Question:medium
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 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:
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Tension grows linearly from \(0\) at the bottom to \(W\) at the top; integrate, giving half of \(WL/AY\).
Updated On: Jul 2, 2026
- \(0\)
- \(\dfrac{WL}{2AY}\)
- \(\dfrac{3WL}{2AY}\)
- \(\dfrac{WL}{4AY}\)
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The Correct Option is B
Solution and Explanation
Use the idea of an effective load acting at the centre of mass.
Non-uniform stress. Because the rod stretches itself, the stress grows from zero at the free bottom end to a maximum at the top support. The average tension over the rod is what matters for the total stretch.
Average tension. Tension varies linearly from $0$ at the bottom to $W$ at the top, so the average tension is \[T_{avg}=\frac{0+W}{2}=\frac{W}{2}.\] Equivalent uniform rod. The total elongation equals that of a rod carrying this average tension over its full length: \[\delta=\frac{T_{avg}\,L}{AY}=\frac{(W/2)L}{AY}.\] Result. \[\delta=\frac{WL}{2AY}.\] This is exactly half of what a point weight $W$ hung at the free end would produce ($WL/AY$), because the weight here is spread uniformly along the rod. \[\boxed{\delta=\dfrac{WL}{2AY}}\]
Non-uniform stress. Because the rod stretches itself, the stress grows from zero at the free bottom end to a maximum at the top support. The average tension over the rod is what matters for the total stretch.
Average tension. Tension varies linearly from $0$ at the bottom to $W$ at the top, so the average tension is \[T_{avg}=\frac{0+W}{2}=\frac{W}{2}.\] Equivalent uniform rod. The total elongation equals that of a rod carrying this average tension over its full length: \[\delta=\frac{T_{avg}\,L}{AY}=\frac{(W/2)L}{AY}.\] Result. \[\delta=\frac{WL}{2AY}.\] This is exactly half of what a point weight $W$ hung at the free end would produce ($WL/AY$), because the weight here is spread uniformly along the rod. \[\boxed{\delta=\dfrac{WL}{2AY}}\]
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