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A split pin requires a force of 400N to shear it. The maximum shear occurs is 120MPa. Determine the minimum diameter of the pin |
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Answer» Stress \(\sigma\) = \(\cfrac{forcce F}{area A}\) = hence, cross-sectional area, A = \(\cfrac{forcce F}{stress\,\sigma}\) = \(\cfrac{400}{120\times10^6}\) = 3.333 x 10-6 m2 Circular area = \(\pi\)r2 = 3.3333 x 10-6 m2 from which, r2 = \(\cfrac{3.3333 \times10^{-6}}\pi\) and radius r = \(\sqrt{\cfrac{3.3333 \times10^{-6}}\pi}\) = 1.030 x 10-3 m = 1.030 mm diameter d = 2 x r = 2 x 1.030 = 2.06 mm |
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