Maximum shear stress:
The state of stress at a critical point of a strained solid is given by σ_{x} = 70 kN/mm^{2}, σ_{y} = - 50 kN/mm^{2} and τ_{xy} = 45 kN/mm^{2}. If the strength of the solid in tension, compression and shear are provided as 120 kN/mm^{2}, 90 kN/mm^{2} and 75 kN/mm^{2} correspondingly, verify the safety of the component.
Solution
Given σ_{x} = 70 kN/mm^{2}
σ_{y} = - 50 kN/mm^{2}
τ_{xy} = 45 kN/mm^{2}
= 85 - 65 N/mm^{2}
Maximum shear stress, τ_{max} = σ_{1} - σ_{2}/2 = 85 -(-65)/2
= 75 N/mm^{2}
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.