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A rectangular block of plastic material 400 mm long by 15 mm wide by 300 mm high has its lower face fixed to a bench and a force of 150 N is applied to the upper face and in line with it. The upper face moves 12 mm relative to the lower face. Determine (a) the shear stress, and (b) the shear strain in the upper face, assuming the deformation is uniform. |
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Answer» (a) Shear stress, \(\tau\) = \(\cfrac{force}{area\, parallel \,to \,the\,force}\) Area of any face parallel to the force = 400 mm x 15 mm = (0.4 x 0.015) m2 = 0.006 m2 Hence, shear stress \(\tau\) = \(\cfrac{150\,N}{0.006\,m^2}\) = 25000 Pa or 25 kPa (b) Shear strain, \(\gamma\) = \(\cfrac XL\) = \(\cfrac{12}{300}\) = 0.04 (or 4%) |
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