11 Solutions 46060 5/26/10 3:27 PM Page 830 © 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 11–1 The simply supported beam is made of timber that has an allowable bending stress of sallow = 6.5 MPa and an allowable shear stress of tallow = 500 kPa Determine its dimensions if it is to be rectangular and have a height-towidth ratio of 1.25 kN/m 2m Ix = (b)(1.25b)3 = 0.16276b4 12 Qmax = y¿A¿ = (0.3125b)(0.625b)(b) = 0.1953125b3 Assume bending moment controls: Mmax = 16 kN # m sallow = Mmax c I 6.5(106) = 16(103)(0.625b) 0.16276b4 b = 0.21143 m = 211 mm Ans h = 1.25b = 264 mm Ans Check shear: Qmax = 1.846159(10 - 3) m3 I = 0.325248(10 - 3) m4 tmax = VQmax 16(103)(1.846159)(10 - 3) = 429 kPa 500 kPa‚ OK = It 0.325248(10 - 3)(0.21143) 830 4m 2m 11 Solutions 46060 5/26/10 3:27 PM Page 831 © 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 11–2 The brick wall exerts a uniform distributed load of 1.20 kip>ft on the beam If the allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 12 ksi, select the lightest wide-flange section with the shortest depth from Appendix B that will safely support the load 1.20 kip/ ft 10 ft ft ft b Bending Stress: From the moment diagram, Mmax = 44.55 kip # ft Assuming bending controls the design and applying the flexure formula Sreq d = = 44.55 (12) = 24.3 in3 22 W12 * 22 A Sx = 25.4 in3, d = 12.31 in., tw = 0.260 in B V for the W12 * 22 wide tw d = 6.60 kip Shear Stress: Provide a shear stress check using t = flange section From the shear diagram, Vmax tmax = = Vmax tw d 6.60 0.260(12.31) = 2.06 ksi tallow = 12 ksi (O.K!) Hence, Use in 0.5 in Mmax sallow Two choices of wide flange section having the weight 22 lb>ft can be made They are W12 * 22 and W14 * 22 However, W12 * 22 is the shortest Select 0.5 in 0.5 in Ans W12 * 22 831 11 Solutions 46060 5/26/10 3:27 PM Page 832 © 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 11–3 The brick wall exerts a uniform distributed load of 1.20 kip>ft on the beam If the allowable bending stress is sallow = 22 ksi, determine the required width b of the flange to the nearest 14 in 1.20 kip/ ft 10 ft ft ft b 0.5 in 0.5 in in 0.5 in Section Property: I = 1 (b) A 103 B (b - 0.5) A 93 B = 22.583b + 30.375 12 12 Bending Stress: From the moment diagram, Mmax = 44.55 kip # ft sallow = 22 = Mmax c I 44.55(12)(5) 22.583b + 30.375 b = 4.04 in Use b = 4.25 in Ans 832 11 Solutions 46060 5/26/10 3:27 PM Page 833 © 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 *11–4 Draw the shear and moment diagrams for the shaft, and determine its required diameter to the nearest in if sallow = ksi and tallow = ksi The bearings at A and D exert only vertical reactions on the shaft The loading is applied to the pulleys at B, C, and E 14 in 20 in 15 in 12 in E A C B D 35 lb 80 lb 110 lb sallow = 7(103) = Mmax c I 1196 c p ; c c = 0.601 in d = 2c = 1.20 in Use d = 1.25 in Ans Check shear: tmax = 0.625 108(4(0.625) Vmax Q 3p )(p)( ) = 117 psi ksi OK = p It (0.625) (1.25) •11–5 Select the lightest-weight steel wide-flange beam from Appendix B that will safely support the machine loading shown The allowable bending stress is sallow = 24 ksi and the allowable shear stress is tallow = 14 ksi ft Bending Stress: From