Question

A square‑shaped body is subjected to only direct tensile stresses \(\sigma_x\) and \(\sigma_y\) as shown. If \(\sigma_x > \sigma_y\), then the value of normal stress (\(\sigma_{\theta}\)) and shear stress (\(\tau_{\theta}\)) respectively are ________.

 

• A. \(\displaystyle \frac{\sigma_x-\sigma_y}{2}\) and \(\displaystyle \frac{\sigma_x+\sigma_y}{2}\)
• B. \(\displaystyle \frac{\sigma_x+\sigma_y}{2}\) and \(\displaystyle \frac{\sigma_x-\sigma_y}{2}\)
• C. \(\displaystyle \frac{\sigma_x+\sigma_y}{2}\) and \(\displaystyle \frac{\sigma_x+\sigma_y}{2}\)
• D. \(\displaystyle \frac{\sigma_x-\sigma_y}{2}\) and \(\displaystyle \frac{\sigma_x-\sigma_y}{2}\)

Hint:

Use the stress‑transformation relations: \[ \sigma_n = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos2\theta, \quad \tau = \frac{\sigma_x-\sigma_y}{2}\sin2\theta. \]

Answer 1

The Correct Option is B

Step 1: Stress transformation formula. For a plane inclined at angle \(\theta\) to the \(x\)‑axis under plane stress, \[ \sigma_{\theta} = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos2\theta, \quad \tau_{\theta} = \frac{\sigma_x-\sigma_y}{2}\sin2\theta. \] Step 2: Substitute \(\theta=45^\circ\). \[ \cos(2\times45^\circ)=\cos90^\circ=0,\qquad \sin(2\times45^\circ)=1, \] thus \[ \sigma_{\theta} = \frac{\sigma_x+\sigma_y}{2}, \quad \tau_{\theta} = \frac{\sigma_x-\sigma_y}{2}. \]