5)
iL
260.42I+, -EI
(6)
5
+,2.¡1y;*- 3t].5 + t.25Mo+ lg4.t7-312.5T: 6.25T ¿--'494.8
or, l.25Mo+78.,125-"i¡13¡r or,
+r25)+5s*-¡e¡'12.5
3.l25iMo
ro
I
lBt
f,
18.24
(d)
2.5
.
(_l
(, ('
(3)
tuleAþe¡,f; D
12s)
(-)
(,
E'
:l
: :'::.,.
zs)l
ry"*;i:, ¿E
.l
l3 : lril3
{;
f.: (,'
I
3t2.s)'Et
L)
(,)
r"r'
f
20kN/
t¡-
{-.
a "'j1i
'12
-sÇo..125[l:' t-'
l4l
FRAMES ., ILUUS:IR.ATM EXAMPLES
0
:'
'#*o {} û
i
1
"
j.STRr{IN:ENERGy'IIETI{OD,
combining Eqs. 2,'4 and 6
''
5.21T- 3.125Mo.- 390-ffi5 + 7.il Tr13.02T - 6.25,r{o = ó51.05
0r,
ot.
3.125
Mo,- 2ffi42= 0
'r
Rearranging Eqs. 7 and E in matril form gives,
l:
{*)
I
f-':, #;ïiÌ-fiä:}
{) {)
(b)
{i
or'
{j
Back substitution in the'vàiues
{r.l= {-ri.'+} of M. ; Mp
ñrrtz tl
:,
glves , ' -:
and ME (Fig. 5.1'4c)
Mc = -'33.E5 kNm hogging Mo = - tB.Z4 kñn hogging
tt
,MÈ
The resutting shear force and bending moment diagrams are
Examph
5.10
RELEASED.STR.
*ttz
IrJ lc.zsJ
l-ì
{,1
'143
B¡*AIT,IPLES
n. I
ùown in Figs 5,l4..diin{ e,
MOMENT
,:l
-t
{: (_;
Analyze the frame showhrin,Fig. 5.r5a a¡rd determine the contribution of axial, siqar ---:'
andbendingdefonnationsinthem.emberforces,
(-,
C (_r
{; (, (,
:.
I 't':
"=I;#.f,.r#.Jå#j
'' ::
(
,
:
SHEAR
(c)
THRUST I NÑL'J
FORCE OIAGRAMS:
DUE 10. APPLIED LOADS
(5.ea)
is staticaily indete,ry,iryte ro a degree 3. The rete¡sed srru*ure ¡s shown in
*,^tlr,Ír:: rrg. )-r)D. rhe strain energy wiil be a fi¡nction of
M"
R*.
M¡,' and l-et.us dr¿w momenr, shear and thrust'di{grams due to ttre apptieå, toañ as weli as each of the redundants' This wiil herp in wriririg the srra¡n energy exprespio¡¡ for, the porral frame. The be¡dirrg, Ítoiitelü:shean.and.ihrusi¿¡ ¿uã-to.Oa,appfieA loads and unir vatues of Íhê redundants-"M¡ M¡; and,Ro , *å'irrór"-¡o rlls 5;tic to f. Axial tension is considered æ positive whiiè axiá compräsién;is,considerei as negative. Let us first evaluare basic inregrars required for wriiing ü,er¿ipress¡oi à;
MOMENT
*q ;;r;r;
,.1,
æ
;fil;
Segment
(
..
:
.' ', "t.'' ,. The stain enërgy due torb€hding, sirear and th¡ust deforqations is given by '' i
strain energy.
t.
.
EI = constant,,Ag ='i00,81, GA. = 50El whereElisinkNmeter.unirs.
Solution {.,;
-l-
.'Forallmembers,
Origin Liimits;
AB A o
-,L
SHEAR BC.
B .,,4a 2L
C.D,
c : .0:: L:'
(d) ':
:
THRUST
FORCE DIAGRAMS DUE
.. ;ì . :'; FU. 5.t5 ,,
T0 M¡=
.':
I
:
FRAMES " ILLUSTRATIVE EXAMPLES rvrD M Y^ a:::Jz + 0
I
I'* - wL'2L^2L
.f,TIATIAT 2L' ôMp 2L' . .,.:r. åM,r ,t'tOmelf;+ =.,i.'.. .1
âRo*
[,x
'ÊliJ:Ì-
ií "',- .*
xâMx ' 2L' ôMo
ôM ôMn i"l
145
ôM
2L'
: -I)
âRe*
r,
v:-.wL*w.-þ*f*o avravrav
' ç),,: .
OIAGRAMS
IUE 10 ü0.=tJ
l, I {
2L ' âMo T:0+ 0 + 0 - Ro,"
L1,.
{'
:
..,.:
AT
âRo*
2L '
Y
-:v
ât : o.
= 0,
aM^
ôMp
âMo
ôRp*
-
-l
I
Mênber C D
M AM
ôMa
û
''
.,,
(1.ì:.iEgfi9E-o¡aGRAM?,0u8:iTo,fg¡t.-l;
âMn
{ I
.: lÍ
1rr,r","1r:Trut'lq*,,*r_dliro t... ,r.,i-ii ,i : .¡t-:1,:!'r:;''.,'
-.:;
0 + 0,
:0, T: -
MA 1,._dse-,g#g,f.,e:g-,1*
:.M¡ -,LRo* +
AM âMp,,
= 0,
V=0+0+qav
,,,
=
RDr x
ôM
=:- l,
: - t +:(
âRo*
RD*
AV âRe*
--qÈ=0. ôMo wL + Ml _ 2L
=-
!Íb +o 2L
raT
ôT
ôMe
2L'
2L'
ôM¡
I
ôRo*
=0
For minimum strain energy,
AU
ôMe
=
0,
AU
ôMo
-0,
AU
âRu*
:0,
146
STRAIN ENERGY METHOD
Let us first evaluate
FRAMËS - ILLUSTRTA,TIVE EXAMPLES
t47
au aM^
u*f[-* (
.-ä'-ro
*Er--ï*' -*,.r]
-*" -Ro"L+n*.)(-r)$
;)*.i
try * J r'a,r- * I n*11 - å *,] * o.
i[-*'.* *. - *...
(- Mn*ro)
(5)
*](+)ft .
o
(6)
'oh
(2)
ïl-"'-#.*l(*)s*o*
ar dx J¡, aMAAE=ll-*,-y.*](-+)#.0. 2L 2LJ( zr)t
L.
{L
i[-*'.#-*l 0-
*r-+þ *ï#)#
il : .'. AU
:
MJ
I
Q)
,t*f
.Ço-mbining Eqs. 5, 6 and 7.
r# -'
[å'^'.Y.å**r'
(+,#)'^.(î
Ma +
(3)
Combining Eqs. l,2and3
âMo or,
(M¡' MJ
- (*
+)#
-
j*r'.#.#-#-#j :,
#)Mo+112R*
+: o
2 t âII frr¡.i s -::-=1"'4"+1M^L+jR*L2-lwL3 âMo L 3 3 " z "^ 3 (st. ?\ (t :I
Ma M. M^ + l."'D.ll:O l00L 100L.200L 200L1'-
3-.or' l;-rbjt^*lî*a*Jto*iv**-¡wI,r=0 ,l
'
Ø)
Now let us
evaluate
vâTI v
âRo*
(s)
STRAIN ENERGY.METHOD
r48
T TRUSS
149
L-2L/
: i(-t^ -xno.)(-x)9r.lf *t.-5-t^ 'EI JtM+y-* J0 âRO* EI {\
'lIvto 24 e¿o tszs I e.en 6.670 24 l{Mo ìl=if 640 ì| |
*
M¡x
2L -
Jr j## Jf
Mo* 2L
:
-
**r)1-r¡$ * i(-r"
[ir^r'*
?Ro*L
A"G=
LRo* +
no^x)(-r-+*)$
,,,,'
EI
.i',' .'
*
o
Ro*!
(r0)
25Et
", A.ECombinihg Eqs. 9; l0 ahd I I t't,
,¡(--': u'
.,
,il
RorL
(l
5OEI
'
'.
'
l)
=
(12)
o
in matrix form,
{'',, :t:'"¿ :
ii,,'lil
,
i
rii
''.-'
ii
I,
(;'.#) (:-#) (;'-#)
',
ii,
r:/-. -, :l l'. .,
it.-. ÍÌ{ l-' l'
i.i ;
,(
),: lf "^ l
I^ -lwL' J t_:wL, J
., J
'
Let L = 4 m,
)
(åi.*,i[:1. -awLa
Symmetric
:
.',
l_
5¿
w:30 kN/ln,
the matrix reduces to
l.
3-.¡
-:wL)
-L-
lL ia 32
L
L:4m, w:
8.¡
-LJ
l. -r --wL-
:
J
H}
30 kN/m,
1.334
.ii
rj"-
L
-?*ro J
the maÍix reduces to
'
.:.¡
1r', i|\
I;L IJ I
1*^r,,j*;Ì.(}r.år)o* -l*r,
Arranging Eqs. 4; 8 and 12
the effect of shear and axial deformations is ignored, the expressions for minimum strain energy are obtained by using Eqs. l , 5 and 9, that is,
I
Ro*
147.3)
,ll
I
o
Lþ-
-,aU ôRp*
''':
[R*J
(rs)
o
{'
(_:
=l-url
i(-n,^X-,)#
-2L ¿* : * 9r * fr ' Et J ARD,* AE o [(-n^-X-r)!I
il
frøoì [-øz] 1t" |
ri , ø, .;i,,"
jr"r"r.'*JR'"L'-i-u]*
: I *o",S. v-+-+ ' deu ôRo* A"G ,o :
-
(14)
rzr.oozJlno-J [5l2oj
lSymmetric
= {0}
(13)
6"667
I
,lfue I *o I 24 llM, l=1640f rzo.oezJ [no.J [s rzoJ z+
{0}
(16)'
Y ÍY:
b
t50
STRAIN ENERGY METHOD
ìi
tPl
P
or,
l"
t
l5l
TRUSS
',ft-)
rJ =
I(n'+u, * u, * ur)t
th -2,#,
(5. r 8)
where, i represents i th member in the statically determinate truss, and redundant meinber.
ji
{-}
j represents j th
' i'l "-
..,,i .:|,::
(o)
'.,.[) .li'j U9
rtit) ',1,
!,i{)t
{ .,|,*
u1 u2 U3 rr,zl\z/ \r),
\1,\
L6
'i
,'i
GIVEN TRUSS
I
L
rin..
(b)
rlL-,r
L
#-I,.'.u,+u2
terms containing
riL;
,.
{
:',
ii'jl (, {'-)
'jl t,ä
:r'
r'.
\-
,,'
1i
',i
{
ilr'-"
ii;-, il \. -' ¡:'
lt
,..
i, I
l''-'
(
3.
ôur RELEASED,TRUSS
ôI
,
.
F'
: F'+ ut + u2 + uj
:
:
F2¡L¡
(5.17a)
2AiBi
.,
)
Strain energy in a redundant
(
1
Total sftain energy U =
will
become zero upon differentiar-icfi
of
u,
and Tt , and therefore equal to u, / T,.
= ffi
(5. I d)
Similarly. U can be differentiated wffi respect.to T, and Tr, and equated to zero. ' From these equations, the values ofT,, T, and T" can be obtained. It can be seen that is ,,Êq. 5.19 I
similar to Eq. 3.5.
Deflections in a Truss
., [åi),=.
=',
(#),=, =0, [#J.=, =o
(5./0a)
Tt'iL¡
member, ¿g. = 'r2AjEi
IaU, *f ij
is the ratio between increments
l(r'*u,*ur*ur)-!Li
(5. r 7b)
. and
(.
(
T,
Castigliano's second theorem cari be conveniently used to determine deflecfions in a indeterminate truss. Let the forces in the redundant members be T, , T2 and T, in the truss shown in Fig. 5.16. Let load P be applied at a joint where defleäion is rãquired. The truss can be made statically determinate by removing the three redundant members. The forces in the truss members can be determined in terms o{T, , Tr, T, and P. If U is the total strain energy ofthe.truss, thèn
where = member force due to external loads n¡ r u2, u3 member forces due to T, , T, and T3, respectively Strain e¡ergy in any member except those considercd rõdundãnt
¿g'
and
t6 Statically ináìterm¡nate truss
If a reaction is chosen as a redundant, the constraint support in the direction of the reactiõn is removed. A force is applied at the joint in ths direction of the redundant reaction. The compatibility condition is that net deflection in the direction of the reaction
Let the force in any member F
T,
o
L5
members of the tniss which has now become perfect and statically determ¡nate. The compatibility condition is that the increase in distance between the joints is equal to the elongation of the member itself. The statically determinate truss cañ Ue anatyzêO.aue to the external loads and forces T, , T, and T, individually.
must be zero
,,i i;1:
Fig.
{:,
ôu' L' * IL' : aI AiEi ArEr
L
t;
i
* u,)'
aU,
.,.}
/ ^,,\ loul l_l =^ \ âP /o=o
(5.20b)
The solution of Eq. 5.20 gives values of the redundants as well as the desired
,.,¡,.:.4,$ection
l
t52
STRAIN ENERGY METTIOD
Example
5.ll
Determine horizontal deflection
ofjoint I of the
r53
TRUSS
ôU f -ôF dx AT J ATAE
truss,shown in Fig.5.17a by the
strain energy method.
ât r' r vv :ln-- .ôF dX J AP APAE
and
The integrals can be evaluated as follows using the values given in Table 5.1
Table 5.1 - Computation of member forces L
F'
Ã
,
ttI ¿o
300 Cm
'(o
lot
(b)
)
Pl
2
30 t¡
F
c, <; I
+1
(c)
(d)
ê
^" lT
:."
I Rr-
.oï
F:
AE
1¡'+u)
t.67
P
0 -
(e)
T
l
0
[(4ûr0.8T)x 1.25+1.67P]
I
t.67
t.zs+
I *t* 1.67 P]
/.
: - 3680 or, T=in member 1- 4 = - 26.93 kN , compression 136.4T
:,.
ii,,
(- l.33)
',
$f;ff),=,: f Li
:r.
oo *
I-JJ
-t30+0.6r+p](-0.6)f .c0.8r)(-o.sl"T -
/ ^.,\ :0 l::l t ôT )r=o
or,
+f
Let us treat member I - 4 as a redundant and is assumed to carry a-force T. By removing this member and replacing with the member force T, the trusi becomei statically determínate as shown in Fig. 5.17b. Let us apply hórizontal load p at joint I to calculate the horizontal displac.ement of this joint. Member forces F ' due to external loads, and u due to lirad P are shown in Figs. 5.17 d, e and Table 5.1.
ùhere,
:
-[40+0.8T+!.33P]
0
(40+0.8T)1.25
50
P
[(40 + 0.8 T)
",'fue
f
T
-
i.33
- 0.8 - 0.8
-0.87
0
ôP
-t
0.6
[3Gi0:óT+P]
i,':l
Solution.
1Iq
)U
-P
AT
[40+0.8r+ l33P](-0.s)
Fig. 5;17
sftainenergvu = f J2
t-4
-,
I
+l
2-4
20
- 0,87 - (0.8T+40)
au - [r2l-{ AT J ATAE o -t + È ? o
l.o
20
2-3
-tio+ o.sr¡
R3t I
t-2 t-3
(30+0.6T)
30
ôF
AF
F =F'+u
u
l*,
0.6
r) +
x 0+ (50 +
fføo+
On substituting the value of T,
r+
1.67P
)
26.98kN
l.6i
x
I
0-8r) + (50 + r) r.67x
{
x
r*
f
I x"E +
+
a ü {:}
{) {:} {_)
lqql
[613e
l. ap /"=o
.'.
O
_ A:
a
+
122.7 E
1 (-zo.la¡1 ,E
L)
'25kN/rn
E
E
0.
E:2x
3m
104 kN/cm2
I
I
l4cm : l.4mm
*,:
jii,.
P 5.5
moments are shown as negatüe.
Analyze the continuous beâms shown in Figs. P3.3, P3.5 and P3.6 using the strain energy method. Neglect axial and Shear deformations. Draw shear force anri 'oenciing moment diagrams. Take EI constant unless specified otherwise.
:
',ié
{:
.:::j. .¡,ii.
B
-r I I
ii'*j
3m
3m
L
I
20kN
I 30kN I
D
c
4m
T
T
I
I
2m
(_.
EI
I
t
(,;
{) (,
T-
T
2l
(,' P 5.3
(
lz J.
B
1m
(",
,1.5,
M", = 74.07 kNnJ
Reanalyze the frame sho*n in Fig P5.3 Uy treating M", R"* and 27.94 kNm., R",: 16.4 kN<-, Mo = - 2g.64 53.37 kNmJ
redundants.
k 20kNl
'
3m
8m
F--------------=
I ;
Figs. P5.l-P5.4
: - 32.53 kÑm, Mr": - l+.al rru*1'"
Analyze the frame shown in Fig. P 5.3 by treating R., , R", and M" as redundants. [R,., = 16.4 kN<-, Rur: 77.4 k^fT, M": - )A.Oq *ñor,
P 5.2
:
() {',
6fit
,
Mto
[R¿, = 4i.86 kN+-, Ru, = 45.30 kltfT]
2m
= CONST
kNJ',
Reanalyze the frame shown in Fig P 5.2 by treating Ro* and Ro, as redundants.
I
I
P 5.1
fiO
+
+ 3m
I .30kN
Ãnalyze the frame shown in Fig. P5-2 by treating R* and Ru, as iedundants. Draw shear force and bending moment diagrams. lit,, = A.7l kN<-, R,,,:
2m
I
I
Figs. P5.5-P5.8
20kN/m
[,.
+ 3.'n
t-l
{. .,
T.
ii.,Tri
P 5.7
{.:
P 5.6
:j¡t,
Analyze the frame shorvn in Fig. P5.l by treating \* as a redundant. Neglect axial and shear deformations. Draw shear force and bending moment diagrams. [R", = 24.43 kN<-, M"n -- - 27.84 kNm, Mu" = 122.1a kNmJ
5.2
þ-----q!----n
sm *' þ-3-m +
.ï
NOTE: In the answer, all clockwise moments are shown positive and anti-clockwise 5.1
T
2A, 3 I
PROBLEMS
{i t-.;
:ry
I55
I
2826
Ç\
{)
TRUSS
STRAIN ENERGY METHOD
t54
50kN n *rr
ì
ùI
I
M" as ñm, Mo" =
Analyze the frame shown in Fig. p5.4 by treating R," . R"_ and M. as redundants. SO.0O kN-+, Roy: t7.95 khrl, U, =- SI.SS teñm¡
[n^:
r1,,, P 5.4
[M,:
Determine force in member 2 -
i,.ìi,,.¡,.,*"h
member:
15 cm2
4 of the truss hown in Fig p3.9. Area of
' [Fz-t:
4.62 kNJ
Take area of top and bottorh chords: , An^alyze the tnrs-s shown in Fig p3.10. cm2, area of web members :70 cmz. Treat Ru, ai redundants. 190 "j,*_,:,, , ,, [Rar= 255.5 kl,fîJ ,_,
,.i,
U *
CHAPTER
STRAÑ ENERGYMETHOD
*
r56
û
5.10
e) 5.
I
I
Çt
û
Fig'P3'll;
lobethe
(a) ió 126'3 kN' F6-î: : îír r-:-l ì: ci ;Ñ,'Fr.: - I I'ts N, F, -o =
Ç)
*
assume mèmbers 2 -5'2-7 the ãreas in cmz' indicate redundants. nigotbt ;ith" mãmbers redundants' ¿is 2'7 2 and Treat meribers 5 Treatmembers I -6and6-3 asredundants'
Analyzethe truss shown in
six
12' Treat member 2- 3 as intemal Analyze the truss shown in Fig P3' [Ft - ¡
5J;
=-
59'94 itN' Rs,= 45'28 kMl']
Figs' P5'5 - P5'6 using the strain energyAnalyzethe portal frames shown in p"t",fnü" ttt" effect of axiat and shear defo'rmations' Take E:200
method' GPa, I :20A
x lOaIna, A= 150 x lOarn2' v :0'25' M¡,: -4t'65 kNm' R^-- s'i3,!N-+J [P5-5: '[P5.6: ni,: tt'a kN-+' Rr:'os'¿ *ttT' tt', = I8'29 kNm' ' Mt" = - 36'9 kNml
{-)
{: {,)
5.l3AnalyzetheclosedframeshowninFig.P5.?bymakinguseofsymmetry.
[M"t,: l-.;
{.
5.14
(-,
{_
(t
-
4'36
kiÑ;,
i,t":
5'26
kNm' T'o¡ :
-
I4'7 kN' T¿"-- - 30'3 kN]
P 5'8 by making use of symmetry Analyze the closed frame shown in Fig' : : Horizontal thrust at B I0 kN M"¡' tÑm 15 through ø unal'- fi¡" =
tensilel
{r
COLTJMN ANALOGY METHOD
121'40 kNl
ó.I
INTRODUCTION
The column analogy method rvas proposed by Professor Hardy Crossjn 1930 for the analysis of statically indeterminate structures having a maximum indeterminacy equal to This method is useful for the analysis of fixed beams, arches and single cell open or
3.
It ís very convenient for the analysis of curved members and nonprismatic members. It is based on a mathematical similarity between the stresses created in a short column section subjected to eccentric load and the moments induced in a member by redundant reaetions. closed frames.
ó.2
STRESS IN A COI.UMN
For a proper understanding of the column analogy method, it is essential to recaptulate the formula for the stresses in a column subjected to bi-axial bending. Consider a column section subjected to an eccentric load P with respect to the tw-o fectangular centroidal ¿rxes x-x and y:y as shown in Fig. 6,1. Thé stresses in the column are assumed to vary linearly with .the x and y coordinates of various fibers. For equilibriurn, the following conditions must be satisfied:
.rF
or,
P-JodA
=o :0
Er4 :
. JodAy :0
or,
{,
f M, : o . jodAx :0 or, P*ì -
í-
( ( (
and
la)
0
{,,'
P yo
(6.
M* = fodA y:Pyo My = JodAx:Pxo
(6.1b,)
(6.
lc)
6.za'¡ (6.2b)
çj j
þt'r
DEVELOPMENT OF THE METHOD
COlUtr,,fN ANALOGY METHOD
t58
159
I
Y¡i i ue, A41l___l____{v,. ,)/ \ø-
{"1
ii
t)
:-_--}1
f.-
(o)
i
{i
GIVEN BEAM
(d) LOAOING ON TOP
OF
ANALOGOUS COLUMN,
l
t:; rJ
/\
i
':
f
SAME AS (b)
-/
zt/
-
v'-
lr---
(
b) FREE SPAN
MOMENT
*1
(e) SECTION OF ANALOGOUS COLUMN
{-';
L
{) :
(f)
MOMENT DTAGRAM
(C) END
Çi
{i
Fig. 6.2
{-_;
Fig. 6.1 A column section under load and stresses
Thd moment area theorem gives
{__,
(l) ,
GeneralfÖrmulaforstressatanysecticninashortcolumnmaybewrittenas: {-_')
_*{",1,
: A-L-tt-f;I'L :.f ',,',. - l.t,l..i*l'l rxry-¡xy r
{.,,'
{) {_)
one
.
(-;
6.3
{', ("
(, (
or
P M" .M'x+:'-Y O=-+ I*A l,
for this beam consists of ttro components : (a) The free span móment diagram due to the applied loads (Fig'6'2b)' (b) End moment diagram due to f,rxity (Fig.6'2c)
(t) (2)
Slope ofthe beam at B relative to the tangent at A is zero' Deflection ofthe beam at B relative to the tangent at A is zero'
,-:
two
(ó.5)
di¡rgram rnust be equal and opposite to the area of the end moment diagram.
(2) '
Thç¡def.lectìon of t(e beam at B with reispect to the tangent at A is equal to the 'momeni of area of inoment diagram between A and B.about B.
!
,
as shown Consider a fixed beam having uniform cross-section and a general loading and a roller A at a hìnge inroducing inFig6.2a. It can be madç statically determinatety diagram moment The bending as redundants, treaied are at B. The moments M^ and M"
The net moment diagram can be ôbtained by adding these algebraically. The compatibility condition requiresthat.:
^ tI M^ds " u: J EI
of
1"..,,, but 0 = 0 due to the compatibility condition hence the total area of the moment ,,:' dlagram between A and B must be zero. It follows that area of the free span moment
(6.4)
DEVELOPMENT OF THE METHOD
:
The slope ofthe beam at B with respect to the tângent at A is equal to a¡ea the moment diagram between A and B,
.r,,,
(6.3)
product both of the centroidal rectangular Írxes are axes of symmetry, then to ofinertia l^, ofthe bross-section is zero. In this case, Eq. 6.3 reduces
lf
(_i
(.;
PRESSURE ON BOTTOM gr^,îÈisî?us coLUMN,
., but
4, :0
J
\v.v.t
EI a
due to the compatibility condition, hence moment about B óf *re rree flfi moment diagram. must be equal and opposite to the moment about B of the end t diagram. Because the two areas are equal, this requires that the centres òf of these two ar€as should coincide. let us consider a column section of uniform width and length equal to length
diagrams
of
r as shown in Fig 6.2 e. If the corumn section is loaded with tne freJspan M. as shown in Fig. 6.2d, the sFess in the cólumn section wilr vary
^diagram from A to B as shown in Fig.6.2f. The stress at âny section in a short column
to uniaxial bending is given by:
*{}
*
COLUMN ANALOGY METHOD
r60
ANALOCOUS COLUMN SECTIONS
t61
ü,,
{} * {) {}
o= Thus, f'or equilibrium in
(1)
{}
their oppositeto't't*oï"niofthïtotal'ot"ulou"ttãtåt"point'sinceareasof and oppoiite to each other' *" diagram the road oiugrum ;na stress coincide must of gravrtY
{._;
\..,l
{.,,
1.,
l/Et of the conespondil;;.;il; importa"t
section it'at *i¿tt' - of the column at any of the height oi tt ,"ut structure. The
Jdil;;iJered
(,_;
{',
i'
Mg
analogous
(c) END
'I
(C) SECTION
OF ANALOCöUS
+1 COLUM N
la,
\l-' I
(f)
PRESSURE ON BOTTOM OF ANALOGOUS COLUMN
SAME AS (c)
Fig. ó.3
course, be consistent'
M,
is plotted on the compression face of the corirpression face is on the top or on the outside in the case of frames, the moment is considered positive and thè load on the analogous column as a positive thrust. The free span bending moment
member. If the
Alternative APProach
determinate by remorring fixed beam can also be made-static,ally
"*j^l[t-
*i.ffi ::ir:::ili"in'[i**"s.n;:;1ff#i'iî,Ä:î,ii:*:!ïi:Tü' iï,:y"itoil:liåÏäü,
deflection of R" and M" are such_.rhat rhe srope-and
the
cantilever at B are both zero'
For comPatibilirY, be equal and diagram of Fig' 6'3b about B should The moment "i";*íã;tn" areÀ'of diagram of Fig' 6'39 about opposite to tr't''iot"nt of f . in-Figs' 6'3d and e' The stresses colyn,'"-thoyl analogous the The loading.and : -on Eq' 6'7
(2)
'
i"
'
nig' ã'3ras given by
given Thgsstressatanypointofthecolumnsectiongivesthemomentduetotheredundants at that section in the beam is
The net bending in the beam at that section.
*orí"n,
byalgebraicsuperpositionofthefreespanmomentandthecorrespondingstressinthe
.olu*n
?*T. t\ I A t\ Jl t\
MOMENT DIAGRAM
the free span moment.'-itt. the samê as those used f.or
tr,"..îr"ï*""ttt"*"
LOADING ON TOF OF ANALOGOUS COLUMN. SAME AS (b)
5* él
of ltisimportanttoknowthattheunitsofstressandhenceofthelrxedendmomentare units for A, I, x and y must'
6.4
section'
STGN CONVENTION
analogy method: should be foltowed in the column The following sign convention
(
leL-Me
"to be some small unknown value'
ioràr u,"u of Fig' 6'3c'
t.,
MOMENT
is equal to
(l)AreaofthemomehtdiagramofFig.6.3bshouldbeáqualandoppositetothe \'/
{)
(b) FREE SPAN
"quui
"ano"t a column Thus'itisapparentthattherelationbetweenthefreeSpanmomentdiagramande-nd ti as simply supportetl is the sale'11in the name moment diagram ¡n u utä*"i""ii"e section und-L:lt" the-stresses createj Ln its
A
(d)
GIVEN BEAM
anv or the total lnPrieo. t:u9,1bou'
column is not
{.,
(o)
*o*
{)
i.; -
RAÈ____!___-ÎRB
:
load'
between the applied fouJ-unà ;;;; column analogt' u t";
(--
å'Y
column
to the-applied must be.equal and opposite The total stress on the sectioh to the a¡ea of the àpposite and u"'"qu"r ,ä.rr-a,ugrurn that is, area "r,n. point should' be equal and
(z) fi3 iîîiXî;
{._;
i_)
a
(6'7)
,.
Thus, sagging moments in the case of beams give a positive load on the aralogous Compressive stresses on the analogous column are taken as positive. Since a compressive stress in the column corresponds to â moment opposite in nature to the applied moment, it means compressive stress on the column section represents a hogging moment. It will be ¡epresented as M, henceforth. The net moment at any section will be found as the algebraic difference of the free span and rbdundant moments, that is,
M =-M,
-
M,
(6.8)
The distances x, y and eccentricity e should be taken with proper signs with respect to the coordinate axes passing through tie centre of gravity_ of thb analogous column.
ANALOGOUS COLUMN SECTIONS logous column sections and the¡eference ¿rxes for various statically indeterminate having three, two or one degrees of indeterminacies are shown in Figs. 6.4, -6.5 lf both supports are fixed as in the frames shown in Figs. 6.4a and 6.4b, then
{"J i
{j {} {.1
{;
FIXEDENDMOMENTSINBEAMSoFUNIFORMCROSS-SECTIQ}I163
COLUMTN ANALOGY METHOD
162
i
through the hinge. The first term in Eq. 6.3 is equal to zero and three moments of inertia viz, I*., I' and I*, are required to be calculated'
The stress at the centroid of the anaiogous column' the retérence axes are taken through centroidal the both or Eq 6'3' tf^Tt one any point in the column ls glven pioduct of inertia I*, is zero and the stress the ot:ii"rti4 u*", axes are principal
j:::::::
p rry ;'gl;;iy
by Eq.6.a.
{":i
t_i
{.)
{) ü
If both supports are hinged , the moment of inertia at both these sections will be zero and therefore, infinite area exist at these two points' The axis passing through these two co, hence points is one of the principai axes as shown in Fig. 6.6. A æ, I*y = 0, and x axis the about areas of infinite of The moment i4ertia immaterial. yis axis location of is zero because it passes through these twp points.' The stress at ariy point is given by the
:
p
third term of Eq. 6.4.
I
f-i-----r
1_:r;=r-
i:
l:
{,.; (q) {"_",
- both.ends fixed Reference axes for'analogous column sections
(-:
Fig. 6.6 Reference axes for analogous column section-both ends hinged
i'
rr",:
I
{-
,
{,;
{,
I I
{)
Iv
-
{,-
;
.
t*.; {_,
¡
t, (o)
l,t
(b)
(c)
sections-one end fixed end other hinged Fig. 6.5 Reference axes for ânalogous column
Ifanyonesupportishingerl,itsmomentofinertiaiszeroandtherefore,widthofthe is infiniæ area of the anâlogous column ut"d at this point and therefore this is the centroid of the
inRnit"' rhe
{--'
"Jd;iä;Jí;;tttlt-p;ïti and is assumed to be
(,
.l^,i"'"u,.ThereferenceaxesaretakenasshowninFig.6'5,obviously,moment. x and y axes passi ,"îäã îi ,t r, in¡nit area at thé hinge is zero about both the
l
v
'ffi
{_,
"on""nt
The following examples illustrate the application of the column analogy metl¡od.
ú.6 FIXED END MOMENTS IN BEAMS OF UNIF'ORIì{ CROSS-SECTION Example 6.1 Determine the fixed end moments for the beam shown in Fig 6.Ja by the column analogy method.
$olution
)
Solution assuming simple beam (Fig. 6.7b)
The beam can be made statically determinâte by treating Mo and M" as redundants. load on the simple beam due to uniform load is applied as a downward load on the of the analogous column because this moment causes crompression on the out side of beam. The pressure along the base of the column is constant in this case. properties of the analogous column are /
of inertia about x
'Area A:
:
I x L: L
: lxL3 - x axis I-. ^12
6^t
Y
L
ë¿) ir 2\
u) ii
r,.ì
164
COLUMN ANALOGY METHOD
ffi (o)
iì
tr'i
"xhj*
GIVEN BEAM
rÍ
wlTl
f-ij
FIXED END MOMENTS IN BEAMS OF UNIFORM CROSS-SECTION 165 the analogous column because this moment causes compression on the inside of the
beam.
-ì
=Itlt AI
M,:pressure lìor point A,
B
u.-
'''
{-t ;Y
Y4 M-: "2 -
Ft\
lì-
tl
*
l-. l--
lx
vr :-v777-V7V77/7V71 lx
{J f-¡
p:1* *LtxL:wL3 326 M:Pe, e:L/4, l/y = lxLz/6
v
T,-
d
.l
o
12
(b)
'iiì
(c)
SOLUTION AS
SIMPLE BEAM
ci
SOLUTION AS CANTILEVER BEAM
_
MA
Loadonthecolumn
,i
r,-)
Ç;
' oe:
:
{}
Mai
point B,
0
: PMv Ã*i
1,.
*lj :Jt:t"' = wL3 n -t
l
NetmomentatA:
M.- Mei= O -ff:
,
{_,
;Ihe fixed end moments at A and B are both troggin!.
fl
(b) Solution assuming cantilever'beam
('
:
-r.,
-+
NetinomentatÉ: Mr-,MB¡:
i
:','
,I}
{.
, -$:
wU
E
5wI]
:i
t2
( s*ûl wL2 ... t=' llY
tz
: - Yz¡6
*r-' (*f r ¿''*l q )(il lxL
wLz
2 -t-( lz)
lx
*!' t2
U
z
ÌÌ:.:' ,
:r:.
{,) l,ì,
û64 lx
=M.-Ms¡:M": 0,
M¡:
(_: '
- wU
6
O: 2 wl] "t=ãwL3 i* , Its eccentricity about O = 0, hence moment about x - x axis : .'. Stresses on the column section are :
{")
|r
6 l6 )\-/
lxL
.Fig.6.7
{.),
¡"
.3 (wú lrt il"t r wLl ll ¡l
Ms: M"- Mi:0 -
wL2 t2
: _ wt] t2
the fixed end moments for the beam shown.in
-+
o. K.
'
Fig. 6.8a. by the column
assuming simple beam (Fig. 6.8a, b)
(F ig. 6. 7c)
The beam can be made statically determinate by removing the support B. The load on as an upward load on the top of
the cantilever beam due to the uniform load is applied
\
l.can be made statically determinate by treating Mo and M" as redundants. diagrams, analogous column section and end moment diagrams
111oTlit
Fig. 6.8b.
FIXED END MOMENTS IN BEAMS OF UNIFORM CROSS.SECTION
t6'7
COLUMN ANALOGY METHOD
.166
þt'-"f(t:) (-)+
M.ry_-.-*dtbIr
Ms
T
il
(d)
SOLUTION AS CANf ILEVEB BEAM
Mi
:,. ,,
i..
M¡
:
{lttr.point B,
,"' (b) SOLUTION .AS SIMPLE BEAM
Fig.6'8
Ms =o
For point A,
' M
=
- l {""
:
Pr e, +
i)'
Prer,
|t' - ",(*r-)
",
:-
*,' *þt"-')'
the centroidal axis' where e, and e, are measured from
',,
=
.j, Mo:
i$l;¡l\otut¡¿n
,"+(M"
-
6a2L+
M,
+arz)=".;(r-+)
: - *^?(Z-r3l YL\ L)
assuming cantilever beam (Fig. 6.5 c, d)
,' fhe
"'t - \i-'u^)
,=
(i-?rv)=i-'r{'-uù =
y=-Ll?, Mi=
i. ?
l:lxL3/12
L;}
Rxe¿ beam can be made statically determinate by removing the support at B and R" and M" as redundants. The free span moment diagram, and end moment are shown in Fig. 6.8 d. The centroidal axis y - y passes through the mid-point (s column section at L2. i:.'..
Mr::
Mo, P: - Moa,
M: pe,
y=.-L/Z
(!-11=-l!-ì ": - \, 2) \ 2 )
\=)
:
il
i'l
ir
COLTJN{N
r68
lr
STIFFNESS AND CARRY OVER FACTORS
NUEIOCY METHOD
t69
i
\J
::
(' 't l'-1
(d)
;lì ¡
,U'
(o) BEAM
a) M"; - 3M, å(L -
,
.
ELEMENT
ì
{r.
vzzzzz--.zzz-,--rzz-]l
"^[-ì>--------
3M"å(L-a)
M¡,:
(e)
I
(',
LOADING ON TOP OF ANALOGOUS COLUMN
I
SECTION OF ANALOGOUS COL UM N
oft
Me:
trl,
For point B,
(l
M, ='0,
t\.
'2
(b)
L
{:
MOMENT DIAGRAMS
lto
OF ANALOGOUS COLUMN
(c.) REACTIONS tN
SAME AS (b)
CONJÚGATE BEAM {
Fig. 6.9 Stiffness and carry over factors or,
\.
Ms: -
""i('-
r;)
M^ / EI and M" / EI diagrams are considered as pressure diagrams on the bottom of the column, the column will still be in equilibrium, The MA i El diagram will be considered as positive while the MB / EI diagræn will be considerèd as negative. In this case, it is conve¡rient to take the width of the analogous column as I / EI instead of unity. Thus, Mo and M, can be found directly as the positive and negative pressures at A and B.
6.7 STIFFNESS AND CARRY OVER FACTORS
. Thqstiffness,of a beám at end A is defined as the momentrequired tc produce a unit slope at A when the end B is fixed. A moment M" will be produced at B when the moment .Mo is aþplied at A. The carry over factor from A to B is the ratio of the momént at the,fixed end B to the moment applied at A under the above condition. \.
Consider a beam AB of uniform cross- séction shown in Fig. 6.9a. If a clockwise moment is applied at A, the beam will deflect as shown in dotted line. The slope at A will be 0o which is proportional to Mo , that is,
_..
Downward load on column
MA *on.-
:keo¡
{
:: i
I :.
or,
stiffness factor
ko
The moment diagrams on the conjugate beam due to s
(
ìÍ
: MA ,^
(6.e)
Mo
and
M"
are shown
in
Fig.
6.9b and the reaction 0¡ in the conjugate beam is shown in Fig.6.9c. Now if reaction on the conjugate be.'.m is treated as loading on the analogous column (Fig. 6.9d) and
L
Area of the analogous colurnn
::r,
Distance of eltreme fiber
=
0r
,
y --.1 L 2 Mo:6=
¡3
Moment of inertia
EI-
P ,,MY
AI
Eccentricity
e:
!2
BEAMS WITH VARIABLE CROSS-SECTION
COLUMN ANALOGY METHOD
170
t7t
^L---:-B-------Ð
v
,^
.'^(:Iä) El Irzer,¡
or,
'1":h l: l'
= fe^
lv 1-r
-TIIJ
SimilarlY,
lY 0B,
iìiil ,ì.Í
zî¡
!:l',
Mir
or,
:.o":
l*4 AIr'
t
' ----91-j--
MA 4EI t. ÃA=ooL -
Itznt
T 'oB
u ee
I
¿
zBr ^ -r"Á
-r-rl-
= þ* '^(Ë)(-i) EI
x
Fig.6.t0
iì;
J
ì: ?: i:l:lj.:
M":b-db=ffi"
;; :
M" = -' lMo
;..i:
are both rt is-fg¡ing' In fact' Mn and Ms The negative sign for fUi *-"1nt ttr." treating hogging moments' respectively' Therefore, clockwise causing ,ugg'ng" and
of'
,,, .,.1.,'
3EI
clockwise moments as Positive' Carry over factor from A to
B'
: C* = Mol ¡,rfo ã
kB:#=
(6.10)
the stiffngss methods of analysis' These results are very'usefrrl in'
Example 6.3
:-
Me=-
DeterminethestifñressatendBofthebeamshowninFig'6"10,andthecarryover factor from B to A.
Gn;1
UB
3EI
Solution
(a) ' Properties
3abEI
jì!fgr$..'$i{l'.F,.|.,-þPkwise
of the analogous column section
A=cc,
l.*a3 l*b3 l, = -l-*-n-,
I.x :
factor C"n from B to A'. Stif,hess factor k"'and òarryover edge of the analogous column' Let us apply a load 0" at the right
(b)
M"='PressureatB:
P,\4 Ã-1-*
,..t cc
":
:. t.l
.t,
moment acting on the beam elemen! qs positive for both ends,
M^ M
-
MB
r:ìili1,8
BnAMËìlry.ITH VARIABLE
a
" cRoss-sEcrroN
Determine the fïxed end momdnts for the beam shoún in Fig. logy method.
6.ltáby the column
BEAMS WITH VARIABLE CROSS-SECTION COLUMN. ANALOGY METHOD
172
= t,
?3r43 + 3x(5.44-l.S¡z
o
133
I eï-c zt-Dtsl !
I
I* :
I
t/t.sEl ì,
1
ET
T
T
(b)
'.2
5.44)2 +
2x(10.5 _s.44)2
and
= 14: EI
ordinate
t-
U
ordinate
2-
b=
: f:StNt
x3- l0xi
BendingmomentatB:60
-
cbcde (d) Ms/Et DUE TO ÞOINT LoAO
ax(6 _
Analogous loading due to uniform load (Fig. 6.1 Ic)
135
r35
EI
=l35unit(sinceEI:l)
:67.5 units
2El
momentatD:60
1
^-
à"i.*
The beam can be made statically determinate by introducing a hinge at support A and roller at support E. The M, diagrams may be drawn separately for the uniform load and the point load for convenience.
4-\
obde (c) Ms/El DUE TO UNIFORM-LOAD
lll.47units.
lJ ^ i*
*
173
x9- l0xf
ordinate
I\¡f 3_ d : ll1
ordinate
4- d: +f l:5E I
:
E¡
z
: fSS¡¡t
67.5 units
:9gunits
Fig.6'11
segmenta-b-1. . p.,:i wâ' (3l-2a)x-:-I
Solution
(a) i, EI
Properties of analogous column
T"he length of the aralogous column section The section of the beam is non-prismatic. l/EI as shown in Fig. 6.1 lb. Let the span of rhe beam. its width is equal to determined as follows: be çan section t"f"*n unity. The propertiet
";;;i; :
"itftt
rl Area A = I x3 +! x6*
,a
*3 :
.: from
A
lox
lOrt2-2x3)x
-_ 3x(4x12-3x3) : 2x(3xí:ñt
l:225units
l'e5m
Sunits
c.g. of the analogous column from A
v:
3x1.5+3x6+2x10'5 =
I
axes Moment of inertia I, about the centroidal
5.44 m
:
äþ
*n*(0, -2,)-z(o'-r,)].
j : +e.s units
{i*_ rl
ri-
''¡
ill
:{
,t
t74
j!__:
c. g.
of2-b-d:-3 fromA =
{,_) ,
{..
-;
Areaofsegment
e-4-d'
(", -,.j
(-'l L,1
-
c.g. of e
(c)
--d
'î¡Ei
point load (Fig'
t388.Sunirs
r',g. of the analogous loading from A
= (225 x 1.95+ 495x6+ 150x 10.05+131.25x2+ l16.68x4.08 + 208.36x6|t3+ 62.5 xl0)/ 1388.8
l5ounits
= 10.05 m
from A
to the Analogous loading due
Bendingmomentat
(3L-za)
= Pt + Pz+ Pj+ P4+ P5+ Po+ Pz : 225 + 495 + 150 + 131.25 + 116.68 + 208.36 + 62.5:
768t.67 : ).JJ ln
il
6' I I d)
r
.':..'.t.Hut moment
3BB.8
of the loads about the centroidal axis
M = 225x_(5.M- l9S; + 495x(5.44-6)+ t50 x(5.44- l0.OS¡+ i:, ,,i.+t, 131.25 x (5.44-2)+ 116.68 x(5.44-4.08)+209.36 x(5.M-6.73)+ ,:. 62.5 x (5.44- l0) ,t,,:, '.: 127 units clockwise r, ,,,',' ' Thu ¡* stress at any section can be determined from the following equation as shown
B =29'17x3 = 87'5kNm
:::lìi;rfri
(
l-' = * =87'5unis .'. : and ordinate 2- b = 43'75units 2s'17 x5= 145'85kNm c Ë"îot"rä".""t" ordinate
i'j \, -'
1..
145'8J ordinate 4- c -- æl -
(.r (.. (.
4
=
Net loading on the analogous coltimn
P
I
" ¡2 rl*I= = loxIt¡*l2-2x-' 1.5 l¿
I
i
t
P3
(d)
6m
--.^2
l1<
BEAMS WITH VARIABLE CROSS-SECTION
METHOD COLUMN ANALOGY
'i't'-J
Bending moment
...
ordinateTordinate
.'
{"
at D = Q25
d
l.lil'tlsbrc e.l. til,:.!!::',ilr
; & * P*My _ 1388.8 !::::= -- l27y = 173.6 + l-l4y -+-----rA- t* 8 nt.47 ,:: TABLE 6.1 Computation of net mirrnents
72.93 units
kNm
=31'25units
6- d :41'67units
: a-l -b, P¿ = :-z xti'5 13l'25 units :2m o*g:ofa-l- b'from A
Area of segment
Area orsegment 2.- b
c.g. of
4,', : l\lryE)"'
:
I 16'68 units
z [email protected]::2'2P+32-b - c- 4 fromA = -x-@:S*nS3)
Areaorsegment4-
c.g. or z
g. oi 7
c
- d-7,'r: (-4?ë)"0=
-d-
- d-
e from
e'
: I x3x41.67 :62'5units 2 A = lOm p-t
:
4.08m
llr
the beam shown in Fig. 6.12 a ar irs eirher end and also Êffy.pver factor from A toB, and B to ^A,. .tltfftrgss -of
208'36units
.b-c- 4 from A = +-ffi+5
Area of segment 7 c.
- c-
: 6'73m 28 6
4.667 from A
'177
PORTAL FRAMES WITH ONE AXIS OF SYMMETRY t76
or.
Mo --b
: f
= -2.et E] e^ L L -" [9-t'o.s3l J L6
... Taking clockwise.rnoment acting
"AB i.27 = ".^:2'91
t
EI
@
A load
fb) ,
EI
6-
or,
Ms
T'
EI
L
ze.sz )
L6
._
n,
: E r 0^ l, * o.rorul
:
4.35
T
oB
(:.i:e")(+.ooz) 29.33
EI
M^
:
EI
_r"";;:setl: _z.st T n [:
EI
Again. using the same sign convention,
EI
[å
29.33
EI o" ^ [a * 8"3.3321
= o"6
EI u^
=
(3.3308X:.:¡)
kB : $58: Mo : pressure at A
MA = pressure at A -PMv M^ = ----u^AI
:
Pressure at B
Fig.6-12
so *(+-øote^ )(a.ooz)
or,
is applied at the ríght edge of the analogous column (not shown)-
oR
Letusapplyaload0oattheleftedgeoftheanalogouscolumnasshowninFig.
MA
0.4
and carry overfactor Cun from B to A
icl
Stffiess factor kn and carry over factor Cnu lrom A to B
6.12c.
K,
on the beam element as positive Tor both ends,
jA
lx43 43 r- ---\2+ !!-1-*4(2-0'667)' I- : ;I - !-+z*(z-aøl)z : 29'33units 2.67 'r 14.22 + 5.34 + 7 '10
(b)
0,
M" :
oA ,oA i-------ôjjg
'
Sti-fnessfacty
eA
.t.
0-7426)
:
7.27
E'
:
e^
e' *
067
"^:11r',
6.9 PORTAL FRAMES WITI{'ONE AXIS OF SYMMETRY Example ó.ó
Thus, kA = 7.27 \l
Analyze the portal frame shown in Fig. 6. l3a by the columnanaloly meihod.
Ms
=
pressure
at B
eA
-6ET
(+.oozoo)(-::rrr)
29.33 -EI
: Ero^^ lt_o.rrl L6
I
Solution The frame may be treated as fixed at A and D and having a moment of inertia equal to D on rollers. The
zero. The fraìne is made statically determinate by placing support
l\
COLUMN ANALOGY METHOD
rTt
i)
PORTAL FMMES WITH ONE AXIS OF SYMMETRY
179
20kN/m
60:7:s
pr *'fag¡lo I el l"x+ lus'lu" Ier t' t * ; ¡Oí :
:
EI
Tl"
stress is compressive throughout and varies
,noggtng.
'Stressat A:
rinearry. Hence, the end moments are
0
B:
(b) X-SEC.
It.9ty
-=r-
_7!.47kNm
C = -7l.47kNm
OF ANALOGOUS
D=0
COLUMN
- The net bending moment diagram can now be drawn by subftacting the above varues from the M* diagram, and is sh'own in Fig. 6. l3d. t
Example 6.7
Analyze the gable frame shown in Fig. 6. l4a by the column analogy method.
Solution
(a)
¡nt -'l (d) BENDING
fron*
MoMENT kNm
Properries of the analogous column section (Fig.6.l4b)
A : 2xlx5+Zx'¿ l^ x ?-2ll : l7.2ll
(c) Ms /Et LoAD Dl AGRAM
y
Fig.6.13 analogous column is shown in Fig.6.l3b. The wi
points A -
D
is given
by
sin
2x5x2.5+7.211x7
:
t7.2ll
4.385 m
6 " : =1'. ?2n= o.rrr, .or o: .r.rr1 :0.832
o
Thg'inclined member BC is shown-in Fig: 6.
l4c.
Let width
=
b
:
r2-
r"zEIEI3EI = .9 x62+2*
l_*61
I
=iof
f o rt sin2e dr 0
0.5552 i. i r;^ : *r¡nre =!x7.2t¡3* 12 2 --" " 12 :-r4.812 unirs
= -2 x2Pë *g =907-'5 3 zEI EI lv!=Pe*, €r=0, M*: Per, er:6m
Load on the column section. P
Similarly
[',
of member
nc
:
ScoUt
=
j
x7.2ttr
"ry
:
l0.Bl
unirs
Moment of iàertia about the centroidal axis of tire frame
Stiess at any point (x, y) on the section
P M" M. o : A+ I-** loy
*
.L
of member BC = 2f bdr (rsiri0)2 = 0
The free span moment diagram is shown in Fig, 6.13c.
.'.
from AE
r¡nits
I,:2 -sJ x lx !-+ (6.4)
2xs
x(4.385
-zs¡z +2x4.Btz+2x(+ x7.2¡)x2.6152
20.83+ 35.53 + 9.625+49.31
-¿
=
115.30 unirs
COLLJMN ANALOGY METHOD
r80
C:L,OSED FRAMES
ã't r lx 5 x 62 + 2* ,r]*
r------t__ r r r_IN/m-r--Ì L 25^k
WITH ONE AXIS]OF SYMMETRY
r8l
7.2t1)x32 + 10.81 = jgt.63units
sl,,tically determinate structure can be obtained in several ways. By introducing a
it('A and a roller at E result in a symmetrical free span moment diagram and thus
ët'ical loads on the analogous corumn as shown in Fþs. 6. l4d. e and L Load on BC Pr
--
: 'l'otal load
E
i
Z25
:
t 0B l
.65 downward
load on'CD, p,
P -- 2163.3
= ì-:l? . 3.7-5 rn , c,6 ol load P, fionr B = :-5i, t6 t6 , ,ceeentricitv along ¡' axis. e.. - 2.5 + (5 - 4.385): 3.1 l5 rn
COLUMN
(o )- 6ABL
x t.ZU
r
X-SECTION OF ANALOGOUS
kl2mJ
^,
i
FRA ME
i..,rtr:cielrtricity along x axis, e*
ffi:
i\,1-:Per
'f'h<: stress at
:
=
2163.3
0
x 3.115 :6738.7 andM..:g
an;, point in the analogous column is given by
PM" = -_+_iv
A I.'
:
!
25 kN,/m
(e) Ms LoAD
DIAGRAM
¡_!_63.i
6738.7 x
t7.2u1lsi
y-
= 125.69
+ 58.44 y
sectlo.n¡ B, c. Drvíll be in compression and A, E in tension due tothe action of the l.riofiìentM.\ alone. The computations are shown in Table 6.2. .,.,
TABLE 6.2 Computation of net.moments
,/y P2
-
4.385
0
12.s.69
0.6t5
0 450 0 0
t25.69
1.6ts
irsorN fruo*N
rsorruf
-
(d ) STATICALLY DETFRMINATE
ìi6.IO
F RAMEf
't'
LOADING ON ANATOGOUS
125.69
t25.69 t25.69
-
256.26 -- 130.57 35.94 t6t.63 269.70 395.39 35.94 t6t.63 256.26 130.57
-
t30.57 161.63 54.60_ 16
t.63
130.57
CLOSED FRAMES WITTT oNE AxTs oF SYMMETRY
.
i:r Example 6.8
COLUM N
Fig. 6.14 Gablb frame
0.615 4.385
-
iiir .',.fi;11tte
the closed rectangular frame shown in Fig.
6.
l5a by the column analogy
COLUMN ANALOGY METHOD
t82
CLOSED FRAMES WITH ONE AXIS OF SYMMETRY
183
' )s3 + i-| x5 x(1.25)2 : 2 x I x :j2x = t0.42unirs 122' .: r*:2 lr212 " :. :+ 2x(t x2.5) x 2.s2:41,67. vnits
l_"
62.5
62.5
Free body diagram of the released frame is shown finalogous column section.are,shown in Fig. 6.15e.
q2.5
F .20kN/m , 2.5m , 2.5m (o) BOX FRAME
(b)
Ms
p:2*f - I *z.sr62'1 _.- 2 +2x2.5x62.! _ __.J+ 62'5"t_rx)x_. 3 2 : 52.08 + 312,5 + t56.25 _ 156.25 : 364,58 (pult)
DIAGRAM
c
çTr-rl4r)
(c) X-SEC.
OF ANAL060US
\
62.5
(d} FREE BODY DIAGRAM
COLUMN
125
M,, = Pe, = 2 x 26.04 x -_- :65.10 ariti- clockwise 2
)i
50
ls0
the
).s
62. 5.
F
in Fig.6.i5d. The loads on
v,
62
My * 0, (=Pe* but e* =
(viewed towards positive x- axis) 0)
E
,.+ :r:.:l
Net stress at any po¡nt
62 .5
+:
ls6.2s Ê
PM= _+-v A I*-
ilitl
364.58
,1,
65.1
l0 10.42' l'Í '',¿ ut:t='.|ë : Stress at A, C = - 36.4s + 10.42 .z::t::.t:
i.
stressat
.'.
D, E : -
Net end moment at
\.
É.
36.45
- E#F
:-
z.o,4kNm
(sagging)
44.z6kNm (sasging)
: - 62.5 + 28.64 : - 33.g6 kNm D,E:-62.t+44.26 =-l8.24kNm
A, C
Alternative Solution
(e)
LOAD ON ANALOGOUS
Let us make the franle statically determinate by providing hinges at each of the four
COLUMN
çorners. The M, diagrarn is shown in Fig. 6.16. It hig. 0.ts Box.frame
p
Solution Let the frame be made statically determindte by making a cut in the member ED ai F. Under such a condition, thg M" diagram is shown in Fig.6.l5b. It is hogging on sides ED. CD. The load is.a pull as there is tension on outside. Properties ofthe Analogous Section (Fig.6. I 5c)
AreaA =
l{ence. load is a thrust.
2x:I x5 + 2x lx 2.5 : 2
lOunits
= :, 5.* t?1 * ? * s * 62s : 2232
l\4 = 156.25 x 1.25 -
104.17
x
26o.42uniits
=
65.1 anri_clockwise (viewed towards positive x_axis)
M, =o Netstress atanypoint
1.25
creates compression on outside face.
*
: ry+t l0
95't u t0.42'
r84
'..,
COLUMN ANALOGY METHOD
,, '., .:t
lv
..
,
PORTAL FRAMES WITH NO SYMMETRY
l8s
-
,ílJ¡1r,1',i
I
',1i',rt:Ç iifÌi'.r': 'l
:'.lr',:rI
:.1i '
I
?gk N
Ms
175 DTAGRAM
(d)
Fig. 6.16 Box frame - alternaty'soltr.:on
A and
c=
-26.04
*
7
The net moment diagram can now be obtained by algebraic addition of the above streqses (end nrontents) with the free span moment values.
J ii v
ò. ¿.
ó.tr
DTAGRAM
65:!
"_!r_2I : _-.85 33. kNm (hogging) t0.42 Stress ar D anci E = 26.04 - 7.gl : 1g.23 kNm (hogging) srress at
Ms LoAD
poRTAL FRAMES WITH NO SYMMETRY
l*-r+(7-Ï)
Example 6.g
(b)
Analyze the portar frame with fixed supports shorv¡¡ in Fig. 6. r7a by corumn anarogy
X-S€CTION
d
{e)
OF
Mr.zEl
ÂNALOGOUS COLUMN
method.
Solution-
(d
!
Properries of analogous colunm ¡Fig.tt. ITbl Taking nìotnents abour the line AC
40t.25 -,r*7x3.5+lx4x7 X=r, _ -,'-: , ---*-''urtr lx7+!x7.+lx4 l4.s =2.7gm 'faking
tìtoments about a horizontal line passing through Â.
I
lx7x3:5+l 2*7*7*l*4*5 i=r@=r4J=4'76n (
(
A= lxZ*1*7+lxi .2 .t*: lx73
-rt +
t. =
62.82
=
I
.
lf )
LoA.0s l¡.j ANALOGOúS C0Ll:l'4N
'' 69
l¿.Su,lit,
{c) 1.. = I x
lxTx(4'76 -3.5)2 +!x7'x0-
2'
f*,
:
REL.EASED STR Fig.6.t7 Unsymmetrical portal frames with flxed ends
i x27821t x4x(7 -z.tt,lt* ) "l*
I x 4 x (7 - 2.78) (s - 4.76;ù+
= 34.21
J,
*t *(3.s
-2.7Ð2
* z x (3.5 - 2.78\ 0 _ 4.76r + txTx(-2.78)(3.5_4.76)
]
="
t4t.4s
t86
COLUMN ANALOGY METHOD
'[hè frame is made statically determinate by removing the support F. The released structure.is shown in Fig.6.r7c. The M, and M/EI diagrams ur"iho*n in Figs. 6. 17 d and e. The loads on the analogous column,are shown in Fig. 6.17 f and are tabulated ¡n Table 6.3. Since the M" diagram causes tension on the outside of the frame, the load
PORTAL FR.AMES WITH NO SYMME-I'RY M*.
lvls
t87
/ EI diagrams are shown in Fig. 6.lgb
c. The anarogous roading is ¡. in"n9Fig.6.18 d, and is taburated in Table 6.5.andSince rhè M, aiug.ur.uu.". rression on the outside of the frame, the load p represents thrust.
represents pull.
TABLE 6.3 Load on the analogous column P
-
Ð,
M*,
:
M*
€"
600.00
-2.78 -2.78 :1.62
1225.00 153.12
Mr' : M, - ¡4*
I
Y-
'¡2
r,,: '
('
ró68.0 3405.5 248.0
2058.0 t543.5
1.262'.24
-
325g
s32t
-
343.0
Ð 3258.5
## : ß71
"
_ tzsl*U
5321
_:
T
r7. a6 frz.as (o) RELEASED S1R.
34.2t2
- AtAs
62.82
247.8 D
= 54.54
+I' .: ut.4s _ ry! : 62.82
TABLF
51.69 415.6
+
S",l
:3s46
/
ez.Bo
(c)
The net moments in the frame are shown in Table 6.4
.
b.). MS DIA6RAM
600
=
î
r"! _
-3.43
-
=
>,5321.5
I'u- y"' t.. :
I*'=r*
M^ = Pe,
-t978.t2
t
(r.
Pe^
M,
e,,
C.!
Ms
/ EI
(d)
LoAD DIAGRAM
Moments at points A, C, E and F
/ / LQAD ON ANALOGOUS COLUMN
Fig. 6.18 Alternate solution
1
Ms
v
Point
m
P,ã
m
,MV
--:-x
ir
M*.
----y"
A
-
2.78
-
4.76
-
475
c
-
-
2.78
E
I
F
-
t75
-1978/14.5 =
-
-
36.41
-
80.27
-
80.27
36.41
2.24
ft
-
36.4t
r
1.76
0
-
Jõ.4
t2t.85
t
M=
Mr-Mi
P
Ix
Iy
!.
TABLE 6.5 Load on the analogous column
Mi
-
21.85
-
t72.OO
-
JÜÜ.70
-
tó.40
80.95
-
135.70
-
J9.30
ðu.y)
+
66.40
óJ.óU
-
78.t
-
e.,
207.80
t.88
415.60
l.0l
2.24 2.24
54.69 900.00 600.00
66.40 78.15
ltó
I
(b)
€*
M, = Pex
-
1.62
2.24
-
2.78
- 390.66 - 419.76 - 88.60
o.75
-2502.00
2.09
-
-2.78
-
1668.00
M*:
Pe,
465.47 930.9: 675.00 1254.00
t22.50
-
87 70
939.92
Alternative Solution
The frame can be made determinate. by .statically M4, R antl R." as 'is .treating redundants, thaf ìs, a hinge introduced at A and a roller ^ Fig. 6. ig a. at r, u, ,Èo*,ï^¡n
M"
M._M,+=
g3g.g2
+
4287.7* 34¿l
t4r:4i
1976.9
f,*fi
COLUMN ANALOGY METT{OD
Mr':
M, - Mr
*
: -
4287.7
-
x
g3g.92
PORTAL FRAMES WITH NO SYIVIMETRY
34'21 62^82
=
-
t89
4800
'l'he nct rnornents in the frarne are computed as shown ín Tabre 6.6
TABLE 6.6 Moments at points A. C,.E and F Point
v m
Ms
P/A
-
C E F
-
2.78 2.78 4.22 4.22
4.76 2.24 2.24
-
1.76
0 300 0 0
I50.2*
08.67*
t50.2 t50.2
08.67 6-s
150.2
6S
M:
M;
M" --i-y
MY
---;-'x
Mr-M
I,
Iv
A
-
X
m
-
t72.sI 81.2
86.4
340
81.2
-
ç3.6
66.4
-
-
COLUMN
-40
-
78.6
66.4
78.4
** -4s9:;#1Ð--r0867, .iy.,6+_!l!)= -
;#=r502.
(b) X-SEC. OF ANALCGOUS
86.4
r725
87.5/
Et
'rhe two altcrnative releasecl ,r.ì,.,u.., give idenrical ¡-esults ivirhin the round-ofl.
ul'rorS.
lìxnmple 6.10 I
.
Figure 6.t9a shows a portar.fiame u,ith hinged supptrrrs moment diagram by colurnn analogl,.method. :
A and D. Draw
bending
(c) Ms/EI
The franle has trvo hingcd suppotts and. theretbre. one dègr.ee . of irtdeternrinacy. The hinges in the anàrogous cor.nrn rviil havq ¡nn,it. area .rh'ough trre poinrs
shown in l- ig.
also.
6-
A
l9b.
A .-
D., rhe axes ,yrlrn ¡,
and
Such an axes
:c
and 1..: a
;;;;;; ;iö;i; #. ffi,
'"rþr"r"ntu ¡rrincioal
axe-s
and,
statical passing
mid se*íon
,1,".;fb;;:';:;
:-"0.
AB, I. =
lbds(Ssin4:5)' 0
t.
t.
lbS-ds(sin45)" 0
BENDING MOMENT kt-,tm
LOAO D|AGRAM
Fig. 6.19 Unsymmetrical portal frame with hinled ends
'""
r*=!a
Similarly, for BD.
t-
f"or segment
(d)
fzsrN
Solution
as
,E
Total moment of inenia
g "663 -
¡" =
blr'
: ll-t
The frame is made staticaily determinate by introducing a roiler at end D. The M, diagram is shown in Fig. 6. l9c.
P: : x87.5 x7 : 306.25 ,2 : 2.47 m en : !.5 sin 45
Total Load
PM o :. _*--*y Alr
=
M"
îY
srnceA:€
,tq0
û {--)
...
tt
COLUMN ANALOGY METHOD
. q:-_-;-y 306.25x2.47
l"j-
(P6,2a):
o:
or'
(P6.2b):Kn:
diagram is shown in Fig. 6. t9d.
TABLE 6. 7 Net moments at Point
(.."
KA:6.%î,
A,, B, C and
M*.u
Ms
v
The nèt bending moment
AB
M: Mr- M¡
0
0
B
4.950
0
c
2.475 0
87.5 0
D
.(
0 32.7.0 16.35
0
:(., :
t,,
6.1
Determine the fixed end moments for the beanns shown in Figs. p 6.1 a treating the beams.as (i) simply supported, and (ii) cantilever,
(P6.la): lf'ln=-
: MÅ :(P6.lc) : Ma:(P6.lb)
!
(P6.td) :
Rn
'- p¡2.
MB
96 0.00725.wL2,
MB
55.40ÌcNm,
MD
ss.3e kN1,
Mc
7.26T,
¡
COon:0.68 .
COui=_0.58
ZI
f \
-
d. Draw bending
momenr
(b)by removing support Din frames in Figs. p6.3a and F6.3b, (c)åy removing support A in frame in fig. lO::A.
PROBLEMS .'
+,
(b¡
Analyze the portal frames-shown in Figs. p 6.3a_ and shear force diagrams by¡:eating: (a) the frames as simply supported,
71.t5 0
=
'. 1 F 6r Figs. p 6.2
0 32.70
K,
43s
1t
(o)
:Mi A
co.ta :0.61,
I 6m
, 4m .,-
kNm
I*
6e+, L
L__-J-2l
D
l9l
co*:0.42, Ka :
6.616 y
The stresses can.be computed as shown in Table 6.2.
t
PROBLEMS
(P6.3a) _
d by
(P6.3b)
= ie,u ll* Mo = t4t.67 1y3, kNn
RDx:, IZkN MA :-147.28
i:
RAr
:14.;s íu1,,un=
Rny
=7.s kN1,
83.33 kNm,
Mo=__72.94kNn,
-
kNm
=1*L2 96 : 0.024 wL2 : 55.40 kl,{m : 332.88 kNm
T I I
8m I I
I
_t_
( '','
+
(ql EI = (o)
C0NST.
(b) EI=CONST.
20 k
(b)
N/m
25 kN/m
(
B 2t
75kN
ìr-r
c
¡*-?-,Fl'5-1.1.5 l. 2m -¡ (c) t"igs. p
( (
t
r/r;l.','" 6.1
Determine the stifftess and carry over factors for beam erements moment of inertia shown in Figs. p 6.2 aandb.
¡---_-JqL_______r ¡d)
(c)
with
variabre
Figs. p 6.3
$ï,,,
,.,
COLUM}{ ANALOGY METHOD
þj
(P6.3c) Rtx
gr;
Mt.t¡ = (P6.3d) Rt.v :
H'i ü, Ë.
-272.4 17
kN'-+,
R,ty = 200 kNî,
Mo:
258.88 kNm,
Rty =g,48 kN|,
M.4=
87.41 kNm,
\""
II,
kN-+,
Mr,
tr
t.st
l
B
.
T
T
(' i
5m
INFLUENCE COEFFICIENT METHOD
c
þ----l3ia----i
\
{
+
I
3
E
5kN/m
(o)
10
kN/m
l_
rT-T-1_ ,1-î-T-r1 F--g!!--t -!L-l
Analyze the frames shown in Figs. P 6.5 a and b and draw bending moment and shear force diagrams. (P6.5q) M^ : -21.78 kNm, Mor, : 17.21 kNm, M<:o : 30.54 kNm,
(P6.5b)
MD : -l I;08 kNm Ru : 12.5 kN-+,
R,cr
:
INTRODUCTION
The use of flexibility matrix provides a systematic method for the analysis of a structure having a high dégree of static indeterminacy. The matrix approacli is more convenient for use on computers. There are two flexibility maüix approaches in vogue:
(i) (ii)
(b)
Figs. P 6.4 6.5
7.I
r
8m
rsrnTm
I
= 207 kNm
n), Ix: 170.9, Ir: 576.75, I.tt : 62.72 uníts MA -- -28.86 kNm, lMu : 329.46 kNm, Mc. : - 32.22 kNm, M,, : -21.66 kNn, ME = - 12.14 kNm
çr
(.
seYen
klììn
(P6,4b) c.g.from ¿ = (5.458 n, 5.14
I
i{.
CITAPÏER
dnalyze the closed frames with one axis of symmeûy shown in Figs. P 6.4a.-b. Draw bendingrnoment, shea¡ force and thrust diagrams (a) by intioducing a cut at the midspan of top girder of each frame (b) by inhoducing a hinge at the corners ofeach frame. (P6.4a) Thurst in AB : - 29.95 kN (compressive), Moment at the nidspan of AB : /,5.80 kNm (tension outside); , Mce : - 6i.8 lNm
1,,.
:-/
61.53
Mtrt;: I4l kNn,
ä{,) f
:.
16.67 kt{T,
Møt
= IÛ|kNn,
system approach member approach
In the system approach, the flexibility matrix is developed directly for the stn¡cture as a whole, whicli is also known as the inJluence coeficient method. This method is good 'for small sûuctures only. In the member approach, the flexibility matrix of the structure isldeveloped through the flexibility matrices of the constitue¡t elements of the structure and using the transformation matrices. The member approach is more amenable for a cgfhputer than the system approach, and can be used for solution of very large structures.
Mcn: -100 kNm
In
th is
chapter only the system approach is discussed. The basic steps of the influence ¿¡s disct¡ssed for the flexibility method in Chapter l.
coefficient method are the same
Each displacement component can be determined by using the Castigliano's second lt¡eorem or the unit load method. Both these methods involve complicated integrations' .and are prone to errors. There is a graphical method which can be conveniently used to integrate expressions to determine the displacements. Once the various displacements are known, a compatibility equation must be written for each redundant in the structure. These equations will be in terms of all the redundants. This implies that the action of one redundant will influence the'displacement associated with the ðompatibility equation of another redundant. Thus, the redundants can be determined by the solution of all the [þepr simultaneous compatibility,equations,
(
t ( ( (
( L {
. Figs. P6.5
of applying each of the redundant loaås, a unit load is applied in the cli¡ection -lnstead dfeaèh redundant. The {isplacements due to the unit loads are the influence coeflicients.
The actuai
displacement
due to a
redundant
R is simply R times the
: -t-,:: '
.
i'r',:'ìtù!1;;
trryi'
INFLUENCE COEFFICIENT METHOD
..
FORCE DIAGRAM
195
displacement due to the unit load. The compafibility equation (1.5 as derived in chapter I) can be reproduced as ;
_
{^-} + tFl {R}
or,
-Á"o*.FR:
7.2
:
{^.}
(7.r)
Â,
STGN CONVENTION
O
In the inîluence coefficient method, right hand side axes system is used as shown in Figure 7. l. The y. - axis is aligned with the longitudinal axis ãf thu member, and .ùe y, - axis is perpendicular to the piane of the paper. It may-hand be either upward or downward depending upon the direction of yz axis. rne right side ruie *ù u" stared -
UPWARO
J. TO
(o)
OOWNWARD
1
TO
PLANE OF PAPER
RIGHI HAND StDE AXES
SYSTEM
xl
as
follows:
O
PLANE OF PAPER
Fingerc of the right hand are pointed towards positive 12 - cuis and are curled towards positive rs-.æis. The directioi oJ'the'thumb gives
pos itwe direc tion
the
of y.,_æis.
(b) (-
,.
Y1
POSITIVE
Xl -
MoMENT
'Y1
sl-__+t3
(C)
( (
R. H. S. AXES
t
F,ig. 7.1 Right hand slde axes system for a member Figure 7'2 a,shovs t{e
ili,,
(
i,lli;,
direction of y, - axis according,to the right hand side rylitive rhe foree tunctions are reprer"ot"d by *, , *rr"i" trtË ;il;*þr i mayöe l,
.:;
,r
(
{.j
Fig. 7.2 Sign convention for force funçtions
..
A statically indeterminate structure.is made statically deteÉniinate by introdueingone or more releases as shown in Fig. 7.3. Corresponding to a giùen release, appropriate biactions or pair of redundants p, are introduced as shown in the same figure. The i subscript i represents bending moment, shear'force or axial frrce, as before.
(' (': (,
,:
POSITIVE,
x3 L---L-- -Jl_ x3 lY¡ r e.HSlOtt XZ -SHEAR (d) POSIT|VE X: -THRUST
I = bgndingmoment, 2 : shear force, and -J = axi4l force.'
f;lå:î
Once a structure is made staticaily determinate, the nexí step is to determine displacements or discontinuities at a release in the direction ofthe redundant or bi-actions due to applied loads as well as redundant or bi- actions. Diòcontinuities can be evaluated 'if the various force diagrams are available. These are classified as :
(D x - diagrams (ii) h-diagrams
Bending moment x, is positive if it causes tension on the positive y, side; shear force x, is positive if ihe positive face of an element moves alông tt pJritiu" y¿ ¡xis
and the
7.3 FORCE DIAGRAMS
negariie sidè iirove.ç ar-o,,g th o¿t^rrr; " axis; the axiar f@e rension in rhe erement. -rn"" poiiti"" i; r*"¡ngl-iu""rî.
x, is positive. if îioì"i in Figs. 7.2b,
x
-
:
:
i.
The force diagrams in a released structure due to applied loading are called x,o diagrams, where the subscript i may be 1,2 or3. The subscripts 1,2 and 3 represent bending moment, shear force and axial force; and subscript o iepresents external loading. If shear or axial deformation is ignored, the corresponding force diagram need not be drawn.
diagrams
i0ø
INFLUENCE COEFFICIENT METHOD
197
-FORCE DTAGRAM i:
:
.
;]4"ì:
lol
'
----l3t-__ -1:
P.=1 I
¡-r'¡--r fl L_+t_J (o)
MoI{ENT RELËASE
(b)
l---f
1
I
Pr=t
SHEAR REL EASE
I
l-+-l
I
.-,.i,,':i.
Þ3=l (c I r¡1Pgtt
,i¡tt''
,,lir, , ' : , whore
RÊL EASE
ll a, :
_
I
degree of static indeterminacy
,if'::i,.'
effect or nexwe' shear and
,Ë': rff":*:":""-'.:"*i.':T:-*,j::T:,ffi3r" can be easily determined as follows ,;Ìl'i... O-eformations
æ
:
:
.¡;1'. llt
,i.,'
(d) Fig. 7.3
,irr-,, .:: I ri..
CUT RELEASE
Release discontinuities
rl¡*:-fhi.¡hr¡.dy¡,1Þ¡ht$l*fhr¡h:rdy, *J-;"c *J
,
J-=EI
.
0.5)
AE
Once the bi - actions or redundants are determined using F;q.7.1, the net force ,,.;:l, ' ,i.lr can be uç lv¡r¡Pulçq computed using uDurË, the u¡s squ4nvl equation : llr the ulç Ju structure uçtul ç ldr sectiogin x at ar any stiçllut ally ,:,-:. l l r_, {x} iì,.,.r ' : (7.6a) '.',.. ' {x} : {x.} + {H}r {p} (7.6b) = xo * HT p i:,,i,.. .,or, ''. :.. ..... . , where {xo} : force values at the section under consideration due to the extemal loads ,,,ìrr.ir.. = {H}r hi¡ values at the section under consideration due to ,,r,tlì.'-,:: ','.' :' unit bi - action {p} : bi-actions ,:..i:]i.:,ii,r.,,::
t.- ài"g..at. : The force diagrams in a released structure due to unit redundant. or unit bi - a*ions are called n:r.þ giug.To that is force i ca¡sed due ro unit redundant j. The subscript i represents thË force fi¡nction and may te ,equal * i'à, * ,. The subscript j'represe¡ts the redundant. Total displacements or disconginuities due to applied loads in the released structure conesponding to bi - actions can be calculated as :'
4" =J (
_i
/',
,bJ
(.,
(1,2)
The deflection due to unit load method can be written as
-acÌions
,
,,
where,
",j,, __ fh¡;h¡¡dy3 J__nI
-
'Ihe flexibitity rnatrix [F] is a square matrix consisting of influence $*
elements,
J
:.'
(7,3)
:
!
Total discontinuity due to bi _ actions in the releàsedr'strucfure co[esponding to can be calculafed,4"
ír (".
7.4 GRAPHICAL METHOD OF INTEGRATIOFI
h¡¡ x¡o dy3
: Jl'Mmds EI : bending moment due to applied loads : bending moment due to unit load
^ M m
(7.7)
Mds represents a small element of the M-diagram and m represents the corresponding - diagram. Hence, displacement  can be rewritten as :
or
coefifrcient
)
a=
Area of the M
-
diagram x Ordinate of the
m-
diagram at thc
cg.of the M -diagrarh
EI
(7.8)
Alternatively,
('
Á:
A¡eiofthe m- diagramx Ordinate of theM- diagram
EI
at the
cg.of the m- diagrarn
(7.e)
r; i-_j (..
.j
{.
:
í, ,)
,
{._;
TNFLUENCE COEFFICIENT METHOD
t98
ln case a structural member is non-prismatic, it should be divided into a sufficient . number of parts so that the M - or m - diagram is continuous and E I is constant in each part. lt should be noted that M need not be a bending moment diagram. It rnay be any forcing function diagram such as a axial force or shear force. Table 7. I gives the values of 8q.7.8 or 7.9 for common shapes of force functions. This table can be directly used for computing displacements once the diagrams due tci various fÒrce functions are drawn for a given structure. Thus, Eqs.7.3 and 7.5 can be evaluated directly using the Table 7.1.
I ...
.L
[:'
Table 7.1 Evaluation of volume integrals
t(',t
I
nY? tf i' ,t ^/\7--+------çrì
m, m, dx
0
-P, o
..72ï *Ë
à>h
(".
(b)
I
Pr-l Lo(Ze
+t
Y2
)
h11
I
L
h2t
(c) Lc(2e+ f
--6
)
l6
h3t FORCE DIAGRAMS DUE TO pl t =
.Yt Pr= I A l' ,, ,n---.,*--€'rE /zr7>
(! + f ) 5
Ld(e+2f
I
cg
Y3
. dg(L+e)
>>þ
/\ 1
(
() (
(.
L
hzz
7.5 ILLUSTRATIVE EXAMPLES
(d)
n12
h3z
FORCE DIAGRAMS OUE T0
Example 7.1
v^=
Develop the influence coef,flroients for the beam shown in Fig.7.4a.
1
t.
+ (
o
i
I
Pr=1
H'< Æ__-___-__^ zÐ
ht¡
hzs
hg¡
/?Ð?
o
Solution
The beam is statically indeterminate
to a
degree
3. It can be made
staticallv
determinate by inftoducing a hinge at A and a rollèr at B. The axes ancl the three bi actions p¡ , p2 and p3 are shown in Fig. 7.4b. Let us draw moment (h, , ), shear ( h, , ) and thrust (h3¡ ) diagrarns due to each of the three redundants (i = l, Z,'3¡ as showiin
Figs. 7.4c, d and
e.
.Ths influence coefücients, that is, discontinuities, due to unit
bi . actions can be evaluated using the graphical integration technique.
(
e)
FORCE DIAGRAMS DUE
Fig. 7.4 Fixèd beam
T0 p3 =
1
i.li
''-
'a
INFLUENCE COEFFICIENT METHOD
GRAPHICAL METHOD OF ÑTEGRATION
201
..'
:t
{n
h¡¡
hrrdv¡
:l:
.J+"dþ.JIr*lþ
iì'i,l
L- I o l' 6ET-rA"c l6i+¡¿"c [t*
{.l.i,,
dY, h, h, dy, t r;îr, ¿r, * ¡ --eE J¡ !iEr *'Jf A"c--J -
:
frr
fn
F:l#-#;"#.ra|õol I
!,r
å
+g*rî'$+o= 3EI A"G
- J¡
hn h¡?dy¡
EI
-L ' le"c :er * r
loo-L-l
i¡i,Lr,
L
tt'ii,'
{¡
¡
hn hr¡dY¡
'1'his is alwrys a symmetîic matrix. If shear deformation is ignored, the h, diagrams , t ìþ6eome zßro, and the influence coefficient matrix can be written
as:
#'l L' '#j
f'h F:l##ol
J pr = o =t,
r
*'J¡hrrhzrdyr,*J¡ hrr h3,dyr
Añ
AEJ
"l:'
h¡r hgz dy¡ * Jlhrrhrrdy., *-JA.G ¡ AE
A"G
c)
,ì;l
L_ :tillt-*ît*i*o='--oni-L4c
-6Er : fzt
I
(ii)
If axial deformation is also ignored, the influence coefficient matrix can be written as:
-eE
,:
tl J-lt 6ErLr 2l
(iii)
Alternative Solution fzz
h¡zdy¡ - J¡h¡z !ZdY¡ EI * Jfhzrhrrdh* Auc 'Jih¡z,AE
[.r
=
(-r fn
(r.
'
U¡-l***t"i-n3EI A"
-
dv¡
hrz
t" !¡ EI ,0:fn
G
L-
3EI
't
{.' (i
(
(, I
hr¡ f33
J¡
r
shear (hrrf and thrusr (hr, ) diagrams due to ea"ch ortr," t¡re" fray ToT"lt !!i¡ ), redundants Q : 1,2,3) as shown in Figs. 7 .5 c, d and e. The influence boefficients are :
LA- G
lxlxL EI
frt r,r, ¿y, * , ¡hrrìrrdyr-'J,' r,r,-AE-
A"G
lxlxl zBl o=ftt
f-
'12
h¡¡ h¡¡ dy¡ !l dv¡ * ¡hzt .hztdvt *¡ EI J A"G ,J,-AE
:0+0+
t_et us
agfn
(_.,
t.
The beam can be made staticaily determinate by removing the support B, that is, introducing a cut at B. The axes and the three bi-actions are shãwn in Ég. z.su.
L =-AE
The influence coefficient matrix may now be writteä as:
f,,
=-_
L
2Et
:
f
'2t
lxlxl L L ---:-=- . lx lxl- -.3EI'AeG 3EI A"G
f""
fT
L
EI
:
0:f¡z
^b
¡a1
AE The influence coefticient matrix may now be written as:
!j., ;¡":. :.\,'tJ
.ir.:tj: I.;..:. ].: ':]:
:
:
INFLUENCE COEFFICIENT METT{OD
:È0å
6¿
¡1^
r--i
l-
Arl ^',
'(-_:
-----i{
roi
r) (.
Yt qr ryz
T- , ---
i.
âA *r--T: I2
.,,
Pt\
l_ ,ro,
R.
-Pl=
'í -
Ð)
flg ,
I _L oI " EI zBt I =l-ãET l-=!= =!= o
i..' tt,'
(bl !
{,
,-
-l
I
ffi :l IL o o +l
":
(v)
I
AEJ
hil
:.:
ff axial deformation is also ignored, the influence coefficient matrix becomes:
o
(
h2t
(cl
h3l
troRCE DTAGRAMS DUE T0 .Pl =
iir
l. \/
4----------.--=1, tPr=l " ;,lZ
l
t(;
1ì '
':#f:-)
r
. It may be seen that the influence coeffrcíent rnatrices obtained by using the two altemative.solutions are different. The stress resultant however, always remain the same-
LIì-----"--=--hn
Example 7.2
o
[---_ll'
!
|
CRAPHICAL METHOD OF INTEGRATIO}J
h¡z
Determine the flexibility influence coefficients due to the combined eflect of bending, in Fig. 7 .6a. Take A,.G = 50 E I, AE I00 E I
shear and thrust deformations in the frame shown
h22
'.
Id)
t,
FORCE DIAGRAMS DUE
t---t"
í'r
hz¡ (.
:
Þ3= |
T0 Pr=1
Solution
0
The frame is indeterminate to a degree 3. A but release is introduced at the support,D to make it státically deterininate and stable. The axis system and the unit bi-actioni p, to p3 are as shown in Fig. 7 '6 c. Since it is desired to include the effect of shear and axial deformations, all the three force diagrams are drawn due to each unit bi-action as shown ih Figs. 7.6 dto f.
ht¡ I
+ h33
(e)
(.l
Fig.
{) ll
(' If
shear deformation is
L' O #J
ignored, the influence coefficient rnatrix becomes:
:
rrrr - ¡ hrr h¡r dy¡ *'J¡ hzr hzr dy¡ -J:-p * ¡ h¡¡ h¡r dy¡J EI A-c
7.5 Fixed beam - alternative solution
-# l* .=l-# #.db 'l'l
{'
The influence coefficients can be evaluated as follows
F0RCE ü¡AGRAMS DUE TO p3 =r
:
(iv)
ll
6 I ' toì+ ' rnr- " lx4+lx-+Ix4l+ jf '|*t.oJ.rhror*ffiror= ErL r
-l
f,, - J¡hrr!?dy¡'-J* ¡hlhzzdv¡ *.J¡h¡rhrzdv¡ EI A"G AE
= (-)+il -'EtL2x4x4xz*a*61*o+ 2J-- o= - 31 Er
...
I,1.-.''11 1. r,
::
,
t.
:
'.'.
i¡.S{l{
INFLUENCE COEFFICIENT METHOD
b
..
AFHTëfi L'fi
ßT-T{ÕT,*
'üF
ÍNTHG RATION
(-
T ¡
t n
F--__q*
(c)GIVEN
--;
m, IJ, l,'l
f---_-_l
(c ) rr¡¿DER
STR.
")- (d)
",-o.iå'*,
ti II li
'-,1
i*-
¡r.iz
ts^-
(e )
|
¿
h2z
r-1 P.=l r
*l
,IJ
Pi-l
.
(f)
idyt htr hzl dyl- , hrz h¡rdy, 7:J AE * Ëf A.G -î--;-
lia*tueeãse] + *JrlÉ+--:
[---¡--r
rÑ
n'
{q
fr¡
g
fn
ft hr¡ hr¡dy¡ . I¡h1¿h1¡dy3 . -- JEIJA.GJAE -T
t
hl:
h?¡
lt= t'
lrl
El
+ r[
.hz¡ dy¡ A"G *'l ¡ l;-
'*[u"o * Ju*s, f,
f,,
h¡r h¡¡ dy¡
hzr
f].0.
o=
- *(33)
fr, : ¡hrz !ldY¡ * ¡hzzhzzdyt *.J¡hnhndyt J EI -AÈJ A"G
EI
h33 h33dy3
t
ttx4+rx4lJ IOOEI'
-
g lg0.l4 IOOEI EI
I r,, fn
F
dy¡
-T
¡I
84
=
Ihe influence coefücient matrix F is
Fig.7.6 Fixed end port*! frame - force diagrams hr r
* x4x4x(a.;G)'o'$]+o+o
I I =-i-[lgo]+'+=EI''sOEI
t
: .ft : :.
!,L:tt-:rl iì,i. r1
:i¡.
(b} RÉLEASËO
FR,XI4Ë
lt
aa;9fiþlë¡¡t:¡oo'E,I and AE,='.100,.E (2I):= 200 E I l,i
tI
I
rzz
lsy*.
f,rl
t,I frrj
irr -28 -33 I F : +l EII 90.86 E4 I I
s
o
reprcsenrs rymmetryl
I
rso.r4j
If shear and axial deformations are ignored, the h, , and h, , stress diagrams will be zero. Hence, the second and third terms in each.of,the -q., expreséions would be zero. The influence coefficient matrix F becomes
.
2CIo,
INF'LUENCE COEFFICIENT METHOD
hr
-2.8
GRAPHICAL METHOD OF TNTEGRATION
-331
tltt s0.67 *+l F - EII raol L$ Alternative Sol¡rtion I
I
The frame can be made statically determinate and stable by introducing a cut in bearn as shown in Fig. 7.7 b. The axes system and bi-actions p¡ to p, are also shown in the same figure. Let us draw moment (h¡¡ ), shear (h2, ) ana thrust (h3 ¡)
eaéh
of the three unit bi-actions
ß:1,2
and'3)
as óhown
lt r-=
ft' ':
I 6 *o * " * r *+l+o+o = -ll -l-it EIL2JEI
cr12
0:û,
LL
in Figs.1'.1 c, d and e. The
(b)
- ErL2lEr J=lo*!^"1=-* :
fzz.
3i3x3* 2+3x+xsl+-l['-'r +h"4x3+ ErL 3 A"c
p^ Pz
\l
rï
r
Þ.
lF't
[f4**lt_¡_¡ /2 + -!
)
I
/7217
(cl
RELEASED STR.
'llI
f,,
/77V
UNDER Bl-ACTIoNS
o
nrr
I
OIAGRAMS DUE
TO
,,,
J
1
L;l+;r[t
x4x2f
fst
-l r ll+x+xa-rj.iãttx4x2f+-!-Lîj t flxol : rL 6 42.86 _ t28. I 3EI 5OEI 2OOEI EI
I
n1t
t--.tl
hzz '
î:ll.;lt a*:I: lQ)
l[sol* f * 8 -eo.l4 EI.'',t00EI lg0EI EI r r33
Pl
T
influence coefÏicients can be evaluated as follows:
frt : o:
207
(f
)
h3z
FORCE DIAGRAMS DUE l'O P2 =l
FoRCE ,D|AGRAMS DUE
Fixed end portal frame - alternativesolution
Í0.
p3
-
1
I
The influence coefficient matrix can be wriften as:
o .|-,, tt
[n o
-16l
F:llErl eo.t4 o L$
If
I
l
-16
F:+l EII eo o
I
42.86)
shear and .axi{ deformations are ignored, the .influence coefücient matrix i can.be
written as:
rl tt
Alternative Solution
II
l_s
I I I I
n.atl
The p-ortál frame can be made statically determinate and stable by introducing tl¡ee Itinges; õne at each support and one in the beam BC. The axes system and bi-actions
GRAPHICAL METI{OD OF INTEGRATION
TNFT,UENCE COEFFICIENT METHOD
pl , pz and p, are shown in Fig.7.8c, Let us draw the moment' shear and thrust diagrams ài¡eìõ unit bí-actions as shown.in Figs. 7.8 d, e and fl The influence coeff,rcients can be evaluated as follows :
r---ãr--l
lr
(ù) GlvEil
!,j.ot',= tl r'6
T
lf l l6l
Pr
.hI
,ln--=r 6¡ri F
'
tìAl-1Ë
I
(b)
r-tr l-!
o¡acirus
+L
hzt
DUE Tô
P¡=l
,l .1 h22
DIAGRAMS
c
0.5
T0 P2-l
hre
(f }
i. |
)
-qi
h23
I
FORCE DIAGRAMS
-
'5.667 0.5
5.679
(,
Fig,7.8 Fixed end portal frame - alternative solution
^r"
*
+' z'i.r
*
å"
EI
= fzt ¡i
AUG AE
II
.
å]. trr[å *
,
* l*
'3r
*
], e.,].#¡.;,. +' ;]
EI
lf I I ol *L-A* r "tJ
r,, = *[åtrxl+2x0.5+0.5+0.s)*os*ff.*#t] [å: ]
1.9856
:,,*[.' "I.#t'o.r*,)]*r!f-i.*".,4t _ --
.
0.t87
T0p
':t
¡",.
