Question

A uniform heavy rod of weight \(W\), cross-sectional area \(A\), and length \(L\) is hanging from a fixed support. Young's modulus of the material of the rod is \(Y\). Neglect the lateral contraction. The elongation of the rod is:

• A. \(0\)
• B. \(\dfrac{WL}{2AY}\)
• C. \(\dfrac{3WL}{2AY}\)
• D. \(\dfrac{WL}{4AY}\)

Hint:

Tension grows linearly from \(0\) at the bottom to \(W\) at the top; integrate, giving half of \(WL/AY\).

Answer 1

The Correct Option is B

Step 1: The rod hangs under its own weight. The tension is not uniform. Measure a distance \(x\) from the bottom (free end). The part of the rod below \(x\) has weight \(\dfrac{W}{L}x\), which is the tension carried at that section: \[T(x)=\frac{W}{L}x.\] Step 2: Consider a small element of length \(dx\) at position \(x\). Its extension is \[d(\delta)=\frac{T(x)\,dx}{AY}=\frac{W x\,dx}{A Y L}.\] Step 3: Integrate from the bottom \(x=0\) to the top \(x=L\): \[\delta=\int_0^L \frac{W x}{A Y L}\,dx=\frac{W}{A Y L}\cdot\frac{L^{2}}{2}.\] Step 4: Simplify: \[\delta=\frac{W L}{2 A Y}.\] \[\boxed{\delta=\dfrac{WL}{2AY}}\]