Question

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:

• A. $100\ \text{N/mm}^2$
• B. $200\ \text{N/mm}^2$
• C. $300\ \text{N/mm}^2$
• D. $400\ \text{N/mm}^2$

Hint:

In pure bending, $\sigma_{\max}=E\,c/R$—no need to find $M$ or $I$ if $R$ and $E$ are known.

Answer 1

The Correct Option is B

Step 1: Curvature–stress relation 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 occurs at $y=c=\frac{d}{2}$.

Step 2: Substitute data.
$d=20$ 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: Conclusion.
$\sigma_{\max}=200\ \text{N/mm}^2$.