0.175 1.580 A"G AE EI
*['. r*!*z-i] EI
I I
_
l lr r r r 6l rf l r*4*2+1*1rgJ ,\"[rx-x4x2+6:e'iJ.æ[-à'à' I 8 2J EI
{
il--11' îr
O.25 _0.09375
1.583 0.208
'r' \l
t/ V\
1ö*
2_
EI sOEI IOOEI
lf 0.5x4¡n ,rL- O 12x0.5+l)x2-0.5*O.S*1,4* r
ll
r! rI
-*è t,
Jå t
.r.,1n -
i;-r;T-:t |'/ l,
--i
tt tt
2 0.25 0.09375 EI A"G AE
UNDER
BI-ACI:ICNS
.hlt
I
(d) roRcE
\t ç [----T
'*L.c',?',tl
P3
'rua'ffT
l-ut +
l-l
(c)
RELEASED 5IRUC f UfìE
I
l./\l tt
3.25 0.208 0.26s: 3.25 0.208 --+ 0.269 3.727 'ç1-*--:-----:-+EI A.G AE EI 5OEI IOOEI EI -+ . lflx4,^ 41 I l- I I xax2l+ .\ I tzx0'5+l)-0'5xt'tJ*+"oL-;";
Erf 6
il ¿'
hL
209
* + x z'+
i' * . ;]
0.667 0.25 0.0937
EI A"G AE
= -ftz f33 : f,,
0.661
EI'
by symmetry
The influence coefticient matrix may now be written as:
.GRAPTITCAL
INFLUENCE COEFFICIENT METHOD
2t0
METHOD OF INTEGRATION
2tt
lt.tzt r.xe -t.saol
: 1l
s.67s
EII
"
Is
o.uu,
I I
ttztl
If shear and axial deformations are ignored, the F matrix becomes:
2
_I_
L3
-r.ssrl
ls.zs 1l F- EII s.667 o.uurl 3.250 J L$ I
rc) x30
Thus, it can be inferred that the influence coefficient matrix.depends on the chcice of releases and axes system. However, the final solution of any süucture will remain
î,2
'l l/
\!1
unchanged.
tr¡ L_J
''l
Example 7,3
A joint is
-7
suspended by three springs as shown in Fig. W. Determind the forces in the springs.
7.9a. It is subjected to
¿\
?tr¡ i-ù
.J
(
ÈJ
l{
'{i'\¿/1,r'
a
vertical load of
,r4
lrv
Solution
(
t
Three springs are meeting at joint O. Hence, the joint O is statically indeterminate by degree Let us introduce a thrust release in the memb.er BO. The released joint and axes system are shown.in Fig. 7.9b. From geometry,
REi-EASED STRUCTURË
l.
sulq:
DTAGRAM
Fig. 7.9 Indeterminate joint
L)
.
-Ll
Ll
Let.us write equilibrium equations under the application of a unit bi-action
applied at O along OB,
sinp
= L3 !'z
"orp:3L3
P,, cos a = p' P,, sino + P:r sinp = I
and,
Let us apply load W at joint O in the released structure. The axial force diagram xro due to extemai load W is shown in Fig. 7.9c. For equllibrium
P,o cos an4 of,
Plo
Qin cr
^ = rro P¡o
:
+
P¡o sin
Wcosp
,¡r([lþ) Wpos
ø : Pro cos p p : '1Y tenston
D_
and
'21 -
cos B
compresslon
cos c[
P¡l :
't"G+ I
pI
compression
The axial force strbscript
g,
't"GlpI
Pr:
tension
pl : l
diagram h' due to qnif bi_action is shown in Fie. 7.9d. The 3 indicates axiar forcã and subscript l indicates *,. ou*u"i à-f redundancy
under consideration. Álo displacemen?at
=
o
in the direction of the release due to extemal load
to-j
;l
.',
,ir
l
t,;
INFLUENCE COEFFICIENT METHOD
212
GRAPFIICAL METI{OD OF INTEGRATION
h¡l x¡o
.{,-)
¡ JAE
T.
¿""
:
is given by
P3s
+
(,,,
wcosp Þ. " L, *o* Y,"or î," ,îro^."+-l "g, - _f -Lrir@lPf^" iin(a+p) " e,e,' - sin(a+p) sin(a+p) A¡ErJ
i
{., t,{.--)
:t
using the geometrical integration technique
P31 P¡
cos cr W cos c¿ sin(o + sin(a + p)
p)
ioltrc
.
213
of p, from Eq.(iii)
(vi)
''
can be substituted in Eqs.
(iv), (v) and (vi).
).
i('': it\-,/
or, l("-
''
j{-..)
:,
.(,;
Âro =
frt
I\þ0,,
t.' ('
20 k N/m
F Lr * Lz - cos2 tt L, I Ltt"G.FIAtEt a\E sinft+Ð A,ILJ f
Compatibilityequationgives, Âto
pl:
+ !,r Pl =
5m .F
r,
(iÐ
"or2
or,
a two span beam ABC shown in Fig. 7.10a using the influence coefficient
J
Influence coefficient due to a unit bi-action
,
{.,
fsin2(a
1.4 (Ð
.
t'
rti :1
l, *' w-cos2 ct. -¡.l r¡ + p) ArEr sin2(a + p) A3E3 ^
I w.ort P
(q)
Á-_¡
ffi
(b)
]:,1,i1ìr':
,
-SPAN
l,
/>þ
REL€ASED STRUCTURE
cï+5c
+:l#
åiî
The redundant force in memberOB is given by
2
/?D7 I
0
*
GIVEN
8m
(.. (._
.
wcos2Ê' L, * wcos2q, L,
(; (;
sin2(a+p)
Pl:
cott sin2
t,. ;
ArEr
p Ll
(a + p)
1c)
sin2(o+F) ArE,
L2
ArEr A:Ez
"ort sin2
o
3.077
(iiD L3
(o + p) A¡E¡
(d
Member forces Force in member
(; (
(. (
AO, Tt Tl : Pro
)
h11 0IAGRAM
is givertby
+P'Pl
_ cosp pr : WcosÞ ,',i;;pl sin(a +þ¡ Force in memberBO, Tz = 0 t, P2¡ P¡, : Pr
Fig. 7.10 (iv)
\
.
x10 DIAGRAM
(v)
Solution
l.
The two span beam is statieally indeterminate by dcgree It can be ma{e starically determinate by introducing a ihrust release at B, thai is, remöving the support B. The x* diagram is a parabola as shown in Fig. 7. I 0c. The h,o {iagram due to a unit bi-action at
i
at the B is triangular as shown in Fig. 7.10d. The cliscontinuities
'}J
t,,r
applied toã¿ and Ure redundant bi-action
i
t:/-\
ordinateofthexlodiagram.at B
_
¡;
¡-\
i \"-/
20x52
c.g. from
A
Ordinate
of the h,,
(l
ji /-\
't: i.--l .t I
to
the
follows:
\
1(., ) /'',. i\
the continuous beam supported on three elastic springs as shown in Fig. ¿lt;.
::' þo/2-g-s/z+- q---{
:
+aO
5)= 1208.34
3.19m
*tl -z " s¡
(O)
BEAù ON SPRINGS
diag.íam at the c.g. of the segment ADB
:
:.
i.a
release due
(-)
1.e63
ã-
,)
AreaADC = ax4?2.5x13=3661'667 J .'.
AreaBDC :
3661.667 -120.8.34
c.g. of area BDC from
-
' \ -;
.'.
A
cf
:
(c) x
ÐIAGRAM
OIAGRAMS
2453'327
366!,667
x6'5-1208'34x3'19
:
RELEASED STRUCTUFÊ 8.13 m
3661-667 -1208.34
: 13 - 8.13 :
4-87 m
the h,, diagram at the c.g. of the segment BDC
:
(-)
I
(d) '873
h31
f,, .lt :
xr'
Ì
c.g. of area BDC from C
Ordinate
215
iLle using ttre influence coeffrcient method.
= 130x5-20 'L
t2' Irxt3_Zx _ z.s(+xtg-lx.s) :
AreaADB ,! \/
I
cañ be evaluated as
htl xlo n-^ : I _ru J EI On,
l/-r ir1. / i
GRAPHICAL METHOD OF INTEGRATION
INFLUENCE COEFFICIENT METTIOD
214 j.ì
STRUCTURE UNDER
.UNIT B¡-ACT|0N pl=
1
DIAGRAM
¿r, J¡hrrhrr EI
h31 DtA6RAM
(e) h
DIAGRAMS
Fí9.7"U Rs
:
0
41.028 Ra
=
0
For compatibility Âro +
cr, This
-
6967-36
+
f'
or,
RB
:
Solution 169-81Ò'l
ís the same reaction as obtained by' the method of consistent deformation in
Examþle 3.2.
The beam is statically indeterminate by degree l. A thrust release is introduced. in the middle spring to make it a statically determinate structure as shown in Fig. 7.tlb. The springs AA', BB' and cc' carry only axial forces, while the beam ABC carry bending
INFLUENCE COEFFICIENT METHOD
216,
GRAPHICAL ÌvIETHOD OF
moment. The bending moment xlo and axial force x30 diagrams due to the external load are shown in Fig.7.llc. The'itructure under the unit bi-actions p, is shown in Fig. 7.lld. Thelending moment h,, and axial force -þr, diagrams due to the unit bi-actions pr : I are shown in Fig. 7'l le.
a
ZtT
in the outer two springs can be determined using the equilibrium
)Fr:O, Rl+Rz+Rr:P IM": 0, R., x 2a+ P, xa-Px 2Rr+R2_1.5P:0
Discontinuity at the relèase in the direction of the release due to the applied load is determined as follows:
n.^: J¡hllxlo ¿n, *ru El -rr
INTEGRATION
(iv)
l.5a
:
.
0
(v)
forces in various'members of the sfiucture are known.
¿n. J¡h¡rx¡o AE
nlyze the beam shown in Fig. 7.12a. due to settlement of support B by l0 mm the influence'coefficient method.
=
+(î.'Ðl I tl *' 1"u"1-gl *?,gf*4*.+,f,-11.+"+ z"''\-t)^1^ 4 J--1n A"q'\-;)- 4" AtE".l-tj
* [; " î"(-î). ;. +. +{;'"(?.
3 PL : - fttpu' l--¿--+-I ArEr Er 196
;).
I PLI
.1.
t. - Io)
;
6m
GIVEN 3 SPAN BEAM
(Ð
I
8 A3E3
ABCD
./\---.---.-?.\-_ /?fu ¿rb1 Ð- -----Ð ';' 5m
J vr-t
. Discontinuity due to the bi-actions in the direction of the bi-actions or the influence coefficient is determined as follows:
(b)
*ivr=t
i,L
RELEASEO STRUCTURE
htfhrt dy" f,, ¡hrrhrr dy" + Jf II = .J AE EI
or.
I L I L I a 2 all+-.l L x-+-+-x-x: 2(l-xax-x-x2 A3E3 2 2) 2 2 2. 3 EI\2 êrEr -l ^2F2 artl.LL- + - + _ f,, : -_-+2 6EI 4 ArE, AzEz 4^383
Thecompatibilityconditionrequires Aro +
fir Fr = 0
(c) h11
(iD
(d) h12 DTAGRAM
:
Fig.
(
llPa3,3 PL,l
( (
DIAGRAM
PL
%Et-sAÃ-sA3%
ot,
=- at,¡
L
¿EI-AA"E, -
( :
L -AA3% ,l L
AzE2
B2 ; reaction in the middle sPring
7.12
Solution
:
(iir)
Tfie continuous beam
indeterminate to degree 2. ,Letus intoduce thrust ¡s_staJiça[y c. The ner vçrrical deflecrion ãt s is l0¡,|_;; ar 19lease,s c is zero. since therearerg extemal-Ioads, xro diagram is zers. The force diagrams due to the bi_ actions at B and c are shown in Figs. ,7.rzc and d. r-et ur the influence coefüciênts. "oöuto .
¿¡ 5up.ports
B and
_
GRAPHICAL METHOD OF INTEGRATION
INFLUENCE COEFFICIENT METHOD
218
r,,: ^nJElflþ¿v, r,, =
=
.jlx3.33a'rs'{]l
[1 x J$av. :- ferLz"
5
x
3.
D
=Y lþ
334 x 1.334 *
T 2.5 m
-n+1a + 3'6) + 2(2 +
t'2)f
+ 1.5 m
+
-L
*e * -z'3'6] -J ]3.'-
_
integration of It is irnportant to note that the geometric
h"
over
h'
:,t:.2
:
::.
-'-lEI =ry r, Lzz-!-'.. = ¡hrzh¡z 0". -rr : []r3.6xr5xr.l+ EI L¡"-'": éompatibility
tFl {p} +{^0}
: {^}
er I
s+.el
e+.s
l[prJ
'
Ìí!r¡?
:
r fss.szz s+.oel{nri.o={o^ot} o
or'
I
+t rt tï tt. ,
RELEASED STR
:::,r,:¡.r.,
it'i
J
lss.stt s+.erlJnr \ _ i+oo
or,
I s+.or
of'
e+.4
l[prJ I o
{;;} i:1i1}and Rc is 35'84 kNJ' This is the same result as À"ir,"¿ of consistçnt defcimation in Example 3'10'
Thus reaction
*t"r"iîtìitã
R"
is 42.46
kN
î
1
AnalyzetheportalframeshowninFig.T.l3ausingtheinfluencecoefficientmethod. l
{
tI
i
(''
('
Fig. 7.13 portal frame
Ignore axial and shear deformations'
Solution
Theframeisstaticallyindeterminatetoadegree2.Itcanbe.raade3tatically system
The released frame and the axes * shown in to the applieq ]9:9 ¡re ihown in'Fig. 7.13b. the tJmeni diagram x,o !uè e, The 7.13d'and Pigs' in: are-"shown bi-actior¡s iìg.-i.ij" ;¿-thore d.re tq the detenninate by removing ihe hinge support
àii*ntinuitv
9.38
1?.7 1
Example 7.7 1
¿,?
n
should be done
in three parts.
requires,
r-T*--_-l tt
(b)
i4.69
EI
219
20 kN/m
Â,0
=.fH"*,
:
20.8
(.f ) MOMENT.;KNM
A.
due to rhe ãpplied loads can be calculated as
follows:
,
"'
=
*l*xaxaxs0+jx160x4,.:.(310.rs0)"o.oj
/
GRAPTIICAI. METHOD OF INTEGRATION 224
= -: f _¿u ^-^ J
ntzxto
EI
¿
The influence coefficients can'be calculated as follows
r,, =
.
h¡rh,
EI
6'*,,
J\þ*,
: :
fll
*!"0] ) J -
*[;
EI
:
x4x4x!*++ara*+l-
3
ErLz
3318
J
-210 +
{4
,{\',;il
:
-25.24 kNm
=
-130 +
{4
,,{;ïi :
20.8 kNm
=
-16
kNm
requires,
The compatibility condition
EI
*4x4x4+!xax+'+]=#
: t,,
frame can be made statica[y-determinate by introducing a rolrer at support E as ñtn in rig. 7.-14a. Tt^e moment diagrams *o unih are shown in Figs. 7.r4i, c and d. xro on the beam. cD ;an be split in a triangié anJparabota as :::?T:l^,9t:ryy 7.14,e for ease in computations. Discontinuiries due ã;öiièã roads can ,nll omputed as follows:
Ig
!ry EI
}+l[ss.r:
[-rtsJ I, sq
Â,0
64 lfp,ì
ll
}=O
106.6?JtprJ
Âza.
[p,ì 137.?0ì .n* tt'l={ u'on }
oÍ'
__
The row vector HT gives the h
,,
+{o
HrP
values at the section under consideration due to the
t.s)
J*P or,
=
*11
ßt.t\ : {s.arl
2
"
+.?x
4ox4.
:l=#
x 30 x l.5x
'4oxq'+fJ
:
I x r.s+ 3o x2.5 x2.75 + )40 -895'42 EI
The influence coefücients can be evaluated as follows:
= 12.74 kNm
x3o x4
: f*P*,
t'
'
unit bi-actions.
(*,)åt: o ,
=
gË
The stress components can be evaluated at different sections i of the frame Using
(xr)¡ = (xo)¡ t
'
Âo + F p = $
[.-:zeol
{
represents the value of the stress component x, at B in the span AC. (x), :s the conesponding stress component in the released stuctures at the samc The resulting bending moment diagram is shov,¡n in Fig. 7.13 f.
8s-333
*q*?x4+4xa"+l= rt2z -J ¿u. : Ilt"1r¿ "-" *r, = ¡hrzhrz 3 ' EIL-"2'J EI
or,
=
¿u"
l3ox4 *Z+lx8ox4
¡
-,i;Ïi
EI
1
ftt=,
* -50 + {0
3760
r Fr : - t I Lx2.5x50x3.167+50x4 x4+)x160x4x4+ r ErLz
.*
221
INFLUENCE COEFFICIENT METHOD
I+þ¿r, ¿ '- ".'-J- Er Er " .=*l*xrx4x ErL2 !+rxt",l=lI 3
xax++
CRAPHICAL METHOD OF INTEGRATION
INFLUENCE COEFFICIENT METHOD
222
Èiii"','''
$ti*.¡Se"þ EI
*, : qg EI
conditioncrequiresthat:
(o)
T
223
4 +Fp:
0
RELEASED STRUCTURE
{,-'i]..['} ,*;]ß}='
I ,J I
t.
1_!
f
:2.
s
17.s 1
I v,,
tnl
/7ÐT
(xr)i
=
(",)å': +
55
V
x1o
(xo)i+H¡rp 30+{o
_.,{?H} :
(*,)3o: o +{l
-.,{?ffi} :
-16kNm
-
25.20 kNm
(b) (*,
qy
)\,*o- :
55
+
{0.5
'{?ffi}
19.40 kNm
Example 7.8 Analyze the continuous beam shown in Fig. 7.r5a using the infiuence coefïicient method. The flexibility difpach spring is fo : f1¡BI, Take e-l:
consi;i
Solution The structure is staticalþ i¡rlÎterminate to a degree 2. Thereleased structure is shown "riä. z. tsu The force diagrams x.,o and xro duà ro the appried ru"a, ,r,ã*" i" 7-r5c,and rhose due to the:bi-actioir iåa p2 afe shown in Figs. 7.r5d and e. It should be noted that the beam ABCD àrr¡es Uåling moment *h"rã-u" tfr" springs BE and cF carry axial thrust. The discontinuities can be evaruated as follows:
in Fig'
p,
Fig.
7.14 Portal frame - alÍernative solution
*"
.i. INFLUENCE COEFFICIENT METHOD
224
,1,0.
Ib¡
dy,
+ h'J¡ f', h'
2.*Lt*?r*r*É*l=llÉ 3EI 3 3 EI 9ET
REL€ASED SIRWIURT.
f,,
due to-symmetry
-Ð,
,Æ---T?,.1 t ,r)r, þ
)75
due to symmetry in the structure.
.J¡hu!¡r EI
^'T-f* GIYEN STRUCTURE
.IQ,&4fSIEAÞMEÍH9Ð'OfIN{EGRATION
3L
I+P
/2 .l
dv¡ + hrr i'., hr,
,,ffi-lt,
+.
+ h' i', h'
#[i'(+ . ?).;( i . ï)l. o.
o
t30 '(c) 1..,r
{"
,,Ð--T".p , l'2
_,
f.---:
-=í-;1.47
(., '{'
L/3\
',
,r+?2
¡rrv
tt,/
Él{;l.{a}' eJtnzl ál
{'_'
iT Ln
{.
: _?p =p,
(;
: Fig. 7.r5
(, {,
Aro =
J*P
ot,
*
h,,
r',
x,o
.',: ,
-,
Auat-yze the portal frame
{L 's
#p
and lrence, momenr
with inclined legs
as shown
at
-=
zlpt
in Fig. 7.16a using the influence
sotff1ç¡.nn*ethod. Neglect axial and shearãeformaqio¡s. Take E I = conìtant Solution .
L3PLL 2 2Et' -x-x-+u _ 23 PL3 24 EI
Jle'frary
is statically.indeterminate to a degree.3.
^ g,ll$'t,:P*n
2
A cut is introduced
at the rdidspan
in 7.16b. The.local axes system is also shown in the sarnð fig;e. I'hc bending nomentlq: x,, due to externafload, and h,,, h,, and h,, dueto-unit -dilqramq bi-actions :aie shown in Figs. Z. l6c to 7 .16i. Discontinuiþ due to the applied load at the point of cut offand along the directions the bi-actions can be calculated as follows: ' '::,
::t'-
:
i :
,i
-' :r' r
'.,1 /'
!'
:
of
,1-.;
f::::.
r ',
..1
226
.
Âro ';t.,
(bì
(a) GIVEtt FRAME tls
i'i: I .-. I
i -: .-
L .. * JI!|} Et
¡hr¡
' EI
rii
¿n.
xt
,
227
88.4 E,I I
=
",!rf(+.;)'
=
25'5Ji :lo' ZEI .i3, I
or','
xro
NreGRArloN
2El .
,
.-
,,.4*=*
_25x sJ-z
dys =
- J\#
fï*
RELEASED; STRUCTURE
GRAPHICAL METHOD Or
= ?:!¿
294.6
EI :
j."trìcient or rhe discontinuities due to the bi-actions ar the point of
i*-.i": Ëllg the d¡rections
.
(.d) ,x¡6 DIAGRAM
-ó '"
/-T*) (1*1 t" )"
f
Yv
)
hlt
r!i¡'.i:!
DIAGRAM
ñ f'
,/
r(^
/-ra1
r..';.
ìr;ir.t'
=
o:
&'
ri-l:l- :,.:,:':
z ( 5x5JtxrJ+o - \ = I\Ti*, á"[-; +i
=
l\þ.er,,
=
)* )"
: ì :i
.
"':
,.. . ,' t'-ì:
.14'.-þ5,::r"
li "'[i "'.19).
EI
(j ) :
s
908.3
(¡) .
?tJa 4!.1 ,l ++" ay, ='þJ-zxlxl+ròx:l,x t+iJì,*r*rfa JEI = EI
h12 DTAGRAM
+ +__a. P3=1 \
.
'
= 5
(s)
of the bi-actions can be'calculated as folrows: ..: ,. ,:,. _
J\T'."
-í;,:'.'
(f
ì
:
h13 DIAGRAM
.'
:
.'l
.l
:.
,, ¡::,':r'i
:. .:: l.ìlìi' ''ì
:i:r.r';::
,-i,rl-,:
'.:l:,'-:
o:frt
-r',1
-r
:
.
.r rh,.h,r . . rrt = :::.:- rl+dv, Er '.:
:
=
2f5
CompatibilityconditioìrÌgives,
MOMENT kNm
''' Fig.'z.16 Portal frame fiíth'inciined lèÈ¡'"'
)
1
*l;xsJlxf5*sfJ*o=I1r EI ?rL¿ A;+ Fpl: o '
I iì
(r,
Í.,. 3)],.
,"INFLUENCE€OEFFICIENT METFIOD
22ti:
.r,
PROBLEMS
22;9
,,1, ':"1{l;i=.
{;ll1}i'ï' o [rtuJ l-rr.r,
ttt'ti[o,i
0 ,ì [ n = { -o.errf [ -r ro,J
oÍ,
The süess
"o,np*á,i'"an
The bending'moment
be evaluatéd.aidifferent sections in the frame.
xl
at
A.in the mefnber 'AB is'given by
(*,)f = (*,0)f +HrP
/-. lcD \^l,f
The row vector Hr
D
gives the hU values at the sections under considerations due to the
unit bi-actions tesulting bending momsnt diagram'is shown'in Pig.7.16 k.
(*,)Îl : -
2s
+ ir-r0-
tt
[0] 1-0.8rrf [ -z.so J
: - +'aotNm
Develop the influeqce qocfficient matrix for the beam shown in Fig. P7.l by
Ignore axial and shear deformations.
,
''. B
,
[']
0Ì -o.8rll =
4A ,, ,- 4fn. . r
4.05 kNm
iì'iì::rjj:lli:i:.r.'
1
[
(*,)3'= o +{l-5
PROBLEMS:
(D introducing a th¡ust release at B and C,.and (iÐ inhoducing a hinge at A pn.d-a !h4rs! f-elease at B.
:
SimilarlY
.,; .,: :
-2.s0
[o
or
J
[ -z.so
J
6m. .'
j, -
;.
,,
:.'l: 'Ì'-- ::')i.:: . i. :
.
Fig. P7.l Redo problem ?.1 -rby ',iñcluding thé axial and shear deformations. Take
I
{-o.t"f =
{¿
A,G
:75
E
I,
and
AE
:
100 EL
4.05 kNm
Develop the influence boefficient matrix forrthe beam shown in Fig. p?.2 by
(i) introducing a th?ust releasg each at B and C (iÐ introducing a thrust release each at B arid D (iii) inhoducing a thrust release each ai C and D lgnore axial and shear
deiòrmations.
""
{r F) ìai j
;[
lü
,
i.-',
.. ."...
TNFLUENCE COEFFIGÍENT METTIOD
230
B
I
3m ,
2l
C
5m
,
¡ f.m
r:
. Ì
231 ..r.,
l
.Þevelop the influence coefficient matrix for the frame shown in Fig. P7.6 öy rake 1{rG Z_|tiul, Sell9.LlSílg. aly. tw9 different released structures. nn'= io et.
D
:
t
1--, ii
Fig.
it'l i
ii_;
7.4
:
,.(j :{"
Develop the influence codfftcient matrix for the beam shown in Fig. P7.3 by (Ð introducing a hinge at D and a thrust release at B (iÐ introducing a hinge at D and a thrust release at C
aAZt D
'(,j
B
I
, 2 , 2 ,'
; 7.5
,
C
6m
Fig. L.)
P7.2
FÍä. P7.6 Develop the influence coefficient matrix for the truss shown in Fig.p7.7 by
P.7.3
(i) introducing a thrust release each at l-5 and 3-5. (iÐ inftoducingathrustrelease eachatl-5 and2-6.
Develop the influence coefficient matrix for the frame shown in Fig. P7.a by
(Ð introelucing a roller at D. (iÐ introducing ahinge at E
Take AE of the vertical members and diagonal memb.ep;0,'.1,-r{.E,ef to¡ ::'::t :ii;rìr i- 31d
bottomchordmembers.
.
:i.'
B
-r
I
¡
l''.
I
1m I I
t.
I'
Fig.
I
7.6 (.
I
P7.4
{':t'¡-.
Fig.
Develop the inflt¡ence coefficient mafrix for the frame shown in Fig. p7.5 by
(Ð introducins-a cut at C (iÐ introducing q hingç a¡ A,an{ .a ¡--oJler"¡! C
and considering the axial and shear deformations. : 2s E I, !,9 t
ot
rake A"G
i
,|,...
( 7.12
i
:
I¡-' 'r,;
'.6n .:',. ':I : r',-¡-|
(
Fig. P7.5
-.-1''.!!
.:,i
P7.7
:låL-) iU, il :
lr',
MULLER . BRESLAÙ PRTNCIFLE
f!.J i
1r.'t
t-:
r':.
i'r
I f
ist
.çr1+qrER
eight
,
J
.ìa'. l"
:i
,,
ì-
ìr-
INFLUENCE LINES
ti
¡t; I
(-)
.. (;
{) (
.:
8.I INTRODUCTION
i(,
to moving loads' Influence lines are required for the design of structures subjected discussed in were arches and Influence lines for staticily detenninate beams, trusses follows: defined'T volume I of this book' InflueDce lines may-be Aninfluencelineisacurvetheordi¡ìatetowhícþatat,ysectionequals acting the ialue of some particulor þrcing function due to:,a unit load particular a ø;which indicate'ine'aten at that section. fnese cnes thà¡passaþe of a unit load function at a given seçtion is afèeed dae:to
.ì
(: t.,
(:
t'
::
across the entire
The
iri-.il'ì:-!iìì ii
in chapter 5 is very helpful
:'
S:2.IÍÍIILI¡EB¡'BRESLAU'P"R'INCIPLf,;'
ì rr' '':'1::: " ',
. ,."'
The lvfuller - er.riuu'princþle proposed ¡n rg6¿':' 87 is the most powerful tool for where the developing.in$ì¡ence tinei for both determinatp and in{etgtminate structufes,. qtatel tm! I-t valid. princiþle of superpositig¡ is : '
(,
,:,
þr
any
þrce quTntity in a structure, ß
represented
. .ttiï'iotiiè::sçl\¿:,þíi in" a"leAt"a shìøie:of lhë:siructu'rëtrësùliing'ftom moving the þice q"oitiq' under consideratibn through;ø':3ßrøll displacement.
:: .-- : ' ,
"
,Fig.g.
, ,. ;:l'i
l'. ¡ir ,:,
--.
l, Irihireincélineforiroryc.iít
ltí,:t
in
In this chapter the influence lines are drawn for statically ;ì: i
frames.
The infuence lines
:ti
r.':liì.iiii.ji:.r:iiir i.;.'
,
\i
I
¡,
Maxwell's reciproóal theorem':diéôussed
indeærminaæ beams and
ld)
I
sPan:
developing the infiuence lines.
eÁo A
_
': ':" influe¡ce ' ,principle :is the basis for -determinrþg lin9.-s for vqrjoug .structures This regardlesi of whether the method is anal¡ical oi experimental'
Consider a frame shown in Fig. 8.1a. It is desired to find the influence line ordinates
ordipate fo¡ a unir load.acfins 4r,.po¡trs.E,l¡d F, is any point in fhe sban rBC. In aecoidanc" lhe rpan with'the Muller - Breslau principlè, the'force compoúent foi which the influence iine ís -desired is removed. In other words, a hinge is introduced åt tlie sirppdrt n,uitr1.*ir,¡¡ Fig. I'lb so rhar a specified rotation -uy b. inrroduced. ¡ ulrudÄluried at E and
,,.,,,lr"rl,'H,:.fil1!jl:_l1,l"etc" IT *l9r,e,,B=i!l AB, and F Fo{
Pornt,,11t
.tþ9pe{9!!ao_91"4-Sqo*LmIlg.qlqf:go1¡3lgd_bJ*yr*¡r,o¿-
is now appried ar A and rh" r.";;ilr ¿"n".rì, shåwn in Fig. 8.r d. *lï:1l_T"T:nr r he rótat¡on Â0o , and deflection Áh, and Arr' can be bomputed by
,.,Jn.!igs-.,8rrç,an!'4¡tganbeeasifysgep,.thst:;...j 0oo : MoAOo or,
::..,-..,::.i:.,:;,..::::::
..:,
ooo
^= aor
M^
a¡y mettrod.
(8.
la)
Maxwell's reciprocal theorem can be appried to sections A and E in Figs. g.l c and d, that is.
t¿ I
{t .'
INFLUENCE LINES
?14
i
ILI,IJSTRATIVE EXAMPLES
235
'':
M^ =&e ao¡
(j {',
(8.2)
The influence line ordinate for Mn due to unit load at E is given by the ratio of displacement at E
to
the rotation at A.
Now let us apply a unit load at F in the span BC as shown in Fig. and e, it can be seen that
0'¿o
:
8.1e. In Figs. 8.1
MA^OA
9isM^ ^ = aoo
or,
d
(8.3a)
Ituj
Maxwell's reciprocal theorem can be applied at sections F and A in Figs. 8.1 d and e, that is,
I
x0'oo= I x Av, MA -'-A :=
(8.3b)
I
\_onF
+ Àoo
lT,"'
(8.4)
I
--A
|
The influence line ordinate for Mo due to unit load at F is given by the ratio of displacement at F;to ,the rotation at A. Sim'ilarly:the influence line ordinates at other sections in the frame can be-determined.
'
HA=,1
--+-i
I
':..
Infiuence line ordinate for Hn
(.
.i
(,;
() (,
'A't¡nit horizonùal forbe öf 8.2
d.
'l
kN is next äpplied
The iteflectió'ris Àh;
,
and
^hF
Âv,
A
thä frarire deflects as shown ' ctrn be öomputed bv anü method.
at
a,äa
In Figs. 8.2 c and,d, it can bp e.Asily. seen th¡¡ i ' lootHoÁho
;
'
,, ,' . I (
is,
I xhoo = I x I
.
(8'59)
Maxwell's reciprocal theorem can be applied to secfions Ai¿in¡t gih,ttrs abbve tþures, that
r:.t.:.
-'./--
lr, þêno td
Let us now calculate the influence line ordinate for a unit load acting at points E and F of the frame shown in Fi& S,24, In accordancç with., the,Mlller - Breslau principle, a shear release is introducJd ai the support A so that a specified displacement may be inüoduced as-shown in Fig. 8.2b. A unit load is ap¡liS g!:_-E,,1p.d. llip, ho{SontaJ displacement at A in the ftame shown in Fig. 8.2c is compirtèd by àny method.
in'Fiç
"
I
".í
Âhu
(8.sb) ìt"'
H^:&e - : i .. . ,114,,..
^hi
Similarly, with respect to Figs. 8.2 c and,gr,it.can be sho¡vn-fhat
(8.6)
i r:
::
t
.':
r-
,.
(e)
) :
.í-.i-
Fig. 8'.2 Influence line for horbonial reaciion at À
:.'
H^Ã^h :
AYF ^.,
(8.7)
É+¡rA_
From Eq, 8.6 it islaþparent that the deflection riu¡ye of rhe member AB is, to some r¡rst9, lhe influence line for Ho .for a horizontal load on AB. Similarly, the deflection th: member BC, is ro soiùé'scale, thèfifliience:tine for Ho for a verrical load on BOi¡' ffts'3¡epes and deflections can be determined by any method.,
glyt:f
8.3 TLTUSTRATIVE EXAMPLES Example 8.1 Develop the influ-ence lines .beam shown in Fig. 8.3'a.
.:: i
for
: Re'- , Ra , Mz- ,
I,
M¡ '".and
v,
in a two span continuous
'
Solution
InfluenceLineForRn'
:
'. -:-
t.i,l
-:
''l::"'' The vertical restraint at support_A is rer4o. ç-d and: a unit vertical load is applied at A. The deflected shape of the beam is, io some Scale, the'influence line for the reaction at A. "i :. :
g)
Y'
II,LUSTR.ATI.I'E XAMPLES
237
l¿,' ål.:
l';,
''''îi:¿
INFT,UENCE LINES.
Ë:.
*i
i:4¡
:,
5!/
Re:
å,' ì,'--'. il.,
, M¡:
anitclockwise
Mz=
i" lr
455
',
1b).
l''-
tsEAM
CONJUGATE
ïL,
L.
-
^2
j
416/ /'J
=
.Tñq
0.406
/J
i(.,' be determined at other sections. The l:iilgll',th.:, influence line ror R" can
f
it,
,
¡
:..
il
i
(d)
DEFLEC¡ED SHAPE
i,(. 1
,{
3B
15 -1;!,1,tj.
(.
CONJUGATE ESAM
".
(
Ir
kN
--
t
Shear at
(
{
(.
i4t
-â : "': r iùf--
--{) îi---JJ '' -\.rj ir ì!.
(g)
(
.: i1i,:.i!j',r
ri--
---------
z{
rlurn -r"
.
:::
i
.:
.:
in the span BC
_:l
Rn
_
=-
8.00J 111,.: ,:
Réacfión
8.0
=
fr
DEFLECTED BEAM AND REACTIONE
i.
(
:-.
;
B
n
.:.tt.
=
2.6'I
38456 '_
-
.-
,-,
¡f
CO,\JUGATE B€AMJ_t-.
,.
Fig. 8.3
,._j
= 21331
'ÏÍNF.LUEN€E LII.IES
ILLUSTRATIVE EXJ{MPLES
tt R. : -x8x8xZ':J
Reacfion
B f :
^ t,K c;-ct
x
-È@@
(d)
nôY oo
o MOMENT
Reaction R^
aa-;ì !' FO
oË
Af
The moment
M,
-cdn'be
o @ o
çioE Q,-..:,..
ë'Ë' .g
;
..f Þl
'Q:
9
EFAç.llo.rN. råT
.9,
å
I
,
.oì--. '.r .-.i ..
@ñ
{Bo o- -o.:
.sci
;
x 16x8'-10.667':53.3331
dcfermined by considering the equilibrium of the span AB.
s. * *8x8*9.:ür:'0,. ¡.2,3
The moments at various sections can be computed.as follows
:
.
I = 53.333 x2 -,- IJ x, " = 105.33 . ,k2 i :,4' '\., I '"_..-Mornent justtotlreleftof2 = 53.333 x 4 - t' x4x4-x:.:ZaZ.C6 23 ' Mömentjusttotherightof2 : -.341.33+202.,66: - 138.66
(:q)',1.!i0ll ENTT.AT F; c
iì
=
2
53.333 @@N
= 10.6671
i.^
'ç
239 '
Mgmen!
at
:
FFO @@ o@6o ooo
1:
a{@ ,--ooõ:ôo ''õ9;o333 '9oooóé :.. 1. :j,,.,r.ij:i;i:r j::- ,-i. :. ,.. :,,-.(íJ!:SHEAR.,ÂT ,; :. r,r , , i: ,,,i .,,, .: t:;:i::.iiì.'i _ 1r_ ,-,:.;;;;,;11.1' .,.,. ., .ti. : .;i ,:-.,,;,,,- ..;,., .:.ì, ,, i, i¡ Fig. 8.4 Influence lines îor 2 _spln ,
{c}
REACTION
AT
B
_.i
Þ$m .
Momentat
:
I
-8
*2-:xlx0.5*?
{., ,
{: ;, ( (,
;,ì
j.
(, (.¡
(ì {'.
lì
;=,2.6V.x4
='-.ie.ZZl
I - ,4'
at left
-,;*,4,. lxi,,,+g,0 :
,.. ;-,,,:l-jt
-
,,..1
|.,
I' L' ordin*',u-i t
i . *# ' I. L. ordinàte drs= 'i*
:
"
,=
0:375
isshowninFig.S.4e.',
:i
:jí:.;ii r.!
1
:.
:.:
:-
lri ax t* !-,=.¡z:o-
,::
0.308
66
341.33
arrighrof2 = -;iä=;:;.;0{0.,6
-! :i:.,!'.. :;--.
,
.:
,:
. .:
-
':.
, ,:ì1
;i":,,': '
herinf,luençe.line :fo¡ sheár.at 2.is: shoqn in pigr f :4 f. ¿{¡S. .ÍÈ.:.ilrj"
.i
.'
i
:
ü.*a'npte,.:4,2,
i
ii
i:'. 1 .:.:
,..i.
1; ..;t - tì':i.l
i:,, i':a t
A shear release is introduced at section 2 and a pair of unit forces is appried as 83.f. re.?crio{¡q are shown in Fig.8.3g. The :l1î1T,lig conlugate Its.deflecjt:d.shape.a1d corespondrng beam is shown in Fig. g.3h. A moment M, is imposed at section
2 correspending thf,i,r]çl¡J1vg dirpracement between.the two'cut portions at lo 8.3g. Lèt'uJ first caiculatè the rêactions iitr,r rãn¡usure beam of
section 2 showii in Fig.
i.:,;r;¡;¡'¡'., ..ir r... ir.; ;.ij..r.ri:j....1::r::,,-,¿.i.;ji;,.,,.¡
i:ìl
-,,
.:
fiolutiqn
:,
i;¡,;. 1:'ir1'i.[û¿tiiïiiJ r.¡,
B Ci'i¡n'd ' âäit ,.!,. .!'. : ':
ahd
---'
Influence Line For Shear at Z
Fig.8.3h.
,:
)î)
x4
..:
r
influence line ordinare ar orher secrions u" "* ott"inàd'à shown in Figfine fo¡ momeltÌat B can.arso be obtained in the same manner
.spr$Iv.tr: ^8,4.d. The influence
of 2
10.,667
t
= - 0.7!i7,, ;.i ;¡, ,
5 :
i 4{. ?? rvJ¡rJ I - 341.33 -
I. L. ordinate at
t r):l;rìÌi .-::: .;jr j. ^,, , ,,1.. ,¡rì . i.r- :,ì , i,. ì, . ,,: i :,.',¡;:;.:j...,t ;.¡¡:.:; _1;j.r The"i influence line ordinateat anypection is computgd by dividing each moment by (or + oz:21.33) as in Eq. B) or 8.4,
(.,-:
(_:
r:
Momentat5
,
Momentat
I ¿:i:i
cotrtiÌiäðüs
idthS¡''ifie principle. A couplés.
method as
2,40;
i,
' r T{N.SI*IEN0E.LINES .¿ ::/i-ii¡::ì.r11:ìi.'..¡::.. - .-l
-
.:
PROBLEMS
24t
PROBLEMS .: Ø.
ø{N
qr
-. l' ,,úl -! o:..-.-(^e,), M
o@ "i,. o
i
È. ci,.,..
'E
a?-.-..-+ ' Nd.o.
tr.?..{
:o.o
.@.
Sketch the influence lines for the moment at A, moment at C and shear at C for the beam shown in Fig. P8.l an'd compute the ordinates at 0.5 m interval.
o
T2 ìi-i- rr..!\'
.'
4A,CB Ë,,:3'
--D,
,2m.3m r-
ó ::,;-oì:,,tdr.si;,.ã-i,,, (b) REACTION AT A
Fig. P8.l
r
9.. -5r r::+.:tú.ì.. (cl _SsjAcTl9N;,êllg
o *'.-g .::F:
ooo (d)
4=
REACTI CN
ó ts: ts o i" 'r.4 -e;l o o ATC
-.. i\.ì -iì::i. :J:t .
@N@
:.;- ;
t:
^@N@ 'îY: ..iÍ ¡ ":S ooô
¡,
Sketch the influence lines for the moment and shear at the midspan of BC for . the beam shown in Fig. P8.2. Hence, determine their values if there is a I uniform load of,intensity 15 kN/m on the entire beam.
j.: .
T-
;): 't i"rü.ìi i
AT
ABC
c
'..l'Tl;,rY.,'0,#t*
ð
¡.-¿m ' +"
:-:'iqi.; T,9.i".
o
à.¡r.:,
.;
:,:.;í,¿ .
åi:¡i:-i' ,,
f ø
ti::?:::
ìi;iiil
¡ Õ
@':
ø",
'¿,,s Ë oi o o È
B.
i¡¡:¿:*.1?¡
s HEARì
AI
2
,
sm
'f
Fig. P8.2 , Develop the influence line for the reaction at B @2 'shown in Fig. P8.3.
ABCD Æ þr Bm + 6m *
:
m interval for
,8m {
'(' 'i
þ- +-
,,,,1:i,,
;,':,.; l
:i( rr,i
.:',i ,
.:
Fig.. P8.4
-=.f 1
the beam
INFLUENCE LINES 8,5
using the computer program given in chapter 14,
for the vertical redction at A, moment at A
BE ofthe two storey frame shown in Fig.
;d
CHAPTER deve_lop the
Uenaing pg.5.
influence lines
*o..nt ;t;;;
nine
,p"n
ARCHES
Eig. P8.5
r( i
t:,{
ìi(
'|.,
_-_1
/ (o
)
Z- HINGED ARCH
,.( 'ii
i
'i',
¡
¡,,(.
lr''ii 1-.' j r,. {'t
(c)
FIXED ARCH WITH SPANOR€L
S
(b)
FIXED END ARCH
(d)'2-
HINGED ARCH
WITH SPANORELS
Fig. 9.f Typical arches
244
ARCHES
TWO HINGEDARCH
9.2 TWO-HINGEDARCH
A two-hinged arch
consists
of fwo verticar support
reactions and two horizontar support reactions. Thus, it.is a staficaily indeterminate stnicture by degree A two_ hinged arch mav be anatvzed.u'ing ;'r;;ãã.¡u¡rry
*f
i"riäJri¿í"i"iio
(Ð method of consistent deformatipns, (ii) sfrain energy method, a4d \ -'
(iii)
Moment of inertia of analogous column
l.
so
far:
at any
:
245
B"
I#
point is given by Eq. 6
PMv AI
column analogy method.
Column Analogt fuIethod
Let us use the column letrrod to determine the horizontar thrust H. consider two hinged arch sho*n ${9sr nrig.-t.za. lt one hinge support andreplacing "*i"'nu¿" ,r"ri""it;;#;;ää uy r"n,o'ing ir *ith;;;ll;.iuppon. Analogous column secrion sriown in Fig' e.2 is 'rh¡
h
in rh";; ¡en'inc.;^;;;;;ypoinr Ioading on rhe analogous.column ;;;;ä;iñ'witt be p. dsÆr.
*-iii'u"
u*.
?¿'
B'?¿t I Px Yqs J^ Er o:¡__+_Ë_ {u.teI. )
a
I*-t'' r ?" ¿s ,ilv--EI
Iv'S EI
rhe
i.
ìlèt sagging moment at any point (xry) in the arch is equal to
l-4 : F*- o : p* -Hy
îa* ----* ,,
nar
RBx
llo
)
rwo
HTNGED
AR.H
Horizontalthrust
lRsr
?ds H: jlf v2
L' (xryI
I
For vertical loads only on the arch,
.
H=
-9! EI
R* = Rsy
Knowing the horizontal thrust; the vertical support reactions
easily evaluated.
horizontal thrust is given by l
Total load t;
(
and
R",
can be
rib may
Fig.9.2
!
\ey
In large span arches, change in temperature and elastic axial deformations in the arch cause considerable change in the value of horizontal thrust. A detailed derivation can be carried out using the method of consistent deformations. Thls method will be used to analyze a fixed end arch in the next section. The expressioi for the
(
I
(e.l)
oi
the column
= f¡"dt
IEI
:. -
:
B
Mome¡t.ofload aboutA _ B axis Area of analogous colurnn
B
= =@
yds [ tt*
BJ. fJ:
iEI
Ju-t# +cI,r A ? r ds ?dscose
lt-el*l T : A. :
n¡
coefficient of themral expansion change in temperature area of cross-section of the rib
(e.2)
246
ARCHES
e : E=
-
inclination of the arch axis with the horizontal modulus of elasticity of the arch material :
the case of ci¡cular arches, direct integration is possible if the msmer¡t of inertia of rib is constant throughout. The analysis ofarches is very senbitive to numerical d.offerrors. Hence, it is recommended that arches shogld be analyzed using at least 6 signif,rcant digits on a digital calculator.
The effect of rib shortening is to reduce the varue of horizontal thruslby abìout 3yo. flat arches o. ø. *1r., tìu¡"ïä i*Ëää"rp*
is geúerally imporrant for very
"t¡¡.
Stiain Energt Method
,ILLUSTRATIVE EXAMPLES
Let us reanalyze the arch using the strain energy method.
9.1
A 2-hinged parabolic arch has a span of 100 m and a rise of 25 m. It carries
B^
u
strainenersy For minimum strain
energy
AU AU AM AH -=__ âMAH
.
In a statically determinate arch, AM
AH
-
: f 1 M'¿s J2 EI A
:
J'
ôH
t,l (,
() (l
,,,,t'
I
l,:
J
'^-
-s')#(-'):o
Intd* J' A
låüil;,.å"'"',Tlitï::'on where, Io =
0:
Vertical
reaction Re :750 kN, RB :250 kN
Ëquation of a parabola with A as origin is given by
4"21*(100-x)= * (r0o-*) v:aonr(t-*)= ' roo' loo'
Simple beam moment
ds
u-.:750x
This is same as Eq. 9. l.
EI
is introduced' It is
assumed,r'ur*o.äì'åit"n.
value of moment of inertia at the crown inclination of the tangent it ,".iion
"t
"
r ar any
t1.
{å
*ith horizontal.
{ a .q
r
in the arch is given by _2
1000
(x-25)
for x
I
50m
for x>50m
(Ð
(ii)
¡r*:25000 -250x
0fi
t
p*
- 20:2
= 750x-
ànd
In the case of paraboric arches, the rib is thicken-ed_near the edges. Thus, its moment y*t* ?rng.rhe span. Ir is rherefore difÍicurr ro inregrare Eqs. 9.1 or 9.2
:*.i"
tt. vo*
H= ¿E-
A
?,¿,
Iosec0
B
!i.li:rìi;r']1
A
J(u-
AH
H: lu.tti
{,r
(''
= JEI f][¿,
f Mds ôM
A
I:
¡ììliÏrr
B
lJEI
ds=dxsec0 and
0r,
,.:.,
B
'in#: iu.,#
or,
dx
'i'i.,'"1
.îB.
(.r
AM
o=
Equation for H reduces to
' or,
(r (r
ds cos
M = F* _ Hy
-:
au_
ìlîffi au
a
distributed load of 20 kN/m intensity of the horizonøl span over its left-half Determine the reactiohs and draw bending moment $agram. Take I: Io sec 0.
lt
and
l,
247
ILLUSTRATIVE EXAMPLES
B
.'. ju-ro*
:
i
:.
50
100
ro-']hå(roo-.110- * ]¡rro.-Lrvu Jirrooo-2s0x)#(r0o-x)dx J "0_ ,o 166.66
x
t05
248
ARCHES
B fv'
i
ILLUSTRATIVE EXAMPLES
a* : =
too
+ [*r(roo-*)'a* lo_10
. =
\
3.33
t00_
is of interest to nofe that the variation of moment in the segment AC is parabolic as by Eq. (i), and that in the segment CB is linear as given by Eq. (ii). The bending diagram is drawn by first drawing the F* diagram and subtacting from it the Hy
ih J{to'.' - 2oox3* *o)a* 0
x l0a
fte F*
itself
.
H I = -!!6'66"105 3.33xld = 5oo kN
if
diagram will be its line of thrust. The Hy diagram will be the arch the vertical scale is selected accordingly. The B.M. diagram is shown in
9.3b.
Net bending moment
M-: and
_ 5x (100_ x) : 750x l9:, _ 500 = 25000 - 250x v : 25000 _ 250x _ Sx1íOO_x¡ _
lvL
2-hinged circular arch has a span of 150 m and rise of 25 m. It carries a uniform load of 30 kN/m of horizontal span from 50 m to 100 m from its left support as shown in Fig. 9.4 a. Draw thè bending moment diagram. Also, determine the effect of rib shortening. The cross-section of the arch is lm x 2m deep throughout. Take E = 2 x lOa kN/cm2.
250x_ Sxz
forxS50m
(iiÐ
25000-750x+5¡z
,nulll
oot, of inflection
;;.'r;;ined
250x _5x2 =
.'Eq.
Y.?Ii*o* (iii) and
q
A
F*-HY
0
or,
by equæing
,0.
If temperature rises by 40oc, determine the increase in horizontal th¡ust and the change in bending moment. Take coeflicient of thermal expansion
,',rl*ïn:ä . ,:T
s.=
12
x l0-6/"C.
30
x=50m
kN/m
sagging momenr in segmenr AC ca¡r be determined by differentiating
equating it to zero, that is,
250x-5x2:0 Maximum negative B.M.
(
Or, X:50m : 25000 -750xJ5+ : - 3125 kNm
5y752
20kN/m
(', (.1
( (o) 2-HIN6ED PARABoLIc + 3125
( I
(-
t" (
b)
-
s212
ARCH
3125kNm
(c)
BENDTNG MOMENT kNm
562 5
BENDTNG MOMENT kNm
Fig.9.4 Fig. 9.3
(a)
, T,lig. horizonfal thrust in a 2-hinged arch
isgiven by
250
ARCHES
ILLUSTRATIVE
,B
fdsr
Jp.yEI A
¡¡-
(e.2)
BB
f r ds fdscosO JY-Er+J-ee
'AA
?a'
2 x l25a
l,a,
^g 1Ï(lzs.o.e*too)2 nde
J¡r.yEI
+qLT
I v-J' EI A
y=R.cos0-(R-h)
.'.
)2h t 4
\4
y = (t}scos0
sí,nc
: #:
q.:
ll.54o =0.20rad. P
p.
=,,Sm
i+* I AE
100)
o.r, sinp =
Simple span moment
Ir*
-
)2x25
# :
36.87"
:
: r#
at point G
g >0 >cl
ln3
r-c3
t2
t2
2 x l}a kN/cmz
0.344x125' 0.0260x125'
H= H:
ds = RdO
.'. J r*vdx :
". zl[tso(tsr25sin0)- t5xt25z x(sina-sine)'?]
(rzscoso- ro0)
125
d0 +"zlTsqT'_rzssin0) (t2Scoso _ 100) x 125d0
=
2 .x 108 kN/mz
lEl
0.6295
/El+ 150/AE 38.12x l0-5 + 3 x l0-7
1651.3 kN ignoring axial shortening 1650 kN including axial shortening
'The effect of axial shortening is to reduce H by 0.079%.
:
!e -
30
x50/2:
M:
750kN
750x 1650y : 750x-- 1650y - 30(x- 5q2/2
, ,
c
.0< x < 50 rn 50< x < 75m
.
where, x is measured from support A.
t
_1,_ = z tzs4^f,
.
l(t 6 - 6 sin 0) -
I0
.R
: fff *,r0, :
.Mgm"nt ai any section is given by
p
x
o.ot2'6:t2s3
,Horizontal thrust is given by,
px at point F p, = 750(75-Rsin0) -30@sina-RsinÐYlZ_ a >0 >0
Simple span moment
I
o do
:
A:b,xD:lx2:2m2
I : "1 :
E
:750(75-RsinO)
t(*l',
r.6cos o+ 0.6a) ùa
'o
'Moment of inertia pf rib
O.S+Zrad.
-
ot : : ,T*"î'S ]lrrr"os AE AEå
Area of crobs - section of rib
: o.u
l25o
EIJ'0
.
*= |.{.n'l+ :ftt*rr'l=-!
¡o.ro5+ o.o67l = 0.344 *
/ 1.5(sin2
a
-
2 sin q, siri
p
+ zJ tzs4
[(s.e -6sin 0)(cos0 - 0.s)Jd0
e + sin, e)] (cos' _
0.
s) d0
. x:20m,y: l2.Z5m, 1 ,, atD,x=50m,y:22.47m, .'. : : at C, x 75m, y 25m, .
,|.
Thc,bending moment diagram is shown in Fig. 9.4c.
L,¡.:, .
@
251
EILJEI
Let O be the origin. (Fig. 9.ab)
and
EXAMPLES
.
tf temperature rises by 40"C,
M
M
: -5212 kNm : 424,5 kNm = 5625 kNm
150
AE
FIXEDARCH
ARCHES
H:
O.6295+12xl}axl50x40
_ 0.7015 :
0.0003812
0.0003812
due to the applied loads
l840kN
idor the released arch as shown in Fig. 9.5b. Let net moment af any section be If B is taken as the origin of coordinates with the positive direciion of axes h in the same figure, the deformations Âx, ay and Ág oi a sma,ll element of the -F oñn bo evaluated using the moment area theorem as follóws:
The th¡ustH increases bY ll.50Vo.
Momentatcrown =
Mc :
:150
x75
-
1840 x 25
-30
253
x25212
875 kNm
Thus ma:
A
horizontal displacement of B with respect to
9.4 FD(EDARCH
rMyds Er
J.
verticaldisplacementofBwithrespedtoA,
Ay = - î$,É "'
derive some basic
B
rotation
of
¡ Mds
B with reslrect to tangent at A,
^0
JEI A *"u or
h
I I '+.'rl
shortening in element ds
horizonøl component
FIXED END ARCH
I l.
DUE TO THRUST
Éå
vertical component B,
; H
(b) ELASTIC
S diagram
-l;
of
of point
O'L =
OO'cos 0
={$"org AE
e.aa)
OO' = OL =OO'sin0 Ay
rotation dQ
Pdt AE
o
= P& *ioe AE
oor
(e.4b)
.
R
\
(d ) ELASTIC DEFORMATIONS OUE TO IEMPERATURE RISE
or, Aþ
Pds
AER
(e.4c)
R > > D (thickness of arch), Â0oo -;;:,, 1,::' 1lillce llofOnnntions due to temperature
(-,
(
OO' =
DEFORMATIONS
DUE TO APPLIED LOADS
(,
{
fo,
of
= oo': or,Ax
ELASTIC DEFORMATIONS
Ax
(.'
(9.3c)
Let us
I (o)
(9.3b)
e
better.physical' relationships fof deformations due to moment, thrusÇ shear and temperature. Let us remove the support B and the fixed arch is reduced to a curved cantilever beam fixed at support A.
T
(9.3a)
B
A f,rxed arch is shown in Fig. 9.5a. There are six support reactions. Hence the fixed arch is statically indeterminate to degree 3. It can be analyzed using any flexibility method described earlier. Let us use the consistent deformation method which givgs a understa4ding of the stucture. Let us fust
A, Áx
Fig.9.5
conlidgr the rereased arch shown in Fig. 9.5 d. If the temperature rises by to and cr is the'copfiltient of thermal expansion, length of the element ds will increase bv
.
254
A,RCHES cr,
SYMMETRICAL FIXED ARCH
element ds :
: and vertical elongation of element ds horizontal elongation of
a t ds cos 0
(e.sa)
t ds sin 0
(e.5b)
o,
Deformations due to shear
Mc
('-
These may be neglected being very small. The net deformations in the arch are obtained by supqrposition ofdifferent effects and integrating aiong the arch:
Âx i+f .l#
"os
e-|ot
ds cos o
þ-\t2
:
(e.8)
IEI
moment at any
Âx=0, Ây:0, and
f$A
^0=0
(e.e)
There are three unknowns M4, [Io and Ro and three simultaneoüs equations. The moment M and thrust p can be o
(., i.
(:
I
.t
l,'
.
.',. '.,-II11 ,
Á{o repiesen*he disptacemenæ of C¡n
Âxo
: ^xB
(e. (9.
fl"y + R.x-
r0b) l0c)
is given as
mn
(9.1
la)
cantilever B.M. about K of all loads between K and C
that
is,
*
^,
Hc
(9.1lb)
-approximation is accephble because the effect of rib sho,rtening is itself very small snd it simplifies the calculations
fte
LThus,
irä;i";
arch CB, then
(9.10a)
values of M and p given by Eq.
âx,r
9.1l
can be substitured in Eqs. 9.6, g.7 and 9.g.
A
r
i{t.*".r*Rsx-m^)yil*fHccoseas - J'f atcosOds
(9.t2a)
c
A
-I
Af¡ = J(tut. * Hsy+
Let C be the origin for each half arch and positive directions of the axes are shown in
for compatibility at C;
Â.Yn
Â0¡
.
is symmetrical about a vertical axis through the crown c. The symmetry is in terms of geometry and not the loading. Let us introduce a cut at the crown C an¿ ttre resulting a¡ch is statically determinate. The redundants at c are Hc , Rc and IVl". These will act on either edd at c such thar the section c is in static eqqilibiium"a, ,holin in rig. e.o. ou" to the extemal loads as weil as these redundants, rhe point c in each half arch wiil qndergo certain displacements. Foi compatibility, the displacements in either half must be consistent, that is, continuity of the arch axis must be mãintained : ,
[f
=
: =-
c and B at K is aheady included through the vatues of at C. For vertical loads only, the thrust p at any section can be taken horizontal thrust.
'
The solution of Eqs. 9.6, 9.7 and 9.g becomes simpler in the case of symmetrical arches having their supports at rhe same level. considerãn arch shown in: Fig. s.6 whìch
lnlr*3 ¡t*", .*d ^xA,. AC and Âx", Âys and ^V¡ ÂQ" in the otheihaif
,
of loads between
calculated.
9.5 SYMMETRICAL FIXED ARCH
.þ
pointK in the portion AC
M = Mc+
"
For compatibility,
(r
Á01
.,,
f Mds
l--
';ll;. Ltz
Fig.9.6 Symmetrical fixed arch
(e.7)
ÂYn
^Q
Hc
ln C
(e.6)
? par 3 : - J? tttx¿r*JO-sine-Jards;in0 ,, ^y .AA
.B
255
ds t
., A$e
A
ds ?I H" tt^ sin si Ods i Rçx-r^)-#*l Er.l-=;'--Jatsinods
(e.tzb)
Ï,-* = J(Mc+H"y+Rsx-m^)3 'ds õtr
similarly in the other half arch, the net moment at any point K' is given
e.Dc) as
!i¡i r;i.:r: ìií 25i6
ARCHES
M = M"+
and
HcY +
\x-
(9.13a)
P =Hc
where ms =
ELASTIC CENTRE
iL¡',:'
m"
(e. r 3b)
cantilevermomentaboptK.ofall loads-between C anci K'.
The values of M and P can be substituted in Eqs. 9.6, g.t and 9.g to get.Âx¡ Âys , and Â{". Substituting these values of displacements in the compatibility .oñ¿iti*i given by Eq. 9.10 yield the following expressions:
ifl :;, i..'. -.: ii,lÍ.:,,
-.
I
Rc:
-,'")** Î,*^ c
Â0e = -Â0s
.i{r.
jì *î ""',ï to' - î ttr,,+la6-=Jatcos0ds
: -ifr" + Hcy-nsx- mr)y$-f ï'-EI B
,
ð
nc
¿
c-os
e¿s
AE
*fot"o,
(e. r 5)
A. x'ds ^f J EI c
&e= -Âx"yields
gives
Jit.*Hcy+R.x-*^)# c
* H"y+ Rçx-mo),
A.
A
c
cc
ît-rds : -J(Mc+H"Y-Rsx-*")EI c
¡' cls I j ,r.J# *.r".i#: J#:J(.o*-")Ë
eo,
c
257
I
ds
(e.t6)
since the arch is symmetical about c, the integrals of similar ñ¡nctions from c to A the same values. For å given loading, Hc, Rc and M" are constant and can be taken outside the sign of integration. The abov" can ùe simplified as : or
c to B have
"{uutiãn,
2McÎy#+r*"i# j{*n *,""¡r#.r".iqp A
î
:
2o,tldscos0 J
c,
Âya:
-
yields,
: atl.
(e.t4)
in a conjugate beam. A similar observation may be made in arches. An column section of the fxed arch is shown in Fig. 9.7. Width of the analogous at any point on the axis is equal to lÆI where I is the moment of inertia,of the at the same point. Centre of gravity of the analogous column is O. The ofO are chosen so that
jl : o= fuA J'EI
Jf* EI
(
1.
(
(x,y)
1
or,
(
ANALOGOUS COLUMN SECTION
Fig:9.7
ARCIIES
Yr =Y-Yo
the three integrals given by Eqs. 9.20 in Eqs. 9.14 and 9'16, we get
Let the
(e.t7)
zM"
(e. I s)
(, (. (
i( I
(
The point
yo
'
¿c
zM.l#.2""
+f¿s
(e.
¿EI o is calred the Elastic centre.
keepìng positivd direction of trr" *ãr by making use of Eqs. 9.17 andS.iS.-
The concept of erastic c-entre is
-;;"
Jtil;Jløîäu",n
iso
c
r,Î* =
Ï,-^
*'")*
(e.ztb)
le)
(e.22) usefur
z
o, rur.. --"-'"oordinaterusefur relations can be derived firsr ro
¡'#=f{,,*,:)# = j+i.*J# =,.j#
[!r
.cJEI
it#=ir,
*,0),#
[l'nn *,n")r$ -c
(e.23)
=
J'"#-*;ig.",j,,s =îr:#."'i#
Al
(9.20a)
u fixed symmetrical arch can be analyzed using Eqs. 9. I 5, 9.19, 9.22 utd 9
ffi
(e.29b)
A
.
*rn"Xr, J('no
erymÞJo
*yo)#
':: ,,i,
'
.
t
r.r
rtnglyzæ the arch
of Example 9.1, if the two ends are fxed
.-,:r.',
, rolúrrçn
(
(
(9.21a)
,å¡.hetituting the value of Ir'Í" in Eq. 9.21a gives
El
(,
: crtL
o
in the anarvsis of frames. The origin of
(
(
=
Ards - EI JlY'
=
(, r.
61
J
[(y--y,)* t' 'c
or;
oî30: *. *'i*]. r'.i
.,ds +, , ds Ï, (m¡ +ms)Yror - YoJ (mn *mn/
A
(.
y,î*. r""[îr,'
? v,¿,
JEr:o c
and
259
ILLUSTRATIVEEXAMPLES
Due to symmetry of the arch, the c.g. should be on the axis of symmetry. point O be at a distance yo below ttre croln C.
(9.20c)
Tha olastic centre of the arch is given
liy
as shown
in Fþ. 9.8a.
-23 -
:.tì
260
. ARCHES
t)¡:,
261
ILLUSTRATIVE EXAMPLES
20kNlm
T--f-Ì_-r-lt_r_tl { t lc
ï",;,
Hc:
'l:.':....,
îlt,,,,ì-Ì,F
.j.,:.t
:*,,'-
A.
Mc : t
I
*-")fr Ã
(e.22)
-HcYo
,J*
ds _ dxsec0 Io sec0
BENDTNG M0MENT, kNm
-50
Fig.9.8
50
f n-9e
J-F,I
Yo:å--: td' JEI
î"9 i. EI l'
¿
100
I
Yo
(e.23)
l.¡
3125
$ìotL Èt
eresl ,li"3e. er f AE
J(*^
(c)
+ms)yr
Lå'',
.
::::il::¡:.rlj,,:.:.'
(o) FIXED END PARABOL|C ARCH
î(*^ -^
0
(e.te)
qs
Yr
J Dr
Tl¡e stress components at thJcrown R.,
Hc
and
Mc
are given by
:
Y-Yo
lx" dx {rooeÇ
j'î
_
|.*t ¿* -l-100 J
--#-
:
8.334 m
?
Jdx 0
:[*--rro), * =ro+, -"]:o
i'-,.-r,'',Various integrals can be evaluated as follows (Fig. 9.Sb) : n
dx J
(mo
_m")xfr
EÇ Io
R.=a..-' .¿l._-¡ x.ds
¿EI
(e.
f-^ ¡ dx luxJI EI^ å -elo
l5Ó.25xl0F.I^ EIo
l5) ds Itr
EI^ J,
il,o.,)l¡1_r.rroì-4' = '\roo
{'
)Eto
27.77
rÊ
EIo
262
ARCHES A
[(.o *,n")S r,o é
, Hc and M"
156.25x105/EIn
z*o'+ro*rd7fr
,"r?BuEr"
.,
'1
(.'. :
or,
() (.r
(r
x:
or'
-
Z"SOIEL- -
14
:
Me
= ,o "{-
l0õ
: 0 or, (5x-
dM': dx
20x
c¿rn
now be eúaluated.
*
-
187.5
-
lOx
fQ x 18.752- 187.5x
o
=0 or, x:l87l5m
t8.75-500x# :
-n57.8tcNm
sagging
!.þ,.g,'drawn as
500
x
8.334
=
the rcactions in a symmefiical fixed ended arch shown in Fig. 9,9a due to kN acting at 20 m from the left support using the cotumn analogy method çgnfre method. The width of the arch rib is 0.90 m constant, and depth varies
zero
i,g¡Bression (i), where x is measured from 'atr:
:3tzskNm
with C
(r)
3l25kNm
as
xt2
origin y' =
r00
hogging = tan
-
sagging
f
0, and
n'S
v'=|¡s
ds
= l0 sec 0 = l0Jt * trtte
:5.r6m
JEI of the arch section at the crown
I
2 -R. x -Hc,y _Mc = 0
crown.
Analogy Method
-rr,r"
tn7.sx50 1500x25
th_e
.
':('.")*
of inflection can be determined by equaring tvÇ equat.ro zero, in rhc teft wx_
0
ions can be done for the right half 4rch. The bending moment diagÍam shown in Fig. 9.8c. The various integrals should be'evaluated .f *ignificant digits on a calculator to avoid numerical errors.
187'5kN
çan be easily calculated.
ä-R"x-Itr
:
o.
H^: =wx-Ra- 2r 100
dx
187.5)x
:5ookN
RB = R" = 1S7.5 kN I HB : Hc = 500 kN..
n'llrîf:'o
v2
500
263
37.5 m.
Ro = 20 x50 - R. = Bl2.5kNt HA - Hc=500kN-+
{-.
(':
:
4'16TxlollPt^
Now the supportreactions
('
or,
dM*
at the crown
z?l?tos/Ero H^ "'c :
(r
-
Eto
Ero
The reactions Rc
(.'
EIo
:Jl;-r.ttoJrro ìr¡" _2780
lt'-.
(l
xo0 l0x2
tl( ,,
i z d,
(.-
EIo
0
Í
A
M. - -L:
4.167xl}s
: f*'-9t :
t'î.'-g. EIo
R. --(-
50
: [to*t d*,: J.
ILLUSTRATIVE EXAMPLES
ô=Tf
-:o.o75ma
ARCHEb ILLUSTRATIVE EXAMPLE
,
265
The computations of the elastic centre are done in a tabular form for
f.i.
The computations of elastic loads and properties of as shown in Table èoiumn seclio.n are shown in Table 9.2. Thus, values of the reactions Ho ,
.È¡d Vo can be also computed at the elastic centre. However, the stresses in an column,section can be determined directly without making use of 'the values Élastic centre as shown in Table 9.3.
Table
9.l
Computation of elastic centre
Ð67. Yo
Table tr-
1,
9J
Computation of loadsu moment and inertia
,
'{ I
;(
l:
r\
,t !it f
(c) ELASTIC
' ;
CEN,TRE
P=
i i
i( '' i ì--. '
-27923.7
33.89
i(',
I \..'
t.,
|
'',
17
(d)
35.83
6.67
Mo
BENDTNG MOMENT¡ kNm
Ër _ no
Fig.9.9
(
The arch is symmehicar -about tåe crown. Let harf the arch be,divided in segmenß AD, DE, EF, FG and five GC. Their centroids are indicated by points 1,2,3,4,
and
P _ A
27923.7
:
52.23
:.
16.69
534.6
M*_355027 :
,W
vo i:-ffi:-42s
:
1380.060 _s.16
m
266
ARCHES
ILLUSTRATIVE EXAMPLES Table
Point'
x
M,
v-
P
M"
Ã
A
50
19.84
3
25
1.09
c
0
3' B
-25
r.09
-50
19.E4
-
267
9.3 Computation of stress in anrilogous column
5.16
I*'
M"
Iy -X -t000 -
86.12
-52.23 :52.23
-t8.19
-52.23 -33I.13
0 106.25
+2t2.5
M=
Mi
M"-Mi 404.1
t76.67
33.89
-JJ.ð9
35.83
-35.83
-170.86
170.E6
î'* Yo=å.
(e.1e)
J# c A. Ít
Réactions at support A are given by
J
HÀ - Ho: 16.69 kN -+ VA = 50 + Vo:50 - 4,.25 =45.75 kN I Mn : - 404.17 kNm hogging
Rc
r
(mn -mn/x
=c
'
A 1.
oS
''
(e.r5)
2fI-9! JEI c
Reactions at support B are given by
HB =
VB
Mn :
Ho
:
16.d9
A¡
kN <-
Vo :4.25 kN I 170.8 kNm sagging
J(.o "c
ft
(Ignoring axial deformations)
A''
(e.23)
2fv,29
å"
The bending momerit diagram is shown in Fig.9.9d.
(iD
**")v,# -Er
Elastic Centre Method
ïhe moment
at ahy section in thç arch can be determined knowing the values of M0, and Vo at the elastic centre as follows:
M=tnt itUo+Hoy;Vox
l
1000
+ 52.23+
16.69
x
+
Mc:0 +
+ l6.6gx (- 5.16) + 4.25 x0: - 33.g9 kNm hogging
52.23
c
(e.22)
.Hcyo
-
19.ß4+4.25 xS0:_404.t7kNm hogging
M¡ = 0 + 52.23
16.69
J{.^ *.")# Vt, : !-- -¡
,J*
sagging moment is taken,as poqitive and hogging moment is taken as negative,
M^:-
A.
x t.09 + 4.25
x25:
t76.6I kNm sagging
I
,l
nl .n'ft 'r
Example 9.5 Reanalyzethe circular arch of Examp le g.Z
I
[¡/
iç
V/
its two ends are fixed.
0
Fig.9.l0
t/*
268
ARCHES
ILLUSTRATIVE
y=R-Rcos0, R= l25m y:125-l25cosO
EXAMPLES
269
LetC betheorigin,
:.
. EI þ
:constant, u.:0.2 radian: ll.54o,
Hc
sinq:0.2, cosa=0.9g
:0.6435 :36.87", sin p : 0.6, cos p = 0.S
F¡
n5
B
=
y,
:
2"4405:1975
:8.36m
###
: :
Et
f(n-n*,e)H, =#f(r-"o,e)ae :
Erastic cenrre
x l}s
80.37
¿Er i Er Er
Î'#
174
*',")fr J{-^ c
Let us fi¡st evaluate the va¡ious integrals used in these equations.
It- ¿r = n¿e
:
from
#
:on:,io,e f0o
c
L€ *lr,.,ro(nsine-r2.5)H
^?.c
F
ffjro"", 0s
32.98xl}s Mc : n rSO3l -
x-.YY__3" oetermtnate.
kN
edo+Í(r2sine- l2)de
: -+#
1975 x 8.36 = 4007 kNm (sagging)
support reactions can be easily calculated as the arch becorires statically
R^ : 750kN1, HA: l975kN+.
I
mA
: toî, 0
mB : mo: mn,
,+, 750 (R sin
0-
0<0< q<0<
t2.S),
n,
ct
involve addition/_subtrqcrion of quanriries of nearly the j :.:: y...rv .rçu!ç, urçrç lü.tt very rarge iluctuatton ""J::::l::.:t^"" :,|.]T:erals nuctuaiion.inlhivarue in the value ;,hr'.iiË;rai; of the integrand ¡;, liåil3i,T*',"^"-11;e'-9î:,:'_i-l"g .. significTr.dieilr considered. liìr.¿*.rnended rhar alt l:f:t*]Ti:.1",ïïber.of be. computed.uq to least 5 or 6 sigiificant digits dn u *l.uiia, in the ,.r'..'., !llu"1 analysis I ..,.;;¡ Of a
p
0, yl = e16.64 - R cos 0 ) = yr when measured downward from the erastic centre is taken positive as Since
Rc
circular arch with fixed ends. enrtc
:
A
Ît(.o *.n")v, ' ds : "i
Ë
J
^.(rrc.enRcose)I99 a
. . Reanalyze the circurar : elastic cenFe method.
[2"rcxz
arch of Example 9.5 using the coruinn anarogy method, and
$olution þ
I
1ì
J'I
\r
() (,
"
zso(nsin0
;,Er
:
('
z
-
r2.5)(l lo.o¿ _ Rcoso)
RdO
f rds
"on,put"Jãr'rf,o*n
Jv'-Er c
Thedistanceofelasticcenrre from C; y0
p
=
|
6rc.æ-Rcoso)z
iEI
Column Analogy Method
The circurar arch is symmetric about the crown. It is divided in 7 segments as shown :.jn Fie. 9.1 la. The elasric.cente i, in Table 9.5.
tl4 x l}s EI A
(i)
Rdo
=
0 *r K- r,
(o.el:-cos0)'d0
''J
4405
= ,î
:
##=
8.387m
The load on the anarogo¡s corumn.and its properties are computed as shown in Tabre ,þ',6, The stresses in the analogous column arä cåmputed * ,rr"..n-lnïJre 9.7. Thus, no! bending moment at differeit sections ¡n tr," uìrrrrun be determined as shown in the
:$ðttl€ table.
ARCHES
270
ILLUSTRATIVE EXAMPLES
9.6 Computations of loads and inertia
Table 30 k
N/m
t0
-
(o)
135190
FIXEO CIRCULAR ARCH t80.395 t158545 2438t .U.146484 t_16ó039t 2_so2.0s t_ s6.s,
: 160.79, I, : 2x 158545 : 2x4381 :E762, P =-33.20xt0s
2x80.395
A Ix
À4 670 5
(b) M0MENT
=-
Fig. 9.tl
x
Nç=
105
A
: -- 33.2x105 l60.tt- -20653
M*=
-
P
kNm
173.7
(-902.05 +902.05) x 105:0
xl}s
173.7
Ix
3t70gp
8762
Table 9.5 Computation of elastic centre
Mr= Segment Point
xt
v'
dy=
m
m
dx
ds:
I,
y'ds
0
Table9.7 Stresses in anatogous.column
clx sec €
tan e
AD DE EF
J
70 60 50
I a
FG
4
40
GH
)
HI
6
IC
7
30 18.75 6.25
Ll.43t 0.676
12.070 258.75a
Point
t5.341 0.547 I 1.400 t74.88', r0.43: 0.436 10.910 I13.84: 10.300
37.62(
1.4t4 0.152 t2.643
t7.87',
A F H
t2.5t6
t.95i
C
6.57a 0.338
3.65i 0.247 0.15( 0.050
10.556
69.38¿
74.329
(ii)
x
Ms
v 16.613
75 45
-
25
0
- 5.850 - 8.380
0.006
P
-A
46875 -20653 14375 -206s3 - 9375 ¿0653 0
¿0653
M *y
M¡
M:
Ms-M¡
I,' -3292',1 -53590 I 1.89 -20640 I t595
16609
-
905t 4Q4,
6705 saeeine
-3734 hossinp
-
317 hossine 4044 saeeinp
Elastic Centre Method can be.analyzed using rhe concept orelasric cenrre. Bending is given by :
.-^"113.y::-"Y:^1f_ry.L nrÕmont at any section in the arch
M:
Ms
+
Mo
+ IIoy -
Vox
,
272
,.i::
ARCHES
INF.LT.JENCE
vo
0
MÂ MD MF
Mc
I*
::=:
1982
L-z
tl: rX
:20653 kNm
Mx :
273
trnit load is at a distance z from A
Sagging moment is täken as positive and-hogging moment is taken as negative.
Ho
LINES FORA HINGED ARCH
an-*l
l¡x
kN
10rx
L
for x >z
L-
B
î
: :-
46875 +20653 + 1982 x 16.613 - 6705 kNm 35625 +20653 + 1982 x 6.954 lt89 kNm 24375 + 20653 + t982 x (- 0.006) : - 3734 kNm 0 + 20653 + 1982 x (- 8.38) = ¿t044, kNm
J
i¡
l.
vo*
i(=).'*.i;t'-.i,*
A
:Ì,r
,íi.
..ã-:ì
=
:ì,::l:r':-
::;ì'
The bending moment diagram is shown in Fig. 9.1lb.
9.8 INFLUENCE LINES FORA HINGEDARCH Influence lines are very useful in the design of arch bridges. iA,n arch is economical because mom€nts are reduced due to the presence of horizontal tlrust. Influence lines for three hinged arch were drawn in section 13.9 of volume I of this book. Influence lines for horizontal thrust, radial shear and moment in a 2-hinged arch are drawn in the same manner. Let us draw influence lines foq a parabolic arch. The following slmplifications are introduced :
ftt
#!,n-'lxz(L-x)ax+
FÏ,t-
xf
xzdx
ro,(t-z)(Ê +:.',-22)
------æ H
nz(r-r)(t!*rr-r') l3 x(tns)hzl
.
L
Moment of inertia at any section
t:
Io
ã[[.t- rf .fj
H= 5L(z 223 roì
sec 0
2. Effect of rib shortening is ignored. Horizontal Thrust
(e.24)
line for [I can be plotted for different values of z from 0 to L as shown in The maximum value of H occurò when L/2 andis eqúal to 25Lll28h.
z:
Nonnal thmst at any section d is given
f tt* Yds
J-¡r-
H:+
Io sec 0
(e.l)
-
lv-dx
J
I:
r,
'J F* Ycrx
B----1).
" y'ds f -Er
since
B,
and
ds cos
0:
'(i'
dx
i\ ¡'l
lr. ri
\cos0* + Hsin0,.
x)
-.r,-.= I| ronzx2(r.' - x)2ox L' L' li ?lar-4y
t..
,
(e.2s)
Radial shear at any section D is given by
line for radial shear is given by the gum ofinfluence line for FI times sin 0.
and the influence line
i
j
Y:-T-
-
T:Hcos0* -\sin0* 0 : slope ofthè arch axis at D V*: vertical shearatD ii ,
line for normal thrust is given by the influence line for horizontat thrust the infll¡ence'line for ver,tical shear times sin 0.
Equation of the parabola wirh A as origin is (Fig. 9.2a) +nx(r_
as
8h2L 15
0=0,
Q = V*
idluence line is shown in Fig. 9.12b.
(e.26)
for
vertical
274
ARCHES
INFLUENCE LTNES FOR A FIXED ARCH
275
(o) TNFLUENCE LINE FOR Ft LINE FoR M0MENT
:yNFLUENCE cos 0
(b) INF.LUENCE TINE FOR SHEAR AI CROWN :j,
:
,ti:ì
;wt'tnrwENcE
LINE FoR THRusT
i!irì:È.ì,
-tt'.' t-
',
iii',1i1 ì:r:a
(c) INf
L-UENCE LINE FoR MOMENT AT CROWN
Fig. 9.12 Influence lines for a two-hinged arch . at crown
Moment
Moment at any section is given as
}È}"TNFTUENCE
À4 : t-
^-,Ïï"" lI
-.FIy o,:lt, tr,",.rny"ry¡ tine for p*,
subnacring from it ord¡nare of the influence rine for H, trre innuelfe Td r¡r¡ç for fihe ror Ä4x tra_ ls isiotaineo. obtaine shows the influence line for Fig. g.r2c M* tú
",
, ,r^:"':"
;;;**'sv'eç
The infrr¡ence rines at any other section can be protted using Eqs. g.24 to g.26. The influence rines ar t"t ouatæirpun åtöffi;ir for moment] nor¡nar thrust rh".* sho*n in rig. e. t: u au¿ radial T t ï';,äì ," check rhe arch 1
sprngmg, quarrer span an! rhe ""ì;. crown. rrrr,e assumed to be safe at all other points.
rib ar rh¡ee secrions
-"r, ¡r."r"-"'rrrl*'ri,înJïl"r"nr,
'
Fig. 9.13 Influence lines for
from the crow! C. Qolrtilever moment mA at a point
¡ç = (x_z)
=0
and,
K,:
Cantilever moment ma at a point
&'
O
:
it is
*;r;;T forcg--at any se*ion of a fixed arch -"; M" . Thus, ir is obvious thar rThe üiá*n ". Td ro¡t uirr," varues of iàifii*.* øiir¡on, oru unüi"J;;il "ro*n. *r[ from 0
a two'hinged arch - at'quarter span
io,:L. Cônsider an arch shown in Fig. 9.14 where a unit load is placed at a distance p
'.,'
9.9 INFLUENCE LII\TES F'OR A FD(ED ARCH The Áfluence rines for momenq rhrusr can be easiry carcurared tno*ing t¡¿ ?ori"l influence lines for rhese forces õrn rt""" ä Hc and I\4" will be calculated
L|NE FoR SHFAR
K
(xry
)
Fig.9.r4
for x>z for x<0
276
ARCHES PROBLEMS
î-o' I mÁXr^.. z
the circular arch if the uniform load of 25 kN/m is spread on.the full.
%
sPan.
Ir,.-¿.#
EI -'
Eq.9.l5giveg Rc = c-
=
f*'A ¿-I EI
't*
-_q__ z i*'S
353.64 W,R¿=433.0k1,1,
0
194.1 lNmsaggingJ
lJ2,
\ caii be evaluated. Due to bufîheir sþ will be opposite. simirarry; iË"'."u¡ìo lì tt" *o"o'tn,,¡elääi*n rvrç can be evaruared. The influence lines for Hc and nrg iír1 t" ;r-r;;ar about rhõ crown. These influence lines a¡e shown will
Mc=
IEI
By putting difr'erent values of z between
symmerry, the same ordinates
277
and
appry when the road ¡,
\... Goo I
so' ,1(." -<Þ'*"'
in Fig. 9. r 5 4 b and c. inuence line ì1, R , rlc and rine or¿irr¿æs råi "øi""r"ro,""*;t;,h;;;;ct¡õn
o*-ü;
hil:r-d:frr,i:,#lJ**
*y
ci,
be
0
Fig. P9.t
If
ternperature
of the arch of problem 9.1(a) drops by 50o C, ¿çtÀ¡ne t¡e
(i) including axial deforrnation, (ü) excluding orial deformation. Take coefflcient of therrral expansion c¿ -- I I x l0-6fC change in torizontal thrust
,(.
TNFLUENCE LINE'FOR
A twe.hinged pà*Uoti" a¡çh has a span of 150 m and a rise of 20 m. It carries a uniform load of 50 kN/m of hori¡ental lenglh over its right half span. Draw the bending moment diagram assuming that the moment of inertia of fhe arch sections I varies as Io sec 0. [H^ = 3515 N, Rn = 937.5 kN, Møimum moments = + 17578 kl,Im at
:
,{
(,
(
(b,
TNF.LUENöe
quarter spansJ.
lrNE FoR Hc
,Reanglyze problem 9.3 usìng the column analogy method if the ends olthe arch are fully restrained.
(
For the arch in probiem 9.3, draw influence lines for horizontal thrust H in the atch and the bending moment at a section at horizontal distance of 30 m from the left springing; Determine their muimum values if
(
(c)
(
INFLUËNCE LtNE FOR Mc
Fig.
('
'(i)
a concentrated load of I 50 kN cfosses the span. (ii) a pair of concentrated loads of 75 kN and 125 to\ spaced I 0 m apart
9.15
crosses
the span.
PROBLEII{S
g'r(a) (
(
:
(
(
;t¡v
A 2-hinged circurar arch is shown in Fig. pg.l- Deærmine the support reactions due ro a unifr,rm road of 25
ñ/,o-;i
ûy.Jeft
hJf
ff"iä:"iffi,}:,;îîtffi:iî;¡Ä;';ä :
[IlÁ
176.85
W
RÅ =
szs
I *n, iu-Jt'07.7 W E- =
,Ë;äril" -*i'ä" ï , s.bending ¿,.e 96.8
k;t{n saggingJ
Reanalyze problem 9.5 if the ends ofthe arch a¡e fully restrained. Detemrine the support reactions in a fixed parabolic areh carrying a vertical load 100 kì.I at left quarter span its span is 200 m and rise is 25 m. Take I = Io sec 0 tH,t = 05.6,3 N, R¿ = 84. I 5 kN, MA = e) I 03 4 kNm hogging at Iefi support,
of
if
!
Mc = I
193
ÌNm sagging at lefi quarter span Mu = J
right support
-
795.5 kNm, hogging at
278
ARCHES Reanalyze problem 9.7
9.8
'
theixis of the arch is circula r.
if
[Ht = 106'AO N, Rt = 84. 1 2 kN, Mt = - 991 .5 kNm hogging at left support, M = II82 kNm søgging at lef quarter span, M, = - 832 kNm hogging
at
right supportJ ?.e(a)
Determine the suþport reactions for a 2-hing
Also draw bending moment diagram. Take A
I
:
IHÅ: II.0 kN, Rn: 38.25 W,Mr: - IJ kNn hogging, Mr= 12.34 kNm M, = I 5. I I kNm sagging, Hr = I 1.0 kN, RB = 1.75 kNJ
sagging,
Fig. p9.2 (b)
.
Reanälyze the circular arch if both its ends are fully restraint. RA= 3!._25 N, Mn: 6.54 lNn sagging, Mc = 6.g6 [H,t= 10.5
W kúm , MD : 13.6 _ kNn segging, ME : Z.iS kl,Im sagging, M¡: 18.54 kNm hoggingJ 9.10(a) A 2-hinged parabolic arch segment shown in Fig. p9.3 is loaded with a
PART 2
hoSSinS
' .
A:
(b)
55 YTETHDDb
concentrated load of 50 kN. Determine the support reactions and draw bending I Io sec 0, Io 025 ma, E = 2.5 106 kN/m!
moment diagram. Tíke
:
:
x
1.0 m2.
IHÁ:77.38 N, Rn.: sg.r kN 1, R, : _ g.t tN J, nr: Mr:80.5 kNm, sa4CingJ Reanalyze the parabolic arch of Fig p9.3. if both its ends are ñ¡lly
IHA:
35.60|N, R,t= 48.35
W1,
Ro= r.6s kN
îJ
77.38 ktrm
:ll;:,
r"rt uint.
?:., :+ì,
50 kN
B
T
5m
1
u5+ 10m ., Fig.
,
P9.3
CTIAPTER
ten
SLOPE-DEF'I,ECTION METHOD-
,.;r.,lo.l
INTRODUCTION .
The slope-deflection method can be used to analyze any statiòa[y indeterminate beam or rigid frame. It is based on the assumption that flexu¡al deformations are predominant axial and shear deformations. The rotational and translational deformations of the ., .., over
ir'rigid joint are treafed as'unknowns, and their values are detemrined using the .r:' compatibility and equilibrium equations. A joint is said to be rigid when the angles
the members meeting at the joint remain unchanged after the application of the " :,.- between Thus, tangents to the various joint
rload. members meeting at a should undergo the same rotation, that is, the joint rotates as a whole. The end moments and shears of a uìember :rlr;:.,:Qan be exiressed in tçtms of-the loads on the span, slopes and relative deflections of the tir,',,: o ends of the memb.er. It will be shown that for each unknown joint rotation or :'.1-:&flection, there is a conesponding condition of joint equilibrium. consequently, there .,r.{e as many equations as there are unknown joint slopes and deflections. Thus, knolving ,, r ;:ir.:,:ll¡s slsp.r and deflections, thé end'moments and shears can be determined ,,
¡,:,;1
(: (,,)
("l
-..,
() \.) (_r
(, () (_ì
The slope'deflection method was the forerunner
of the modem stiftess matrix it is now
,.:.'analysis. unfortunately, it is not apenable to computer programming, hence r:-.;¡sed only to understand,tbe physical behaviour of simple problems.
DEVELOPMENT OF SLOPE-DEFLECTION EQUATTONS
;i:.
consider a member AB of a continuous beam or a rigid franie as shown.in Fig.l0.l. ::fite member gets deformed as shown in the same figure under the application:of the .,,:r[g¡ds and deformations of the continuous.beam or the fr¿me. Let the end ¡otations be 0o
.,o, er both in the clockwisé directions. The end B sinks by  relative to the end A suçfr that fhe_ noember rotates
clockwise. Let the net end momenis be Mo and M", again both.
êlockwise. Thus, clocktise rotations and clochvise moments.are considered to
be
i¡osltive. It is possible to express the two end moments acting on the member in terms dto two end rotations relatlve deflection  and the loads dn the rnember.
of
$'t-' 280
SI,OPE DEFLECTION METHOD
DEVELOPMENT OF.SLOPE.DEFLECTION EQUATIONS +
ffL
$$i
281
+
i
Ë,1
iil
(o) Mao
+
BO
T Â
I $t
*['^, i"i-""';.+]
fli
*['^,+":-'u,T'!]
Yez
(tC.2a)
:'
(r0.2b)
of Eqs.l0.2a and 10.2b gives,
Msr
fl:(,'
-*0e
:
+ó^
T A
(r0,4)
,L
'r
(.
(' a
(10.3)
Fig.
l0.I
Development of slope-deflection equation
The total deformations.of this mernber may be separated into two parts shown in Figs.lO.la and b' The condition shown in Fig.l0.1a is called the fixed end condition in which the moments M.o" and Moo are capabre of maintaining zero slopes
A
il
Ë
with loads acting on the span AB. The condition shorvn in ni!.to.to is called "t the ftee end condition in which the moments Moo and M"o are çapable. of maintaining the jdint
deformations 0a, 0u and  withour any iòãas actirif on the ipan AB. Thus,
*[å"^,i"IL - ]"*1"å] :^
282
BEAMS
SLOPE DEFLECTION METHOD
o4
MÆ=Ms¡=
6EIA
-E-
(r0.5)
l:': '
Substituting the values of end moments in F4.l0.lc and
+f
Mras
M¿=
Mrnn.iþ*.g"-l]
e^
Mn:Mrn¡*Seo*
or,
M" = Moo
Thus,.
if
fr" _
Dmw free body diagalo of eaôh member and draw shear force diagrarns. force, bendii,rg momenr diagrams and the deflected
llj 1"*
shape
structure.
(10.6a)
' trlro dcflected shape of the centre rine of the sfiucture herps in understanding its Hfiilïj."3:Í:är;:fl *t*1":,^11euenàinq,;;;i-.sr"*.*ä",rop".*o m as carculated are plotted at trre appropñaæ supports or joins by drawing r in the proper diirecrions- me points or inn""tiìi, uç p.oË,J, nä ur" .moinenr diagram. The elastic çurve or the deflected ,tp"i"í-¡"Ë*öl"rø fte supports or member joints and points
6EIA
. ?þ"*ze,_.!Af
(r0.6b)
known, the end moments using Eqs. 10.6- Tlpicar varues of rhe fxed fetqyinea loadings are shown in Ap¡endix È.
in any member AB can be drä",;.,î
ze the piopped cantilever bearn shown in Fig.l0.2a using the srope-defiection Þr.aw bending moment diagram and deflectedihaf.
