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Contents
PART ONE
INTROOUCTION TO STRUCTURAL ANALYSIS ANO lOAOS
Introduction to Structural Analysis
2
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1.2
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loads on Structures
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PART TWO
ANALYSIS OF STATICAllY OETERMINATE STRUCTURES
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EqUIlibrium and Support ReactIons
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6
Dellecllons of Beams' Geometnc MethOds
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Plane and Space Trusses
7
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Dellecllons of Trusses, Beams. and Fra'...... Wor1l-Energ, Methods
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Influence lines
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Beams and Frames: Shear and Bending Moment
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'lomenl
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M"m;:nl Diagram, I(,~
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AppUnbon of Inn..net Unn
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13
Meltlod of Consistent OeronaatiOfts-FOfC8 MIUtod
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Analysis 01 Synunetrk SIrUctUm
14
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Three-Moment Eqaation aad the MettwMI et least WOft
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15
PART THREE
11
.
InnutBCe LinK tor Stamally Indetemllnatl! SlNl;turn
'"
AHALYSIS OF STATICALLY INDETERMINATE STRUCTURES '"
IntrodueliOn to StatlCall, Indetemllnate structures
111
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Siope-Oeflection Method
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Preface
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(, .don II II I
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Part One
Introduction to
Structural Analysis and Loads
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Introduction to
Structural Analysis
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1.3 CLASSIFICATION Of STRUCTURES
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CHAPTER 1 Introduction to Structural Analysis
•
lh.tI till) t\\ll or !l1t)re of the b.I"1l.: larut~ S
I)-pes descn
'
1-.., c('l1lhlnl'J In .1 elg1I ~
struLturc., ~uch
as a
III Iht: 10 II l'\\IO' 01,1) I~
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C , I ' n1 "I I C' ,truClUTe ... functional requirements.
he
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IIlg l)f t 1,rU~l'
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II
II
Tension Structures
The memba of ten
1011 stTUl.:tun:s aTC
tht' .tclion of e\lI:mal IlMd . Becau~
10nn"
O\l'T
subjected to p~re
lension
the tensJle sin:... !'.
IS
thstributed
the: cro ,·....:ctional area.. of members. the material of
a ..(r~ctue
I ' utili/cd in the most efficient manner. T e n ~ i o
t
lomJXhcd of 11r.:\lbk 'lcd l..tblc... are fre~u ntl}
e~pl )ed
to U
bndi!c and long· p.m roof Bt.'Cau~
01 their flexlblhty, cables
negligible bending liJlne' and can dc\e1op onl} h:nsion. Thus
c\tcmallo,Hb. ,I c.lble .ld0ph a \hare that enabl~
It to 'upport the
b\ ten,ik forl:cS alone. In other \\ord':i. the shape of a cable c
a~ the lo.ld, .Kung on it ch.lJlgl: A\ an el\ample the ,hape~
Ihal a
cable rna) a"uml' under
t\\O
AG. 1.4 SU',pen')lon Bndgc
different loading: conditions are shown
Fig. 1.3
Figure 1.4 ~\oh
a familiar t)-pe of cable strw.;ture Ihe WJ:
hrid!jt In J "u"pcm.lon bridge. the road\\a} i., suspended from Iwo
cable... b)- me"n, or \crtu;al hangers. The main cables pass over a
of lO\H~r.,
Jnd are anchored mto solid rock or a concrete foundabOD
their ends. Be-cau.,c su"penslOn bridge., and other cable structures
.,tifTnc,., in latcral directions. the) are elbitp~us
to wind-induced
Fig I 5,. Bri.KlIlg or .,tiffening systems arc therefore pro
lations c~(
to reduce such o\clllatlon,.
Bcside., cahle ...truclurc\ other examples of tension struclures lOci
\ crtical rods used iPi hanger., (for e>.ample. to support balconies Or ta
and membrane .,tructurc\ su(,;h as tents.
AG. 1.5 Tacoma Narrows Bridge
O\4,;illating before Its Collapse in 1940
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Compression Structures
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Compression structures devd P mainI
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acllon of external loads Two common
c fum and ar h Columns are
compressive loads as sh "'II m Fig. I
subJocled to lateral loads and or m
called a
oIumn
An arch I a curved structure w th
verted cable
F,g I
b
upport brid
and long
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t .. MALmeAL MODELS
line Diagram
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Two I ~
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Loads on Structures
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SUMMARY
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ClW'18I2 Lua on _rea
Bl I dm~\
m,d (Jllu, Smu tim \ (SEt ASCE 7-02) II· Manual
I'Ll l)tdmltml Sr t llfIC dllom lor H,ghn,a Br
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and Jnlf "l(IlIon(l/ Blllldmt/ (mit )1 Sj
Although the load requirements of most local buddtn.
generalh based on thllSC of the nallonal codes listed herein 1
rna' co~tam
addltll1nal pn.l\ IslOns \\arranted by such re
dlu~ns
as earthquakes tornadoes. hurncanes, heavy snow and
Local bUlldmg codes are usualll legal documents enacted 10
pubhc "elfare and safcI} and the ~emgn
must become
famdiar "nh the bUIlding code for the area 10 whIch Ih struetlll
be budt.
The load descnbed on the codes are usually based on put
enee and stud, and are the m",;mum for which the anoUi
structures mw be designed. HO\"e\er. the engineer must decidIi
structure I to be subjected to an) loads in addition to those
by the code and If so, must deSIgn the structure to reslSl the
loads. Remember that the engineer is ultimately responsIble for
TULl 2.1 UIlfI WElIIl11 Of
COIlITRUCTIOII 1IA1DIAU
Malenal
Alummum
Bnl:k
C om:rete mnfon::cd
Structural teel
Wood
1
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150
490
40
6
i
design of the structure.
The objective of thIS chapter is to describe the types of
monly encountered in the design of structures and 10 introduce
concepts of load esumation. We first describe dead loads and
cuss Itve loads for buildings and bridges. We next consider the
effect or the impact. of live loads. We describe environmental
indudmg wind loads, snow loads, and earthquake loads. We IlIWl
discussion of hydrostatic and soil pressures and thermal
condude with a discussion about the combinations of load
design purposes.
The material presented herein is mainly based on the A.S
dard Mimmum De>ign Loads for Buildings and Other Slruc
ASCE 7-02), which is commonly referred to as the ASCE 7
and i perhaps th most widely used standard in practice SUIllII
tent here IS 10 familiarize the reader with the general tOpiC of
structures many of Ihe details have not been included. Need
the complete provisions of the local building codes or the
Slandard must be followed in designing structures.
ExIlnPle 2.1
2.1 DEAD LOADS
Dead loath are gravlly loads of constant magniludes and fbed
: " : : ' permanently on the structure. Such load CORStSI of
truetural sy lem Itself and of all other rnatenal and
• The numbcr'l
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brae:
BeD
ref1
t llcms 11s1cd 18 the bibtiop'apby
be purtlwcd lrom the American
Drive Reston Vuguua 201914400
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SlCTION Z 2 ltf'I L..-
2,2 LIVE LOADS
Uve loads
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FOR IUft-OIflGS
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C!lAPIUI 2 Lo.da GIl St,uclUl"
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live loads for Bridges
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CllAPTER 2 lDlGllNl SII"'IIl...
2.3 IMPACT
2.4 WINO LOADS
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the procahLM d
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II wmd pc,-J I JllJ Ihc d) IlJOB" pr" \lr'" '''<.incoJ "n .1 tiJI I/'!.',\\mJ n"w "hL,h "In be "hl,n!lc<! b Jl'jtn~
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CHAmR 2 LOid. on Struclu"'.
1ECTIOIl2A _
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BUlldmgs rcprc<;cntmg loy. hazard to
fallurl.:. uch.1'; agncuhur.ll and minOT h'rage lal:lhtu:
\11 hUlldmgs oth~r
than those 11 tcl.l
In
(utcgonc 1 111 and IV
Building rcprc$CntlOg d ub-.tantlal ha/.ud t human hrc:
,
f.
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-;;:: .. .-"'t ...
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In
of failure. ~u.:h
as those "here more th n )00 people
f.u':lhtle\ \\Ith I.:dpal,;lt
congregate in one area: da~·cre
-':<l
the
II
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gTe-,Hcr than 150: ",hooh "jth
c(111cgc" y,nh capal.:lt)
~rcal
emergenc) tre.ttmcnt or
C<lpall!)
greater than :!5fl
than SI.Ml hO<lpltcil without
~urgc y
fUl:lhue but \\oath
patient capacit) greater than SO; Jail,. ro"er stations and
utili tic.. not c....cntlal in an emcrgcnl1. and hUild O~<;
l.:ontaining haLardou!> and explo~I\C'
material
E'i...cntial facilitic'i. indudmg. ho"pltal fire and pollet
,tatlon'i. national defen-.e facihtle'i and cmergenC) shelten
commUOIcatlon center.... po"'cr ~ta i(lns
and utlh ~S
requircd in an emergen')
1\
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Souru' ·\daplcd "ilh permi~
IOn from AS( E 7.t1~
\#mlnl/lm J)
from ASCE 7·(J2; for further tnlOrmdlllln the I;ompleh: le'),!
)O~1 9 }(l' mth
TABLE 2.4 EXPOSURE CATEGORIES FOR BUILDINGS FOR WINO LONIB
c
Exposure
Urban and suburban areas \\Ith closet paced
obstruction) of the size of single mil h
or larger This terram must pmallln the
up'"'lod dlrec.1:lon for a distance of at t
2630 ft 800 m or 10 urnes the buddin
height whichever I greater
Appll s to all bwldangs 10 "'bK:h ••1""'..... B
or 0 do not appl)
Flat unobstrUL1ed areas and w
out Ide humcane-prone regl
must f"\ II In the up" nd di
dlsumcc of at lea I 5 000 ft I ".. m
,.... til< bUlldmg bcigh. wlu<
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2,5 SfKIW LOADS
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2.6 EARTHQUAKE LOADS
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2.1 HVDROSTATIC AND SOil PRESSURES
2.8 THERMAL AND OTHER EffECTS
2.9 LOAD C(lMBrNATlONS
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CHAPTU Z LIlIU on SlrUCIIJ,"
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SUMMARY
PROBLEMS
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Part Two
Analysis of Statically
Determinate Structures
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Equilibrium and Support
Reactions
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3.1 EQUILIBRIUM OF STRUCTURES
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CH'PT(R 3
EquU,boUl1llnd Support RelClKln$
$ECTIlIH 3 I
lqulto"""", 01 ~
IS 3.1
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nd ""u-
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lJl<; wm
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on m hal
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anJ 1 , n I'
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to'", ru/!. lh"fl" "r'''' thee JlI '11 [<>nidl
t'ud.... ,
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Atlernatlve Forms of Equations of Equilibrium
of Plane Structures
pr,'"de
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Concurrent Force Systems
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CIlIoPTER,
(q"llobllUllllrtlI S"PlHlfl Re.cIlOftI
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3.3 TYPES OF SUPPORTS fOR PlAJlIE STRUCTURES
3.2 EXTERNAL AND INTERNAL FORCES
.
External Forces
3.4 STATIC DETERMINACY, INDETERMINACY, AND INSTABILITY
Inlemal Forces
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CItAJ'T(R 3 Equlhbrlum and SuPl'ort lIf:.cl,ons
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SUtic Determinacy ollnlernally Stable Structures
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ram See Fl. ) 19{b
Del
The frame
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Internally
_cally determinate
" &actfcms
A
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2
EF
0
IS
0
A
A
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2
840k
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J 200
ISk
C4lJltidlrinl the eqwllbrium of portion AC we wnte
EF
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A
A
0
0
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~EMA
125
ill 200
By
80 408
408k
0
EF
A
0
225
A
0
Ion
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the 1ructun:5 hO\\ll
I.I ....... U
nat or
talK II
(nnmatc If the structure 15 stahcally ind.......
n 11\ (hen detcrnllOC' the degree of external
md
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(e)
b
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CHAPTER 3 Equlllbrium.nd Support Reactions
3.1. tbroUlh 3.•1 Dctcrmml,: lhe
lor the trm:lurc, ho""n
m
TI:.tctton
•
,II the urpons
2 klfl
25k_
I I I I I I I I I I I
70 k.
FIG P3.10
... '11
5m
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b(n ft
pj~
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1-1- - - - 10ft -----ll
4 al6 m =:!4 m
.... P3.11
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rtUUIIJJt
-
r-'5 ft -t'----50 ft - - - - I
.B
a~
24 k
6a12Qft=1201l
3.1' Th~
\\clghl or a ~·Il\1g
Irolh:~.
mo.. mg
peed on a Otam hndge: 1$ modded it a m
1.11 tnbuted load a ho\\ n In f Ig pl 15 ~limr. :cD
pre IOn.' ftlr the \crlu.:al nl~tcar
al th u
01 the po IlIOn 01 the tr IIIe') ~,l. mea un:d b
,.md rIot the ¥raph hi \\ mg the . . arlatlon
FIG. P3.17
nf,,""'••
fUnl1l0n
(ll \
8k
24 ft - - - -
.... '113
,
L"_
-'1
-". 25m
.... P3.15
....
A
24 k
FIG P3.14
AG. P3.12
A
I
~- - 4u20ft=Wft- - ~
.... '118
8k
24 k
ft
-t
15 ft
15 It
!41llU
....
m-jI---5 m - + - J
I I I I
10k
,j
B
R
.......-_.
•:
I
.....'
I
I ft
Hill
ft
L-_ _ AA
I ft
r-I~
•.-~Ift
•
2 5 f t - + - - 2J
... PUI
•
...PUII
...........
-
.
F
T
·
,
. ··'f.r
I:~>'
"
. . ':'
~
."
,:'
.
~f
_.:~
shown m Fill 4 12
11M
SECTION U
CHAPTER" P1,ne Ind Spice Trusses
II" ,,\IIK llr',I.ll-lIill\ of tru~,e
.... ,i' bOlhncees
Thr.: hr ~ \..'C'nJIUtll1 . M ~l.:h1
If III
21 -- r. the '.'u., .. ,.. dclinncly ta
dod ,uthucnl III thl.: ~1ICS
•
d
r
• th' rem llllllll! 1\\-0 l:on Ilion .... lor ,talle del
,,:all} un"t;.lhk Ilo\\CHf. t.:
• _
'l..
•
,
ld mJc:ll..'rnlll1.lC\ (m ":'J J . •lre nCl:c\sary
Stag 0
c etenninlCV, Indetermtnacy, 1M InItIbIItty of PIIne T . -
I
minJI..'\ m
T' al
..
I
nol ... ullKu:nt t(111l11h l n
-
,
In other \Hmh. thc...c (\\0 equation, .. imply
, .'
-,
.
·I n -mha and rc.lclums I' "uffi.uent lor stabiti
w, that the mOil h(r \' I l.:
,
- -oy mform.llIon regardmg their (lTnml/(m 'Iff
The) d (1 ntll pro!' I d c.: ....
_
Ill numl:x'f 01 mcmhcr... and e'\ternal reach
Iru.... rna) h Jloe.1 'U II h. t.:
,
•
hut ma~
1111 be un...tabk" du\.' 11..1 Improper arr.lOgemcnt 01 members
m=17
)=10 '=3
m+r=2J
",_17
)
If)
' ..
2
'"
mt,<,q
' I
(a) Sldlleally Delemunalc
StatICal) IRdct rm
or e"lema! upp..lrt...
_ ' .
\\ e cmph.l Ill." th.ll In ,,'rdcr for the Crltena lor ~1.li·
~e( ml
and mdelcmlmaC\ ;l ~l\(:n
b~ Ell." .Ll and 4.4 to be \·ahd. the lrUII
O1u..t be !>tabk an"d act a a. lOgIc rigid bod~
under a general system
'oplanar IllJJ \\hen atl'lchcd to the "upport-.. loternan} ~table
tfUllel
~U"l
lx= upportcd b~ at lea t three reactIOn ... all of v. hlch must be
ther parallel nl1r concurrent. If a tru,~
l~ mlernall} unstable then It m
be .. upporteJ h~
reactions cqu,l1 in numhcr to at least three plus
nu01lxr of cll.uallom of comhtlon I.l + t, and all the rcacUons mu t
neither par.llld nor concurrent. In addition. each joint. member
portlon of the Iru..... O1u"t be con"tr.lined against all po!'>~ible
rigid bod,
mO\emenb 10 the pl,tne of the tru"". either by the rest of the truss or
external "urpart If a Iru" contam" a sufficient number of mem
bUI ~eht
are not properl} arranged. the lru'" i'l said to have aiticalfi
For ..orne Iru...~e"
it nhl) not be oh\ious from Ihe drav.ing~
"'hether
not their members arc arr,tnged properl). Howe\er. if the member
rangemcnt is improper. it will become e\iden! during the analysis of
tru..s. The anal)"is of .. Ul..:h unslahle Irusses will always lead to meOD
tent. indetcnnmatc. or infinite results.
m=16
J=IO ,=1
m +'< 2J
(d)
",=26
Un"lable
J=
15
,=4
(e Staucall Determinate
E
nr=1O
E
j=7
,=3
nr+r<2)
(0
- 4
".=11
,-)
J=7
m+ ,.2
Un~table
19) StatlcaJly
DetemunalC'
E
2.4tlpm~
~
~
@,
- ":~
_
CIa .. if} each of the pl,lnc tru.. ~ .... hlmn in fig. 4.15 a" un"tahlc statlcall
~tal mre
or statical!} ml.!ctcnmnate If the tru" i.. ,tatll.;all) mr~tedni
then enimrk~d
the degree of 'fati!." indctcnnin<ll').
Solution
m=
(a) The tru..s ho.... n In hg 4 15 a, contains 17 members and 10 Jom
upponed h) J r al."l101i I hu m ,
2, Sml.:C' the thn."C reactions
ther parallel n r l rn.:ulTtn' an d th e m('mlll.'rs
•. 01. thl: truss are properl) am.....
I
It
I
tdlll.;ally
(b) hlT
un table
no
I' )
",=16
1=10
,=4
".+,=2)
".-19
J=12
r=S
m+r-2J
FIG. 4.15
B
... -13
F
8
'" +
hi Stabcally Delcmunale
~A
III dod,
(c) for thl lru m 21 J 10 and, 1 Ikcau..e m , 2
lattt.:all) mdetemllnate with the degree or tallc Indeterminacy ,
2
4 It should be ob\lOt15 If( m fig 4 I Iii C that the lru S \;onlalPJ fI
membe lhan reqwrtd tor tablht
j=7
,=4
m+ ,=2)
(h) Statically DelmninalC'
dclermlnate
thl tru
10
t) Slabcally DetmniIWC
-
-...
~
,,
,,
,,
,
,,
,,
CI)
(b)
BecaUlll
tioD,
the
EF
JC
component of FA• •1 zero the secoad
0 can be ..usl!ed only if FAc il also zero.
The IOCOIId type of arrangement .1 shown 1D PIt
_
of three members AS AC and AD COII_lili
joiDt A Ole that two of the three members AS IUl4
9tI ...... from the liIure that 8UIIlC tbm IB DO
IppIiId to the jOIDl to MI,_ the ~ m o c
JiIIIilIDD equaltoD E F. 0 can be satisfied only if F.
1"
.\. _I,
CIW'lIR 4 PIlAt .... $pICI TIUSSIS
6
.
8
{)r.I\\.1 Ircl.:·llVU.
.
hlfl,.',""
.
in
.
arH)\\"
lorce' h\ .lrnl\\"
1ECT1OIl4.5 _ _ ........ T _ " . . _ , , _
th II'T,lIn of lhe selected joint. ShOWID
•r
h' ..
pullin
l ' ,t\\.I\ from t l.: JOint and com
. ~,
.
.'
pu..hlll11 1010 lhe JOIOt. II 'IS usually
F
e
. I"' llnklH)\\11 member forces to be tensile
I .
Ih' unklH'I" n [lm:cS b\ app ymg the two
t.:
r-: 0 A
..
. m MU Ilion.. \ ..... F
0 Jnd L f ::::.
posItive a
l<l 3 ..,UI11C 111.:
n..
b.
A
.
LA'ICrmIl1t..:
.,'
anumember
force -mean.. that the mcm be"
. r I,S In tension as
're'"
a
neoatiH
ans\\cr
indicates
that the
as...ume d \,ht:..
e
IS 10 .. omru·c..,ion.
-'
If at lea... one of the unkno\\ n torces actmg at the
Joint
be ~ltncie\Oo
I"
24k
lOk
411
•
in the horizontal or \ertical ~ir t on.
the uno
dctcnnincd b) satlsfymg the two eq
cquaUons b) inspection of the joint on the free-body
F
G
the truss,
If all the desired member forces and reactions have been
mined then go to the next step. Othern ise. select another JOIDl
no more than t"O unknoy,.m. and return to step 6.
8. If the reactions "ere detennined in step 5 by using the equa
eqUilibrium and condition of the "hole truss, then appl
maining joint equilibrium equations that have not been u
far to check the calculations. If the reactions were com
appl) iog the joint equilibrium equations, then use the eq
equations of the entire truss to check the calculations. If the
has been performed correctly. then these extra equilibrium eq
must be satisfied.
7.
-t
A
B
A, = 0
FIG. 4,19
Detennine the force in each member of the Warren truss shown m P
Solution
The Iruss has 11- members and 8 Jomts and
reactIOns Because m r 2'J an d the reactions
.
and the JDl:mlIla
are properl} arranged It IS statically detenninate.
StallC Dt'lemu/UJ()
y 3
tru
Zeru-Fi r Ut!mber It ca be
members CG FG and G
n seen from Fig 4.19{a that at
and (G I not S H a r e connected of y,. hich FG and GH
zcr r. rce ecn~me
no external load IS apphed at jomt G maD
F(
0
Fr m the dunen510n Of
lopes of 4
rJl
F the truss we find that all IDchned
wn m Ig 4 19 a Th f
I shown on FIg 419 b A
e n:e-body dIagram of
not be lime r
a Jomt with two or fewer unknIC'''....- ...
Mot be found \\ Ie I
J n1 G ha DI t\\O unk
l;a u ate the suppon reac
nOWn force FF and F. H actmg on It
lO
b
collmear
F~
they cannot be dermn ned
0
A
by the method ofJomts.
48
24
04=36
Rea ',ons 8 usmg propom
b
12k
It-110ft
12
....
SECTIOIl' S ""ysIs of ,.... T..- by'" _
of _ _
'IS
F
•
th u "no"n
fl..lrl.'C
actmg at the
•
tal or \ertl.,;al dlrecUllO the "on~ u
med ~b sal! I. mg the w,o w
n.pCC1lion of the Jomt on the Ill.::" body dlit
fllrl'C and n.: ~noltc
rune bc:cn
tiP Omen' 1~ sek.OCl anothlo:f J I
U known.. and return to h:p t.
..
d t mnned an !\h:P 5 hv usmg th cqu
and nd lion of the" holt: ~urt
then ~prl)
th:
nuum
lOt cquihbnum ~nl."auqe
th.ll h.t\\," not he..:n UII}
f; r to kceh~
the Iculallons It the fe.Kllon \\l'fe (omputaj
appl mg the JOint cqulhbnum equalon~.
thl.:11 u thl.: cl.jUlhbri
7.
mber
the
equa
ODS
All 4.19
ofth entire truss to \.·heLk the l.:JkUl.ltilllh It the al\J
the~
l',\lr.t cquihhnum cqu I
been perf< nned correct)) then
mu I be 1I jed
h
R, (',,,
B
'pT pion
"
E....pltU
D:teMnlnC the
In e ch member of the \\ ,lrn:n Illh'" hm\-o
the m hod of J 101
: ;" ~ : 1;
rc
The
a.,anlll'd.
In
o
hg ~ 1
l
.:&
\1)
••')
,,,
o•
F
lrur 2}haBndl' the
member-;
.tnJ 8Jntnh and is uppo
rcal:tlOn'i .md th m.:mher
It I
statIcally deh:nnlOatl:
th..tl at JOlnl
(j
RlC'mbcr (G
F.
o
4
r
Ans
4 ..
4 ..
(b)
20
JF
5~1
F
1 1
1
4 F",
• "a. __ ....-
81. . 4.1
II
I
20k
2Ok. 20k
~6al20ft
20k
= 120ft
10k
------1
(a)
A
A-~Jf_+
Example 4.lI
I
20k
A
20k
20k
20k
(b)
'----x
0-
-
A
t
SOk
60ft - - _ 20 ft
B
10k
+
10k
20 11-1
20k
«
7'.......,. ._
.,1. . . . . .'. . . .
• _.&. _.....From F 424{a we ,an obse'f'Ve that the hon onlal section ao p8SIIDt
the three members of tnt rest FJ HJ and HK also cuts an additioDal
Fl thereby rete 109 fi ur unkno\\ os \\ hl&:h cannot be determined
equanOll of eqwbbrium Tro
ueh as the onc hemg conSidered
the member> arranged 10 ,he fonn of ,he I.".r K can he anal)'llOd by
CUf\'Cd around the rmddk Jomt Ilk section bb shown ID Fig 424(& 't
the kuI bon of upport reaction \\ \\ III usc the upper portlOD IKN.
InISS abo\ .e<Uon hb for analYSI The free-body dlagram of !hi
shown 10 FIg 424 6 II can be .... 'ha' allhough section hb hu
merobe FI1J JK and HK fo= 10 members F1 and HK can be
by summmg moment about pomt K and I. respectIvely bccaUle tile
action of three of the four unknown pass through these pomts We
fore first compute F A b) consu1enng section bb and then use sectiOn
F
_bb
M
L
_ _ 4.7
N
tertnme F
1Ion
4.7 ANALYSIS OF COMPOUItD TRUSlEI
and F
hb U 109 FIg 424(6 we wnk
25 8) - FHa 12 - 0
_aa
F
......
td
FHa
-1667 k
FHa
1667 kN (C)
The f.....body diagram of 'h. portion IKNL of the
sectJ.on aa IS shown In Fig. 424(c). To determine FHJ we sum
about F whICh IS the pomt of intersection of the lines of action of F,
S.
1Ion . .
Thus
o
50 8) + 16.67(12)
25 16
FHJ
FH
By IIIIIIJIlIIlI
62 Sk
62Sk
C
m the horizontal dim:tion we obtain
3
625
5
o
EllIImpll 4.1 0
.......... - -
·.,.. --.....-
The degree of stallC mdetermmacy IS given by
(III
,)
2}
The foregomg conditions for staUC detenmnacy and IDIdc1ernllinl
necessary but not sufficient conditIOns. In order for these en
vahd Ihe \fUSS must he table and act as a single nllld bod
general s stem of coplanar loads" hen It is attached to the u
To analyze lallcally delerminate plane trusses, we can
method of JOInts whIch e senllally conSIsts of selectmg a jom!
more than two unknown force aclmg on it and applymg the
hbnurn equallons [0 delermme Ihe unknown forces. We repeat
eedure unlll we oblam all deSIred forces. This melhod IS mOlt
wben forces 10 all or most of the memhers of a truss aR: desired.
The melhod of secllOns usually proves 10 be more con
forces 10 only a few specIfic members of Ihe truss are
method essenllally mvolves culling the lruss inlo two pomona
109 an unagmary sectIon through Ihe members whose fon:es 111'1
and delermmmg the desired forces by applying Ihe three
eqwhbnurn to Ihe free body of one of the IWO portions of the
The analySIS of compound lrusses can usually be expedited
a combmallon of Ihe melhod of joints and the method of
A procedure for the delerminatlon of reactions and member
space trusses IS also presented.
(b)
•1l1li._
..... 40'
.., ....... U
(d)
the plane _
shown as
statioall indeterminate
If the truss 15 statically indetenninate
degree of static indctenmnacy.
(I)
/ll,PU
•
RG. P4.3
...NA
PI.
CHAPTER. Pllne .nd space Trusses
'12
1 hu
111 th(; Illl
lhl.: uniformly dl Inhut d I
n ll) t , \l.1I\ mlth:d hv. Ih
Ih~
uu
purim
<I
dm
on lh
onl,; ntrated I lad 1
Illint
1~ l
lO,~'f=7r
-.
12 It
1,
IHt
'm
~=~ ~ ;:{
B
I
r-- ~5
'm
I 7
ft--t- ft
C
+I
7
FIG.
ft - - 25 fl - -
P4.31
AG. P4.29
I
Jm
15 IN/m
L
1
Jm
fIG.
_I A ~ = = ~ = = ~ 1 ' f \
P4.25
4.28 Determine the force in each member of the
"urporling a floor deck as shown in Fig. P4.28. The
1\ )1pml~
dctrop u~
on floor beams which. in turn a
neeted 10 the jomts of the lru"s. Thus. the urnform!
tnbUlcd loadmg on the deck i, transmitted by the
hearn'S as l.:onl.:cntratcd load .. 10 the lOp joints of the
30'
lOk
1-----
6 aI 1611=96 II
JOk
JOk
Aoorbeam
-----1
Deck
Roof
Purlin
4.32 1IInuP UI Det nmne
Identified by x of th ,
s«tions
2at8m=16m----
FIG. P4.3D
I
J
K
15 kN/m
B
40kN
6a1Sm=30m -----
l2"+2 m-I-2 ...1-2...\.2...\.2 m+2 m+2
F
A~kf~ ±
FIG.
BCD
IJO k
IJO k
1 - - - - - 4a13O t=\
II
I
G
H
I
P4.28
each ",,:naI1Il'
supported
whICh IR tum are attached to the jOints of the
4.21 MIl 4.3D Oetennme the force
root lru 'OJ hOl'n. The roof IS ylpmI~
10
L
-
1M
CllAPlBl 4
PIa.. ami Space Trusses
Pi,
,.
G
I I.
III It
'1 (.:= ~ := ~ = = l'[)
\11 k
10 II
III ft
11111
III ft
HI II
.... 1'4.33
FIG. P4.36
J
b~=;":iFr(H
15k 15k
~=bdlE
4 at
~5
25 k
20 ft =xo II
bm
k
.... P4.34
"'OlN
- - - - b II 20ft = 120ft - - - -
4at4m=16m
2ULN
FIG. P4.39
FIG. P4.31
1m
.....1I--40kN
--"'E
1m
60kN
J
1m
2m
3m
3m
3m
G
m
.... P4.3li
6aI4m=24m
FIG. P4.40
l!i1i
CHAPTER 4 PI,ne Ind Space Trusses
r-
"9'
4
m-j
SOkNIM
SOkN_
SOkN
N---r
A
4m
100 kN ~.Z=)r;:
L
---.l.
:{'p.;1'5-l1t
100 kN l/.it~:
I
4m
-t
D
-+
4m
F
4m
A
~d",c
. . . . . 4.7
1
.............. D<te
(he tru h en
All PUI
P4.44
AG.
:'\0 k
_~O
J
II:
K
9 fl
9f1
20 ft
t
20 ft
9ft
9ft
)Ok
40 k
4 at 15 ft = 60ft
9ft
9 It
Yll
FIG PU2
9tt
9 f1
AG.
P4.46
9 f1
FIG. P4.43
,
8
C
D
E
-S m+sm-!-6m+6m+Sm+sm
FIG P4.47
20ft
All.
N.4lI
I
_' ' "_4~
-,
J__A
-,
J
~c
20 ft
10 ft
If
k
~tA
.HAI
e,2t~
0
,
IOftrlOft
~t.
C
5ft
II,
,
3m
)IUdli ~ 2lJtN
H
~ ~ 2lJtN
-+
3m
Sm
~ .:a_ ;~
y
,
I
60kN
A
tAy
sm-/
c.
et t.,.
C.
I-IOft-1!Al- 6ft
-...
ID
momber
1
or
I- Oft-~6ft
...
•
CHAPTER 4 Pl,ne and Spice Trusses
180
c
...-=~5f~=: ~
t
5
Beams and Frames:
Shear and Bending Moment
5.1
5.2
5.3
5.4
5.5
5.6
c. 10--rfI
C
, /
------,l-- - \
10 (t
•
/
•-l'liftt.
_-l'ift-,
,., 5t
C
jj.'
Axial Force, Shear, ~na
gnl~ eB
Moment
Shear and gnl~eB
Moment Dagrams
Duailtat,ve ~te lf D
Shapes
Relationships between s~aoL
Shears ~na
~neB
ng Moments
StatiC Determinacy mret~nI
nacy ~na
lnollabllity of Plane Fra es
AnalysIs of Plane Frames
Summary
Problems
r
Sf!
36 ft
A ....'
A --"
IA,
:
,
_ _ Ie; 11----15 f t - - l
Elc=\allon
fIG
P4.55
B
IB
Sf!
Sf!
I
uLu_
x
)
Plan
Unlike trusselol, considered in the preceding lhapter "hose m e ~
always subjected 10 onl) aXial ~cro
the member llf nglo fmID
beams rna)' be subjected to raeh~
fOTl:C-. .!nd bending moment a
as ax.ial forces under the action of e\temal load The detemun I n
these internal forces and moments stre.,s re ultant I D e r) f r
design of such structures. The obJecti'c of Lhi chapt r I t prese
anal}sis of internal forces and moments that rna) de\d p to be:
the members of plane fnlmC) under the ac.:tlon of c pi
r
f
('xlemal fon:es and coupl~.
We begin b) defining Ihe Ibree I)'pe of sue
force!'l shear forces and bendmg moment that rna
sections of beams and the members of plane fi ames \\ n t di;"w"
conslrul:llOn of the shear and bendmg momffit d
m b the m,'th,ld
of sections We also consider quahtatl\e deft(Cted h J'C'$ f
Ihe relauonsh.ps between loads he rs and bendmg momenls.
lion we dC\elop th PfOL--edures for Lon tnk;Un the he r
moment dJagnims using these relau n hi
f
II 'At P
sificauon of plane fn,mes as tauuU del nm t loo."erm,nale,
un table and the anal SJ of lallLB ly etanore:I~
111
112
CHAPTER 5
Beams and ff'lmn: Shelr and Bending Moment
IfCT1ON5.1
_ _. _. . . . . . . . . _
III
p
I')
C
i!
A'- !l=,d =~
A
"
!l.LL.UB
'{a I
(.
P
B,
Ibl
tu=D=o
QIC"'('ltB
P
S
Ie)
'" 5.1
B
(d)
5.1 AXIAL FORCE, SHEAR. AND BENDING MOMENT
Internal forces \\crc (kilned In Section 12 as the forces and couples
ened on a ponion of the '!Iructure by the rest of the structure. Consider
for esample. the Simply supported heam shown 10 Fig. 5.1 (aj. The ~
body diagram of the enure beam ~i depicted in Fig. 5.I(b), which sh
the e\ternal loads. it'! well as the reactions A \ and AI, and B at
ports A and B. respccti\c1) i\'i di'!cu'l'icd in Chapter 3. the support
actions can be computed b) applying the equations of equilibrium t
tree body of the entire beam, In order to determine the internal fo
acting on the crOS:l . .tclion l)f the be.lln at a point C, \\c pass an I
oaf) section {{ through ( . thereby. cutting the beam into two parts
and CR. as shown m hgs S.lIe, .tnd 5.I(d). The free-body dIagram
the ponion -Ie Ihg. 5 Ifc sho\\:l. in addition to the external loads
uppon reaclion\ acting on the portion A C. the internal force Q
and \./ exerted upon portion -IC at C by. the remO\ed portlon
tructure ote that \\ nhout Ihe'Se internal forces. portion AC I
equJlibrium osI~
under a general coplanar "y stem of external
and reaclions three intemallorces (1\\0 perpendicular force com
and a couple are nCf.:e ary at it section to maintain a portlon
beam m equihhnum. The t\\O internal fnrce components are
onented in the dirCt:llon of and perpendicular to. the centroidal
of the beam at the secllon under l:On ideration. as sho\\ n m FI
The lntemal force Q In the dirt.'l:lion of the l:entroidal axis ofthc
The mtcmal aXial force Q at an) section of
but oppo lie in dUe1:t1on to the algebra!\; urn re ta
in the din.'Cllon parallel to the a\IS of the beam of a th
II n und r
support reaction.. acting on clther side of the
Using similar reasoning \\e can define the hear and bendmg
mcnt as folloy.s:
The shear S at an) 'ieClion of ~ beam I eqUdI In m,alll""Ld<butt :o~
direction 10 the algebrall.: sum resultant
the Io;:onlpooc,"s II
perpcnd",;ular 10 the a\IS of the beam of al
reaction admg on clther SIde of the SlXII
The bendlOg moment \I at an 5((11
but oPPO lie In dlft\:tlon to the algeb
cenlrold of the ..TO !tCC1.Ion t the bca
of all the ('\1 mal I ds and uppon a
§«tlon
Sign Convention
The SIgn com mUon commonl) used ~ r the ax
bendlDg moments' dep.C1Cd ID F,g'
.gn convenllon ..h h. flen rd rred t
,pana,,,
Ihal II ),elds the
me posit" or no
C""''t'ftl'ioft.
...
_
-...-_.s
.. -.wAaialForoo. -
. . of !be _ _ c:oasilImd for c:omputiDl the mlllnlll
pcIIilM dinlcliODl of tbc mtcmal fon:es acting on !be
_
.... OIl ..a. Iide of tbe IllCtiOD are shown m FJI. 5
From a computalional vteWJlOUIl however .t IS I U~
_ t to expreIII this IIlID convention m terms of the
In,"'"
IIIId 1ll&dI0DI Il:liDll on tbe beam or f1'llllHO member u
to 52 d
indicated m FIS. S2(b) the
t!tRIIIdIrfd ID poIll wMn tho ex,mrat [orca lit:
",.dIM:e IIIIUimt or hmJe 1M Ierumu:y 10 plitt 1M _ .
CHAPTER 5 Bums Ind Frames: Shear and Bending Moment
.... 1IondInt _ _ ....
SECTION 5.2 -
til
Example 5.3
EIlII.... S2
Ora""
55.
•
A
60k
FIG. SA
A
Solution
.
~
1"" hh ' '..' ~-lg
:-, (
nalh un"upport..'J pomen B
.')"u
PUtl~f'
To ,nollJ computmg reacuons,
\\hICh
I" 10
we
th..: nght of the !>Cetion bb
180 k·ft
I ~
' -_ _ 6m
LIO ft -----!---IO ft
LJ
I
10 f t - L IO ft-l
(a)
the mh:mal force
Sh(ar Con lJ..:nng the cxlt:miil forcl.:
d
•
A
A _ 46 kl
}
-J-
b
E
,
b
I
b
acting
\\ nte
').
':!O ...
S
HO k
46
. ote that the 500 ~k
m coup!..: dtlC" not h.l\t.: any effect on shear
Buuli"y
Con..tdcring (he counlt:rc1ock\\isc moment
yoc
.\[omUIi
Yo rite
,0
A
E
8 L.....,.,,......;C,,-
E
-14
\I
SOO
\/
.140 kN m
,(14)(,1
]40 kN m
The reader rna) r.:hcr.:k the re.. uh.. hy .. umming forces and momen
tlon A B 01 the beam after computmg the reaction.. at support A
5.2 SHEAR AND BENDING MOMENT DIAGRAMS
Shear and bending moment diagram"! depict the \anattoDl
quantities along the length of the member. Such diagram
trueted b> gni~u
the method of '>e tion~
described in the
tion. Proceeding from one end of the member to the other
from left to right section arc passed. after each succesu
loadong along the length of the member to detennine !be eq
pre ing the hear and bending moment in terms of the dis
seclion from a fixed ongm The \ alues of shear and beodiDI
determined from these e~uations
3re then plotted as o,rdiDIlllll
the po ilIOn Yo IIh respect to a member end as abscIssa to
hear and bendmg momenl diagram This procedure
the follo\\ong e ample
-]4
(d) Bendmg Moment Diagram (t·ft
(e) Shear Diagram (k)
FIG. 5.5
171
Cle .&1 _
... ~
SMar and BendInD Moment
III:lIa.I.2 _
. . . . . . . - . . ....._
Sht r D ram To detc:nnlm: the equatlon for ~hear
10 segment
beam \\e pa a sa:hon aa ,It a dl~ta",:e
\: from sUpJXlrt A as
Ii 5 b Con ld ~n
the free tx"ld\ to the len of this section W obtam
s
.u; k
10 ft
lor 0
As this equation andlcah."S the !l;hcar IS tnasO'~
at 46 k from an
dl lance to the th~n
01 pomt -I to an mfinlteslmal distance to
pomt B At pomt A the shear s~acrm
abruptly from 0 to 46 k,
hne IS drolwn from 0 to 46 on the shear dIagram Fig. 5 5 c at A to
ehange TIus IS (0110\\00 b) a honzontal line from A to B to Indicate
hear remams con tant an this segmenl
exi b) u mg $CCtlon b hg 5.5 b
m segment BC as
S 46 60 -14 k for 10 ft < , s 20 ft
The abrupt change m shear from 46 k at an mfiniteslmal distance to the
to ~ 14 k at an mfimteslmal distance to the right of B is shown on the
agram Fig. 5 5 r.; by a \ertlcal hne from +46 to 14. A honzontal
IS lhen dra\\n from B to C to mdlcate that the shear remains constaDt
\alue throughout this segment.
To detennme the equations for shear to the right half of the
convenient to use another coordinate "(I dmx:ted to the left from the
the beam as sho\\on in Fig. 5.5(b). The equations for shear ID 8egmaa
DC are obtamed by consldenng the free bodies to the right of secti
ce respectively. Thus.
S
2"(1
aod
S
2"
10 ft
for 0 S "(I
54 for lOft
x, S20ft
Ikndrng M fMnl Dra(Jram Usmg the same secbon and """. . .
p1oyod pn:v1OusI for compulong shear we detenmne lhe foil_
bending moment In the four segment of the beam For segment A
46
forO
10ft
IBe
M
46x
60
10
14
M
for 0
F'OI ._",1 DC
M
54
M
from which
13 25 ft; thaI
from end E or 40 13 25 26
FIB- S S d
I Example 5.4
Solution
Rea
These equatIOns mdlcate that the shear IDcreases linearly from
20 II; at an mfimte Imal distance (0 the fight of D; It then dropI
34 It at an mfiDlleSunal distance to the left of D; and from there
lmearly to 14 k at C This mfonnatlon is plotted on the shear
shown m Fig 5 5 c
M
M
10
I. nJ
See Fig 5 6 b
B
CHAmR 5 Bums Ind Frames: Shear Ind Bending Moment
SECTIOII U
h
, -.:~;8+LT,
..L....I.-JI.-..!.--l
a
h
...-'- B .- 60.75 l.'l
,
I
hm
- ~-m
(bl
'"
-1725
(e)
,
,
-W.75
Shear Diagram (kN)
75.5
5.3 QUALITATIVE DEFLECTED SHAPES
Id) Bending Moment Diagram (kN
fIG.
rn)
5.6
The \alues 01 S computed from Ihe\oC ~noilauqe
arc plolted to obtam
diagram ho\\n In Fig 56 c The pomt D at \\hich the shear I zero
from the equation
s
from \\hleh \
636 m
(' .
;)
60.75
o
Q ......
DIn, .......
m
>
_
~
1IOftIIIII
60k
180k ft
2k1ft
F
(I) Beam
'
c) Qualillbve Deflected Shape
5.4 RELATIONSHIPS
..... .......-_............ ..-.
automallcalll Y sat~'
menl pp ymg
fled me" no hon onlal force are
qU~'bnum
equallOn E F. 0
we'"
.. d,
S
EF.
0
dS)
dS
0
S
Dl\1ding by d, " .. " nte Eq 5 I) as
dS
d,
.. dx
..
m "hich dS d., represenlS the slope of the shear dUl&J1111
5 2 can be expressed a
ope of shear diagram
ta poUlt
&XIS
To determme the change m hear between points A
of the member see fig. 5.8(a)), we mtegrate Eq
to obtam
f
dS
S8 - S.
=
J:
w dx
in which (S8 S.) represents the change in shear
and B and "'~J
dx represenlS the area under the diltri
gram between points A and B. Thus, Eq. (5.4) can be
c:IwJae m &be r between
AandB
ApplYIng the moment eqwlibnum equation to tho
beam element shown m FIg 5 8 b we wnte
EM
0
M
wdxdx2-S
tIS
By ncgIecting the terms containmg second-order
dM
dl8'"
Sdx
which can also be wotlen as
dM
dx
S
- _•• the slope of tbc bbelldillil
ted as
.'AU
178
dong Moment
CHAPTER 5 BHmslnd Frames: Shear and Ben I
COU
SEClION 5.4 IIetItJoaoMPI_ ~
les or concentrated Moments . .
P
'Ix'l\\c:c:n the 10,luo,; dod shears
I" III. ,Inl! . 5, )~I
arc \alid at the
I'h .<'>, .'
h.p•. :'\.1 I mll ~
.... 'lIlo.:ntr.ltcd 1110menlS, the relatl
f . 'lil,ie.. III Lt.
b E
\ltlwugh ,he
l~ r
,'"~ 1 l It .dr I
arrlll:.t1ilHl l' U
Ixm.ling JnlHllCIHS as gi\cn.) qs S
l\H'c:n thl.' . . hc:.If' ,lIlL!
h P lin'" .\, Illustrated In Example
ltd It ,ut:
(..
.
pit' the bcnlllllg mom"nt chan
.
pt.1ml 01 .trr It.
th' 111.1l:!l1itudc 01 lhc moment of the
h) all .tnWlIlll ('qu.d w \,~ ('llll'" ida the equilibrium of a ddli
the lx.tlll of Fig. 5.8(a) b} passing
dcri\t' Ih" m)~l .n t dr
:i.ln
.II''!.:
.
not \.1 ' .
I' tlit)1l (11 .1
.
lllU.
ment th.tI I l\(llah:
I.' 1<lIKe, to the left and to the right
mfillltC\lffi.1 ul'
.
-,ec:t1onI .1 I 1 n 01. the coup ,_c 1/ fhl' frce-bod} diagram of
(If ,tpr Kalu ~I
5 s J \prhmg the moment equilibrium cq
I"
"htl\\ n m fig.
.<
•
\\ rile
t ~
.\1
\7
( \I
L
1/. - 0
+ ,1.1/) - 0
d.\/
.1/
\\ hKh can be . . ,.!ted ~a
change in hending moment al the
pomt of application of a couple
magnitude of the
moment of the cou
Procedure lor Analysis
The follo"mg step-by-'tep procedure can be used for cons
<hear and bendmg moment diagrams for beams by applylll'
going relaliomhip, bet"een Ihe load" the shears, and the
moments.
J.
2.
Calculate the tropu~
reactions.
ConMrul.:t the shear diagram as follO\.\ s:
a. Detenome the ,hear at the left end of the beam
centrated load 1\ applied at this point the shear
pomt. go to step 21 b). Otheru i,e. the ordinate of t
gram 8tthi, pomt change, abruptly from zero to the
of the concentrated force. Recall that an upward
the hear tu mcrca'ie. \I. hcreas a downward fo
shear (0 dttfcase.
b.
Proceedmg from Ihe poiOl al "hieh the shear w
the pre\lOU, top to"ard the right along the length
Idonllfy the next pomt at "hich the numerical
3.
- . . ..........
•
and nunlDlum value of bend,ng moment 0CClII'
the shear IS zero At a pOInt of zero shear if the
from posIDVC to the len to negative to the rip
bending moment d,agram WIll change from
of the pOInt to negative to the nght of It; that
mmnent WIll he max,mum at th's pomt. Con
of zero shear where the shear changes from
to posItlVC to the nght the hendmg moment will
For most common loadmg condition such
loads and umfonoly and linearly clistnbuted I
zero shear can he located by COIISIdenng the
shear diagram. Howevcr for some cases of
loads, as well as for nonlinearly dlstnbuted
bel
ery to locate the pomts of zero shear by
JftSSIODS for shear as Illustrated in Example 5 4Co Determine the ordinate of the bending moment
pOIOt selected m step 3(b) (or just to the left
acts at the POlDt by adding algebraically the
shear diagram between the prevtous pamt utd
rcntly under conSIderation to the bending
VJOUS pOIOt or just to the right of It, ifa couple
.. Determine the shape of the bendmg moment
the prcvtous POlDt and the point currendy
by applYJng Eq S.8) which states that the
moment diagram at a poInt is equal to the Ibt8t
e. If no couple IS acting at the point under
proceecI to tep 3(f). Otherwise, determtne the
bending moment dIagram just to the nght of
mg ....braically the magnitude of the 1DOlMi!
to the bending moment just to the left of dw
bencti... moment diagram at !hi pomt challlt
amount equal to the magmtude of the JDOIIIIlIi£
t If the pamt under COnSIderation .. not mal'
of the lam, then return to tep 3(b
diagram has been completcd. If dw
ClIIried out cornctIy then the value of
the rig1n of the ri&btlllCl of the t.am
the """"'4 enon.
JlI'OC*Iure can he IIIlld for
...... bypra-'U11 cr-
Tbe
....-
cad,
i~ - DIl O :eht y1 anr
...... by .. c oed,...
the JlflIClIlCbue
be
112
CIMPTER 5 IeIms .nd Frames; Shear and Bending Moment
lECTlOIIu .......Oi........ _ ~
. . . . . . . . - . ••_ _
•
SOlution
R
ttl"
~
r
(I
n
(10)
I~
'::,r
to n
12
In' A
Potnt 8
(I
()
IJ
IJ
P
P
18 k
lt1 ll1
24 k
P Inl(
IJ
Pin, D
IJ
240
40
a po 111\1;• up" <lrd I.:om:cntr,tted fOTl.:C of 18-k
Int ., ':c~OIS
. h h aT diagram
ad at pomt .J I C I;
Im;TC.I..e..
18 It
abruptl) from () to
pomt
P ml B The ,h~ar
SB L
I
Ju..t 10 the kft
tlf
POint B i~
gi\en b)
_ ,"", • _ an:a under the load dlagr.tm bety,ecn just to the right 0
jU'ot tn the kft of B
h ·"pl'" /"
and'
"just to the I
m",)tc!>u~
' R" arc llsc.d to denote
"
h· h h
•.)U"1 to t h
·
hi"
rcv\t'Cll\ch
A..
no
load
"
apphed
to
this segment
e ng.
t·~
•
heam.
s.
I
18
+0
- 18 k
Bl'Caus( a ncgali\c (do\\l1\\ard) com,;cntTatcd load of 12-k magmtude
point B. the ..hear .IU'I to the nght of H .,
S..
18
12
6k
Po;", C
S(
I
Sa
H. ..
area under the load diagram bet\l,ccn just to the nght
jn'll to the lell of C
I Example 5.6
5,,_60_6k
.~(
It - (, - J()
~4
Draw the shear and bendlDg m
shape for the beam b wn ID F g. S I
k
Purnt D
24+0
S/l
It
24
+- 4~
-24k
- 0
The numencal \alue of hear computed at points A. B. C and D
to construct lhe shear diagram a ho\\n in Fig. 5.9 c The shape
bel\\een these ordinates has been C'Stabh hed b) applying Eq S 3
lhal the slope of the hear dlagrdm al a poinl IS equal 10 the load
thai polDl Because no load IS apphed 10 lhe beam between thac
s1 pc of the shear dlagnuTI IS zero bel\\een Ihese points.. and the
n is of a tien of homontal hn a ho\\ n in Ihe figure ole
diagram cl
I e return 10 zero JU I to Ihe nght of the nght
beam lO<bcatmg that the anal 1 ha been earned oul colnCtl
or
A
k
114
CHAPISt 15 Bel.... and Frames: She.r end Bending Moment
70 kN
IICTIONU ft ••
I
~h
r
D
I•••• ~
a
.
m
P, nt A
p, tnl 8
4m
om
(a)
70 k
m
H4:.620~
~
p, "" (
I
B
-\ ::; 70k '
a..ding Mommt D
tb)
,
Zero slope
70
I
\
M
~
B
(e) Shear diagram (kN)
A
m/8
P, Inl C
A
-620
(d) Bending moment
diagram (kN m)
A
I
EllImple 5.7
... 5.10
(ej Qualnallve Deflected Shape
CL: M,
M
06
. .IJDII
0
200
0
MA
620 k
m
Me
M
.00
M
-200
•
10 IlN/rJI
_
air
'M' _1IId _ _ 1... 1I...lng Momon'
P ml D
\Ill
O'i 1to;
1'168
47247
472 68
IICT10IlU
472 47 kN m
nl"
,,,
1111_
0.21 ~ 0
The hendlng moment diagram IS sho\\ll m Fig. 5.11 (d 'The
diagram between the Mdmates Just computed has been based on
thai th lope of the bending moment dlag-m m al. any ~ m t
I equal
at that plint Just
10 the nghl of A the shear IS positive and
of the bending moment diagram at thiS poml. As we move to the
the shear decrea
linear!) but remains positive) untd It beco...
Therefore the lope of the bendmg moment diagram gradual)
becomes Ie steep but remams posl1l\C as \\c move to the ri&ht
It becomes zero at E
ote that the shear diagram in segment AE
the bendmg moment diagram In thl segment is parabohc or a
,un because the bending moment diagram IS obtained by IRtegra
diagram Eq 5 II Therefore the bending moment curve will
degree higher than the correspondmg shear curve.
We can see from Fig 5.II(d that the bending moment
maxunum al pomt E where Ihe shear changes from paslbYe to the
live 10 the nght As we move to Ihe nght from E. the shear bc<:oa~
and It decreases hnearly between E and B. Accordingly the slope
moment diagram becomes negauve to the nght of E, and It
uously becomes more steep downward to the right) between E aDd
left of B. A pOSitive clockwise couple acts at B, so the bendina
crea
abruptly at thiS pomt by an amount equal to the magmtudl:
ment of the couple. The largest value (global maximum) of the
over the eonre length of the beam occurs at JUst to the nght of ..
no abrupt change or dlscontmuity, occurs in the shear diagram •
FInally as the shear In segments Be and CD is constant and
bendmg moment diagram m these segments consists of straight bDII
or
bve sJopes.
A_ _
Quolilotwc /kfler"d Shope See Fig. 5.II(e).
Draw the shear and bendIng moment dIagram Ind the q.uoH'-II!
....pe for the beom shown In FIg. 5 12 I
See Fill 5 12 b
EF
EM
I 3 6 2
2
B
0
0
0
SO k
.1..
- . .. .1
.
190
CHAPTER 5 Bumllnd Frlmu: Shear Ind Bending Moment
1Itl-.,. _.-. - ............_ ...
RClION U
II
til
hathelr
pomt F.
T0 l~at
II
c
Ih pamt f mflectl
hon fot bcndmg m ment 10
upport pomt B hg 5 12 b
16..\ k
.\h r l>ull}r. un
1/
\,
0
\
L
II
"II
Ii
\,
L
S
•
f' nl .,
Plml R
Poml (
Paml
7.~)
50."'
IR
."
'~
k
20
I
9
2
~t
0
The ,hear ,h.tgram I' ,ho\\n In f-i£ 5.12(c) The shape of the
t"fen the ordinall:s just l,;omputed is obtained by appl) ing the condi
~Iopc
of the ,hear diagram at an~
point i'i cqual to the load In
P0ln!. F'or e\ample a'o the load ~tli'en
at A is lCro. so is the slope
diagram at t. Bet\\een ~ and B. the load mtemilty is negative and. I
Imearl} lrom ll:ro at I to - ~ k· ft at B. Thus. the slope of the shear
negatl\e in thi.. ..egml:nt. and It dccrea\Cs (becomes more steep
from A to jU'ot to the left of B. The rest of thc shear diagrdm is COD
using ,imllar rea..oning.
The POlllt of lem ,hear. E. i, located by using the similar tnan
the 'ihl:ar diagram bcl"l'Cn Band C
To facilitate the con..trucllon of the bendmg moment diagram
the \ariou'i ~gmen
... of the 'ihear diagram ha\e been computed aDd
m parenth-.~
on the ,hear dlal:!faffi (Fig. 5.12(c)). It should be n
areas of the par,tbolit.: pandrel.., A8 and ('D can be obtained by
mula for the area of thi, lJ.hape gi\cn In AppendIx A.
II
Point B
II.
0
Poml £
Solution
~nOltca! R
(See Fig. 5 13 b
20 10 5
1/,
Pomt (
1/,
In6 21
Pomt [)
\fJ)
-18 , 18
0: II:
0
C 10 - lOll 15
0
C
LF
20 10
250
.00
A
72
n
k·ft
72· 17822 - 10622 t-f.
11412
M
18 k-ti
()
The wpe of the bendmg moment diagram bet\\een theIe
tuned b) ~ng:
the condlhon thaI the !llope of the bendmg m
an) pomt I equal to the r.hear at that point The bending momeat
con lfUeled I shown In "II!! 5 12 d
It can be seen from Ih IS fi gure that the maximum De
ment OCCUR at POlOt B h
\\ erea the maximum poslUve bendinl
AIls
Draw Ihe shear and bendlDg moment diagram and th qlualitalli.. della;;led
shape for the beam shown in Fig 5 13 a
A
1/ ,
B
IExample 5.9
&ndlll!J \fun/tnt 1>/lJIJfum
Pomt A
1931 fI
Qual".,
lk ,n! Shup< A Qlua'lltativc ddlornd
hown m Fig 5 11 e The bendm@ momm
beam I bent c nca e upward In tb
n
ment I negative m segment AF and GD
In these segments
9k
6
249 ft and
from 1lihKh
21 3 k
2" .' + ~td
SD
f)
2
or
IR k
1 12
D
Po
A
B
0
t
-
IIIU • • _
. . _ _, .... 1IIndInlI ~
20
~m5 ol- + - m
1#
100kN
I
1m
B
.,
E
-500
I
lOOkN
20kNIm
j , , , 'l'"="c
',-i
C =250kN
B
A
----:D
(b)
fIG.
5.13
P, urI C
P, urI D
The
S
L
S
R
20(10
SO
ISO
250
SD L
100
0
S
100
100
R
ISO kN
100 kN
100 kN
0
diagram II shown In FIB 5 13 c
&oding M
p,
,j
Po
B
P,
IE
Po
D
'DID ram
M
SOOk
SOO
0
62
m
SOO
0
kN m
62
SOO
0
m
'IN
CHAPTER 5 lei.... lnet Frames: Shu, Ind Bending Moment
IECTIOflU _ 1 1 1 , , _ ~
... 1../11
'-""'-""'-'-'--r.,
P 'nl
- . .........
". ._
n
.4
Th
t1I ~
"It
ar dl
m
,n
Bend" II
(.Il
f! n TTl=rpPTI '9
~
,
H
C
, =:!..J I..
R ="7~
I..
P nl A
It
,F.
It
p,
l.J1l
24
44
'44
P .,8
D
D =24k
Pin' F
It
"'0
II
PI
,I»
'44
\44
The bend n m ment d
(/wI
'"
I
iN
Sha
f,
r
Example 5.' 1
-24
Draw the hear and bending moment dJ
shape for the tatlcall mdetenntnate be m
al;Uon dctcomne<! b) usmg the proccd
r
lennmale beams presented In Pan Three (th
J"
(() Shear Diagrilffi (k)
J 1lIfI
I
f I
8lkft(IA
1m
1----18"
Jt8
154k
I
•
OXI~
Id l Bendmg Moment Diagram (k.ft)
fI6.
( ) ()uahlall e Detlc\''tcd Shape
I I I I
5.15
I ft
l:;t) k
Ik
IECTION 1.1 ..... Da..
' u:..'111.101101.........,. . . P $ 2S, ......" -
,.
,.
D
... 5.15
(d) QualJtatl\e Deneclcd Shape
old
~/B : . c:~ %
Solution
Regardless of "hether a beam is staucally determinate or iJulletelllli.
the support reactions ha ...e been detemuned, the procedure for
shear and bendmg moment diagrams remains the same. The shear
moment diagrdms for the gl\en statically mdeterminate beam are
5 IS band (c. respectively. and a qualitative deflected shape of
B:
~
~
shown;n Fig. 5.15(d).
