A wire of circular cross-section of diameter 1.0 mm is bent into a circular arc of radius 1.0 m by application of pure bending moments at its ends. The Young’s modulus of the material of the wire is 100 GPa. The maximum tensile stress developed in the wire is ____________ MPa

To find the maximum tensile stress developed in the wire, we can use the classical beam bending equation:

$\frac{M}{I}=\frac{\sigma }{y}=\frac{E}{R}$

Where:

• $\sigma$ is the bending stress.
• $y$ is the distance from the neutral axis.
• $E$ is the Young's modulus of the material.
• $R$ is the radius of curvature.

The maximum bending stress (which is tensile on the outer fiber in tension) occurs at the maximum distance from the neutral axis ($y_{\text{max}}$).

For a wire of circular cross-section with diameter $d=1.0\text{mm}$, the distance to the outermost fiber is:

$y_{\text{max}}=\frac{d}{2}=\frac{1.0\text{mm}}{2}=0.5\text{mm}$

Given values:

• Young's modulus, $E=100\text{GPa}=100\times {10}^3\text{MPa}$
• Radius of curvature, $R=1.0\text{m}=1000\text{mm}$

Now, we rearrange the bending formula to solve for the maximum tensile stress (${\sigma }_{\text{max}}$):

${\sigma }_{\text{max}}=\frac{E\cdot y_{\text{max}}}{R}$

Substituting the values:

${\sigma }_{\text{max}}=\frac{(100\times {10}^3\text{MPa})\cdot \left(0.5\text{mm}\right)}{1000\text{mm}}$

${\sigma }_{\text{max}}=\frac{50\times {10}^3}{1000}\text{MPa}=50\text{MPa}$

Thus, the maximum tensile stress developed in the wire is 50 MPa.