the moment diagram, Mmax = 30.0 kip # ft Assume bending controls the design Applying the flexure formula Sreq¿d = = Select W12 * 16 Mmax sallow 30.0(12) = 15.0 in3 24 A Sx = 17.1 in3, d = 11.99 in., tw = 0.220 in B V for the W12 * 16 wide tw d = 10.0 kip Shear Stress: Provide a shear stress check using t = flange section From the shear diagram, Vmax tmax = = Vmax tw d 10.0 0.220(11.99) = 3.79 ksi tallow = 14 ksi (O.K!) Hence, Use kip kip Ans W12 * 16 833 ft kip ft kip ft ft 11 Solutions 46060 5/26/10 3:27 PM Page 834 © 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 11–6 The compound beam is made from two sections, which are pinned together at B Use Appendix B and select the lightest-weight wide-flange beam that would be safe for each section if the allowable bending stress is sallow = 24 ksi and the allowable shear stress is tallow = 14 ksi The beam supports a pipe loading of 1200 lb and 1800 lb as shown C A B ft Bending Stress: From the moment diagram, Mmax = 19.2 kip # ft for member AB Assuming bending controls the design, applying the flexure formula Sreq¿d = = Select Mmax sallow 19.2(12) = 9.60 in3 24 A Sx = 10.9 in3, d = 9.87 in., tw = 0.19 in B W10 * 12 For member BC, Mmax = 8.00 kip # ft Sreq¿d = = Select Mmax sallow 8.00(12) = 4.00 in3 24 A Sx = 5.56 in3, d = 5.90 in., tw = 0.17 in B W6 * V for the W10 * 12 widetw d flange section for member AB From the shear diagram, Vmax = 2.20 kip Shear Stress: Provide a shear stress check using t = tmax = = Vmax tw d 2.20 0.19(9.87) = 1.17 ksi tallow = 14 ksi (O.K!) Use Ans W10 * 12 For member BC (W6 * 9), Vmax = 1.00 kip tmax = = Vmax tw d 1.00 0.17(5.90) = 0.997 ksi tallow = 14 ksi (O.K!) Hence, Use 1800 lb 1200 lb W6 * Ans 834 ft ft 10 ft 11 Solutions 46060 5/26/10 3:27 PM Page 835 © 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 11–7 If the bearing pads at A and B support only vertical forces, determine the greatest magnitude of the uniform distributed loading w that can be applied to the beam sallow = 15 MPa, tallow = 1.5 MPa w A B 1m 1m 150 mm 25 mm 150 mm 25 mm The location of c, Fig b, is y = 0.1625(0.025)(0.15) + 0.075(0.15)(0.025) ©yA = ©A 0.025(0.15) + 0.15(0.025) = 0.11875 m I = + (0.025)(0.153) + (0.025)(0.15)(0.04375)2 12 (0.15)(0.0253) + 0.15(0.025)(0.04375)2 12 = 21.58203125(10 - 6) m4 Referring to Fig b, Qmax = y¿A¿ = 0.059375 (0.11875)(0.025) = 0.176295313(10 - 4) m3 Referring to the moment diagram, Mmax = 0.28125 w Applying the Flexure formula with C = y = 0.11875 m, sallow = Mmax c ; I 15(106) = 0.28125w(0.11875) 21.582(10 - 6) W = 9.693(103) N>m Referring to shear diagram, Fig a, Vmax = 0.75 w tallow = Vallow Qmax ; It 1.5(106) = 0.75w C 0.17627(10 - 3) D 21.582(10 - 6)(0.025) W = 6.122(103) N>m = 6.12 kN>m (Control!) Ans 835 11 Solutions 46060 5/26/10 3:27 PM Page 836 © 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 *11–8 The simply supported beam is made of timber that has an allowable bending stress of sallow = 1.20 ksi and an allowable shear stress of tallow = 100 psi Determine its smallest dimensions to the nearest 18 in if it is rectangular and has a height-to-width ratio of 1.5 12 kip/ft B A ft ft 1.5 b b The moment of inertia of the beam’s cross-section about the