Strenghth of material Solution

Strenghth of material Solution - Strength of Materials: An...

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Strength of Materials: An Undergraduate Text Graham M. Seed Saxe-Coburg Publications, 2001 ISBN: 1-874672-12-1 Chapter Solutions For more information about this book contact: Saxe-Coburg Publications 10A Saxe-Coburg Place Edinburgh, EH3 5BR, UK Tel: +44(0)131 315 3222 Fax: +44(0)131 315 3444 Email: buro@saxe-coburg.co.uk URL: http://www.saxe-coburg.co.uk G.M. Seed, 2001. All rights reserved. No part of this document may be transmitted, in any form or by any means whatsoever, without the written permission of the copyright owner.
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1 Chapter 1 Solutions 1.1 The cross-sectional area, A , of the bar is ( 29 A = = - - p 50 10 785 10 3 2 3 2 x x m . The axial stress, σ , due to the axial load P =30kN is s = = = - P A 30 10 382 3 3 x x MPa . . 1.2 From (1.6) the axial strain, e , is e E = = = - s 382 10 210 10 18 2 10 6 9 6 . . x x x From (1.3) the axial extension, δ , is d = = = - - eL 18 2 10 05 91 10 6 6 . . . x x x m 1.3 From (1.16) the in-plane strains are [ ] [ ] e s ns e s ns xx xx yy yy yy xx E E = - = - 1 1 , Re-arranging the first equation for σ xx s e ns xx xx yy E = + and substituting into the second equation then σ yy is found to be [ ] [ ] s n e ne yy yy xx E = - + = - + = - 1 70 10 1 0 28 60 0 28 40 10 54 2 9 2 6 x x x MPa . . . Substituting σ yy into ε xx and re-arranging for σ xx [ ] [ ] s n e ne xx xx yy E = - + = - + = - 1 70 10 1 0 28 40 028 60 10 4 3 2 9 2 6 x x x MPa . . . 1.4 From (1.23) the bulk modulus is ( 29 K E = - = - = 3 1 2 210 10 3 1 2 0 3 175 9 n x x GPa ( . ) 1.5 With E = σ / ε , σ = P / A , ε = δ / L and δ = α ( T ) then P is given by ( 29 P EA T L = = = - - a 120 10 71 10 11 10 100 0 75 125 9 4 6 x x x x x kN . ( ) . illustrating that a compressive axial load of 125kN is required to cancel out the extension due to thermal expansion. 1.6 The compressive stresses of each bar are ( 29 ( 29 s p s p copper copper steel steel P A P A = = = = = = - - 150 10 25 10 76 150 10 37 5 10 34 3 3 2 3 3 2 x x MPa x x MPa . The contractions of each bar are
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2 ( 29 ( 29 d p d p copper copper copper copper steel steel steel steel PL E A PL E A x = = = = = = - - - - 150 10 05 120 10 25 10 318 10 150 10 0 6 200 10 37 5 10 102 10 3 9 3 2 6 3 9 3 2 6 x x x x x x m x x x x x m . . . The total contraction of the composite bar is δ = δ copper + δ steel =420x10 -6 m. 1.7 From (1.31,1.32,1.36) the in-plane strains ε xx , ε yy and γ xy are given by ( 29 ( 29 e s ns e s ns g xx xx yy yy yy xx xy u x E v y E u y v x = = - = = - = + = 1 1 1 2 0 ,
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3 Chapter 2 Solutions 2.1 Figure Sol2.1 illustrates a circle of radius R with an elemental strip of thickness d y at distance y from the x -axis. The area, d A , of the elemental strip is therefore dA xdy R y dy = = - 2 2 2 2 The area of the strip is therefore given by, (2.1) A dA R y dy A R R = = - ∫∫ - + 2 2 2 The integration is assisted by making the substitution y = R sin θ , d y = R cos θ d θ ( 29 A R d R d R R = = + = + 4 4 1 2 1 2 4 2 2 2 2 2 2 0 2 2 0 2 0 2 2 cos cos sin / / / J J J J J p p p p The first moment of area Q x is, (2.2) ( 29 ( 29 [ ] Q ydA y R y dy y R y dy R y x A R R R R R R = = - = - = - - = ∫∫ - + - + - + 2 2 2 3 0 2 2 2 2 2 2 3 2 / Similarly, it is found that Q y =0. Therefore, as expected the coordinates of the centroid are x c = Q y / A =0 and y c = Q x / A =0.
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Strenghth of material Solution - Strength of Materials: An...

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