,
*a.ã.""r, ä;"
.
r03
statically indeterminati stn¡oture, two conditions must be satisfied (a) compatibility condition, and lOi moment"quitiUriurn. a
compatibirþ condition
(
that Jeqyires
(bI
:
angle^
*"h;;ã
"ñîä;
il ffi; ilå
ofjoinrs rhar .ue can undergo a rotation. The condition of 'n*y moment equ'ibrium is satisfielb¡, making the qm of rhe end moments of ail memb_en ,neering äoäro This provides as many'near simultanuo"r t¡, nu*¡"r"oru'iããr"',e,s and Â,s.
"-ft";*,
il;1"il
[
,"ro.
fr" procedu¡e for the sorution cf ary ve¡rluÃsvsù veqrr beam or non-sway -frame is steqlr¡sæp --¿ continuous as follows : . ' step 2 -
write themoment equilibrÍum ' - slopedcflection equations
Step 4
-
-
Å,s.
end moments.
PL,
8
Equations
Mpec
*
tDr
î
rPto+oc)
PL , 2ET T * -i-o"l
-:
step 5 ' substitu'æ the values of 0's and Á's in the slope-deflection equations tò get the member
-
I
equations. Fo¡ the end moments, substitute the developed in stç 2 in terms of 0,and Â,s.
Solve the linear simultaneous eouations for-O,s and
,llYll'
\Z )\2 ) --7-
PL
Determine the fixed end moments due to the appried loads on any member. write thq slope-deflection equation for both ends of each member.
step 3
.,
^ rao_ -
"q;tio;ü
I
DEFLECTED SHAPE
Fig.l0.2
between the members meeting at a joint before the appricarion of fhe road must reirain apprication of the toad. Trris condition is sarisfied by assuming rhe sa¡ne í;ù"ãle-ilr a, members mpetins at a joinr: Thus rhere wif ä
step
POINT OF INFLECTION
EQUATTONS OF EQUTLTBRTUM
At any joint of
(
of inflect¡ons,
NNAMS
L2
0^,O.andÂ-¿¡p
of the
Shape
. +0" #
M^:
i
'r.
283
*-
PL E
.St"
0¡ =0beingafixeôend
SLOPE DEFLECTION METHOD
tsEAMS
285
(c) Joint-Moment Equilibrium Equations
Mnc
!+!
EMc (d)
':- 0 or
Mc¡ =
0
¡^ --32EI - PLz
or
,{"c
Back- Substitution
816
PL
:-
106.67+
Zff rrr" *r"¡
:-
106.67+
0.5
EI eB +
0.25
EI 0c
106.67+ 0.25EI 0B + 0.5EI ec
3PL
l6
o
(Ð
L/2
Free span moment at
.'. Net momentat
i
M",
PL
M^^:Mce:
I
L/2:
=
lL , andMomentaf, Ll2duetocontinuity 4
+ -+
=
The bending moment diagram is.shown in Fis.
yb.
S
=
-
iiqo'
i, 1 jltlj1¡!l.3 : ,', i-.Yçå !)
russine
Fig.3.lb.
0.4EI0o.+ 0.8EIoB
The deflected shape is shown in
,', tdtl Matri,
+
+ 0.5Er0s + 0.25EIec _ rc6.67 : EIoB + o.2s EIoc - 65 = o or, ' 0.25 Elon + 0.5 EIg. + 106.67 : 0 41.67
0
(iÐ (iiÐ
Solution
ESKmplel0.2
Y
tr+'o span continuous beam shown
in Fig.l0.3 using the slope-deflection
^urrr"a method. Draw bending moment diagram and deflected shape. Take
EI:
lo.s 0.4 erf o.+ 1.3
constant.
L
POINTS OF INFLECTION
þ- sT, 'F
8m
(b)
DEFLECTED SHAPE
.r:
I
Solution
(a) Fixed End Moments
Mpsc =
-
(b) S lope-Deflection I MÂB { MÂB
Mna
: :-
:-20x-
wL2
t2
t2
--41.67 kNmandMn¡n=41.67kNm
4t.67 +
[i;
.,r'
ir,
Aø"f - Substitution
("CJ ür.ÌüiÌ L-zoJ.J)
tlir nubstituting the values of 01s in the srope-deflection equations
M* = 0, Ms¡ = __ .
:
kNm, M"" = l22.5kNm and Mcn : 0 ïtrÊ bending morflent diagram is shown in Fig.3.2d and defrected ,r,up" i, shown in rrlg; lo.¡¡. l22.S
r'- -'ll ìl.Lìl:':!ì
R2
20x:-: - l06.67kïm t¿
Eqwat ions
Mres +
',ïl{r}
....,,;..".::.-;...
1.. {$
t2 -
0.2s
d
(a)
Mres:-
0
$tze^ 0.8
+ os)
EI eA +
41.67 + 0.4 EI
0.4
and
Mr"" =
l06.67kNm:
- "f '' 'ft¡lårvze the continuous beam shown in Fig.lO.4a ',. Ënd{ry shear force and bending rno**, àãg.u_r.
using the slope_deflection method
End Moments
EI 0¡
eA + 0.8 EI
0¡
)<3 =-'#=- too*þ :-
6Z.skNm = _-Mrco
SLOPE DEFLECTION METHOD
286
BEAMS
100kN c
ËÌ$
ôr
r:*çfl
i
j!
:1
Éë
rl
M¡u .; ì'
Ë|' ,..
!&
$5.
'. '..,i.-.,'
,
äi
(b)
f,;
#¡ i. ?ì
roo
BENDING MOMENT KNm '
s1.5 s1.5 lo:,9 10.9
I
^r-5rj'}-6DÈ+--ÌE 39.1 60.9 16.35 16.35
$¡., $¡,
#t'..
1.82
i:,-,
ü
,
97.75
(c) FREE
Ëi,
1.8 2
BODY DIAGRAMS
í,,
2oD) = 0.5EI ec
sQ
Torr+
op) =
(d)
r6.35
POINÏS OF INFL ECTION
+
0.3334 EI
,FI Equations
A being
a
simple support
Môn=0
(iíÐ E being a sfunple support
0gives, g.3F,I0a + 0.48I0"
-
62.5
Ogives, 0.3334810o+ 0.666?EI0E
:
0
,
(e)
'' (,
0.4 t.8
DEFLECTED St-tApE
fl)
0.5
Fig.l0.4 (b)
0
S ûíp e-D eflection E qu attuns
Mec
0 0.5
t.6667 0.3334
*iJlil lï1
leeo*e") :-62.5
Mc.c
62.5
+ +
S{zeo
Srt^
*e") = 0.8 EI0n +
+20ç) = 0.4 EI 0r +
(r)
(ii)
equations may be rearranged in the matrix form :
ii. lir',
OE
?(0D+20s)=o.3334EI0o+O.6667EteE.
*0,
'q
i:,
(
fu u il ibrium
0.6667 E[ 0o
(iv)
=0 0.BEt0"+ 62.5 + Elgc. 0.5 E I0o= .g {$,'* lOn + 1.3EI0c + 0.5EIoD + 62.5 : 0 * 0gives, 0.58I€c+ EI0o + 0.6667 EI0o + 0.3334EI0.:9 L,o. + 1.6667EI0D +-0.33348I0É: o
sHEAR FORCE, kN
'
li
+
+ EI0p
0gives, 0.4EIeA+
1i't
¡i
hl¡ú *Mamant
=
lirr
li!'
ç 0
*0,
,
ii
o" + 0.5 EI oD'
Trr.+
i,..Mun +
fiiÌ' ll
**
= EI
xQ
IÍ^r ', li$ !.,/ ll
287
-
62.5
Elfl" +
62.5
0.4 EI 0B
0.8
Ë:l=f'::,:i'}. tå:l [ ]l;i 1u' :VËlues
of eA, 0c, go and 0, in the slope-deflection
equations,
üt-,
ir
,
BEAMS
SLOPE DEFLECTION METHOD
{
i-.
f
i{-'
Mec = O,McA
f
fr,
Mo¡ :
f
ir,
ÍXr
: -
54.51kNm,M"n
10.9 kNm, tpo :
Free span moment
fi'
:
54.50kNm, Ivfo. _:
-
l0.9kNm,
289
AFr
ç^ Ø
¡, 5 ', 1"1
-OO
Pl : nj in span -4 AC at B :
kNm. The bending
(o)
momenr
diagram is shownin Fig. 10.4b. Shear force diagram can be drawn by first calculating the support reactions. The free body diagrams of each of the three spans are shown in Fig. 10.4c. The free body diagram gives, R. 77.25 kN and RD 18.17 kN.
i{l'-''
ji,t,
:
: -
The shear force diagram can now be easily drawn is shown in Fig. 10.4e.
'fl,, i;o l:..
68.85
in Fig. 10.4
(b)
,t'
Esrlmple 10.4
68.85 LI Af--__é,s_\ a^ \ \l I fr-l " P----Jo 13.77 ßl7 zd.z zl.z z3t ,.rr, (c}
\-/
Determine the reáctions in the continuous beam due to a vertical settlement
$lt-'
of l0
mrn
at the support B as shown in Fig. 10.5a. Draw shear force and bendiirg moment diagrams. Take EI : constant, E : 200 Gpa, I = 200 x l0{ ma.
ft,
FREE BoDY oIA6RAMS
Solution
fli,
BENDTNG MOMENT kNm
(a) Fixed End Moments
t' ir, Ii
There are no loads on the beam. tlence fixed end moments due to loads in each span rre 7Ê.fo-
íi
(b)
ì(.
it, (J
S I ope-Defl ec t i on Eq
Mes:
T(æ^+eo-iì!.01) : orEr0^+
Mse=
?(" +20r-it)
Msc=
T(ur+e.+ {!!)
(, (-.
t-'
(, (' (', (,ì {
)
(d)
uati o ns
0.4Er0B
SHEAR FORCF, kN
POINT OF INFLECTION
- o.0o24lr
:0.4Er0^+ s.sEreB- o.0024tlt
(e)
: rr0" + s.5Er0. + 0.00375Er
+(r"+20.+i*) : 0.5Er0" + Br0.+ e.6s375Er Mco: Totr+ 0o) : 0.66678 r 0c + 0.3334 E I eD
DEFLECTED sHAPE
Fig.l0.5
Mcs=
Moc=
'FT
fO.+
20¡,) :0.3334EI0r+ 0;6667EI0o
(c) Joint-Moment Equilibrium Equations
Mo" : 0, M"o + Msc:o
A being
Mc"
+
MDc
IMo:
Ivl"
0, =
EMc. = a simple support
Mcp:o 0, D being a simple support o0,
0,
gives, 0.8 E Ion +
0.4 E I0B
gives, 0.4EI0,q+
l.g,EI0"+ g.5EI0"+ 0.00135EI =
-
O.OO24E
I=0
gives,
0.5EI0s+
1.6667
EI0c+
0.3334EI'0o + 0.00375EI:0
IMo : 0, gives, 0.3334 EIo" + 0.6667 EI0D = 0
0
294.
SLOPE DEFT,ECTION METHOD
Thc four simultaneous equations may be rearranged in the manix fonn
[;
lo
Lo
0.4 1.8
0.5
0.5
1.6667
0.33341
0
0.3334
o.oeozJ[eoJ
0
substituting gives :
r. | = l=o.oorzsf
I o
]
[o.oors I
I
l_o.oozzl
lr"J
Io.oorr J the varues of 0A , eB , gc
= Mrr= MAB
O,
M"^ =
-
and
00
in the srope-deflection equations
n** 44 kNm, M"o = _ 44kNm, and Moc : 0 TheÊeebodydiagramgives, Ru 41.97kN, and t{c:3j.5kN =-
f' 'I
l' \...-
.
6g.g
kNm, M"" = Or.,
The bending moment diagram is shown in Fig. r0.5b. The shea¡ force diagram can be d¡awn if the support reacrioñs are known. Thp ã"e ¡"¿y ãi"er;is shown in Fig, 10.5c. The shea¡ forpe Fig. 10.5d. {ç.g can be easily drawn as shown in -'Þ' The deflected " shape is shown in Fig. 10.5e
r0.5 FRAIVIES : NO SIDE S\ry-A,y A frarne tends to deflect horizontally under latera! loads or unsymmetriial loads unress it is restrained to do so. Typicar frames that ;;;;;;;;;;;ä;
(
(' ;'. i (
{,
l
l:"i=i-o'ooorf le"
('
o ..
vertical
horizonrafly, that no side rt"uy, *e. ihown in Figs. ló.0" a d. In addition, if a is, lvitf frame is symmetrical with respect to its geometry, boundary conditions, and vertical loading, ir will nor undergo side sway as'shown ií rigr. ié.0"-*¿ f. The foilowing t"*^Wilþßtate the analysis of such frames.
, i¡:i
Fig. 10.6 Typical frames with no Side sway
Mrse
:
lM.t7 kÑm
89
is a-cantilever span. It is statically determinate. We do not write slope-deflection for such spans. The end moment at B in the span BC is 15 kNnn, anticlockwise. 'pquations
,l
Mns
:-Mrm *\Ð-eto+on)
<4Yrc,xÁøq{o.s
\/' Analyze .Y
-
the frame shown in Fig. 10.?a using the -'vrv slope-deflection sv¡¡vwlrv¡ method and draw
bending moment diagram.
Mse
Solution
(a) Fixed End Moments
wr¡¡g
: _ wl] n
=-50"1
=
to4.t7
to4.t7
*
15 øI (g")
since oa
=-lo4.l7kNm
i''Uu'
:
o
__,\
ê
+{u,lrqrjl;]Il 5 -'lJ-*)z **^)¿
0.ff(20"+oD):EroB
(
I
291
; lß;l [iiüi,l
¡.r^ì
or,
FRAMES : NO SIDE SWAY :
= T,,,,
since 0o =.9
v'
-
292
SLOPE DEFLECTION METHOD
293
FRAMES : NO SIDE SWAY
bodv diagrams are^sh:111;tilç: The resulting bending moment and free i; drawn. The deflected shape is shown in easily now U" d shear force
il i**t
10.7d. ,1,
""r
\"
^^*r{q'å"
slope-deflection method and draw the frame shown in Fig' l0'8a using the moment diagram.
m
t-
I
D
75kN/m
(b)
(ol
.6
\r' 1.2j* fF=T-t-T-Jn J¿s 1/' \t r11.26 r08.71 131
19.3
4.28
,
DIAGRAMS
F
INFLECTION
,
-5m ^
++e"
¡t) u)
lrzt.z
t
(di
DEFLECTED SHAPE
:
Fig. 10.7
'.
(c) Joint-Moment Equilibrium -Equations
(, I
.'rl
i,
f
IM" : ,
\..,' .,
"{ ' (,)
or, or, or,
() ,(,
M¡a + Msc + Msp = 0 104.17 + l.6E_!!]"*_:!5_t EI.O"_; O
2.6EieB =
-
BENOING MOMENT KNM
89.17
^ UB =-=;-34.3 EI /
Hoggingmoment
.'.
sO"t : + qS.Z\= n**rt
atLl|inthespanAg atLl2 = O.Stl¡O.O
Net sagging moment at
¿??
-
//,/'.ño'"'ù j
- l3l.60kNm, M"n :49.30kNm; M"": Msn : - 34.3 kNm, Mo, : lZt5 kNm Freespanmoment
ãt40 o <4'3?\
POINTS OF INFLECTION
MAB:
''
ii'
0, or,
(d) Back-Subsritution'
(ì
.
I
POINTS OÊ
1123.71
(
.(
B
0'5m
l"'
(.
i
ls lts tc Bì tul'
2l
1.2s z+\t7.ts
(c) FREE BODY .:
MOMENT kNm
Ll2 = +
=
YL2 =
66.26 kNm
15 kNm,
(c)
DEFLECTED SHA'PE
156.26 kNm
Fig. r0.8
"u,/----
SLOP.E DEFLECTION
METHOD
FRAMES: WITH SIDE SWAY
Solution
0.8EIEe+ 5,22EI0" + l.l4EI0ç- 150:0 Mc + Mce : 0 II\4c : 0, oÍ, l.l4EI0s + 3.88EI0.+ 0.8EI eF +301.45:0 0.8 E IOc + 1.6 E I0F - l2.B =0 f,Mr. : 0, or,
(a) Fixed End Moments MÈns
: - wL2 V:-75x-
Mrsn =
52
l56.25kNm
Mrcr = -
:-
(-,¡
(;
(,
Mrrc =
t.6 0.8
þ)
'('
Mo"
=
Mrn¡
156.25
Mso
156.25
Mi"
: :
o+
o
t.t4
3.88
o.s
Mrc : -
+ oÐ
:
r"¡
Mes
f,,
e0c+0") = - 4.8 + ¡.6Eroc +
i
1.6 E
I
eA
+
0.8
M¡e + Mec +
EI 0s _ 156.25 =
Msr = 0
I
l5o
I
lle. | l-:or.+sf
0.8 r.oJ[e.J Irz.r
ll9.80kNm, Mcr =
0
FRAMES
:
J
-
333.06 kNm,
ll9.80kNm, M."
:
M", = 52.04 kNm,
0. shape in
Fig. l0.gc ,
WITH SIDE SWAY
i practice there a¡e many
I.34EroB + o.67eroE
l2.s + 0.8EI0" + l.6B¡9,
= .0, or, IMs = 0, or, ÐUo
[ rse.zs I
l bending moment diagram is shown in Fig.l0.gb and deflected
frames that undergo lateral displacements under vertical
s:âlone or under lateral loads. There will be additional joini translation at each storey frame in addition to joint rotations. Additional equilibrium equations are requireà for each joint translation. Typical frames that can undergo lateral displacements
in Fig- 10.9. A fr¿me can undergo lateral sway if thãre is unsymmery with tp geometry as in [ig. 10.9a or unsymmetry with respect to loading as in The frame shown in Fig. 10.9c can have two independent lateral joint at the top of the two columns. The additional equiiibrium equalionJ are by considering the shear equilibrium in any storey pf the frame. In the six frame of Fig. 10.9e, there will be six additional shear equilibrium equåtions.
'.BEI'F
(c) Joint-Moment Equilibrium Equations ¡l t-
=
= 0, M"o = 280.98 kNm, M"" = -
M." :
eE
lt
lf e" |
Eaèk-Substitution
0.67EI0" + l.34EIeE
. "y,
lfeoì,
f::ì[]iíiJ, lå:l=hilf"
+ l.6EI0A + 0.EEI0s + 0.8EI0A + l.6EI0s
To
(iv)
- r2.8 kNm
= _ 306.25 + 2.288I0" + l.l4EIec : 306.25 + l.t4Ele" + 2.28EIOI
McF = - 4.8
l
4o*l e x3 -2):
'M". - Mrac+'fprr*r"¡
MEB
{,
-
+.3E$lrro*
MAB
Msr
'('
=
Slope-Dellection Equatiora
:
t,
a).
o
t.t4
o o
5.
o
0.8 s.22
o
: - 4ox I e*2-3) =- 4.8kNm #e^-b) L-
- #oo-
(iiÐ
306.25 kNm
:
(, ',
(ir)
Mot¡ix Solutíon
j2
: - wL2 n:-75xMrcs : 306.25 kNm
,c
-
156.25 kNm
Mr¡c
tj
=
295
(Ð
l.iord". to analyzn a fra¡\e with side sway, it is essential to be able to draw its shape- The.following rules are very helpful in sketching the deflected shape to relation between the.lateral displacements of various mÞmbers meeting at a
SLOPE DEFLECTION METHOD
296
FRAMES: WITH
STDE
SWAY
T
LP3
L3
nn
u"i',
l, riir,
(c)
297
I
(d)
H1
(b)
(e) Fig. Rule
I
I I
10.9 Typical frames with side sway
Rule 4
The frame undergoes small displacement. Axial deformations are not permitted.' The joints are not allowed to rotate while angle between members at a rigid joint remains unchanged. only displace in a direction perpendicular to The member ends
Rule 5
itself. The disønce between the member ends does not decrease due to
Rule 2 Rule 3
;;;
curvature of the member.-
Hr+H3+H5+Pr+ Pz+Pr:0 where,
Hr=
M,
+
Mj - PraLl
Pzb
,
H3
(10.7)
=+
-*H4 (c
4
Hg
)
r-\
tat
F4{
i
t
L1 I
Consider a single storey. frame sh'own in Fig. 10. l0a. The free body diagrams of the columns are shown in:Fig. l0.l0b and that of the girder in Fig. l0.l0c with' respect to shear. Thère are thiee columns with six joints and each column end will have certain shear. The shear equilibrium equation can be written as follows :
ll
Hr
I
J-
(ò)
DEFLECTED CURVE
Fig. 10.10 Single storey frame with side sway- free body O4"grorn
298
FRAMES
SLOPE DEFLECTION METTIOD
,, _ Ms+Mu-Prc 'L3
299
Â, are small and axial deformations are neglected. since the top of thi columns are 6 and l0 will have the same horizontal -¡ Â, , and the points 4, 8 and 12 will have the same horizontal displacement ,âr. Each storey of the frame is cut through an imaginary prane as shown in Fig. r0.l lb. lI::. -" two unþòwn lateral floor displacemenrs l, aná Â, , and two addirional shear
in the same horizontal line, the points 2, ¡l-
Alternatively, The shear.equilibrium equation or simply the shear equation may be written as
I{z+Hn*Hu:o
r+ =
1VITH SIDE SWAY
in the same ílgure. At the first floor it deflects by ¡ while at the secohd , to* , its total yay !s Â, and.net sway is e - Â,. It is assumed that displacements Äl
rr<--
where, Hz =
:
(10.s)
M, +M, +pr(L¡
-a)+rr(r., -U) Ho: Ll
M5 +Mó
:
ll3(L3 -
I
L2
H¿+He+Hrz*Pr:0
(10.9a)
Hz+It+Hro+pl+pz=0
(rO,eb)
f
c)
H4
: ¡4j-14{
Hs
M, +M, L4
upon simplification, Eqs. 10.7 and l0.g turn out to be identical. Thus, the shear equation helps determine the lateral sway A at the level of the beam assurning there is no axial deformation in the fr¿me. consider a two storey-two bay frame shown in F'ig.
l0.ll a.
Pzi
I
I
Í ,FI_N
'! l"
I L1
i
=
&#fu
the frame shown in Fig. l0.r2a using the slope-defrection method and draw ifinalre l moment diagram and its elastii curve.
12
'*l
/^ :=
ll
r7
..*H4 *-_ H3 P2 I ....*Hz
Sl ope-
-
50x
* =
-l50kNm, Mrc¡ =
t50kNm
l50kNm, Mrr" = l50kNm
Deflec ¡ i o n Eq uat i ons trsymmerricar with respec*o
-_Hg
_Hl2
nH7
Hll
I
'-Hs
-
geomerry, roading and boundary condirions, --T".3i: the lateral sway may be taken as zero. Neveftúeress, thã sway
is taken ns non-zero
later turns out to be zero.
*rvrÄB
= o+#(æ^.r"-#)
t-*r,
(b)
Fþ: l0.l
",,
2
L1
+
Hu:\i*
The_follow_ing examples illusfrate fhe application of the slope deflection method for analysis gfgeneral frames.
Its deflected shape is
,oz,l
Fl
ï
nr: Y'JJ[z,
I Two storey frame with side sway- free body diagram
=
0.898I0o + 0.45EIeB
-
0.29EIA
0 + 0.45EI0o + 0.89EIeB
-
0.29EI^
SLOPE DEFLECTION METHOD
300
50 kN
FRAMES: WITÉI SIDE SWAY
tY" = o, or,
/m
Msn + Msc
:
o
0.45,EI0r + 2.89EI0"+-ere¿- 0.2g9l^
IM. :
301
-
150:0
(ii)
+ Mcp : 0 EI0" + 4.89EI0c+ O.45EI0o + g¡gE- 0.29tto.-': g $o. ,: 0' or' 045EIec + 0.89EIgD - 029ElL:0 IME -- 0, or, Mrc + =O "u, * tt9n 2.89EIeE+ 0.45EI0F - 0.298t + IJQ : I MrE 0, or, 0.45 8I0r + g.39 ! I0r : O.zg9l L : 0
0, or,
Mca + McE
(iii) (iv)
(v) (vD
dquafion gives,
(vii)
\ Matrix (b)
BENDIN6 MOMENT kNm
Fig.l0.r2 l5o +
Msc
+(2eB
o.+s o 2.8e I 10.45 io.ar
+oc)
l0 Erl 0
l5o + 2EIeB + Elec
Ma" Mpc
150+EI0"+28¡6" 150+2EI0.+B¡9, 150+El}r+2EIeE
MDc
s
=
Mce
*
Æ!(2s^ *e. 4.5
\
r,t
t
i"(Y
I
lo l0
f l.r+
- 0.29EI^
:
0 + 0.45EI0o + 0.E9EI0c'- 0.29EIA 0.89 E I0r + 0.45 E I0E - 0.29 E r^ 0.45 E I eF + 0,89 EI0E - O.2gEl L
M"o
Mrt Mer
:
0, of,
0.4s
o I
0
0
0
0
4.8e0.45
0
1.34 1.34
I
0.8e
0
o
2.89
0
0.45
1.34
1.34
0.89EieA+0:45EI0B-0.2gEI^:0
eA
-28i:49
0B
'56.34
0c
(c) Joint-Moment Equilibrium Equations
MAB
l 0
o
o 0 o o
'0
eD
0
eE
-56.34
0F
28.49
(Ð
^
0
-o.2e.l[eA
] -0.2e lleB I -0.2e lloc I -0.2e lJeD | =
0.45 -0.2e ll eE I o.8e -o.zr ller I 1.34 -1.74ll ) ^
-!Aì 4.s )
0.89EI0o + 0.45EI0c r';'À
Solution
I EI
0
r50 '0 0
--t50 0 0
FRAMES: WITTI SIDE SWAY
SLOPE DEFLECTION METHOD
102
(e) Back-Substitution
= l4B : M""= Mns
0,'Msn 206.34
:
37'32kì'Im; Mac = --37'32kNm
kNm, Mcp
:
0, McE =
37.32kNm, M". = -
-
206.34 kNm
37.32kNm
andc.
,/
,.
Flgrftple7078
\--/
.
,/
-.:
Mco
0.8EI0c
0.4EIec -
IM" :
bending moment diagram and elastic curve.
0, or,
0.5EIeB
5m
Msn + Msc
:
0
+ l.8E l)c- O.24EIlf :
(Ð
0
(iD
0.
lhown).
I
o
0.24EI^
There are three unknowns, hence there should be three independent equations to determine them. Theihird equation can be obtained using the shear criteria. (Fig. not
ï
z
being at a fixed end
0.24EI^,
l.SEIeB + 0.5Elec- 0.24EI^ + 20.84 = EMc : 0, or, Mcs + Mcp = 0
AnalyÉ the frame shown in Fig. 10.13 using the slope-deflection method. Draw
J
0o:0
s)
5 l.
(c) Joint-Moment Equilibrium Equatiors
,/
E
zBr (ze. * e^ - 1A'ì
o
MDc
The free body diagrams and the bending moment diagram are shown in Figs. 10.12b
*
Mcp
303
Re*+RDx+1Ox5:0,
i
'
:
8m -.---Ðl
(o)
(b)
PORTAL FRAME
l.2EIeÈ +
ELASTIC CURYE
t.zEIec- 0.96EI^ + 125:0
(iii)
'!di) Matrix Solution
Fig.
10.13
Solution
Erlo.s
52 Mr^B:- wL2=-l0xn= " Mrsn 20.84 kNm ,;it(-, \--.
.
r.a
f
(a) Fixed End Moments
-20.84kNm
(b) Slope-Deflection Equations
MAB =
i.
rt í
.ita
't.,rl-.
il. l:
.
Msc
:::)t .1',.1
rl
2ß.54
ji:.
r,i¡ 'rli:,
ií
20.84
r.
l
i:l
=
Mcs :
0+
+ 0.4EIeB + 0.8EIeB
- 0.248I^, - 0.24EI^
I
Back-Substitution
0o=0beingatafixedend
iii.,Back substituting the values of slopes and deflection in the slope-deflection equations the values of end micments.
MAB
: i-(208+0ç) EI0Bt 0.5EIec zBr2t)
0.5EI0B+ElOc
r.8 -o.z+llecf=l o
{i} {:nl+
- 20.8!. ?(rt^.t"-+)
: Mso :
- 0.241[eBl [-zo.a+l
Ir.z r.z -o.e6l[^J l-rzs)
,.j
:,
o.s
lɡmt,
: -
57.75kNm,
Mc"
22.39 kNm,
RD* = -
I0.58kN
- 14.37 kNm, Msc : 14.38 kNm M.o : - 22.38kNm, Moc : - 30.Ag*î
Mse =
is thè sanie resu'lt as obtained by the strdin energy method in Ex. 5.7.
*.
304
DEFLECTION METHOD
SITOPE
FRAMES
/' tO.9
Exagfle
,/rrrur" "f".;;?;
supports A and F' 10.14 shows a portal ftame with hinged gf the frame' curve elastic u"naing *ot"ni áiugrutt and
T
Member CE
T
t I
I
I
4m
'';
4m
B
Mre
I
,
Mc,
::
H¡ A
. 3'5m . 3.5m
l'-{t¡--
(bI FREE BODY DIAGRAM
,
+ 2Ebr\ ---::=(Zgc +0r)
!gIe. * 1pro" 7'7
7"7"
I"
-
2Ebr\ +..-(9c
1ere.*-9pre,
Solution
.
43.75'
43.75
M¡c
Fig. 10.14
(a) Fixed End
305
)a6 7^7"49
Hç 3m
SIEE SWAY
2F.r / ¡¡l + #l eo + 20ç- --7)I 7 \"
73.47
Mco
: WITH
43.75
+ 20)
+ 43.75
Moments
P
ab2= t:"74x32
M'o' = -
#
ì\Í ¡v¡FcA
p a?b
Mrcr = -
SOxf
= MrEu =
43'7J kùrlm:
û
Mrsc
: -55'rokNm
:
= ls*4'7t!3 :
?<3
o.!(ze".r.-+)
MEI,
-
43,75kNm
ù
0=MF'E
*
lPro Tt. -
o.?(t,.tt,-?)
MFE
=
EI0E
?3.47 kNm
=
fe,+EIo¡-fEra
-Moment Equilibrium Equatíons
(b) Slope- Deflection Equations
Mnc:
Me¡nber AC
Vce+Mcp:o
(ii)
;,M,çc+Mpr=o : o ,MFE
(iir)
Mnc
= :
or,
M¿c
Mrec
_
55.10
. 2|(rø^. t. - T) '¡çt ( + -+ll20o
7\ )
+
0.
(r)
o
(iv)
_;3^\
6--. 4 +irlec-Ëu^ = |ue.
t)
I
-ss'lo
(v)
t4EIeA +
0.2857
EIec-
0.1224EI^
-
55.10:0
SLOPE DEFLECTION METHOD
-
Mce + Mce 0.28578 |
g
0, gives 1.7
l42E I 0. +
I
0'57 14 E
^+ MEc + M¡r = 0, gives o.57t4E I 0.+ 2.1428E I 0r+ 0.5000
Mre =
FRAMES: WITH SIDE SWAY
OE
-
+ 29'72
0'1224 E I
'o*r gdto.to
: 0'
(iD
^
E I eF- 0.375 E I À+43'75
:
,
10. 5a shows a portal frame with fixed supports A and F. Draw the shear '6r"and bending moment diagrams and elastic curve of the frame. I
2
0
i.i8
¡s.¿s
0, gives
t ¡
3
: 0'5 EI0u + BI0F - 0'3750E1Â
6.1
lzr.re
(iv)
0
\t
l./
28.82
36.r 128'82
The shear equation gives, (Fig.l0. l4b)
Moc + Mc^
HA:-|7.nF-4
36.1
I
- z5 x 1. and
HE
=
MEr
t
-Yrr.,o
Mre
I
E 0.0408 E I e^ + 0.0816 E I 0" + 0.25 EIoE+ 0.125 E I 0F- 0'1112
+
:0
53.3547
form These five simultaneous equations may be arranged in the matrix
l^
rs.r ,:f{a'n
1,,, .,,
(v)
,r.tt I
(b) FREE
:
BODY DIAGRAM
Fig. lû.15
0.28s7 0,
fo.szt+
0.57t4 0.s714 2.1428 1.7112
lo.28s7
"rll0
0
o
',
00.5 0.0816 0.25
lo.o+os
i l;;il[::l [ ;i:å ] -]:,'1,111å:l= i" f j
0
0.5
I
Fixed End Moments are same as given in
I
-o.rrrz.lIa
0.125
0o :
l-sz.tsn)
0: %
Ex. 10.9.
due to
fixify of support
-Moment Equil ibrium Eguat ions
0, : EME 0, ÐM..=
oÍ,
iighear
{i} {r;ql. This
is
0,
and
A
in fte slope deflection equations
MAc =
0, ì4cn = 2.175 kNm, ì4r = -
Mpc HA =-
172.85
kl'Im, Mrn
3l.E0kNG-' Hr=
-
: -
Mcn + Mc¡ =
M¡c + Men :
condition is (as in Ex. 10.9)
0
(Ð
b
(iD
:
H¡,+Hr+75:0 IM. : 0 gives,
(d) Back-Substitution Substituting the values of 0¡ , 0c , eE , give the values of end moments.
.or, or,
2ll7 kNm
172'85kNm, Mo, =
:
t.tt42Bl0. + 9.5714EI0e- 0.1224EI^ + 29.72:0
(Ð
IM, : 0 gives, o.57l4qttir+ z.t+ztEI0E
-
0.375EIÂ + 43.75 =
(iÐ
0
0
43.2okN<-,Ro:.0.62kN 1,
&:
49'38kll 1
in Ex' the same result as obtained by the consistent defonnation method
3'l
t'i
0.1224
El0" + 0.375810E- 0.2225 EI^ +
45..4812
=
simultaneous equations mãy be ananged in the matrix form
0' :
(iiÐ
SLOPE DEFLECTION METHOD
308,
FRAMES: WITH SIDE SIù/AY
;
c
o.s7t4 -0.'224.]|.ecl -zs.tz | ) Erlo.s7t4 2.1428 -0.37s lje,l= 1 43.7s |' 0.37s -o.zzzs)lti' l-+s.+tn) L0.1224
* lr.tr+z
{::}=*{ ;:'::',1 t^ J l23e.e57|
oft
309
x
l-\
l.*l
(b) FRAME UNDER SWAY
(c) Back Substitution Srlbstituting the values of 0. , 0, and
 in the slope deflection
equations gives
Mec = :86.8 kNm, Mce : 39.45 kNm. Mcp : - 39.46 kNm M"ò' : 66.24kNm, Mpr : -66.24kNm, Mre : -78.11kNm The values Fig3.
of
reactions can be obtained from the free body diagrams shown in
l0.l5b.
H¡ : - 38.90kN<-, IIF :-36.1kN<-,Re= 2l.l8kN1, & : 28.S2kN1
(c) FREE
BODY
19.46 Take
E:
200 GPa,
I:
l0{ ml.
'
Solution (a)
300 x
;
S I op e- Defl ec ti on Eq uat i ons
The members AB and DC sway to right by an amount x (Fig. l0.l6b).
.
'tçt ( "îlrr^+0¡-
Mrs M"o :
?v\
;)
:
0.28s7EreB
-
Due to the settlement of support D, memþgr BC sw?vs
2lbr\ / 1Ï!/lZOr+ 7
(d)
: +(ro*rrr-+l " 7) o.sTt'EreB- 0.t224ltx 7 \^ clock*i""
hv
v:l0mm. J. Mo.= Dç
17.5 5
0.t2248rx.
[
Ac-
?v\
i7) l:
l.l428E I g" + O.57l4E I0ç - 0.2449I 1y
BENDING MOMENT' kNm
Fig. 1ò.16 Frame with support settlemenr -
:91t" +20ç-]{):o.r,o E r0" + l.l428Er 0r- 0.244eEty " 7) 7 \"
zEl(^^.^
3*1,
=î[2oc+0o- ;)=rI0"+
s.5EI0D- 0.375EIx
,,/
,
SLOPE DEFLECTION METHOD
310
25
Given (b)
Jo
Mc" +- l&o_-:-
ur¡j4 ^r
I
+ I
(Ð
M¡n+ Msc = 0 HA *.
4m
'0D
int-Moment Equilibr íum Equations
5m
(ii)
0'
I
Ho---ll.
Y¿e.1!g!À * Mcp t Mpc
:6
A1* A2 k---.{
T
+ 20o
Y
. AZ l. --{
kN/m
1*\ - Ð=0'5 EIec + EI 0o -0'375EIx "r.= ?(.0. : 0'004 : = l0 mm = 0'01 m' an4 tÊt (
3ll
FRAMES: WITH SIDE SWAY
(
b)
FRAME UNDER SWAY
(iiD
IMp : 0 gives, x 0'01 x'EI = 0 l.7l42Elg"+ O'57l4EI0c- O'l224EIx - 0'2449 EMc = 0 gives' 0'00045EI = 0 0.5714EI0"+ 2'l428EI0c- 0'375EIx Shear equation gives,
or,
O.l224EIeB + 0'357E|øc-
0'22248Ir+ 0'0015EI =
be arranged as The three simultaneous equations can
0
:
(c)
(d)
BENDING M0MENT kNm
0.s714 [o.ooz++el I r.tr+z z.t4zß -o.l224lfeBì 4.37s 11t" i='l o.ooo+s I o.s7l4 I 4.37s o.z2z+ )lx ) [ 0'0015 L-0.1224
Fig.
ELASTIC CURVE
10.17
J
End Moments
or,
fe"ì
ootolrl
of
0s , 0. and
x
Mro¡
in the slope deflection equations
47.55kNm, Muo =
-
l9'45kNm,
zs\t- : t2
Mr.cD: 'ú:-
t';l=+tln';l
Substituting the values
MAB : -
[o
Msc:
:
133.4 kNm,
'Mi."r :
133.4 133.4
kNm
kNm, Mrps
:
133.4 kNm
Deflection Equations
l9'47kNm frame is symmetrical about its centre line with respect to geometry, vertical loads conditions. Hencp, lafgral,sway will be .ze¡o. Neverthéless, it is taken as at'each floor level to illustrate the point (Fig. 10.17b).
.
$(zec.+ os- ?), n 0.4EIeB - O.24EtLl o
: o, cbeingarixedend
FRAMES: WITH SIDE SWAY
SLOPE DEFLECTION METHOD
312
0.8EI0B
Y".
* n v'-4
M"o
2Et
M""
(reo* 0¡ - 3^¿l :EIeB +
[-"o'-o 0.5EI0B + EI0o
MAB
:
- 0.2481\ 4)
-
-
0.375
æpirt^+0p)
t33.4
+
Mon :
133.4
+ 0.75EIeA + l.5EI0D
M"t
z1ßt).^^ : 133.4 + "-ï)!!per+08)
Mes
133.4
Mo, :
0+
0E )
-
Mon*Men _^
l.5EI0A+ l.5EI0"+ ¡.5EI0o+ l.5EIeE - t.5EI^2 :0
or,
(vi)
-133.4
+ l.5EI0B + 0'75EI0E
'Ï(rr"+0s- +):EI0o+
0'5EI0E
-o:lslr^z
:
o
- ?þre+0rr+)
0.4EIeE
-
0'24
= 0'8EIeE
i;:;
tlo
I0D+EIeE-0'375EI42
-
lor.s
o'248t^l
0.75
0.5
'l o.zs
L
0
0
0 0.75 2.5 0.5 0.5 3.3 0 1,2 1.5 1,5
J.J
0
0.75
t.2 1.5
-0.375.l feA
133.4
-0.24 -o.rzs | | e" 0 -0.:zs lJe"
133.4
4.24
-0.375
;0.96 0
|I
-133.4
eE
-133.4 0
i, l[î;
0
EI^r
:---- \ !y:)-t! .-: - 0'375EtL', - 133'4:0 2.5EI0A + 0.5EIe" * 0175Èi4:IMn - o, or, W-. *;; ""t)
194.66
It" (i)
0.5E,I0o+3.3EIeB+0.75ETqi:-o74-EI4--0'37581L2-133'4:0(iD
or,
(v)
0
^r
(vi) can be rewritten as :
M*+M"o 44
:
0.96 E I
+ 0.75EIeB + l.5EteE
IMa = - 0, or,
EMp
u
-
(c) JointMoment Equilibrium EÃuatio'1s
or,
_^
Matrix Solution
Mro
MFE
M"a + Mr.
î-
l.2El (gB+ Ëq.
= -t33.4 + l'5EI0A + 0'?5EIeD
Mra
55
EI^2
0.37581L2
$oo =-
Mer
0.5EIeA
+
3i3
= 0, or' ÚlÞ1¡
I r8.7
le"
jto ^
ltt
I
-t94.66
EI
-iì r.z ,0
lo'
0
[^z
(iiÐ 2'5El0o+ 0JEI0E- 0'375EI^2 +133'4 = 0 - = EMe = o. or. M-rva-Mu, EF L_ ^-. 133'4=0(iv) 0.75EIeB+0.5 E l0o+l'i Èiq=oz'rEI^r -0'375 EI^1+
0.75EIeA +
the otlter ' one correspoqding to the upper d'torey a¡$ There will bef two shear equations' : corresponding to the lower storey'
(v)
(vi)
values
of
slopes and deflections
: , 15.80 kNm, Msc = 31.65 kNm, : 72 kNm, Mes = 84.67 kNm. : 84.67kNm, Mop: - 84.67kNm, = d03.66kNm, Mer = - 3l.65kNm,
in the slope-deflection
MBr =
-
103.66 kNm,
=
- 84.67 kNm, Mep= - 72 kNm, Mrp= * l5.g0kNm, Mno
moment diagram and the elastic curve are shown in Figs. 10. I 7c and d.
FRAMES WITF{iSLOPING LEGS
SLOPE DEFLECTION METHOD
3t! Example 10.13
-
by the slope-deflection method' ÃÁalyzethe box frame shown in. Fig'10'18
I
t20 kN/m
Fig.
+
ing
10.18 Box section
Eqs. (í) and
'*ú52
Mrpp
(b)
8 : Zox:- :
= ;
4l-67kNm
Slope- DeJlection Equations
:
'Yo. :
Mroc
+
T(rt^.t. -l)
- 62'5 -'$("o*0. i)
Due to
--'-
M¡r
:
Men
:
o*froo+2gs)
symmetìY, 0c =
+';þtr* -
0n 0
(c) Joint-Moment Joint
A,
^= Equilibrium Equations Mec + Mne =
0
(i)
+ 4EI e-:
5
"
o
fo'rm
(ii)
:
l-2e.3J
Substitution c Mec = - 33.86 kNm, M* :
Back -
Mro :
-4
at B
'" : looxl : : Þr
momentatmidspanofED¡bending
33.84
kNm, Mro
: -
1g.24 kNm,
18.23 kNm.
l25kNm
nl] : s2 : 20x!-
= -,
62.5kNm
moment diagram is-qhown in Fig. 5.14 e.
\ilITH SLOPING I¡EGS
nlyze the frame shown
its elastic curve. 41.67
o
4¡.67:6
(ii) in matrix
lerJ
,f.RAMES
MAE: o+'fi
:
ft"Ì= EtI {rs's1
sÞan moment
Mnc
e-
Ïl{3:i ={J';2,}
"fii
(a) Fixed End Moments
--- 8
2EI
,{
Solution
P ab2 Mrnc: -T
4r.67
0.8EI0Â + 2.4810.+
D
2.5m
+
e. e^ 5^2.5^2.58
¿llg^ + 4EI e.+ 2.5 ^ 2.5. "
t
2.5 m
2.5m
* 4EI
+ 4EI
2.4Ete^ + 0.8EI0p- 62.5:0 Ioirit El, Mpe + MEo = 0
T
,.
62..5
315
in Fig. l0.l9a
using the slope-deflection method. Also
r" - l) slope-deflection niethod, axial deforàations of the member are ignored. Thus, AB, Bc and cD can not deform axially. The joint B will deflect horizontally l$right. member BC beiúg horizontal and cD being inclineil, the joint c will .The ,pcrpendicular to both BC and cD so as,to take the final position at ó". c' c" is tdlcular to BC, and cc" is perpendicular to cD as shown in Fig. l0.l9b. The sjoint deflections may be written as follows :
'I
FRAMES WITH SLOPING- LEGS
SLOPE DEFLECTION METHOD
316
1\l ,'NI :
tú", A
(b)
þ--lr--F-l-g----l
:
o+
317
Irr"+0.+ .-ò I Tf +l l.)
0.8EIeB + 0.4EI0" + 0.t68EIô FRAME
ln ì
swAy
McB
'NDER
o.4EIeB + o.8Ete"
* o.toario
,etztl,( ,-L I M.o 0*-#|'20r+0o-+l ,{¿ r : Mpc: (c)
FREE BoDY
=
45",
l.t2
E I Oc
-
E
CC': ôcoscr: en': =fi \J'
t...,.,
ô
.,
¡7
Muo
,+l o+ ?lrt^+os-*t + 0.4EI9b
. Momentabout B giúes; Ho
EI
õ
:
Mes 1
Msc
(ir)
MBA
MomentaþoutCgives, Ru
:
Moment about C gives, Ho
: M.o+M*-Ro5
t
Mça
5
Rc:
Ro
(Ð
0
IF* = 0 gives, Ho = H" : IF, = 0gives, Re:Rs
)
- 0.168 0.4 E I0o + 0.8 EI0B - 0-168EI õ
0'8 EI0n
:
A8
(a) Stope-Defleetion Equations
=
9
body diagrams of the frame members are shown in Fig.
. :'I
MAB
:
0.56EIOc-0.24EIô
0.4EI0" + l.g2BtÛc- O.O72EIõ
,, p¡.rftee,
-.( "r
I E, since 0o
IM" : '0, or? Mso + Msc : 0 l.6EI0"+ 0.4Elgc:0 IM. ': 0, or, Mce + Mco = 0
i ¡
cc" = ô.
c'c,, = õsincr =
)
0.'24 E
Joint-Moment Equilibrium Equations
DIAGRAM *ìu*to
nig. f O'tl frame with an inclined leg s.
l.
J=nt
10. I 9c.
SLOPE DEFLECTION METHOD
3r8
FRAMES WITH SLOPING LEGS
Shear equation
or,
0.48E'Iec- l.l52EIõ+
M"n =
Mcg+Msc +25:O
M*+M"o- + Moc+Mco
zEbù( Msc: --¡^s+oc-*î*J
5
125
=0
1.34 E
I0B + 0.67 El0c +
Msc
1.34 E
I0"
Mc"
0.67EIeB +
McD
0.67 E
I0"
ùbc
0.34 E
I0c +
(c) Matrix Solution
'{*r i*, -{:al8j}={å'}
.r
iiÌ "häÌ
25kN g
2t
Back - Substitution
Mrs = Mca
:
,r'þffi:
-
18'97
kNm, . Mn¡ : - lg'4 kNm, M".
2l.58kNm,
Eton + 0.34E IeB _ 0.167 EIô 0.34EI0x+ 0.67EIeB- 0.167EIô 0.67
F[tnl;î:Li
:"s5
319
Mco
: - 2l'59kNm'
,ment diagram is shown in Figs. 10.
MDc
+
0.678I0"+ 9.3339¡6 l.34EI0c+ 0.333EIô 0.34EIeD- 0.t67EIô 0.67
Et
eD
-
0.t67 E I
ô._
c.
36m
19.4 ll.{m
I
l-
24.03 kNm
(b}
6m
f9 d.
force and bending morfent diagram
+
0.333 E I (2ô sin 30)
FRAME UNDER SWAY
for a portal frame
Solution 'lr
Under the lateral load, the frame will sway to the right. Its deflected shape can be drawn using the rules cited earlier arrd is shown in Fig. 10.20b. The displacements of members AB and BC at joint B are as follows :
: C C,' B-E': Cc':
(d)
BB,,
B
B": ô cos 30, B'8" = õ sin 30
FREE BODY DIAGRAM
The displacement BB' is perpendïcular to AB, while B'8" is perpendicular to BC- The displacement ofjoint C is identical to that ofjoint B.
(a) Stope-Deflection Equations
o
BENDTNG MOMEHT
Ito
Fig. 10.20 Two hinged frame with inclined 1-
tNm
tegs
Moment Equilibrium Equations
MAB 0, 0.67EI0o+ 0.34EIOB- 0.167EIõ ItVl" : 0, or, Mse + Msc = 0 0.348t0^+ 2EI0" + 0.67 Etoc + 0.16zE Iô : b
=
0
(i)
(ii)
FRAMES WITH SLOPING LEGS
SLOPE DEFLECTION METHOD
,320
IMc or,
: o,
M.".
o
I-
'
" =0 0'167EIô 0.678I g.-+ 2EI0t+ 0'34EI0D: 0, 0'34EIoc+ 0'67EI0D - 0'167EI6 = -._
i*o
(iii) (iv)
0
AB
0.34 0 2 0.67 0.67 2 0 0.34
lo.at 10.34
EII
l0'20c' the frame is shos'n in Fig' The free body diagram of Member
32t
o
L,:.
Taking moment about B
-1.33
lr} ',;i,ï'JI';j ,l,li'fl{å:}
-1.33
[-,å,j
;-=61-l\'1ëlAC 3J3 R¡= Rs, Hn:
BC
Hs
Taking moment about C
Ël 4ïÅ
Msc+McB
i, =-
- Substitution
6
Rs+Rc=9 or,
Rc=
M"c * lrlr"
= 0, Me¡ : - 32.5 kNm, MBc : 32.5kNm, Mcs : ' 32.5 kNm, McD = - 32.51ñm, Mo. = 0. MAB
.t, 6i
1
i1
Taking moment about C
CD
Mco+Meç-RP3 ,, "D3J3
moment diagram is shown in Fig. 10.20 d. 1
the fxed end, frame shown in Fig. l0.2la using the slope-deflectión
It
__--À
D¡aw the elastic curve.
Shear equatio
I\
HA+HD+25:0
ôr v,,
M*
+ M"¡ + R^3 3J3
(MÆ+
or, or,
M*) -
so'
* M"o+MDç-Rp3 +25=o
6'm'
3,t3
r(*+r*)<*-*
Moc)
-
r(t*#*).
(c) Matrix Solution
3
(b)
(o)
BENDING MOMENT kNm
zs"6= Fig. f0.21 Fixed end fiame with inclined
=0 (M¡s+ Msn) - (Msc+ M.s) +lvfcn+ 75'f3'
0.34EI04- l'33 EI0B- l'33 EIec+0'34EIeD-Elô+
D
legs
ì
129'90
=
0
(
as in example 10.15,
0=
0" being fixed ends, rhe joinr-momenr equilibrium Eqs. (ii) and (iiÐ
5:?'EI0^ +
0.67
EIe^ + 0.16?Er =
o
FRAMES WITH SLOPING LEGS
SLOPE DEFLECTION METHOD
322 0.67
EI0u +
Shear equation
or, or,
Mos
is
}Eløc+
0'167
Ho + HD + 25 =
EIô
:
Ioad wL is resolved intò two components : along the member and at right to the'member (Fig. 10.22c). The perpgndicular load per unit length is 0 / L sec 0 : w cos2 0: The fixéd end moment is equal to
0
0,
+ Msn - (Msc+Mcs) +
trdcp
+ uo. + 75v5 =0
t2
(c) Matrix Solution
ofr
(d) Back-Substitution
l2
25 kN
/m
'î t:{\i]{it}= { ,!,,} iiÌ "{ xi!
M¡¡ : - 20.29 kNm, M"o = - 22'50 kNm, Ms.' : 22.50 lù'{m, McD : - 225 kNm, Mo..= Àác" :
-
0(Lsec0)2 wl]
w cos2
-EI0B- EIo" -l'33EIô+129'90=0
"þ':.¡
6m
22'50 kNm
6m
(b) FRAME UNDER SV/AY
20'29kNm
vB
I'
Example 10.17
t-''
Analyze the gable. frame shown in Fig. 10.22a using the slope-deflection Draw bending moment diagram and the elastic curve'
t l.
-ùsø
Yt
Solution
wL cos
o
The fxed end moment in a beam with a horizontal span of lengfh L when
x"
ñ* 4Mac
,/
"'Vl:-
-
M Mag
i
I
,,ro
(d) FREE BODY DIAGRAMS
-¡I
5m
at C" and C"'
(a) Fixed End Moments
I
I lvVA
disþlacements. The joint C cannot have t horizontal displacefnent due to Thus out of three joint translations, onfy one translation is independent, that A possible"elastic'curve of the gable frame under the given BB'= DD' : and D'DCC'i' are drawn so that B ^. The,par4llelograms 10.22þ, Fig. shorlrr-r in
ICC"C'=90":ICC"'C' ' c'c"' : c,c,,, - al BB' : Â =.^lsinO: CC"
Hac
ô j*.Ha
This gable frame has 5 rigid joints A,B, C, D and E' Since the method ignot"r the axial deformations, joints B and D cannot have
!'B!C:Draw perpendiculars
r
25 kN/m
fflTrfD
HBA
MtBA
-/
fr,I
DC"' aie equal-to BC and DC, respectively. B'C" and D'C"' so that they intersect at C''
,
o) GABLE FRAME
The bending moment diagram is shown in Fig' 10'21'þ'
and
32?
I I
130.81
AL AND CONJUGATE
,
BEAMS
(f
)
BENDTNG r40MENT kNm
Fig. 10.22 Gablé frame under symmetric loading
pReuPS wITH sLoPlNG LEGS SLOPEDEFLECTION ME'THOD
324
25kN/m, L = m,
=
if
w
(b)
SloPe'DeJlection Equations
6
MAB
Fixed end moment
o ..
=
equation is given bY
25xl = t75 kNm
o'z¿ E
IÂ'
0o
?þ^+20"+ +)
75
(i)
-
:
Ho+Hr:0,
r^: %#*
= o+?("^*0"*+J: o'o''é" ì
Msn
=
l.9Ei0" - 0.l76ElL'-
325
and
Hr:
Mes + Mse + Mnp + Mep
-8..^,^,1 end' 0' A beingafixed
:
Mee;Meu
0, This
leads to a
trivial solution.
IMc : 0 gives, H"" x 4 + 150 x 3 Msc- McB- 150 x 6:0,
= o'IEIoB + 0'24EI^
_ rr ^'Bc
Mu" +'M"" +450 4
B, the shear equation is
:
H".-Huo:o = 7s Tia@"- å.Í^) -
=-75+ M,""
of,
75+
McD =
. ,lt,
or, .
.
Mcn
-
-
ä,t"
75
* '#rroc
0.?5Â)
+
=
-
75
* p1e"+
=
-
75
-
oD
+
0.l76EIeB - 0.304EI4 + 112.5 :
= 75 + o'75
0'555
-
0'416EI^
these equations in matrix form
"[_,i
^)
î,
- 0"
]='
duetosymmetY
(iÐ_
0
r--
:
ï,'Jf]{';}
be seen that the coefficient
,t *ffioc+zøD+o:sL)=7s -
MDc
EI0B
c
0.754), but 0u= Jl3 0.555Ète"* 0'416EIA ,)tI
+ 0.48EtA
f
= -'15 + t.toEIOB- 0-416EI^
Ma"
"
+ffieeB-075^)
=
{,1j,}
matrix is symmetric.
;. EI0a = 77.933 and EI^:415.185 - Substitution
M¡s
l'l0EI0"+
0'416EI
:
130,81 kNm,
Mcs: -
Mnt = 16l.99kNm, Msc :
-
16l.99kNm
54'46kNm
Check
a¡t /
Yo"
ert\
= o - ?["p+0r -+)= - o'8EIoB- o:48r^
Mro = - o'4gIgs
-
o'z+Ell 0B
(FiS' I0'22d) (c) Joint'Moment Equilibrium Equations solve two independent equations are required.to There are two unknowns and
B,
M"o + Mr" =
0.8EIOB
+ 0'24EI^ -
Joint
or,
displacement at
0
75
+ l'l0EIeB- 0'4l6EIA = 0
joint B can be checked using the conjugate beam method
1x 161.99"s-l
22EI
x
130.81
I x 130.81 *5 *2,.s-I 2323
*5=
77t?.,5
x t61.99*s'
clockwise
I - 415'186
o. K.
o.K.
SLOPE DEFLECTION METHOD
'
327
FRAMRS WITH SLOPING LEGS
Similarly, it can be shown that 0. : 0. The bending .moment diagram is shown in Fig. 10.22 f. The elastic cr¡rve was shown in Fig. l0-22b.
T
Example 10,18
4m
Analyze the gable frame shown in Fig. 10.23a using the slope-deflection method. Draw bending moment diagram and the elastic curve.
+ I
5m
I
Solution Due to the horizontal loading, the horizontal deflection at B may not be equal to the horizontal deflection at D A probable deflected shape is shown in Fig. 10.23b.
Let
BB'1 A" and DD':
(û) GABLE
H "Yti' M", Å#, lB
The parallelograms BBC"C and DD'C''C are drawn so that BrCr and'Ctr'D' are equal to BC and CD, respectively. Draw perpendiculars at C" and C"' to B'C" and C"'D' so that they intersect at C'. Thus,
: BC : CD (Deflections being small) t;; 'anti-clockwise cosec0 (As sway C'C" (^B - AD) = = 2 ^") î length B'C = B'C" and C'D' = Crt Dr
c' ctrr
- t¡" -
^o)#
( .,:.
( (
Mr"c =
t- = Mrc¡: l3'3akñm - Y= "''on"', --'oî
0A
(b)
FRAME UNDER SWAY
lve
X--0" l"o
\
re?
Hoc
T*o.
_ 'wL2 -
,-.- * : - l'x t2
vtEw
Dt.
I
clockwise sway
Mms =
E
ENLARGEO
"li" ïrW"
xsc BC
(a) Fixed End Moments
(
FRAME
Ào
hl 'oE.li "D.
-20.B3.kNiir, Mran = 20.B3kNm
'lfo¡.
0 = 0, being fixed ends.
(b) Slope-Deflection Equations
I
FREE BOOY DIAGRAMS
l"
I
(
'¡ i
MAB =
(,
( :1.,
(
('
Mnn
- 20.83. ?(rt^- t"- +)
: -
20.83
=
20.83. 20.83
+ 0.408108 - 0.24EI^B
T(t^. rt"- +)
+ 0.80EI0B- 0.24EI^B
(eI M/EI
(d) M /Eldiagrams Fig. 10.23 Gable frame under,wind loading
DIAGRAMS
SLOPE DEFLECTION METHOD
328
M". : or,
-/--\r + o. - ¡(¡"-¡o)fJ"ìl ,_-'lY\20" nEL""","., 2Jß [4/J
-
t3.34
: -
13.34
+
l.ll
IMo = 0
EIeB +0.554EI0. +0.208 E I^B -0.208EI^D
Adding these two
:
o+
ze
trrl
f
;#Lrr,+
op
)
l.ll EIoc + 0.554EIeD -
Hsc
¡(r":¡o)f J"ll MDrj o .':g)lo. ¿{r'L + 2oo - -z"lE-l. )J
s \-
(
,")=0.4EI0D-0.24EI^D
B, Joint C, Joinf D, Joint B, Joint D, Joint
(; (', 'l
(, ('
Msn + Mac = 0
(D
Mcs + Mco =
0
(ii)
+ Mor = 0 H". - H"o = 0
(iii)
MDc
HDc +'
Ho" =
.8 for the member CD gives,
(xi)
AB
for the member BC gives,
Mnc+ Mcn + lQx4x-2-Vc*6-Hcx4 =
0.208ElLD+ 7.49:
(Ð
2.22EI0.+ 0.208EI0o+ 0.5548I0" + l.9l EIeD - 0.208EI^B - O.O3}EIAo : 0 J o.olzEI0B- 0.208EI0D+ 0.20EI^B- 0.104EI^D- 55:0 0.208EI0" + 9.932EI0o + 0.104EI^B -- 0.20EI^D'+ l0 : 0
(ii)
0.554EIOB+
13.34 :6
:
ler 0.s54 o 4.032 -o.2o8lIeB I [-2.+s'l I 0.5s4 0.554 o o lle"I f-ß.:c| 2.22 rtl| o 0.554 l.el -0.208 --o.o32 l{ eo I = { o I -0.032
o
L-0.208.
0
|
(v)
0
(x)
-30
o
(iv)
The expressions for H"n and Ho, can be obtained from the free body diagruns of and DE in Figs. f0.23c.
IMs = 0
Msc+Mcs-Mpc-Mco
,ins in the matrix notation
There are 5 unkirovms in 0", 0. , 0p , Âs and Âo. The moment equilibrium provides threáequations and shear equations provide the other fwo equations :
(
(
.:,¡,:,
= o.BE teD- o.I4EI^D
(c) Joint-Moment Equilibrium Equations
il
0 or, H"" = Hc -
ffie five simultaneous equations can be w¡itten as : ;+, l.9lEI0s+ 0.554EI0c- 0.0328IÁ" -
0.554EIOc+ l.llEIeD-0.208EIÂ"+0.208EI¡
Mo, o. f(reo + 0p +) i5\ Mro = 0+ ?E(e^* 26._ -
Hc =
-
Hc
^
or,
for the member BC gives,
0
Eg
(ix)
I
HBc+ 40
0.208EI^B + 0.2088I^D
(viiÐ
Mnc*Mcs+80-MDc=Mcp
0
I
.
t2
'H
3(^" Jl3 ll ttr*['î)] ^"
-
(vii)
Mnc*Mcs*Moc+McD+80
Hc=
13.34 +0.554EI0" +0.11EI0c +0.208 EI^B -0.208EI^D
H.x4-Vct6 = O,
Euations,
vc
3(o"-o)f!îTll 1-." *,r"Y* 2",!t3 \4/J 2Jt3 lD
Mco
forthe member CD gives,
Mpc+ Mcn +
r= ,o * :
329
FRAMES WITH SLOPING LEGS
be seen that this is
E.I 0" : EIÂ":
-0.208 o.zo -o.røllanl -0.032 -.o.to4 o.2o J[^DJ
r
I rr lo I
Ðrmmetric square matrix. Its solution is
68.938,
EI0ç =
613.46,
EI
- 45.0{5,
^D= - Substitution MAB = - 140.43 kNm, M"o : -
EIeD
I
]
:
= 87.49
454-693
:
0,
(vi)
7l.25kNm,
M". :7l.22kNm
(iiÐ
(iv) (v)
330
FLEXIBILITY AND STIFFNESS
SLOPE DEFLECTION METHOD
Mc" : Moe : -
34.55
tNm, M.o = -
39'13 kNm, Mto
(e\ CompatibilityCheck (Figs.
10.23
: -
d and
34.55 kNm,
Mo.
:
39.13 kNm
MATRICES
33I
Member DC
74'13 kNm
:
ec
e)
oo
-,+"#',zJß-
),t#rfr
Member AB Using the moment - area
theorgm,
3
: !92-l
e-
EI
-
613'4 E
I
454.70 4 x-----7 +r-143.45 CCU+CrrC'sino:'-='iEI. EI 2JI3
:
534.27
EI
as calculated using the left and right hand sides of the frame is the same. It
^cxthat the calculations are corréct. proves
,,I,0.8
(Jß) _ tzs.tt (z x¿.ltr _" m\ st tz ( I + \v.-/ Er
T1.;
-\ - zJß"+T ) Yl;"zJß)
143.63
EI
Acx: Cc" - C'c"sin0 : Member DE
'' 2Jß I ..39.13 ... ñi xlx2Jl3 ..n xz{tJ6î. --:-X--x¿r,t.1 3 32281
34.55
EI
^a"
¡rn¡aiti.,- aaa-(positive means +^ to -t^L+\ right)
45.16 . :-:rf(ne8ativemeansanti-clockwise) uc:ua- 48.07- 128.37 *. 62.27 EI ,, EI C: C!': CrCr - C"Ct
_
C'Cz
143.45
LL
Moment - area theorem gives,
48"07
(negative means anti-clockwise)
I Er22El
(positive means clockwise)
Mentber BC
_
'
r:; 87.s: ^ ¿att5x-
: #"(i",) - +#(i) - *#,",
ÂB
,'
EI
C'C"' = C* Cz
: 1* ¿ *(r¿0.¿t -7t.2s)- ? .*,.['t " EI ,t)
o^
45.?
-
,
FLEXIBILITY AND STIFFNESS MATRICES
ple 10.19 Develop-the flexibility matrix and stiffness matrix for the beam elements with respect the degrees of freedom shown in Figs . 10.24 a, b and c.
,,-2
'A
,-
# +f "(#"): +#
I ..74.13 . I 39.13 ô vD :=, 87.5 "t t" El tx-x5
. I 74.13x5l?.sl_ t,.3g.l¡*s ao=i" el-l¡ ) 2 Et
(positivemeansclockwise)
(s\_ls+.to
,_^_,1,_._
..;J=Ë(positivemeanstoright)
I = CONST.
E
'otrT¡z r., 4A zEI
2
J' ,.1 ,l>;¡------aÐ, -1;_J_:
,. . E
3
!-
¡3 rfr¿
='CONST.
(c)'
Fis.t0.24
I = CONST. (b)
FLEXIBILITY AND STIFFNESS MATRICES
SLOPE DEFLECTION METHOD
Table 10.1 Stiffness coefficients for a beam
Solution
rs,
force successively at degree The flexibility matrix can be developed byapplyìng a unit at all the .coordinates' displacements the of fre'edom (d.q.f.) each and evaluating
Similarly, the stiffness. matrix can be developed by giving a successiely ar each d.o.f. while restraining all other degrees determining the forces required at all the d'o'f'
un^it
of
Rotation 0n at A when for end is
-displacement freedom, and
^t
B
(a) Flexibility Màtrix - Fig.l0.24a '1, and the dcflection and slope at B can be calculated with Apply a unit force at d,o.f. the help of standard results given in Appendix B,
ô,,
- öB H=* (
ôr, : ot=m=m
+
ô,,
- õB Y=L zBl 2El
ôzz
:
MLL EI
(c) M/Et
l7-----I-l
EI
*E:to ú12 i¿ I
=
l r' L'f I ¡e¡ 2ltl ll:-!l
I t'
L'
L,EI (a) Stiffness Matrix - Fig.
EI
|
LOADING
,{E!ê¡ L
The desired flexibility matrix can be written as:
[Flz*z
CONJUGATE BEAM
:=--.--_lH
Now apply a unit force at d.o.f. 2
oe
b)
t lzú
srrrruess
/
2_Elea L
lo€ler
6EI L3L2 6L
F is a symmetric matrix.
10.24a
Apply a unit deflection at d.o.f. l, and the forces produced in the beam at end B can , be determined with the help of standard results given in Table 10.1:
(d) STIFFNESS
COMPONENTS
Settlement ô at B when for er'd is fìxed
Mg
)
J
11,,.
coMPoNENTs
Rotation 0u at B when for end is fixed
3L2f
6El5
ú
B
1q"
l*o zei
e"
' îT- " ^ 6E_r eB il'/--L2
M
(b)
¿lt
r-L
\l
r
es
eele^
T:-u
6EI5
i
SLOPE DEFLECTION METHOD
334
Table
l0.l
"e
9¡
r--
3!r
eB
E;4 Iti (o)
¡
(o)
[¡-' Itl f.tzel l3Ef zltltxt I r' [F] x [Kl = I ^ L I l6EI lt:
Settlement õ at A when for end
<::ø --!l
-Ma
335
Stiffness coefficients for a beam (contd.)
Rotation 0o at A when for end
^\./l
FLEXIBILITY AND STIFFNESS MATRICES
"Bl
3Er5
en
itr
L-ru EIJ L û
[r I oli:til
6EI
t
4Et L
o. K.
LOU
Since the product is an identity matrix, it can be concluded that the two matrices are iinverse ofeach other.
F'lexibility Matrix - Fig. 10.24b
Apply a unit force at d.o.f. l,'and the deflection and slope at all the d.o.f. can be ined using the moment-area theorem or unit load method or the standard results in Appendix B.
Krr
k.zt:
t?Et
It
^ = o"
MLL Òr : ua:3EI = 3EI
t2Bl Lr-
MLL òzr : o":6EI =
So"=-Ë
ôr, : 0
k,,: 6TIe"--6EI --t¿ û "B- -T ,Kzz 4EI^ o"= 4EI L L
MLL - 0e: 6EI 6EI -=-
The desired stiffiress matrix can be written as:
IKlz-z;=
| it,
LÛ
(axialdefonnation)
apply a unit force at d.o.f . 2,
Now apply a unit deformation at d.o.f. 2
l!!I I 13
6EI
,;u:
6t : 0 (axial deformation)
npply a unit force at d.o.f. 3,
_ 6Erl
12
ô,, = 0 =ôr,
I
q; I
xisasvmmetricmatrix
LJ
Multiplying the flexibility and stifftess matrices
:
.PLL o." JJ --.
AE -=-
AE
matrix can be written
as:
SLOPE DEFLECTION METHOD
336
337
FLEXIBII,ITY AND STTFFNESS MATRICES Itiplying the flexibility and stiffness matrices :
le 3EI
lFl
¡*¡ :
L 6EI
Lo 6EI
(b) Stiffness Matrir -Fig.
dl ol
'li 06 ,')
Lo
=utt
3EI
2
FK = whererz
: llA
Lo
ol. AE
0
I
:'Apply a unit force at d.o.f.l, and compute deflections at all the d'o'f. using
10.24b
theorem- Fig:10.25 a.
M¡: +go L^L = +
k,
Ms:-!e^=
k¡, = 0
(axialforce)
o. K.
lo Loorj
Flexibility Matrix - Fig. 10.24c
Apply a unit deformation at d.o.f. l, while restraining all other d.o.f. The tbrces produced at all the d.o.f. can be determined using the conjugate beam method or the standard results given in Table 10. l.
kr, :
*
ftool I ol : I
f--'\| - ----
+
â^,le cl v
.,
i..rl
t
L
^r-
I
E¡ (q)
Now apply a unit deformation at d.o.f. 2 and restrain all other d.o.f.
krz
:
kzz
: Ms: fe" = +
Me:
î^
-9r" L"L = -ry
-----)
t------\-
I
_\_
-----
krz:o
(c
)
Now apply a unit deformation at d.o.f. 3 and restrain all other d.o.f.,
Fig. 10.25 Moment area diagram
I
.
kr¡ : 0 : kz. and k." : 4E
The stiffrress matrix can be written as:
L
axial
[Kl¡*s
ô"
:
4Et zBt L -t 2EI 4EI LL 00
ôtt
L
2.
x-L
L2
2Et
=
a
L3
3EI
-
=Ò¡,
tt] 5--x-L 2EI 3 =
0
4E
tt]
2EI 3
5L3 _-
6EI
unit force at d.o.f. 2, Fig, 10.25b
L
r
the
L L EI 2 -x-=-
ôtt
339
FLEXIBILITY AND STIFFNESS MATRICES
SLOPE DEFLECTION METHOD
338
I} 2EI L: 3EI
F:
3L= 3t] : L *T zEl Et
ô¡z Now apply
a
unit force at d-o.f. 3,Fig. l0'25c
I:L 2EI EI 5L3 3Û BL3 6br 2Er ãEI
ûL2û zEI EI
i$iffness Matrix -Fig.
'
c òt¡
:- 3L..fr.+¿r.')l =-st: zlr"lL"2L)3 aEI
2L EI
EI
10.24c
a unit displacement at d.o.f. I (Fig.l0.26a), and compute the forces at all d.o.f. moment-area method or using Table 10.1.
i,t] ^ òz¡=ffi
i*(#)*!tzt) = å*
(o)
l*' t -"i
=ffi I ". r
ntt
kzz
I
\ *rrt "N+4v krzl
( b)
Now apply a unit force at d.o.f. 4, Fig. 10.25d
^U.L : Òr¿
2El,
ou
:
EI
kz¡ I î_/Fk_4 ,,r¡l -Yø I'"
Rtl
k33
t"
K43
.
ò¡¿
2L ^ : 2r], ooo: pi EI
The flexibility matrix can now be written as:
(c)
(d)
Fig. 10.26 Deflected shapes for stiffness coefficients
k qz
PROBLEMS
SLOPE DEFLECTION METHOD
340
k¡r
1o,
6EI- 6EI.k?t : - ,r U"+ ç äc:0
.
6EI
l.À34
l.zet* lz¡ll" '= 24Et = (l-r -E-l-
l-
K¡r
i{ho
341
t]
^ t-o"=
4El -
4El
stifftess matrix can be written
L as:
tzBt
À.rrr
.:
.k¿r:
^ -r;or=T
Lt
6Et
6EI 24ET
t: Now apply
a
unit rotation at d.o.f. 2 (Fig. 10.26b)
K:
=0
=
ko, fe. = +
6EI
kzt:
.K¡¡ = Now apply
a
lzEI ^ t2Bt o"=-T._ Lu
- po"=-Ë:n* IZET _F_
.R2¿
=
2Et
-t"= +
.3.
6EI
4EI
*; be seen that the product F x K is an identþ matrix. !¡ÍË l Ti PROBLEMS
É*; Analyze the beams shown in Figs. P3.3 - P3.7 byfhe slope-deflection method. Draw shêar force and bending moment diagrams and the deflected shapes. ' Solve problems 3.3 and 3.4 by the slope-deflection method.
¡
i!'
Analyze the beams shown
in Figs. P4.2 and p4.3 by the
slope-deflection
method,
.
Solve problem 4.3 by the slope-deflection method.
tI
Analyze the frames shown in Figs. PlO. I - p10.4 by the slope-deflection method. Draw shear force and bending moment diagrams and draw the defleoted shapes.
unit rotation at d.o.f. 4 (Fig. 10.26d)
k,u: ffe"= T
2EI
EL-EL
Nbw applya unit rotation at d.o.f. 3 (Fig. 10.26c)
,Kr¡
lzBl
3 -2 L'L-L
lEr ...oc =
k32 Se. =-T
0- 8EI L 12Et 6EI
' ¡i
Solue problems 3.6 and 3.7 by the slope-deitection merhod.
Analyze the frames shown in Figs. P5.2, P5.3 and p5.4 by the slope-deflection method.
SLOPE DEFLECTION ME THOD
342
CHAPTER
l0 kN/m
l5 k N/m
r
T
I
6,D
im
I I
MOMBNT DISTRIBUTION METHOD
I I
þ2t+
Pr0.t
elbvmJ
¿t
',
Pt0.2
T
T
6m
I
4m
I
+
I
3m
DEVELOPMENT OF THE METHOD
:
llhe moment distribution method rvas proposed byfProfessor Hardy Cross in l932jfor
lm analysis of staticatly indeterminate beams and E¡-namer
insar=ite*tiu"
l0 kN/m
immediate attention.
30kN
;|
c
,ft, is necessary to measure the capacity of a member to rotate when a moment is i$nblied at that end. Let us consider two cases :
t.
50kN
(Ð A beam fixed at one end,
ì: *¿ (ii) A beam hinged at one end. iÈ : ßeam
Fixed at One End
,1:..i:: Let us consider a beam fixed at one end and simply supported at the other end as ,i:i,rl.chown in Frg. I l.la. If a clockwise moment M is applied at the end B, it is required to :i ,ffna the rotation at B and tfie moment produced at the fixed end A. This can be done by Pt0.4 Fige.
pt0.l - pl(L4
.1." using tåe conjugate beam method. The moment produced at end A will be such that the i' mtation at.A will be zero. It is, therefore, obvious thatthe mom€nt at A should also be r.iclockwise so that the slope at A is zero. The shear at A due to the MÆI loads is
I0.8
Analyze the gable frames shown in Figs. P6.3c and P6.3d by the slopedeflection ¡rethod. Also draw the deflected shape
1*1ML-?MoL:o v^ ^ = .3 2Et 3zBl
10.9
Solve problem 6.4 by the slope-deflection method.
MA=l
10.
l0
Solve problem 6.5 by the slope-deflection method
¡
fiod and is very usefirl when the degree of kinematic indeterminacy is very large. The ion method would require solution of a large number of simultaneous for such a problem. In the absence of a digital computer, it was very )me to solve these equations. That is why, the moment distribution method
Trlte slope 0 at
B is equal to the shear at B due to the MÆI loads,
(11.r)
344
MOMENT DISTRIBUTION METI{OD
DISTRIBUTION FACTORS
345
nioment M causes-aslope equal to ML/3EI at the same.end,of a bean,when the hinged. The stiftress of a beam hinged at A and simply supportêd at B is equal
is
v-"
with Relative End Translations
M¡-
z\MA
Let us consider a beam AB whose end B sinks by  as shown in Fig. I l.lc. It will a moment at each end whose magnitude can be determined using the conjugate method. The moment at end B due to the [,f/EI loading will be equal to the  as shown in the development of the slope- deflection method. Thus,
(sJ
:ó'E-x¿\
;lFI
IM
Fig.
tl.l
(l l.6a)
Basic conditions
pnd of the beam is simply supported;.the moment atthg fixed end will be given
.zML tMAL_ ML 32Er 3zyt 4Eti
i^ itt
I 1i or,
M
4EI
0
L
I
I
L._
The moment M causes
fixed. If the value of this
(l r.3)
I
__l
a:lgp:rYll4ll lqption
arjhe same end
oll,hçu. *li"o
the far end is
uni{r t-hen the,v.atue. or u,ré!úirea tò produce this ottúuti ,tø*--n rl,¡rÄ*il;;;;* nxed at A
is
ïÍl,"31îl:jig1l"jr:n * ',
M
(tt.2)
and slmÞty süÞported at B is èqual.to 4EI|L.
.=
,-' lPb tv_
(r l.6b)
DISTRIBUTION FACTORS concept of distribution factors is essential 1o understand the moment distribr¡tion Four members mçt at joint O as shown in Fig. I1.2a. The joint O is rigid, that angles between the members meeting at O remain unchanged. Let the members be at A, C, D and the end B is hinged. The length and moment of inertia of the are shown in the same figure. Under the application of a moment M at O, the
will deflect as in Fig. lt.2a.
llis
desired-to-*no¡¿-wrat¡lrt!4lJ[-o-lugnts
ll_1!::4:.{l¡"_ygqul_4rnÞe¡_at_o.-Asrheioiu!-Q_þ¡þ_i_!_.theggles A Beam Hinged at One End
Let us consider a beam AB hinged at one end and simpry supported at the other end as shown in Fig. I l.l b. If a clockwise moment M is appliàd at ttre en¿ B, it is required ro furd the rotation at B. The moment at A wilr be zeroúeing a hinþd Til;ìqp" e ;; B can be determined using conjugate beam method.
E _ML 3EI
or,
M
c
!uoo
.I1, Ll
(M
12,L2
""ã.
l¿r L¿ (l1.4)
\tI \
13,L1
B
.î# (b)
3EI
L
(r r.5)
Fig. 11.2 Joint equilibrium
'\
Mog
Ê
are
146
MOMENT DISTRIBUTION METHOD
SIGN
between the various members at o remain unchanged after the application of the moment M. The summation of the absolute stifûresses of all the rn"ntüår, meeting at o will be a measure of ihe resistance ofjoint O to rotation. It is obvious that each member will resist a proportion of the total applied moment such that IU: O at the joint. Let the
T9T9nj. tlul:d
*"1 9v
Moo
oD be Moo ,_yo" vi" trr,.""a ü|o , respectívely aT shown in T:*1".r^94,.o_B,oc Fig. I1.2b. L.r..,þç rotarion createdâ o ¡î'gI . morirenf l.9gli¡ed to produce a rotation
Moe
r
1Ir,0.7512 13
$oc
: i+r N{o. : l, L4
ìVf
: d. ìvt
L2
L3
I¿
M: d.M
L4
r4
Moo
t
(r r.7)
. 4EI4 L4
Moo: Mo" : Moc , tøoo , ,!. : o'15 12 .
L1
J3
L¡
.14 'L4
Moo: M
L4
I_rr-rÏfi,JM: LL, L2 L3 l
d4M
(il.il)
Lo
quantities dl , d2 , d, , and do , etc. are called the distribution factors. Thc tion factor for any member at a joint is equal to the relative stifrness of the i'divided by the s-um of thdrelative stiffnesses of all members meeting at the joint.
(l I.8)
In fhe above expression absolute values of the sriffness have been considered. lt would be more convenien,, if they may be reduced to lowest terms by a common divisor. when such a division is made, the resulting.numbers will be related ¡n some constant -;;¡;;;; ratio to the corresponding varues of the absorute stiffness, and are cull.J us 'stilitnesses. when a frame is composed of prismatic members it ir;urt;;;se valuos of, l/L for the relative stiffness of each membei. Thus, relative stiffnçss oía member whose far end is fìxed is r/L, andthat of a member whose far end is simpry or hinge
LI
, I¡ Jt* 0.75 l2 T-1---
Ll
L3
Mon + Mo, + Mo. +
14l
IrL3
Mo,
ol'o
347
Ll,*-fr-.ú.r.J
0 in each of the fouì memb-e-rsis-gìven-by-Eo.fi.iã,.ì i.ï,
thus, Moo: Mo, : Mo. : tøn, ,$L ,ill, , 4llt ' L¡ L2
Also,
L)
: *lte Lr
Moo :
i
0,7_1-12
and.
Jhat is,
CONVENTION
help qf distribution factors, it is very convenient to calculate the bending in various members meeting at a rigid joint under the action of an
'the created
moment at that joint.
CONVENTION moment considered at the end of a member will always be the moment the joint rpport exerts on the member. If a support tends to rotate the rqember clockwise, will be consldered as positivq otherwise the moment will be considered as
(r t.e)
(il.
t0)
Il
Moe :
M:drM
lixomple
llil
,$n¡!Vze
itrothod.
il,
the propped cantilever beamShown in Fig. 11.3 by the moment distributlon
MOMENT DISTRIBUTION METHOD
148
BEAMS AND FRAMES WITH NO SIDE SWAY
349
A and C are zero. The total balanced moments are obtained by all numbers in the corresponding joint columns. ilfddtng i,.l.',' lancing moment at
i',f ,
Llï
,
The resulting moment diagram is the same as obtained by the consistent deformatibn
Ll? Aualyze the two span continuous bea¡n shown lrlbution method.
Fig. ll.3
.50
kN
in Fig. ll.4a by the
l5kN/m B
Solution
(o) Ftxed End Momenß
Mrec = M.co
:
(ol
p(tlz)(rlz)2
PL
-_--_}-
.8
. p(t-lz)'?(rtz)
l-
.PL
-------------ã-
U
8
(b) Distribution factors
-Ì\
!T"9"
(b) FIXED
END MOMENTS
58.60
89.8¿
_Itng_Eg4Þglq m9g!!93!igg!Ç-ttç_tpe!r9nþþ_c_e Jnoment wilt remain -r¡0.-jò¡nt¿¡üiriu¡¡mÍl
corúFeliTóE_-L_selq9-þ_ir! !!ence, ãßniuullon facìõi is
¿
(c)
ct!_eþl3tia9¡nern*;'s""d-Otq.{en9!_aj{gþap1øó1ere,æfr¿çg j-oirr js_ær.o, (c) Moment- Distributio,n Sable I I.l) Table Joint Member Distribution factor Cycle I FEM (
BAL Cyele2
{
il.
I \
CQ
r8.lL-
ll.1
_
A AC 0 0
-
PLl8 0
r_In
io.:ì
BENDtN6 MOMENT, kNm
Fig. I1.4 Two span beam
of $or4eq! ¿¡striþuti-oa,_tl-rc_cSgqþLq!9gkyr,sg_USlqçú_at_ç q!qry_-jy_er_¡¡9!q9qq-9f ll2 l_-!_LJÐ. 1_;_ PLIII cssnrer-clockwise ar A. The ll$ges 1þq seqond cycle
(dt
i:.
0
End c is simply supported, hence the net bending moment at c will be zero. The initial locking moment at c is PLl8 (clockwise). Joint c can be released by applying an anti clockwise moment equal to - PLl8. Since support A is naturally fixed, it can readily carry the entire moment assigned or carry-over to it. This joint need not be released.
-
1.
1.0
0
-
-7r_!t5.2I
C
+PU&
3PLn6
LOCKINO MOMENTs
CA
PU8,
_ PUI6
BAL Total
(
I
I"pyd.Moments (Fig. I I.ab)
Mres
:
-${=-58.6okNm
MFBA
- *
5o(3-)-'(Ð
:
+35.l6kNm
moment -
l4¡,1 + ¡f¿l-a¡¿¿-c't
¿
"àI+.
ppo,* È1
t\
tt
BEAMS AND FRAMES WITHNO SIDE SWAY
MOMENT DISTRIBUTION METHOD
350
ra'e
35r
Table 11.2 Span
BC
t^2
M.". ,: - 15xî:
-
l25kNm
: + 15* lo' = + 125 kNm
Mr.s
t2
(b) Distribution Factors Stiffrress of member
4Et
4E(4I) -à'="'
4Er
4E(6I)
AB = î:
BC: î: Stiffiress just to the left of support DF at joint A in
sPan AB :
A:
-lL
-
t22.5. ,+ 6.69
i=:2.4Et cc
'
:
o
-
i
oF
at
SPan
joint B in sPan
joint c in span cB
DF
at
(c)
Moment Distribution
I
X stiffiiessof membersmeetingat joint B
2El 2El + 2.481
DFatjointBinspan
btifftess
l.16
+ 0.63
l
Bc :
:
##ffr:
:
0'4545
+ 115.28
0'54s5
1.00, since there is only one rhember meeting at joint
c.
and clockwise There is a counter-clockwise locking moment of 89.84 kNm at B release the locking to srder In I1.4c. in Fig. as shown c at kNm of-125 locking moment kNm is applied at B .or"it at B, a clockwise moment equal to 0.4545 x 89J4 :-40.83 :49.01 kNm is applied at x in span BA, and a clockwise momentequal to 0.5455 89.84 counter-clockwise a applying by released is n iå tp* BC. The locking moment at C *o¡r¡"nt equal to 125 kNm at C' itself since end C is simply supported. This completes the first cycle of moment distribution. The clockwise moment 40.83 kNm applied at B in span BA induces a carry over moment moment equal to ll2(40.83) at A in the same span. Similarly, the clockwise (49.01) lD to equal moment over a carry induces BC in span B at applied kNm 4g.01 = applied at G 24.50 kNm ui C in ttt. ru¡¡" ipun. The counter clockwise moment 125 kNm in the same span induces a counter-clockwise tnoment equal to l/2(125):'62.5 kNm at B + are the new set + 24.50 and 62.5 20.4 (co) to equal BC. Th: carry over moment L cycle. joints frfst the of the end at lock all to required of fxed end moments
second cycle, the joints B and C are released for the second time. The countermpment of 62.5 kNm at B is released by. applying a clockwise moments of 34.10 kNm at B in spans BA and BC. Clockwise moment of 24.50 kNm at C d by applying a counter-clockwise moment of 24.50 kNm, at C. Joint Ä is rfixed, hence, need not be released. This completes the sèçond cycle of moment
over moment of + 14.20, - 12.25 aúd + 17.05 kNm are the new set of fixed required to lock all joints at the beginning ofthe third cycle. The above be repeated for as many cyclet to achieve the desired accuracy.
tâl moments are obtained by adding all numbers in the corresponding joint -The free span moments are superimposed on the support moi¡rents. The net diagram is shown in Fig. 1 1.4d.
MOMENT DISTRIBUTION METHOD
352
(d)
BEAMS ANN FRAMES WITH NO SIDE SWAY 20kN,/m
Check
The moment distribution method is an iterative method. It is essential to cÏreck the accuracy of the final result. íIhe first g¡ççk is to see if t!¡e algeb¡aþ sum of all_the end p_glg$1p9"!!Itg_e! e jg_in! iq Tero_ 9r_4ej exç€pl_ar a fixe_d, supporr._ ,This check.is on the ioint moment equilibriuml
F_st + 3 +2,F2,F3r+2+2{
"f- -
M^
:
Mrns
+f
{reo*
)Êt (0n+ Ms : Moo *î
r,1.67
0
e")
I
b)
FIXED END
MOMENTS, KNM
12.87
.i\
20s)
Solution of these two equations give
(c) L0CKING M0MENTS kNrn 17.7 7
--*-3?.1
oA=
3ET/L (rra"
oB
-
rr¿oo)
-
ã:
j(rr¿o
ii.
- n¿.*)
20.1
.t;i:a
3ErrL
.jr;ii
These equations may be rewritten as:
:i,:i:
i104.18
ittt
ffi., Slope at nea¡
end
:
Change at near end
-
--(Change at far end) .,
Usag tlie above equation, the compatibitity condition can be satisfied. The slopes at the end of each member meeting at a joint must be equal to.each other. The slope at a
fixed end slioul'd be equal to zero.
[]i '
þl
It [] / I \-t,
The total moments obtained at the end of cycle g
Ært.1-ll l
)/ i\ ìt
.
FREE BoDY DTAGRAMS
62.33
if-li l/110
(e) Final Moments
,la t1.
1û3.65
13.1
inoreased.
1ome,nts._lhese
l
(e)
:
'ï.,Î)-Jî'iiyïk;i
53.67 16.33 16.6
It cau be seen in Tàlle I 1.2 that the slope at the fixed end A is ¡early zero. The slopes at joint B are nearly egugr. For a uetter u".*ury, the number of cycles should be
,'
¡:; !t
17.77
(tt.l2)
3ET:IL
fl:
(d) BENDING uovEÑr kNm
;ii,;1
I
ff:
18.61
f^
66.3 9
ååf'
ffi
1.67
t
There can be another ilidependent check on the calculations. This makes use of-the conditions of continuity of slopes at each joint. The slope-deflection equations for a span AB may be written as :
ffi
353
are
the end moments or support
moments and the fiee span moments are superimposed algebraicalry as
snoryþ¡Åra\ /ãrì '&j/
Analyze the'continuous beam of Fig. ll.5+by the method of momËnt distribution. " Draw the shea¡ force and behding momãnt diagrams and skerch the deflected;*"ä;;.
W (f)
'
Fig.
SHEAR FORCE
KN
11. 5 Three span beam with over hang
57.67
MOMENT DISTRIBUTION METHOD
354
BEAMS AND FRAMES WITTI NO SIDE SWAY Solution
{yfr*ea
Distribution
nna Moment (Fig. I I.5b) 20 x52
Span
AB
t:*
Span
BC
Mrsc
n
Mrcp
t$
Span DE
¡ffiistributton
stffiess
M*o : + 4l.67kNm
4l.67kNm,
a-¡2 6oxl ^í =
M"cs = Span CD
: -
-
28.80 kNm
+43.20
l2 -'¡
+23.47
60x- )'=+43.20kNm )-
+
7or2'^5': 7'
5ox
!4
7t
Mrpc
rcr2'15 + 5ox t:2
MDE
80 kl.Im
+11.74
=
-
91.84 kNm
: + 79.60 kNm
+
0.17
+
-
0.10
-
0.18 0.18
Factors
kes
: qÐ :
2.408
r,
kBc
:
ry
:
3.zoEt
A : 0, since joint A is,a fixed
DFatjointBinspan BA
=
+
-
0.14 0.19
-
0.01
+0.015
kc"= gg) =3.438r I
DF at joint
6.71
end.
0.4286
ffi:
\
BC:1?op_r=0..57t4 5.65E
DF arioint
c in span
cB = cD
#ñf
I
:
0.4826
tltP-] = 0.5174 = 6.638r
The cantilever span DE is not a ,suar member. It can be removed by imposing a fixed end moment ar D in th¡ span DE equar to g0 klrr*. rÀ-n;..ìrì.g"i"" since it acts in anti-clockwise däection. Now there remains only one .".u.. ãi¡oint D. Hence Cisfibution factor at D in the span DC is 1.0000.
moments are written in the first line of cycle l. The locking moments $ fn.support are shown in Fig. ll.Sc. The joìnt B is released b! applying €r clockwise moments equal to 5.52 and 7.35 kNm at B in the spans BA and BC, efively. similarly, joint c is released. Joint D has an unbalanced moment of - 0.40 A clockwise moment equal to 0.40 kNm is applied at D in the span DC. Span DE tvjl hang whose moment equilibrium depends on the loading in the same span DE ly. lf:tltere is no load in the span DE, the moment atjoint D will be zero. [Ê¡¡{l{-e-d e¡rd
!t,{tg3elna cycle of moment distribution, the carry over moments are applied at i A. B., c and D as shown in the table. They are the new locking moments. These 4.8, !ft$ nre ngain released in proportion to the distribution factors at the r:spective
joints.
