07 Solutions 46060 5/26/10 2:04 PM Page 472 © 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 •7–1 If the wide-flange beam is subjected to a shear of V = 20 kN, determine the shear stress on the web at A Indicate the shear-stress components on a volume element located at this point 200 mm A 20 mm 20 mm B V 300 mm 200 mm The moment of inertia of the cross-section about the neutral axis is I = 1 (0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4 12 12 From Fig a, QA = y¿A¿ = 0.16 (0.02)(0.2) = 0.64(10 - 3) m3 Applying the shear formula, VQA 20(103)[0.64(10 - 3)] = tA = It 0.2501(10 - 3)(0.02) = 2.559(106) Pa = 2.56 MPa Ans The shear stress component at A is represented by the volume element shown in Fig b 472 20 mm 07 Solutions 46060 5/26/10 2:04 PM Page 473 © 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 7–2 If the wide-flange beam is subjected to a shear of V = 20 kN, determine the maximum shear stress in the beam 200 mm A 20 mm 20 mm B V 300 mm 200 mm The moment of inertia of the cross-section about the neutral axis is I = 1 (0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4 12 12 From Fig a Qmax = ©y¿A¿ = 0.16 (0.02)(0.2) + 0.075 (0.15)(0.02) = 0.865(10 - 3) m3 The maximum shear stress occurs at the points along neutral axis since Q is maximum and thicknest t is the smallest tmax = VQmax 20(103) [0.865(10 - 3)] = It 0.2501(10 - 3) (0.02) = 3.459(106) Pa = 3.46 MPa Ans 473 20 mm 07 Solutions 46060 5/26/10 2:04 PM Page 474 © 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 7–3 If the wide-flange beam is subjected to a shear of V = 20 kN, determine the shear force resisted by the web of the beam 200 mm A 20 mm 20 mm B V 300 mm 200 mm The moment of inertia of the cross-section about the neutral axis is I = 1 (0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4 12 12 For … y 0.15 m, Fig a, Q as a function of y is Q = ©y¿A¿ = 0.16 (0.02)(0.2) + (y + 0.15)(0.15 - y)(0.02) = 0.865(10 - 3) - 0.01y2 For … y 0.15 m, t = 0.02 m Thus t = 20(103) C 0.865(10 - 3) - 0.01y2 D VQ = It 0.2501(10 - 3) (0.02) = E 3.459(106) - 39.99(106) y2 F Pa The sheer force resisted by the web is, 0.15 m Vw = L0 0.15 m tdA = L0 C 3.459(106) - 39.99(106) y2 D (0.02 dy) = 18.95 (103) N = 19.0 kN Ans 474 20 mm 07 Solutions 46060 5/26/10 2:04 PM Page 475 © 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 *7–4 If the T-beam is subjected to a vertical shear of V = 12 kip, determine the maximum shear stress in the beam Also, compute the shear-stress jump at the flangeweb junction AB Sketch the variation of the shear-stress intensity over the entire cross section in in in in B in A V ϭ 12 kip Section Properties: y = INA = 1.5(12)(3) + 6(4)(6) ©yA = = 3.30 in ©A 12(3) + 4(6) 1 (12) A 33 B + 12(3)(3.30 - 1.5)2 + (4) A 63 B + 4(6)(6 - 3.30)2 12 12 = 390.60 in4 Qmax = y1œ A¿ = 2.85(5.7)(4) = 64.98 in3 QAB = y2œ A¿ = 1.8(3)(12) = 64.8 in3 Shear Stress: Applying the shear formula t = tmax = VQ It VQmax 12(64.98) = = 0.499 ksi It 390.60(4) Ans (tAB)f = VQAB 12(64.8) = = 0.166 ksi Itf 390.60(12) Ans (tAB)W = VQAB 12(64.8) = = 0.498 ksi I tW 390.60(4) Ans 475 07 Solutions 46060 5/26/10 2:04 PM Page 476 © 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 •7–5 If the T-beam is subjected to a vertical shear of V = 12 kip, determine the vertical shear force resisted by the flange in in in in B in A V ϭ 12 kip Section Properties: y = ©yA 1.5(12)(3) + 6(4)(6) = = 3.30 in ©A 12(3) + 4(6) INA = 1 (12) A 33 B + 12(3)(3.30 - 1.5)2 + (4) A 63 B + 6(4)(6 - 3.30)2 12 12 = 390.60 in4 Q = y¿A¿ = (1.65 + 0.5y)(3.3 - y)(12) = 65.34 - 6y2 Shear Stress: Applying the shear formula t = VQ 