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A bolt having a diameter of 5 mm is loaded so that the shear stress in it is 120 MPa. Determine the value of the shear force on the bolt. |
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Answer» Stress \(\sigma\) = \(\cfrac{forcce F}{area A}\) = hence, force = stress x area = stress \(\pi\)r2 = 120 x 106 x \(\pi\) \(\left(\cfrac{5\times10^{-3}}{2}\right)^2\) = 2356 N or 2.356 kN |
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