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© 2008 by R.C. Hibbeler. Published by Pearson Prentice Hall, 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. 6-45. The beam is subjected to a moment M. Determine the percentage of this moment that is resisted by the stresses acting on both the top and bottom boards, A and B, of the beam. Section Property: B 25mm ' / = — <0.2)(0.2')-— (0.15)(0.15 3 ) =91.14583(10-') m4 "» : • 25 mm 150m m Bending Stress: Applying the flexure formula Afv £" M(O.l) = 1097.143 M 91.14583(11)-*) M(0.075) Resultant Force and Moment: = 822.857 M For hoard A or B F = 821857M(0.025)(0.2) I + -(I097.I43M - 822.857Af>( 0.0251(0.2) = 4 800 M M' = F(O.I76I9) = 4.80A/IO.I7619) =0.8457 M — | = 0.8457( 100%) = 84.6 * Ans 6-46. Determine the moment M that should be applied to the beam in order to create a compressive stress at point D of <TD = 30 MPa. Also sketch the stress distribution acting over the cross section and compute the maximum stress developed in the beam. M 25 mm 25nu Section Property: / = 1(0.2)(0.2')-— (0.15)(O.I5 5 ) =91.14583(10'') m' Bending Stress: Applying ihe flexure formula Mv c=T M0.075) M = 35458 N m = 36.5 kN m Me / 36458(0.11 •=40.0 MPa 91.I4583(10-») Ans 266 25 mm © 2008 by R.C. Hibbeler. Published by Pearson Prentice Hall, 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. 6-57. Determine the resultant force the bending stresses produce on the top board A of the beam if M = 1 kip • ft. Section Properties: IA 0.75(10)(1.5) + 7.5(l)( 12) + 14.2S(6)( 1.5) 10(1.5) = 6.375 in. .5') + 10(1.5)<6.375-0.75) 2 + -M6)( 1.5') + 61 1.5)( 14.25 -6.375)' = 1 196.4375 in' Bending Stress: Applying Ihe flexure formula <To = Myc / 1000(12X6.375-1.5) = 48.90 psi 1196.4375 MyD ~7~ : 1000( 12)16.375) 1196.4375 63.94 psi The Resultant Force: For lop board A F =-(63.94 + 48.90) (10)(1.5) = 846 Ib Ans 6-58. The control level is used on a riding lawn mower. Determine the maximum bending stress in the lever at section a-a if a force of 20 Ib is applied to the handle. The lever is supported by a pin at A and a wire at B. Section a-a is square,0.25 in. by 0.25 in. 20 Ib 20(2) - M - 0 ; M = 40 Ib in Me 40(0.125) / ~ £(0.25X0.25') 15.4 ksi 272 © 2008 by R.C. Hibbeler. Published by Pearson Prentice Hall, 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. 6-93. The beam is subjected to the loading shown. Determine its required cross-sectional dimension a, if the allowable bending stress for the material is cranow = 150 MPa. 60 kN 40 kN/m 2 !* ' -2m- -1m- Support Reactions: As shown on FBD. Section Properties: 4«V««frotoil . ^ £vA ^ X h) «+ H r«) ( f") " " r 1 I \ /• 1 V 5 1V / = — 10) -a + 0 - 0 — a — a\2 U >/ 2 b A 12 I ^ 6 ) 5 37 . Internal Moment: As shown on the moment diagram Allowable Bending Stress: The maximum moment is A/,,,, = 60.0 kN m as indicated on the moment diagram Applying the flexure formula 150(10') 60.0(IO')(a-ia) a=O.I599 m = 160mm Ans 6-94. The wing spar ABD of a light plane is made from 2014-T6 aluminum and has a cross-sectional area of 1.27 in , a depth of 3 in., and a moment of inertia about its neutral axis of 2.68 in4. Determine the absolute maximum bending stress in the spar if the anticipated loading is to be as shown. Assume A, B, and C are pins. Connection is made along the central longitudinal axis of the spar. 801b/in. D 2ft 3ft- 10 li/. AM Note thai 2S.8 ksi < ar = 60 ksi OK 291 -6ft- © 2008 by R.C. Hibbeler. Published by Pearson Prentice Hall, 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. 6-98. The wood beam is subjected to the uniform load of w = 200 Ib/ft. If the allowable bending stress for the material is <ranow = 1.40 ksi, determine the required dimension b of its cross section. Assume the support at A is a pin and R is a roller. .irrrrrm V(lb) Allowable Bending Stress: The maximum moment is M mll = 5688.89 Ib ft as indicated on moment diagram. Applying the flexure formula M,.,c l.40( 10') 2844.44(12X0.75*) b = 4.02 in. Ans 6-99. The wood beam has a rectangular cross section in the proportion shown. Determine its required dimension b if the allowable bending stress is cral)ow = 10 MPa. 500 N/m 2m r — 2m I , fm 7#> Allowable Bending Stress: The maximum moment is Wm<1 = 562.5 N m as indicated on the moment diagram. Applying the flexure formula ; ^N\ -zso MCA/-*-; 10( I0 6 ) = 562.5(0.75t) b = 0.05313m = 53.1 mm * L, Ans 294 © 2008 by R.C. Hibbeler. Published by Pearson Prentice Hall, 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. *6-120. The composite beam is made of 6061-T6 aluminum (A) and C83400 red brass (£'). If the height h = 40mm, determine the maximum moment that can be applied to the beam if the allowable bending stress for the aluminum is (<railow)ai = 128 MPa and for the brass (cr aUow ) br = 35 MPa. k 50 mm ] 150mm Section Properties: For transformed section. Allowable Bending Stress: Applying the flexure formula Assume failure of red brass Eb, lOl.O(lO') btl, = nbii = 0.68218(0.15) = 0.10233m Me (<T-"°")br = £7 . _ lyA 35( 10') _ 0.025(0.10233)(0.05) + (0.07)(0.15)(0.04) 0.10233(0.051 + 0.15(0.04) M(0.09 -0.049289) 7.45799( lO-«) = 6412N m = « . 4 l k N - m (controls!) Ans Assume failure of aluminium = 0.049289 m Me 4M-|5(ft«mj)(a051) +0.10233(0.051(0.049289-0.025)2 7.45799(10-')] •|-J2(0.15)(0.041)-HO.I5(0.04)(0.07-0.049289)! = 7.45799(10-') m4 6-121. A wood beam is reinforced with steel straps at its lop and bollom as shown. Determine the maximum bending stress developed in the wood and steel if the beam is subjected to a bending moment of M = 5 kN • m. Sketch the stress distribution acting over the cross section. Take £w = 11 GPa, EA = 200 GPa. 20 mm 4- — M = 5 k N - m 20 mm 7 = —(3.63636X0.34)' - — (3.43636)(0.3)' = 4.17848(10"')m' Maximum stress in steel: _nMc, _ 18.I82(S)(10')(0.17) '"' "" ~ 7 4.17848(10-') 3.70 MPa Ans Maximum stress in wood: 7 = 0.179 MPa 4.17848(10-')' An » st ) = n(<rw)-m, = 18.182(0.179) = 3.26 MPa 1\ F I rt 3U Mf. 309