5.2 Elastic Strain Energy

The rotation associated with the moment is
Example
Section 5.2
3
2
∂U
PL 5M
0L
Δ C = = +
(5.2.28)
∂P
12EI
24EI
∂U
θ C =
∂M
0
2
5PL 2M
0L
= +
24EI
3EI
(5.2.29)
■
Consider next the beam of length L shown in Fig. 5.2.12, built in at both ends and loaded
centrally by a force P. This is a statically indeterminate problem. In this case, the strain
energy can be written as a function of the applied load and one of the unknown reactions.
M
A
A C
B
Figure 5.2.12: a statically indeterminate beam
First, the moment in the beam is found from equilibrium considerations to be
P
M = M A + x,
0 < x < L / 2
(5.2.30)
2
where M A is the unknown reaction at the left-hand end. Then the strain energy in the
left-hand half of the beam is
L / 2
2 3
2 2
1 ⎛ P ⎞ P L PM AL
M AL
U = M A x dx
2EI
∫ ⎜ + ⎟ = + +
(5.2.31)
⎝ 2 ⎠ 192EI
16EI
4EI
The strain energy in the complete beam is double this:
Writing the strain energy as U U ( P,
M )
0
P
L / 2
L / 2
2
2 3
2 2
P L PM AL
M AL
U = + +
(5.2.32)
96EI
8EI
2EI
= , the rotation at A is
A
2
∂U
PL M AL
θ A = = +
(5.2.33)
∂M
8EI
EI
Solid Mechanics Part I 189
Kelly
A
M C