Chapter 12

12 Solutions 46060 6/11/10 11:52 AM Page 883 © 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 •12–1 An A-36 steel strap having a thickness of 10 mm and a width of 20 mm is bent into a circular arc of radius r = 10 m Determine the maximum bending stress in the strap Moment-Curvature Relationship: M = r EI however, M = I s c 1 c s = r EI s = 0.005 c E = a b C 200 A 109 B D = 100 MPa r 10 12–2 A picture is taken of a man performing a pole vault, and the minimum radius of curvature of the pole is estimated by measurement to be 4.5 m If the pole is 40 mm in diameter and it is made of a glass-reinforced plastic for which Eg = 131 GPa, determine the maximum bending stress in the pole r ϭ 4.5 m Moment-Curvature Relationship: M = r EI however, M = I s c I c s = r EI s = 0.02 c E = a b C 131 A 109 B D = 582 MPa r 4.5 Ans 883 12 Solutions 46060 6/11/10 11:52 AM Page 884 © 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 12–3 When the diver stands at end C of the diving board, it deflects downward 3.5 in Determine the weight of the diver The board is made of material having a modulus of elasticity of E = 1.5(103) ksi B A 3.5 in in C ft ft Support Reactions and Elastic Curve As shown in Fig a Moment Functions Referring to the free-body diagrams of the diving board’s cut segments, Fig b, M A x1 B is a + ©MO = 0; and M A x2 B is a + ©MO = 0; M A x1 B + 3Wx1 = M A x1 B = -3Wx1 -M A x2 B - Wx2 = M A x2 B = -Wx2 Equations of Slope and Elastic Curve EI d2v = M(x) dx2 For coordinate x1, EI d2v1 dx1 = -3Wx1 d2v1 = - Wx1 + C1 dx1 (1) EIv1 = - Wx1 + C1x1 + C2 (2) EI For coordinate x2 EI EI d2v2 dx2 = -Wx2 dv2 = - Wx2 + C3 dx2 EIv2 = - (3) Wx2 + C3x2 + C4 (4) Boundary Conditions At x1 = 0, v1 = Then, Eq (2) gives EI(0) = - W A 03 B + C1(0) + C2 C2 = At x1 = ft, v1 = Then, Eq (2) gives EI(0) = - W A 33 B + C1(3) + C1 = 4.5W 884 18 in 12 Solutions 46060 6/11/10 11:52 AM Page 885 © 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 12-3 Continued At x2 = ft, v2 = Then, Eq (4) gives EI(0) = - W A 93 B + C3(9) + C4 9C3 + C4 = 121.5W Continuity Conditions At x1 = ft and x2 = ft, (5) dv2 dv1 Thus, Eqs (1) and = dx1 dx2 (3) give - W A 32 B + 4.5W = - c - W A 92 B + C3 d 2 C3 = 49.5W Substituting the value of C3 into Eq (5), C4 = -324W Substituting the values of C3 and C4 into Eq (4), v2 = 1 a - Wx2 + 49.5Wx2 - 324Wb EI At x2 = 0, v2 = -3.5 in Then, -324W(1728) -3.5 = 1.5 A 106 B c (18) A B d 12 W = 112.53 lb = 113 lb Ans 885 12 Solutions 46060 6/11/10 11:52 AM Page 886 © 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 *12–4 Determine the equations of the elastic curve using the x1 and x2 coordinates EI is constant P A EI d2v1 dx1 = M1 (x) M1(x) = 0; EI EI d v1 dx1 L = x3 dv1 = C1 dx1 (1) EI v1 = C1x1 + C2 (2) M2(x) = Px2 - P(L - a) EI EI d2 v2 dx2 = Px2 - P(L - a) dv2 P = x - P(L - a)x2 + C3 dx2 2 EI v2 = (3) P(L - a)x22 P x2 + C3x2 + C4 (4) Boundary conditions: At x2 = 0, dv2 = dx2 From Eq (3), = C3 At x2 = 0, v2 = 0 = C4 Continuity condition: At x1 = a, x2 = L - a; dv1 dv2 = dx1 dx2 From Eqs (1) and (3), C1 = - c P(L - a)2 - P(L - a)2 d ; C1 = P(L - a)2 At x1 = a, x2 = L - a, v1 = v2 From Eqs (2) and (4), a P(L - a)3 P(L - a)3 P(L - a)2 b a + C2 = C2 = - Pa(L - a)2 P(L - a)3 From Eq (2), v1 = P [3(L - a)2x1 - 3a(L - a)2 - 2(L - a)3] 6EI Ans For Eq (4), v2 = B x1 P [x2 - 3(L - a)x33] 6EI Ans 886 L 12 Solutions 46060 6/11/10 11:52 AM Page 887 © 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 •12–5 Determine the equations of the elastic curve for the beam using the x1 and x2 coordinates EI is constant P A B x1 L Moment Functions Referring to the FBDs of the beam’s cut segments shown in Fig b and c, P(x1) = a + ©MO = 0; M(x1) + a + ©MO = 0; -Px2 - M(x2) = M(x1) = - P x And EI M(x2) = -Px2 d2v = M(x) dx2 For coordinate x1, EI EI d2v1 dx1 = - P x dv1 P = - x1 + C1 dx1 EI v1 = - (1) P x + C1x + C2 12 (2) For coordinate x2, EI EI d2v2 dx2 = -Px2 dv2 P = - x2 + C3 dx2 EI v2 = - (3) P x + C3x2 + C4 (4) At x1 = 0, v1 = Then, Eq (2) gives EI(0) = - P (0) + C1(0) + C2 12 C2 = At x1 = L, v1 = Then, Eq (2) gives EI(0) = At x2 = P (L3) + C1L + 12 C1 = PL2 12 L , v2 = Then Eq (4) gives EI(0) = - P L L a b + C3 a b + C4 2 C3L + 2C4 = PL3 24 (5) 887 x2 L 12 Solutions 46060 6/11/10 11:52 AM Page 888 © 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 •12–5 Continued At x1 = L and x2 = - dv2 L dv1 , = Thus, Eqs (1) and (3) gives dx1 dx2 P P L PL2 = - c - a b + C3 d AL B + 12 2 C3 = 7PL2 24 Substitute the result of C3 into Eq (5) C4 = - PL3 Substitute the values of C1 and C2 into Eq (2) and C3 and C4 into Eq (4), v1 = P A -x1 + L2x1 B 12EI Ans v2 = P A -4x2 + 7L2x2 - 3L3 B 24EI Ans 888 12 Solutions 46060 6/11/10 11:52 AM Page 889 © 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 12–6 Determine the equations of the elastic curve for the beam using the x1 and x3 coordinates Specify the beam’s maximum deflection EI is constant P A Support Reactions and Elastic Curve: As shown on FBD(a) L Slope and Elastic Curve: For M(x1) = - x3 d2v = M(x) dx2 EI P x EI d2y1 dx21 EI y1 = For M(x3) = Px3 - = - P x dy1 P = - x21 + C1 dx1 EI [1] P x + C1x1 + C2 12 [2] 3PL EI d2y3 dx23 = Px3 - 3PL dy3 P 3PL = x3 x3 + C3 dx3 2 EI EI y3 = [3] P 3PL x x3 + C3x3 + C4 [4] Boundary Conditions: y1 = at x1 = From Eq [2], C2 = y1 = at x1 = L From Eq [2] = - PL3 + C1L 12 C1 = PL2 12 y3 = at x3 = L From Eq [4] = PL3 3PL3 + C3L + C4 = - 7PL3 + C3L + C4 12 [5] Continuity Condition: At x1 = x3 = L, - dy1 dy3 From Eqs [1] and [3], = dx1 dx3 PL2 PL2 PL2 3PL2 + = + C3 12 2 From Eq [5], C4 = - B x1 Moment Function: As shown on FBD(b) and (c) C3 = 5PL2 PL3 889 L 12 Solutions 46060 6/11/10 11:52 AM Page 890 © 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 12–6 Continued The Slope: Substitute the value of C1 