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Maximum shear stress:
The state of stress at a critical point of a strained solid is given by σx = 70 kN/mm2, σy = - 50 kN/mm2 and τxy = 45 kN/mm2. If the strength of the solid in tension, compression and shear are provided as 120 kN/mm2, 90 kN/mm2 and 75 kN/mm2 correspondingly, verify the safety of the component.
Solution
Given σx = 70 kN/mm2
σy = - 50 kN/mm2
τxy = 45 kN/mm2
= 85 - 65 N/mm2
Maximum shear stress, τmax = σ1 - σ2/2 = 85 -(-65)/2
= 75 N/mm2
All the stresses are inside the strength limits of the solid and thus, the solid is safe.
Factor of safety in tension = 120/85 = 1.412
Factor of safety in compression = 90/65 = 1.3846
Factor of safety in compression = 75/75 = 1
Here, maximum tensile and compressive stresses are well inside strength limits, maximum shear stress has reached the strength limit and thus if the state of stress is proportionally increased the solid will fail in shear.
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