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ch01-03 stress & strain & properties

01 Solutions 46060

01 Solutions 46060 5/6/10 2:43 PM Page 13 © 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. 1–22. The floor crane is used to lift a 600-kg concrete pipe. Determine the resultant internal loadings acting on the cross section at G. 0.2 m 0.2 m E B H C 0.4 m 0.6 m G F 0.3 m 0.5 m D 75 A Support Reactions: We will only need to compute F EF by writing the moment equation of equilibrium about D with reference to the free-body diagram of the hook, Fig. a. a+©M D = 0; F EF (0.3) - 600(9.81)(0.5) = 0 F EF = 9810 N Internal Loadings: Using the result for F EF , section FG of member EF will be considered. Referring to the free-body diagram, Fig. b, : + ©F x = 0; 9810 - N G = 0 N G = 9810 N = 9.81 kN + c ©F y = 0; V G = 0 a+©M G = 0; M G = 0 Ans. Ans. Ans. 13

01 Solutions 46060 5/6/10 2:43 PM Page 14 © 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. 1–23. The floor crane is used to lift a 600-kg concrete pipe. Determine the resultant internal loadings acting on the cross section at H. 0.2 m 0.2 m E B H C 0.4 m 0.6 m G F 0.3 m 0.5 m D 75 A Support Reactions: Referring to the free-body diagram of the hook, Fig. a. a+©M F = 0; D x (0.3) - 600(9.81)(0.5) = 0 D x = 9810 N + c ©F y = 0; D y - 600(9.81) = 0 D y = 5886 N Subsequently, referring to the free-body diagram of member BCD,Fig.b, a+©M B = 0; F AC sin 75°(0.4) - 5886(1.8) = 0 F AC = 27 421.36 N : + ©F x = 0; B x + 27 421.36 cos 75° - 9810 = 0 B x = 2712.83 N + c ©F y = 0; 27 421.36 sin 75° - 5886 - B y = 0 B y = 20 601 N Internal Loadings: Using the results of B x and B y , section BH of member BCD will be considered. Referring to the free-body diagram of this part shown in Fig. c, : + ©F x = 0; N H + 2712.83 = 0 N H = -2712.83 N = -2.71 kN + c ©F y = 0; -V H - 2060 = 0 V H = -20601 N = -20.6 kN Ans. Ans. a+©M D = 0; M H + 20601(0.2) = 0 M H = -4120.2 N # m = -4.12 kN # m Ans. The negative signs indicates that N H , V H , and M H act in the opposite sense to that shown on the free-body diagram. 14

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