Question

A body is subjected to a tensile stress of \(1200\) MPa on one plane and another tensile stress of \(600\) MPa on a plane at right angles to the former. It is also subjected to shear stress of \(400\) MPa on the same planes. The maximum normal stress will be

• A. \(400\) MPa
• B. \(500\) MPa
• C. \(900\) MPa
• D. \(1400\) MPa

Hint:

Remember the principal stress formula: \[ \boxed{ \sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau^2} } \] The '+' sign gives the maximum principal stress, while the '−' sign gives the minimum principal stress.

Answer 1

The Correct Option is D

Concept: For a two-dimensional state of stress, the principal (maximum and minimum normal) stresses are given by \[ \sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^{\,2}}. \] The larger value gives the maximum principal (normal) stress.

Step 1: Write the given stresses. Given, \[ \sigma_x=1200\text{ MPa}, \] \[ \sigma_y=600\text{ MPa}, \] \[ \tau_{xy}=400\text{ MPa}. \]

Step 2: Compute the average normal stress. \[ \frac{\sigma_x+\sigma_y}{2} = \frac{1200+600}{2} = 900\text{ MPa}. \]

Step 3: Compute the radius of Mohr's circle. \[ R = \sqrt{\left(\frac{1200-600}{2}\right)^2+400^2} = \sqrt{300^2+400^2} = \sqrt{250000} = 500\text{ MPa}. \]

Step 4: Calculate the maximum principal stress. \[ \sigma_{\max} = 900+500 = 1400\text{ MPa}. \] Hence, \[ \boxed{\sigma_{\max}=1400\text{ MPa}.} \] Therefore, the correct option is \[ \boxed{(D)\;1400\text{ MPa}.} \]