4.4 Strain Energy in an Elastic Body
111
dU =
1
2
(σ x dydz)(ε x dx)
=
1
2
σ x ε x dV
(4.34)
where
dV = dx dy dz volume of the element.
If an elastic body of total volume V is made up of such elements, the total strain
energy U is obtained by integration
U =
1
2
v
σ x ε ν dV
(4.35)
Taking σ x =
P
A
and ε x =
δ
L
where
P uniaxial load on the member
δ displacement due to load P
L length of the member,
A cross-sectional area of the member
We can write Eq. (4.28) as
U =
1
2
P
A
δ
L
v
dV
Therefore,
U =
1
2
P.δ since V = L × A
(4.36)
Next consider the shear stress component τ xy acting on an infinitesimal element
in Fig. 4.2c. The corresponding deformation due to the shear strain component γ xy
is indicated in Fig. 4.2d. In this case, the force τ xy dxdz acting on the positive y face
does work as that face translates through the distance γ xy dy. Because of linearity,
γ xy and τ xy grow in proportion as the element is deformed.
The strain energy stored in the element, when the final values of strain and stress
are γ xy and τ xy is
dU =
1
2
τ xy dxdz
γ xy dy
=
1
2
τ xy γ xy dxdydz
Therefore,