neutral axis is (b)(1.5b)3 = 0.28125b4 Referring to the moment diagram, I = 12 Mmax = 45.375 kip # ft sallow = Mmax c ; I 1.2 = 45.375(12)(0.75b) 0.28125b4 b = 10.66 in Referring to Fig b, Qmax = y¿A¿ = 0.375b (0.75b)(b) = 0.28125b3 Referring to the shear diagram, Fig a, Vmax = 33 kip tmax = Vmax Qmax ; It 100 = 33(103)(0.28125b3) 0.28125b4(b) b = 18.17 in (Control!) Thus, use b = 18 in Ans 836 11 Solutions 46060 5/26/10 3:27 PM Page 837 © 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 •11–9 Select the lightest-weight W12 steel wide-flange beam from Appendix B that will safely support the loading shown, where P = kip The allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 12 ksi P P ft From the Moment Diagram, Fig a, Mmax = 54 kip # ft Mmax sallow Sreq¿d = 54(12) 22 = = 29.45 in3 Select W12 * 26 C Sx = 33.4 in3, d = 12.22 in and tw = 0.230 in D From the shear diagram, Fig a, Vmax = 7.5 kip Provide the shear-stress check for W 12 * 26, tmax = = Vmax tw d 7.5 0.230(12.22) = 2.67 ksi tallow = 12 ksi (O.K!) Hence Use Ans W12 * 26 837 ft ft 11 Solutions 46060 5/26/10 3:27 PM Page 838 © 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 11–10 Select the lightest-weight W14 steel wide-flange beam having the shortest height from Appendix B that will safely support the loading shown, where P = 12 kip The allowable bending stress is sallow = 22 ksi and the allowable shear stress is tallow = 12 ksi P P ft From the moment diagram, Fig a, Mmax = 108 kip # ft Mmax sallow Sreq¿d = 108(12) 22 = = 58.91 in3 Select W14 * 43 C Sx = 62.7 in3, d = 13.66 in and tw = 0.305 in D From the shear diagram, Fig a, Vmax = 15 kip Provide the shear-stress check for W14 * 43 , tmax = = Vmax tw d 15 0.305(13.66) = 3.60 ksi tallow = 12 ksi‚ (O.K!) Hence, Use Ans W14 * 43 838 ft ft 11 Solutions 46060 5/26/10 3:27 PM Page 839 © 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 11–11 The timber beam is to be loaded as shown If the ends support only vertical forces, determine the greatest magnitude of P that can be applied sallow = 25 MPa, tallow = 700 kPa 150 mm 30 mm 120 mm 40 mm P 4m A y = (0.015)(0.150)(0.03) + (0.09)(0.04)(0.120) = 0.05371 m (0.150)(0.03) + (0.04)(0.120) I = 1 (0.150)(0.03)3 + (0.15)(0.03)(0.05371 - 0.015)2 + (0.04)(0.120)3 + 12 12 B (0.04)(0.120)(0.09 - 0.05371)2 = 19.162(10 - 6) m4 Maximum moment at center of beam: Mmax = P (4) = 2P Mc ; I s = 25(106) = (2P)(0.15 - 0.05371) 19.162(10 - 6) P = 2.49 kN Maximum shear at end of beam: Vmax = P VQ ; t = It 700(103) = P C (0.15 - 0.05371)(0.04)(0.15 - 0.05371) D 2 19.162(10 - 6)(0.04) P = 5.79 kN Thus, P = 2.49 kN Ans 839 4m 11 Solutions 46060 5/26/10 3:27 PM Page 868 © 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 z •11–41 The end gear connected to the shaft is subjected to the loading shown If the bearings at A and B exert only y and z components of force on the shaft, determine the equilibrium torque T at gear C and then determine the smallest diameter of the shaft to the nearest millimeter that will support the loading Use the maximum-shear-stress theory of failure with tallow = 60 MPa 100 mm T 250 mm C 50 mm 150 mm A x 100 mm Fz ϭ 1.5 kN From the free - body diagrams: T = 100 N # m Ans Critical section is