into Eq [1], dy1 P = A L2 - 3x21 B dx1 12EI dy1 P = = A L2 - 3x21 B dx1 12EI x1 = L 23 The Elastic Curve: Substitute the values of C1, C2, C3, and C4 into Eqs [2] and [4], respectively y1 = Px1 A -x21 + L2 B 12EI yO = y1 |x1 = y3 = L 23 = PA L 23 B 12EI Ans a- 0.0321PL3 L3 + L2 b = EI P A 2x33 - 9Lx23 + 10L2x3 - 3L3 B 12EI Ans yC = y3 |x3 = 32 L = P 3 3 c2 a L b - 9La Lb + 10L2 a L b - 3L3 d 12EI 2 = - PL3 8EI Hence, ymax = PL3 8EI Ans 890 12 Solutions 46060 6/11/10 11:52 AM Page 891 © 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 12–7 The beam is made of two rods and is subjected to the concentrated load P Determine the maximum deflection of the beam if the moments of inertia of the rods are IAB and IBC , and the modulus of elasticity is E EI P B A C l d2y = M(x) dx2 L M1(x) = - Px1 EIBC EIBC d2y1 dx1 = - Px1 dy1 Px21 = + C1 dx1 EIBC y1 = - (1) Px31 + C1x1 + C2 (2) M2(x) = - Px2 EIAB EIAB d2y2 dx2 = - Px2 dy2 P = - x2 + C3 dx2 EIAB y2 = - (3) P x + C3x2 + C4 2 (4) Boundary conditions: At x2 = L, = - dy2 = dx2 PL2 + C3; C3 = PL2 At x2 = L, y = 0 = - PL3 PL3 + + C4; C4 = - PL3 Continuity Conditions: At x1 = x2 = l, dy1 dy2 = dx1 dx2 891 12 Solutions 46060 6/11/10 11:52 AM Page 892 © 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 12–7 Continued From Eqs (1) and (3), PI PI PL2 cc+ C1 d = + d EIBC EIAB 2 C1 = IBC PL2 Pl2 Pl2 c+ d + IAB 2 At x1 = x2 = l, y1 = y2 From Eqs (2) and (4), IBC PL2 Pl2 Pl3 Pl2 e+ c a+ b + dl + C2 f EIBC IAB 2 = PL2l PL3 Pl3 c+ d EIAB C2 = IBC PL3 IBC Pl3 Pl3 IAB IAB 3 Therefore, y1 = Px1 IBC Pl2 PL2 Pl2 e+ c a+ b + dx1 EIBC IAB 2 + IBC PL3 IBC Pl3 Pl3 f IAB IAB 3 At x1 = 0, y1 |x = = ymax ymax = = IBC Pl3 IBC PL3 IAB I Pl3 P e f = e l3 - L3 - a bl f EIBC IAB IAB 3 3EIAB IBC IAB P e a1 b l - L3 f 3EIAB IBC Ans 892 12 Solutions 46060 6/11/10 11:53 AM Page 1023 © 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 *12–132 The beam is supported by a pin at A, a spring having a stiffness k at B, and a roller at C Determine the force the spring exerts on the beam EI is constant w A B Method of Superposition: Using the table in appendix C, the required displacements are 5w(2L)4 5wL4AC 5wL4 = = 384EI 384EI 24EI T Fsp (2L)3 Fsp L3 PL3AC = = yB – = 48EI 48EI 6EI c yB ¿ = Using the spring formula, ysp = Fsp k L C k L The compatibility condition requires ysp = yB ¿ + yB – (+ T) Fsp k = Fsp = Fsp L 5wL4 + ab 24EI 6EI 5wkL4 A 6EI + kL3 B Ans •12–133 The beam is made from a soft linear elastic material having a constant EI If it is originally a distance ¢ from the surface of its end support, determine the distance a at which it rests on this support when it is subjected to the uniform load w0 , which is great enough to cause this to happen w0 ⌬ a L The curvature of the beam in region BC is zero, therefore there is no bending moment in the region BC, The reaction F is at B where it touches the support The slope