) )
BEAMS AND FRAMES WITHNO SIDE SWAY
MOMENT DISTRIBUTION METHOD
3s6
357
The iteration process is continued till the locking moments become suff,rciently small.
At the end of cycle 8, the total moments are obtained by adding respective columns.
) )
) ) J. /l
(d)
5
Check
The moment equilibrium at each joint is satisfied. Now let us apply the check on continuity condition at each joint as disçussed in Example 11.2. The change in moments at the near end is shown in the first line, the change in moments at the far end is shown.in the second line. Their difference is shown in the third line. These values divided by the corresponding -value of 3El/L. It can be seen that E I 0 value at the end each member meeting at a joint is the sarne. This shows that the moment distributi
-:v¡.lENT kNm
(o)
method has converged.
Moe¡
BD
s2 : : 20x-
Mreo
:
AB : 0, A being a fixed end stifftess of member AB : 2Il5 : 0.4000 I
ve stiffiress of
member BD
so12\_5
:
BA: 171.43 kNm under 70 kN load at
The total moments obtained at the end of cycle 8 are the end moments. moments and the free span moments are superimposed algebraically as shown in Fig. l 1.5d.
0.2500
I
span
: DB : BD
0.4000I
:
0.6154
0.4000I + 0.2500I
I -0.6154
:
0, D being
a fixed end.
0.3845
In the case of frames, it is usual to arrange all the joints in alphabetic order. The ftlcmbers meeting at each joint are written under the joint as shown in Table I I .5. Since
,.'
l,
-
i.iJhere may be
tt.l
more than two
members meeting at a
joint, the carry overs are not
between adjacent columns. The carry overs are always from near end to far the same member..
Analyze the frame shown in Fig. I l.6a by the moment distribution method and d¡aw moment diagrams.
¡The iterations stop at the end of cycle 2 since there is only one unknown joirit rotation
.,
Solution
in the given rigid frame. A linear equation of one unknown can be soived directlv
., without any iterations
(a) Fixed End Moments
Span BC
jointB in the span
'ÞF at joint D in the
The free body diagram of each span is shown in Fig. I I .5e. The resulting shear diagram is shown in Fig. 11.5f.
Span AB
:
171.43 kNm under 70 kN load
_
'-,,
Il4
.;'.,:,Member BC
îv{ <-2 : 70x::-: + 50 x ":- :
M2ìc¡: to1512 +
:
is over hang member and it can be replaced by imposing a counter .Çþkwise moment at B equal to 15 x I : 15 kNm.
J
Exaryple
Mrpu
l_atjoint A in span
62.5 kNm
3x2
77
o:
Distribution Factors
Mosc:60x-:72kNm Mlocp
ll.6
Fig.
(e) Bending Moment and Shear Force Diagrams The free span moments are as follows:
_)
- --49.3
50 kN./m
equilibrium, the net moment at each joint mu'stbe'zero.
50x52 Mnes : - -æ- : - l04.l7kNm, Mrsn: + 104'17kNm Mr¡c : - 15xl: - 15 kNm
ì,í¿t)'t'cnec*
tne joint moment equilibrium is satisfied at joint B. The compatibility check is also The slopes at fixed ends A and D are zero. The slope at joint B in the spans BA ,_-Ë,þ.g1n.
i
':,:aind
i
BD is equal.
tn"
bending lnoment diagram is shown in Fig. I1.6b.
MOMENT DISTRIBUTION METHOD
358
BEAMS AND FMMES WITI{NO'SIDE SWAY
359
Table 11.5
A
Joint Mémber Relative Stiffr¡ess I/L
DF Cycle I
BD
0.4000 0
0.4000
0.2500
0.6154
0.3845 0 - 34.30
-t04.17
BAL
0
CO
-27.43
104.17
-
54.87
Total Moments
0.2500 0 0 0
-15 0
-
17.15
-
17.15
-
17:15
0
-131.60 -27.43
-
54.87
-
34.30 34.30
(change in
-27.43
-
13.71
-
8.57
moment Difference
0
-
41.16
EIO
0
-34.30
Change
D DB
BC
^0
BAL
Example
BA
FEM
Cycle2
in
momeil t/2
Check
B
AB
'
49.30
-15
- 25.73 - 34.30
(aI FRAME STRUCTURE
7. tl
17.t5
A
11.2s
0 0
.71
B
1.79
11.25
ll.5
Analyze the frame shown in Fig.
ll.Taby
the moment distribution methoò:
35.62
t
8.57
Solution
-(a) Fixed End Moment
:- àstNrn,
Span
BC
MFBc
Span
EC
Mrrc : - 40xl'5\,?'52 = 4'
=
- 50x'+
Mrcn
(b)
: + 25kNm
BENDING MOMENT Fig.
kNf
ll.7
23.44kNm
joiirt D is hinged, either the regular stiffiress factor 1: 4EIIL) or the modified factor (: 3/4 x 4EllL:3EW) can be used for end D of membãr cD. They will to identical results with a slight changc in the moment distribution. In the present the modified stiffness of member cD will be used and the moment distribution s is shown in Table I 1.5. ress
Mrcr : +
4oxl'52
-x2'5 = + l4.lokNm 4'
'Ihere is no extemal load on spans AB and cD, hence fixed end *om"nt are zero on
these spans.
joint B in span
(b)' Distribution Factors
BA =
Relative stifkress of members
kes: ot'å kcp
=t.3334E1,
: ,u"T:
l.2oooEr,
t-L ÀBc =
Kcp --
1t"T
BC
= 2.ooooEr
4Ex+:
l.5oooEr
C in span
=
u'E: ^h
t.3!34F.1 t.3334F,1 + 2.0000E
'-'9909-tl:0.6000 3.33348t 2.00008I
@:0.4255
0.4000 i =
f-) (_.,
MOMENT DISTRIBUTION METHOD
360
l; (i
BEAMS V/ITH T.TNEVEN SUPPORT
rI.5 l'2oooEl :0.2553
cD: CE IDF
Check
I
1.5000E
:
4.70008 I at
joint
Il.6
Analyze the beam shown in Fig. I l.8a due to sinking of support B by l0 mm by the method. Take E: 200 GPa and I : 200., x l0-ó ma.
moment distribution
0.3192
A8C0 m ,. sln jjBãroË' s'
C: I
DF at joint D is not applicable since modified stiffuess of member CD is used. DF atjoints A and E are zeror ,since they are fixed ends.
The moment distribution process is shown in Table 11.5. It can be seen that since modified stifftress is used for member CD whose.f¡ end D is hinged, no carry over moment is transferred to end D. In this case 6 cycles are required to achieve a reasonable accuracy. Both the joint ñoment equilibrium and continuity of slopes conditions at each joint are satisfied.
A
(a)' Fixed End Moments ,/
Mpes :
BA
Cycle I
0 0
0.4000
\ (
FEM
BAL Cycle2
CO
BAL
co
Cycle 3
BAL
(.
Cycle 4
CO
BAL
(
Cycle
CO
5
BAL
co
Cycle 6 (
BAL Total
(
Change in
moment
l/2 (Change moment ) Difference EIO
irr
+l
+
0 1.6:
0
0.3i 0 + 0.1( 0 0.0i
+ +
0
0.6000 25 0.0( T I 5.0( 8.3t
0"
-
+ 5.00
+
I
J.JJ
+
0.61 T
+
0.2
+
0.01
+
: +
-
0.0,
"+
14.21
+ -7.1
+
14.21
+
+
3.5(
0 0
+
4.91
1.6(
7.1
7.1
c BC
0.9( 0.5: 0.3r 0.1( 0.0(
CB CD CE 0.4255 0:2553 0.3192
+25
0
-l 6.6, - 9.9t + 7.5( - 3.lf - 1.9 + 2.5( - 1.0( - 0.6¿ + 0.4t - 0.21 = 0.1: + 0,I( - 0.0; - 0.0¿ + 0.0:
+ 14.l(
-
t2.4t
D. DC 0 0
EC 0
-23.4¿ 0
-6.2¿
-
2.4(
-
0.8(
-
O.li
16.0( 10.6;
-
15.8t 10.5t
- 10.5f 12.7
0
-
t.zt 0 0.4( 0
0.0t
r
l.9l
10.5!
+ 6.35 + 5.29
6EIA : . Mo. = Mrcs :-"î : 150 kNm \, Mrco = Mroc:o
E
+ 0 - 0.0i - 0.04 - 0.0: + 0.0i - 0.0i - 0.01 - 0.0 0 - 14.2: + 14.5( - 12.71 - r.7l 0 - 3 1.3! + t0.7: - r0.i - t2.71 - 15.8f 0 - 7.9: - 5.2! + 5.31 - 3.9t - 6.35 - 7.9:
+ 10.6{ + + 10.6f f
MFB,a
+(\
l\
6Er,A _ 6(200x l0'- x200x10-)-o.oro _ :--e-:--=_e6kNm
ll.5
B
AB
DF
Fig. ll.8
Solution
The bending moment diagram is shown in Fig. I1.7b.
Joint Member
.
CONTINUOUS BEAM,
(c) Moment Distribution
Table
36t
BEAMS \ilITTI UNEVEN SUPPORT SETTLEMENT
Exámple
4.7000E I
SBTTJSVEWT
0 0
o
"
(zoo x lo*6 x 2oo x
lo{)x (-o.olo)
42
. llj:_lld D are simply supported. The.distributionfactors may be compured using eltrer regular stiffriesses of members AB and cD, or their modified stiffiresses.
:0
Let us first compute the distribution factors using the regular stiffiresses of members
¡..:il18 and
CD.
k*:; I
:0.2r, k".= I :o.2st, i4 joint DF at A in the span AB : I.O aq:ioint B in the span
BA
= 0.21+0.25t ==!''!-
öu : joint C in the
span
k
0.25t 0.45I
cB : -= =o'ttlr, 0.251+ 0.1667 t
I
":i
ó
= 0.16671
=0.4445
:
0.5555
:0.6000
362
Joint Iuember DF Cycle I FEM
(.
( I
-
co
+
BAL
t
Cycle 4
co BAL
ii
Cycle 5
CO
BAL Cycle 6
co BAL
Cycle 7
lì
co
Cycle 8
co BAL
¡r;ilf"T"g
-øiffias
\;
Ë (¡
CD
0.6000
0.4000
-96
-
+
t¿.uo
+
24.00 48.00 1.34
+150
-
- 90nn - 60.00 0 - l).uu 0 30.00 + 9.00 + 6.00 + 30.00
30.00 45.00
-
1.66
0.67 ó.00 + 0.67 - 461 2.34 0.34 2.34 + l7¿ 0.97 T t.ll +
+
U.lb
_ +
0.20 0.24
- 0.12 0. + ol) + 0.14l3 U - 69.80 +96 + 13.10 82.90
+t38. t7
+
26.20
+49
-
4.50
21.80 36.33
4.25 2.17
+
+
1.32
t.39
+
-
36.24
I
1.08
I
't
t"
0.88
0.50 0.50
43.9s
the modified stiffnesses cD. rhe modined ,,,1Tîi_:r;F;"using of spans AB and ;ffiö.rhar irs f".;_ilñ;ed
{¡j1
ïif
or simpry
r,*i:ääñ"i,TÏå#:ilîi:i*1:{fJï,'ï""¿r"ii¡"ìiîi;i";ïi,ba,ancedby BA. The net mome¡
span
,ha,,he,"",,'*- ;;
Table
0.12 0.12 0 0
BC
=
GÈiih
BAL Cycle2
4J.95
Cycle 3 Cycle 4
Cycle 5 Cycle 6 ,
A.6250
-48
+150
-
63.7s 50.00 + 31.25
-100.00
-
3
+
10.62
+
15.62
+
6.64 J.¿l)
-
10.41
38.25 18.75
3.99
+
-
1.95
CO
+
BAL Cycle 7
-
co
BAL Total momèñi-
0.42
0.20 0.04:
+ 0
7)"s
r I,ll
CO
BAL
0 0
+
co
0
0 +150
CO
BAL
=r.oEr
0.3333
U
+
BAL
0.6667
+t50
0
CO
BAL
CD
0
-
CO
BAL
87.90
Cycle 8 0.3750
MFEM
0.o?
oy.t
l
+
0.69 u.54 0.34
0 0
-
0 0 50.00
+
10.63
0
0
s.2t
3.32 2.21
+ I.I¡
+ 1.62 1.08
-
+
0.34 0.23
+'0.t7
0.07 0.06 0.04 + 69.77
-
+
0
+ 21.25
-
DC
t.88
+ 0.lI
-
D
CB
+t50
+48
0 0
BAL
i,il: =
+96
co Cycle I
BC 0.62s0
-96 -0
-96
BAL
:r.:r1.
BA:
*;m;
C
BA 0.3750
FEM
?ilíi;:¿l¡:li
8I
B
AB
DF
1,1,
0.5
:*i,*rfu
was shown in Fig. I0.5b. Moment distribution using modified stiffnesses
ruemoer
U
-
ll.6b
Joint -l--Ã
+ +. 0.23 _ 0.23 t
;.''åiî*i*i*"iÏy
tr fixedendmomenß. thåmoment distribution cycres are assumed point in rhe usuar mannerto begin from this Th" b"";;;';;;;;;å,u*.u,
0.Bg
1.00
-
The distribution factors are computed
liill
3.00 3.00 2.84 2.94
0.44 + 0.46 0.25 0.21 u.12
+ 0.ll
muomentdistribution process using regurar and modified stiffnesses is shown in
,,
DF atjoint B in the span
'¡
+ +
177 t.42
363
Modified Momeft Distribution
now ::r_T. AB compure rhe disrriburion factors using the members rnodified sfifihesses and CD. of
3EI u ]EI = ^"o = -I- 6
(
5:67
+ -
65.9t
u*=# =+ =o.6Er k¡"=# =+
(
+
1.50
oÁ9
Let
('
-
2.92 2.65
t,50
0.75 0.66 0.34
0
+ 15.00
u.
+ + 0.33 - 0?o - 0.35 u.¡8 0.15 + 0.17 + olÁ + 69.80 + 43.95 - 80.20 --t06.05 )J.02 - 40.10
-
0
5.83
,"#Ì
D DC
+ 150
+ + - 0.87 l.t0 0.55 0.43 + o{< + 0.52 + ¡
+ ÈTõ"-
CB
0.5555
+
BAL .L .....;.
ÞL
u.+¿+4)
+ 12.00
BAL Cycle 3
BA
r.0000
+96
co
(c) Moment Distribution
C
AB
-96
SETTLEMENT
cB : ,==t:ott-:0.6667 l.0EI + 0.58I cD : 0.3333
DF at joint C in the span
1.0
A
BAL Cycle2
(
t,
BEAMS WITT{ UNEVEN SUPPORT
0.4000
Tabte ll.6a Moment distribution using regular stiffnesses
ri;,
ü
CD = DC =
DF atjoint D in the span
ffi
l
MOMENT DISTRIBUTION METHOD
0.1I
0.04 0.03 + 43.95
+
-
0.54
+ 0.ll
-
0.06
+
0.01
45.95
0
FRAMES \ryITH SIDE SWAY
MOMENT DTSTR.IBUTION METHOD
II.6
superscript I refers to the first stage and 2 refers to the second stage.
F'RAMES WITH SIDE SWAY
The frames with side sway were analyzed using the slope-deflection method. The fixed end moments were computed due to an arbitrary side sway  and using additional shear equilibrium equations, the magnitude of side sway was determined. It is not possible to repeat this procedure in the moment distribution method which is iterative in
?.
Fig. ll.9a. Its true deflected shape is shown in Fig. an imaginary support is provided at the level of the beanl (s) to prevent
Considerthe frarne shown in
I.9b. In stage l,
fimt stage, all lateral motions are restrained, one at each
the side sway. This problem can be solved using the moment distribution procedure as discussed earlier. In stage 2, an arbitrary lateral displacement  is imposed at the level of beam (s), and the fixed end moments are computed as before. These moments are written in terms of the arbitrary displacement Á. The extemal loads are withdralvn from the frame for stage 2 analysis. The moment distribution procedure is carried out for the arbitrary values of the sway.
ion is performed for this case.
I moments
are computed in terms of Â, and a moment distribution is performed. 2b, a lateral sway of Â, is imposed on the first floor while the tàp floor is from any lateral motion. Fixed end moments are computed in terms'of Â, and distribution is performed. The principle of superpositiòn requires that
Ql*k,eí'+kzelb+q:0 Qå + ,.-{..n09*
9r Td Qt
computed as:
k,e3'
l2-t ^ iil'i
;,:
i t:
( I
ML,*
Ll
length BC
L2
length AB
ytt'tl
HAI
2
vilr,
Single storey frame with side sway
i::
,'
r¡i
of stage 2.
ML=* Nråo
"ril * LtL2LL,L2I
,'t
STAGE
The true situation is obtainid by superposìtion of the results of stage I and 2: T\e equation of shear equilibrium is written assuming that the correct rorn"-nt, in stage 2 are k times those obtained due to the arbitrary sway Â, The factor k tells us how much to
scale the moments
l*i*.
..,
SÏAGE I
(
:o
Lt'bj
6Et
ll.9
+p,
Irl" * lnbo .* vri, + [¿b.I
+k
Fig-
+kzez2b + p,
are the storey shears in rhe top storey and first storey, respectivery,
Q,,
.
floor. The moment
In stage2a, a raterar sway of displacement Â, is on the top floor while the other floor is prevented from any lateral motion.
#{
Érí
the shear equation provides the value of the factor k. The scaled moments of ûre then added to the moments of the stage I to obtaln the actual moments in the The true sway Á' is equal to k times the arbitrary sway Â.
a two storey portal frame shown in Fig. I l.lOa. It is a typical situation in there is more than one degree of freedom in side sway. The key to the solution is the principle of superposition. The actual equilibrium and kinematic state of the ,is considered to bo the superposition of the three stages shown in Figs. 'e I I . lOb.
nature. The moment distribution method is cbrried out in two stages: I
365
*
r.f
pr3"=tvråt
*
Mic
+ Måo:l
STAGE I =o
Fig.
ll.l0
STAGE 2 o
Two storey frame with side swâv
STAGE 2 b
MOMENT DISTRIBUTION METHOD
366
k,
FRAMES.WITH SIDE SIü'AY
.
and k, These are the correction orscale These equationS are now solved for 2a and 2b to obtain the correct moments. of stages the moments to factors to be applied The sum of all three cases give the correct moments in the actual structure.
The following examples illustrate the application
of the
Factors
ofrnembers AB
moment distribuiion
procedure to the sway problems.'
atjoint B in the span BA
and sketch the deflected shape.
àt
Fê{
F4.t
joint B in the span
:
'EI
""
6
-
BC :
e(zr)
_
nu6 Etl6+Erl4
8
=
0.4
Ptla : o.a
EIl6+Er/4
will be the distribution factors at joint C. The moment distribution for the fixed end moments is shown ü Table ll.7a. The values of Ho and Ho are
,Èame
by considering the free body diagrams of the columns.
l i-,
-'
rr _ M* + M"o _ He:
l:!,ìt
ff
þ2-F---6m (o)
--4
(b} FRAME UNDER
.
il
SWAY
nD-
É¡
iiit:
L-
13.57
+27.18 _
6.7ekN+
6
""
=
fi'i
,
Table
,
ll.7a
/ f, l'
+
18.75
tt.25
åt
-
+ 16.87
-
10.t2
(c)
I
BENDING MOMENT' kNm
i(
Fig.
i('
+ l.6E - 1.00 + 1.50 - 0.90
ll.ll
Solution
( (
(
)..-Ñloment \_/'='.:
-
This frame has only one sway, that is, to the right. If this sway is restrained, the fixed end moments are given by
(..
Mrsc (
Mres (.
+
distribution for the applied loads with side sway prevented
50x2x62
:-
ir-
M*o i
0,
- SoxT x6 :+ 1g.75 kNm Mrpc= Mr"o : o
56.zskNm,' Mr."
+
13.57
0.15 0.09
MOMENT DISTRIBUTION METHOD
1
II. Moment distribution for arbitrary sway Âl (6.79 Fixed end {-..
moment Mres
: - q9 : t;
,
Mrsn
6ElA'
(. Assuming
EI
: -
---;-
l6.67kNm
100 kNm3
^' Mrco :-
1_'
=-
Similarly,
(,
There is no external load on the frame in this case. The moment distribution is shown in Table I 1.7b.
(_l
Table Joint Member
A AB
Cycle I
FEM
-16.67
BAL Cycle2
0
BAL Cycle 3
co BAL
Cycle 4
CO
-
CO
BAL Cycle 6
CO
1.00 0
+
BAL Cycle 5
-
0.30 0 0.09
0
+
CB
0.4
0.6
0.6 0
Total moment Corrected Moments Moment (l) Net Moment
-14.t0
-
2.0
0.6,
+
0.06
-
0.02
0.18
4.74
-
+
+23.30
8.83
0 10.0
+ + -
10.0
+ 5.0 5.0 3.0 - 3.0 - l-5 - 1.5 + 0.9 + 0.9 + 0.45 +. 0.45
-
-
0.27 0.14 0.08
0.02
+ I1.54
3.88
+ 27.t8
+
-
0.27 0.14
+ + + 0.04. +
l1.54
+t3.57
., côlumns. T" values of Ho'and r
the
+
0.03
BAL
{
BC
CO
H^'=
0.08
III.
+ I1.54
+
+
-
+ 15.66 + 19.54'
3.88 27.18
+23.30
- t6.67 + 6.67
3.88
t I
-16.67 0
il,ì:l
4m
J J.5J
2.0
+
0.6
+
0.06
-
0.02
-
-
0 1.00 0
+
0.30
_
0.09 0
+
0.03
0.18
t1.54
) + k(Ho'+ H¡') :
D
T
'-a
2¡
T 4m
B
I
t A
, 3.5m 3.5m l.-------!tF---_l
0
Fig.
11.12
distribution for the applied loads with side ¡ray prevented. ':ftame has
-t4.10
only one sway, that is, to the right. If this sway is restrairied, the fxed
- 4.74 - 7.83 - 19.54 -12.57 3.88 15.66
:-
55.1kNm, À4rcn:
75x42 x3 72
. ?{r e_ 50x=_ :_43.75kNm:_ Mrr. 7'
.Kec =
Correction in sway
(He + HD
-
,^,., 4'27kN+-:Ho'
The shear bondition gives, I
++ EI
3m
0.4
0.04
-'0.02
D DC
CD
Ho' are computed by considering the free body diagrams of
- ft¿.lo+tr.s+) \ l)=- -
=
the portal frame shown in Fig. I l.12 by the moment distribution method.
I
{
EI
ii.fhe mo.enrs obrained in faþle I l.7b can be corrected by multiplying with factor k i:ÈhO*n in the same table. Thètotal moments obtained in Table ll.Tacan be added fo the corrected,moments of Table I l.7b and the net moments are obtained. moment diagram is shown in Fig. ll.Ilc. The elastic curve is shown in
l6'67kNm
C
BA
- t6.67 + 6.67
: k^' : 0.336. g
ll.7b
B
DF
^
+ k(-4.27'4.27) : 0, or, k = 0.336
ìiil.l lb.
Lr
#:Mrnc
-3.92)
sway
6 x 100 zz
|¿ LU
369
FRAMES WITH SIDE SWAY
o
*ffi ,.a.'.:.t
4Et
l-
n f,¡"ged end, modified .
stiftress
:
f "+
= o,42BBr
=
73.47 kNm
FRAMES WITTI SIDË SWAY
MOMENT DISTRIBUTION METHOD
tL'
4EI 4g(:tl h": # r"'4= l.l43EI, k--: --- : s¡ : ]'f I : 0'75 E I F being a hinged end, modified stiffitess
.:
-
in
span CA
0.428EI + l.l43EI
cE : DF at ¡oint E
in
span
A AC -55.1
BAL
+55.1 0 0
MdFE
M BAL UU
BAL
-
t5.46'
+
3.54
co t.70
co
+
BAI. Cycle 5
^' Mrre E
-
CE c.73 43.75
-
0 43.75 41.81
-
CO
+
BAL Total moment
0
+
- 13.13 + 9.59 + 6.30 - 4.60 t.44
0.38
CO
BAL Cycle 6
0.27 73.47
27.55 0 +101.02
BAL Cycle 4
EI
+
EF 0.40 0
EC 0.60 43.75
F FE
:
6x100
:
-1,2.25 kNm
72
0.19
1
+ -
1.06
0.69 0.50 0.
0.04 87.63
+
-
r6
0.12 87.63
+
-
43.75 26.2s
-
20.90
+
+ -
+
0
-
17.50
: Mrrr=-lE]A' = 42
i+*
moment dist¡ibution is shown in Table I1.8b.
Table 11.8b 0
12.60
+
8.30
4.80 2.88
-
r.92
+
0.92
230 1.38
0
-12.25 +12.25
+ - 0.53
+
0.32 0.25 0.15
+ 10.3t
a.2t
ï -
o.lo 10.31
0
The moment dishibution for these fixed end moments is shown in Table I l.Ba. The values of Ho and H, are computed by considering the ñee body diagrams of the columns.
+162.t6 HA
Mn"iM.o -75x3 -87.63 -75x Lnc777
= M."o
100 kNm3
+
CO
BAL
Cycle 3
:0.60
c +
72
11.8a
CA
DF
FEM
6EIA'
Mrnc
.'.
(
Joint
Cycle2
0.27
moment.
6EIA'
Table
Cyclç I
:
end
0.73
l'l43EIEC = l. 143 + 0.75)E I EF : 0.40
Mç¡qLer.
sway Á'
(
0.4288r
=
- _ 2.5g kN<_
ll:: Moment distribution for arbitrary
4
DF at joint C
lo.3l
L¡ -flr
+172.47
=
-
37.5 kNm
c
The values
of Ho'and Ho
III. Correction
i':,where
HF
-o.85kN<-,
:--
I 1.34 4
=-
2.84 kN<-
,Ner He' :
and H. = -
!
i'-
I
2.58 -
19.62
:
k:
14.30
and final moments are shown in Table I1.8b.
+
{!ë e I + + )
f"tro. : *
all
I
rFr ""'
\
=
a
: "e vg
:
4
x200x
106 x
14'30
2J-4-.
L
300x lC-6 x 0.004 4
e
: + 240kNm
+ 120 kNm
Dßt¡ibution Factors
: l.l4EI - 4Bx2l 7
14'30 x"0'85 = -31'78kN+-
x 2'84: -43'22'kN<-
These are nearly the same values as obtained by the slope Example 10.9.
- deflection method, atjoint B in the
span
MES WITH UNDVEN SUPPORT SETTLEMENTS
tt.7
I)'' i on- t4
o-o}c,
0.01 m whích tries to rotate the beam BC in clockwise direction.
l_Mr.o
l:
r*uy n,o*"ntr
Â
t"
in swaY
The shear equation gives,
The coi-ected
se
FRAMES WITH UNEVEN SUPPORT SETTLEMENTS
are given bY
H;: - +:
c4,u.^ì
*-tb,^C'{^c;¿w-
MOMENT DISTRIBUTION METHOD
tqz
{ro
:ÞF at joint C in the span
m4. TakeEI=constant.
BA
-
BC
=
CB
CD
o'57 : 0.57 + l. 14
0.33
O.67
r'14 : - l.14+l = 0.47 \
0.53
Table ll.9a
T 7
I t:
'
,orr"
î
,
'îîi *< Fig'
+
- 13.04 + 8.73
ll'13
Solution .
I.
+ +
-146.94 98.45.
"LL
+
4.37
+
0.ll
lVlomept distribution due to yielding of supports when side sway is prevented
(a) Fìxed End Mome4ß
'1
: {Mrm
:i,
in, I "-^FBç
+
0.71
+
0.07
l.16
o:Mru¡' --6EI L2
6 x 200
x
l\
l0- x2
x 300 x 72
l0-ó x (0.01) Total moment
(l
0.10
0.2t
120.û
12.66
FRAMES WITTI UNEVEN SUPPORT SETTLEMENTS
MOMENT DISTRIBUTION METHOD
374
end moments is shown in Table I l'9a' The The mqment distribution for these fixed the columns' ã"t"rmined from the free body diagrams of reactions Ho and ffo
The values
'
"u"'ù"
_
HÂ
:
108.52+234'26
=
85.70 kN-+
.
Monnent distribution -q for arbitrary sway
(a) Fìxed EntI Moments
Mrse:
Mpas
M"oc :
Mrcp=
where. it is assumedthat E I
u.;loo:-
ry--
in case The frxed end moments so obtained are of the same order as obøined
I'
The
moment distribution is shown in Table I l'9b
Teble
-
l22.sa
BAL CO
CycleZ
+
20.24
-
16.40
BAL CO
Cycle 3
BAL Cycle 4
co
+
CO
BAL Cycle 6
CO
-
1.46
+
0.16
BAL Cycle 7
co
-
0.13
BAL Total
I.. {
Corrected moment (2)
Moment ( Net moment
-
-
40.4(.
' 32.8(, + 3.59
1.79
BAL Cycle 5
+
n2.5e
lIìl 8.34 73.1
+
25.J4
-
47.79
+
0.47 375.00 82.IC + 198.75 + 176.25 99.38 .|" 41.05 19.30 21.75 66.58 JJ.Z9 l0:87
+ +
-
+ + + +
8.82
3.64
5 .90
1.93
l.7t
1
1
+ 0.96 0.64 +
+
0.78 0.52
+ -
0.08 0.03
+
73.13
+
50.ó3
-
19.9t
0.0J
J
+
1
- 375.( + 8.1
+ 15.65 +
7.28
0.2(
- ll4.l4
-
0.53
0.61
2.92 0.32
D DC
CD
CB
BC
0.33
FEM
Cycle I
c
B
BA
DF
Hu:
--2A2.75-288.93
:-
33.21 kN<-
=
-
122.92 kN<-
of the horizontal reactions at A and D must
(Hn + Ho) + k(Hn + H¡'): 0 (10.85 +85.70)+k(-33.21-122.92) CorrectswayÅ =
kÂ':0.61E"
1
r'
be
=0 or, k=0.618
1000 :0.618 * EI
1000
200,.300
= 0.0103m
are shown in Table I1.9b.
I.8 SYMMETRY AND ANTI-SYMMETRY Many diffe¡ent types of modifications have been suggested by various researchers in
i:the moment distribution method for faster convergence. Use of symmetry or anti-
11.9b
A AB
-t18.34
by
; Tþe corrected sway moments and final moments
 = 1000 kNm3
Joint Member
- l14.l4
H;
The shear condition is that the sum eqnal to zero,
:'or, :,, .,
t22.5okNm
fl'--9!:-375kNm
ry:
are given
10.85kN_+
7
H¡ : II.
25.34+50.63
of Ho'ahd Ho'
375
I
17.64
2.95 1.56
+
1.39
032 0.17 0.26
0.t4
9.6: 7.83
14.14 + 202.75
irdiagram will be symrnetrical. The conjugate beam for span cD.is shown' in Fig. I :,¡'ì he slopes at the ends C and D are equal to
l.l4c.
e:tML2ET or. ¡4=2Ete L
+
O.7[
-
Therefore, the moment per unit rotation at either end, equal to 2EUL. T\e usual stifftress of the beam is EIIL.
0.ut
to one-half of that of the usual end is
the stiffiiess of the beam relative stifhess of a
and the carry over moment
-288.9:
-
70.54 + 125.3C + 12534 178.5( 108.52 + 108.52 +234.2( 50.6 55.70 16.78 16.78 19.91
+
¡i,' Consider a five span beam shown in Fig.l l.l4a. It óarries symrnetrical loading and ¡i e deflected shape is also shown in the same figure. In the central span cD, the moment
0.E6
0.12
202.75
considerably.
-
0.l5
+
;¡symmetry is one sueh modification. When 4 structure is perfectly syrnmetrical with{ :.:Iespect to geometry and boundary conditions about its cenfie line, the applied loads may 'cause a condition of symmefiy or anti-symmetry in the bending moment diagram in the ¡'central span. If either of these conditions exist, it is possible to modi$ the stifftress of ìi:the central span that will permit the moment distribution to be performed for only onei¡half of the structure. 'The convergence becomes faster and the efforts are reduced
-
+
consider a five span beam shown in Fig.
ll.l4b. It carries an anti-symmetrical
and the resulting deflected shape is also shown in the same figure. In the central
ll -l i):'
MOMENT DISTRIBUTION METHOD
ï',|(t
SYMMETRY AND ANTI-SYMMETRY
/' I
!fr
377
nxay{en.rc
\-/
I "(} ()
Analyze the portal frame shown in Fig. I i.l5a by the moment distribution method.
ii
f')
(q)
SYMMETRY
(b)
ANTI-SYMMETRY IN CENTRE SPAN
lN
CENTRE SPAN
I
(; (") i,
Mr--rM s ¿---AD
Ç,;
,! ij
ï 4m
f>-=
(._)
f--¿¡.-¡n * z a
(d)
'
¡t-
t.. ì*t ! i.
i
, ( b) BENÞING MOMENT
(o)
{-T-Tlf}
Fig.
KNM
11.15
Sólution 'The frame is symmetrical about its central line with respect to geometry and loading. There will be no side sway. only half of the frame need be considered for moment disribution ä'
\4
IIgfSYMMETRY 'Jpr1N Fig. l!.14
(a) Ftxed End Momenls
Tl- SYMMETRY
Mrsc:
span CD, the moment diagram will be anti-symmetrical. The conjugate beam for span CD is shown in Fig. I t . l4d. It can easily be shown that
0:+-! Thus, anti
6EI
or, M:
-
t++
40x5x22
=
--ir--
-
57.t4 kNm
;-
Mpc"
:(h) DistributÍon Factors
68ifr
j
l$tiffitess of
in the bendi
column
lhe moments on the two srdes oÏ tie beam wlll ry*--
AB
T=''
BC
I 4EI 2L
either side of the beam
at joint B The frame shown in Fig. I l.l4e represents a symmetrical case, while the frame shown in Fig. ll.l4f repreSents an anti-symmetrical case. In both these cases only one-half
frarne need be considered. The following examples illustrate the applications of symmetry or anti-symmetry in the motnent disnibution method. The same concept can also be coileniently used in other methods'of structural analysis.
in span BA
BC
=
:
I I +0.57
I 2 4.64
0.36
moment dishibution is shown in Table I1.10.
*
4EþII
-i-.
.=
0.57 E I
...:l::iii::fin ,, ir..'¡'
'l
MOMENT DISTNBUTION METHOD
118
SYMMETRY AND ANTI-SYMMETRY
Table l1'10
ì
(b) Dist¡ibution Facton
AB : 4E]
Stiffuess of column
-
Stiffrress of beam
DF at joint B
The resulting moment frame can be written by symmetry' The moment in the other half
Example
L
Modified stifftress due to end A being hinged =
57.14
+20.57
diagram is shown in Fig'l
379
l'l5b'
in
span
span
ll.ll
3
4Et
-x4L
: - 3El 4.5
0.678
4EI _ 4E(3r) :zBt þú= tø, L 0'67 :o.zs BA: O.67 +2 . BC : 0.75
The moment distribution for theç fxed end moments
Table
AnalyzethetwobayframeshowninFig.l0.l2abythemomentdistributionmethod
is shown in Table I I.l l.
ll.ll
and draw bending moment diagram'
Joint Member
Solution
A BA
DF
Theframeissymmefiicalaboutitscentrelinepassing.tluouq|'t{recentralcolumn will bc no side sway' to g.ot"fr--"t *af as. loading' Henc¡' there
Cycle I
Moreover,therewillbenomo-"ntinthecentralcolumnduetosymmetry.Hence, l l6' consider the frame showrr in Fig' l '
Cycle 2
with reppect
50
kN/m
iíi::':.,;lTh"
I
A
il{ffese
ll.16
One-hatf frame due to symmetr.y
(a) FÍxed End Moments Mm"
Mrcs = +
*f
j1
50x6" : _ l50kNm t2
12 150
kNm
0.75 150.00 + t 12.50
CB
+
+
0 0 0
+
37.50
-
37.50
+
150.00 56.25 0
206.25
McE Mec :Mr. :-
Mc" = -
M".
206.25 kNm
=+37.50kNrn
Mne'=-37.50kNm
are the sanre values as obtained
in Ex.r0.7,by the sloie-deflection method.
iil.1t¡ì
i.l:È 'i:,'
;',:.
,
37.50
BC
moment in the other half can be found by symmerry, rhat is,
4.5 m
Fig.
CO
Final momenl
T
I
+
BAL
q SYMM. c
B
Ð.25
FEM
BAL
c
B
AB
:.Ç;/Analyze the two storey-single bay frame distribution method and d¡aw the efastic curve.
shown
in Fig. l0.l7a by the moment
il .t.' r:.:ì:lil
SYMMETRY AND ANTI-SYMMETRY
MOMENT DISTRIBUTION METHOD
380
in
soan ag : J i' 0.75+
:
0.57
1.0
!t:.
To carryout the moment distribution process only half the frame needs ifconsidered ag shown in Table I1.12.
T
4m
to
bê
Table 11.12
+ 5m
I
-t33.40
+
40.00
'..'
Fig. 11.17 One-half frame due to symmetry
+ +
FÍxed End Momenls
rr - - *t : -6È\: TYTFAD 'r.--9/ n
t33.4kNm
= Mo"
+
M.oe :+133'4kNm:Mttt (b) Dístributìon Føctots Stifrhess of column
4Er _.48I _ 0.88I
BC
final moments are nearly the same as
L'
beam
DistributionfactoratjointB in
span
BC
=
0.41
0.23
',.^naryze the box frame shown in Fig.r0.lga by.the moment-distribution method.
o'8 **ro*gJ5 tn
span BA: o**ñoJs
ln
span
Distribution factor at A in sPan
AD =
10.12 by the slope-
Ínmnlp I I 11
in
BE=
in Ex.
l:::;i'_
1"481 ,'L*¿E(lI) = 2L'28 \, : stifûre.ss of beam-BE
Âi-r
obtained
method.
4
of
1.45 0.83.
+71.81
1EI = p¡ Stiffuess
7.4t 4.22
0.75
0.8+1.0+0.75
#ïi
SYMM.¡t
-o.75EI
=
I
T
o'31
2.5 m
:0.39 =
1,
0.30
20
, :0.43
I
l.lE
kN/m
2.5m
I
One-half frame due to symmetry
SYMMETRY AND ANTI-SYMMETRY
METHOD MOMENT DISTzuBUTION
x82',.
"
ll.13
Table
Solution
in Fig. I l'18
Cycle
i00x5 : _
Pabz PL tlrYrFAc g l] ilt:lTYTFED
I
BAL 62.5kNm
20x52
12
12
=
0 41.87
0 27.92
4t.67
BAL
2.32
co
Cycle 4
0.26
0.79 0.53
34.17
34.17
member
AC AE
= =
AE and AC =
fhe poÉal frame shown in Fig. I
I
l.l9a.
-
14.û3
-'
6.90
-
1.55
4.68,
-
3.13 2.35
-
r.57 0.52 0.35
-
18.52
0.78
+
0.17 18.52
Il2l 2.52 -'-x-
0.4
A',
0.4+0.2
5
'
(b) FRAME
UNDER SWAY
hA.l
B
0.2 O,4+O.2
Stiffiress of member
I AE= L5
Modifred siiffrress of member
ED
T
E
5m
z¡
I
o
(o)
5
II
2.5
Fig. I l.t9
I.
l2l = t"T 5
=
0.67
:
0.33
0'
'frame is symmetricâl about its ce¡tre line with respect to geometry. However, it of anti-symmehical loading with respect to the beam BC. I{ence, this frame analyznd, by the usual moment-distribution method.
a case
Carry over factors betweenAandE = 0'5, betweenAandB =
Draw the bending moment diagram.
B,
2.5.
Joint B
F
13.75
t2l
-x25
0.67:0.33
DF at
-
20.93
1.05
co
-
-
4.69
0.52
BAL
Modified stiffiress of member
-
7.01 1.57
BAL Cycle 5
-
13.96 9.36
4.60
Total moment
sPan
0.33
62.50 20.63
CO
41.67 kNm
Joint A
DF at A in the
0.67
BAL
(b\ DÎsttlbutíon Facton
Stiffiress of
0.67
CO
Cycle 2 Cycle 3
wl-z
-
FEM
ED
EA
0.33
DF I
AE
CA
Side
Sincetheframeissymmefricalaboutitscentrelineconsiderone.halfoftheframeas shown
A
Joint
(ø) Fixed End |lloments
383
between
EandF:0
ThemomentdistributionforthesefixedendmomentsisshowninTablell.l3.The by symmetry' The .{È: moments in ü," ."muining one-half frame can be written directly +ï final end moments are nearly the same as obtained from slope-deflecfion method in Ex.l0.l3'
distribution when side sway is prevented End Moments
f,M¡*=-
*t] n
=-î 10x52
=-20.84kNm,
Mrun: + 20.84kNm
column
DF
BC
BA :
span
at B in
sPan BC=
=
offi
lr
==l'o;;:
0.8 + 1.0
-
Cycle 3 '.
( '':
t,
-
co
(,'
(r () (
(,
BC
0.56
9.17
CB 0.56
-
0.72
-11.67
DC
CD
Cycle 4
0.05 +10.90
lhe horizontal reactions Ho and Ho
+
-
-10.90
are given
by
Hn
.:
HD -
l:./sqarrjq - 2s : 7t
+
(-;U
BAL Total moment M'
1.28
Corrected moment 0.45 0.25
+10.56
+t3.44 6.72 3.76
+13.44 + 6.72 3.76
1.88 1.05
+ 1.05
2.77 the. free
î
-
2.96
+
0.83
-
0.23
+
0.07
-
0.02
0.52 0.29 0.15 + 0.08 + 0.04 0.02
1
0.04
-19.86 -31.78
CD 0.44 24.0 +10.56
-
-
+
-
-
-
+ +
0.52 0.29 0.15 0.08 0.04
-
0.o2
-
24.0
+
5.28
-
l.4E
+
0.41
2.96
1.88
+
DC
+'0.83
-
0.23
+
0.07
-
0.o2
- 0.ll + 0.04
-15.75
+t5.75
+t5.75
-25.20
+25.20
+25.20
-2s.20
-19.86 -31.78
+10.90
-10.90
-
+ 2.77
+
-14.30
+14.30
-22.43
-30.40
-15.75
M' + 0.20
+
+ 0.12 - 0.07
- 0.ll
UU
Cycle 6
+2.57
+ 2.77
M 0.10
+ 1.38
Net moment
ü..tn.
-25.78 -57.56
horizontal reactions Ho'and Ho'
, : -:::::---::::: -19.86- 15.75 : )
body diagrams of the
H^'
columns AB and CD,
)5
0.56
0.41
BAL
0.91
-
+
CO
0.44
CB
0.56
1.48
BAL
5.83
+
-25.78
5.28
BC
0.4+ 24.0
-
BAL
-
0.36
BAL Total moment,
-
+
co
3
Cycle 5
+ l.2a
CO
Mfi) .,
BA
Cycle
D
+ 1.63
BAL Cycle 5
c
0.44
-
D
C
B
BAL
+20.84
-
co
Cycle2
ll.l4a
CO
Cycle 4
is shown in Table ll-14a.
4.58
BAL
t'
(
-20.84
BAL
24.0
ll.l4b
BA
AB
FEM
Cycle I
Similarly'
end moments is shown in Table I 1.14b.
A
Joint Member
I.
24kNm
Table
B
AB
BAL
|,
\
A
co
Cycle2
-
Mrcp =
DF
BAL
:
=
i!iLr *oni"n, distribution for these fixed
0'56
Table
(,
Mroc
0.44
The moment distribution fo¡ these fixed end moments
Joint Member DF FEM Cycle I
100 kNm3
fixed end moment so obtained is of the same order as that in the case
4EI 4EI L5 49x2l 4El : -_ = EI L8
AB
beam
that E I Â":
is assumed
(b) Distribution Føcton Stifhess of
385
SYMMETRY AND ANTI-SYMMETRY
MOMENT DISTRIBUTION METHOD
384
iii;i-Ihe
shear condition requires that
fu
1.38
given by
7.t22kN<-
the
2.77 +22.43
sum
:
Ho'
of all horizontal forces must
be
lS'
27.e8 kN<-
+ HD) 1 k(He'+ Ho') + 50 : 0 ç27.98 +0.83) + k(-7.122-7.122) + 50 : 0 (HA
2'77
!t'38 = 0.83 kNe
þfi:
5
II. Moment distribution
under arbitrary side sway Â'
The fixed end moment due to a side sway
'he correct side sway
..'.--EIEI .i'''.
Â'i-sequal to :
6ELA' _ 6EIA' _ rr Mres:_ - -E-: - -,z-
6x100 _ =
1
is,^given by
Â
:
k
Â'
:
k
i,-.The corrected moments are shown in Table
-
24kNm = Mrse
ll!¡nud in Ex. 5.7 by the strain-energy method.
:,'
l-100:
or;
k:+
1.60
160
ll.l4b.
These are the samevalues as
o.K.
t,-j I
'l
{'
SYMMETRY AND ANTI.SYMMETRY
MOMENT DISTRIBUTION METHOD
ì-
387
,t.. ("
,')
(-..'
the Portal frame of ExarnPle load of 20 kN at B.
{^Ayrt
ll.l4 if it is.subjected
only to a horizontal
conect sway is equal to Â
Solution
t": t)
Theframewilldeflectinananti.symmetricmode.Thefrxedendmomentsduetothe for the H;;;; ,ft" "í"*"" distribution will be carried out only external load are l'20' ""r". one'half the frame shown in Fig'l
;Jit;ä;t
^'ã
lè
Cycle I
I
i'. Fig. f l'20 One half frame
iltt
- -T-
BAL
due to anti-symmetry
f:.
r!:
I:
at B in
r
bearn BC =
'
(-
the
sPan BA = sPan
BC
=
The moment distribution is .
written
Mcp MDc
4El L
4El )
-
B
BA
BC 0.65
-24.00
+
8;40
+ 15.60
- 15.60 - 10.40
+ t5.60
4.20 0
-
-
t7.80 l1.90
+ 10.40
4EI
L. , -x 4E(2t) = l'5 î o'8 : 0.35
= l.5EI
End Moments
-- 6EIô' . =u6"
The shea¡ condition mqy be
-
as,
t(gi*Ho¡+20=0
l.t6
is symmetrical abôut its center line with rpspect to geometry, but not with ifô the load. The beam deflects in an anti-syrhmetric mode. There are no fixed ments due to the external load. Therefore, one-half frame will be analyzed for an ;.!way Á' as show¡ in Fig. I1.21. The deflected shape was shown in Fig; 10.20b,
0.65
ll.l5.
= Msc : + 10.40 kNm Ms¡ = - l0.40kNm : Mn¡ -- ll.90kNm
10. I 5.
0.8 + 1.5
shown in Table
:
method.
rEIô' =
':
I
EI
-m
the portal frame with inclined legs shown in Fig. 10.20a by the n¡oment
o.8EI
Modified stifttess due to anti-symmetry
DF
I
=-24kNm=Mox
Mrpc
ofcolumn AB =
Stiffiress of
24.00
+
co
Total moment Corrccted moments
Ntu
6EIA'
:ta :
66.8
in the other half fr¿me can be written by making use of anti-symmetry
''
-Stiffitess
-
FEM
BAL
þ),Dßtribution Factors
.:\-,_i
AB
0.35
Cycle2
(a) FÍxzdEndMotnent
t/'.
x-100 = EI
DF
5m
=
0.668
0,668
A
Joint Member
c'ü.
(._)
Mrcp
k:
corrected moments are shown in Table I I . 15.
T
and
kÂ' =
or,
Table 11.15 20 kN
MErs =
-
o,
SYMM. rè
lt,
.tl
*[_!Z'ri!lg) * zo :
'.:i:
16.67 kNm 100 kNm3
6
x 100
;--:-l6.67kNm
SYMMETRY ANÓ ANTI-SYMMETRY
.
389
MOMENT DISTRTBUTION METHOD
388
ll.16
Table
lc'
25k
svt'lu
¡:t
T
F_3m
FEM
Fig.
Mrsc
Cycle I
t: 24x100x0.5
-_-..-.;:-_-
:
+
Stiffiress of member
-
3EI I
:
-
62
33.34 kNm
0
-
0 0
-
- 8.34 - 5.00
+33.34
-20.00
13.34
+ 13.34
32.50
+3250
sway is given by
: ,ET2EI ô cos 30o
k ô'cos 30"
:
2.436
*
100
*
'13
-
210'96
Mcs = Msc:+32.50kNm Mco = Mse:-32.50k1Íl
o.5EI
o.K.
tt.t7 the gable frame shown
L
inFig. l0.22aby
the moment distribution method.
Modifred stifttess of member BC due to anti-symmetry
le
=,.'(0"ä") :'u"' DF at B in the ¡Pan span
BA :
0.5
0:5+2
=
25
4m
+
is shown in Table I1.16.
5m
The sway equation is
or, or,
HA+HD+25:0 kIMse- (Mnc+M"s)+Mspl+ 75$ = 0, asdìscussedinEx' k[-13.34 -(13.34113.34) -13.341+75{3 :0
^?
V:)4?.6
kN./ml
,T
0.2
BC = 0:8
The moment distribution
m
correctéd moments are shown in Table 11.16. The moments in the other half
{EI
BC
0
Mod.FEM
.Total moment Corrected moment
(b) Distrìbutìon Ftrctots Stiffrress of member AB
+ 8.34
BAL
24Elô_sin3oo
BC 0.80 .+33.34
16.67
CO
3m.l
ee(ztX)o'tin:oo) _ --
1
BAL
f 1.21 One half frame due to anti-symmetry
6EI(B'8" + c'c')
BA 0.20 t6.67
-
t6.67
BAL
I
+
B
AB
DF
3/3m 600
A
Joint Member
10'15'
I Fig. 11.22 One half frame due to Bymmetry
Ë
[\' l,' Iri-' Ét r E\J F,l ft
MOMENT DISTRIBUTION ME'iT{OD
390
SYMMETRY AND ANTI-SYMMETRY
(
Mruo
Solution
Ëï'
line with respect to geometry. and loading' The frame is symmetrical about its centr€ .¡f,*" *ìfi U" a symmetrical side sway. Its deflected shape and free body diagrfuns may ,Th" ryoln"nt distribution mar ue 91{.10 out for one be seen in Fig.!0.22b, B*lo.rz. and for sway in Fig. 11.22 in rwo parts : for non-sway morients,
Etr'
moments.
FÇ; F¡"',
il ,:t'
jr,
i,',"
!f|rh; I.
i1..,
ii,
wL2 12
f,r Mpcs
:
25x62
l(
I
I
ir-
(assuming
EIÂ
= l00kNm3)
(Â'coseco). -g " rano= 16 L'sc
Mpcs
r¡kNm
-
=
-
:
0.416x100 =
AB =
+
-
DF at B in span BA : in sPan BC :
,, : + l.l I = =o:1 0.8
Cycle.l
r.rrEl
Total moment M' corrected mornent M' Moment M Net moment (M+M,)
ll'l7a'
Table l1.l7a
i!, Joint Member DF FEM Cycle I
A AB
CO
BAL Total moment M
c BC
+ 31.5
-
+
75-0
+ 75.00
43.5
+21.75 .a
+ t5.75 0
+ 15.75
CB
0.58
0.42
BAL
+ 31.5
II. Moment distribution due to arbitrary sway Â'
15.75
+ 130.59
+ + +
CB
0.42
0.58
24.0
41.60 + t0-21
- 41.60 + 5.10
-
- 36.50 - 151.33 + 96.75
7.39
31.39
+ 130;14 + 3t.50
+
BC
161.64
31.39
-
t30.14
-
t61.64
-
-
31.5
Hr+H¡=Q = - He , hence this leads to a trrvial solution.
3t.s0
-
54.5E
(i)
B, the shear equation is
HBc-Hs,A-o I forces H""
(iÐ and H"o
due to non
- sway momgnts are given by (Refe.r
+ 96.75
l¡c
=
Msc
+ Mçs +450
4
(a) Fìxed End Moments
6EI{= 6x100:+24k1.{m 5. -
+
c
B
BA
sway equation is given by
B
BA
3.70 0 + 27.70 + I14.84
BAL
shown in Table
24.0
+
co
Cycle2 0'42
is
+
FEM
riun n.flb.
tl.l7b
A
BAL
0.58
The moment distribution for these moments
Cycle2
4t.60kNm
AB
DF
qn(zÐ
î:ffi:
:
4l.60kNm
Â,cosec0 =
#
Joint Member
=0'8EI
c -= 4EI member B^
Mrne
-
momenr distribution due to the arbitrary sway moments is shown in
¡(
,
kNm,
Table
Stiffiressofmember
i, l(
=
12
+75 kNm
iL ii,' lr..'
24
6E(2r)^, 2.1ß 6 __ _ =-6Ñ*-¿ =-;JE=,EI^'
fixed e¡d moment are given by' The frame is resbained againstsway, and the
Distrìbulion Factors
1,
+
Moment distribution due to applied loads
ij,j
,.,
Mrsc =
lu,nã rr,o*n
I/'', íor'
:
391
H¡n =
M¡s +Msn
)
=
= d2S.8l lcÌ.I+
tLii¡l
=
e.45 kN-+
ffir
syvuErny
3gz
er.rD ANTI-SYMMETRY
393
"tott*tDISTRIBUqPIIMeTnoP
Thehorizontalforcesduetotheswaymomentsaregivenby:
ffi
,, - Msc+Mcn nsc-4
$I. t,''
rr - M*+M"n nBA-5
it
1.,
-31'39- 36'50 4
=-
tã.SZtN
= ll.82kN -27.70!31'19 5
shear eq; (ii) gives Substituting these values in the
128.81- kt6'97 -9'45
k=
or,
-kll'82:0
4'146
Table net moment (M+M') ar9 shgwn in The corrected moments M'and the method' Ex'10'17 by using the slope--deflection These are the same uutu"' * in
$[ li
ll
i,, l,i
l ll ll
Example
tl.l8
in Analyze the Vierendeel girder showñ anti-symmetry
fig' t t'Zla
ll'l7b'
+ (b)
SYMMETRTCAL LOAD
by making use of symmetry and
75kN rt-
, ll
D
c
B
r:l
I
2l
21 A
lll lii'lì
I
¡
I
I
I c' Fig.,,ll.2l
2l
2l
I
(cl
D,
T 4m
,l
Vierendeel girder
Solution
TheVierendeelgirderorframeissymmefticalabouttheverticalaxisbuttheloading a symmeftical
into two sep'arate system¡, one is un-symmetrioal. Let the load be broken system shown in shown in Fig. ll.23b and another an. -ii-ry.tetrioal
(iÐ
inclined members,
¡,:.¡äD
ì
for- given
loads' This
approach
"o*puøtionul
TheViere¡deelgirderisastaticallyindeterminatestructurebutanapproximate if three.hin.go.T" can be u¿opt.ã. It will liecõme shrically determinate
-AVri,
p-.i*J ;äËt";ü;ul1,h.;;rp*s
introduced in each
tn. characteiistic behaviour of the girder is maintained by
of the chord members andmid-heighr of the verticals. determinate structure. However, such an The Vierendeel girder u.."nr¿s a statlcally : with girder is not suitable fdr Vierendeel analvsis
.LOAD
Vierendeel
cliords of widely different stiffiress,
girder
;ontd.
non-prismatic vertical members, and loads applied away from node points.
Fig'll.23c;obviouslythesumofthesetwosystemsisequaltothe.givenloads.
ts that "q"A considerably' efforts
11.23
(r)
system
separate systems it
I ANTI- ÇtMME TRICAL Fig;
Thdrefore,accordingtotheprincipleofsuperposition,thesumoftheresultsforthetwo reduces the
l'.
(c
moment distribution methods have been developèd fôr the analysis of girders. The substitute frame method can ôe used for the analysis of parallel girders with chords of different $iftress in the panels, while the top chords of any panel are of the Same section: The fixed end moments are from the panel shears. The stiffrress of the verticals is take¡r as six times of (= I / L) and the distribution factors are determined. The. Carry over factors
-1. :.
The restofthe mom€nt distribution procedure is as usual.
Symmetrical Loading Only one-half of the frame need be considered.
SYMMETRY AND ANTI-SYMMETRY
MOMENT DISTRIBUTION METHOD
394
Table
Member Stifness
(-t
J\
I
.T ksc
í'r
Joint Member
-- 6x2 7 = 3I = kcc'
kAB
MemberAB,
DF
= i=O.25
1r'
l)
I
l-r ("',
,,
Distribution Factors
Atjoinr B in
|':
atjoint B in span.
(''.,
it is cut in trvo halves by the line
of symmetry.
rì
at
span
joint C in span
"ok
- ãtjj}
=
0.08
CBI
0.25t = 0.074 0.25I+0.125I+3I
=
t-
+ 33.34
+ I.38
+ o.22 + 2.67 - 37.5 + 36.01
0.22
-
0i,037
37.06
+ 0.11
+
1.49
0.25I
=
0.25I+3I +O.37St
in span CC'
0.827
in span CD
0.104
End Moments Roactiqn RA,
=*
I2;5 kÎ.f 1
(\ ;
ì(r
?I
in span CC'
3.375r
inspan CD :
.,
(, ('
l-0.E89-0.074 =0,037
Fixed End Moments
'Reaction Ro
.'.
=
37.5
kN,
Fixedendmomer¡t =
-
Shear in panel CD Fixed end moment
0
= =
Shear in panel
37.5"
BC =
37.5 kN
1 * 1 = -37.5kNm
22
II
(ì
k* = 3I, 'vu24k^
=ltl
k". = 0.25
momenb in the
The:
DF
0.92
FEM
BAL Cycle2
co
Total M'
M Ner(M:+ M)
ll.l8b
B ¡t¡4.
BAL
=o.r1sl
The stifûress of member CD is taken as 1.5 times because it is cut in nvo halves by the line of anti - symmetry
JañrMember
Cycle I
1
(-ì
*d th;.r.r'"ö. T¡ble
Anti - symmetrical loading
Member Stiffness
The moment distribution procedure is shor¡n in Table I l.lgb.
erlralf have rhe sa¡ne,"gnitua"
0
The moment distribution procedure is shown in Table l,l.l8a. The moments in the remaining half frame have the same magnitude but opposite in sign. Case
(ì
('l
37-s
CD
0.889
',
(,,
(,
CB
o.92
BCI =
c cc' .
CB 0.074 - 37.50
Distribution Factors
'At joint C in span
:
('
total M
BC 0.08
+ 3.0 + 2_78 - 2.7E - 3.00 +
34.5
+ 2.56 + 37.06
BAL
The stiffiress of member CD is reduced by 50% because
:
+
co
Cycle2
hD=;";=o'r2sl
"11'
-
FEM
BAL
tl
i
0.92
Cycle I
ll.lEa
B
BA
395
+
-
+
u.5 o.7g 10.71
+37.06 + 47.77
Itc 0.08
- lz,5 + l-o +
0.ü6
-
0"07
to.7l -37.06
- 47.77.
c cc'
CB
0.069 t2.5
-
0:86
-
1.0
-
t4.29
+ 0.069
0.827
-
1.0.34
CD 0.104 25.0
-
l:30
+ 0.827
+0.t04
-
+ 23.80
9.5,1
-37.50
+ 36.0t
-
+
+26.50
+25.29
sr.79
l.4g
396
PROBLEMS
MOMENTDISTRIBUTIONMETHOD
as shown in Table I I ' l8c' The net moments in the half frame a¡e computed
Table
397
PROBLEMS
ll.l8c
Ãnalyze the beams in Figs. P3.4 and P3.5 by the moment distribution method. Draw shear force and bending moment diagrams.
n.2
lt.l8b) M'
Analyzn the beams in Figs. P4.l by the moment distribution draw elastic curve and the bending moment diagram.
method
and
Analyzethe frames in Figs. P3.8a, b, c and d by the moment distribution method and also draw the elastic curves.
Fig'l l '23d' The net bending moment diagrqm is shown in 26.3s
I
t.4
Analyze the frames in Figs. P5.4,P5.7 and P5.8 by the moment dishibution method and draw the bending moment diagrams.
Analyzethe frames in Figs. P6.3a and b by tþe moment distribution method and draw the bending moment diagrams.
Analyze the frames shown in Figs. PlO.l, P10.2 and P10.3 of problem 10.5 the column is pinned at support D in each case.
(d)
Analyze the box frame shown in Fig. P 6.4aby making use of symmetry.
NET BENDING MOMENT kNm Anialyzethe frame shown in Fig. P6.4b by making ur. of
i
Fig. 11.231 Vierendeelgi¡der,
I
I
11.9 COMMENTS
: :
ON
THE MOMENT DISTRIBUTION METHOI)
analogy method. Some additional comments are as follows
2. (l
:
It is better to take all calculations in the table to at least four pþces of decimal for better accuracY
The moment distribution can be stopped at the end of any cycle (that is, after balancing the moments). It is desirable to stop it when the unbalanced moments become quite insignificant.
3. It should be ensured that the numerical sum of the balancing- moments is exactly t.
4. I
(: (. I
5.
ryrrOry.
contd.
of The moment disfibution method is one of the most powerful tools for the analysis this chapter in corsidered structures All thc hand. by statically indeterminate structures consisted of prismatic members. The method is equally applicable to a structure consisting of curved members or non-prismatic members. The difficuþ lies in the determin,lion of fixed-end moments, stiffttess and carry over factors for such membeis. For curved or nôn-prismatic members, thcse values can be determined using the column
l.
¡f
equal to that of the unbalanced moment in every cycle' A-consistent practice should be used while rounding -offthe values. It is desirable to use nearest even values. The check on the moment distribution works assuming that the fixed end moments or the as computed are conect'. The check will work even if the fixed end moments distribution factors are taken wrong.
Analyzethe frame shown in Fig. P6.4c using the moment distribution method.
Analyze the gable frames shown distribution method.
in
Fig.P6-3c and
d
using thê .moment
DEVELOPMENT OF STIFFNESS MATRICES
399
CHAPTER
twelve
.L*
u,
\t
,í-2",
DIRECT STIFFÑESS METHOD.2D ELEMENTS
i
,'I
(rul
þ'
ro)
I2.2
Fig.
-ú3
(b)
Truss element in local axes
I (ur +0,ur:0) Let the end i undergo a displacement equal to
T2.I DEVELOPMENT OF STIFFNESS MATRICES The concept of stiffiress method was developed in section 1.3. The slope-deflection method and moment-distribution method fall under the system approach to stiffness mcthod. The dircct stiffness method falls under the member approach. This is a very computer friendly and the most powerful analytical tool devetoped so far. A right hand aJ(es sy$ern as shown in Fig. l2.l is used in this method which makes it quite easy to
Force at end
(!¡r = 0, u, *
IP,
j
u,
displacement, p,
is equa!
P3
=
4o : {Eu,
to
AE
Lut
0)
Now let us restrain the eird i and impose
.u" Jtraln € =
a displacement equal to
u,
at the end
-L Frr
StreSSO
P1
*X
Right hand axes system
L
P,=
Consider a pin.ended,truss or a bar element i-j as shown in Fig. 12.2a. It can carry only axial force and undergo axial deformation. Only two coordinates are required to
defineitsdeflectedshapei'-j': ulandut. u,isthedisplacement of endianduristhe displacement of end j. Lct us define an orthogonal system of axis x-y such that the longitudinal axis of the truss element lies on the x-axis and y-axis is perpendicular to x-axis an shown in Fig. 12.2a. Let the axial force and axial displacement be positive in
the direction of positive x-axis. The stiffitess matrix of the truss element can
= 66 = ".*2
Forceatendj requiredtoproduce u, displacemenÇ
TRUSSEI,EMENT
:
required to produce
ium, the force at the other end
Y
developed as follows
i
Pz=-fr=
Fig.l2.l
while the end j is resüained.
L
understand.
9r
u,
Strain € : $, I is member length L' Stresso:g.= Eut
be
;.1. ¡¡.
(ut *
0, u,
*
Pz=
-fn
0)
. }rly'hen both displace¡nents P,
:
Pr:
AE L
p, = Ao
T l(-
(u,
arg imposed, the resulta¡it forces are
-
u2)
ur + uz)
:
j,
DEVELOPMENT OF STIFFNESS MATR¡C.ES
DIRECT STIFFNESS METHOD.2D ELEMENTS
400
I ne
P=K^
or
Y1
aPl
{r} l+ fl t;}
or
Ki,z=f[-l l]
40t
(r2. I a)
u2l'\u3 I
u, ._t-t-
.lu5
lit
u6\¡
u¿
x
|
(r2.rb) (12.2) P3
A= E Case 4
:
area
of cross-section of the prismatic truss'element
modulus of elasticity of the material
ltt
Let us consider a general case where the ends of the truss underg,o
displacements u¡ , u2 r u, and uo âS shown in Fig. 12.2b. Only u, and u, axial force. Eq. 12.la can be rewritten.as :
Ët
til
*l; -11
will produce the
P3
TT
: 1 lËÌ
P,
E¡
(b)
u3.
fi 3----
: : :Jt:i
æ ï
P,
whqre, P, and P, are the forces at end i, and
P6
and Po are the forces at endj,
tul
Pg
EI
BEAM ELEMENT The stiffness matrix of a prismatic beam can be generated using any of the followlng methods:
. . 2. 3. I
-
Solution of basic differential equations Any flexibility method Conjugate beam method òr moment-area method.
Let us use the conjugate bearn niethod to generate the stifhess coefnicients. Consider a prismatic beam shown.in Fig. 12.3a. A 2-D bearn is ãS$¡med to lie in the x-y plane. lhe translations u, and ú, take piace along x-axis and y-axis, respectively, while the rotation takes place about the z-axis. The degrees of freedom perjoint are three. It has a total six degrees of f¡eedom : four fanslations u1 , u, and uo , u, and two rotations u. and u,,. The size of the stiffr¡ess matrix is 6 x 6. Let us impose each of the six displacements individually and compute the forces produced in the beam.
F'ig.
12.3
Beam element - development of stiffness
I u, +0 As derivèd in thè:case of a truss element
Pr 'L =4Eu, kn= AE L
and P4
and k,,
:-
*,=-fo,
'AE L
(i)
(iÐ
402
DIRECT STIFFNESS METHOD.zD ELEMENTS
DEVELOPMENT OF STIFFNESS MATRICES
Case2 uz+0
l$hear at end i is equal to slope at end
to u,
is imposed along the positive y'axis while restraining all other displacements as shown in Fig. 12.3b. The conjugate beam and the MÆl loading are also shown in the same figure.
A displacement equal
t.:
( nr z
Deflection at end
i:
ot,
Po
P3
lll"'
,:
(iii)
beam subjected to forces P2 , P3 ,
P,
and Pu
eP = ff*1
(iv)
can be
lo
:
IZET 6EI . . \z= T = &2, x22= -V-
uJ
ano
ksz:
ry
us
(v)
(vi)
i
:0
or
Pz =,2Pa
pj
'1,,.,1.o
uo
u4
,
u5
and
and pu.
6EI
(vii)
uu imposed individuålly at
:
,orkon=+ or
k,o
=
-f
(viii)
P¡=-#q,orkr= # -n-
A displacement (rotation) equal to u, is imposed about the z axis while restra¡ning all other dþlacemertts as shown in Fig. 12.3c. The conjugate beam and the M/EI loading are also Shown in the same figure. The rotation u, again produces forces P2 , P3 , P5 and Pu. The deflection at end,i with respect to end j is zero, that is,
-- ,-
follows
or kur=
-gt]
:--pus,, Pz: Pr+Pu __l2EI
Case3 u3 *0
zBt 3 2Et 3
to as
= - P4 = - 19"r,
P , o: P¡
forceatiduetoaünitdisplacementatj
RLLRL2 _ _-r-x-+-o-X:L
n:io
P
:
Moment of M/EI loading about end
L
-Prj Pu - 6EI .. ^
evaluated. The final values are
Pr lo
k,, =
Or rX¡¡= - 4EI
ln the same manner, member forces due
lj
Ps=-Pz=-ff".t
where
,
ork23:T z-:7lur1' Pi:.-p1 = -#ur,or .r$= -f6EI p.
Po
By cdnsidering the equilibrium ofthe
The stiffness coefficients are
1åI Ut,
considering the equilibrium ofthe beam subjected to forces p2 , p3 ,
or
:
(clockwiserorarionis+ve)
o:?EJ,--zÛr *.u: ur' or xor: -Il-
Ue
.'equal to u,
I P?r. zL P^L Ll l-^::x=+:-xll:-uz L zBt 3 zBt 3l
t,=
p^ Fr-=
moment of the MÆI loading between i and j, an¿ ¡t sl¡oul¿
or
=
P¡ :
P.L r-lr
P1 PÁ: EI -ã+? -u, r,
j
lP3L.lP6L_^ 2EI 2ET
i
l-::x:-L+:ï*; l_ = - 0¡ = -u3 \ 2Er 3 zBt 3)L
The displacement u, produces forces Pz, P¡, P, and Pu. Since the change ofslope between the encls i and are zero, the area of the lvI/EI diagram must be zero, that is, the total shear must be zero.
403
i'
(ix)
or
.
Kr<
tzBt
_ l2Bt . lzBt Ps: -P2 ç us I, or Kss = --:iLzEI zEI Pr= J- u6 ' or r¡o = -L4EI 4EI Po= ;uor or Xø: l, 6EIuo or K26: 6EI Pz= '. It T
L'
(x)
ELEMENTS DIRECT STIFFNESS MIETHOD-2D
404
6El
It is interesting to know some of the properties of the stiffiiess matrices developed in
as follows arranged in matrix form Equations (i) to (x) can be AE AE
Rl Pz
0
L
rzBl
--;-' L-
0
6EI ---;'
0
P3
P4
P5
0
6EI
4El L
0
0
I2EI
0 lPo
----';'
6EI 6El tEl -----------;-
6El
-v
2El ZEI
0
L
L:v ----46El 2Er t]. L
--4
0
lzBl
r
AE
6EI
previous section.
:
0
L 0
--4' L"
v
_AE L
0-
0
0
ur I
!¡r
The stiffrress matrix represents a set of equations.. relating member end displacements mernber end forcês in equilibrium. Thus, gach column of the stifftess matrix must ff the equations of equilibriurn. Consider column-5 of Eq, 12.5.
I I
U
ü3
L
0
0
lzBl ----
6EI
-F
6EI ''-';-
4El
L-
-v
L
(t2.4)
0
R¡
ü4
t2Er
Pzi ü5
P¡¡
^4j
116
fpiì _= [K,, i"i,l fgiì oÍ, tr;Ï,., L*j;ï*;;Ju.uluif.., matrix is of size 3
l2El
-r-
Psj
of Pt, to Pu, Pl; + P,¿j = 0 + P¡j : 0 Pz¡ ¿t
l'
For E F* .: 0, : 0,r Eor XF, --Y Pt¡ ÐM,:0, p",+Lp., For tM,:0. s¡ +
Ifaxialdefonhationsareignored,theforce.deformationrelationscanbewrittenas:
6EI rzL
P3
Symmetric
4EI
L3
(12.6)
J 0'a' u3^
freedom are shown in Fig' The corresponding degrees of
"rÀI
P6¡=
OK OK OK
U¡
oÈl tzgr' tíH -T --iT eel zgt -q -FLt]
Pj_
6Et -T
On substituting the values
x 3 and sub-wctor is of size 3 x
ryI r
L3
6Et _T 0
D .
Poj
Each sub
40-f
PROPERTIES OF STIFFNESS MATRICES
6EI --;- va' or k56: --F
Ps-=
PROPERTIES OF STIFFNESS MATRICES
-
"r;h
axial deformations Fi,-. 12.4 Beam element d' o' f' without
stiffness matrix of any element or any structure is always symmetric about thc : k¡ i . It is obvioub in Eqs. 12'2, 12.3, l2-4 and 12-6. Symmetric , that is k¡ ¡ can be prõved úsing the Maxwell's reciprocal theorem. This is a very useful . In a computer, only the upper triangular matrix or only the lower triangulat need be stored.: It saves considerable memory requirement and computer tirne.
stiffriess matrix of an element is singular, that is, its inverse does not exist. This explained as follows with respect to Eq.l2.4:
first three rows represent the member forces at end i, while the last three rows the member forces at end j. Because of static equilibrium, for any specified set
F*¡
: -F*¡ , and
Fyi
: -Fr¡
406
DIRECT STIFFNESS METHOD-2D ELEMENTS
TRANSFORMATTON OF
ut :
of a 6 x 6 stiffrress matrix given by F'q.l2.a can be multiplying the third row by (-l). This means that row by obtained from the second
It implies that the
f,rfth row
rows 2 and 5 are linearly dependent. The same is true for rows I and only three independent equations in a general 6 x 6 stiffiress matrix.
4. ln fact, there
similarly, it can be shown that some of the columns are linearly dependent. ln a 6 x 6 general stiffiress matrix, tJ¡ree columns are linearly independent. This is due to the presence of rigid body displacements. Consi,f,er the rigid body motion of,
\':\¡:^ with all other displacements zero. Substituting these in Eq. 12.4 gives, 0
0
R
-E-
P2
6EI _T
U 0
P4
^+
TzE,I
--e-
P5
6EI
T
P6
0
tzBl
0
6EI -T
0
0
Thus , equal displacements at the two ends, that is, rigid body displacements, iesult in zero force. Hence, columns 2 artd 5 are linearly dependent,
or,
{k}z:-{k}s
Linear dependeñcc between rows (or columns)
of a matrix is a condition for
singularity. R ig ìd B o dy
L
o':{",
u2
ur u5 t*}
?
us substitute this displacement vector in Eq. 12.4 and calculate member force
¡3 (say):
P:=
o.#u,*T(-;*)+ o-1r*. T(*;")
same is true for any
of the member end forces.
The force-deformation equation developed so for have been written in terms of local forces and displacements. In practice, various members in a strucfure wiil be differently and each member wi¡ have its own axes system. rr," ¿ir""titirtn.r, ro be.wrinen. ¡n terms oi a y ùrstt I T-'^rtT::' .l1ili"^ y:ir." a relarion beru¡een trrsrocal ai¿ srouur T the locat and glóbal membr, :"1,1ï:::an{ and disptacements nre vecror quanrities, we "na Ël"J the resurr ro eirher rorces or ¿irprur"Ã"nt .-ðãîr¡¿", .*r:gt- ""9 Tplr rh,e rocar axes is shown in"" axes. rhese two i""l ä""' å, rc e. _::"*inare coordinates ofpoint p are :
If an element is subjected to a rigid body ilisplacement or motion, no forces should develop at the ends of the element. Consider a beam element shswn in Fig. 12.5 which
;;;;'ñ"iy*.,
:.:li:Ì
!li,t!.:::-J.ll1l1Tpr"1
àirpiã";ñ.' ö'i#i*n d";il;rh;rä'i""ri"* "*
t*" 3."'1"t.::lg: l?l
iñ
D Ísplacement
(x'
y)
byx-ñ.J;;d;ãr,"* ;", ;; i*;
l'*:i:.1,r-,tl
in local coordinate system,
aird
(x' , y') in globar coordinate system.
has been displaced to a new position withoút any change in its length or relative rotation of its ends. The end displacements have the following relations :
{'l
1': iu,, t-+t
P(xry) lxt ,ytl I I I
jur
l+|
Fig. 12.5 Rigid body displacement
=,
TRANSFORMATION OF COORDINATES
^=
-il-
u<-u.
,The dispiacement vector can be written as:
0
L3
6EI P3
0
tzBt
lzBr
¿OZ
u-1"
u¡:uo
are
COOR.DINATES
tl
\
x
I
LOCAL
I I i
x'6L0BAL
Fig. 12.6 Trênsformatiirn of coordinateaxcs
ELEMENTS DIRECT STIFFNESS METHOD-zD
408
that
be seen By simple geometry' it can
TRANSFORMATION OF COORDINATES
409
DEFORMED
Ê
POSITION
, 1 --'Tl"í .-' .uí \ -.f 'l*
and Y:-x'sinO+Ycos0 In matrix notation
or where,
f"Ì : [vJ
[*:i':]fi]
{ô}
t¡'1 {ô'}
[À ]z *
AlternativelY,
I z=
coso
l-sine
sinO
Iz./'T
(12.7a)
(t2,7b)
I
or whêrè, []rl
l,
tö) = :
Fig. 12.8 Displaced position of a truss element
f",l : f coso .sinelfu,'] lurj l-sin0 cos0J [ur'J
(12.8a)
"o'eJlvJ
(12.8b)
{",}:f
is' is an orthogonal matrix' that
tlltÀlr: ill
or'
Iirft
=
[]'lr
disptacements at the two ends can be considered simulaneously, and the hip can be expressed as :
uo represent shown in Fig' l23 ' ut" u¡-'^l::^1d Let us consider a truss element displacements represent uo' and u3' ui *rtif" ' '-u¿i ' displacements ulqng tlr".tJcll-uxes' ifáp" of the truss element is shown in J"fo.."i . æ
'ilr
flil
i,i
[¡,J
I
{^.}
ir i,l 1,,,
=,
1-.u\
l
= : :
,
(r2.e)
{i} Lî' 'î' *, JJ{i}
"\a:,,
i'
lr.
coso sinel{u:'}
tuoJ l-sine coseJ [uo'J
rotation matrix
Truss Element
i.{'
POSITION
(b)
t?tlr {ö}
-,':;
.INITIAL
I
(o)
*ì : lcoso -sinollxl fsine
"åT í,4
cosOl
ln matrix notation _
tt',|
"7W -_yt ..
(t2.7c\
x'= xcosQ - ysinO !: xsine + ycos0 f
I
=
llrll{:;:i
tRI
{^*,}
(l2.l0a)
.
(t2.t0b)
member deformation vector at end i in local system member deformatiog vector at end j in local system member deformation vector in local system rotation transfr¡rmation matrix for the complete element
i
fi, :Lô
o iú Fig. 12.7 Local and global displacements ii
a truss element
ol
il
prime indicates the respective vectois in global system.
(l 2. I 0c)
TRANSFORMATION OF
DIRECT STIFFNESS METHOD-2D ELEMENTS
4IO the element.
or, {P.}:
[R]
{P.'}
systems,
'
be written as
P.: RP.'
'
=
{P*'} : {P.'} :
Ro*o
because
tRl-t
il.
:
'ir,
,1,,:
'ii! :i i'r
(12.1
lc)
R=
being orthogonal matrix'
lc'
tRl or' K.' = Rr K.
R
(t2.tzt
conesponding Thus, member stiftress in gtobal coordinates can be obtained from the 12-12. The given Eq. by product matrix triple the using coordinatei local in stiffiress given by Eq' global coordinates,is transformation of end displacements from local to are These l2'l0d' given Eq' by is loads or forces 12.lOb, and the transfqrmátion of end method' stiffiress direct of the the backbone form which the most significant relations
iir''
Element
Beam
l;,1.-
in Consider a beam element shown in Fig. 12.9a. It sho.ws the six degrees of freedorir global system. Its displaced position is shown in Fig. 12.9b, It is obvious that out of thc
i(
lr-' L
0i 0
0
0l ol 0 0l -----------I
oi o o
I ..9.. r: 0 0 0. 0 : cosO sin0 0 00 0 i -sinO cosO 0 00 oi o o rJ
(r2. t3b)
fï
i
(t2.l3c)
i1.",
12.l0b, l2.l0d utd 12.12 derived earlier for a truss element are also valid for a element except that the size of these vectors or matrices witi be 6 x I or ó x ó
¡.,Eqs..
lof4xlor4x4.
ELEMENT LOAD VECTOR
tTïryl
the sriffiiess meihods incruding the órope-doflectron .of and thé ¡traractelilig moment-disrri6ution method is that t¡.r" rnu"rtt¡r;1onI';;'""" between loads and displacements. The toads acting across the
ir l;
cos0
|
K.i tRl {4'} tRlr t K.I [R] {^'"'}
tRlr t K* I
(t2.t3a)
rJ
|
tRl-r t
: [R]r
tK-'l =
:
t K'nI [Rl {^m'}
Comparing Eqs. 12. I lb and l2.l
f cose sin0 0l Àr*¡ = | -sine cos0 0 |
-sin0
Letus substitute Eqs. 12.10 b and Eq' 12'10 e in Eq' 12'l Ia [R.] {Pm'}
:
cosO sin0
(t2.r rb)
in globalsystem
components
complete rotation matrix for the beam element is given by
(12.1 la)
in local system
4I I
the z'-axis coincide. Hence thd need no transformation. Thus, the rotation matrix l,
Lo o
(12. I 0e)
l2.lb) can also be written in local and global
{P.} = t K* I {^. } {P.'} = t K.'l {A''}
tional componelts u3 and uri
(r2. rod)
or,
The force-displacement relation (Eq.
l,' ,jr' Td lj ar end i, only thë'.first tw,o translational r,nccd be-transformed. The local z axis and global ;lhree displacements
load vectors at each end ol' A similar expression may be written for the corresponding
{i} ti illtr+}
COORDINATES
.
ir
l\.
rp* ,;;i
by an equivalent nodal loadin! which satisfies rwo criteria. Ëiot, th"
loading must cause the same structural responsp as the original
lqøingl aì¿
reference coordiiate sys,"* rr,'"=eqììi,ä'""i,i"i"ì lil l"* 9: 9"T.þ.1" "¡ine ¡h_e æe opposite ro rhe fixed end forces as ihown in nig. iz.to.'m" nirJliä fo,. the joint displacemeñts *g últ r:prelent rhe ."ätio¡r. ", rrr" rq"i"J*t ìoa"r
topresent ¡ent the actions. actions. The fixed end forces can be conveniently cårcuìued using analogy method.
t;
l¡ f\ i
slo¡ìe deflection and the moment distribution methods, crockwise monent was *Â-i+i. '^ r- ¿L ^ r: , as positive. In the direct sniffness method, anti-clockwise ,no,,,"nt . The positive forces or adions act in the direction orthe positive;;r:-ñ;;:'f*
i;l
\'
ii
i;-;i"" ;
il
irr
in Fig. 12.104 rhe element road vecror ror't'" l2..l0c can be written as :
lr-h^o*n
lr. i:
l
]r,
l¡ i,
t
i', rl
ilr
(q)
(b)
Fig. 12.9 Disptaced position of a beam element
eçiuui"irlàïJrsîi#;
: -{P"}r,*o = {0 -I: 2t2 -wLz 0 -*L2 end forces for typical loadings are shown in Appendix C.
wLz _.-( ttx.6 -
12
UTF.ECT STIFFNESS METHOD.2D ELEMENTS
4t2
ASSEMBLY OF GLOBAL MATRICES
413
T.-l '1. 3'l r
DoF
I
(o)
wt2
2
12
5
1g
¿wl
(b) FIXED END
(c)
12.5
6
z
FORCES
ç-
t"
(
(; (
.,'
(, :
(
I
l{ 5 6 3 MEMBER
.
(
MEMBER
,(d) ,l
# # # / /
1 2' 3 ,1 5
STRUCTURAL
be
determined knowing the connectivities, that is, node numbers at its two ends. The global stiffness matrix of member is 4 x 4 and is shown in Figs. 12.l lb, c and d
(
123
2
Fl ,F ?.+ ¡
Let us consider a pin-jointed frame shown in Fig. l2.lla. The joints or nodes numbered sequentially, ftom I to 3. Similarly, the members are also sequentially. The degree iif freedom per node is 2, hence the total degree of freedom the frame is equal to = noof nodesx d.o.f. pernode. : 3 x 2 =6' Hence, size of global stifttess matrix of the structure is 6 x 6..The locations the contribution ,of member stiffriess matrices are to be placed dépend upon the
node numbering. The locations can
't
4+-D0F
(c)
For a given structure; stiffiress matrix of each membç(can bq computed'in local axe system. The stifftieSs matrix in global axes system can be'computed by making use Eq. 12.12. The next step is tó generate the stiffiiesÈ matrix of the complete strucl ttris is also called globa! stffiess matr,x of the structure; It basically means that contribution of stiffiiess öf ãll'members rneeting at a given joint must be added and this process must be continued for each joint gr node in the structure. Thus: there i a need to determine the locations where the,elements of a member stiffnesq:matrix aie be placed in the globai stiffness matrix of the structure.
of
5 +-DOF
(b) MEMBER I u1:2t
lÞ--¡ r. , r
.'ZFFNE,S^S MATRIX i
I
NODE
EOUIVALENT NODAL L0ADS
ASSEMBLY OF GI,OBAL MATRICES
2 ,5
. Fig.
K-
MEMBER
6
1
(e)
l2.lf Assembly
of K - pin-jointed truss
3
DIRECT STIFFNESS METHOD-zD ELEMENTS
414
I
I
I
I
J
2
J
2
I
2
J
j:
ASSEMBLY OF P AND K FOR TRUSS
iL:
node number at any one end of a member node number at tlre other end ofthe same member
The location vector of mernber
I
is given as
:
:
DO
l00M= I.NEL
LM(3) = LM (4) Do 200 t: t,4
rr: P(r
JJ = LM CONTINUE CONTINUE
I
The process of assembling the global stifhess matrix can be easily carried out by visual inspection. Once this process is clearly understood, it is convenient to wrlte a subroutine to compute the locàtion vector LV or location ryaffix LM and assemble the global stiffness matrix knowing the global stiffness matrix of each'element. A conceptual
p[ii]
(J)
+ = pe
[i]
;
for(¡= 1;j<=4;j+)
{ )t)
i=lmül;
stif[ii] [i] += st[i]
a
[]
;
plane truss
.
Now let us consider a rigíd jointed assembly shown in Fig. l2.l3a. There,are threc tlvo members. The d.o.f. per node is 3. Flence; the total d.o.f. of the structurc liiis ¡ ,.3:9. The stiffness mahix of member I is shown in Fig. l2.l3b. The lôcatisn -: vector of member l'is given as : i:'nodes and
13 : 3xi : 3xl: 3,J3 = 3xj : 3x3= g t2:13-l:2, t2:J3_l:8 Il:t3-2:1, Jl=J3-2=7 2 3 7. S e} {LV}r:{t
12.llcandd. Thestructuralstifftressmatrixisshownin.Fig. l2.lle. Thefirsttworown and columns represeni the contributiotr of node l, the nçxt two rows and colunrns
are to be placed.
Ð + PE (r)
for(i=l;i<=4'i++) fo¡(i=l;i<=4 ( ii=lni[i];
Fig. 12.12 Program segments of assembly of K and P for
Similarly, the contributions from each d.o.f. of members 2 and 3 are indicated i0 Figs.
I
n[3]=lm[4]- l;
srlF(II, JJ) = STIF(lI,JJ) + ST(l,J)
=2xi = 2xl:2, J2 = 2xj:2x2=4 Jl:12-1:3 ll:12-1,=1, .{LV}r:{l 2 3 4}
contributions of member
P(r
n[]=lm[2]- l;
r
LM(r)
r) =
(m);
2l = 2* nodi [m] ; 4l = 2i nodi [m] :
DO200l=1,4
12
(
TRUSS
2* NODI(M) LM(4):2+ NODJ(IVi) LM(l)=LM(2)-l
:
represent the contribution of node 2, and the last ùvo rOws-.and columns rePresent the coïtr¡bution of node 3. The coneiponding nodal'd..o.f. *" ãtto indicated along the row and the column. Thus, each element of each membér 4aü'ix will go to the conesponding location in the structure matix. These elements are added algebraically in the global ótiffness marrix. The shaded pp'çtions in Fig. l2.lle show the locations where the
{m=l;m<=nel;m#)
LM(2) =
=2xi : 2x3: 6, 12 - 2"j : 2x2:4 Jl=J2 l=3 ll:12-l:5, 4} 3 6 ilvlr={5
'
i
CALL TRUSS (M)
:
In Fig. 12.l lb, the first two rows and columns represent the contribution of thç node I j (node l), the last two rows and coiumns represent the contribution of the node (node 3)' (d;o.f. l), and global xd.o.f. of contribution the rePres€nt colu¡nn ih" nrrt ro* and first (d.o'f. y-d.o.f. 2) ofglobal contribution the represent column second and row the second global at node l. Similarly, the third row and third column represent the contribution of x-d.o.f. (d.o.f. 5), and the fourth row and fourth column represent the contribution of globaly-d.o.f. (d.o.f. 6) at node 3-
node number.at end
LM
location vector (4x l) PE = element load vector (4xl¡ P = structure load vector ST = ele¡nent stift¡ess matrix (4x4) STIF: structure stift¡ess mat¡ix
12
The location vector of member 3 is given as
total number of truss elements
NODI:
12:Zxi = 2xl= 2, !2 = 2xi = 2x3=6 Jr=J2-l:5 Il=r2-l=1, il'vlr =:{l{ lt t2 Jl Jzl 2 s 6l The location vector of member 2 is given as
415
subroutine is shown in Fig. 12.12. The array ST represents member stiffness nratrix which is computed in subroutine TRUSS. The array STIF represents the structural stiffiress matrix. The DIMENSION statements are not shown for the sake of claritv.
ConnectivitY
Member
wher€ i =
ASSEMBLY OF GLOBAL MATRICES
The first th¡ee rows and first three columns of the matrix correspond to node i whi¡. i!:ì[he last three rows and the last three columns conespond to node j. The nodal d:o.fì arc rl,rlso marked along the rows and columns as shown in Fig. ti.na. The structural i:tstilïness matrix in Fig. l2.l3c shows the locations where the contributìon of mcn¡ber I í:.,.will be located. Thus, the procedure of assembly of global stiffens marrix of the rigidgl,iointeO frame structure is the same as that of the pin.jointed tfuss. The program segnìent i¡'.1.shown in Fig. 12. 12 can be easily used for a framed structure by changing the indices
:i..'
ì;:,Appropriately. :ilì; t:
'
ASSEMBLY OF GLOBAL MATRICES
DTRECT STIFFNESS METHOD.2D ELEMENTS
416
'J #
-F 3'l
4t7
The elernent global load vectors can also be added to generate the structt¡ral loarl in the same manner as the stiffness matrix. The location vector computed earlícl ides one to one correspondence between the element load vector and the structure vector. The loads applied directly at the nodes must be resolved or input along thc I axes. The nodal loads are read directly into the global load vector ofthe structulc the element loads are assembled.
{P}
: {P"} : {Pn,} :
{ (ol
{ I
237 1
,/1,/l/
T
[P]r*,
1
,/1,/l/
t_
(.
{Pq}*q,}
(t2.t4a)
{P,} _ {P"}
( 12.
t4b)
loads applied directly atthe nodes
fixed end forces due to the loads applied
oR the members
:eonsider a truss assembly shown in Fig. l2.l4a. There are 4 nodes. The structurc vector will be of 8 x I size. The equivalent nodal forces are shown in Fig. 12. l4b. global load vector can be assembled as follows :
(b) K-MEMBER
\,
:
i
IP'
0
l*
9 0
1r
wl
I
= lPo
v:
l¡;
0
IP,
-w3
2
(
+ t. .
(
I (
..1
3
,/lr/
t /t/t/ (c).
STRUGTURAL K
@
(o) -
MEMBER
1
{
Fig. 12.13 Assembly of K - Rigirl -jointed frame
APPL IED LOADS
(b)
EOUIVALENT NODAL LOADS
Fig. 12.14 Equivalent load vector - pin jointed truss segment shown in Fig. 12.12 assemble5 the gtobal load vector t load vectors {PE}¿, r.
