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**03** Solutions 46060 5/7/10 8:45 AM 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. •3–13. A bar having a length of 5 in. and cross-sectional area of 0.7 is subjected to an axial force of 8000 lb. If the bar stretches 0.002 in., determine the modulus of elasticity of the material. The material has linear-elastic behavior. 8000 lb 5 in. 8000 lb Normal Stress and Strain: in 2 10 s = P A = 8.00 0.7 e = dL L = 0.002 5 = 11.43 ksi = 0.000400 in.>in. Modulus of Elasticity: E = s e = 11.43 0.000400 = 28.6(1**03** ) ksi Ans. 3–14. The rigid pipe is supported by a pin at A and an A-36 steel guy wire BD. If the wire has a diameter of 0.25 in., determine how much it stretches when a load of P = 600 lb acts on the pipe. B Here, we are only interested in determining the force in wire BD. Referring to the FBD in Fig. a 4 ft P a+©M A = 0; F BD A 4 5 B(3) - 600(6) = 0 F BD = 1500 lb The normal **stress** developed in the wire is s BD = F BD A BD = 1500 p 4 (0.252 ) = 30.56(1**03** ) psi = 30.56 ksi A D 3 ft 3 ft C Since s BD 6 s y = 36 ksi, Hooke’s Law can be applied to determine the **strain** in the wire. s BD = EP BD ; 30.56 = 29.0(10 3 )P BD P BD = 1.054(10 - 3 ) in.>in. The unstretched length of the wire is wire stretches L BD = 23 2 + 4 2 = 5ft = 60 in. Thus, the d BD =P BD L BD = 1.054(10 - 3 )(60) = 0.0632 in Ans.

**03** Solutions 46060 5/7/10 8:45 AM Page 11 © 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. 3–15. The rigid pipe is supported by a pin at A and an A-36 guy wire BD. If the wire has a diameter of 0.25 in., determine the load P if the end C is displaced 0.075 in. downward. B 4 ft P A D 3 ft 3 ft C Here, we are only interested in determining the force in wire BD. Referring to the FBD in Fig. a a+©M A = 0; F BD A 4 5 B(3) - P(6) = 0 F BD = 2.50 P The unstretched length for wire BD is L BD = 23 2 + 4 2 = 5 ft = 60 in. From the geometry shown in Fig. b, the stretched length of wire BD is L BD¿ = 260 2 + 0.075 2 - 2(60)(0.075) cos 143.13° = 60.060017 Thus, the normal **strain** is P BD = L BD¿ - L BD 60.060017 - 60 = = 1.00**03**(10 - 3 ) in.>in. L BD 60 Then, the normal **stress** can be obtain by applying Hooke’s Law. s BD = EP BD = 29(10 3 )C1.00**03**(10 - 3 )D = 29.01 ksi Since s BD 6 s y = 36 ksi, the result is valid. s BD = F BD A BD ; 29.01(10 3 ) = 2.50 P p 4 (0.252 ) P = 569.57 lb = 570 lb Ans. 11

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