5.2 Elastic Strain Energy

But θ A = 0 and so Eqn. 5.2.33 can be solved to get M A = −PL
/ 8 . Then the
displacement at the centre of the beam is
Section 5.2
3
2
3
∂U
PL M AL
PL
Δ B = = + =
(5.2.34)
∂P
48EI
8EI
192EI
This is positive in the direction in which the associated force is acting, and so is
downward.
Proof of Castigliano’s Theorem
A proof of Castiligliano’s theorem will be given here for a structure subjected to a single
load. The load P produces a displacement Δ and the strain energy is U = PΔ
/ 2 , Fig.
5.2.13. If an additional force dP is applied giving an additional deformation d Δ , the
additional strain energy is
1
dU = PdΔ
+ dPdΔ
2
Solid Mechanics Part I 190
Kelly
■
(5.2.35)
If the load P + dP is applied in one step, the work done is ( P + dP)(
Δ + dΔ)
/ 2 . Equating
this to the strain energy U + dU given by Eqn. 5.2.35 then gives PdΔ = ΔdP
.
Substituting into Eqn. 5.2.35 leads to
1
dU = ΔdP
+ dPdΔ
2
(5.2.36)
Dividing through by dP and taking the limit as dP → 0 results in Castigliano’s second
theorem, dU / dP = Δ .
P + dP
P
Figure 5.2.13: force-displacement curve
In fact, dividing Eqn. 5.2.35 through by d Δ and taking the limit as d Δ → 0 results in
Castigliano’s first theorem, dU / dΔ
= P . It will be shown later that this first theorem,
unlike the second, in fact holds also for the case when the elastic material is non-linear.
5.2.5 Dynamic Elasticiy
Δ
Δ + dΔ