at support A 1 3 2 c = c 22252 + 1502 d 2M2 + T2 d = c p tallow p(60)(106) = 0.01421 m d = 2c = 0.0284 m = 28.4 mm Use d = 29 mm Ans 868 B 75 mm y 11 Solutions 46060 5/26/10 3:27 PM Page 869 z 11–42 The end gear connected to the shaft is subjected to the loading shown If the bearings at A and B exert only y and z components of force on the shaft, determine the equilibrium torque T at gear C and then determine the smallest diameter of the shaft to the nearest millimeter that will support the loading Use the maximum-distortionenergy theory of failure with sallow = 80 MPa 100 mm T 250 mm C 50 mm 150 mm A x T = 100 N # m Ans Critical section is at support A s1, = sx s2x ; A + txy sx s2x ,b = A + txy Let a = s1 = a + b, s2 = a - b Require, s21 - s1 s2 + s22 = s2allowa2 + 2ab + b2 - [a2 - b2] + a2 - 2ab + b2 = s2allow a2 + 3b2 = s2allow s2x s2x + 3a + t2xy b = s2allow 4 s2x + 3t2xy = s2allow Mt Tc a p b + 3a p b = s2allow c c 4M 2T ca b + 3a b d = s2allow p p c c4 = 16 s2allow c = a = c p M2 + s2allow p2 12T2 p2 s2allow (4M + 3T ) b (80(106))2(p)2 100 mm Fz ϭ 1.5 kN From the free-body diagrams: (4(225) + 3(150) ) d 2 = 0.01605 m d = 2c = 0.0321 m = 32.1 mm Use d = 33 mm Ans 869 B 75 mm y 11 Solutions 46060 5/26/10 3:27 PM Page 870 © 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 11–43 The shaft is supported by bearings at A and B that exert force components only in the x and z directions on the shaft If the allowable normal stress for the shaft is sallow = 15 ksi, determine to the nearest 18 in the smallest diameter of the shaft that will support the loading Use the maximum-distortion-energy theory of failure z C F¿x ϭ 100 lb in A x in 12 in Critical moment is just to the right of D T = 1200 lb # in Both states of stress will yield the same result Let s s 2 ; a A 2b + t s = A and s + t2 = B A4 s2a = (A + B)2, s2b = (A - B)2 sa sb = (A + B)(A - B) = A2 - B2 s2a - sa sb + s2b = A2 + B2 + 2AB - A2 + B2 + A2 + B2 - 2AB = A2 + 3B2 = s2 s2 + 3a + t2 b = s2 + 3t2 4 s2a - sa sb + s2b = s2allow s2 + 3t2 = s2allow‚ s = Mc Mc 4M = p = I c pc3 t = Tc Tc 2T = p = J c p c3 (1) From Eq (1) 16M2 p c c = a 12T2 + p2 c6 = s2allow 16(2396)2 + 12(12002) 1>6 16M2 + 12T2 1>6 b = c d = 0.605 in p2s2allow p2((15)(103))2 d = 2c = 1.210 in Use d = in Fz ϭ 300 lb 10 in in M = 220572 + 12292 = 2396 lb # in sa, b = F y ϭ 300 lb D in Ans 870 in E B y 11 Solutions 46060 5/26/10 3:27 PM Page 871 © 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 *11–44 The shaft is supported by bearings at A and B that exert force components only in the x and z directions on the shaft If the allowable normal stress for the shaft is sallow = 15 ksi, determine to the nearest 18 in the smallest diameter of the shaft that will support the loading Use the maximum-shear-stress theory of failure Take tallow = ksi z C F¿x ϭ 100 lb in A x in 12 in Critical moment is just to the right of D T = 1200 lb # in Use Eq 11-2, 1>3 2M2 + T2 b p tallow c = a 1>3 2(2396)2 + (1200)2 b = 0.6576 in p(6)(10 ) in Fz ϭ 300 lb 10 in in M = 2(2057)2 + (1229)2 = 2396 lb # in c = a F y ϭ 300 lb D dreq¿d = 2c = 1.315 in Use d = in Ans 871 in E B y 11 Solutions 46060 5/26/10 3:27 PM Page 872 © 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 