5.3Application to Deflection Problems 121Fig.5.7
Deflection of a simply supported beam by the method of complementary energy.
and
∂C
∂PC,f=
∫
Ldθ∂M
∂PC,f− (^) C=0(iii)
Assumingalinearlyelasticbeam,Eqs.(ii)and(iii)become
(^) B=
1
EI
∫L
0M
∂M
∂PB,fdz (iv)(^) C=
1
EI
∫L
0M
∂M
∂PC,fdz (v)FromAtoB,
M=
(
3
4
PB,f+1
2
PC,f+wL
2)
z−wz^2
2sothat
∂M
∂PB,f=
3
4
z,∂M
∂PC,f=
1
2
zFromBtoC,
M=
(
3
4
PB,f+1
2
PC,f+wL
2)
z−wz^2
2−PB,f(
z−L
4
)
giving
∂M
∂PB,f=
1
4
(L−z),∂M
∂PC,f=
1
2
z