Question

Seven bolts are used in the connection between the axial member and the support shown in the figure. The ultimate shear strength of the bolts is 320 MPa.

If a factor of safety of 2.5 is required, determine the minimum allowable bolt diameter required to support an applied load of F = 1225 kN.

Answer 1

The minimum allowable bolt diameter required is 33.3 mm.

Step-by-step explanation:

1. Convert the applied load from kN to N: F = 1225 kN = 1,225,000 N.

2. Determine the ultimate shear strength of one bolt: S = 320 MPa.

3. Determine the required bolt force: Fb = F/7 = 1,225,000 N / 7 = 175,000 N.

4. Determine the minimum allowable bolt area: Ab = Fb / (S * FoS), where FoS = 2.5 is the factor of safety. Thus, Ab = 175,000 N / (320 MPa * 2.5) = 219.3 mm^2.

5. Determine the minimum allowable bolt diameter: d = sqrt(4Ab/π) = sqrt(4(219.3 mm^2)/π) = 33.3 mm.

Therefore, the minimum allowable bolt diameter required to support an applied load of F = 1225 kN with a factor of safety of 2.5 is 33.3 mm.