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01 Solutions 46060 5/6/10 2:43 PM Page 57 © 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. *1–84. The assembly consists of three disks A, B, and C that are used to support the load of 140 kN. Determine the smallest diameter d 1 of the top disk, the diameter d 2 within the support space, and the diameter d 3 of the hole in the bottom disk. The allowable bearing stress for the material is 1s allow 2 b = 350 MPa and allowable shear stress is t allow = 125 MPa. C B d 1 A 140 kN d 2 d 3 20 mm 10 mm Solution Allowable Bearing Stress: Assume bearing failure for disk B. (s b ) allow = P A ; 350A106 B = 140(103 ) p 4 d 1 2 d 1 = 0.02257 m = 22.6 mm Allowable Shear Stress: Assume shear failure for disk C. t allow = V A ; 125A106 B = 140(103 ) pd 2 (0.01) d 2 = 0.03565 m = 35.7 mm Ans. Allowable Bearing Stress: Assume bearing failure for disk C. (s b ) allow = P A ; 140(10 3 ) 350A106 B = p 4 A0.03565 2 - d 32 B d 3 = 0.02760 m = 27.6 mm Ans. Since d 3 = 27.6 mm 7 d 1 = 22.6 mm, disk B might fail due to shear. t = V A = 140(10 3 ) p(0.02257)(0.02) = 98.7 MPa 6 t allow = 125 MPa (O. K !) Therefore, d 1 = 22.6 mm Ans. •1–85. The boom is supported by the winch cable that has a diameter of 0.25 in. and an allowable normal stress of s allow = 24 ksi. Determine the greatest load that can be supported without causing the cable to fail when u = 30° and f = 45°. Neglect the size of the winch. s = P A ; T 24(103 ) = p 4 (0.25)2; A u f 20 ft B T = 1178.10 lb : + ©F x = 0; -1178.10 cos 30° + F AB sin 45° = 0 d + c ©F y = 0; -W + F AB cos 45° - 1178.10 sin 30° = 0 W = 431 lb Ans. F AB = 1442.9 lb 57

01 Solutions 46060 5/6/10 2:43 PM Page 58 © 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. 1–86. The boom is supported by the winch cable that has an allowable normal stress of s allow = 24 ksi. If it is required that it be able to slowly lift 5000 lb, from u = 20° to u = 50°, determine the smallest diameter of the cable to 1 the nearest 16 in. The boom AB has a length of 20 ft. Neglect the size of the winch. Set d = 12 ft. u 20 ft B Maximum tension in cable occurs when u = 20° . sin 20° 20 = sin c 12 c = 11.842° d A f : + © F x = 0; -T cos 20° + F AB cos 31.842° = 0 + c ©F y = 0; F AB sin 31.842° - T sin 20° - 5000 = 0 T = 20 698.3 lb s = P A ; 20 698.3 24(103 ) = p 4 (d)2 d = 1.048 in. Use d = 1 1 16 in. F AB = 22 896 lb Ans. 1–87. The 60 mm * 60 mm oak post is supported on P the pine block. If the allowable bearing stresses for these materials are s oak = 43 MPa and s pine = 25 MPa, determine the greatest load P that can be supported. If a rigid bearing plate is used between these materials, determine its required area so that the maximum load P can be supported. What is this load? For failure of pine block: s = P A ; P 25(106 ) = (0.06)(0.06) P = 90 kN Ans. For failure of oak post: s = P A ; P 43(106 ) = (0.06)(0.06) P = 154.8 kN Area of plate based on strength of pine block: s = P A ; 25(106 ) = 154.8(10)3 A A = 6.19(10 - 3 )m 2 P max = 155 kN Ans. Ans. 58

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