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ch01-03 stress & strain & properties

01 Solutions 46060

01 Solutions 46060 5/6/10 2:43 PM Page 9 © 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. 1–15. Determine the resultant internal loading on the cross section through point C of the pliers. There is a pin at A, and the jaws at B are smooth. 20 N 120 mm 40 mm 15 mm + c ©F y = 0; -V C + 60 = 0; V C = 60 N : + ©F x = 0; N C = 0 Ans. Ans. A C B +d©M C = 0; -M C + 60(0.015) = 0; M C = 0.9 N.m Ans. D 20 N 80 mm 30 *1–16. Determine the resultant internal loading on the 20 N cross section through point D of the pliers. 120 mm 40 mm 15 mm R+©F y = 0; V D - 20 cos 30° = 0; V D = 17.3 N Ans. C +b©F x = 0; N D - 20 sin 30° = 0; N D = 10 N Ans. +d©M D = 0; M D - 20(0.08) = 0; M D = 1.60 N.m Ans. D A B 20 N 80 mm 30 9

01 Solutions 46060 5/6/10 2:43 PM Page 10 © 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. •1–17. Determine resultant internal loadings acting on section a–a and section b–b. Each section passes through the centerline at point C. Referring to the FBD of the entire beam, Fig. a, a 5 kN b B 1.5 m a+ ©M A = 0; N B sin 45°(6) - 5(4.5) = 0 N B = 5.303 kN Referring to the FBD of this segment (section a–a), Fig. b, +b©F x¿ = 0; N a - a + 5.303 cos 45° = 0 N a - a = -3.75 kN +a ©F y¿ = 0; V a - a + 5.303 sin 45° - 5 = 0 V a - a = 1.25 kN Ans. Ans. A 45 C 45 b 3 m 1.5 m a a+ ©M C = 0; 5.303 sin 45°(3) - 5(1.5) - M a - a = 0 M a - a = 3.75 kN # m Ans. Referring to the FBD (section b–b) in Fig. c, ; + ©F x = 0; N b - b - 5 cos 45° + 5.303 = 0 N b - b = -1.768 kN = -1.77 kN + c ©F y = 0; V b - b - 5 sin 45° = 0 V b - b = 3.536 kN = 3.54 kN Ans. Ans. a+©M C = 0; 5.303 sin 45° (3) - 5(1.5) - M b - b = 0 M b - b = 3.75 kN # m Ans. 10

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