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ch01-03 stress & strain & properties

03

03 Solutions 46060 5/7/10 8:45 AM Page 16 © 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. *3–20. The stressstrain diagram for a bone is shown and can be described by the equation P= 0.45110 -6 2 s 0.36110 -12 2 s 3 , where s is in kPa. Determine the modulus of toughness and the amount of elongation of a 200-mmlong region just before it fractures if failure occurs at P=0.12 mm>mm. s P P 0.45(10 6 )s + 0.36(10 12 )s 3 P P When e = 0.12 120(10 3 ) = 0.45 s + 0.36(10 -6 )s 3 Solving for the real root: s = 6873.52 kPa 6873.52 u t = dA = (0.12 - e)ds LA L 6873.52 u t = (0.12 - 0.45(10 -6 )s - 0.36(10 -12 )s 3 )ds L 0 0 = 0.12 s - 0.225(10 -6 )s 2 - 0.09(10 -12 )s 4 | 6873.52 0 = 613 kJ>m 3 d = eL = 0.12(200) = 24 mm Ans. Ans. 16

03 Solutions 46060 5/7/10 8:45 AM Page 17 © 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. •3–21. The stressstrain diagram for a polyester resin is given in the figure. If the rigid beam is supported by a strut AB and post CD, both made from this material, and subjected to a load of P = 80 kN, determine the angle of tilt of the beam when the load is applied.The diameter of the strut is 40 mm and the diameter of the post is 80 mm. 2 m B P A C From the stressstrain diagram, 0.75 m 0.75 m 0.5 m D E = 32.2(10)6 0.01 Thus, s AB = F AB = 40(103 ) p = 31.83 MPa A AB (0.04)2 e AB = s AB E = 31.83(106 ) 3.22(10 9 = 0.009885 mm>mm ) s CD = F CD = 40(103 ) p = 7.958 MPa A CD (0.08)2 e CD = s CD E = 7.958(106 ) 3.22(10 9 = 0.002471 mm>mm ) d AB = e AB L AB = 0.009885(2000) = 19.771 mm d CD = e CD L CD = 0.002471(500) = 1.236 mm Angle of tilt a: = 3.22(10 9 ) Pa 4 4 s (MPa) 100 95 80 70 60 50 40 32.2 20 0 0 compression tension 0.01 0.02 0.03 0.04 P (mm/mm) tan a = 18.535 1500 ; a = 0.708° Ans. 17

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