is zero at this point and the deflection is ¢ where ¢ = R(L - a)3 w0(L - a)4 8EI 3EI u1 = w0(L - a)3 R(L - a)2 6EI 2EI Thus, 8¢EI R = a b 9w30 Ans 72¢EI L - a = a b w0 72¢EI a = L - a b w0 Ans 1023 12 Solutions 46060 6/11/10 11:53 AM Page 1024 © 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 12–134 Before the uniform distributed load is applied on the beam, there is a small gap of 0.2 mm between the beam and the post at B Determine the support reactions at A, B, and C The post at B has a diameter of 40 mm, and the moment of inertia of the beam is I = 875(106) mm4 The post and the beam are made of material having a modulus of elasticity of E = 200 GPa 30 kN/m A 6m Equations of Equilibrium Referring to the free-body diagram of the beam, Fig a, + ©F = 0; : x Ax = + c ©Fy = 0; A y + FB + Cy - 30(12) = a + ©MA = 0; Ans (1) FB(6) + Cy(12) - 30(12)(6) = (2) Method of superposition: Referring to Fig b and the table in the Appendix, the necessary deflections are (vB)1 = (vB)2 = 5(30) A 12 B 8100kN # m3 5wL4 = = T 384EI 384EI EI FB A 12 B 36FB PL3 = = 48EI 48EI EI c The deflection of point B is vB = 0.2 A 10 - B + FB(a) FBLB = 0.2 A 10 - B + AE AE T The compatibility condition at support B requires A+TB vB = (vB)1 + (vB)2 0.2 A 10 - B + FB (1) 36FB 8100 = + ab AE EI EI 0.2 A 10 - B E + FB p A 0.04 B + FB 36FB 8100 = A I I 36FB 875 A 10 - B 8100 = 875 A 10 - B - C 1m 0.2 A 10 - B C 200 A 109 B D 1000 FB = 219.78 kN = 220 kN Ans Substituting the result of FB into Eqs (1) and (2), A y = Cy = 70.11 kN = 70.1 kN Ans 1024 B 0.2 mm 6m 12 Solutions 46060 6/11/10 11:53 AM Page 1025 © 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 12–135 The 1-in.-diameter A-36 steel shaft is supported by unyielding bearings at A and C The bearing at B rests on a simply supported steel wide-flange beam having a moment of inertia of I = 500 in4 If the belt loads on the pulley are 400 lb each, determine the vertical reactions at A, B, and C ft ft A ft ft B For the shaft: (¢ b)1 = (¢ b)2 = 400 lb 800(3)(5) 13200 A -52 - 32 + 102 B = 6EIs(10) EIs By A 103 B 48EIs C ft 20.833By = EIs For the beam: ¢b = By A 103 B 48EIb 20.833By = EIb Compatibility condition: + T ¢ b = (¢ b)1 - (¢ b)2 20.833By EIb Is = = 20.833By 13200 EIs EIs p (0.5)4 = 0.04909 in4 20.833By (0.04909) 500 400 lb = 13200 - 20.833By By = 634 lb Ans Form the free-body digram, A y = 243 lb Ans Cy = 76.8 lb Ans 1025 12 Solutions 46060 6/11/10 11:53 AM Page 1026 © 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 *12–136 If the temperature of the 75-mm-diameter post CD is increased by 60°C, determine the force developed in the post The post and the beam are made of A-36 steel, and the moment of inertia of the beam is I = 255(106) mm4 3m 3m A B C 3m D Method of Superposition Referring to Fig a and the table in the Appendix, the necessary deflections are (vC)1 = FCD A 33 B 9FCD PLBC c = = 3EI 3EI EI (vC)2 = (uB)2LBC = 3FCD (3) 9FCD MOLAB c (LBC) = (3) = 3EI 3EI EI The compatibility condition at end C requires A+cB vC = (vC)1 + (vC)2 = 9FCD 9FCD 18FCD c + = EI EI EI Referring to Fig b, the compatibility condition of post CD requires that dFCD + vC = dT dFCD = (1) FCD (3) FCD LCD = AE AE dT = a¢TL = 12 A 10 - B (60)(3) = 2.16 A 10 - B m Thus, Eq (1) becomes 3FCD 18FCD + = 2.16 A 10 - B AE EI 3FCD p A 0.0752 B + 18FCD 255 A 10 - B = 2.16 A 10 - B C 200 A 109 B D FCD = 6061.69N = 6.06 kN Ans 1026 12 Solutions 46060 6/11/10 11:53 AM Page 1027 © 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 •12–137 The shaft supports the two pulley loads shown Using discontinuity functions, determine the equation of the elastic curve The bearings at A and B exert only vertical reactions on the shaft EI is constant x A 12 in B 12 in 70 lb 180 lb M = -180 x - - (-277.5) x - 12 - 70 x - 24 M = -180x + 277.5 x - 12 - 70 x - 24 Elastic curve and slope: EI d2v = M = -180x + 277.5 x - 12 - 70 x - 24 dx2 EI dv = -90x2 + 138.75 x - 12 dx EIv = -30x3 + 46.25 x - 12 - 35(x - 24 - 11.67 x - 24 + C1 + C1x + C2 (1) Boundary conditions: v = at x = 12 in, From Eq (1) = -51,840 + 12C1 + C2 12C1 + C2 = 51 840 v = at (2) x = 60 in From Eq.(1) = -6 480 000 + 114 880 - 544 320 + 60C1 + C2 60C1 + C2 = 1909440 (3) Solving Eqs (2) and (3) yields: C1 = 38 700 v = C2 = -412 560 [-30x3 + 46.25 x - 12 EI - 11.7 x - 24 + 38 700x - 412 560] Ans 1027 36 in 12 Solutions 46060 6/11/10 11:53 AM Page 1028 © 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 12–138 The shaft is supported by a journal bearing at A, which exerts only vertical reactions on the shaft, and by a thrust bearing at B, which exerts both horizontal and vertical reactions on the shaft Draw the bending-moment diagram for the shaft and then, from this diagram, sketch the deflection or elastic curve for the shaft’s centerline Determine the equations of the elastic curve using the coordinates x1 and x2 EI is constant 80 lb A x1 EI d2v1 dx21 = 26.67x1 dv1 = 13.33x21 + C1 dx1 (1) EIv1 = 4.44x31 + C1x1 + C2 (2) EI For M2 (x) = -26.67x2 EI d2v2 dx22 = -26.67x2 dv2 = -13.33x22 + C3 dx2 (3) EIv2 = -4.44x32 + C3x2 + C4 (4) EI Boundary conditions: v1 = at x1 = at x2 = From Eq.(2) C2 = v2 = C4 = Continuity conditions: dv1 dv2 = dx1 dx2 at x1 = x2 = 12 From Eqs (1) and (3) 1920 + C1 = -( -1920 + C3) C1 = -C3 v1 = v2 (5) x1 = x2 = 12 at 7680 + 12C1 = -7680 + 12C3 C3 - C1 = 1280 (6) Solving Eqs (5) and (6) yields: C3 = 640 80 lb 12 in For M1 (x) = 26.67 x1 C1 = -640 v1 = A 4.44x31 - 640x1 B lb # in3 EI Ans v2 = A -4.44x32 + 640x2 B lb # in3 EI Ans 1028 B in in x2 12 in 12 Solutions 46060 6/11/10 11:53 AM Page 1029 © 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 12–139 The W8 * 24 simply supported beam is subjected to the loading shown Using the method of superposition, determine the deflection at its center C The