Answer
For steel, $ 1.67 \times 10^{6} J/m^{3}$
For brass, $ 7.44 \times 10^{5} J/m^{3}$
For aluminum, $ 5.48 \times 10^{5} J/m^{3}$
For titanium, $ 2.22 \times 10^{6} J/m^{3}$
Work Step by Step
Given:
refer to the given table (steel alloy, brass alloy, aluminum alloy, and titanium alloy)
from Table 6.1, the modulus of elasticity for each alloy is as follows:
Steel alloy = 207 GPa
Brass alloy = 97 GPa
Aluminum alloy= 69 GPa
Titanium alloy = 107 GPa
Required:
moduli of resilience for each alloy
Solution:
Using Equation 6.14:
$U_{r} = \frac{σ_{y}^{2}}{2E}$
For steel,
$U_{r} = \frac{(830 \times 10^{6} N/m^{2})^2}{(2)(207 \times 10^{9} N/m^{2})} = 1.67 \times 10^{6} J/m^{3}$
For brass,
$U_{r} = \frac{(380 \times 10^{6} N/m^{2})^2}{(2)(97 \times 10^{9} N/m^{2})} = 7.44 \times 10^{5} J/m^{3}$
For aluminum,
$U_{r} = \frac{(275 \times 10^{6} N/m^{2})^2}{(2)(69 \times 10^{9} N/m^{2})} = 5.48 \times 10^{5} J/m^{3}$
For titanium,
$U_{r} = \frac{(690 \times 10^{6} N/m^{2})^2}{(2)(107 \times 10^{9} N/m^{2})} = 2.22 \times 10^{6} J/m^{3}$