5.5 STATIC DETERMINACY. INDETERMINACY. AND INSTABILITY OF PLANE FRAMES
As defined in SectIOn I 3 rigid frames, usually referred to
frames are composed of straight members connected either
moment-resistmg connections or by hinged connectlons to
configurations The members of frames are usually COl_*'!
j01D1S although hinged connections are somellmes used
prevents relative translatIOns and rotatJons of the member
nected to It so the jOlDt IS capable of transmllllng two
components and a couple between the connected memben.
aetlon of external loads the members of a frame rna
ubjected 10 bendmg moment shear and axIal tension or
A frame, cons,dered to be slalkally delermlNll
mome", shear a"d a 101 forus '" all 1/ member. IU
lernal rearl/OII ra" be del ""''''ed by "''''9 lbe eqI-m-1i
"1On and rondlllon
Smce the method of anal)'Sl p.......led In the foll'.....l...'
be UIed onl I anal u tallcally determlDate frames, tt
the studenl to be able to recognIZe tallcally detcl'l1liDale
proc .. l,ng WIth the anal ,
fIG. 5.18
b
-
_mil . . . . . _
.............
_1
member end forces but in oppoSlle dnectlons in accord
Ion s thIrd law Th analySl of the frame mvol~es
the
Ihe magmtude of the 18 member end forces SIX per JIIelq
three uppon reactions At A, and Dr· Therefore the
unknown quantllleS to be determmed IS 21.
Because th entire frame IS m eqUllibnum each of
and JOlnlS mu t also be m eqUIlibrium s shown m F...
member and each jomt are subjected to a general ""PI8uii
forces and coupi wh.ch must saIl fy the three equatillQl
num EFx -0 EFr -0. and EM O. Sinoe the
three members and four JOint mcludmg the two jom
uppons , the lotal number of equations available IS 3 3
1bese 21 equdlbnum equallons can be solved to calcu1a
known The member end forces thus obtained can then ..
tennme 8X181 forces, shears, and bending moments at
along the length of members. The frame of Fig. 5.I6(a
stallcaUy detenninate
Three equations of eqUIlibrium of the entire frame aa
could be wntten and solved for the three unknown ralClilil
and D y . However these equitibnum equations are II8t
from the member and joint equilibrium equations and do;
any addillonal information.
Based on the foregomg dIscussion, we can develop
the static detennmacy indeterminacy, and instabiltty of
frames COntalOmg m members and j joints and supporfllf
ber of) external reacttons. For the analysis, we need to
member forces and r externsJ reactions; that is we need
tota1 of 6m r unknown quanlllles. Sinoe there are lit
JOInts and we can wnte three equatIon of equiltbrium
ber and each JOint, the number of equilibrium equa
m J Furthermore If a frame conlalDS mternal
na1 roI1ers, thae mtcrnal condillons provide additioDal
can be IIICd m COIIIunCllon WIth the equilibrium eqlq::::::
the unkoowna Thill, if there are e equations of Cl
the tota1 number of equallons eqwllbrium equatiOla
CODdition vaiJable IS 3 m J e For a frame, if
equal to the number of equ&tiOllS, \bat
hQ.
6m
r
m
J
....-................
hi"""
JOml lbe number o( equaUODl
_ ..., ... tbc number of mcmben rncelll1ll al tile
....lIIIPle, conJider tbc binpI Joml H o( the (rame
biqe C8IlII I traDJIIUI momenl the m......"'.
three memben EH H and HI meetIng at tile
M
0 M
0 and Mill 0 HI_ _
H
III
ale IlOt IIldepcndcnl m the
that if any
ale
tisfied aIonI wllh the = ~Il t ncmo
H tbc mnauul1ll llIuaUon will aub
tile binpI JOIDt H pnmdca two iDdcpendcnt eq
similar
IIJIII, .1 can be shown that all
pn:Mdiel tbc llIuabODI of condiuon whose numblr
_ben meetJl1ll at the Jomt I
. . . . . . AppI'lIICb
altalDlli"" approach that can be UJed (or delID!lJtlli
indcIcrminacy of a (rame IS to cut enoUBh
by
nc nn"IIl'ry 1leCU0DS and or to remove
IaIdcr the _
Illltic:a1Iy delInnmate The IOl\I1
and tenIII I'CIIrI1D thus removccl equals tho
an example consider the (rame sbowtl
M
......_ ..c B_-Ho-Q
s
A
D A
......- .....c
-
ClW i8i I
. . . II1II F.-: _
Ind IIndlng M....nl
=
I.n
".-5
=6
r=8
~
=0
3m.r> lJ+~
Slalicall loclelenninare • =s)
frd.mes shown In Fig. 5.20
mdeterminacy.
I
Vmfy thai each of thCdP
nate and dctennme Its g.
or~talc
....
...
m=4 ;=4
r= 3 e, =0
3m.r>3)+t'
(b) Staheally Indetemunate (. =3)
IS
sta
m=6 J=6
(e) Statically Ilnde....
Hinge
..
m= 10 Jz9 r=9 e =S
3m+T>3J+t
d Staticall Indelermi.... • = 7)
• ..
,110,
• = 3 (number of girders) = 3(4) = 12
(e)
j
= 3 (number of giJden)
(I)
II&. 5.20
SlIIulIon
See FIg S 20(0 through f
U
,
3m.,>
_IL'... OF PUlE FRAMES
The foUowlDg step-by-step procedure can be used for
member end fon:es as well as Ih. shean bending DIClIIIIiIi
fon:es m memben of plane Iatically determinate f1nm-.. '"
I. Check r. r IaIU: detennmacy Usmg the proeed
ptIle:eding lIIlClion detennm. whether or DOt the
................
..
IIECIIGII U
2lII
ClIAPTER 5
....... end FnIm11: $hI1f Ind Bending Moment
- . .. _
....
y
~-_x
T
T"~-
I
15 fI
x
10ft
18
--'--
-
x
42
42
C
IB
360
JOh- ---1
(a'
lB~
C
8
D
A,- A
t
-- D
Y-~t
Y
L
t
18
42
18
(d)
Dr
Ay
C
(bl
r~
18
B
-sf
M:N:-AO
B"
CB'~ 8
B:'
42
CBe
r
Miff.
8
C
• 8rc
C:c
8rAO
x
8
360
C
8
Be
c:'4,t
MiD'P
Ii
D
D
C
A
A
(I) IIeadiDI Momenllliql-. (l
(e) Shear Diagnms (t)
Cr
rt+8:'
8
MAO
•
C
B
8
Ar A
:-
42
-18
AAO
r
r:o
I! A
D
A~
(J) A>ial Fon:e IliIIn- It)
IB
...121
... 5.21
(e)
and FnImes: ShNr "net
Bendl-... Momenl
......
~hT
tnd f r,
If m
of the fr,tme •• rc hown
In
"Ig
bod d agram, of .all the memben Bnd
,', f'l.anl'<.e
I
L..'<t1O the: computatIOn of In
- D both (,("hleh hale anI} three unko
'~nl "i~l,
or al 1"101
t 0101 ~
A
S 21 d
Sumlarh b\ .Jppl)mg ~
f!
ben are shown 1ft FIg. S 21 g
IS k
o \\f' obtam
~ ltd ~
memben of the frame Fig
From the bendin ':~= . : "
S 21 f we observe thai the
bend concave to the left and conca"c upw rei
b
menl dnelops ID member CD It don not bend but terniU
bve deflcetcd shape of the frame obtained by
In t
the three memben allhe jOlnb i shovin In Fig S "I h
lhe deftecllon of lbe frame al support A 1
Due to":. ~ ,:= ch~
JOint S deflects to the nght to B Smcc the axial defol
neglected and bending defonnabOns are
umed t be
only 10 the honzontal dlJu:tlon and JOint C deflects
joml B: lhat I SS
CC
ole thai lhe cUfUlu of
SlSlent With their bendmg momml diagrams and thai the onlPna
lween memben at the ng.:d jomt B and C ha c been rna
mcd
A:'
d of . f ' and
no,", kno":D mernbIr
\femht'r .fB \\llh the u~f
f'd \foil uhkh can be detennmed by
.fB ba Ihm unkno",ns B r
r an
~
'~ F
(J'\"F,
OandE\(,t-·
us.
I
p)mg
~
Th
s(,
,
18k
\fi B ::..360k-rt
-18k
B t"
J IJfI 8 Proceedmg De\! 10 Jom! Band com.idenng its equllibnum
obtam
o
B,
\lemhI' BC . elll
+ - [F, -
0
. [h
0
·C[,\f,
()
\f:C = -360 k-ft
~'omldcng
the equilibrium of member BC we wnle
18
360
2(30)
213011(5). 42(JO)
+ cf' == 0
+ .\ftC ~ 0
C:c
0
C:c
42 k
M!C
0
0
\Iemhi'r CD Appl)lng.
cflJ
L Fl
-42 k
0 and
L F)
.\f,(D
-= 0
Draw the shear bending moment and ulal force diagram
deftecled shape for the frame shown In Fig S 22 a
Dd
Solullon
IN, rmmQ)- m 2) 3 r 3 and
and lbe frame I geometricaUy stable III tall II
SIal'
- 0 in order, we obtam
3)
~
Sinct' all unkno\l,n ~,"rol
and moments ha\c been detennined we chcc:k
computatIOns b) u mg the third equlhbnurn equallons for member CD
EF
A
: ~
\//) - 0
Jo"" D Cbeclung computation
42
42
0
Slwar Dwqra11lJ The x coordinate 5)' tems sel«ted for the three melD
of the frame arc shov.n In Ftg S 21 d and the shear d18gram for the <:~I l"
co
bUSIng Ih cru~ p
descnbed In Section 5 4 are depicted
S 21 truckd
c
BnuJing \I nt D (lranu The bendmg mOment diagrams for
mcmbenofthcframeare OVtlltnflg 521 r
AIls
IExample 5.14
Joillt C Appl)lng the three equilibrium equations. we obtain
(CD
AIls
Quo/"G"
I k
.f "
Dill ranu From the ~
we observe that the axial ~
comprewve WIth a ~
tant magmt\lde r
gram for lhl member I a stought hne pa
a hOWD 10 f II S 21
Similarl 11 n be
forces In members Be and CD are al
42 k l'espectl\cly The uUlI fora: dl3
.
(OfL'C§ elther,j, J
~ from ih free-bod) diagram
h mol -4 u"t.:an
J In' -l lk'pnnl11g. t }l "
',t lthe n"hl
O.fmu,tall
e llith J. magnitude of I
In order to satl f\ L 'I
of 1~ k to the left. Thu..
to balam:e the honzontal rtlU.. Uon
.f:'
'ui 11,
2
0
Boca
lenD
qualitative
:no
aIAPJEfI 5 ...... _
IEl:TIIII U
FnmIS: Shelr ,ReI Bending Moment
x
I 6 klfl
I I
0
-,,,,-_
c
8
D
(
sf·
-E I11-o'lc....LI_ITJ
1.6 kilt
.e
1Sf'
.11.
("
25
S
8:(
R
V
IJd./ll
Shear Di,acn.... ttJ
c
B
180
c
8
A
(d)
11m
2H
D
.no
t\
~niJ B
y
25 _
rh, uta
- ..
A
'fJ
(J) AJ<iaI Fa... DO...... (Ik
... 5.22
A, •
x
Mument Diagram (kAt)
Ax
410
24
(el
c
111
1ECTJOI5.I .....,." _
CHAPTER 5 ...... and Frames: Shear and Bending Moment
H~
338
"'-
C
,'(P
D
60
It
x
611
B
\ __-\
~I
60
60
A
E
(8) Bendinl Moment Diagratll$ 0lN m
C
12
-96
B
D
-384
314
48
"'5.23
((, Shear Diagrams (kN)
fit 5.2:3
Contd
(i) Qao~"
DoOocIId
2lS
• -,
218
JOUII
R
dd'
Do
...
/Ii
48k
.
n
[F
0
• E
II
[F
0
,,'
1I
12.$ 2
()
~1)
I:
..ax..s
0
0
\
.f
'r - 4RkN r
12k
UfOnm" DE
12 8 ~
'
'-
E
48k
C[ I( -0
60
JOint
1: F,
{I
£,
(J
£1
-12
12
\ftmhn
D Appl 101 the three eqwl bnum eq
m
12 5
[F
L.~
0
£
0
12
[F -0
48
48
Checks
Checks
0
IS« hg. 5.:_~lc
End JQrH
Joint A 8) arrl)ing the I,."qudliom. of equilibrium
\\c obtain
L F,
-=-
0 and
L F y 0-
\Iemher AS Cono;.idenng Ihe equIlibrium of member AB. we obtain
,
B"
B"
I
12kN
.wi
4RkN
B
-
-60 kN· m
loint B Appl)mg the three equilibrium equations. we obtain
Bft
B"
I
-12kN
48 kN
\':( - 60 kN· m
Shrar and lkrulmg Moment DIQR1anJJ See Fl.
\ftmha BC
[F, -0
48
[1',
0
BC
eI
0
liB
0
1241
• [
60
JOllfl
124 2) ... 12 ,
ef< :..
AXIal Fore DIagrams The equauons Ii r axial fI
frame are
-12 kN
Member AS Q
C·<
I
0
Member 1K' Q
Member 'D Q
0
Ch8cIcI
Member DE Q
C Conssdering the equlhbnum of Jomt ( 'ol.e detenrune
(D
,
12 k
cctJ
0
Vnnht-,CD
124
[f,
0
[1'
0
0'
0
[ 110
0
\/0
0
0'"
x
or p
\/ p
-12 k
SUMMARY
48k
60kN m
Ans
••
21'
CHAPTER 5 Beaml and Frames: Shear and Bending Moment
("If (hc mcmt,,:r 01 .ill ,Ill: l Il'rn.11 Ill,ll.1o- .lOd re.lcllono,; .Icung ,,In either
, 0 I Ihl' Xlllln. II'l "'"'
"" "I,'f II W ~ Pll«III\C. \\hen the c'tema.' forr-..
IUl'
. ' _
IlnJ to rnxhlt:c 1",'11 j(ln, The ,IK\lr .H.IIl) ,ct:t!on oj a m~bcr
IS equal
In magnlludl but llPPll Ill' '" Jlredllln. W the al~cbr.':
o,;um of the
",'omJl'.lnlnl in thl' dirl'\:lil11l p.:rjX'l1l!lcul.tr to the 3:\10,; 01 ~ht
member of
.111 Ihe lam"t;~:e
hid...md fl'a",·ltlll1 al"ling on either Side 01 the lieC1ion
\\ l' l.·on'IJa it hl he rt1 ItI\ e \\ hen th" e:\lernal force:-. tend to push the
J'l'rtJon tlf the nll'mbcr lln the kft of Ihe ...et:t1on up\\ard \\ IIh respect to
Ihe portion on Ihe nghl of the ,edlo!1. The bendmg moment at any sec..
u(ln llf J mcmtx'r I" equal 111 magllltudc. but l1rpO.. lte III direction, to the
algebrah-' urn of the mllmcnh abl1uI the o,;txtioll of .ill the eXlcmalloads
and reactlt'n ....tl.:ting on eilher Idc of Ihe ,no1lc~
\Vc consider it to be
p0,iu\c \\hcn [he c\h:m.lllon..:!,;' and couple"l lcnd to bend the member
Wned\ c UP\\ .lrd. l'au...ing C(lmpre ..ion m the upper fibers and tension In
the lo\\cr tibe~
at the '-Cl:UI.m
Shear bending moment. and .I\ial force diagrams depici the vanation.. of Ihe..e quantitic'\ alonf! the length of the member Such diagrams
can be l.'On"ltrul;led h) determining and plotting the equations expressing
these '\treo,;, re"lultant... III lenn'l of the d"tancc of the :section from an end
of the member. Thc con"trul;t10n of ..hear and bending moment diagrams
("an be. con'\lderabl) e\pcdlted by apply mg the followmg relationships
thal eXltit bet\\cen the loads. "lhearti, .tnll bending moments:
..lope of '\hear diagram
at a pomt
intell"llty of distributed load
at that point
(5.3
l'hange in tihear bct\\een
points A and B
area under the distributed
load dIagram between
point\ ., and B
(5.5
change in shear at the
point of application
of a concentrall'd load
magnitude of the load
slope of bending moment
dIagram at a point
change In bending moment
bel\\een POints -4 and B
hear at that point
area undcr the "lhear diagram
betv.cen points A and B
change In bending moment
at the point 01 applicauun _ magnitude or the moment
of II couple
of the (,;ouple
(5 12
58
510
514
A fio:lme IS l;on Idered to be
bendIDg moment Bnd a\ial lore tatl(,;al.l) determinate if the
external reactIOn can be d
e ID all Us members as well a all
clennmtd b
h
num and condUlon If a plane f
) uSI.ng I e equations of equil
I uPPOned b) r react
rame contams m members and J JOID
Ion and h
.
as t' equallons of condition theJIlf
Jm
lm
r
r
J
iJ
3
~
e
Ihe fr..m
I
tahcall u tab)
""I.
the fram , tatoe II del nn
3m r 3J e the frame I tatlcaU md t on
The degree of ta
tiC mdetennmacy I gl\en by
3m
bend~
r
l
procedure for the detemunall n of m mber end r.
moments and aXial force In the members
detennmate frames is presented in Section 56
P
~n
111
CHAPTER 5 ...mI.nd Fro....,. SbN, .nd Bending Moment
I~k.:
I I I I I I I i l l 1111
A
Hnge
===t
B
I-----L.---I
4m-1i~
FIG.
... P5.9
r==r
p
(1
m
fW
A
L
!
I
B
L-_
fI6. P5.18
10m
PS.12
40k
1
fI6. P5.19
20ft
12 k
FIG.
24ft
.,b===!B:="""'----.... C
•
Aprrrq
I
RG.
L
Sf{
10 f•
B
I
fIG.
PS.2O
P5.25
20 kN
=~! -" :! - "
IV
20kN
+
1m
./.-
3m
3m
FIG. PS.21
PS.15
5 m----4-------- I m
A
1i====;)M
I--L_----.,I
I
100 k-ft
~-C!_ID
t"'ii'"
A
6ft
fIG. P5.27
c
6ft
6ft
fIG. P5.16
B
10kNlm
fi r
af
the resWu g
ldiqram
B
fIG. P5.17
I-15ft
20kNim
ISk
... P5.11
30ft
fIG. P5.2II
air;
A~
--L---I
:f--rrrrrr-r-rr~
I
PS.14
"'-PS.l0
fIG.
fIG.
PS.13
7m
---+----7m---l
.....u
. . . . . . . U1
qualitaliw
c
.. _I _.. __.. --.-
I
20.
60
;.--!l-----J.. -..;,t..
...
-
A
V
D
1--10 ft--+--6 ft----l-4 ft-J
m
1 m----
20L
1CE
V
--I-- --I-B
1--5 m
10 ft
5m
5 m--l
20.
, •
75kN
20.
-L~J!- l - ,.4
D
~S
ft.J.--
I51,--1-'0 ft-l
2S kNIm
4.',;-,,,---,--,':'::',::u:I:9:
SJ<
15 kill
I
A
8ft
8ft
8ft
6 m ---+---6 m,·- --1--
8fl
,.,.1.31
fIG. P5.37
50kN
'OOkN
10 ft
-I-
3m
J klft
24 II - _ .
3m
fIC.
... PUI
I
12m--1-6
A
•
"PI.AO
-+_
I-f_ _
JOft
......
t
2 kJft
B
"'PUI
A
B
--.+-_
3 kill
9 ft --i- - - - - _ 24 ft
1--11 11'-+'':'
10 kNlm
P5.36
A
~
.... P1L41
AF~b6~;.J .u .u .u . r. ;~
'0 --+-- J m
40k II
30 ft ----+---10 ft
RG. P5.43
50kN
"'PU2
.. PI.M
10k
"'1'U7
!-I
1----3
3Sk ~
2.5 klft
flC.P5.36
... P5.30
IJ -!: !- L ~
A~C
D
:b.
-
BCD
I
1-_ _1
...._ _1
•
B
C
•
60kN
I- 6 m --l- 6 m -I--s m
'50kN
600kN m
M1k
- -~-
A~
7Sk-ft
11======8
'50.N
_ +-
_H_'...:.
...
..
+ ':'-1--3511--1
Pi
Hinge
soctI""U
J7 ...... 5.71 Draw the shear bendmS moment and
I I ron:c dIagrams and the qualitatIVe deflected ha~
Ii
.1\]<1 I'
....
r
the: frame !>ho"n.
12kN1m
~
4m
4ft+4fI~
4ft
A
~
IS ft
15 ft
ISl!ft
I
BmmC
D
25 k
(a'
(b)
I~
Hinge
o
OS klft
...
1 S m + - - - s m----l
"'P5.58
JOk
10ft
.
I
c
(d)
(e)
Sm
fIG.
P5.57
AG,I'5.55
150kN
AJJ
I
C
B
.1.4
1-9 m-+-9 m-+-- a-l
12ft-+--1 ft
... P5.8I
1II.P5.53
60k
A
1=:Hi~-9rI
90kN
J.
(a)
8
(b)
1 - - 2 0 f t - - + - - - 2 0 f t_ _
25 kN/m
(e)
(d)
1 m
S m---<1--
.......
6 ft4-6 ft--l
-
--_.......
-'
B
I ft
I ft
A
.......
1----20 II - - - . I
ISkNIm
c
D
9m
.....
1---10m---+--SllllilJ
SECTION 1.1 DI""..... ' ............. Dsn •
8
Deflections of Beams:
Geometric Methods
6.'
6.2
6.3
6.4
6.5
6.6
D,Nerenlrat Equallon 'or Beam Deflectron
Drectlnlegrallon Meltlod
nOJ~sol epuS
Meltlod
Moment·Area Method
Bending Momenl Dragrams by Parts
Conjugate Beam Method
Summary
Probtems
J
Strut:lUfC
like ,III \lthcr ph} \11.:,11 bodic!'>. defonn and change shape
\\h"o suh)ltCtcd to fon.:c\_ Olher common causes of deformations of
'JtrUl.:lUn:, mdude Icmpa,llurc changes and support settlements. If the
deltHJndllons t.I1\.lrrcar .md the 'itructure regains its original shape when
the action . . l.:.1U mg the deformations are removed, the deformations are
h:nncd
deWit ,/(/omwflofh.
6.1 DIFFERENTIAL EQUATION FOR BEAM DEFLECTION
The permanent deformations of structures
are referred h) ..... mt/mlll or ,,!mli£ cJelormutions. In this text \\e WIll
Ilx:us our alh.:nUon on l/nwr t/tlllic tIt1uT/1wtimlJ, Such deformations
\011") Imearly "lth dpphcd loads rfor instance. if the magnitudes of the
load altlflg on the lrul.:ture ar.: dl)Ubled. its deformations are also
doubled. and so forth Rel:alllr\)m Scl,;lion 3.6 that in order for a struc.
lure to rc.: pond Imearh 10 Jpphed load'\.. It musl be composed of lInear
cia til. matenal ilnd II mu t undergo lam~
deformations. The pnoclple
f UJlt.iTk.'SHlon I \ahd lor ut:h trudures.
h r rno I trueture e;\l.:\." (ve deformations are undesirable a
they ~am
lmp.ur the tructurc s abiht~
10 '!oCl"\e its intended purpose
f or example a high n bUlldmg rna) bt= perll.'\:tl) safe in the sen that
the II )v-able Ir
re
d
ft
a n I,",eec ed )CI useless unlX"Cupied If It ....
Ie\:
I I) due (0 \\md l.:au 109 uack in the \\alls and window
51 ruclUre
re u uallv de
d
h
Ign
0 I at their deflections under DOrmal
d
~;m
n III n \\111 not e l.:eed the aJlo\\able \alues specified tn buiJd.
221
•
lECTJONu
CHAPTEIII DetIectionI of Beams: Geometric Methods
6.4 MOMENT-AREA METHOO
--
P,
M,
.
on the beam. The slope and
g lIlJl\ idu. l~
to ""h.h \'1 Ihe I(lad .ll.:tdm · .1 kllJ C,1O Ix l,:oll1pulcd by using elt". d t e l -il HI 1\ I d u . ·
_
ICT
dl,;lkxllI.m Ut' l l . l
h ' d' 'Tlocd prC\10U'iJ) or one of the 0'''-I In mel llU l....
~
Ihe Jlftx:t mh:gr.t It
b' 'Ill ~ d l o n '
Aho. man) struclural tnDl
·
J III .. u -.t:qUt;
- '
eo
Illet hi..'d dI"l.-U 'l
II nlUlI of S(I.'/ ('Of1\trlU fUm p. ubhshed by .L_
. - h ndh:ll1k o!
,1
_
'UlI:
nC\:nng a "
~
I (QmI TUl 1/ n wllta," ddkClIon formulas for
""til TIl all Inwfllt: ,: ..'1 f,~ lo.ld...and \Upport condltion\. ," hich can be
t'leam.. fN \.trI(lU ., rx. -h fonnula ... for ,Io~'
and deflections of ....._-d tor Ihl'" rurpo"C: ..s m:
. .
.-ms
u
. 11 Illd .tOd ,upport conditio"" are gl\en IDslCle
for \,.lmt' l,.'omnll1n ~t pt: \. l •
,
.~
.
_
'·,h. bOl1k for clm\t'Olcnt rekren\,;t'
the tronl l:l" er 0
..
P,
wlx)
---
A
I--x
m
Beam
I /Ix I
--------
-
T.....'"B
The mot'n;~re.1
method for computing ...Iopes and def1 ctio~s
of beams
\\3" de\dopcd 1:1) Charlc... E Greene m 187J, The me,thod IS based 011
t\\O thcorem .... called the mOn/tIlt-area ,I"ore",.\'. reiatlOg the geometry
of the ela...lic cune of a ocam to
\f EI diagram. \\ hich is constructed
b) di\ iding Ihe ordmate... of the bendmg moment diagram by the ftexural rigid II} £1. The method uhle~
graphical mterpretations of integra)
imohed in Ihe \DIUlion of the deflection dllTcrentJal equation (Eq. 69
in term... of the ilreas and the moments of arcas of the .\1/ El diagram.
Therefore. II is more comcnient to use for beams with loading dlscoo
tinuitle... and the \ariahle EI as compared to the direct integratlOD
method de-.cribcd pre\iou ..l)
To derive the moment-area theorems, consider a beam subjected
to an arbitrar} loading as sho",n in Fig. 6.4. The clastic curve and the
.\1 £1 diagram for the beam arc also shown in the figure. FOCUSing our
attention on a differential clement d\ of the beam. \\ e recall from the
pre\iou... noltc~
.IEq. (6.10)) that dO, which represents the change m
slope of the e1a!lllc curve mer the differential length dx. is given by
1"
dl/
If
EI dl
611
Note thai the lerm \1 £1 dx repre\ents an infinitesimal area under
M Ef dIagram. as ho\\nin Fig. 6.4. To determine the change ID slope
bet\\een (\\0 arbitrary pomts A and 8 on the beam we integrate
6.11 from A to B 10 obtam
'
rr
dll
or
,
o
~d<
'M
J El dx
MDiapm
_ _-'--..J EJ
fIG.
6.4
in wblcb fl. and fI, are Ibe slopes of tbe ela tiC curve al pomt A and B
respectIvely. wltb respect 10 tbe axis of tbe beam m tb und ~ rmOO
(borizontal stale fl lU denotes tbe angle bet\\een tbe t ng n t I
elasllc curve at A and Band
\I £1 d< repre n th
undo
M EI diagram between polDIS A and B
Equallon 6.12 represents Ibe malbern IIcal
momtnl-a"o Ih "m wblcb can be stated a ~ II
f:
The chan
In 51 pc bet\\een !be
la
polDl I equalt tbe a... under ,he \I EI ~ : , :
proVided that the elastIC curve I
Un LU
SECT10IIu
CHAPTtR 8
DeflectionS of Bums: Geometric Methods
_
d, llO I I1Jll , .
lrllm a roml 81' gl\l;ll t'l~
ek'l11l n1
dO
dl
"here J the dl 1.101,.·..: In1m
ft.!1 mIl) Eq "I ~ \Idd:lo
.
n1C'mcnt cJ n~let
granng Eq.•'
bc:J.m \\e: obtain
necl~
61
8 t' [he ckment lh Sub-.tllution of
.11
ote Lhal Ihl.: ~;h
ProeedUJlI for Analysis
t) thl,.' unddomll.:d aXIs of the
........ qxllllU.lfl
J I
l
(U\~)
,",
6
on the dm:.h~t!ln
side of Eq. 6.14 represent
• 1m 11 .trC.1 u"rf\: pendmg to til: llbout B
an ... (\\(l arhllraf) point -f and B on
.
r:
d\
"r
As<
f'1
,U
Hit
fit ,d,
,E/
in \\hich ~'H
fl:prc..enh the IIl1/e/l'II.';O/ dlrwtioll of Bfrom the tan
at 4 \\ hlch ... the deflection of rOlll! B In the dlr:~t1on
perpendl
to the undcfonncd ,1\1'" l,f the beam from the tangcnt at pomt A
\1 f1jrdr Tepre.. "nh lhe moment of the Jrea undcr the M £1
gram bctwt.-cn poinh 4 itnd B about pomt B
the mathematical expression of Ih
Equation (6.15) hn~erpc
ond m01lUnr·/lHU Ihl oft m, \\hil.:h can he stated as folloYos.
r:,
Tht' lan~gl
dC\lalion In Ih~
dtf('(tl\ln pcrlll:ndil.:ular to the undefi
alll of Ihe tlcam tIt a POint on the (I,...u!.: ("une fwm the laD Dt
or
cIa tiC tune Jt .molher pt1lnll" equal {{l lhl: moment
the area
\11:/ diagram hel\\.een (hI; 1\\0 pl1mt.. Jboul Ihe pomt al ",hi h
tlon I d m:J pr l\u,k"11lhal the cia Ih,; I.un I l;llntmUtlU between
pomts
It is Imponant 10 nole the ord~
of Ihe subscripts used for
6 15 The fir,t !lu~npt
denutes the pt.)!nt "here the devlaUoD
temuned and ahout "hl..:h the muments are c\aluated where
ond ubs4;npt denote the poml \I, here the tangent 10 the elastIC
dra\l,n Also In the dlstanl.:e an Eq. 6.1<; i alway taken
tl\e the IgD or ~B
I Ihe same as thaI of Ihe area of the M
gram bel\\ecn A and B If ,h area "f 'he If Ef dlagram beltwe_
and B 1 positne then ~B
IS also po Ilne and pomt B I
direct
h
the po5ll1
IOn t e tang ot to the cia tl4; cune at polo
ICe
'rna
_
-
6.-58ENDIGMO:=T~A';RSBYP
IECT10ll u
-
.,
ClIAl"IER • DeRictiN. ...ms: GeOmetric M...oda
.
..... - - .
IpplKallon of Ihe moment
J n'
'(tll'O .•
Ihl: prc.."'(c ~)I
of the ;In:.'" and mo e~ts
of areas
_.1
nH11\L'''' ~l .turm ',l
m It \\ ill ~ ..hown In the ~lof
h
01('( lOU I
.
\I f I dlagr.1 .
,.
d fl
1.." the
lh lld for detcnmnmg
\anou" nolrthlO'
I"~
• be 1m me
-.
h e eCliona
'-__
ah
-I n that the (cmJug :- < I" thc:.. e quanuues. W en a ~
-.ex 10
mputatu1n
0
.
.
f
d
1
m
.....
.......tm.. •11"'0 n:lju :' \.'t . 01. loa J ... ... m.:h as a combmatlon
. 0 lStI'i
..ubJccted to Jltfl.'renl ~t ~cr
dc{ermin.Hllm of the properties of the
ut·d and COl1l.:Cnlralcd loa
h' 'ombmcd effect of all the loads,
t:
due to I I: r..:
oded b
..ultant \I 1::/ tlJagrJnl.
Th- difficult\ can be 3\01
Y con tl'UCt
ba:ome d fomlldabk t.l"~
,...
.
\ . . Jllu..tr.llw
Ilk
Hill
~C
Jk
JOft
Jk J
~
10
1
klft
.;i I c::r:=Q!
c
+
.... Ie
4k
16k
B
I
t
I
3D k
30 k.
A
A
t
I ID ft
225
169
8
C
A
+
II
II
120
120
fa
(b,
C
46kt
h6k
I
(e)
JOft
I lOft
225
c
,,....---(r-
120
d)
Ill. ..9
Ft6. 6.10
DID
........ "
.....
210
k Energy Methods
CHAPTER 7 Deflections of Truues, Beams, Ind FrlmtS: Wor-
lECT10Iu ....... _ _
III
p
s/
7.2 PRINCIPLE OF VIRTUAL WORK
b) John Bernoulh
1 11/ rli. \\Im.h \\.1' lI1trnduccd
' l .
,
rl,,1
11111\ 11l.11 hllli lor mal1\ prol.)ems of t
I ~I"
pro'lu
J pt.l\\!;
".
.
.
Xl~n
\\C ~ t u d )
1\\0 lonnul.ltl(,lOS of Ihl Prm.
tural nlIX h.mll. I n Ih IS 'l.:
••
n Ipl( nll/Twal ~/l\pt(
nlUlt\ lor riC/ill 'kid,
l,.lplc nam I) Ull prl I
"II
r
'lor
dt
(tlfl/whlthod/(
\ The latter fonnula
'
th.. p'It/' 'I' al (IT IIi..
,
h'
r,ll,'\\
m"
,,-'1.:11\111\ 10 dc, clop the lilt thud of'"
uon l U'CU 10 I \; II.
eo
,
••
'11 IuJ,·n.:J
III ~ one: of the mo,' general methods
II ,,.. \\IIK .. 1"1 l,.(
...
The
flllt If
I
I
IINl/!1
dctaffiuung delkl:twn
(If
1---.---1
I------
p
L~l
1 he pnm.:lpk of \ mu••) lh... pl.KC!l1t'nh for rigid bodies can be stated
1011('1\\ :.:
If a ngld boJy
I' In
sl
,truelun:
Principle of Virtual Displacements for Rigid Bodies
~"i,:td
~"IUlh num
under a ~ ,tern of force.... and
w .m> sm.llI \ ,rlu.11 }do - l~n
.tncl'TI - ~alp, d
~ht
If
II
I
Dunng the \ Inual transl tion~
A
work done b) the forl.:cs ~I gi\cn by
II
Pli
and
II
Ali
Pli
( li
A
and <\
P Ii
-
LI'
t\
..
A
c _
lniliaJ
equdibrium
P
position
sl
JL~A
p
f
Ii
AG.7.2
see FIg 72 c The vlnuaJ work. don
Virtual rotation (J can be expressed as
W.
P a
By subsUlutmg Eq
78 through
total virtual work. done as
Because tbe beam
EM
c
c
If
of the beam the
I
'"r1UaI
JI' ,
II
c
b
by thl" c\1t:mal fon.:!':' I' I~ro
II
p
c
\ lrtual work doae
rhc term I irl/1al 'Impl) means imaginary, not real. Consider the
he.tffi ..ho\\n in hg. 7 1\<1' The free-bod) diagram of the beam i shown
in fig. 7.2(h; m \\hich P, and P, represent the components of the
ternal htd P in the.\" and y dlrCl:tions. rcspecti\c1}.
0\\ SUppO'iC that the beam is gi\cn an arbitrary small virtual
~dob· ign
di ..placemcnt from it-. initial equilibrium position ABC to
another ro..ition I' B'C' as 'ihm... n m fig. 7.2(c). As shown in this figure
the tOlal \'irtual rigid·hody dhplaccment of the beam can be decom
po!\Cd into translations !i. and 6., in the.\" and y directions respecti\el). and a rotation (J ahout point A. Note that the subscript L IS used
hL're to identify the displaccment'i a.. \Irtual quantities. As the beam
undergoe.. the "irtuitl displal:ement from position ABC to poilU
4' B'C'. the f(,lrce.. actmg on it perfonn work. which is called
uork Tht: total \irtual \\ork. H . pcrfonned by the e"ternal forces
mg on the beam can be d:.M'"crp~
a'S the 'Sum of the \ irtual work. Iv.
and It donI:' during tram1<ttion.. in the \" and y directions respectl
and thL' \irtual \\ork It' . done during the rotation: that IS.
_
L
tberefOl<
EI'
10
Eq
t\
I'
LB
IIEC11OIIl.2 1'ItnoiIIIo .. _
_
_
ClW'TER 7
'qulhlmum under a \ _nual S) ..tem of Ii
I;
ll, :l,.fcd l(l an' Tn.11l fl.'al dc:lonnation (:ODII
ndlllfl'iO)U'"
to 1,;( U
J
unull\ lt1nJlllll11\ (11 the: ,trudure then the
llih the UpplS .• n .. on
.
d
lh- \frlU.11 c'\h:mal I"Tee.. Ian coupl
IeI \\ori. done b
I;
.
.
. I JI pJ.Kl.:m.:nh and rol,lIll)O" '''' I,.'quallo the
thTl u h Ih. n.: I h;m..
.
d
, J n· bl Ihl. 'lrtual mtcm.l! 100.:':" an couples
wI ml ~rnal
\\ 1[.. 1..' I;
.
I'm I dl'.plal'cmc:nt .md rotatIOn,
thh ugh III,," n: III l.
'
I t h' teml rirluaJ ,.. a......ociated with the foren
It • J t(lnna
d
pc
bl
trul,;tuT
I ,n
In thl .. IJh.'mCll
c
mdll...lle Ih.lt Ihe: force ... ~ h:nl l' arhHrJr) and doe.. not depend on
action \.JU 109 ,hI,.' n:.t1 Jdoml.lU()O. _
. _
.
the twoI , Ihe: \ ahdit\_ 01 thl" pn.nc.lplc. consider
.,
,
r \.1 d cmlln, tr ,It.:
".
I
h""n
,n
I·i.'
i
1.1
The
tru"s
I
~
m
equlhbnum
unOOtbe
mc:m~r
ru.. v
e."
•
J~lon
llf a \ inu.1I t:\(l:rnal fon':l: P a~ ,hlm n, !he fr~.lx ~
diagram
10101 ( of th~
tru , i, ,ho\\Jl In hg. 7.J(b). Sl~ce
JOIO.t ( IS In equilib\ irllial c\h:n1al .1IlJ mternal lorce, acung on 11 must satisfy the
rium. th~
10110\\ lllg
(\\l)
I:l\ulltbrium equation,
-L '
\~f
F.
0
()
P
F'H co,lh
F,
U 'Ill
F;B( ~)(soc
- 0
O[ + F , B( sin
(h
0
7I
m \\hlch F l( and F H( rcpre,cnt the VIrtual internal forces in members
.J( and 8C n:'~l.:t \c1y
and fh and fh denote, respectively the ang1
of mdin<llion of Ihc~
member, with respect to the horizontal FIJ.
7 J(a)).
No\\. lei u, a"i"iume that joint C of the truss is given a small real
Ji,plal.:cml:lll .:\ 10 Ihe right from its equilibrium position, as shown
Fig. 7 ~I.t
011: that the deformation is consistent with the support
E.,u I
8,
8,
~
7.3
11
or
W
A JOdlcated by Eq
stde of Eq
~
IJ
Thus Eq
1~
F
&e05O
In .... hleh the quanllt on the left-hand ide rep
.... ork Iv. done by the \Inual eMernal fo
real external dl placement A Also real n th the
& cos IJ are equal to the real Internal dl p1acernen
members AC and Be respectively we an C fK:11Kk that the nighl-hand
Side of Eq 7.15 represents the "Inual mternal.... rk JJ d
"irtual internal forces actmg through the real mternal d p.Ia,:em,en".
lhat is
w
w.
which is the mathematical tatement of the pnnclple f \lnual Ii tee Ii r
deformable bodies.
It should be realized that the pnOClple of Inual ~
here IS applicable regardless of the cause of real deli mat
deformations due to loads tempe:ntlure chan
or n t
be determIned by the apphcalton of Ihe pnn. pi How,.,·e,.
mallon must be mall enough so that the Iftual Ii
an magnitude and dlrecllon while pcrfi rmlO t
In
though the apphc tlon of thl pnnclple lD thl t t
structures th pnJlClple IS ltd re rdl
la lie or not
The method f Irtual w
fon:es ~ , deformable bodlCS
rewntten
(bl
fIG.
F, 11
1~
m
ClW'TEII7 Deflections
-
0' Trusses. Beams, and Fllmes: Work - Energy Methods
II/will
\lrlUal ('xh:m.a1I~C
)
'" ( ftal (; h.-mal 1.11 r1al'(ment
~- -
p
fl'\{(m
D
(~\
\lrlual mternal force
-
n.',.1 mlcmal dlsplacem
o
P,
Member}
(; _ FL
M
B
r r " Illd d/\pJaftll/tllf are used in a general
In \\ h11: h I hI: I cnn s "
.
r~'pc l \
cl), ~ o t e
that because
and mduJe mlmlcnh .mJ rl)t;j1~n,
. nd 'no'oJ..:nt 01 the ,lctlons. t.:au".mg the real defonna
'trtU.1 I ron,:cS.lrI.: I l,:1~
durin"
the c,,"premo
tlon and fl.-mal n 1.\'nsllnt
•
.
eo the rc.1.1 deformation,
""I
the ('''\ll.:mal.mJ Internal \lfto.11 \\ork In I::.q. { 18 do not contam
l.
f,H,:tor I .:!
\ ['I.
(a) Real Syslem
D
Member)
F.
. .
IIldll:atc. the method 01 \Irtual \\-ork empl
1"0 ...crant1c ,)sICm ,I \Inual forc~
'i),h:m and the real ~}s_tem
ofl
or other
H
~ Il\
that cau...e the dc/ann,tUon
dk'\:h
(0
be determmed To
terminI.' the dcRL"(llllll ,or slope) at an) point of a structure a
fon:c >~ lem i, '>Ch.."\:lcd '0
the lml\ unkno\\n in Eq.
"ark ~cthod
to Ix used
and frames are dr:\c1opcd
Vln
B
thai the dC''ilrcd ddlcl:tion \ ~r rotation WID be
71M The exphcH expressions of the Virtual
for computing deflections of trusses be
in the rollov.ing three sections.
(b) Virtual System
FIG. 7.4
W.
LF
. By equaung the Virtual external W rko Eq
2 t
mlemal \\-ork Eq 721 in accordance '4uh the pnoclp
forces for deformable bodies \\e obtam the r. II WIn e p I
method of \Irtual work for truss deflecuon
When the defonnauons are caused by external load Eq
be subslltuted 1010 Eq (7.22) to obtain
7.3 DEFLECTIONS OF TRUSSES BY THE VIRTUAL WORK METHOD
To de\dop the cxpn:ssion of the virtual \,,'ork method that can be uaecI
to determine the delb::tums of trusses. consider an arbitrary statically
determlllate tru a!i !iho"" In Fig. 7.4(a). Let us assume that we
10 determinr: the \ertlcal dcfle<.:tion. 6. at Joint B of the truss due
PI and p, 1 he truss is statically detenmnate
gi\en extcrnalod~
axial forces in Ib member.. ('an be determined from the method of
described prc\ iou...l) III Chapter 4 Ir F represents the axial force I
arbitrary member J e.~,
member CD in hg. 7.4(a)) of the truss
from n/( han/( of n/lt'rl ~
the axial dcfonnation c) of this I'IICIIlber
gl\en b)
1A
!oJ,
fL
At:
1 A and 1. dcnote re rectI' el) Ihe length c:........oec1ia1'*'
area and modulu 01 c1asllut} of member J.
T() determllle the Vertical ddla::uon d. at joint B of the
Ittl a \Irtual y t econsl
r n 'IlIlg uf a unit load actmg
. at the
m the dlrecuon 01 the d
d
d
fl
. Fig
eire e ethon as sho\\-n m
In \\ hu:h
,..N'·_
Because the destred deflection A IS tbe onl UollO
valu can be determmed by solvmg thl equauon
Temperature Chang" and Fabrlcalion EmIlI
The e pressl n or the Vlrtual w rk method
qUit sen I In the sense thaI II n be
ftecIJ os d 10 temperat= chan
~ b,rica.ioD
elfeet r. r which !he member
1de
can be
luated be haud
The
I deformalion
chan 10 temperat=
n 10 Eq
19
n
-
_7 _
.
....,.a ........ ' -
.... III - . ........... _ : W....-En.rgy M....od.
H r
lIlt I I n I (J \ T1,.1 lal:lhtatc the compulallon of the
defl lion the real and \1rtual membc'r lon.'CS are tabulated alonl
Jcngths L and the lanot('.S-"~orc
~aer
-t of the members a hown III
71 The modulus of ('la5111,;11\ E I!'> the same for all the members 50 II. .""'.'
ROC mcluded In the: table
ote that the same sign convention I used Ii
reo..) and \Irtual S) tems that IS In both the fourth and the fifth colUlllDl "'''''..-..
table ten lie fon.-es are entered a" posit" (' numbers. and compreSSIve €I
negatl\( numbers Then lor eal:h member the quantlt) F, (FL A II COIIIIP.
and It!> \alue IS entered In the sixth column of the table. The algebraIC 111II
the entnes In th '501 Ih fo:olumn L F. Fl A is then detenmned and I
recorded al the bonorn of the sixth column. as shown. FinaJly. the deIirId
flectlon.1 IS detennmed b) appl)mg the ,irtual work expression Eq
shown In Table 7 2 ote that the positi\ e answer for .1G indicates that
deftects to the nght In the dlrecuon of the unit load.
TABLE 7.2
Member
AB
CD
EG
AC
CE
BD
DG
BC
CG
A (in ')
L In
192
192
192
144
144
144
144
240
240
4
3
3
4
4
4
4
3
3
,n
n
,n
84m
F (k)
60
0
20
60
0
15
15
-75
25
I
0
0
I5
0
075
075
-125
125
~7.
......-,_.....
CHAPTER 7 DefttctiOItI
o, Trusses. Beams, end Fremes·. Wortl ...~grenE
Methods
1ECTIOI7A a.IUlI••• II . . . '" lie
Ell lillie 75
.
Lkt
i:
t
"',
'Ii.
tbe l;n
{f ''0 (l
mt:th\x! f
II ddl\.
In (I('
Iu,,"
I t01l11 J)
I""f! . nJ m
he
11l
\11 the
r
f f
Iru
\'0
fl.-I
h.'"n
III
In
100
hg
shon
11I1III'" ~
•
Tlill 7.1
MembeT
tu.!I" ri..
CF
f
E:F
r
4
20 ft
lD~ ~
I
.l al O~
I
h - 60 h -----<
.n
In
'a)
E -O.-lm. F
4h,
7.4 DEFLECTIONS OF BEAMS BY THE VIRTUAL WORK METHOD
It'll Real Sy\tem - 8
Ik
IIG. 7.9
IC) \
Inual S)stem - F Forces
Solution
kat S m The re-dl S} lern ..onw'ls of Ihe I.:hanges
members ( f and E.f rth tru a ho"n In FIg 7.9 b
llfun/S 1m The \
lUI di
In
the leo
I
Inua )' tem Call! I Is. of a J -k. load applied
rtl.:tI nat J Int 0 a hov.n 10 "Ig 7 9
The
F In mcmbe (F d
. I.:
necessa
I n
an Ef cotn be ed Ily computed by uSlOg the
'-;;;'Jj
In
Ana
•
_ .& 7
Da""_" nu-. ..... and Frcmu: WOIk-EnIl1Y Methods
"'·_- - A"; .I~'":
SoI\IlIOII
I nd \Irtual \ t 1~
rc luw.n til Fig 7 B{b) and
n from Ilg
I' a Ihat the Ill.: ural rlgldlt)' £1 of the
B b) and (c mdlcates
abruptl) at pOint 8 and f) \1Sl,', Fig
and \,rtual o .dlO~
an: dl \,"IIOUO"", at p'-,mb ( and D respecti
quenll th \an tlon (If thl: quanllt) ( " \' f.I \\1" be dllCOlltin
Bend D Thu the beam musl ~ dl\ Ided mto four scgmen ....
and DE. m I,:h gmcnl the quanllt) \' \I £1 will be COIIti
erof~
can be mlcgrated
~
\ coordmal sclC\IW for deternllnmg the bending m ~
are ho\\n In hg
I' band 4.:
ole lhat In any particular ..._ _
beam the me coordmatc must be u!oed to ",nle both equations
equation for the real hendlng moment \1 and the equauon (or
bendJn, moment "
Th equation for \1 and \I for the four
the beam dc:lenmned b\' u 109 the method of sections are tabulated
76 The drft lion at D can no" be computed b) applymg the vartlW
pression 11\ en by Eq
'0
The
........
E••",pl. 7.9
am be
,
1
Af
HIli
,
3011
EI
E
I
V V
EI d'f
1~
r(;)
):(f~+
1 kN .1.
~f
75, d,
(:)(75, dT
J:(H
+Td)~ ,57
2 193 75 k
£/
2
m'
Therefore
2 19375 kN m'
EI
75
2,193.75
200(300) ~ 0.0366 m
.......
366 mm 1
.1.
TABU 7.'
-';I
Sqmmt
AS
A
o3
EI
Be
A
3 6
2EI
D
A
6-9
2EI
ED
E
0-3
EI
75
."
.._,....-
......_
, , - . _ .... _
....._ _ MoIIHIdI
6 SOIl k
£1
6S01lkft
£1
019410
7
1IF f U - BY THE VIR1UAL WORK METHOD
ApplicabOD of the virtual work metbod 10 determine tbe
8octions of frames IS Slnular to thaI for beams. To determiIIi
bOD
or rollbOD 8 al a poInt of a frame a virtuaJ UDit
couple IS applied al that pornl When lbe virtual sys1eID
the defonnabOns of the frame due to real loads, tbe
work performed by the unll load or tbe urnl couple II
W.
I
porbOD of the frame may undergo axial
In addibOD 10 the bendiDa defonnallons, tbe total virtual
done on the frame IS equal to lbe urn of tbe mternal virtIW.
bead!QI atId that due to axtal defonnabons. As diacuallll1J1
ma lOCbon, wbell the real and vlnual loodin.. and tba
Elan COIItilllIOua over a seament of lbe frame the virtuaJ
due to beadml for that seament can be obtained by
.....,my JI, M EI over the lenatb of the seament Tbo
due to beadial for the ent,,,, frame can then be
the work for the iDdividual &eamenll, that 11,
EIMM
dx
£1
~
W.
eht~fi §
',<.FL/AEl,
axial fI
F UJ4 F. dose to tbe nil
and the uiaI rilidity A£ tile _ a l
tile
!bat
Tbua, the
u discuaood m
due to axial . . .
mtanal work clue
&epICItt
lie
JIIae\
__,.
.
.
.
_
.II/__.
.
.
__
•
-E_~
Ikft
G------::o!D
I
B
A
I,
~cr(
I
cr)
I
I~
I
lli
I
C
EII_7.11
1•
-l
B
I •
. . n.
AJ
I
(c) VUlUl1 S)'IIem - At
D
III. '.11
•
awrrEII 7 DeflL1IoftI of TruIIII . .ms, and Frames. W0rk-Energy MethodS
TAIlE 7.1
(
B
- y
(
~
'-Y
4
~-t!.'t',
-:.~
H
rh
H
' 0
TIl<
k;H
•
4 ISO kN m
£1
~
4
ElIlllJlle ~,7 •.11.'2:
-t>
M
IS. 711
d
10 mm
m
m
An
--:-_-:-_-:::-:-:::-i
312
CHAPTER 7 DrttIIcttons of Trusses. Beam•. and Frames: Work-Energy Methods
lIl:naIIu Dollica". Of " - " tis
_
: lift
1.!!-l.,,1=,T~_
19 !=:r~
c
B
£:29000
I 1000 ...
A = 35 ..
IHI
n
t-
I() ft
-Jo,.
-A
.1
IOU-
/l
116
I
FIG 7.17 (l;ontJ)
115
107
tJ
(<)VutualSy ....
'MUfTO
5
B
B
A
167_
I
I 5
M717
1.
27S
125
b) k
ISm
AI F
-D
III
. W k-Energy Methods
or
ClIAPTEl1 7 Deftections of TrullleS. Beams, and Frames.
lI11e l,ln tx- ddl?fllum'd b~
fle.,:lll,m It Illtn' ( (l t 11I t r.
o:prellllllhnh\ly
10li
,-.
I \
w
(
II
('I'"
~o
If "
1
f/
[ \
2
I h7 d\
,
I~
o -
Ik
or
the \lrtual
""I,
Ju
1
I~
.lpr ~ Og
11 67"1:
~
)~
Strain Energy 01 TI1IIIIs
01\ !..-It
If
9. r'5 I.At 1
D
EI
6=t.l:.
AI:
510S
9.rSII:!
+- -- .
'\5 29000
29.000 1.000,
OI)()()()5
0.04655
o,n466 ft
0559
... Member}
In
ConSIder .he arbitrary truss b n tn FI
a load P wblcb IDCr<a graduall fr
lhe structure 10 deform as shown tn I fi
enng linearly elasllC tructures, tbe deflect
of applicallon of P tncreases linearly tb
I
cussed tn Section 7 I see Fig 7 I c tb
tern
dunng tbe defonnabon 4 can be expressed
p
Sote th.tt the magnitude of the il'\lal defonnation tenn i!'o negligibly smaD
compared to that of the lxndmg dC'limnation tenn
7.6 CONSERVATION OF ENERGY AND STRAIN ENERGY
Before we t.:an dc\clop the next method for computing deflectio
it is nccc..sary to under~tand
the concepts of conservabOD
energy and ""tram energy.
The (ntryr of a ,lrUl:tun: can be 'iimply defined as its CQpacr
doiny 1I0,J,; The lcnn ,train tnat/y ., attributed to the energ t
(ru("(ure hm h((uu\{' 01 ;H t!t'/ormutiun. The relationship between
~lruct s.
\\ork and train c:ncqn of a strUl.:turc ~i d~ab
on the prrnc,plt
{'rtatum of tturql v.hit.:h can bt:' 'tated as follo\\s:
-,_<.
w
AG.7.18
To develop the expreSSIon for IDternal w rk or tra n
truss let us focus our attention on an arbitrary member
CD tn Fig 718 of tbe truss If F represent tbe
member due to lbe external load P tben d. twed
ax,al defonnallon of Ibis member I 8' en by
FL
Internal work or stram energy tored lD member t
The tram energy of !be enlJJe
SlDlpI
stram..,...... of all fils membeTs tUKI
be ...~ ......
The \\otl peth rmed on an ela tiL Mrul,;lurc in equJlibnum by ~i IdJO1
gradually applu.-t.I external lot(~'i
is equal 10 the \\ork. done by
t tC(' or th tr.tln energy hm.:d 10 the Iructure.
4
ThiS pnnciple Can be mathc:maIlCall} expre!'osed as
w
beca
IV
each
FLtllJU>l.
~.!
••
.............. -
,
... - " " - " " ' "
3'8
k Energy MethodS
CHAPTER 7 Deflections of Trusses. Beams, and Fr.mes: Wor-
amount dP th n lhe me
111 ram
apphcatlOn or dP can be \Hlllen a
en r
7.7 CAST/GUANO'S SECONO THEOREM
I
n
11
Ihl
,
8ECTIOII77 - " _ ' . _ _
SC... tl\111
IIOIlS ,11
• 'r
\\C 1,.011 luI:
trudun:
1111
..
cnl'rg~
mCllllxl for dctenninlDl
.
mdhoJ. \\Im:h l.:an he .lpphed only to
<llhHhl.:f
IL,
rhcorcm Lan ~
ll\r 1m
r I\
~I"
1__
dl
-
\ I Imll.dh rrc,cntcd 0) \ l>crto (astlgha
.:Ifl\ ell III,. slru"llln:
\.
.
I
L
,
'
I • 1 1\\ n \' (t.l\llylul1JO \ \t'{ mil 111l1lr( m C
I -, !OJ 1 ... OlllilWIl \ 1..1 l
•
- .
",
I
\ hi -11 l" In be u...cd to c",I,lb1l,h equatIOns of
ghaOl\ fir I t H,'MCm \ l '
_
_
'-.
,
I n 11 Cllll...1JacO In tim, tc'\L C .htl[!hano
hhnum 01 Irultun:s
II'
and the total tram energ)
l
..
"\lotredmth
l
dl
l
tatcJ.I 1.. ,11(1\\
l:
turl:
IJS1II; leu ...
thl.' partial dcri\<lIi\1.' of the \tram
n 'rrhcd fllrcl.' ,)T couple .
i... cqu.d
to the di p1acemen
" I t h (I;
l
'
c'"C
rolaU, n ,1 f Ie
h I
l . . .or. .,'ourlc aillm!~ 11 hnc of actwn
In mat!ll.:mal1cal hlml. thi thcorcm c.m be" ';Ialed 3!'o:
il
II
t,ft
,tr.un cneqn. ~
defkl.:tion of the point of apphca
P In the dlrcction of P:; and 01 -= rotatlon of the pomt
apphl,;.ltion of thl' couplc\l in the direction of Xi,.
To pro\C thl'> theorem. col1,idcr the beam shoy. n in Fig. 7 20
beam i, subJcl.:lt:d to c\ternal loads PI .~P
and P,. which increase
uall) from .fcrn to thell' final \alues. causmg the beam to deflect
,ho\\ n in the figurc The strain energy (L stored in the beam due to
external ,\Ork It ) pcrfonncd b) these forces is given by
10 \\
hid l
of thl,; ~:,lf'
in y.hid \1 ~\ and \ .Ire the deflections of the beam at the pOlO
applil.:allOn of PI p, and P, .~lcvit pscr
as shown in the figure
7 51 i indil.:atcs. the ,tralll cncrg) L IS a function of the external)
and l.:,m be C\pn.:sscd as
J
PI
,~P
P,t
I
- dP, dlJ..
2
•
Since dP2 rematns constant during the additional deft u n .\
Eq
poinl of apphcallon, the lenn dP Ill.. on the ngllt-hand de
7.55) does not contam the factor I 2 The term 1 ~ dP d6. rep •
sents a small quanut} of second order so It ~an
be neglected nd Eq
(7.55 can be "ritten as
.JIP~
2
By subslllulIDg Eq 7 51 IDIO Eq 7 56 we oblaln
U
a ume that the dcf1cClIon tt.: (If the beam at the polDt
plication of P i~ to be determined. If P: IS increased b~ an IDfinil......~i'
dPIJ.
l
OY.
and by <qUlllng Eqs 7 54 and 7 5
t
we wn
l
P,
P
P,
Ill. 7.2l1
P
.... \ 'II \II
_i __
J~
P,
r
or
{
P,
which
lite madtema
Ial
enl
IJ.
C:"';Iiano's
320
CHAPTER 7 Deflections of Trusses. Slims, and Frames: Work-Energy Methods
Application to Trusses
To dc.:H"k'p tht: l"prt: ](In (,I ·on~ilg t'aC
. . ".:cond theorem whM;:h
he used fo Jl'lt:nmnl" the dd1cctll'll' 01 trw·....e" \I.e . . uh!>titute Eq
for the lram cllcrg) l I of tru"se mIll tht: gt:ncral c\pression of
~hano'"
..n'onJ thl'tXCI11 for Iktkdil1n" as gi\cn b) Eq (7.50 to 0
\
0\ .. Ihe flJ.rtl31 dl'n\<J;t1\e tF 'p
~,.
tF tP. the c:\pression of
llghanl) ~'\:ond
thn)f~m
fM tru "e Cdn be \\ nllen a!>
0
and
,,(,!)
FL
,p U ..
\
L.-
The- foregoing e\pre ".on i!> ".milar In fonn to the e"presslon
Eq. P~3)0
As illustrated by
method of \lflual \\ork for ,~>!urt
. . ohed e\ample... at the end of Ihl' ~ction.
