Question:medium

A steel wire of $20$ mm diameter is bent into a circular shape of $10$ m radius. If modulus of elasticity of wire is $2\times10^{5}\ \text{N/mm}^2$, then the maximum bending stress induced in wire is:

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In pure bending, $\sigma_{\max}=E\,c/R$—no need to find $M$ or $I$ if $R$ and $E$ are known.
Updated On: Feb 18, 2026
  • $100\ \text{N/mm}^2$
  • $200\ \text{N/mm}^2$
  • $300\ \text{N/mm}^2$
  • $400\ \text{N/mm}^2$
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The Correct Option is B

Solution and Explanation

Step 1: Relationship between curvature and stress in pure bending.
For elastic bending: $\displaystyle \frac{1}{R}=\frac{M}{EI}$ and fiber stress $\sigma=\frac{My}{I}=E\frac{y}{R}$.
Maximum stress is observed at $y=c=\frac{d}{2}$.

Step 2: Data substitution.
$d=20$ mm $\Rightarrow c=10$ mm, $R=10\ \text{m}=10{,}000$ mm, $E=2\times10^{5}\ \text{N/mm}^2$.
\[\sigma_{\max}=E\frac{c}{R}=2\times10^{5}\times\frac{10}{10{,}000}=200\ \text{N/mm}^2.\]

Step 3: Final result.
$\sigma_{\max}=200\ \text{N/mm}^2$.

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