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01 Solutions 46060 5/6/10 2:43 PM Page 71 © 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. 1–106. The bearing pad consists of a 150 mm by 150 mm block of aluminum that supports a compressive load of 6 kN. Determine the average normal and shear stress acting on the plane through section a–a. Show the results on a differential volume element located on the plane. Equation of Equilibrium: 6 kN a +Q©F x = 0; V a - a - 6 cos 60° = 0 V a - a = 3.00 kN a+©F y = 0; N a - a - 6 sin 60° = 0 N a - a = 5.196 kN 30 Averge Normal Stress And Shear Stress: The cross sectional Area at section a–a is A = a 0.15 b(0.15) = 0.02598 m2. sin 60° a 150 mm s a - a = N a - a A = 5.196(103 ) = 200 kPa 0.02598 Ans. t a - a = V a - a A = 3.00(103 ) = 115 kPa 0.02598 Ans. 1–107. The yoke-and-rod connection is subjected to a tensile force of 5 kN. Determine the average normal stress in each rod and the average shear stress in the pin A between the members. 5 kN 40 mm For the 40 – mm – dia rod: s 40 = P A = 5 (103 ) p 4 (0.04)2 = 3.98 MPa Ans. A 25 mm 30 mm For the 30 – mm – dia rod: 5 kN s 30 = V A = 5 (103 ) p 4 (0.03)2 = 7.07 MPa Ans. Average shear stress for pin A: t avg = P A = 2.5 (103 ) = 5.09 MPa (0.025)2 p 4 Ans. 71

01 Solutions 46060 5/6/10 2:43 PM Page 72 © 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. *1–108. The cable has a specific weight g (weight>volume) and cross-sectional area A. If the sag s is small, so that its length is approximately L and its weight can be distributed uniformly along the horizontal axis, determine the average normal stress in the cable at its lowest point C. A C L/2 L/2 B s Equation of Equilibrium: a +©M A = 0; Ts - gAL 2 a L 4 b = 0 T = gAL2 8 s Average Normal Stress: s = T gAL 2 A = 8 s A = gL2 8 s Ans. 72

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01 Solutions 46060 5/6/10 2:43 PM P

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Magazine: ch01-03 stress & strain & properties