Ihe procedure for COlD
When the effect of aJUal deformatIons of the membe
g1e<:ted In the ana1l"ls Eq 7.62 and 763 red
t
defla:l1on.. . b) (.tstigliano'" -.ccond theorem is abo similar to that
\lrtual \\ork method
A
and
Application to Beams
Bi substituting Eq, (7.44) for the stram energy (L) of beams In
of Castigliano's second theorem (Eq. (7 SO
general exp~'ts.on
obtam the lolJo\\mg expre.. . . . ions for the deflections and rotab
spcctl\el). of beam...:
.- JI ")L'ltl\"
,\fz
OJ)
I
and
(/ _L
("'i
{/ - --=
t'.\1
or
I'
0
\I'
-dx
2EI
8
PrtlCldure for Analrsls
A stated prevIously the procedure ~ r
ture by Casugliano's second theorem I suml
work method The procedure essentiaD In I
t,
and
L
8
1
(,
M) Md
,\I El <
J24
k Energy MethodS
1IC1IOII77
CHAPTER 7 Dettectionl of TruUlI, 8.8ms, 8nd Fr8mes: Wor -
Solution
l 1111 the
llWld,natc IW\\l1
Nnllmg R111m lit In the I1t' 1ll.1
Il~
III
1/
The
p:!0111 JI;n\.IIl\
\I "llh
('I
rc
r\
pcd III
r I~
al R
l
II
C t1ghano's :\:ond thlort'1tl
nll\\ be Jcnl,th~
J.
'l\Cn h~
El
8
~b gni~lp a
Fq. 7 (10
.t\
the ex,on"';'.
foIlO\\5.
~
D
gl\t"11 b\
,r
IWO
AI
c
,1/
The Jelk
1 Sk/l,
""(h\ \\C \\ nlc the equal10n
ew.......'..... .,......
-
I
kill
D
t
4Ok_ B
E 29.oooUi
I 2JOOm'
A
40- A
30ft
t
(aj
6S-!!
6.S "
480ck'•i
PL
.lEI
-l
40- B
Delenmne the Hllation of joint C of the frame shown in Fig. 7.23(a
ghano's -.econd theorcm
I •
Solution
This frame .... as prc\iousl} analylcd by the virtual work
7.10.
No external wuplc is acting at joint C. ,,"here the rotation IS desired,
apply a fictitious couple .'01 ( 0) at C. as shown in Fig. 723 b) The
dmates used for dctenninmg the bending moment equations for tbe tbnI
ments of the frame are also sho"n in fig. 7.23(b) and the equauoDl Ii
temlS of (j and i \I i \I obtained for the three segments are tabulated
712 The rotation of Joint C of the frame can no\\ be detenmned by
\I
0 to the equations for \f and I \f I q and by applYing the ex
Casughano' second theorem as gi\cn b) Eq. 7.65.
fl,
4O_
... 7.23
TUII712
.
AB
El
o
00129 r.ld
="~=
~
00129 rod
'-
6.S-1
'd~ ) CIL
6487 Ik·n
U
(\l
•
CHAPTER 7 Deflectionl of Trussel, Beams, and Frames: ~grenE-koW
Methods
1ICnotI7......
• lIw· ...... lIw ......
·'WIlD?
... a.......
EumpIe 716
D
.l5D
----------r
Dt:: n'fl~t.li
I
£=ZOOGPa
-t - I.ZOO mm·
4m
.~5
P.-
,4 )~P34.0+lP 51-(
-15+043P
1
B PI I
L
=0)
P,t = 84)
4m - . . . . -
10+ P,
(a)
7.8 BETTI'S LAW AND MAXWELL'S LAW OF
Max II
aI
MaxweU m 1864 pia aD unponan
mdetennina.e stnJctum 10 be 00t1lido:red
MaxweU law WID be derived here
Bm /a which was praeuted
(b)
FIG. 7.24
Solution
~lhT
Iru..., \\.1 pn:\Iou I) an.l1)zcd b) (he \Irlual \\ork method in Example
stated as foU
For a Imearl C be
tbe
ron:< and coup a<tmg through the de,romlatio
ron:< and coupla I equal
the
TAItl 7.13
I
m
Memhc'r
,F
,F
F
{"'PI
IkN)
rkN kN)
t'P,
rkN kN)
(i"'F/t'P I )FL
(kN m)
OA]
04]
-061
84.48
0
0
0
0
._---
4B
B(
4/1
BD
CD
4
15
15
]
566
4
P, OAW,
04]P
061P.
~"'2k-
P
15 -
5
(J
I
0
0
0
0
71P,
1
071
Ee)FL
£A
k
£A
7Y708
m
0
5m
Ans.
~
For PI - 0 and P2
8448
through the deformation due to lhe P
P,
P,
',
l---+-
'---+
-- . ---
'L---+-
(a) P Syslrm
..
kN m
7Y708
2UI) II) o (K1I2
332 mm
flf,MH32 m
(hi QSYMeID
Ans.
p.
I I
~AEC;)fl
~.
IIECIPROcAL IIEFI.ECnOU
w.
or
.,
_
CIA lSi 7 _ _ ... - . ....... Inti FrI_: Wort<-EnlllI' Me\lIodl
25 b
25 a
ext "e assume (h,H the beam "ilh the Q forces aetia&
IS ubJf1:led III the deflccuons causc.-d by the P
B\ equaling (he \ IrtUOI) external work to the VirtuIl
"ork "' obtam
.lfQ Ifp d<
I
i
EI
o
oung thaI the nght.hand SIde of Eqs. (7.66 and (767
we equal the left·hand Sides to obtam
~
L n'Q - L
I
+
A
I
I
kd-
J:
L
Equation 76 represents the mathematical statement of
Max\\elrs 13" of reciprocal deflections states that
elastic structure ~hl dejfeclwn 01 a POInt I due to a un" lotIII
poml J IS equal to the deflectIOn 01 j dut' 10 a unll load at I.
In thi tatement the terms defierI/on and load ""' UIed
eral sense to Include rotalion and couple. respectively
prevIOusly MaxweJrs law can be considered as a special
law. To prove Maxwell's law consider the beam shown ID
The beam I separately subjected to the P and Q systems,
the umt loads at points i and j. respectively, as shown m
and b As the figure indicates. f, represents the deftcc:tiOJl
the unit load at J. whereas fj, denotes the deflection at ) d..
load at i. These deflecuons per unit load are referred to a
e/fieumlS. By applying Belli's law (Eq. (7.68)), we obtatn
1(/0) - I(fj,)
J.l.
I
or
f,
(b Q S ~
... ,.8
Qt\1')
I
fj
which tS the mathemaueal statement of Maxwell's law
The reCIprocal relallonshlp rematns valid between
caused by two un,t couples as well as between tbe~;:
rotauon caused by a urnt couple and a unit force n
:=:
In Ibis chapter we have Ieamed that the work doIIe by
~OPla=oent.1
or rotaUon OJ of I
of I hoe of acbon IS gIVen by
W
r
Pd.1
........,._.......
.;
d~: = :~_
........, .........
~
theon:m ror
IlDearI
4m
F
... P7.3, PlA7
CMAPTBI 7 Deflections of Trusses, Beams, .Ad Fr.mes: Work-Energy Methods
---- 4
l()()kN
$KIlon 7.4
7.2O.M 1.21 l~'
1'-4~fl
thl \lrlUal \\orl... mcthod to de
lhl IClfll' .I1lJ ddk.."'~i(:
at r~'lt
HoI thc beam h
- ~C!: = 8~= r
12 II
2. Ult
f
a
G
b'l 10
r-
10m
/
,..
tal dd1ectt n .11 JOint II 01 the
nF
P- 1 due ((.. a temperalure lOlrca l f
c beTs BD DF nd Flf l se the mdhClJ fli 'lr
=loO"lanl
=21.).{)OO I..
I
= 3.fX)() ~:ni
2 kilt
I
'~c
P7.20, P7.S1
RG
!
15f,
I",
~,it
1 reqUired for the beam 'ioho\\n. so thai It maxunum
deflcl.:tion ~eod
not exceed the hmil of I 360 of lhe pan
length (I.e. 5:~am6
L 360) Use the method of \"1r1ual
work.
P7.2I, P7.S2
7.22 through 7.25 Usc lhe \irtual y.ork method to
mine the deflcl.:tion at roll11 C orthe beam shown
p
A
==::::::8,===....:1 C
L
2
2/
a=6~(10
RG.
If
RG.
L
T
30k
~
1
All
8ft
1
I
E = conSlal11
I
15 It
fIG. P7.29,
30k
P7.57
tf
8f1-t-8f1=1
L=24f1
£1 = constant
E. = 10.000 U.
A
~,/-1Iiqe_.:
1---16 II
le4000m. 4
HKlkN
A
l============t------
~. (f;~=IO:!t_m
8
hm
f = con!>tant = 70 GPa
I = SOU ( Iff') mm4
P7.23, P7.54
A
8
. . .~-.
i---6m-!--6m--I
1 - -_ _ L-12 m - - - - I
3m
1
~
fIG.
8 II - 1 ~ 8
fIG. P7.26
P7 .22, P7.53
P7.17, P7.I'
the rucal deflection at Jomt (j 01 the tru
P 16 II member Be and CG are 0 5 m. too
method f rtual Yo lTk
!~_ :
1.28 tllrough 7.28 Detemune the smallest moment or lOer.
8ft
~
~A
. . ._ _
1-:1 = .:on:-.tanl
E =70GPa
1 = 1M (lot') mm4
8 ft
_
k
AG. P7.2S, P7.56
m~
AG.
6f.--I
=constant
£ = 29.000 ksl
1=3.500m 4
, JF=-.:~
H
'IF= ~Di
1
FIG P7.24, P7.SS
£1
iF~
5m
21
f = con tant 250 GPt
I = 600f I W', mm4
t.1
1
•
D
,m
H
fII. P7.16, P7.18
ft.
IOOkN
fIG P7.21
EC
... P7.311, Pl..
E./= _ _
. . . . 7..
E. - 200GPa
7.J1_ 7.11
1
336
CKAPTER 7 Deftectionl of TnllIeS. Beams, and Frames'
Work-Energy Methods
c
Pc
Il sc Ih \lfltl,ll \HlTk method t d
1.3 I H (II the If,lIne ho\\n
III 1,'111
1l
Ihe \ Iflllal "'l~rk
method I de nnl
1
lilm
1.
at
Jomt
B,ll
the
fram h n In FI
\
J.. t ..,-
8
PJ 1
'm
20kNim
,m
£1 =con,lant
15 (I
£1=1.: n lanl
F = 290111J I..
I _ ~OCIm."
£ = 711GPa
I
t
AG.
=1.03011((ilmm"'
~I
P7.33, P7.60
5 m+1 5 m - l - - - S m _ _--I
P7.31, P7.59
~
7.34l'-.c the \Irtual \\ork method to detennine the
of JOint D of the frame sho" n.
c
fl
P7.31, P7.l12
--.
.-fFdb~=zJ,
B
~O'
1 kif.
I'~ ~ E
8
T c-
D
2/
tIl_
3m
-Wk8
mual
D
Lift
rk method to dc:temune the honInt
C f the framC' sho.... n
c
D, -~
c
ilt
Z3.7P~
I
7.38 L\C the virtual work method to detennmc lhe rotabOD
of Joint D of the frame sho\\n.
I
ft
Hlft
I
I
18 ft
E/ = (on'laOl
E = lO.nuO bi
I =l'I.Ib() in."
o~
AG. P7.36, P7.37
7.35 L;,c the \lrtual \\ork method to detenmne the
zont•• 1 deflection at jomt £ of the frame shown ID
P7 ~3
15 ft
~
E1= COn!itanl
£ =200GPiI
1= S()()( lOb) nun"
II&.
2/
Sm
1/
200
D
E
/
- - 1 - - - 1 6 ft -----I
P7.34, P7.35
fIG
P7.38 and P7.61
'--
A
I
_,.., ...11I._ _11I'-- _
.... _
-E"""'---
_
15kN1m
c
1
3m
20ft
C
+
u
D
U
65_
kN
1:
3m
·A
- 3 1 0 + - 510
D
fIG,
1----,1011----1
EI - '
P7A3, P7.44
. . . . . 7.7
.ooobi
,AI ....... ,AI Use Casngliano I :;~"
..."A1
lermme the honzontal and vertical 0
ftectlon at Joint B of the trusseIi shown m
•
c
,.II Usc Casbghano's second tbeoJaa
honzontal deflecuon at JOint E of thI
P78
,.It ....
,.u
Use Casbgll8DO IOCOIIll
mme the slope and deflection at point B
in Figs P7 20 and P7 21.
20ft
,.13 ....... , . . Use Casbg!iano'
detennme the deflcctlon at POlDt C of
Figs P7 22 P7 25.
,.I, ... , . . Use Casbgl.ano
A
D
--+--IOft
"
mme the slope and deftcction at poiDt
ID Figs P7 29 and P7 30
, . . U C'SbgllanD S secood ~lI_obt
mtlcal defIect,OO at JOIDt C or tbo
P71
, . U C bgllano second ~ =
borizoolal dcftection at Jom! C
P7 3
ft
,.It U Castigliano second a~cobt
talioo of)DIm D of tbe frame
~
...
_..._-
ob med b
mph sUMIIlutmg Eq .
a
o~\'"
L
L
The mfluencc Ime for 5. fig
2 e shows that the
zero "hen the umlload IS located at the left suppon A oftt.
the umlload mo,e from A to B the shear at B decreases
It becomes
L "hen the umt load reaches just to the left
As the unll load eros
pomt B the shear at B mcreuea
I
L It th n decrca
hnearly as the umt load mo
until It becomes zero \\ hen the umt load reaches the ngllt
Influence Une for Bending Moment at B
When ,he unllioad IS located to the left of point B (FIg 8
pressJon for the bending moment at B can be convemend
uSlOg the free body of the ponlon BC of the beam to the
ConSldenng the counterclockwise moments of the external
reactions actmg on the portion BC as positive m accordaDcl
'altl Ign com: nllon
(Section 5.1). we determine the beDd , .
at Bas
Ms
C (L -
aJ
0S x Sa
When the umt load is located to the right of point B we
body of the ponlon AB to lhe left of B to detennine M.
the clockWIse moments of the external forces and reaction
portion AS as posItive we detennine the bending moment
Ms
as < S L
A (al
Thus the equa',ons of the IOftuenee hne for M. can be
M
s
{C (L
A a
a
0SxSa
aSX L
Equat,on 8 5 IOdtca... ,hat 'he segment of the iD1I_
betwoen potnts A and BOx a can be obtained by
onIi..tes of the gment of the mftuenee hoe for C
by L a Also according thIS equation the segmeal
Itoe for M. between pomts Band C a x S L can
mulbplYlO8 the oro,..... of the segment of the iDII_
and by a The iDlluenee hne for
\hili
......, lines for A and C IS shown m Fig. 8 2
of this iDII......, line 'n tenns of the JIOSIbOn of the IIDiI
obtaiDed by ubstitubng Eq 8 I and 8 2 mto Eq
'0
:;::"S C
M.
SECltON 8.1
lnftuenct Unn tor Beams and Fmnes by EqulIlbrtum MetbOd
CIW'1tR 8 Inftuence Unes
£IIIIlPII '.1
b.
Solullon
f' _ unol II the desired IOllucot;:
d I 1m td
nl ..
e ~nt
I l.: n~ tOld th~
ulllucncc hn r. r
momc:: t b) u m th mllu n hn II r upper.
bel pr
mg'Althlh.. coru tnl.. lllnof ntnn
r bending m mcnt Jt a J"llOf l n lhe tflJl.:tun.:
mflucn.. hnt; fll( IIlhe reJl.{1 ns lm III (the Idl
I h rumt undl;rt,;on Idcr.tuon -trt.: J\dllahle OthCN
.'
d \'0 the qu ero nnuence lines IN n;,h:1l0ns ,,) 109 the pr
du de nix'\! In the prc\lOUS h:p \0 tnl u 'n~e
1101: f(l" Ih she,
r bcndlO£ IOl1meni .11 a ["llOt ~ln lh..: .. trudure 1;.10 hl: con tnll:tcd
s follll....
a. PI.1cc the UOIt load l)O the ...trudur..: ,II .1 \arl.lhk IX"'llon :t f
the t /r flf Ih.. POIllI under nlO... IIJerathl n.•lUd del :n i ~
the (\
pr.. II'n hlf the "he.lf 'M bcodlllg IlHlO1..:nl If th..: c~l[euifm
linl"'S l,)r 11 th\' fe.\I,:lIoth arc kn\mn. then It i~ u... u.t1I) (ome·
01.. 0111' u the pllrtillil of Ih,: erult~
tll the rillhl nr Ihe l"'Jlnl
hlr JctcrnulIlntt the c\prc......ioo for ~hcar
(or \'lending mome01
\\ hl.. . h "1111,;llOI.II0 terms im 01\ iog: nnl) r.:l\tin~,
The ..hear C'
lxnJmg moment IS ,,:olhukred to Ix po.. iti\,: llC e\ha~ n
III ¥i.
l.urd.tOC \\ llh Ihe hal'" S III j om. tlt,o" t:... tuhh hlo:ll In Sl.'1.:tll\
.:;;1
2~gI
.. t pi cc the untt 10<:ld to thl,; rt /11 Ill' the pOIOI uOOa
I. n uh:ralll 0 and ddennlOt: th.... t: pre ... ,on fllC the !>hcar or
~ntd eb
m ment If the IOllucn..:c ~.loh
for .111 the rc.KllOns are
k "" then II IS ~laus
com .. ntenl to u thl' ponion
the
(TU ture to th
I of the f'\)int for dO::h.:nTImtng tb d", m:d e
on \\h h \\111 conl,nn teml lmolvm ~ (lolv rca..:UQns
P
c. (f e pre
n for Ihe shear or bcndlOg moml:nl l:on
rm n\; htng ooh reactIons then It IS ~1l.ren
Impkt
l.: n truC1. the mfluence hne f('lf shear or ~ n d m
moment b'l
comb ng the gm Db of the C.:.II.1.IOO IDflurnc lin 10 tlC r
et \\ h these e pion Oth\"f\\
ubslltul 1 cxp
10 r. f the aC' on mto the c presslOns f r the heal
be dl m m nt
nd plut the (I; uluog e p
100 \\hKb
be 10 terms ooh 01
(0 ohw.m (hI: mRuenl hnr.:
d. R pc t tel' until all the deslced mtlucnl.... Itn for shc.a
bend n m cot.. h.. heen detennmcJ
[_-7.---=-==-"
l,~
L·
'f
"
JIUIlCLI
(
r(
(
l
b.
(
r
... L3
_ _ L' ......
CIW'IEIII ""'...... Una
{
,
rh~
nRuen
hnc
CIr \
I II
L
f B od dd mlln' lh~
r
II
~
,
(
hcm n
III
I
ft
.............., , .
:!t1 rt
12 ft
•
I~
J'
Ie
U br I \\ pl.lt": th..: umt load at a po Ilion
bent.hng mt ment ,II H b:- u mg the lree bod
II n II th beam to tht': nght 01 8
o
(
II
\:::. I:!
ft
lx~
the uml I ad I It Il,.'d 10 the nghll'f B. and \\1: use the free body
pom n of th beam to the left of B to dclcmllne \f B:
I:!~
"
12ft-sxs:!Oft
Thu the equ.lllon of the nflucm:c hnl' Illf \fa arc
~
II,
L:
1f2I; :co\~I1
h
,
I~
12fts\:s;20ft
The ml'luenl"C hne lor \/" .....ho""" in !-ig. IS 3(1'
FIG. 8.4
Ora\\ the mttuen.:c lines f(n the \crlical reaction and the reaction
",upport A and the ..heM and bending' moment at point B of the cant
sho\\n in r if!. X4'aJ
The IDftllellOO tine
ht}Iunt Line
Solulion
Infillllll l' l.int
lor A,
M
12:.,1'
0
The inft
A -1_0
A
Examp118.3
The mllue-oLe hne for'" 15 bo" n In fig S,4(c
Injfulnlt'Lm /or \1 f
>L·II,O
\/~
If\-O
1/ ,
I ,
x
The mfluence hne for \I "hKh I obtained b) plouing thl equa
m FIg: 4 d A 11 (he ordmate of lhe Innuence line are nepb
(hat the n of \1 ~ for all th nohI~
of the unn load on the be... II j
unterd k i n , ad of dockwi a IOnia 11) a umcd
den mg th equatl n ( f th mOuen!,; hne
line
M.
354
CHAPltll. , _ UIlOI
IECTIaIII.2 -air
Hon
D
,+-----,,'/\
E
F
4 --
m
S m ----t- 5 m
,.
G
.
I.............. " •
-.............
EG
-
8
--+-- " m
c ,-----:""""o:"""'=--+__-!:-....:..!
D
05
IkN
c
I
D
A
_I,
I
{dJ Influence Line for B (kN/kN
,
E
G
F
8I1_8
I
A
OHJ3
B
b>
(kNJkN)
(ellnfluence Line for A. and B~
05
c
o
o
D
F
-0.5
(el Influence Line for A
m
(0 Influence Line for SE(kNIkN)
(lNk~J
m
AG.8.7
Ans
Influent I' I II/(' for B,
Lf
()
H
I)
A
,) 10, - 05
B
The inftut=nl.:e line lor B
10
hm.. n in hg. l:(7(dl
I
L
Inffutn ( I III Iflr ~ \\ e '" III U'C the cqualion of condition
M
hctennme the e:..prc Ion 1M ~
t~rh
\\c platt the UOIt load I the
d
mge 1.
• 3
.
that
I
,
on Ih fl!,!ld part ('1
"L
\1'
I
I III
•
01
the frame
to obtam
/I
",
J•
6
10
<
"
:(I 5 I~)
III m
8.2 MULLER-BRESLAU'S PRINCIPLE AND QUAUTATIVE INFWENCE UNES
IEt11oIII.Z
CHAPTER 8 Influence U....
F
Released structure for MD
25
I0
0"----,1B
A
1.0
i~o
C
£'"""'""""J
F
o
c
o
A
-05
Influence line for SB(kIk)
A I
"
ICli'---_----::i4:i:---F
r----'B
o when
C
t
£ =0
1 k is at just to [he len of B
5,=
I k when I k is at just to lhe right of B
Influence line for Mn(k·ftIk)
Ik
~_-L'_-r;.E_
A 1_ _
c
D
MD =2.5k-ft
f
£y=OSk
(I)
(el
AG.8.10 (eonld.)
Solution
It,tiUI ntl Llllt jor..f To detcrmine the general shape of the Inft
for A . "e remo\c lhc n:slraml (:orre!iponding to A, by replacmg lhe ftIIld
pon at A b) a rolll;r guide that pre\ents the horizontal displacement
tatlOn at A but not the \crtil:al displacement. Next pomt A of lhe
structure IS gi\en a mall displaccmenl 6. and a deflected shape of lbe
dra\\n as ho\\n m fig 810 b Note that the deflected shape IS co
the upporl and contlnully I;onditlon of the released structure. The end
beam \\hlch I altal;hed to the roller gUide cannot rotate so Ihe poI1lil1lU"
must remam honLonlal in the displaced configurdtion. Also pOInt E i........
to the rolkr upport; therefore It cannot displace in the \'erucal diRlCllOlL
th portion ( f rotates about E as ho\\n m the figure. The 1\\0 rigid
AC and (F of the beam remam straight m the displaced configurabOD
late relatne to calh other at the mternal hmge at C which ;~.:1tlmrep
lion The hape 01 the Inftuence hne IS the same as the deftected shape
leased truclure a ho\\ n m ~ Ig 8 lOb
B) recogOlzlng that A
1 k. .... hen a I-k load IS placed al A we
.. alue 01 I k k for the mftuence-hne ordmate at A The ordlna
and f are then detenmned from the geometT) of the Inftuence bne TIle il~I;
Ime (or A thu obtained IS ho\\n m fig 8 10 b
m
_&-.
-
CHAPTER' InOuene. lin..
Ex..... 8 8
~
hn
.........
I III
"
I
A
fllnge
II.
J)
f
B
4m
4m
10m
.4
"
'-1-';r~
4m
'
. -~L:iJ)<5;'
F
JlOO"
Rd~·J
t"d Beam tor A
I0
~-:'o
,
-; :- ~O
_-::--
BCD
8.3 INFLUENCE LINES FOR GIRDERS WITH FlOOR SYSTEMS
£
(h)
o
J)
o
(
Rdl'a ed Beam fur C
A
-04
A&. 1.11
("
In the prcYIOUS section w con den:d t
were ubjected to a moving wnt I
Innuell(t" Lillt" lor A, (kNIkNl
Inl1u nee lme for ( IIt ILN)
F
In mo.1 bndges and bwldmgs there
are nol ubjected to lave loads dlRCll bu t
tnmsnulled vIa ftoor franung tern T
bndges and bwldings were de!aibcd
re'p«:tivc' Another cumplc f the fnlnon.
hown tn FIg 8 12 The deck f the brid
" which are suppofled by
by the ,. r Thus an li
gardlcss f where they
Ioca
"""""nlnlled or distnbuled I
as ootICCtIlnlled I
applied I
tIoor beams
T
and
-
CHAPTER 8
Inftuence Linn
IfI:TllIII u IIIIIaonco Uooo .. _ _ ... _ _
F
rbt m<;
G der
"
~
r::L
W-
,---- -1---
L
/\--"::;-
;
b
/
Aoorbeama
(e) lnnuencc Line for F
I
I
/
Q
I
r
---~
"--
Slnnge
"
I
I
Q
r-----
.J
I
I
I
L
.J
- ----
-....... Girder
b
-
x~',!_
lot .F
~
o
Br=r
(2US) - •
F8 =--
F
C
riD
E :..
Deck
d
floor ma~
-
Gmkr
Strmger
.....13
In
m
II-.
IIECl1OIlI.3 1nII_ Uooo lor
CHAPTER 8 Influence Lines
I.' S'111'
- • , In,tt:.ld. the "UinpCf!l. and the
\'llh..:r , Jt'pldl' III
J
Ih II th..:.r lOP l...dgc, .Ire e\en With
1
IIIIllll'
\\'1
•
I
h
II
N .lin.. ,Ill U II I I\ Pi 1'\\1:1"11.1111
I" '
It tht.: .... lIne 1c\C ;,IS t e top " I ' ~
.
olhd .llld In: ('It Kf I ,
d
the: glrd..:rs see I l~
Thu .h quail
an
.h 10O
F
8 1_
I
4
L
F
Influence Lines lor Reactions
L
A
'1",,-"- for Ihe ,.crtlcal rcaclton A
I Ill' 1Il11UI,:IlLl
The C"-IUatlon ('I
1,. •.•
h
rr l)lI1g tht' eqUilibrium equations
f lao hi:: ddcmum."U V.I
L
L
l' •
• 1/
,~
.L
\f ~
0
F L
1\
0
\l
I I
L
I
0
A
0
F
,
L
L
equations
The mfluence 1lIle" (I bt <II'ned b\ rlQuing these" d are
. sh
I
ole Ih.lt the"" IIlfluence lme~
are I entlca to
,
d (,,'
Flg:.. I.t h .10
d beam to '" hleh the urut
#
for the n:::u.:1I1111 01.1 "'l1p~
applil'd din..' dl)
..,upporte
Influence Un. far Bending Moment at G
Influence Line for Shear in Panel Be
"uppose that "c "i.. h to construct the influence lines for shean
(J and JI which.ire located III the panel BC as shown m
R. n(a). When the Unit load j, located 10 the left of the ~nel
po
the ,hear at an) point wllhin the panel BC (c.g .. the pomts G and
can be dcs~ rpxe
as
'C\!.
pomls
SHe
F,
O$x~
I
Similar!). 1,1, hen the unit load IS located to the right of the panel
the ..hcar at any poilll \\ithin the panel BC is gi\cn by
M
I.
")
I.
4\
-
1+-
L
F L-a
, L a
L
o
L
2L
5
L
When the umt load I loca.ed wIthin the pane
8 13 d the mOmeDt of the fan:< F. exerted n
beam at B about G must be IIICluded 10 the "I_we
mOmeDt at G
-
A
M
When the UOlt load IS located to the right of the pan I POlOt C
bendmg moment at G 15 gIVen by
L
5
When the Unit load IS located 1,1, nhin the panel Be. as shown
8 13 d th fOfl.:e f B exertcd on the girder by the floor beam at
he mcludcd In the e'pre \Ion for shcaf In panel Be
The mfluence bne for the bendmg m ment t pol t G
in the panel BC FIg, 8 IJ a can he nstructcd by
procedure When .he unit load I located t .he len f t
the bendmg moment at G can be expressed
B
............
\ L
L
2L
5
4
(I
0
0
«I
0
DO
When the un,t load I located to the letl of C
moment at C I gIVen by
rll-
3
M
Sx
When the unit load IS located to the right of C
MAcn
nUl
(I
~2)
L(~
the equanon of the Influence hne for M.
M
l~C)
A
c~)
~
~x
L
x
The iD8l1C11CC line obtained by plollUlg these
1
that this iD8ucncc hne IS ideQIk:II
IIIDIIlCIlt
c:orl<ljlOillljDg beam WIthout
........ Iar . . .
_
. . i. . . .1.lul .....
Plan (deck not shown)
.....17
connected at their ends to the JOints on the bottom chorda
longlludmallrusses Thus any li\e loads e.g. the wetght 01
regardless of »here they are located on the deek and wbe!lla
concentrated or dlstnbuted loads are always tranSlD.ltted to
concentrated loads apphed at the Jomts. Live loads are tl1lll_111
rooftrus In a Imllar manner. As in the case of the girder
the tnogen of the floor systems of trusses are assumed
supported at the" ends on Ihe adjacent ftoor beam . ThUi
hoes for trusses also contain straight·line segments between
To Illustrate th construction of Influence lines for uu.....
the Prall bridge truss shown in Fig. 8.18(a). A uml (I.k
from left 10 right on the stringers of a ftoor system attached
tom chord AG of the truss. The effect of the unit load ..
the truss at Joints (or panel points) A through G where tU
are connected to the truss Suppose that we wish to dill
lines for the venical reaclIons at supports A and E and
force In members CI CD DI IJ and FL of the truss
IntIU111C8 UIIII for R••ctlons
The equallon of the Inftuenee hnes for the verti<:aI ....._
can be delermined by applYlJlg the equilibrium equa
EM
0
A60160xO
Ix
E60
0
- ..
,
Influence Unl for Force In Dllgonl' Mlmber DI
1br e pressJons for Fn1 can be determined by ~
Fill 18 h and by apply,"g the equlhbnum equaliOD
of the two ponlOn of the truss When the wut load II
ofJOint C pphcalJon of the equdlbnum equauon I: F.
porn n DG of the cruss yields
LF
0
4
SFm
E
0
FD/
12S£
When the I-k load IS located to lhe right ofJOIOt D .....
LF.
0
A
4
SFD/-O
FD/
I.2SA
The sqments of the mOuence hne for
FOI
bctwcea A
tween D and G thus ronstructed from the mllueace IiDoI
n:speclI...t are shown 10 FIg 8.1 g I . The ordiDa...
then c:onnccted by a straIght hne to romple.., the in1I_
as shown 10 the figure
Innuenee Une for Force In Top Chord MIIlIbIr IJ
By c:onlldenng section bb FIg 8.18(h)) and by pllCllll
tint to the left and then to the nght of jomt D we oblItia
exPreallons for FIJ
+<LMo
FIJ 20
E 15
0
0
075£
FII
A 45
<LMo
0
F, 20
0
Fu
225..1
The iD1I_ line for F thus obtained I
310
CttAPTER. Influence Lin..
L
1 - - - - - 4 paMl
I
m ~ 3~
IIlCT10II u -... ~
.\
II
II
m -----j
(e) S~clion
bb
F
0
F
o
C
•
D
LM
o l.:U
(f)
66£
ted to II1e ngln
F
A
0
£
a
Ior_
Influence Line for FUti (kNIkN)
12
A
4
F
0
-2A
24m
m
1 kN
I-- ,----.j
(b)
A 7-"!~
1 (I
B
F. 12
C
A,
~o
FHL.
(g)
A
D
c Influence Lme for A IkNIkN,
A 16
_J
~ O!5
A C E
d Influence Un< I
E kNlkN)
R&.1I.2O
1661£
"
o
~7
A
C
0.208
......... r
D
-0417
F HL
= -I 667 £) - I 2S
• -0 833 E
o
£ 16
12 F
When the l-kN load I to the ngbt of D we blaIR
Seclion aa
f.
10
J
SFH 8
Fl'tI
Ih, Influence Lme for F HL (kN1kN
o
.
-
CHAPTER 8 Inftuence Unes
B
SUMMARY
1«1-----£1 ... constant
J:, .. 2 ~ . O h i
1
500 in. 4
(3)
Ik
Aj-~iB
X
I
x~
(b) Real Beam
B
A
[
)5
£J
M Diagram ( D
k-ftlk)
(c) Ei
15
PROBLEMS
£1
(d) Conjugate Beam
SoclI_ 8.1 _lid 8.2
'.1 llnugh ... Dra\lo the influence line for vertlCal reactIOn .. at support!l A and C and the shear and bending momem at POint B of the beams shown 10 FI P8 I throusb
o
Pl<4
fIG. 8.22
fIG. PI2,
(e) Influence LIRe for Vertical
DeflectIOn at B (M)
;m
FIG. Pl.l, Pl.59
B
AI'F"=~D
... 1'&3
PUI
CHAPTER 8
Influence Unn
1,:=R~;i-P
n
r "
f
nd bendm
'.15 Dr,l\'· the ..,t1u\"n\:c hne for the
TTlll Dll"l1t .11 pomt /) 01 the he.tm ho.... n In hi P8 I
r \l,:IUlal n:.J~l on')
nd bcndm morncnl
rs "and I' b
I~
011
n
R
FIG
c
n
f
'm
Xm
m
P8.6
1.7 Ora" lhe mtlut.:n\:\· hnc\ f()T thc \crtll.:al rc:.u.:tion\ at
uPI' rt -I and ( thc hcar at JU\I to the right (If ., and
the bendm ' rnomc:nt at POint B of the Ix.tm \ho" n 10 hg:
c
f
AI. P8.1.
.A
'm
-:\m-4
•
FIG.
3m
Sm
Sm
'.12 Draw lhe mflucnce line.. for the ..hear and bcndIDr'
moment at pOInt B and the ..hear.. ju.. t to the left and
PS.16
'.17 Draw the mfluence line!> for the "ertleal reactions al
'lupports A and B and the shear and bendIng moment al
point D of the frame sho"n in Fig. P8 17
Ihe right of .,uppart C of the beam sho"n in Fig PBI
~
B
.
C
LP
4ft
6ft
"C===";D;,,,==_-iEi-
f r the hc:ar and bendmg 010otlle er beam ho\\o In hg pg
. . . Draw the mtlue
upports Ii 0 nd F
B
IIiDae
\
B A-.~
'.13 Drd" the mfluem:e line.. for the vertical ,..rillll!l
upport -I and E and the reaction moment at u
the beam htl.... n
lD
6m
·A
FII. ..... PUl.
fig, P8.I3.
11 ft
R
~m
fIG. 1'8.19
1M ft
FIG. 1'8.12
Hinge
5m
F- - r
IS ft-+--15 ft
12 ft
20ft
FIG. 1'8.7
Ih I ftuenlX lin
3m
'.11 Dr,I\\ the mfluem:c line.. for the !l.hear and
mom,nt at POIllI £ of the beam ..hown in Fig. P8 10
B
:b..(
------t-I 0 ft ~
Hi.
PS.l0, PS.ll
A:c:
1--- ~1
E
E
D
401
6m
P
A.Li"
/)
C
H
'm
AG.
FIG. PI.8
B
1.10 L)r.lll Ihl' mftucnct' hnc.. for Ihe hear and
momt.:nl at pomt ( and the ..he.IL jU')1 to the left and
Ihc n1!ht Ill' \UpJ"'.lrt D of the beam \ho" n in hg P8 I
FIG. P6.5
m
f
,.16 Dr.I\\ Ih, influcnl,,'e hnc fur the n I r U
at
lIrfltlrt I .lOd F .md the: hear and hendm mom nl at
roml 1> of the Irame hO"11 10 118 P8 16
PS.9, PS.58
H
R
Oft
It
4m
U Dr
h
R
hnl.:
FIG.
In
-
ilnd be dm
P8 1\
,1. Llr.I\' thc mllu..:n,.;c hnc for th
. ,,·ut ,11 r oml Hoi the hcam ho\l."
f\11 111
FIG 1'8.4, 1'8.61
'm
he
I.' 1)1,1\\ 1h... 1I111ul·IKe [1I1C\ Ill!" thl" \Crll\:.a1 reaction
l 1l,:(lvll l1l\)lllt.:1l1 ,II UppMI I ,nul Ihe ')hcar and
lllonleni .11 1lI'~,
Hill tl\l.' \:.\Ollk,er hcam ..hown ID
D
4m
FIG. PS.I3, 1'8.14, 1'8.15
4m
12 ft
FIG P8.17
1.11 Ora" the mfluence hnes for the
[C"dlun moment al upport A and
n},lment al point C of the frame
6111
P&22
6
au
_
CHAPTER 8 Influence Unea
lh~
InftUt."T1 hnt.' or
l:. nJ (, • f lhe: ~nl.:>t
\ertKal re
ho\\ n In
Hinge
,
'Oft
.,
.J
--IOift
\
JIL
B
6m
hG
F16 PB.35
Hmge
E
D
C
F
I
--+115ftl5fl
-II---i
AG. PB.31, PB.32
I.ZI Ura" the ml1uenl,;t: lines tor the ~hear
..ho"n in
mom t t pomt !> (It the ma~
and bending
~IF
moment at pOint! of the beam ..hown in Fig. P&,:!4
LZ7 Draw the mtluenle lines for the reaction moment at
upport of and the \crtll,;al rCdl:tlons at ~tropU>l
A D. and F
1t Ih bcdm htl"n m hg, Pl\17
Hinge
1=,;CiD~".E*Fr:lG
PI.270 P8.28
81
....
A.
20 It
IOft--l-IOft
FIG.
8.32 Ora" the IOfluence hne... for the shears and
moment'! at poinh Band E of the beam shown
PR.)I
8.33 Dravo' the influence line'! for the vertical reacti
,upporh A B. G. and /I of the beam sho\\n m FtB PI
A
8
JJO
3m 4ml.mjJ
II&.
L
10 ft
I
. . . Ora
10ft
upports A B (
poInl E (the
D
E
F
G
Hop
PB.36
P8.1'
La Ora" the ml1uence hne.. for the ..hear and bendmg
A
10 (t
-0)
1.31 DI m the mfluem:e line· for the reacllon mo....
.Illd Ihe u'rtKal r...action' at '!upporh -I and F of the
h,,\\n III } l~ PS JI
F
.
1.30 Dr;!\\ th... IOtluence hne, for the ,hea~
and
mC'mt.'ntc; .tt JX'JnIS ( and F of the t'leam s.hown
p,
E
10 (t
AG PB.29, PB.30
(j
lHI-
•
,
/)
10 (t
'.37 Draw the influence hne~
for the reaction momenl at
support A. the vertical reactions at support A and F and
the ,hear and bending moment at point E of the frame
nwoh~
In Fig. P8.17.
T
8~= ;iC~=. ; D
..........;E;'-'_"!iF::..
·A
\--4m
8
4m-f-6m
6m
II&. P8.3lI
Hinge
IOJ
fIG. P8.33, P8.M
A
10M Dray, the mfluence hne for the shean aud
moment 011 POints ( and E of the beam sit
PO)
. . . . . . . . Draw the mfluence hnes for
and \-erllcal reat.1ltlns Oil uppons It and B
~o",n
In hp Pll 3S and P8 36
5m--1-- 5m
5m---I--5m
FIG. P8.37
1.31 Dra\\- the mfluencc hnes
Co
r the
upport.-4 the .. ertical reactions t upports A
j1;:=XC:::::1:f =i=l
.......
..
CHAPTER 8 Innuence Lines
J)
~,;: ~
n
1(
n
I I
hIm
Ilh gtrlkr \\Hh th~
r
n I(
fl{'l,
I
G
f
n ,Jnd
(elll
8
\
5m
F
F
D
8
J
10 fl
AG.
R6 '8.41. '8.42
l,:
'I' Ihe
,m
,m
ll'ift-90ft---- -
~G
~
T
In
;.1'p==F
I r the ra~h
In pand B(
nd
f he rdt:r "lth tlle 11.)0)1 ) tern
~.
o
~~ 1 2 f t
m9= ~la, enap~
4pane1 811511 =60h
AG.
PS.46
"'- P8A3
.... Ora" Ih~' mflucnU' Ime Itlr lhe hear in panel cn and
the bendmg m ment 1 /) 01 ~ht gIrder "llh the floor ,,) ,1em
\\llmh'P844
A
i,
.....
(
8
i
n
1
i
(,
t
i
'; panel III h m -
~±
G"
H
8
C
2.
f- ~
A
"R
C
- - - - 3 panel!lo at 4 m - 12 m
AG. PS.47
'0 m ----_--1
L
AG.
J
P8.54
4fr
H
"""
8ft
L
PS.51
~A~
"'- P8A4
T~
L
M
N
0
P
8
C
0
E
F
E
F"
G
IAI"",,- a.u 1>,
cd
l
Ihe I ftuencc Ian Ii r Ihe for
In
n
f the tru ses ho\\n m
I Om! uaJ t the boll m
! - - - - ~ panels al 16 fI = 64 ft - - - - " ' \
A
I
3 panel at 4m 1 m - - - - - I
AG. P8.55
G
- - 6 panel!. at 30 ft = t 80 ft ------I
"'- PS.48
1.53 . . . . . 1.57 Dm\\- the m8uence Ii
Ii r the
In the member... Identified by an
of the
01. figs. P '\"\ PR, <;7 LI\e loads a
transDll ted I
chord, of the IruSoses
4m
"'-1'151
-D
"'" PS.52
_1.4
-I
_
D
panel, at 20 f1 = 80 (I - - - t
K
24 m
D
1 - - - - - 6 panel, at 16ft = 96ft----_
~G.
; - - - - - - 6 pane 114m
10 ft
PS.50
I
Hmgt
- ~- ,
F
10 fl
PS.49
PS.45
panel £F 8nJ
rh :trde' \\ Ith the lloor ,. ll,:m
he,ll
10 ft
4m-+4m-+-4'D-j
402
CHAPTER 8 Influence lines
,
8
Section 8.5
8.58 Dr,1\\ 111l' :\ln~u1 l
hn..: I(lr thl.' \(TIKal deft
pOint BIll Ihl.· 1..1ll1Ik"..:r tx',lm ()J Pr('hll:m
~l nst. I l
\ll. 'Ig px 9.
8.59 .Itd '.60 Dr,1\\ thl.' lllllul:lH:c hnl.' lor Ihe \
fk twn t r0ll11
\)1 ~ht
~lpnI
'UPporll:d beam
I ms
and 8 2 U
Clllht.1ll1. ~c
h
P8 I and
Application
of Influence lines
RespOnse at a Pam
n
el
A6 P8.57
116ft
-ISU
9.1
8.&1 PI'''' thl.' mtlul'nl.'l: hnl' for the \l:rtKdl d
pc nt J) ,If 'hl: !'learn :'ff P(ohkm SA I:J 1,;0
h' P ..
9.2
esn=~
9.3
ralue=L ~:.
9.4
~
Load
eular LocaIJon Due to a SIOgie Mov ng Concentrated
at a Pameular Local on Due to a Unltormly Dstnbuted Lye
Locabon Due to a Sones 01 MovlOg
Absolute Maximum RespOl1se
Summary
PrOblems
Hlq/nml Bridqe SlIhjc'cll'l/to
lIariny Loads
s~ofthe
nos oepartmentof Tr
tl·
In the preceding chapter y,e learned ho\\. to I,; n trul,;t mfluence lin
In thh chapter y.c con Idt'T
the application of influence lines In detcn i ln~
the ma\lmUm \",lu
of response functions at particular loca1n~
In Ir\1l;ture due to \i nable loads. We also discuss the proctdurcs for c\aluatm the a lut
maximum \ialue of a response function that may occur an here m a
structure.
for various response functions of struc re~
9.1 RESPONSE AT A PARTICULAR LOCATION DUE TO A SINGLE MOVING CONCENTRATED LOAD
As discussed ID the precedlDg chapler ea h rdin
a i,n,8",,,.:e
gives the value of the response funcllOD due ( a n c
load of urnl magnitude placed on the uuelure al the I
ardmate Thus. we can state the f; II wing
I.
The value of a ...ponse funcl,on d I -1l'"11.de
load can be obtained by mulnplymg the mOl
the ordinate of the response funcl n 1D8uel1C
the load
..
CHAPTER 9
Application of Influence Lines
S£CTlDN U
1\' lh:llllllllll: thl..' 11M l1llUll1 pO... itlH' \aluc (11,1 rcspon
,
2.
1/
C
D
B
P
r t 'lJmum
p
A
c
B
I
D
n of Lo.1d P fl r MaXimum
a
eMs
FIG. 9.1
'S1Il~ t.
fi
load must be
.11 thl' II,. .llhlll of the 1l1.1\llllUI11 pI'... lll\C ordmatc of the
funClIlll1 11111UCIh.:1' hnl \\ hCIC.1 tll dl·t 'n l1 ~
the max.lmum n
\dlm: of thl rC"1 '11 l lunditln the: IlMd lm~n
be placed at the
tlt'll (lIth\..' 11l.l ,"Will !ll'g.ltl\C ordllldh.'. 01 the lIlflucnce hne
JUl.' h' ,I
111\\\ Ill!,! I,.l,.lnl.:I,.'1l1r.lh:d 10,IIJ the
r:::==-
(('l1Slder fl'r l' ampk .1 1x:.1111 uby.:ctcd 10 a mo\ing concea
h.lad of lll.lgmtudl P .1 ShlH\ n 111 J If! 9.I\a 1. Suppose: that we
Jdammc the bt:ndll1£ mOlllent .It B "hen the load P IS located
dl,Un(1; :\ Inlm the left uPfX)fI I The IIlfluencc hne for \t. &1
Fig i). I .1 h.1 .m llnhn.Hc \ at t~c
PO"I~lon
of the load P ;" ~ :
that a umt 10;10 rl.hXd .11 thc po Ilion 01 S'~uacP
a bendmg
\f
,BI.:C.lU'l the prnh:lplc of ~ure p0'oil n
I~ \alid the I
mac.nitudl: P mu~t
I.:au~
a ocndmg moment at B. "hich IS P
largl: ~a that l.:.lU cd h~ thl: IO<ld of lImt magnitude. Thus the belllll.'
nll)ml:nt at 8 dUI: tl.:) the lo.ld P 10;; .\1 8
Py
I: t UPP{hC that our ohJl:di\ c:: i... to detcrmine the maximum.
ni\1: and the m.l\imum negatl\e bending moments at B due
10.tO P From the mflucl1l.:C:: lll1c for .\1 8 (Fig. 9.lla)) "e observe
ma\imum po ilne and the maximum negati\e influence-line ard
OCl.:ur at rx)inl'o B .1I1d D. re'opt:cti\e1} Therefore. to obtam the
mum POSltl\C bending moment at B, "c place the load Pat POlot B
shown in Fig. 9.1(0) and compute the magnitude of the maximum
iti\c bcnoll1g moment <to;; .\1 8
PYH. whcre YH is the inftuenceordinate at B (hg. l).I(a}). Similarly. to obtain the maximum ne
bending moment at B. we place the load P at point D, as shown m
9.I(c). and compute the magnitude of the maximum negative bendidI.
moment as .\1 s
Pro.
IIooIIonIo "'" .......... LocatIon D"'10 I ~
DIIlIIt_LM!oId
H
4m
-
4
14
n
b Inti
\OkS
~,g;- j :i£ -
-<!4'; D_ 5- ~
Hmg
FIG. 9.2
H
'c
Eumple9.1
For the beam ~ho\ .n
In hg. 9.2(a determine the maltlmUm upward ...1di"j!J;i
s.uppon C due 10 a 5()-l. conf.:cntrated Il\e load.
Solution
Influt'nCl! I InC 1 he InflUcnLC hne for the \ertical reaction at u
thiS beam 'las prevlOU Iy con\truf.:ted in Example Its and IS showD
9:! b Rt\:all that ( \\.;1 a umed to Ix fklSllI\C in the upward di
\:On trUl:tton of lhl mflucm:c hnc
R tJl tllm at ( To Oblain the maximum poll
conl.:entratcd Il\c I()ad \l,e place the load at B
Yther the rnalumum po IIl\e ordmate 1.4 k kS of the mflucocc
B) muJllpl IR~
the mdgmtude 01 the load h) the \alue of thl
\fa lnlwn (pit rJ
( due
(0
the 50·"
t nmne the maximum up'l<lfd rcaL110n al ( as
\0
)4
70 k
70 k
9.2 RESPONSE AT A PARTICUlAR LOCATION DUE TO A UNIFORMLY DISTIIIIUTED LIVE LOAD
SECflON '.2 ..... DItIi It • hi
CHAPTER 9 Application of Influence lines
\\hlIT I I the mllu":l1li:' 11l1l.: MJIIl.Itl· ,It \ \\hit:h i.. the POlOt
... u n I r.h SI1l1\\ n III Ihl' llgure 10. ckr...-mllne the tOlal
mt."lm..:nt at R Jue t,l thl' JI Inhut...· J 1~.IJ
l ,~ m
\ II to \ h
glJt~
• i.I. 9 1 tx't\\1,;l n 1111.: l 11I11It III l ath1~
~,.mp : le 9 2=-
$
"lie I.ICItIen
Ow" ............, 011*7 " ...... LIIJ
;: ;: - ~ ;= = j
(
\Is
I
\1
\c/\
II
J
Id\
Hn
J
~'n
III \\hllh th..· IIlh:gral
1.1\ repn;-.('nt the ,Ire.' under the 5<I:molltli
thl.: mlluc,:oll'lme \\hl("h Jn'~r; l,c:o
to the loaded portion of the
Thl MC' I "ho\\o J .1 haded MC'.1 lJ1~ the influent.'C line for M
_
("J
An 1111111111 ffik
.tJ.
A
B
1---
5/
A
4m
14
lJ
ell
A ""0
1---<-021
4m
fqu.ltl(ln l).2 .11 (1 mdK.ItC" th.It the bending moment at B
Ix maAlmum po ~,l\itI
If the uniforml) dl·.tributcd load is placed
th(1'1,;' p\',rtwn" c,lf the ~,tnl
\\ here the mfluencc-Iinc ordinates
tilt.' and \ Kc.: \CT.•1. hom fig. 9.J(.1 \\e Gill sec that the ordiD8I1;~'
the mtluenlC' line for \lB are po,it1\c bct\\een the points A and
negatl\e bc.:1\\I..'en ( ,lIld n. fhereforc. to obtain the maximum po~tr(.
bending moment at B. \\e pl.ICl' the uniromJ~
distributed load
A to C a... ,ho\\11 in hg. 9.J(b). and compute the magnitude
mi.l\imum po.... me bt:l1I.hng mom..:nt <1'
.\1 8
II
bl lnflue
(area under the influence line bet\\een A and C)
", G)
CD ''iV,.ngemcnt fUn formly
o tnOOted LI e Load \I for
M
urn Pili M
B
B
91.
1 0'<1-<
(O.7,t)( 181
037,,,
15kNm
1' 8 L
c
B
Similarl) to obtain the:: maximum negati\e bending moment at
place the load from C to D. it ...... ho\\ n in hg. 9.3(c). and com
magnitude (~f the maximum negative bendmg moment as
·\18
\I
area under the influence line bet\\een C and D
1I(~)J:!5L
~aB
on the foregomg discu~on.
In
0.125" .lDL
\\e can state the fall
The \alue of are ponse function due to a umfonnly di tn
apphed o\er a ponlon of 'he structure can be obtalDed
pl}ID@. the load inten It) h} the nel area under the com.
porh n of Ihe re ponse fundlCn mftuence hne
2. T detemllne Ihe maximum po!'OlIl\e or negall
re pon funl,;hon due 10 a uniforml) dlstnbuted Ii
I ad mu , be placed 0 er Ihose ponlOnl'i of the tructure
rdmate (f (he re ponse function mftuence hnc
ne lI\e
I.
H
c
Hinge
FIG. 9.4
o
H
.,
SECTION 9.3 .......... II ........... ~
ClW'TER' AppI_ of Inti..... Un"
• l:e 11Il('.lre £X' llnc. \\hcrcas the 20-k
the (rdmah: (,1 t h ,",,11&:1"
.• d d I I pI l' d (" .. r (he .. nlln: kngth ol the beam
JI,tn butL-u ..3 II Il'\ •
JX»111\C
~nhmcb
ml1m of
\ta lmum
I ( I'
po IIlH" \II
~IH'n
m
The
h~
9(1
~oC)
2
"' :0:
9 2
G)
J
-2)
8
w} kN· m
\l \ mum \t atl
The loading ~lI'"
BlnJlIJ!/ \fun! II' or (
obt".l.In the maximum ncg.ttl\C bcndmg mllment at CIS sho\\n 10 FIg. 9
maximum Dc-gam \1(.\ glH~n
b)
Ma'imum
Degall\.:' \I,
.... II ........ -..00_..1..
__ &.-
- 9U -2
20[
t-
40(~)
G)
-180 k
3
J)
-2 +
2
G) 9
2
m
FIG. 9.6
9.3 RESPONSE AT A PARTICULAR LOCATION DUE TO A SERIES OF MOVING CONCENTU
LOADS
As discussed in Section 2.2. live loads due to vehicular traffic
way and railway bridges are represented by a series of m
centrated loads ",ith specified spacing between the loads (see
and 2.3). Influence lines provide a convenient means of anal
tures subjected to such moving loads. In this section we dlSCUM
IOfluence line for a response function can be used to detemune
\alue of the response function for a given position of a sertII
centrated loads and (2) the maximum value of the response fi
to a senes of moving concentrated loads.
ConSIder for example. the bridge beam shown in Fig 9
that we wish to detennine the shear at point B of the beam
wheel loads of an HS20-44 truck when the front axle of the
tated at a dIStance of 16 ft from the left support A as showa
UTe The Influence hne for the shear at B is also shown 1D the
distances between the three loads as well as the locahon of
are known so the locatIOns of the other two loads can be
hshed Although the mfluence-hne ordmates correspond.ina:
can be obtamed by usmg the properties of the SImilar lrillllll'by the IDfluenee hne It IS usually convement to evaluate
nate by mull1plymg tbe lope of the segment of the inIIl...._ ' .
the load I located by th di tance of the load from the
41'
el2
SECTIONU
CHAPTER 9 Application of Influence lines
{
1
..r
'011---
.ft
3ft
(.j
'~+-
c
B
\ b) Influence tIDe ... ;r,.I_
-0
8k
u,.
(lk
5k
c
S.
101.151.:
51.:
I
C
~-15,
ext the entire senes of loads IS moved to the left b 4 n t pta the
second load of the senes the IO-k load. at the locallon of the maximum
positive ordmate of the mfluence hne as bawn m FIB 9 d The he r
at B for thiS loadmg position IS gIVen by
S.
IOk15k
)~3(
18 Sk
12 tt
R"
8 20
86
(3~)
10 20
)~3(
IS 17
IS 567 k
5k
c
20 (t
hear at POint B due to the senes ot four 4.:oncentrated loads
the figure 1 he mDucoee line It,lT 8~ is !ibo", n in Fig 9 (b
that the load n moves from nght 10 lefl on the beam we
from these figure thai as the series mme from the end C
t "aed pomt B. the he-dT at B Increases continuously as the
The
n of load I then mo ed funher I
third load of the senes, the 15·k load us\
The shear a\ B I now gIVen by
k
)~(
12
.16
CHAPTER 9
Application of Influence lines
lECT10IIu _
•
Iladmg f't
'I
,.
I
ng fklMllOn 1 I t!f ~
*4t::::===dr_·__.. . :. ;.
~
00
J::!
(1_)
'0
1~
:!80 k T
SI.
J2k
16k
ih::::iD
E
2
1-,01,+-15 ft
1
D
---I
oft
"7
1::!6t1
')
B
C
- - • p.tn< =tf~ a
Mo:d
d
Ig: t} s
16111(:)
I ,.T I
•
Itll)n
8l
A~t:='"-_
cb
J
I
2011
8k
Ali 9.8 contd
H
~2Oft-I1
- - - - flOft - - - - - - I
r
OJ
•
A
- - 4 - - - - 40
15ftl--l_
I...o.ain Pus
4
For loading position 4 Fig 9
T
B
'I-T~J
,
1
,
16l
~(
12k
I
I
I
• In
: 32k
I
B
I
n - l - - 20 [1----+-15
(:1
Loading Pu
,
--~
f1--1
I
I
8l
I
10ft
15~ ,,-l--I
I
20 I. --1---25
II-----l
d I loading PositIOn 2
Thus far we have COIISIdered the maJUJJlUDl l<Spon..
a parI
10 tJOn 10 a truetun:: 10
sect
,"'...."
delCmUue the
I max ... a1ue. ~ 01 :, f~
"
IDWIII1IIIl
32k :
,
1-10
9.4 ABSOLUTE MAXIMUM RESPONSE
herein
I
2l
Ans
44H T
oa:ur
al anbeams
IocaUODCXJDSidered
through uf
upponcd
m athis•
, I
,
"16k
E
A
Ilion J
------T---
~t:=W4i
i ill
cbb;,J)
~20fl_+ur'i
Maxunwn F
I
I
I
be used to develop ~.: ~
1Jl'
SIngII CaiWiib.1Id lIId
CoaIida:-~,-:maeb' ~:
'a:~
r.:=
.18
CHAPTER 9 Applicltion of Influence Lines
,
.;J"'=~
IEChOIU
Similarly the lllaXunum negab e
~.,*=
-0--<
I
..
~
.
rna unum negah
7
Pc
bear
4
L
a
(~A
I
L-- a
~
A~C
-r
L-Q-
J
(e)
Influence I me tor Shea• ..It
SeLl..lon a a
,
En\dopc of Ma.lI:imum B ndUlI Mc....:\:
- Single Concentrated load
0(1 .L)
t---a
L
()
rr(L-
ar
()
L-a-j
(I.:' Influence Lme for Bl'ndmg \toment
at SectIOn (I a
(l
J Envelope of Ma\imum Shears
Di"tributed Lood
.k_-=:::::::::::::"',':-L"
()
,
p"
L
L
-f
fd I Emelope of Ma\imum Sheaf'; -- Smgle
Concentrated LUdd
(g)
Envelope of Maximum BendlOg
- Unifonnly Dlstnbuted Load
maximum bendmg moment
pa( 1
Uniformly DlltrlbutId IOId
exl let us de\enDi.. the aboolu.. IDlIXImUID
menl ID the 5IDlpi supported beam of Fig 9
and c, ft:' .~1e\I:'CJ
RCl.:all that the\e mfluence lines were 1m
,e1op<d In S<xllon 8 I fig. X.2Ie) and (fll.