z •11–45 The bearings at A and D exert only y and z components of force on the shaft If tallow = 60 MPa, determine to the nearest millimeter the smallest-diameter shaft that will support the loading Use the maximum-shearstress theory of failure 350 mm D 400 mm 200 mm B A Critical moment is at point B: M = 2(473.7)2 + (147.4)2 = 496.1 N # m x T = 150 N # m c = a 1>3 1>3 2 2 2496.1 2M2 + T2 b = a + 150 b = 0.0176 m p tallow p(60)(106) c = 0.0176 m = 17.6 mm d = 2c = 35.3 mm Use d = 36 mm Ans 872 y C 75 mm Fy ϭ kN 50 mm Fz ϭ kN 11 Solutions 46060 5/26/10 3:27 PM Page 873 © 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 z 11–46 The bearings at A and D exert only y and z components of force on the shaft If tallow = 60 MPa, determine to the nearest millimeter the smallest-diameter shaft that will support the loading Use the maximumdistortion-energy theory of failure sallow = 130 MPa 350 mm D 400 mm 200 mm B A The critical moment is at B M = 2(473.7)2 + (147.4)2 = 496.1 N # m x T = 150 N # m Since, sa, b = Let s s 2 ; a A 2b + t s = A and s 2 = B a A 2b + t s2a = (A + B)2 s2b = (A - B)2 sa sb = (A + B)(A - B) s2a - sa sb + s2b = A2 + B2 + 2AB - A2 + B2 + A2 + B2 - 2AB = A2 + 3B2 = s2 s2 + 3a + t2 b 4 = s2 + 3t2 s2a - sasb + s2b = s2allow s2 + 3t2 = s2allow (1) s = Mc Mc 4M = p = I c pc3 t = Tc Tc 2T = p = J pc3 c From Eq (1) 12T2 16M2 + = s2allow pc pc c = a = a 16M2 + 12T2 1>6 b p2s2allow 16(496.1)2 + 12(150)2 p ((130)(10 )) b 1>4 = 0.01712 m d = 2c = 34.3 mm Ans 873 y C 75 mm Fy ϭ kN 50 mm Fz ϭ kN 11 Solutions 46060 5/26/10 3:27 PM Page 874 © 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 11–47 Draw the shear and moment diagrams for the shaft, and then determine its required diameter to the nearest millimeter if sallow = 140 MPa and tallow = 80 MPa The bearings at A and B exert only vertical reactions on the shaft 1500 N 800 N A B 600 mm 125 mm Bending Stress: From the moment diagram, Mmax = 111 N # m Assume bending controls the design Applying the flexure formula sallow = 140 A 106 B = Mmax c I 111 A d2 B p A d2 B d = 0.02008 m = 20.1 mm d = 21 mm Use Ans Shear Stress: Provide a shear stress check using the shear formula with I = p A 0.01054 B = 9.5466 A 10 - B m4 Qmax = 4(0.0105) c (p)(0.0105)2 d = 0.77175 A 10 - B m3 3p From the shear diagram, Vmax = 1484 N tmax = = Vmax Qmax It 1484 C 0.77175(10 - 6) D 9.5466(10 - 9)(0.021) = 5.71 MPa tallow = 80 MPa (O.K!) 874 75 mm 11 Solutions 46060 5/26/10 3:27 PM Page 875 © 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 *11–48 The overhang beam is constructed using two 2-in by 4-in pieces of wood braced as shown If the allowable bending stress is sallow = 600 psi, determine the largest load P that can be applied Also, determine the associated maximum spacing of nails, s, along the beam section AC if each nail can resist a shear force of 800 lb Assume the beam is pin-connected at A, B, and D Neglect the axial force developed in the beam along DA D ft ft A ft Section properties: I = (4)(4)3 = 21.33 in4 12 S = 21.33 I = = 10.67 in3 c Mmax = sallow S 3P(12) = 600(10.67) P = 177.78 = 178 lb Ans Nail Spacing: V = P = 177.78 lb Q = (4)(2)(1) = in3 q = 177.78(8) VQ = = 66.67 lb>in I 21.33 S = 800 lb = 