beam is made of A-36 steel kip/ft kipиft A B C ft Elastic Curves: The elastic curves for the uniform distributed load and couple moment are drawn separately as shown Method of superposition: Using the table in Appendix C, the required displacements are (¢ C)1 = -5(6) A 164 B 2560 kip # ft3 -5wL4 = = T 768EI 768EI EI (¢ C)2 = - = - = M0x A L2 - x2 B 6EIL 5(8) C (16)2 - (8)2 D 6EI(16) 80 kip # ft3 EI T The displacement at C is ¢ C = (¢ C)1 + (¢ C)2 = 80 2560 + EI EI = 2640 kip # ft3 EI 2640(1728) = 29 A 103 B (82.8) = 1.90 in Ans T 1029 ft 12 Solutions 46060 6/11/10 11:53 AM Page 1030 © 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 *12–140 Using the moment-area method, determine the slope and deflection at end C of the shaft The 75-mmdiameter shaft is made of material having E = 200 GPa Support Reactions and B A M Diagram As shown in Fig a EI 1m 1m 15 kN = (1) B = ΗtC>A Η 7.5 a b (2) R + c (2) d B a b(2) R EI EI 5.5 kN # m3 EI = (1 + 1) B 7.5 1 a b(2) R + c (2) + d B a b(2) R EI EI + c (1) d B a b(1) R EI ΗuC>A Η = kN # m3 EI = 7.5 a b(2) + a b(3) EI EI = kN # m3 EI Referring to the geometry of the elastic curve, Fig b, uA = ΗtB>A Η LAB 5.5 EI 2.75kN # m2 = = EI uC = uC>A - uA = = 2.75 EI EI 0.25 kN # m2 = EI 0.25 A 103 B 200 A 109 B c p A 0.03754 B d = 0.805 A 10 - B rad Ans and ¢ C = Η tC>AΗ - Η tB>A ¢ = 5.5 a b EI EI = 0.75 kN # m3 = EI LAC ≤ LAB 0.75 A 103 B p 200 A 10 B c A 0.03754 B d 1m kN Moment Area Theorem Referring to Fig b, ΗtB>A Η C = 0.002414 m = 2.41 mm c Ans 1030 12 Solutions 46060 6/11/10 11:53 AM Page 1031 © 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 •12–141 Determine the reactions at the supports EI is constant Use the method of superposition w A wL C L3 - 2(3L)L2 + (3L)3 D 24EI ¢B = ¢C = D B L 11wL 12EI = Due to symmetry, By = Cy By (L)(2L) ¢ BB = ¢ CC = 6EI(3L) C (3L)2 - (2L)2 - L2 D 4By L3 = 9EI By (L)(L) ¢ BC = ¢ CB = 6EI(3L) C -L2 - L2 + (3L)2 D 7By L3 = 18EI By superposition: +T = = ¢ B - ¢ BB - ¢ BC 4By L3 7By L3 11wL4 12EI 9EI 18EI By = Cy = 11wL 10 Ans Equilibrium: a+ ©MD = 0; Ay = c + ©Fy = 0; Dy = + ©F = 0; ; x 3wLa 3L 11wL 11wL b (L) (2L) - A y (3L) = 10 10 2wL Ans 2wL 11wL 11wL + + + Dy - 3wL = 10 10 2wL Ans Dx = Ans 1031 C L L 12 Solutions 46060 6/11/10 11:53 AM Page 1032 © 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 12–142 Determine the moment reactions at the supports A and B Use the method of integration EI is constant w0 A B L Support Reactions: FBD(a) A y + By - + c ©Fy = 0; a + ©MA = 0; w0L = [1] ByL + MA - MB - w0L L a b = [2] Moment Function: FBD(b) a+ ©MNA = 0; -M(x) - x w0 a xb x a b - MB + Byx = L M(x) = Byx - w0 x - MB 6L Slope and Elastic Curve: EI EI EI EI y = d2y = M(x) dx2 w0 d2y = Byx x - MB 6L dx2 By w0 dy = x2 x - MBx + C1 dx 24L By x3 - [3] w0 MB x x + C1x + C2 120L [4] Boundary Conditions: At x = 0, dy = dx From Eq.[3], At x = 0, y = At x = L, = From Eq.