Suppose that \lot:' \lol",h to ut:'termme the absolute maximum
the beam due to a ~mgle
mo\ing concentraled load of magDltude
discussed in Sel.:l1on 9. I. the maximum posithe shear at the "':OOD'
IS gl\en by the product of the load magnitude P and the
poSltl\e ordmate J
u L of the influence line for hear
a a fig 99 b . Tbus
di5lnbuted
live loadorof
m\ellSl shear
•
maxunum pclOIbve
negab
~1t \C
shear
p( I
~)
5
The envelope of maxImum bending moments constructed by pIonlng
Eq (9.5 i••hOWD ID FIg 99(e It can be .... that the aboolu.. maxI
mum bendlDg momenl ocam at mIdspan of the beam aDd has magmlude PL 4
FIG. 9.9
maximum
~)
I
the~bam.:
the
by pbu:iDc the load over the porb • f
the shear iDftuoaoe line Fi 9 b
,..
multiplying the load 1II\CII5I
Ioodod porbOD of the beam
the
maxunum DOOiitive
b
~:E
SfC1IOIl U
CHAPTER 9 Application of lnftuence lines
- ",---+--". "--"1
Th~
\:O\dopc 01 m \Imum hl.:ar due w a unift ' mlly dl!~tnbu ed
Jwd t.:on tru..ted '" p](1wng Eqs. I} 6 and 9. l. ho\\n In FI n.-'-''"
II t.:an Ix
n that Ih .tbsolulC rna lffium rl.~h
dc\dops al secl " 'IJ .~
In Idc tht: SUppOI
dnd ha magmtude" L -:
To ddcmllnt: the I.: pre Ion for the ma\lmum bending m
xXLL n a'a \\(' mu!llpl) thl.: 10.11.1 mtcnslt))I b~ the area oftbe
mg moment mthll.:n"c hnl.: Fig I} <} t.: t() obtain
rna Imurn hendmg O1OO1l.:nt
" u
-,-
I
L
,.-l"---!
a
AG 9.10
The ffi\d,,\pt' of maXlmum bendmg moments due to a umforml
tribult:d Inc load constructed b) ploulng Eq. .~9
I
hown
9.9 g It \:an be seen Irom thl cmdopt: that the- absolute ~ : ~
hendmg moml.:·nt OCt.:uTS pi mid pan of the beam and has m
II L'
Series of Concentrated Loads
The ab~oluh"
~uh.tcl,; d
m.l\lmum \.lIuc of .1 rc pon ..e lundilln m any 1b lC "'~
to a serics of IlltlVlIlg cOllccnlr.ltcd IOtuh ur any
loadm!! condltioll l.:.tn he ddcrmlilcd from the em-elope of ~!
of the re'pon~
lunuioll. SUl'h an cmc!opc l.:an be: co
by e\alualmg the 1ll.1\lmUm ".t1uc~
of the: csnor~:e
fundlon at n
of poinh along thc ht~ncI
01 the :erul"Ut~
by gm~u
the pr
de~crih
III SCI,:tion.. tJ.1 thTllugh 9J ,md b) plotting the m
\alues BCl:iJu..c of th ct'lhlde:r,thlc amount t'll:tlmputatlOnal
\ohed. Xl:tpt for :1IO~
o"nnpk trul:lurcs th~
anal} IS of absol
irnum rl: ponse IS u u,tll) flI:rlormcd u'lIlg l'omputer In th f ~ =
Sf't:uon: \\-f dN'U Ihl.: dlrl.: t method Ih.1I an.. (t1mmonly emn
d~{(:n me
thl; "bs(llute m.tXlmum rt,ch~
and hcndmg momeD
pI) upported hc.tm uhJeded [0 .l sc TIes 01 111m mg l:oncent
As In th 1.:a5e (If In 'k t.:onl,; ntr.lled md umforml dilllriblil
I.,ad the absolute ll1a mlUm he'H In ,I ylpmI~
upported beam
senro: 01 mmmg' nl,;C'nlraled load al\\-.l tll.:lUrs at section
Ihe uppon I rum the mfluent' line lor. hear at an arbl
au 01 d Impl upporled heaIn htl\\- n In • Ig 9 9 b w
n
order to de lop the maximum p 1 IU I: he u at th
u
place a man load of lhe TI d po thle on the portl n
fur y"hlch (he Inllucnw Ime I po Itl\e and a fe\\ load
po~1I i
th ponlon \\h n: lh mflul:nc hne I n g3tl\C MOTeO
U U I
hJlil-d ttl\\ard the Iclt uppon 01 the L .•,
the
max.lmum
h
"",om
po 11 e
ar \\ll1l ntmuouslv mcrea
beca
and the rna Irnum ordlnale I the po lt\~
roruon of the inllIul_
In ria
Voh rc
Ih
I thl: neg lI .. e ponlon dec
ut rna Imum po tl
h ar Vol II tX:4.: ur \\ hen the ",,:licle
I ted
oc
JU It Ih n hi f Ih I ft uppon 4 l Ing a
\aluc~
I
IIocl
_
11...._
Lit
II can be sh
.L_
L_
n Utilt tall; absolute rna Imum N"020
section located J II .L_ I f
._~
t1cd L_
0 U~
e I of Ihe nghl uppon C
po
SlOce the location f the ab!.ol u ~ ; , : m
knov..n UIl;am
the...--a
y·~ure
f r compuung maximum
d
ue to a senes f concentrated load de loped ID
emplo ed 10 delemulI< the magn tude of the a luI
Because the Influence hne for hear Just, Ide the left
caito the IOftuence hoe for reactJon at the left uppon t
convementl used ~ r detemumng the magmtude f the a
mum hear
To detcmune the locauon of the absolut ma m
momenl conSIder tbe slDlply upponed beam ubjected I
sencs of mO\ 109 concentrated loads P P and P
h
don ted
P
9 10 The resultanl of lhe load P P and P
hown m tbe fi
I located al the dIStance from the load P
bendmg momenl dIagram of the beam con I Is of tral t I ",gmalls
between lbe load pomts regardless of the postuon of lbe I
absolute maJumum bending moment aceu under De of the 1
sumlng that the absolute maximum bendlOg moment aceu under t
load Pour objectl\ie is to detenmne Its position fr m the nudspan
the beam a shown ID the figure. By applymg lbe eqwlibnum cq I n
~ M.
0 and usmg the resultant p• •lISlead of the md Idualloads I
the eqwhbrium equation \\e detemune lhe ventcal reach nAt be
~+
AL
P.(~
C
If.
0
)
0
A
Thus the bending momenl under tbe load P
M
A
(~
)
I
bemWmum,
I
P
P
CHAPTER 9
-
Applicahon of Influence lines
Ihe ma'tmum
hilI.. al Ih I.. ptl.. llh1ll III lhl: Ill,ld To Ode:rmllle:
.
.1
ft
n:
Iloll,e
Illlll..'lltlll
uue
to.1
lOgle movi
or 1\ (\I1L~
\.1 1UI: \1 •
h I
n10 t tx' pl.ll:l..'d .11 I e _
rx:.ltlon
of thc
l..I:nU.llr.:u• Il'1\ I lh'I.: Illd
I.
.
, ,rdUllle of lhe: :e~lJ .p':n
lum:t1(lIl lOflucnce
JX'SIII\I.:' or negltl\1.: t
'.
,.
I
luncllOIl due to a ullIi(lnn. ) dlstnbu
rh I.: \.1 Iu... 1I a , t.: ,n.1l~ .•
In 01 Ihl' ,lrudun: elll be l,ht.lIned by multi
4lpp II l I ,n cr 8 ['l.' r" ,
IhC' 1000Id Inkn Il~ b> {hI.: Ill..'t .m:;.! under the .corre...pondmg POrtion
n:pnl:un..
I "'n
1l mlhllncc hne . To ,",."Iennllle the ma\lmum posj
ncgatl\(· 'I IUl.:. 01 a re rt1nX lunctllln due to a u.mfonnly di
11\ c I,'ao the load mu t he pl.ICl:O r:.~o
Iho-.e. portion of the
\\ h..' n; thr.: ordmale III thl.:' n:'JXln c tunctlon mfluen",-e hne are P C l . ~
I.)r ne:1!.tll\ e
m.tXlmum ,alue (If a re,pon..c function at a panicular I-·de"",
in a ..tructuTt" due to a xne... of TTIO\ lIlg concentrated loads can
tt'mlined h\ uc..:e 1\ d) pbl"mg ca..:h load of the "Cne.. on the
at the loca"tilln of Ihe rna Imum ordin.uc of the re:"!ponse fUOCli
flut'nt:e line b\ l"omputlng the \alue of the re..ponse funcllon for
po ilion of the" "l'rtt:... Ihfl1ugh algebric~I)
sUTTITTI.ing the ~roduetl
lo.ld magnitude... and the respcctl\C mf1uence·lmc ordmates aad
comparing th~
\alue... of the respon!'>e function thu.. obtained to
mine the ma"imum \alue of the respon!'>e function
In Imp!) supported beams (a) the absolute maximum shear
op:"! at ...ecuons JU'lt II1sidc the supports. (bJ the absolute m
bendmg moment dut: 10 a smgle concentrated, or a unifonnl
tributed. Ii\e load occurs at the beam midspan. and (c) the a
maximum bendll1g moment due to a series of moving concen
loads occurs under one of the loads near lht:: resultant of the loads
the midspan of the be..lm i!oo located halfway between the load and
resuhant
The:
6 I <lr the I'>t:.lln III mkhlH~I
10 de renm t
, and neg"dtl\C hcar tAd th m tmu
u\C bendmg ml)rncnt at pomt (d I
""
d of 1)0 k a umlorml d l"bul
~H fIl and a uOllonnly dl trabuled d d I
,.
p>SIU\
1 ror lhe hl'dnl of Problem 8 2J de mil
t. IU\C and ncgdtl\e he nand th ma Imum
r:Ju\e bendmg: mOffit:n 011 POint D d
a
load \)1 )0 k .1 umfonnl) dl tnbuted It J
,"J.l uOllonnl) di tnbuled dead load of I It
tOO
'0''''''01''''«1
il\C
Sn" 11.1 • • 1.2
1.1 I- r the beam f Problem 8 4 detennme the maximum
e bendlDg moment .1t poml B due to a 15-k ..on.
trate<! h i d
be4m f Problem 84 deternul1C the maximum
n I upport,of due to a l-k ft umlonnl
I d
m f Pr blem 4 delCnmne Ihe maximum
I poInl 8 due 10 a l·k it umf< rml dl
d
SA for lht: beam of Problem HS determme the
POSltl\C and ncg-all\e hear and the maximum poll
negatl\C bending moments al point C due to a co.......~
h\c load of IOU l. a umforml) d ...tnhuted hve
l m dnd.1 umlormJy distributed dead load of
• .5 f-or the t:dntlle\t:r beam of Problem 89
dcla1""
the
miUlmum ul'"ard \ertl",al reaction and
\.ounterdock"'lse reJr.:tlon moment al support
,"Onctntr4h:d Il\e load of :!5 Ii;. a umforml dil ri~.
load of ! k rt and a umforml) dlstnbulCd
oSk It
... Pl.13, Pl.17, Pl.ll, PI.22
t' 1or the beam of Problcm 8 29
delellillne lhe ma,im,um
e\l h~
ami neg4tl\e hear!> and the ma mum posI
IlC\!.lIl\e ocndmg moments at pomt F due to a
h\~
IO.ld of .to k.•1 umfonnl) dlStnbuted h load
dod a UOlfonnl> dlstnbuted dead load of I k fl
t,9 f'or the tru...s of Problem K47 determine the rruwmum
ten,lle and compre......l\e a tal force In member eN due
concentrated Ii\C load of 30 k a umforml dl Inbuted li e
load of ~ k fl. and a umfonnly dlstnbuted dead load of
I l. ft.
9.10 For the tru.... of Problem R.50 determme the rnUlRlum
comprc.. si\e axial force in member AS and the maximum
tensile a"ial force in member EF due to a concentrated live
load of I~O
kN a umforml) distnbuled hve load of 40
kN ,'roo and a unifonnl) distribuled dead load of 20 kN m
9.11 For Ihe tru..s of Problem 8.51 detennme thc maximum
(cn.. ilt: and comprcs..i\'e aJOial forces In member JJ due to a
concentrated live load of 40 It a umfonnly dl tnbuted live
load of .. k 1ft. and a unifonnl) dtstnbuted dead load of
2 k ft.
SIclIo.9.3
PROBLEMS
•
9.12 For the beam of Problem 8 2 deternunc the mwumum
and bending momenl at POint B due to Ihc
JXhlli\e raeh~
lceh~\
loads of the mo\ing H20-44 lruck shown In FI
M.l:!
.m~; bi
18lN
! - - - 43
72kN
m---l
FIG PI.1Z, PUll
1.13 For the beam 01 Problem 8 I ddefmine .... O I _ ' ~
po ItI\C hear and bendmg momeol I poIDt 8
~
I-l5h
... Pl.14, Pl.ll, Pl.18, PI.23
12k
W
120ft
... Pl.15, PI.21
12k
cb!:;~
---+---
k
. . . . . 'IN' • ""
•- : ~ : " - .
Cll elI~
D"I UJ~:; 5 '":
~.
.. It»...
boom due
L-.
momeot
m
Ibe wheel
Ioods
., F... P9 12
duebeDdinI
to Ibe wheel
Ioods
moment
In
., F'I P9IS
&II Determine the absolute rnaxiD...HiI
• IS-m lon8 SImply upponed boom
three movlftg concentrated loads sbDwD.
t.D D<ltlllUll" the .bsolult
111lWIIl_
.60-1\-10.8 SImply upponed beam due
mOVIng concentra.... load shown
42lI
ClW'TER 10 AnI'ysI. 01 S",",,",", Structures
1E1:YD11o.t Ii
10 1 SYMMETRIC STRUCTURES
Reflection
The d
OIU n
lJ(
I
lh'7;-'==-'1i
I'II.lll.2
A
lh R
n
II n hout A I
al AduaJ 51
I)MI
... 18.1
c
11IIC S1IUC\tIIII
t ... I •
•
Q)
CltAPTEll10 Analpl. of Symmetric Structures
,
'iii"
1KTIOII10 1 Ii
I w
I
Lf': = iB~l = ~
CE====fiB=_ _
L~-o;
£
..,----+---;"'IT-i--I
I, -\
E I. ~
L
L
E.l.A
Ellt
o
Reflection
Be;tm
(al
I
•
..-:m It - - - - t
.4
-
8f!
16 fl
I
B
1011-+-1011-1
EIt _ _
",
F F r.~ E D ~ = Hmge
le~nJHI
B
A
•
30 ft ------I
I
"'-
•
28 II
...
21
E. -\ =- con..tant
£. A = constant
Beam
ReflectIOn
L
M
N
N
M
H
J
K
K
J
F
G
lAI E
15 II B C D
-'- - '" 25 II - L:: 25 II .,
G
F
AI
( b)
15 II
-t-
ItI
15 II
,
-t-
I
I
lAl
W
E-COIUtant
0
0"
2511
I
W
E
Ret1ectioa
Frame
FIG.
E. A = conMant
E. A ;: conslanl
\~urT
Reflecnon
(e)
•
C
•
ilO ft---j
o
0
-,
ilO II---j
'
21
I
21
12 h
1
A
B
~
E.A
""10.3 E
fS rom I.OC
cun
I
B
20 h-------l
0---2011
E.It= _ _
l&nl
Frame
ReflccbOD
ld,
Ro_
ft~1
E A .COI"""
10.3 (contd.)
If)
L
Q2
CHAPTER 10 AnI'p'o.1 Symme1rtc Structures
.Ie
t---L
L£. 4 - con unt
... 10.4
trUl:1UfJ.1 )mml'tr\ lor lilt: pUrp(l l' or.tIl .tnal) ,is, II is 0'..,.....
l'lln IdeI' the: \ Il1'tr~
l,r ol ~
thl''''C ,trlll:tur,11 properlles that
dic-d Oil I't:sult ... llf Ih.1t p.trll(ul.tr t~p.:
of .tnal) ... is In other
IflKlme l.tn ~ l:llll ldl'rt:d 10 !x ~Ilmenc
for the purpose of
\ I'" If Its ... Irul'lural rnlllt'rttl' th.lt h.l\e .In el'tect on the resul
anah "'1'" drl' \ mmdm:
(:on hjt:r for l' J.mrle the ...t.tticall) dctemltnate lruss su
\ertlc.i1loalh.•l'" h \\n in hg. lOA. \\e C,tn '>Cc from the figure
~cl-'m t )
llf the tru... Il.e the dime:n ... ion... of the truss and the
Ill.:nt of tru ... r~m'l 1I
and Ih materi.ll and cro...s-s«tional
E .tnd 4 an..: )mmelrl( \\llh re"'pcct to the .\ a,is bUI the
\ 101all' ... ~ ~rt:el m
lx-c.w,< Ihe 11Ing.:d ",uppon at A can exen both
IonIa I and \t.'rlI(JI re.Klion . \\herca ... the roUer suppon al C can.
~lno
a \t.'rtJcal re,lclllm. HO\\eH'f. the tru s can be coosl<iered
... ~ mmetrll: \\ hen ... ubJl'Clcd 10 \ t:rtil:al load onl) because under
load. the hOrizon I,ll reaction al the hinged suppon will be
.,
01, therefore. It \\ill not ha\c any effect on the response
member <1'\1.11 force: ... and ddlectiom) of the truss. This truss
con"'ldercd 10 be ,) mmClrlC \\ hen subjected to an) horizontal
hO\\eHT.
£ample 10.1
The tru .........ho",n in bg IO.5(a) ~i to be ,malyzcd to detennine its member
fon.:C\ and deflections due to a general s) met~
of loads acting at lhe JOID
the tru,.. he wO'\ldcrcd to he ...) mmclric for ~uch
an an lysi ~
s
G
H
ABe
D
~ l-~l
F
4 panels '11 25 II _ 100
E , - ,on 'ant
fIG.
~It
lD.5
lao
E
ft--'
~
A
25
8
C
It..l- 25 It ~
E. A
D
25ft..)..
=constant
(b)
Solution
\\e c,m see from 1-1.. I(J 5 b L_
th
c
tndl the dlmen Ion the arrangemeat
e matenal and l:TO lanU t~
propt'rtlc £ and A and the lIII...rlI
gl\en trus are all mmetn
h
the m mber ((I of the tr l: \\U ropecl to the \ertlcal aXI
aXI
U
Thu ~hI
Iru IS s)-rnrnetnc wllh
ClIAPTER 10 "'"I,... 01 Symmetric Structuri.
,
SECTION 10.2 "'IISble MIl
,
.
I
\[1j';-"I~fPTJ
-Q---<--Q-
c
1--0-_-
Loadmg
Rcn
RefleCl10D
Loading
b
Id I
• --i
-'I-
~ It~
~.I'
-,-,-\,~
s
I-~l
-
:
i
"fl:r:r--.x-..
Rel1ection
Loading
Ie,
,
Loadmg
,
~
A~f
I--a--+---a--l
C ......
----a--_--I
f----a-----;I
J-~_l
Ie
_ C 4' . . . . ' & •
"
_,,----11-- a--l
--1---0
Ai,,,,
Rcflccli...
0)
''',--,--
",'
I
D
P
:;==rtJ
I
B
'u:r==
-",----- a- - of~
Loadmg
Reflection
II)
AG.10.9
s
.<
w
"
cD:r:r--eB, oo::::c:cuA
c
"1--,,-I-a--1
I--o--l--a --l
Loading
Rcnecbon
leontd."
a--+--o---I
W
cgob
o,freflecti...
(d)
Antisymmetric Loadings
A loading I con Iden:d 10 lx' J.ntl\)mmctnc v.ith respect to an
plane if the ncgatl\l' 01 the n:flcl.:llOn of the loading about the
lllal to the loadmg it -II
M I-~l
~
.<
-
M I - - - '1 - - - I
M
i
~
~-+'t!
I-----a-----I
Some e'i:imples 01 antis) mmetrit: loadings are shown in Fig 1
each loading ~ac
the reflection and the negative of refteeuon
t---
r--_-----"I:
n
pi
~t,- -I ;B
-----<
Rcllection
Loading
1-;------1
W
£
----a_
R
b
--I
.,~o- + -
n dlnn
B
,. 1---0----0- -
Loading
I )
FI5. 10.10 E mpl
.l
IA
nil )- rnm
ttll:
loadmg
.. 10.11 (......./
M
M
1---'1
'81----+---
CHAPnR 10 Anllysis of Symmetric Structures
~':X C 2 < ZL
IECTl0II1D.2
c10 ~' . :U
.... IIIt'.1101••_10
C'-IIOO.1o
blaC.
$I
kif'
--- T
"m
~"'l~ -'
~
j j j
81
0
0
000
40 -10 -In
0, O. 0.'
6nctm41maGnnl
(til
mg
~
Half Loadmg
-lO O~
80 ~O
0.' oS 0, O.
~{)
Ie
ld IS) mmetnc Loading Component
FIG. 10.12
~"(I
HITm (ric 1.(J 'I1~
( mpOllUI! The anti~}m etric
componan
loadmg I ,lbtained h~
..ubtractmg the ~} mmclric loading component
10.1211.1 from the IOlalloadmg hg, 10.12(<11) and is shown in Fig 101
FlG.l0.13 (eonld.1
Solullon
~ION
that the ..urn of the .. ymmclric and antisymmetric componen
to the gncn loadmg
I:~ ; ~
and S
tbemm
halfInloadl.B
Loud",are Cbowo m",FITheI I b
loadlDg about the UI 19 drawn In Fig 10 I
The
lbe Slyen loadlDB I detennined by addlDB the b If I
reOeotl•• FIB 10 13 • as showo ID FIB 10 I d
Eumple 10.5
:\ beam is ..ubjel.:ted 10 Ihe loadmg ..hown m Fig. 1O.13(a). DetenDlRe
metn( and antlsymmclrl( components of the loading lAith respect to the
,,> mmctry or the 1x,Im
,
I
lJJItlDtm
-
Q
20 It --+-10 ft-t-IO ft-4-- 20 ft---l
(a)
G,.. rn loadmg
,
IS.
10.13
X.l~
I
2st ft
Q
I b, Half Loading
Example 10.6
50
CftAPTER 10
Ana'ysls of Symmetric Structures
2.
. Il'('1
Sc
Ll
h "Ul
., LI" '
I II til' thl.' "tfUl.:Wn:',. on either side of the
"lll\"I"', JilL' l.:W"-'CCIIOI1.t1 i1r~
... and morn-
U"
l I • ' h
"-q
'·,h" l """,.",(\1' tIll.: 'llh'lrw.:tufC. \\ teh
are located
l
1'",_
• . ,. ","'1,'(n ...lhlUIJ ~.lh r('JUl.:c:d b) h.llt. \\hereasfuU
[ hl .1\1 (\
"'- .
.
II I
be
l,r t he..c pn.... l hllUIJ h.: u'l'.d hlf.1 O( leT. mcm rs.
11110
..., mmctrll,; and aOII;.Y1",.,....
1'1\ ell IlllJlIlg
3. Dl'l.-("llll pI"\.: ,h l ' c
' .
'
...
,,-,""'1.:1
to
the
.t'\I'
01
,
((lmf'\ 111.'11 .. ""h .. I'~
,
_'i,mmctrv
.
• of the stnlrho_
- - ......
'
U'!OI:! the pnx'cJUTl' dt'~nh.-J
111 Section 10.2.
-t. [)c,;mllfl tilL' fC pl'n'L' of the ,trlll.:lUrc due to the 5}mmetnc:
"Il1IT1L'lf)
. .
inertt,.
(1
elP," ~
-:==--_-:-
10:.:,9:---
OclCmunc lhe ~
I,',..,·"
Ill!! l.onlptlncnl.1
fllI!O\\
.
"
10
~ft
.
\1 c.II.:h Jllllll ilnd end 01 Ihe ,ub,tructure. "hleh IS loc:a
a.
of )mmctr: .Iprl) rc,lramls to prc\ent rolaltoa
JX'rJX'ndil:ular 1O the a I~ of ~ mmetl). If there
hmgL'.1t uch a Jomt or cnd. then 0~1).
the deflection but
rot.ltlon h"uld lx' ~r . . tr<lIlled at that Jomt or end.
b. .\rpl) Ih~ . .) mmetril; I:omponent of loading on the sUb,tnl:\1"'~
\\ilh the mac.l1llUde . . of the concentrated loads at the
s\mmetf) n:duced b) half.
the ,ub,truclUrc to determine its response.
c. el)an~_
d. Obtam the ...)mmelrll' respon-.e of the complete structure
fleeting the re.. pon...e of Ihe substructure to the other Side nI ....~·
a\ls of~)
mmetr).
5. Determme the rc.. pon-.c of the structure due to the antJsiYlTlIIIIlldl/!.
loading component as 1'0110\\ ...:
8. AI each Joint and end of the substructure located at the uil~!f"
...ymmctry. apply a rc:-Maint to prevent deflection in the am""".
tion of the axis of symmctry. In the case of trusses, the
forces III members located along the axis of symmetry will
zero. Rcmmc such members from the substructure.
b. Apply the antisymmclric component of loading on the
structure \\Ith thc magnitudes of the loads and couples alP i ~
at the ~ixa
of ~ym etry,
reduced by half.
c. Anal)lc the substructure to determine its response.
d. Obtain the antisymmctnc response of the complete strut
rcflectlng the negall\c of thc response of the substructure
other side of thc axis of s}mmetl)
6. Delermme the total reSp()n~
of the structure due to the gIVen
ing ~b gni~opmrXJUs
the s)mmetric and antisymmetnc rei
obtained in "tep, 4 and 5. rc'pc:t;tl\e1i.
the
<1\\'
dcfl~(twn
The foregomg procedure t.:an be: applied to statically dcterminatt
\l.ell as mdctennmate ~)me(rit.: struc(e~.
It "ill become o,bvi~j
sulhequent chapters that the utihJ'ation of structural symmetry
erabl~
redu~,
the I;omputatlonal etfon required in the anal
cally mdetermmate Mructurcs
==========J
1ECT1OII11.4 ProcIdIn f8r ....... at " ••"'" ......
f
I
2H
-
G
H
JOk
12k
.t panch al 20 fl _ 80 fl
(a) Gi\'cn Tru~s
•
I
I
----J
18 k
Jlk
and LoadIftB
c MembcrF
,
~v
411
JOk
I
18k
S)'I1IIII<ll'lC ~
Due
Coa_..
H
G
18 k
JOk
18k
(b) Symmetric Loading Componenl
,
F
~
I
,G
6k
(cl Anli~)'metrc
Jlk
6k
loading Componenl
ISk
(d) Sub'lrUcturc
H
With
... ,...
Syrnmetne BoundarY CoDdi1JODl
6k}k
Forces Due" AaIi y.-ux I.-liDc C:"'.....
,-
(g) _
I3 k
6k
A ~ = I~
~k
b _
F64
24k
G64H
JOk
I k
~
IE
k
Forces Due
...~1~
qc 0t
15
1
U
$10
':---::-
I~
A_IO
15
Member _
~1~
10-qC
~1
OI~
15
15
Due tD,Aatioym_1rI
1010
0t- -'~D:-,.;j
15
I'
456
CHAPTER 10
AnalySiS of Symmeh1c Siruciures
Solution
"L I 11.
1I1 11,1): "'~01
I
Thl
b
N.I l ~
I h kf h.lI!
\mllldrK \\Ith II.: 1''1\."\:11\1 the \crtlcal
th... 0..:.1111 1\ ckded lor ,malYlil
"j
.
\ I
I
\fllmdnl,.
n:
I hI.: uhslrlldulC'>
U
fI', the an.lpi~
J'Il1nsc .In. hO\\11 In I-Ig 10 ~
llf the ~ymet
d .lI1d ('1 re peell\ I
-
k
,
-'-
ExlmpJe 10.12
Ddanunc: thl' ut'! trudUI 1M the .lI1:~
1 of the ~
n: r o n ~
thl.: ~laJt.w
indeterminate frame; \l(h~
"r
12"-
I
-'-
-'-
A,Rb.ymllllClnc looohni Cempno...
mmelm: and antill1lll"iii
n lI1 Fig 10.23 a
I k It
I I I I II
I
Sft
I I IbTdl I I
I I
2. klh
I I
12"-
_L
I·-+
-+1
I S ft
I I I II I I
I
k-
I S ft
-
-
_L
-''I - 25 ft --t--- 25 It -----t- 25 f1---l
E, I. A =c()n~ta
(3) Gi\en Structure and Loading
,.-.-.-r-hl.:ft~_
11 "
3.
• _
k.
I
I I
, 2 k ft
Solution
'I
S ",m"
and c
-3k
I I I I I I I I l
, 2 II. ft
I I
-'fIG.
10.23
I I I I I I
-'-
-'-
t--l75It_
I b, S}mmctnl Luadmg Component
(d) SublrtIUc~
for Anal
of Symmemc Response
FIG. 10.23 (eonld.)
SUMMARY
- L....
and
An"u'''....,.,
C'nnlj_'"
.------,_+.
•
..
CltAPTER 10 An.lysli of Symmetric Structure.
.\ IO,ldinc: ~I n'll IdcTl;d It' Ix: :cirlm~,
\\ uh respect to an
It rlane If ~ht
n:tJC:dll'l1 l,f the IllJdm!,! about the axi\ is Identteal
loading It~lr
\ !o.ldlOf! ~l Cl' l1 ~:cr h.H
to Ix anu,,}mmetnc WIth
tlhln .1'.'" In 11 plane If the- nc:g.ItI\C olthc.n:llcl'1Jon of the 1oad1lll
the 3\1 I Idenlll,...i1 to the: )O,lomg Ihdl An) general uns.YttUllllll
loading c.m be dc(,:ompll cd into ..) mmctm: and antis} mmetn
nenb \\lth n: p...xt wan 3\1
\\ hL:n J \mmc:ITll: qrudure I" .. ubjected to a loading that
mctnc: "ith rc:'l""Cl to the: \,tfm:tun:', a\i~
of s)mmctl) the res
the ,trw.;turc I...ll'l) ') mmetTic Thu.. \\c can obtain the response
entire 'lructure b) .ma1) 7mg a half of the ..truc-ture. on either side
a ,.. of ,)mmetr) \\llh .. )mmetric bound'H) ,,:onditions; and by
mg the computed T6pon,c .ll;\out the axis of ~ mmetr).
\\ hen a }mmdril..' ...tructurc- i...... ub.ll.'Cted to a loading that
..)mmetric \\Jth re'lpcct to thl..' ..."~:nutcr
~IXa
of s)mmetry. the
of the ~truc e
i... al 0 anti,) mmetric. Thu~.
the response of the
erutc ~
can be obtained b) anal)7mg a half of the structure on
...Ide of the aXl~
of ...) mmctr). \\ ith anti symmetric boundary COI>dilii
and b) reflecting the ncgaU\c of the computed response about
of symmetr)
The re"'pon:-.e of a symmetric structure due to a general unsoy""",.
loading I.:an be obtained by determining the responses of the
due to the s}mmetril.: and antisymmctric components of the
metric loading. and b) superimposing the t\\O responses.
60
k'l
A
(,()
kN
f>()
kN
30
kN
18
Ie ID IE IF
H
/
J
r---- 6 panels at 4 m
K
G
~m
DHinpE
r---t-'-
..
E. A =constant
F1G PIO.3, PlO.18
soco- 10.3 .nsI1DA
10.1& ttwougfIl0.ZO Detennme the force
10
each member
.1I--1-l
A
01 the tru'..c..... ho\\n in figs. PIO I PIO S by ubhzJDI
3m-+-Sm
,trudural ) mmc{T)i.
fll.PllIJI,PlO.21
24k- C
FIG. PIO.4, Pl0.19
'm
PIO, I PIO.IS with rC'lpect to the ax.is of symm
..trUI..{urc.
fII.
Pl0.7, Pl0.22
5m
2H E
I:! ft
5m
Ht
4ft
FIG
A
PlO.5, PlO.20
1-1'11
Eo/A
80
lUI ......... Il1.D Determan< the member eud
the fr.lmc .. loho\\n an Figs PIO 6 PIO
U
f. A - ron tanl
fl&.
PlO.l, PlU8
FIG. PlO.2 InsI Plo.17
'm---+
E./A-_
........ 10.1 _10.2
10.1 ......... 10.15 [)l:temllnc the :,\mmClnc and antirometnl..
rnponem
f the loadi~gs
..hoy, n in rig..
F
=24 m =----1
E. A =con tant
PROBLEMS
-
""
JO
kN
tural \)mmt:'try.
fII.
PI.... Pl0.23
III
_1'18I11 -.... ......_ _
... I .
35
1
I
0
m
...........
15m
E.IA=~
... PlG.13 ... Plo.2l
C
B
...............
2t1ft
20ft
10k J
L"T
K
I
Hlft
10k G
H
3k1ft
,t
:Mil
10k 0
E
F
A
B
C
E
I
I
m
I
1---3Oft
30ft
.. PlG.14, Plo.2l
o
........
11-+-2011
B
•
11
Introduction to Statically
Indeterminate Structures
11.1 Advan1aQes and Dtsadvantages lnlleterm
11.2 AnalysIS of Incle1ermlnate Struc:.
nate Structures
Summary
es
SF/ncr Harhour, Australia
v
glIal Vis
In Part Two of this tex.t. we considered the anal)
1
of tatlcall) deter
minale structures. In this part Chapters 11 through 18 we focus our
allention on the analysis of staticall) mdetennmate tTUCtUres
As discussed pre\ iousl) the support rea,lIon and mtern I f, n:
statical!) determinate structures can be detennmed from th equat
of equilibrium includmg equations of condluon If an H
er
indelerminale slructures ha\e more support reacli n nd r mem
lhan reqUired for static stabdlt) the eqUlhbnum equall n I oe a 0 t
sufficient for detennining the reaction and mtem I r.
f
h
strw..'tures.. and must be supplemented b\ add II n I I n nilu
on Ihe geometry of dcformallon of tr\1l.'1ure
These additional ",Iallonships "htch a t nned .he ,·..,polihiliIY
("(}lid", ns c:murc that the connnult) of the dl plllt.'tITlC'n
throughout the structure and that the trueturc v
gether For example at a ngid Jomt the deftecti
the members m«tlDg at the Jomt must Ix the
an mdetenmnate trUCture
ID 01\:
III
ddi
arrangement of members of the uuelurc
nal propertlCS ueh as cross-sectIonal a
of .1as1lCl' .'e which to tum <!<pend
CHAPTER 11
Introduction to Statically Indetennlnlte Structures
1EC'hON111 ......
is therell
' I!!"I ( 11 •III indch:rmin.llc structure
strw.:tun.: 1 hC Ul
.
)
'nl"nnCI
\dh.
,
~
d
r
.
'
the
(rclatl\e
SizeS
lHll 111 an IIl:r.1 t 1\1.:
..
h ofthc
." .In.:, IIll 1',11,
I. . •Is,Unll..:d ,lI1d u ..cd to anal}lC
.
h I e structure
.
Ihu..
lhl
IIIlt'd
.m.'
u..ed
lO
fe\
I"C
t
IIlh:rna I lm.. e
l.
. .. e member
"
. , ... Il....
asswned
Te'\) ~u , menh>t:f
'"'" ,In." 11llt dose to those Inltlally
.
" Ilal,....
" ..U1 U 101'e- the I,tlcsl member sizes. The Itcra
slrw.:IUfC I' Il,l
of an anal
tmue... unll Ithc me.:'nl·""
\'\; ,iln t'l.l cd on the stlu~er
e '''DIIld,
$
e
1I.,:h1l: A• • •
ml,.· ~1
to Ihl)~
.•"uIned lor that an.i1~"I.,
Despite the !i.....,n:g(lll1g lhtlkult) m. gm~ lscd
L
£/
...
mdetermma
lUre". a great maJorll) of ,trul:lure, bemg ~Ul t
today are sta
determmate: I\.)f C'\.impk. most modem remfon:ed concrete
afC 'itaticall) imktermin.tlc. In this chapter. . "c dlsc~
some or
portant .IJ\ anlage .lnJ di ad\ antages of ~ndet munate
strUCIIUri
lX)mpared III Jeternunalc ..tructures and mtroduce the fun
concept-. of the anal) .." of indetermmate structures.
-'.........
11.1 ADVANTAGES AND DISADVANTAGES OF INDETERMINATE STRUCTURES
... 11.1
The ad\antJ.ges of staticall) indeterminate structures over det
structures include the following.
nol ncc:essanly collapse and the loads WIll be Ico~ubirtsde
lacenl ponlons of the ,Iructun: Consider for
detenmnate and indetemunate beams shown m FI 1
specIIvely Suppose thallhe beam are upponmg brid
way and that the nuddle pteT B I destroyed when a
1. Smaller '~enrtS
The maximum stresses in statically I
nate structures are generally lower than those in comparable
nate structures. Consider. for example, the statically detennmate
In_.
indeterminate beams shown in Fig, 11.1 (a) and (b), respecti
bending moment diagrams for the beams due to a uniformly dis
load. Ii. are also shoy. n in the figure. (The procedures for
indeterminate beams are considered in subsequent chapters. It
seen from the figure that the maximum bending moment and
quently the maximum bending stress in the indeterminate beam
nificanlly lo\\er than in the determinate beam.
2. r~ta G
fes ~n itS
Statically indeterminate structure
ha\e higher stiffnesses (i.e. smaller deformations), than thoac
parable determinate structures. From Fig. 11.1. we observe
maximum deftectlon of the indeterminate beam is only one-fiftIl
the determinate be-dm.
3. f~d'" J~R
Statically indeterminate structures If ~lp n
SIgned have the capacity for redistributing loads when certatD
ponlOns become overslresscd or collapse in cases of ovcrloadl
eanhquakes tornadoes. Impact e.g. gas explOSIOns or vchidc
and other ueh events Indelerminate structures have more
and or suppon reactions than reqUired for static stability
or member or suppon of uch a struclure fails the enbre sU1111l1l11i
Agj
if
S_s1ly~Bam
Inoaasl\ "'"
if
r~A
... 11.1
A
B
•
•
CHAPTER 11
Introduction to Statically Indeterminate Structures
Internal
IlCTION 11.2
,(~nlh
AIIoIroIo aI h,d".i_'" '11 •
A
B
4
B
EA I
(a
B
a) Stath.alh Determinate Beam
lbl SIJtically Indelemunate
Beam
F
al~T)AE
a
a.l.T L
Slab all Detennirwe B
11'"==========
B
-F
a
T
AG.ll.3
E. A. I
Ihe: "l.ltlcalh~. detemlinate beam is sUpported
ram IOto II. Bl ,... ",'u~
iu.. t tht: ...utfillel1t nUll1ha of reactl )n~
reqUired lor ~ta 1c
stabil
·remo\.llllf urrl)ft 8 \\ill l·au ..e the entire ~truc e
lO: collapse aslllll,.
in hI!. II.:!'J . Ho\\c\a. the indeten1l1nate beam (Fig. 11.2 b has
Ol it:h er~a t\
In the \crticdl direction: therefore. the slructure wiD
nc-ce :.arih collar"\.' .tIld rna) remain ~labe,
c\en after the support B
failed. \ ~uml!
that the he.lm h.h been de...igned to support dead
onh m C.l"C of ;ueh an .tceidel1l. the bridge will be closed to traffic
pie; B i" reraired and then \\ ill be reopened.
AG 11.4
lb, SlalocaJl Indctmni.... Beam
from deforming axIally by lhe fixed uppo
temperature change AT a c rnpressl c a I ~
F
• dT AE develops In Ihe beam as hown In Ihe figu",
fabncatioD erron are SImilar to those of temperature ha
rnmate and mdetenninatc structure
of ... tatlcally mdetcnninate structures
The mam ~egaln\dJi
delenninatc . . truclurc'l. an: the follo\\ing.
I. Strene\ Due W SUppOI·t Settlements Support settlements cia
cause an\' strc:-.'lC'l in determmate structures: thcy may, however
..,ignillcal;t . .tres.. es in indeterminate structures. which should be
mto account \\hell de.. iglllng indeterminate structures. ConS1der the
tennmate and mdeterminate beam shO\\.-'1l in Fig. 11.3. It can be
from Fig. 11.3(a) that "hen the upport B of the detenninate
undergoe.. il ...mall settlement tJ. H • the portions A B and Be
beam. \.\ hkh arc connected together by an internal hinge at B
rigid bodies \.\ithoul bending that I .... they remain straight
~Ires
. . de\dop m the determinate beam. Howe\er, when the
ous mdetennm,ltc be.lm of Fig. II 3(bJ IS ... ubjected to a Sll1ular
tnem il ~
it bend. as ho\.\ n in Ihe figure; therefore, bendID8
dc\elop in the beam.
2. Streur, Due to TrmprrQture Clrllngt's lI11d ~irb"F
like support s.:lllcmcnts the'<C effeels do not cause stresses m
natl' tructures but may mduce \ignlhcam stresses in indetemu·'-'......
(on Iller the dctennmate and mdetcrminah: beams shown m F
It r... . n Ix s«n from hg. 11.4 ill that "hen the detenninate beam
Jected to a umll..'mn temperature mcrease tJ.T it simply elop"'~
the aXial defonnatl\)n gi\en b~ () 1 tJ.T)1. Eq. 724
de\dop In the d tenninate hearn. smce it is free to elon
e\er \\hen Ihe indetenmnatc: lx-dm of Fig. 11.4 b whICh
me"".
11.2 ANALYSIS OF INDETERMINATE STRUCTURES
Fundamental Relationships
Regardless of whether a structure IS statically determmate r mdetermnate its complete analysis reqUires the use of three pe f Iall
shIps.
•
Equllibnum equations
•
•
Compatibility conditions
Member force-defonnauon relauon
The eqUlhbnum equauons ",late Ihe r.
Its part
eDSunDg that the entire structure
w
In equillbnum. Ihe compoubihty condiuon ",I I
Ihe strueture SO thaI Its ,anoDS parts fill
defonnaUon ",lau os. which In' I e Ihe material
proper1JCS E I and A of Ihe members..: ~
Iween Ihe fon:es and displaeemen
Ihe~
In tbc anal
of III
detenm
blainhe~1§P
of equilibrium
a firs.then
usedIheI i~.rebtfm
fo
f Ihe trueture
Ihe companbili
ndi
.,. em
di pJaeemen F r eumple
10
If
'SInD.·
!OUt
M".1
e71l
CHAmR 11
1K1IOtI1t.2
Introduction to Statically Indeterminate Structures
",
.
12 II
f. -\ = (on lant - 20.000 k.
16 (I
DeI{lrmcJ hapc
5ll0k
'"
Substilutlon of Eq
or
F. ,
F.
48F.
500 k
(e)
Ib'
FA,
c
The member aXial defannatlons can now be compu =~ :' f <-~.
these values of member aXial forces Into the member f(
relallon. Eq. 11 10 Ihrough 11 12 I obtai
<lA'
<l
0136 ft
1629 In
and
Fmally by uhslltuung the values
Ibe oompallbilil oondlllOO5 Eq
dlSplacemenl of JOlOl A a
A
<1011\2
500k
(d,
FIG. 11.6
tn:lme;, alp~h a;,ol1 re\
'\ 01 Joint .1• The d"J~p Iacement d·lagraDI
h
. ~
0\\0 In 19 ,11.6 CI gm u~sA
the dlsplal..-ement A to be
Yo nle the l;ompallbllity l,;ontJition\ a,
IS
i)lB
J-rc
i)w
"
1\ nj~
(J -
0.8.-\
10
m
en
CttAPTtR 11
Introduction to Statically Indeterminate Structures
12
h' INl:1.: IllclhoJ .... ,\ hu.:h ~na
.prc'Cnted In
'rmng l,"yu.ll llm ", 1 \:
"
I
'r,11\ ((\l1\l,"I1\Cl1t lor .Ina )/lIlg 'mall st
ler... I ~ and I ~
,IfI," I!cl1l;· .
.
1.'
~I
I ' k\\c:r 1.'\0......... mcmber... alllLor reactlo
\\ Ith I k\\ (l'uUllu,tll'" .I;
'.
, I I hl\ 1ht·.. C' mcthod" .ln~
also u't:d to den
requm.'d It'lf "t<.HIL .1'1 •
.1
. , .1 'j' nlill on fI... I.I!ltlll' llL'l'dl.'d to ul,"\t~lop
the displ
member loru:-ut: l'f '
"
,)~
F(
l
"
mc:tht1u....
11' 'I,plal:....ment method... ,m: consll.kred
throul:!h,
I
11.: u
~
-~
r 1emenll,"u
('n u
large Jnd hlg ~
10
Approximate Analysis of
Rectangular Building Frames
Chant-
...- .
' m ·tht)lh Me 111('r" ,),tcmi.ltlc can be easil
Th c.; I.:
t:
. "
. '1mputcr ,Ind .~m
thr.:n:lore prdared lor the anal
rl'dundant tfw.:lun:.
12.1
12.2
12.3
12.4
SUMMARY
In thl I.-hapla \\c h.ne It:amed th.1t the ad\ anta,ges of statical
tenninate ,lrw,;turl,".. O'er dl,"h:rmmate ...t~_erut;.lU
Include smaller
mwn "tn~
t:". grt:atcr :.c"~nrlu
... and redundancies. Suppon sett!
temperature ch~lng"'.
and fabricauon. errors may m d u ~
51
..tn: ,e" in mdt:tcrminatc . . trUl.:ture". whIch ...hould be taken IOto
\\hen de igning . . uch . . tructurc .
imohes the usc of three fun
The anah"I" of ~trUl.:tun=
rdatlon.. hip.,: ~ equilIbrium equation". compatibility condittolll,
member force-deformation relation,,_ In the analysis of inde
~truc e.,
the equilibrium equation... must be supplemented by the
~til b ar
condition... based on the geometry of the deformation
"tructure. The link bc.:t\\-een the equilibrium equations and the
patibilit) conditions is c.: . . tablishcd by means of the member fl
deformation relation", of the structure.
The methods for the analysis of indeterminate structures
classified into two categories, namely, the /iH("(' f/Ii'xihility) meth
the dill'lac(,lJ1l'''' srilll/f_'.' I /IIethods.
AssumPtIOns for ApproXImate Analys,s
Analysos for Vertlcal Loads
Analys,s tor Lateral Loads-Portal Method
AnalYSIs for Lateral loads-CantIlever Method
Summary
PrOblems
Sr. Louis Gate1my Arch and Olel
Courthouse
Jeff
N
:lI1 Na'"
,beans 01' Mel'lOl
PtServe
The analysis of statically indetenninate strudures uslDg the force and
displacement methods introduced an the preceding chapter l;3.n be con·
sidered as Hact in the sense that the compatibilIty and equlhbnum
conditions of the structure are e,,-aetly satisfied in uch an anal
Ho\\c\er. the results of such an exact anal}sis represent the ac 081
structural response only to the extent that the anal 1Ical model of the
structure represents the actual structure. Expenmental result ha e
demonstrated that the response of most common type of ruct
under senic.-e loads can be reliably pmlicted by the fnrce and d p
menl methods provided an accurate anal}t1cal model fthe troet
used in the analysis
Exact anal) i of indetermmatc truetUfiS IR I
of del1ecuons and solution of simultaneous equations
time conswmng Moreover. such an anal
depends
cross--secuonal area and or moment of merna f
of tbe trueture Because of these ddlicullleS
at<d "
analysIS the prehmonary design nf mdetermm Ie t
based on the results of appronnwi ana
In "hi h
In
are nmat<d by maJul18 ""rtain a wnpn os about the : ~ =
SIZeS
and or tbe dlStnbunon of forces bet
n the membe
thereby avo,d,ng tbe neeesstty of computmg dellecti
01
IECnoN 12.1
CHAPTER 12 Approxllnlte Analysis of Rectangular Building Frames
_... 1119...... tor ..................
h ." rn)\(" hl ht: lJlIIte l"omcnu:nt to use
'rrn'l:\lm.ltl.: 111.1.
I
'
f rro1 xl \\hen ('\aal a lern.tll\e deSigns
plannmg r h.1 t: l~
J\.:
.
Th
., 111I Ih.'J for rd.lI \~
c~.·l)nOm .
e results
...trudurc ar~
lI"lJ.l II \ 1.:. •
In 11,(' he u cd 10 (',ornate the Sizes of
PH) mMIl: an,. I\ l'i (,:. •
I
.;.. 'f "l'('Jed 10 Iill[\.lh: lhe. C:\iK.'t ana }SIS. The
tnll.:tur.l I mlnh...:
..
.
I
' J(' u:m II I' ~r·.\l..
n" t.: ~l
n.ln
.In: then fe\ I,,"-,d I.tcr.tll\c}.
, .usmg the
- . "\ an ,I,..c hl tfmC at their fin,ll deSigns. Fun
lIcec .. 1\ C e .Il ....
n
"1\\" IIlth 1 I
(\meumc
Jrrr,,\I......
'.
. . . u..cd to . roughl) check the
of (,.'\.KI .m.ll) r \\ hll.:h dUt..' to 11 l"ompIC\H) can be prone to
hoalh.
ren(l\~ti !
III
n."Ccnt ) car... there 11,1'" nc~
an Increased tendency
and fClfohtung {,Ida .. trUl.:tun:.., Man} such strueturea
,lfUdcd prillI' to 19(i(). inl.:luding man) hig .ri~
buildings. wa.
hmed okh on thl.' b.I\, of Jpr o'-Im<lt~
analySIS. so a knowlceIIe
u~dl'L
(amh'ng of .lrrrlnmhlie method, used b) the original des.i
)l au~
helpful in a rem.Hallon undertaking.
.
l nlikt.· thl.' l',-,Kt .~dohtc1I
\\ hll:h are general In the sense that
can be applied (0 \ arious I) ~cp
of ~truc res
subjected to vanoUl
in!:! .~noitJd c
a \pecilic method is usually required for the appro
an:lys.i, of .t p.trtlcu!ar t) pe of struclUre for a particular loadiDJ.
example. a different approximate method must be employed ti
anal)sls of a fecti.lngu!ar frame under \ertical (gravity) load than
the ~is)lan
of the samc frame subjected to lateral loads. N
methods ha\e been de\eloped lor approximate analysis ofindete
structures. Some of (he more common approximate methods pe
to rectangular fmmes arc presented in this chapter. These methoda
be expected to yield re..ults within 2()'~/o
of the exact solutions.
The objectives of this chapter arc to consider the appro
anal)sis of rectangular building frames as well as to gain an
standing of the technIques used in the approximate analysis of
tures in general. We prescnt a general discussion of the sun
assumptions nccessar) for approximate analysis and then con ider
approximate analj-"is of rectangular frames under vertical
loads. Finall). \\e present the 1\\0 common methods used for
proximate anl~s.i
. . of rectangular frames subjected to lateralloadL
12.1 ASSUMPTIONS FOR APPROXIMATE ANALYSIS
As discu sed In (haplers "' through 5, stati<.:allv indetemunate
ha\e .more uP.JX>rt realtions and or me b r~
than rcqwred
tablht): therdore. all the realtions and internal forces lOci
me o~
f~
uch truetures cannot he determined from the
.equlhbnum. The eXl;e s reactions and internal forces of an
ar'c f eIerrcd to as rt dUlUlant\, and the Dum
dmlTlale IrUl.:ture
.
undants I e the dltfercn(,.'C bet\\een the lotal number of
the number 01 eqUlhbnum equations is termed the degr
un"""".
Assumptions about the location of Points of InftectIon
478
CHAPTER 12 Approximate Analysis of Rectangular Building Frames
1lCTIOIl12.2
.J 1"1(ICI.'Jurl." Il'".
.1ppI0'\1I11atl'
ana
I). SIS of n""'.....,.
. )
.
, 1.11111.:. ~lIb'CJ
Il) \l'rtH.:.lI
t!!r.l\ll)
l
o
.
d
~
Involve ,,'"-.
. , .
\ l'(lll1l1hlll I~
lal hull J log
thrn.•1 ,ulllplh1n"
lut (hI." heh.1\
1M of_ c.lch
girder of the fram
_
.
III Ul1lfllrlnl) lh,tnbulcd _loath", as sh
•
I' ,
Il,' '·'I:l:-h!.ll.!\ JI.lgram tlf .1 t) pH:al glTlh:r DE of the Ii
.-1'
- - a
1.:.
J I
. h
I
hO\\1l In I "It! I __
"" ......v hlml lilt: JdlCl:I.1." ·da.re 01 I. e girder '1Ql1d1lt.
_,
In thl.' figun: \\e (lb~nc:
th.!t t\l~
mtkl:tlon POints C\lsl near both
oj the rcJl~
Jht.' I: mllL'cULlIl poml dc\do!, becau-.;e !he columna
thl. IJ'.I(('111 ~lrJc
(tl 1n ,.~h:J
to the end ... 0'. gIrder DE offer partial
'trJIIl! or rc l'I.IIKe .Ig.tin t rt1t.tlIon b) c'\crtln.g ncgatl\(' moments
and \tUl at thl.' r~dI!
end .. D .Ind E re~pcl.:t\1)
Although the
IlX:3tllln of the mflccti,m ni)~
... dc:pend' on th.e rclall\c stlffnesses of
fr.lme 'f.,lbm~
.md I."W be d~teml nC'd
onl) from an exact anaIIYltL ••
~an
e,tabh h the regll..ln...dong the girder In which these points III:
I.att.:d 0\ namll1l11,!! ~hI
l\\ll extreme lo·ondillons of rotallonal restraint
and (d J. If the girder ends ~
the i!ird"er cnd 110\\ n In I Ig. I ~.:!c)
to r~lt.I
.•1" m the ca..e loll' a ..impl) .. urrorted girder (Fig. 122 c
l'~ro
h<.-ndlll1! m()l11cnh and thu.. the mtlection poinh would
Ihe other eXlrcme. if the girder ends \\ere com
at the elld .. nO~_
fixed <l!!Jm..t rotation \\C C,1Il .. hm\ by the exact analysis presented
"ub-.eq~ nl
ch,lpt~r
.. that the mtlection points would occur at a eli
of O.::!IIL from each end of the girder. as illustrated in Fig. 12
Therefore, \,hen the gIrder ends are only partially restrained agaunst
tation (fig, 12.2(b) the mtlection points must occur some\\here WI
a di,tanl.:e of 0.211 L from each end. For the purpose of appro
anah ..i,. it i.. ('Ommon practice to assume that the inflection pomts
loca;ed about hall"'\ay between the two extremes that is, at a d.
of 0.1 L from each end of the girder, Estimating the location of two
making Iwo snoitpmu~a
about the behaVIOr
flection point-. ~c'\loi
the girder. The third a\\umption IS based on the experience gained
subjected to \erticalloads
the exact anal) ..~c of rectangular frame~
\I, hich mdicate\ that the axial forccs 111 girders of such frames are
\ef)- ..mall. Thu\. in an apprm;imate an lysi~.
it is reasonable to a
that the girder axial force\ are Jero.
ILI!..'r .1 fnll1H;'
, I..
2.
The inflection ~)lI1t'
arc 10l.:a1Cd .it one-tenth of the span from
end of the girder
The l,!lrder axial force IS lcro.
The effect 01 IhelM: )IiJpm~
II1g a\\umptions is (hat the middle
tenth of the p.m 0.8/" I 01 each girder can be considered to be
uppoftcd on the tv.o end ponion!'l of the girder each of wbich
the Icn,gth equal to nne-tenlh of the gIrder span 0.1 L). as shown
12,2 e
ole that th girder arc nuv. stalicall} detemllnate
end force and moments l.:<ln he dctemllned from statics a sboWIl
....,...ror_
~
CIt
H
G
Pl
• '."J
lIl'I!';L
l
To ,ummarizc Ihe foregoing di""us..ion. in the approximate ~ : . ~
uf a rectangular frame subjected to \ertical loads the following a
lion .. are made for each girder of the frame'
I.
w
l"t.:
w
D
E
A
8
~_-L
F
c
- - - --L
----1
fa) BUilding Frame
L----
~C)
(b) TypIcal Girder
~
w
~
---
(c)
-
I
Simply Supportod Girder
fll.1U
I
-
-
I
CHAPTER 12 Approximate Anllysis o' Rectangular Buildino Frames
I
,
,,
,
,,
,,•
_L
_
L
_-1---
--.,,,
,
,,
,•
,,
,
,,
,,•
,,
,
,,
,•
,,
1',- ,
,,
,,•
1'-
L
,
--1---
1ICIJOII123 ....,. ......... . . - -_ _
,,'~
•• •
-l,
,•
i
I
L
-----j
" Intlection Point
(a) BUilding Frame
--.-!
P,-
---+•
2
-+•
P,-
.'-
.L
•L
-+•
-..
2
-L.
z.
(b) Simplified Frame
"-nnn
"-nnn
- -- -- s
IS 12.5
I.
(c) Equivalent Series. of Portal Frames
statical!) delermmale because II IS oblained by inserting onl 14
hm@:e I.e one hinge 10 eal.:h of lhe 14 members IOto die
frame \\hlCh IS IOdelenninate 10 Ihe 18th degree. Thus the
mdelermmaq of the imphfied frame of Fig. 125 b I 1
mlenor column
h
CHAPTER 12
Approl mite AnlfysiS of Rectangullf Bulldmg Frames
1EC1'IDI1U - . . . . ....... 'DIIiI . . . . . . . .
a.
b.
"
Ih
1/
Q
RG 12.6
--Cl----<c__
'+
II
t'!C.rlldlr
(
)
-
-
CHAI'TEII 12 Approximate Analysis of Rectangular Building Frames
- t-r::jZ
I:':.
-
~
,
"
;<- C ~
~
~
,r
1EC1JoII1U . . . . . . LIWII' II. .............
~T1
~
-
f-
\b-'
~
'"
:. ~-ff"
_
(:
f0~
""tiJ
..
~ '"
a
'"
~
v
-g
",
-"
c
~
0
."
0
"'l
c
E
:;"
v
~
~
-
~
-
0
§"
~
"
'"
~
z
S
~
~'-V:.
~
.
~
,~p*
~
v.
~
:f:
Z
v
0
~
-
EF
0
4
F
0
12
EM
0
60
M
Q
12 1
-
-
CHAPTER 12 Approxl_1e Analysis of Rectangular Building frames
IICI1OII12A AooIroIoIor'-nl' ••, :
\\t,.' .:.m .lbll ~ c
that 10 order t
f Ig I'Sll
~,
.~ .
"
(II thl 1,'lrdcr \,.'nd moment M H m
1 lll1enl ll.JluhhnUill
Th
If
'"
I I '·~-I
ll']umn l."nd llHl1l1l."nl,
H
I k
mJ \IN'(NI I,) I 1<,;
I k . us
d
.
I
11'11 IlIl Illtnt (, bUI .1 r.; \)( "ISC lrectlon at the
l(luntenk,I..\\ISC
t In'( I
Jll'r IIll \)1 "'lllt (,
glrJlf
-J
(,/I
h
10 Jt:tenmne the ',fda hC,lf \, II \\(' (,:oml cr. t e moment equi
the kfl half III ~lr&
(,1/ I H'lll the frcc-hod)" diagram of girder
·'t the he IT !tlrn' " 11 t~um
act downward
WI
,h
"8
\\"lIOX'"
k
11411 It \oan dl.'\l·]llp ,I "Xluntcrcloc WI
m
tude (II ,
- "'.
h I
•
/
••
'th
Internal
hlllgc
to
h,llam.'C
tee
ockWi
mtuuc' I aouu
"
I Thus
-
/
/..
t ....
\/ H
60
I ,
The I lal JOTle h'.'.lT .lntl Ilhlmcnl at the nghl end H can now be
mg the three cl.jUlhhnum CqU.I!lll I1S to the fr~
bod) of girder
'2 e '\ppl}lng L: F l O . \\C oblam QNe.• = 7 5 k . From EP
obtain 5
I k .md to l'ompule \fH&. \l,C appl} the eqwhbrium
~lpra
.. L\/u
II
15
\IH ,-+130=O
12.4 ANALYSIS FOR LATERAL lOAlIS-CMTuvER MEnIOD
\In
otc thallhc guder end m\lmcnh. \hll and JtHC , are equal 10
hale the same dm:("tion.