{pl
4t8
DIRECT STIFFNESS METHOD-2D ELEMENTS ASSEMBLY OF GLOBAL MATzuCES
similarly, consider a rigid jointed plane frame assembly shown in Fig. 12.l5a. Thc equivalent nodal loads are opposite ofthe fixed end forces. The elemeniload vector (6
I
x I ) can be transformed from
I
2
local system to global system using Eq. l2.t0d. The net nodal loads are shown in Fig. lz.lsb. The structural load vector .un b" urrr.bled as follows : I
P[,r = {-p, p,
tr
4lq 3
J
-p3 ip4 -Ps -P6 i-P7
P8
Ps
iPro
-&r
oio o ol
1
(o) Fig. 12.16 a Node numbeiing in a truss Let the member stififrress matrix in global system looks like
l-*
xx
XX K: l,. xx lx
lx xx
(o)
APPLIED LoADs
(b)
structural stiffness matrix can be assembled making use of the location vectors as before and is shown in Fig. l2.l6b. The empty cells in the matrix show 2ero
Two x's in a cell show that two members meeting at the node under are contributing , whereas, five x's in a cell show that five menrbers
The program segment shown in Fig. 12.12 can be easily modified to assembre thc element load vector generated in the element subroutines into the structurai loao uector. Thc final assembled structural stiffitess mahix, and load vector are related similar to -Eq. 12.ll.
= 4,*n Ân" r where n : total structural d.o.f. Pn* r
(r2. I 5)
matrix.
I¡r other words, the above equation is singurar.
The shape of the structurar stifhess matrix depends considerabry on the order in which the joints or the nodes of the structure are numbered. Let us ooÁid", a prane truss shown in Fig. l2.l6a. -
,
,(
.l
i\
:
total nodes : 6 stiÈfness manix = 6 x 2:
D.O.F. per node (
size
of structure
n number of non-zero off-diagonal elements in the structure rnatrix plus onc for rgonal term' Thus- due to symmetry, onry the upper triangurar matrix neecr bc
in a computer algorithm.
numbering of a structure is rlone sr¡ch that the band width is mininrum f-or laum computer solution. This becomes even more important for a large slruclurr r the number of nodes may be 5000 or much more. Á u-¿"J ,uni* ?Jri,n*,, ¡,, 12.17. Let us renumber the rruss of Fig. 12.16 as shown in Fig. r2.fga. 'r'hc
ring is done width-wise. The strucfural stiffrress matrix can be assembted as
EFFECT OF NODE NAMBERING :a'
at node 2 are contributing to the nodal stiffrress matrix. All the ,on-zero of the structural stiffrress matrix are located within the marked area. Thc I matrix is said to be a banded matrix. The hatf bønd width is defined as thc
9lte node
The properties of structural stiffrress mahix are identical to those discussed for the member stiffness
tl".
x indicates an element of the structure matrix. EAUIVALENT NODAL LOADS
Fig. 12.15 Equivalent load vector - rigid jointed frame
/',
*l
= 2,
12
in Fig. l2-18b. The harf or semi-band width is g against r0 obtained earrier. ¡n a rh solver, such as, the choreskey merhod onry upper triangurar matrix need be as shown in Fig. l2:r8c. The eremerrts on the diàgonar are shifted to the tirsr tl* respective rows. The storage requirement in iig. . t z. l is n x n L" gf where n tl d.o.f' of the structure- The storage requirement in Fig. r2.lgc is n x m where nr Ê semi-band width. The saving is enormous in a large-probrem. A'sparse nratrix {ome time lead to an iil-conditioned set of equationr and the soluiion rnav bc
tt
¡6^+
.
ASSEMBLY OF GLOBAL MATRICES
DIRECT STIFFNESS MEI'HOD-2D ELEMENTS
,120
i.
The program segment shown in Fig. 12.12 can be very easily modified to skip lhc clements of the lower triangular matrix and shift the diagonal elements to the lirst i\column.
þ ,1 + 2 + 3 + a + 5 + 6 { xx
\ {
xx \ x
x
x
x
I
f
x
x
x
I
x
x
x
x
2
xx x xx
x
x
x
x
x
x
x
x
3
x xx
X
x
x
x
x
x
x
x
x
1
x
x
\
x
x
5
x
x
xx \
x
x
6
xx
x
x
x
x
x
x
x
X xx x
X
x
x
x
x
x
X
'l
x
+
x
x
x
x
x
x
t
xx
x
+
x
x
xx
7
xx x
x
X
x
x
x
x
x
x
x
x
x
I
x
23/,56789 '
i..
x
X xx
l0
Fig- l2.l6b Stiflhess matrix of a truss
x
x
9
x
x
10
X xx x 11
,Fr-F3,F4+
h-l T ñ I xx
xx lt x
X
+
x
xx
x
x
x
x
I
\
x
x
x
x
2
N
xx
x
x
x
x
3
x
x
x
x
1
xx x xx
x
x
x
x
x
x
5
x xx
X
x
x
x
x..
x
x
6
x
x
X
x
x
7
x
x
8
x
x
9
x
2
x
x
xx
12
+
x
x
X
x
x
x
x
x.
x
x
x
x
x
x
x
3
+
X xx
1
{ }
HALF BAND
\..,...i
t
,
+ 5
.,
+
'i
(., (
t 6
.r
\
xx x
xx x
x \ xx xx \
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
1 s 6 7 I
Fig. 12.17 Banded stiffness matrix
Fig.
12.f
8
rr x
xx
9 t0ti
(b) Stiffness matrix
il
i,
x
WTDTH
(t,
42t
Node numbering in a truss
x
I0
xx
II
x
x
12
I2
422
DIRECT STIFFNESS METHOD-2D ELEMENTS
I_?
ILLUSTRATIVE EXAMPLES
II.LUSTRATIVE EXAMPLES
1231s678 I
x
x
x
x
x
x
o
o
2
x
x
x
x
x
o
o
o
3
x
x.
x
x
x
x
o
o
I
x
x
x
x
x
o
o
o
5
x
x
x
x
x
x
x
6
x
x
x
x
x
x
7
x
x
o
o
x
I
x
o
o
9
x
x
x
r0
x
x
1t
x
t2.l .For the four spring system shown
Fig.
o o @ þut þuz þug
/
x
/
Fig.
\
ruo M
=
nodal d.o.f. are shown in Fig. r2.20b. The stiffiiess matices of each of the four gs are as
follows:
c
l, NEL
Itor {m=!;m<=nel;m++
I
LM(2) = 2* NoDr(M) I LM(4) = 2* NODJ(M) I I LM(3) = LM (4) - I I Do2oor=1.4
[lm[3]=lmtal- I
DO200J=1,4
I F( LM (J). LT. I r ) coTo2oo
l' 200 t00
JJ=LM(J) - II+ I (il , J J) = STrF (r r, J J) +
sTrF CONTINUE
ST (r, J)
goto CONTINUE;
K's =
r.*
)))
l
Kt2 =
qlt'l? L-I 4)
k.I ,'f r -rl¡.
L-r
U3
I
Kt¿
fJ4
:
ho
I r -ll¿t.
I
L_I U5
{ )
+ r; stifliilffl +=st[¡]ûl CONTINUE: { printf ("ByE',);
(
23
34
;
lfor{i=l;i<={;ir-r ii = lmtil ; l{ p [ii] += pe[i] ; I for(=¡;j<=4;j++) if(lm[i] < ii) { JJ:LMtJl-Il
,
Fig. 12.lg Program
(m);
t2
r -t1,, L-l tl2
: u,l
lnpl=z llm[]=lm[2]-t;
P(rÐ = P(rÐ + PE (r)
:
TRUSS
K'r
llnt2l=z t* nodi [m]; nodj [m];
rr =LM(r)
.
(
": [.] il
12.18c
I CALLTRUSS (M)
(,
(,
12.20
this case, the local and grobar axes coincides. The stiffiress matrix of a spring is
FORTRAN
luu
(-r
F*u¿ . þuu
AssEMBLyoF nexoeD
(,,
(,.,
o
@
(b)
Fig. l2.l8c.
(l
\i
12.20a, construct the structural stifftess
:trzg:gtEs
Stiffness matrix The program segment shown in Fig. r2.r9 stores the K matrix in fhe form shown in
(,
in Fig.
Z-iYlV\tvì1+-.1¡ú¡W\-@ 7k1k2k3il;
12
(,,
423
DIAGONAL ETEMENTS
e stiffness matrix of the structure or the system can be assembred with the help on vectors as shown along with thc member stiffnesses.i Thrr" ur" per node is
l.
of structural stifftiess matrix
of
S nåd.r, un¿
: 5x5
424
DIRECTSTIFFNESS METHOD-2D ELEMENTS
o -kr fn, k, + k, . -n1 ,*, : l -n'' | | I
=kz o
Four important points may
l.
Let us enter the data and store various matrices in the LOTUS worksheet in the cclls as indicated.
It
Member
lz k, -k3 -t, k, + kn -ko la -k4 knj5
k,
be noted
L A E
+
13
'
K R Rr
2.1
Each degree of freedom carries the confribution from all the.members meeting that node. Thus, locations (2,2), (3, 3) (4, 4'¡ cury the contributions from the springs meeting at the corresponding nodes.
2.
3. 4.
The elements of the member stiffiress matrix are added algebraically in the stiffness matrix at the cells identified by the location vectors.
System matrix presence of the
Prepare a template for generating global stiffiress matrix LOTUS worksheet.
of a2-D truss element on
lnclination
K:
o -AE L 00
(.
In cell D6, enter the formula : + 86 * (øPV\S} to compute the value of 0 in radian.
Step
3
In cell 88, enter the formula : i n¿ * SSß3 l.n cell D8, entei : - 88 In cell Bl0, enter ' : - 88 In cell Dl0, enter : + 88 In the remaining 12 cells of 88 to El l, enter
:
In cell B 13, enter the formula : @ COS (D6) In cell Cl3, enter.the formula : @ SIN (D6) In cells Bl4 and Dl6, enter : - C13 In cells Cl4, Dl5 and El6, enter : + Bl3 In cell El5, ente¡ : + Cl3 In the remaining 8 cells of Bl3rto El6, enter
0
Step
4
(, (: í'
f
R
cose
-l-sin0
l0
L0
:
0 (zero)
Step
5
In cell Bl8, Clg,Dz|and E2l, enter : + 813 In cells Cl8 and E20. enter : - Cl3 In cells Bl9and D2l. enter : + Cl3 ln the remaining 8 cells of Bl8 toÐ2l,enter : 0
Step
6
The data matrices K, R and Rr are now ready. Let us compute
AEo L
The rotation mafrix R is given as
:
2
---
00
Bt3.El6
cell Bl8.E2l cellB23.E26 cell B28.E3l
Step
AE^AE v
0ût
R
B 8.El I
Enter the value of L:300 cm in cell 83, A : 15 cm2 in cell 84, E: 12000 kN/cm2 in cell 85, and 0 : 30" in cell 86. Only the numerical values are to be entered as shown in Fig. 12.21.
Stiffness matrix in local coordinates is given as
LL
K
B6 D6
I .
LoruS template :
: L, Area : A, Modulus of elasticity : B of member at node i with the global x-axis :0 degree
Length
K
Product Rr
B5
Step
Solution The fóllowing data and matrices are required to prepâre a
Product Rr
cell'B? 83 84
cell cell cell cell cell cell cell
'[he computational steps are as follows
K is singular, that is, its inverse does not exist because rigid body displacements.
Example t2"2
no.,
0 in degree , 0 in radian
:
The structural stiffness matrix is symmetric and bairded. The semi-band width js
DATA, MATRIX MULTIPLICATION Command
sinO 0 cos0 0 0 cosO 0 -sinO
425
ILLUSTRATIVE EXAMPLES
.r2345
RrK
:
/DMM - rangeof fìrstmatrix: Bl8.E2l
,iJ
rarige of second matrix : B8.El I range 9f output matrix :B'23.E26 This gíves Rr K" Now let us compute Rr KR, again using the D M M command
:
using thc
42(T
ILLUSTRATIVE EXAMPLES
DIRECT STIFFNESS METHOD-2D ELEMENTS
/DMM
Step
7
is shown as
A 2 J
4
follows fol
6
E
'
The following data and matrices are required to prepare a
Length
AE
l5
L
30
Theta'1r¡=
600 0 - 600 0
0 0 0 0
0.86602s
0.5 0.866025 0 0
K
0
lt
600 0 600 0
0 0 0 0
l2 t4
t\'
t5 r6
(
t7
:.' (,
t9
¡:
23
RI
-
0.5
0 0
i8
0.86602:5
lRl
r
-
20
,0
2t
0
0
22
25
26 27 28 29 30
3l
519.6152
0
300
0
-5t9.6t5
0 0
Rt*K
-
Rt*K*R
300
450 259.8076 - 450
-259.807
259.807(f
0.5
0.866025
-
0.5
-
-
0 0
300
-259.807 450
259.8076
0 0 0.5
0 0
300
450
Momentof inertia
temprate :
= I, Modulusofelasticiw
_-
L'
AE AE
_AE L 0
0
L
IZEI --:cL' -.:6EI 6EJ ._ 2ET --;VL
t2Et
T
. 6EI --..'...-
UL
4EI
rotation matrix Ris given as
0.866025
5t9.6ts2
-
A,
Lorus
0.866025
-5 t 9.615
1s9.807 -r 50 temnlate pla
0 0 0.5
0 0
150
32
Fig.l2.2l sample
-
0.5 0
0.5 0
24
0 0 0.866025
Area =
t2Et
-6EI AL
0.523s98
-
: L,
,, lnclination ofmember at node i with thé global x-axis 0 degree = *$tiffriess matrix in local coordinates is given as
300
Theta:
l3
(
D no. I
C
12000
IO
(
Solution
element
L: A: E:
8
I
(,
= Rr KR.
)
.i
(.;
worksheet.
:
B 2-D Truss
I
i.
Prepare a template for generating global stiffiress matrix of a 2-D beam element on a ¡ LOTUS
For other truss members, cells A2 to E3l can be copied to 433 to E62 lòcations using the COPY command, and data can be change
The
{
Example 12.3
range of first matrix :823.826 range of second matrix : Bl3.El6 range of ouþut matrix : B28.E3t This gives global stiffrress mafrix K'
_
427
-2s9.807
-r
50
2s9.8016 I 50.
R:
where À=
f
cosO sin0 0l
l-f'
.î'
i-J
ur enter the data and store various matrices in Lhe LoruS worksheet .L:l in the cells
' tnotcated.
Member no.,
L A I E 0 in degree 0 in radian
K
cell cell cell cell
82 83 B4
85
cell 86 cell 87 cell D7 cells 89. G14
flv ii.) iËi-',
DIRECT STIFFNESS METHOD-zD ELEMENTS
428
R 'Rr
Ë¡'.
cells
iä)'
Product Rr
i'i'--'
Product
iil,,.1
I
Step
,fc,
R
Enterthe value of in cell 85.
j'r;
i(:
Rr K
Bl6.G2l
range of ouþut matrix : B.37.G42 This gives global stiffrress matrix K,
cells 823.G28 cells 830.G35 cells 837.G42
The computational steps are as follows
':;,L.'
,ì1,
K
ILLUSTRATIVE EXAMPI-ES
=_
Rr KR.
For other beam members, cells A2 to G42 can be copied the COPY command, And so on.
:
L:400cm incell83. A:55
E:20000 kN/cm2 in cell86,
and
0:
cm2 incell 84, I = 8600 cma 270" in cell 87' Only the
is shown as follows
:
numerical values are to be entered as shown inFig. 12.22' Step
2
* In cell D7, enter the formula 1 + B7 @ PI/180' (:4.712388). in radians to compute the value of 0
Step
3
In In In In In
,i('
ji,
cell 89, enter the formula cell cell cell cell
:i M*
B6lB3
formula 1 + 12 * 86 * 85/83^3 formula : + 6 * 86 * B5/B.3"2 l, enterthe Cl Dl l, enter the formula : * 4 * B6 * B5/83 Dl4, enterthe formulà: + 0.5 B4 * Dll
C 10, enter the
Now copy these values with appropriate signs in the remaining cells of B to
:'r
G14. Elsewhere enter
a zero
value. This gives K in local coordinates.
816, entertheformula : @COS (D7) In cell Cl6, enter the formula : @ Sn'{ (D7) ln cells Cl7,Elg and F20, enter : + 816
Step4 Incell
il
IncellFlg,enter í + Cl6
In cells Bl7, and E20, enter : - Cl6 ln cells DI8 and G21' enter : I In the remaining 26 cells of 816 to G2i' enter This gives R
:
i{
i{'
Step
5
Step
:
0
:
0
+ 816
Incells B24andE27,enter : + Cl6 In cells D25 and G28, enter : I ln the remaining 26 cells of B;23 to G28. enter
.( (
In cells B23,C24,E26andE}7,enter In cells C23 andF26, enter : - Cl6
:
This gives Rr
6
The daø matrices K, R and Rr are now ready. Let us compute Rr K.using the DATA, MATRIX MULTIPLICATION command :
/DMM
-'
range of first matrix : 823.G28 range of qecond matrix : B9.Gl4 range of ouþut matrix : 830.G35 Now let us compute Rr KR, again using the D M M command
:
/DMM nnge of first matrix : 830.G35 range of second matrix :
Bl6.G2l
Fig.l2.?2 sample LOTUS template
to
A44 to Gg4 using
430 12.
7
DIRECT STIFFNESS METHOD-2D ELEMENTS
BOT.JNDARY CONDITIONS
BOUNDARY COI{DITIONS
Knowing {A¡ }, unknown
It has been shown in sec. 12.2 that the stifihess mtatrix of a member is singular and it contains rigid body displacements . The same is the situation of the structure stíffness matrix. The rigid body displacèment can be eliminated from the stiffrress matrix by introducing appropriate boundary conditions. A structure may consist of roller supports, hinge.supports or fixed supports. The purpose of introducing boundary conditions is kr restrain the d.o.f. corresponding to these supports. Sometimes, we like io specifo certain displacements and wish to compute the corresponding forces produced in ìhe siiucture. Yielding or settlement of supports is another commonly encountered situation. I'hr, boundary conditions can be specified in different manners.
l.
i,g
r.,
i;.1{
år' íì:
,lt
Zero Displacements
-
il¡
i¡¡
i', I
P.: Zero Displacemenß
Il:
where, ,\
I, = P2 ^r L2
:
i (
= = :
KÅ
(r2.t5)
^
or
supports.
: IKrr] {Âr} + [Krz] {^z} {Pz}: [Kzr]{Âr}+ [K22]{^2}
(l2.l7a\ (t2.t7b)
Eq. 12.l7a gives the unknown displacements,
{^r
} = [Krr ]-t {pr } _
{Pz
Kzr
I
(or,
[ Kil
]-r l Kr2l {^2}
(l2.l8a)
P"
=
K" Â.)
(r2.r8b)
(l2.l8c)
{Ar }
In general, a load vector consists of two components : loads applied directly atthe nodes, Po loads applied on the members, that is, equivalent nodal loads, P".
At active nodes
*} :
{P.r
}-
(12.19a)
{P"r }
At inactive nodes, the load vector further consisis oftwo components
(a) loadsapplied
:.
directly at tlre nodes or through the members connected to such nodes,
that is,
{P,
.(b)
*}
=
{Poz
}-
(r2. teb)
{P¿ }
reactions from the active degrees of freedom, that is,
{Pz} :that is,
= [Kzr]{Âr} + [Kzz1{¡;}
(t2.t7b)
support reactions corresponding to the load vector {Pz*} . The net support reactions are given by
-
[Prt] ar equal
and opposite,
(t2.29)
{P.}:{Pz} -{Pz*}
The limitation of this method is that it is not convenient to number the nodes in the manner. Moreover, the partitioning. shown n Eq. 12.17 is cumbersome to program and complicates the computation of member forces.
3. SpecifiedDßplacements . Let us consider the following set of equations : [x,, K,z r,rl |.ô,1 l'r,I *,, r,, rr, l{0, = 1., I |
LK' K' rrrl Lo,
j| |.rrj
(12.2,1a) '
Suppose õ, has a known support displacement value equal tq õz *. Eq. 12.21a can be modified to.read :
-
:
{pr}
and
e2.t7)
known load vector unknown load vector corresponding to support reactiðns unknown displacement vector corresponding to active d.o.f. known or zero displacement vector co¡responding to active d.o.f.
F.g. 12.17 can be rewritten as
K¡I I {Pr}
(a) (b.
; fn.
Number ail support inactive d.o.f. at the end
fnì fK' iKr2ll^rì-} {_'l:1""i"""|'i iPrJ lra irrrj[arj
or,
I
(
-
and
'(t2.t6)
The unrestrain nodes or free nodes are also called as the active nodes and the corresponding d.o.f. are called, active d.o.f,, The constraint nodes are also o" inactive nodes and the corresponding d.o-f. are called inactive d.oJ the "uil"d forcedeformation equations (Eq. 12.15) can be rearranged so that all the active d.o.f. are arranged together and all the inactive d.o.f. are arranged at.the end.
(
(
K"Â"
Krr I {Âr }
{Âr
{P,
,
2.
or,
load vector {F2 } can be computed using Eq. l2,l7b.
0
{^z } {Pr
Deleting rows and columns
The effeci of inactive d.o.f. can be eliminated by deleting the corresponding rows and columns from the structural stiffness matrix, load vector and displacement vecto[. K, and P muit be compacted and linear simultaneous equations ìan be solved for the ^ unknown displacement vector. This method is very convenient for hand computations but is very cumbersome to program. The compacted force-displacement relationship is written as :
l
f(. l:
If
43t
l¿, ; ïl{:i} {t -
K,, ôt
ôzt K¡z ôz
:Ì
(t2.2tb)
Modi$ing the K and P as shown in Eq. l2.2lb effeetively rernoves the i th row and column.of the K, Â and P. All the quantities on the right hand side are known. The joint displacements can be found which yield ôz : ôz *.
432
DIRECTSTIFFNESS METHOD-2D ELEMENTS
BOTINDARY CONDITIONS
Program STAP
A
srructurar Anarysis program
tz utd 13 w'r
chaprers "
be presented
(srAp -3D) based on the theory iir chapter r+. i hJ;;;;;;;r..
for(m=l;¡4=¡sl;m+ deveroped
to introduòc
zero displacements as well as specified displåcements. 7,ero Displacements
DO200l=1,4 rr = LM(r) P(r
oi;;;;..f.
total number of active equations. jrre^¡roqram segment shown a relation between an active nodal d.o.f. anã numbers. "quution
P(r
I) + PE
lmt3|:
id[][]
;
lm[a] =
id[][2]
;
for(i=l;i<=4;i+) { if (lm[i] <= 0 )
gotoCONTÍNUEI ii =lm[i]; p[ii] += pe[i];
(r)
r F ( LM (J). Lr. r r ) GOTO200' JJ=LM(J) - II+ I STIF(ll, JJ) = STIF(ll,JJ) + ST(I,J)
;oth";rü;;
in Fig.
I) =
DO200l=1,4
óf,freedom. Ail rþstrained d.o.f. are stored as zero, while the unreshained d.o.f. are numbered sequenriatty from l. Thus, ir gives rhe totat number
\
rF (LM(I).LE.0) GO TO 200
degrèes
(
NEQ:
po
60
N:
NffiF
sti{iilffl
r).cr.otcoõ'
rF(rD(N,
neq=0:
ro1!:l;n<=numnp;n#) IOf ln : Ì'
n (=n¡l¡f.
lD(N,r): 60
,l '. í i.\
?o
û^f^
I
^Dtl
)¡ 'I
)
Displacements
=i
The specified displacements at the given nodes are imposed on the structure through use of boundary eiements. A boundary element is simply a spring. it may be a translational spring or a rotational spring. Its force-deformation relation is given by
mi@ffii;iÈir
CONTINUE
I
or Fig
12'23 Rerationship between active d.o.f. and equation numbers
This information is very useful while assembling the global structural stiffrress matrix
and rhe road vector. The program is
d'o'f in STIF a¡rd p if it rF
is
insrrucred,J;Il"*"lä;ffiäïra
iestrained. This is dãne rhrough rhe srarement
(LM(r). LE.
0) co To
The relevant program segment is shown
200
nFig.l2.24.
l {
t
fa¡ (i
ID(N,Ð,Ì= l,l.fDõF
:
printf("BYE");
.LABEL2.
CONTTNTJE
;
-
idlnltil
ID fN.D = 0
stlilil
Fig. 12.24. Program segment of assembly of K in U.T.M. and banded force with boundary conditions ^
ñB-ffi
NEQ
WRITE ('t,*) CN, 50
r
lrJ
GO TO 60 70
- r¡
coNTtNUE):
+: I id[n][i]:n"q:
NEQ=NEQ+ I
+=
{
goto xy (50,20) ; print("Press any key") getch( ) ;
neq
r,ìJpoF-
if(m[j]
NTINUEI:
O
DO 50 ry: l,
{
goto CONTINUE2
FORTRAN
c c l¡EQ : equation number c NUMM: rotal nimber oñõG C NDOF : totat d.ojìeriode
;l
for6=¡;j<=4;jr-r)
CONTINUE CONTINUE
12.23 estabrishes
(m);
TRUSS i = nodi [m] ;
LM(2) = rD(r,2) LM(3) = ID(J, l) LM(4) = tD(J,2)
An array named ID (identity) is generated as the,nodar data alongwith the boundary pondirions are read. A boundary cóndition has to be ,p""in"Jãroïg'äacn ot-trre six degrees offreedom per node or¡ôint. A restrained degree offreeaom-i, lpe.ined as r, while an unrestrained degree offreedom is rp."ifird as zero. Thus, the array ID stores the node number and the information abouieach of its six
{
CALLTRUSS (M) I : NODI (M) J = NODJ (M) LM(1) = rD(r" r)'
rrr
particurnr.
p : kô f,ortranslationalsprings rn : k 0 for rotational springs
(t2.22a)
(t2.22b)
, The stiffness of the spring is usually specified to be a very high value say l0¡0 . At a given node as many boundary elements or springs may be specifïed as the d.o.f. of the node. The force-deformation equations are modified similar to Eq. l2.2lb so that the desired displacements are imposed;
,.
The reaction
displacement.
in
the ,spring gives the force required
to
produce the specified
434 I2.8
DIRECT STIFFNESS METFIOD-2D ELEMENTS
I
|;;ä,fr'ï:'::ï,î ,,:trliï:""1i'ï:o:11,::i'::llr,f ;:"Ë"ï,*iä*#;ff ïå:"ii:,iTi
12.20 whichis a some whar H::1ï::":.1] ï1lrrrg of Eqs uuul¡uary elements. In case or fl,gid supports, the dispracement "'" is ,p"riti"o ,;;;" t:t9' The boundary ". placed in the direction elements are placed 'the directinn of ôr rha support .,,;^^* d.o.f. r - ¡ An infinire ùu¿, ro usç
fi
:iii:î*.
12.9
435
rhe reaction
i",r,.,p.,,u'rilJ
ro.t"*ïif#í
# ö;#ï:ili:,ïì:i;lr'",:;i,i;
zero
INCLINED ROLLER SUPPORT
over come this problem, the roller support is replaced by a boundary element. lt the_same restraint as the roller support. Thus, the spring shoultl have flnite axial stiffrress normal to the inclined surface. It would offer nó ."rIrtun." normal ¡lts own longitudinal axis. The length of the spring is of the same order as other !ffibers in the structi¡re.
ggp or{1
.
,n"_i"l"r:f tl"^.:q91e the spring should have infinite axial area and -lf rof inertia. If the joint ofy.rieid, the structures is pinned, only its area need be infinite as in Fig. l2-25c. Examples 14.2 and 14.4 explain the use of the boundarv
The dispracement of aþint musLbe described in.terms of independent components of dispracemenr wirh respe* to the grobar rhe ser of equations.
' {P,}: IK,l{4}
*"iàã.åtre
suMMARy oF DrREcr srrFFNEss rvrnrriot
would be singular.
T::.1;,1ï",îüîï; :::"j,:1:r":q:l' ;;;;äÏ
ä
fl
in"
o
*o,ucem enr are
principles involved in performing the direct stiffrress matrix analyses have been I above. A flow diagram showing the inportant tasks to be accomplished by a
rh e
program is shown in Fig. 12.26. For those interested in thè details of ing a source listing for the analyses of 3-D framed structures is availabJe on a and its salient features are discussed in chapter 14. The basic steps are
:::'1?',iå':":tr"Ji; î:ii,:,"n::*t-":::*-*îiöJd;1üdLiifi r i! n ís,i in;; l; ;"* ä :ä,i: ä:?i ilii:i :ffiå ::',ll,nî :ïn::: üijïT;,' jli""'åfffåå'J.i'::it::l j.îTî::,5 j];äru;äilä'ff ";,i Íï3,1ffi parallel oriented
11
J,l-.'"tf¿rffiJ*.
displacement components'are retated-to
"u"r.
àit,",
;ilir'i#i,.
ized in the following
:
Idealize the structure and establish global axes.
{.
Number the nodes so as to obtain a minimum band width.
::
Identifu the type of members, that'is truss element, beam element or boundarv elements. Number the members in each category.
!. I
I
compile the basic data : nodal coordinates, material and member geometrical properties, member connectivity (i and j nodes).
I
\
X (6LOBAL) :
SUMMARY OF DIRECT STIFFNESS METHOD
SUPPORT REACTIONS
X
O
(o)
(b)
(
(GLOBAL
ì,.Ettt )
t
calculate element stiffrress matrix in local coordinates vector P*. calculate rotation matrìx
?,,
Ç *d
element load
and R, and get global member stiffriess and member
load vector.
R
Kmr :
P_ SPRING
A=ø
fi;l R'IÇR
': R Pr'
(l
(t2.t2) (t2.t0e)
Assemble the structural stiffness matrix and structural load vector. Superimpose nodal forces, ifany.
Introduce boundary conditions to.eliminate rigid body d.o.f. and take care specified displacements, jf any.
)/
r''
Fig. 12.25 Inclined roller supports
2. I 0c)
of
solve the system of linear simultaneous equations and get the unknown nodal displacements.
P" + Kr^r
(12.t6)
436
DIRECT STIFFNESS METHOD-2D ELEMENTS
ILLUSTRATIVE EXAMPLES
l0 Extraðt
437
element end displacements and transform into local cóordinates
Â. :. R Â_'
Read control data nodal coordinates I
t
(12. I 0b)
Solve for member end forces using
P-*= P.+P" P_* : K_ ,* P"
or,
(12.23a\
(t2.23b)
^_
These equations may be written in the local or global coordinates as appropriate. 1,2 Superimpose the solution of kinematically indeterminate structure to get span moments.
l3 Evaluate
the support reactions by considering the static equilibrium ofboundary nodes.or by using boundary elements. -\
(12.20)
Read nodal load data
.14 Frint outjoint,displacements and member forces.
stiffi
Assemble global maüix and load vector
ILLUSTRATIVE EXAMPLES 12.4
$¡Anelyze the three bar assembly using the stiffness matrix method as shown in ïs,12.27a.
,.lfiet the origin of the global axes be at node 4. The nodal d.o.f. are shown tlß..12.27b.
Nodal data x (cm)
Node Compure
-
I
*.rb". *ããir",
2
4
airpiã"rãñliãiffii
Print nodat and member forces in locìl axes
v (cm)
173.2 0 300 0
300 300 300 0
Member data Member
Connectivify
ieometric propertier
Material
Direction cosine
properties J
cm2
Lcm
E kN/ cm2
c
4
I
6
346.42
2
4
2
J
4
6 6
42420
8000 20000 8000
- 0.50 0 0.707
='cos0. Fig. 12.26 Flow chart for a.general purpose stiffness matrix program
A
I
I
J S : sin
300
S
0.866
I 0.707
Member stiffness matrix in global system is given by
I(n': Rt ÇR
(t2.12)
ln
438
DIRECT STIFFNESS METHOD-2D ELEMENTS
ILLUSTRATIVE EXAMPLES
^
439
sinol_f c sl cosoJ l-sin0 L-S cJ
_f
coso
-c2 -csl 1., cs s2 -c: -t'I *.' : +El
A2tE2 L2
I symm. 78lz
st
j
-60 -34.64 60 I 7
K_":llz+.u103.92. 60
s+.a+ _60
I L i'-
rì
7
,_l ,2 -
[o | I
-103.921 8 I I
t03.e2J 2
83478
0 o 0 17 lsø.sts s6.s7s 4oo:*:rl: K.''3=| se.szs 0 013 $ 4ooj4 $ L I
I
L
56 -s6.s7s -s6.s7sfI 7 I -s6.s7s .t -56.57s I 56.s75 56.5751I 5 s6.s7s J 6
There are a total offour nodes, and the structural d.o.f. = 4 x 2: 8. The stifftress be assembled since the location vectors are indicated on the member stiffnes$
trix can
I 34.64
+10.45
23
7
-ó0
-34.64
103.92
60 0
I
60 lr
-t03.e2
0
0
013
400
0
-400 l!
56.575
.
56.s.75
-s6.575
56.s75
-56.575 (34.64
+0+ s6.s7s)
12
-56.575 -s6.s7s
l)
(-60+0
16
+s6.s7s)
I, |
(r03.e2
or,
. liill
Fig.
12.27
Symmehic
+a00+
|
l8
s6.s7s) J
440
DIRECT STIFFNESS METHOD-2D ELEMENTS
ILLUSTRATIVE EXAMPLES
44t
Boundary conditions
ül:0:uz=u¡=u4=u5=u6
.
The structural stiffitess matrix reduces to 2 x 2 by eliminrating ihe first six rows and
Area
columns,
Lo¡C-veeter It is a 8 x I
the pin-jointed truss shown in Fig. 12.28a using the stiffrress matrix method. of diagonal members : l0J2 cm2 and that of the rest : l0 c¡n2, ancl
Instant.
vector
{P}: {P"}+{P,}
pr:{o o
I
(12'l4a)
o o o o
o_p}r"e
Eliminatüigthe first six erements corresponding to the boundary condftions:
T
=
L
', {-:},",' {-,iJ The force.deformation relation
gives p.
I
= K. Â"
ol-lsr.zrs -2.+zs][uì r..ì = / : {l-roo/ l-ro* so'a'iJ{,1} iï} {ffff;}
-" ""
*
(b)
The member forces can be determined by using the respective force-deformatlon
relations : (Fig. 12.27c)
{P.'¡ = [K;] {^'.i For-ce in member
1l2.nb)
I
_60 _34.64 ] lst.e+to3'e2 60 JP'f:l 34.64 lP, I I |..,
[p,i t
$
60 lfl-o.ooozl tß.e2lJ-o.r7so[ _
,ili,Jl
I ro.+sl J
-rr.ro I
:i:1m,|
(c) FREE
Force in member 2
Fig.12.28
'A ;;',
Let the origin of the global axes be at node
iinhown in Fig. 12.28b.
',:'::
Force in member 3
|.p,ì [r I -r lp-l sa'stsl I - l -r
1.rl:
[r,J
r
t
il{-í1 lïíil
BODY DIAGRAM
Nodal data
l.
The nodal degrees
of freedom
are
DIRECT STIFFNESS METHOD-2D ELEMENTS
ILLUSTRATTVE EXAMPLES
443
Member data Member
Connectivity
Geometic
Material
properties
properti€s
'K
Direction cosines
L
l2
I
J
A cmz
t0
Lcm L
U
I
a
E, kN/cm2
I
cos0
sin0
E
0o
2
J-
t0
0
+0+0.5 0+0+0.5
2
I
L
E
900
J
J
L
E
0o
4
I
I
4
0
L
E
900
0
5
I
I
J
6
2
4
l0 l0 t0J2 t0J2
I
0+l+0.5
4
0
Global stiffness of a member is given
LJ'
E
450
LJ2
E
1350
by IÇ' = Rr IÇ
K;:fl
L r^'. l2
::i,
-f
=
I
| L
' ol,
t/J2 t/J2 -t/J2 t/J2
78
5
R
ll o -l roEl o o
Tl
r
L
vector
l: (
(,
(r
K.s
:
I
fo.s o.s -0.5 -0.51 0.5 -o.s -o.slz
rThere are-a
K,o:
tuL
0
-l
0.5
-l 0
:, lj
åil; ols 016
(0+ o-0.5 l? (o+ I +0 s)J8
I x,l vector
0.5
0 0
p
0}
P'
o
0)
P. T K.Â,
(12.t6)
35 1.5
0 0
0
=
lOE
L
P
-0.5
o.5l
0.5
-o.s l +
3
I
Symm.
0.5J
I
i;,.'
fn¡s system of simultaneous
=g.
The stiffness matrix can be assembled since the location v€ctors are indicated on the member
7
u5
1.5
PL
l0E
8
0
0
-o.5
0.5
l¡¡
t.5
0.5
-l
0
u5
t.5
0
0
u6
-0.5
u7
t.5
ut
equations can be solved using the Gauss elimination 0.5
u3
u6
4x2
6
1.5
0
8
0.5 -o.s I z
:
0
force-deformation relation gives
7
total of four nodes, and therefore, the structurar d-o.f.
is a
8
f0.5_ -0.5 toJ-zel
It
pI:to o
'Js
34
L'L "' llil
rcJlsl
{.5
the 1,2 and 4 row elements corresponding to the boundary conditions:
I:
L
OJ6
0
u+
Pr ={0 0
, lo.ol o olrK,4=toEl o Ll o -tlz olz
1256. i:'(
o
t2 7
ol7
0
0
-0.5
Eliminating rows and columns corresponding to these d.o.f., that is, 1,2 and 4. the global stiffiress matrix reduces to 5 x 5.
L
ols ols
-{.5
I
conditions
ul:0:ü2:
Ll
6
7
0.
(l+0+05)
l
'OJ4
6
-0.5 -o.5
(0+l+0.5) (0+0+0.5) (l+0+0.s)
34 5 6 io o o ol3 x*, = loEl I o -tlo
olr , l' 0o -r0 0lz K.¡=lonl
5
0+l+0.5
Sr]o,o
34
34 -t0 00 t+0+0.5 0+0-0.5
lr, cs s2 -cz -csl
Krl
_lOEx
-0.5
U7
2.0
Ug
0.5
444
DIRECT STIFFNESS METHOD-2D ELEMENTS
The member fòrces can be deterntined oring
ti"
II.LUSTRATIVE EXAMPLES
corresponding foice-deformation
lr,
relations.
: K- = +l I
K,'
Force iymember 6 (Fig. 12.28c)
,
{P.*}= {P_}+{P"l or,
{P,n*}
=
4L l2
il
')
J
:1"'
2400 9600
-
800
-2400 800
6
240012 .l
4800 2400
| |
3
s
e600J 6
conditions
ur:o:
il
The f¡ee body diagram of member
4r:
6Ll
ffhis is also the structúral stiffiress matrix.
-0.5 _0.5 o.5l[0.5ì f!ì lo.r jPo[_ toel 0.5 0.5 -o.sli o Ior :_ - Tl 0.5 -o.sllz.of roe ,f3i| 't It[ o.s] [o.sJ [',j L
.
-t2
'lLs
02.23a)
[,K', ] {^', } + {0 }
6L
445
u,
structural stiffrress matrix reduces to 2 x 2 by eliminating the fTrst and third rows
6 is shown in Fig. I 2.28c. There
Isooo K_:l '
compressive force equal to 0.707P.
14800
4s001 e6ooJ
Eiarnple 12.6
fr,l
Analyze the simply supÞortgd beam carrying a uniform load by the direct stiffness method shown in Fig- 12.29a. Neglect axial deformations.
tv I l0
+.----gt
(t2.t4bl
I
'
.4\ .
kN/m
":lll=,.")-{p"} lP' Io'J
15
is no load acting directly at the nodes. Therefore
--.'x
¡
p":0.
he equivalent nodal loads due to the uniform loád acting on the beam are shown in 12.29c. d.
-¡: :,ro 30 30 -30} - {* 2 -*} tzj L2 +t2 +
(b) NODAL D. O. F
.
load vector in view ofthe boundary conditions, 1
FIXED END
FORCES
"" (d)
{;,J relation for the structure gives,
EO. NODAL
LOADS
Fig. t2.29
Solution
I
:
Let the origin of the global axis be at node L The nodal degrees of freedom are shown in Frg. 12.29b. Since, there is only one member, its K in local coordinates coincides with that of the global coordinates. K is a 4 x 4 matrix si4ce axial deformations' are to be ignored, that is, u, : ü¿ : 0
p.
= K,
^.
: l?::1 i:ggl 1",} vr f-o ooozs} ;. f",l = oeooJ J f+roo luu/ l"ul | o.ooezsJ
{ ¡o ^'^'i [
slopes at the ends are given by
v^ : wL3
24Et=
lox
63
rl;14ñ
=
0.00625rad
o.K.
DIRECT STIFFNESS METHOD-2D ELEMENTS
446
ILLUSTRATIVE EXAMPLES Support reactions
: Let us rewrite
Pt' :
the member equilibrium relation
:
20
K't Á''
447
kN/m
362s I looo 4B0o i 2400 -24001[-o.ooozsl ¡ p*' : +aoo s600i 2400 -24ooll o.oourrlu | j - ll -o lz I iÃoo ziöö 8ôo :ð'óo L-2400 -2400:-soo s00J[ 0 J5 P.:Pz-P*z Ç, Â, -.Poz _ P"z
(12.20)
24001{-o.ooozs} _ f¡ol= fsol l*,Ì = | z+oo -aaoo)lo.ooozs/-o*iml l*,J- l-z+oo troln*
+ul uzrt\
r\lu3
ua
lus u6-f-r
Member änd forces
/
{P. *} : {P'. } + {P';} in local or global coordinates {P,+} = [K'n'J{^'r}+ {P'"}
faoo 24oo
p.*:l
I o I
-8oo
z4ooj
,oo
-.,4ooll
t
[ ¡o ]
L_¡oJ
REDUCED D. O. F.
If'/--E--\i r a--4.,
(
;-l.j ;l
e600Jt0.00625J
I
(c)
(d}
-tr--
FIXED END FORCES
{¡}
Example 12.7
22.13
Analyze the non-prismatic beam subjected to a uniform load as shown in Fig.12.30a. Neglect axial deformatisns.
17
.s0.11\+ |
MEMBER END FORCES
Solution Let us assume a node at the point where moment of inertia changes.. The d.o.f. shown in Fig. 12.30b. The origin ofthe global axes is takei at node L
¿3.3 9 are
Nodal data
(',
,.
Node I
x(m)
v(m)
0
0
.2
4
0
3
7
0
(S)
BENDING MOMENT kNm Fig.12.30
448
DIRECT STIFFNESS METHOD-2D ELEMENTS
Member data
l, Connectivitv
Ueometric properties
Member
Material
Direction coslne
properties
I
A
L
m2
A A
J
I
I
z
2
2
J
c
m
E kN/m2
4
E
3
D
I I
0 0
written as follows:
vt : n'- _v =* _
Fl-t,
4Û
-6L
2'3
o.:zs I Erl 0.7s = Ç,
-0.375 _o.7s | L 0.75 t
i
K.' =EIl
4"
I
3.334 J4
v€ctor is no load on member
l.
The equivalent nodal loads due to uniform load on
2 are shown in Fig.t2.30 d and e.
-
Po-P" directly at the nodes, therefore,
.Po
o
=0
30 15 30
-r5Ì
load vector
P.
lz
0.37s 13 4.7s 2J4
: P"_ P. : o_
=
{::} {_il}
relation for the structuré gives,
:"[-i:i,;1,ä]{:;}
3456
o.a+s Çr=E tl o.eæ 1.334
-o.os3l3
L-0.0&3
$l'
I
o.sz
*t] wL -y4l: ,o ,: pr-' = fo o *L 2 t2 2 t2J |
I
L* 2r] -6L 4t] l
I
3
is no load acting
I
tz
t,,,
P
$I
449
2, 5 and 6 rows and columns are to bc deteted to obtain the reduced Kt2 *2
S
Since the local and global axes coincides, and Lhe axial deformations are ignored, the d.o.f. mayrbe reduced as shown in Fig. 1r..30c. The member stifihess matrix can be
lr, EII 6L
EXAMPLES
ILLUSTRATIVE
$ l3
P,: Iç^,
f";l: (-r)Jrz.t341 lo¿J u ls.tz+J
.r,
lt
4.44s -0.667 0.445 ' It lo.en 0.667 -0.667 rs4l6 |
The system stiffiress matrix can be assembled as follows
:
{Pm
t234 0.375 0.75
K':EI
20
-0.375 -0.75
-0.75 0.375+0.445 I 4.75+0.667 2+t.334
4.Ms -0.667 ô.US 0.667 0.6;67 -0.667
+
{Pe}
'lj
[r,ì | o.:zs lp,l: I o.zs {P'*}, : 4.7s 0.37s
andur:0=ue
$
2
t; ls
ß34J6
1*i l-o.rr, [poj,
lo.zs I
-o.zs
forces in memb-e¡ 2 can be determined.
Boundary conditions
ur:0:ur
}
6
{P,,,*}
- rx ì t\,, ì
rP
I
*{o} 2
l{.'1
{ffi}
DIRECT STIFFNESS METHOD.2D ELEMENTS
f-zo.rul
(P,*)z
froì [rsol
Nodal data Node
: l;l?,Tl. ; i={ * J
[-zs.rJ
[-t'j
i [-r:.rJ
The net member forces are shown in Fig.r2.30 f, equilibfium. Example
ILLUSTRATIVE EXAMPLES
g.
x(m)
v(m)
I
0
0
2
L
J
L
0
_L
The foriei at node 2 are in
12.8 :
Analyze the propped cantilever beam shown rnetnod.
in Fig.l2.3la by the direct stiffness
matrix in global system AE
L 0
l"'
0 +
Þ I
len
(c)
AE _L 0
0
-T
6EI
TL
4EI AE
00
IZEI 6EI _-F-E
.6EI -:-.;VL
2EI
L
^ tzBl 0 L6EI n v-.:
4El
L-L
a4 ¿J
Fig.l2.3l
There are two þpes of elements in this structure: beam element and truss element. -. The nodald.o.f. are shown in Fig.r2.3lb. Ler the origin orthegtobaï*i, u, arnode r.
where.
AE LL,
can be assembled as follows
:
,2F.7
ILLUSTRATIVE EXAMPLES
DIRECT STIFFNESS METHOD-2D ELEMENTS
452
67
I
453
I
AE
t 0
il
t2Et -t-
0
6EI
AE
K_ -L 0 0
4EI
TL
I
t,
00 6EI -t2Bt _T _T6EI zEI -'=VL
(+.') 0
0'
(r2gt
l.
2Er\
l-r-.-r-J . 6EI -TL
l5
4EI
l'
00
0
2Er -r¡-
0
o
Å,1' TJ8
spring carries a compressive force equal to P/8 as shown in Fig. l2;31c. mole 12.9 Analyze the continuous beam shown clect axial deformations.
Boundary conditions
.
ür = 0
:
u,
: ¡,
and
If axial deformation is ignored,
u¿
us consider the global axis at node
url0:u, =
0.
nFig.
l.
12.32a by the direct stiffrress method.
The d.o.f. are shown inFig- 12.32b.
Nodal data
The corresponding rows and columns can be
eliminated.
56
to -u:lt *.' =gl r L-6L 4u 16
Node
x m
I
0
.2
4
)
m 0 0 0 0
9
lt
4
Load vector
{P}:
{P"}
- {P"}
Element data ì
Fixed end forces dúe to the point load are as follows
(o
P"':{o - 1,2+
I
:
I
+E o +2 -+I q o} J
I
n
2
J
J
J
4
fpì l-: P: {_ll ^r IPL ITJ
I
Itz'
$I
l-12 -6L
tz
er|eL 4Lz
I
-lL,
P. = K,
Â.
L
Geometric
Material
properties
propertíes
I I 2l I
J a
Member stiffiress matrix is given by:
The reducçd load vector.due to the boundary conditions is
The structural forcç-deformation relation gives,
Connectivity
Member
There is no load directly atthe nodes, hence p = 0
I I
I
ut zt] 4Ll
L(m) 4
E E
5
L
)
E
ffil
r
DIRECT STIFFNESS METHOD-2D ELEMENTS
ffiËr{
a.,#_fnr*,g
#F
Ëfi ãïi:l Fi'l
ILLUSTRATIVE EXAMPLES
of
(q
åii
)
''l' (b)
itj(,
(c)
(d)
i,,
-0.375+0.48 l+1.6
. --0.192 0.48
-0:48
0:192+
I.5
0.80
-0.48 +
1.5
--1.5 1.5
I ! I
*.
2.0
-t.5 l.0
1.5
-t.5 2.0
I l
ul=0:u¡:us
FTXED END FORCES
reduced stiffiress matrix is given by
246 EA. NODAL
j 1.6 +
conditio¡is
LOADS
f
Kr=EIlo
i,,l
I
| L
I
(
7
t'o
10.5
ç.
ir i:
.0.50
:
¡t'..
1lY í
þzs
t
z'n-_l r;-ø--f -E--
.-.
ii( '
j'
1.0
i
REDUCED D. O. F
f-r'\,r
iir.-,
$
375
;
iÍí"
1l'
t8+5
l8?s 4.375 0.1875+ 0.192 I -oi
Efl.il
{å,
78
fî
CONTINU0US BEAM
,
$it
K
global I
r,äl
2
'
455
8
$
lz
2.6
o.8o
0
l4
3.6
l6
-r.5 l-0
t,
1.5
-1.5
z.oJs
i;llts semi-band width is 3.
J.) t-
.,
$
K_,:Efllo.rtzs o37s l.o
_o.rt75 _0.37s
j{,
Only a 20kN load is applied direcqly at joint 4.
ll
...
Iz
K*z
,r,(
(i ,
: t,l
r.6 :.::^_0.48 4.1s2 o.ts2
0 0 0 0 o -20
$
l¿,
1.5
-1.5
I
are
0}
:
-'î:-- l0x4':20kN P,=P,: wL "22
I
x., = u,f ;:.; ;, I fs l_r.5 _r.5 L o.4B o.Bo _0.48 r.a]e [ ;; ;;
PJ:{0 u'
Fixed end forces on member
34s.6 '
P=Po-P"
,,
o.rt75 l¡ o.sts _0.37s o.5o I , olo I
l
ri(
- flhe fixed end forces and equivalent nodal loads are shown in Figs. lz.3z c and d.
1234
(
j¡
Load vector
Fig. t2.32
(,
'ì(.
BEND|NG MOMENT kNm
ls
I' P
2.018
p:-Þ:**Lt:* '2
1ox42
{,-}
DIRECT STIFFNESS METHOD-2D ELEMENTS
456
ILLUSTRATIVE EXAMPLES On Ìvlember 2
,, -r'
or
:
ob. L''q
:9kN, po- l5x2x32:
er: ff
.
5,
The fixed end force vector is given by 20
FJ..
P"=
il
ili,
t3.34 2
20+ )+9
29.0
- -10.8 14+ 13.34
i:(
6.0
)
-7.2
6
0
0
7
0
0
I
-l
P':P" -P":
\i
(l
I
relations.
The
I
l_o.rn
joint displacements
L
I
or
[-n.:+l I z.s+l
o.rez IllI
slf f,., r.s z.o p*, _
?
J,5*l I
"
can be determined using the system force-deformation
,'^
o
l,ool IOJ I L
o
tr. o
ff;l
lr0.15l "\, o4 lt2.6el " r,ç [ -oo J ",,1 ub
I
o
u
z.e 3.6
u(
,/
o
-1.5 1.0
AE :8000kNm2, EI =20000kNm2.
;ffi1
t
Solutlon Let the global axis be at node I . The d.o.f. are, shown in Fig. 12.33b.
g
ú'x
12.32e.
Analyze the inclined frame shown in Fig. 12.33 a using
0.8
l'-
'r/
|_1.5 _1.5 ll-.zt.ot r.5 ll_tot.+t | Lt.r r.o -r.5 z.oll-rr,.ot inFig.
$
Iz.lr.|"lt o]
Y*Y
lr.o
lo.s
Lå.Ì
o.4s o.8o _0.+t r.sll_zt.
The bending moment diagram is shown
17.2 f =sll
=
Member 3
í
0
_0.48
sll
.
i
Pr: K.Â.
il
ri
-20.10
Liit #"::',:,llil lo.sz. o.¿s t.6
2
I:6
7.22
{
t Solution Vector
lr
..' :
I P*, =
2.:544l
I I
I
il
li
flÌ
writing the member force deformation relations,
{P*, } : {P.}+{P"} : tK-l{^.}+{P".}
The reduced load vector'ts gr given by is g
!'
;ii
can be determined by
J
2+l0 -7.2t
t -13.334rìI
*/
'forces
2.54 4
+0 6+(
.iii
(
-
tti,:uu*
I
13.3 ,.34
fti ['.i
20
=
+l0.gkNm, pr=
P^-- l5xlx3:-7.2kNm -ó
-
[::ì [ ,il;il i"'f !\-awl tïl lïiil
,t,:T,t,=-$
,,:'u!' Ú
457
method.
DIRECT STIFFNESS METHOD-2D ELEMENTS
ILLTJSTRATIVE EXAMPLES
459
123456
@
s.20 s' . It i rzr;3.60 l8o4.8o lz | -t' 2BBo 3840 l6ooo Kt-t : ll + 153.60 2880 r7ts.zo 1s.20 -t7 lq | 53.60. -1804.80 -3840 -153.60 r804.80 ls 2880 3840 Booo 2B8o -3t40 reoooJo
r.sEr
I
(b)
Ir L
0.0. F.
56789
4
I
ii
f ,r. ot
(d) FREE BODY
LOCAL AXIS
$'.|4
DIAGRAMS
rm stiffrress matrrix can be assembled
with the help of location vectors.
y conditions
ii
.
Fig.
12.33
Nodal data x Node (m) 2
il
ul
v
J
0 J
Kf
5
ireooo
=l
¡rf
Member
-:a+o L 8000
2
2
2
Direction cosines
Geometric proPerties
Connectivity J I
I
I
2880
|
Member data
if
AE kNmr
trt
L
kNmr
m
8000 8000
20000 30000
)
0.8
4
1.0
C
:
0.6 0
ironoo
| |
2880
where
K.
RT
K-
R
I 7ts.zo
$
+zeoo
-153.60 2880
1804.80 +
8000
(12-t2\
vector
10242.3
2880
13035
lo
I
15
OTOOOJe
.r/
f=Po-P" ijfliwaá an¡l f^'^õô iñ l^^ôl ^^^-,{:--+-.
16875
s l¡
data in and R, are given by Eqs' l2'4 and i2'13b' Substituting the member
Eq. 12.12 gives
8437.50
-3840+
37rs.2o
-:s+o -153.60
Member stiffriess in global coordinates is given by
K'.=
4'6
S
L
'
\
ta4x4matrix
(m)
I
J
:0: u, and ur: Q: uB: uc
rsponding rows and columns must be deleted in the global system matrix.
0 4
1
ir,
)
B437.so It lo 16825 45ooo X'-r: | 0 lu _200 ooo2ooooþ | l0 -s437.s -t6r7s o t437.5 ls lo t6r7s 22soo o -t6t7s +soooJl
/*-g't'S (c)
zoor
Di^ tt at^
l' 14
lS
16000+45000J6
lt
460
DIRECT STIFFNESS METHOD.2D ELEMENTS
ILLUSTRATIVE EXAMPLES
: {P, P2 P3 P4 P5 Pu } where, P,':0: Po, Pz: Ps = | : tO t* - : +_PL : + 20x5 +l2.5kNm = _po, p¡ P-r
end forces (Fie. 1233d) member end forces are given by
fo.r 4.6 o; lo.u o.B o
4.49
-6.0
-6.49
4.07
8.0
12.o7
i
P'=11 g lj - rä: I 0 ;o'8 | io.u 0.8
;0
L
Pr*
=
-12.50
12.5
0
0.49
-6.0
-5.5 I
4.07
{i:}
8.0
3.93
30.24
-12.5
t7.74
5.50
0
-rE.93
l5
-3.93
-17.72
0
-t7.72
-5.51
0
-5.51
8.93
0
l&93
-58.0
0
-58.0
0
0
{P"}
l0
,il|il ..r.?:1-
0
t0
0
-t2.5
[il
Pz* =
I
Member 2 There is no load on the
:
{P,o*¡: {P*}+ {P"}: tK_]{^.}+
This local vector can be transformed in global coordinate system.
Ëil iiill
46t
member, ...P": 0
5.50
Therefore, global load vector is given by
:H
-6.0
6.0
8.0
-8.0
\?,1
-12.5 '' ò.0 '
-6.0 8.0 -12.5
l=
¡, Anatyze the portal frame shown in Fig.l2.34a by the direct stiffiress method ing the axial deformations. TakeE:200GPa r!300 x l0{ ma,
=
100
x l0{
m2
5
-23.0
l?,1.
0
0_
0
0
.0
0
:î
T
il
9xl
4rn
1
Reduced load vector in view ofthe boundary conditions is given as :
P.t Solution
l.m
: l-t2.5 6.0 -23.0 t2.5ltx4
vector
I
Pr: K. Á,
Â":
or,
^,
=
I
K'r. Â,
{-o.oo:se 0.oo27s -0.00580
(ol 0.00178}
F'ig. 12.34
462
ILLUSTRATIVE EXAMPLES
DIRECT STIFFNESS METHOD-2D ELEMENTS
Solution
l.
axes are taken at node
The d.o.f. and member numbers are shown ín
Node I
v(m) 0
J
0 0 7
7
4
I
5
2
Element data Connectivity Member
I I
a
2
2 4
J
ffi
z
I(ma) x
>,
4
where
l0-6
x
300 600 300
200 200 200
100
Direction cosine
c
E(kN/mz)
100
R
, ut = 0= uz =u,
S
106
0 I
0
0
I
Lo i ÀJu,u
5
-7347 -2099
0 0
.
0
-2857 14
0
-7347
0
17t43
5
12
0 285714 1347 0 678
2857t4
ifii
Kt- 2 --
4198
0
14694
-2.85714 0 0
lsÌ
342s616
ìj:.,
luil*t
ls
+
342886
6Bs7t
0 -14694 34286
le
285714+11250
l7
0
4Ì98 +-500000
22500
-14694
¡
l8 68571 + 60000J96,6
0 289912 7347 14694 102857 0 -285714 0 0 -4198 -14694 0 14694 34286
296964
0
504198
22500 -14694
128571
4198
The member stiffness.m4trix reduces to a 4 x 4 matrix as giyen.þy F,q.12,6. The same the
i rasult can be obtained by using the following boundary conditions and dropping
9
contribution of AE/L of elements 1,2 and3.
l4
ül=0-u2=u3l uro= 0 : ulr : utz ut = ur: 0; ur:u,, =0 u+ : üz = ldteral sway '
l;
28s714
0 0
9
:{ll). Ignortng q4ìal deformations
68571
00 4t98 -14694 t46i4 34286
:
14
$
0
K.i =
l3
2099
0:ulr :ürz
287813
Ir
34286.
7347
Lorus
6
$
285714
4
9
.0
14
0 -4198 14694
:lll:l
0
and u,o =
r)
2099
ffil
22500
8
I (
uu
software on a PC.
Kt. l=
30000
7
500000
system stifttess matr-ix can be assembled ignoring the rows and columns to the. above d.o.f. It reduces to a 6 x 6 matrix as shown below
I
'[he element stiftress matrices can be easily generated on a spread sheet using
[,
0
t1250
0
0
Considqring axial deþrmatiqns
RT K_ R
2
ffi,
Material property
0-4
100
=
22500
-500000
-22500
Element stiffrress matrices in global coordinates are given by
K'6r6
iì"'
7
J
-11250
t2
60000
conditions
I
I
ll
0 '0
= --22500 0
Geometric properfy
L (m) A(mz)
J
K'.3
l0
s 500000
0
x(m)
I
12
1t250
Nodal datà
J
ll
l0
T!. global -. Fis:12.34b.
463
I,
-14694 68s7tJ9
i:,ili
if
.u.rurn stif,hess stiffrress.mat¡ix redrices to ae 3 i x 3? mafrix mafrir as shown below. Th" system matrix reduces ;illt:a¡" :
$
fzogg+ttzso K,
I
=
l,
I
-
9
6
4
7347
34286+68571
zzsoo
34286
14 lrzt+s =ln+t le
68571+60000J9 L22s00
102857
P*r =
,rrt rJr",.
34286
Load vector
P:Po-P" Member
I Pr
#Íi
p¿
gfii
' 75.x3 =
: î
-32.rn{,.
= - 42.9 kN,
P¡ po
:
+.'
- -
75 x4x32
+55.r kNm
:f:= 75 x-!' x3 7'
P*z
-
: - 73.4 kNm
Member 2
#r
,r= +25 = pe,
Îl
!r.. *. 4l: * l. üt L-.'
= +A#
= +43.75kNm,
ps:
-32.1
-41.0
21.7t
0
2t.71
30.15
55.
8.93
I
-4t.s
.!l'?1. -34.0
-21.71
0
-21.71
32.31
-73.4
-41.0
'8.di
34.0
0
-3.17 -3.0
25 43.<15
-34.0
0
3.17
25
-19.2
43.75
-42.g
55.1
25(-73.4+43.75) O
25
-34
[üt] I
|
:::l
28
I]ii
Ëfil
34
-28
|
62.85
l-øz.ss)
-43.75kNm l0 mm and rotates by 0.004 rad,
-43.75 O
0
method.
I
0lr*rz
= 300 x
l0{
ma.
Since Po is zero
(i)
If axial defornrations
are considered, the reduced load vector can be written as
T
:
4m
1l
$i I ìi-
73
and P*r =
28.17
The load vectors were calculated in global coordinates. The system load vector can be assembled as follows :
p,r: (-x-32.1 0
#l ñ1
po
465
ILLUSTRATIVE EXAMÞLES
DIRECT STIFFNESS METHOD-2D ELEMENTS
p'r = {42.9
(ii)
-25 29.65 0 -25
43.751t*6
Ifaxial deformations are ignored, tlre reduced system load vector can be writton p''r
= {42.9 29.65
=lomm I
43.751t*3 -
= 0.004 RAD
Solution vectors
I.
If axial deformations
are
considered II. If axial deformations are ignored
Ëill
0.0038
rI
-0.000076
Â,:
I
as:
0.00013
Â"=
0:0037:
I o.oorzl+
Fig.
10.000r416
[ -0.00035J e
-0.000056
'l
12.35
he stiffness matrices of the frame members can be developed as in Example l2.l ilf6$A"l d.o.f. are shown in Fig.l2.34b. The boundary conditions are as follows :
;j:
-0.00034
ul = 0 :
uz
:u3
ulo = 0, ün : - 0'01 m, ut2: - 0'004 radian
Member end forces Member end forcei can be determined using the relation
P*r= P.+
?m
h
ll¡ilillleadto
:;: '.
P"
This calculation can be easily done using LOTUS, and the forces are as follows
:
a 8 x 8 stiffrress maûix.
flr¡xìol deformations are ignored, the boundary conditions are as ''': ..
follows
:
l.
ILLUSTRATIVE EXAMPLES
DINSCTSTIEFNESS METHOD.2D ELEMENTS
466
üz=ü3 ul= 0 u2: üs: 0 us : lt': - 0.01 = urr* u¿: u7
467
0.0103
-
say
0.001639
-0.00157 -0.01
This leads to a 5 x 5 stiffiress matrix as shown below
fzogs
Ç
-0.004
:
4
ll
+ttzso
$
,to, =l| zzsoo l0 22s00
12'
l:
342ß6+6857t
'', 34286 68571+6oooo -14694 -14694 1,000,000+4198 0 30000 0
I
L
The conesionding force-deformation relation cari be written as follows
[j l':::i 1:l=lî"
'j
$
.is the same
,li,
'1l\nalyze the bracket shown in Fig.l2.36aby the direct stiffircss method.
t::'i
6o0ooj r 2
:34286 128571 :,'14694 -14694
ltz l l-4
:
,
10
1t004,198
..0 ¡ooo0 0
Since the di5placements:u11. and rearranged as follows :
u¡,
tfil
60000
are known' the above equations can be
40 kN
I
(a -22500uiz'
Ii
l469aui¡
t.
1r+oe+úi,
il:-
tL
lti
o,i
il'
0o
lt+a.s+l I tt+t ), -ze.s+ [
: lzzsoo
t02857
34286
I J
,ll
L
o
o
128571
00
ftrl+e
| -o.or
BRACKET
(
b)
0.0.F.
102857
I ro;o ] [-o.ooa
li;
lo
a .'
urz
il,
i[.
7347
| =,1";oo 34286
' un.,
ili. I
)
Fig. 12.36
Irzzdo
-:OooOuir¡
I
'lil
l,3 : I = 8600cma, J = 4300cml Member2 : I:5100cm4, J:2500cm! E:20000kN/cm2, v = 0.30
Members
t02857 j.-
trj lzzsoo:
/i1
result as obtained by the slope deflection method, Ex.l0. I l.
bracket is similar to a grid structure in which th,e loads act perpendicular to the of the struçture. The stiffness matrix of a grid elernent and its transformation
','ffiis
;,1{fi
Nodal data
00 00
Node I
x(cm)
I _l 0 21300
12857t
;J{ü
J
4
300 0
v(cm) 0 0
400 400
,
ILLUSTRATIVE EXAMPLES
DIRECT STIFFNESS METHOD-2D ELEMENTS
468
Elen¡ent data Member Connectivþ I
J
cm4 I
I
2
J
za
J
J
4
3
kN/cm2 20000
0.3
100
cm 300 400
20000
0.3
8600
300
20000
0.3
8600
4300 2500 4300
5
le t,
48125+2293334
I
= 2341460 -
lu
19.+76.45
0o
0
90"
=95.45
0o
3825
I
1020000+ I 10370
=
written using Eqs.l2- 12 as follows The member stiffiresses in global coordinates can'be
-48125
:
56
2
,r
$.
Þ
E
I
Inclination
with x-axis
L
I ctt{
J
s67
Material properly
Geometric Property
469
l7 I I
I 130370
48125+2293334
o
0
ls
= 234t460
I
te + 76.4s
0
11467
-3825
-19'
=e5.4s
le
I
I 10370
Kl:
-l
0
2293334
0
-11467
76.45
0 I 10370 -t1467 0 -76.45 0
0
10370 0
1146667
0
1t467
6789
)
4
Kz:
0
48t25 0
te
510000
0
382s
0
48125
-3825
0
t0
Ks:
0
-l
10370
0 0
,r.*]2
11467
there is only one nodal load, the reduced global load veclor can be written as
rl={0 0 -40 0 0
16
:or, Âr, : { 0.0039
1146667
'
10
ll
P*.
I
110370
1t467 -76.4s o
2293334 11467
76.45
7
P'r :
8
2
9
: p.+
'I
12
76.45.
-11467 0
0.0093
- 1.89 0.0039
0.0012 "-0;20}
end forces
2293334
0
p"
2
{-434.5
'4
Yir : 1- azn.s
Boundary conditions
-
456
J
1t026
434.5 -3ig}
38.06
56
7
389.0 -
1.94
-
434.5
follows
ur : 0: u, :u, and
:
ulo =
O:utl
=ut2
as shown below The reduced system stiffuess maEix can be assembled
P.r : :
l0 {_434.5
ll
- g77
12
l.g4
7
434.5
-,38.06)
8? - 389,0 t.g4)
3
The boundary conditions are as
Â,
solution algorithm in dou6le precision may be employed because there are äome very large numbers in Kr.
$
0'
i P. = K,
il. n suitaUte
17
18
-11467
0}r*o
¡lhe solution vector can be obtained from the equation
l02o0o0 0 4812s o -19 -352s 0 l9J9 t27
ll
:
v.ector
ls
11a370
0
i;lroad vector
2293334
$-l4
r020000 3825
li
89
:.
,'
389.0 - r)41
ILLUSTRATIVE EXAMPLES
ELEMENTS DIRECT STIFFNESS MI]THOD-2D
470
The member stiffiress matrices in global coordin¿tes can be developed using LOTUS rpread sheet.
i.j:
ExamPle 12.14
A portar frame hastwo pin-ended
^J;;"
Itqf*.:*lll l'"ì*î" stirfrress method' ft;;;;; r*#bt using theãirect
ineig' r2'37a'
matrix for beam elements
l2
5
Kl:
o
4t
'1.
(bl
4t
o.0.F
Fig.
Kz: 12.37
""d;h;; ft.
Beam tYPe
I
I
J
t2.75
0
2ss0
00
-12.'15 1ss0
25s0,
340000
2000 0
12.75
0
2550
-2000
00
0
-12.75
0
2550
0
-
400
Geometric ProPeffY
L
A cm4
)
55
8600 3400 3400. 8600
cm 400 400 400 400
,
2
J
AO
J
J
4
4U
4
4
)
))
I
I
J
t0
565.7
2
J
)
l0
s65.7
Vfaterial Propert) Inclination E
with x-axis 0
l0
.0 0 l0
0 0
800
lo
2000
0 400 800
)
lt ls
I
4
5
1720000 6
ó80000
v(cm)
cm¿
Truss tYPe
0
0
K¡:
2750
6450 0
$
1
0 400
4
2000
x(cm)
ConnectivitY I
860000
Node
J
J
6450 32 0 0
""il1l}"
""nt", data Nodal
2
Member data Member
1720000
8
0
and truss type' They used in this frame : beam type There are two types of members rhe d'o'f' tr';1@'¡19,1Î"1"Ï"can be joined at a singre 2. nods at or*," global axis is taken
2
4
-2400
Solution
i: J*#ïft:äïï
0 2750 -6450 0 -32 0 0 -2750 -64s0 0
'l
I
s
32
I
471
17
12..75
-2550
ll
l8 680000Je
L2
7
$
H
I 680000
9
l0
2000
-2ss0
.340000
l2
0 0 l3
12.75
l1
-2s50
680000 t2
l4
l5
l0
$
-kN/cm2 20000
90"
20000 20000
00
20000
270"
20000 20000
JIJ
0o
0
rq:
6450
-32 0 645A
2750
u
0 1720000 0 -64s0 32 0 2750 -2750 0 0 860000 -6450 0
t2 t3
t'i 1720000
l5
Égg
ffiì¡
;
ffili
472
DIRECT STIFFNESS METHOD-2D ELEMENTS
ILLUSTRATIVE EXAMPLES
473
Stiffness matrix for truss elements
1278781314
. lt77 x, : I w t77
=+40kN, po: P8 : +40kN Pe : p5
$lt
$lr
lrtt 177 xr: , lz l-ttt t77 177 | l -nt -177 177 lt l-'177 ll3 ntlt
ffi
l-rr, ffij
8
-177 177
wþ
lw
If axial deformations are ignored, the boundary conditions
ffii
follows
are as
I*!' Ðl
ffr,
li:= i' : ut¿: Ulr =
$¡
Ïiî 0
tii P':1;1
:
IoJ as shown
below
K.:
Ë[
t77 -177
tfi
0
6450
743
6
1720000 +
ln
1550 340000
12.75+177 + 177
2550
0
34oooo
t¿33333.
-1.726
0.000057
as
follows
0.1 13)
:
2ev.r4
8
1.94 x 105
9
-743
t 298.14 *, : I _rnr.,o
306
4
+12.75
-2550+2s50 :å3333.
P": Iç^"
:
.
-226
680000
I li:
r
-80 l* o ls za,søo
l-za,øøet]
member force vectors are
+177
u50
üå
1+o++of=
ber end forces 4
32+32+ 177
fÍi ,t
t2
6
4
[,u,\uu,]
Âr, : {0.0394 -0.113
:
$r $l'
15 l+
¿6,666716
force-defo¡mation relation can be written as
5 matrix
-26,6667kNcm
reduced system load vector is given by
: il
The system K reduces to a 5 x
prz:
P=Po-P"
üz:u¡ ur:.0: üz: us= 0
ffi.Í..''
-26,6667kNcm
Ps= +40kN,Ps: +26,6667 kNcm, P,,:+40kN,
Boundary conditions
ffi)
=+2ó,6667kNcm
3
-n7 -n7
The system stiffress matrix can now be assembled with the help of location vectors. It is a 15 x 15 matrix.
ffi)
+ +=*20x!002 t2 t2
t2
-3
x
f .l-2e8.t
.:J-ill IrrzJ
105
, !,r
äi¡
ll)'
'
| +ta
¡!
frrr
*:.i
:
. x,i
üi
',
or
x,: | 0
ù,
flj .l
'&;
l,i
:,
..
i
ffj,
,
,gi.
:A
r
tn¡ss is supported on a roller support inclined at 60" with the horizontal as shown in 12.38a. Analyze the truss and determine the movement of the roller support. Take E, t l0 cm2 , E = 15000 kN/im2 i
24o,oooo
-2550
389
I o 34oooo o looto o 25so
'J:n,,,,,j
::1f.he
Load vector The load vector in global coordinates for each member can be written as follows
.&i
ffifi
letso
:
roller support may be replaced by a very stiff spring which will not pennit any perpendicular to the inclined surface. However, node 3 will be,free to move to the inclined surface. The d.o.f. are shorvn in Fig. l2.38b.
DIRECT STIFFNESS METHOD-2D ELTMENTS
474
ILLUSTRAI]VE EXAMPLES 4
3
K¡:
T
2m
s
*t92 92.7 ',.7
t228.653
l-t-288. 72
l992.',1
I
Ã92.
Ç,+
T .3m
I
6
lz288.8.72 I
I
5
t [+
92.'l17t -lgt
It
6
$
I
0.t62
I
t- 1,87 ,87
t.(08
t
lr
l. 87
-0.t.62
1. 082
Fig.
I ls
28. i8.t63 t3Jl6
F
K¿ = l0lo x
'll3
288. 8.72
-12lt.t63
) I .si si 082 t--l .08:
475
1-
t'
l6
r
-1. o82 o. 62 J8
12.38
Nodal data Node
x (cm)
v (cm)
I
0
.'
0 200 500
4
1019.60
2
inclined roller ar support 3 was repraced by a infinitery stiff axiar spring. The are as follows :
y conditions
500 300 0
ul =0:u, and ur:0=u, will be no axial deformation in the spríng, hence dispracement normar to the I surface will automatically be zeroi The reãuced system stiffness matrix can be :e
ConnectivitY Member
GeomeÍic
Material
Direction
properties
properties
cosine
I
J
L
A
cm
cm2
I
I
5
583.1
l0
2
I
2
538.5 360.5 600
l0 l0
J
2
3
4
3
4
l0ru
I
Kr
I trs.rl
=l
113.54
J S
0.857
0.5r5
o.37t
0.928
0.833
-0.556 -0.50
0.866
4
2
188.93
t13.54
34
t;
68.23)6
ó
(288.72+38.34
']
= 327.t)
$
(-192.7 +9s.9
(128:63 +239.88
= -e6.8)
=¡6g.5),:
_288.72
l:92J
t92.7
''|j
68.23
-188.93
nl
C
.
6
-r 13.54 l-r t:.s+ -68.23
|
E kN/cm2 5000 5000 5000 5000
:
loilows
as
Element data
-12g,63
(288.72+188.93+ l.87 x t0to¡ (-t92.7 + tl3.s4' (t2ß.63 + 68.23 + , 0.62x I0ro) -1.082 x
l0r9)
vector loduced system load vectorcan be written as :
prr- {50' 0 0
0}
vector
$ It t7
I rt.r+ x, : I os.ro -:s':+ |
239.88
-95.90
38.34
L-e5.e0
-239.88
9s.e0
I
vector
is given by
lr' = K, A"
th:.:,u* few very
large numbers in [K], a suitable scaling scheme should be Irll o o'rarn the correct sorution of the.four rinear simultaneous equãtions.
13
239.88)4
Â,
^r
: {0.1658 0.0435 -0.0149
-0.026}
cm
PROBLEMS
DIRECT STIFFNESS METHOD'2D ELEMENTS
476
The computational effort required to compute deflections is much more than that
Memb er end forces
required to compufe forces at the ends of members.
using the member force-deformation The member end forces can be determined relatior rs as follows :
I s.zt
P'
I t'+t
I
-'
I
=1_5.7s1'
f-
t
[*.:,
The chances of making mistakes are much more in computing deflections in,.say, the unit load method or strain energy method. The chances óf making mistakes, say, in the slope-deflection method or moment distribution method are relativelv
f+38.sì
0.53ì
=1,'rl;il'
l-t.+t)
l-zs.sl
P3
={
l-38.8
J
less.
} |
l;:
[+zs.eJ
i:
- 46'65 kN' ate - 6'74 kN' Tht : net forces in the three truss members the rollet of movement The spring will be zero' respec tiï"iy. fft" axial force in the 28' 35 kN, and
In the flexibility method, once the member forces are known, additional In computations become necessary to determine the slopes and deflections. on the contran/, in the stiffrress method, deformations precede the computations of members forces , with the exception of moment distribution, method which is iterative.
METHODS COMPARISON OF FLEXIBILITY AND STIFFNESS
been discussed in Chapters 3t 4' 5:6 andT' Thr I various flexibility methods have in Chapters l0' l and |2. Let us first discussed u,"n while thestiffness methods r,uu, methods : summ arize the main steps of these tw'o
l
Step I
óffiin--
th- d"g.ee of
indeterminacy, identiff
obhin
static redundants and
released structures (statically
determinate). 2.
Compute displacements and áirections ofthe
at the
releases due
pomts to the
LrelerÏnlne ulË uçË¡çe
indeterminacy, identiff the degrees ol
freedom and obtain
4.
each member along the selecfed degrees oifreedom due to the aPP.!9q-!939!:each C-õmputeTrces at thg ends
of
words,
develop the flexibiliry matrix'
other words, develoP 'the
Impose compatibility condltlons-anq
impo¡e ,q,rîiçitm . conditions
õõmæã"mòGE.ces
l.
Ease in
bY using the
a method for a given problem
computations 2'
the following problems'talæ E ¿ 20a Gpa and I
and
stffiess marrices and load - lhe spread sheet. Alternatively,
depends upon
the tbllowing factors
:
working through the various methods in each Based on the e4perience gained while : that general, in category, it can be concluded,
(l)Thechoiceofredundantsisverycrucialinanyflexibilitymethodbecausethe the other hand' in any stiftress method' computationaf tmotttì"p"nO upon it' On
may be
.!:,.
the beams shown in Figs P 3.3, P3.4 and p3.6
obtain the disPlacements
Compute member rorces DJ¿ ustllË trrc force-deformation equations.
I = 150 x 10-6 ma unless specifìed vectors may be generated using'Lorus
the solution of linear simul¡aneous
(a) lffi.t'åi
liìi
ár'
Accuracy
there is only one restrained structure'
vise.
"quàtioi, ed using the programs given in Appendix A. verify the requhi using program diseussed in Chapter I4 and øvailable on afloppy.
stiffness
matrix.
equations of eQuilibrium. The choioe of
PROBLEMS
member due to unit displacement along the selected degrees of freedom' In
obtain the redundants.
).
It is difficult to write a general pu{pose computer program for flexibility methods, whereas, stiffrress methods are very computer friendly. The direct stiffrress matrix m.ethod based computer programs ar.e now used extensively even in small design offices.
restrained
delrtttnqtÙCmputeEãdãforces at the ends of
Compute displacements at the polms and áirections ofthe releases due to the
unit redundants. In other
a
structure (kinematically
applied loads. 3.
Physical concepts become quite clear while working with any of the flexibility methods. On the contrary, stiffrress methods are mechanical in nature.
StifÈreis Method
FtexibilitY Methods
Most elements of the influence coeflicient or flexibility matrix are non-zero, whilo, matrix is usually banded. Thus, the number of arithmatic operatione required for the solution of linear simultaneous equations is far less in thostiffness method than that in the ftexibility method. stiffiress sf
Both the flexibility and stiffrress matrièes are symmetric about the diagonal.
suppol t is 0.03 cm down the sloPe'
t2.12
477
g.r
asspmble the stiffiress mâtrix, with and without axiar deformations and mark the various degrees of freedom in the strucfure. (b) assemble the load vector, (b) (c) evaluate reduced structural stiffrress matrix and load vector after introducing (c) the boundary conditions, (d) solve for structural displacements and determine member forces. (d)
Dett the support reactions and member forces due.to a vertical l42 Determine
;ì.
i
orz of 7.5 mm at support B of the beam
in Fig. p3.4.
settlement
o", Analyze the frame shown in Fig. p3.8 c and determine the support reactions.
iL¡ n-, Draw shear
force and bending moment diagrams.
478 l2-4
DIRECT STIFFNESS METHOD-2D ELEMENTS CHAPTER
For the trusses shown in Fig. P5.7. and P5.8
(a) (b) (c) (d) (e)
(f)
thirteen
number the nodes so as to $et the.minimum band width, simulate the supports using boundary elements, assemble the global stifftress matrix, assemble the load vector; evaluate reduced structural stiffness matrix and load vector, and solve for structural displacements and determine member forces.
DIRECT STIFFNESS METHOD- 3D ELEMENTS
12.5 Analyze the closed frames ¡hown. in Fig. P5.ll and P 5.12 making use ol synimetry and taking appropriate boundary conditions.
12.6 Analyzethe circular arch.shown in Fig. P7.3 by dividing the arch axis in ten clemerits. Tâkeú E :26000 MPa,t A = 0.20 mz,and I : 0.67 x 10-3m4. 12.7 For the frames shown in Fig. Pl:0-l
(a) Mark the various degrees of (b) (c) (d) (e)
an
beam
STIFFNESS MATRIX. TRUSS ELEMENT
dPlO-z
freedom on the structure ignoring
deformations and assumÏng supportD is hinged, introduce boundary elements, assemble the globai stif,frress matrix, assemble the load vector,, and solve for structural displacer4ents and evaluate support reactions.
the axial
'
*:up"d... resist onty axiat force. The deformation afthe ends a¡e ,j,T:::fTïr.ft lations in the x, y and z directions. with three possible degrees of frcedom at each Fig. t3.t the member stiffrress matrix is rh; ly in the same manner as in the case of a "f ";;;i;;";.-i;;;Ëï;,Ë; ptane truss erement discussed in section and looks as follows
:
AE L 0 :
IÇ'o
=
0
"'ÃË"-'.'
-T
0
0
0
0
0
0; ^'AE u::t
'----!---..--.
0:0 0i0 ll¿
('!'v j'?j't
Fig.
l3.l
3-D truss elemenr
DIRECT STIFFNESS"METHOD'3D
480
STIFFNESS MATRIX. BEAM ELEMENT
ELEMENTS
BEAM ELEMENT r3.2 STIFF'NESS MATRD(as'ltrown''in nie' tl'2" six degrees of freedom ner-join1 hæ element its stiffness beam 3-D A per.elem.ni. ln order to de¡ive ¿"srä, twerve section' total cross are the of There "rî;"dom ,tt" piintipal axes T-rio'J"" ,"'""t*l¿å *ittt mafiix, the local o", ur" Hence bending and shear
iliì:ï:;;ä; ; i;dd;;ffi;:|";;*Ï;;í,:f:î"Ï;
T":#:,ii*T3:1ffJ;Hffi
:låJï't;n-*Ji--*""',i"t,r'ecoordinatelYstem'Foran' Now' the forces auoue simptifrcation is invaiid'
arbitrary choice of benoü;ä;ñ" can.be arranged in srx,.T;,i;ïffi;;;;;tilered
follows (Ð
(iÐ
(iii) (iv)
independent of each other
:
forces shear forces shearforces bending moments
Pt
and Pt
P,
and Pt
P,
and P,
P,
and Ptt
bending moments
Po and Pto
axial
as
corresponding displacements-are ul to ur2. Each force can be impcised on thô and the corresponding reactions at the ottler end and displacements can be íned in the same manner as discussed in chapter 12. Thus, a 12 x 12 stiffness ofthe beam element can be generated. Such a formulation does not account fbr deformations which may be impgrtant for deep flexural members,
ieniecki ü9681
used the differential equations relating the force and the. deformations in each ofthe above six sets and derived a general 12 x 12 matrix for a beam element. This accounts for the shear deformations about the of bending. The derivatión of the stiffrress matrix is not presented here but final expression is given in Eq. 13.2 a to d. The Structural Analysis program D presented in thehext chapter makes use of the same stiffrress matrix.
K,r",,
' *"1 f*" :f .,... r ..... . l_Kr, :
(v)
(vi)
(13.2a)
I
Kzrl,,,rz
is a 6 x 6 matrix and is written as follorùs
Prr'urr
I
I Psr
us
f
P10,u1o
Pg' u9
234"56
L
SYMMETRIC
l2El.
r-16;l;) rzBIy
0
LnñÐ
0
0
Pl2,
ulz
:
EA 0
/e,u,
48t
(r3.2b)
GJ
L 0
6E!.
--;7--L'(l +
,r,",? ,,/ errul Fig. 13.2 3-D
beam element
crrj
(+*o,)u, -(j;d;
STIFFNESS MATRIX- GRID
DIRECT STIFFNESS METHOD- 3D ELEMENTS
482
t _EA
-L
0
234 000 -l2Bl,.
Kzr =
0 L3(l+a.)
',C;J 0
0
-t
GJ
0
-6EIy
lm-+a,)
(13'
l9îIlit
(,
0
9EI,ooo
Tr-ã)t
-crll
is ciefïned as a frgrgsrructure wirh rigid joinrs whose members and
-ñ,"îr"r*ïi""ä;í#1,
gul is, alcng z_axlr. ¡-r*"-l,:" :lJl.:*:*r", the In rhe case át z_o
::h:,P1". if ,j:,i:T1î
(z-o,\Yt,
t?(t+ar)
radii ofgyration
i:STIFFT.IESS MATRIX. GRTD ELEMENT
=Krz
0
:rr
7
I
L
0.
l2El"
rn'-"J
l2EI,
Fa;Ð CJ
lo
ll
:.
o ''a;Ð ,nt,ib 6E
I'
o
L (+ 0
+a")el,
T;ÐL ,
(+
* or)e t"
0
ELEMENT
gi In these equations, arandarare shape factors afongthe y and z axes and are
follows
:
Fig. 13.3 Grid element
'
l2gl. r¡,, "Y : _GA.yL2
and
sz
=
lzBly c^_L:T
'or'."1fr(f)' z+(r+v)*[])'
,Ë';;i""il:
s*J;","d;ì;ö; il,; iii;,,j,lï i"::î;T earlier. The srifftess marrix of agria:" elJm"nt ts gi;;yä.;î;
SYMMETRIC
9
Kzz =
:. * J;',î;,"ä;rffiffi ;i,";of ",' i. s¡.iu¡v, îo una u, jiïg-*,-_T1 li,i""lJjüî"""j,,,T1i:::: "t. "ía * *d..y axes and i
. uu iepresenr t *rluiiã" aJ"g f:rj.j::]:-l-":111 ,i#.:,:å1':":^:f:"-T ",*
ll
t0
EA
illi
b;;;;;;,'#;;;
11uc1ure. " were designared in rhe order of rranstarion rùr4!tun {efor.n:alions i¡r rhe order of rotaiions abour rh" ;, y;;;axes at "X1"1,:::.t1:ns^lîdrhen case of a grid member, a sriehtty differenr scheme end,
f,:*-"1":::-"t il f." inli g. :: ï.:'"_ f 'lryf .e"1o'1n 7
483
lyl.L << l-andr" lL<< l. thana, and q.z maybetaken aszßro,thatis,shear nations can be negiected as ís the caíe with slånder beams.
-68t,
0
-l2Bly'
00 00
ll
0
l3(r+ar) 0
Íy,Íz :
6
ELF,MENT
STIFFNESS MATRIX. SHEAR WALL ELEMENT
3D ELEMENTS DIRECT STIFFNESS METHOD'
484
GI* L 0
f(=
L -6Ely
----;-' V
:9r"
0
0
rL L 6EIy
,1L
L
hl* ^t
4ElL
0
L
'l*
2
I
48_5
3
I
I
I
I
Gl*
4Ely
----;v
L .0
----- -lzBly L3
YL L YL r]
v
I
I
tzBly
',7777
I
(o)''ELEvATtoN
a
--
Ir
Ji-,
I
el
'r,[*,
(cl
(b)
(d )
Fig. 13.4 Shear wall element :Îhe stiffiress matrix corresponding to the coordinates in Fig. 13.4b is obtained by erting the flexibility matix, that is,
of freedom has a of nunberinq'of th" degrees The change in the sequence
nnt
advantase.rhegrid'"åi""iïä;äå:::*:*iî*i"""1åi,i:t:"ä":l:
iffi
':-f
.##'iäË;;':rll'n:::::::'lhi;JiJ;i;"1åï,J*""'
manix rouows the sarne
I tlh1
K=
o'":il"å'i"iuTtr'^.Ji"'"*Î:::."S"1î1îîîï:"fgT;:""t is given bv Eqs' 12' I 3'
L ht-
can needs i:5::"1-l-?"HÏl;fi:T i-p u"u* erement' åHi i'i"ltä' iö"f '"dil:: of freedom' "r"T'üirrt'iÏff restrain undestrabledegrees It may be noted ürat tfr3'iifft':ï
,",*"i'åiJiiä?'iï'ã.åiåî-il;"r::l:l*ï-',"j.iìï':::å:iä: H:*":""J*:ifi'"åiïläiì;#;;;;J*Tì"::îi*'"'ff enorÍnous as compareq '" Hîfrin'fiFig. two adiacent floors shown ^'l'::.'::,;'T;;"; *"lt.b^:y""i,,11. ä""it-r'ar** Consider a shear oi, l?.4b. The flexibility matr Tiiå'ffff i';:"1,äï;:*ffi ;ñ:y.:"*,ii*.11*i""iit'Y'-ii at end A can be written îXi'Ïl'::i:iäffi;fiilî;";ffi;"';¿icarte¿
If
oeÏl
rn
6'J2
the eleinents óf the stiffr¡ess matrix of Eq. 13.5 and by considering equilibrium, corresponding ro rrrc coordinates in Fig. 13.4c ¡*¿ åevoloped. .{tnr.:Ju*ð the strtlhess matrix of a shear wall element as shovm in Fig. 13.4d can be wri'tten
"*
I
I
EII
$
h'
6EI
r = ---L (l+a)
----;-
h"
. * o)If ,h
(¿
-t2E I --?_ h'
h-
'[t*;Ëål | ,tt
2
lzBt ------;-
6EI
shear deformations are included'
(r3.sb)
h,GA"
\ryALL ELEMENT STIFFNESS MATRIX- SHEAR as a vertical de"t'Ï1r]"]a;:i*Ïf;.i:-#1#3 A'shedr wall mav be treaæd
n=
(13.5a)
'
I3.4
(13
I
I
l2Bt C : -=--
øken to
l2 n lznz lrrlt
h2 ,tt j
(++a)nl
fiìã)l -orr
tri'ifJl"ü: ffi:ffii,iH'å;i,.ï; öi¿ "r"**'
,
-6Er I
-6EI
ht(z
' - o).EI 'h
thê axial deformations are included
:
I 2
(r3.6)
tzBt
-F-
+h'
'3
(¿*o)El 'h
4
j
STIFFNESS MATRIX. BEAM WITH RIGID ENDS
3D ELEI\dENTS DIRECT STIFFNESS METHOD-
486
EA
AE
h 0
G;;)r,(;)"-t
-AE h 0
0
12El 6EI
K6*6 =
s
t_
:;'l:
-lzBl
c-")F 6El
G*"tr''
o
(!1s)rt
i
trÐ-r
i 'lin
ol
(13.7)
h
-6Er
0
lzBl
G-")F'
6*a)h'
h
G-.")F
(z-o) Bl
FT;T
-6EI
ÊEIl"*
El'ì S':)
\ SU -EA o L -2Er l)-
0
(l3e)
/zett" ,, eI'¡
-;SD
(¿+q) Bl
c;;I
ffi.#.r,) 1 sr -sr'/ (2Btt.t, Isú -
o4
,
o laþ.41 \ sti sl: )
r,
4(t" *to)*tr)
(zerc', | EII (--+ stl frEl1Þ* SE SI:] \ 4lro *u,)
BEAM I
id""tü;ih
WITII RIGID ENDS 13.5 STiFFNESS MATRIX'BEAM
.t"., *"* are usuallv connected bv b3a'l,i1^11j$ll'r:ti5:':J:i#å"1; 'i" *Ìl:i'i:H"ïöä'äïi;ä;J"'1;'i*:"*mi*î,:Îî1ii:ïi:i:: ;i:i"*til Ï"#""ili'oää"Ji:. ,r'" iä" "r " l"-iil'"îtå'i*ri
**':ïll lä:l"",iJi'i'i':iÏii j:ä""ffi".Ë***:';'*:î,J1J'JH:'J,iîi;;ffi ffi ïi#ilié"ïff':i-#'"':'hËJ;;1*'::-1"":P,^*f *ï'J'J,ff ffi äil'i"'tr'J:ïJffii'i'åÏ'ui"'ìiî¡"*'1,*"9::*iî'5ii:.ï.îålli;iäT 'å:"liåä1",iï: trffi :i"LH'ìå:i"ä'tùiËÏ;'"9:i{**:::::*r*iïffi Â* at A and B are rerated to the
o
sr
l{-
stiffiress matrix becomes are ignored, that is cr = 0 the In case the shear deformations element. the matrix in Eq. 12.4 for a beam
SHEAR WALL
(s)
i¡
where,
t6
:
A at
{^}=tRltA*} or'À=RÂ* t"o l0 R= 0l 00
l,l
I
SHEAR WALL
Ïii;J":ji;":i:tri;:Ë.""ï:ï;ãl'pì""**"" A' and Bl by geometry as follows ä¡rpl""-*r""
487
(13.8a) DEFLECTEO
(13.8b)
jz
aauz
I'ig.
