Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

720
ENERgy METHOdS
Since the volume of the element is dV = dx dy dz, the strain-energy density in pure
shear is
1
u = 2
τxyγxy
(17.7)
For a linearly elastic material, τ xy = Gγ xy and the strain-energy density can then be
expressed solely in terms of stress as
or strain as
τ xy
2
u = (17.8)
2G
γ 2
G xy
u = (17.9)
2
The total strain energy associated with shear stress can be found by integrating the
strain-energy density [Equation (17.8)] over the volume of the member:
U
xy
∫ τ 2
=
2G dV
(17.10)
V
Equation (17.10) can be used to determine the strain energy for bars in torsion, as well
as the strain energy associated with transverse shear stress in beams.
Although Equations (17.3) and (17.7) were derived for σ x , ε x , τ xy , and γ xy , additional
strain-energy density expressions can be derived for the remaining stress components in a
similar manner. The general expression for the strain-energy density of a linearly elastic body is
1
u = [ σε x x + σ yεy + σε z z + τxyγxy + τ yzγyz + τzxγzx
] (17.11)
2
17.3 Elastic Strain Energy for Axial Deformation
The concept of strain energy was introduced in the previous section by considering the
work done by a slowly applied axial load P in elongating a prismatic bar by an amount δ.
If the load–deformation diagram is linear (Figure 17.4), then the external work W done in
elongating the bar is
W =
1 2
Pδ
A
L
FIGURE 17.10 Prismatic bar
with constant axial load P.
δ
B
P
and since the strain energy stored in the bar must equal the external work, the strain energy
U in the bar is given by
U =
1 2
Pδ
The prismatic bar shown in Figure 17.10 has a constant cross-sectional area A and modulus
of elasticity E. When the load magnitude is such that the axial stress does not exceed the
proportional limit for the material, the deformation of the bar is given by δ = PL/AE. Consequently,
the elastic strain energy of the bar can be expressed in terms of the force P as
2
PL
U = (17.12)
2AE