Next the end a"lion" for girder HI are computed. The equili
tlon" ') F t ~ 0 and L \I 0 arc fir"t applied to the free body ofJ
11.8(f} to obtain the a'iial force QHf ~ 2.5 k - and the moma
15 k-fl ) at the left end If of the girder. The shear SHf - 1.5 k IS tbaa
b} dl\idmg the moment \tlll by half the girder length. and the three
equatlon~
are applied to the free body of the girder to obtam QIH
Sill
1.5 k 1 and \1 fH
15 k·ft .:> at the right end f of the guder
12.8([ ).
All the moments and horizontal forces acting at the upper nghtJ
no\'. kno\l.n so \I.e c<tn check the calculations performed thus far by
L Fx - 0 and L .\f ~ 0 to the free body of this joint. From the Ii
gram of joint I sho\l. n in fig. 12.R(f). it is obvious that these eqwb
tlons are indet:d satisfit:d.
The end a"tions for the fir~t.soy
girders DE and EF arc
similar manner. b} starting at the left joint D and workmg acrou
The girder end a"tlons thus obtained arc shoy.n m Fig. 128 f
Column A \/UI Fon e We begm the computation of column
the upper left Joint (i from the free·bod) diagram of Joml G
128 e \l,e absent: that the aXial force m (olumn DG must be eq
site to the shear 10 girder GH Thus the aXial force at the upper cod
DG I Q n I k By applymg L: F,
0 to the free: body of
obtam the aXial force at the 1000er cnd of the column to be Q
the column DG I subjected to an a:tlaltenslle force of 1 k ADa.I
Temammg second tory columns £H and FI are detenmned
ldenng the equlhbnum of Joml 1/ and I respectively ~:
~ rces fi r the first tory column AD BE and CF are c
equlbbnum con lderatlon of Jomt D £ and F respecbveJ
thus obtamed are hown m Fig 12 8 r
Centroidal axis
,
'~,
I
I
-
/
I
- 0:
r;,
/0 ,
I
Tension ew...
FIG. 12.9
/
/
f
I
,,
,
I
I
/
/
RIG""
_
CHAPTER 12 Approximate Analpis of RectanGular Building Frames
502
CcnlfPlJ 1'1 \.tllumns
H
1. fl
•
(
8
~
I"
r
Q
'0 ft
Q
H -
Q(Xj
0 I.:!'i
Ie) SCl,:llon ad
CcnlrC11d of Ctllumm
r-\
1bb711-l
Gi, __~_- ,- ;Hr-~_-;l
H
T
]OL-r-
~D!.-
,0 L _
-'1rE~_j
-I
f
F
-
A
QBI
QM)
+
12 ft
~
0.125
8ft
..L
QAD
(d) Section bb
h) Stmrhlll:d Frame:
HV-
I- 15 It -1- 15 It-l
Q(iH
-
6,R~
-
~
M'H~18.9
5~1f
G
':
M
11(;
= 1.26
~189
.
SHG
= 1.26
y
,,i
L _ _- x
(e
FIG. 12.10
QHG
..
CHAPTER 12
-
Approxlmlte An.'ysis of Rectangular Building frames
---lH:.-_ _- ; '
1 0 , -':-
kD------+.E,...----j F
~, -
~
15
095
11;;-
(8' Support ReactJon..
FIG 12.10
d F elm '~hT
IInphfi..:J lraml: obtained b) insertlng
at midpOint
I all '~ht
'f~m .t
(If th..: gi\ cn frame. I" shown
Q
.\unp
bln~
I ~ I l'l
M.
To l-ompulc J\.ial forces in the column oftbe _ . , .
tory of the '~mart
"t' p.l .in lrn<lgm.ll'} -.ectlOn aa through the IDtcmal
at tho: bh~l't J m
01 tolumn f)Ci. Elf and 11 a .. sho"" in Fig 12 I
free-body Jj:tgram llllh(' pt1rtilln llf tho: frame abo\e thi.. no1t~
is shown
OI.~
r.: Bel.aux the ..et:tlon CUI!< the- ~nmulo:.r
al the Internal hinge&., 0111
nal hear:'> and J\lal fllfl:r.:: but no Internal moment.. act on the free body
point "he~
the l'l,lumn.. ha\t' been cut. " ....uming that the cros~ectlODI
of thc wlumns arc cljual. \\e Jetl"mlinC !he kx:ation of the cenlrold of tile
columm from the left column OG h) u..mg the relatiomihip
(. IU11111 4
jJl~),
I{O) + 4' ]01 t
~A
~
:q
Q£"
26,67 QDG
Qff
~; ~
"'-H
Qo
0.125Q1>('
() 875Qo"
By swnmmg moment!< about the Ielt Internal hinge J.
L
SUMlJtuD~
fq
If)
lJ
III 6
Qw]O
lAC lAnte
Q" 50
I and ~
blam
60. ()
h(~21
1(1
II 875Q
50
QIX.
0
I 26 k
Subs lUual Eq
0
and 4
b1a a
60
66 k
The columa axaal forces arc hown m FI
diagrams of aU the members and JOlD
1(50)
- - - 26.67 fl
The later.t1 load .. art' ,u.:tmg on the frame to the right. !lo the axial force
umn D(j "hir.:h is to the left of the centroid. must be tensile. whereaa
force .. In the columm 1:'11 and f1 located 10 the right of the centroid
wmpre....i\e a.. ho" n in f-lg, 12.1 U(c). AI..o. since the axial forces m the
are a....umed to be line.lfl) proportional 10 their di!ttances from the cell
rdat\(ln~h p)
bet'Al"cn them C.lO be c'!ltabh'ihed bv means of the Similar
shl)"n m Fig l~_I()v.
thai i...
•
lAC
B I1U1lmi,aI m"""mls
0
G"der Sh ar and Momml KnOWIng hunn
can now be computed by OO1I5Idmng eqwbbnum m
JomlS Starttng at the upper ltfl: JOint G we
L F r - 0 10 1he free body of Ih,. jOint F
S" I 26 k .ttb< Idl end of 8mler GH TIl<
detennined by muluplyilll 1he bear by ba the .",Ir1en~ih:
It
508
CHAPTER 12 Approximate An.lysls of Rectangular Building Frames
'ra'••
lid .\/1 af I \ \ ilh the f:mkr llll,.lnienh now kn
bI: dd...·nllllll·d l'l\......~ nll1'ldenrlf: mument .....
......1 Ilimn nlt1lll.. n I, ... on
.. ull,
lOin!) HI,; 'mom' JI lh... s.......·,md ~rlh'
and .Irpl~
tIlg ~
\1
II to (he' fi
end of column
( ' . I' III C 1\ (ltol.un 'h..· Imllileol .It the uprer
J0m',rle-~
.
tx: \I
I~ 'I) l·lt· fhe heir al tht.: uppt..·r . ·. nd III culumn DG I
puted b\ l!1\1 ...hn' \I ~) h lit the t.:11lumn th~le
thai 1\
(
1'11/1. 1/f\
""/11
115 k
llul "I, mu I. d h1 tht: nl!hl 0 th.lt it can tle\c1op a c10ckwl
. .-c me l'lmllt<;r... kll:k\\l--e end ml'ment "up, The ..hear and moment
10" . . T . ·. nJ narc lh... n ddermw.. d h) .Irpl\ 109 the equlhbnum equations
Oand""'" \/ UtothelreePlxiyoft:t1lumnDG -.eeFig.12.IOe
Q
moml;;;' nd !oheJr'o lor column '-"If .tnd fI an,' I,:omrmed In a similar
thaeaft..· r [ht.: rnxedure IS rcpe.lted to detemllne the momenl and
th...' fir 1- I ,t} ...·0]umn.. -l n Bf. ,jnd C1' '>t:e Fig I ~ IOlIl ,
0""
to m.lj,~
Gmf. T A \101 / OTt I \\ e bcgm the computation of girder aJ.laI ft
the upper lefllomt (, \prJ) 109 L F, -- 0 to Ihe free-body diagram of
hlmn 10 bg I~.JU
e "e hnd the a\l.lI force in girder liH to be 68S k
pre. Ion. The i.I\ial force for girder III i... delemlined ...imilarl) b) consideriq
eqUllibnum of J0mt /I i.Ift....r \\ hlCh the eqUllibnum equallon L FK
plied to the free ~ix<b
of the right Wint I I(l check the calculation The
force' for the IIr 1-..IOT) glrd....r.. DE .md EF are then computed from
librium COiNder,ttl0n of JOin I f) and F. In order. The axial forces thus 0
are ,ho\\n 111 FIg. J:!.IO.f).
RW<f/flf/\ The force' amJ monU,'nb at the lo\\er ends of the first
umn,.fD, BE .1I1d CF repre'ent Ihe rcat:tion.. at the fixed supports A B
re'~di\l).
,1'0 .. ho"n 1I1 I ig, J2.10(g).
apply the
Chc'(klnfl CmJll'urlllio!l\' To check our computations, we
equilihrium equatlom
+
''0
LE,
'LF,
L \II
W(2ISJ
(0
PROBLEMS
Section 12.2
12.1 through 12.5 Draw the approximate shear and bend·
ing moment diagram.. for the girders of the frames hown m
Iigs. P12.1 through PI2.5.
the free body of the entire frame (Fig. J 2,10 g
0
10
0
20
95
15
5.54
7.5R + 0 95
+ 6.63
0.04
~
30 kNfm
0
0
IIII1
III11
E
f)
0
20116)
15 ft
F
4m
75.
1)
7 SRI 50 • 1202 - 0951201
+ 44 3
6m
6m
AG. P12.1
SUMMARY
In Ihis chapler. \\c hale learned that In the approximate
sLaIIl;all} ~hanjmret dm
lruclun:... t\'-o I} pes of simplifymg 8IlUIapi
are commonly emph}}ed: I assumplions about the locatlOD
tlon ~lm)(p
and 12 a ~umptjon
about the distribution of fo
111.,,2.2
117
5eclIon 12.3
n j;=!=='==!,dd=!::1s\=g,Jd=;/F
T
I
l.~
4() .-,
A
X()_~D= =iEr_- ~F
12.1 ","",,1112.13 Do:h:mllno: the arp o~imale
sl'le rs lind m(lmCnl for all the memben of lhe
Shll"O mIlt; PI26 thn.)ugh PI2.13 by uslRg the
nu:thlx!
Xm
,N
" 2
Ir
G
T
H
I
12 II
8m
_-rr£=====__.;;F
12 II
D't-~"_
f
E
.i. '-A
c
8
B
... 'lUI, 'lUI
f- 8m-+ sm~
fIG.
-
CHAPTER 12 Approximate Analysis of Rectangular Building Frames
_
-;:cF=~
P12.3
2OtH--;/
FIG P12.8, P12.16
D
: k..:tl
I
(J
I
I
I
I
I
I
40tH
I
__
K~_-rJ,.
~
.,..
-t.E~- :F~- G -~H
4
6._
H
10ft
.BAt
I
n
I
I
I
I
I
81"
A":'-
I
f
[
~
I
1--8 m
Ii=:'l-~F
I () It
40k
~
c, ~
0<
6 m,-+-,--' DR, mm----l
... Pl2.12, Pl2.2lI
16 It
-,~l
:W ft
FI&.
(,
P12.4
~ d:;I r=i~ O lt" /m :db ; J
I
I
T
I
I
I
I
I
:I::I:I:i;:I:Jc'
"O:lIN:"mi:rj
J
I
I
I
I
snk-if=====_4E;..
D
I
£
f
--}-24II--i
I
15k-i---_I--OO+---..
J
K
E
8m
fJ
ft
~"
FIG. P12.9, Pl2.17
/-,-
H
I- 24
~
--+-
_
....1 T
30 kN_-i H~= G.r _
50 kNI'!:" -_~ =.,I _ Oli
+
4m
E
F
4.
~
C~ • .1.
1-6 m-+-6m-l
8m
A - L-
B
12m
20 I.
N.'l2.5
FIG.
'l2.7, 'l2.15
FIG. P12.10.
P12.18
S-
A
F
8
1-- 3011
... Pl2.11, 'l2.21
--+
G
D.........
IIfCT10II 11 1
13
-...- _
....... _ _ II •••,......
. _ . ."
-...-...-
-
I»
111
Method of Consistent
Deformations-Force Metho
13.1
13.2
13.3
13.4
Structures wIth a Single Degree of Indeterminacy
Internal Forces and Moments as Redundants
Structures with Multiple Degrees of Indeterminacy
Support Settlements. Temperature Changes. and Fabncatlon
Summary
problems
13.1 STRUCTURES WITH A SINGLE DEGREE DF INDETERMINACY
Enora
In thl3 chapter \\c sludy a general formulation of the force (fleX!
method called the method of ((111.\';.1"1£'111 t/e!lmnClfio"" for the anal
statically indeterminate !'>!ruclUrcs. The method, which was IOtrod
b\ Ji.lmc'i C. Ma'(\\-cll In 1864. essentially involves removing enoup
straints from the mdetemlinalC structure to render it statically
nate. ThiS determinate struclUfe, \\hich must be statically stable
rerred 10 as the prinwrr S!ru('/u/'t'. The excess restraints removed
the gi\cn Indeterminate . . truclure to convert it mto the detenmna
maT) struclure are called r('t/wuJwlf rl'.\lraint!i. and the reacUons
lerna I force~
a~ociled
\.\-lIh lhc\C ~lniart~c
are termed redundan
redundanh are then applied a... unknQ\\ n load~
on the primary
and their \alue!'l arc dctermmed b) gniho~
the compatibility cq
ba-.ed on Ihe condition that the deformatIOns of the pnmary
due to the combined dTcct or the rcdundanls and the gIVen
loadmg mu!\t be the same as the ~noitarfed
of the original
nate tructure
Smce the independent \ariable.. or unknm.. ns in the method
Mslenl dcformallon arc the' redundant forces (and or momen
must ~ determmed before the other respon~
charactensh
pla('cmenls can be c\aluated. the method is deit ~ad
as afi
In thi" chapter \\e fin,t de\d()p the anl}~is
of beams. Ii
tru s.es \\llh a lOgic degree of indetcrminaC) b) usmg the
510
A
0
To determlDe the m1undant C by ng ~; :
we rem e the roller support C from t ill
It IOto the detenDlDate cantlie' r bea
I rmlnote beam I reti ned I
the 1';:~ P
Iben pplicd as 0 unl<o
I
n
gIVen
Iemalload p
ccanhedctcmUned
unl<n
load C
mg
same
the
I of
the
upport
2k
US(
the prinwy
C
I
10
,:00'_
512
Method of Consistent Deformabons-Force Method
CHAPTER 13
IIfl:IIlIIII.1
"* =_;O-'~J
II
__~-,' 5= !~_
..
I ft
p- J2"
tc
------I
10 ft
--;
> - - - - l . _ ~ ft - ~ - :
E- ~O l k.
.... ~",215=,
I
a
lndet:U'a~
., I
~
, -.;;:-
"
I
---- --- ------B
"',
COD\'eDlenli
by I pen
f tho pnmary beam due
t
ft<ct,on due to tho
1
VldUl1l on tho beam thaI I
e
---
A
.1
'
0
<1
10
10 ExltmalloadlRg and Redundant C
p= -'2 k
-E- B.EI~=-_Al!f
,
AI
- '20
(r:
--------------- :Feo
" =32"
IPrimlll) ma~B
c ~-
Ie
B
1k
Ik
dl Primary lkam Loaded hti~
M
I
120
Bendmg Moment Diagram for Primary
Beam Due to External Loading (k·ft)
=OJ in.
xC
20
+
[ _.,-~OI;: = -
:it
A B C
Bending Moment Diagram for Pritury
Beam Due to Unil Value of C (k ftIk)
'00
-E- I, -A-. !BI~-
_ _4:oi
A
(
kl'l
'"
RtdundAnt C
32k
A -
~1
Subjected to External Loadinl1
________
0-(. IA
L
=
+
=21) k-ft
1.1
A
__---"C
".fll.
I
2k
=10 k
(0 Bendmg Momenl Diaanm
for Indetenninale Beam Ut.ft)
II&.
13.1
which
l1co and <1 repment '": ~I
end C of tho pnmary beam due t II
C
I'b Prinwy Beam SubJ~cted
". ._"
~;:f.
B~am
p= l2 k
"
of the )lnJJIary bea due
and lhe redundanl
m be t
rnmate beam
ppon C
lhe IOdelenm..1e beam
Prt1JllIry beam d
tho
lhe rodundant C m I
be
o
\
a
_
dant C <a<h actmg alone n tbe be m
used 10 denote the deflectIOns <1
nd <1
first ubscript C ,nd,cates the Iocau n f
subscript 0 I used 10 IOdieale that <1
loading, whcr<as the second ubscnpI (
lhe redundanl C Both of Ih deftecll
If Ihey occur '" the duectlon of lhe redundanl
be upward a hown '" FIg 13.1 b
Since the redundant C I unkn wn It I
:~I
D
<1 by first evaluatlOg the deft<ct,on al C due I a~um:te
dundanl C as shown m FIg 13 1 d and then III
lion thus obtamed by Ihe unknown maglll
f lhe
<1
C
1
III
CHAPTER 13
Method of Consistent Deformations-Force Method
lt. nc \~
huJ r l"IU l" .1 l.nt\1I ~d
Ot'ncdlon of A(() at end
hll\\11 III !il'~
I \.I(\"· ~:.IlJS
the: dd1c,:tlon <It C In the original
minah.: ;na~
I It:ll.\ the n:duno.IIH foro:C\ must be of
h131:.!niludc: ((l pll h lhe: t,"no ( b.ll:)." 11110 Its onglllal position by
In!:: ~dn
ur\\ilrd ddlt'I,.'llt1 11 llf \{ I .It em.! C of the primary beaaL
e,;IUJh.: rhe: dkd l.lf ( llll Ihe ocJrn. \\f' l:omputc the ftexibJln
I.:ienc j; { \\ hh.:h I the: ddkdllln at C due to a unit \alue of the
Janr fig I J I J
Sm(c ,Uperptl.. llWn IS \ alid. deflection I
prorx.rtlOnal {("I load: thaI 1'•. I f a unll load cau~
a deflection
then a load lell HOlt: a" OlUl.:h \\111 t.~uac
deflecuon of IO/ce Tb
up\\<lrd rcdunJ.ml of magnitude C cau"c:-- an up\\ard defledioa
C II, at cnd C tlf tht: pnOlar) be.lm. Smce (he up\\ard
(r. cau't'd b~ lhc redundant C mu,,1 be equal 10 Ihe d
dellL'dlon j, () duc 10 (hI." c\tl."fnal load P \\C \Hile
C fcc
pre\lou,l)
Since the primar} be.lIn I ... "taticaJl) determinate the dellleclliI
.1 l () and Icc can be computcd b) either u"mg the melhods p
dC""-Tibed in Chapter" 6 and 7 or by u.. ing the beam-deflection Ii
gi\en m"ide the front cmer of the book. By using the beam-dcllalll
fonnula.... \\C determine the deflection at end C of the primary beam
to the e\temalload P(12 k) to be
5(.12)(20) ,
48(3().OOOj(512)/I44 ~
-0.25 ft
(see Fig. 111 (el) in "hleh a negatl\e sign has been assigned
magnHude 01 6( () to IIldlCalc that the deflection occurs in the do
direction that 1\. in the direl:tion orp~itc
to that of the redu
Simllarl} the fll:\lbiht) coefficient !e ( is evaluated as
j"
~e
oj
lEi
J(lOO(){)
(512,
1-14
0025 ft/k
hg, 13.Jld B) \uh,tituting the c.\prc...sions or the numerical
and f« mto [4. 115, \\e dctenninc the redundant C
.1({J
(5/'/) (lEi)
C
4Xt./
The po\iu\e ans", r I) (.
I'
~
16
P _ 10 k
...
(r
m d It.'<lte:> that our IOllIal
assumpb
the up",ard dlrcruon of ( "'.I~
corre("t
"' Ilh (he rcacllon ( kno" n the three rC'maining reacUo
be
delcrmmed
by a r p)
I 109 tel
h h rl:e e"lulhbnum
..
hod
equations
} of the mdetermmate beam hg. 11.1 e)
I,;
0
A
M.
0
32 10
32
A
0
10
0
1020
.,=,
A
111
k
0
After Ihe redundant ( ha • _
other response characten lJcs of u=>
mputed
~:, ; ~er
the beam
n Ih be
:PIOYJD.g uperpDsUlOn relatIOn hips ImJlar In 3f< nn t
perpo IUon relaUon hlp expressed 10 Eq 13 4 Th
can alternatIVely be determlOed by uslOg the UperposUI
see FIg 13 1 a c and d
A
An
A
A
0
A
C
0
+A
C
32
1I.a + lI. c C
I 10
22 k
320
ole that the second subscript 0 i
eXlernalloading only Fig 13 1 c wherea the
nd u
notes reactionS due 10 a unit value of the redundan C FI 13 I
Slmdarly Ihe bendlOg moment dIagram ~ r the on term
beam can be obtaoned by supenmposong the bendong 10 ment dl
or Ihe pnmary heam due to external loadong onion the bendi
ment diagram of Ihe pnmary beam due to a umt value r redundan
multiplied by the ..Iue or C The bendong moment dIagram r r I
detennmate beam thus constructed is shown 10 F18 13 1 r
Moment as the Redundant
In Ihe roregomg analysis or the propped canul.. or FIg 13 1 a
we arbltranly selected the vertical reacUon at roll r uppon C t
redundant When anal) :lng a truetuT b /h m
!ornuJt,ons Itt' can choose onr suppor', a,1I n , ml Tiki
menl
(21))'
......... Degree .. In....'
0
~ M.
""rd. '\ote that Eq 11.6" <qui\aknt to Eq,. (134) and (13 5
L
LF
LF
~ L II.
.1 c o
m \\hld both ddlcctlon ... ftc and \((). arc a:>:>umcd 10 be JJOSI
5PL'
48£1
IECT1OIll3.1 .-...-
as tht
redundant prOf 'ded thai Ih , mo
I
re. tratnl from the glt n rna terminal 1m
sITU turt that sIan 0111 Jetem,mat and t
CODSldenng agam the propped cantil
which t redrawn on Fig 13 2 a we can
Iha
traml comspcmdmg 10 the honzontal
labcall unstable Therefore A cannot be used
H _ e.ther or the two other
t ppon
the m1tmdanl
Let us COOSIder the anal
or the beam
menl M
the m1tmdant The actual
arbitran usumed I be coon lock
obtaJD the pnmary beam. we rem
the =:. ~
end A by replaCIng the fixed suppon
bi
Ih
~: ' (
iI
51.
CHAPTER 13 Method of Consistent Deformations-Force Method
IEC1tOIlIU Duc1urIo _
p= J:! l-
I
c
--__ --------B
_---- 1I.
tC
8. = 0
"± t'I.
If.
A,
I
rlOlt
r---:--L
The lopes
and f
an be
delleclton formul wide tbe fr •
PL
16£1
10,ft--=j
10 ft -~
E = .10.000 k"', 1= 512 m. 4
L
=
3£1
fa) Indetenmnate Beam
p= 32 k
'd>=0~
9140
I
-----
I
8
----
...... '
A",= 16k
...... Dogroo" ..
_ ......._ _
'1
.i.
t
ote lbal a nesal. e gn has been •••,,_...
because thi rotatl n occun ID the cl
to the counterelockWi ducetl n a
13 2 a By ubsulutlDg the numen
compaltbihl equallon Eq 13 7 w nte
o0075
00000625 M
from whICh
C.o= 16k
120 k fI
(b) Primar)' Beam Subjected to External Loading
+
..... ------- ...... -,
=~A
0
--
M
I k-fl
B
.........
"
C
t A'A = 'RiI k
FIG. 13.2
(c) Primary Beam Loaded with Redundant MAo
Fig. 13.2(bJ. :'<Oote thaI the "mpl) ,upported beam thus obtained
call) determinate and ~tablc.
The redundant Jf~
is now treated
kooy, n load on the primary beam. and ih magnitude caD be
from the compatibilit) condl1ion that the slope at A due to
bmed efftXt of the ex.ternal load P and the redundant M... must
The primal) beam I!I subjected separately to the ex
p -:. 32 k and a UOIt \alue of the unknown redundant M,t
Fig. 13.2 band c respecti\el}. As sho"," in these figures
!tenh the slope at end A due 10 the eXlemalload P whereas
the flexlbilit} coefficienl thai i~. the slope at A due to a unit
the redundant \f.... rhu the epol~
at A due to .\lA equals 9..
Because the algebraic sum of the slopes al end A due to 1b8
load P and the redundant .\1 ~ mU!lo1 be zero. we can e prell
paUbllll) equauon a
PracedUfI far Analysis
Based on Ihe foregomg diSCUSSIon we
by-step procedwe for the anal)'$l of tern U
WIth a SIngle degree of IDdetennlnacy
I. DetenntDe the degJee of indeternl\lllCY
lhe degJee of indetemtIDlICY area
lWe IS mternally indetenttJDllte then end
The
of mternall iodetemt
~: "
WIth mulbple degrees f inde
..
117
518
CHAPTER 13 Method of Consistent Derormillons-Force Method
I - I,', the: rc:dund.1Il1 \\ III imph
IX JlI\C' 111.ll!ilI 1Ull
, • d " •1\ l:Mrcd, \\hc:r....l ....1 neg,lll\e
ume:
,
113 II \.1
nlluJt' \\ III mO ......lll: Ih.ll the dClu.11 ~n'(>
J.
4.
that the
value of the
1\ oflflO!\lte to
.1' um J lmua~
R,'mo\ t:: Ihl re Ir.lIlll corrr.:,pnnding 10 the
b
h
J!1\\.'n lIllkll'mllll.lll ,tfuetore 10 0 lam t c
tructun:
8. Dr.J\\ a dl;Jgr.lm of Ihe ~amnp
...tructurc ",ilh only the
nat huhn!! .Ipplled 10 it. Shh.:h a deflected \hape of the
lure. and h(m the ddkcllon tor slope. at the point ofa
tlllO and In !he' din.'ction of the n..-<lunda"t b} an a
",mho!.
. C\L dr.l\\ •• lh.lgram of the primar) structure WIth
umt \alm.' of the redundant applied to It. The uml fOlQl
moment O1U,1 ~ applied in the positl\e direction of die
dunJanL Sketch .\ deflected ,hapc of the structure. and
<111 ~tanpor t<
s)mbol the flcxibilit) coefficient represen
ddk\:lion (or ,lope) al the point of application and ID the
rcxllon of the Lnadu~r
To indicate that the load as well
rc'pol1se of the qructure is to be multiplied by the red
\hO\\ the redundant pre(,;edcd b) a multiplication sign
to the diagram of the \tructure. The deflection (or slope
IOl.:ation of the redundant due to the unknown redundant eq
the flexibility coefficient mulliplied by the unknown rna
of the redundant
5. Write the compatibility equation by setting the algebraic sum
deftecllom (or slopes) of Ibe primary struclure at the locatton
redundant due to the external loading and the redundant
the gl\cn displacement (or rotation) of the redundant support
al.:tual mdetcnnmate slruct ~.
Since \\e assume here that
arc un) icldin!!. the algcbrail: sum of the deflections due to
lerna I loadmg and the redundant can be simpl} sel equal to
obtain Ihe compatibility equation. The case of support m
is l.:on idered in a subsequent section.)
6. Compute Ihl' ddk·(:tion" of the pnmary structure at the I
the redund.tnt due to the external loading and due to the
of the redunddnt A detllXuon is con,idered to be posmYe
the arne s.c:n'tC a, that a sumed fOf the redundanl The
can be determmed h) usm~
an) of the methods diSCussed
tcrs 6 and 7 for beams \\oIth l:onstant nexund rigldlly EI
all} comement 10 determine (he~
quantities by uSing the
formula ~!I\cn
In ide the front l:O\ef of the book w ~
flecllon vf trusse and fram~,
can be comemently com
u Ill@: (he method of \lftual \\ork
7. Sub IJlute the \alue of ~noitce1ld
(or !lolopes
II mto the !,;ompatihiht) l'quatlon and sol\C'
redundant
b.
520
CHAPTER 13 MethOd of Consistent Deformations-Force Method
_______....n
\/..
\ =
o-f- I,
I
8 =~
\ =
1ECTlCJII13.1
....... __ 11 ..............._"
lIL
M
and
2EI
t
ID which the IlC b e gn for 4
mdlcal
EI
that th
:' ~I
••: : ;
downward dlrectl D that I oppooJltc t the Up" rd d
redundant B
ldl50pport ReactIOn for Indelenrunate Beam
\la mhuJ.
0
IUdtmdant 8
I
UbsblubDg the " ,..."""
I.. mto the compaubilit equation Eq I
ML
determl
o
2EI
\/.
L
8
AIls
The positive an wer for B indicate! that ur m
ward direction of B was rrect
.,
'"
Shear diagram
R II
The mIlalDmg mu:ttons f the indetmnl
detenmned by superpoilbon of lhe reatbons f thc
external moment" and the redundant B hown In FI
",
pecUvely
8
+
A
A
M
Bending momenl diagram
(e) Shear and Bendmg Moment Diagnum
FIG.13.3 cOld.
L
I
0
)lI
- -Ie~)
lI. - lI-
lor Indetenninale Beam
AIls
A
2L
A
M
2
LC2~)
II
3M
2L
AIls
M
2
Ans
The reactions are shown m Fig 13 J d
Solution
Dt'f/fl'(
A, A \f~.
Shrar and Bendmg Moment Diagranu By wung the reactl ns the shear nd
bendm8 moment diagrams for the mdelenDInale beam re nstructed The
diagrams are shown in Fig 13.3 e
Ana
oj Incit (( rmin{/('y The beam is supported by four
and B, It ig. D.J(a): that is. r 4. Since there are only
librium equations the degree of mdclerminaC) of the beam
IS
equal
Primun &um The \crtkal rt:J.l,;IIOn B al the roller suppon B
be the redundant. The sense of B i) avmmed 10 be upYtard as
11' a The pnmar: beam obtained b) remming the roller support •
gl\cn mdctcnninatc beam i hmo,n in fig. 13 31h) Note that the
le\cr beam IS tallGtlly dctennlOalc and stable. ~ext
the pnmary
JC\:ted separatel} 10 the external mumenl \1 and a unit value of
redundant B as ho\\,n 10 fig 11Jrb and c) respective)
hgurc. 6. 0 denotes the defta:tlon at 8 due to the external moment
,•• denotes the tlell.lhJ1l1y l;octhl;lent repreloentlOg the dcfteeuon
•
umt value 01 the redundant 8 Thu the deflection at B due to
redundant 8 equals JuS
(ompatlhllJl Equal, n The del1l'Ctlon al !lUpport B of the
mmate beam I zero SO the algehral... urn of the deftecuons of the
at 8 due to the eternal momenl \f and the redundant B m
Thus the l:Ompattbl1lly equation l:dn he "ntten a
6"
JaBS
0
lkfi llum 1 Prrmar Sturn 8y u 109 the beam-deftectiOD
btam lhe ddkctl n 6s( and Iss 10 he
IExample 13.2
DetcmuDC
the reaenon and draw the
shear:~ : ;cb~a
the bcom shown In Fill 134(0 by the method
the ractlon moment at the fixed upport be the
IIICT1CIIll1.1 - . - ..........
liliiii"'"011'1...._
11
BlFe=!==!:d....lU
I
I
crt
I
I
I
\
\
A.-_
20ft EI
A
30 ft
Indctcnninate Frame
(3)
=
I
t
I
I
I
4H
I
I
I
.~ :
,
.A
(b) Prinwy Frome Subje<1ed '" E>IaMI LoadiDB
M_
+
B ----------------I
=-~
C_ 1
,,
,,
,
,
Ik_ A
~
Ir
(c) l'I-r Frome SSuldljj"".'ID lJai1_oIIPW
........ A:,-· ..,..._ .
FIG. 13.7
I
ClW'TfR 13 Method 0' Consistent Deformations-Force Method
~
UH
IO-)~I
uH
Tangeot at R
12 ..
B .'__
j
I
I~;:r
:0 It
-"Ia.z _ " ' - _ _ ."'.'__
~12~
c
,_0_",,-, -;:._._-_._:0_-_:_·.-11....
_~o
A
ft
r-
714
9~-'14
gtK
B
1:../ - l'lln,tant
l) ~ & d
,a) Inddl'mllOale B~am
'0 l
I
16
I k
andSuppon _ _,for 1,. . . . . . . ._
12 k
A
A,'F."J,I
Cli: ' .= E:f5~;=9,
........... _.. __ .!L..
A_,-
.. .....
,/
TangenlJust
left ot R
Tangent JU\'
right of B
III
_OJ
10
B
B
5719
-Iliapms for - . . AS and Be
(bl Primal) Beam SuhJected 10 External Loading
+
I kAt
A11'F.==r==!f=;
....... __ pl!J- ...
'~" B §,f ~
A
C~,: =
f88R
-----------
B
Tangen! JU\{ to
left of 8
5719
....
fiG, 13.8 (eonld.)
Tangent ju!>t to
right of B
(el Primary Beam Loaded .....Ith Redundant M B
,
"0
"
"
!lQ ~
U
,
!lIUl
EI
EI
(d, ConJugale Beam for EAtemal Loadmg
l-
EI
, i ,
flI.I3.lI
' "i7
fJ
£1
Ji-C
,
ii
Ie) ConJugltlt Beam lor Unn Value of RedundantM.
(b -1Ill MomeDl Diqnm I CoaIin
Bam
ft
-
536
CHAPTER 13 Method 01 Consistenl DefOrmations-Force Method
\\hllh Iss .mJ BR dlilotc thl' . hlfX'~
.It the ~nds
B ~f the
the: nghl '\ri~tn
(l[ thl' i:'l".IIll, fe pc.:dl\c1)
due to the umt value
In
n.'.dund.tnt \f B
.
The: lomp.lllbllll\ ('I.jU.ltl(lll I' ba-.cd on the reqwremenl
IlllX' of thl' d. Ul,; ,:Uf\ l ()f the JI:.tu.11 lI~ctm:a
beam IS COD.
at H. lhat I then: ,... nl' l,:h.lIlgc of .. lope Irom t~uJ
to the left of B
III the nght (If B. Thc:n.:h1fC. the algcor.tlc . . um ot the angles be11W1l.
tanglnt at Ju .. t (0 the: left and at JU"1 Wlhe fight of B due to the
IO.ldlOg .100 the fedund.tnl .\1B mU'.. t ~ zero. Thuo,.
OBfl
+- fss
\la -" 0
I
\\hll.:h l,ln be l,hcd for the redundant bending moment M.
change.. of ,lope" (}so
and f 88 n::1 ha\ c been evaluated.
after
SIOl:C t\ll:h IJf lhe "pan.. of the primary beam can be treated
"Impl) .,uppMted beam. the ..lop..:') at the ends B of the left and the
~nar
C.ln be c,I"'II) computed by using the conjugate-beam method.
conjugate beam ... for the e\ternal loading are ..hown in Fig. 13 8
calling that Ihe "lope at a point on a real beam is equal to the
Ihal point on the core~pndig
conjugate beam. y,e determine the
and OSR at ends B of the lert and the right ~pans,
respectively
rJSI
4CO 'tf-~
t"_
(}SL
, 7
"'/
, d
an
OHR
533,33 k-ft'
£1
Thu" from Eq, (13,8), we ohtaln
420 + 533.33
()HO rd
953,33 k-ft'
£1
£1
The nc,ihility codlkicnt ISBrr:.J can be computed similarly by
conJugate beam for a unit \aIUt~
of the redundant .\la shown
1U(el, Thu,
IBBL
V
667k-ft k-ft
FI
and
IBBR
c-
10 k-ft'fkoft
£1
hom Eq. 13.9. y,e ohtam
fBB
f881
f88R
6,67
+ 10
£1
B~
16,67 k-ft'fkoft
£1
"'DII!'"
Ub!>tllullng
the \aluc:s of II BOrd and f BBrellOO
'I the ~
.
I, I() J y,(' dctemlinc the magnitude of the
equation fq.
\f11 as
~53
33
fT
or
118
5719k-ft
Intlmally Indeterminate Structures
As Ihe foregomg dIscussion mdtc:ate lructures Wllh a ngle deg
f
Indetenninacy that are externally Indetennmate can be an I
b
lecttng either a reaction or an Internal force or m ment
the redun
dant However If a structure IS Internally mdetenmnate bu ternaU
detennmate then only an Internal force or moment caD be used the
uch
redundant because lbe removal of an extern I reac\I n
.tructure WIll )'teld a .labcally unstable pnmary lruclure
Con'lder for example the truss hown ,n FIg I
conSl.ts of SIX members c:oonected logether by ~ ur JO
ported by three reacbon c:omponcots 1b
4 4 the degree of mdetermlllacy f the I
6 3 24
I Because tbe three reac\I
the three equal100 of equiJibnum 0 the
ternaJl inde\ermiDale \0 the finl de
tba
member \ban required for internal stability.
To lIDI1
the \rUSS, we must
memben \0 be the reduodaoL Su
tba
the dia
member AD
be the redlUDdanl.
spood"'l F.
\ben mooved ~
\0
the
member D
no Ioopr
d$
When the priJJlluy
II deli
and gap
member AD
..
ClIAP1'EII t 3
~
.~nI-Fo:e
01 eonll._....
_
Method
,~, -
D
/l;C;===7.ir
q-
C
p
+
A.~=B
(b) Primary Truss Subjected
(a Indetemunatt
Truss
to E"temal Loadmg
-FoForces
D
O.. erlap =fAD
A.D
/~
A
(c) Primary Truss SUbjected to
Umt Value of Redundant
FAD -"AD Forces
fIG. 13.9
mm'
the actual mdetennmale truss we conclude that the
£. D must be of suffic,ent magnitude to bring the ends of Ihe
lIOns of member AD back together to close the gap To
feet of F' D m closmg the gap we subject the prmuuy
value of F' D by applymg equal and opposite unIt .,,,alIOlIdI
pomons of member AD as shown m F,g 13 9 c
ote lba
sense of the redundant fAD IS not yet known and II arbi
to be ten de with the unit ..,al forces tending to elo....
of member AD a shown m the figure The umt value
the pnmary truss and causes the ends of the two po
AD to overlap by an amount I.D AD as hown In FIg I
erlap m member AD due to the ....I force of ma,piitullllI
f
F.
lD
CHAPTER 13
Method of Consistent Deformations-Force Method
JfC ~hI
1\,pc .i1 thl; end 8 \-.1' the left and the
am rl; p..·dl\cl\ due hI lht: ntt:m.ll hladmg 8
,\1.: \'1:11
•
lin
m'
l'i 10
'·UI
r..?" "
I" I fl
60 III
~E
U
:'00 k, ·m
EI
,
--~-
1M
~I
21
Thus
•
The Ill-\Il'llllt) ... oefhl..h:nt I
fig 13 10 ~ \\t: "an 'o("C that
fBI
In
tSIlL
IIIBR
\\hlCh
IRBI
10
'\.nm
,lEi
FI
10
,... - JE(21) - ~
and
1.67 m
Thu'i
In
EI
Ifill reI
167
EI
5m
EI
(1/ th" Rllllllltlalll By !Iub!ltituting the values
Into the compalihililj equation (Eq. (I)), we obtain
\JIIIJllitullt
IBfirel
1 125 • ( 5 ) \/•. ()
U
U
-225kN·m
Rt'(HI/II1J!j The fon:cs at the cnd .. of the members AB and BD
bt,:,tm can no.... be detcnnim:d by app1)ing the equations of eq
to the free hodle, of the memocrs ho\\n 10 Fig. 13.IO(d By co
equilibrium of member .-18 \\c obtain
tlOUOU
t.
GJ 15
(~)
8"
Slmllarl
111
15 1lI
)~ 1C
'2.5 kN
C25 )-97
10
5k
for member Bn
l'i IfJ
D
15 III
(~')
•
(~J)
(21~)
~
CI~)
1275 k
825 k
B) \.:On ad nng the tqulhbnum 01 JOint 810 the \er11cal direction
H
H'
8"
975
1275
225k
CHAPTER 13 Method of COnstlten1 Oeforma1ions-Force Me1hod
F
t
T
~
A
,
,
V
18
\01.
4H
1EC1llIII1U ~ _
I •••
, .....
TMLllU
.. It
n1
In
AB
BC
CD
EF
BE
.\ pand at IS ft _ 54U
£-\ -
... Mol
CF
AE
Cln~at
SF
CE
allodelenmnale Tru ..
DF
E
F
k
F.
£.."i+---I/.__-:&.~=o: D
30
I
BI
2625
45k
CI 2625
.10k
4(1
I
35
Next the ftexlbdily cocffiClent
exprallon see Tab) 13 3
lb) Primary Tru.. ~ Subjected to
External Loads-- FoForce'i
F
t
t
o
o
(; PriIlW) Tru s Subjected 10 URn Ten lie
Force 1Q Member CE. -"c
fIG.
13.11
computed
n
+
E
I
Fone
I
A =40
..
CHAPTER 13 Method o. COnsistent Deformations-Force Method
13.3 STRUCTURES WITH MULTIPLE DEGREES DF INDETERMINACY
Th(' rot,thad 0t ("\)0 .. 1 I('nt Jd('1rnl.ltion .. \~J eloped in .the ""eadil..
uon .. Itlf gnJZ~I.d
..tllIl'Wf(' \\ llh iI. "mgle ~cfged
of md~te'rm iJu. ,
('.1 I" lx' (' r..:nr.kd w tht.: .m.tl) ~I 01 ..~erutc
\\ uh multiple
Inde;('nmnal..·). C\.'n ider. for n.lmpk. the four· pan c~ml1nUO I
Uhll'\:ted III .1 ~lmrof u
dl ..lnhuted Ill.ld ". iI ho\ ~
In Fig I
Th~
~am
I" "uPflI.'1rlcd ~b
1\ "upp<.lrl reactIOn ..: thus I1s degree
tenmn.\(\ ) equal to b _, 1 To .mal)7(' the beam. we mUll
three Upptlft fe.ll:ll0n ...1" rcJund,mh. Sup o~
thaI \\e select the
c.tl rea("lion .. 8 ( .•lIld D ,11 the mtenor supports B, C and
p.?Cti\d~
to bI.' th(' redund.mt . . , The roller ...upport... at B. C and
then reOlO\ ed ff(lm the gi\ en indetemllnatc beam to obtain the Ita
detennmate and ,table pmn....) ~.tm
a......ho\\n in Fig. 13 12 b
three redund.mh .tTl: no\\ trc<lted as unknown loads on the
beam. ,md their magnitude . . can be detennined from the com
condition.. th.1t the deflection .. of the primary beam at the I
B. C and f) of Ihe redundant.. due to the combined effect of the
e.\ternal Io.ld \I' and the unkllQ\\ 11 rcdundanb B\'. CI' and D m
equal to lefO. ThiS i\ uac~b
...e the deflections of the given inde
beam at the roller supports B. C. and D are zero.
To C\lablish the compatibility equations. we subject the
b<am ,cparately to the external load If (FIg. 13.12(b)) and a untt
of each of the redundant> B,. c.. and D, (Fig. 13.12(c). (d). and
specli\d) A....hown in hg. J:~.12(b).
the deflections of the
beam at points B. C <lnd n due to the external load U' are dena
t!1J(J, j,( (J. and j,f)(). respectl\d)' Note that the first subscnpt
flectIon j, indilatt:-. the location of the deflection, "herea! the
subscript. 0. 1.. u--ed to 11ldicate that the dcfleL:tion is due to the
loadmg. The flc\lhility cocllkicnt.. representmg the deflections
pflm.lI} beam ~ud
to unit \aIUt:'i of the redundants are also
u..ing double sUb..Cflph .I'i \ho\\ n In Fig. I ~.12(c)
through e
~ubscnpt
of a Ilexihillt} coelhl:ient denote... the location of the
and the s....-cond suh Tipt i'cta~ldn
the location of the umt load
the detlcl:tion. lor e,ample, the fle;ubihl> coeffident j; B
dt:flect!on at POJllt ( of the pnmary beam due to a unit load at
Fig Ill2 c \\ herca JB( d~note'i
the deflection at B due
load at C FJ~
1".12 dl and so lln. Altcmalhely a ftexlbili
Clent I ma~
al 0 ~ Interpreted .1') th~
deflection correspondiDI
dundant i due to do umt ,alue 01 a redundant j; for example
the ddk"Ction corrt.: ~nid op
to the redundant C due to a UDJt
the redundant B I-Ig 1'\ 12 c IBf denotes the deflection
mg to the redundant B due 10 a unit \alue of the redundant
on. A deflt\:lton or flexihlht} (;Oetlll:lent at the locatlon of a
con Idercd lO be po III\e II II ha the Same sense as Ihat assUJDDCI
redundant
-----0
(b)
...... __ ..
Primary Beam Subjcdcd to ExtmW Loodt
+
A
//-----F-h----F------t;.r-tB
:A
C
D--..::.':%l
....
I
«
Prinwy Beam Loaded WIth ReduDdanI B
+
d Priawy Bam Loadcd
th - -
+
_
FlG. 13.12
..... Loadc wil~
.R....~lI ...t e...... D,
B
1lCI1OII1U
CHAPTtR 13 Method 01 COnsistenl Deformations-Force Metttod
f lXU"'lIll:!
'nllll ll at
llur..It L:
ptlll1t
B of the primar) beam
I I
d
we
the t:".tcrna 03 IS A
I
the Jdki.'lll1n .It III
,-.
:
I", 12( )
1:-' ISb S (} Ig . . -'
c the
I ~ ,., b thr.: d...· lkl'lllll1 dlll' to
. -.
(.1' , I t L~IJ
.md the ddleclion due to D
JUI,.' l\) ( I ~ / B {
,1£
.
d
h
the'hl!,II lklkX"tlon .It B. ue to t e('combined
I hu
lu!I 1 I , c
I
I
'
and
all (ll the. rcdundanb .1'0 d.lao -+ .JBBB
01 me ("\.h:m,1 O,IU
..
h d ·t1eclltll1 01 the: .lctu.11 In etermmate bcaaa
tspD ~ml.:
let:
, " '
f
!'oum 0 the
",'rl'
\\ C' -.et the ali!cbrau.:
LU 2.1 at upro rt B t I
·
~.
. h
• 'am at B cqu.a1
to 7ero [0 obtam
the com
01 t e pnmLIr) ~
.
•
N'llli
due.
_
10
n,
B t (.eC
"DD, - O. :"exl. \\e fOCUl
.I H()
BIl,
b . II I~ .
__
Ie-nil on at p..ltnl C of the pnm,lr) beam; b) alge falea y ............
.md the redundants
and by
"
t ("u'
IIlxuon.1
U L: to the c\tem,llload
_
• •
eqUJlIt1n.
the ...um equal W zao. \\e obtam lh~
second comJ>a:ubihty eq
\
t (oB
!tee ,,,,0. o. SImllarl). b) setllng equal
the algebraIC '-om of the ddlecuon'l of the pflmary ~ a m
at D
the extemal load and the redundants. \\e obtam the third com..
equallon. \DO t (o.B.
IDC C t fooD,
O. The three compa
equalion~
thu~
ohtamed MC
.IB01"
(e.B t
f" C
Procedure far Analysis
Based on the foreg 109 diSC Ion \Ilo
by tep proe du~
for the anal
of truet
51 tent deformauon
1.
1.
leDD, = 0
.1'0 + fc.B. t .Icc C. + fcDD, = 0
:\00 t
hmO, •
.lDCe!,
+- JnnDy
~
0
Since lhe number of compalibility equations is equal to the
of unknown rcdundants. these equations can be solved for
dundan!>. A, Eqs. (I) 15) through (13.17) indicate. the campa
equations of struclUr ~
with multiple degrees of indeterminacy
general ((Jupln/. m Ihe "iCn\C that each equation may contatD more
one unkno,", n redundant. The coupling occurs because the
at the location of <t redundant may be caused not just by that
tlcular redundant (and the external load) but also by some or
the remainmg redundants. Bc au~
of such coupling. the com
equations mu~t
be sohed \imuhaneously to determine the
redundant".
The primaf) beam IS SliltkaH) detenninate. SO its dcftectiom
the external loading a... \\cll a, the flexibility coefficients can be
b) u"ing the method, dl'l.:u..\Cd pre\ 10llsl) in this text. The toW
of deflections indudmg ftexlbiht) coefficients) involved m a
compatlbllit) equationl> dt:pend'l on th~
degree of indeterm
structure hom Eq
1315 through 13.17) we can see that
beam under (ol13lderatlon. \\ hlch is mdeterminate to the third
the compatlbllit) equatluns cOnlam a total of 12 de6ections
flection'S due to the eXlernalloadmg plus 9 flexibility C~oeficnt:'=
e\er accordmg to \I'HIU'1/ \ /em oj ru ,procul deflections
JUI ISf JDB IBO and 1m 1<D. Thus. three of the
effiCient can be ohtamc:d by the application of Ma well
3.
4.
II
111 ."
..
550
CHAPTER 13
MethOd of COnsistent Deformations-force Method
.~
\\ nh.'
1.0Illp.IUhlhl)
danl h\ ~t lng.
lqu.IIH1n for
the 10c",lIon
of each
th\.' .llgc:hr.IK 'lUll {If the dcllectlons or slopee
pnmaT) tf'l l ·{un.: due: III tht: ntanal loading and each
dunJant l·~U.I
to (he "mmn dl pJ<Kcment. (or rotation at
J n • I x:alwn l'll lhl' .Idual mdctemllna.h: '\tructure
n:Pl-ml~'
numba of (omp.ttlbJlll) l'qUJUOIl thw. obtamt..d must be
thl nurnlxr l)f n:dunJant .
6.
7.
<-ompute the ddll'I.:llllIlS and the tlc:\ihhllil) cOhefficlents
In (he l:1.mp.;luhilil) CqU,IUl'lh b) u,mg. t c mel ods di
\lllU !) In thl' Ie t ,lOLl b) the, apphcatlon ~f.
Maxwell la
l:ipfl'l(.ll ddk,IIOIl
\ ddk"l.:lIon (or f1c\lblht) ~ C l D
J(xation of ..l rcdund.ult I com,ldcred to be positive If It
.tme 1.'0 ..('.1 lhal a.....umcd for the redundant.
Sub..Ututc the \.IIUl:' of deflections computed in step
the comr.ltlhilit) cquatiom. and sol\e them for the
redund.lnh.
8. Once the redundant .. ha\c been determined. the other
charal:teri,til:' (e.g.., reacuon\, shear and bending moment
gram ... and or member forces) of the Indeterminate structure
chtlu<tled either through equilibrium considerations or by
pm.ition of the re"'polhcS of the primary structure due to the
nal loadlllg and due to cal.:h of the rcdundants.
Example 13.8
lXtermme the fI:"J(:tlOn, and dray, the ,hear and bending moment
the ·e~rht
pan contlnUOU beam ,hm.. n in fig. 13.I3(a) using the
con I tl.'nl deformations
Solution
D.
n ('
oj Imlt tl rmlllill
\
I
2
Prl11ldrl Blilm The \crtll.:al rcactlOm Band C at Ihe Intenor
and C re pcdl\"I) ar~
~k"l cd
as the redundant.:. The roller
<l.nd ( arc thl'Tl remo\ed to obtam the pnman beam shown m F
I the pnmary he.un I !>UbJCdcd !iCparate'" to the 2-k ft exterllll
the UOit \alues 01 lhe redundant Band (' a' ~hO\ n
In Fig 13 t
d r f'l"(tl\ I)
(mlp,mhlhtl Iquau ns Slnt,; the detle1.110nS of lhe actual
beam dt uprom B Jnd ( .ne ;ern \\e set equal to 7Cro the algebraiC
del1t\:tlOn at pomt" 8 and ( rc fIt.'Ctivel) of the primary beam d
external I ad and t"dl,;h 01 the redundant to obtam the compabbili
A,
IlthB
Jse
C
()
1\1
1/ liB
he (
0
CftAPTfR 13
Method of consistent Deformations-Force Method
""f11
IICTIlIIlIU .......
••
10 ,1111...._.
20
16
II I Ie
8
e
F
tJ
fi
-
B
-20
e
44k
Shear diagram Ik)
Ans.
20
~F
0
16
60
4444D
D
I k
SJwar and Bntding Vommt Dio ,
The
diagram of the beam are shown m Fig 13 1
of
and
Bendmg momenl diagram (k·ft)
As.13.13 wntd.
(0 Shear and Bending Moment ~margiD
erally develop at the mtenor supports of conbn
moment diagram I usually po$Itl e over the mIC1d1e
bendmg moment al a tUnged uppon al an end of me beam m
pe
I generally Regatlve at a fixed end uppon Also the
for Continuous Beam
ment diagram lS parabolic for lhe pans ubjected to : = ~ : ~
and It conslSts of hnear segments r. r pan bjected t
actual values of the bending moments of
depend n
!he loading as weU .5 on !he lengths and lie> I gidi
conbnuou beam
D( 1ft ( lion \ IIJ rhl P"murl' Blum By u...mg the beam-deftectlon €I
obtain
29'333333 k-ft'
£/
3.555.556 ft'
f(1
----
E/
Example 13.9
3111111 fl'
E/
By appl)int:! \{dx,\\ell's la\\
foe
1.111 III fl'
EI
\I"llnJtudl' if the Rt'dllIJdunl B\ ub'Stituting the values of the
and Rexlblhty \:oeth\:lent of the pnm~ry
beam just computed mto
blbt) equation 'F:q 1 dod 2 \\e obtain
2911)1 133 • 15555568
3111.J1IC
0
291Hlll3
J 555 556C
0
JIIIIIIB
:~ : nemo
bendlDr. t th
to th
shown 10 Fig 13 13 f A shown lD thi figure neph
Themshapes
the SImilar
hear
beams.
general are
Ans
CHAPTER 13 Method of eonststent Deformlbons-Force Metttod
IEC1lGI IU . . . . . _ ..... - . . . . . . . . . ."IS
,
m
1) ~3
p
U
tI
A
EI
ft ~I ...~.
A
.----1...•• _-i:o __-
-_,~
.-p
o
la
lnll
,1
..~- ~-_
82500
111.
Imwtanl.:0U
and 1.
1 tnr (
f
If~
~I
\\
kN
;Q.~I
I hl.: rern.llnlng f .. lh~ ~
III 11(1\\ he lktcnnmcd b
three .. qualJlln ot yUJllh(lUlIl III Ih.. Irl.:C hl)dy nl Ih\: md I mll
l ~,
.
0
I~O
"
I.'
I~
<
\I
"
1'0
\I
\I
\
, nd&
\I
I.
,
411714
c
...
I
I
Primary ..... Loodod with RedundIlII'"
+
0
A~-"'
~
r
I. 10
k
I
~ 'i:--------- ·-···-7"fA"t:=>c
r;;
0
120
.....
+
b 1ft
R,,11 flOII
~ -I
•
t ",
I
b Primary ..... Subjected to Exlcmal Lood
"'111 '-"XI
H 111(
Suhm' I &.I
=-~Cl7
El-_
EJ
1
~-_!"'
I---L
H1Hm
1'1111(
b_---\
m
11 I. f
I
Primary ..... Loodod _ , t
,. ,111110
~
1115
II
510
CHAPTER 13
Method of Conilitent Deformations-Force Method
ApT~E
"
-
IIIl:nalIIU ••_ _
,,
,
'lusw_ ...............,..,.....,
'
~(
~L-l+I
L
I
OLE
I
£1 = constant
(a) Indetennmate Beam
.
0*
C
lie
I
C.
---_ ... -
.
E
D ......... --- ...
I
I
D,
Eo,
(b) Substructure for Analysi
w
( f-:.l!t'D~LE
2..L'
IC
--l!O____
------
2wL
--
Am
(e) Primary Beam Subjected to Extemal Lood
+
L
( Ic
II
______y-----tl111
---]
<too
D
E
I
(d) Primary Beam Loaded with ReduodaDl D
+
2L
( h :--=--:';--::.--:....--!-.d-_ !ED/~
.........
.-.'.'
C
... ,3.,8
II
Ie) Prinwy Belm Looded with Red'mcI...
FlG. 13.16 (eontd.
-..---10<0_--
_ ..-.._D$
IEl:I1aIIIU
-, ,
_
c "'l
,
A
-
D
t,
I
II
D.
I
D
JOII--i
Indetcrmuwe F.....
II
~"
I.050k·1I
mind Ie lru...,> ':.111 he uClcmullcd
1050
In
In
Llble 13.4 and Fig
I
I
I
'xD
I
l~k
•
H
,3.,3
"
(e) Prinwy Fnme Subjocted 10 UOI' Value of , ....... D
Determine the rCdllums and dr,,,, the he.lf and bl:ndmg moment
the fr.tm h \\0 In fig Il IS OJ by ~hI
mcthlJd of l:omistent defi"onDal'"
+
Solution
Dt r
Ind( It milia/,.
P,mar Irum
,
The real.:lIon
0
\ B
\
:!
f)
and D) at the hinged su
JOtfl~A
kxted 4 the redundant I he: hmgce.J uppon () I then removed
rnman kame h I n In I It; 13 I h
Cll.t the pnmary frame
Tald) h the h::mitllo.ujlng dod the UOit \aloe of the redluotW.. 1
D
ho\\o In Ilg I' l~ bland d re pectl\ Iy.
if/II I "
ling that the honLOntal and
L1 al mdeicrmmalc: tram I the hmged upport Dare
(mpul
I Ih
rnpallbl II
I
quail n
.- .-'- "
It
FlG.13.18
d rn-y
.- ,,"
I
,,0
\
\
C
\
Dr
.
-
flier
I
I
~
-
I
D
c
--- --fD'l.DXI/
& ......
60
+
B
The member rorcc\lhus obtained <Ire ,ho\\1l
r
I" I
60
Ib) Primary F..... SubJccted 10 Ex..maIl.oodu\i - Mo _
the n:maming members of the I
USlllg the \upcrpoMtion relationship
~h
~
.0
IOk-
\ft mhcr 4 'wI Fon l" The fon.:c\
117
£J._
I
I
_
,.......
..
r
IEcnOll 13.4 "_1011111..._a"_ lemp11"11ur1 CII.g••,
.....
CHAPTER 13 Method of Consistent Defom1.tions-Force Method
570
A
SUPPORT SETTLEMENTS, TEMPERATU'!E CHANGES, AND FABRI_CA_T_ID_N_E_R_R_OR.:.:S:-_.J
A
o
13.4
I
o
Support settlements
Thu, f.u :1\ah~"
um"dat.·d th~
.m.I1)"':-o of structures with unlYieIdliIit·
d'...... u -.('..1 111 Ch.lpll:r II. :-oupport mo\ements due to
SUpp"l rt ,. \
. . ..
fllundalwn and the IIkl: m.t) mducl: significant. s~ert
In external
In their d e s l g n : ~ f
detemlln.Hc ,truclure .md must he clnsder~
~t lC'men s.
hl \H~ 'r
do not ~\t.h
.111) dftXt on the stress condi
~truc
th.1t ~na
mtern.i1I) etanimr~.hed
but externally de
Thi~
lad of etk,;:t 1:-0 dUI; III the fad that the settlements cause
:-otrud e~
III displ.H.:e and llf rot.lIe as rigid bOOies without cIIIuqlilii
their sepah~
The OlC'thod of consi...tent de.formati~ns,
as developed
the prel.'eding .snoit~
can be ea:-oll) modified to mclude the effect
trop u~
settlements in the anal)sis.
.
