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A tube of outside diameter 60 mm and inside diameter 40 mm is subjected to a tensile load of 60 kN. Determine the stress in the tube. |
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Answer» Area of tube end (annulus) = \(\pi\) \(\left(\cfrac{D^2}{4}-\cfrac{d^2}{4}\right)\) = \(\pi\) \(\left(\cfrac{(60\times10^{-3})^2}{4}-\cfrac{(40\times10^{-3})^2}{4}\right)\) = 1.5708 x 10-3 mm2 Stress \(\sigma\) = \(\cfrac{force F}{area A}\) = \(\cfrac{60\times10^3}{1.5708\times10^{-3}}\) = 38.20 x 106 pa = 38.2 Mpa |
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