Question

Question 2 A stone which has a mass of 800 kg rests on a pad at A and a roller at B as shown in Figure Q2. G represents the center of gravity of the stone. The pad at A is fixed to the ground and has a compressed height of 30 mm, width of 140 mm and a length of 150 mm. The pad at A is made of a material with Youngs Modulus of 7 MPa and Possions ratio of 0.35 The coefficient of friction between the pad and stone is 0.8. Determine the horizontal displacement of stone caused by shear strains in pad, before stone begins to slip. (35 Marks) 0.4 m 0.3 m B 1.25 m1.5 mA Figure Q2

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Answer 1

E = 2G (1 + 1) [E young's modules G = modules f rigidity = parsons ratio] 7 times 10^6 = 2G (1 + .35) G = 7/2.7 times 10^6 = Component f weight f stone on pad F_1 = w times 1.25/2.75 = 800 times 9.8 times 1.25/2.75 = 3563.64 force f friction causing shed strain = F_1 = 3563.64 times 8 = 2850.90 Shear strain = F/A = 2850.90/150 times 140 times 10^-6 = 1.3575 times 10^5 Now theta = F/A/G = 1.3575 times 10^5/7/2.7 times 10^6 = .05236 If l be displacement theta = l/r = > l = theta times r [r is height f r Rad A] l = .05236 times 30 times 10^-3 l = 1.57 mm