Forty bolts are to be selected for fixing the cover plate of a cylinder subjected to a maximum load of 990 kN. If the design stress for material is 330 N/mm², what is the diameter of each bolt?

Concept:

Design of bolt is done by,

\(Strength\;of\;bolt\;material = \frac{{Maximum\;load}}{{N \times \left( {Area\;of\;cross\;section\;of\;bolt} \right)}}\)

where N = No. of bolts supporting that load.

Calculation:

Given:

No. of bolts = 40, Maximum load = 990 kN, Strength of bolt material = 330 N/mm2

\(Strength\;of\;bolt\;material = \frac{{Maximum\;load}}{{N \times \left( {Area\;of\;cross\;section\;of\;bolt} \right)}}\)

\(330 = \frac{{990 \times {{10}^3}}}{{40 \times \frac{\pi }{4} \times {d^2}}}\)

d² = 95.5 mm

d = 9.77 mm

d ≈ 9.8 mm

Hence the required diameter of the bolt is 9.8 mm.