Con..ider. for example. <t tv.o·"pan continuous beam subjected
uniform I) di~trbue
load \I. as ..ho\\n in fig. 13.19(a). Su~
the ..upports B .and C of the beam undergo small settlements A. and
respecti\el). a" sho"n in the figure. To analyze the beam,
the \ ertical reactions 8. and Co to be the redundants. The sulJPOJ'tl
and Care remO\ed from the indetenninate beam to obtain the prj....;;'
beam which is then subjectcd separately to the external load w and
unit \alues of the redundants OJ and C, as shown in Fig. 13 19(b
and (d) respccli\ely. B} rcali/ing that the deflections of the actual
dcterminate beam at supports 0 and C arc equal to the settlemen
aod ,1c. rC5pectively. we obtain the compatibility equations
.&:
I I I I
A
I I I I I}
C A _-D~
8 ABO
--------
(b Primary Beam Subje<ted 10 E wnaI
we ca.ii"
A,'
Ji
+
--------
----i-- -.
n
............ D
(c) Primary Beam Loaded With Redundant 8
+
A
._-r;~·,
Jl'
0
8
Ct
I
v.hich can be sohed for the reduodants BI and C 1 • Note that the
hand sides of the compatibility equations (Eqs. (13.18) and 13 19
o~ longer e,qual to ~ero.
as in the case of unyielding supports COIwll
III the pre\ 10US sections. but are equal to the prescribed values of
ment ~t supporb 0 and C respecti\ely. Once the redundants ba
detennmed by sol\ ing the compatibility equations. the other
char cle~stlc
of the beam can be e\ aluated either by eqwbbri
superposition.
Although suppon settlements are usually specified WIth
the undef~
~it on
of the indeterminate structure the 1IIl. . . .
of uch dISplacements to be used in the compatibIlity equalJOlll
be measured from the chord connectmg the defonned postti
supports of the pnmaf) structure to the deformed postllOD
dundant
suppon A ny uc h support d.splacement IS consic\eIWd
.
poslt1\e If It has the same sense as that assumed for the red
8
FIG. 13.19
d Primary Beam Loaded willi JtedundaDI
A
C
1'_-,,._...--
. . . .,,1_1II1II...". 1II1II1.....
,
•
............ AM
"
t
---_ ....
t
c
CGmp1111IiIiIy eqoB.,
1..B.,+locC _4.
laB +1cr: C
.. lUi
-"at
tho _bbility equabODS instead of tho •
..... Ac This IS bocause only tho c1ispl.. !MiI"
cause
.0 tho beam In olber words, if tho
wooId have oett1cd ..Iber by equal amounll or by
deformed poIIbODl of all of the supports would iii
Ihon tho beam would mnam slraisht WIthout bon....
would cIooeIop m the beam
_001
DeImnine tbc
and draw tbc ....... and bondiaI
the three-tpon CODbDUOUI beam Ihown ID F... 13 1
dimibutod load and due 10 the support :~;.ItnoalO
and ~ ID .t D U.. the method of _
d
MM_
..........
...._lyaMIyzod m Eump!t I
ruction••'1 the roller 1UPJlCl'.:.:.-""
.. IJ.
JIl1IIWY beam
....
.)l(Irjl.... ""....."" Tho
~
upport 1OlIII.....
_.-conned
"can bepClIIIIOII
_ that the :....: ...
~:
of the
__
C aad E relalift 10 the c:bard
thai
04m aad
578
CHAPTER 13 Method a. Consistent Deformations-Force Method
(
t"i lUI mm-)
[)
OX
E
"
1'(IO~l
TlIIU lUi
M
GPa
~O(l
06
(
•
AB
CO
A(
BO
AD
Bc
..
0-
I
•
0
((:1 Pnmal) Tru.... SUbjected to
Ten de Force in Member AD
Fon:e!'o (kN')
a Inddemunah: Tru .
lOT
t
0
8
m
C
•
C
D
32.067
2·t.05
At =O¥~_-:=
-r
A
__~
32.067
F
The member ~of
o
It'll ~mnP
o
B
Tru\ Subjected
to Temperature Change..
SUMMARY
fiG. 13.23
numenr.;al \aluc'i lIt these quantillC\ an: tabulated in Table 13 6
1;; deh:mllncd to Ix
.1 WI1
192 rnm
'\41)0
C\t the flexlblhty l:oethl.'lcnt
Uln see l ..bh: 13.6
pre
Jto
An 1\ computed b) u~mg
the virtual
0,0479 mm
\111 nlIllllt vf lIlt Rlc/mu/mrt lJ) ub-.tituung the \alucs of '&"DO'''''' Jr.
mto the compatlblhty equation I::q
I '12
I
lJ 1I479 flO
f.
\I.e ubtam
0
40084 kN T
thus btained
F
Sf
•
_
l'llhcr
~n'(l
thn'\l ~h
\,j
_13.1
13.1 ........ 13.. ~IC"
--
101) l
I
I
8
(
3m
m
E
I
= ,on~(at
'g
FIG. P13.2, P13.6
U.5 tIInIIIIII 13" Dc:tennme the reactiom and dra\\ the
h r nd hendmg m lment I.hagram for the beam\> hu\\n
In f I
PI I PI '\.1 h~ usmg lhe: method of conM"tent de~ rmat n
lett the re ell II momenl at the h\e:d uppon
t be the redundanl
13.13 through 13.25 Determine lhe reaction and draw the
~hear
and bending moment diagrams for the structures
shown in Figs. P13.13 Pll2S using the melhod of con Is.lent dcfonnations.
FaP13.16
I
'12h
C
L+L
----+- U2
£1 = consrant
I
12ft
12ft
t, =29.000 k!oi
EJ-_
1=1500in-'*
RG. P13.9, P13.30, P13.50
fiG. P13.13
--"C
I.S klft
B
~
SOk
120 k.."
" l~dIb
c
A
- + - - - 6 m ----I
£1 =-
£=<00",,",
Fa P13.17
SOk
- . rD =~bJ B
6m
lJ
AG. P13.12, P13.33, P13.51
the redundant
1l:'ri'
l
12m
/
I
10 m
2~m
1
£ =200 GPa / = SOOf (06J nun'
ft
13.1 tIIroUlh 13.12 Determine the reactions and draw
. . hear amI b<nding momenl diagrams for the beam
m '·ig... , PLl9 Pl112 U'iing the method of con
fl)nnatlon .... SclC\:t the reaction at (he interior support
EI-l"omtant
... P13.3, P13.7
A~
)m
"'---L--I
i---bm
A~
FIG. PI3.4, P13.8
FIG P13.1, P13.5, P13.49
m
... P13.15
AG. P13.11, P13.32
25 kN/rn
1--10
- - - - cO It
£1
I_l ~"f1(lt m ~
"
£/
l
-' kltt
D
~UOGPa
,,
60k
~
I k
8
PROBLEMS
m n the readlOn and dr.l\\ the
bendl
m m~
t dlagr.lm for the Nam h. \\n
PI' I PI'.1 U 109 (he methl'ld 01 ...on I It:nt de·
tl
I! h
dl n ,lI the: roUe:r urp.:ln 1(-' he
the red ndan(
SOk
or
l"3l.'h of the rcdunJ,lIlt
60l
50 k
lillo.' ".1lkICrmill;tlC ...tructure ca~
be ev
CLjuihbnum «'Ill Idcr,ItIOn" or b~
~UJlCrpoSUl0n
ollhl' r l . r~
,trudun.' due to thc c\tcmal loading and
n:"!,('1n"e t:h.lr.ll.'tcri ...Ul.'"
= lb=: ~
PI.'"
CHAPTER 13 Method of Consistent Deformations-Force Method
B
10 ft
con tant
~- -tfO2
£1 = COOItanl
£J = conSlallt
FIG. P13.10, P13.31
__
FIG P13.14
... Pl3.1•
111
_
CHAPTER 13 Method of Consistent Deformations-Force Method
PI UI.
10k
I "Ill.
E
20 fl
\
.
18
'\0 It
E/
10k
=l'n~ta
1-_ _ 20 II - - - - - - j - - 15 f l - - l
FIG P13.22
'pane
fA
/;./ = I,;on tant
fIC.
FIG. P13.24
=
tant
j--151! - + - - 1 5 f l - - I
2k1ft
C
8.
D
C
P13.27, P13.52
.D
~
1L~
~A = d;, : = ~(
1011
8
""l
I tilt
90kN
c
• A
E
~J
15 tl
I
£1 =constant
2 kfft
A
FI6.
P1U8
F'G. P13.25
D
r
IS h
13.26 tbrough 13.21 Determine the reactions and the force
in each member of the trusses shown In Figs PI3 26
PI129 u~mg
the method of consistent deformanons
E/ = ctln~a
FIG P13.20
3 kllt
Ltnl
m
,.
8
1----20
5m
30k-~.f:
A
fIC. P13.21
A 6
D
AG. P13.19
E. =: CUll
• fl
29000
f&P1129
60
40k
rt --+--1
1----2 panel..! 8 fl
5fl
2J------j
EA
E =conslant
FIG. P13.23
FIG.
P13.26
16 fl---1
constant
..... ,3.1
'UI ....... 'U1
1«tJDg Ibe
Ibe
UPdaD
-
1M
ClW'TBI13 M....od .f Con"stent Deformations-Force
Method
'11'..
S{lI..N
1J.M . . . . . U . . ll<
,
mm
30k
15 k
I
I
h
-~
5m
EA
.-.....I'j$..
5m
16ft
B-1
, I:h====a'
I.:! It
•
FIG P13.39, P13.54
EJ
B
FIG. PI3.36
k
-
8ft
It- f1
-
flI.
and dIa
13.3
A
45 Determine, the reactions
13.37 UIroUIII13.
oment diagrams for the
rJ Ix-ndlOg' pm Pll.45 ll'.ing t he m ethod
hear
an
hl.)"n 10 f 1£'_ PD..
.
E:- .. tXlOll
Pl3.34
I~nl
6ft-jt----
deformatlOn~.
flI. Pl3A3
501.
:!5l~/m
FIG P13.40
B
Am
8m
1001...
50k
E=70GPa
A~CjfE
RG. P13.37, P13.53
Am
2k/fl
~30
B
I
50k
D
I
2kift
F~
fl-+15 ft+15 ft+15 ft+15 ft+-30ft-j
£/ =constant
!Ok ft
FIG. P13.41
--.---
3m
f..A = con lant
flI.
Pl3.35
r-~B
===iCr--OD;J
25 kNIm
I klft
21<1ft
A
B
10 ft
c
+75kN
B
EI
Jm
IfIfl
10 I,
.J
E.I :;: c:onstant
fIG.
D
I Jm
Pl3.38
A
EJ
9m
RG P13.42
_'l3M
,,_III
D........._-1~I =
.........
21
... PlUB
•
11M Sol Problem I 1
_ I of 30 mm al oupport D
.PlUI
~
11.11 Sol Problem 139
PI3 9 aad • IOItIemenI of I
B
1U1 SoI1e Problem I I
PI I aad tho oupport lOlII_oi(l
at" aad 40 mm at
1. . SoI1e Problem I
PI
aad tho 1UpJlOJ'I.aIo.....llaC
aad I ID atD
IrN
-
a••
60tH
m
1. . So... Problem 131'1'7"'_
PI
aad tile IUpJlOJ'IIIIII.....irJ
mmatC
11.14 Sol Problem 13 39 fI
PI3 39 aad tho ouppon oettI_1IIM
3IDatHaad IolltO
14
Three-Moment Equatl
the Method of Least W
14.1 DIIIvIIIalI of T111l11-Momenl Equation
14.2 AppIkIlloll of TIIn1t-Momenl Equalioll
14.1 IIlIl/Iad of LRSl Wurk
SUmmIry
PrubIems
:clI t~= -
In this chapter we COI1Ilder two alternate ~
lIcxibility method of anaJy... of talically 1D'''1lii
the I ~
nDi~
and 2 the IMthod
The thrcc-mOlllCDt equation wbich _
Clapeyron ID 1857 PfOVIw a con_ tool
beama The thrcc-mOlllCllt equatlOD NJIi
to,
COIIIllBlibility c:oadilioa that the slope of the
an _ r IUpport of the COIItiDuoua
IX
: : :..
~
momrnIs-the
IDd
at the two bcmdm,
~.
~
at the lIIleri«
....
arc tnaled
~a:the md, !" '
l1IaI
JIIlIied
at
III
.......bhtllty cqua
the
i;~' ~
........IS_'_...,_ ..-.. oIl.1at .....
,.
w
B_
(Ie_ _
,.
w
.
w
CHAmR ,. Three-Moment Equation and the Method of Least Work
11C!UJ11141 . - . . - 01 _ _ .......
upport
I.. ')
/.1
..,. 24-£f
II'
rquullUn
Ulan 01
w oblam th
ral
~
f I
th,
-
.",,,,,,nt
1'1 A I
6H
~
In \\ IUl.;h the ..ummJIJOIl '1lg~
h,n e been added to the first term
nght 'Ide.. of the ..!,; equ,t1I~ln
. "tl th<1t multiple concentrated loads
arr1it.-d hl e,tch 'p<.111 in..tc.uJ of a 'lIlgk concentrated load as
In Fig... 14. J J and b I'M . Implicit) ). \ ... ~uo nlt oc
beams us,ualJy . .
loaded \\uh unif('rml) dl tnbutcd lo.td, o\er entire spans and _ _~.
trated load... the ellect' of onJ~
the ...e t\\O types of loadings generally
con ..idered In thc three-moment cquallon. Howe\er. the effects of
t) pt.' ,,"If load.. l.<tn bt: mcluded. 'ImpJ) b} adding the expresu
..It.. . pc.. due to the:-e IO<ld .. to the nght ..Ide, of Eqs. (14.6a and 14
The ...Iope.. 0 , and O,~
of the left and the nghl spans. res'''''Clht&;
due to ..upport -.elllemen". can be obtained directl) from the deft
po.. itltm of the ... pam. depICted in Fig. 14.l(c). Since the sen1emen1l
as umed to be mall. the l)lopes can be expre!\sed as
I
(I ,
c\
c\, - c\
C\.
L,
I
The ...Iopes at cnd.. (" of the left and the right spans. due to red.._......·
support bending momen". ,Hg. J4,l(d)), can be detennincd COJllII'c(
nientl) b) u...mg the bcam-dcncction formulas. Thus,
OIl
II, ~
.\[ L,
.W, L
6t:I,
3£1,
- +--
\1, L,
H,L,
31:.'1,
6£1,
---- + - -
in which \.f,. \f and \f, denote the bending moments at supporU
and r. re\pecti\e1). As shown 111 Fig. 14.1 (d), these redundant
moments arc considered to be po\iuvc in accordance with the
cOllum/ion that is, when causing compression in the upper fiberl
tcmion in the lo\\-er fibers of the beam
81 sub (iluling Eq,. 14.6) through (14.8) into Eq. 145
the compatibility equation as
4
m \\:hlCh M
bendmg moment at uppon
heIDS COll5,dered If M
bendIDg mo
to the left and to the nght of
pec\J
E
L L
length of the pan 10 t
fl and ,
livelli
moments of menta f the pan t t
of c r pcctlvely, P P concentrated load a I
nght span rcspec\J el) k or k
ra'
f th d
from the left or nght support to the pan len Ih
distnbutcd loads applied 10 the left and the nght pan
A
settlement of the suppon
under con uieratl
settlements of the adjacent suppons to me left and t the n
spectlVely As noted before the support bendmg m men
ered 10 be positive In accordance with the beam ( I I 'ion-lIlllt
when causmg compression 10 the upper fibers and ten 100 ID the I
fibers of the beam. Funhennore the external I ads and uppert
tlements are considered positIVe '" hen ID the downward dl
I n
.ho"n in fIg. 14 I a
If the moments of inertIa of two adjacent pan r a Dtmu
beam are equal (1 e I
I, 1 then the thr~m
ment equa
pllfics to
PL'A 1- A', II I.
6El- - t 24El
L .-
4
C\,
c\
L,
1/ I
If l_ + ~,!
61:.'1,
3EI
+ M,L, _ 0
1£1,
61:.'1,
B) simphf)lI1g the toregomg cqualton and rearranging It to
terms contammg redundant moments from those involVlDJ
cttAPTER ,.
Three-Moment Equation and the Method of least Wont
\I
\I
4\1
H'/~
... P I k
f
4
"
I
k
Off"
f
The: flln: 'lllllg Ihn':C:-l1llllllc:nt c:qu.ltlon Me applicable to n
urMrt
( .100 r, 01.· oj l'llllllO.UOU' be.am pro
.
, I\e:
...Il:\'~nl(
t·
.
Ihere .w: n,l :eIlJU tnl(~IO
. !i.m:h .1' mternal hmge.... In the beam
the kit urrort .1l1d tht' n1!lll uprx1rt r
14.2
APPLICATION OF THREE-MOMENT EQUATION
-The follo\\ ~m
~b·ret'
tinuou lx.iITI ~h
I.
2.
3.
4.
5.
6.
-...tep prOl.:cdure can
Ihe three·moment cqu.illon.
Sdecl the unkno\\ n bendlllg moments at all interior supports
bc:am .t... the: re:dundi.lIlh
B) tn.'.ltlllg cach Illlcnnr support successive!} as the In
ate ",urr0rt ( \\ ritc a thret>momcnt equation. When wntIDI
equatllln', it ...hould be re,lil/ed that bending moments at the
end support!'> .Ire known. For ...ueh a support with a cantil
hang. the bending moment equals that due to the external
actlllg on thc cillltile\er portion about the end support. The
number of three-moment c"Iuations thus obtained must be
the number of redundant support bending moments. which m
the onl\' unkno\\ m, III the,e equations.
Sohe thc ,),tcm of three-moment equations for the unm
pon bending momenh
Compute the span end ,hears. "'or each span of the beam
a fn:e-bod) diagram ,ho\\ II1g the external loads and end
and bJ appJ) the: noitauq~
... of equilibrium to calculate
fon.:es at the t:nd.. of tht: nap~
Determine ,upport rca tlon~
b) con..idering the equlhbnwa
uppon joints of the ~am
If so de Ired. dra\\ !.hear and bending
~am
by u..illg the hwm H"" nJllt tnliml.
Fixed Supports
The three-mument equatulll . a 1!l\Cn b} Eqs. 149 thro
y,ere utrl\ed to atl Iy ,hI." I.:ompatiblilly condition of slope
the IDlenor upport 01 contlnuuus beams These equaU
-
CHAPTER 14 Th....-Moment Equation and the Method of least Wont
,,,
,
I
, =-
- ' 1 U ''fa .... II TIll. 7
IoIUl1an
I
_B~
'0 oJ
......
•
"'(}!l
B
= h:!.1
(ttl Span End Momem<' and Shears
\(1
-1
:!1H..
k
2.5 klft
.Q..B
= tl
B =62.2k
Id Suppon Reactions
32.6
10.4
D
A
-9.6
Shear diagram (k)
F,
B
Bill ••
Bendmg moment diagram (k.-ft
III. 14.3
(d) Shear and Bending Moment D'I;·, - -
_icoI.__ ..,......
CHAPTER 14 Three-Moment Equation and the Method of Least work
ED""" 14.2
l"ktl.:nmnl.: tht: rc'h.U n h,r lhl.: t,.t'Il11I1Ul'U' lx.lm hl\\\n In Fig 14
the umlOnnly Jl lflhutw )l' d .mJ du\: tlllhc ~ur l t
:'>CttlemC'nLs of 10
'0 mOl at R ~n mOl ,il ( .wd·1O mm II n L'il; lhl.: thnx·momrnt eq
10m
10m
IlJm
II
='
,-on lanl
1-7IX)(I tlmn '~
E=lOOGPa
1~ 5
)1""
rr:D:::i:D')-
'I
8
A - 1'H5
I
llbA
~8-J
(
I
B,
o::iII::o')
1
8
1152\:ft:J I152
8'" = 1"1 5
~)1.2
II""
C
14512\Ji:~
e"' = IX)"
B,H' = 11".4
=~7.9
l~. b
~C-J
2.15~
195.1
I
C =
kN
(("....t-.;30L-.....,m..l....J:;;
(
C;IJ= \95.\
]7~.
(b) Span End Moment' and Shear,
:\OJ..N/m
A =0
~')
L
I
, = I]~
H
I~J
8 =
9 'N
C
t
= nXHr<
I
D = 104.9 kN
(\:) SUpplll1. R..:'adl<m
RG.14.4
SoIulion
R( dundalll
and (
The bendmg moml.: ~tn
\fBand \f( at the 1D.t~"
.JTe the redunddnt'i
Of Jomt B 8) consldenng the lIII(IIIIln
re pectl\ely and ubstltutmg 1
10 m £
lhw.'- '10m nt Equatton
{~md,
and (d
2.00 10
m I
tUO III
mOl"
700 10
m" M
Iflmm OUI m -\
As "Omm 005m A
002mdndP
p, (J Intol:q 1411 "('"Ole
~
...
r..:' pet.:tl\ct
~l
k
600
CHAPTER 1. Three-Moment Equation and the Method of leasl Work
1ECl1CIII14.3 _
It ~
_
..
The bend..
-
IOf!
-
-
Inll
10ft
...
II
II
EI ::: con..t,ml
I j, InJ.Clcnmnak Hcam
II
-
~OI1
IOU
lOft
(b) Equnalent B(am 10 he \n.II)/eJ by Thr\'C'-Moment EquJtion
9 ,
','\ I
2475 2K.75
1--"'1'":T~,"f9
.. ..
142 5
J.;12 S
~:T-' " =;r ,.
90
I.H kilt
=53.5
(t:)
Y7 ~
b.·ft
14.3 METHOD OF LEAST WORK
25.25 18
W
I
90
2025
B,
AIls
AIls
__....;'_-..J-l_,(
I ! !
..,bB
IOtt
k
AIls
-I" k
(}
9
90
C, = 43.25
Span End Mnmcnh and Shear,
).M
C
"'.Itt
j--......l.._-4;.L.LLJ...Ju...J.....L.I....L.1.-1-ci:i..1..J
c
I
·n.2S"
o
1t'---, I '- B. ~
t
--_ ....
B,
FIG. 14.6
FIG. 14.5
Equa
Thrf:' . \ttln nt l;quutlon at Joint B SImilarly,
upJXln A B and ( \\ \\ nle
"I 20
2. \.1,20
45 211
,\()
I 2 I
\/( 30
beam
102
CHAPTER 14
Three-Moment Equation and the Method of Least Work
the Jl,·tku~'I
.tl
the J1i.llllt of dprli~at on
ilrrl ) mg CI'.llgll.IIW
"l, '\:~ Ild
of tht.: rt.:dundanl B
Ihr.:orcm. \\c can \\ nle
,L o
,8
It hlmld be: rc.lh/t.:J Ih.ul:q. I·U~I
r~p csent
lion In the dirl,,"l.'thln \ll redundant B,
the compatibility
and It I.:an be solved fl
redundant
['I 141.\ IIldil.:atc . the first partial deri\ati\c of the ~;' :. ~;
\\Ith re fll..'ct to the: rt.:dund.lnt t~um
lau~e. b
to zero. Thi
that fllr the: \alul: 01 thc rcJund.lIlt th,ll saustie... the equation
.\
erg~
or
hbnum and n1rnp,ltihillt) the tr.lin energ) of the structure IS
mum ,)r rna IIllum, Smo: for a linearly elastic structure there
m.n.lmum \ ,llue of "tram encrg~
becau"c it can be increased ill
b\' inl.:rc,1 ing the \ alue of the redund.tnt. we conclude that for till
eul~\
of the n..:dund.tnt the slrain energy mu"t be a minimum Tbia
du IOn t kncmn as the: primiplt ol/tmt lI"ork
The moqlliwdt'" oltht rUl/llldalll5 ora stutically indelermmQ
tlln mInt ht \/1( h tltat tltt' .Hraill Ult rqy .lwrt'd in th(' \trucl r~
mllm ,. t (Itt i"(t mal II ork dO1/{ i.\ fht' /t'mt
The method of least \\ ark .IS described here. can be easily ex
ttl the an.ll},,!s of slructure" \\ith multiple degrees of indetemuD8CJ
struclure i~ ind~tenm ate
to the mh degree, then n redundants
kxted. and the ...train cncrg) for the structure is expressed in tenns
kno\\ n c"ternalloading and lhc IT unknov.'n redundants as
tdeliaill4ii
{
nu, R, . R, ...
. R,,)
whil.:h II' rcprc..,cnh all the known loads and R I • R2 ..... R" denote
redundant ... .tx~
thc prinCiple of least work is applied separa
each redundant h} partially differentiating the strain energy ex
(Eq. 14.1411 \\ith respect to each of the redundants and by set
partIal dcri\atl\c equallo lero; that is
III
tl."
tR 1
tL
t R~
,L
,R
0
0
0
v.hKh repc~m
d ystem of IT Imultancous equation
n:dund.mt and t:an be sohcd lor the redundants.
1 he procedure lor the an ly~is
of mdetenninate truetunII
method or lea t \\ork IS diu trated by Ihe following examp
&04
CHAPTER U
Three-Moment Equation and the Method 0' Least Work
_1403
1(11 nno
900nB
0
lUll 14.1
(rum" tll...i t
r;F
0
1~
'0
~
0
~Jf
0
~_
1:\ b.
H
T, detenmm' the
the eqUlhbnum ~qu.:I on
_II -
n: II. two
flg I·t
n;ma ~
,
IS
If, 10
II
...
...
()
IS
If, In 15
_~OJ
()
If,
"
...
...
Eumple 14.5
F
[Xh:nmne thr.: reddl\)lh fOT the I\\o·...pan continuous beam shown ID
b) the mt"thod of bJ t \\ork
,,
A,=O~£
AI =245 -0.5 B)
,_se_
SOkN
30 kN/m
1l
W
!
To de1mn1
D
I ~ xljD,=13'
tB,
!--IOm--+I-Sm-+-Sm
AG.I4.8
£1
245
The bl:am i
A
8 1 , and D
I,• I
\I
'E·- ih
A,l,:C rdmg to the pnnclpk
(lllt'ast "ork.
d
,8
I',-8
(,II)
If
lid, ~ O
Bel re \'oe ...an bt.un the equation lor bending moments M
the feat:tlon at the UplX'Tl ... and n of the beam in term
Appl 109 the three equlhbnum t:quallon v.e \\nle
LF
0
AB BC and CO The
I
£f
urported b) four reaction!'>. A
L
the cquau
... - 0
ID
Fig 148 and
of B are tabulated ID Table 14 I
"the den
With respect to B are naluated. These dcri t
ofTabl.141
By ubsUtullng tbe expresstOD ~ r \/ nd
Solution
Sioc:e
By
o
o
equauon are shown
=constant
onl) thTL'\." cqullitmum equiltlon.... the degree of indetenmnacy of tbI
equal to I Ld us !>elcd the re.H.tlon B, to be the redundant. The
the redundant "III be detennined by mimmi/ing the strain :-~'Y1len
with re p..:... 1 to 8
The tram t'n rg)' of a beam !>ubJet:ted onl) to bending is e·xpI""'• •
B
608
CHAPTER 14
Thl'ft-Moment Equafion and the Method of least Work
TABLE 14.3
\ loo,Jm.llc
(B
(
B<
BD
(
~
o
" 4
I~
41
~
ORT
T
I
I: 1(': If) 'd~
L
",.:cordmg: to thl: pnnl,.'lrk of lea,t \\ork
d
·T
\~
(iF..) fL4£ .
....... (T
PROBLEMS
11'
lEfth
(T
£1
sectkHll4.2
~
1••1 tfWouIli 14.1 Detennine the reaction and dra the
,hear and bending moment diagram fI r Ihe beam hO\\'ll
in ~giF
P14, I through PI48 usmg the three-moment
equatIon
0
The c\pn: ion, l\lT the: tx-ndmg f\~tncmo
and the axial forces F m
the redundant T and Ihl'IT dCmali\c\ nlth respect 10 T are tabulated
1~_3
8 ~ ~ul:
{HUlin!! '(~cht
c\prC"lOn, and dcmati\cs into Eq (2 we
H
0811 OST X) 12 I
f
' 12
400
12
lf2 0.6(\
4
m
£1
fill.
SOk
I(T)(IO)(I2J'
08
I
41( 15, . 0.6T\ - 2.4 TJ d,]
n
Pl4.4, Pl4.11
£
Jl..
0
T
12m
1
E= consWit
E. cOlUtanl
FIG. P14.1
fill. Pl4.5,
,
Pl4.12
,
15k
SUMMARY
In this chapter. v.e ha\c studied 1\\0 formulations of the force
it}; method of analj\i... of statically indeterminate structures,
the three-moment equation and the method of least work
The three-moment equation represents. in a general form,
patihilit) condition that the ~lope
of the elastic curve be COD
an inh:nor UpPOrl of the ~uonitc
beam This method whicla
d~u
for aniJl}zing ~uonil(
beams subjected to external
uppon ~t lem nh.
mvohc~
treatmg the bending moments
rior and an} fixed suppon!'! of the beam as the redundanta
moment equation i thcn applied at the location of each
obtain a Sc:( 01 (:ompatihJlil} ClfualJons which can then be
redundanr hcnding momenl
The pnnciple of lea t \\ ork tales that the magnllUtk
oj un mdt'tammute true fure mllJl he me h Ihal lhe
lured m Ih IrUl/urf" I a nummum To analyze an andete
tUTe by the method 01 least v.ork the str-tin energy of the
fir t expre sed In lenn of Ihe reoundants. Then the paruat
2S kN/m
~c
A.a:
I
I
10 m
1
£ 200 GPa I
=
20 m
2I
I
=SOD( 10') nun'
B
lOk
C
:i/O
D
tIOf~±21
£.29000
FIG. P14.2, P14.9
fill. PI4.&,
Pl4.10
lklft
A
D
Jallh
£ ....-
FIG. P14.3
I&Pl4.7
k
£
10ft
OliO
.000
I
F
....
III' ='4
?_ ........ _of..-_
nR......r~ .....
ft-+-IIDft--i--IOft -'----:20 ft---------<
-.PMI
_ ......
.... Iooding shown ID Fig PI42
IOmm.tA 65mmatB
'4.'4 Ashown
beam
Be a
fixed
IS upponed by a
ID FIg. PI4 14 Dtl&_
cable by .... method of least wort.
.... method of least work See
.... method of _
work See
RKbOaI aad .... fon:e ID each
ID
Fig PI4 13 _
mem.
!he method of
- - - - - 1 5 f t - - _...
£=~
fIB. P14.14
CHAPTER 15 Influence Lines tor Stattc.lly Indelennlnate Structures
612
. . . . 15.1 ................ _
\!though ,lily of thc- mcthC'd, of ana.l) si~ of ind~ten I ate
pn: l'nlc-d In P.lrI fhl\'1.' l'.lIl be u'c-d lor computmg the ORtiDlllI
mfluenl'C Iinl.". \\C- "ill u,c the method of COfl'i.lstent deforma
eu d~
m Ch.lpta I J for su("h .~e-)pru
Once the innuence linea
determin.ttc tructures h.I\1.' lxen constructed. the} can be UIed.
,amI.' mannCT .1' IhlN." t<'lr detenninate struct re~
discussed m
In thIs ,..'harter. the pro("edun: for constructing innuence hnes Ii
calh indC'tenmnatC' lx.lOh .lnd tru,~
.. IS de\eloped. and the a
of \fulkr-Bre".tu· principle for comtructmg qualitative loft
for mdetenmn.ttc ~.Ims
and frames is discussed.
.._
-- - - - - - - - - - - - - - - - - - - - - - - - . . . . ;
15.1 INFLUENCE LINES FOR BEAMS AND TRUSSES
r-x-l
'!LL
t
X
B
on:
t
A
B
C
oa:
D
t
C
la, Indetenmnale Beam
A.Jl\ X
B
C
~
b Pnmary 8Qm SubJ«1ed (0 Urnt Load
+
COlhtder the COl1linuous beam shown in Fig. 15.I(a). Suppoae
\\bh to dr.I\I. the mfluence line for the vertical reaction at the
suppon 0 of the beam. The beam" subjected to a downWard
concentrated load of unit magnitudc. the position of which 18
the coordinatc \ measured from the left end A of the beam a
the figure.
To de\elop the mfluence line for the reaction B), we need to
mine the e.\pression for Or in terms of the variable position x oftbe
load. Noting Ihat the beam is statically indeterminate to the first
we !tClect the reaction 8 1 10 be the redundant. The roller suppon
then removed from the actual indeterminate beam to obtain the
cally determinate pnmary beam ,hown in Fig. 15.1(b). Next
mar} beam IS \ubje<.:tcd separatel) to the unit load positioned
arbitrar} point .\" at a dlstan<.:e x from the left end, and the
O. as shown m fog. 15.1(b) and (c). respeclively. The expreSSlOll
can no\\ be detemlined b) usmg the compatibility condll1OD
deflection of the pnm,,,)' beam at 0 due 10 the combined efti
ex.ternal unit load and the unkno\\n redundant B must be
l.ere. Thus
J.. t JBBO - II
o _
la.
fBB
\\hi4:h the lle\lbilny 4:oenlClcm /8' denotes the deflection
mary beam at 0 due In the unll load at X (Fig 15 I b
fl ;Ubillt) coetfklcnt IBN denotes the deOection at B due to the
of the redundanl 0 fog. 15 IIe) .
\\ c \:an use Eq (I:' I for 4:onstructing the inftuence I
plaCing the unit load sU\:cc:s hel)' at a number of pos1uons
beam C'\alual1ng f tHo lor ecI\:h position of the unit load and
\alues of the ratio fB~
ISH Ho\\evcr a more efficlcnt
equill
In
dllnn
fIG.
IS-I
InlIUnCi U
hodllliU"""
.... Sb'UClUIII wIlIt IIJIlIph D.......
111
.s._ &-......
'IJ II'. . . .'1117............
x
t
._--
t
B,
b PriIU)''''' t i l ' ' ..
+
-------",
B
I
PlimorJ .... Loodod willi
616
CHAPnR 15
lnftuence lines for Statically Indetermln.te Structures
I •.
_ I L l ........ UOOOllr_ . .' -
x
c
R
L
3m
D
3m
R
EI = con..lant
la) Indetenninate Beam
Prj
..... Subjected", U.. Lood
D
1S
=
J kN
8
8
A
C
e I Inftuencc Line Ii
x
IkN
fBX
!
8
lIII
A
(hI Primary Beam Subjected to Unit Load
8
I=J S
c
1.56
+
8
I kN
... 11.3
D
Ie Primary Beam loaded with Redundant B
FIG. 15.3 contd
D
I
618
CHAPTER 15
Influence Unes for Statically Indeterminate Structures
---========
8
._cn__~,.51:-IO
Example 15.2_
IRl
=n==:llMo:...
:.::...:T:r:~-.I1t
..
I. B
Sm.. by 1.1 \\dl' b\\ ~)f rl'\.'lrrocal d..:tlcl.:ti(lIh. fBl - 'HI \\e place
load at Btl" thl' rnnl.lf) Ix.lm f 1£ 15 d(~
d~l.
compute the detleoliel"
pt.'ml ~ through r t'l} U 109 thl.' bcJrn-Jdlel.:llon lormula gl\CJ1 mside
\.o\<."r \llth\; bl,1{lk 1 hu .
fBA
JBB
:\tH." k' m' k.
£/
I ..
:!.B k' m
EI
fB'
I••
JBD
lOB
IB'
ftB
k.
m' k.·
1:!6 k
£1
'HN m k:'<
from which
EI
"
"hlCh the e\la~Cn
"go", indICate that Ihc'\C deflectIons are in the
direction. Note that the flexIbIlity coeflicient
in Eq. (I) denotes the
Ipo,iliH' Jeflection of the primar} beam at B due to the unit value
dUDdant B, Fig. 15..J)c(~
~acr h,\
the del1cction f88 represent!> the
(negative) deflectIon at B due to tile ntemal unit load at 8 (Fig. 153 d
In
iSB
I..
TABLE 15.1
lml
L ad
at
A
H
(
n
l
Influence Lme Ordmates
\1
kS
m/k~)
I ,
I ,
o "'llJ
I"
"1.\6
" 148
lJ,,J4
"
"
1"8H
+
or
243kN·m 3 kN
-£1
The ordinates of the influence line for B) can now be evaluated by a
Eq. (I \ ,>ucl,;c,>"i\c1y for each position of the unit load. For example w
unit load is lOl.:atcd at A thc \ alue of 8 1 IS obtained as
i..
J.,
164J _ I 5 kN 'kN
241
'.
The remaining ordinate of the mfluence Ime for 8\ are calculated m
manner Thc~
ordmates arc tabulated m Table 15.1. and the IOftuence
8 issho"n In hg. l'U(c)
InflU! n((' '-me Jor .\I( \\ Ith the influence Ime for B known the
of the mfluem:c line for the bcndi~
moment at C can now be
placm~
the unit luad ,ucre I\d) at points A through £ on the
beam and by gm~u
the com: pondmg \alues of B computed "",""'lIlfI
namplc. llS deph.:led m hg 15 1 1 J "hen the umt load IS located t
\alue 01 the rtJlllOn .11 B IS B
I 5 II. kN R) con",idenng the eq
the Iree bod) 01 the purtlon of the beam to the left of ( \\'C obtaUl
II,
1.5kNmk
The alue of the remammg ordmatcs of the mfluence line arc
Imilar manner Tht:SC nrdmate
for \(( I ho\\n m hg IS 3lg
Slllce ~
f
D
lD iU:COrdanc:e
th M
on the pnmary beam Fog 154(d and
through F by USlnI the cooJUPte beam
In Fig IS 4(e from whiCh v,e obtalo (he
fDA
fA
fOB
JBD
r. t
0
I
£1 86 10
G)
G
10
122
CHAPTER 15 Inftuence lines for Statically Indeterminate Structures
IECTJoIll11 ......... u.. "" __ .. T _
TABLE 15.2
Load
.t
r
n
A
0
O.·N~
O.60~
n.
o,~q
£
F
t''i
~70.O
0
k
The alue r the
Imllar manner TI.
d
ordog r I
r
I~
1I1a are Ii
lor the shear and bend
tngm
lie
faPCCliv
U
0.397
o 746 (left
0.154 [right
0
O.O7:!
0
0
O.W3
10
0
10
0.079
0
k k
M
,k k,
0
0.095
0.119
I0
B
C
D
o
0603
lmt
~
Example 15.3
Draw the ml1uencc hnes for lhe ....~IO "
Fig IS Sa
The llrdmate, of the mfluence hne for n, l:dn no" be computed by
Eq. I ... ec~u
... ~le\I
for eal.h po Ition of the, unit load. For eumple
unit load i... dela:x~I
at B. the \alue of D 1'1 gl\en b)
D
108
I pp
816667
1680
0491 k k
The r e m a m ~
ordmate... of the Influence line for D, are computed m I
manner. Thc\C" ordmate' are tabulated m Table 15.2. and the Influence
D, l'1 .. ho.... n in F-ig, 15.4(t'
With the mfluem:c line for D
mfluence lines for the rem.llning reactions can no.... be detennined by
the equatiom. of eqUlhbnurn. For example. for the position of the UDI
point Bas ,ho....n In ....g. 15.4(g). the value of the reaction D" has been
be 0.492 k:'k By applying lhe equilibrium equations. "e determme the
the reaction, A I and F. to be
SGIU11an
The beam I IIIdetemunate 10 the
' ' ' ' Lin< fo, R-.Jundan, D and G The
and G for an arhilrary posulon X of the uml I d l;:
the compaubdny equation sec Fig 15 5 b Ihr ugh
Inl/IIt'f/( (' Lmn lor A. Will f
tCL:II,
()
,1,(50) t 1(40)
+'L:r
0
0603
0,492
+- ~f
F
~
()
-0.095 k k
The \alues (If the n:maming mfluenl.:c line ordmates are computed
manner. Tht:'iC ordmatc!ii Me II ted 10 Table 15.2, and the influcDCle
and f' afC ho\\n in hg. 15.4 h and (I re"''''''i.:ti\e1).
Intlunl/t I Inc lor .\( unci \I( The ordinates of the intluence
hear dnd hendmg moment at C (an no\\ he c\aluated by pl3ClDl
llOCC ~1e\
at romt A throu~
F on the mdetenninate beam and
gmdnop~r oc
\aluc of the rcactlnns computed prevIOusly F
hown 111 • Ig ~4.51
.... hen the umt load i Ioc;ated al pomt B tbe
reachon ue A
fJ IlfJ3 k k· f)
() 49:! k k. and F
009
Idenng Ih~
eljulhhnum of the frL"C yj~ub
of the portion of the
of ( \\C obtam
fo,
Ip
f"
f.DD
D
G
0
GO
fDx
ltD \loe place lit unlll
beam (Fig IS 5 e and compute the deftectlon al pam A Ihr
the beam-deftection fonnulas gnco IDSI(1e lite front c \ r
Smce by Maxwell slaw
fDA.
(A.D
fpB
I.
0.492(20) - 0
A,=0603kk
nd
D and G 81 the roller uppons D and G ...~«tlvc y
The IIIftuence hne ordinat \Ii II be
ua
lItrough G h",~
on Fig 155
0
•
. , '&11 .... II u.: far SIItIcIIIr In. I ........ 8trUCtWII
I
~(
AI
A
I kN
x
5m 5m 5m 5m 5m 5m
D
G. ,
I
15m
J5m--'-'
._Beam
EI - '
Pn-r Beam 5ubje<1Iod 10 UIII\ Lood
(1)
+
--_oJ!.
f oo
- ~9 : '"-A 1
G~D
IkN
(e) Primary Beam 5ubjeclod 10
'edund... D
+
A'O:;-fRlI._"G~
xG
I kN
d) - , . . . . SoihjorPd III
• '
d 'G
IkN
I
A
C
•
II
.... '.
,
G
~
D
~.D
bd.
826
CHAPTER 15 Influence lines for Statically Indeterminate Structures
TA8LE 15.3
l 011 l\\ d
,,
/l
cit
(,
(
J)
t
f
G
10
0
0
4
B
o O.l~
o ~8
(I fl""
OIl6'\
1.0
0
{I Ij'\ I
O.~2
o ';~5
1I.5l\2
1.11
0
O.K04
o .1Rb
o
-0.159
-O.J~7
o
Th\,.· n.'111ulnlng ardm,ih.' (If the Influence line, for the redundants are
m .. Im.lar manner Tht''oC (Irdmalc are t.lhulated in Table 153 and
enl'c hnl.' lor D ,IOJ G ,m.' hlmn In FIg. I S."lgJ and (h) respectJ
{I/flu nd Lim \ lor ~ amI \/ t The ordinate!> of the influence
reactIon 1;.111 no\\ he dch:nnincd b) placing the unit load
al POlO!'. ., through G on the indetcnnmatc beam and b} applying the
of eqUlhbrium. "or e\ample. f()r the po"lltion of the unit load at 8 FIIthe value, of lhe rC,H;llon, () and G, h,nc been found to be 0 228 kN
~nimJ :n
-(J.()32 k'W kN.
B) con,idering the equilibrium of the
yl~Hi :('.Il J Cr
dt:h:nmnl,.' the \ alw.:, of the reactions A. and .\/ ~ to be as follows
o
.1
I + 0.228
0.032 - 0
A, - 0.804 kN kN
II,
1[5)· 0.22R( 15)
0.032(.10)
0
M, - 2.54 kN
and\t~
The \.tlue of the remaimng influence line ordinates are computed m
manner. Thc'\oC ~'.rtanido
an: h~tcd
in T abk 15.3. and the influence I
arc hO\.\El in Ilg. 15.5(J) and {k" .yle"itcp~er
Eo..... 15.4
Ora" the mfluenl.:e hne lnr the furee, in ~rebmc
BC. BE and CE
ho"n m • Ig 15€l 1a l,v load arc tran..mith.'d to the top chord
Solution
I tnlcrn<ally mdetermlllate to the fir t dcgree. We select tile
the diagonal mcmhcr ( F. to he the redundant.
The truss
Ff t
In
bljlu£nfe Lme lor R( JUtlJUIII ft. t: T{) determine the mftueoce
.... pla a uml I ad Su!.:!.:e I\dy at Jumh Band ( of the t
po Ihon of the unit kldd "e apply the method of COl S sten 1I~:
\lmrut tht' alue 01 II
The J'lnmitr) (russ obtained by 1'1
(t. I ubJt:Cted fl<trat Iy tn the URlt luad at Band C as shown
_...-_.....,.....---._.... .
630
CHUTffl1S
Influence Lines for Statically Indeterminate Structures
LIlloo., _ ....INoI...... " " ' -
IECTION 15.2 0 llUw 1nII_
On~c
,I qU.i1IL.ltI\\,.' e~l',Ut1lI
hne for a ...tructural response
h.a IX.'l.'n l.'on IrUl.:tl.'d, It (,tn Ix: lI"Cd 10 dt..'X.'idc ""here to place
load to m.I\lmize Ill\: value 01 the re"'pOIl-.e function. A eli
\i."(t1on 9 ~ the \alue' l,f .1 n;"poll"C function due 10 a undi
tnbul\.'d hH' 10.ld h m.1XlmUIll po Itl\C or negatl\c) when the
plao:'d O\a Iho~
POrll("!J1 01 thl' tructure "here the ordinates
rt." Pl"lJl '" fundlon lIl11ut."n(c Ime .Ire POHJ\C 'or negati\c Beca
ml1ul'nl..'l,'·linc Mdmall,' tend to dlmll1l ...h rapidl} with di
thr.: JX,int of .1rpIK<ltllll1 of thl' re"'on~
function. Ii\e load p
than Ihn.'l' '1'.10 kngth.. ~a"
from the location of the resPOIlle
generJII} ha\e a negliglPle ctJeLl on the \alue of the response
\\ ilh the 11\ e·load p;.ltlern kml\\ n. an indcterminale anal
...trudurc ('an he pc:rformcd to determine the maximum value
Pl"lJl..c funl:tlon.
8
D
Qual
A_
•
or
•
Innue
111
£
l .. l
8
8
l,
U>od.
8
D
8
£
F
EulllJllB 15.5
Dra\\ quahtatl\e mflur.:l1l:e lin!':) IOf the \Crll(;al reactions at suppon
the o..:ndmg mllmr.:nt .It rom! B. •miJ the shc.:Jr .:Jnd hcnding moment
of the four-...r,tn ulI1linUlIUs he.lm ,ho"n in Fi!!. 15.7(a) Also _L._.u
r<lngr.:mr.:nl 01'.\ ul11lonnl) dl,lnl1uled iJo"""ard h\e load n I
ma\lmUm rChlll\C' re.t<:llon .. at 'Urporh I and B. the: maximum nep
lng moment at B thr.: lll,l\lmUm neg<ltiH.' ..hear al ( . and the maximum
bending mon1l'nl at (
Amnaemem of Live Load ror Mnunum NcphYe M
Cdl
•
Solution
IIIJluc/J/( I inc lor I fo iJl.'tcfmlOe the qualilatl\e influence
\r.:rtk<ll re.actiun .~ at ?>uprnrt of. \\-c femme the \-crtlcal restramt at A
<tl'tu<ll ma~
and gi\l' the rclc,I'oCd beam a .. mall dl'oplacement In Ibe
8
C
D
E
F
Quahtabvc InOuence LIM fur S
E
C
Arronpmcnt of (j,", lood for MuimwD NepO
1<1
.:---:8- C~D:-";F
Qua1iIIIi... - Lme
H
()
E
Quaht<tllV Inl1uen\.(' Lme (or A
C
~ofLI
AG. 15.7 (I.:untd.
II&. 157
An n em nt of LI\C' 1 ad for MaXimum Posiove A
Ib,
"
I.-dforM"'==
F
832
-
CHAPTER 15 Influence Lines tor Statically Indeterminate Structures
dir..."ct! n
,f I
rill,: Jdkl;l\:J sh·lI'll.: 01 ,Ihl: n.:lc.I'CJ beam th
h' I"" t'l hn~crp;n
Ih lo'Cn .11 ~j;h
l~ the 1IlI1ucn,,"C hoe 1e
U\C nllueru.:e ltoe 1M I
l'l Ih.ll the Jdkdcd ,hare 1:0;, con I teat
uppcrt onJltllm 01 the: rek d Ix •• m: th.ll I pomts B D E and
TI,; e ..1 beam "hlch an.: ,1I1....:hI;O III wlkr upr~l t
do notdisplace
the roSlt \~
Tll max.lm.
,.dUl'lll
t _ the 11\..: I{lad II
IS
placed
fB
oJ DE f till: o.:am "here thl' {lflhn,ltc'o of the mfluence line
IX' ~\I
a sho\\n In f-I l'i "7 I:l
In. I
,8 The quahtatl\c mflucm:c Ime for 8 and the
l,r Bare determllled lD
ho\\o In 1"1£ l'i'" (
III
.uT'a11{!('lllcnt for ,he md lmum Pl)!<.lt!\t' \ollue
nunner and
In
aT\;
I
B
A
_....
•
To determine the 4uahtatl\c mfluence
gnld ~
moment Jot B \'t! msert a hm~('
at 8 in the aClual beam aad
rdea..oo ma~
I mall rotation In the JX' ItI\C dlfct:llon of M.
the rortl n to tht.: left (If B ulunh:rd\'l(k" 1'< and the portion 10 the
clcx:k"l ' a.. shen\n In Ilg 15'" d Thl: ddk'\:ted ..hape of the rdeued.
thu ('Obtamed f\'pn:!oCnt 1hl: qU'Jhtatnc mf1uencc Ime for \I"
To cau the ma'lmum 1lI.'gati\e 1x'nding moment at 8 we place
load II <J\ .. r pan IB RD. llnd Ef ('If ~h(
beam "here the ordina
influence Ime lor \/. Jf~
nq;..ltl\\.: as h(mn m Fig. ~.51
d.
U 'f
I
,,,
lnflu 'Ie, LUll fi T S( The quahl<ltl\C mfluence line for S( IS de1..._
cultml! the auu,11 ocam at C !nd h) \I!~ Ill!! the relea'Cd beam a
dj,pla~emnt
In Ihl: po Itl\1: dIrectIOn of S( b) mo\ing end C of the
of the be-... m dl)\\ n\\ ard and I:nd ( of the right portion uP'Aard
Ilg, 15.7 e
To obtam the nl<lXllnum ncgatl\e hear at C. the Ji\e load IS
nap~
DI:. and the portion BC of the "pan Bn of lhe beam. where the
the influence line for S .lfe neg,Hi\c.•I~ ~ho"
n in fig 15.7(e).
In/lu(/1C't' Lm!'
lor \/( Thc quahlati\e influcm:e line for the
moment at C and the 1I\c-Ioad arrangement for the maximum poIIh
\Ie arc !.hov.n In I ig. 15.7(f).
~
A
_L..
_L..
..
_L..
L..
Quahtab
Quahwsve Influence Line feX'MA
1111
L..
Influence line ,
I I I I
W
w(
.'(
,1111
IIII
_L..
InJiu nl( l.lII( /lIr \1" The quahtattve mflU('n","t' hne for the
ment <1.1 -II
o"n In t-Ig IS 8 h
ute that IDt.:e the mernben
connected tOgJ.:lhcr hy ngld Jomt th ongmal angles between
mtersa:tmg I a l0lnl mu I he mamlamed to the deflected shape M,.... 1iII
blam the hlaXUllum po lint: hendmg moment at A (he hve load
o\er th
pan of the frame v.here tht ordmat of Ihe mfluence
~
""'Il A
~
FIDPlll5.6
Solution
....
~
~
•
W(
W(
I I I I
1111
A
Ora.... quahtatl\e influence lmes lor the bending moment and shear at
Ihe bUilding frame ho"n In I-Ig, 158(a Also. ~ho"
the a r . ~ :
unifonnly dl InhUll:d do.... n..... arJ 11\1.' load II that \\111 cause the 11
Iti\c bending moment and Iht: mu l~ .m
ncgati\c !lohear at A
-
..
d •
-
Arrangement of Uve 1.oId for MaJ,unum PoIibve MA
(11
... ,5.8
134
CHAPTER 15
Influence lines for Static.lly Indeterminate Structures
,
SUMMARY
In 11ll'S •.: h.lph:r \\t: h.ne lh,cu'...ed lIl11llf..'m:e ~cnil
for statlcall
mlll.ll.. tru.. lUl"C'
'hl' rn'o..'dure for (.'on,trudmg such 1R8
b\ thl,.' Illl..'1lwd l,r con I lelll JcfMmalll\ll'" c"i'C'lltially lovol
;rudmg the IIl11UCIlCC: linc lor the n:Jundanb h) plaCIng a
U(CL'~
I\d~
,It ,I r'xt~nu
l,f r<'l11h along the length of the struc:tult
tllf c:.Kh l~
ItJlln 1\1 the lIll1t lo,ld (omputmg the values or
dund.lI1t h) ~lr
Ill!): tht: melhod of con ... hh:nt dcformatlo
u...mg thl,: mllul,.'lKC hnc . . for th.c hna~l udcr
and. b} applYIng the
11011' of c:qUllit'tnum. detammll1g the mfluence Ime~
for other
funclIon of the ...tructllre
balu.lllon 01 the dclll'Ctlon Ill\ohed in the application
method Ill' Ll'n ... I...(('l1t dcfl.'mlJlion can be con. Iderably expedited
mg \1J.\\\c1r... la\\ ofrL't:ipn.:x:,l! .~noltcen d
The procedure for
iog qualn.niH intlut:ncc line... for mdetenninate structures by
Bre,lau· principle I ... prc'Cnted in Section 15.2.
'\Cr = ~B ", DF=JE
801---1-401
401
- 1--1---- 3
401
E con tun
"'''5.1
FIG P15.5
15.' DrJw the IIlftuen 11
port .Ino the fOfl:C In m m
ITU ho"n tn hg Plj 6 LI
bf,llJllm choro 01 the truss
E
F
... ,,5.1
PROBLEMS
B
SoclIon 15.1
15.1 Ora\\ thl,.· II1flucnlc hoe for the reactiom at the ... up·
port nd thc hear <tnd hendmg mllment at point B of the
beam ho"" m Ilg. PI51 Determine the influence line or·
dm t OIl '-m Hllcndl'i Sded tht" reaction at ~uprot
C to
he th redul1l,b.nt
8
c
£/ = con~ta
AG. P15.3
601
11m
15.4 Draw the influence line... for the reaclion at
POrt!. and the ~hcar
and bending moment at pamt
beam ,ho"n 111 Fig. PIS.4 Detcrmine the influence
dinate... at lO·ft intenal...,
£/:;: cun tanl
FIG. P15.1, P15.2
A
.il
BCD
;;Y;;
rt ~
l
I
ne the.: Influence Jines C r the real:tlOm: at the
r the be.am f Pr. blem I j i b ) !<Iccung the morpon 4 t be the redundant. s....e Fig PI Ci I
the mfluen lin
r the rC"dl:Uon at uppon {
d bendmg moment at pomt B of the beam
•
I
FIG. P15••
15.5 Dra\\ the inftuenex Jine~
for Ihe reactJODI
ports and the ~hear
and bendmg moment at poiDl
D
A
B
E
601
_I
L - 2paneI1l8m-16m
:;,
£1 = constant
C
... "5.10
T
1-
I--- 20 II --41-10 11-1-10 It
15.2 De
B
15.7 Draw the inftuem;e lines for the forces tn memben Be
and CD of the truss shown m Fig. PIS 7 live loads are
transmitted to the top chord of the trus
15 It
A
£A:;: con tant
AG. P15.6
B
A
C
l---- 3 panels ailS ft - 4S ft
hO\\11 III fig: PIS.3, Dctermme the influence hne
at 5-ft mtcn ,i1...
£.4=_
FIG. P15.7
15•• Ora\\ the influence hnes ~
the
Be BF and CF of the truss shoWD ID F PI
load, are tr.ln"mllted to the bottom h rd the
... "5.11
-
., ,& 15
LCI UI8II far S1I1Ic8IIJ' ............. ~
....
C
A
~-
L ----+_
11I.,.5.15
11.'. Draw qualn.bYe inlI_
ment and shear at POlOt A of' Idle:'===
FIJ- PIS 16 Also show the ..
dlstnbuted downward hve load
posIbve bendmg moment at A and . .
shear at If
000
... ,.5.12
..... ,u
11.,. . . . .
A
,I.,. CIa
~
11I.'15.16
11.17 For the bwlelin. r..... """'"
mine the arrangements of a unif'onaIy
ward live load w that will cause the
bendIng moment at pomt ..4 and till
beneli.. mOlnent al po"nl B
..--:--:....;.-;;--1IIr
B
C
B
-+--L
I~-". ;.-, -, -~
....,.
1--- L
--+-.-+-. -~
L --'+--L
"''11.17
-
CllAPI8l
I.
51a11t-DoIIectIOnthe fundamentals of this method proVtdes a valuable intral~
matnx tltrues method \\ hleh fonns the baSIS of m
\\are currently used for slruclural analysis.
We first derise the fundamental relallonsbips 1I01lII.
apphcallon of the slope-deflection method and tbco
concept of the slope-defleclton method We cOllSldcr the
the method to the anall" of conltnuous beams and PI Dill
of the frame In \\ hleh joint ticlOslatIons are prevented
sider the analy I of frames with joint translatlOns.
--
r
•
r
16.1 SLOPE-OEFLECTIOI EQUATIONS
When a continuous beam or a frame is subjected to CIl1le1llll
temal moment generally deselop at the ends of Its inClividai
The SIQpe-d(lIt!( llon equatlOnr relate Ihe moments at the
10 'he rolations and di.fplacemen1s of lis ends and lhe eo
plied to the memher.
To derive the slope-deflection equations, let us focus
on an arbitrary member AB of the continuous beam
16.1 a). When the beam 's subjected to external loads and
tlements. member AB deforms, as shown in the figure and
ments are induced at its ends. The free-body diagram and
curve for member AB are shown using an exaggerated
16.1(b). As indicated in this figure, double-subscript notabOO
member end moments, with the first subscript identifyIng
end at which the moment aclS and the second subscnpt
other end of the member. Thus, MAS denotes the moment
member AB, whereas MS A represents the moment at end B
AB. Also. as shown in Fig. 16.I(b), OA and Os denote
rotatIOns of ends A and B of the member with respect
deformed honzontal) position of the member; <1 den
translation between the two ends of the member m the
pendicular to the undeformed axis of the member, and the
notes the rotation of the member's chord (I.e the mm,1i
necting the deformed poslt,ons of the member ends due
translalton <1 Smce the deformat,ons are assumed to be
rotation can be expressed a
<1
L
111 member end moments end rotallOns QN/
hurd fUiallOn ar, po lllve wMn colUt/erd
_....
~A
:----------'"'i
JW
T
A
•
lei
_
•5 •
71117_.-"
By subotilUbllll Ihi cquabon mlo Eq 16 Sb ad
cquabon (or AI we obtain
M.
2EJ
L1JJ
and by subotilUlillll Eq 166a mID ather Eq J,... ",IIUI!
obtain the expmllllOll (or M••
. . . . . , DI I
__
,.
_
. z- A~C
-'r_·~J:=L,
(I FIUd Bam
I
,.
I
: (b) Simple-Bam 1IeadinI....... "hI
I
au
(e) Jllud.Ead
"=_..
