Question 2.17: A rectangular parallelepiped of linearly elastic, isotropic ......
A rectangular parallelepiped of linearly elastic, isotropic material is subjected to general triaxial stress σ_x, σ_y, σ_z, as shown in Fig. 1. Determine the volumetric strain, ε_V. Assume that ε \ll 1 for all three coordinate directions.
Plan the Solution We can write the change in volume in terms of the change in length of each of the three sides of the body, which can, in turn, be expressed in terms of the extensional strains ε_x, ε_y, and ε_z.We can use Hooke’s Law, Eqs. 2.38, to relate these three strains to the three stresses.
\epsilon_x =\frac{1}{E}\left[\sigma_x-\nu\left(\sigma_y+\sigma_z\right)\right]+\alpha \Delta T\epsilon_y =\frac{1}{E}\left[\sigma_y-\nu\left(\sigma_x+\sigma_z\right)\right]+\alpha \Delta T Generalized Hooke’s Law (2.38)
\epsilon_z =\frac{1}{E}\left[\sigma_z-\nu\left(\sigma_x+\sigma_y\right)\right]+\alpha \Delta TThe basic equation for volumetric strain (dilatation) is
ε_{V} = \frac{ΔV}{ V} =\frac{L^*_x L^*_y L^*_z – L_xL_yL_z} {L_xL_yL_z}where the L* lengths are the after-deformation dimensions of the body, and the L’s are the corresponding pre-deformation dimensions of the body. For uniform strain in each coordinate direction,
L* = (1 + ε)L
so
ε_V = (1 + ε_x)(1 + ε_y)(1 + ε_z) – 1
= ε_x + ε_y + ε_z + ε_xε_y + ε_yε_z + ε_zε_x + ε_xε_yε_zSince all three strains satisfy ε \ll 1, the squared and cubed terms can be dropped in the preceding equation, leaving the approximation
ε_V= ε_x + ε_y + ε_z (1)
From Eqs. 2.38,
\epsilon_x =\frac{1}{E}\left[\sigma_x-\nu\left(\sigma_y+\sigma_z\right)\right]\epsilon_y =\frac{1}{E}\left[\sigma_y-\nu\left(\sigma_x+\sigma_z\right)\right] (2)
\epsilon_z =\frac{1}{E}\left[\sigma_z-\nu\left(\sigma_x+\sigma_y\right)\right]Finally, Eqs. (1) and (2) may be combined to give following expression for the dilatation ε_V:
ε_V = \frac{1 – 2\nu}{E}( σ_x + σ_y + σ_z) Ans.
Review the Solution From the above answer we see that the change in volume per unit volume is of the order of magnitude of the strain, and is therefore small, as we would expect. It is interesting to note that, because of the
(1 – 2ν ) factor, a tensile stress causes an increase in volume, which is what we would expect to happen, but only if 0 ≤ ν ≤ 0.5. As noted earlier, most materials have a value of n that falls within the range ν = 0.25 to ν = 0.35.