122 CHAPTER 5 Energy Methods
FromCtoD,
M=
(
1
4
PB,f+1
2
PC,f+wL
2)
(L−z)−w
2(L−z)^2sothat
∂M
∂PB,f=
1
4
(L−z)∂M
∂PC,f=
1
2
(L−z)Substituting these values in Eqs. (iv) and (v) and remembering thatPB,f=PC,f=0, we have, from
Eq.(iv),
(^) B=
1
EI
⎧
⎪⎨
⎪⎩
∫L/^4
0(
wLz
2−
wz^2
2)
3
4
zdz+∫L/^2
L/ 4(
wLz
2−
wz^2
2)
1
4
(L−z)dz+
∫L
L/ 2(
wLz
2−
wz^2
2)
1
4
(L−z)dz⎫
⎪⎬
⎪⎭
fromwhich
(^) B=
119 wL^4
24576 EI
Similarly,
(^) C=
5 wL^4
384 EI
Thefictitiousloadmethodofdeterminingdeflectionsmaybestreamlinedforlinearlyelasticsystems
andisthentermedtheunitloadmethod;thisweshalldiscusslaterinthechapter.
5.4 ApplicationtotheSolutionofStaticallyIndeterminateSystems...............................
In a statically determinate structure, the internal forces are determined uniquely by simple statical
equilibriumconsiderations.Thisisnotthecaseforastaticallyindeterminatesysteminwhich,aswe
havealreadynoted,aninfinitenumberofinternalforceorstressdistributionsmaybefoundtosatisfy
the conditions of equilibrium. The true force system is, as we demonstrated in Section 5.2, the one
satisfyingtheconditionsofcompatibilityofdisplacementoftheelasticstructureor,alternatively,that
forwhichthetotalcomplementaryenergyhasastationaryvalue.Weshallapplytheprincipletoavariety
ofstaticallyindeterminatestructures,beginningwiththerelativelysimplesinglyredundantpin-jointed
frameshowninFig.5.8inwhicheachmemberhasthesamevalueoftheproductAE.
The first step is to choose the redundant member. In this example, no advantage is gained by the
choiceofanyparticularmember,althoughinsomecasescarefulselectioncanresultinadecreasein
the amount of arithmetical labor. Taking BD as the redundant member, we assume that it sustains a