_ 1 1 of the area IIIIdcr !be IIJDple-bam "Ii
about !be eoda A lUId B are liven by
CHAPTER 16 Slope-Deflection Method
1t1
L
0.
can be expressed
"'I + ( f-E\1s. - FEM ••)
-~
{l
It' J~
.md ~1.6
manzed .1
i:lrc similar in form
1/
JEt
L 1/
1/
(l
r/I + ( FEM
~EF
)
II
1662 161
!('1pt.'-d.:ftcct!on equations gIVen
the m('lJ,fled
BecJu
D7 •
4£1
.loJ Ih... modified slo[X'-Jc:tkl.:ti,)!1 ...~noltauq
\f IS
IICnaIlt.2 - ~ o I " ' 1
L
- IFEM fB
1/
3~rJ':~ t)1
1.5k1ft
S,,= 1338
B
lOk
1613
32 5
I
C
18
d , _ ElId M........... Shean
in \\ hll..:h the "ub~npl
, rcler" to the ritllillv connected end oftbc
here the moment \/ A ,Il'h and the ~ubrjpt
h idenlifies the II
of the member lhe rotallon 01 the hinged end can now be wn
~
, ,, r/I
II,
L- 1FEM , )
4t:1
'
392k.~
I)
lOk
pntr ml B:. ~ :- ilr - "'D
,
,
,
I
IlJ8k
327Sk
1877k
49k
Ie) Support Reactions
A~4.1
To illU'"tr,tte the basic com:cpt of the !'ilope-dcflection method
the three-span continuous beam sho".n in Fig. 16.3(a). Altho
1 ~ kilt
III k
---""
98
--20ft
B
~1lI"-r l/)ft+-
(A~_.
..... - ' O : l ~
B(./"
10
,
EI c \;onslanl
E =19.000 k
I
"-D
49
16.62
1=500in"
k(~_S)f
1511-----J
197
I
8eIdillI- DiqdID (k-ft
'bl
fIG. 16.3
Jll,lU
4k ft
_
•
MIl
,- ..
CHAPTER 16 Slop.-O.flection M.thod
.............
,tru(lure .K1U 1~
!.:l)n\lsb of ~l 'lI1glc l:(lnlmuous beam between
A .1Il0 D, for tht,.' purro",,, l~1 anal},is It ,is. consIdered:
\.:omfl4.'1scd (If thn.'C' rehm~
., 8. B( .InU CD. ngldly COD
lOll1h f. B. C .InO D 1(\(,11"0 .11 111(,.' supporh of the structure
\Urr('1rt
I.:Ol1lll1UOU 1x,Im h.... 1IC~
di, ulc:d 11110 members and
the unknO\\I1" lerna I rc,ll:LJom, i.ll.:t onl} at the JOlOts.
the
j 0 1l1ta.
Degrees of Freedom
With the loint l()(';i.Itions
nOll
c..tablished
\\C
identify the unla~ol
dept.-odenl dl'.placement-. If,tn,kuion.. and rotations) of the J01DtI
,lruclure, Tht'.e unkno\\11 jOint displaccmenh are referred to U
qrees of !rlrd(mr of the ..trueture. From the qualitatl\e deflected
the continuou.. beam ..ho\\11 in hg. 16.3(a), \\e can see that nODe
jomb can translate Furthermore. the fixed joints A and D aumaw
tate. \\herea .. J0lnl'! Band C arc free to rotate. Thus the COD
beam ha.. t\\O ~ergd
of freedom. 0B and Oc which represent
know n rotation .. of joint~
Band C. respectively.
The number of degrees of freedom IS sometimes called the
kint:matic ind(!lt'rnlinac:r of the structure. Since the beam of FIg. 1
ha~
two degr ~
of freedom. it ~I considered to be kinematicaUy
mmale to the second degree. /\ structure without any degrees
dom i~ termed kinemalically dt'laminate. In other words, if the
~tnem calp
of all the jOlO1S of a structure are either zero or kno
structure is considered to be kinematically detcnninate.
FEM.
FEM••
L
50 k-ft
12
50 k-fl
or
For member BC
FEMsc
PL
3020
The unknown joint rotations arc determined by solving the equa
equilibrium of the joinh thaI arc free to rotate. The free-body
of the members and joints Band C of the continuous beam are s
Fig. 16.3(b). In addItion 10 the extemal loads, each member I. IU
to an mternal moment at each of its ends. Since the correct sen
member end moments are not )oct kno\\iIl. it is assumed that 1M
menh at the ends of all the members are positive (counte
In accordance \'-ith the slope·denection sign convention adopted
diagrdms of lhe JO
preceding section. Note Ihat Ihe ~dob-erf
the member end momcnh actmg in an opposite (clockWise diJllCIiili!
accordance \'-ith . c\'-ton\ la\'- of aClion and reaclion.
Because the enllre structure is in equilibrium. each of Its
and JOints mu't also be In equilibrium. By applying lhe mlOJlllSll.l
hbnum c4u.tion, L: .11. () and L: At, _ () respectively to
bodies of JOints Band C \'-e oblam the equilibrium equatlODS
II. A
+ II.,
= 0
\.((8
t \It I'
0
75 k-ft
T--8-
FEM • - 75 k·ft J
Equations of EqUilibrium
50 k·ft
or
r
k
75k ft
Note that In accordance Wllh the no1lce l~
the
counterclockWIse fixed-end moments are consIdered lO be JlOIIII Smc:e
no external loads act on member CD Its fixed-end m men are
thalli
FEM D FEM
0
The fixed..,nd moments are shown on the diapam f
Fig 163 c
The s1ope-dellectioo equations for the lluee members
unus beam can now be wnnen by usmg Eq I 9
rotariom
upporlS f the _bDunUS beam transIa the
rd
1hree members are zero I e
6
poria A and D are fixed. the rotabOOS
169 for member AB WIth A the
obtam the ~
equa
trIIcture In
2E/
O
by coosideriII8 B as the near end
I5ll
ClW'T£R 16
Slope-Deflection MelIIod
21:1"'0
0
:0 - II
~lf . mS
b) gm~lrI.
~EI
If
20
2FI
II •
~O
and for mcmtx-r
I),
IICTJaIIIl2 ..... e....,. .... ..,..,DD,IIIRIooI._
- 50
Be. \\c obtain
B UbollluUn lho: n me: I 01
SIlO 12' ft'
and I SIlO 'n
aode. be
+ 0.1 EIII, + 75
.lI/E~O
75
8,
8
75
+ II.
:!.fJl
O.'~EIf}
50
16.9) for member
flj.
:!.tJB
01
000\\ rad
mu
o
r
Member End .....,11
., I
-2EI -Ie
'L7£11
.0•"
I,
'21:.1
- 0"
15
0.133£/l/,
15
\Io(
~'.:
The momen at lhe ends r th Ih mc:mbe •
can now be determined by ubo!lIUltng the nn
and £18 rnto the s10JlH\eflect equa
Eq
M.
M,.
0 I
\08 46
SO
02
108 46
SO
10846
01132
These lope-ddledion equations automatu:all) satisf) the com
condition, of the '1rUl.:lufC. Smce the member ends are rigidly
to the adjacent joint. the rota l n~
of member ends are equal to
lations of the adjacent joints. Thu~.
the (J terms in the slope
MB(
02
M.
0218382
01
49 1 k·ft
or
equation... Eqs. 16.1R)) reprc\Cnt the rotations of the member
'" ell as (hose of the jomb.
M
Joint Rotations
To dClcmline the unknown joint rotations 08 and tic. we SUbstl
slope-dene'Clion equations (Eqs. (16.18)) inlo Ihe joint eq
equalions (Eqs. (16.17)) and solve the resulling system of equat!
multaneously for O. and II, Thus by substituting Eqs. (161811
116.18c) 1010 Fq. (16.17a), \\e obtam
50)· (0.2EI0. ' O.IE///,
+ 75)
71
k-ft
10846
+0
and by substltutmg I'qs II ~.18d!
I Ellie
D
026718382
491k.ft
Moe
0133 18382
24 4k·ft
and (16.1 Xc) into Eq. (16 I
,IIIEIfI.
75) _ O.2~7EI
0
or
Olf.///. + O.4~7f10(
Soh 109 Eqs. I1>.1 d~
"-e obtam
·75
and (I h.1 %
(Eqs. 16.17 If the solution
should be satISfied.
IS
UO,
18l~2
k.ft
5
correct then the eqwbbriwn equall
-71 7
71
0
491
0
Member End Shean
The member end momen" )USI compuled
diagnun of the memben and) 10" ID Fig 16
the ends of memben can now be de1enniDed
of equih"bnwn to the ftee bodies the mem
M
0
S1mulldneously for £/9.
1fJ8 46 k-ft
1.10.
I k
or
Note that a positive answer for an end moment indICa thai
counterclockwise whereas a negative answer for an end momen unpl
a clockWise sense.
To check that the solulton or Slmullaneous equalt"'" Eq 161
has been earned out correctly the numencal values of member end
moments should be .ubslttuted into the )0101 equiJ.briwn equa on
0
··25
EJ
49 I k-ft
491
1I.4EIfI.
beam
392 k-ft
or
IO.~Ef(
II
00018 rad
+ 0.1£///. - 75
,l f£~.O
cn.
(O~EIO.
r£
de
..
o
392
S
20
152
....,. ..
CHAPTER 16 Slope-Deflection Method
~91
~/
II
II
.
.Ioi sc
I'
.\
(I
" 1/
0
~F
k
F
\'CR
_
n
D87 k
Il
'92
II
S(n
49k
II
So<
49 k
The: fOfcgl1mg mcmlxr end ,he.u can. altern i\l~I)
---
I 5 20
127
L"
I
JO
4
0
55
18 7 I
5
244
0
Chedcs
be evaluated
b~
~UJ"l :rp. ti m
or ('no hear.. due: 10 the c:\lcmalload and each of the
cont! ~Hl'(m Kn
~nu:.lt
~d1t.ra ·
t l l1 the ffiemocr. For example the shear
at cnd .-1 of ffieml'ler -tB i, gl\cn b)
1 5120
.19.2
717
20
20
I
13.38 k 1
In "hleh the fin•• tcrm equah the ~hear
due to the 1.5-k/ft unifonnly
di..tribuh.-d Illad. "herea.. the ~ond
and third terms arc the shears due
to the 9~ ..:!-I..·ft and 71."'·k-ft moments. respectiH:I). at the ends A and 8
of the memher
Support Reactions
Shear and Bending Moment Diagrams
With the suppon reacUons known the hear and bendlDg momtDt d~­
grams can now be constructed In the usual manner by U51Dg the btam
sign com nlion descnbed In Section 5.1. Thr shear and bendmg m
ment diagrams thus obtained for the continuous beam are shown 10 Fig
16.3 f and (g) respectively
From the frce·body diagram of Joint B in Fig. 16.3(d), we can see that
the \ erlical reaction at the roller support B is equal to the sum of the
shears at ends B of members AD and BC; that is,
B
S.,+S"
~
16 2+16 3~ 2.75kr
Similar!} the \crtlcal reaction at the roller support C equals the sum of
Ihe ::\hean. at end~
C of members BC and CD. Thus
c - S" + S<o -
13.87 ,49
= \8.77 k f
The reacuon::\ at the fixed support A are equal to the shear and moment
allhe end A of member AB; that is.
A
S .. -13.38k
W.
II .. - 392 k-ft )
Similarl} the reactions at the fixed support D equal the shear and m0ment al end D of member CD. Thus
D
Soc
\10
\10<
49 k I
244k-ft)
The suppon reacuon are shown m Fig 16 3 e .
16.3 ANALVSIS OF CONTINUOUS BEAMS
Based on tbe d,SCUSSIon presented on tbe precedmg tI n .... procedure
for Ihe analySIS of conlmuOUS bealll> by the s1ope-<lcftecboo method
be summarized as foDows
I. Identify the degrees of freedom of lbe true\we F rolalio",
beams. the degrees of freedom CODSI
.... unknown
theJOIDIS
1. Compute lixed-end moments For each member
evaluate the lixed-end moments due to the
the expemons g",en inside the bock ""' of the
3. ten:Iock
In the
lixed-end
of
supportmomen
senJemen"" de
chotds of memben adJ8COllI 10 the uppo
!be re
e uanslabOD be\WeCIl ....
member IengIh
A L The ho
:~
thebee:
I
the
I
-
1fCl1lII1U ....,." C1C,
ClW'TBI 18 SI....·DeneeU.. MelIlod
-
_
Ex-" 18.1
I k
Dt:!emllne the rt3.l.:1lllnS mJ dnt\\ the "hear and hendmg moment dUlgranu Ii
tht 1\\0- pan l,;(l111IRUOUS ~am
shm\" In ftg 165(a by the IOpe-deftceuon
m hod
Solution
Ikq
f',f dfmr From Fig. 16 'i(a. \\e can see that only Joml Sofdle
beam IS free to rotate Thus the structure ha.. onl) one degree of freedom wtticb
1 the unkno,," Jomt rotation Os·
J.£nd"", ntf B) usmg the fixed-end moment ('xpres Ion given in_
SIde the back co\er (If the book \'oe C'\aluatc the fixed-end moments due to the
BeD
Fi
e temalloads (or eal,:h member
FEM.
Pa~
648 k-ft ,
L
FEMs f
Pa
L
FEM B(
"L
12
FEMcs
150 k-ft ;J
~
- 43.2 k·ft
2 30
12
- 150 k-fl ,
or
or
64 8 k-ll
or
-43 H-ll
or
18k
I
M~IA
~
Bl )
+150 k-ft
ole that in accordance with the slope-deflection sign convention the counterclockWIse fixed-end moments are considered as positIVe, whereas the clockWiSe
3S6( l
984 27.57
I
(lBl)
Bl ).01.5 :It
A
816
1
984
Sumlarl) by applymg Eq
equation
•
18k
2£/ 20
30
M,
'
2£/ 0
30
'
ISO
ISO
Be. we obtam the
0133£/0,
A
4
B
27.57
lopc:-deftecliOD
ISO
00667£/0. - ISO
4
•
It
,.,1U
C
t
t
3741k
81n
2
2 kill
D
d~-
Equ,librQlnt Equal on The free-body d18graJD of JOint B 11 showD m
thaI the member end momenb whICh .... asawned 10 be
counte lock
directiOD on the ends of the members must be .ppIied ID
16 b
I
3,n.lI(
64 8
432 -0.16£/0.-432
169 for member
10.5(
27.57
(e) Member End MomenIs mel Shan
Slope-DeflectIOn EquatlOIH To relate the member end moments to the unknown Joint rotation, 08. we write the slope-deftection equations for the cwo
members of the structure by applying Eq. (16.9). Note that smce the suppon A
and C are fixed, the rotations 0.,4 - Oc O. Thus the slopc-deftectlon equation
for member AB can be expressed as
25
rm+mm
IB
Cl)
B =3141
Chord Rotations Smce no support settlements occur, the chord rotabons of
both members are zero; that is'''AB "8e O.
2£/ 20
1
18k
fixed-end moments are considered to be negative.
0.08£/0.
M
b
-150 k-fl
2£/
25 0.)+648
rm+mm
(I B
1)
M
MMC1B1)
)14
I43k
til
CHAPTBl16 Stope-Deflection Method
IECTtOII1U """"' .. Clnrnul
__
OpJ'ltl II\; .. kld\\! Jlrl'dll'll llll tho.: ~rI
Ixxl) ..,I thl,; Jl1lhl In uccl)rdance
W1lh
CUll'lI s lhmJ 1.1\\ 0, .lprl)lllg thl' Ilwml'nt l.'qUlhhnum equation
M.
to thl: 11 ...' ( bl~
of l(llllt H \\ .. oht.lIn thl: eqUlllhrlum equalll,"
t:
~Bf\
0
\11l(
J
Inl R
I t " ToJ d.:lI:nmnc the unKno" n lomt rotalt..ln.. ". \.\1: SUhstllulc
Inp.:--dd'k-ctHln etjUJtllm f q
~ .\Od .' I mto the cqUlhhnum equation
(q 5 III (lbt.;,un
th..
o 1UE/OII
.p"
IIIMI II
• 150 I
0
A
I (j(,X
fwm \\hh:h
18 fl
h
£1_
\It milt r Fnd \I, III nlf The mcmlxr end momt'nh can no\.\ be computed
hy uh tltuUng the numl'nL.11 'alue 01 HOB bad mto the lopc-deftechon equations hi I through 4 Thu
:' . b-IX
M~.
\/1B
OIlX
\1 ~H
(I
\1 il(
o U'
\/l II
O.ll6b71
Ib
'645 i
1015 k·ft
1'0
1015 k·ft )
JM 5)
or
If
B
I74J k·ft
I SO
101 5 k·ft;
174Jk·ft;
or
~otc
Ihat it J'llhlli\c an...\\..:r for an cnd moment indicates that ih o;ense is coun.
h:n.:ln':.... \.\I'oC. "herea.. a negati\e an,,"er for an end moment imphc3 a c1ock\\lSC
-.en....: Sinl:e the end mumenh .\f8~
and .\IBe are equal in magnitude but oppo.
'>Ite In '\C n~.
the cquihl:lOum equation. \1 8 I +- .H 8(' - 0, is indeed satisfied.
.\lo1/ha Em' Shc·t", The member end
,>hear~,
obtained b} considering the
equilibrium of each member. an: !'Iho"n in Fig. 16.5(c).
Supporf RUJctionl The ceactionl> at the fix.ed l>UppUrb A and C are equal to
the force" and momenh at the emh of the members connected to the3C jOints. To
determine the reaction at the roller support B, \\c con!'lider the equJllbrium of the
free bod) of joint B In the \ertlcal direction (see Fi8. 16.5(c)) to obtain
B
SS4
9.K4
SIIC
The ..upport reaction acc ho\\n
in
+ 27.57 ~7.41
tb Free· Body Diagrams of J n
3 klft
20~ rJ2
6J
18
0
)741 -2 30) + 32.43
0
27
20.7
<; E 1/(
J5b
81b 55
1845
374\ 30
230 15
174 J
C -471
Member End Momen and Shean
3kJft
CheCkS
0
0.2 ",0
I
21
B =477
63k
CheCkS
Sh rand B ndrnq \lomn" DUll/ram The shear and bendlDl moment diagram can 00\\ be .,;on tructed b) usmg the Mum SJyn cona: "'10" delaibed
II n S I Thne diagram are ho\\n In Fig. 16 S c and f
AJtS.
201
c
Ans.
EF
1~
21 7p~r q20
k1
fig. Ib.5(dj.
Bonde
3 klft
20.7 27
Eqwllhrlllm (ht'Ck To t:heck our l.:alt:ulation.. of member end shears and
upport reactions \\e ap l~
the equations of equilibnum to the free body of (be
entire trUl:ture Thu !iCC hg 16.5(d
8.16
Beam
356 k-ft ":
432
.'\M5J
Coo
4
tk
...1"
J kJft
1 ~D;')I
(CC.CI::
02 Ie
-
-
CllAPlElt1. _alii 0.............
. . . . . . . . . . . . . .1.
•
21
-21
(e) Shear Diagram (k)
51 3
_
•••. 0.1 II IE/Ie
4
102
M,1U
-102
(0 B,ndillll Moment Diagram (t II)
Ioluaan
IHgru of Futdom
(JB
and (Jc
FI td End Momenls
FEM.
324k-ln
or
324k II
48 6 t-Il
or
48 6 k.fl
~
81 k-Il ,
or
81k II
or
81 k.fl
486kll'
or
+48.6 k.fl
324 k-II
or
34k
SECTION 11.3 AnaIyoiI 01 _ _
R( tollOn To delcnmnc the unknown j~mt
rotation Os, we substitute
lhe slope-deflection equatIons Eqs. 1) and (2)) mto the equilibrium equation
Eq. 3 to obtain
JOIllI
187.5
0.lE1 08
+ (0.6EltJ s --+-- 300)
10kNim
- 0
-
! - - 6 m - + - 9 m_ _
or
EI • constaftl
0.9E10 8 - -1125
(a) ConIlQUOUS
Beam
lOkN
from \\hu::h
EU~B
- - 125 kN - m~
\20 kN m (
\#emh , End \fommt The member end moments can now be computed
b) substituting: the numencal \alue of £/88 into the slope-deftection equations
Eqs. I and 2 Thus
\18. -OJ
WBe>
06
lOkN
fb StabCally Detenninarc
CanliIC\'ct Portion
225 kN .mAns.
or
1875- -225kN-m
125
125 +300= 225 kN m~
Ans.
30 kl'l
'[em"" End Sht:tJn and Support ReQcI;om" See Fig. 16.7(c) and (d)
52 5 - 15(20) + 225 - 60 + 82.5 ~ 0
0
mNk021~B
Ai
Equlllhrrum Cht(k See fig:. 16.7(d)
+ILF
Checks
(c) Stabcally lnde1mninate Part to be Analyzed
+(LMD~O
-525(20)
+ 15(20)(10) -
:-:-_-:1D
IC
225(10)
+ 60(5)
~
Checks
0
IEumplI1U
(d) Free-Body Diagrams of Joints B and C
Detennine the member end moments and reactions for the continuous beam
shown in Fig. 16.8(a) by the slope-deflection method.
6873472
SOlution
13(....,...---::-) (IBI)
Since the moment and shear at end C of the cantilever member CD of the beam
can be computed dlfcctly by applying the equations of equilibrium (see Fig.
16.8(b) It IS not necessary to include thiS member in the analysis. Thus, only the
mdetennmate part AC of the beam shown in Fig. 16.8(c), needs to be analyud
Note that as shown m thiS figure, the 120-kN . m moment and the 3O-kN fora:
exerted at jOlOt C by the cantilever CD must be included in the analysis.
FEM
...
FEMe.
-~
12
675k
m)
BI 275;;; 275 I B
CIllO
687
I
3472
.
By' 41.59
5527
6.87
13
~7 DI"- rj;L l~! . 1r4!J-._
I
0
675k
IA
687
m)
or
or
675 k
+675 k
m
m
I
... lU
I
4159
OSupplldIleo<""'"
120
C • 8527k
30
FI'C J·£nd Alomml
FEM6A
(DJi]j]m) CW) ~'-(
(e) Member End Momemsllld Sbean
lkg"e~
of Freedom From FIB 16.8(c) we can see that jOlOts B and C are
free to rotate Thus the structure to be analyzed has two degrees of freedom
\\hlCh are the unknown Jomt rotations 86 and 8e ·
FEM,."
552730
~127
lOkN
IC
30
..
ClIAPmI'8
SECT.'•.3
SIope·_ MolIlod
It> Q) t\l nwmhc:r "B lind Be
'EI
\I
tl
0
211
\I
'F!
\I
9
-= ~_
n3F10 s
[J
n
~
0-U4£/88
07 '
o ~2n!l{
• t>1
'i
~m
'1./
9
\I. "
~=~
lM~1I0.
"
0
~1 6: .5 -
........ at ConInllll . . . .
(''' 5
0
(I ~ £ l n 8
11444EW(
l'tl S
£_7 CPa
4
C""tinuous_
r WI! fi/I II n{ 8~ l..-on"lI.knng the mornen! I.."quihbriurn of the fl'C't
I:o...\(hc of l(llOts B .J.nd ( fig 16 Md ~c
obtain the equilibnum ~quatl n~
Equl
()
B
6
). Inl
through -t
R
n Suh<.tltutl\ln
the ..Iopc-deflection equatlonll
mill the equilibrium equatlom lEq". 15) and (6) }ic1d..
1011
I 111£108
\If
,
o 222EIl1c
Eq\ 12
- -675
fbi Owrd Rolan
Due t Suppon Sdt ~
(7
(8)
8) ..ohing Fq
EW,
7)
and II( ..imullaneousl), .....e detenninc the .. alues of ElfI s and
llI.-
10
ElO H
4125 kN m 1
EIlJ{
-97.62 kN m 2
(el free-Bod) Diagnutlll of Jomt Band C
\I.mlta End \lOlllt'fl/\ The member end moments can no\\ be computed
by lIuhstltuting the numem:al .. alue.. of £/8. and EIlJ( into the slope-deflectton
equation.. l-.q\. I throug:h (4))'
311
41.25,
,~I\
(J
\l~
() 667
412';
It,
0444
412S
-131 kN m
275 k.
0222
m
97.62
or
137 k
m) Ans
or
275 k
m)
675
Ans
27 S kN m
\I
O'l"ll
412S
120 k
m
() 444
9762
675
or
120 k
m)
te that the numemal .. aloe of \I"... \I"
nurn equall n Eq
\1
m
EnJ 'a~
f. ull , unr C
d Member End ttl
MS.
MS.
and \Ie" do salls! the eqwbb-
S and 6
and Support Rt'Q lrom Sec fig 168 e and f
Ii The eqwhbnum equations cha;k.
AnS
M.1U
n and
117
-
C1W'TB116 $Il1jlt-Delloction Molhod
SECTION 1U . . . . . of ewlt
.>.
105
It'
-
1
C
8
A
Til . . . . .
n
M.
2EI 8.
-IN \7
She-dT Diagram (kN)
(I)
2£1
\I
91
See FI
9
\I
Jom, RotatIon. Substltutlon r the ~Olpe-<Icft:i'>n
mto the eqUIlibrium equall
o
(8) Bendmg Moment Diagram (kN· m)
lU cold
40
O.
By solving Eq
SolutIon
Filed-End Mom('nt~
moments are zero.
8.
cate the chords (not the elastic curves) of the members in the deformed positions
Because the length of member AB is 8 m the rotation of its chord is
0.02
8
-0.0025
whIch the negative Sign has been assigned to the value of "'AB to mdlcate that
Its direction 15 clockWise as shown In Fig. 16.9(b . Slmilarly the chord rotation
~'"
0 ~2
M.
MIA
CD
"011
0
Equations Applymg Eq. 169 to members AS
WC\\inte
2EI
8 O.
00075
and
AnI
or
91 k
m
Ana.
M,=
m
or
56 k
m
An&.
56k
S6k
m
AnI
-18k
m
An&.
MD
An&.
and Suppor 1IIoc
See F... 16
I:
Be
m
m
Efoi/ibr- c
l/lco
Sl 'fN-Dejk
AnI
91 k
M"""" EM _
From Fig 169 b we can see that
98 kN m
101",- 91 k
101
_ 0 0025
0002 rod
Member End Moments To compute the member end momen
lute the numencal values of (J. (J and El
70 800
56 000 k
the naht Sides of the slope-deftecllon equation Eq I throup 6 I
In
Be IS
00005 rad
o
Since no external loads act on the beam, the fixed-end
Chord Rotalwln The specified support settlement is depicted in Fig
16.9{b). using an exaggcr'dted scale. The inclined dashed lines in this figure indi-
000 5
9 and 10 Simultaneously we detcrmme
Degrers of Freedom Os and fie .
for member
<qua,tio",
through 6
-9K
fI&.
o
M
0
236)
4
II
1fl:IIOlI1U _ _ .. c.Coo"",,
F
_
1ft
: : 1U
Determm the member end moment and reactions for the three-span contlnu
ous beam hown In b 1610 a due to the umformly dlstnbuted load and due
to the uppon nlem nt of In OIl B J In OIl ( and .. In at D. ~U
the lope
deflection method
AI!-.UJL . 4~ .J:rb...
~.& !. l~
B
~20ft-l1
ft
£ 29.000
Solution
FI
C
liB and He
J-End \I
l!sc
m nit
B
FEM..
FEMe
FEM..
FEM.
FEM/>(
FEM
1_ 7.800 lD 4
c.m;...... 1lcam
lH r
I Fr.
"" Although all four Jomts of the beam are free to rotale
" \\111 ehmmate the rotations of the Imple supports at the ends If and D from
the an h s b) usmg the modified slope-deftection equations for members AS
and (D respecmel) Thu the analysl \\111 mvohe ani) two unkno\\n JOint
rotatton
220
12
_ 66 7 k-ft
~
or
J
or
0.0026
1Y
(el Free Body DiBgmns of Jomls B and C
the hard rotation for member CD
1 38
.
41 38
I
81 79
B_I2317
000313
II EqIlQI
o
,.,11.1'
000039£1
100
41 79
I
C_6219
Cd MembcrEnd- andSheaB
IS
OI5£IB.
808 41 79 204
808
o::=IT=o)
(I I) ( o:TI=o) CW) (o:TI=o
lA
BI7 24~ IB
CI
IC
I
000365
100
-f~
B
in which the negative Sign has been assigned to the value of !/IA' to indicate that
Its direction IS clockWise as shown m Fig 16.1O(b) The chord rotation for
member BC can be computed In a Similar manner by usmg the settlement of
upports Band C From Fig 16 10 b we observe that the relative settlement
between the end of member BC IS I tn
tn
0875 In. 00729 ft and
0026
-
-66.7 k-ft
41 38 81 79
M
to
(b Cbord Rotati... Due to Support _
66 7 k-ft
00521
20
.[H
}
c
66 7k-ft
(hord RllallOns The specified support settlements are depicted 10 FIg.
1610 b usmg an exaggerated scale The mcllned dashed hnes In thiS figure m·
dlcate the chords not the elastic curves of the members In the defonned POSItions. It can be seen from thiS figure that since support A does not settle but
upport B settles by In the relative settlement between the two ends of member AB is In
00521 ft Because the length of member AB IS 20 ft the rota·
tlon of the chord of member A 8 IS
Similarl
0
C
4
HCTIOlIII.3 AIlIlyIlI ol _ _
172
CHAPWl16 S1opo-_" Metllod
'.,
'f1
If
u~
If _
o "I'tl
111 flO,
~/.I
"-
.. J
20 _I
o It/II,
In
II
0
Equ I J,
11m
j, Int
0
.:'11
II ,
through ~
1{
II,
3
66
(\ ("\(111 f.I
n ()().lbS I
i
"" 7
n:'/HI
n lMIII f.I
unOlll
IOn
"f>'
OI5UOt,.
Ion
O lM~71.'
•
Ans.
Equall" St.'\: hg Ib.101.:
\lBi
'f B(
\fl'
\/ 1 0
0
6
0
-
Ro at III 8~
uh tltutmg the ,lope-deflection equdtion Il:q
1010 the eqUilibrium «Iuall0n\ I Eq"_ 5. and (6). ",e obtam
o 'SElfl.
Sub~ti ul ng
EI
equatton.. )' Il:h,h
0.1 EIfI{
0.00149£1
o I E1f/ B + II 'SflfJ(
0.00063£1
I
:!I).lMXl 1",KUO) I:!' ~.rt
0,1£/(11/
By gniho~
• t>tl 7
i)I)()lhS
+ ~:U
J13
~ into the right ~ides
().3SE/fJ( =-
be
EJO II
£If"
-2.307.24
PI
39,2 0.1~
(KI
,
A
- 1,131.57 k-fl'
"'tn/her End .\lOInol/\ To compute the member end moments, \\c sub'ititute the numem:al \alues of Elfl, and E/O e back into the ~Iope·d nection
equation\ IEqs f1 throug.h 14)) to obtain
427.7 k-ft )
-427 7 k·ft
\.IS(
421 7 k-f1 ....
Ans.
"'<'8
!10K k·ft ,
Ans,
Ito
KOK k-ft
808 k-ft )
I-- L - - \ - - L - - \ - - L --l-- L--I
Ans.
\lilt
or
P 'l~B- --.,-tJ ';g:q
£1 = constanl
Ans
\#l'mb.., E"J Shran and Support R('dcl;uns See hg. 16.IO{d and (e Ans
EquIlibrium (nt" Ii The equilibrium equation~
check.
\'. e pre\ IOU I) analyzed lhe l,;onUu~
beam consuJered ~ h
m Example 13 14 by the method of L-on\1 lent defonnatloos. Theoreticall) lhe slopedefla..1.lon method and the method of consistent defon ation~
should yteld
IdentICal
ults for a gIVen structure The small dlfferellCe$ between Ihe mulls
determined he and those oblallled m Example I '\.14 are due 10 the round-oft'
errOR
,
{
6.268.81 ~tf·k
0'
HI
II
FEM
of the abo\c
Eq"'. 7) and (8) Imultaneou,Jy. we determine the "alue\> of ElO s and
EIfI( III
If
(a> Continuous Beam
,.,
Ap-ib'le
I-- L
I
L
----j
111) One-Holf Beam with Symmetn< IIouDdUl' Condi-
... 11.11
E
L
m
RCT1OIlI6A .....,...., _ _ IIIl ei_Of
Of
CltAPTfR 16 Slope-DenllC1lon Melllod
"l. l\L
.,
~
2
T
1
r
12
"i
"2L
c
8="L
DI'
,
.L-
.L-
"2L 12
12 "2L
12
2
,'o)(kb
(c:ilJ)
1
" I) (181) (el', 1)(lcl)(,
81 .e!!!...
~
18 cl .L cI'~ 1 .Lo
L 12
16.• ANALYSIS OF FRAMES WITHOUT SIOESWAY
nL "L
HL .... L
.. L
12
I
D
L
12
E:I'
;1.
"21.
D,="L
Cd) Member End Moment and Shear.>
,
,
,,
CI" 1 I'
•
~I'
.L
.L
T
.L
T
T
.L
-T
I
I
.L
T
(c) Support Reactions
.L
2
24
24
.L
-T
.L
-T
A
V,
.L-
-If
wL 2
.L'
.L'
E:
.L
.L'
)12
E:
.L
.L
.L
T
T
I
I
I
D
C
8'
.L
~
D-
8
A
24
c &
D ~
\/ \l, V
..L
-If
..L
-If
wL2
-If
wL 2
24
~
E:
wL'
-12
(8) Bending Moment Diagram
(0 Shear Diagram
R6.16.11
coold
The shean. at the ends of member AS are detenruned by considenng lhe eqUIhbnum of the member
The shear and moments at the ends of member Be can now be ~natbo
by reflectmg the corn pandmg responses of member AB to the oght of the
aXis and the member end moments and shears on the nght half of tbe beam can
be ddenmned b) reftectmg the correspondmg responses on the lefl half to the
other uJe of the au The mc:mber end moments and shears thus Obtained are
!ilhO\\n In Fig 16 II d and the support reactions are given In Fig 1611 e
The shear and bendmg moment diagrams for the beam are shown 1R Fil
16 II f and g respectl" I)
AnI
A thl eumple silo.... the uuhzatlon of structural symmetry can C()IlSider
ably reduce the: mputatlonal effort reqUired In the analysiS The beam coosid"'
ered 10 this
mple Fig 16 11 a ha three degrees offrcedom, 8. 8 and
Herby kiD' ad aDtase f the lrUcture s symmetry we were able
IInma all the de
of freedom from the analySIS
PI
1KT1CIII1U . . . . ,, _ _ '.111'_"11
ClW'IEIIl' ......._ - . "
c
I
80'\
\
B
(a) Frune without Sidesway
~
D'H~T-r.:1
E~
I
I
I
I
I
I
I
I
I
I
B
(b) Frame with Sidesway
D
I
I
I
-~­
,
E
,
,
1
,,
Axis of
......- symmelr}'
\
\
,
\
I
I
I
I
B
£J=con
... 11.12
(ej S)'IlUDCIIIC
tant
Frome SUbjcclallD S)'IlUDCIIIC
Loading - No Sidesway
JOIDts prevents one JOlOt translation ID Its axial directioo. The Dumber of
independent)OlDl translal10ns
I then obtained by ubtractiDS from
the total Dumber of _ble translal10ns of I f.... JOID" the Dumber of
translalJons _
by the supports and members of the fnunc We
ean .mfy our <:onclUl1Olll about the frames of FIgs. 16 12 a and (b
."
m
IlCTIOIIIU ....,... 01 , .._ _ ......,
CHAPTER 16 $IotIe-Ooftection Molhod
f
In
dEnd M
nr B umg the ~xif
r fthe book \II; bl
Id the back
40 20
FE"',
HM
FEM
lhl Fr«·Bod,. Dlagr m~
- 6~
C
I~
21 2
21._'/
~I_
I
L
! lifl
I!'-:D~
\.I C
~
'2 15 68_~
21l
LID I ~
2265 L
'-V -.
J! '5
I~D:-'
36.86
69.21
~1I'i9
---h
193
21.2
1.45 -
D
ext
.2 klft
-'--"-22 65
IE
vtC
"nle the
It,
23.14
k ,
k
9*
69.21
(C I Member End Momem... Shears and Alr.iaJ ~ecroF
23.14
-145
9.7,+6921
(d) Support Rcacuons
.2E/
(1
20
.,
100
o IE/U
100
2EI
20
it'D
2EI 0
20 (J
\fDB
2EI(2H
20
D
ltD
2£ 21 12l1( +HDI
+ 1;0
0267£111
o )'\3EI
It
2£ 2J 20D + Dc - 150
o I HEW
02
02£1
"
It
B
h«lt~epO
\114
1,4;- B
"'- 11.13
FEM
k
AC BD and CD and Eq
4011._
18S_ A
k-
I
of JOint!> C and 0
276~
C -
2 )(,
FEM
FEI
1:11864 \.1::1."::1
-1.,205.7
115.9
27.65
100 k-
I
(al Frame
k t
FEM
FEM
E=- !Y.OflO II.
100kft"
100 II. ft ~
4
nd m
In
30
"l
100 - 0 2E/()
100
0,1 fJOp
0.2£/fll)
£1
JO
(ISO )~I
ltD 0
lib
EqIMlI
By appJ)lR the mOlllC'n
'[; MliJtw0 ;~
em: bodies of on' C and D F 1
eqwlibnum eqUibODS
\t
It
An.
-
_,U ....,.....
_
_
11 . ; Da. 1', _
DEC
0
~
u--
tn.
=~':ti;
B
1--30 ft---+,--30 ft
;;'29.000)(8011)/(\2 1' k
t ml_l_
A
E=29000ks.
B'
(b) CbonI RIMa.ODS !lbe 10 SIIJIIIOR SettIcmco"
a) Frame
MDC(:f;)MDE
..v
MDt
£1
c) Free-Body Diapams of Jomts C and D
M
46
C
•
~
854
411
411-:-[;,1
.,...
54~1
L
467 254
::-: ' \ 4 II
C:-
De
467
76.2
floh f :-~£
T_I
467
48
-...j..-I069
46
JO
2.54
-48
t
254
M
721
--h 92
~S48
C _411
0.69-
A -411
069- B
M
o
541.8 +54.8
4'-Y
4~
46
4
721
d Member IilId MomeIIII _1IId Axial FoI<:ea
.:+
4.6
... ,..,.
4.6't'
7 1
SuppoR-
11.5 MALYSIS OF FIWIII WI11I
-
-
ClW'TER 16 S/opt-Dellection Ml1!lod
1ECnoI1U AooI,oio "' _ _ .'.'_
---L---
whICh the ch rd
11 n
be negative beca se the ar I k
FI
rotahan can be expressed In t nIlS f the
S1denng the do placemen, diagram f) I
16 16 b SInce CC I perpendicular I AC hi b I
fll wIth the vertleal CC must make ,he same ngl
tal Thus from the diSplacement diagram f J In C
In
(dl
Re.l&.16 conld
frame to an arbitrary horizontal displacement .6. and draw a qualitative
deflected shape of the frame, which is consistent with its support conditions as well as with our assumption that the members of the frame
are inextensibJe. To draw the deflected shape, which is shown in Fig.
16.16(b). we first imagine that the members BD and CD are disconnected at JOint D. Since member AC is assumed to be inextenslble
joint C can move only in an arc about point A. Furthermore since the
translation of jOint C is assumed to be small, we can consider the arc to
be a straight hne perpendicular to member A C.
Thus In order to move joint C horizontally by a distance A we
must displace It in a direction perpendicular to member A C by a dlr
tance CC Fig 16.16{b so that the horizontal component of CC
equals A ote that although JOlOt C IS free to rotate Its rotation IS Ig·
nored at thiS stage of the analySIS and the elastic curve A C of member
AC IS drawn With the tangent at C parallel to the undefonned direction
of the member. The member CD remains horizontal aDd translates as a
ngId body IOto the position C D) with the displacement DDI equal to
CC as hO\\ol1 In the figure SlOce the hOrIZontal member CD II a
sumed to be InexteDSJble and the translallon of Jomt D IS assumed to be
small the nd D of thiS member can be moved from Its deformed poll.
lion D only In the vertical direction. Smlllarly smce member SD IS also
can see that
CC
cos
Next let us consider the dl placement diagram
It has been shown preVlou I tha'DD I
and parallel to CC Therefore
DDID
DO,
and
DD
at
DD
n
DD
A
t
..
-
680
CHAPltR 16
Slope-DelIlCtion Method
~H
~Ol U "hu
h.J' Ib,171 thn)ugh
the- l:h(lrd r,11.1I\('11 III the three m c ~ r
lh.2Q) inh\ rq fI6.:!h)
... ltl {":Iln ... 01 \
(lhl ctln
"I.'
~
""
Ih lOot
l:\""P.
1-.
\
VBn
1iI,0
I. co... p!
\
I Itan
P
SoIU11on
Dr,
,hUJ.~1
161 b
- tan
Pol
1~
l d(6I.~1
01
al .. in
By ~ol\ing
obtain
fli + 0]
= tI~
cos
II]
~in
II:. = L
cos 11'1
Eqs. (16.3la) and (16.3Ib) :)imultancously for
"I -
a~
L
co, Ii, (tan
P, + tan /1,)
i.
- ------ -- -----cO,/I,(lanp" tan/I,)
411 kN
Jil<;
The: h1fcgomg c prc"lon, of l:hord rotations can he u-.ed to "Ole
the I, ~-d k'\. uon
cqu.ttll'n lhcr b~
relating member end nWmC:h1s
to the thrt."C unkno\\ n jOint di"placemcnh Oc OD, and .,\. A.. In the case
of the rcdangular frame ... c"n'ldcn:d prc\iou"J}. the thn.."1: eqUlhbrium
t."quatlon, nl..X"C' ~ra,
for the ..noluI~(
of the unkno\\ n Join! dl'iplacl.'mcnts
can he c..tabh hc.."d b} ...umming the moment'<. aCIlIlg on Joint" C and D
and b} ... ummll1g the hMil'onlal fon;c" acting on the entire fr,lme. Ho"_
C\ef. for framc!'l \\!th IIh':)IOCd kg . . , it I' usuall) more comcnicni to
e..tabli,h the third equihbrium equation b) ~ u m i n g
the moment-. of
all the force' and I.:ouple~
al.:tlllg on the entire frame about a moment
I.:cnter O. "hKh i~ 1(l(.lled at the intcr-.cction of the longitudmal axe:.. of
the t" 0 indined mcmbel', a~ ,ho\\ n in Fig. 16.16(d). The location of the
moment center () I.:an be determined b) using the condit on~
(-.et I-ig.
(1~. I'
I
C
T
401tN
0
c
I
0
5m
7m
8
J.'~ -fL
-
8
a°'4..Ju
£/ =confotant
(a)
Frame
(el Free-Body Diagram of the Entire Frame
(l6.Jlbl
tI]
and
d!.
\I.e
~C' ----,
,-
(16.12.)
C
,-- .
~ID'
I'
b21.~ (
,,/
I :
I
I
Sm
lI'AC
J ,'
,,
,
'1BO
8
,,
Multistory Frames
The foregomg method can be e:\lended to the analysis of multistoJ)
frame subj«ted to sides\\ay as illu!l.traled by Example 16.12. Ho\\e',er bo.:ause of Ihe I,;onslderahle arnounl of compulallonal effort 10\ohed the analysis of such structures today IS perfonned on computers
u)lng the matnll. fonnulalion of lhe displacement method presented 10
D~
'--;;00;:---,--...,-"""'1
I :
I '
I :
I '
Once the equilibrium cquatioO\ have been e!<otabli!<ohed. the anal)sis can
be completed m the u'ual manner. as diS(:ussed pre\ iousl).
(hapter 18
"
A
lb) QualilabVe
Det1ected Shipe oftbe Frame
FIlL 111.17
d) F....Body o;.,r-
C....... AC .... 8D
-
-
IECTI0II1U AnIIyIIa of _ _ •• Wi,
I, d f.nd \I
Ide th ha k
rEM
2k
fEM
411
fEM
fEM
4
k
294 k
m
fEM
H
HM
(It rd IW
~
If
5.8- B
~7
~
0"1
El
I
16.47
I "EI
A _5.8
146~
2£1
23.53
5
Ie) Member End Momcnb. Shean. and Axial Forces
IfDB
II,
40kN
c
I
£J~
2£1
5
o14E~
04EI
'1
"Inv
o 24E/~
08EI
2EI
211
7
flo
2£/ tJ
7 I ,
2HD
~
392
OS'IEIfl
026£1
294
0 286EIH
0" lEI
4
D
£qulllbrlum EqUOtlo1U B} considering the m ment eqwbbnum f J n
and D ..l ie obtam the eqwlibnurn equations
B
0
\I,
Y'
0
s
-5.8
In which S
16.47
s
o
and S, repmcnt lhe shea
stw-.11 m Fig 16 I
ImD
~6.41
umns F'lI- l I d
23Sl
and
<0 Suppon Reacti....
... 16.17 amid
\J
To estabb h the lhird eqwbbrium equation
equatlon L FtOto the free bod of the ent
'-f.-I77
5.8_ 04
W'4
S
CHAPTER II
_
III
:n3J ft /
,,
,
,
,,
,
,
,,
,
,
,
,, "
,
,
,,
o
-f1j';"- - - 1 ~
26.61 ft
/
,,
,
,,
c
x.:.'884
D
1762 9 1.7
.
leI Member
16 ft
-
End_
MIDI
i~
$BD -
D
t
1---20 ft---
(c) Free-Body Diagram of the Entire Frame
B
A
1764-
'Y
2
911
88
I)SupponIteoclioas
M
rh-SDIT
DI
D
I
16ft
s_s,J
'fMID
""1.,,
.,
1IC1IIlI1U ......... _ _ ....
' _
(d) ......Body Diqnms or CoIUlDD1 AC ODd BD
...,1."
COlI'"
-
..
CHAPTfR 16
Slope-Deflection Mertlod
1lC'l1OII1U - , . . . " ' _
./\
\
,,1
f
'ti
\I
'ti
't/
'!.H
1/
o Ot"''i \
I.
't/
\I
20
2fl
10
1/
o 021.an \
1I1'J1"F/O
II OblCi \
I.
onlHIH/\
II ~I "
II 1)6"<;; \
'11
1/
111118f1'/\
t11l1fl
II 1)(,"5 \
'11
1/
II
o 021-1U \
II ~51 )/
•
11.1 nfl,)
~
Q
l (loY'5j
02t-W(
, I
II,
.1 II oY', \
o 2.ElOn + ()
11:1fJ(
1111111//\
~l1 )(O
,
FqUllthrwm f.quCJIWnI B) con ldcnng the momenl eqUilibrium tlf Joml
dnd D
tht eqUilibrium cqu.Ullm..
"I: ~)hla,"
\I l f
\IllS
"'!
\lrD
0
,toc
0
,
The third equilil:\num equation i c'labh,hcd by .. umming the momcnh of all the
fon.-.: and couple.. actmg on the fn:c body of the emire fr<lme abnul poml 0
E·••di/,,,urn (
"" hil:h j.. IOl:atcd .tl the inlc!"'C..'dion of the longilUdinal axe., of the two l"olumm.
a.. ,h('l" " in fig. ItI.1X(L:}. Thu..
T\L
\If{ - \f( (53. B) • Jl s /)
()
\10
S80(42.67). 10(26,67)
In .... hir.:h the ,hear.. al the lower end.. of the columns can be expre..sed in ~mrcI
column end moment a, ("oCc Hg. 16.18(d))
SHU ~
and
B) !>ubstitutlOg ~cht
~erpx
.\1Hf}
0
of
Det nmne the
det1lXllon f J I F
+ .\fIlS
16
defteclt n m thod
IOns into the third equilibrium cquation. "'"c obtam
9
Sub)titution 01 the slope-deftection equation) Eq~
IOto the «jUlJihrium cquation\ (EW" (7 through (9)) )Ield!i
Jmn, DI pIal t'ffl'"
through 6
"4E/IIr
8)
I\lOg Eq
0 I E/ll o
U 11JH(
II 45J:..Jfl o
1I711:JiI(
0877f:/H o
HI through (12
l),lKJ"'5£/ ~
OI2EJ~
0.183£h\
0
10
0
II
800
12
Imultaneou!il} ","e dctennine
HOc
66 648 k-ft-
t/flo
12"i 912 k·ft
Elf>
=
5233 6k·ft
I Example 16.12
T
$'
-
1M
CHAPTER 16
Slope-Deflection Method
P,
SUMMARY
In thl' ,h.tph:r. \\co h.ne stUJI..:J .1 d,t lI.:,d fonnul.t110n (If thc 1.11 pia
ml;nt ..tlllncss Illdhod. ,.tlh.'d Ihe slop..:-ddll,.'\:tlon mcthod. fur thc anal
I'~
of be.tm .tnd ff.Hnc Thc method I' ha-.ed (10 the: ..lope-deflection
eqll.Hllln
\I
~EI
_'II
(I
1
1
.ljI
~ FF'I
'''"
169
\\hu:h relate the ml,mt.'nh Jt the: emb of a member to the rotation and
hnemc~alr id
of 11 .. end, and the e'\ternalload.. applied to the member
The rrll\:t.'durc for .tool!) .. I ' <:-"entiall) im·ohe.. tl) idcntif)ing the
unkno\\11 Joint di'plol,emcnb ldcgn.-c.. of freedom) of the structure, 2
for each memb..:r. \\riling slope-deflection equations relating member
end moment to the unkm)\\n Joint dlspla,ements; (3) establi..hing the
equation.. of equilibrium of the ..truclure in term .. of member end moment..; 14) ub..~nitu
Ihe ..lope-dcflection equations into the equihbrium equatiom and ,ohing the resulting s}stem of equations to deter.
mine the unkno"n joint displ.tcements; and (5) computing member end
mOffic:nt .. b} .. ub,tituting the \alues of joint displacements back into the
slope-deflection cquations. Once member end moments have been C\al·
uatcd. memhcr end shear.. and axial forces. and support reactions. can
be detennined through equilibrium considerations.
IOOkN
Af=~
-
1"1 ...... IU Del "mne the re t.1lons dnd draw the
r d be ng m men! dlagr m for the beams fooho"n
f
PI I PI 5 b) u ng the I pe-deftectlUn method
I k
(
lJ
/---1-E.
f
6m
2/ - - - = constant
"'-PIli
25 kS/m
A
8m-----l-8m
fI
8m
E=70GPa
£
1= I.lOO I Hi'l mm"
2kift
25 kNlm
A
I
10 m
,
£
FIg
fIG.
Pt6.5
I
2~m
I
(
8
I
10 It
£1
15 It
15 It-l
15fl
1. =29000kl
fI=""......
=200 GPo 1=500( 1()6) mnr'
III.Pl8.10
... Pt8.2,PtU
20ft
1= 165010"
~
1---15 It--I-
20 k
15ft
,
f!! I ! III ' 1 ! ; ~
£
D
GPa /
"'- Pt6.ll, Pl8.15
AG. Pl6.4, P16.7
10k
I
... Pl1.1
ID
I
fIG. Pl6.3
PROBLEMS
1... Sol\c Problem 16.:! for the loadmg loho""
PI fl .2 and a settlement of I in. at support B
I
_9m--+-om+6m
A~C
.-,1.3
UXH
III.PtI.11
I
11M
CHAPTER 16
Stope-Deflection Method
IH
I
A.....
B
'T- o 'r51t~lIOJ-
-I
\0
~tH
I
....e
t
/l
~J
F
__
T
/)
3m
+N
II c I.:on lant
f.• 20() Gp'
3m
~
1
E
I
....
~9nOk
1
I
1
"
\11 kN/m
2'\ kN/m
~
L
1 c 4fXl( 1061 ~
1=:'\.onOm"
Pl6.12, Pl6.16
fill.
FIG. PI6.17, Pl6.21
•
IfIm
~
III
1M ""il
Be/}
6m+ bm- i-6m-l bm~
£1
fIG
""
EI -
alii II}
2 lift
B
T
= cetnstaJII
Pl6.13
J
1='~!:;C,*>
A
1~em=I
2:
1--1--+- 20 II
B
10ft 10ft
;:l=In~f.1
J1O/}
f1~
+
:A.
F
G
11.21 Sohe Problem 16.17 for the I dl
Plo.I"7 and a -.ettlement of 50 mm at urpon
ronslant
=29.000 ko
I = 3.500 in. 4
£1
f.
=:
15 I.
--'-----+--1-5
40"
15 .JfI
P16.lts and a 'iCulement of"
RG. P16.18, PI6.22
tn.
dl
n
FI
I
I I
I I
I F-
/}
C
lO ft
~
~
I
4Ok-
30 It
15
B
+
' ..17
+
2n1l
' - 1...
5 ft
21
, . . . 0.:
I:
fill.
Pl6.19
= I.:on lant
PllI.25
-r--
30 It
\0 It
£1- comlanl
f.= 1O.000k I 1-3oooin'
H.
A
c
FIG. Pl6.23
......,U
II.J4 ........ 11.31 Detenn'" the ..... b<r end
.tnd rCactlOn$ for the frames shown ID FI
b} gn~
(he slope«ftectlon method
PI6 -4 P
flI. PI&2I
III,
.1
D
1
A
A
IOU
-t:='!~9
E = constant
II&.
101t
169 for the loadmg ho"n m fig
I
mot l .. '\mmiu upponl
k
I f.
D
err-I
U kif.
IIII
10k
---, B
3IVft
PlB.24
upron A
2 klft
10ft IOfl
Pl6.1.
fill.
18.12 Sohe Problem 16.18 for the loadmg h
201.--1--1--1
E =: confootanl
f-
n
11.23 Determine the member end momenl and r II n
for the frame in Fig. PI6.~J
for the loadmg ho"n and t
,upport settlements of I in. at A and 1 IQ at J) ~
the
nol c tfed- p ~
method.
.
- - 2I----!-fill.
PlB.20
D
100k·ft
15ft
4Uk
~
lin tant
...
~ '-
i
1 r--
8
25k
-rlllllllil
:!Ok
C
-
(,
0
12 It
Moment-Distribution Method
16~
ft - ' + - - - 1 2 ft----I
4~
'l8.27
17.1
17.2
17.3
17A
17.5
81
J
£1= con lanl
fl5.
17
3 kI(1
C
4ft-j
160
ft
fl5.
8
'l8.30
18k-
I~
2/
C
2J
~I
0
ft
-+1
15 ft
The Empire State Building,
Veil' York
Proto eoortesv of Bethlehem Slee COfP(ll'al
_L.!!
.1L
1 - - - 2Oft--+-15 ft--l
I 15
/
/
--3011--
£1 = constanl
fl5.
MetIlOll
Problems
£/ = conslanl
I 5 kJft
Definitions and TImlIllOlogy
Ilasic Conceot of tile Moment ~
AnaI'jIis or Continuous Beams
Analysis 01 Frames _
Sidesway
Analysis or Frames with Sidesway
Summary
E= constant
'l1.211
fl5.
'l6.31
18 kNlm
2m
...L..
"
1---5 m---+-3 m--l
_PI...
-
no
CHAPTER t7
_ 1 7 1 Dsne."
Moment-Distribution Method
the "lllutUlIl III 1n1U1t mulUS el!u,IIIOlls. \,here.,s 111 the case of fram
\\Ilh sid \\.H. Ihl.: llumllC'1 (11 slInult,Uleous equallons mvolved u ualt
equals the num1"ll.:r 01 lI11kpc:ndl.:nt JOIIlI translations.
The moml.:l1t-(,i!stnbulllll1 I11dhod " classified as a dlsplacem nt
method and from.1 IheorcllLal \ IC\\fKlllll. It is \er) similar to the slope
defk'1.:tlOn method C(1l1sl0ereo 111 the prL"(.;eding chapter However unhk
the lope-defleclloo methtlO III \\ hu;h all the structure's equthbnum equa
lion are sallshed Slmult.wellu...l} III the momenHJlstnbutlon method
the moment cqUlhbnum equ.H1ons of the JOI~ts
are solved Iteratively by
uccessi\eh coosldenng the mome-ol eqUlhbnum at one JOint at a tune
\\ hlle the ~mal g
Jomts of the structure are assumed to be restramed
agamst dlsplaL"'Cmenl
\\e fir...t deriH~
the fundamental relations necessary for the apphcation of the moment-dIstribution method and then develop the ba IC
concept of the method. We next conSIder the applicallon of the method
to the anahsls (,If continuous beams and frames without sidesway and
finall) di.-.c·uss the analp,is of frames \\ ith sidesway.
fit
opplied
. . fa I lSi
fit
Q?-
r
if)
t
fit
I
oppIied
Gtf-t
sA
I
L
£1
b Beam w
fI6. 17.1
FuEndHin
17.1 DEFINITIONS AND TERMINOLOGY
Before we can dc\clop the moment-distribution method, it is necessary
to adopt a sign comention and define the various terms used in the
anal}sls,
Sign Convention
In applying the moment-distnbution method, we will adopt the same
sign convention as used previously for the slope-deflection method:
M
The bendmg tiffne
K of a mem r
be applied at an end of the member ,
Thus by setttng 0 I rad ID Eq 17 I
bending sttlfness of the beam of FIg I
4£1
K
Counterclockwise member end moments
are considered positive.
L
Smce a counterclockWise moment at an end of a member must act m a
clockWise directIOn on the adjacent JOInI. the foregomg sign convention
imphes that clotli.ulse ~tnemo
on JOlnH are considered po~it,ve
Member Stiffness
Con Ider a pn matlc beam AB which IS hmged at end A and fixed at
end B a sho\\n m Fig 171 a . If we apply a moment M at tbe end A
the beam rotates by an angle 0 at the hmged end A and develops a m0ment M. at the fixed end B a hnwn ID the figure The relationshiP
bet"... the apphed moment M and the rotatton 0 can be estabhshed by
K
K
4
1
L
"'
n,
au lSi 17
_ _-II • ..,... .......
-"17,2 ..... ~ o I
....._I...1II1117.7
... _ _
then the eli tnbuuon facto" for the ends B of membe" AB BC and BD
are gIVen by
OF,
Xac + X'D
005
XII<
_ 0051 -04
Xac + X'D 0.1251
OFac
OF,
KBA
X'D
+ KB(
0051
KBD
o 1251 -
0.2
M.17A
These elislnbuuon facto" ondlcate that 40 po"""'t of the l50-k·ft 10
ment applied to joont B I exerted at end B of member AB 40 _ t a
end B of member BC and the remaonong 20 percent at end B of member
BD. Thus the moments at ends B of the three members are
M,.
OF.. M
-OF, M
M,
M,o -
OF,oM
04150
-60k-ft
or
60 k·ft
04150)-
6Ok-ft
or
60 k·ft
02(150)
30k-fi
or
30 k-ft
Based on the foregoing discussion, we can state that In general the
eliSlnbutioo factor (OF) for an end of a member that I ngIdly COD
n<ct<d 10 the adjacent joint equals the ratio of the ",Iau>e bendIng ~
ness of the member to the sum of the relative bending stilfnenes of B1l
the members framtng 1010 the Jomt; that is,
OF
x
X
Funbermo", the moment dlStnbuted to or resISted by a rtgIdJ
n<ct<d end of a member equal the dlstnbuuon factor for that end
the D<g8uve of the moment apphed to the adjacent joont
FIud-End MomenII
The fiud-end 10011I<II\ CXJlIIaoi,
c:oudib
weD
Uve eli placcments of member
the
k
the book for con_t refeRIKlt.
: : ~ ibu
pan
-
bon method the effects of ont tranIIab
8IId
are aI taken onto lI<COUDI
Contliderthe
.oetIIlomeat A
8IId FEM
d
FM
on which FEM • and FEM 0 dono
the!alive
bon A between
I
the ma8Jllludeo weD the eliRlCbODS
are the same It can be seen from F g I
~ ~ebt
then the two fuu:d.."d momen act 10 the
placement
a chord
rotabon
on
to
maonW.C8U1OS
31'0 s1DJlO1
t the
IW ends
the
chord IOtabon due to relab :.a :~
Fi J 4 b then both fuu:d-end ]
(poaiu diroction to pmeal the
17.2 BASIC CONCEPT OF THE MOMEIl-llflTll1IUJIlIII a l a
_17----
1Il:1IDI17.2 . . . . . . . II "II
, .. - .....,
he
~
106
1 II" •
_
I
,L_~I
106
0
,01Ac
B
,f Balanctng JOJnl 8
AXed-End Moments
(8 Balanf>mg Jomt C
("111111111)
A
911
FEM
30k
15k fi
245)
C) (C
1
(B
BI7187171
149
01
491
FEM.
or
FEM
Sk·ft
FEM
SA
Be
CB
I os
+50
,- 6.3I - 41
- 0.4
- 0.3
+38.9
... 17.5
05 I
+75
-125
+161
8 1 I - 81
+ 14
• - 07 - 07
+ 09
• - 0.5 - 0.5
+0I
- 0.05 1- OOS
50
• -125
-71.8 I +71.7
-~
~-
or
FEM.
