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ch01-03 stress & strain & properties

01 Solutions 46060

01 Solutions 46060 5/6/10 2:43 PM Page 65 © 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. *1–96. If the allowable bearing stress for the material under the supports at A and B is 1s b 2 allow = 1.5 MPa, determine the maximum load P that can be applied to the beam. The bearing plates A¿ and B¿ have square cross sections of 150 mm * 150 mm and 250 mm * 250 mm, respectively. 40 kN/m A A¿ B¿ B P 1.5 m 3 m 1.5 m Referring to the FBD of the beam, Fig. a, a+©M A = 0; N B (3) + 40(1.5)(0.75) - P(4.5) = 0 N B = 1.5P - 15 a+©M B = 0; 40(1.5)(3.75) - P(1.5) - N A (3) = 0 N A = 75 - 0.5P For plate A¿ , (s b ) allow = N A ; 1.5(10 6 ) = (75 - 0.5P)(103 ) A A¿ 0.15(0.15) For plate B¿ , P = 82.5 kN (s b ) allow = N B ; 1.5(10 6 ) = (1.5P - 15)(103 ) A B¿ 0.25(0.25) P = 72.5 kN (Controls!) Ans. •1–97. The rods AB and CD are made of steel having a failure tensile stress of s fail = 510 MPa. Using a factor of safety of F.S. = 1.75 for tension, determine their smallest diameter so that they can support the load shown. The beam is assumed to be pin connected at A and C. Support Reactions: B 4 kN 6 kN 5 kN D a +©M A = 0; F CD (10) - 5(7) - 6(4) - 4(2) = 0 F CD = 6.70 kN a +©M C = 0; 4(8) + 6(6) + 5(3) - F AB (10) = 0 A 2 m 2 m 3 m 3 m C F AB = 8.30 kN Allowable Normal Stress: Design of rod sizes For rod AB s allow = s fail F.S = F AB ; 510(106 ) A AB 1.75 = 8.30(103 ) p 4 d2 AB d AB = 0.006022 m = 6.02 mm Ans. For rod CD s allow = s fail F.S = F CD ; 510(106 ) A CD 1.75 = 6.70(103 ) p 4 d2 CD d CD = 0.005410 m = 5.41 mm Ans. 65

01 Solutions 46060 5/6/10 2:43 PM Page 66 © 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. 1–98. The aluminum bracket A is used to support the centrally applied load of 8 kip. If it has a constant thickness of 0.5 in., determine the smallest height h in order to prevent a shear failure. The failure shear stress is t fail = 23 ksi. Use a factor of safety for shear of F.S. = 2.5. A h Equation of Equilibrium: + c ©F y = 0; V - 8 = 0 V = 8.00 kip Allowable Shear Stress: Design of the support size t allow = t fail F.S = V A ; 23(103 ) 2.5 = 8.00(103 ) h(0.5) 8 kip h = 1.74 in. Ans. 1–99. The hanger is supported using the rectangular pin. Determine the magnitude of the allowable suspended load P if the allowable bearing stress is (s b ) allow = 220 MPa, the allowable tensile stress is (s t ) allow = 150 MPa, and the allowable shear stress is t allow = 130 MPa. Take t = 6 mm, a = 5 mm, and b = 25 mm. 75 mm a a 20 mm b 10 mm Allowable Normal Stress: For the hanger (s t ) allow = P A ; 150A106 B = P = 67.5 kN Allowable Shear Stress: The pin is subjected to double shear. Therefore, V = P 2 t allow = V A ; P>2 130A106 B = (0.01)(0.025) P = 65.0 kN P (0.075)(0.006) Allowable Bearing Stress: For the bearing area t P 37.5 mm 37.5 mm (s b ) allow = P A ; 220A106 B = P>2 (0.005)(0.025) P = 55.0 kN (Controls!) Ans. 66

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