[4], dy = dx By L2 - C1 = C2 = From Eq [3] w0L3 - MBL 24 = 12By L - w0 L2 - 24MB At x = L, y = 0 = By L3 - [5] From Eq [4], w0 L4 MB L2 120 = 20By L - w0 L2 - 60MB [6] 1032 12 Solutions 46060 6/11/10 11:53 AM Page 1033 © 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 12–142 Continued Solving Eqs [5] and [6] yields, MB = By = w0 L2 30 Ans 3w0L 20 Substituting By and MB into Eqs [1] and [2] yields, MA = w0L2 20 Ay = 7w0 L 20 Ans 1033 12 Solutions 46060 6/11/10 11:53 AM Page 1034 © 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 12–143 If the cantilever beam has a constant thickness t, determine the deflection at end A The beam is made of material having a modulus of elasticity E L w0 x Section Properties: Referring to the geometry shown in Fig a, A h(x) h0 = ; x L h0 x h(x) = L h0 B Thus, the moment of inertia of the tapered beam as a function of x is I(x) = h0 th0 3 1 t C h(x) D = t¢ x≤ = x 12 12 L 12L3 Moment Function Referring to the free-body diagram of the beam’s segment, Fig b, M(x) + B a + ©MO = 0; w0 x a xbx R a b = L M(x) = - w0 x 6L Equations of slope and Elastic Curve E M(x) d2v = I(x) dx w0 x 2w0L2 dv 6L E = = dx th0 3 th0 x 12L - E 2w0L2 dv = x + C1 dx th0 Ev = - w0L2 th0 (1) x2 + C1x + C2 Boundary conditions At x = L, = - 2w0L2 th0 (2) dv = Then Eq (1) gives dx (L) + C1 C1 = 2w0L3 th0 At x = L, v = Then Eq (2) gives = - w0L2 th0 A L2 B + 2w0L3 th0 (L) + C2 C2 = - w0L4 th0 Substituting the results of C1 and C2 into Eq (2), v = w0L2 Eth0 A -x2 + 2Lx - L2 B At A, x = Then vA = vΗx = = - w0L4 Eth0 = w0L4 Eth0 Ans T 1034 12 Solutions 46060 6/11/10 11:53 AM Page 1035 © 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 *12–144 Beam ABC is supported by beam DBE and fixed at C Determine the reactions at B and C The beams are made of the same material having a modulus of elasticity E = 200 GPa, and the moment of inertia of both beams is I = 25.0(106) mm4 100 lb/ft a A B C D E a ft ft ft in Section a – a + ©F = 0; : x Cx = + c ©Fy = 0; By + Cy - 9(6) = a+ ©MC = 0; 9(6)(3) - By(4) - MC = Ans (1) MC = 162 - 4By (2) Method of superposition: Referring to Fig b and the table in the appendix, the deflections are vB = By A 63 B 4.5By PLDE = = T 48EI 48EI EI (vB)1 = = (vB)2 = A 42 B wx2 A x2 - 4Lx + 6L2 B = C - 4(6)(4) + A 62 B D 24EI 24EI 816 kN # m3 T EI By A B 21.3333By PLBC c = = 3EI 3EI EI The compatibility condition at support B requires that A+TB vB = (vB)1 + (vB)2 4.5By EI = 21.3333By 816 + ab EI EI By = 31.59 kN = 31.6 kN Ans Substituting the result of By into Eqs (1) and (2), MC = 35.65 kN # m = 35.7 kN # m Ans Cy = 22.41 kN = 22.4 kN Ans 1035 a ft in Equation of Equilibrium Referring to the free-body diagram of the beam, Fig a, a 12 Solutions 46060 6/11/10 11:53 AM Page 1036 © 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 •12–145 Using the method of superposition, determine the deflection at C of beam AB The beams