5.2 Elastic Strain Energy

Section 5.2
5.2.10. The element deforms with small angles θ and λ as illustrated. Only the stresses
on the upper and right-hand surfaces are shown, since the stresses on the other two
surfaces do no work. The force acting on the upper surface is σ xydxdz
and moves
through a displacement λ dy . The force acting on the right-hand surface is σ xydydz
and
moves through a displacement θ dx . The work done when the element moves through
angles d θ and d λ is then, using the definition of shear strain, Eqn. 3.6.1,
( σ dxdz)(
dλdy)
+ ( σ dydz)(
dθdx)
= ( dxdydz)
σ ( 2d
)
dW = ε (5.2.16)
xy
xy
and, with shear stress proportional to shear strain as in Fig. 5.2.9, the strain energy density
is
u = 2 ∫σ
xydε
xy = σ xyε
xy
(5.2.17)
Figure 5.2.10: a volume element under shear stress
The strain energy can be similarly calculated for the other shear stresses and, in summary,
the strain energy density for a volume element subjected to arbitrary stresses is
( σ ε + σ ε + σ ε ) + ( σ ε + σ ε + σ )
1
u = xx xx yy yy zz zz xy xy yz yz zxε
2
Using Hooke’s law, Eqns. 4.2.9, with Eqn. 4.2.5, the strain energy density can also be
written in the alternative and useful forms {▲Problem 4}
1
u =
2E
μ
=
1−
νμ
=
1−
2ν
Solid Mechanics Part I 186
Kelly
xy
xy
zx
(5.2.18)
2 2 2 ν
1 2 2 2
( σ xx + σ yy + σ zz ) − ( σ xxσ
yy + σ yyσ
zz + σ zzσ
xx ) + ( σ xy + σ yz + σ zx )
E
2μ
2 2 2
2 2 2
[ ( 1−ν
) ( ε xx + ε yy + ε zz ) + 2ν
( ε xxε
yy + ε yyε
zz + ε zzε
xx ) ] + 2μ(
ε xy + ε yz + ε zx )
2ν
2 2 2 2
2 2 2
( ε + ε + ε ) + μ(
ε + ε + ε ) + 2μ(
ε + ε + ε )
xx
yy
λdy
Strain Energy in a Beam due to Shear Stress
zz
dy
y
λ
xx
σ xy
θ
dx
yy
θdx
zz
σ xy
xy
x
yz
zx
(5.2.19)