Five bolts are used in the connection between the axial member and the support.

Question

Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 270 MPa, and a factor of safety of 5.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 425 kN.

Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 270 MPa, and a factor of safety of 5. 2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 425 kN.

Explanation / Answer

Allowable stress = 270/5.2 = 51.923 N/mm^2

So Force = Stress * area

Force = P/2 ( double shear case ) = 425,000 N / 2 = 212,500 N

212,500 = 51.923 *( 5* area of each bolt)

area of each bolt = 805.9259 mm^2

pi*d^2/4 = 805.9259

d = 32.28266 mm

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