Answer
$E = 7.12 \times 10^{10} N/m^{2} = 71.2\ GPa$
Work Step by Step
Given:
aluminum bar 125 mm long with square cross section 16.5 mm
pulled in tension with Load 66,700 N
elongation - 0.43 mm
Required:
modulus of elasticity assuming deformation is entirely elastic
Solution:
Using a combination of Equations 6.1, 6.2, and 6.5 and considering that for a square cross section, $A_{0} = b^{2}_{0}$, where $b_{0}$ is the edge length:
$E = \frac{σ}{ε} = \frac{\frac{F}{A_{0}}}{\frac{Δl}{l_{0}}}= \frac{Fl_{0}}{b^{2}_{0}Δl} = \frac{(66,700 N)(125 \times 10^{-3} m)}{(16.5 \times 10^{-3} m)^{2}(0.43 \times 10^{-3} m)} = 7.12 \times 10^{10} N/m^{2} =\ 71.2 GPa$
