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ch01-03 stress & strain & properties

03

03 Solutions 46060 5/7/10 8:45 AM Page 6 © 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. *3–8. The strut is supported by a pin at C and an A-36 steel guy wire AB. If the wire has a diameter of 0.2 in., determine how much it stretches when the distributed load acts on the strut. A 60 200 lb/ft Here, we are only interested in determining the force in wire AB. C 9 ft B a+©M C = 0; F AB cos 60°(9) - 1 2 (200)(9)(3) = 0 F AB = 600 lb The normal stress the wire is s AB = F AB A AB = 600 p 4 (0.22 ) = 19.10(103 ) psi = 19.10 ksi Since s AB 6 s y = 36 ksi, Hooke’s Law can be applied to determine the strain in wire. s AB = EP AB ; 19.10 = 29.0(10 3 )P AB P AB = 0.6586(10 - 3 ) in>in The unstretched length of the wire is stretches L AB = 9(12) sin 60° = 124.71 in. Thus, the wire d AB =P AB L AB = 0.6586(10 - 3 )(124.71) = 0.0821 in. Ans. 6

03 Solutions 46060 5/7/10 8:45 AM Page 7 © 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. •3–9. The s–P diagram for a collagen fiber bundle from which a human tendon is composed is shown. If a segment of the Achilles tendon at A has a length of 6.5 in. and an approximate cross-sectional area of 0.229 in 2 , determine its elongation if the foot supports a load of 125 lb, which causes a tension in the tendon of 343.75 lb. s (ksi) 4.50 3.75 3.00 2.25 A s = P A = 343.75 0.229 From the graph e = 0.035 in.>in. = 1.50 ksi 1.50 0.75 0.05 0.10 125 lb P (in./in.) d = eL = 0.035(6.5) = 0.228 in. Ans. 3–10. The stressstrain diagram for a metal alloy having an original diameter of 0.5 in. and a gauge length of 2 in. is given in the figure. Determine approximately the modulus of elasticity for the material, the load on the specimen that causes yielding, and the ultimate load the specimen will support. From the stressstrain diagram, Fig. a, Thus, E 1 = 60 ksi - 0 0.002 - 0 ; E = 30.0(103 ) ksi s y = 60 ksi s u>t = 100 ksi P Y = s Y A = 60C p 4 (0.52 )D = 11.78 kip = 11.8 kip P u>t = s u>t A = 100C p 4 (0.52 )D = 19.63 kip = 19.6 kip Ans. Ans. Ans. s (ksi) 105 90 75 60 45 30 15 0 P (in./in.) 0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 7

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