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**03** Solutions 46060 5/7/10 8:45 AM Page 18 © 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. 3–22. The **stress**–**strain** diagram for a polyester resin is given in the figure. If the rigid beam is supported by a strut AB and post CD made from this material, determine the largest load P that can be applied to the beam before it ruptures. The diameter of the strut is 12 mm and the diameter of the post is 40 mm. 2 m B Rupture of strut AB: P s R = F AB A AB ; 50(10 6 ) = P>2 p 4 (0.012)2; P = 11.3 kN (controls) Ans. A 0.75 m 0.75 m C 0.5 m D Rupture of post CD: s R = F CD A CD ; 95(10 6 ) = P>2 p 4 (0.04)2 P = 239 kN s (MPa) 100 95 80 70 60 50 40 32.2 20 compression tension 0 0 0.01 0.02 0.**03** 0.04 P (mm/mm) 3–23. By adding plasticizers to polyvinyl chloride, it is possible to reduce its stiffness. The **stress**–**strain** diagrams for three types of this material showing this effect are given below. Specify the type that should be used in the manufacture of a rod having a length of 5 in. and a diameter of 2 in., that is required to support at least an axial load of s (ksi) 15 unplasticized 10 copolymer P Normal Stress: 20 kip and also be able to stretch at most 1 4 in. 18 s = P A = 20 p 4 (22 ) = 6.366 ksi 5 0 0 flexible (plasticized) P 0.10 0.20 0.30 P (in./in.) Normal Strain: e = 0.25 5 = 0.0500 in.>in. From the **stress**–**strain** diagram, the copolymer will satisfy both **stress** and **strain** requirements. Ans.

**03** Solutions 46060 5/7/10 8:45 AM Page 19 © 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. *3–24. The **stress**–**strain** diagram for many metal alloys can be described analytically using the Ramberg-Osgood three parameter equation P=s>E + ks n , where E, k, and n are determined from measurements taken from the diagram. Using the **stress**–**strain** diagram shown in the figure, take E = 30110 3 2 ksi and determine the other two parameters k and n and thereby obtain an analytical expression for the curve. s (ksi) 80 60 40 20 Choose, 0.1 0.2 0.3 0.4 0.5 P (10 –6 ) s = 40 ksi, e = 0.1 s = 60 ksi, e = 0.3 0.1 = 0.3 = 40 30(10 3 ) + k(40)n 60 30(10 3 ) + k(60)n 0.098667 = k(40) n 0.29800 = k(60) n 0.3310962 = (0.6667) n ln (0.3310962) = n ln (0.6667) n = 2.73 Ans. k = 4.23(10 -6 ) Ans. •3–25. The acrylic plastic rod is 200 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change in its length and the change in its diameter. E p = 2.70 GPa, n p = 0.4. 300 N 200 mm 300 N s = P A = 300 p 4 (0.015)2 = 1.697 MPa e long = s E = 1.697(106 ) 2.70(10 9 ) = 0.0006288 d = e long L = 0.0006288 (200) = 0.126 mm Ans. e lat = -Ve long = -0.4(0.0006288) = -0.0002515 ¢d = e lat d = -0.0002515 (15) = -0.0**03**77 mm Ans. 19

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