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**03** Solutions 46060 5/7/10 8:45 AM Page 28 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •3–41. The stone has a mass of 800 kg and center of gravity at G. It rests on a pad at A and a roller at B.The pad is fixed to the ground and has a compressed height of 30 mm, a width of 140 mm, and a length of 150 mm. If the coefficient of static friction between the pad and the stone is m s = 0.8, determine the approximate horizontal displacement of the stone, caused by the shear **strain**s in the pad, before the stone begins to slip. Assume the normal force at A acts 1.5 m from G as shown. The pad is made from a material having E = 4 MPa and n = 0.35. 0.4 m B G 1.25 m 1.5 m A 0.3 m P Equations of Equilibrium: a +©M B = 0; F A (2.75) - 7848(1.25) - P(0.3) = 0 [1] : + ©F x = 0; P - F = 0 [2] Note: The normal force at A does not act exactly at A. It has to shift due to friction. Friction Equation: F = m s F A = 0.8 F A [3] Solving Eqs. [1], [2] and [3] yields: Average Shear Stress: The pad is subjected to a shear force of V = F = 3126.69 N. t = V A = 3126.69 (0.14)(0.15) = 148.89 kPa Modulus of Rigidity: F A = 3908.37 N F = P = 3126.69 N G = E 2(1 + v) = 4 = 1.481 MPa 2(1 + 0.35) Shear Strain: Applying Hooke’s law for shear g = t G = 148.89(1**03** ) 1.481(10 6 ) = 0.1005 rad Thus, d h = hg = 30(0.1005) = 3.02 mm Ans. 28

**03** Solutions 46060 5/7/10 8:45 AM Page 29 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 3–42. The bar DA is rigid and is originally held in the horizontal position when the weight W is supported from C. E If the weight causes B to be displaced downward 0.025 in., determine the **strain** in wires DE and BC. Also, if the wires 3 ft are made of A-36 steel and have a cross-sectional area of 0.002 in 2 , determine the weight W. 2 ft 3 ft 3 0.025 = 5 d d = 0.0417 in e DE = d L = 0.0417 3(12) = 0.00116 in.>in. s DE = Ee DE = 29(10 3 )(0.00116) = 33.56 ksi Ans. D B C 4 ft W A F DE = s DE A DE = 33.56 (0.002) = 0.0672 kip a+ ©M A = 0; -(0.0672) (5) + 3(W) = 0 W = 0.112 kip = 112 lb Ans. s BC = W = 0.112 = 55.94 ksi A BC 0.002 e BC = s BC E = 55.94 29 (10 3 = 0.00193 in.>in. ) Ans. 3–43. The 8-mm-diameter bolt is made of an aluminum alloy. It fits through a magnesium sleeve that has an inner diameter of 12 mm and an outer diameter of 20 mm. If the original lengths of the bolt and sleeve are 80 mm and 50 mm, respectively, determine the **strain**s in the sleeve and the bolt if the nut on the bolt is tightened so that the tension in the bolt is 8 kN. Assume the material at A is rigid. E al = 70 GPa, E mg = 45 GPa. A 50 mm 30 mm Normal Stress: s b = P A b = 8(10 3 ) p = 159.15 MPa 4 (0.0082 ) s s = P 8(10 3 ) = p A s 4 (0.022 - 0.012 2 = 39.79 MPa ) Normal Strain: Applying Hooke’s Law e b = s b = 159.15(106 ) E al 70(10 9 = 0.00227 mm>mm ) e s = s s = 39.79(106 ) E mg 45(10 9 = 0.000884 mm>mm ) Ans. Ans. 29

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