12.0 in 66.67 lb>in Ans 875 in in s B MA = Mmax = 3P P C in 11 Solutions 46060 5/26/10 3:27 PM Page 876 © 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 z •11–49 The bearings at A and B exert only x and z components of force on the steel shaft Determine the shaft’s diameter to the nearest millimeter so that it can resist the loadings of the gears without exceeding an allowable shear stress of tallow = 80 MPa Use the maximum-shear-stress theory of failure Fx ϭ kN A 75 mm x 50 mm 150 mm 350 mm B Fz ϭ 7.5 kN 250 mm Maximum resultant moment M = 212502 + 2502 = 1274.75 N # m 1 3 2 21274.752 + 3752 d = 0.0219 m 2M2 + T2 d = c c = c p tallow p(80)(10 ) d = 2c = 0.0439 m = 43.9 mm Use d = 44 mm Ans 876 y 11 Solutions 46060 5/26/10 3:27 PM Page 877 © 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 z 11–50 The bearings at A and B exert only x and z components of force on the steel shaft Determine the shaft’s diameter to the nearest millimeter so that it can resist the loadings of the gears without exceeding an allowable shear stress of tallow = 80 MPa Use the maximumdistortion-energy theory of failure with sallow = 200 MPa Fx ϭ kN A 75 mm x 50 mm 150 mm 350 mm Maximum resultant moment M = 212502 + 2502 = 1274.75 N # m s1, = sx s2x ; A + txy sx s2x ,b = A + txy Let a = s1 = a + b, s2 = a - b Require, s21 - s1 s2 + s22 = s2allow a2 + 2ab + b2 - [a2 - b2] + a2 - 2ab + b2 = s2allow a2 + 3b2 = s2allow s2x s2x + 3a + t2xy b = s2allow 4 s2x + 3t2xy = s2allow Mc Tc a p b + 3a p b = s2allow c c c Ba c6 = 4M 2T b + 3a b R = s2allow p p 16 s2allow p2 c = B = B M2 + s2allow p2 12T2 s2allow p2 (4M2 + T2) R 4 (4(1274.75)2 + 3(375)2) R (200(106))2(p)2 = 0.0203 m = 20.3 mm d = 40.6 mm Ans Use d = 41 mm Ans 877 B Fz ϭ 7.5 kN 250 mm y 11 Solutions 46060 5/26/10 3:27 PM Page 878 © 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 11–51 Draw the shear and moment diagrams for the beam Then select the lightest-weight steel wide-flange beam from Appendix B that will safely support the loading Take sallow = 22 ksi, and tallow = 12 ksi kip/ft 1.5 kip и ft A B 12 ft Bending Stress: From the moment diagram, Mmax = 18.0 kip # ft Assume bending controls the design Applying the flexure formula Sreq¿d = = Select Mmax sallow 18.0(12) = 9.82 in3 22 A Sx = 10.9 in3, d = 9.87 in., tw = 0.19 in B W10 * 12 V for the W10 * 12 wide twd = 9.00 kip Shear Stress: Provide a shear stress check using t = flange section From the shear diagram, Vmax tmax = = Vmax tw d 9.00 0.19(9.87) = 4.80 ksi tallow = 12 ksi (O.K!) Hence, Use W10 * 12 Ans 878 ft 11 Solutions 46060 5/26/10 3:27 PM Page 879 © 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 *11–52 The beam is made of cypress having an allowable bending stress of sallow = 850 psi and an allowable shear stress of tallow = 80 psi Determine the width b of the beam if the height h = 1.5b 300 lb 75 lb/ft B A ft ft h ϭ 1.5b b Ix = (b)(1.5b)3 = 0.28125 b4 12 Qmax = y¿A¿ = (0.375b) (0.75b)(b) = 0.28125 b3 Assume bending controls Mmax = 527.34 lb # ft sallow = Mmax c ; I 850 = 527.34(12)(0.75 b) 0.28125 b4 b = 2.71 in Ans Check shear: I = 15.12 in4 tmax = Qmax = 5.584 in3 VQmax 281.25(5.584) = It 15.12(2.71) = 38.36 psi 80 psi OK 879 11 Solutions 46060 