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A bar of thickness 20 mm and having a rectangular cross-section carries a load of 82.5 kN. Determine (a) the minimum width of the bar to limit the maximum stress to 150 MPa, (b) the modulus of elasticity of the material of the bar if the 150 mm long bar extends by 0.8 mm when carrying a load of 200 kN. |
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Answer» (a) Force, F = 82.5 kN = 82500 N and cross-sectional area A = (20x)10-6 m2 , where x is the width of the rectangular bar in millimetres. Stress \(\sigma\) = \(\cfrac FA\), from which, A = \(\cfrac F{\sigma}\) = \(\cfrac{82500\,N}{150\times10^6\,pa}\) = 5.5 x10-4 m2 = 5.5 x 10-4 x 106 mm2 = 5.5 x102 mm2 = 550 mm2 Hence, 550 = 20x, from which, width of bar, x = \(\cfrac{550}{20}\) = 27.5 mm (b) Stress \(\sigma\) = \(\cfrac FA\) = \(\cfrac{200000}{550\times10^{-6}}\) = 363.64 MPa Extension of bar = 0.8 mm Strain \(\varepsilon\) = \(\cfrac XL\) = \(\cfrac{0.8}{150}\) = 0.005333 Modulus of elasticity, E = \(\cfrac{stress}{strain}\) = \(\cfrac{363.64\times10^6}{0.005333}\) = 68.2 x 109 = 68.2 GPa |
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