13.5 Beam with rigid ends
ri !./ $ì
¡'
488
Ê:'ij
e.'i $i.,'
Þ.t
Ï,'.
=
;¡r,..
ù.. ..9: ,,
As
h":
lIr
=
equivalent shear
2l(t
half width of wall/column at one end
half width of wall/column at other end clear sPan of beam
-rÍarea, v =
*l1L:l^:^1":^t*
tiTT:.tT e the internal * 9,: :i:"1"i":::fJ*ï:ì ^r. ¡ôncr'nr å'nu'vr'o^"1:":d'"litlJ:-T:::'-'::"::-:':iii äÏ:Jäi'il;ä'#;;ffi :'.::::;ffi i;Jä;;;;u't"år*ut"¡alrherero¡e'.:t'*l jio"::::,:'i':å j:i ffi ;;i:::ili'ï".ïä.";ri:r;iîT=,::i:'^'l.oîi**"î'lï;;i:iï: rhe stiinress matrix of a stepped member can be :::ÏLiïä;:;;;;ry;;ã.
''; i
,
obtained in two manners
s... l
l.'
n:'
:
analogy method is used to obtain The conjugate beam method or the column then obtained by inversion of the is the flexibility .;;i*. Th" stiffiress matrix
flexibititY matrix'
ill
2.Thesecond-"tt,o¿treatsthesteppedmember'asaone.dimensionalstructure in series and analyzes it by which consist, oi prir-uti. ."*b.r. connected acquired in
¡l
ù
Í,
segment. Thus, the knowledge already ,t stiffness ir "a.h " the previous cnapter is sufficient' beam shown in Fig. 13'6 using the Let us derive the stiffness matrix of a stepped ; ,""on¿ approach. There are a few simplifieations
ffi
lìl
I
r.' l|
l.
k¡z: -k' = kzl
(13.r rb)
t?l l-11 \A^E)¡
fne¡ = stidess of i rh segmedt ITJ, : n number oftotal segments simplifies the analysis due to reduction in size of the stifhess rnatrix. stifftress matrix of any segmþnt r between joints / and / + 1 is
tzBl -î-
i
6EI ÉEl
kt= ....t:
i i
'trE.i""'-'-6EI ----;-T
.........-...1=......j..
I?EI
--í-
-,=f
L-
i
zEI --6EI L,L ;
4EI
:
lr.: ;
::' I
LKr*t., : þ*
:
[Kl*
r, r]
1-¡* rJ
r
We can now adsemble the stifhess matrix of , say, first two segments :
H. l
fi..
f"li ":: i II |.; i "" i",,J,.,
* =l*rr iKrr*K" iK"l
hi 't:
Fig. f3.6
StePPed beam
(t3.t?a)
:
6EI rlLi
i where, [Kr. r* l ]
å, 't
þ,-
(13.1 la)
-
l¡..
I
-î-J, r-
The entire structure lies on a straight !ine' the coordinate transformations'
(13.10b)
å(#)
2.Theglobalcoordinateaxescoincideswiththelocalaxesatendi'eliminating
É:,
(l3.l0a)
u
poisson's ratio
,1 is expected othi-end. At certain sections, thã member
:
'='å(#),
kn =
strelelh required ùuv¡rõ¡¡' luË r-çtluulu often the Quite otren Qulle "f
fi.
to axial deformations are
P:
be
analysis because forces devetoped ar rhe ends due
;;;;ä;;in","r*r,n"
+ v)
13.6 STEPPED MEMBERS
ill
\l
L:
= l*ZU
,-
=
tu:
kz=
lir
t!' ,t.
t"
kr=
where',
)-i' J:_ l t'1
ttii
489
at the ends can The analysis of the system due to axial displacements
,ki
''r-
STEPPEDMEMBERS
3D ELEMENTS DIRECT STIFFNE{S METHOD'
(r3.r2b)
.¡"
þ:t
i
F:ll
ì
,;| Þl: u' 't,
I ,r
F:Ë, ì
¡*,r'.
i
þ¡.7:,
I
Eiu
t
F.r-¡.
l'
t'
Each sub matrix
.
l:
,,
F::'r:':t:
[:rl
I
þji..,.
h:':lì I ,
K.¡-r,.¡-z K¡-r,¡-r
.
:.li
Þii
,
l'll: [oJ |
þ;-i,,, li
I
^
:
$i ,
.where,
:
Fi..i,, F',ii'l
li:llr
It is similar to, (for any member)
P¡'
.
i
+ u¡ T - k¡i u¡ \.¡
I
I
Thus, Eqs. 13.16, 13.17 and 13.18 completely define the stiffrtess rnatrix of a stepped member in.local coordinates. Thp calculation of fixed end foúces follows the same
Example 13.1 Determine the stiffrress matrix of the stepped beêm shown in Fig. 13.7a.
(l
3.
l4a)
Thus r
.
ñ
¡'l
r ìt:., ffi;¡i ',, l,
,
(o)
K¡-r,¡-z K¡-r,¡-r
tb) Fig.
{r,l
[P,i,
IÇ,,t K'tu,
(13.16)
rjlp,J
13.7
force-deformation relation is gii,en by Eq.13. 13,
(13. I 5)
tlft'ì := _l*t -L-t
,h
/1,,
Ktrr K-tu,o
The cross-stiffiress matrix lq i md the direc th obtained by the equilibrium of the member'
If A:900
.4m,4mr
:
,
3. I 8a)
(13. r 8b)
,
suþ¡natrix Kr,, isa2x2sr¡þ¡natri: K*u isa(2j-4)x(2j-
Pr = [Klr lr¡¡ : [Klr -
(l
procedure.
The solution of this equation is
,
F_...i,
þ,.¡i
j ki¡ uij - h¡i k,j o jt
fp,l=Lr,,, [K,.' ix,."lf^ l*ì,,,Jlr {i'} to.l
rl
'i
b¡
kj¡= -h' jkt' k.i¡=-h'jk,j
uj-l uji
final matrix becomes
,
o
It can be rewrittèn in the partitioned form ás
ßl
..lt;
K.¡,¡
j-t,i
' Kir [n'ì l*" 1...11...:"' lol I*" :Krt : Kzt Kzz Kz¡ {ol=l ttt l,ll
:
.
l
Kj-,,i
local coordinate axes atj.
Pii : -
13.13)
uj,=o
column' Thus, eliminating tþe last row and last
F.'i ,' .::l
(
K
.
'
$.:1¡,r;,',
491
The equiliþrium equation of the member can be written as :
,"
ll2
Kz¡
rons aÏe Ifthe element is fixçd atj, the boundary eondit
r
*,'l
j'in
ul
[P¡'
Fill,
MEMBERS
(13'17) Pji = -hi¡ Pi¡ stress matrix relating the force vector from end i to end hi ¡ :
size'
0.
I
l0
þ,'
of 2 x2
Kli Klt Krz I(zt Kzz
t-;
.
Þl:1,,
l':.l
IÇ,
is
Inthis,n*",.**stiffuessmatrixoftheentirestructurecanbeassembled.The as shown below : 13 '6 ' ltis a tri-diagonal matrix segment numbers *" ,r'o*n ìn:Hg' uij Kii K¡l
F.i
Þr;ii
STEPPED
ELEMENTS DIRECT STIFFNESS METHOD'3D
490
;
the other end can be
l'l,l= l5::iÏir *,, K,i9l[:ii] l*,,
1.,1
i
iJ", I
[u,J lo iK¡, K,,jtu,,J I
nrr,
6Er,
",,= | ,tÈ1, .i'i,
LT;
I l,
I¡ J
-rzrr,
I L3t
eBr,I L2, I
L-?'
L,
f
x,' =l
-är, ,i,, l= *''' J
493
STEPPEDMEMBEP.S
METHOD- 3D ELEMENTS DIRECT STIFFNESS
4g2
...
"l= L3r
t+:.ry #.w1 4El, 4EI2
Krr = Kirr+Kitt = -erl, | r*,11?0, = 0.002, 400'
+lj, =ry 400"
I
*onr, L Ú, L-z "'- " ) - o.*,+ =zl!!900 = tzo
#='a#
ff Ki¡ = [oal'/ ,lr!, =
kj,
o.oor,
= 0.4,
: - ù,¡ k,, : -ljå, T]["är" ,lï,]
direct stiffness component at end j
k¡¡
=-
k¡¡ h
ii o"t".-in"
K-','= [;.H'lå,
If
end
j
where
= K¡
8¡'SJttJ-]
r(, Ki,', ,.,, =
- tlr K,ir
[-:;ï
- ¡*"*' ,.t* l-
,lrï"'ifii' #l
l'za
I on.r:i
ts i without the axial component Direct stiffuess matr¡x at end
,q,:
l3'l I'
9oo
^E the axlal component is stiffness matrix at end i including ki¡
o oI fr.sso'oo2l8 l'02
=lo f_ o
t.oz
method' Take Lr
:Lz-
400
cm,.Ir:64000
,i13,]
ôrr :
r
-
-
40!
*?x
,t*ooo'rt, 3
400+
s6x
lgg,
106
¿oo
|
58l87J
cma'
l3.l
using the
andrt=2t,'
40014001!-600ìl " fooo* 3 \4oo+sooil
Etz L ft + !, r.6671
= -3x2x64x103 -qrL91- * 64xl0rL 3 õrr : 2500'17 ñ^. -zt:
:
J
lx400x{9 * 990*+oo = .4ooo * 24xt0a z*oq7tú eo;ic z"'-- EIr EIa (-)
4'37s
deflection at d'o'f' fiiow appty a unit force at d.o.f. 2 and compute
u,,=å= _ùry-=r'.35 " l'5x9oo
The direct
:
['^ï' #J r:xï' ;;îl ri';,"
by Eq' The axial stiftress is given
)
and compute Let us restrain end i of the beam, and apply a unit force at d'o'f' l ' (Fig' 13'8 a'b) theorem, deflections at , d'o.f. I and 2 using the moment-area
Kr¡
:;î][;i,',ï
L-0.726
--o;zo] 34s.27
whose inverse will Let us first determine the flexibility matrix of the stepped beam the desired stiffuess maftix.
at end i is given by Eq'13'16' is fully restrained' stiffrress
\i
fo.ooztt
the stiffrress matrix of the stepped beam of example
fhn;;;;;;;;frtcient
,
_
i']
lrl
L*¡i
*,, = få?å +,o;'å¡,813^X1=f,'9?f ßÅ1
I
final stiffness maFix is given bY
*:ll" ä-t;]
-r.oz
+232.53)
is given by
¡r = l^;älì' ;;:å][i
EJz = rco
*'l = f:':?t
= f-o.oozts L 0.726
t î'tto' , 2x64x,ot
I
and2,
xl}a
: {-l Err ãn*io, 409 * 1q = 0.00e37 -lqo--- * 64 "22 x lo' ^ = 1S 2x644lo'^ EIr Elz = --
li,, :
200 x
--T *
69¡,.
4qq
Erz=
4s7s
F = fzsoo.tz-o.ooe:zJf '
f,::zs
*
24
4'37s
TR,ANSFORMATION MATRIX. 3.D TRUSS ELEMENT
3D ELEMENTS DIRECT STIFFNESS METHOD-
494
l'vaoo
1
+
¡._-
aoo cm
Õ_ ,
,{
-r
2x64000
.:
F¡-'
1/El2
1/EI1
3.125
64000
3.12s : f rsooo.ooe37j
1
lr.rzs
l-o.oo2l8 4.7281 \ : l-r.rrt 34s.7ol
The crosS-stiffrress terms k,, can be computed by considering the equilibrium of the as given by Eq. 13.17 and 13.18. The results are:the same as in Exirnple 13.1.
800
100
EIz
(00 cm
,
F:
1
=
1
100
ffi"' 100/ El
80000 : f *400x 400 + 600 + 240000 2 Elz EIr"¿00 - 64000 2 x 64000 ôu: 400x 1 400x
ô¿¡
495
o,K.
Elz
tol
$
-
LoAos
(b)
*
LoADs
7 TRANSFOR]VIATION MATRIX - 3-D TRUSS ELEMENT
a space truss is assembled from the various elements :the member stiffness matrices. The member stiffness matrix:must be expressed in of the structure or global system of coordinates before the stifftess can be
f:.'The systgm stiffness matrix of
1' v
The two reference
system,s
of coordinates are shown in Fig. l3.9. 'Also
in this figure is a general, vector A with its components along the global axes x', y'
+ 800/E I
10A
1
/'El2
x (LOèali 1/EI1
$
rcr
(d)
LOADS
A,/ å
4
LQADS
T1
Fig.
F-r=K:
or,
(ii)
lo.oo2l8 +l.olsl
l+r.olt
( GLOBAL )
13.8
o.K.
5sl.7BJ
a unit force at d'o'f' 3 and Now let us restrain end i of the beam, and apply d) (Fig' l3'8c' 4' and 3 the deflections at d'o'f'
"otp*"
u,, =
+
!00*ass.þ*.
#-ix400 _ 64xlo6 * n61l9i, x 1.555 = l5oo x 64000 2x640A0 åx400x 3
fr-q(ffi#)] Fig. 13.9 Local and global
axes in 3-D
xt
TRANSFORMATION MATRIX - 3-D TRUSS ELEMENT
METHOD'3D ELEMENTS DIRECT STIFFNESS Ð(es x A alo¡rg'one of the local components of vector
496
To obtain the necessary to add the
v'and' z components t*iäoä oi*' x-axis is given bv äïiä";;;ivectoi A'alongthe ' x ; x,cosxx' + Y'cosxy' + z'çosxz'
where, and
497
rfirst rotation consists of a rotation p about the 2' axis as shown in Figs. 13.9 and 13. l0a. ithe direction cosine terms in Eq. 13.19 a can be readily computed as :
y orz,it ' ' Thus, tht Ð(rs' that along
cos9 o sinÊl I Àu:l o I ol l-sinP 0 cosPJ
:
cosineofanslebetweenaxesxand{ cosxx' = cosÏl = axes x andy' : y' ;;; = i'; cosine otanãte between cos x : ;;i;; ãi"'ãr" between axes x and z' * o = ï:'lX "o' be written as :
(13.22a)
t:"t:]-L'"T y = fäsyx'+ Y'cosYY'+./cosYz) +lcoszl
z components of Similarly, the y and
and
"=.';;Li*¡cosiY' as : written in the matrix form *t *'l[.ì l*ì fcos xx' cos xY' cos lnf = I cos ¡x' cos YY' rz'l{v'f
be These equations caq
t;l
o'
cos
ü ti i
or'
or
l"o'
zY'
cos
(l
ill']
A = ?"4 J, Irl¡ iüru'¡ ----"-.i""tion ' ' cosines 12, mz, and
l] , Ï1
n,
are
:ru;i ;;
zz')lz')
.=[i
i
it gin"n
bv:l-l:l:?
Ko*o:RTo*o K6*6 Ró'o bY Eq' l3'l
,1r..,;
'
Asysremaricprocedurer",.,hi:l*11'-":..:l^*,*1îîïïåî:îrrl,,i.t;'riii
iää"-'d;îT:i'":'1":,::':"1ï*îffi::l rur.r¡v' separately three each other. Let li:Ì;,äî:;'få"liii#i'i:il;#Ë,ïîf;î'":ï":r::-ff the with :î:;:åå,:ri 19*:" svstem l'-2''3' whi'Jh coincides axes tocat a an¿ :;;;;';v;:;
xt
:
that i matrix can be written as berore' (13.21)
an expression for the 13'20' into Eq' 13'21 yields Substituting Eqs' l3'1 and for a space truss' ,tifftr.r, *otti* in global system ,.- E
,"å"äiiäli"ilåffi
r't, lltt
for the y-axis.
;l
1ffi""åffffi:ilj'rt#:i#.ï;itîíÁ",'
where, K is given
ROÍATION ABOUT 3,
T: z-axis. ãi'ection cosines ^^cineq for the z'axis' th; :l-^^.,^-
9l:-"i:3^TiÏr:*i:t:?l n roi u space truss is orthè rorm "rffî:*Tit"äi:'fi;,"Tä;;ää;il'äË* (13 :I
(b)
ROTATION ABOUT 2,
(l3.l9b
be d:tu*T-"-d Tr,"r"'¿ir"oion cosines mav
where each element
(o)
3. I 9a)
3,r,
zt
(cl ROTAIION ABOUT
1,/
Fig. 13.10 Transformation of axer Now let a rotation cr takes place about the 3'axis as shown in Fig. 13.9 and l3.l0b. rotated coordinate system is given by double primes. Thrr angle between the I " axis the plane
l' - 3' is a: The direction
Lo
| ":'"
= l-smq,
lo
cosines are given as
:
sincr 0l coscr 0
0ll
(t3.22b) I
498
¡p eleupñrs
DIREcT STIFFNESS METHoD-
TRANSFORMATION MATRIX - 3.D BEÄM ELEMENT
third r¡tation 0 take place about the axis l" as shown in Fig. l3.l0c. The rotated coordinate System is given by triple primqs which also represents the local axes of a element in space. The direction cosines are given as : Now let
a
o ol f., : fr cos0 l0
sinO
L0 -sin
0
TR.ANSFORMATTON MATRIX - 3-D BEAM ELEMENT
of a 3-Ð beam element is identical to the gencral in 3-D truss element. In a truss element, therç are thtod of freedom at a node, whereas, in a beam element there are six degrces ol'
The transformation matrix nsformation matrix used (13.22c)
I
cos 0J
The final rotation matrix is given by the triple product
per node. The resulting stiffhess matrix ând the transformation matrix aie of thc '12 x 12. For a beam element, the transformation of the nodal displacement vectors ve the transformation of linear and angular displacement vectors at each node. :fore, it requires the transformation of total of four displàcement vectors, two aÍ of the two r¡odes. The transformation or rotation matrix can be written as :
;
I : i,, Ào Ip
(t3.22d)
For a truss member, the orientation of the principal planes is immaterial. Hence, the
thirdrotationmaybeignored,thatis0:0. l"
499
-2"-3"
represents the local member axes x-y-z as shown in Fig. reduces to
lroool
írxesasshowninFig. l3.l0b t3.9. The final rotatiõn matrix
*,r-tr:
I = L.LF
lf ol ;
lo
f cospcosa sina sinBcosal : I l-cospsinø cosq, -sinpsincr L -sinÊ 0 cosp
(t3.2s)
L),,,,,
where À is given by Eq. 13.22. The individual elements of this rotation matrix
(13.23)
I
o
;l
are
te complicated. Therefore, a¡r alternate approach is used.to determine the direction
I
All
elements of this matrix happen to be in terms of the direction cosines of the x:axis, which are also the direction cosines of the member itself. These direction cosines may be " determined from the coordinates of the member as (Fig. l3.l and p¡g.'ù.siroflo*i,
/, :
cosrr
where,
X: _X: : :i:"i, *,
: ,#,
,llìzI\2t\2 u:
nr = cos r, =
{\x¡.-x¡l + (y¡ -y¡) + ,n.
surÞ:
and
= cosy2
sin cr
=
m,
\.:-.,)-
À=
cosp:
cos cr,
-/.rmr
E;î -nt
^n;":
(ß.24a)
(t3.24b)
I
e.ncs of a beam etement and L, a poinr noiysj, r."l1h-" kl,:T:.ï'.d:ll1*::" the local x-y plane as shown in Fig. 13.l i. ry9 Their coordinates (*,
,' z¡)
and
r given by l1
+
nr2
Ir2
+
nr2
=
lr2 + nr2
, yt , 4).
,
ï" The direction cosines of rocar x-axis arong
,
l.i
,
+ !.)
lt' + nr'1
lx¡t Y¡t z¡) (l3.lec)
h
r------:-
ltlf + n(
Fig. 13.11 L - node of a 3-D beam element
z,i)
tit; il;,n
(t3.24d)
nt
.J
(*t
(tt.24c)
,,!lr2
ml Therefore
3=
,.T!e nine direction cosine elements forlthe matrix rr,r, Eq. 13.19 are usually rlculated from the global coordinates of three points. rwã points are the t*o end, ol' beam element along ;the loc-ar.x-axis. and third point ir tui"n any where in the x-y 'I plane, where y is one of the principar axes of the cross-seðtionar area of the
,'¿{
:
erement
TRANSFORMATION MATRIX.3-D BEAM
DIRECT STIFFNESS METHOD'3D ELEMENTS
5OO
x¡
cosxx'
-X;
: lt = T,
Yi -Yi : -f, ^t
- = trr -
zi-zt
(13.26)
-L
ELEMENT
5OI
Y*:mln3-ntm3 Yy : nr lt - Itnt y, = lt m, - m, /r
Thedirectioncosinesofthez-axiscanbecalculatedfromtheconditionthatany to the plane formed by any two vectors along trre z-axis must be perppndicular no* I to J' along the x-axis' and L could be in the local x-y plane. These two vectors three vectors can be written as: these of poi* L. rr,ur, ,rr" cross-prdducl
u""*ri,
i
;ofr,"¡,, ñ
+: + + Z XxL +
Z =z*i+zyJ+zzK:
of,
I î*, iYi l*, -
Y¡
-
(13'27a)
z:-
;l
(t3.27b)
"il
l*r-*, yl-Yi 4-zil
'r= ffi, ^,
:'ù' ": : A
z'
(13.27c)
- Y)(zí- +) - @- 4)0r- Y¡) zy = (z¡ - zi)(x¡ - 4) - (x¡ - xi)@¡- 4) z, = (x¡ - xi)Or - y¡) - (y¡ - y')(xr- xi)
where,
4=
-and,
lzl=
(yj
cosines of the y-axis can be calculated Once the z-axis is established, the direction orthogonality' of using the similæ condition
'
Thus, where,
i1
and
++'+
(13.28a)
Y:XtxZl
i,
are unit vectors along the x and z-axes' respectively'
(13.28b)
Therefore, /, : ä, -,
Yv
lvl '
"r=#
(r3.28d)
y*2 +yr2 +yr2
by using-the following expressions and solving simultaneously
ry+inz2+ nzz:l l, l, + mzmr * nrn, : l, l, ¡ mrmt + nrnr:
0
:
(t3.2e)
0
conditions of orthogonality.
often the point L is selected as a known point in the structure which is placed on
y-axis, although it could be any point in the local x-y plane. The only is that this point L should not lie on the local x-æcis. If this point does not lie itself, its bounda¡y condition is given as restrained in all the six-degrees . Thus, as many t-points may be selected'as are required to define the of the beam elements used in the stucture, without affecting the storage or the band width.
I. A. (1967) Shedr Wall lnteraction, Portland Cement Association, Chicago.
(r3.27d)
,*z +zr2 +zr2
=
:ly, once the z-axis is established, the direction cosinès of the y-axis can be
arE again the
global coordinate axes x', y, and where, i , j and k are unit vectors along the by arcgiven z axis respectively.'Ño* the direction cosines of
cos",('=
lYl
(13.28c)
i, J. S. (1968). Theory of Matrix Structural Analysis, McGraw Hill Book
SALIENT FEATURES Routine
CTLAPTER
fourteen
ffi
503
The
program makes use of the dynamic dimensioning concept, which allocates to various variables in a single COMMON array A depending upon the of the given prbblem. The program checks the memory requirement with available size of the COMMON array A. In case, the memory exceeds, a message is and the execution stops. It can be restarted by increasing the size of anay A the variableMTOT.
STAP-3D COMPUTER PROGRAM
This program employs a total of eight files on the hard disc during the various stages ion. These files are oRen¡d and closed in the main routine. I
STAP
It is, in fact, the controlling routine for the entire program. It
reads the control data, the named COMMON statemenß, and calls the core subroutines. COMMON A consists of control parameters, whereas, COMMON blocks B and C are meant working arrays whích may vary in different subroutines. The important variables are
I4.I
INTRODUCTION is a general purpose program for the analysis program is ue'am oi tottt tñese elements. The
program) srApj3D (structural Analysis
of any struchry"
.onrir,.fli *r; ;t
element can be easily included in its element organized in' such u -*oä tf'* tty new matrix method as developed in chapters 12 library. The program is based on thé stiffiress typical test proþlems with input and ouþut are and 13. The user lnrt u"tioir along with of the STAP program is available on a provided in this chapter. ftã fOn:fneN-iittitrg floppy.
tit'3:11iîliî-:i H:
vr Prof' r wr¡llwr¡ bv .,rt*urdrrJ written originally
l'wirs.o1 universitv-of ltián on a Personal computer' The for^exect modified bean has frameiomputers, it ^, ---,,^- TL^ -¡ar¡am ia
FoYl1I::iïlî*,;T::il"ü'; ilË:;":'#u'ïilîñi"eu-r"r¡"*oft a user-.specifid.* f']:-1i-"111ïi"1i" from iead ã"''i' ffiücti'";ffiËiliñ" ll ili'ffi ;"iäiiriå,*n*;;;;;"sr"*readsthËinputdara,c:**r.:I:ÌTg_19,1 the .¡ft"-uto re-r-un and if the input data is correct öä;riil,lr.'ã¡".
"*-examine
CORE
.MBAND
band width
.INELTYP
number of element types total dynamic memory, number of equations number of load cases execution mode number of blocks Due to limited dynamic memory size, the equations are stored in blocks in a file. They are called blockwise for fr¡rther processing,
.MTOT
,MQ LL
MODEX NBLOCK
as and when required.
DUMP DUM
:
dummy or working variables dummy or workiúg variables ind_icator for starting location of different arrays in the dynamic storage array A.
FEATTJRES
The program is divided in
I.
number of paramete¡s number of nodal points
NUMNP
1Nl, N2, N3 etc.
program with the execution option'
I4.2 SALIENT
.NPAR
SE'MENt
NODES a number
of segments as follows
:HI"'"".ï:X'
2. ELEMENT segment -
:
STAP. NODES, ELIB, LOAD, GSTIFF,
subroutines TRUSS and TRUSSI subroutines BOUND, and BOUNDI
reads the nodal data and $enerates missing nodal data,
ifnecessary. The boundary ions are also read herein. It generates one anay tD (identity) which consists of tion numbers of all the active degrees of freedom. It is very helpful in locating any error or instability in the structure.
ELIB This is the elernent libtary subroutine and calls the desired elements. At present there three elements
:
Truss, Beam and Boundary. These
will
be discussed a
little later.
Ll' Frl þ-i. g\-.
504
ine PRINTD
I
Subroutine L'AD
Ft,
or both' or at the structure level either the element level The specified data' be may element The loads tiãt"""sponding are ïrã"Ji" and toa¿s svstem tevel global "blg*irl,rJ;ì ür il ttre The element il,ro"eñrh" ;;ñ; roads revel struch'e "r" read in this subroutine' D. These are t'he elemlnt level : case A' B' C' and at loads of or any There can be four cases toua' live load' temperature' J*". t" live n*""¿î'i-å'""' load' tr'" 9t1ã s¡ecified through e's' c and D to dead b"i"Jil a'sig ã in a n"]ìnîî'i"" user combined The "ri*î*"*ruil"".f, can be oth", toã¿
::It extracts the nodal displace¡nents and prints them
at
Hj{i
å) F-i.
t:
F, þ!i l
"uor". load or any imposed
r
öä';;
f+,,
[;
þi
load case
,ftlt"
fou¿."i
subroutine extracts and prints the member forces in the local axis for each . It calls the element library for transformation of element forces from global to
TRUSS/TRUSSI
"*"t
is a main subroutine which checks the memory requirements and calls TRUSS I . latter routine reads element data, generates missing data, if any, and generates stiffness matrix and load vector.
1ou¡
'corresponding to each
"u"' B' c and D for structural "i;;;*;ã 11' irätrtiJi"* and q2a ; loads' "rt*iäã';;;' for concentrated load case 2 are q" ' e22' Lzt tT:y:t l'o; qr, are I o,r;;;r;;ä 1o' : load case ' mày ue written as follows
subroutine FORCE again calls TRUSS to extract the element forces in local
+9lz B' +Qrr C' +9t+ Dl +Pl =9rt'Al I Structuralloadcase + qro D' '+ P' 2 = gztL2 + gzz}2 + bzt Cz +qro D3 +Pz 3:O' Or+ grz B¡ +q¡¡ C¡
Fí...
"o*åii'ul'iont
BEAM is a main routind which checks the memory requirements and calls BEAMI. called by FORCE, it helps extract the element forces in local coordinates.
BEAMl/STIFF
Þ;1.,
kl,
:
'
LL= qu.,ì Ar¡+
þir þ,r
where, 4,, B, ' C¡ ' Di 9¡
þ,i,,',
Br¡+
at element
ft 9r-1,: Cr¡
+
+ 9l-l,l Drr Ptr
to level in local coordinates'
to LL' and (i | -'"t"Á".,, bád multipliersI to1l-t-) concentrated loads (i =
Pi
' =
LL
=
j
:
þi,, Fri,'. þ:ii.'l þi il..
f¡l,
ine BOUND/BOUNDI
is a spring element for determining support reactions and imposing known on -the structure. It is also used to model an inclined roller support. The t data is read in BOLINDT which also generateS the stiffiress matrix. A boundary is defined by a single directed axis through a specified nodal point by a linear stiffriess along the axis or by a linear rotational stifÊtess about the axis. It be al.igned with the global directions to avoid ill-conditioning of the stiffrtess
I to 4)
important variables used in the program a¡e as follows
Subroutine GSTIFF
: :
ä.","J; *,*",' matrt ."Prylt--t*,,J:"'ï*"å1,i5;:"ff;åtiå
,,"#i*"ii,T."iJ"'Jii:ä.".-Jüïî;':n"1'3.*n:-';:"H;Hïi"i:i î,:ï:'."i"Í"i'J"iïi'ffi :'3:iiTï#'.:i#iï$:r$ïî:,iiåiÏ:"'#:$ or the specirred boundary Ë;'1Ë ::iå1-Ï;"ff "ffï#ïï'ä'";;äi matrix "::"*:g: Ïläil'i.'i ;;'*: ;;ded svmmetric stiffness It
for the nodal elimination method and solves is based on the standaid Gauss
displacements.
any, and generates transformation
STIFF. be
load cases' total number ofstruitural global axes are computed and of the element aiong the The loads due to self weight
Subroutine SOLVE
if
for each element. The stiffness matrix and load vectors are generated in the
added seParatelY'
þt'.,
reads element data, generates missing datA
;-'f"*edinglobalcoordinateswithintheprogram'
.' (i=ltoLL)
p..r''.,
f+i,
= loaids-
gr.l-,2
for each load case.
ine FORCE
-"i *:lt*pm':i:i:l ¡*;"* l, 2,""""""""' $iËi"ffii,l;få;,ïfä"-i""¿ it' Thgre ar'1no multipliers
and so on, the toad
F¡i.'
505
SALIENTFEATURES
PROGRAM STAP.3D COMPUTER
: ST, ASA: Y,Z,T
TTYP ME
=
:
location matrix array for an element which depends on its connectivity maximum number of stress or force çomponents of the current element degree offreedom oflhe current element max. degree of freedom of any element used in the library member stiffness matrices x-, y-, z- coordinates and T-temperatur€ arrays
member force aray material type number of elements in a given group
USER'S INSTRUCTIONS
506
STAP.3D COMPUTER PROGRAM
NUMMAT : = SF = RF : EMUL
The program accepts data in any set ofconsistent units.
number of materials in a given group fixed end force in local coordinates for the four load cases A,B,C'D' fîxed end force in global coordinates element load multipliers for self weight of an element along x-' y-, directions for each of the four load cases A, B, C and D.
Conrrol Data (415)
z
, Colurpns I 6ll _
J
Total number of nodes (NUMNP)
l0 l5
of different element types (NELTyp) Numberof different load cases (LL) Number
20
Execution mode (MODEX) 0 : Sòlution
I¿.3 ADDING NE\ü ELEMENTSTO THE LIBRARY
I:
A new element (one or more) can be easity added to the current library. It will require d'o'f' some changes in subroutine ELIB so that anew element iS called. If the maximum of the new element is 12 or less, the control COMMO-N blocks A and B as shown in subroutine TRUSSI remain unchanged. Otherwise, a suitable change is necessary in COMMON block B. There will be at least two,subroutines per element : a mâin routine and a working routine. The element geometry,.material property, connectivity, loâd etc' will be read in, stifttess matrix and load vector generated and saved on a ftle in the working routine. The band width is computed by calling subroutine BAND residing in
arguments.
Nodal Data (7 I 5, 3 F I 0. 0J5, F I 0. 0)
Columns
E
while executing
the program on
3. 4. 5.
no.
I. Number II.
I-5
Totalnumberofproblqms
7t-80
a
ifte coordinate axes correspond to the right hand thumb rule. ÌNodal data lines need not be in ascending order. If lines are omitted the nodal data f.orthemissinglineswillbeautomaticalIyg9n*ut"ausirigKN.
N is the increment to be added to the prwious node number. KN should be given on the last line of a generation r.qu"n"". ïhe lines so generated have the same boundary as the first rine in the "ondition, l€eneratlon sequence.
I - 80
Probtem Title
NOTE:
l.
means no generationdata of the last node must be given at the end
f.KN:0
Truss Beam
element : element :
Boundary element
Heading Line (20A4)
Columns
Boundary condition for rotation about x-æ
.
Data
of Problems (15)
Columns
1 ; åii,","". ......displacement in y-direction ......displacement in z-direction
56 -65 66 -70
::.The
DATA PR.EPARATION
Boundary condition for displacement in x_direction
36 -45
Personal Computer: The nanres ofthe input and output files. Do you want to print.generated nodal data : (YN) / Enter Y'or Y for Yes; N or n for Do you want to print ID matrix : (Y,N) The Identification matrix is helpfirl in ciiecking the generated nodal load vectors for each load case and in debugging, Do you want to print final nodal loads : (YN) Do you rvant to print nodal displacements : (YN)
Nodenumber
46-s5
I4.4 USER'S INSTRUCTIONS questions must be answered
I - 5 6 - l0
ll - 15 16 -20 2l -25 26 -30 3t -35
There are sufftcient comment cards in the current elerñent library and should bc helpful in developing new elements.
The following
Data check only
Load Case Headings ( I 5A4) ive as many liires as is the number of load cases. Columns I ' - 60 Title for load case I
the corã segment of the program. The location indicators Nl, N2, N3 etc. are to be appropriateiy def,gred in the main element routine as weli as in the core segment routine STAP. The nodal coordinates, ID array and material properties arê stored in array A which has to be carefully transferred to the {vorkìng element routine through the calling
l. 2.
507
The user may also indicate his/her name, units employed and date.
:
Give its data only if required. Give its data only if required.
Givc its data only if required.
'odal Load Data with respect to global øcis (215,6F10.0)
Columns
I -5
Nodenumber
USER'S
STAP.3D COMPUTER PROGRAM
508
- l0 ll - 20 2l - 30 3l - 40 4l - 50 5l - 60 6l - 70
6-10 lt -'15 16 -20 2t -30 3l - 3s
Load case number Load in x-direction Load in Y-direction Load in l-direction Moment about x-axis Moment about Y-axis, Moment about z-axis
6
KN is equal to node difference which must be constant for
2. 3. This set of lines, if
LoadCasps (4F10'0) Element Load Multiptiersfor Structural WIL '"'G;; defined in line type III. ; many lines *'tt " ní*n". of load cases load case A element I - 10 Multiplier for
dT= 1
II to VIII' where and J.
3-D TRUSS ELEMENT
DATA Set VI-l
A. Control Data (315) Columns l-
II 6
5
õ;;
I - 5 6 - l5 16 ' 25 26. - 35 36 ' 45
Columns
C. -
properties (NUMMA
;i";
I - 5
fit or initial
be
I
speoified using an equivalent
dT: r/crL
q':
dT: - P(øAE) coeffofthermal expansion, A
modulus of elasticity.
V/eightPerunit length Coèff' of ttrermal exPansion
Membernumber
prestress can
nodal data lines for nodes
prestress = P (tensile) which is releásed after the member is connected to the rest ofthe structure, then
Cross-sectional area
Mer,tber Pata (415, FI0'0J5)
_T*r
2
are the temperatures specified on the
wherè
I - iO I ' 20 2l - 30 3l 40
Columns
I
Modulus of elasticitY
+ve y-direction' Line 2 : As above for gravþ load in ¡þe +ve z:direction' ¡þe in load for above iravity lin. ¡ , As load case. ;i;;; , Multiplier otit .r*át loads to be considered for each element
D.
and
+T',
If initial
Materialnumber
Element Load Multipliers (4F10'0) i : Multiplier of gravity loads in'the +ve x-direction' Element load case A Columns Element load casP B I Element load case C Élement load case D -'
KN
change in temperature. If member is too long by r, then
B. Geometric Property Data (15' 4FI0'0) or material property' line is required for each different geometric
-
T'
The initial lack of
TYPeI
l0 Number of truss elements (NUME) l5 Number of different geometric/material
given generation
The data for first element and last element in the series must be provided. Automatic generation should be used only if all the elements to be generated in a group have the same material and geornetric properties. The change in temperature in a member used to calculate thermal loads is given by:
Multiplier for element load case B Multiplier for element load case C Muttiplier for element load case D
lines type IX. For nut problem, darry, repeat
a
*r : I¡ + KN
Ii
J¡*r : J¡ +
values.
' I I ' 20 21 - 30 3l - 40
t
KN- musf be specified on the first line of the generation seqûence. Member number to be generated must be one greater. than the previous member number. The node numbering is done as follows i
lóad case sequence ' The lines musi be in node number and loads only need be given' concentrated non-zero The daø for nodeshaving u"í, *"tt be terminated with a line. containing only
Columns
509
Node I Node J Geometry identifióation number Reference temperature for zero stress Member generation code KN
sequence.
NOTES:
l.
INSTRUCTIONS
Set VI:2
area
of
5
Type2
l0
Number Number Number Number
6li 16-
20
2t
25
t5
of beam elemenrs (NBEAM) of differénr marerials (NUMMAT) of geometric properties (NUMETP) of fixed end forces (NUMFIX)
Mdterial Property Data (15,3F10.0) Give as many lines as the number of different materials. Columns I - 5 Material number : 6 - 15 Modulus of elasticity i: 16 - 25 Poisson's ratio .,
cross-section and
3-D BEAM ELEMENT/
Control Data (5i15) Ìl:i... Columns I -
':
:
E:
youngs
(15'6F10'0) Geometric Property Data
Give as
,*y
tin",
Columns
local geometric PfoPerties about
ä-ti,à-';;ú of different
axis of a member'
I - 5 6 -15 t6 -25
26 -3s 36 -45 46 -55 56 -66
Weight Per unit volume'
26 ' 35 C.
USER'S INSTRUCTIONS
PROGRAM STAP.3D COMPUTER
5IO
t- 5 6-15
Geometric Property nuniber
Axial area
Al
associated
il;;;; . 26 -35 "ï;;;;;t
with shear force in local2-axis force in local3- axis
16 -25 26 -35 36 -45 46 -55 56 -66
Áã associated '"i't"tt"* about local l-axis. inertia iãttionuf
36 -45 local 2-axis 46 -5s liåä"ti "r ¡tt"rtia about local 3-axis about 56 -65 fntãm"nt of inertia
5l
F.E. Force in local 3-axis at node I F.E. Moment about local l-axis at node I F.E. Moment about local 2-axis at node I F.E. Moment about local 3-axis at node I
Fixed end force number F.E. Force in local l-axis at node J F.E. Force in local 2-axis at node J F.E. Force in local 3-axis at node J F.E. Moment about local l-axis at node J F.E. Mome¡t about local 2-axis at node J F.E. Moment about local 3-axis at node J
NOTES: must not Any value except shear areas
l. 2.
Shear area rn"" uãî"r"ã"J
,
ir
be sheal deformations need
for shape facrors: using the fouowing expressions
:nl.r:lä,1i1äilir'iåd for rectangular
be
31o' lnon'"ero¡ oniv
sections
:
l0(l
Thermal forces may be input through fixed end forces. Member Data
I - 5 6-t0 H -15 .16 - 20 2l -25
+ v)
Gl"-rlrt 6(r + v)
circulêr secttons
where
4. ,
36 -40 ,41 - 45 46 -50
principal directio¡s of amember' Local 1,2 and3 axis indicate
""'ðofuÁn'.' I - tõ
i - zo 2l - 30 3l - 40
Element load case A
Elernent load case B Element load case C Element load case D
NOTE:
loads but fixed end forces due to gravity This program does not comPute reactions'
ii::ri#iï{""ìrÍJri;fi,l'3ln
in rocar axis' Skrp u",ou" set or nrxed end rorces
there a¡e no fixed end forces'
t.-
5
6-15 t6 -25
Fixed end force number I f.E. For.. in local l-a
Member number Node I Node J Node L Material number Geometry number F.E. Force set number for load case A F.E. Force set number for load case B F.E. Force set number for load case C F.E. Force set number for load case D Member generation code KNi
KN is equal to node difference which must be Gonstant for
a
given generation
sequence.
KN must
be specified on the first line of the generation sequenie. Member nunrber to be generated must be one greater than the previous member
number. The data for first element and last element in the series must be provided. Automatic generation should be used only if all the elements to' be generated in a group have the same material and geo.metriç propertie¡. The details for speciÛing the L-node were discussed in section 13.5 and Fig. 13.4, Set VI.3
tPio'ol
Line I Columns
5t - 55
tl,t
Line2:n' uuol"" for gravitv 1"4 i" iL:1-ji::tti""' +ve ;i;; ã , A, ubone for !'anþ load in z'direction'
'
3l - 35
v = Poisson'sratio
r
l.
26 -30
C..t+ñ
bl.ElementLoadMutltipliersforGgvityLoads.(4F10.0) direÓtion' gtuJiq.r"ua Line I : Multiptier rtt il*"]x-
(l Iß)
Columns
BOUNDARY ELEMENT
Data (215)
Columns I -
.
6-
5
10
Type 3 Number of boundary elements
Member Data (51 5, 3F I 0:0)
,'Cslumns I -
5
6-10
Node N at which the boundary element is placed Node I , far end ofthe spring
I
STA?-3D COMPUTER PROGRAM
512
fl
15
t6 - 20 2l - 25 26 ' 35 36 - 45 46 - 55
: :
tó
2.
2:0 poins
loads at all panel pointi ofthe bbttom chord
rlIlrI 2 0 0.ìr
:
rot¿tion i5 sPecified Member generation çode KN
3
t5
l6
I
t29 I 20000.
N' ,
Thefollowingexampleswillhelpdevelopanunderstandingofthevariouscapabi of the program STAP-3D-
l4.l
AnalyzetheWarrentrussshowninFig'14.lusingthetrusselementinSTAP-3D.
. 0. 0. 0. '0.
2 3
Fig.
@ loo'32oo
l4.l
cm
Warren truss
Sotution in The node númbers are . shown .in circles and element numbers a.e shown l4'l given in Table are properties ms.nber boxes in Fig. 14.1. The
Tebiè Member no.
Ito14 t5 -2t 22 -29 Modulus of elastigity
l4.l
Member geometric þrdperties Area, cmr 30.78 5.68 I 1.36
E :20000
.Remark
2ISA 100 x
100 x 8
lSe 5ox 5ox6 2tSA50x50x6
kN/cm2
given earlier and The input datp prepared in accordance with the instructions
complete input and output are shown in EXI4I'FII- and EXl4l'DOC'
0. 400.
tll
I 400. I 2800. I 400.
I
I
0.
I
r 3200,
,..
I
l
2800.
3
30.78 5:6,8-
20000.
I l136
,0.,
0.
U.
0¿
20000.
0.
0.
0.
0.
.q.
0.
û
0.
0.
J lt 7t3 fr 8 15 l6 4 92 t4 t2 l4 . t5 2 2t 14 l5 22 I 2 f . 232 9 256 269 t0 28 13 l4 29 .14 t6 .2 I t0. 32 5 l. 52 7.2 9t 92 ll 2 13r 132 152 00 0:-0. 0. 0. J
I
l0-
IIr
l4
Specified displacement along member axis Specihed rotation about member axis SPring stiffness
I4.5 ILLUSTRATIVE EXAMPLES
Example
I
loads at only four
3:t:"Hii":åpecined
to The positive direction of the element is from nod-e,I If spring stiffriess is input as zero' it is taken as l0'u '
ì
Analysis of a Warren Truss units are kN+m Example 14.l
NOTE:
L
ILLUSTRATIVEEXAMPLES
I
Code for disPlacement 0 disPlacement not sPecified" I -displacement is sPecified
I
]
EXI4I.FIL
t0. 10.
I 10. 10. 2 0'.
20. J
30. 3
30. 30. J
r0.
15. -50. -25.
-25. -100. -25. -25. -75. -25. _25. 0. 0.
0.
400:
5t3
ILLUSTRATIVE EXAMPLES
STAP.3D COMP'UTER PROGRAM
514
EXI{t.Doc *
*È* * **ttÚ
ttú*t**
** **Í t** *** * i* t ** * * ** ** *tt * ii**
******
** * *t * * * i*t*
STAP'3D S'TRUCTURAL .AI{ALYSIS PROCRAM Example 14'l kN-cm a¡e units Truss Analysis of a Wanen
rr*t*l
t
tt*l**lt***+***l***
+*tta*t*lt*+*tltt
*rf ****t**+*+llt*++*
*
= ELEMENTfiPES NUMBEROFLOãDCASES = SoLUrIONMODE(MODEX) -
NUMBER OF
+* I I
lS
0
0
I
2
0
0
0
0
J
5
4
0
o
0
0
4
5
6
0
0
0
0
o
0
0
0
0
0
0
0
78 910
6
I
813 9
2 o
loads æ only foui poins loads at'atl panel poins of the bottom chord
DATA
BOTJNDARYCONDITION
t1 19 21. 23 25 27
ii: ló
0
0
0
0
0
0
0
0
0
0
0
0
l8
0
0
0
0
20
0
0
0
0
22
0
,0
0
0
24
0
0
0
0
26
0
0
0
0
28
0
0
0
0
0
0
0
0
29
.TIìMP
OFTRUSS MEMBERS .000
NOD.{L POINT COORDINATES NODE BOI,JNDARYCONDITIONCODES xYz NUMBERXYZXXYYTZ .000 .000 .000 Itlllll 400.000 .000 400.000 200llll 400.000 ..0ffi, -.000 00lll 3 800.000. 400.000 .000, 00llll 4 s00.000 .000 .000 o0IIII ) 400'00-0 -'000 1200'000 I I I 0 0 I 6 700lltl¡200'000'00p'000 E00IIIt1600'000400',000'000 900llll1600'000'000'000 r0001¡¡12000'000400'000'000 0 0 I I I .l 2000.000 -000 .000 ll c 0 t I I ¡ 2400'000 400'ry ^'0oo t2 130otlt12400'000'000'000 t400llll2800'000400'000000 .000 .000 t500III12800'000 .000 .000 160llltl3200'000
:29
OF D'FF. MEMBERS
.000 .000
TYPE
AREA
E
.000
WI/LENCTH
.000 .000
I
20oob.oQo
2
20000.000
TEMP
30.780 5.680 l r.360
.000 .000 .000
.000 .000 .000
LOADMULTIPLTERS
AB .000
x-DlR
.ooo
:Y-DIR
.000
z.DtR
.000
TEMP. COEFF.
20000.000
GENERATEDNODAL.DATA
zz
t2 t4 t6
15
l0 I,t t2 t'13 t4 t5
CODES NOIDAL POINT COORDINATES KN , Z Y Y Z XX YY 72 X NUMBER X rlllltl:000'000'0000 t I 400'000 400'000 '000 0 0 0 I 'l 2 l4-00IIII2800'000400'000'000v 3001111400'000'000\0000 150oIIII2Eoo'ooo'000'0002 160lltll320o'oo0'000'0000 NODE
0
16
ïi,',.' NODAL POINT TNPUT
YY
0
**+*+****+*
EQ.I : DATA CHECK
LOAD CASE NO 2
xx
0
')
EQ.O: EXECUTION
NO I lS
z
0
*** **l **i**
5
NUMBEROFNODALPOINTS
LOAD CASE
ËQUATIONNUMBERS N x Y
TEMP
c
.000000E+00 .0000008+00 .0000o0E+00 .000000E+00
.000000E+00 .000000E+00 .0000008+00 .000000E+00
D
.000000E+00 .00m0oE+00 .0000008+00 .0000008+00 .0000008+00 .0000008+00 .000000E+00 .0000008+00
.000 .000
I
J
TYPE
TEMP
.000 .000
I
I
3
2
3
)
.000
J
f
.000
4
7
9
.000
)
t6
I il
lt
.000
.0Q0
.
.000 .000 .000
BAND
LENGTH
WIDTH
I I
l3
l5
t3 t5
t6
2
4
l l l r | r I t I
.00
2
400.000
.00
6
400.000
00
6
400.000
.00
6
,t00.000
.00
6
400.000
.00
6
400.000
.00
6
400.000
.00..
J
400.000
.00
6
400.000
515
517
ILLUSTRATIVE EX.AMPLES STAP-3D COMPUTER PROGRAM
5t6
if ..:r.
ltl
6l Bl lol 12 l4
ro4
ll i28 lo t3 14 '12 152 164 176 I l8 19 lo 12 20 14 2l 22I 23'2 244 2s6 269 nll zB 13 29 l4 6
NODALLOADSTNBLOCKNUMBER I
.00
6
400.000
.00
6
400.000
6
400.000 I
.00 1
.00
6
400-000
I
.00
6
400-000
2
.a
.00
4
400.000
3
52 72 9' lt 132 152
.00
4
400.000
4
.00
4
400-000
5
.00
4
400.000
.00
4
400.000
.00
4
400.000
.00
4
400;000
.00
)
5ós,685
.00
E
56s.685
.00
E
s6s.685
.00
8
565.685
.00
4
5ó5.68s
.00
4
565.685
.00
4i
565.685
.00
f
565.685
2 2
a1
53 t3 93
ro
3
tzj 143 163
TOTAL NUMBER OF EQUATIONS
I}ANDWIDTH A BLOCK NUMBER OF EQUATIONS fN NUMBER OF BLOCKS
LOAD CASE
2
J
)
I
5
2
2
1
9
I
9'
2
ll
2
t3
I
IJ
2
t5
2
7 8
ilo i. ll
i;; :i
rz'
13
t¿
'!,
i it: ., í18 :; li" il. it
15
ló
t9 20
i¡:l
. :-iì,,
2t
)', 23
i24 APPLIEDLOADS
RZ RY RX .000 .o0o 10.000 .000 -25.000 .000 .000 -50.000 .000 .000 -2s'000 '000 .000 -25-000 .000 .000 -100.000 .000 .000 -25.000 .000 .000 -25.000 -000 .000 -75.000 '000 .000 -25.000 '000 .000 -25.000 '000
MX '000 '000
'000 -000
'000 -000
'oo0 '000
'000 '000 '000
26
My .000 .000 .000 .000 .000 .000 .000 .000 .000 .000 .000
ltiz .000 .000
LOADCASE I
ELEMENT LOAD MULTIPLIERS
¡BCD .000 .000 .000
'000 MA
-100.00000
-25.00000
-75.00000
-25.00000
.00000 .00000 .00000 .00000 .00000 .00c00 .00000 -2s.00000 .00000 .00000
25
.
-25.00000
.00000 .00000 .00000 .00000 .00000 .00000 .00000 -25.00000 .00000 .00000 .00000 .00000 .00000 .00000
17
:
=29 =8 =29 =l
-50.00000
.00000 .00000 .00000 .00000 .00000 .00000 .00000 -25.00000 -00000 -00000 .00000. .00000 .00000 .00000
o
'28 29
.0Q0
.000
NTS AND ROTATIONS
.Q00
LOAD
,.000
CASE
.000 .000
.t 2
.000
I
.000
2
.000
I 2,
S'TRUCTURE
l2 10.00000 .00000 .00000 .00000 .00000 .00000 .00000 -25.00000 .00000 .00000 .00000 .00000 .00000 .00000
6
::,: :li
NODAL POINT LOADS
NODË NO. 2l
LOAD CASE
TION NO
I 2 t
XY .00E+00 'ooE+oo .00E+00 '008+00 8.902E-01 -l.64lE+00 .000E+00 .008+00 .008+00 '008+00 6.0928-01 -¡.2g,6f,+00 .969l+00 .008+00 '00E+00 ¡96+00
9.ò828-01 ó.660E-01
.0008+00 .000E+00 .008+00 .000E+00 .0008+00 .00E+00
.00E+00 -008+00 -7.486r.-02 -l.64lE+00 .0008+00 .00F.+00 .00E+00 .00E+00 .008+00 .¡69f,+00 -1.6248-02 -l.ll8Et00 8.1228-01 -3.1258+00 .000E+00 .66g+00 .6¡g+00 .ggp+00 5.523E-01 -1.998E+00 .000E+00 8.t08E-02 -3.2848+00 .000E+00
.00E+00 .00E+00 .008+00 .00E+00
'00E+00 .699+00
ll
2
8.122F.-02 -2.130E+00 .000E+00
I
6.563E-01 -4.083E+00 .0008+00 4.5488{l -2-690E+00 .000E+00
')
l0
I 2
2.6638-01 4.2418+00 2.031841 -2J34F+00
.000E+00
4.7ilE-01 -4.670E+00
.000E+00
3.3308-01 -2.926F+00
.000E+00
t
4.807E-01 -4.6?08+0q .000E+00 3.330E-01 --2.9268+00 .0008+00 2.g24--Ol -3.979E+00 .0008+00 2.ll2B-01 -2.690E+00 .000E+00 6.951E-01 4.1728+OO -000E+00
I 2
I
4.630E-01 -2.734F+00.'000E+00 l.4g4E4l -2.930E+00 .000E+O0 1.r37E41 -1.998E+00 .00õE+ü) 8.?38E-01 -3.1239dÐ .000E{0
2
I .,
4
-1.
i¡ iir,l Ìri.;
5.848E-01 -2.130E+00 .000E+00 7.472E.{i2 -1.540E+00 .669E+00 5.686F.42 -1.206E+00 .0008+00 l.0l7E+00 -1.540E+00 .000E+00 6.823E-01 -l.tl8E+00 .000E+00
2
3l
ilri
2
il,
I
1ìi
7
frl
.TRIJSS
.0008+00
2
2
::ll
ILLUSTRATIVE EXAMPLES
STAP.3 D.COMPUTER PROGRAM
518
I
.000E+00 .g¡9f,+00
1
.000E+00 +.000E+00 .000E{ù
.0008+()0
.00E+00 .00E+00 .00E+00 .00E+00 .00E+00 '008+00 .00E+00 .00E+00 .008+00 .00E+00 .00E+00 .00E+00 .00E+00 .o0E+00 .00E+00 .00E+00 .008+00 .00E+00 .0oE+00 .00E+00 .çgg+00 .00E+00 .00E+00 .00E+()0 .00E+00 .00E+00 .00E+00 .008+00 .00E+00 .008+00 .00E+00 .008+00 -96g+00 .00E+00 .¡99+00 .00E+00 .00E+00
'008+00
STRESS
FORCE
.00E+00
.008+00 .00E+00 .008+00 .00E+00 .008+00 .00E+00 .00E+00 .ggf,+0o .008+00 .00E.{-00 .008+00 .00E+0Û .008+{0 .008+00 -00E+00 .00E+00 .008+00 .00E+00 .008140 .008+00 .008+00 .008+00 .00E+00 .008+00 .00E+00 .008+00 :008+00 .o0E+00 .008+00 .00¡l+0(l
*.
CASE
I
I
I
2
) )
2
J J
4 4
I I t I .,
3.73622 I15.00-l 2.84277 87.500 3.73622 I15.001 2.84277 87.500 7.t47s3 220.001 4.87332 150.001 8.93442 275.001 6.09165 187.501
9.259tt
28s.001
f
I
)
a
6
I
6
2
7.
2
I I
,l
)
6.09165 187.501 1.79730 240.001 4.87331 150.001 3.89865 120.000 2.84276 87.soo 3.89865 120.000 2.842'16 87.500
9
I
-7.14752 ,.20.001
9
2
4.87331
-1s0.000
t0
I
-8.93441
-275.001
I
2
-6.O9164
I
-10.72t29
-330.00t
2
-6.49775
I
-10.72t29
-200.00r -330.00t
2
-6.49775
-t
87.50r
-200.001
¡3
.l
t3
2
-6.09t64
-r 87.50t
l4 l4 l5 l5 ló
I 2
-7.79729
-240.00t
4.8733t
-t50.000
I
.00000
.000
2
4.40r4¡
2s.000
I
-9.683 l6 -6.60215
t6
MEMBER ACTIONS
MEMBER LOAD
l0
It
2
jzss.oot
-9.25929
-55.000
l7
I
-9.68316
-37.s00 -s5.000
tt
2
-2.20074
-t2.500
l8
I
.00000
.000
't8
2
.00000
.000
t9 t9
I
-7.92260
-45.000
2
1.20075
20
I
lt.gzzsg
-t2.500
20
2
-6;60215
-45.000 -37.500
2t
I
.00000
2t
2
4.40t41'
25.000
.000
22
t
))
-t3 î7 t57
-148.493
2
-123.i44
23
I
-10.89297 13.07t57
23
)
1.78070
88.389
24
I
6.84702
77.782
24
2
4.66843
s3.033
25
t
6.84699
77.782
25
2
t.55614
17.678
26
I
?.602t0
63.640
26
2
r.s56l6
17.678
)7
I
5.60212
63.640
27
2
4.66843
28
I
14.93892
28
)
29
I
29
148.493
53.033 ,
169.706
7.78069
88.389
-14.93892
-t69.706 -t23.744
-r0.89296
519
t4.z ,
Analyze the ùanen truss shown in Fig. 14.2 using the truss and boundary elements in P-3D.
ILLUSTRATIVE EXAMPLES
STAP-3D COMPUTER PROGRAM
520
0.
0.
0.
0.
0.
0.
0.
0.
q.
0..
0.
0.
3 t5 t6r 4 l4 3 15 23 s 93 t0 t4 163
,l ::, 8 ¡.. o
it.
@rY
,:t¡4
l3 l5 ?
t2
¡'l:
i::l 5
i¿r
l4 I
GLO,BAL ATES o
elements Fig. 14.2 Modelling of supports using boundary
Solution
,23
2
25
6
26
9
28
IJ
,29
l4
52t
10. 10.
10.
I 0.. 20.
20. 30. 30. 30.
J
Theboundaryelernentsareusedtoevaluatesupportreactíons'Thenodeandmember gloup al'e the same figure' The members in each numbering for the truss a." ,ho*n in unrestrained'or free as treated are 16 and I Ño¿. numbers numbered from
t on*arã.
.|hespringshaveinfinitestiffiressandthespecifieddisplacementinthedirectionofeach spring is zero. The r"n6r,
lengths' spring is äf th" tum" order as other member shown. in EXI42'FIL and EXI42'DOC' Three
ãr"*rt
The inpuf data and part output are astrisks
***
here' indicate that only selected output is reproduced
irl I
t8
!ìrI 6
l9
.:l;:,
¿
r
0 0 0
t0.
t.
0.
2
-25.
:l
I
i:::.<
2
-50. -25. -25.
.9
I
.9
-t00
2
15.
2.
-25.
I
2
-t). -25.
2
-25.
t
are kN-cm Btample Warren Truss wirh boundary element' units
l4'2
J
192-20 loads at onlY four Points chord loads at all pancl points ofthe bottom
I
I
i0
10. l 400. r 2800. I 400. I 2800. I 3200.
I
0
0.
0.
400.
0.
0. 0.
0. 0.
400.
0.
r -400.
I
00. 00. 00.
tlt 1
EXl42.FlL
l00lll 2o0lll lll l4 Slll I 15 i600lll lltll l7 I I t l8 llllt 19 I2gt t 20000. 30.78 2 20000, 5.68 3 20000. 11.36
I I I
t7
-400. I 0. t 3200. . -400.
*** *****
** * * ******
*
i *
q9*
0.
***É*********1t
r{*+l*t rrrr**+**++rr+frr+rf*+***********t**** OFNODAL POINTS
0.
t******
a* * t * * t * * ù** * *t*t
ANALYSIS PROGRAM STAP.3D T¡uss wiih boundary elernent, units are kN-cm Example
=19
OFELEMENT TYPES OF LOAD CASES
MODE(MODEX)
=0
**t t ****
14.2
r* ***+** *r+****
u*******
ILLUSTRATIVE
STAP-3D COMPUTER PROGRAM
522
E
I..AMPLES
523
EQ.O: EXECUTION
EQ.I : DATACHECK LOAD CASE
NO I IS
I.OAD CASE NO 2
loads at atl panel poins of the bottom chord
BOT'NDARY ELEMENTS
.fì i¡:.. '.:t: r
IS
NODE 'NODE DEFINING CONSTR. DIR.
i[,
NNI t11 l18 16
¡fl
Table 14.2 Member geometric properties
=3
NUMBEROF ELEMENTS
il'
ilThe node nurnbers are shown in circles and the member numbers are shown in squara ffes. The member properties are shown in Table 14.2
loads at only fow points
CODES
19
SPECF
SPECF. STIFF
DISPL
ROTAT R
KD I
KR
KN
U
0
0
.008+00
I
0
0
.00E+00
I
0
0
_008+00
.00E+00 .00E+{x} .00E.r{G
CONSTR.
I
1.00E+10
2
1.00E+10
J
A cmz
l,3A
66.7 78.46
2
:'
I;_" cm4
I" ClIt4
Remark
s37.7
13630.3
622.t
20458.4
ISMB 350 ISMB 4OO
"
Table 14.3 Fixed end forces Member
**t
Reaction
Momeng kN cm
Remark
KN
CONSTRATNT FQRCES
35.
I
NUMBER
I,_, = J cma 32.4 46.9
The member local axes are shown in Fig.l4.3b. The equivalent nodat loads in the members are shown in Table 14.3 with respect to Figs.l4.3c and 14.3d.
NUMBER 1.00E+10
Bearn
LOAI)
FORCE
MOMENT
CASE
2
ll l2 2l
-.10000E+02
-.r7108E43 .t0500E+03
))
3l 32
50. 50. 26.5
.00000E+00
'.00000E+00 3 .00000E+00
.87500E+02
.000008+00
.120008+03
.00000E+00
.875008+02
.000008+00
2042.00
35.
-
end i
-2042.00 4 I 67.0(anticlockwise) 4167.0(clockwise) -2992.5
The input data and complete ouþut are shown in EXI43.FIL and
Example 143
; / i.,:
,:,.i ii
@@ u-xooe
load
Yl t-
Ir 20
20kN/m
3-Sm
-r- l'5 -l-
5m
|
(o)
CONTINUOUS BEAM
I' l l'l
iilll ')1,
"Ì
lil.,;i
(b
I,
'2 .2
ùrî-rúi*l+"
l:rll
I
tt lt tl
3
,
cm Example t4.3
ll0
l:'i
-,;'
beam kN and
)
MEMBER
(c)
LOCAL AXIS
,- l'5 -r- 2m J
4
I I I
6l 24 l 2r000. I 66.7 2
78.46
I
II l2'
l
r
I r r r I
t0.
o.
0350-
0-
0 850.
0.
0
1350.
0.*
0
1550.
o.
t0.
t00.
J
0.
'0.
0.
0.
32.4
537.7
r3630.3
0.
0.
46.\
622.t
20458.4
0.
0.
0.
0.
0.
0.
0.
0.
0. 35.
Fig. r4.3
I
35.
0. 0.
j
2782.5
28t5
Analyze a three span continuous beam thown in Fig. 14.3a using the beam element of STAP-3D
end
0.
2M2.0 -2û42.0
EXl43.Doc.
ILLUST'RATIVE EXAMPLES
STAP-3D-COMPUTER PROGRAM
524
20.
,
0
0.
50.
4167 .0
50' 3
0.
I]EAM GEOMETRÍC PROPERTIES
4161.0
0.
l. I 2 '6 6 2 2'3 4 6 3'3 4456 0. 5 10. 00 0. 0. t.
I'LEMENT AREê. SHEARAREA
278i2.s
0.
0.
0.
26.5 I
I .l
I 2
I
1)
I
l3
I
I
0.
0: '
0.
t23 66J00 7E.460
ryPE
-2992.5
28.5
-4000.0
r* * ir* t++*****
*
A -DIR
*+
*+r
**
rr**t**
STRUCTURAL ANALYSIS PROGRAM S'TAP-3D . continuous beam kN and cm Example l4'3 t++***t**t*f +rt*** ***tt*f**1** + * * l r + t*++*+*t*r
NIJMBER OF ELEMENTTYPES = =t NUMBEROF LOADCASES SOLUTION MODE (MODEX) = NUMBERoFNODALPOINTS
irl ì
t+*+++*t***+ f*
*
*t +** ++**tt+*
*
li;. li,,
I .000 .000 .000 .000 .000 .000
FORCE
t* I *
6
I 0
EQ.I:DATACHECK I IS dead load
I]ËAM
LOAD CASENO
NO
,i'
INERT'l/.1
32.400 537.700 t3630.300 46.900 622.t00 20458.400
.000 .000
B
,C
D
.0000008+10
.0000008+00
,000000E+00
.000000E+ú0
.000000E+00
.0000008+00
FIXED END F€RCES IN LOCALCOORDINATES
ryPE NODE **rt+i+**t*++
INERTIA
FORCE2 FORCE3 MOMENT
35.000 35.000 50.000 50.000 26.500 28.500
.000
I MOMENT2 MOMENI3
.000
.000
2M2.000
.000
-2M2.M0
.000
.000 .000 .000 .000
.00p
.000
.000
.000
1992.500
LOADS.
LENGTH
.000 .000 .000
.000
4167.000
.000
-4r67.000 2782.500
EQ.0: EXECUTION
..
',,
t****it****t**+***
AREA INERTIA
J
.000 .000
.0000008+00 .000000E+00 .000000E+00 .0000008+00 .000000E+00 .000000E+00
X-DIR
0.
t** ****t **f***+*
SHEAR
III-EMENT LOAD MULTIPLIERS FOR GRAVITY LOADS
EXl43.DOC *
s25
NODAL POINT INPUT DATA NODE BOUNDARYCONDITIONCODES
NUM.XYZXXYY I
iìi, J
4 f
ó
ITI¡{EE
lllll 0ltll 0llll 0llll 00ll:él lllll DIMENSIONAL
BEAM
I
l
I
NODAL POINT COORDINATES
ZZXYZKN'TEMP .000 .000 .000 r 350.000 .000 .000 0 850.000 .000 .000 0 0 1350.000 .000 .000 0 t550.000 .000 .000 .000 [email protected] .0c0 I
0 0 0 0 0 0
.000
J
J
4
4
.000
AL NUMBER OF EQUATIONS
.000
:DANDW¡DTH. ,ñutr,lesnop EeUATIoNS rN A BLocK
000 .000
OF B,LOCKS
NUMBEROF BEAMS NUMBËROF MATERIALS
I
LOAD
NUMBER OF CEOMETNC PROPERTY SETS NI.IMBER OF FIXED END FORCE SETS
2
CASE
MATERIAL
. r
S POISSON S MODULUS RATIO 21000. '00000 ,YOTJNG
WEIGFTT DENSIT,Y .00000
350.000 500.000 s00.000 200.000
=9 =5 =9 =t
POTNT LOADS
4
3
ELEM
ABCD 1000 2000 3000 0000
000
NODAL
ELEMENTS
.,
2
NODES MATL CEOM JLNONO 26tt 3612 46Il 56¡t
t
APPI,IED LOADS
RX .000
RY RZ .000 .000
MX
MY
þa
.000
.000
-4000.000
ELEMENT LOAD MULTIPLIERS
A B rC r.000 .000 .000
D .000
!1'
STAP.3D COMPUTER PROGRAM
526
NODALLOADSINBLOCKNUMBER
i.
I
it
I
r 2 3 4 5 6 7 8 9
-
Ii
',
!.t.
ri
@
Fig.
1384.50000
2992.50000
I
.00000
oontinuous beam with boundary elements kN and cm Example 14.4
9210
.00000
-4000.00000
I I | I | l
.000E+00
load
llrl 200t 301 401 5t 6lll TtIl Sttl 9 I, I 24t2 I 21000. 0. t ao.t 0. 2 7E.46 0. 0. 0. 0. r 0. 35.
.000E+00
.0008+00 .008+00
.00E{0
'008{{)0
.000E+00 -4.191E40 .0008+00 .00E+00 .008+00 -3.49E 03 .0008+00 .000E+{n .0008+00 .00E+00 ' .008+-00 -6.98E 04
.0008+00 .0008+00 .0008+00
,:;
,:tl
,
t::,
\;
a
.000E+00 .000E{{)0
.0008+00 .00E+00
.00E+00
5"16E'04
.000E{û) .008+00 .g6p+00 -4.498-M .000E+00 .000E+00 .008+00 '00E+00 .00E+00
BEAM FORCËS AND MOMENTS
BEAM LOAD
NO. NO. I I 2 3 4
AXIAL SHEAR R2 RI
.9699+00 .000E+00 .000E+00 .000E+00 .000E+00
14.4
EXT44.FIL
.00000
L" ;,, j
tr
.00000
CASE
¡t:,' i,:'r
@
E
.00000
NODELOADXYZXXYY ii:.
@"
-2125.00000
NODE DISPLACEMENTS AND ROTATIONS
6 5 4 3 2 I
Íx7
1x
CASE
EQUATÍONNO
ILLUSTRATIVE EXAMPLES
@@
SHEAR TORSION BENDTNG
R3
MI'
2.37¡þ+01 .0008+00 .0008+00
4.129F+i1.0008+00 5.0698+01 .0008È00 I 4.931E+01 .c00E+00 2.525F,+01 .0008+00 I .000E{0 2.9758+01 .0008+00 l000E+00 -1.3688-07 .0008+00 .000E+00 7.368E-07 .000E+00
BENDING
M2
.0008+00
M3 l'308E+03
.000E+00 .000E+00 -3.5llE+03 .00081-00 .0008+00 3.5llE+03
20.
Solution The node a¡rd member numbering are shown in the same figure. Nodes 2,3 and 4 are unresrrained nodes with zero displacement in the ver'tical direction. The input and part output are shown in EXI44.FIL and EXI44.DOC
I
I I I
t I
I
10.
3
0.
0.
0 850.
0.
0 1350. 0 t550.
0.
0.
10.
t00.
l
350-
-200.
t
850.
-200.
I
1350.
100.
0.
0.
.
0. 0.
0.
0.
0.
0.
0.
32.4
537.t
46.9
622.t
0.
t3630.3 2M58.4
2M2.0
:2042.0 0.
4167.0
4t67,0
26.5
0.
6
I
ll
t
I
J)
I
I
2
2
)
3
J
4
6
I
l3
5
6
I
I
I l r
0 0 0
J
0.
J
28.5
4
0.
0 350.
50.
.000E+00 .000E+00 -4.0008+03 -0008+00 .0008+00 4.0008+03 .000E+00 '000E+00 -4.0008+03
Analyze the continuous beam shown in Fig. 14.4 using the beam and boundary elements of STAP-3D.
I
50.
3.165E+03
Example 14.4
I
35.
.000E+00 .000E+00 -3.165E+03
.0008+00 .0008+00
I I I r I I I l r
I
2782.5
-2992.s
3
2 J
E
4
9
.f
I0.
0
0 0.
0.
00. 00. 00, -4000.0
ILLUSTRATIVE EXAMPLES
STAP-3D COMPUTER PROGRAM
528
'l¡(
EXt44.DOC *
** **
****
****
**
****
**
** *t * **
**
***** ** ***
**
** **
**
** * ** * *****
*
*******
*
***** ** t*
STRUCTURÁL ANALYSIS PROGRAM STAP.3D
l5 kN
continuouíbeamwith boundaryelements kN and cm Example 14.4 *** +* ** t+**t*** **+t*****+È** +t*rt r*t+**r r****ú *r r+ r*r **** * tt ++*+**t+i
*
++*
23 L-NoDE
19
tl**
ts
=9 -a
o o
:I =0
NUMBER OF LOAD CASES SoLUTTON MODE (MODEX)
å
o o
EQ.O: EXECUTION
t5
E
tr
@
E
E
BOUNDARY ELEMENTS
NUMBEROFELEMENTS
:
@ I
3 L
NODE NODE ìii N 2
iji' '1
DEFTNING
CODES
CONSTR. DIR.
iiì.
.
5
ii,
4
ti.
rr
NI 710 810 910
KD
KR
SPECF.
SPECF.
DISPL
ROTAT
KNDR 0 .008+00 '.00E+00 0 .00E+00 .00E+00 0 .00E+00 .008+00
-
22
CONSTR,
, 500 cm f-1.------{,f
NUMBER S t.008+10 I 1.008+t0 2 t.00E+10 3
(o)
@ 400
5-STOREY FRAME
t8
@
[õ]
12
ø
8E
7
i*t
@
,Fm
t0
EQ.l : DATA-CHECK,
Ø
É.t
13
@
2l
rz lìîl
tiil
E
20
f5l
t6 NUMBER OF NODAL POINTS NUMBER OF ELEMENT TYPES
529
9,
E c ft-l
E
EI 6
E x2 2-€Þ- z .-l---2.il|, t3
NOOE
ts
@
I
3
t+
sm
, (b¡
13 tI (m l¡
NODE AND MEMEER NUMEËRS
ttt CONSTRAINT FORCES
NUMBER LOADCASE
I .2 3
I 1 l
FORCE
MOMENT
.9¡.99(f,+02
.6¡gggg+00
.745598+02
.00000E+00
.297508+02
.00000E+00
Example 14.5 Analyzn a six-storey two bay frame shown in Fig. 14.5a using the beam elentents of
STAP-3D.
Fig.l4.5
-l
For column members (13 -24), the local 2-2 axis is along the global x-axis, the local axis is along the global y-axis and the local 3-3 arís is along the global z-axis, that perpendicular to the plane of the paper. Node 22 having (=500, 0) coordinates and restrained is the L-node for all the above column elements. For columns 25-30. the of moment of inertia are specifiéd"in Table 14.4. The L-node remains same as 22.
Solution The node and member numbering are shown in Fig. 14.5b. The columns 13-24bend about'their strong axis and 25-30 bend about their weak axis. The member properties are shown in Table 14.4.
.
The member axis orientation is specified through L' node on the memb'er data line. Foí beam members, the local 2-2 axis is along the global y-axis, the local l-l axis ls along the global x-axis, and the axis of bending (local 3-3 axis) is perpendicular to the plane ofthe papier. Point L is chosen in the x-y plane but not lying on the local l-l a¡ii¡ of any beam. .Thus, node 23 having (0,2500.00) cooïdinates and fully restrained is the L node forall the eight beani elements.
Table 14.4 lllember geometric properties Element
Rectangular
Aiea cmz
rr-,
r350
cma
Ir-,
cma
sectiorr cm
Beam l-12 Columns l3-24 Columns 25-30
30x45 30x35
t01250 78750
227800
1050
35x30
1050
107190
78750
107190
The dead and live load intensities are 20 kN/m and l0 kN/m, respectively. The fixed d forces for beams of different spans are shown in Table 14.5.
ILLUSTRATIVE EXAMPLES
STAP.3D COMPUTER. PROGRAM
530
531
Table 14.5 Fixed end forces Fixed end force set
m
kN
t *L' t2
I
kN
storey-two bay frarne liN-cm Example 14.5
"t
I
)
Dead
2
5
Live
J
4
Dead
50 25 40 20
Live
4
4t66.67
+ live load
2083.34 2666..67 1333.34
+ live load + 0.5 x earthquake
of 15:456-1978 code specifies the'short term rLodult¡s of elasticity
-
where o"u
:
ol'
Due to creep, the rnodulus of elasticity gets reduced depending upon the age ol' concrete at the time of loading. It is, therefore. desi¡able that the modulus of elasticity shpuld be modified appropriately.for creep. for use in the structural analysis. The effective modulus of elasticity of Ml5 grade concrete at 12 months age is given as follows
where 0
:
l+l.l
:
10500
Mpa =
0ll
20
oir
6
1050 kN/cm2
2l 3l 22r 23t I' 230 r 1050. l r350.
i;i
The frame is analyzed for two load cases
T:i.
Case
iiil
l-
:
500. 0 900. 900.
I 850.
350. 1850.
10.
0.
¡ 500.
0.
1900.
0.
I -500.
0.
10.
2500.
0.
0.
0.
l.
10125,0.
t.
787s0.
t0719 0-
3
1050.
0.
0.
l.
t07190.
78750.
0.
0.
0.
0.
0.
0.
0.
0.
0.
10.
30.
4t66.67
4166.67
2Aß34
25.
-2081 .34 0.
0.
0.
2666.67
-2666.67
40.
20.
:
4 + L0 x Element load Case B
Element load Case A + 0.5 x F,lement load Case B
The nodaL loads for case 2 are directly added to the corresponding element loads within the program. The input data and'part output are given in EXI4S.FIL and EXr4s.DOC.
1333.34
zoì. 4
ó
t9
l2
20
l3 t8
l6
f
The structural load data is specified as follows
25. 40.
I
I
is
2
24
t7
25
J
30
rE
i
220. 220.
227800.
0. 0.
50.
40.
Element load Case
I 850.
0.
20.
The dead and live loads are specified through the fixed end forces on tåe element data. Element load case A reprdsents dead load and case B represents live load. The lateral loads are specified through the nodal load data for load case 2.
x Case 2- 1.0 x
350.
50.
Dead load + live load
1.0
0 500.
0.
Case2- Dead load + 0.5 live load + lateral load
Case l-
350.
1050.
creepcoefficient
but not the member forces. In most cases, we are interested in the member forces.
ltI tll tll tll ltl 134
00.
2
It is interesting to note that modulus of elasticity value affects the frame displacements
ili
II ll
MPa
5700'Jl5
ll
)
characteristic strength ofconcrete at 28-days
E* : E" l+0
I 10. l I l | I I I' I I
40011 l9
concrete at the origin ofthe stress-strain curve as follows:
E.:5700./o"¡
20
23'l
-
4
Clause 5.2.3.1
*L. 2'
Load
Span
523r 2023t 623r 2t23t 422 t9221 522 20221 622 2t221
11200 tl2 i; 34 134 I
z 1
I
2 2
I
3 J
-t333 3
J
.34
STAP-3D COMPUTER
532
FIXED END FORCES fN LOCAL COORDINATES
l0 220. 13 . 220. 16 220. t9 2 15. 00 ll.0.0. 0.5 l.
,'rVpf
NOOE FORCEI FORCE2 FORCE3 MOMENTT MOMEN'T2 MOMENT3
.000 .000 .000 .000 .000 .000
ll
0.
0.
.oóo
EXI45.DOC Ë
**
+* r+ * *
****r* t*t+***+** Êt*t** ***t * +****+*** ********t*
*r**È* **+***r+++++**ttt
= NUMBEROFELEMENTTYPES = NUMBEROFLOADCASES = SOI.UTION MODE (MODEX) =
íNO
23
r. 3 ',4
I 2
:s t.6
0
EQ.0: EXECUTTON
EQ.I:DATAdHECK LOAD CASE NO I IS dead + live load LOAD CASE NO 2 lS dead + live load
ti':
+
BEAM
ELEMENTS 30
MATERIALS
: NUMBEROFFIXEDENDFORCESETS =
I
S MbDULUS 1050.
"13
I 4
POISSON S
WEIGHT
R.ATIO
DENSITY
.00000
.00000
BEAM GEOMETRIC PROPERTIES
ELEMENT ,ARI]A
il,.
SHEARAREA SHEARAREA
TYPEI23I23 1350.000 .000 | 1050.000 .000 2 1050.000 .000 3
INERTIA INERTIA
1.000 l-000 1.000
INERTIA
101250.000 227800'000
78?50.000
107190.000
107190'000
787s0.000
c
AB
i1:i
X-DIR
'iÌ,,'
I
Y-DIR Z.DIR
.000000E+00
.000000E+00
tL
47 710 ir. 15 13 tó .t0 t7. t3 ¡6 16. 19 r8 25 l9 58 .20 8 ll 2t ll 14 22 't.. ìi: t4 l1 zl t7 .2o 24 .: ?{ 6 .3 69 26 27 28
I]T,EMENT LOAD MULTIPLIERS FOR GRAVITY LOADS
),
il
.000 .000 -000
IJ 45 78 t0 lI ¡3 l4 16 17 19 20 )o 89 12 lt t4. 15 .t7 l8 20 2l
{4
3
NUMBEROFGEOMETRJCPROPERTYSETS
YOUNG
l0
ä: il
NUMBEROFBEAMS = NUMBEROF
7
I I
0.5 + earthquake
ìr 'THREE DIMENSIONAL
NODES
BEAM I
NUMBEROFNODALPOTNTS
D
.000000E+00
.000000E+00
0000008+00
.000000E+00
.000000Ei00
-.000000E+00
oooooðE+oo
.O966ggg+00
.000000E+00
.000000E+00
nnn .000
2o.ooo
.ooo
.000 20.000
It+
STRUCTURAL ANALYSIS PROGRAM STAP-3D Six storey- two bay fiame kN-cm Exampte l4'5 $+t++*****+**i****Ë+++'**+*+*+++****+*+*++*+*t****t+ll***tt+*****+****l*+*t*+ttlt
MATERIAL
533
ILLUSTRATIVE EXAMPLES
PROGRAM \
.29 30
sri
t2 15 t8
15
18
2l
.000 .000 .000 .000 .000
.000
fìnn -000
OnO 000
.000 .000 .000 .000 .090 .000 .000
.000 .000 .000 .000 .0Q0 .000 .000
ELEMLOADS MATL GEOM LNONOABC I 2'-0 I I 23 llt20 23 lll20 23 llt20 23 llli20 23 23tlt20 11340 23 11340 23 11340 23 23r1340 11340 23 23t1340 12000 22 12000 22 12000 22 22r2000 0 2 d 'o t 22 0. 0 .0 2 I 22 0 0 2 0 l. 22 l2000 22 12000 22 12000 22 22t2000 12000 22 13000 22 0 0 l 3 ,0 22 13000 22 0 0 0 1.. 22 13000 22 13000 22
(t7O 416(\ 4166.670
4166.670 2083.340 -2083.340 2666.670 -2666.670 1333.340
-1333.340 LEbIGTH D
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
500.000 500.000
s00.000 s00.000 500.000 500.000
400.000 400.000 400.000 400.000
.400.000 400.000 350.000 300.000 300.000 300.000 300.000 300.000 350.000 300.000 300.000 300.000 300.000 300.000 350.000 300.000 _ 300.000 300.000 300.000
300.000
ILLUSTRATIVE EXAMPLES
STAP-3D COMPUTER PROGRAM
534
@@
BEAM FORCES AND MOMENTS
LOAD AXIAL Rl NO. NO. I -l -098E+01 I 1.098E+01
2
2
I
_2.t778+00
|
2.1778+00
2
3
|
7.147E+Ol 7.853E+01 2.834E+01 9.666E+01 7.433F'+01
3.084E-02
-t-6108+01
7.567E+Ol 3.69$f,+01 8.812E+01
_1.239þ+00
x.5738+Ol
l.6l0E+01
2.
7.4278+Ol 4.369E+01
4
I
4 5
2 I
1.359E+01
9.596F-02
7.321lË+Ol
-9.596F-02
6 6
2 , 2
' .000E{0
5.0318+01 .000E+00 7.4698+0l .000E+00 7.3661+01 .000E+00 7.lg4B+ol .000E+00 5.707E+01 .0008+00 6.7938+0l .000E+00 7.3t4E+01 .0008+00 ?.6868+01 .000E+00 5.77gF.+Ol .000E+00 6.72tF'+01 .000E+00
1.499f,+0t -1.489E+01
_6.592t+00 6.502E+00
5
.000E+00 .966þ+00 .000E+00 .¡gQf,+0o .000E+00' .000E+00 .000E+00 .0008+00 .0008+00 .000E+00 .0008+00 .000E+00 .966f,+00 .000E+00 .000E+00 .0008+00 .000E+00 .000E+00
9.ll7E+00 -9.1 178+00 2.576E+01
-2.5768+01 3.222E+01
-3.2228+01
.699f,+00 .0008+00 .000E100 .000Ë100 .000E+00 .000E+00
M3
.000E+00
4.5 l3E+03
.000E+00
_6.279þ+03
.000E+00
_4.9968+03
.669f,+00
-l.2l9E+04
.000E+00
5.339E+03
.000E+00
_5.672i+03
.0008+00
-2.3298+03
.000E+00
-1.048E+04
.000E+00
5.6391+03
.gggp+O0
-5.267E+03
.000E+Ò0
ó ¡l
6
dt
_8.836E+03
'
5.8578+03
-4.964E+03
.0008+00
l.l8óE+03
.0008+00
-7.283E+03
.0008+00
6.2548+03
.000E+00
4.7228+03
.0008+00
3.0748+03
.000E+()0' .000E+00
-5.790E+03
.000E+00 .000E+00 .000E+00 .000E+00 .000E+00 .oooe+oo
-5.2488+03
.0008+00
-5.1 l6E+03
.000E+00
E
úr
-5.7llE+02
.000E+00
8-t3lE+01 ..000E+00 .000E+00 ..0008+00 7.6798+{1 .0008+00 .0008+()0 .000E+00
-1.359Et01
BENDTNG
MI
R3
.000E+00 .000E+00 .0008+00 .0008+00
1.2398+00
3
TORSION BENDING M2
R2
-3.084E-02
2
I
SHEAR SHEAR
BEAM
I
53-5
(o)
SPACE FRAME
4,316E+03
2.7U8+{3
M2= 438
'5
| /,=,,,,
Example 14.6 Analyze a thfee storey-one bay space frame shown in Fig.l4.6a using STAP-3D.
UJ
\-/
Mr =
Solution The node and member numbers are shown in the same properties of the.members are given in Table
figure. The
432.3
-/o
\¡í
geometric
l4'6'
Table 14.6 Geometric properties Section cm Members 30x45 Beams l-( 30x40 Beams 7-l 30x30 Columns l3-2¿
Modulus of elasticity E". =
Þ
A cmz
0.195 0.170 0.141
1350
:
[,_, cma: J I, , cma 101250 237000
1200
183600
900
tt4200 MPa
äi¿ =-10s00 Poisson's ratio v = 0. 15 Torsionalinertia : p x3y (wherex
90000 67500
:
L,
la cma
f1-,,,,
227800
ló0000 67500
1050 kN/cm2
| (b)
FREE BODY DIAGRAMS Fig.14.6
",
=
,or.o
STAP.3D COMPUTER PROGRAM
536
The vah¡es
of p
ILLUSTRATIVE EXAMPLES
for different aspect ratios are shown in Îable 14.6; The L-node
various members is shown in Table
l4'7'
Table 14.7 LNode for space frame elements
I
2
J
4
6
l0 l2
6t7 I t7 t4 t8 t6 lE l0 19 t2 t9 l4 20 16 .20 2t7 417 6l 8l l0 18 12 t8 149 169
2
(--2,0,12) (-2,4, t2)
(s,
-2,
12)
I
4.
t0 t2
6 t
l5
J
l6 l8 l9 2t
partial ouþut Thus, a L-node need not be a fully restrained node' The input and shown in EXl46.FIL and EXI46.DOCEXI46.FIL
Three storey - one bay three dimensional frame kN'cm example 14'6
110
tlll Slll I glll
500.
0.
500.
350.'
500. I
I
t3 l3 t3 l3 t3 I3 l3
20.
z
10.
J
I
t
10. I 20. 10.
25.
20.
25.
0.
l0
400.
350.
t2
400.
1050.
r 500.
400.
0.
5û0.
¿å00.
350.
500.
40{}.
1050.
lll
tTtll 18 I I 19 I 1. 2olll l3 224 1 1050. 0.15
l
r-200.
I
I
l-200.
400.
1200.
l
I
I
t
l.00.
t200.
I
I
r
--200.
t,2n0.
NUMBER OF NODAL'POINTS NUMBER OF ELEMENTTYPES NUMBER OF LOAD CASES SOLUTION MODE (MODEX)
0. 0. 6750 0.
1350.
23700 0.
10125
227800.
1200.
18360 0.
9000
r6000 0.
3
900.
I t420 0.
=20 =l =l
=0
EQ.O: EXECUTION
EQ.I ; DATACHECK CASE NO
r
*trt**r****i***ir*r*
ANALYSIS PROGRAM STAP-3D
0
2
** *r+* t
storey- one bay three dimensional frame kN-cm example 14.ó
r200.
I I
500.
****r+r****** r ****** *rtËr*****r***tr
050.
400.
t4 t6
I IS
lateral load due to wind
DIMENSIONAL BEAM ELEMENTS
6750 0.
-
0.
0.
r3
1050.
I
6
0.
IJ
t5
1t000 ll tl tl t2 t2 t2 t2
350.
4
t3
9
il
22
4
0.
10.
2
ii
1
24
4
lateral loãd due to \¡/¡nd
i:
)
I
I
20
8
t3
537
NUMBEROF BEAMS NUMBEROFMATERIALS
NUMBEROF GEOMETRIC PROPERTY SETS NUMBEROFFIXEDENDFORCESETS
.
=24 :l =3 :0
*
*+***
**
**
**
ILLUSTRATIVE EXÂMPLES
S'TAP-3D COMPUTER PROGRAM
.538
MATERIAL
I
S MODULUS RATIO 1050. .15000 YOUNG
S
POISSON
L POINT LOADS
WEIGHT DENSITY
i. NODE
'00000
LOAD
NO.
CASE
2
I
BËAM GEOMETRIC PROPERTTES
!
ELEMENTAREA,'SHEARAREASHEARAREAINERTIAINER.IIA'INERTIA
TYPEI23l3 1350.000 000 I 12c0.000 000 2 900.000 .000 3
-000 '000 .000
237000'000 t01250'000 227800'000 1s3600 000 90000'000 160000'000 I
14200'000 67500'000
67500'000
J
I
J
t
4
I
4
I
LOADS ELEMENT LOAD MULTIPLIERS FOR CRAVITY
x-DtR .000000E+00 ..0000008+00 .000000E+00 Y-DIR .000000E+00 .000000E+00 .000000E+00 .000000E+00 Z-DIR .000000E+00 -000000E+00 .000000E+00 '0000008+00 ELEM LOADS I]EAM NODES MATL GEOM ABCD NOIJLNONO 0000 ll t2617 0000 ll 23717 0000 ll 14817 0000 4t014l8ll .l 0000 I 18 s ll 15 0000 6l2tól8ll 0000 2 | 7 .2 l0 t9 0000 2 19 I 3 ll .19 I 0000 2 g I 4 12 0000 1061420t2 '117152012 0000 0000 t28162012 00.00 13l2l7'13 0000 14231713 0000 3 I 17 4 15 3 0000 3 I I 6 5 .. 16 0000 3 I 6 7 'l t7 0000 3 I 18 7 I I 0000 3 I .tg 9 l0 18 0000 3 I l8: 20 l0 ll 0000 2tlll2l8l3 0000 3 I 22 13 14. 9 c000 231415913 0000 241s16913
ABCD .000 _000 .000
'000000E+00
NUMBER OF EQUATIONS
-11
42
BANDWIDTI{ NIJMBER OF EQUATIONS IN A BLOCK NI IMRFR OF RI,C)CKS
=i
APPLIED LoADS
RXRYRZMXMYMZI 20.000 .000 .000 .000 .000 2s.000 .000 .000 20.000 .000 .000 .000 .000 25.000 .000 .000 20_000 .000 .000 .000 .000 25.000 .000 .000
.000 .000 .000 .000 .000 .000
.000 .000 .000 .000 .000 .000
ELEMENT LOAD MULTIPLIERS
AB^CD
,TOTAI-
539
LENGTH
M FORCES AND MOMENTS
500.000 500.000
LC,AD
No'
500.000
NO. I
500.000 500.000 500.000 400.000 400.000 400.000 400.000 400.000 400.000 350.000 350.000 350.000 350.000 350.000 350.000 350.000 350.000 350.000 350.000 350.000 350.000
.000
1
I
AXIAL SHEAR SHEAR RI R2 R3
TORSION BENDINC BENDINC MI
.
