What is the safe tensile load for a M36 × 4 bolt of mild steel having yield stress of 280 MPa and a factor of safety 2.

Concept:

Designation of bolts:

∴ M 36 × 4 means the nominal diameter of the bolt is 36 mm and pitch is 4 mm

The load P lifted by a bolt with failing is given by,

\(P = \frac{\pi }{4} \times {\left( {{d_c}} \right)^2} \times {σ _t}\)

where dc is the Nominal diameter of the thread in mm or m, and σt is the tensile stress of the bolt in MPa.

Calculation:

Given:

Tensile stress σt = 280 MPa ,FOS = 2

Load on the bolt

\(P = \frac{\pi }{4} \times {\left( {{d_c}} \right)^2} \times \frac{σ _t}{FOS}\)

\( P = \frac{\pi }{4} \times 36^2 \times \frac{280}{2}\\ \Rightarrow {P}=142502.976 =142.5 KN\)