Question
A 0.5 -m-long steel rod with a 1 -cm diameter is stretched in a tensile test. What is the work required to obtain a relative strain of $0.1 \%$ ? The modulus of elasticity of steel is $2 \times 10^{8} \mathrm{kPa}$
Step 1
The formula for the area of a circle is $\pi r^2$, where r is the radius. Given the diameter is 1 cm, the radius is 0.5 cm or 0.005 m. So, the area is: \[a = \frac{\pi}{4} \times (0.01)^2 = 78.54 \times 10^{-6} \, \text{m}^2\] Show more…
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