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