M2
M3
6.077F-05 7.t55E+00 -t.754E+00 4.3238+02 4.385E+02 L7898+03 -6.0778-05 -7.155E+00 1.7548+00 4.3238+02 4.385E+02 1.7898+0j 3.5108-04 4.936E+00 -2.7t0E+00 -2.6848+02 6.7748+02 1.234E1.0J . -3.5t0E-04 -4.9J6E+00 2.7t08+00 2.6E48+02 6.7748+02 t.2348+03 2.7998-04 t.9t2E+00 _2.9558+00 _l.ult+oz 7.3898+02 4.7808+02 -2.)lssrq+ -l.9t2E+00 2.955E+00 1.0478+02 7.3EgE+02 4.7EoE+02 -1.3028-04 -7.t548+00 _1.?4lE+00 4.3128+02 4.352E+02 _t.7898+03
l.t02E-04 :7.t54E+00 t.74tE+00 4.312F'+02 4.352E+02 _t.7898+03 2.6128-04 -4.9358+00 -2.699}+00 -2.6858+02 6.7468+02 _t.234I1+O3 -2'6128-04.=¿-Ò¡sg+oo 2.699E+00 2.6858+02 6.7468+02 -r.2348+03 4.020F-04 -l9l2B+00 -2.9458+00 _l.M7E+02 7.361F.+02 4.779r,+02 4.020E44 l.9l2E+00 2.9458+OO t.041F'+02 7.36t8+02 4.779t'+02 1.2468+01 -3.70t8+0t I.947E+00 -2.558E+02 _3.906}+02 _7.4048+03 -r.2468+01 3.70rE+Or -r.947F.+00 2.5588+02 -3.88rE+02 1.2508+01
-7.400E+03
-2.378E+0t Z.SS7E+00 -1.726F'+02 _5.l2BE+02 4.7568+03
-1.250E+0i 23788+01 _2.557E+00 .t.?268+02 _5.t0iE+02
_4.7578+03
1.2498+0t -9.006E+00 2.7788+00 -6.l:,7}+01 _S..S7OE+02 _l.80te+O¡
-t.2498+01 9.006E+00 -2..77i8+00 6.t77r'+ot _5.5438+02 _t.80tE+03 6.EllE-03 -1.227F'+01 t.g47E+OD -2.558E+02 _3.906E+02 -6.EilE-03 t.2278+Ot _t.9478+00 2.5588+02 _3.881E+02
_2.454E+03
5.9958-03 -8.713E+00 2.557F'+00 -1..t268+02 -5.t288+02
_t.tile*ot
_2.455E+03
-5.9958-03 8.713E+00 -2.557E+00 1.7268+02 _5.l0lE+02 _t.?438+03 4.79tl43 -3.5t2E+00 2.t788+10 _6.177F'+01 _5.5708+02 _7.0258+1)2
4.79tE-03
3.5128+00 -2.7788+00 6.t77F'+Ol _5.543E+02 _7.024F'+02 -5.5808+01 -7.2838+00 3.0t3E+01 3.943F'+02 _5.889E+03 _t.398E+03 5.580E+01 7.283E+00 -3.013E+01 -3.9438+02 -4.ó56Et03 _l.l5tE+03
ILLUSTRATIVE EXAMPLES
STAP-3D COMPUTER PROGRAM
540
54t
_9.936f+02
:Z.sg4E+Ol -5.3368+00 1.934E+01 3'464F'+02 -3'l8lE+03 _9.719F+02 -3-588E+03 2.594E+01 5.336E+00 -t'934E+01 -3'464F'+ù2 í'ssle+oo .l 8l9E+02 -1'4379+03 4.3268+02 -7.094E+00 -'2.778r'+00 z.llty+oo -9.551E+00 -t 8198+02 -l'9068+03 -5.3988+02
-14 t5
i.ãnot*o
_¡.3999+03
consider a single storey portal frame as shown in Fig.l4.7a. It is subjected to a point of i0 kN at Lhe mid span of the beam. The relative stiffriess of the beam and columrr
ed
given by
7'400E+00 3'943E+OZ -¡'5679+03 -3.359f,+01 -7.2339+00 _t.l5lE+03 -3'q438+02 -l'0838+03 7.2E3E+00 -7'400E+00 3.8508+01 _8.936E+02 f'4648+02 -9'3928+02 -5.33(f,+00 5'653E+00
t6
-¡.9g7g+01
t7
i.nott*t
5.336E+00 -5.6536+00
039E+03
_9.739F+02
l'819E+02 4'3419+02
4.326E+02
-3'4648+02 -l
2'9508+00 -5.424f'+OO -2.778r.+OO' -5'97E8+02 5.424E+00 2.7788+OO -2.95gg+00 -¡-31p1+02
l8
5.580E+01 7 '282t.+Oo
l9
3-0078J01 3'935E+02 -5'8778+03
-3'9359+02 -4'6478+03 -5.580E+01 -7.282F'+00 -3'007E+01
2.594E+0t 5.3358+00 t'935E+01
20
3'464E+O2 -3'184E+03
-3'46/]E+02 -3'589E+03 -2.594E+01 -5-335E+00 -¡'9359+01
2l
23
.9.9359+02
-?'464F'+02 -l'0398i03 -1.907E+01 -5-335E100 -5'654E+00 2.778F-+OO 2'9508+00 l'8i88{-02 -4'3468+02
t.398E+03
F--l!O ,{, ZSo
l.l5tE+03 9.935þ+02 4.3258+02 5.3975+02
::::ä',i*äff"*".'"î-"r-r""rã¡¡Tavb€o'ry'T'*1"::L11T::'-i:":':ff:
äåÏ:'ï;"'i;h*k''d¡;"1:g:ttlo:'f "lf 1"^Í;,Ï,"*TJiîåì"TJi; lilXl,;,"i jl;ii"",ï'",**ö¡ì;ìFo:Jr"j^î:':j::îti.:"ff of the ìiil each $",ff illlåå""'"""il;ij"l* in local rñember axes. Net moment along :
M*¿ Mrr+ M,o3 432.3
Nl
M":
-
M,ut
-Mrr'
2455 + 1083 +939.2 -- 0
M,, - M,ot a Mro3+ M,z3 nsg + 255.8 - ll5l -893.6 Mrz +
M1o2
*
M,ut
*
M,r
:
o
t
438.5-390.6 -394.3 +.346.4 = 0 while subscript indicates the The superscrþt indicates the local axis' t-
E
MB WHEN FRAME
z I
IS HINGED
z
l¡J
M (b)
In a 3-D problem, the interpretation of results =o:::J1.et":::t: i.5':ll;"lni ;"ä:äïä;;;'ìàãiã* u""^se th e '" em ber r::'î ;": fl :"::",i n Jl
global axes is as follows
Itlf*iu*n
53976+O2
o =
,l
E
(o)
FIXED END
";
"
ueam
ISMB 3OO I = 8600 cm4
9.739E+t2
9.7398+02
-t'81ßË+02 -5'9779+02 -5.424g+00 -2.7788+OO -2'950E+00
(Vl)
.l.lslE.F03
-¡'43Sf+03 7.095E+00 2-778F,+OO 9'5488+00 !'8188+02 -l'906E+03 -¡'31$p+02 -9.s48E+00 -i.ónit.oo -2.77$fi+00 3'9359+02 -l'5078+03 7 7.2828+00 '4028+OO l.ssoE*or
5.424E+00
^-
-5.398E+02 ¡.3991+03
4.3258+02
.7'4O28+OO -3'9358+02 -t'084E+03 -3.850E+0I -7.282r.+00 '1.9078+01 5.3358+00 5'6546+00 3'4648+02 -9'3959+02
t-
t231 RELAIIVE ST,IFFNESS,
(c)
K
INFLUENCE OF RELATIVE STIFFNESS OF BEAM
FORCES
Fig.l4.7 re
column ha3 a span of 350 cm and consists of ISMB 300 section. The beam has a
of 500 crn. The moinent of inertia of the beam is varied such that the relative varies as 0.25,0.5, 1.0, 1.5,2.0,2.5,3,3.5 and 4.0. The data of the frame is for srAP-3D and the progr¿rm is executed for different values of relative The point load of l0 kN is replaced with the equivalent nodal forces as showu 14.7b. The values of momenr ar A and B are plotted in Fig.l4.7c for difrerent of the relative stiffnesses. It is seen that the inoment produced in the frame es as the
relative stiftress of the beam increases.
any case, the joint equilibrium has to be maintained. The distribution factor at B in BC increases with the increase in beam stiffiress which results in decrease in the produced in thejoint.
problem is repeated by taking the portal frame as pin-ended. The momenr ar B is plotted in the same figure. It is seen that the moment is consistently lower than in the frame with fixed ends. The decrease in the moment produced is ãttr¡uuted to lecrease in the frarne stiffiress.
number. 14.8
Example 14.7'
of beam and column on the moments Illustrate the effect of relative stiffrreqs
the effect of relative stiffness of beam and column on the lateral deflection in
STAP-3D COMPUTER PROGRAM'
542
Solution
NON-LINEAR ANALYSTS : MATERIAL NON.LINEARITY
earthquake loads. 10kN
1
J
TSMB 3OO t = 8600 cm4
1 þ
(o)
500cm
-a
PORTAL FRAME
l-2 € g.
;
r.0
5 w
o.e
9 J
t! 0.5 l¡, o -t O-t, É rrj 0.2 t-
There are two different forms of noq-linearity in structurai analysis. The first is due non-linear material behavior and is usually referred to as material rion-linearity. The is geometric non-linearþ which is caused by large deformafions leading to iable changes in the geometry of the structure. In a structure, there may be either non-linearity, or geometric non-linearity or both. In the previous chapters, it assumed that the material remains rvithin the elastic limit and that the deformations in the structure a¡e smäll.
If
stress-strain relationship of the material is non-linear, the behavior of the structure becomes non-linear. The displacements may still be considered small so that the ic relationshipç remain linear. In the present chapter, the mefhods of analysis of non-linearity in structures consisting of tiuss and beam elements are discussed. classical theory of plastic analysis is discussed first to prepare the necessary back The modifrcations in the stifhess matrices of a 2-D truss and a 2-D beam )Rts are then suggested álong with the solution algorithm. The geomeüic nonity will be discus$ed in thqnext chapter.
STRESS-STRAIN CURVE OF STEEL The uniaxial stress-straiú curve óf steel shows three distinct regions : elastic, yield and
hardening as shown ín Fig, 15.1. At yield the material flows and the strain ¡sesfrom €y to €sr atnearlyconstantstress..Beyond er,,Strainhardeningof the ial takes piace. For Fe 250 grade steel, the strain values are as follows :
ro
RELATIVE STIFFNESS, k
(b) INFLUENCE OF RELATIVE STIFFNESS OF BEAM .
INTRODUCTION
Fio- l4-8
Yield stress' o, = 250 MPa yield strain e, : O.tZ"lo strain hardening er, : 1.5'olo maxi strain e^o : 25%o modulus of elasticity E =2 x 105 MPa
NON.LINEAR ANALYSIS : MATERIAL NON-LINEARITY
544
THEORY OF PLASTIC AI.IALYSIS
545
Stage
The fibre strains as well as the stresses increase linearly with the load as long as the remains within the elastic range as shown in Fig. 15.3a"for a rectangular section. moment is given by :
vt vt
Mo ly M:SO S : elætic section modulus o = stress in extreme fibre
STRAIN-HARDENING
l¡J cE
tt/l
€y
€st
€
mqx
STRAIN
(r5.r) (1s.2\
€>>€v
Fig. 15.1 Stress-strain curve of mild steel
The idealized stress-strain curves are shown ìn Fig. 15.2" Figure 15.2a perfectly elastic curve, Fig. 15.2b shows a rigid plasiic curve, while Fig. 15.2c shows perfectly elastic-plàstic curve. There is no strain-hardening. The theory of analysis Ítssumes aperfectly eiastic-plastic sfiess-strain curve for steel.
þt
tJl
vl
.Þ ,v,
l=
ELASTIC
ul
-
PLASTIC
ln
ELASTIC PLASTIC
STRAIN
STRAIN
STRAIN
(o)
(b)
(c)
e
equaló €y, the stress
{noment
M;
pLASTtC
and the section deveþps yield moment, that is,
:...
associated with. the first yield is called the yield mgment.
Pla;tic Stage increasing moment; yielding will take place at,thÞ çxfteme fibre. As the yielding will progress towards the intériot of the beam ãs shown in is I,fent inoeased, ¡.15;3b. Taking moment oflorces about the neutral axis :
M:2fo,
+I-",++] :"'-l; #l=",1;
:
The stess-strain relationship is assumed to be perfectfy ehstic-plastic as discussed
earlier.
The deformations are assumed to be small. There is no axial load on the beam.
Let us determine.the momenfs in a simply.supported beam under increasing lateral l
load,
o : or,
Nt: o, s
Plane sections remain always plane and nonnai to the axis of bending.
3. 4.
(c)
æ,
¡-
The thecry of plastic analysis is based upon the following assumptions
2.
ELASTTC (b) ELASTO-pLASTtC
t¿,
RIGID
I5.3 THEORY OF PLASTIC ANALYSTS
L
(o)
ttl
l¡J æ,
-.{
an
vt
¡¡J
.É.
b
hus, moment
#l
(15.4)
capacty M is greater than the yield moment Mr; and its value increases
the decreasd in yo. Stage
ting will progress towards the interior of the beam until the section is'fully as shown in
Fig. 15.3c., At this point,
NON-LINE.A,R ANALYSIS : MATERIAL NON-LINEARITY
546
Compressive force Tensile
force
and because C is equal
MOMEI.IT-CURVATURE RELATION
: o, A"
(l s.5a)
T = o, A,
(15.5b)
C
to T for equilibrium,
A" =4,
(15.5c
HINGE ¡_ ,l.SPREAO' .
\i'!
The moment tvç associated with yielding over the entire depth of a beam is called plastic moment. force x |ever arm
Mo
bdd '2 2
bd2
MECHANISM
Z
where
(ls.7 Fig. f 5.4 Plastic hinge and a mechanism
plastic section modull.rs
A"y¡ + A,y2 yt,yz
:
full plastic hinge will develop only under .ideal conditiorrs. The factors'which
( 15.
the forniation of the plastic hinge.are
c.g. ofthe compression and tension areas about the neuftal
,
"-
Mo .t
M.,
PLASTIC HINGE AND MECHANISM
Consider the simply supported beain of Fig. t5.4a loaded by a concentrated load the mid span. As the load increases beyond tlat causes the first yielding, y proceeds not only towards the centÌe of the be4m, but it also moves out. .The beam collapse when the centre section becomes fully plastic. This section witl keep rotating at constarit moment and the venical deflection will also keep on Thus, the beam will transform into a mechanism consisting of two links with a hinge in the middle. In all there are three hinges in a simply supported beam: two hinges, one at each support, and one plastic hinge. within the span; Such a plastic hi has no additional moment résistance. The plastic hinge is not a section but it is the z of yielding near fhe section of full plasticify. The shaded porrion in Fig. 15.4b shows plastic zone. The length ofplastic zone depends upon the shape factor. Greater.is ratio, the larger will be the lengtþ of the plastic zone. For flanged sections, the value varies from l.l0 to about 1.16.
:
Local buckling lateral-torsional buckling axial force, and
axis
The neutral axis under the plastic condition divides the section i¡to two equal that is, Ao = 4. The ratio of the plastic. moment MD. to the yield moment N4, is the shape factor f, thus
I5.4
(bl
(15
Zo,
,Mp=
",Il'
(o)
= O.,t4
:",+: r.sr'ç
Also,
-
s47
shear.force.
With the formation of
a plastic hinge in a structure, there is a redistribution of internal causing other sections to reach their full strength and develop plastic hinges- This goes on until the structuie forms a collapse mechanism. That is, the deflection ori increasing without any increase in the lqading. The loading associa'tgd with the of colþse is called the ultimate load Tlne ratio of ultimate load to working load called load facfor. Thus,
Load factor =
A
ultimate load working load
(15.e)
mechanism is a condition of instability. Mathematically the degree inacy of a stn¡cture at the verge of a mechanism is negative.
of
statical
MOMENT-CURVAT{,RE RELATION an el¿istic section, the
curvature
{
is given by
o,t€ - ---"Ry y:
distance of ihe fibre from the,neutral axis
(l5.l0a)
548
NON:LINEARANÀLYSIS: MATERf,ALNON-LINEARITY
e: R
PLASTIC ANALYSIS
strain at{he fibre
:
If a structure has a degree of statical indeterminacy equal to c.;then it will require o. to be a statically determinate structure. Therefore, just one more hinge will be
radius of curvature The moment-curvature re.lation is given by
icient to convert it into a mechanism. Therefore, minimum number of plastic hinges to convert a statically indeterminate structure into a mechanism is equal to
Meô =-=Mv9ûv For an elasto-plastic reciangular
,
M
section
: ",|:-+fl ,L2 d.
are two methods for analyzing beams and .frames when using the plastic
ì
arso
o,
=
J
-a
:
0Jd
St¿tical oi equilibrium method. Mechanism or virtual-work method.
Yon = 10t = 0t d nd 02 2þ
Yo
setting,
M
My=
or,
if,-llq.ì'l 2L 8\0/l
(rs.l
The equations l5.l0b and l5.l0c are plotted in Fig. 15.5. A straight line is obtai in the elastic range, while a curve is obtained in the plastic range which asymptotic to the line when M becomes equal to the plastic moment. '[his shows when the section becomes fully plastic, it undergoes infinite curvature. Thus, i rotation can occur at a section under the firll plastic moment, and it can behave as plastic hinge without application of any additional moment.
r
1'5
:E
= F
z
t.o
l¡J
o
Method method consists in óonstructing a bending moment diagram in which M < Mo at section in the struÈntre, such that a collapse mechanism is formed. There is'no' to analyze the structure using the stiffiress or flexibiliry methods. The various steps
I
Rernorè redundant forces such that a staticallv determinate strucfure is
2 3 4 5 6
Draw a suitable bending montent diagram at faildre due to the applied loads. Draw a suitable bending momerfi iliagmm at failu¡e dire to the redundant forceò. Superimpose the trvo moments diagrams such that a mechanism is forméd. Compute the ultimate load using static equilibrium equations. Check the moments to see that M
M exceeds M^P anywhere, repeat
.
Mo every where in the structure.
steps 2 to 6 so that a plastic hinge is formed at the
of maximum bending moment and the other basic conditions are satisfiêd. there may be more than one possible bending moment diagram. In such a
draw all possible bending moment diagrams, check various conditions and the collapse loads. The minimum collapse load among them is the true load, The statical method gives a lower bound on the ultimate.load.
Fig. 15.5 M-0 curve for a rectangular section
15.6 PLASTICANALYSIS
t5.l
Any plastic anaþis must satisfi th¡ee conditions
l.
staticequilíbriumcondition :
2.
Plastic moment
:
The structure is under static equilibrium
the ultimate load for the fixed beam shown in Fig. 15.6a. The beam is ic ivith plastie moment capacity equal to Mo.
coHapse.
condition : The 'moment no where moment capacity M,
3.
consideration
Mechanism
must exceed the
of the
section
.
condition : There are just sufticie¡t mechanism forms.
hinges when a collapse
plastic
:
2. Number of hinges required to form a degree of statical indelerminacy a, Two hinges can be assumed to form at each support and one at 2+ n. The bending moment diagrams are shown in Figs. 15.6b and 15.6c. The equation at the mid-span section is written as :
:
l:3.
550
NON-LINEAR ANALYSIS
:
PLASTIC ANALYSIS
MATERI.AL NON-LINEARJTY
w...t]
+ N/L= ----!-M-PP8
"
l6MD
-uL" It is obvious that no where
lVf
t-Mo. Th" beam mechanism
A
is shown in Fig. l5:6c-
o.
.:-ï-TVB '---¿ ?F---t (o)
Mp M
---\. r---\ (b)
It;
Få'l
B
Fig. 15.7 Continuous beam : statical method
(c l
ReactionRo
Fig. f5.6 Fixed end beam: statical method
.
Shearforceat
Example 15.2
Mechanisml- SpanAB Support A is simply supported. Hence or¡þ two plastic hinges are required to beam mechanism. One hinge will form just to the left of support B.in the span AB moment capacity of the span BC is higher. Let us assume that the other hinge will at a distance x from end A and not at the mid span.
- : +-+-Y. 2LL x=
for maximum moment to occur at x
;_+
Maxirnurn moment at x
l
Since there a¡e three spans, there can be th¡ee independpnt beam mechanisms. bending moment diagrams due to the applied loads assuming each span to be a si¡ supported span is shown in Fig. l5:7b. The moment diagram due to thg rgdundan! moments in each span is shorvn in Fig. 15.7 c. The two moment diagrams superimposed as shown in Fig. 15.7d.
t\/f W-Mo : rrr 2L
0
.,.
Consider the continuous beam shown in Fig. 15.7a. It has a plastic moment of Mo over the exterior spans AB and CD, and 1.5 Mo over the interior span Deterinine the collaPse load.
Solution
(d)
M,*: ,
(i +)ti +) #(i-+)'
For plastic hinge to form at
Mt
x,
-l\Ço
,:
ùç
ly_lt_w*ltlll_"0ì \2 L 4 2L)\2 w) (w Mo llr Mo I f+- rrjt.r- * )
55t
552
or'
NON-LINEAR ANALYSIS : MATERIAL NON-LINEARITY
*r(!- "t"l'
to
2
:
Ùrg
settine -wL
'or,
-
4m2
.'.
w
or,
W"
PLASTIC ANALYSIS
\2
m.2m
rExample 15.3
wL)
A portal frame is shown in Fig. 15.8a. Determine its céllapse load.
'm) = fl--l' \2
wL
oÍ, m:2.914
tl=0
l2m
-,
/z
or0.0g5g
w,,==Y+:os43MP -.'L 2.914L M.
553
E
M.
l-1,65t'
r*r*
x-
0.5L-0.0858L:O.4l4L It will be seen rhat m = 2.914 gives x :. - 2.414L
the collapse load
v/" =. lr.65lL
and \ :
(b) which is inadmissible.
0.414L
Hence
-
(o)
wL/2:OL
)
i¡r
wL/2-OL).
Mechanìsm2- SpanBC It is a cortinuous span having statical indeterminacy equal to It needs ttrree plastic hinggs t9 form a mechanism. Two prastic hinges wiil'form just outside me supports B and c where rhe moment capacities yo Td nor 1.5 NL. Th" other hinge wiil form at the midspan. The equilibrium equation is'given by
2.
ï.
NÇ
or,
*
Ls Me =
w :
w"
1.5+ l0 Mo
:
*î
3;ï I
- -_Mp : e.æ-'-
,0 2
(c
)
(dt
(ii)
Mechanism3- SpanCD sga.n needs only two plastic hinges to form a mechanism. one hinge will form at end C and the other at the location of the point load. The equation of equiliirium is given
!i¡
¿ts :
3 M^+ P :M_: 4--p
3WL
-
g
'
or, w =w"=T{r:6.67yt The ultimate load is the smallest of the three collapse loads, that is,
w"
=.
6.67 l{e-
L
Fig. 15.8 Portal frame : staticat method ciÐ The frame is stâtically indeterminate to a degree l. Letrhe horizontal reaction at E be redundant and equal to Q. The vertical ."ariion, at e and E are given by : 'the
)))
PI.ASTIC ANALYSIS
554,
NON-LINEARANALYSIS:MATERIALNON-I'INEANTY
| and.n' Jht yi"i:|tÎ:tl1t *:":;}:t 'ilï'.ï,*;i@ï*":TïïH:,i'iT,i:,[:';'#;i:"i:
rhere are natural hinges at the suppoß
*Jåi"#;''ffi ll'X'lli'lüi;iJi";äñJ*'!iu'""r'*1T*'.'iï11;"-îl¡;3:Y:;1i|åTåi' hin$es rorm at c and D :åiffi:ä",iä;i;;ilì-;:.1.:'." "':ï:i1'^y?"1'î*" - iá* the equilibrium equafions at c and D' ó;;. iö "t*ti " v/L 3 At c î-;at ^, -2M,
or, Q=Qu='+
AtD +=* Eq. I .'.
gives,
otì
Y =r*,* f(zr"r,) w =t"=r+
Check
ttrr
MoinentatB: M" = î
-
:
QL
3
5lnç
(o)
- 2 M,
span
AB' the
as
shown in
rig'ls'8i'
The equilibrium
M":+-åot = 3Mo-r:5N!
Hence; the collaPse load Mechanism Method
;älffi,¡j-ffists
is Wu =
(o)
'+
method.
(d) BEAM
SWAY
({
) swAY
o.K.
equations in assuming a mecìanis4 and writing the requilibrium. r-, ¿L^ -.1¿!*^+l^¡¡l load ultimate the for solveg are These equations
.,"*':iìlit'iii"i¡üi"å"nr
J0INT
L
w:wu=.+ = 1.5 tnL a zMv
l
¿^Mo
Eq. (i) gives,
c,
BEAM
;{e
(c
Me ArB Y-ot or, W = (aL*Nç)i AtD + = Mo oÍ, Q =Qu=
(b)
assumed mechanism is
equations are :
Mornent at
POSSIBLE POSITIONS
NOT O.K.
tv! in the
D Let us assume two hinges f.orm at B and
Check
cr,
OF I{INGES
tnL
incorrect'
/
NFLASTIC.
Select possible independent and combined mechanisms. The independent mechanisms are known as beam mechanism, swøy mechanism or ioÍnl mechanism as shown in Figs. 15.9a-f They may occur independent of the
3.5 rvrp
:
Since moment at B can not exceed
:
2
failure of the str.uch¡re as a whole,and may lead to partial failure of the structt¡re.
: l'5 Mp .a 2Mn
,
method is very convenient when the structure is highly indeterminate. The various are as follows : I Determine the degree oi statical indeterminacy ct. and the total possible locations of the plastic hinges NPLASTIC. Therefore, number of independent mechanisms
(c
)
CoMBINEo
(-h) COMBINED
.Fig. 15.9 Mechanism$
PLASTIC A}'ALYSIS
NON-LINEARANALYSIS: MATERIALNON-I'INEARITY
5s6
Ms-
3 Step 4
Mp
.
(sagging) (hogging)
vinual work equation may now be written
-
Step
Mp
Mc= -
Thecombinedmechanismisformedwiththecombinationorsuperpositionof g' h' The correct ;;;t ;echanisms as shown in Fig' l5'9 the beam ""d *""t'uni'ts are those which satisf, the other two basic mechanism conditions for the plastic analysis' - by solving the equilibrium equations' Determine ttre låwest ultimate load Mo everlwhere in the structure' Check the moment to see that M S
ilnrernal work
::
Themechanismmethodgivesan,upperboundontheultimateloadorcollapseload.
as.:
done :
yo (_ 0) + Ms ( ú + O) + Mc (_ O) +M M:0 l0+ ó) óì + Me M ,t O Yr91 Mo(e+ l.sMe0 a 2.5Mpó n IvçSs= 5Me0 =
:
workdone
:
22
The average deflection in the span AB is (0 +
:
Externalwork done
of the smrcture' l. A dotted line is drawn at the.inner side individually' drawn ;. ntipossiute mechanisms are angles are taken as negative' 3. open angles are tak;;; positivè and closing si¿á ¡s taten as positive, and that causing Momenr causing trnri*'on the dotted 4. ' ""ö;";s¡on ¡s ãrcn * negative'
(iii)
2W¿ + O.¿WA .f 0.6W-A
Sign Convention
for,ttreanalVsis of stn¡ctures The following sign convÞntion is used
557
:
2.5 W^ :
Ly2 :
2.5 W( 0.6L0
A,
)
/ 2_ using Eq.(i)
Equating intemal and external work done,
2.5W(0.6L0):5
w:
5 Mp _ loMp L 3 L
w"
1.5
ExamPle 15'4
l5'l0a by the mechanism metht¡d' Analyze the fixed beam shown in Fig'
Moo
us check the moment at the
mid span
Re: Y + zffx o.ó : l:7w and Rc = 1.3w bending moment at-the mid span
- Mrt l.3W x 0.5L
Mo
-
Me + o.os 0.75MP
irig. ts.fO Fixed end beam : mechanism method Solution
the collapse load is
hilqes to form
is 2' It requires The degreerof indeterminacy of the beam. -tl¡ee. twó supports, while the third hinge will the of ateach *iriio,rn hinge mech4nism. one rand rotation at C : 0' Then at A : 0 form at B under trr" poiniïJ.=iet rotation deflection at B (i)
"positive
...
if it
causes teirsion on the dotted face'
M
Mu
(h,oeeins)
=
Me
|
*o*, o.K.
l0 M.
3.L
----l-
a
'L= ABx0: BCx$ 0.4L0 = 0.6L0 (ii) Hence, 0 : r.5 I slope S at C is also negalive béing closing The slope 0 at A is negative, *9 tft"moment is taken angle' .Tlie ."gi;;. ï'";lope (0 * Oi o:t s it'positive being an opening as
.w"
<
xfrrao
w l!ì' - 2L\2 )
Reanalyze the three span continuous beam'
of Example 15.2 using the mechanism
I -spanAB (Fig. l5.llb) Two hinges are required to form a mechanism in the span AB. I-et one hinge forms at istance x from A and the other hinge forms at B in the span AB..
x0
:
A
:
(L
x ô= ' L-x o
- i)0
bygeomerry
(i)
NON-LINEARITY NON-LINEAR ANALYSIS : MATERIAL
558
PLASTIC ANALYSIS
559
2.414L or 0.414L
(v)
\ first value is inadmissibte. The ultimate lbad is given by
w : w.. " 2 - span
2
xl.4l4 0.414 x
M^ lL 0.586
M11.65 ' L
(vi)
BC (Fig.15.l lc)
workdone
:
MoO
+ 1.5M0(e+ e) + I!ç e : 5Nç0
: workdone l.5W^ .)
t.5W+O
intemal and exteirnal work'done.
0.75wLe = 5NtO M^' w:wu=6.67i3 - span
(vii)
CD (Fig. 15.l ld)
It requires only two plastic hinges to form a mechanism. One hinge will form at C and other at the section of point load.
method Fig: 15.11 Continuous beam : rnechanism Internal work
M, ( e +- O) + Mo
done =
: ",(*)'.H'=",(:Ï)' w.lAì * y(r_*)A
External'ivorkdone L\2)
Lì
: .w^ = lw*e 22
*e: 2 a::f 2'qoa
Ö
lworkdone:
.
(iii)
work
2
or,
*=
=
I.
W - *; t
w
: wu = 4t.67+ L
(viii)
collapse load is the smallest of the th¡ee ultimate loads, that is,
(iv)
Forminimumcollapseload,differentiatingWwithrespecttoxandsettingitto'7ßfo: gives
1
ãMoO
*1, : J I'a,o
",(iÏ)o
ffir,
'
0):
internal and extemal work done,
usingEq(i)
Equating internal and extemal work donq'
lw*e 2
done:
+ Mo(0+
MoO
0':o/3bygeometry
W"
:
M.
4.67----!-
15.6 Reanalyze the portal frame shown in Fig.15.8a using the mechanism method.
dw dx Of'
f+ZLxz - L2:0
Degree of static indeterminacy C and D, respectively.
: l.
The locations of possible plastic hingos are three
þr' þ4i
560
Numberofindependentmechanisms
þgål
=
Equating the
3-l = ?
oneisbeammechanismandtheotherisswaymechanismasshowninFigg.Js:t:?::j:q The ultimate loads f¡otiil The combined mechanism is shown in Fig. l5.l2d.
þq'
',JtS.tZ". the various mechanisms
Flj,
are as
fo[óws
:
..
þf
"T*#,fl'
Fi FI.
(b)
"xte*at
BÊAM MECI{ANISM
w.i
and inæmal work done
.z Yo: but and
to(-o)
+ MB(e) + Mo
(-0)
+ MB (0)
MA:0=M¡ MB : M, Mo :-MP
Mp
Mp
561
PLASTIC ANALYSIS
NON.LINEAR ANALYSIS : MATERIAL NON-LINEARITY
w_^
: %s + Mpô :3Nçe
or,
;tt
or,
w:w':6i
M.
(iÐ
Combhed Me'ch1nßm
Wt
þA4
ffii.,r
þA*{
ffii,.
or,
: L0 ,þ ^ 0 =20
by geometry
Equating the extemal and internal work done
w i:-Â+WA' = Mc(e+e) +
wi,
Vto
(-O-S)
2
ffii,'
(cI
ffiiÌ
Fig. 15.12 Portal frame: mechanism method
and
BesmMechøn¡sm
ffii:
lg:
þ{,. Mi,'',
2
ffii,' Mi,'
...
^tbYg"o,n"ttY
Equating thepxternal and internal work done
wÂ
Pqi
where
SWAY MECHANISM
but
=
MB(-e) + Mç(o+ e) + Mo (-0)
Ms = - Mo: Mc = 2lv1þ
or,
Mo
: lateral sway disþlacemeit at the level of the-bçam ^ : vertical displacement of the boam ^' : 2M, sagging Mc Mn =-Mo hogging w + rvfe(e+o) = 2\(o+o) +À+wÂ' v' 2
ry.#:4Mee*3Nço:lwoo
w w"= 1iM^
or,
(iii)
''
= Moo * 2Me(o+ o) + w*e 2
ffii ffii: t:
of'
ffi
Swøy Mechønßm
þã.
Wri.'
or,ô-29
Meo :6Mee
Hence, the collapse load is equal to the lowest among the three loads
wt+
W:Wu=
:
t+
Example 15.7
^
=
L0:
'bv i o geornetv
Determine the design mbment in the two bay frame shown in Fig. l5.l3a using the mechanism method.
NON-LTNEARITY NON-LINEAR ANALYSIS : MATERIAL
562
PLASTIC ANALYSIS
563
:
Mr(-e) + M; (2 e) + M4 {-e) Mz=-Mp M, =2M0, Ma,:'2Mp
6x2.5 0
:
lse
T I
5m
7 NÇ0
or Mn = 2.14 kNm
(i)
2- Beam 6-7-8
I I
Ivt
I
I
:
2.la kNm
(iD
J-Sway (Fig. 15.l3c)
I I
l-
(cI
(oI
Fig. SWAY MECHANISM
1
)e
Ml :-Mr: Mlo--M¡ M2: Mo = À4 :-Mr
CI e
.(bI
o
BEAM MECHANISM.
3 Ç, the arrows represent direction of rolation
Opening angles are taken as positive and closing angles are taken as negative. moments are taken as positive" that is
PORTAL FRAME
tg
5. I
ing the external and internal work done
¡
(d) COMBINED MECHANISM
:
3x50: M,(-e) + Mz (0)+ Mro(-0)+Mr(0).+ 150: 6NLe M, :
Ms (0) + M8 (-e) (iiÐ
2.5 kNm
uscombinemechanisms
I and3 (Fig. 15.l3d)
ile combining these two mechanisms, the plastic hnge at2 will disqppear.
(c)
(f )
COMBINEO MECH¡NISM
workdone :.3x50 + 6x 2.50
MODIFIED ÑIECHÀNISM
I work done :
z0
(9)
OOMBTNEO MECI{ANISM
* 4ln!e + 2Mee.+
: ll Mp0 Mp:2.73 kNm MODIFIED MECHANISM
300
M, (- e) + M3 (20) + M4 (- 0) + M5 (0) + tvt, (- e) + Mo (0) + Mto (- 0)
= MoO
(h)
:
tits combine mëchanisms 2 and
Moo+ lvçe
\
*
M;e
*
Moo.
(iu)
3 (Fig. l5.l3e)
tVhile combining these two mechanisms, the plastic hinge at 8 will disappear. There
plastic hiirges at the top middle joint. Let us rotate the middle joüt clockwise by 0. The two hinges at 5 and 6 will disappeár and a dew hinge will form at 4 as in Fig. 15.13 f.
Fig. f 5. 13 Two bay portal frame : mechanism method Solution
.'.
6, :
Degree of statical indeterminacy = Num¡er of independent mechanisms
These are : Beam
mechanisms:2,
Total no of nodes in the frame 4
Joint mechanisms:
l'
ten nodes' A plastic hinge can form at any one or more of the
Mechønism
I-
Beam z-à-¿
(rig'ts't¡U)
Equating the external and internal work done'
:
l0
Sway mechanism
workdone =3x50 + 6x2.5
:
I
e:300
workdone :4(lqpe) + 2Mo0*zlv\QÐ +
$20)
12Mee
Mr=
2.5 kNm
combine mechanisms 1,2 and
(v)
3 (Fig. l5.l3g)
PLASTICANALYSIS
MATENAL NON-LINEARITY NON-LINEAR ANALYSIS
565
:
564
^ ''' lit; *t¿ ã î,,, disappear and only one hinge îill :î:"ï":riJ-ry'üTÍ:';tiä: *îhT,li'^{i3""^.". h. 15.13 h' T: Fis. r5'r3 *îu" 20 as shown in Fig'
shown-in is trtuw¡r "' mechanism rs rhe combined mechanlsm The
u*o*,
mple 15.8
l:
t'
"t";;il;ö;äi;åäi"'i't'ts"; ::î}il:îi1 Externalworkdone work
done
";t or,
Ãnalyzæ the gable frame shown in Fig. l5.l4a subjected to determine the design plastic moment value.
=45Ø =3x50 +6x2'50 +6x2'50 + lvfe(2e) : e) + z Mo (20 + 2Q + 20)
7.skN
o
= 17Me0 Me = 2'6 kNm
The largest value of IvÇ
=
30
(vi)
@
@
o
2'73kNm based on Fig'15'l3d'
þlm + sm
toIT"".orodrawB'M'diagramcorrespondingl'Ïi"::*:'-*lJ:,iri-"li;i'"i
,"i:ff 'å::ffi-'Ï;Ïtii"ï^$î-ï::ï¡:,;3i"?"ï:"ff at 7 rocations onrv' that is' ro!ã" t"o* ::i,iå'Ji*i*'rÏi:T: ;äl"o*ffi T;,#:ti:::'it %l ::r::ïå (vii) lvii' Mr - Mro = ME: -2'73kNm M5 - Me = +2'73kNm ' M3 : 2x2'13 = 5'46kNm
o
of'
M6(-0) + M', (20)+ Ms(-e) = M6 - 2M? +'M, = - 15
or,
M6
Span 2-3-4
of'
:
2M7
=
o
BEAM
(d
6
x250
)
bI
OOTTED L¡NE. tENSIoN S¡oE
frl
EEAM
(iI
SWAY
sL
'rff: ll Iì
,"t
+ 3m
I
(f)
t\
H+
(9'
I FRAME
',0) kNm
o. K. than or equal to the conesponding
frT:-i.t,|tss Thus., the moment enerywhe'e.in.'nt 2'73 kNm' rn"'"roí", the design moment is
+ 2m 3m
4'75 kNm
.ZMo
ì
+
FRAME
VM
I 5m
ft
-!
I
15
FREE
BOOY
DIAGRAM
2m
(viii)
0.12
\.T','
I
.{
|
r.ot la17.35
@..4.1n
+
Middle Joint z Joint mechanism
rn#;;;;;;;-iqu,
I
3m
Beam mechanism
M4+M5 M6 Me i - 2'73 kNm'2 Mo + t2'23): ... Eq.(viii) gives,M, = i C2'73
+
3m
@#@
e
-12'23
lvlz-2M3+Mo =-¡5 tvl: - 2(5.46) +(-5'46): Mz'= l.38kNm'Mp
7.s
T
Beam mechanßm
-
T
2m
tiY
e
of static equilibrium' Let us write the equations
z
Ito
and wind loads nnd
(o) GABLE FRAME
= -5'46 kNm
M4 Span 6-7-8
I
3(Mp
gavity
Fig. 15.14 Gable frame : mechanism method
566
PLASTIC ANALYSIS
MATERIALNON-LINEARITY
NON-LINEARANALYSIS:
Net horizontal load in top storey
Solution
SwaY
:
3 Desree of statical indeterminacy plastic hingeà = e "ipossible 6 techanisms
ili"itãi"."t s"a;
ryPe mechanisms
SwaY måchanism
=
Frame mechanism
=
4
ó
"i
= 0'371
Extemal work done
f = lo x
: Mo :
Internal work done
L5e-l
l-|- )
4MP
= -
7.5 'J Øv '
(l)
kNm
Intêrnar
-itî
:
(
\^ (, ., s.rss)s : [/')x T"2 )" l.ro,( #)'s
4
Externarworkdone =
done 1
y:1.670
x,l0 = t25.6le
=
2.785
:
2A-B9xz.Se +
=
2x
30x2.50 + tO*
iI 2 ,
Mo
:
7.34 Me 0
(6)
I in Fig. 15.14 e M7 (- e) + Ms (0) =
mechanism ó is not possible.
,
('o"f)e -
("'+)t
: n:s
(5+2)g = 37 or
y = 7/3e:234e
vertical load in top storey: (7.S + 15) cosô
Extemal workdone
20.89 kN
2.785xto
a'325
same as'beam l-2-3 lVIechønßm 4 - Beam 7-8-9
¡¡*
+
+lo,.l€*11
2) = 1500 - 52.230 +2.785 0+50.1 e: 150.650 : Internal work done Me [0 + 20 +. (0 + y) + y] : 8.68 Me e
= 1.875 kNm
(Fig'I5'lae) Mechanism 5 - Sway Mechanism
:
:2x30 x250 + z}.ggx(-2.50)
4Mo 0
N! = -¿ = Mp
(5)
'Let us combine the frame and sway mechanism as show¡ in Fig. l5.l4g. The centre nitation 0 is shown in the same figure. The sway of right leg is given b/
=6.85 kNm
l7'1
or,
kNm
7- Frame Mechanism (Fig.15.l4g)
MechanismS-Beam5-6-7
Intemal work
27.4g
==n' ^'T
9.59
t I\,f3 (e) + 38.355 0 (Mechanism 5) Mr:M¡-Mz+Me-38.355 - - t7.l - (¿17.1) + t7.t - 38.355: -Zt.255kNm > l?.t kNm NG M, (- 0)
3-5 = 5'385 m
Exþrnalworkdone
4 Mo e
us check the moment at section
(Fig' l5'14 d) Mechanism 2 - Beam 3-a-5 Inclined lengttr
2.785x 30 = 3g.3550
[0 + 20 + (0 + y) + T] Mo: l7.l kNm
Internal work done
O
l'875
2.785 kN
- Frâme Mechanism (Fig. t5.laf)
Extemai workdone
(Fig'15'l4c) Mechanism'l - Beam l-2-3 1.50 Average disPlacement =
' ts*
: Mo :
50-37 or
*u ;ä;ä;" ;*rili
;-
:
The plastic hinges form at 3,'5,7 and 9. The centre of rotation 0 is shown in Fig. 5.14f along with the deflected shape. The sway of right leg is given by
and cos 0:0'928
TheinsideofthegableframeisshownbyadottedlineinFig.l5.l4b.Itwillheþin * *.n as hogging and sagging moments'
determining positive
ó
:2x10,.3*
done
: I
*'ä;ïiloiL"ìion Theframeissubjectedtouniformlydisftibutedloads.Thetotalload.oneachmember ü"*¡ttt 3-4-5 and 5-6-7 with the sin O
7.5) sin
A':30
Internal work
horizontal is $, that is,
-
2
l-
is apptied at its mid-sp;:
(15
External workdone
:
inã"p"nd"nt
:
567
",
=
t;iit
=
17.35
\2
kNm
(7)
NON.I.INEARITY NON.LINEAR ANALYSIS : MATERIAL
568
NON - LINEAR STIFFNESS MATRIX ANALYSIS
5.8 NON . LINEAR STIFFNESS MATRIX ANALYSß
Çheck 3 in Fig' l5'14 e Let us examine the moment at section
Mr (- e) + M, (0) +
'M7
(- e) +Me (0) =
3S'355
0
'Mo
in the member 3- 4 Let us also examine the moments
'
The analysis der:eloped in section 15.6 gives an idea of the collapse load only that too simple beams and frames under static loading'. For a plastic design, it is desirable to
(sway mechanism 5)
M¡: Mr +Mi-Me+ 38'355 = -13'695 M3 : (- l7'35)+(-17'35)-(17'35)+38'355
the deflections and rotations in the plastic hinges, in order to get a clear ing of the behaviour of the structure. Such a non-linear analysis due to ial non-linearity can be easily carried out using the direct stiffrress method earlier in Chapters 12, 13 and 14.
o. K.
A non-linear analysis of a multistorey structure becomes even more desirable in tle of a severe earthquake. Such a severe force is expected to cause significant uctural damage resulting in loss of structural stiffness: A linear analysis is cleady not
5'
B'M' diagrams' (Fig'15'lah) Let us infioduce a cut at 5 and draw
M¡
icable in such a-situation as it cannot account for changes in structural properties, of forces.due to the forrnation of plastic hinges, nor it can give an idea the location of plastic hinges and amount of rotations. The main object of a nonanalysis is to arrive at a satisfactory and economical design ,which limits the imum rotations to the rotation capacities of the respective sections and produces a uniform distribution of plasticity throughout the structure.
: - 30x2.5'+ 7.s x ry: - 54'8'kNm
Mr : - 30 x2.5-10 x l'5+7'5sinO(4)+7'5cos0(2'5)= -
Me = -
M? =
3o x2.5
+ 6 * I!1= -
61.46 kNm
34'61kNm
x2'5: 30 x2-5+l0x 1.5.+15sin$x4+ 15cos$
-
At9. -5V+5H+M 5V + 5H + M At7, -5V + 2H + M
At l,
2.94 :
Determine the member forces at the service loads from an elastic analysis, and
(Ð
61.46: -Mo 34.61
The latest codes on the design ofsteel structures and reinforced concrete structures are on.the Limit State Design Philosopþ. The main problem is'"he determination of forceq. at the limit state of collapse. Apparently, there are two possibilities :
2'94kNm
ContinuitY analYsis gives'
magnifu them by appropriate partial safety factors. Carryout a deailed non-linear analysis.
(ii) (iii)
-Mp Me
obviously, the first approach is very convenient but no-where near the true behaviour. design codes, however are currently based on the fust approach. The second is quite involved and needs a very careful evaluation of material properties, ion of input data and interpretation of the results. That is why, it is confined to ly special structures, such as, very tall buildings or unsymmetrical buildings.
But M: Mo: 17'35 kNm Eq.(ii) and (iii) give .'. = 34'7 H=1.01kN and V=0'42kN
3H +31.67 = 2Mo
or,
ere are
i
'
M*
=
= M, =
"
17.35+0.42cos0x 17.35+0.389x 17.35 + 0.764
+
x+
+ l'0lsin$x +
0'3?5x l'888
+
0'696
(
7.5
\ xz
lta"J;-1.
xz -
r 30cos0l x2
53s5
dx
= 0 or, x:0.40m
use
2'585x2
M*':
l7'35kNm=
Mo
of initial stiffrress and tangent stiffiress, respectively. It is clear that
the
stifftess although it involves more computational in the overall analysis. .;The incremental method is again an iterative scheme in ich the load is appiied in small increments as shown in Fig. 15.17. It is a very scheme for nonlinear analysis of structures having material or geometr.ical non O:K.
Thus,theplastichingedoesnotformatthevertex(section5)butat0.40mawayfrom section 5 in member 3-4-5'
"
gence is faster using the'tangent
x2
and
:
These methods make use of eíther the initial stiffness or the tangent stiffrress of the as shown in Fig. 15.15. An iterative scheme is illustrated in Fig. 15. l6a and b
J,
For maximum moment
dM*
two basic methods for the solution of a rionlinear problem
Iterative methods Incremental methods
LetuswriteanexpressionforM*inthemember3-4.5takingtheloadsasuniformly distributed (Fig' 15' l4i)
or,
s69
Method with Initial Stiffness ¡
The salient features ofthe iterative method will
be explained with the help of load curves shown in Figs. l5.l6a and b. Let 'us f,irst consider Fig. 15.l6a which !
NON. LINEAR STIFFNESS MATRIX ANALYSIS
. MATERIAL NON LINEARITY NON.LINEAR ANALYSIS
57t
:
510
road-denectiol behavtour represents load-deflection
ï::::ï:.,
tî"Ä| "t'å or I similar lniian""tion n-,ï"flî:fit"*iff it þ' ol"*t:.*,11",
a toaq tl :-llj, ?lTl;:fliffi1":rîãî.,.** äå"ï'o*o*':::l"S-:ilJl"i?:T:::;'"ää"t-*"T'T:'f:*ingthe conesponitins to
geñerated rf the T'*ï:"";iì;'uif'oq"ni"' be i--^-,^, can u' tl",ìs*r nroqerties, -fl1l;. ;;f;;ü;Ñes ffi:iffi;'åes of each member tiwr¡ "an or ^r conespond matenal "r¿ tne -"Yil,::,"':"'l"i:ç;io:'åïil;ï;"ï"hection behaviour ot the strain li*in ¡.ttuuiour rv' step' th::tä"îH:i#ålï.ïäi"î is iter singre. jii;.ï a . il outaine¿ obtained iterative i;d ì,ß
ror each
älil;i" ;pì ::ï'j", b, as shown rn rr
to.Point
STIFFNESS
äP,
load
p^
=>. ' Ëõ p,
Hã z
:'ä:'iìÏ:i;.;'#üioi
(rJ
t
ONLINEAR BEHAVIOUR
pt
Ár'
NON-LINEAR
.
fANOENf ST¡FFNESS
STIFFNESS
o o
ATA
AZ A3
A1
DEFLECTION
Fig" 15.17 Incremental method for nonlinear analysis
J
The ordinate a, b, represents the error in load for the structure. The correction in the load vector is carried out by considering the load-deflection equilibrium for each member, thãt is,
INIf IAL STIFFNESS
OEFLECf ION
Step
2
SteP 3 SteP
ft"..19{
4
is referred
vector matrix as well as load
Thuserror dPm(l) :0
as
:
lntroduce
chaPter 12' ur vr¡qPrv-¡-.' -' . discusseq in ãir"urt.d equations. ot llneal simultaneous system Solve the
,:.p.
Âul : ô- in the
ifmemberremainselastic.
p" - pul if member becomes inelastiç.
(iii)
is obtained for each melirber.
where
structu¡e does =KJ1 -^^" the structure"since the in poini nodal ea:l b, instead of \ to b' for deflectìon A at r --^-+ Jv .^ vv!ç¡'¡'nhr¡inert "ithe:",:l#;tå#äffi;;ä.""'*p""¿'to ' dlsplacenrvtrr ^ "onerpbnd. not remain linear' Po
in Fig.
(ii) is obtained by substituting
member
to-T tttj
e*tur" fe sf-u]f co-no 111#;rlii" the stirfiress boundarY
ordinate Pur
equilibrium equarions of each member computed using the *ä¿lR.¿'ltiffness matrix taking into account formation of one o. *orã prastic hinges in the
:::i:::i:j:::1,ÏilÏËi
I
mafix SteP
p. =ç-ô;
and tangent stiftnls rr'rr Initial Fig. !l$' 15.15 '"' stiffness g-lobal rber and assemble the marrlx or ""'" "'-"'"p,"rti stilfr' stirrness stiffness This rhis elastic.stiffrress erastic . î or 1,13. rate the carcu Calculate ", the'yuîlf matrtx or ttt" fr ressmatrixof stif ,ï:*::1"'"'il,l1.tlr*.structure' stiffrress ":",";î;ñiiffners matrix of the structure'
:
P,
:
yield load for the member
The Load dP.(r)
¡r called unbalanced road vector for a member. simir-ar unbalanced load vectors are calculated for all members of the structure. Finally the unbalanced load vector dp(t) for the structure is assembled.
l5.l6a'
The second iteration is canied out using dp(r) in Eq. (i), that is,
dp(r)
:
Kr d
(iv)
^(r)
The global stifhess matrix of the struetures K, remains unchanged.
o o J
7 5f IFF NESS
fhe total displacement
+d^(r)
^ This conesponds
TAN6ENT SIIFFNESS
by %bz OEFL ECf ION DEF L ECT ION
(b)
I
mcthod for non-linear analysis
to ordinate
br.
The error in load for the structure is given
Repeat step 5 and obtain dp (2) for the structure. Now repeat step 6 and step 7 till the error in load (iþr resistance) falls within an acceptable level.
r¿^rñ+¡r,o
is now equal to
NoN-l-ngnn
ANALYSIS
:
HYSTERESIS LOOPS
MATERIAL NON-LINEARITY
573
simplification is suggested through the iterative scheme shown h Fig. l5.l6a which employs a constant stifftess. In a given load increment, the iterations are carried out using the tangeirt stiftress matrix of the structure computed at the beginning of the load increment. The tangent stif,frress matrix is. updated at the beginning of next load increment. This is known asthe modified Newton- Raphson method.
*lmru,-ro*t*ml:g***rw
Choice of a Method
whole.
wfth Tangent Stiffness fterative Methdd as shownìn -':^- ''!rile of a stn¡ctures
Now the obvious question is which is the besl method of solution of a nonlinear system.'The answer cannot be given unequivocably as the algorithm,most ef;ficient in one case may be divergent in another. The stability and accuracy of-an algoritlim depend upon the degree of nonlinearity in a given problem. Nevertheless, th,. incremental method with tangent stiftress matrix in sufliciently small steps can be recommended. The tangent stift¡ess mahix is reasonably well defined and can be easily calculated for discrete structual systems. At this sage it is necessary to mention.that direct solution algorithms for the solution of linear simultaneous equations are faster than indirect numerical iteration algorithms. Thus a nonlinear problem is solved at a series ,of linear problems. The problem is further simplifieC with the existence of powerful linear
t:i-ll;tÍl;"Otåi'l
con,i¿",.*,"t*:*lî:Ï"1#î:$r:,'#:liil:iJi"tr'""ã'ectdenection u" e-".l"Iul]^-
o -^¿-;v e' :ffiilt;oãirving steP carried out r*r'Ë is iäî'.iirt""* TTt::tr'n" The second iteration Step 6 n:::r*"i1:*::^:ï;äi'Jitî1.î,'iå:,:'**'i't':ll,Ïri cispracement to aacisp curves can
U-*ïïïi;:'åii'äff #õãing
Jiìä;;;
the ,ri stnrcrure. ^t
-involv115T;:lli't#;;*rrãrponaingto stiffrress mr using uP¿ut"¿ ¿p(t) where Solve
Kt' = foidA
:
Kr d^
O)
at bt ' matrix of the structure *ng,"ni stifftress
(l)'
computer progrrims which can be conveniently adapted to non-linear solutions.
I5.9 HYSTERESIS LOOPS
is also known as iterative scheme
The elasto-plastic M-$ behaviour shown in Fig. 15.l8a, b is-valid for monotonically
Rcsrortheanarvsisr;ma'ääîTå'"j,#i:l*gff,**"n,liiîïJl Newton-Raptso"\"!!#irräåii^ã"¿,,tterefore assemblv of tanc:1':^1tlÏi:r., each
iteration' NeveÍnel
Incremental
;;;;r.nce
is
usuallv faster'
Method
and displacemet ¡rv small increments
rntheincreme",qi:'ïi',",1f
*t l""tJ;ä; i, .i*i* äiji:.iîIlj$*åf":'"J*î'H;3l,iiti tottt"m c,ffi",n.nt.l is estabrished. The ;"b remains consu *-t'ïïil;; îqÌill:",he stiffrres.* rangent member urrurtc *X tt-iJ'"h;;;:utî tt un", is considered ,o, "u"t is updat the load increment
is evaluated'
orthe necessary' Rest ot eacr' begirinin8 the At
stç'. tliï-"'""iotrot?r".liil-.. n;
during the load the the stnrcture for
rncrwg
th"
step. The 1"u¿""r"*"ntal
a, unqlan:ed loads
I
,!, l,; total number number í';,ri"d, the totar ïi' ì; ïii.i'"i";'f"d' deflection îlii:ji ,:tiri;:ï" l'"",ä#"ì äï"',1-":1':: ,:'ioi":ï" î,1¿ii;"i olurnU"rofloadsteps. collapse
üililÏli;uuù'ãà"'"'i"" ?,!',;ihcorresponos t" till
to the n itËrations is equal
'"ïî:îi:ii.",:Ti**1,:tL.:tn**,t**l*åîlj'rî,'ffi merhod. rhe unbalancï*onoo may be seen as i
;;;;,
lrttre stin
"otnbÏ11:i"1t-iffii requires'pdl*Ii efTorts' io'n"' T".T::T:':T"i.ili* tl9:,tt".t"1"r.lii| rr'ri computattonal orrig'
road step'
Fis. l5.t6t,
^lu
a beam element are shown in Figs. l5..l8a and b, respectively. In Fig. l5.l9a, l-0 shows unloading from a maximum stress level ofjust below theyield level.
road vecror
st îr :.::: sma* hence negrigrile, rf the load ";ia ¡1 *l_u_l?,ir"¿. ThemodifredtancentsriÏ;;ïtg.ls.rz increment, Thus, in each roao ,ne procedur i'-::-1ï orb¡ , bz
repeated
increasing loads. When the structure is unloaded, the beha'¿iour is invariably assumed to be elastic until further reveise loading produces plastic yielding. The loading and unloading cycles may be continued on the structure. The force-deformation (or moment) relation is known as hysteresis loop or model. The area of the hysteresis loop ,gives the energy dissipated during each cycle. This behaviour is again idealized as an ic behaviour for convenience. Typical hysteresis models for a truss êlement
ro sreater
Mp
/TDEALTZED
My
t-
zl¡J
SI
R
AIN - HARDENIN6
:E
o
-o CURVATURE
(o)
Øv
Øu
CURVATURE
(b)
Fig. 15.18 ldealized m-$ curues for beams
HYSTERESIS LOOPS
: MATENAL NON:LINEARITY NON.LINEAR ANALYSIS
574
M_ I forl < Mp
loading takes as shown t"q elastic be to unroadins rhis i:::;:î3":i"1"îo',ÏT,Jå::il"JI'"TÏ*n This unloading i:T-ilirú"d fne vrcru beyono or ':':;;;; above ptastiè süain' It occuf: the prastic ãs as tne óru." totn a stress elongatton o-D residual I 5- I et; o-D The Fþ l5'l9b; in Fig. 2-3path as ourrtrË irrv traçe.s traçes the same
üä;'";
level; O-H represents Plas
0.15
(r5. r2)
Py
;.'""-:bou",î.::ti::"*il'åi':JL"*" :I"i},l::':"ï " ili; löl ilä''r"'r "^ .; Jitnilrrv, ?å;:fi :*.:':9:îl:ï:#':å vi rd beyond the vierd ¿ú is bevond äïr'" ;;rãadine pk:1 tl"::,::i"yi:"j[:h; n-ä A:; *r'iÉú :l'iil;;' ä'"rv ""t:îgig"in::J,1"fi,ï:fi.ii:"i'ïä"í; stratn-l-.',^ su4¡u,irci** plastic plas'ç F. represents from F'
;il"ï;; [,ît"tr*,
575
M
e
trSM;
^.-^¡- ,,.^n unloadingfrom "p"riunloading
Mv*
* P
> 0.15 Py
MOMENT
3'
SHORTENING
(O)
TRUSS ELEMENT
'ffi-1
i/ ,v '_/1"; (b)
,,/cunvrrune
RAMBERG_OSoOOD
l'
CURVE
(o) STEEL
(b) beam element loops - (a) truss element Fig.15.19 Typical hysteresis The
curve and as Ramberg-Osgood in Fig 15.19 b is known hYsteresis toqP.sh own
equatton expressed bY the
ffi;'-J,*À;o""*0,î1'1"å*:ï"J5i::îîäïl:Hä1"" i:Ë'ru;ii'i'iäï' ã"u;"¿ing hvsteresis moder' ""otT*llí"i,Hi|å"i'i"'.i|
r""il;;;;;;;";'.n:i*r^"::l',':'v^,:'.;:,f ï:t:îff lîLïr?i#;lî'":''ii as elastrc' ¡ì¡¡J
*.'1""*äT'lli?":,::ii?#;'"r,ï:"i::xl:yi;:"îïîi:L'T'i'ii;.ffi
'¡w¡Esr"' the columns are freated l"t"r"oo. Inelasticity in beams ar€ Ëravru columns because äî;;il;;",secorumns*:ry"Y]::u":,X::?,";;:;îîiriotu*ndàsignphitos collapse
collapt:lTl:IJj"Íl.i* ïlîXllLl;åil#iJi"itËtilii:Ii''ì."'iîf ;lXr:ll**ål'J:3'Ji Tfyl:g"i#är¿ï:"i,i!ï"i;ri;;ií":r :#ilä;;;crete ;:üäääffilaxial
4partial or local oro column is reducec i, p,"r".'.ã rol interaction cqrves ror steer axialÏorce prcsctl ur ¡t' ¡rr---.21au âro axial column steel column' For oa ¡r*-' b. D' rur upon the amount of un¿ and f S.ãf .r.^.,,. in Fiss. l5'2Ia -- ^-^ ut",tl.:]Li:,.rigs' columns torc" relation is givenbv
;.";drî
lMt BEAM
(b)
R.C. BEAM
Fig. 15.20 Idealized hysteresis loops for beams
i
In non-linear analysis, it is absolutely essential to keep track of the sequence of ding and unloading as the solution is path dependent. The tangent stiffness.of a ' depends upon its current state of foree and deformation. the ¡ncremen-ø of analysis with very small load steps is almost invariably used. The change in stiffness depends-on the axial force-axial deformation characteristics truss members and moment-çurvature or moment-rotation relationship for beam or members: The moment curvature reration is required.to get the måm"nt-.oiut¡on the momenr-curvarure retation is simitãr to ttrat strown in nig. å"1:::l,b_".l.Ts l8a. The small curve near the yierd moment makes the derivation,rrr,"tr.ä""i_
ttion relation quite complex. Therefore, rcrng any
it is desirable to idealize it without appreciable'erro1. The elastic-plastic moment-curyature or the elastic _
moment-rotation relation is frequently qsed in any non-linear structural analysis as in Figs. l5.l8a and b. rh.e srrain-hardening srópe in Fig. I5.tgb,"f-u.,ur."" n2%oto 5o/o of the erastic slope for steer ur
*ilr *
"on...,"
beams.
members, rhe moment curvature retation exhibirs complexiries
f:f:t'::l9"T.:t: during the inirial loading srâges due to cracking- of concrere una li"riing or rcement. The moment-rotation relation is again idealized .trrl¡,ç-plastic for
*
_jËæ*-*: slcl
MEMBER STIFFNESS MATRIX : MATERIAL NoN-LII"¡E4'RITY uoN-LnenR ANALYsIs
)/o
2
I
A
T 0
K=-
EI L
cutrves
for columns
I
a mu n-linear analYsis of non-trnear,i::ï; .,rns are made masç for the i" 'l:: partition partl or assumptions assumptjc bearing rv.e any load "rrti., The following wrtnout artr ---^ ranalyzecl without is o-atrrzed o.TJf*u
'Ï'
ft"t"d oiitui"t'"t* "iJ
plarre after bending ' Plane sections remain beam element' only at the ends of a n ot*ti" hinge can form 't'i,ig. ttuuing u zero length'
2.
The moment-curvaturerelation elasto-Piastic' ¡"-* aho erorv relatton ls Ílrsu tri"i"i-rt
4.
.OnS deformations
5.
The
6.
F'or a Fig. l5'l9a'
0
t2 ----;. L.
6
0
-6
-0 I
I
^'
(ls. r 3)
0
o+
6
U
0
2 J
L
-6L
4
4
.EI K=--
00
+o V
_A I 0
5
0 J _E
0
L
A I
Itisa 0
0 0i4 :I
-1 v; oio IL! oio
0 J
r] L
0 J
I
I
2 J
4
(ls.l4)
.0
-t J
J
is bi-linear or elasto-plastic'
is neglected that is' o-delta effect - - small' assumed to '- be are 4rJurr¡ve aIv
¡
--r.riôn relation force-axial displacentent axial the element, mrss
is
given si.
MATRIX r5.TT MEMBER STIFFNESS 2-D BEAM ELEMENT
'
0
iL "i"'ã"-"""
L
0
undetrgo infinite rotation' A plastic hinge can
J.
4
L'L 6.
A I
;;t
etc' rtuit"uttt or lift wells
1.
L
0
I
ASSUMPTIONS
:
U
-T
When moment at any one end of the beam exceeds the plastic moment'capacity of that end, a plastic hinge is formed. The beam has one end fixed and the other end hinged. If the plastic hinge develops at the left end, the stiffrtess matrix is given below :
(q) SfEEL COLUMN
15.IO
6
5
A
R. C. coLUMN
?=+,+=+ Pt' tY "'-"' Fig' 15'21 M - P interaction
12
00 .:t2ó
_T
0
b)
0
L-
0
(
0
I
0 - 'rt"
4
J
behaviour elemerit under elastic : below and is reproduced -^:^ L:-^ôG\ is oiven uo eq'iZ'¿
of a 2-DThe stiffrress marix
prismat:-f3
Analogous
to Eq.l5.l4, the following equation gives the stiffrress matrix for
a
prismatic beam with a plastic hinge at the right end J, that is, rows and coulmns 3 and 6 in Eq. 15.14 must be interchanged,
:
NON-LINEARITY NON-LINEAR ANALYSIS : MATERIAL
I A I 0
o=î
0
J
J
E
I
4
A _T
f
6.
0
'l'
0
00 33 --;vL -. 00
k" = 0.1%
013
J
0 J
0
0
E 0
(rs.l5)
;r
I5.I2
K' :
ol
becomes null.
reducestozerowhileiti,ut*t"yieldlevel'However'incompression'itislikelyto is reached unless the member has a very low buckle as soon as the critical Uuckiing load again becomes zero after it buckles. lf the slendèrness ratio. The i"*U"r stiñess iucking' its behaviour is similar to that in member yields in .o-pi""ion 'instead of tension.
of a truss element in tension and There is a significant differenòe in the behaviour
compressionunderreversedcyclicloading'Theunloadingfromthetensionor elastic and the energy is absorbed, whereas' the compression yield tevels is *roln"¿ to be to be along the original path, and no assumed unloading from the u""r.ìi"i level is energy is absorbed.
-'l
(rs.r8)
rJ
Rr
kR
(l.2.t})
here,
LM
(l) to LM (6)
previous structural stiffness matrix. new structural stiffness matrix
of Ç are the same ad those of K. Thus, there is no need to ble the entire stiffness matrix for all the elements. Instead, the stiffness matrix is for only those members whose stiffness matrix undergoes a change. If more one member develops plastic hinges, the same procedure is repeated. Again, the procedure is repeated in each load increment.
Rest of the elements
The procedure remains the same for a truss element except that four elements of l( need modifications corresponding to.the four d.o.f. of each element undergoing a se of state.
:
(ls.r6)
its stiffness matrix is If the member yields in tension or buckles in compression mafix, that is,
u'=
= K = Kn : J
reducedorwithdrawnortheultimatestrainiSreached.Theaxialstifûressofthetruss
l0 0l r [o o]
L L-' "t ïl_,
K. (t, J) : K(t, J) + k'n (r, J) - k, (t, J) I : LM (l) to LM (6)
Theaxialforce-axial,deformationbehaviourofatrusselementisshowninFig. keeps on increasins^until the load is 15.i;;. Ë;ñù; ;; t*;r"" rh;axial etongation
AE
t
The structural stiffness matrix can be assembled as discussed in Chapter 12. When a plastic hinge is developed, the stiffness of the structure is reduced. The new stiffness matrix of the member in which the plastic hinge(s) þas formed can be generated using Eqs. 15. 14 to 15. 18. Let the new stiffriess matrix of ihe member be k'n and the previoui stifûress matrix, that is, prior to the formation of a plastic hinge, be k' in thl global coordinate system. The location vector for this member is already known whibh gives the locations of the elements of the stiffr¡ess matrix of the member under consideration in structural stiffness matrix. Let the location vector be given by vector LMu * , - Now elements of k' which were used earlier have to be replaced with the elements of k'n.
2-DTRUSS ELEMENT
,']
¡p I
MODIFICATTON OF THE STRUCTURAL STIFFNESS MATRIX
olu
the beam the stifftess matrix plastic hinges are formed at both ends of
-,:i[],
579
The global stiffness matrix of a member is given by
t5
matrix is given by For an elastic prismatic member the stiffness
VECTOR
A null matrix is likely to introduce numerical erro¡s in the solution algorithm. Hence, it is usual to set the post-yield stiffness to a very small value, say
ol'
0
.L
0
if
_1
0
J
EI
Finally,
)
INCREMENTAL DISPLACEMENTAND LOAD
a
null
5.I3 INCREIVIENTAL DISPLACEMENT AND LOAD VECTOR equilibrium equation at i th load step can be written
\U¡ , at
(r5.17)
(¡+l) th load
=
P¡
step, this equation
0!
as
(15.20a) would be
+ aq)(U¡ + ÁU)= (P, +
(rs.20b)
^P)
581
UNBALANCED LOAD VECTOR
NON:LINEAR
580
MATENAL ANALYSIS : i
gnorin
NON.LINEARITY
give¡ 4e g very smalr terms'
j.,i subrractins i lrr:.**, ä:"äitibrium equation: "* "ï;'"11";îd *,"lJ,liiiiil;Ï A: IqAY
Mi ts
(15-20c)
zt¡J
-
where
and
load ÀJ = incremental
ot = **"'n""iins
Knowing K
Í. o 2
vector
steP' and A? in a load
Mi-t
'
vector. ,
)lacement increllJti"iï:"ffi*'"ìi"'"'**'
Mpi
dispracement
can b
0i-t êy
^U
ROTAf ION
vectorforamember,","""""",uinut",Ãõcanbeobtainedas:(l5.2l) ,n" ,r"r","""r1;u*="i:;:"*", rn"*b"'
that in local coordinates'
^j rheincrem:I1,
î,;;i li: ;i;f
aican
matrix orthe using the stirfiress determined be
is''
= k ai _= kRÀu
(rs.22)
j::ï#:::'iffi""i:,äi',i:i:"ffiåå'JÏåï:îiJ'ï:#fi'1:
i'i'TJJi"i
iä*
step Fig. 15.22 Unbalanced moménts in a load equal to.pe it predicts'a moT^"1.t:r:t::::: tnstead of preäicting a moment increment ' momenl In oider to avoid errors unbalanced the pn is .
ä;;il;;"órno^"n, tXlij:i:' ; ;;;;i;; fr o' u""u*u lutin g equ bri um ulbalrestore 11c¡¡ :]::,,i:Y Because of equilibrium. step ro ^? ;i"Jïui;iilffi*Jn"xt byload tÎ"0': this eliminate To pn' the actual member fo... i, lt'í TmPgrary Pi9 iIi
rou¿s ro, inè'nexr iiå ;: ffiä "àî.ãi"ìi. only' period o..*itttd to act for a short
unbalanced moments at the ends i and
LOAD VECTOR UNBAI'Arrupv "-14 UN"NLANCED ,,.,N the moment at is confirmed when *"t"1' :' or a Prastic hiif ", D everopment : plastic momelrr-ov**t-l*rl ¡î:::,:":: the remains unbalanced' -"i:Ïrtt:i:"iäåJåi "oourt therefore' outvrcu* th€ rnembers that *::: i:T:::iJlî: Mo' and *::"j""j""jll,uti|Jå '""1 ',".äit than Mp, in.all those sreater moment rr"*r *""ä 'o'n"nts
î\il:.åiiïÏ::** f i .,"T:ï,":'*"i'":åJj]ËÏ'rËî::î,iliä;;r"'ü-tî'*ff x:'"::; rheroadvecrorcorrespolj'::,::"T;;:iiiä'iiiiä!'f jfu:*l'fi
'åï:iï:iï::
a*å:f -i**i"+ :Ji:'"i'lli"'i"ääffi;canbedeveroped rt[l1iT'#t*i*"ileålåli*i**"1.'**rru*xiÏ:iï:: follows
pn¡
:
Pnj
:
a
îldjlt;":::îiiïïir:ïr':î'':ï:i'iiJi{j#,'J:î:Ëlçri'r*ff iri'i,,i"i.Ïi*f ll*1";*î1,.,"1f ¡ru'rv'--' ;ffl,Ïffitui""ìîo-'atpointBandthe in Fig' trzz-as shown
;;ÑexceedMo'
.
rhus'
Linear momentincrement,
on,
incrementpo Nonlinear moment
=='T"
fl"'
as
follows
:
pni
.t
(15.24)
Pnj
shear force at end
'
Pnsi
'the
incremental moment' The
will always have a positive sign' The proper sign
is
: O"t*ffi
Pnj :
Pnsj
= \vo'l-lv'-tl lr f
(1s.23)
by
l*,1-lt,'i
;;;;;;btained
member'-': Ï: :::1,,i:i:"i^ï :iä liï'ïînü:fl{ iiH"' T:l;
."olï,iiil:î,äî','1ilöif ü'Ë;":îyk+,ËJttffriil*:t*l'-'"i:': ii:i:i
are given
¡rra,¡-lrraorl
:
:":::",
j
road srep. The fictitious extemal load
unbalanced moment should have the same sign as
Ëftiî",i"":fr as
ê¡
:
Pni
M¡ .. xläl l"'jI i is given
*
(15.25)
bY
l
Pnr
--;-
, ., Pni + Pnj : (-'L
05.26)
members, it is assumed that unbalanced axial toads ur{uto,because in flexural in loeal coordinate system' ,l load carrying capacity is never exceeded' Thus,
NON-LINEAR ANALYSIS
582
:
DUCTILITY
MATERIAL NON-LINEARITY
583
T5.15 STEP.BY-STEP INCREMENTAL ANALYSIS METHOD 0
ln the incrêmental method, load is applied in a number of steps. The basic assumption
of the process is that the displacement varies linearly during each load increment while the properties of the structure remain constant during this intervà|. The analysis procedure çonsists of the following steps:
Prsi
Pl=
(t5.21\
Pni
0
Step
Pnij
I
Pnj
follows i the global coordinate systern as This can be transfornred nto
2 Step 3 Step
:
4 Step 5 Step
(1s.28) Step
:
= l"o,l o, set lv¡l = I"o,l
(1s.2e)
6
Update the structural stiffness matrix and load vector and repeat steps 3 to 6, the total load is applied or a mechanism forms.
I
ln case of a truss element, the total axial foree is compared with the tension yield load or bucking load depending upon the nature of the axial force. In case, the axial force exceeds P, or P"., the unbalanced load vector is computed and rest same as for a flexural element.
M, urd M, remain unchanged' TheunbalancedroaûvEçru:::#äl*uuluil"edloadvectorisaddedtothe rhe unbalanced load vectdr for. the *ructur: ,1^]i ,1s511:::. il'åå:ii" li: fr1 ioad vectors for all membetl:,, .Lt^r,o., ir nrastifi cation has' occurred, jiT"1'.Tl'll'i:iii:'"î,i1l"",l"lilll
""u"iã"J il"i:'å:ïL'å1i1""äiil::i'il:"':äï
néxt load srep ro set the incremental {isplacemenr
vector.
unbalance.i":j"t::,:,Y-,tfÏtii:î"::,îi#tff'ii"1: the axiar rorce is kept constant .",ili1T,'ff;T,i;tiï:;"ri;#. i;;'ther situation. rhe procedure to carcurate the
in the moment' and:the changes are made
P
uHglt¡Hce
l--_*|
UNBALANCE
Ï----'{
\--ti+t
YIEL D SURFACE
-)(Il,'.'-å.' (o)
Compare the lotal moment developed at each end of a member with its plastic moment capacity. If plastification occurs any where, calculate the unbalanced
.
The sign of
M
Determine the incremental and total member forces in local coordinates.
till
Pn¡ + o, set ltvt¡l
"
Solve the linear simultaneous equilibrium equations to get the incremental
load vector.
\
member are reser as developed at the ends of the Now the actual moments
:îï",il":H,;:i*'llih';;i-i"r,t
Assemble the stiffness matrix and the incremental load vector for the structure.
displacements.
= - pnr, sin 0. - . , p,nl p:;; = irn.¡ cos 0 p'n¡ = pni. ^ P';; : Png sin e pns¡ cos 9 p,n5 = p'no : Pnj
Pnj +
Determine elastic stiffness and equivalent nodal loads for each member and transform in global co-ordinates.
I5.I6
DUCTILITY
Non-linearity is measured in terms of ductility. It is defined as the ratio of the maximum deformation .to the yield deformation. The deformation may be strain, curvature, rotation or displaeement. A strain based ductility depends on the material ,properties, while curvature based ductility depends on the shape and size of the crôsssection ìn addition to the material properties. A rotation based ductility also includes the of member length and membér end conditions. when the definition of ductiliw is with respect to displacement, the entire configuration of the structure and loading also taken into account. The various definitions may be written as:
Curvature ductility,
Fo= lgh*
(r5.30)
0y
Rotational ductility,
Ite
:
lel I rmax
(15.31)
t,
l^
l!
r
t0-l | rlm&\
ol
0, (b)
in M - P interaction curve Fig. 15.23 Unbalanced load
ofthe procedure is the
Displacement ductility
l¡¡
r rmã : l¡l
é"
if m- 0 relation is bi-linear (15.32) (r s.33)
584
. le l;,* le, f =
Â, =
to
I5.:I7 ILI.USTRATIVE EXAMPLES
0, : Yield curyature e; = Yield roøtion e; = Plastic rotetion
where
for The yield rotation yield anti symmetric
A= 'v
A sinele
storey portal frame shown in Fig. 15.26 a is subjected to incremental lateral
ttou¿-p. "plo, thê load-displaceñTent cur-ve and the sequence
pf plastic hinging.
Also,
I
Yield disPlacement
when-tlrl d'efTl:s 1:J:tation Fig'l5'24' in *o*"nil"fo u* 'tto*n
a beam is
"';îä;
Example 15.9 (15.34)
lo,
58s
ILLUSTRATIVE EXAMPLES
: MATENAI- NON-LTNEARITY NON-LINEAR ANALYSTS
t"T':":
is subjected
and is equal
to
:
tsMB 150 (15.35)
M"L 6Fl
NON -TINEAR
z
t
500cm
.
d ? o
r
LOAD
roo
SIEP =
10 KN
J
(o)
'-'y
¡.--,-
in a beam element ßig' 15'24 Yield rotation
LATERAL DISPLACEMENT' cm
lfastructureisductile,itwilltendtodeforminelastically.lntlrisprocessitwill
,:'ll ,.uilloîiä',*=::l':åiiï""'iïiJifflJ':Jff Ï::l-t"'ï:"Ï'.'i:iffia brittle
t9-dgform on ,tt" the occupants b.f"r;^;;tü;.. Fig' to deform'as shown in sienals sufficient wamlnf o'itä"''"n¿'ir'L'"ro"' predict the to* to very able is has .tir.tu." e "i"-ìií*tánalvsis 15.2s. tr faits suddenry
;;;'',i"".
"*i.,"
:":':"'
"to;;õ;ftv
färj ä;;r"g å"*,,i'v**n:T:1:"1";'o:iii¿d"fi*ïi:Tiäilrkfu can as a steel structure
nì*îtri*ì*ole
or
:'^ilÏ'Ï:Ïîä:'ll
forces' to resist severe earthqtrake
(b)
SEOUENCE
(c) L0A0-DISPLACEMENI
OF
CURVE
PLASTIC HINGING M
M
EF
Up
1.0
l.o
ts
z
E
o =
zt¡J
o =
t¡J.
2.
0
5
(d)
15
10
ROTAT ION
MOMENI-ROTATION
20
0510 RoTATtoN e./oy
ê/Av CURVE
-
COLUMN
(e}
MOMENT ROIATION CURVE
sËau t¡J
L)
É
" Fig.l5.26
o l!
DEFORMATION
behaviours Fig. 15.25 Ductile and brittle
, The geometry and yield properties qf the beam and column are shown in Table l5.l. lateral load in steps of l0 kN is applied at B a¡d the frame is analyzed using the analysis program MINA-2D. The program prints lateral displacements, plastic at the ends of each member as well as the location of ihe plastie hinges. The
587
: MATERIAL NON-LINEARITY NON-LINEAR ANALYSIS
586
sequenceofplastichingeisshowninFig.l5.26b.andtheload-displacementcurvein displace¡nent of 2'51 a lalteral loui oi 140 kN and pie. I S.ZOc. The frame i- ìkï
",i'.
Und., increasing lt"d;tht
:
"p," collapses at a load 200 kN' ¡Åe -of
Table
l5.l
Mp EEAM = 0.5 Mp
'
Member ProPerties
lsMB
(o)
COL
300
S-EQUENCE OF
PLASTIC HIN6ING
Themoment.rotationcurveM.0forend.AofthecolumnandforendBofthebeanl -:i:::'::Y: curve in Fig. 15.26c- rhe strain':
il:'å"J:ri"i"ï
;î
LOAD STEP
i;'t*
lll lÏÏYïîl i:î;: äiX*ï: Lìíä;;" the road-disptu""'"nl
d are shown in Figs. I s'2 6
hardening stope is data for these members'
^."
-
IO KN
il;i;: îi.Zøi"r,it" column. This is consistent with the input
the plastic of a weak girder-strong corumn frame. rn order tö study the behaviour is. that column, the of to 50Zoof that ,.t
*"ff|äji¡ryä*
"
i*,oì,
Mps :
"quul
0.50 Mpc
in FigJ5'27 a and MINA-2P: lld. lh" results are shown The frame is reanalyzed using of 2'4 cm' displacement lateral and upiíu load of 130 kN 15.27b. It remains and the frame "rur,¡l ¡, J"i"r a displacement of 51.8 cm Fina'y, at a lareral l""d;,iöiff-¡,.
Ib)
LOAD-DISPLACEMENI CURVE Fig. tS.27
collapdes.
rn the frarne
t of Fig- t5'26 a the plastic mcT:lt ':fi'j:v :TT :1T,':Ï l,' It can sllies' n t521a' y":.0 i¡,r{" rrame or Fig.
"r,Ï"liiïiäi,i,i,ii;r;;i
moment capacity effect of further reãucing the plastic seen thar there is no ,ignii,";ni frame' ductile a is frame the either case' the beam îromll%o to 5ôt"' ln
ISMB 400 in order The beam section is increased to
^to ::îO1.t*"*i,T.tli is 1.80 ti prasric moment capacitv of rhe beam. _,rotJ:*::äJ"iffir*t.n. n" i' unãrv"'¿ und':'-" l'Ïll'.ïîîiîîl ]T :'*,:i1-; ãf that of the column' rh;';;;' much smar! corr"ps., at a load of leO kN bur at a
ilffi:,Iiä"ilï;;;;fr","e 3'2 cm' Thus' a strong girder-weak displacemeñt
of about
¡z o.
Mpg - I.E Mp6
ô o J
SEOUENCE OF PLASTIC HiN6ING
NON
- LINEAR
LOAD SIEP=IOKN
o r z t1
column frame is a britt
OISPLACEMENT, cm
(bt
frame.
Fig.
15.2E
LOAD
-.ÐISFLACEM€Nr
CURVE
ÑOU-I-ME¡RANALYSIS:
588
ILLUSTRATIVE
MATERIALNON-LINEARITY
EXAMPLES
589
Solution
ExamPle l5'10
:î''"Ï'Jiïi
ú.t' uu ¡our of the i# ;;": iii"u,loJff Ï";":;-obayrrame::T-*",i,ilåJi;3ï;"1î,ff A rwo storey-tw"
93J;î: ;ili""i"t i'I;lï.' ll-^, road ,^., tirr rilt coilapse' collapse. iit.llïi'ñäåiäiiut"'ut lateral and incremental
10
2ookN
100
kN/m
8j -35 kN q
100
83.3 5 9
3
(b) EOUIVALEN'f NODAL
LOADS
The geometry and yield properties of the members used in the frame are shown in Table 15.2. The yield properties are calculated in accordance with IS : 800-1984. The gravity and lateral loads acting on the frame are shown in Fig. 15.29a. Thp uniform gravity load is replaced by equivalent nodal forces as shown in Fig. 15.29b. The lateral load of 100 kN and 150 kN is applied in I00 equal load steps, that is, in increments of I kN and 1.5 kN at the first and second floor, resqectively. The nonlinear analysis is canied out using the Monotonic lNelastic Analysis (MINA-2D) program developed by the author. The salient features of such a non-linear program have already been discussed in sections 15.8 to 15.15. A detailed discussion similar to that for the STAP3D program is beyond the scope of this b-ook.
The storey shear-latèral displacement behaviour is shown in Fig. 15.29c. The in the same figure. It can be seen that the frame is linear upto a base shear of 85 kN. Elastic analysis predicts a maximum lateral displacement of roof equal to 6.25 cm, whereas, the inelastic analysis predicts a sèquence of plastic hinge is also shown
displacement equal to 29.75 cm. This is a ductile frame.
Table 15.2 Member properties
(o) LINEAR
Member NON L¡NEAR
FIRSI SECOND
ú. <1 501 l¡l T
rn
td
o ?
Compression
Buckling
A
I
Load,
cm2
cm4
moment capacity + Mo; kN-cm
Plastic
56.26
8603
+ Ió300
P",, kN
6
FLOoR,À
z.t
Tension
yield Load Py, kN
FL00R' A2
Beam
ISMB
3OO
Outer Column
I
I
l4l8
-126
66.71
r3630
+22240
t?60
-1004
92.27
30390
+ 38825
ISMB350
lt
lnner Column
ioo
ISMB
(n
DISPLACEMENT
(c) Fig. 15.29
4OO
GEOMETRICSTIFFNESSMATRIX.TRUSS.EI.EMENT
591
The elastic and geometric stiffrress matrices are determined for each element at the beginning of each iteration and assemblçd to form the system stiffiress matrix. Except the,solution algorithm, rest of the procedure is the same as disgussed earlibr,in Chapter
CHAPTER
sixteen
12.
16.2 GEOMETRIC STIFFNESS MATRIX - TRUSS ELEMENT
NONLINEAR ANALYS
äËo*rntRrc
Consider a 3D-truss element shown in Fig. 16.1. dssuming linear gla¡tic behavio-ur, the strain-displacement equation can be written as :
N
'N-LTNEARITY
.:(*)'.;(#)'
a
(t6.2)
'.å(')' negtecting the higher orrler term
llÏrmltinff o ,.?lnîo î"*i"
(r6.3)
Let us first consider a 2-D truss element for which
t""":^:itfft'ru:tffif"'n;;iîÏi
{1611assump'fion T;;:;;"re rapidlv than predicted l"n""ti*_,
for a 3-D tn¡ss elèmeút:
#.;[#)'.i(*r)'
16.I INTRODUCTION that the deformations so f1r' i1 was assumed of sup-erposition was ln the previous chapters discussed principle i, *tv,irte structujf,l#ätî:-rr,u, deveroped in the äir-,""un, varid. rn practical ,"r*r] "* is no more varid L mav
åîîi:i.$:::::
* compared l*lt ' to $ôx ia*/
T '
í9rì' ôx Z\Ax)
r.' w:0,
9s *
(16.4)
deflËctions increase yPfY be found that a load tr'ere t'"åäìi"tìv "rtere -may 1" "'¡1t"î"à" bv a linear solution;
:i;ü';il;jF'""e:#,'i:ii'i,""'ïft
"ï'îff
"J'$i6:mr,'ty:ä'Täii't
iutgt tpun :"0t1t",:lt:åaciry deireases witn coniiìrît"r carrylng cal
å"l:*lli"l
¿lso falls within
ul T' ia
where load
äåïïîä;:=':r:ll-"*:iä*r**:J:ril'ï"ïîuil1#::ii,::l relations betweetr stress
t.
/
(u2,v2rw2
7Wl
this book.
(
with the
:pry"Ï:L"-:,::",iåî*':n:"î::lliå;l#ìi""il i"ii^:ïn:i*t**n:",'**:miilïlåiii, iff ,."i'",ï'"'"iï:."ff lhe i,*01o.",,""
increases
u1r v1r w1
)
)
cnanBç on îïî;"ä:ï:ï'",1ÏJJJJ"::'"ÏT;:';:ü;ËI-;;ôfxr;n*:'Ï1p:iff íhe ,tifüterr matrix depends carrreu uul "' """';rì;å i*ru,i". methods or incremental; be can o{ of unutv'ìt:lnb:.*ttt-t-d^":lj'11"1]-"¡-ned analysis ç¡Ùrrw¡
Äät;iltt""t
Sucir a nonlinear
äìir"¿t.
c""metric non-linearity may
"^[i-t":
non-linearitv with material non-linearity ttäiji"ã *f,n additional involve not no"nuo'ue t*,. does
i:rï: iÏ"ffii in the *.u;** -"Ïi*ï" formulations' *ilti::,#'ffåî"'o' äå'liì"'"'p"rated
i:.i,iî
ï'"J"i:l"
components: element consists of two truss element or a beam The stiffness matrix oFa
¡6 lvhere, and
= lÇ +K,
K^ =
fì :
elastic stiffncss matrtx geometric stiffrress matrtx
(16.1)
Fig. 16.1 3-U trúì element The displacements u and v vary linearly 4long.the:bar,
leigth. Let (u1 ,v¡)
represent the nodal displacement components for nodes i
andj, respectively.
Setting,
.x
and (u2 , v2 )
GEoMET,RIC STIFFNESS MA1fRIX -.TRUSS
NoN¿INEARITY NoNLINEAR ANALYSIs:GEoMETR'IC
sisz
l"l:L o
tt
'
[n;J
'' = EâTI4, =
, IL*.
-rlerl-l* a,.l.*J +\axl I
(16.7)
and integrating 16'7' neglecting higher order terms , Substituting Eqs. 16'6 in Eq'
92L
(u,2
-2u, u, + u22)
*L
tu,
- *,). ='
. #zt]
(uz
-
ur )
(vr2- 2vrvr+
v'])
(16'E)
F, even for relatively large displacements
u=
ffi
+ fi(u, - u2)2 f
(vr
-
Yz)"
(r6.e)
aU EA (u, _ _ Pt = I \-, ã;;: AUF pr:#:i(vr-vz) P¡= P¿= i
99
I:I -l;0 L0
The Slobal stiffiiess
mlrix
K'=
dYt
:
,
(l6.l3a)
llr
ol
Lo
ÀJo*o
is given by
:
I
.
The direction cosines are determined at the beginning maintained constant through the iteration.
K,=|9
0 0 0:0 0 0 Oi0 -l 0 0; I
0 0 oi
oi ..
(r6.10)
(l6.l3b)
Eq. 12.7c.
ur) -¿'
=
L
as,
RTKR
where R =l À
written
is
11e.tz"\
-ljo*o
r 0 0i-r
+ u2) Pt-u, t y=|{-u,+nr)
\
I
of each iteration and
For a 3-D tn¡ss element the corresponding expressions can be written as
oYl
ottz .
ol
1
Using Castigliano's first theorem,
'
ffi ffi;
FA-F-\i
(l6.r2b)
1i9:ll and Kr=rl9 Llo o!o ''o'l
where,
Eq.16.8 reduces to
lìll{l}
I il
[0, tit o;o j
yields,
U=
+lj,i.i
:..
aufavl2
4l i
å :l
6
(t6.t2a)
'
= ;it"',.4"=fJ"1o*
"
"
(16.6)
equals
; ;l
K =IÇ+K,
Thus,
ôu -ur + u2 ôxL ô, = -vl+v2 ôxL elastic material The strain ençrgy for a linear
[,^lå å
=1'L;: ltl
(r-e) o ejl"'l
by differentiálion,as 'The displacement derivati,vecan be obtained
'-0
tl;l
(r6.5)
:
593
In matrix notation,
fu'ì
_tt Eol]''f {t'l } =l[(r-e)'o
ELEMENT
t.