are made of wood having a modulus of elasticity of E = 1.5(103) ksi 100 lb/ft a A B C D E a ft ft ft in Section a – a Method of superposition Referring to Fig b and the table in the appendix, the deflection of point B is ¢B = 600 A 83 B PLDE 6400 lb # ft3 = = T 48EI 48EI EI Subsequently, referring to Fig c, (¢ C)1 = ¢ B a (¢ C)2 = 6400 3200 lb # ft3 b = a b = T 12 EI 12 EI 5(100) A 12 B 5wL4 27000 lb # ft3 = = T 384EI 384EI EI Thus, the deflection of point C is A+TB ¢ C = (¢ C)1 + (¢ C)2 = 3200 27000 + EI EI 30200 lb # ft3 = = EI 30200 A 12 B 1.5 A 106 B c (3) A 63 B d 12 = 0.644 in T Ans 1036 a ft in Support Reactions: The reaction at B is shown on the free-body diagram of beam AB, Fig a a 12 Solutions 46060 6/11/10 11:53 AM Page 1037 © 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 12–146 The rim on the flywheel has a thickness t, width b, and specific weight g If the flywheel is rotating at a constant rate of v, determine the maximum moment developed in the rim Assume that the spokes not deform Hint: Due to symmetry of the loading, the slope of the rim at each spoke is zero Consider the radius to be sufficiently large so that the segment AB can be considered as a straight beam fixed at both ends and loaded with a uniform centrifugal force per unit length Show that this force is w ϭ btgv2r>g A B v r Centrifugal Force: The centrifugal force action on a unit length of the rim rotating at a constant rate of v is g btgv2r w = mv2 r = bta bv2r = g g (Q.E.D.) Elastic Curve: Member AB of the rim is modeled as a straight beam with both of its ends fixed and subjected to a uniform centrifigal force w Method of Superposition: Using the table in Appendix C, the required displacements are uB ¿ = wL3 6EI yB ¿ = wL4 c 8EI uB – = yB – = MBL EI uB ¿– = MBL2 c 2EI yB –¿ = ByL2 2EI ByL3 3EI T Computibility requires, = uB ¿ + uB – + uB ¿– = By L MBL wL3 + + ab 6EI EI 2EI = wL2 + 6MB - 3By L [1] = yB ¿ + yB – + yB –¿ (+ c ) = By L MB L2 wL4 + + ab 8EI 2EI 3EI = 3wL2 + 12MB - 8By L [2] Solving Eqs [1] and [2] yields, By = wL MB = Due to symmetry, A y = wL wL2 12 MA = wL2 12 Maximum Moment: From the moment diagram, the maximum moment occurs at btgv2r pr the two fixed end supports With w = and L = ru = g Mmax wL2 = = 12 A B btgv2r pr g 12 = t p2btgv2r3 108g Ans 1037 ... B 12EI yO = y1 |x1 = y3 = L 23 = PA L 23 B 12EI Ans a- 0.0321PL3 L3 + L2 b = EI P A 2x33 - 9Lx23 + 10L2x3 - 3L3 B 12EI Ans yC = y3 |x3 = 32 L = P 3 3 c2 a L b - 9La Lb + 10L2 a L b - 3L3 d 12EI... publisher 12 14 Continued From Eqs (1) and (3), PL2 PL2 PL2 + C1 = - a b 64 128 16 C1 = -5PL2 128 v1 = v2 at x1 = x2 = L From Eqs (2) and (4) PL3 5PL2 L PL3 PL2 L a b = - a b a b + C4 768 128 1536... Eq [2], C2 = y1 = at x1 = L From Eq [2] = - PL3 + C1L 12 C1 = PL2 12 y3 = at x3 = L From Eq [4] = PL3 3PL3 + C3L + C4 = - 7PL3 + C3L + C4 12 [5] Continuity Condition: At x1 = x3 = L, - dy1 dy3