12(65.34 - 6y2) = It 390.60(12) = 0.16728 - 0.01536y2 Resultant Shear Force: For the flange Vf = tdA LA 3.3 in = L0.3 in A 0.16728 - 0.01536y2 B (12dy) = 3.82 kip Ans 476 07 Solutions 46060 5/26/10 2:04 PM Page 477 © 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 7–6 If the beam is subjected to a shear of V = 15 kN, determine the web’s shear stress at A and B Indicate the shear-stress components on a volume element located at these points Show that the neutral axis is located at y = 0.1747 m from the bottom and INA = 0.2182110-32 m4 200 mm A 30 mm 25 mm V (0.015)(0.125)(0.03) + (0.155)(0.025)(0.25) + (0.295)(0.2)(0.03) y = = 0.1747 m 0.125(0.03) + (0.025)(0.25) + (0.2)(0.03) I = (0.125)(0.033) + 0.125(0.03)(0.1747 - 0.015)2 12 + (0.025)(0.253) + 0.25(0.025)(0.1747 - 0.155)2 12 + (0.2)(0.033) + 0.2(0.03)(0.295 - 0.1747)2 = 0.218182 (10 - 3) m4 12 B 250 mm 30 mm 125 mm œ QA = yAA = (0.310 - 0.015 - 0.1747)(0.2)(0.03) = 0.7219 (10 - 3) m3 QB = yABœ = (0.1747 - 0.015)(0.125)(0.03) = 0.59883 (10 - 3) m3 tA = 15(103)(0.7219)(10 - 3) VQA = 1.99 MPa = It 0.218182(10 - 3)(0.025) Ans tB = VQB 15(103)(0.59883)(10 - 3) = 1.65 MPa = It 0.218182(10 - 3)0.025) Ans 7–7 If the wide-flange beam is subjected to a shear of V = 30 kN, determine the maximum shear stress in the beam 200 mm A 30 mm 25 mm V B 250 mm 30 mm Section Properties: I = 1 (0.2)(0.310)3 (0.175)(0.250)3 = 268.652(10) - m4 12 12 Qmax = © y¿A = 0.0625(0.125)(0.025) + 0.140(0.2)(0.030) = 1.0353(10) - m3 tmax = VQ 30(10)3(1.0353)(10) - = 4.62 MPa = It 268.652(10) - (0.025) Ans 477 200 mm 07 Solutions 46060 5/26/10 2:04 PM Page 478 © 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 *7–8 If the wide-flange beam is subjected to a shear of V = 30 kN, determine the shear force resisted by the web of the beam 200 mm A 30 mm 1 (0.2)(0.310)3 (0.175)(0.250)3 = 268.652(10) - m4 12 12 I = Q = a 25 mm V B 0.155 + y b (0.155 - y)(0.2) = 0.1(0.024025 - y2) 250 mm 30(10)3(0.1)(0.024025 - y2) tf = 268.652(10) -6 30 mm 200 mm (0.2) 0.155 Vf = L tf dA = 55.8343(10)6 L0.125 = 11.1669(10)6[ 0.024025y - (0.024025 - y2)(0.2 dy) 0.155 y ] 0.125 Vf = 1.457 kN Vw = 30 - 2(1.457) = 27.1 kN Ans •7–9 Determine the largest shear force V that the member can sustain if the allowable shear stress is tallow = ksi in in V in in in y = (0.5)(1)(5) + [(2)(1)(2)] = 1.1667 in (5) + (1)(2) I = (5)(13) + (1)(1.1667 - 0.5)2 12 + 2a b (1)(23) + (1)(2)(2 - 1.1667)2 = 6.75 in4 12 Qmax = ©y¿A¿ = (0.91665)(1.8333)(1) = 3.3611 in3 tmax = tallow = (103) = - VQmax It V (3.3611) 6.75 (2)(1) V = 32132 lb = 32.1 kip Ans 478 07 Solutions 46060 5/26/10 2:04 PM Page 479 © 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 7–10 If the applied shear force V = 18 kip, determine the maximum shear stress in the member in in V in in in y = (0.5)(1)(5) + [(2)(1)(2)] = 1.1667 in (5) + (1)(2) I = (5)(13) + (1)(1.1667 - 0.5)2 12 + 2a b (1)(23) + (1)(2)(2 - 1.1667) = 6.75 in4 12 Qmax = ©y¿A¿ = (0.91665)(1.8333)(1) = 3.3611 in3 tmax = 18(3.3611) VQmax = = 4.48 ksi It 6.75 (2)(1) Ans 7–11 The wood beam has an allowable shear stress of tallow = MPa Determine the maximum shear force V that can be applied to the cross section 50 mm 50 mm 100 mm 50 mm 200 mm V 50 mm I = 1 (0.2)(0.2)3 (0.1)(0.1)3 = 125(10 - 6) m4 12 12 tallow = 7(106) = VQmax It V[(0.075)(0.1)(0.05) + 2(0.05)(0.1)(0.05)] 125(10 - 6)(0.1) V = 100 kN Ans 479 07 Solutions 46060 5/26/10 2:04 PM Page 480 © 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 *7–12 The beam has a rectangular cross section and is made of wood having an allowable shear stress of tallow = 200 psi Determine the maximum shear force V