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?
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A5 kN/m²
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B10 kN/m²
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C15 kN/m²
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D20 kN/m²
Answer
Correct Answer: 15 kN/m²
Explanation
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:
Identify principal stresses: sigma_1 = +20, sigma_2 = −10.Compute difference: sigma_1 − sigma_2 = 20 − (−10) = 30 kN/m².Compute maximum shear: tau_max = 30 / 2 = 15 kN/m².Report the result with correct units.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