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# ch01-03 stress &amp; strain &amp; properties

## 01 Solutions 46060

01 Solutions 46060 5/6/10 2:43 PM Page 63 © 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. *1–92. The compound wooden beam is connected together by a bolt at B.Assuming that the connections at A, B, C, and D exert only vertical forces on the beam, determine the required diameter of the bolt at B and the required outer diameter of its washers if the allowable tensile stress for the bolt is 1s t 2 allow = 150 MPa and the allowable bearing stress for the wood is 1s b 2 allow = 28 MPa. Assume that the hole in the washers has the same diameter as the bolt. A 2 m 3 kN 1.5 kN 2 kN 2 m 1.5 m 1.5 m 1.5 m 1.5 m C B D From FBD (a): a +©M D = 0; F B (4.5) + 1.5(3) + 2(1.5) - F C (6) = 0 From FBD (b): a +©M D = 0; F B (5.5) - F C (4) - 3(2) = 0 Solving Eqs. (1) and (2) yields F B = 4.40 kN; F C = 4.55 kN For bolt: s allow = 150(10 6 ) = 4.40(103 ) p 4 (d B) 2 d B = 0.00611 m = 6.11 mm For washer: s allow = 28 (10 4 4.40(10 3 ) ) = p 4 (d2 w - 0.00611 2 ) d w = 0.0154 m = 15.4 mm 4.5 F B - 6 F C = -7.5 5.5 F B - 4 F C = 6 (1) (2) Ans. Ans. •1–93. The assembly is used to support the distributed loading of w = 500 lb>ft. Determine the factor of safety with respect to yielding for the steel rod BC and the pins at B and C if the yield stress for the steel in tension is s y = 36 ksi and in shear t y = 18 ksi. The rod has a diameter of 0.40 in., and the pins each have a diameter of 0.30 in. C For rod BC: 4 ft s = P A = 1.667 p = 13.26 ksi 4 (0.42 ) A F. S. = s y s = 36 13.26 = 2.71 Ans. B For pins B and C: t = V A = 0.8333 p = 11.79 ksi 4 (0.32 ) 3 ft 1 ft w F. S. = t y t = 18 11.79 = 1.53 Ans. 63

01 Solutions 46060 5/6/10 2:43 PM Page 64 © 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. 1–94. If the allowable shear stress for each of the 0.30- in.-diameter steel pins at A, B, and C is t allow = 12.5 ksi, and the allowable normal stress for the 0.40-in.-diameter rod is s allow = 22 ksi, determine the largest intensity w of the uniform distributed load that can be suspended from the beam. Assume failure of pins B and C: t allow = 12.5 = 1.667w p 4 (0.32 ) 4 ft C A w = 0.530 kip>ft (controls) Ans. Assume failure of pins A: F A = 2 (2w) 2 + (1.333w) 2 = 2.404 w 3 ft B w t allow = 12.5 = 1.202w p 4 (0.32 ) 1 ft w = 0.735 kip>ft Assume failure of rod BC: s allow = 22 = 3.333w p 4 (0.42 ) w = 0.829 kip>ft 1–95. 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 size of square bearing plates A¿ and B¿ required to support the load. Dimension the plates to the nearest mm.The reactions at the supports are vertical.Take P = 100 kN. 40 kN/m A A¿ B¿ B P Referring to the FBD of the bean, Fig. a 1.5 m 3 m 1.5 m a+©M A = 0; N B (3) + 40(1.5)(0.75) - 100(4.5) = 0 N B = 135 kN a+©M B = 0; 40(1.5)(3.75) - 100(1.5) - N A (3) = 0 N A = 25.0 kN For plate A¿ , (s b ) allow = N A A A¿ ; 1.5(10 6 ) = 25.0(103 ) a 2 A¿ a A¿ = 0.1291 m = 130 mm Ans. For plate B¿ , s allow = N B ; 1.5(10 6 ) = 135(103 ) A B¿ a 2 B¿ a B¿ = 0.300 m = 300 mm Ans. 64

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