Question

The state of plane stress in an element is shown in the figure.

The maximum shear stress in the element is ______ \(\text{N/mm}^2\).

• A. 0
• B. 5
• C. 10
• D. 25

Hint:

Both given faces have zero shear, so sigma x and sigma y are already the principal stresses.

Answer 1

The Correct Option is B

Step 1: Read the stresses from the figure.
The element carries a normal stress \( \sigma_x = 20 \text{ N/mm}^2 \) (tensile, on the vertical faces) and \( \sigma_y = 30 \text{ N/mm}^2 \) (tensile, on the horizontal faces), with no shear marked on these faces, so \( \tau_{xy} = 0 \).

Step 2: Note that these are already the principal stresses.
Since \( \tau_{xy} = 0 \) on the given faces, \( \sigma_x \) and \( \sigma_y \) are themselves the two in-plane principal stresses, with \( \sigma_1 = 30 \text{ N/mm}^2 \) and \( \sigma_2 = 20 \text{ N/mm}^2 \).

Step 3: Apply the maximum in-plane shear stress formula.
\( \tau_{max} = \sqrt{\left(\dfrac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} = \sqrt{\left(\dfrac{20 - 30}{2}\right)^2 + 0} = \sqrt{25} = 5 \text{ N/mm}^2 \).

Final Answer:
The maximum shear stress is half the difference of the two principal stresses. \[ \boxed{\tau_{max} = 5 \text{ N/mm}^2} \]