Mohr’s circle / stress transformation: A body is subjected to two mutually perpendicular normal stresses of 20 kN/m² (tension) and 10 kN/m² (compression). What is the maximum shear stress?

Introduction / Context:
When plane stress involves two perpendicular normal stresses with no shear on the reference planes, principal and maximum shear stresses can be computed directly. This is a core application of Mohr’s circle and stress transformation equations.

Given Data / Assumptions:

• Sigma_x = +20 kN/m² (tensile).
• Sigma_y = −10 kN/m² (compressive).
• Tau_xy on the reference planes = 0.

Concept / Approach:
For plane stress with perpendicular normal stresses: maximum shear stress tau_max equals half the difference between the principal stresses. With sigma_1 and sigma_2 as the principal stresses, tau_max = (sigma_1 − sigma_2) / 2. Here, sigma_1 = +20 and sigma_2 = −10.

Step-by-Step Solution:

Verification / Alternative check:
Draw Mohr’s circle: center at (sigma_avg, 0) where sigma_avg = (20 − 10)/2 = 5; radius R = 15. Tau_max equals the radius, confirming 15 kN/m².

Why Other Options Are Wrong:

• 5 or 10 kN/m²: result from taking half the absolute values incorrectly or from using sigma_avg.
• 20 kN/m²: corresponds to the larger principal stress, not the shear radius.

Common Pitfalls:
Forgetting the sign of compressive stress; using absolute values without considering direction; mixing up average normal stress with shear magnitude.

Final Answer:
15 kN/m²