Question

If a member is subjected to an axial tensile load then the plane inclined at \(45^\circ\) to the axis of the loading carries

• A. minimum shear stress
• B. maximum shear stress
• C. maximum normal stress
• D. minimum normal stress

Hint:

Maximum shear stress on a 45° plane is a fundamental result in axial stress analysis.

Answer 1

The Correct Option is B

To determine which stress the plane inclined at \(45^\circ\) to the axis of an axial tensile load carries, we need to consider the stress transformation equations for a plane inclined at an angle \(\theta\) from the axis. When a member experiences axial tensile load, normal and shear stresses on any plane are given by the equations:

Normal Stress (\(\sigma_{\theta}\)):

\(\sigma_{\theta} = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2} \cos(2\theta) + \tau_{xy} \sin(2\theta)\)

Since only axial load \(\sigma_x\) is present and \(\sigma_y = \tau_{xy} = 0\), then:

\(\sigma_{\theta} = \frac{\sigma_x}{2} + \frac{\sigma_x}{2} \cos(2\theta)\)

Shear Stress (\(\tau_{\theta}\)):

\(\tau_{\theta} = -\frac{\sigma_x - \sigma_y}{2} \sin(2\theta) + \tau_{xy} \cos(2\theta)\)

Given \(\sigma_y = \tau_{xy} = 0\), the equation becomes:

\(\tau_{\theta} = -\frac{\sigma_x}{2} \sin(2\theta)\)

For the angle \(\theta = 45^\circ\), \(2\theta = 90^\circ\), and \(\sin(90^\circ) = 1\), thus the shear stress becomes:

\(\tau_{\theta} = -\frac{\sigma_x}{2}\)

This analysis shows that for a plane inclined at \(45^\circ\), the shear stress is maximized (\(\textbf{maximum shear stress}\)).