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  # ch01-03 stress &amp; strain &amp; properties
  
  ## 01 Solutions 46060
  
  01 Solutions 46060 5/6/10 2:43 PM Page 39 © 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. 1–55. Rods AB and BC each have a diameter of 5 mm. If the load of P = 2 kN is applied to the ring, determine the average normal stress in each rod if u = 60° . A B u P Consider the equilibrium of joint B,Fig.a, : + ©F x = 0; 2 - F AB sin 60° = 0 F AB = 2.309 kN + c ©F y = 0; 2.309 cos 60° - F BC = 0 F BC = 1.155 kN C The cross-sectional area of wires AB and BC are A AB = A BC = p 4 (0.0052 ) = 6.25(10 - 6 )p m 2 . Thus, As avg B AB = F AB A AB = 2.309(103 ) 6.25(10 - 6 )p = 117.62(106 ) Pa = 118 MPa As avg B BC = F BC A BC = 1.155(103 ) 6.25(10 - 6 )p = 58.81(106 ) Pa = 58.8 MPa Ans. Ans. 39
  
  01 Solutions 46060 5/6/10 2:43 PM Page 40 © 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. *1–56. Rods AB and BC each have a diameter of 5 mm. Determine the angle u of rod BC so that the average normal stress in rod AB is 1.5 times that in rod BC. What is the load P that will cause this to happen if the average normal stress in each rod is not allowed to exceed 100 MPa? A u P Consider the equilibrium of joint B,Fig.a, B + c ©F y = 0; F AB cos u - F BC = 0 (1) : + ©F x = 0; P - F AB sin u = 0 (2) The cross-sectional area of rods AB and BC are A AB = A BC = p 4 (0.0052 ) = 6.25(10 - 6 )p m 2 . Since the average normal stress in rod AB is required to be 1.5 times to that of rod BC, then C As avg B AB = 1.5 As avg B BC F AB A AB = 1.5 a F BC A BC b F AB 6.25(10 - 6 )p = 1.5 c F BC 6.25(10 - 6 )p d F AB = 1.5 F BC (3) Solving Eqs (1) and (3), u = 48.19° = 48.2° Ans. Since wire AB will achieve the average normal stress of 100 MPa first when P increases, then F AB = s allow A AB = C100(10 6 )DC6.25(10 - 6 )pD = 1963.50 N Substitute the result of F AB and u into Eq (2), P = 1.46 kN Ans. 40
  
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