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ch01-03 stress & strain & properties

01 Solutions 46060

01 Solutions 46060 5/6/10 2:43 PM Page 31 © 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. •1–45. The truss is made from three pin-connected members having the cross-sectional areas shown in the figure. Determine the average normal stress developed in each member when the truss is subjected to the load shown. State whether the stress is tensile or compressive. Joint B: s AB = F AB = 625 A AB 1.5 = 417 psi (C) s BC = F BC = 375 = 469 psi (T) A BC 0.8 Ans. Ans. 4 ft C A AC 0.6 in. 2 3 ft A BC 0.8 in. 2 A AB 1.5 in. 2 500 lb B Joint A: sAC œ = F AC = 500 = 833 psi (T) A AC 0.6 Ans. A 1–46. Determine the average normal stress developed in links AB and CD of the smooth two-tine grapple that supports the log having a mass of 3 Mg. The cross-sectional area of each link is 400 mm 2 . A 20 B E C D + c ©F y = 0; 2(F sin 30°) - 29.43 = 0 F = 29.43 kN 0.2 m a+©M E = 0; P cos 20°(0.2) - (29.43 cos 30°)(1.2) + (29.43 sin 30°)(0.4 cos 30°) 1.2 m = 0 s = P A = 135.61(103 ) 400(10 - 6 ) P = 135.61 kN = 339 MPa Ans. 30 0.4 m 30 31

01 Solutions 46060 5/6/10 2:43 PM Page 32 © 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. 1–47. Determine the average shear stress developed in pins A and B of the smooth two-tine grapple that supports the log having a mass of 3 Mg. Each pin has a diameter of 25 mm and is subjected to double shear. A 20 B E C D + c ©F y = 0; 2(F sin 30°) - 29.43 = 0 F = 29.43 kN 0.2 m a+©M E = 0; P cos 20°(0.2) - (29.43 cos 30°)(1.2) + (29.43 sin 30°)(0.4 cos 30°) 1.2 m = 0 P = 135.61 kN t A = t B = V 135.61(10 3 ) A = 2 p 4 (0.025)2 = 138 MPa Ans. 30 0.4 m 30 *1–48. The beam is supported by a pin at A and a short link BC. If P = 15 kN, determine the average shear stress developed in the pins at A, B, and C. All pins are in double shear as shown, and each has a diameter of 18 mm. C 30 P 4P 4P 2P 0.5m 0.5 m 1 m 1.5 m 1.5 m B A For pins B and C: t B = t C = V A = 82.5 (103 ) For pin A: F A = 2 (82.5) 2 + (142.9) 2 = 165 kN t A = V A = 82.5 (103 ) p 4 ( 18 p 4 ( 18 1000 )2 = 324 MPa 1000 )2 = 324 MPa Ans. Ans. 32

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