5/26/10 3:27 PM Page 880 © 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 •11–53 The tapered beam supports a uniform distributed load w If it is made from a plate and has a constant width b, determine the absolute maximum bending stress in the beam w h0 L –– Support Reactions: As shown on FBD(a) Moment Function: As shown on FBD(b) Section Properties: h - h0 h0 = L x I = S = h = h0 (2x + L) L h30 (b) a b(2x + L)3 12 L 12 (b) A h0 2L B (2x + L)3 h30 L bh20 = (2x + L) 6L2 (2x + L)2 Bending Stress: Applying the flexure formula s = M = S w (Lx - x2) bh20 6L 3wL2 (Lx - x2) = (2x + L)2 [1] bh20 (2x + L)2 In order to have the absolute maximum bending stress, ds = dx 3wL2 (2x + L)2(L - 2x) - (Lx - x2)(2)(2x + L)(2) ds = c d = dx bh20 (2x + L)4 x = Substituting x = L L into Eq [1] yields smax = wL2 4bh20 Ans 880 h0 h0 L –– 11 Solutions 46060 5/26/10 3:27 PM Page 881 © 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 11–54 The tubular shaft has an inner diameter of 15 mm Determine to the nearest millimeter its outer diameter if it is subjected to the gear loading The bearings at A and B exert force components only in the y and z directions on the shaft Use an allowable shear stress of tallow = 70 MPa, and base the design on the maximum-shear-stress theory of failure z 100 mm B 500 N 150 mm A 200 mm 150 mm x I = p p (c - 0.00754) and J = (c4 - 0.00754) tallow = Aa sx - sy tallow = Aa Mc Tc b + a b 2I J t2allow = M2 c2 T2 c2 + 4I J2 ¢ 100 mm b + t2xy c4 - 0.00754 4M2 4T2 ≤ = + c p p c4 - 0.00754 = 2M2 + T2 c p tallow c4 - 0.00754 2752 + 502 = c p(70)(106) c4 - 0.00754 = 0.8198(10 - 6)c Solving, c = 0.0103976 m d = 2c = 0.0207952 m = 20.8 mm Use d = 21 mm Ans 881 500 N y 11 Solutions 46060 5/26/10 3:27 PM Page 882 © 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 11–55 Determine to the nearest millimeter the diameter of the solid shaft if it is subjected to the gear loading The bearings at A and B exert force components only in the y and z directions on the shaft Base the design on the maximum-distortion-energy theory of failure with sallow = 150 MPa z 100 mm B 500 N 150 mm A 200 mm 150 mm x s1, = 100 mm sx ; A + txy s2x sx s2x ,b = A + txy Let a = s1 = a + b, s2 = a - b Require, s21 - s1 s2 + s21 = s2allow a2 + 2ab + b2 - [a2 - b2] + a2 - 2ab + b2 = sallow a2 + 3b2 = s2allow s2x s2x + 3a + t2xy b = s2allow 4 s2x + 3t2xy = s2allow Mc Tc a p b + 3a p b = s2allow c c c ca c6 = 4M 2T b + 3a b d = s2allow p p 16 s2allow p2 c = a = c M2 + s2allow p2 12T2 s2allow p2 (4M2 + 3T2) b (150(106))2(p)2 4 (4(75) + 3(50) ) d = 0.009025 m 2 d = 2c = 0.0181 m Use d = 19 mm Ans 882 500 N y ... region CD, V = 500 lb Thus qallow = VQA ; I 500 (140) 600 = S– 1366.67 Use S– = 11. 71 in S– = 111 2 in Ans 857 11 Solutions 46060 5/26/10 3:27 PM Page 858 © 2010 Pearson Education, Inc., Upper... = 42.0 in3, d = 13.84 in, tw = 0.270 in At x = ft, V = 11. 56 kip t = 11. 36 V = = 3.09 ksi 12 ksi Aweb (13.84)(0.270) Use W14 * 30 Ans 841 11 Solutions 46060 5/26/10 3:27 PM Page 842 © 2010 Pearson... Vmax = 11. 25 kip tmax = = Vmax tw d 11. 25 0.23(13.74) = 3.56 ksi tallow = 12 ksi (O.K!) Based on the investigated results, we conclude that W14 * 22 can safely support the loading 847 11 Solutions