(h) Fmal Member End Moments (k-ft)
AS
-SOkft
DC
CO
10429
-75
+322
63
+ 27
• - 41
+ 18
. - 04
+ 02
03
+ 01
om I
-491
+491
+428
• +21.4
8111nc1n1 JDInI C
__ ,,__ 8
+ 36
•+
1.8
+ 23
• + 1.2
+ 02
• + 0.1
+ 0.2
- +:10
0
r
k·
k
r
k·ft
CIW'IEII17
_
1ICT1OII17.2
tnhutetJ moments DM c IJ and DM c n to de\eJop at ends C ofJncm.
B( and ([) \\hll;h can be C\.t1uated b) multlplymg the ncpU
unbalanced moment I e
7~·k-ft
b) the dlstnbuuon factors DF
and DF
rcspecmcl) Thus
DM{.
O"~9
OM
0 571
+15
COF,,(OM cs )
COFcD(OM<D)
_
M
75 - +42.8 k-ft
\
= 2(+32.2)
+\6.\ k-ft
SImilarly the carryo\er moment at the end D of member CD
puted as
COMD,
"_
+322k-ft
These dlstnbuted moments are recorded on hne 2 of the mo
dlstnbutlon table Fig 17 5 a and a hne is drawn beneath them
mdll:ate that JOIOI ( IS no" balanced. ote that the sum of the three
moments above the Ime at jomt C IS equal to zero Ie
75 32
41
0
The dlstnbutN moment at end C of member Be mduces a
o\er moment at the far end B fig:. 17.5(d)). which can be dctenniDad
by muluplying the distributed moment by the carryover factor of the
member. Smce JOint 8 remams clamped. the carryover factor of member
Be IS Eq. 17.13)). Thus. the carryover moment at the end 8
member BCls
COMBe
-eooeo.t
\
2 (+42.8)
I
com
+2\ 4 k-ft
These carryover moments are recorded on the same hne of the moment
distribution table a the distributed moments, with a honzontal arrow
from each distributed moment to its carryover moment as bowa ID
Fig \7.5a
The total member end moments at this point m the anal
depleted on Fog 17 5 e It can be seen from this figu", lhal JoiDI
now 10 equlhbnum because It IS subjected to two equal but opposite.
moments Joml B however IS not 10 equdibnum and It needs to
balanced Before we release JOlDt B an lmagmary clamp I applild
JOlDt C In Its rotated pautiOR as shown m Fig 17 5 e
IIaIInclng Jalnt B
JOInl B 0 now "'leased The unbalanced momenl al this J
tained by umDlIDg all the moments aCIIDg al the end B : : : :
AB and BC which '" ngidl connected 10 JOIDI B From the
distribution table I
1 and 2 we caD see thaI there
-:lO....<I~
lixed-end _ I I end B Df member AB whelal the
member BC I subjected to a 75 k.n fixed-end
16 I k II carry r momenl Thus the unbalanced momen
o
OM
OM
Rlllnclng Joint c
With JOint 8 n w balanced we can
I
table hDe 3 that due 10 the carryo r elf<
balaDced momenl at Joml C Recall thaI the 10
zontal hne at Jomt C were balanced pre 1
Th
again and dlstnbute the unbalanced moment t
and CD a FIg \7 S
OM
OM
s
0429
10
OS I
10
These dutnbuted moments are
nIed
eli tnbubOD table and
baIf these ~ : m
ends B and D of memhen Be and CD ..
table. JOIDI C then
ped.
_
SElmON 17.3 Anllylls of ContI_ _
CHAPltA 17 Momen1-0ilbibution Method
rrocl,; ~ The.; Imal 1J'~m
r ~'nJ
nwmcn.... .Ire {)btained .by algcbralcall
umm1l1g thl.: ('nlnI.:S 111 e,<llh ll,lumn III the m{lmc-nt-dl ..tnbution table
The hn tl mom!,;nt thu."i ollt.l1ned .tre n..-cordl,:d on hne R of the table d
. h"'
an
are ho\\n on the fn.'.t.'-bt.lJ) dlagr.m1 (,I l L' mcrnl'Crs In Ftg 175 h
Uh.' that the final moml.'nt satl I) the equatlono,; 01 moment cqudlbnum at JOlOt 8 and (
\\ nh the mc.:mhcr end momt.:nts "'no\\ n me~r
end shears and
upport reaLlIL.m t:an 00\\ he Jcolennmed h) cotl"ildenng the equlllbnum
of the free hodu: of thl' mcmh..:r and JOlllt of the continuous beam a
dtscu sed 10 Sl~t1in
16..:! The ..hc,lr Jnd bendmg
. moment diagram ca n
then be construdl--J In the usu.1I manner b) u..mg the heum Uin (on nI
"
m
Thu the balanCing of Jomt ( mdul:e th foUowm
ment at end ( (f members BC and CD re ('ICCII I
~he
OM,
0429
75
OM
0571
,
l' 2 k f.
4_
k-ft
four dl tnbuted moments are ret: rded on hn .,
tnbullon tabl and a hne I dm\\-n ben ath them- r
\Ioloth of Ihe table t mdlcate that all the Joant re n \\- Nt n
In Ihe ne t lep of the anal) t the carl) 0\ r m me t tha
at lhe far ends of the members are computed b mullJpl 10
tnbuled m men b the carf)O\er fact rs
I
see fig 1o.)
Practical Application of the Moment-Distribution Process
In the ft.lTegomg dl"il,;u ion \\c determined the member end moments b
SucLe....I\c!) halannng onc Joint of thc structure at a time. Although thi
approaLh pro\ldc a de.trI:r 111 19ht mto lhe basic concept of the moment-Ji tnbutnm procl,; s Irom a practical \ic\\-poinl. it is usually more
f,;om enu:nt 10 usc an alternati\ c appro.lch in \\. hich all the joint of the
erutc ~
th.lt arc frr:t.' to rotatc arc h.t1anced simultaneousl} in the same
!'Itep. All thc carr)o\cr moments that arc induced at the tar end of the
memlxrs <lTC then computed "'lIllUltancously in the folio\\. 109 step and
lhe process of b<11ancing the Joinl"i and carl)ing o\er moments I re
pealed u.ntil the unbalanced momenb .It the joints arc negligibly small
To Illustratc this .t1tcrn<1ti\c approach. consider agam the three·
span conlinuous beam of Fig. 17.5(a). The moment-distribution table
used. for carr}ing out lhc u)mpulations is sho.... " in Fig. 17.5 I Th
prc\lllusl) computed di...trihulion factors .tIld fix.ed-end moments are rede ro~
on the top and the tirst line rC!'Ipccti\c1y of the table as shown
It1 the lIgure. The moment-dlstnbulion process is starled by balancmg
Jomts Band ( I rom hne I 01 the moment-distnbutlon table Ftg
175 I \\.c can • that the unbalanced moment at Jomt 8 IS
L M.
50
7'
25 k-f1
A dlscu sed pre\ IOU Iy the baJanLing of Joint 8 mduces dl tnbuted
moment at ends B of Ih members 18 and BC \\-hll.;h can be eulu ted
b multlpl}mg Ih nc:gall\e of the unbalanced moment b) the dl Inbulion factors Thu
~BMD
05
'5
125 k-fi
OM,
o.,
.25
12 5 k-f1
Joml ( I then halanLeo m a ImJlar manner From hn I of the
mom n1-d1 tnbullon labl \\ l,;an
that the unbalanced momenl t
JOlOt ( I
UM,
75 k-ft
These carrymer moments are recordro on the nex.t hnl; Ime 3 01 Ih
moment-distnbution table With an mdmed arro\lo pomtmg trom each
distributed moment to It!\ carr)o\'er moment as shown m Fig I ~ I
We can see from hne 3 of the moment-distnbulJon table that due to
the carryover elfet:t, there are no\\ +16.I-k-ft and 6 '\-k·ft unbalanced
moments at Joints Band C. respecti\ely. Thus Ihese jomt are balanced
again. and the distnbuted moments thus obtamro arc recorded on hne 4of the moment-dlstnbution lable One-half of the dl tnbuted m ments
are Ihen carried o\'er to the far ends of the members hne 5 and lhe
pnx:t:ss IS continued unttl the unbalanced moments are n II Ibl m II
The final member end moments obtamed b) algebralcall urnmm the
entnes 10 each column of the moment-dlstnbutton tabl a rec rd
on hne II of Ihe table Fig. 17 5 I
ole that th
final In In nt
10 agreement With those dctemuned pre\lou I)' m Fig I
a nd 10
Section 16 2 by the ,Iope-deflcctlon method Th mall d ffi n
t\\-cen the re ults obtamed b} different approaf,;h are d t the uDd
olferrors
17.3 ANALYSIS DF CONTINUOUS BEAMS
Based on the d.
dure
r.
r the anal
n p
0
ntal .. the p
beam b
nbDUOUS
mc:thod <an be ummariz<d a foil "
ClW'TER t7
E'
I'"
n_od
IECTIOtI 17.J ....,..." Cc:ll SJ•• _ _
17.1
IXlcmlln Ih m mtx-r nJ nHlmenl hlT lh(' I\Hl.~pa1
(ontmuoU5 beam
I nhlmcnHh Inhulll.m method
II
h
anFgI
<I
\\l1ll
E/
~.JI= ,B¥~b !"J. ~C
A
con tant
lin
2 k. fI
rIOft+-,5ft-!
f---- _.1<; tt
--+-
I 0545
Di tnbuuon fa.:lo"
- 43.2
+64 8
Flxed-end moments
2 Balanc Joml 8
3 Carrymer
-29 1-
J Fmal moments
+35.7
4
58.2
0.455
JO
F M
I
ft---f
I
+150
-150
- 48.6,
.... - 243
101.4
+101.4
-1743
(al Contmuoulio Beam and Moment-Distribution Table
18k
r: I
:'\
Hft
rrllll 1I1T11-\
471)~
357\.r-,A--'----:Bl)101.4 101.4\,IB
(b) Fmal Member End Moments (k oft)
FIC. 17.7
Solutio.
COM
This beam '" as pre\ lOusl)' analyzed in Example 16.1 by the slope--dcJlection
method.
Df ,,,hutiull Fm ton Only jomt B is free to rotate. The distribution
ratto
at thiS JOint are
OFB ..
DFBc
K..
K.. KB<
0.545
KB<
K, .. + K B(
Ie that the urn of the dlstnbution factors at joint B IS equal to I that
DF.,
DF.
0545
C/IeCIII
0.455
The d. lnbUllon factors are recorded In boxes beneath the corresponc:liDa
ber ends on top of the moment-dlstnbutlon table as shown m Fig I
FI
End \I ", nI A wmng that Jomt B IS clamped against rota
calculate the dn~e.uf
m menh due to the external loads by USIDB the fiI<llklol
moment presSion gJ\ n In de the back cover of the book
FEM.
648 k·ft'
or
648kft
COM.
4
111
ClW'18I17 _ - _ -
E
51117.2
Det rm e the member oJ momenl for the thn.'\:"-span I,;ontmuou
F 17 .a b lhe moment-t1t tnhullon mcthod
OF
In
_"D
OF
lllfl
1=
Di
I
buonF
2B
---=8""1
C
4 B
80ndC
--=IO
5 CaR}
B
1-0'
Cony
Balance J lOts Band C
9 Can) r
I Balance J Inl 8 and C
-01
II Final moments
+21.1
....
0,1
+81
-16 ., -16 :!
+8.1
-4 I
· -4 1
+20
1 ()
• -I 0
+05
• ~O 2 : -02 _
+01
-005 -005
+l:!4
I F
18 fi
18 ft
1 It
05
-70.2
·iud··EndM
I
05
81
+16 :!
81
+41
20
+10
• -0'
+02
05
I
+486
+162
324
FEM
-+20
6
+1.0
48
+05
+02
+0 I
+0.05
+005
70.2
+70.2
kfi
_
01
+702
FEM
+8.
+41
FEM
-217
(a) Contmuou!o-Beam and Momenl·Dlstnbullon Table
3 k1ft
)orJI: ~.;r
ml
3 klf1
3 klft
70 {OIIIIIIIIJ) 70.2(OJD:l::o::r:::=.'"\
A
fB
8 1702
et702
Ic
DIJI7
OM.
OF
UM.
05
324
16
OMB('
DF BC
VMs
05
324
I.
OM
OF.
UM
o
OM
OF
fbI Fmal Member End Moments (k-ftl
fI&.
17.8
Solution
Tbl
anal zed P
beam wa
IOU I)'
In
Eumple 16 2 by using the
deftecb n m mod
ad
D
Uti
the beam re f
05
OF
IC
05
k fi
a
fi
100
m
FEM
m
DIF.,(-UJ;t.1
DM
OF
tIMe
•
III
;817 ..... II tl ...........
30kN
ID
10k m
C
B
mcp
4m-l
9m
6m
l20k
(b) SlIIk:oU
CanIiJnw
ConIinuous Beam
CD
CB
1
615
+120
S2S
oS
+
S
-120
+120
d FiJII1 Member End M....... (kN m)
,.,1710
AI
DF
o
FEM
m
FEM
m
-
.M
,.17
III
-
A DII ...............
n
pm'I
beam
nal
.. method
usJy
In
Example 16 S by UIIDI
At JOInt B
os
OF
os
te
OF.
os
OF
os
MO#IWft A qualitatl\ic deftected shape of the con
damped ....... rotabon aod ubjc<tod to the opeciljed
.........., I depicted m FII 17 11(11 IISIIlI an exallF'lltod scale II
..... from \hi figure that tbe mabvc settlements for the _
.......
4
4
0
m aod4
0
By IUIDI the
moment prauOllS, we _
.. the 1IJe.......
moments due to the uppon settlement to be
Fi.
wnll all _
_-end
FEM.
FEM",
lOS tN m
FEM
FEM
lOS tN m
D
FEM
0
Mommf D rribll/ron The moment diltnbution I carried out in the
manner u shown on the momenl-di tribUhon table 1D FIS 17 11
FIMJ M""",n
See the moment-di mllubon table aod flJ. I II d
AM.
aodm
""7.11
CHAPTER 17 Moment·Distribution Method
IICTlON 17.4 AINItpIs of Frt.-I ......... lld.
J}I II' I l l " /,JI1
FE\!
Dt
l'
," 1
FEM
IS
7«1
H\!
FE\!
H\!
\1 luml R
i,
141
IfM
4
'I ~n
sn
H
l(~
cO
11 . 0
tl5'"'1
~o
til momcnt-<li,,,nbuloon
-\t JOlOt (
/ :!11
--1/ XI'
1 :!O
DI,
.\1
Of
At JOint
'I
!'If)
+ 1 :!Ol
~O
0571
O.4:!9
17.4 ANALYSIS OF FRAMES WITHOUT SIDESWAY
n
The procedure for the anal
f fr
that for th anal I of continuo
so.:t1on HOVoe\cr unhke the \: ntmu
rna) be connectoo to a Jomt of a frame
(0 record the computauon m uch a
n
Whereas some engmeers hke to r« rd th
lations directly on a ketch of the fram
the
fonnat for such purposcs. \\e VoIII use a tabular
illustrated b) the follo.... ing e:\amplt:
DF'I)(
FHtd-l.nd \lOIIIC1lt1 -\ quahl.lti\e deflected \hare of the conti UO ~
beam
"lth all Jomt, dJ.mpcd J.gam,t rol.llion and \ubJCClcd to the \JlL"\:iflL'd nop u~
IoCUkmenh I depICted in l i~
171:!(b) u,ing an e,aggeraled "Calc. It 'an be
'en from thi hfure th.lt thc rdati\c IoCttlements for the three member, are
j.u
In . ..\//(
I~
~ in and .'\(1)
1~ -- ~
in_ B) U\ing the fhed·
end·momenl c'prc"ion\. \\c dctermine the li\cd·end ml)ments due 10 the ~U\
port ,<Hlement 10 he
=-0
l
6(29IlOO)(7.800)
G)
(20)1( 12)'
~
1.227.2 L·ft
G)
) (08.7 1~ 92(6
J)il~ f2(
-
I Example 17.7
Delenntne the member end momcn f, r the
ustng the momenl-dl Inbutl n me hod
1718.) k-f,
SOlutian
-
This frame w anal cd 10 E'tampl
H~D
AI
tnt
The fiAcd-end mom<nt due to the 2- .. It extemalload are
2 20
12
2
66 7 k·ft
rhus. the t tal fiAcd-end momenb dut to (he combmed elfC\:( of the
I d and (he upron ttlemenh art
4_
OF
66 7 k·fi
OF
C'
tcmaI
OF
A
_ ,&""
IDII ............
B -
1--301I--J-3O11--1
E 29.oooksi
• F1ame
Carry.....
- .. - .
•
~
•
•
C4
CD
10.429 o.s71
100
+130
_ - 21.4 - 286,
+100
1- 10.7
DB
DE
0.4
-130
0.3
0.3
+130
r
I
I
~
. - 143
243
+ 52
+~
1- 07
-+ 03 + 04
02~
I + 92.1
~
-115.9
I
Ian
I
+ 75
-182 - 182
I
BD
I
1
-130
_+130
~
H
,- 14
+
ED
- 122'
+ 7 ....
+ Z.6
L_jowJ
I
DC
+1159
I - 1.1 I-
I1
-0.6
-+ 02
- 0.1
-186.4
- 01 - 01
-194 +2056
0
7.
CHAPTER 17 Moment-Dlstnbution Method
SEC110M 17.5
...,... 01 ffIIIIII .... "'~I_ '
17.5 ANALYSIS OF FRAMES WITH SIDESWAY
Thu tlr \\1.' h.ne \,.'\)n u.lcred the ,1ll.11),i" of "trw.:turc ,In \\hl\,.h the
tran latit)n, t1f Ihl' It\lI1h \\en: either 7ew or kno\\11 .1'0 In the (use of
In thl' 'oCl.:Ul)1l \\e appl) the morncnt-dl"tnbuhon
sUpJ'l)f( 'o(ttkmcnl
mcth\)d h) .l1la!\ \' Ir.lIlll..' \\ h\)'I: 10111" mil) undergo both rotations and
IIdn I.ltll,.)n lh,lt h.l\C: nnt l'x'1,,'n pre-.cnlxd. ·\s dl, "U'.~d
in Stxtion 164
,ul.:h frame arc l'()mnwnl) referred to.1 frames" IIh 'Ide'" a\
COIlMdcr lor e ample the rectangular frame ~ho\ n
III hg. 17,14 a
~ quahLJII\e detlC\'h:d ,h,lpe of thl.' frame for an arbitrary loading I
also ..ho\\n III the figure u mg an c\.ilggerated ",ale. While the fi cd
Join" .f and B \)1' the frame Me complete!) restralllcd again..t rota lion as
\\cll a tran,I.1I1011. thc joint' C and D are free (0 rotate and trdn..late.
H\)\\c\er. ,ml.:c tht' mcmhcr of the frame are assumed to be inc\tenslble
and the defonnation, .tre " ....umed to be small. the joints C and D displace b) the same .lmount. j, in the honLOntal direction on I) . as ..ho\\n
in the figure.
\I
r- \-1
cl I I I I I I I liD
~-I.
I
-
..
-
I
aJ Actual Frarllt' " Moment..
\--,\-1
I
,
I
8
C
\
I
I
"
A
------1)
I
I
p_ I
•
1IIIIIIIIIrolkr
c
-8
I
I
I
Imaginary
I
I
I
I
+
I
I
p-
I
I
I
8
A
8
IbJ Frame ",ilh Sidc\\\3)' Pre\cnted
,
D
I
I
A
I-~ \
~
MRMomenh
I
I
I
I
An Important que tlon that an
m Ih
IS. ho\\ 10 delenntne the member end m,.m,,,,"
s.tde \\a under Ih a
the frame undt:rg~
Sml.'C the moment-dl tributlon method cann I be
pUle Ihe momenls due to the kno\\n fat rail ad R
reel approach in \\hlCh Ihe frame I ubJ t d t
joml trans.latlon!to !.:aused b) an unkno\\n load Q a
and in the direction of R as s.ho" n m fig 17 14 d
joint tran lalion !to \\e detcrmme the rclatl c t an
cnds of each member and \\c calculat the member
in the same manner as done prc\'IOU I. ID lh
ments The fixed-end momen thu obi med
momenl4slribuuon process to delerm n Ih
\/Q caused b) the )et-unk.no\\n I d Q 0
I m
MQ ha\e been dctemuned die magmlude Q c
apphcatlon of equd bnum equal
With the I d Q and the
d Ired momen \I R dUC' 10 lhe Ia
eo" b) mol pi 109 1/ h the
I
I
I
I
I
I
I
I
8
A
(d I Frame Subjected 10 an
Arbitrary Tram.lation.:1
""Momenb
.... 17.14
II
~: ;
Ie) Frame Sl.lbJeCied loR-
MuMomenh
R
II
hod of ,.aa!,'1is
.n_ s~
.•
I IIII••
"""
B
A
lUod-Il!lId Moo_ Due '" Knowa Tru.I...... ~
BD
fa 17.15 \""'.....,
+ 02
-45.4
IJl Mc= •
-
t
4.4 tD
-
44
Q
.1'
,& 17
.1
I p' DIILINItDn
MIIbod
C
3Ok-·r~
~ )8S
T
D
I k ft
172k-
8S9k
(f) Support
-.1717
f
_00
'-'lit
tJOlnt D
DF
049
DF.
051
Mem/Nr End Momm' D.. ,o ... Arbllrory SitkSWQ A SiDoo
loads at< applied to the: member.; of the: frame the member end 1IlC.....
the: fnun< restrained apln t S1desway will be mo To detcrmino
end moments M due to the 3O-k lateral load we subjOCt the liame
tI8Iy known borizontal translabon A at jomt C. F'8'= 17 17(b
tab.. deflected shape of the frame WIth all jOlOts clamped apIDat
subj<cted to the honzontal dl placement 4 at j0101 C. Tbe JIIOllOl\tJN
strucIIIl& such deftected hapes was discussed In Section 165 0IIl
the frame mcmben are assumed to be mexteDSible and defmma
umed to be _II an end or a member can tnmslate only 10 the
pendicular to the member From thIS figure we can see that the
Iabon 4. between the end of member AC 10 the diroctlon pe,rpeadilllllii
member can be expressed 10 tenns of the Jomt
4.c
translaUOD
'44~
CC
Au
I 254
Similarly the relab.. translabons for member.; CD and BD ...
4
D,D
A..
Tbe
24
3
13 4
3
II due to the: reIati..
FEM
FEM
FEM
FE
34
4
14174
I 202A
_1IllDa....
782
CHAPTER 17 Momenl-OIs1ribution Method
Analysis of Multistory Frames
The fl1r.. . g..'lI1g rnll.:l..'UUll..' L,ln Ix C'\tcnul.'u to the analYSI"i of t
\\ Ilh l1luhirlc dc:gn:c-.. III fn:I.'J"'1ll of 'IUCS\\,I} redi~no'(
the ~t
n'(ungul.H fT.II111: ,11(1\'11 in hg 171S(al. The momenl-dl tnbu
an.II),! l'l Ihl IT.lIllC'" ,Jcnr~:
I,lul 10 three paris. In the fint PUt,
sl"k,\\.ty 01 both 111lM of Ihe IT.IIllL·.• \ prc\cnted hy addmg Ima
Tl,lIl..... at the: tlllN Ic:H:h.•1,h"ml1 10 hg. 17.J8{b) Member end
mcnh \In that dc\cll)P 10 th1" fr,lmc due to the c\tcrnalloads are
~.hup
~b
Ihe- monu.'nl-lh,tn!:luuoll pnxe,... and the restralnmg fo
<lod R at the II1Mgmar) supporb arc C\alualed b) appl)ing the
Ulln' of cqUlhl:mum In the dno'("~
part of the anal}sis the lower
of the frame I' al"'.1\\l.'o It) displace h) a kno,," amount AI while
sldc,\\a) of the upper floor ., prc\cntcd. .a~ shO\\n In Fig. 1718
t1\t."d-end momenl' l.:"au-.cd ~b Ihi!'> di'placement are com pUled and
tribulcd to l)btam the member cnd momenb .\fQI_ Wilh Ihe member
pE
mtni~'1
p-
-j
I
I
I
-R
"
- - - Ff'
I
.1 1 , I
I
---
I
_
E
I
"
pC
"
mnn
p_I'IIIIII'-R
J)
C-- - " D
I
I
I
!I.e..
B
A
(b) Frame with
(a) Actual Frame M Momenh
SUMMARY
Sidesway
Prevented M o Moments
Ffi
E
+ (I
1\\
)(
I
C
f"
"
I
D
011
E
I
I
+C2
C
","17.11
(C) Frame SubjeCted 10
Known Tran lallon.1 1_
"'01
Moments
,~
D I'
I
I
!I. '--
I
I
-Q
I
I
~e.
I
I
X
I
I
A,
Q~F
1,--
!I.
I.-
(d) Frame Subjected 10
Known TransIliO~
"'02 MomeRU
184
CHAPTER 17
--
Moment·OlstribuUon Method
ThL' 'ls~a1I.
(ll If.llne, \\uh a ,'lOgle degree of freedom of!l
I L.tfriL't.I (lut in t\\ll r.lf". In the' hr,t p.ut the ~Ide!\"ay
IS pre\'en
the add ilion of an Inla!!lOar) rllller to, the ,tructuce. Member en~
mcn" (hat de:\d(lp in thl' n:.. tr.t1ncd Ir.lme:. due: to the external
an: ~.hupml(:L
b\ the mOnle:nt·dl"tnbutlon pnll'es,; and the re"llllilliJII
forcc R .It the Imagmal') roller I' C'\.lluah:d b) the apphcatlon
l-quatton, of e:qutlihnunl. In the 'Ccond part of the anal} IS 10 calCU1lIo
the memhc:r mllmcnh due 10 the I""lrce R applied in the 0PPOute
lion the- ..tru,,:tun: i....lllo\\C"d to displace b} an arbitrarily ...,. .....
J..:non" amount' and the member moments and the corre''P<1Ildi1ll
force Q at the IOL:ati(ln of Rare e\aluated as bc:fore. The actual mea.
end moment an: detennined b} algebraicalJ) summing the m
computed in the tiNt part and R Q times the moments computed m
second part
On...'(' mcmber cnd momenh arc known, member end shea
ber aXial fl)rL"C,.•tnd 'mppon reactions can be evaluated throUgh
librium con..ide:ration,.
17.llIInItIIIl17.14 Dete"m
"hear .lOd bendmg mom ot d.
100kN
17.1 ....... 11.1 Detemune the reactillO\ and dra\\, the
h rand bendmg moment diagram.. fl)T the beam.. lOho\\, n
n FI
PI I P175 b) u..mg the moment-di..tnbution
method.
~9m-+61
I
2;"i It
:c:
8
EJ = con
E
o
2U 11---1- IUII-l--Il 11--\---11 II
FIG. P17.8
~
Ok
FIG.
I Hilt
(
III.
P17.2, P17.8
~
IlLPl7.13
\
E
[J- con tan,
E=70GPa 1= 800 (I ()6j mm'
8m
E= 70GPa
E _ 000 kn
pnmwrn:l
D~A
25 kN/m
\ant
20ft
60kN
FIG. P17.9, P17.15
21
8m
15 ft
20 kN/m
1--8m---+-8m-+4m+4m
+
FIG. P17.3
IlC. P17.1
II
20ft
EI=
E = constant
I
c
IlL Pl7.12, Pl7.1'
I I kill
A
~
1--- I -----1-
10k
IRk
I--
~
l
AI
100kN
n the
14 h
method
PROBLEMS
. . . . 17.3
P178 PI
\n fig
A~
1= 1.300 (10'1
~
Il k
8
IllL Pl7.14
C
1--1111----+--1011-1--1011
£1 =constant
rom'
IlC.
P17.10
P17.4, P17.7
3lk
1 kill
o
8
I
10m
I
I
20m
21
~
-10ft
EI
E = 200 GPa 1= 500(1()6) IIIlII'
1= 165010 4
1II.P17.5
10ft-+--
FIG. P17.11
_
'.17 -,-DilL. Ian ........
30kNim
Deocnninc the mombcr end mom
: ' .:~
~
PI I
PI
c
D
•
.."
... "
II
.......
9k-
lIk-
E
1
21
e
21
1
16ft
A
30ft
•
ft
ft
E=C"lall....
... Pl7.J1
,
,
CfW'TER 18
IIC1lllIlll2 .......
Introduction to Mltrix Structur.1 Anllysls
'- __
111
IT]
Fr
(t'll Anal}tical Model and
lal Aduul Tru
Degree.. of Freedom
RG.18.3
18.2 MEMBER STIFFNESS RELATIONS IN LOCAL COORDINATES
In the matrl' stiffness method of anal}"is, the joint displacements of the
structure are determined by solving a system of simultaneous equations
\\ hich i.. e,pressed in the form
P - Sd
L
EJA=con
b Fnme Member - l..ocal C,ooli....'
(18.\
in \\hich d denotes the joint displacement vector, as discussed previously: P represents the effects of external loads at the joints of the
structure: and S is called the structure .'itiffnes.\' matrix. As will be diS-cussed in Section 18.5, the stiffness matrix for the entire structure. S, is
obtained by assembling the stiffness matrices for the individual members
of the structure. Thl .\1i!f;/l'S\ mlllrix !t)r a "U'mher is u.\ecl to eJ:.prefiS the
forte\ al the l'ncl..
Ihl' memha as jimctions oflhe t!i'iplacemt'nts of Ihe
memher\ {'lid,. Note that the terms forces and di\plac('menh are used
here in the general sense to Include moments and rotations respectively
In this section. we deri\e stiffness matrices for the members of plane
frdmes. continuous beams. and plane trusses in the local coordinate
systems of the members.
1.,=
or
-IE ~
",::;: 1
J:
1r:.;~- Lb
I =0
I =0
I-
_ _--<
L-
+
Frame Members
To establish the !!otiffncss relationships for the members of plane frames.
let us focus our attention on an arbitrary prismatic member m of the
frame shm... n in fig 18.4 a . When the frdme is subjected to ext mal
loads member In detorms and internal forces are induced at at ends.
The undefonned and defomu."d positions of the member are sh
In fig 184 b. As Indu;ated In this figure three dlsplacemen
translations In the '( and ) directions and rotation about the _ axl
needed to l;ompletel) speclf) the deformed position of each end
•
11Il.1U
..
• . _.....
_ .........II1II,...
•
k
k
k,
k
ko4 ko
k k
k.,.
-===
1be .um- ...... eD1J, k ClIII be evaluated by
'nher IlIPIJIIteIy to umt val_ of each of \be IIX eDd d
1be _ber eDd fonleo required to ClUIIC \be individual UDlt
_
.... tIeD del mined by liliiii \be prmctplcs of mc:dwmcw
taiaII 8IId de slope ..........00 equaUODl Chapter 16 8IId by
de eqationl of equilibrium 1be member eud fOlOllll t1lIII
a:ptwwt the .,m. . eodk:ienta Cor the member
Let us evaluate \be IIilfDeu coeIIicienll COliespond"'l to
nIue of de d.tpJ·ceml!ll\ '" at eud b of the member u I1toWIl
II ~
ota that all otber ditplacemcnll of the IIIlIIIlber &Ill
callilll from metIumIt: of "",,,riah that the axial dcfOl'lllllIioll
'.IIber CIIUIed by an axial force (2, .. 1IW11 by '" Q Luul..•
IInliDe the force k, that mUll be applied at eud b of the -lfiij;
II
to caDle a dltplacament"
I to be
I
k
TIe
Ell.
L
at lhe far ead of lhe.memb.eqailillrium:
710
CHAPTER 18
Inlnlduction 10 Mltri. Slruc1Urll Anllysls
Pr("l4":,,cdmg In the "'iln~
m.lI111a. the ,~tiflnc!'"
coefficients correa
109 10 the lImt JI'r1a":\.·l11l'nl IIf'
I .In: lound to be (fig, IRA h
o
"
jLl
klL-
AL'
()
()
12
oL
0
EI
0
6L
4L'
()
i
.n'
0
0
12
6L
-I
0
0
l
I
0
0
6L
~L2
2F!
k
"L
0
I
L -
oF!
L'
'"
0
I
()
-12
-6L
2L
0
()
.1L'
oL
Truss Members
-6L
-6L 4L'
12
ote that the ,th column of the member stiffness matrix conSIsts of
end roce~
required to cau..-.e a unit \alue of the displacement u while
r~hto
displaccmenh are zero. For example. the second column of.
st~l noc
of the ... ix end forces required to cause the displacement u
a"ho"n in Fog. 18A(dl. and so on. From Eq. (18.5), we can see that the
stiffness matrix k IS symmetric: that is. k,f = kii. It can be shown by
ing Betti's law (Section 7.8) that stiffness matrices for linearly ew
structures are always symmetric.
Q
ote that Eq 18.7 I obtaI ed fi
is because the members of tru
and. therefore the member fixed
Continuous Beam Members
Since the axial deformations of the members of continuous beams subjecled to lateral loads are zero, we do not need to consider the degroes of
freedom in the direction of the member's centroidal axis in the anal
Thus. onl} four degrees of freedom need to be considered for lhe manbe", of plane continuous beams. The degrees of freedom and the conesponding end forces for a continuous beam member are shown 10 F
18.5
£,1
L
0?-'
Q4' "4
L
£1 = ,:onstanl
'" 18.5
Continu
U'i
Q
Q,. "}
I
beam member -local coordmate
FlG.18.6
k.
•
1
....
• _
10
ate
onented In dlne-t
tranlform tbc
111m
relatioDa
the am_ I«a1 coordinate ystem 10. Cl'llIii1_IIDIN
:
=~ 1be:
amber ...ftbeos relaUODI m aIobIl
then CXlIIIbiDed 10 establilh tile ~: . "
In tbiJ IllCIIOIl, we diacusa tile 1
aDd end dilplacemenll from IccaJ to
the _ben of pIanc ftamea, ~: = '
' : : : : : Coordiaal8 tranlformauon oftbc It
Q
Lid m the r. DowiDIIICbOD
711
CIW'ltII18._ .. Mml _ " , Anl'ysl.
The matrix T I l.:.tI1 _,boo ddlne the t.ransfomi~'
displal:ements from !l')(:,tl hl glooal coordmates; that
&C1llII1U Cow. Ii TIn.
of member
end
IS,
TT u
18 18
Continuous Beam Members
TN
When anl~zig
continuous beams. the member local coordinates
oriented 0 that the positi\ e directions of the local x and .I axes are ~
same as the poslli\c directions of the global X and Y axes, respective)
Fig. 18. . This orientation enables us to avoid coordinate transfor_
mations esua~xt
the member end forces and end displacements m th
global and local coordmates are 1he same; that IS.
·
•
,. = u
F-Q
(18 19
x- _" :'> "~·
Truss Members
Consider an arbitrary member III of the truss shown in Fig. 18.9(a . The
end forces and end displacements for the member, in local and global
coordinates, are sho"n in I'ig. 18.9(b) and (cJ, respectively. Note that at
each member end, two degrees of freedom and two end forces are
needed in global coordinates to represent the components of the member axial displacement and axial force, respectively. Thus, in global coordmates. the truss member has a total of four degrees of freedom VI
through .., and four end forces, F, through F" as shown in Fig. 18 9(c
)
y
IQ".,
Lx;')
L.
.J:ra:----:....
,. _ :=~ri
b
o
®t
.....
,......---....
n
x
o
Q""I
(a) Conunuous Beam
Lx
F
Y-'
F4 • l4
II
bf 0
F,.l,
0
F
v)
) Member End Force llRd End Displacements
In Global Coordinate
fl&.1U
(
Q)."
(b I Member End Force. and End Di p l in Local Cooniinales
)
Y
0
1),-x
Q".,
Y
\
(b) Member End _
aIld End Dispbcanents
m Local Coonlinata
fll,1U
The tramfonnatlOD main
by elpresslDg the local end ~
F as FIg. 189 band
or to maW ~ on
_
718
CltAPTEA 18 In1nHIuca.. 10 Motrla Structurel Molysls
IICl1OIIl" " -. . . . _ _
18.4 MEMBER STIFFNESS RELATIONS iN GLOBAL COORDINATES
f
B\ "'lOg the: Ilh:I1ltx-1 t1tlnc: ... rd.l1iom. in h,xal coordinates (Section
lX.2 and lhl.: tran h,ml.I!lllO ~n)ltadr
(Section Ut-,). \\e can now de
Hillp thl' '1JITnt: ... rd.Hh'" fllT memb\:r.. In glohal coordinates
Frame Members
To c:~tahl s
tht- mcmocr tiITnC' , rdation In global coordinates, \\c first
,ub,tjtu!" thl:' ,uITne...' relation in local coordinates Q ku + Q Eq
I .4 inti) the fon.:e transfom13tion n:latJons F ~ T 1 Q (Eq. (18.17 ) to
obtain
F
T'Q. TT ku+Q,) ~ TTku+T'Qr
(1823
Then. h\ uh,tituting the displacement transformation relations u If
E4 IS.14 Into E4 IIS2.1) "e detennine the desired relations be-
t"ccn the member end force.... f, and end displacements. "', to be
(18.24)
Equation (IR.24) can be comcniently written as
F - Kv + Ff
(18.25)
K .. T"kT
(18.26)
F, .. TIQ,
(18.27)
"here
The matrix K is called the memhn .~·lifJne S
matrix in global coordinates
and F is the memht" fIxed-end foret· rector in glohul coore/mates.
Continuous Beam Members
As slaled pre\ iously. the local coordinates of Ihe members of cooare oriented so that the positive directions of the local x
tinuous sma~
and J axe are the same as the positive directions of the global X
and } axes respecti\oely. Thus no trcmsformations of coordinates are
needed. and the member stiffness relations in the local and global coordinatt'S are the same.
Truss Members
The uffness relations for truss members in global coordinates are ex.
pressed a
18.5 STRUCTURE STIFFNESS RELATIONS
--"
.
.
.
.
.
..n_
Ill""'" IlIIItiOlII
the:~OI=Dt ~
eJIP-the
themember
ml loads
end ~ P m termStoof.nd d,opl.
ber trnou roIauons m s10bal coon!iDltel
oodinI toll B wnlJtll Eq t 8 25 ID oxp81lded fCll1ll
obtIiD
from whicb we dotonD1DO tbo oxprosllODS for f _ at
manbertobo
F.
. . . . . . . . . . . . I II AI.'"
pm d m
D be convemClldy
p
p
p
COII._
CIW'TER 18 In",",uctiDn 10 Mllrix Strocturel Anllpls
2. [\,tIUctlC:- ~ht
,trw.'tun: ,l1lrn..:" matn\
h'T PlOT
~.Kh
<,per.illOn'·
a.
b.
mc ~r
1IECftOII1U "II
S ano fhed-joint force v
th..: ,lruct\lrc, pcrfonn the roll owlQ
(~f
") dOh
.
.
(ru ...~
go OITce11) to ,h:p -( l. ~ cmlsc. compute the
mc:mba titlnc . . main 111 10c<11 .selt<m~ro c
k. Expre I
of k 10f the "~m':t1l
llf fr,lme' .-and continuous beams are given
10 Eq' 1:.5' and (I~.61
re'p<",'i,elj
If the mc ~r
j ... ,ubJected (0 cXh: n.lal load '" then evaluate Its
fixed-end hlft,.. c \l'ctor in lt.:,~k
COllTd mates. Q f b) using the
fllT
Computer PnIgram
pre"ll)n... l'tlf thed<nd ,moments gJ\.cn Inside ~t
kc~
cover of
.Iflrl) mg tlte equatIOns of equthbnum see Ex
ampb 1~.2
,md . I:~
.. .
For horizontal members \\Itlt the local -'" aXis positIVe to the
right i.e.. 10 tlte "um: din.x:l1on a'S tlte global X axis) ,the memthe book and b~
c.
ber ,tilIne" relatJon~
10
the local and global coordmates are
the ",me i.e. Ii - k and F
compute the me b r"~
tran~fo mati n
Q,): go to step 2(e). OtheFWI
matrix T by using Eq
21.~
d.
e.
Dctcrmmc the member slIffncss matrix in global coordinates
Ii = TTkT IEq. 18.26)), and the corresponding fixed-end fon:e
'ector F, -- T'Q. (Eq. (18.27)). The matrix K must be sym·
mctric. For trw~se .
it is usually more convenient to use the
expheit form of Ii g\\en in Eq. (18.29). Also. for trusses. F 0
Example 18.1
the
1
Fig 18 \I a by.he rna
DctenDlDe
Identif, the member'> ,tructure degree of freedom numbers and
store the pertinent elements of K and FJ in their proper pos'.
tions in the structure stiffness matrix S and the fixed-joint force
3.
4.
vector P,. rc,peeti,e1y. by using the procedure described 10
Section 18.5. The complete structure stifTness matrix S obtamed
bj assembhng 'he ,tifTness coefficients of all the members of the
structure must be s>mmctric.
Form the joint load ,eclor. P.
Determine the unkno\\ n Jomt displacements. Substitute P P and
S into the structure 'tiffn«s relation,. P - P, = Sd (Eq. 18.41
and sohe the resuhing s>stcm of simultaneous equations for th
unkno\\ n joint displacements d.
S.
Compute member end displacements and end fonoes. For each
member do the follo\\ ing:
a. Obtain member end displacements in global coordmates,
from the Joint di,placement'. d. by using the member s s\nlO"
ture degree of freedom numbers.
b.
Detennine member end displacements in local coordlOates
usmg the relallon hip u T, (Eq. 18.14)). For honzoD
members \\ith the local \" axis positi\e to the right u v
Compute member end fore.., in local coordmates by USIng tIte
relallonshipQ ku Q, Eq. (18.4 . Fortrusscs Q
0
d. Calculate member end forces in global coordmates by
the trdnsfonnation relationship F TTQ (Eq. 1817 F
e.
Slru lUI Stiffn
WUI
Member I A shown In Fig
begiDlung JOlDt and J lOt 3 the
18 IJ we dete_;nnm<::::=--:-__--::--:
L
r
I
fIr _ _
-
_,U . . ._ ...........
ClW'TER 18 IntnJduc1lon 10 Ml1rix S1rVCluril Anllpl.
Thus the Jomt load \«IM
l~
Q ..
50 ]
p
[ 8b b
1, nt D/\ m m ,,1\ The "it1lfn(".. ~ rel,lIlons for the entire truss can be
pressetlas Eq
41 \\lIhP
0
P
B} subsillulmg P from Eq
expanded fonn as
Sd
1 and S from Fig. 18.11
I.7H!
[ 417b
50 ]
8b b
C
\IoC
wnte Eq
,
417.6][<11]
T Q
d,
1.b44.3
•
B\ sohmg these equation.. simuitaneousl} .... e delcrmine the joint dlsplatemems
to be
dJ
00434
In.
or
d
0.0434].
[ 0.0637 m.
£"',
\lemher
Dl\p!acem('nh alld End Forct'~
Member I The member end displacements in global coordinates, " can be
obtained b} simply comparing the member's global degree of freedom numben
",ilh the structure degree of freedom numbers for the member. as follows.
[:i]
'1
14
~ ]~([
2
o
o
0.0434
-0.0637
d2
in.
4
Note that the structure degree of freedom numbers for the member 0 0 I
are written on the nght side ofv as shown in Eq. (4). Since the structure de
of freedom numbers corresponding to 1'1 and (.'2 are zero this mdlcates tha
t 1 = l - O. Similarly. the numbers I and 2 corresponding to v and..
peeuvel) mdicate that t
d l and 14 d. II should be realized that these
patJbdny equations could have been estabhshed alternatively SImply by a
In pection of tbe hne dlagrdm of the structure Fig. 18.11 b . However the
of the structure degree of freedom numbers enables us convemenlly to Ph.1ID
thl procedure on a computer
The member end dl placements m local coordinates can now be determi
by usmg the relallonsh.p u T. Eq. 18 14 with T as defined ID Eq I
u
06 08 0
0
06
o
~043 ]
00637
B usmg Eq
18 1 we compute member end forces
ID
local coordin
By 1liiO' !be ....1i""oIIip •
1ocI1 coordina
be
ID
..~ !be member end
lOla
pQ U:
....-.q"
T Q,
"'''_no'''-''
_lU
ClW'T£R 18 1_ _ 10 Mltril51rUC turol Anllysll
BOlIN
24kN1m
\ldI'Ibtr 3
Q
6m
4m
10m
(~)
m
I
EI ........
T.
u
y
l n (~
u
0
11
Q
1
0 lJ.I~4
0
Pd
IT]
- n 0637
(1)
%\-X
4
(b AnoIytiaI Model
~'
14<0, _:
:][~043
-
[-:~
k
04+04
2
02
02
04+08
\
Thus thealtlalf f'L'e1O membcr 'IS f-ig 18.11 d
5=£1
629Jk T
F
Q.
Pol,
Q ITJ
kg
Q
I
Q
(a Contmuous Beam
T/Q
629~]
F
62.93
l
-6~
Il
93]
6293 k
o
Ana.
EqUilthrlllm ChUA Appl)mg. the equations of equilibnum to th
of th enllre sirudurc hg 18 life \I.e obtain
LF,
Lf,
0
11
6293
100 (OS 60
0
17 H
6927
100
0
It
11
6927 15
6293 20
In
60
Ch8dll
Chldll
100 eo 60 20
\ 39", 0
C/lIdlI
h .,1111.2
lXt omn the reactl nand tbe m mber end forces for the thnlMlplll
llOUOU be m ho\lo n In Fig 18 12 a hy usmg the matnx. tiff
IlaIulIon
D r
f Fr dt nr Fr m th analyttc.lll model of the beam
I I b
'"" that th lnu;ture ha two degrees of freedom
\\ hi h are the unkn n r tall n of Jomt ., and 3 rnpecb I
membcT I 1
rdmale
tem are chosen
that the poslU eli
L
I =
-200
r
848
jJ 1:200
2
(d) Sb'UCQIIe Sbffncss Matrtl and fixed-JOint Force Vi tor
Support RUI(/IOIU As shO\·\n In hg 18 Il{c) the reactions at the
joints 1,2 and 4 arc equal to the forces in global coordmates al the ends
members wnnected to the jOints.
007 ",0
~0 2+ 5I1-(J
2
~@beD<
4593
\891
30 lIN 176114
6\ 09
176.33 24 IINlm 15388
15338 7694
CW&HP ?m~
\223
Member End
1\77
4616
-
_
CHAPTER 18 IntrodUCtiOn 10 Matrix Structurll Anllylll
1IC1IIIII1U"
8~
U
109 the m~lOhe:r
end
r~me
Il n~
~ ro
Q
Q
OOb
0011
00';
(UII.::!
04
(106
00<>
O~
1I~
0
f,
rdatl~n
ku
Q. Eq. 184
we
~a
EI
[
Ol),;
01
1
006]
[
-006 EI
0012
-006
0
(I
]
0
[
15409
0-1
lK91 K
m
-1598 k
[ 6109.
l"to ~-I
m
~
\Inn... 2
n
..
r
2:
14
k.
Q
d
192:\5
I
Q
[ 0012
0,06
Q-
F.
[~] ~ b[-I'~09]
I
2IO ~
0.06
[
0,06
-0012
006
04
-0.06
0,2
-OOb
0,2
0012
-006
-0.06
04
I221kN
176.83 kN m
117 HN
1[
154.09
00
19235
1
1
(~1
The support react
1891
AnI
4,98
153.8K kN m
Mnnbtr :\
(:'1~
•
I..
Q
k.
Q
(f1 I~
(t'1
Example'••3
Q,
(
F
II
(J
b~(
024
0096
024
024
08
024
04
0096
024
00'lb
024
1
04
19235
024H
0
024
0
08
0
an:
80
18339
•• tor . . . .
•
..-. ..-.-.-
I
4
I
I
9
101
-60
but die 6nt ....- .
tam:
ClDOIIIdo8l
II
lIM
API'£tlDIX B Review 0' Motri. Algebra
,
[
tUlpJ~
rl)\\
~
(1)
I
(I
.1
rol\ .' h)
~.
(1)
Tl'"
I
(I
~(lW
(I
~
1
0 0
~lrit
]
~.;]
~
IB '1e
.15~
~(lW
~
[ (1)
\tul1p~
X~
15 .1Ild ,uhtr.ll.:t It from n.m 1 then multiply
rlm ... b\ 'I'
" ,Ind ,uhIT.ICI II lrom flm 1 lhi,) lc1d~
in t
(II
[ (I
0" ldL~
"II
60
\I X
\<>47
I
- 7705]
Th
~X \I
18.2 111
6
1 II X and ,ubtract it from ro\\ I; then mul.
b) An
1.0-1'"
Jnd \ubtral:( it from TO\\ 2. This yield...
~ b) -I.,
Tl'\\ .'
[ o(~ ~
~
0
2. and
.\3 --
,
I
.
-~] 6
lg ~.BI
6.
PROBLEMS
Matrix Inversion
The Gau!)~·Jordan
~cmi
elimination method can also be used to detcnninc the
... of ...quare matrices. The procedure is similar to that described
(B 24
[~
Bj muillplj 109 ro" I bj A"
obtam
-2
-4
3 and subtracting
B250
It
from row 2 we
:
15]
10
5
B
(
1.2 Determine the matrix C
2A
B If
A -
[ 12
-8
15
1.3 Detennine the products C
A
The augmented malrix. is gl\en b)
-8
7
10
JI If
A
1.1 Determine the matnx ('
pre\ iOlisly for ...oh ing !'>imuhaneous equations, except that in the augmented matrix, the cocftlcicnt matrix is now replaced by the matrix A
that is to be imerted and the vector of constants P is replaced by a unat
matrix I of the same order as the matrix. A. Elementary TOY, operdtions
are then performed on the augmented matrix to reduce the matrix. A to
a unit matrix.. The matrix I y,hich y,as initially the unit matrix. no\\
represents the imeNC of the original matrix A.
To illustrate the foregoing procedure. let us compute the imerse of
the 2 2 matrix
U
Bll'cn ....
SecIIon 1.3
IA Determine the prod_ C
2
5
<aandD BA
-
,
6 4
5
I
4
6
B
AI
1
D BA
4
4
8
-
..... - •
JpM"Np
x-euu& ...
D
8'_1 ReItrDjldl
140
iii
v-cw:&... · "
AG.
CA
JOlOt
x..._
..........
........ E...... l
I--
iii
v..... iii
CHR .. ' ..... E...-...
I
III.c.eM
Coordmat s and Supports Screen
_III
-~
... CoI
-
_
_ I X C Camp_ _1e
~
lift
BIBLIOGRAPHY
20fl
_,-
__
fl
-J'-_
~tf)(3
E I A are lOon tant
E
:!90k.IA=Oan~
1• .fSeE SlUndarJ \f"wmml Dc I 1'1 Load ~ r 8lI
ami 0Iht'1:
2(KH Sfl AS( F ..Q2 Amer:an
St'ICict)
.'tru n ~
01 CI\ II
Z. Arbabi F
~h:Gra, ·HiIL
Engmeen. \
ll991
11. H,b
mlll
IfgtDla
StruilUTIJ! Anah,
cuu/ &ha
Nc\\ York
3. Bathe. KJ. and Wil\On. E,L 1~76
'"m" al \llhJds
In Finilt' Elmlem Anal.nif. Prentice Hall Englewood Chff
J~
•• Beer. .P~
.\fcltt'riuls.
and Johnston E.R. Jr. (1981) \Iedwniu IIJ
McGraw-Hili. New York
5. Betti. E. (1872) II \'uom Cimt'l'lw. Series:! \ ols. 7 and 8.
I. Boggs. R.G. (1(84) Elt'menlllfl Strudurl1l AnahJH Holt.
Rinehart & Winston, New York.
7. Chaje!>, A. (1990) StrUt lural Anall
Hall. Englewood Clilfs. N.J.
I
1L Laible J P 198
Y, mston New York
2nd ed Prentice
L Col/oqUIn! on /lillon' oj Structure 1982 Proccedm
International AS!>OClatton for Rndge and Structural Engmeenng. C ambndge England
••
gnitub r ~ D
.s~orC
H. (1930) 'AnalySt of Contmuous Framn by
fixc:d-End Moments ,. Pr
dill
I
Am('rlcun Soden of ell ,I En III r 56 919 928
11. Elias Z M.
Anal, If. Wiley
e\\
York
11. Gere J.M and Weaver W Jr 196 AI
A
fi r Ullin ers Van ostrand Remhold t:W York
11. Glockner PG
19 I
chamcs.· J urnal of 1M 5:"'..·...,1
89
..
nq ,
,7,
M. Smith Ie 1988 S", r.
ew York
Americ:oll
A
I
..
Ho".,OOd,
•
"."""'U_A
•
WaDI CK
McGraw-Hill
.t. Well HH
I911J [n'mHldia
ewYork
loS
1989 AIIlII "0 SlI'II<lIIn
CIiuIIaJ/ aNi Modmo
York
1.1
StntcIllnJ
II,
II
&11
III
An
lJrotb 2Dd cd. Wi
U
1.7
&11
&11
III
U
1.11
ua
CIIAPTEIl3
1.1
U
U
U
U
Ion
ioU
...
..
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UI
t
~
, b Indeh:mllnalC I
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l n taNe I,. lktcmunalc
I.
It
....
110
1
u,
,)
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100
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1 16
U3
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....
<.
2 L
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Itl~
l
r.
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fRO
("I,
fB(
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t
' H
'\h
to "'4
l
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Rhl" 1
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f( ()
1'\ 81 II.
~8
3' k IT
f
:tol
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kN.
m. ON
4()
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s~
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w·
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5.9
U3
5.11
511 k tT
6
Ul
U3
~
Ft.G
k
14 k
U7
m
50 k Ie).
5.13
C; FR<, - 5 k (TI;
5.15
S
5.l7
S
It L
\hiL
for 0 \
(L 2J. Bending Moment
(-or
(L 2) x
J:
Bending
Moment
\fIt' L)L
S II(L
.1.\2) 16L); \(
\Ix(L~
.-.:') 6L
(-or 0
\
J m: S 20; \{ 20.\
36 k-ft h C)
68.94 k (': Fj(,-
45 '4 kN T
U7
r. H
5.11
"
5.2l
4167 k
C:
"ur J m
10
&A,
- 6 17 k C' FDJ
204kTF
(L 21:S- P12; M - P.\'/2
\: ... L: S - P12; .\1 - P(L - rl 2
2\')/2: .\1 - II".\'(L - .\')/2
\"
6 m: S - 0; \.I 60
(-or h m
\
9 m: S - -20: \I
-20,' + 180
"orO
\" 7 m' S
-60; \I
60\
For 7 m
\
14 m: S ~ IO( 1 - "( • \I
1625 k C' FHI
T F. - 0 14 k C
5.25
U'
.If
U5
U7
5.4lI
245
For 0
\: nOI
For 20 fl
If
For 0
3 J3 If
For 0
20 f1: S
- x
40 - ,
75. If
,,' 2
,
\:
525
l
S
"' If
III k S
'\ "k S
\0 L
k·lt,
R.
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k
k
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k
Il
I
66
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'\8
L
S, If Ss
So 0; ~/.\
I
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-1.350
.\1 D
II
US k
2 7lMi k.
kN m..
Ss
01
\I
4
k
m
36k S
L ....: -27 k; 58 '"
27k;\1 8
\I(
If
94 kfl
at 21 ft from A
S'R
1095 kN. S'L 375 k
S.
·112.5 kN: SCI
-1845k
S
120 k
If
So L =- 60 kN;
\I s 441 kN 01
-450 kN m
St R - SSt =- -8 k; 5 sIl 2383 k' S
- 21 17 k; Sf" '" - So 0; \18 -80 k·ft \I(
.\1 0 :,,:;; -40 k·ft; +.\Im;l~
109 3 k·ft at 2589 ft
from A
.\'" R - 30 k; Ss - S( L - -30 k; S( If
So t
·-50 k; .If,
30Q k-ft: If"
800 k·fi
h\.llll<l\ = 225 k-fL at 15 ft from A
s.~ If ..= SSL - 33.33
kN;
S8 If. Sf) L
-66.67 kN; So If. Sc L
266 67 k S It
316.67 kN; SF L
1667 kN;.If. 3J3 3 k m
.1(" _
333.3 kN m. ifF -1666 7 k m
S~ R = 7.5 k; S( L
22 S k. S( It 22 5 k
S, L - -75 k; S.. 4 17 k
10 83 k
SF R - 10 k; If,
225 k·fi If
50 k.fi:
+ \I~
28.13 k-ft at 22 S ft 10 the left and t
Ss
-=0
II(L
II)
1"0 k·ft \1
211 k \it
\:
1:!5 k·fL \Is
\I(
75 kAt
:!()ll .. m
R6bk •. \f
5U".
\I f
.\8
\1 B ()
<!B
<!t Os 44": kS. Sf -o-17.h k
\f,
x"'5 '\
... , m;
S8
. R944
kN,
It,
~b!U
10., m
O( Qs-O; St - -50 k~;
\1 4 '\0 kN 01;
'\8
h~5
kN; \I" -- -150 kN m
Q4
Qs S8 tl;
S4 ~ 120
kN;
\1 4
.'btl kN 01; .\IB =- I:!O kN m
Oc - 7 ~ kN; S, 12.99 kN; .\1, 77,94 kN 01;
Q, S, 0: .\f. - 8768 kN m
forO
J or (I 2)
711 k
I
\B I
" I
S I
.lOk
m
.\ ~ It. -
....
..
9n kN
\f II
8C1
13113 k-ft
~
<!,
( . F
....
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It
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4.55
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~
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..N 'i
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the right of (
S~
R - 125
k
1875 k ; SOL
SF<
125 k
If
3125k
U7
5.n
m .ISmfromA .... F
a a 3 m b S .. S
S, L -100 k S
S
450 k
m.lf
450k
1.1
be
I
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kft
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."
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-••
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II
k C
Member AE Q
Uk T) S
50 k ft
Member CG Q 96 k C S
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Member EF Q 125 k C S
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k !of
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k !of
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MemberJK Q
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OIAPn:IIIJ
11.1
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5 6k
M
225kN m
M
576k
m
648k
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m
125 k; M
250 k ft
15k; M
IU
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15 k; M
k;M
t
k;M
t
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Member AD: Q
125k-ft
MemberBE Q OS 25 k. AI 250 k ft
Member EF Q 125 k C S 125k AI
I25k-ft
Member AD Q 1667 k C S 15 k AI
l20k ft
Member CF Q 1667 k (T)' S 15 k AI
120 k·ft
Member DE Q-lOk(C.S Ill3 k !of
160 k-ft
Member HI Q 15 k (C' S 333 k; M
4Ok-ll
Member AD Q 1266 k (C. S 1206 k
M 72 4k-ft
Member CF Q 115Ik(T S 824 k M
49 43 k·ft
Member DE: Q 7.55 k (C). S 995 k M
99 Sk-ft
Member HI' Q 12 I k (C' S 246 k !of
1845kft
Member AE Q 918 t (T) S 397k;M
31 75 t·ft
Member CG Q 229 t C S 1478 k; M
118 25k ft
Member EF Q 16HC)S 56Sk;M
84 S1t-ll
Member JX Q
49tc)S 141 t
141k ft
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222 k II
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