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ch01-03 stress & strain & properties

01 Solutions 46060

01 Solutions 46060 5/6/10 2:43 PM Page 55 © 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. 1–82. The three steel wires are used to support the load. If the wires have an allowable tensile stress of s allow = 165 MPa, determine the required diameter of each wire if the applied load is P = 6 kN. A C The force in wire BD is equal to the applied load; ie, F BD = P = 6 kN. Analysing the equilibrium of joint B by referring to its FBD, Fig. a, 45 B 30 : + ©F x = 0; F BC cos 30° - F AB cos 45° = 0 + c ©F y = 0; F BC sin 30° + F AB sin 45° - 6 = 0 Solving Eqs. (1) and (2), F AB = 5.379 kN F BC = 4.392 kN For wire BD, s allow = F BD ; 165(10 6 ) = 6(103 ) p A BD 4 d BD 2 d BD = 0.006804 m = 6.804 mm Use d BD = 7.00 mm For wire AB, s allow = F AB ; 165(10 6 ) = 5.379(103 ) p A AB 4 d AB 2 d AB = 0.006443 m = 6.443 mm Use d AB = 6.50 mm For wire BC, s allow = F BC ; 165(10 6 ) = 4.392(103 ) p A BC 4 d BC 2 d BC = 0.005822 m = 5.822 mm d BC = 6.00 mm (1) (2) Ans. Ans. Ans. D P 55

01 Solutions 46060 5/6/10 2:43 PM Page 56 © 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. 1–83. The three steel wires are used to support the load. If the wires have an allowable tensile stress of s allow = 165 MPa, and wire AB has a diameter of 6 mm, BC has a diameter of 5 mm, and BD has a diameter of 7 mm, determine the greatest force P that can be applied before one of the wires fails. A 45 B 30 C The force in wire BD is equal to the applied load; ie, F BD = P. Analysing the equilibrium of joint B by referring to its FBD, Fig. a, : + ©F x = 0; F BC cos 30° - F AB cos 45° = 0 (1) D P + c ©F y = 0; F BC sin 30° + F AB sin 45° - P = 0 (2) Solving Eqs. (1) and (2), F AB = 0.8966 P F BC = 0.7321 P For wire BD, s allow = F BD A BD ; 165(10 6 ) = For wire AB, P p 4 (0.0072 ) P = 6349.94 N = 6.350 kN s allow = F AB ; 165(10 6 ) = 0.8966 P p A AB 4 (0.0062 ) P = 5203.42 N = 5.203 kN For wire BC, s allow = F BC ; 165(10 6 ) = 0.7321 P p A BC 4 (0.0052 ) P = 4425.60 N = 4.43 kN (Controls!) Ans. 56

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