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Question
If the yield strength of steel is 2.5 × 108 Nm–2, what is the maximum weight that can be hung at the lower end of the wire?
Solution
Clearly, tension will be maximum at x = L
∴ T = μgL + Mg = (m + M)g ......[∵ m = μL]
The yield force = (Yield strength Y) area = 250 × 106 × π × (10–3)2 = 25 × πN
At the yield point, T = Yield force
⇒ (m + M)g = 250 × π
m = π × (10–3)2 × 10 × 7860 << M
∴ Mg = 250 × π
Hence, M = `(250 xx pi)/10` = 25 × π = 75 kg.
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