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ch01-03 stress & strain & properties

01 Solutions 46060

01 Solutions 46060 5/6/10 2:43 PM Page 23 © 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. 1–35. The bars of the truss each have a cross-sectional area of 1.25 in 2 . Determine the average normal stress in each member due to the loading P = 8 kip. State whether the stress is tensile or compressive. 3 ft B C A E 4 ft 4 ft D P 0.75 P Joint A: s AB = F AB = 13.33 = 10.7 ksi (T) A AB 1.25 s AE = F AE = 10.67 = 8.53 ksi (C) A AE 1.25 Ans. Ans. Joint E: s ED = F ED = 10.67 = 8.53 ksi (C) A ED 1.25 s EB = F EB = 6.0 = 4.80 ksi (T) A EB 1.25 Ans. Ans. Joint B: s BC = F BC = 29.33 = 23.5 ksi (T) A BC 1.25 s BD = F BD = 23.33 = 18.7 ksi (C) A BD 1.25 Ans. Ans. 23

01 Solutions 46060 5/6/10 2:43 PM Page 24 © 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. *1–36. The bars of the truss each have a cross-sectional area of 1.25 in 2 . If the maximum average normal stress in any bar is not to exceed 20 ksi, determine the maximum magnitude P of the loads that can be applied to the truss. 3 ft B C A E 4 ft 4 ft D Joint A: P 0.75 P + c ©F y = 0; -P + a 3 5 bF AB = 0 F AB = (1.667)P : + ©F x = 0; -F AE + (1.667)Pa 4 5 b = 0 F AE = (1.333)P Joint E: + c ©F y = 0; F EB - (0.75)P = 0 F EB = (0.75)P : + ©F x = 0; (1.333)P - F ED = 0 F ED = (1.333)P Joint B: + c ©F y = 0; a 3 5 bF BD - (0.75)P - (1.667)Pa 3 5 b = 0 F BD = (2.9167)P : + ©F x = 0; F BC - (2.9167)Pa 4 5 b - (1.667)Pa 4 5 b = 0 F BC = (3.67)P The highest stressed member is BC: s BC = (3.67)P = 20 1.25 P = 6.82 kip Ans. 24

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