If a bolt in single shear is loaded with a shear force V = 25 kN, and the bolt diameter is 20 mm, compute the shear stress.

If a bolt in single shear is loaded with a shear force V = 25 kN, and the bolt diameter is 20 mm, compute the shear stress.

• A. ≈ 31.8 MPa
• B. ≈ 120 MPa
• C. ≈ 79.6 MPa
• D. ≈ 50 MPa

Answer: (C)

In single shear, the shear stress on the bolt is found by dividing the applied shear force by the cross-sectional area where the shear plane passes through the bolt. For a circular bolt, that area is A = πd^2/4. With a diameter of 20 mm, the area is A = π(20)^2/4 = π×400/4 = π×100 ≈ 314 mm^2. The applied force is V = 25 kN = 25,000 N. So the shear stress is τ = V/A = 25,000 / 314 ≈ 79.6 N/mm^2, i.e., 79.6 MPa. Since the scenario specifies single shear, this is the correct stress value. If the bolt were in double shear, the area would double and the stress would be about half, but that’s not the case here.

In single shear, the shear stress on the bolt is found by dividing the applied shear force by the cross-sectional area where the shear plane passes through the bolt. For a circular bolt, that area is A = πd^2/4.

With a diameter of 20 mm, the area is A = π(20)^2/4 = π×400/4 = π×100 ≈ 314 mm^2. The applied force is V = 25 kN = 25,000 N. So the shear stress is τ = V/A = 25,000 / 314 ≈ 79.6 N/mm^2, i.e., 79.6 MPa.

Since the scenario specifies single shear, this is the correct stress value. If the bolt were in double shear, the area would double and the stress would be about half, but that’s not the case here.