0
0
0
0;
0
0
0
0
0.Q,,,
0
qr.o
'o
:
are
.
NONLINEAR' ANALYSIS:GEOMETRIC NON.LINEARITY
594 and
x,
o, =ï 0 9 0 0-l 00
0 -l
ol,
j
o o -tl
0:
'0 I
ól
-1
0i -l
;
ô
0
0
(r6.r4)
ol
Geometric non-lineai pröblernSrcân be analyzed using one of the several numerical techniques. Theie are basically three broad categories of the numerical algorithms : l. Iterative method 2. Iricremental method, and 3. Incremental cum iterative method. Iterative method
In an iterative method,..ê structurejis f¡1lly loade.d uslng an initial v¿lue for the stiffrress. Because of this, the equilibrium is not satisfied; the total unbalanced load is used in the next iteration to compute additional increments of the deflections as show¡ in Figs. I 5. l6a and b. For the stiffiress, either the initial stiffiress or the tangent stiffrress at the beginning ofeach iteration (Fig. 15.15) can be used. In general, the convergence is faster if tangent stiffness:is ,used, although, it increases:1þe.,programming efforb considerably since the stiffriess matrix is computed, transformed and assembled in each
lJu,o
The transformation matrix is given by Eqs' l3' l9c and l3'20' 16.3 NOI\'-LINEARITY OF' CABT,E SUSPENSION SYSTEMS
A cable is a truss elemen! that is, an axial element. However, under compression it to always becomes slack and looses its stiffiiess. Therefore, cable systems are designed non-linear remain under tension. CableS do not exhibit linear-elastic behaviour but rather Another defQrmations. to large dué caused is nonJinearity Furtirer behaviour. elastic salieiit'feature of cable suspension systéms is the desirability of prestressing the cables' The modulus of elasticity of a¡r unstretched cable varies widely. Under presfress, there is a higher rnean tension in the cables which results in a more constant value for the
For geometric non-linear,problems the Newton-Raphòon iterative technique is very ffective. The basic steps *jflr-respect to Fig. 16.2 are as follows;
,
NON - LI NEAR RESPONSE
modulus of elasticitY. The.equivalent nodal force due to prestress in a truss or a cable element is given by
:
-cos0l
p"=
-sine
595
16.4 NON-LINEARSOLUTION ALGORITHMS
01
000 010
NON]LTNEAR SOLUTION ALGORITHMS
I
ô o J
(l6.l5a)
Po
P
"o,e I sineJ
Similarl¡
the equivalent nodal loads due to thermal expansion is given by
0
DEFL €C
-cos0l
p"=
-'*r l Ec[t'A cosO I
sineJ where
I
iì
t: A= E:
coefficient of thermal expansion change in temperature area of cross-ssction of the element modulus of elasticity of the material
a Â3 or , ..-o.t" .. ..
:
T
ION .
l.:,...fttj.r.r.'r.....
Fig. !6.2 Itgr.ativç,¡^lelhod for
(l6.l5b)
Computethevalueöf^iÈomKÀi
r:
j:,-.,"ì:
.-
a gepmptçiçql:ngnlinpar- probtem
:p.:
.;" i' j;: ,'l
Compute rhe unbalanced load vector R, uiing thd iiiitìai stiffness. at A, and determine, Â, using the
Comqule the,tangenfstiffrregf equilibrium equation
KÅ : P-R
I!
(16..16) Compure the,uñbàhn"céd;load vectór R, on the basij obtàined ii,æp á çjftVatqe¡ Repeat stçs 3 and 4 until the converge-nce is achievefu;,ûiaiii,-the vdlues of R
597
CONVERGENCE CRITERIA
NONLINEAR ANALYSIS:GEOMETzuC
596
NON-LINEARITY
Thenumberofloadstepsdependso¡tlenalureoftheproblem.Therefore,itisnot * optinu¡n load size in order -to achieve r"ï-*u.r computational
possible to speciff
Incrementa! method the total roas ln this type of alSorithm, -Durinc,lh"
experience gained by the application of !m"i"o"y. 'if,i, "* oniy be determined from :' problemC' of this method to differdnt types
r rr
is the value of stiffitess ;;';;"orargorithm'$"'"?lî*-'^i:,Tf$ä;r"-"|f ',ii*ffä'li:";ltJlïii",ffi ":'i the in
,h#"'ï;iä.-16.3.
The basic steps with respect to Fig' 16'3 are as follows
rru;*"1'q"iiì"'*ii'iJff
Step
Tp]::ît1"^:^:*?:-*, is round a$ qn ,ncrement *ru;f'*¡'iJ:T'iiiii"å:i"i":'"1Ïft *i::tåäJ¡JÏï:iî:"i'ii: tÏ'i"ä;iá;aPrecewislrinearprobrem' that is,
4- r
i,
dPi
is given as maEix' The tcital deflection the tangent stiffiress
a,:ao * .
whefe'^''is
(r6.18)
n
!aat i=l
n: the initial deflection and
point A'on the Apply a small load p' assume initial stiffiress t:nd locate R load unbalanced gives the response curye. This ' iteration till. point A Iteiate using the tange-nt stiffiiess matrix at.A' in the first
' I Step 2
(t6.17)
I(-, dÅr =
:
Step
3
Step
4
is reached.
point A, and Àpply the next load incremeit, assume tangent stiffrress'K, at locate point B'on the responbe curve' Repeai steps 2 and 3,iitl atl ttte load'increments'are exhausted'
Itisimportanttgnotethattheloadineachincrementalstepneednotbeequal.
(i-1) load step' no' of load steps upto
rò.5
CoNVERGENCECRTTERIA
In
u"r[* tiur U""n ;;u*ó;;;"¡ u ,"r¡"t of numbers, measured by
,Ï
,, :r,
degree of using any iterative scheme, computation is terminated when the desired achieved. If single numbers are used in the iterative scheme, the
...""""'r x* to a number i is simply'
Jay x¡ , X2 r X3 :
_:,
(r6.re)
lim¡*-lxr -,xl = 9,
':
,i:
...,.1
where k represents the k th iteration.
o o J
The rate of convergence C is determined by
Í
lh*-o l***t-'11 = ç lxr
t
x¡¡x2¡x3r-.exn
method
is reacheû within each load step tttl'coî;"tt"""e is fasler' r"ad is ahä s{nall and the convgrgence
'
,,--
.:
Thismethod.requ.iresgreatercomputationaleffô'rtsbecausediestifhessmatrixhasto : are' however' two advantages
"v"it"äi'i'"t-"itíi'Ï'"'" of the str-ucture can b€ ql?:-elastic. o¡ phy¡ic'q¡ P,¡operties chancg.T $ny st!trnÎÌ
:
different iterations, the cg¡vergence is said to occur
Îil]l;"--
.o,o
^f
If if
(t6.2t\
'
tim¡*-ffi="
:
(t6.22>,
(a)
Infinite-norm ll*ll. =' rno
: fx:
that is, the absolute largest value.
(b)
First norm
¡'r
t '
:
'
,
'.".
Any one of the following three veçtor norms is genetally used ,
be computed in every
th9
'
Order ofconvergence p, and the rate ofconvergence C are calculatéd'as
R.d;ï;itt"d
fo¡in
ãre the vectors ir¡
limu-.lliu -,tl[=-o
the conv erg€nce very and iterative techniques improves out A combination of incremental ls'u*.¿ in incåments' 'ft'eË¿tioni arer'cárried the: small' fast. The Newton is If thç lqad step
easily accóun:ted
l).'
While working with v€ctors, the coúveigence is definèd through vector norms.
DEFLECT ION
lncrement¡l cum Iterative
1a¡ '-'
(16.20.
xl'
wherep rep,resents the order of convergence (p >
problem for a geometrical rìonlinç¡r Fig. 16.3 Incremental 4ernoã
"ïi"i*tt"J
-
(16'23aì
¡1" ;'.i 4ì{,,..
t:*liå,,¡ ,:,:l;:....,*
r.r::,¡i::rt
NON-LINEAzuTY NONLTNEAR ANALYSIS:GEOMETRIC
598
CABLE-3DEROGRAM
n
ll*Tl, =
(c)
f,l*,l i=l
(r6.23b)
ILD
:
IT
=
NEQ
in any iteration' that is, zum of all the a-bsolute elements
IMA'I
Second norrn or Euclidean vector norm
NOPE
(t
^\/' ll*ll,=[It.,t'; =J*'*
(16.23c)
AK EST
iteration'
RI
Inlargedisplacementproblems,theconvergencecriteriaisusedonthedisplaéement be tess than a specified percentage The norm ortneiispla"e*ent vector ihould
vector.
say0.0l% or 0.001%'
.
analyze a large' displacement a.floppy. Ii-enplovs avallable.on is ttt sourie listing rin n rr, pro5ìu, -n"*torì_nuphson "i"*¿n6. The program w4s loading. "'r¡"g incrèmentar with technique
to
iterative irr. p.of"ràr of Civil Engineering, University of Roorkee, written by Dr. p. N. C"iiof", the follolvine subroutines : and later moditìed by the author. The program 'srtpu, csnsists.of
GEOMBC' soLVE' UPDATE and
Main Routine subroutines for various It calls GDATA, ASSEMB, SOLVE, UPDATE and FORCE a file and the results from. is read data rouä ffi ]|J ir"*iont for each load step. The are five labeled There file. a into are displayed on the terminal as well ai'written
FIVE' COMMON block TWO is èortm,iOÑ blocks : ONE, Two, definedinsubroutineconrewhileCoMMoì,IblockFIVEisd:.f,,.":dinqubroutine variables.' CON4MON ASSEMB. COMMON block ONE conll-sts.of iniegpr c-o¡trol THREE, FOUR'and
.f d"ñ;;ri"r"d "r¿li-¡"ift as coordinaies, connectivity, mäterials, conditions and specifred displacements. The remaining coMMON
bloctc TWO òór¡sirts
tng,h, uoun¿ary
: :
R.:¿
The main roi¡tine is a controlling blocks consist of various working variables and anays. ':' routine for the entire ftrogram. "
.
.
The important variables are as follows
:
NP = total number of nodes NE = total number of elements NB : total number of boundary nodes' NDF = number of degrees of freedom per node NLD =' number of load stePs NMAT = numberofmaterials NIT = number of maximum iterations
, ,. . r '.i
l
global load vector or displacement vector global concentrated load vector = total displacement vector : total force veÇ-ter . ,., = fraction oftotal load in á load step
subroutine
GDATA This routine
reads control data and geometric data including
m¿terial properties and initial prestre,s-s.data: The generæed ¿at¿-¡s wiitten into the ouput fi
la
ASSEMB It assembles the structural stiffiress matix AK and load vector I and calls each ele¡r-¡ént ône'by òñe. concentrated loads are read by calling subroutine )ATA. Boundary conditions are arso introduced in.'the stiffrrêss ,utri*. ine
FORCE.
structural stiffness,matrix
= elementstiffrressmâlii* i'
TDIS
:.
cunent load step current iteration number of total equations current material element nodal connectivitu band width member length initial prestr€ss in a member prescribed diSplaceùentr '
: : :
PS
PD
thatis,squareroot.ofthesumofthesquaresofalltheelementsinany
rr¡"ï, épnre, nsserr¡e' LDATA,
= =
IBAND
ALI
r-=-
16.6 CABLE-3D PROGRAM program'is wrìtten in FORJRAN A çomputer -
:
599
LDATA It reads
the concentrated load data at various nodes, and fraction
of
total load to be appried i1 a road step. It arso upda{eq ¡he road vecto¡ wåen -si.v9n for each load step. The initial prestresi load vectoi rr is i stored in array Rp and load vector is stored in arrav Rj!.
srIFM It generates the stiffiress mahix incruding rarge deformation t of an element and road vector due to initiar prestress irr ãachiail. These are in global coordinates through necessary'transiormations. .rn" st¡rrnes, lat
rhe stifhêss ,'::::1.'::l,r-"13:îgïtî_*sformarions. lar matrix in banded form_
¡" mgtrix is ,,o."a io upp",
utine GEoMBc It modifies the assembled stiffrress matrix Rl for prescribed displacement.
AK and totar
road
cails ir_rwice: In rhe firsr cail, ir rriangutarizes ,t^""llll,Tl:lain.program symmetric banded matrix. In the second calr, it performs the back substitution and incremèntal displacements ar,-g re¡rne{ in arralr Ri which wæ.gligaqly,a load anay.
UPDATE From the incremental displacement, it calculates the total updates nodal coordinates and prints the results foi eart, it"rut¡oo oi u
load step.
rbroutine FoRCE From the incremental:dispracements it calculates member forces current load step as well as tgfal memþer folc.g at the end of the current load step.
NONLINEAR ANALYSIS:GEOMETRIC
600
V Prestressing Force dqta
INSTRUCTTONS
dn a personaiil the progrut ""'" '-"Ytt-Ï: executing while questions must be answered following The ComPuter:
16.? USER'S
,
rhe
2.
q:lTi:älåîTJlll,Ti, *-'*"f ror enter print code
names
: Print I itt-ï.
Please
"'
'
Columns "
'
I
20
Prediess force in membgf
1l
(N
. ., -:
Û:free; l:restrained
-9
II
Matgrial.lroperttes
(15'
e*ttl:
i:::
\:
: r
t -5
26
i. l-oad daø for
NoderymfJ,
ligtrt
-f
6. -10 . ....,..:, :. ll -15 -20
Load in.z-direction
(8F10.-0),,
'
þ-e
:
glveq' ,
Total load is multiplied by a factor given{Èeach'lbad step' Give as many values as the number of load steps' Columns trana Ithumb
rule'
(4I5)
cótumns I
Node number Load in x-direction Loãd in y-direction
the last nq.dg mgst
VIII Load Multipliers
ttrè
-5 - 15 . 25 - 35
NOTE:.
'' -'
6 ' l5 x-coordinate'' 16 - 25 Y-coordinate 26 - 35 'r z:coord¡natÊ'ì:
tti l.'The'c*ràinateaxes correspold
x - d.o.f. y - d.o.f. z- d.o.f. Specified displacement in x'direction ' .....,........y-direction z-direction
(15' 3F10.0)
l6
NOTES:
ilt,.Menb;erData
VII Load Data
.
(15' 3F I 0'0) ilt"Nodal' Còordìnaies
't,,
-10
6
';
cotumns
, y - d.-o"f. IF NDF:3
ll -20 2t-30 3l - 40
2FI0'0)
5 ' . Colurnns' ---- "I:6 - 15 Area '^ ' .:..ll
.
-8 -9
Columns I
for each material proper-ry" Ome line is required Matçrial nurlr-ber -:
x - d.o.f.
-10
60
2 ; fot 2-D analysls 3 :'for 3'D analisis
..- r .
l
Boundary node number Boundary condition foi displacerñeñt
:
(6/15)
.. '
(215,i F I 0'0)
Columns I
Total numbçr ofnodes 'P) 'LE' (NE) Columns I - 5 'LE'40 ' 6 -lo +;äiïä;;'ïr "uur" "É*ents(NB) 'LE' boundary:todes of number .2O . ll - 15 Total stePs (NLD) s0 Ï'í ''" "iload '20 ""lU* materiats-(Nvr¡r)'LE -21 äffi;; ;l ;itr*"ni node (NDF) ' :' 2l " lòättã"-"e *""dom ner
'
andsoon
, ff- ,
IFNDF =2;
in the input file méntioned must be made available data following The
Contol Døta
:
values per line'
Frestress föice in mèmber
date etc'
DATA PREPARATION
I.
I
- l0
{u¡ñeprogramaccei:î*îA:ìffi ;::"J,"ï:"*'ntunis' or '----
4. Maximum number
1-
(SFI0'0)
This daø is required for each member' Provide
VI Boundøry Conditions
0 : do not Pnnt: and 3. The title of the Problemndicate his/her name, units emptgyed lnurçcrw (a) The usef may fllso
60r
.''.USER'StrNSTRUCTION'S'
NON.LINEARITY
Member number Node I Node J number rnrii¿t¡ut ia*tif icatiön
I
- l0
20 ll : .30
2l 3'l'
-
40
4r 5l
6Ï .,.70',,7t :' 80
MultiPlier for load steP I MultiPliqr for lqSd ¡1eP 2. ..:..............,....'..-.....'.......;4
.
NoN-LINEARITY NoNLINEAR nN¡lvsís;cEoMETRIc
6ö2:
I
LLTJSTRATIVE: EXAMPT,ES
16.8 ILLUSTRATIVE EXAMPLES ExamPle
16.l
r
"
It,hag a saggr-2-5,.m and canies a trl]¿"¡"oiU of'thë loaded
A cable is suspended across a sp-,?rr.,o!,?1,,*..D"*..i"; i1, i¡g, ir +à verrical load of 100 kN n]lÑ; prosram cABLE-3D' tf ;;ä,ù;rh" "*- *-iÐ- tit. iäa load is applied in orre step (ii) the total roãã it upitr;t¿ in five steps'
lr00
Eiq = 12000 kN
l:10
(o)
1,2
tZ r¿. t6 t'__ll
1.4
i .
NO OF IIERATIONS
E'l¿
I
r.o o.s
d
0.6
Fþ. ló.4 INPUT DATA
<
9 o
(c)
.
lIJ
ñ I
(Through Key Board) EX I6I.IN EX I6I.DOC single cable under a point load kN-m t. J
LOAD IN 5 STEPS
lt.&
(Through File EX f óf .IN)
t
2-3t,
NO. OF
IIERA
32
.' ).:ì
t:
Load aPPlied in one steP
.).:, :
.
.
212.5
.
'
2.5
3 25.0 :.,
;ir
i
i
giien as-i2000 kN' The area of Five iterarions are ípecifìed. Thé:üâitiê'ôf'Ei{'it of elasticitv I cros;-*ctio; ortr'e ca¡ieir-utuio*ilv q$é"e 11tul9 fo"lprp'.modufy no.,sigtificance' v{u¡s.lrav1, individuà] Théir tÑZmz. of the marerial is raken *-fiOtiil deflection is ,ho*n in Fig. 16.4b. In the fir¡! iteraii:l The defrecrion-ireration -th" to 0'88 m in an'fl,co¡verges and subsequent cycl.es
";*; is 1.37m which reduces ¡i the second
12000
l
Solution
(i)
2t
¡l
:IONS
(b) ,,:'Fig':'lÓ'4r
cable was 19i.2 kN. Inp!¡t data and ouþut the 5th iteration.. The final tension in the and EXl6l'DOC the nrosram CABLE-3D are shown in EXI61'FIL
z
180 J 150 c;
of
lt 22
2 J
0.
Iil
3u
-30. 2
100.
0.2
0.2
contd.
,1,¿_ñ.
0 5
0,
o
..
NON-LINEARITY NONLINEAR ANA'LYSIS:GEOMETRIC
604
Node no x2 .000 3 .000
''l
rr+*l *tt**+**+**+*ft#II***I**i*****,tt+r+r*+I*Ú*t*I**I*It*t+*t+lII*tI* ' 'Program for the a¡alysis of a cabl€ system li¡ê*t¿n:Raphson iærat¡on technique
u'¡ofi'Ul¿
rr**tr*tr++r***r*a***a.*r,*r****ii*******rr*+***r.it***+*s*t
****tú*t
LOAD
MULTIPLIER
ADDITTONAL NON ZERO LOAD lN LOì,D SJpP t00.000_.
.L8.60 = .LE.40 :
¡.000
2
.LE. 5 =
Number of difrercnt materi4s
'
3
MEMBER
2
MEMBER forces at-the BEGINNING of the bunent iteration
2 ,1
In the beginning ofthe first iteration, ofeach toad step, it assumes ze¡o membe¡ forces
'1. Node
AREA
MODULUSOF
.1008+{l
.1208+{5
no
Total Di.qplacement ¡n cur€nt load step
.0000E+00 .0O00E+00
.t3ElE+01
.00008+00
.00008+00
.0000E+00
.00008+00.
.o0o
y-
.0000E+00
)
.1250F+02
.3881E+0t
.25008+02
;00o0E+0o
step and ¡teration = load
TYPE
No.¡J ¡l2l .731
MEMBER
Member Pr€stcss Member. Pres¡¡ess Member Prest¡ess s
NODE no I, 3
BOUNDARY condition
ll ll
2
2
a
Member
3.21078+A2
Prpstr€ss
Node
no
Total Displacement ¡n cune;t load step
'0000 SPECIFIED DISPLACEMENT
x-
.000 .000
Y.000 .000
z'
." I 2 3
AT
I
MEMBER forces at the BEGINNINC of the_cunsnr iteration 3.2lO7E+02
I
.0000
.0000E+00
UP.DATED COORDTNATES AND DISPLACEMENTS
ELEMENT NODAL CONNECTION MATERIAL
t
z-
I
2's00 '000
z_
.00008100
Final Coordinates
x-
.ooo 12.500 25.000
.. y_,
.0000E+00 .00008+00 .0000E+00 .t38lE+0t
Node
Z4ORD
Displacement in cunent iteration
, x-
.x-'y'z-
I 2 3
elasticitY
X.CORD Y'COßD
1,
I
3 : for a 3'D analYsis Max. number of iterations
I 2 3
¡
. load step =. l and iteration = |
=l
Numb€r ofttoad steps
.000
UPDATED C@RD¡NATES AND DISPLACEMENTS AT
.:.
Tótal numÚer ofboundarY nodes
I
.000
and
Total number ofnodes ToÍal number ofcable elements
Number
.z-
l
kN-m single cable under a point load
MATERIAL
y100.000
Engg'' Dr. PN.Godbôle, P¡of' of Civil Dr. Ashok K' Jain, Prof' of eivilEngg' UniversitY of Roorkee, Roo¡kee
Degree offreedom Pcr'node 2 : for a 2'D analYsis
Load
Step
Program $ritteh bY
NODE
605
SEMIBANDWIDTHISEQUALTO 4
EX 16r.DOC
r
ILUSTRATIVEEXAMPLES
x'
Y-
.0000E+00
.00008+00
.0000E+00
-.4389E+00
.00008+00
.0000E+00
z-
':
:
:.,:.
Displacement in cur€nt iterat¡on
x'
Y'
".00,008+00
o000E+o0
.,00008+00
-0000E,r-00
.94?08+,{0 . .00008+00
z-
þi,i þ{$i
,¡,.noo
þfii
I Node .
þÉÊ,
l' 2
LLUSTRATIVE EXAMPLES
607
NONLINEÆR,AUe¡:vst'$øeo-rdlRlcl'¡oN'LrNEÀRlrY UPDATED COORDINATES AND DISPLACEMENTS AT
x-
-125gç+O2' '3442E+Ol :00008+00': .25OtE+07 .
Fqi
þfi
' and
load
step =
MEMBER I 2
t 2 3
l.9l20E+02 l.9l20E+o2 Total Displacement in curent load step
x-
i i, :.r :
J
:
J-
.0000E+00 .0000E+00 .0000E+00
I
j .lììi,:;:. ),Yi'',
niðptaðement in ÈúÈeiï ircraiiitíù Zt,:, Zi:.,,:t.,.:' :ì-::..:-!C?ri:,.tì,:r': ;;:'!,Ìjry:ii¿
:,: |,t.)
.00008+00
.
-8869E+00
¡çg0B+00
,., r.
Li:!:irl--i: !:{i:i::¿ììi '11'ì'll1 ?l')
Node' -ì
i::-: ' ..' :.
j;. j1i
x-
":
'
r.
.
.0000F,.'00 '0000E+00 .0000E+00 -.5513E-01 .Q000E+00 .0000E+00 -,1;::Ì,i :, irj! z-
.00{nE+00 ,0Óu)E+00 ,',;.rrr ii37g+0t ' í--oooot*oo
.t
jr:' 1¡liiîi'i:.r:i,1ìi.ìiií
-8867E+00
.00008+00
.34t3E-05
.0000E+00
.0000E+00
.00008+00
t'
:.:
¡",. .-t'.
:
r\ii.,:.i-|
sfeP = iteration
and
@i.
Wi
,
t.gtzsÉiòz
i...:,. ¿ ij.t .0000E+00 .0000E+00 .0000E+00 .88ó7E+00
I '2
Bri
3
i::
.oooog+oo .:-
::
":.
.oil'oog+oo
-ì. 1 'j'ìi
r'jj i:Í;
1l
kffiiítiË*o
2 J
yi:i,,"ì:,,:iiõ008+00
i,,15¡.iÉ+'oz iil::ir:'Tír878+01
:zioor+oz
i"' "".oooog*oo
z-
.3387E+0t .0000E+00
I
t9r.l99 .r91.199
Member Forces due
the cunent
r9l.l99 l9r.l99
Load applied in five steps
The load is applied il five equal steps of 20 kN each. The deflection increases with the increase in load. Within each load step, these were .five iterations, tn the first
iteration of each load step, the deflection was largest which reduced and converged in the 3rd iteration shown in Fig. 16.4c. The final deflection was 0.96m. The final tension in the cable was 207.9 kN.
aiíl :{:
:aii t:t
ûo
load.step only
.
In this problem the Newton-Raphson iterative scheme is more efficient than that with incremental loading. The latter scheme overestimates the deflection by gyo. niôpt Jcement in '''''"
x-
"ííiÉnT
y-
-.243?E-03
.0000E+{0
.0000E+00
¡
j
i :.:Ì17ü. il;
iít'uiffi
t'':-'::
: -::'''
'.0úoog+oo .0000E+00 :, i.
'1'
ofcurrent
2
(ir)
z'
.ooooE+oo
ì!:1 liì
:r:::i'.li-iì i, 1'¿
-T :'.*,.;
"
A cable is suspended across a horizontal span of 100 m with.unequal supports as s]rown in Fig. 16.5a. It carries two point loads of I00 kN and 200 kN. Determine the final deflected shape of the cable if it carries an initial prestress of 20 kN, 20. kN and 30 ' as shorvn in the same
figur". iuk" en
:
lsooo
Solution This problem was very sensitive to
it,t'.:tí""t'
-,.1:irj::l¡rirl eil-r'¡.i'ltt.:ì'-.Í"riIit'i¡i¡.
rìrj,::iiìaiÌr
Example 16.2
kN ín the three segments 11
Final Coordinates
Node
.l
z-
Total Membe¡ Forces at the end
::: i; +-.iìÌ ti-- : . .:il.i..ìtil:Í1]i. ì
Total'Disþlacement in óunent.load step
x_
y'
rj'lr'
lì1:!l
:
a
MI,'
;
4
l.9t26E+O2
no
::'11
I
l
Nride
J
MEMBEÈforces ai the BEGINNING of q¡..r¡el¡liirFralion
MEMAER
z-
.0000E+00
:
i
load
I
load step
'.,1
".i3sôÊip r' I.zíooe+oz
y.00008+00
Final Coordinates
Member
Final Coordinates '.ir' 1:ì y-
x.0000E+00
.0000E+00
x'.00008+00 .t2508+O2 .25008+02
2
.;;lr,i:li?:ìì ¡:-ii :¡i': l: :li
,;.1:.
Displacement in cument iteration
z'
,.
Total'Displacæment in oúirent load step
.0000E+00 .0000E+00 .o00oE+00 '
kFi
Hlil
MEMBER forces at the BEûINNÍNG of the current iteration
2
Node no
,
ffii,
ffi{i
I 5
I
3
2P4?3F.+o.2.,',,'2'04938+02:
2
Fj'.'
:li::'
..
MEMBERfoÎc¿satthèBcGÑNÑe-tntiøi¡'¡$¡¡øÉiàfr'"r'iri':;
:
ffi{i
. :í.lt
Node
Noderio
þt,l
ffili
MEMBER
.
ii''¡':i
þ$i;
H{i
I
l
iteration :
x:
ffiüi'
load steP = iteration =
and
AT UPDATED COORDINATES AND DISPLACEMENTS
þ{il' þqil
: 'i 'r. '''"-.'
' .0000E+00' '0ooor+00''
þr,çl
þ¡ir
'j- i): -li'.ii:':; :'r;;'iiìir:'::tlt z;.:.'.,.1 -.
FinalCoordinates y-
(i) the amount of initial prestress in the three cablé segments, and (ii) maximum number of iterations.
iN.
608
ILI-USTRATIVE EXAMPLES
NONLINEAR ANALYSIS:GEOMETRIC NON-LINEARITY
-
609
Solution
T 7.s
l-
This is,a 3-D cable net. There are 12 nodes and 12 elements as shown in circlog {nd boxes, respectively in Èig. tO.Oa and b. The area of crois section of cable elements 4, 6, 7 and 9 is 60 cm2 and that of the rest of the elements is 40 cm2. The modulus of elasticity is 20000 kN/cm2. The load is applied in l0 unequal steps. The complete input data and selected ouþut of the program CABLE-3D are shown in EXI63.FIL and
EXl63.DOC. The solution is quite sensitive to ri¡e number of load steps and mæcimum number of iterations in each load step. The final vertical deflection of node 4 is 1.03 m ftom the initial position.
-.r - __r'
FINAL/
0.482 m
(b) Fig.
8.31 2 m
-'i le0.813 m
16.5
A mechanism is formed if there is no initial prestress in the cable. The solution diverges if the maximum number cf iterations are specified as 5 or 6. The final results were obtained by using a maximum of l0 iterations. The total load wæ applied in five equal steps. The final deflected shape is shown in Fig. l6-5b. The deflections of the nodes 2 and 3 at the end of each step are shown in Table 16. I
, tn this problem
.
the Newton-Raphson iteratiúe scheme with incremental loading
proved to be quite efftcient
Table 16,1 Deflections at load points
u,,Jn I
2 J
4 5
Node
Node 2
Load Step
u,m
Vrh
t.469
0. 70 0. 75 0. 72 0 66 0. 60
3.581
- 0.094 - 0.13 - 0.100 - 0.084
0.692
0.0733
0.590
1
-
3
v,m 1.r65 0.854
Deflections - 0.482 4.770
0.843
Iq)
FLAN
1.669 1.225
0992 0.845
8.312
(b)
ELEVATION
Example 16.3 A cable net is suspQnded across a span of 300 m as shoún in Fig. I 6.6. It is subjected to a load of2000 kN applied equally at nodes 4, 5, S and 9. Each cable is prestressed to 50 kN. Determine the final deflected shape of the net.
Fig.
16.6
NON-LINEARITY NONLINEAR ANALYSIS:GEOMETRIC
ó10
ILLUSTRATIVE EXAMPLES analysis of a cable net kN-m Exl6.3 Total number ofnodes Tofal number ofcable elements Total number ofboundary nodes Number of load steps Number of different mater¡als Degree offreedom per node 2 : for a 2-D analysis 3 : for a 3-D analysis Max. number of iterations
DX I63.FIL
.LE.60 .L8.40 .L8.20 LE.s
12128102 I 0.004 20000 0000. 20000 0000' 2 0.006 3oo' I loo. 2200. 300. 200. 3 300. 100. 4 300.
= t2 = 12 = I =10 :2 =3
.a
=10
5 200.
6
100.
1007 200. 8 2009 t00. 200l0 200. 100t I 200. 100. 12 1C0. l19l 2210 3891 4.9102 5103I 69r22 .7 l0 ll 8712'I g t2 ll l0-ll4l ll 12 6 5 t2 ll 50. s0. 50. 50. . I lll 2. llt 3 ,lll 4 lll 5 lll 6 lll 7 tll I lll
MATERIAL
NODE I 2
f
ó
r00.000
4
2 I
I 50
50. 50.
50. 50.
50
50.
50.
x-coRD 100.000 200.000 300.000 300.000 200.000
2 3
.000 .000 100.000 200.000 200.000 100.000
I 9
l0
lt
l2
ELEMENT No. I
10
2 J
4
-500.
-500. 0.1
0.15
0.15
0'l
0.1
0.1
+**+****l******:it** ***+l*+* * t* *****++t*** t* *****t* + ******** r*+i***i** ***
*rr+**rtt****l+****t**+*+*t*i+*******
***********+*'+f*
Program written bY Dr. P.N.Codbole, Prof. of Civil Engg'' and Dr. Ashok K. Jain, P¡of. of Civil Engg' UniversitY ofRoorkee, Roorkee
5
0'l
6
I 9
t0
ll
Program for the analysis ofa cable system usiig the Newton-Raphson iteraiion technique
'
.400E-02 .2008+09 .6008-02 .200E+09
2
-500. -500.
0.05
MODULUSOF elasticity
I
9
ll t2 0-05 0.1
AREA
Number
-20. -20. -20. -20.
l2 i+**+**
*
Y.CORD Z-CORD
3oûìoo
300.000 200.000 r00.000
.000 .000
100.000 200.000
.ooo .000 .000 .000 .000 .000 .000 .000
200.000 -20.000 200.000 -20.000
t00.000 -20.000 r00.000 -20.000
NODALCONNECTION MATERIAL
I t9l 2t0 89r 9102 l0 9t22 l0il2 712 t2il2 il4I t26t il5t
J
TYPE I
3
I
I
**t**l*t**+ Prestress Member Prest¡ess Member prestress Member prestrdss 50.0000 50.0000 50.0000
2 6 l0
50.0000 50.0000
3 7
50.0000 l I
s0.0000 4 50.0000 I
50.0000 :
t2
50.0000 50.0000 50.0000
6il
ILLUSTRATIVE EXAMPLES
612NoNLINEARANALYSIS:GEOMETRICNo.I{-LINEARITY Node
NODE BOTJNDARY
SPECIFIEDDISPLACEMENT
x-
no condition
.000 .000 .000 .000 .000 .000 .000 .000
lll lll lll lll llt
I 2 3 4 5 6' 7 8
Y-
111
lll lll
SEMIBANDWIDTHISEQUALTO
z'
.000 .000 .000 .000 .000 .000 .000 .000
.000 .000 .000 .000 .000 .000 .000 .000
27
Load Node no x- y- z' 9 .000 .000 -500.000 l0 .000 .000 -500-000 Ir .000 .000 -500.000 12 .000 .000 -500.000
LOAD
MT'LTIPLIER
Step
I 2 3 4 5 ó 7
.050 .050 .100 .150 .150 .100
ß 9
l0
a 3
4 f 6 7
I 9
l0
il
t2
Total Displacement ¡n current load step yz-
at the end
load I 2
.100
J
STEP: I
.00o .000
49.029 49.029
-9.806 -9.806 -9.806 -9.80ó -9.806
:44t4845 ..44t4E.45
9
l5
l5
63.5
63.515 62.227 63.515 62.227 62.227 63.515 62.227
it3.5 rs 62.227
l0
63.5 r 5
t2
63.515 63.515
ll
63.515
63.5 15
63.5
t5
62.227 62.227 ó3.515 62.227 63.515
63.51i
.ADDITIOI¡AL NON ZERO LOAD IN LOAD
-.971
-5.388
-.971 -.971 -.971
-s.388 -5.388
.C00 -25.000 .000 -25.000 .000 -25.000
I
l0
9 l0 lt t2
.000 .000 .000 .000
i
63.515
-9.806 -9.806 '9.806
STEP:
2
.000 -2s.000
-5.388 UPDATED COORDINATES AND DISPI-ACEMENTS AT load
steP =
and iterat¡on
=
2
l0
n' .00{10[+00
.00008+00 .0000E+00
.00008+00 .0000Er-00 .0000E+00 .0000E+00
.00008+00
-3575E45
.3702845
.4438E-05 .7961F.45 .3577E.45 .7985E-O5 .79858-05 '.3692845
dueto the cuffen¡ load step only
.000 .000
UPDATED C@RDINATES AND DISPLACEMENTS AT
steP = and iteration =
) 6 8
49.029 49.029
ofcurrent
step
63_5
4
load
.7960E-05 .,14388-05
.t000E+03 .¡666t+03 -.2005E+02
.
y'
.0000E+00 .00008+00 .00008+00 .00008+00 .0000E+00 .0000E+00 .0000E+00 .00,00E+00 .00008+00 .00008+00 .0000E+00 .0000E+00 .00008+00 .0000E+00 .0000E+00 .00008+00
Member Total Member Forces Member Forces
.t00 .too .r00
.000 .000 49.029 49.029 .000 .000 49.029 49.029 -.971 -.971 =.971 '.971
Displaoement in cuFent iÍorûtiÕf!
x-
x-
.0000E+00 .00008+00 .0000E+00 I .00008+00 .00008+00 .0000E+00 2 .0000E+00 .00008+00 .0000E+00 3 4 .0000E+00 .00008+00 . .0000E+00 5 . .0000E{0 .0000E+00 .0000E+00 .00008+00 .0000E+00 .0000E+00 6 .00008+00 7 .00008+00 .00008+00 .00008+00 .g¡¡9t+00 .0000Ë+00 I 9 -.2s82Ê42 .2619842 -.5424E41 .2619E42 .26t9E.42 -.5424E.41 l0 .2619842 -.2582842 -.5424E.41 II t2 - -.2582842 -.2582E.42 -.5424E.4t Final Cobrdinates Node yzxI .1000E+03 .30008+03 .0000E+00 2 .2000E+03 .3000E+03 .00008+00 3 .3000E+03 .2000E+03 .0000E+00 4 .3000E+03 .1000E+03 .0000E+00 5 .2000E+03 .00008+00 .0000E+00 6 .1000E+03 .0000E+00 .0000E+00 7 .0000E+00 .1000E+03 .0000E+00 8 .0000E+00 .20008+03 :00008]{)0 9 .1000E+03 .2000E+03 -.2005E+02 t0 .2000E+03 .2000E+03 -.24058+02 ll .2000E+03 .r000E+03 -.20058+02
12
ÀDDITIONAL NON ZERO LOAD IN LOAD
I
no
613
NONLINEAR ANALYSIS :GEOMETRIC NON-LINEARITY
614 Node
no
I 2 3 4 5 6 ?
-0000E+00
.00008+00 .0000E+00 .0000Ë+00 .0000E+00
.00008+00
8
.looon*oo 9 . -.2583E-02 .2585842 r0 l
I
x-
x-
.00008+00
.0000E+00 .0000E+00
.000ÓE+00 .oo00E+0ó -0000E+00 .ooooE+oo
.0000E+00 .0000E+00 .0000E+00 .0000E+00 .0000E+00 -00008+00 .00008+00 .0000E+00
.2s70E.42
.2s8s842
.0000E+00 .00008+{0
.25?0Ê42 :.2583E42
.0000E+00
.0000E+00
.00008+00 .00008+00
at the end
load
I 2 J
4 6
I 9
l0
It
12
ofcuncnt
-.6s23E4s
9
-.5388E--01
-.8951E-{5 -.8951E-{5 -.9rgSE-{s -.3r48E-{5
-.53918-01 -.5395E-01
-33ç0E-O5
l0
)
.0000E+00 .0000E+00 .0000E+00 .0000E+00 ..0000E+00 .0000E+00
6
ll
t2
2 J A
6 7 8
9
t0
il. t2
-.5164842- -.5164842
x-
.00008+00 .0000E+00
.00008+00
'
.r
000E+03
.20008+03 .0000E+00
.0000E+00
0008+03 .00008+00
.00008+00
I
.0000E+00 .00008+00 .99998+02 .20c0E+03 .2000E+03 .99998+02
.10008+03 .2000E+03
.0000E+00 .0000E+00
.20008+03 -.20228+02 .2000E+03
-.2022F|02
.99998+02 -.2Q228+02 .99998+02 -.2022F+02
Total Member Forces Member Forces at the end ofcùr¡ent due to thè current load load step only.
253.20t
I 2 J
4
247.765
t23.596
J
253. I 0s
t26.t7l
6
247.765 247.765
123.596
253.105
t26.t7l
t0
il
247.765 253.201
126-l7l
253.105
t2
253. I 05
I 9
=
3
t26.t7t t26.t7t
253.t05 253. l0s
t26.l7 t
t23.596 123.596
t26.l7l t26.t7l
ADDITIONAL NON ZERO LOAD lN LOAD STEP 9
l0
¡l l2
.000 .000 .000 -000
.000 .000 .000 .000
=
4
-75.000
-7s.000 -75.000 -75.000
UPDATED COORDINATES AND DISPLACEMENTS AT load
step = 4 = l0
and iteration
.0000E+00 .0000E+00
-.131
lE{4
.t058E_04
-.1927844 -.52878-05 _. t9058{4 -.1 l30E-04 -. t l30E-{4 -.4755844
z-
.3000E+03 .
-.t3l0E-04
step
63.515 63.419 63.419 61.941 63.419 61.941 61.941 63.419 61.941 63.515 63.419 63.419 STEP
y-
0008+03 .3000E+03
z.00008+00
.0000E+00 -0000E+00 .0000E+00 .00008+00 .00008+00 .0000E+00 .00008+00 .0000E+00 .0000E+00 .00008+00 .0000E+00 .0000E+00 .0000E+00 .0000E+00 .00008+00 .00008+00 .00008+00 -.5285E-05 -.t927844 _.t9038_{4
Final Coordinates
.2000E+03 .30008+03 .3000E+03 .2000E+03
I
Y-
.00008+00 .0000E+00 .0000E+00 .00008+00
.0000E+00 .0000E+00 .0000E+00 .0000E+00 .0000E+00 -0000E+00
-.1070E+00
.0000E+00 .0000E+00 .0000E+00
.
Member
Forces
x-
-.st49842 .5¡5lE-{2 -.10698+00 .5121E42 .5t21F.42 -.t069E+00 .5t5lE{2 -.5t49F42 -.t069E+00
l0
.s36sE-O4
-i6524E-05 -.6898E-04
I
UPDATED COORDINATES AND DISPLACEMENTS AT
and
-.33ór E--05
.0000E+00 .0000E+00
efF;
Displacement in cur¡ent iteration
y-
.00008+00 .0000E+00 .00008+00 .0000E+00 .0000E+00 .00008+00 .00008+00 .0000E+00
Node
-000 .000 -50.000 .000 .000 -50.000 .000 .000 -50.000 .000 .000 -50.000 3
':0000E+00
-.3t48E{)5 -.9¡988{5
.00008+00
.00008+00 .00008+00
x-
I
-.s39tE-01
.0000E+00
Total Displacement in current load step z-
2 J 4
8
.0000E+00 .0000E+00 .0000E+00
no
due to the cunent load.steP onlY
step 127.029 126.934 126.934 124.t68 126.934 124.t68 124.168 t26.934 124.168 : 127.029 t26.934 126.934
steP iteration =
.0000E+00 .0000E+00
.0000E+00
ADDITIONAL NON ZERO LOAD IN LOAD
load
Node
.0000E+00 .0000E+00 .0000E+00 .0000E+00 .0000E+00
Member Totat Membe¡ Forces Member
9 l0 rr ' 12
.0000E+00
.9999F+02 .9999F+A2 -.20llE+02
.
.00008+00
..00008+00 .00008+00
t2 \. -.2586842 -.2586842 Final Coordinates Node yzx¡ .1000E+03 .30008+03 .0000E+00 2 .2000F+03 .3000E+03 .00008+00 3 .30008+03 .2000E+03 .0000E+00 o .36969+03 .t000E+c3 .0000E+00 5 .20008+03 .000cE+00 .00008+00 6 .1000E+03 .0000E+00 .0000E+00 7 .0000Ë+00 .1000Ë+03 .00008+00 I .00008+00 .20008+03 .0000E+00 9 .99998+02 .20008+03 -.20llB+02 tq .2000E+03 .20008+03 -.29¡¡f,+02 I I :20008+03 .99998+02 -.201lE+02 12
curent iteration yz-
DisplacÆment in
Total Displacement in curment load step .yz-
ILLUSTRATIVE EXAMPLES
,.1-:,rijj
,"i..ii t.,.,t, ''::rr:T:l:,':.i
NON.LTNEARITY NONLINEAR ANALYSIS:GEOMETRIC
616
no
Node
l i ã i i ã ; á ö
step Total Displacement in currcnt load zYx-
.ooooe+oo .ooooe*oo .óoooe*oo .ooooe*oo .õoooe*oo
.oóoog+oo .ooooE+oo
'¡99Qf,+00 'o0ooE+oo
.ooooE+oo 899!.99
-0000E+00 '00008+00 .ooooE+oo '6s¡6s+oo .oooot*oo .ooooE+oo 'ooooc+oo
.ooooe*oo .9¡¿68+oo .óoooE*oo .ooooE*tlo
t0 i; i;
'66gQn+oo
.00008+00 .0000E+00 .0000E+00 .0000E+00
-.4198E4s -.6979845
-.6984E-0s
-'¡532¡+00
Final Coordinates
x-
Member
z-
Y-
steP
440-979 440.884 44o.g7g 43t.728 440.884 4it.s7t 431.725 440.884 431.871 440.979 440.884 440.979
4 J 6 7
I 9
l0
ll
t2
l0
ll
t2
load
l0
ofcurrent
step
due to the curent load step only
187.778 187.778
I
627.i27
r
2
I 86.443
+87.874
J
627.327 627.422
183.964 187;178
^
6t4.262
182.533
)
184-lo7
6
627.327 614.405
I
183.964 187.1?E 184.107
7
6t4.262
l 82.533
8
627.232 614.405 627.327 627.232 627.422
I 86.348
9
l0
r8i.778 187.778
u
i87.874
T2
= :
DISPLACEMENTS AT UPDATED COORDINATES AND 5
.I
at the end
.000 .0ô0 -75.000 .000 .000 -75.00Û .000 .000 -75.000 .000 .000 -75.000
load step = and iteration :
.5206E-0s
no Total Displacement in cumrent load step Displacement in current iteration y' xzyxzl.0000E+00 .0000E+00 .00008+00 .0000E+00 .0000E+00 .0000E+00 2 .0000Er-00 .0000E+00 .0000E+00 .00008+00 .00008+00 .00008+00 3 .0000E+00 .00008+00 .0000E+00 .0000E+00 .0000E+00 .00008+00 4 .0000E+00 .0000E+00 .0000E+00 .0000E+00 .0000E+00 .0000E+00 .0000E+00* .00008+00 .0c00E+00 5 .00008+00 .00008+0û .00008{{)0 .0000E+00 .0000E+00 .0000E+00 6 .00008+00 .0000E+00 .0000E+00 7 .0000E+00 .0000E+00 .0000E+00 .00008+00 .0000E+00 .0000E+00 .00008+00 .0000E+00 .0000Ë+00 8 .0000E+00 .0000E+00 .00008+00 9 .t373E-04 -.1034E-0s .¡333E{4 -.7593E.42 .7s96842 -.t558E+00 l0 .7636842 .7636842 -.1558E+00 .t722E44 .t722844 -.t642E.44 II .7596842 -.7593E{2 -. t558E+00 -.t0348{5 .t373Ë-M 3338-44 12 -.7594E42 -.7594842 -.t558E+00 .7576845 .757s845 .4516E-44 Node Final Coordinates y' xzI .1000E+03 .30008+03 .0000E+00 2 .2000E+03 .3000E+03 .00008+00 3 .3000E+03 .2000E+03 .0000E+00 4 .3000E+03 .10008+03 .00008+00 5 -.2000E+03 .0000E+00 .0000E+00 6 .1000E+03 .0000E+00 .00q0E+00 7 .0000E+00 .10008+03 .0000E+00 8 .0000E+00 .20008+03 .0000E+00 9 .99978+02 .2000E+03 -.2053Ë+02 t0 .2000E+03 .2000E+03 -.20538+02 I I -2000E+03 .99978+02 -.2053E+02 t2 .9997F.+02 .99978+02 -.20538rd2
load steP onlY
IN LOAD STEP ADDITTONAL NON ZERO LOAD 9
.4599E-05 -.8247E.46
617
Node
Member Total Member Forces Member Forces
Total Member Forces Member Forces atthe end of cunent due to the cunent load
J
.0000Et00 .00008+c0
.¡9¡95+00
.0000E+00
.7632842 -'766s842' -'ls82E+oo
.46068-0s
ã .¡ooop*o¡ .2ooog+03 'ooooc+oo o .3969g+03 .¡669þ+03 '6666f+00 i .2qg6E+03 .9699f,+oo '69soP+oo ã .iooos*o¡ .oodog+oo .66¡$P+oo ! .6¡ss¡+oo .loooE+o3 oooo!+99 à .ooooe*oo .26ggf+03 'ooooE+oo n .sÐgge+o2 .2000E+03 -'2Q378+02 ín .2000E+03 .2000E+03 -'20378+02 ii .2oooE+03 .9998E{2 -:0378+oz ii .ósesE*oz .9998E+02 -.2637s+02
z
.¡ggg5+00
-.4206E-05 -.7622846 -.4297E.45 -.2271E.45
-.tøogs-tz -:66sr¡2
I
'00008+00
-.429284s
'7675q-42
.t000E+03 .39668+03 '0000E+00 2000E+03 .3¡6gp+03 '00008l{0
I i
.00008+00 .00008+00
-.ioese¡2 1632r¡2 -'lsg2E+oo -'¡531tr+00 .7675842
Node
'ooooc+oo
Displacement ¡n curent iteration z' vX: .6[sgg+oo .ooooE+oo .0000E+00 .g6g69+00 .0000E+00 .0000E+00 .0000E+00 -0000E+00 .00008+00 .0000E+00 ;0000E+00 .00008+00
ILLUSTRATIVE EXAMPLES
86.348
186.U3 t86.443
I r
82.s33
82.533 8ó.348
t86.318 l8ó.443
ADDITIONAL NON ZERO LOAD tN LOAD STEP 9
l0
ll
t2
.000 .000 .000 .000
.000 .000 .000 .000
=
6
-50.000 -50.000 -50.000
-50.000
UPDATED COORDTNATES AND DISPLACEMENTS AT load
steP = =
and iteration
6
l0
ILLUSTRATIVE EXAMPLES NON.LINEARITY NONLINEAR ANALYSIS:GEOMETRIC
618 Node no I J
5 6
Yxoóìos+oo .6[6¡s+oo .ððoõs-óo .ooooE+oo .ðóóoe*¡o .ooooE+oo .ðõóóe-óo .ooooE+oo .óóðós*oo .oooog+op
.óóõóe*oo .oooog+ob .õóoóe*oo .ooooE+oo
8
9
.00008+00 ,0000E+00
:ooooE+oo .0000E+00 -0000E+00 .0000E+00
.ooooE+oo
'ooooe+oo 'oooog+oo
.69ggt+O0
.õóóóe*oo .ooooE+oo ;'102!E-19 "ooooE+oo -.so/ÉJn-{r- .5043F.-02
-.3178E-{5
-'1e2sþ+oo
.3240E-{s
.sõiãi-oz .sólál-oz -so42ç-42 -.íí:iel-.oz -s036rq2
ll
.3179E-O5
-.30758--05
--r028E+00
x- ,.
Y-
no
I
.ooooedo
.00008+00 .0000Ef00 .00008+00
2 J
.¡969f,+00
4
.6666[+00 .0000E+00 .6¡699+00 .0000E+00
f
.0000E+o0 .0000E+00
6
.00008l{0
.00008+00 .0000E+00
.69¡9f,+00 .0000E+00
.0000E+00 .00008+00 .0000E+00
.3240E.45 -.12848-04 .3179E-05 -.12858--04 .3177F.-05 -.1284E-04 :'1204E44 -.3075E-O5
8
.00008+00 -.503 t E-02 .s008E-o2
t0
il
.¡¡ç6f,+03 .369Qf,+03 '96¡6f,+00 'ooooE:Èoo
tz
J
4
) 6 8
9
l0
ll
12
[N ADDITIONAL NoN ZERO LOAD
9 l0 tI 12
I J
123-'187
4
123.78',1
lzl-021
6
123j87
7
l2l.02l
I
r2r.02l
9
123.787
t0
l2l
LOAD STEP
l2
7
AND DISPLACEMENTS AT UPDATED COORDINATES load
steP = =
7
l0
Member Fo¡ces due to the current load step only 123.215 t23.1 t9
t23.tt9 120.449
t23.tt9
855.875 8ss.732
120.449
874.t38
l23.tt9
t20.449
120.u9 t23.215
t23.tt9
t23.ll9
ADÐITIONAL NON ZERO LOAD rN LOAD STEp
=
.000 .000 -50'000 .000 .000 -s0 000 .000 .000 -50'000 .000 .000 -50.000
and iteration
step
855.875 874.233 874.138 874.329
il
'021
121.692 123.787 123.787
Forces
ofcurent
814.233 874.233 874.329 855.732 'E74.233
2
t23.692
.oooOe+O¡O
.3000E+0j .1000E+03 .00008+00
load
9 t0 rI t2
.000 .000 .000 .000
.000 .000 .000 .000
=
E
-50.000 _s0.000 _50.000 _50.000
UPDATED COORDINATES AND DISPLACEMENTS AT load
step = I = l0
and iteration
.00008+00 .0000E+00 .0000E+00 .0000E+00 .0000E+00
-.5020E-02 -.totsE+00 .t743845
.2000E+03 .00008+00 .00008+00 .1000E+03 .0000E+00_ .00008+00 .0000E+00 .t0008+03 .00008+00 .00008+00 .2000E+03 .00008+00 .99968+02 .20008+03 _.20738+02 .2000E+03 .2000E+03 _.20738+02 .2000E+03 .99968+02 _.2073F!1t2 .9996E+O2 .99968+A2 _.20738+02 Member
.0000E+00 ..0000E+00
-.5031E{2 -.t0l8E+00 .5t80E{5
.3000E+03
at the end
.
.0000E+00 .0000E+00
Final Coordinares
MembelTotal
Ioad steP onlY
751.019 751.114 75l.2og 135.283 751.1 14 735.426 73s.283 751.019 735.426 7sl.0l9 751.019 751.209
I
.0000E+00
.20008+03 .3000E+03 .00008+00 .3000E+03 .20008+03 .00008+00
Mg¡ber Forç1. Member Total Membe¡ Forces to the cunent due of"utt"nt àit¡e
2
.0000E+00 .00008{-00 .0000E+00
x-
.00008+00 .0000E+00 .00008+00 .00008+00 .00008+00 .5032842 -.l0t8E{-00 .7290845 .50088{2 -.tOtsE+00 _.1t73E_{5
-.5020F-02
Node 2 3 4 5 6 7 I 9 t0 II t2
z-
steP -
.00008+00
.5032F42
t2
.00008+00
..0000E+00 .0000E+00 .0000E+00 .0000E+00
-0000E+00 9
Displacement in cuÉent iterâtiòn
z'
Y'
x-
x-
.ióóoe*ot .366¡r+03 'ooooc+oo i 'toosg+ol .2669f,+03 'sg6¡f+fi) ; ¿ .lóooe*os -loooE+o3 'ooo0E+00 i .ãóoos*o¡ .ooooE+oo '0000E+00 ã .ióoor"o¡ .66s6f,+00 'ooooE+oo i .ôóõos*oo .1e¡sP+03 'ooooE+oo -26sss+03 i .õoòõÈ-oo '.sgsil*oz .20008+03 -'2063F+02 ö -'20638+o2 io .iôôoe*ol .2ss¡f+03 .ãóõóe.ot .ssslg+o2 -'2s53s+02 ii '.sggir¡¡oz .s9979+02 -:!063p+02
"nd lbad
Total Displacement in current load step
Lt0008+03
Final Coordinates
Node
t
x--
z'6664f,+oo 'oooor+oo 'ooooe+oo 'oooog+oo
.so38c-02 --lo2eE+oo
l0
Node
Displacement in cunent iteration ZV.0000E+00 .0000E+00 .ç9¡99+00 .0000E+00
load step Total Displacernent in currrent
619
Y-
.00008+00.
.00008+00
z.0000Ë+00
.00008+00
.00008+00 .0000E+00 .0000E+00 .00008+00 .0000E+00 .0000E+0O .0000E+00 .00008+00 .00008+00 .0000E¡¡0 .0000E+00 .00008+00 .5t798{5 -.t t97E-O4
--.1l74E{5 -:41468-i4 .7290845 . _.1197E44
.1743845
.2t44F.44
ILLUSTRATIVE EXAMPLES
NON-LINEARITY NONLINEAR ANALYSIS:GEOMETRIC
620 Node no
I 2. 3 4 5 6 , I g r0, lI Ò
Displacement in cuilent iteration
step Total Displacement in curÍent load
.oóooe+oo
.00008+00
.ooooE+oo
.0000E+00. .0000E+00
.0000E+00
.00008+00 .0000E+00
.00008+00 .00008+00
.00008+00
.0000E+00
.0000E+00 .0000È+00 .0000E+00
.00008+00
.00008+00 .6699þ+00
.¡¡6gE+00 .00008+00
x,
Z-
V-
x-
'
.0000E{{0
.Oóooe+Oo
-0000E+00 .00008+00 .0000E+00
.0000E+O0
.0000E+00 .0000E+00 .0000E+00 .0000E+00
.00008J{0 .00008+00
-.4293846
.0000E+00 .00008+00
-.4998842
.50llE_{2
_.t0088+00
.5006E-02 .5ol lE-02
_.4998Ë-{2
_.1gggt+O0
.1280E-{4
_.4gg1B_42 _.1008Ë+00
.1124844
.1097E45
.50058__02 _.1009E+00
-.4993842
V-
,
Node
Z' .96¡9f,+00 .6gg6p+00
.0000E',0 .9¿665+00 .0000E+00
.128'.E-04 .1100E-45
-.4250846 .lt25E-04
4
.00008+00 :00008+00
6
.0000E+00
8
x-
Y'
ll
.5004E-02
12
-.4966842
r 2 3 4 5 6 7 I 9 r0 ll t2
z-
t .¡9¡9þ+03 .3000E+03 '0000E+00 t .2¡¡98+03 '3000E+03 'gggSþ+00 I .ioooe*o¡ .29¡61+03 'oooog+oo ¿ .ãooor*ol .1996f,+03 'oooog+oo i .iooos*os .¡66ss'r00 .00008+00 ã .roooe*o¡ .ooooE+oo 'g66sl+oo i .96gsg+oo .¡9¡¡5+03 'osgsf,+oo e .õoooe*oo .2oooE{3 'ooooE+oo n .qÐg5g+tz -2000E+03 -'2083F+A2 io .zoooe*o¡ .2000E+03 -'2083E+02 ii .26sss+03 .Ð96g+02 -'2083r+02 l) .nsøg*oz .9996B+02 -'2083E+02 Mernber ';-"--
Total Member at the end
load
Forces
ofcurrent
steP
.5004E{2 .4966F42
y-
.99958+02 .2000E+03 .20008+03 .99958+02
.2000E+03 .2000E+03 .9995E+02 .9995F+02
Total Member
Forces
step
I
I I t 8.660
l I18.660
J
J
t22.547
4
I I 18.85 I 1094.770
4
996-876 9'15.609 '
119.877
f
l l t8.660
)
99ó.685
r22.452
6
9
975.752 97s.609 996.590 975.752
tt9.877 ttg.877
6
t0
996.685
095.057 1094.770 t I 18.565 1095.057 I I 18.660 r r 18.565 I I18.851
996.590 996.E76
t22.547
LOAD ADDITIONAL NON ZERO LOAD IN 9
t0
il
12
STEP
I t0
ll
t2 9
.000 .000 -s0'000 .000 .000 -50.000 .000 .000 -50.000 .000 .000 -50.000
AT UPDATED COORDINATES AND DISPLACEMENTS 9 load
step = and iteration =
l0
r
9
-
.0000E]{)0 .00008+00
.00008+00
-.9991E-{l
-.4915E--06
-.99908-Ol -.99918-Ol
-.1l88E{4
Merrber Fo¡c¿s duc to the curent load step onlY
t21.975 t21.975 t21.975
tt9.162 12t.9?5 I 19.305
tl9.t62 tzt.975 r 19.305
t21.975
t2t.975 12t.9?5
ADDITIONAL NoN ZERO LOAD IN LOAD
9 r0 II t2
.4444E.46
-.1200E-{6
--20938+02 -.2Ù93F+02 -.20938+02 -.20938+02
ofcunent
2
t2
.00008+00 .00008+00
.t0008+03 .0000E+00 .0000E+00 .0000E+00 .1000E+03 .0000Er{)0 -000cE+00 .2000E+03 .0000E+00
122.452 122.452
ll
.0000E+00 .0000E+00 .00008+00
z-
99ó.685 996.685
122.452 122.452
.0000E+00 .0000E+00 .0000E+00
.3000E+03 .10008+03 .0000E+00 .2000E+03 .0000Ë+00 ,0000E+00
load
ttg.877
.0000E+00
.0000E+00 .0000E+00 .0000E+00 .0000E+00 .0000E+00
.t000E+03 .30008+03 .00008+00 .20008+03 .3000E+03 ;00008+00 .3000Ef03 .20008+03 .0000E+00
at the end
122.452
.00008+00
-.4980842 -.4966842 :.Ð9lE-01
I
I
.00008+00
x-
Final Coordinates.
K-
Member
Member Forces due to the curent load steP onlY
.00008+00
.00008+00
.m008+00
.4966E42
Node Final Coordinates
y.0000Ë+00 .0000E+00
.00008+00 .00008+00
-.4980842
l0
.10ó8E-44 .1394E-04 .10008-04
.0000E+00 .0000E+o0
.0000E+00 .0000E+00 .0000E+00 .0000E+00
.00008+00 .00008+00 .00008+00
.9
.0000E+00 .0000E+00
.00008+00
.0000E+00 .0000E+00 .0000E+00 .0000E+00 ,.0000E+00
) J
-.l39lE-O4
y-
2
.0000E+00
Displacement in current it€râtion
Total Displacement in currcnt load step zx-
l.
.00008+00 .0000E+00 .0000E+00 - .0000E+00 .0o0oE+00
no
.000 .000 .000 .000
.000 .000 .000 .000
STEP: l0
-50.000 -s0.000 -50.000 -50.000
UPDATED COORDINATES AND DISPLACEMENTS AT load
steP = =
and iteration
l0 l0
621
.0000E+00
.4487846
z1
.0000E+00
.0000E+00 .0000E+00 -0000E+00
'.9736845
-.1¡88E=-04 .937584s
-.487tE--06
-.tt74846
.97698-05 .r
13lE{4
NONLINEAR ANALYSIS:GEOMETRIC NON.LINEARITY
622 Node
no
I
i ã ¿ 5 ã i I s io II t2
Total Displacemont in curment load step zxY-
.0000E+00 .0000E+00
'00008+00
.ooooe*oo .ooooE.
'ooooE+oo 'oooog+oQ '0000E+00 '0000E+00 '6966P+oo
.ooooe,*oo .ooooE+oo .ooooe*oo -0000E+00 .00Û0E+00 .q996f,+00 .oooos*oo .ooooE+oo .ooooe*oo -ooooE+oo 'oooop+oo -0000E+00 .00008+00 '0000E+00 -.+g$e--oz -4947842 -'9899E-"0r
.4s78l-42 ¡978r¡2
-'9898E-{1
-.4967842 -.4966842
-'9899E-{l
.4947F.42 -.4947842 -'9899E-{l
Displacement ¡n current ¡teration
x.0000E+00 .0000E+00
.00008+00 .0000E+00
.00008+00 .00008+00 .0000E+00
v.00008+00 ..0000E+00 .0000E+00 .0000E+00 .0000E+00 .0000E+00 .6gggB+00
.6g9gg+00
.00008+00
-.33E7E--05 .90028-.05
-.9088E-{5
-.909EE-05
-.3377845 .2366F4s
.2360E45
-9008E-Os
Appendix A
z.0000E+00 .0000E+00 .0000E+00
.00008+00
PROGRAMS FOR SOLUTION OF LINEAR SIMUI,TANEOUS EQUATIONS
.g96gB+00 .0000E+00 .00008+00
.6966f+00
A.T
:.1446844 -.4ll2E-.o4 -.t438E-{4 .1271E44
Total lvlember
Forces
ofcùnt load step
at the end
L
C C
I
240. l 58
l2l.49E
t240.063 t240.254
121.403
4
12t3.646"
I 18.876
l
1240.063
t2l.4o3
6
121i.646
C
I I 8.5E9 I I 8.876
8
9
t213.646
l0
I
240. t 58
121.403 I t 8.589 121.498
12
t239.967 1240.254
t21.403 t21.403
N IS SIZE OF MAT¡ìIX TO BE INVERTED OR SOLVED M IS NUMBER OtI R.H.S. VECTORS IAI lS THE COEFFICIENT MATRÍX. SIZE (N+M) * N = I X IS SOLUTION VECTOR B IS R.H.S. MATRIX
L-
CI.IARACI'ER *24 DA'I'FIL COMÌvtON A(-50.60)
WRII'E 1*.*¡ ' Please entcr name of'thÈ ofrtput fìle' READ (*,50) DATFIL OPEN (UNIT=6, FILE= DATFIL. STATUS='NEW') wRI-rE (*.100)
wRtTtì (+,r-i0) READ (*.*) N.M WRITE (ó.100)
12t.403
1213.646 t239.967
ll
L
Member Forces
I
GAUSS ELIMINATION METHOD FOR SOLIJTION OF AX=B WITH PARTIAL PIVOTINC written by Dr. Ashok K. Jain. University ofRoorkee, Roorkee
C]
due to the current load steP onlY
2 5
ELIMINATION METHOD
C
Node Final Coordinate zyxI .10008+03 .3000E+03 .00008+00 2 .2O0oE+03 .3000E+03 .00008+00 3 .3000E+03 .2000E+03 .0000E+00 4- .3000E+03 .1000E+03 .g00gf+00 .2996¡+03 .0000E+00 .00008+00 6 .1000E+03 .66968+00 .0000E+00 7 .0000E+00 .1000E+03 .00008+00 I .00008+00 .20008+03 .0000E+00 n .gg95B+02 .269!f,+03 '-.2103E+02 t0 .200tE+03 .20018+03 -.21¡3f,+02 ll .200tË+03 .99959+02 -.2103E+02 t2 .gg95B+n2 .99958+02 -.2103E+02 Member
GAUSS
wRn'Ë (6.200) N.M Nl=N+l NM=N+M WRII'E (*.*) ' Please enter coefficient matrix rorv-wise READ (*.*) (A(l.J),J=l.N),1=l.N) wRITE (ó.300) DO l0 t=r.N wRITE (6. 400) (A(l.J),J=l.N)
WRITI 1*.t¡ ' Please enter R.H.S. vectors column-wise (r.r) (A(r,J),r: t.N),J:N l,NM)
R€AD
wRrTE (6. s00) DO 20
l=l.N
wR|TE (6, 400) (A(l.J),J=N t.NM) CìONTINI.JE
CALL GAUSS (N,M) wRrTE (6. 600) DO30 l=l,N wR¡TE (6. 400) (A(l.J).J=N ¡,NM) 30 L
CONTINUE
'
ù
625
APPENDIX
[.¡
624
U
50
t00
L,I 150
(_)
(,,
t,) rJ
200
APPENDIX FORMAT (A) FORMAT (' Program for the solution of equation Ax : B 'l y * '. using Gauss-Elimination method with partial pivoting * 'wr¡tten by Dr. Ashok K. Jain, University of RoorkeeY) FORMAT (' Pleaie cnter matrix size : .LE. 50'/ * . 'No. of R.H.S. vectors : .LE l0 7) = '.15,/ FORMAT ( 'SIZE OF COEFFICIENT MATRIX { 'NO OF R.H.S. VECTORS = '.15"/) FORMAT (/5X,'COEFFICIENT MATRIX A Ð FORMAT (8810.4)
100 400 500
FORMAI'(isx.'R. H. s. MATRIX B'/)
600.
FORMA-I' (/5x.'
600 500
CONTINUF]
100 C
CONTINUE
c c
Back-substitution
CONTINTJE
DO 700 l= I -tvf DC 800 L=2.N
NPI:N+l
II=N-L+l fPl:ll+l
SUM=A(ll,NPr) DO 900 J=lPl,N
sot.l.lrloN MATRIX X 7)
c STOP
900
SUM=SUM-A(ll,J)*A(J.NPI) CONTINUE
800
CONTINTJE
704
CONTINUE RETURN END
A.2
GAUS-S -
END
t_,
(.r
SUBROU'TINE GAUSS (N.M)
coMMoN A(50.60) C
c .¡-''t
A(ll.NPI)=SUM
C
Pivot selcction
L
NM=N+M
JORDAN METHOD
DO 100 l=l.N 1....
IPI=l+l
'1
i) (.) (
j./
C C C
AÁ=ABs(A(l.l)) K=l DO 200 J=l.N
200
lF (ABS(A(J,I)). Gr. AA ) THEN K=J AA=ABS(A(J,r)) END IF CONTINUE
c
c C
f
*
C L
r'
Rorv interchange
c
L
C C C
rr.'(K. NE. r) THEN
J:I.NM AA:A(l,J) DO
3OO
(_
A(r.J)=A(K,J)
Cì
A(K.J)=ÀA
L L
CONTINUE END IF
3OO
rF (ABS(A(!.I))
wRIl'Ë (t.l
c
.t-l'. IE-I s) rHEN
I IOO FORMAT (' DIVIS¡ON BY ZERO
I
L L
STOP
END IF' Tri-angularization
C
AA=1.0/A(l,l) DO 4OO J:I,NM
A(l.J)=A(l,J)*ÄÂ CONTINUE DO 500 J=lP I ,N
AA=A(Jd) DO 600 K=lPl.NM ^(J.K)=A(J.K)-AA+Á0.K)
** * **
********
+
**
**
**
***
*tt
****
t * * **
+
*
+
l* + ** * * * +* ** * *t+**
EXECUTION TERMINATED 7)
+
+* **
*t*** ** *+
N IS SIZE OF MATRIX'TO BE INVERTED OR SOLVED M IS NUMBER OF R.H.S- VEC1ORS COEFFICIENT MATRIX. ITS DIMENSTONS MUST BE CALCULATED t,Al rs THE .IHE RELATION N r (N+M+N) FROM X IS SOLIJTION VECTOR B IS R.H.S. MATRIX [C] CONTAINS INVERSION OF [AÌ lF ONLY INVERSION IS DESIRED' GIVE M=0 lNv='1. PROCRAM COMPUTES SOLU'IION OF R'H'S' ONLY INV= 0. PROCRAM COMPUTES INVERSÍON ONLY INV= I. FRóGR.AM COMPUTES SOLUTION OF R.H.S. AND INVERSION BOTH
C
100)
(
l
CAUSS-JORDAN METHOD FOR SOLUTION OF AX: B AND/OR INVERSION OF MATRIX A writteu by D¡. Ashok K- Jain. University of Roo.rkee. Roorkee
DrMËNSION A(20.30)
CFIARAC'TER *2.I DATFIIWRITE (*.*) ' Please enter nanìe of the Output tìlc '
RËAD (*-50) DAl-FlL OPEN
(UNIT=ó.flLE: I)ATFIL ' STA-IUS:'NEw')
E (*.2000) READ (+.+) N.M,tNv
wRll
r'¡úRITII (6.450) wtìt tE (6.500) N,M.lNV
NI=N+I NM=N+M
NMI=NM+l NMN=NM+N
APPENDIX
626
APPENDIX
c
C
READ R.H.S. MATRIX IF SOLUTION IS DES¡RED
TO
1000
wRrTE (ó, 550) (A(¡,J),J:Nl.NM) CONTÍNUE
L
WRITE (*.*)'PLEASE ENTER COEFFTCIENT MATRIX ROW-WISE' READ (*.*) (A(l,J),J=¡.N),¡:l.N) WRITE (6.2400)
C
I=l.N
STOP
wRITE (6. s50) (A(l.J).J=l.N)
END SUBROUTINË GASJOR (A.N.NMN.INV) DTMENSION A(20.30)
CONTINUE
c
rF (rNv.GE.0) THEN
L
NM=NMN-N NN4l=NM+l JJ:NMN lF (¡NV.LT.0) JJ=NM
L
C
CONSTRUCTDIAGONALMATRIX
C
DO l00 l=},N NMI=NM+f DO 200 J=NMI,NMN
200
100
L.
DO 500 K=l.N RAKK:r./A(K.K)
A(I.J)=o.
A(l.NMl)=1.
Kl=K+l
ENDIF
DO 600 J:K
600
C CALL_GASJOR (A.N,NlülN.lNV)
.
IF (INV.NE.O) THEN
wRrTE (6.2600) DO 700
l=l,N
'
wRITE (ó, 550) (A(l,J).J=N l.NM)
800
7OO CONTINUE ENDIF IF (INV.LT.O) THEN
,
cl-osE (r-iNrTd) FORMAT (A) FORMAT (' Piogram for the solution'of equation Ax=B',1
.
* * +
I
'
INVERSION CODE
=.1. ONLY SOLUTION'./
A(l.J)=A(l,J)+SAIK+A(K,J)
REI'URN END
4.3
CHOLESKY METHOD
c,-** * *
t+*
**t
+
**
t+
**
** ** ** t ** * *t*+*
*
***+
****
tt ** * *** +***** * *t ** * *f t** **** *
C
I using Caüss-Jordan method 7 ' written by Dr. Ashok K. Jain, University of Roorkee'/) '
'NO OF R.H.S. VECTORS
DO 700 r:r,N rF (r-K).NE.0) THEN SAIK = -A(t.K) DO 800 J=Kl.IJ
C
i
L
5OO FORMAT ('SIZE OF COEFFÍCIENT MATRIX
A(K'K)=1.
ENDIF
ENDIF WRITE (6,2800) DO 750 I:I,N WRITE (6, 550) (A(l,J).J=NMI,NMN) CONTINUE
* *
.JJ
7OO CONTINUE 5OO CONTINUI]
STOP
50 450
I
A(K.J):A(K,J)*RAKK
C
C.
750
I I
2200 2400 2600 FORMAï'(/sx.',SOLUTION MATRTX X 7) 2800 FORMA]'lsx.'TNVERTED MATRIX C /)
IOOO CONTINUE
ó50
t written by Dr. ,{shok
:
READCOEFFICIENTMATRIX
c
DO 650
I
by Causs Jordan Method '/ K. Jain. University of RoorkeeT I Please enter size ofcoeflicient nratrix : .LE. 20 7 no. of R.H.S. vecto$ : .LE- l0'/ indicator INV : -l Solu. onlyi/ : 0 inversion onlyT I Solu. anfinvirsion'/) FORMAI'(/5X.'INPUT R. H. S. MATRIX B 7) FORMAT (/sX.'rNpUT COEFFICTENT MATRTX'4
:DO 600 I=l,N
C
* * * * * +,
=0.ONLYINVERSION'./
550
wRtTE (*,*) ' PLEASE ENTER R.H.S. VECTORS ROW-WISE', IF (lNv.NE.0) REeo 1*.*¡ (A(1,Ð.J=Nl.NM),1=t'N) wRlTË (6,2200)
600
'' '
: I. SOLUTiON AND INVERSION BOTH',/) FORMAT (8Fr0.3) 20tJo FORMAT (' Program for the solution of equations Ax=B'/
c
rF ([NV.EQ.O) GO
* *
627
=',15,1 =',15,1
=',15./
C C
CHOLESKY METHOD È-OR SOLUTTON OF Ax = b wr¡tten by Dr. Ashok K. Jain, University ofRoorkee, Roo¡kee
c c*
***
+
+* * *
+****
f,
**
+
*
t
*t*
+* * * * * f
+
**
*
Ë
*+ È* * * *
+
** * *+*+***
c
i
solulo¡l
oF postlvE DEFINTTE EeuATIoNs
+*
** *** ** * t* f +*f * *
\,....:
#
APPENDIX
628
N = NUMBER OF EQUATIONS TEBMS PLUS I OFF-DIAGONAL NON-ZERO M = NUMBER OÈ VNXIVUT"I TRTANGULAR UPPER MATRIX' IIALF BANDED A = POSITIVB DEFINITE
C
c C
SYMI\,IETRIC MATRIX B = R.H.S. VECTOR
L
L
C'
rF (IND.EQ.
,:i¡íì
L rüVRITE
C
cALL SOLVE2 (N.M.A.RA.B) wRrrE (6.3s) WRITE
FORMAI'
tile' (*.*) ' Please enter natnc ofthe output
ìrvR|TE (*,45)
wRlrE (t.50)
;
mafix A ' row-wise 7'Please enter coefticient semi band width is 3'l and 6 is lf size of matrix
*
follows :'/ ' then feed the data ãs
FORMAT -
60
(
'
* ¡ + , + * t
+**r/ *+.xtl +++tl +xttl
(//
l0X.' Cholesky melhod for the solution of positive'i
* +
lOX.'dcfìnite symnretr¡c banded nìatrix Ax: b'/ l0X.'rvritterr by Dr. Ashok K. Jain' Univ- of Roorkee'/
* * * *
IOX.'FIALF BAND WIDTH l0X,'INDICATOR FOR SOLUTION
l0X,'
t0
FORMAT(/25X,'
25
FORNIAT (25X,'D*L TRANSPOSE MATRIX
35
FORMAT(/25X.' Solutiorì vector')
40
FORMA'r'(A)
45
FORMAT
(
+ *
' I
* *
' '
.
,rr1*,*¡'Pleaseenterrighthandsidevectorb'' FORMA'r
WRITE (ó.10)
l5l=l.N
wRfrE (6.i) (A(l'J)'J=l'M)
STOP
CONTINUE
END
WRITE (6.20)
c -ir I
1 I
L
L+D*L TRANS;OSE FORM DECOMPOSE MATRIX A ¡NTO
C
CALL SOLVEI (N'M.A.RA) WRITE (6.25) DO 30
l=l.N
wRt tE (6.t) (A(l.J):j=l-M)
: DECOMPOSE
vector')
)
Progranr f'or the solútion of posit¡ve defìnite and 7 .symmetric banded matrix Ax = b using Cholesky method?
Written by Dr. Ashok K. Jain. University of Roorkee7 by shilìing the diagonal elements on to the first'/ colurnn in eaclr row. The bottom right triangle 'i
(
'
Please enter no. of rows :
senri-band
t
WRITE (6.i) (B(l).1=l'N)
I
shoulð be filled with zerocs. '/) 50
WRITE (6.5) N,M'IND
t5
b
EQ.
Only the upper triangular 4atrix need be supplied '/
* ' *007) .READ (È,+) (A(l.J).J=l,M).1=l'N)
DO
='.151/
20
**0'/
READ (*.*) (B(t).t=l.N)
=',l5ll
A=L*D'U "//. lOx,' EQ.0: sot.u'iloN oF EQUATTON'.i4 FORMA-TV23X.' A MATRIX')
1*,*¡ N.M.lND
wRtrE (+,60)
(6..) (B(l).1:l.N)
CLoSË (LiNl'I=6¡
+24 DATFIL.
srATUs ='NEw') iu*tT=u'FILE= DATFIL'
gE4¡
SOLUTION OF EQUATIONS
c
READ (*'40) DATFIL
oru"
l) SroP
'r€
IMPL¡CIT REAL+8 (A'II'O'Z) 00)'RA( I 00) DtMENStoN A(l 00'25)'B(l CHARACTER
1:
61c
APPENDIX
l-r
'
width
indicator fbr solution
.LE.
: .LE.25
:
100 7
'l
'l
A:L*D*U 'l
EQ.
I
F.Q.
0 : solution of equations'/)
: DecomPose
sutlRou'l'lNE soLVE I (N.NSBW.A,RA) C
.c t,
SUBROUTINE SOLVEI REDUCES THE'A MATRIX INTO L,L TRANSPOSE FORM VEC-TOR RA IS A LOCAL
VECTOR
NSBW IS NUMBER OT- OFF-DIAGONAL TERMS + I
tMPLICIT REAL*8 (A-H.O-Z) DTMENSTON A( I 00.25).RA(
I
00)
'
t '.,,,t
o c.J
APPENDIX
630
APPENDIX
63t
JA=2
{_,'i
LB=I+NW
NA:N-I
o
rF ((LB-N).GT.0) THEN
NW=NSBW-l DO 50
I=I,NA
LB=N
:
ENDIF
LA:l+l
1.,
(-r
tB=t+Nw
DO 90 J=LA.LB
CONS=1./A(l,l)
c(J){(J)+A(t..rA)*Ct JA:JA+l
RA(r):coNS
M:l
(._,.
CONTINUE
90
c(N)=c(N)/A(N,l)
rF ((r,B-N).GT.0) THEN
r)
LB=N
C
ENDIF
c
DO 50 K=LA.LB
L.
M=M+l
BACKWARD SUBSTITUTION NB=N+l
I
LA=NW
JC:I FUN=-CQNS*A(l,lvf)
DO 75 l.=l,NA
JA=M
l=NB-L
DO 50 J=K.LB
CI:C(I)
A(K.JC):A(K.JC)+A(l.JA)*FUN
M:I
JA=JA+I
LB=l-l rF (r-Nw).LE.0) THEN
¡g=¡i+l ]U
.
LA:LB
CONTINUE
RA(N)=1,/A(N,l)
DO 75 J=l.t,A M=M+l
RETURN
c(1.ß)=cì(Lts)+ct*A(LB.M)* RA(t.B) LB=l-B-l
END
/) SUBROUTINE SOLVE2 (N,NSBW.A,RA,C)
CONTINUË
C
RETURN.
C L
L
-
ENDIF
c
SUBROUTINE SOLVE2 SOLVES THE EQUATIONS FOR THE GIVEN RICHT HAND StDË IN THE FORM OF MATRIX C. ANSWER IS RETURNED IN MATRIX C.
c
END
4.4
SUCCESSIVE OVER RELAXATION METI{OD
IMPLICIT REAL*8 (A.H.O.Z) DTMENSTON A( I 00,25),RA(
00),C( I 00)
c+i C
C
c
I
FORWARI) SUBSTITUTION
-c
**
+
**
*+*+*
***
+
****
****
*
+
****
**++ **
*******
****
** ***
****
***+**
SUCCESSIVE OVER RELAXATION METHOD FOR SOLI.JTION OF SIMULTANEOUS EQUATIONS by
C
NA:N-l NW=NSBW-l DO 9Q-l=l,NA
LA=l+l
cr-c(r)/A(l.l)
c(l)={l CI=-CI
C
Dr. Ashok K: Jain, University ofRoorkee, Roorkee c* * **++ t +* *****t+ + t* t* ***+++** * t ** * *****+***+++* * **
**
+
*
t
***
c .. DESIGNED TO SOLVE A MAXTMUM OF 80 STMULTAÌ,¡EOUS C. N = SIZE OF MATRIX C A: COEFFICIENT MATRIX C B=R.H.S.VECTOR C X-SOLUTION VECTOR C EPSL,OÑ = AbS. MAX. ERROR
**
***
EQUATIONS
Èi
() I
APPENDIX
612
(-'
w =OMËCA=OVERRELAXATION FACTOR
(: C
MAXIT = MAX. NO OF rIERA'I'IONS
ä
633
APPENpTX
c
coMMON /^/ A(80.8 I ),x(80)'N'MAXIT'EPSLON'W
FORMA'T (A) ITERA IONS') FORMAI' (s)U'SOLUTION DID NOT CONVERCE IN "I4:' FORMAT (5X.'NO. OF ITERATIONS ="t5//(lPEla.6)) FORMAT ( 'sotution of linear sinrultaneous equations us¡ng'/ r * Suciessive Over Relaxation method'/
LOGICAL FLAG WRITE (*.+) 'Please enter name of thc output
CLOSE (UNIT=O
l0 20
L
CìHARAÇTER*IO DATFIL
40
INTECERCNVRGE
50
DATFIL
READ (+.IO)
+
file'
' by Dr Ashok K. Jain, Unive¡sity of Roorkee '/)
Sl OP
'
OPEN (UNIT=6.FIIE=DATFIL'S'fA'l'tJS='NEW')
ENI)
wRlrE (*.50) SUBROUTINE READAT (FLAG)
wRrrE (6.s0)
coMMoN /A/
c
A(80.8 I ).X(80).N.MAXIT.EPSLON.W
LOGICAL FLAG
FLAG=.FAt.SE.
CALL READAT (FLAG) CNVRGE =
rF (FLAG) GO TO 2C WRITE (+,+) ' Please enter size of matrix '
I
20
READ (*,*) N \ /R|TE (+. l0)
IO
FORMAT
CALL SOLVER (CNVRGE.I'TERSi lF
: r50 F
(cNVt€E -NE. l) co'l'o
ITERS:ITERS.I WRtTË (r.40) lTtlRS'(X(l)'l=l' N) cAli- PRN1 (l'IERS.FLAC)
wtìlTE (*.100)
100 FORMAT(
t *
'
130
READ (+.*) tF
(lcH
' Please enter code for calculations'i
' '
NPLUSI=N+l rF (FLAG) GO TO 30
WRITE (+.+) ' Pleasc enler the cocflìcient matrix rorv rvisel
l) THEN
READ (*,*) ((A(l.J).J=l.N).1=l.N) WRITE (+,*) ' Please cnter r.h-s. vector B'
I=l'N
READ
x(l¡=6.s
C
CNVR(iÊ=l CALL SOLVER (CNVRCE.ITERS) lF (CNVRGE .NE. l) Go TO 130
30
I;IERS:ITEßS-I WRITE (+.40) ITERS,(X(l)'l=l'N)
(t,*) (A(I,NPLUSl),1=l.N)
RETURN ËND SUBROUTINE SOLVER (CNVRCE,ITERS)
coMMoN /A/
ELSË
NPLUSI= N+I 160
'
CONTINUE
CALL PRNT (ITERS.FLAG) (;o TO 150
co ro
ITERS
A(80.8 I ).X(80).N.MAXTT.EPSLON.W
= l*. i'!ì:
CONTINUE
ENDIF
I30
' w= | nrcansGauss-Seidalmethod'l) (*,i) MAXIT. EPSLON. W
2:stopcalculations'/)
CALL READAT (FLAC)
tz(t
CONTINUE
RESID = 0.0
WRITE (t.20) MAXIT
DO20
WRITE (6.20) rv{AXrr
SUM = 0.0
GO'tO
DO30
150
','
' 3. ovcr rclaxatiotr paranìetcr y
READ
FLAG=.TRUE. DO 120
' Please enter the l'ollowing data :7
' l. nraximurn riunlber of iterations '2. maxinrum absolu¡e errori/
l:reoalculate'/
lcH
.EQ.
(
+ * +
l=l.N l=l,N
:
APPENDIX
634
_--__,
-
CONTINUE
APPendix B
-
SLOPES AND DEFLECTIONS
'rEMp = ((t-w)+x(r))+(w *(A(|.NPLUSr) - SUM) / A(l,l))) RESID = AMAXI (RESID, ABS(X(I) - TEMP)) X(l) = ¡sYn CON'I]NUE
S.
No.
Case
Slope or Denãct¡on
ÍTERS: ITERS +I rF (RESID.CE. EPSLON .AND. ITERS -LE. MAXIT) CO TO
i:l
l0
'.1,
â,
lF (ITERS .GT. MAXIT) CNVRGE = 0
e^:
I
RETURN
, L/2 , L/2 F..-..._tn-l
END
pr
2 = oB,
ä
^L3
^c:
*rf
SUBROUTINE PRNT (ITERS,FLAC)
coMMoN /A/ A(80.8 l ),x(80).N,MAXIT,EPSLON.w
a _ pab(2l-a) 0B: ^ 6EIL . pa2b2 " 3EIL
LOCICAL FLAG 2 rF (FLAG) GO TO 40
lvRlTE (ó.20) (A(I.JN:1. N).1=1. N) FORMAT_(/r X.'MATRTX Å /i(sF r 2.4)) WRITE (6,30) (A(I,N+l).1:l.N) 30
FORMAT (/l X.'MATRTX B'/(5F r2.4))
40
CONTINUE
3
Yuþ7
wL3.^^= 5wL4 o^:oo" " 24El' - 384EI
wRtTE (6.50) EPSI-ON,MAXIT.W,ITERS
50 FORMAT (/ . * * ' * * 60
IsX.'MAX. ABS. ACCURACY DESIRED
= '.F10.6/
l5X,'MAXTMUM TTERATTONS DESTRED IsX.'OVER RELAXATION PARAME'IER
='.I5l
IsX,'ACTUAL NO, OF ITËRATIONS
wRiTE (6.60) (x(l),1=l.N) FORMA'I ( I X.'Solurion vector X
4
eA:
='.F8.5/ ='.r5l)
7(E I 5.4))
L
.
N{L ,
oB:
I!
MM
5
Á]\c/-
{\a
o Mt] oo=ZEl=Us'Ác=,, -ML
lLlz +L/2 1
RETURN END
B
6
8
-",)
6EIL
eB=
J4, ^B:l¿
o":
tL3',
oB:
Y!,
^": = ^B
14 M4
APPENDIX
636
Appendix Ç
9
c
:. L/7 . l#+l
-{*, A
l0
. Slope
Case
S. No.
L/2
oi Defbç!þn
^ 5wlj=vs, ^ aç^ -tüy'L3 uo: 60EI xgr (where
,
W:
total load )
p 5'.rr PL3 Lc:ImFl
B
c
lL/? 4
Lt
,2
'/
4
ll
t\ :-
Pal
b3
¡ña vv
3EIL3
A :-
A
l3
_
wLa
-nwt]
n:YoLt =1"1!-¿=¡"¿ 6EI 6EI I2EI
ì\
.-. F.
.
re) - cr(lr- - rc) -:å IzL'LI "'(¿l -
A
l4
+-4 ,razat\
B
B
l5
l6
*tj t2
384EI
A
12
FIXED END MOMENTS
pt] ao: Ã, ¡c:
7PÊ
wherea+b=d,
b+c=e,
ZOSeI
r^=#,o-J-Yx-1r--x)2
*u ¡ : ** (L' -3Lxz +2x2 A= ¿ggl'-* 48EI
þ o+
b {
"f'-';.i(Ð']
";þ-,(;)]
APPENDIX
II\DEX F 26
Fixed end moments
t6l
Flexibility matrix Flexibility method
4,41,92,
Force method
107,193 4
fxed end
243 2s2
influence lines
272'
G
two hinged
244
375
ì
637 196
Gable frame
t79
Gauss method
3E
GeométriC non-linearity Geometric stiffiress 400,480, 576 Global matrix
590
t97
483
419
Grid element
s9l 412
ßa 435
H Hinge, Plastic Hysteresis loops
546 573
594
over factors analogy
168 157
I Idealised stress:s-lrain cun es Inclined supports
544 434
572,596
t2
Incremental method Indeterminate Stn¡ctures Influence coefücients Influence lines
t96
Initial stifhess
193: 232 570
7,279, 343, 398
Iterative method
569,595
deformation
4t criteria
offreedom inuities method
597
I
583
Jr Joint mechanism
555
257
K beam
:rboundary
lgrid :
truss approach theorems
480,576 435 483
398, 57E,591
Kinematic indeterminacy
L Linear simultaneous equations 38
l0 l13
12
M
INDEX
640
Maxwell's law Mechanism Moment distribution MullerBreslau PrinciPle
N Newton-RaPhson method
Non-linear analYsis P Plastic analYsis Plastic hinge Plastic moment Programs, computer
52 546 343 232
Statical method StePPed members
SteP-bY-steþ method Stiffness coefficient Stiffness method Strain energy method
SuPPort reactions 572 mechanism SwaY, 548,569,590 SYmmetry 548,569,590 T Tangent stiffness 546 Three moment equation 546 matrix Transformation 500,598,623 Truss element
549 488 583 168
398 107
434 555
375
570 92 4O7,495 398,578
s9t R Redundants Released stfucture Releases
s Shear deforntations Shear wall element
Slopes and deflections Slope'defl ection method Statical indeterminacY
79,149
Trusses
42 42
U
195
Unbalanced load vector
110,483
Virtual work method
580
v 484 635
w
279
Work
1l
Work
sheet
115
108 36
il 1l i
,3
':i
:
::.,
'.€
'..r1
r.ã
: .:..rl
.'i.:,.::., liji 1;
