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

Unit Volume change
Consider an infinitesimal element with dimensions dx, dy, and dz (Figure 13.14). The initial
volume of the cube is V i = dx dy dz. After straining, the lengths of the three sides become
(1 + e x ) dx, (1 + e y ) dy, and (1 + e z ) dz, respectively. The final volume of the cube after straining
is thus
The change in volume of the cube is
∆ V = V − V
f
i
V = (1 + e ) dx(1 + e ) dy(1 + e ) dz
f x y z
= (1 + e ) dx (1 + e ) dy(1 + e ) dz − dx dy dz
x y z
= [1 + e + e + e + e e + e e + e e + e ee − 1] dx dy dz
x y z x y y z z x x y z
= [ e + e + e + e e + e e + e e + e ee]
dx dy dz
x y z x y y z z x x y z
z
dz
y
O
dx
563
gENERALIzEd HOOkE’S LAw
FOR ISOTROPIC MATERIALS
dy
FIGURE 13.14 Infinitesimal
element.
x
If the strains are small, we can neglect the higher order terms (i.e., e x e y , e y e z , e z e x , and
e x e y e z ) to find
∆ V = [ e + e + e ] dxdydz
x y z
The change in volume per unit volume is called the volumetric strain or the dilatation e. It
is defined as
V
e = ∆ = ex + ey + e
(13.21)
z
V
Note that the change in volume is not dependent on shear strains, which change the
orthogonal shape of the infinitesimal element but not the volume.
Next, consider again the generalized Hooke’s law equation (13.16):
ex 1
= [ σx E
− νσ ( y + σ z )]
1
ey = [ σ y
E
− νσ ( x + σ z )]
1
ez = [ σz E
− νσ ( x + σ y )]
If we add these equations together, we find that the dilatation can be expressed in terms of
stresses and material constants as
1−
2 ν
e = ( σ x + σ y + σ z )
(13.22)
E
Since the dilatation e represents a change in volume, it is independent of the orientation of
the element considered. It follows, then, that the quantities e x + e y + e z and σ x + σ y + σ z are
also independent of the orientation of the element.
An interesting fact about Poisson’s ratio can be observed from the volumetric strain.
Suppose we consider a material for which the volume doesn’t change, regardless of the
intensity of the stresses acting on it. In other words, the volumetric strain is zero (i.e., e = 0).
In order for Equation (13.22) to be equal to zero for any possible combination of stresses, it
must be true that
1− 2ν
= 0 ∴ ν =
1
2