Elongating

#648 / Mechanics of Materials Easy

In the figure shown, a 4 inch diameter by 2 feet long aluminum rod is placed under tension by a 40 lb force. If the rod’s strain is 0.001, how much did it elongate in inches? What is the percent elongation?

Expand Hint
Strain is the change in length per unit length.
Hint 2
$$$\varepsilon=\frac{\Delta L}{L_o} $$$
where $$\varepsilon$$ is the engineering strain, $$\Delta L$$ is the change in length, and $$L_o$$ is the original length.
Strain is the change in length per unit length.
$$$\varepsilon=\frac{\Delta L}{L_o} $$$
where $$\varepsilon$$ is the engineering strain, $$\Delta L$$ is the change in length, and $$L_o$$ is the original length.
$$$0.001=\frac{\Delta L}{2ft\cdot \frac{12in}{ft}} $$$
Solving for change in length:
$$$\Delta L=(0.001)(24in)=0.024\:inch$$$
Percent Elongation is:
$$$\%\:Elongation=\left (\frac{\Delta L}{L_o}   \right )\times 100$$$
$$$=(0.001)\times 100=0.1\%$$$
  1. 0.024 inch
  2. 0.1%


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