Page 205 - Mechanical Behavior of Materials

206 Chapter 5 Stress–Strain Relationships and Behavior

Only two independent elastic constants are needed for an isotropic material, so that one of E, G,
and ν can be considered redundant. The following equation allows any one of these to be calculated
from the other two:

E
G = (5.28)
2 (1 + ν)
This equation can be derived by considering a state of pure shear stress, as in a round bar under
torsion in Fig. 4.41. Recall from elementary mechanics of materials that the state of shear stress, τ,
can be equivalently represented by principal normal stresses on planes rotated 45 with respect
◦
to the planes of pure shear. Similarly, the shear strain is equivalent to normal strains as shown in
◦
Fig. 4.41, also on 45 planes. Let the (x, y, z) directions for Eq. 5.26 correspond to the principal
(1, 2, 3) directions. The following substitutions can then be made in Eq. 5.26(a):
γ
σ x = τ, σ y =−τ, σ z = 0, ε x = (5.29)
2
This yields
2 (1 + ν)
γ = τ (5.30)
E
From the definition of G, the constant of proportionality is G = τ/γ , so that Eq. 5.28 is confirmed.
Measured values of the three constants E, G, and ν for real materials will not generally obey
Eq. 5.28 perfectly. This situation is mainly due to the material not being perfectly isotropic.

Example 5.2
A cylindrical pressure vessel 10 m long has closed ends, a wall thickness of 5 mm, and an inner
diameter of 3 m. If the vessel is filled with air to a pressure of 2 MPa, how much do the length,
diameter, and wall thickness change, and in each case is the change an increase or a decrease?
The vessel is made of a steel having elastic modulus E = 200,000 MPa and Poisson’s ratio
ν = 0.3. Neglect any effects associated with the details of how the ends are attached.
Solution Attach a coordinate system to the surface of the pressure vessel, as shown in
Fig. E5.2, such that the z-axis is normal to the surface.
The ratio of radius to thickness, r/t, is such that it is reasonable to employ the thin-walled
tube assumption and the resulting stress equations, given in Fig. A.7(a) in Appendix A. Denoting
the pressure as p,wehave
pr (2MPa)(1500 mm)
σ x = = = 300 MPa
2t 2 (5mm)
pr (2MPa)(1500 mm)
σ y = = = 600 MPa
t 5mm
The value of σ z varies from −p on the inside wall to zero on the outside, so its value for the
present case is everywhere sufficiently small that σ z ≈ 0 can be used. Substitute these stresses
and the known E and ν into Hooke’s law, Eq. 5.26, which gives