HW5+Solutions - HOMEWORK PROBLEMS Week 5 Due Friday 14...

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HOMEWORK PROBLEMS Week 5: Due Friday 14 October 1. Figure 1 shows an assembly for measuring the tensile strength of the wooden specimen AB . The specimen has thickness (perpendicular to the figure) of 10 mm and width d = 15 mm. If the wood fails in tension at a stress of 800 kPa, determine the force F at which the specimen will fail. If the wood also fails in shear (parallel to the grain) at a stress of 200 kPa, determine the minimum dimension h if shear failure is not to occur before tension failure. h d F F A B Figure 1 Solution h d F A B F Figure 1.1 Figure 1.1 shows a free-body diagram of the specimen cut in the middle, where the cross- sectional area is A = 10 × 15 = 150 mm 2 . The force F is perpendicular to the cut, so there will be a tensile stress σ = F A = F 150 1
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and if F is in N, this will be in N/mm 2 = N/m 2 × 10 6 in other words, MPa. Since tensile failure occurs at 800 kPa, we conclude that F at failure is given by F 150 = 0 . 8 or F = 120 N . h F F/ 2 F/ 2 Figure 1.2 Figure 1.2 shows a free-body diagram of the specimen split along two lines parallel with the bar, leaving two small rectangles of wood behind in the left hand grip as shown. In this case, there are two cut surfaces, each of area h × 10 and they will share the load equally, giving a shear stress τ = F/ 2 h × 10 = F 20 h .
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