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ch01-03 stress & strain & properties

01 Solutions 46060

01 Solutions 46060 5/6/10 2:43 PM 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. 1–42. The pins on the frame at D and E each have a diameter of 0.25 in. If these pins are subjected to double shear, determine the average shear stress in each pin. 3 ft 3 ft 500 lb A 3 ft B C Support Reactions: FBD(a) 1.5 ft 1.5 ft 3 ft a +©M E = 0; 500(6) + 300(3) - D y (6) = 0 D 300 lb E D y = 650 lb ; + ©F x = 0; 500 - E x = 0 E x = 500 lb + c ©F y = 0; 650 - 300 - E y = 0 E y = 350 lb Average shear stress: Pins D and E are subjected to double shear as shown on FBD (b) and (c). For Pin D, then V D = F D F D = D y = 650 lb z (p D ) avg = V D 325 = p A D 4 (0.25)2 = 6621 psi = 6.62 ksi = 325 lb Ans. For Pin E, F E = 2 500 2 + 350 2 = 610.32 lb then V E = F g z = 305.16 lb (t E ) avg = V E = 305.16 p A E 4 (0.252 ) = 6217 psi = 6.22 ksi Ans. 29

01 Solutions 46060 5/6/10 2:43 PM Page 30 © 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. 1–43. Solve Prob. 1–42 assuming that pins D and E are subjected to single shear. 3 ft 3 ft 500 lb A Support Reactions: FBD(a) a +©M E = 0; 500(6) + 300(3) - D y (6) = 0 D y = 650 lb B C 3 ft ; + ©F x = 0; 500 - E x = 0 E x = 500 lb 1.5 ft 1.5 ft 3 ft + c ©F y = 0; 650 - 300 - E y = 0 E y = 350 lb Average shear stress: Pins D and E are subjected to single shear as shown on FBD (b) and (c). D 300 lb E For Pin D, V D = F D = D y = 650 lb (t D ) avg = V D A D = 650 p 4 (0.252 ) = 13242 psi = 13.2 ksi Ans. For Pin E, V E = F E = 2 500 2 + 350 2 = 610.32 lb (t E ) avg = V E = 610.32 p A E 4 (0.252 ) = 12433 psi = 12.4 ksi Ans. *1–44. A 175-lb woman stands on a vinyl floor wearing stiletto high-heel shoes. If the heel has the dimensions shown, determine the average normal stress she exerts on the floor and compare it with the average normal stress developed when a man having the same weight is wearing flat-heeled shoes. Assume the load is applied slowly, so that dynamic effects can be ignored. Also, assume the entire weight is supported only by the heel of one shoe. Stiletto shoes: A = 1 2 (p)(0.3)2 + (0.6)(0.1) = 0.2014 in 2 s = P A = 175 lb = 869 psi 2 0.2014 in 1.2 in. 0.5 in. Ans. 0.3 in. 0.1 in. Flat-heeled shoes: A = 1 2 (p)(1.2)2 + 2.4(0.5) = 3.462 in 2 s = P A = 175 lb = 50.5 psi 2 3.462 in Ans. 30

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