that can be developed in the cross section of the beam Also, plot the shear-stress variation over the cross section V 12 in in Section Properties The moment of inertia of the cross-section about the neutral axis is I = (8) (123) = 1152 in4 12 Q as the function of y, Fig a, Q = (y + 6)(6 - y)(8) = (36 - y2) Qmax occurs when y = Thus, Qmax = 4(36 - 02) = 144 in3 The maximum shear stress occurs of points along the neutral axis since Q is maximum and the thickness t = in is constant tallow = VQmax ; It 200 = V(144) 1152(8) V = 12800 16 = 12.8 kip Ans Thus, the shear stress distribution as a function of y is t = 12.8(103) C 4(36 - y2) D VQ = It 1152 (8) = E 5.56 (36 - y2) F psi 480 07 Solutions 46060 5/26/10 2:04 PM Page 481 © 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 7–13 Determine the maximum shear stress in the strut if it is subjected to a shear force of V = 20 kN 12 mm Section Properties: INA 60 mm 1 = (0.12) A 0.0843 B (0.04) A 0.063 B 12 12 V = 5.20704 A 10 - B m4 12 mm 80 mm Qmax = ©y¿A¿ 20 mm 20 mm = 0.015(0.08)(0.03) + 0.036(0.012)(0.12) = 87.84 A 10 - B m3 Maximum Shear Stress: Maximum shear stress occurs at the point where the neutral axis passes through the section Applying the shear formula tmax = VQmax It 20(103)(87.84)(10 - 6) = 5.20704(10 - 6)(0.08) = 22 MPa Ans 7–14 Determine the maximum shear force V that the strut can support if the allowable shear stress for the material is tallow = 40 MPa 12 mm 60 mm Section Properties: INA = V 1 (0.12) A 0.0843 B (0.04) A 0.063 B 12 12 12 mm = 5.20704 A 10 - B m4 80 mm Qmax = ©y¿A¿ 20 mm = 0.015(0.08)(0.03) + 0.036(0.012)(0.12) = 87.84 A 10 - B m3 Allowable shear stress: Maximum shear stress occurs at the point where the neutral axis passes through the section Applying the shear formula tmax = tallow = 40 A 106 B = VQmax It V(87.84)(10 - 6) 5.20704(10 - 6)(0.08) V = 189 692 N = 190 kN Ans 481 20 mm 07 Solutions 46060 5/26/10 2:04 PM Page 517 © 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 *7–60 The angle is subjected to a shear of V = kip Sketch the distribution of shear flow along the leg AB Indicate numerical values at all peaks A in in 45Њ 45Њ 0.25 in Section Properties: b = 0.25 = 0.35355 in sin 45° h = cos 45° = 3.53553 in INA = 2c (0.35355) A 3.535533 B d = 2.604167 in4 12 Q = y¿A¿ = [0.25(3.53553) + 0.5y]a2.5 - y b(0.25) sin 45° = 0.55243 - 0.17678y2 Shear Flow: VQ I 2(103)(0.55243 - 0.17678y2) = 2.604167 q = = {424 - 136y2} lb>in At y = 0, Ans q = qmax = 424 lb>in Ans 517 B V 07 Solutions 46060 5/26/10 2:04 PM Page 518 © 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 •7–61 The assembly is subjected to a vertical shear of V = kip Determine the shear flow at points A and B and the maximum shear flow in the cross section A 0.5 in B V in 0.5 in 0.5 in in in in 0.5 in y = ©yA (0.25)(11)(0.5) + 2(3.25)(5.5)(0.5) + 6.25(7)(0.5) = = 2.8362 in ©A 0.5(11) + 2(0.5)(5.5) + 7(0.5) I = 1 (11)(0.53) + 11(0.5)(2.8362 - 0.25)2 + 2a b(0.5)(5.53) + 2(0.5)(5.5)(3.25 - 2.8362)2 12 12 + (7)(0.53) + (0.5)(7)(6.25 - 2.8362)2 = 92.569 in4 12 QA = y1 ¿A1 ¿ = (2.5862)(2)(0.5) = 2.5862 in3 QB = y2 ¿A2 ¿ = (3.4138)(7)(0.5) = 11.9483 in3 Qmax = ©y¿A¿ = (3.4138)(7)(0.5) + 2(1.5819)(3.1638)(0.5) = 16.9531 in3 q = VQ I 7(103)(2.5862) = 196 lb>in 92.569 7(103)(11.9483) qB = a b = 452 lb>in 92.569 7(103)(16.9531) b = 641 lb>in qmax = a 92.569 qA = Ans Ans Ans 518 0.5 in 07 Solutions 46060 5/26/10 2:04 PM Page 519 © 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 7–62 Determine the shear-stress variation over the cross section of the thin-walled tube as a function of elevation y and show that t max = 2V>A, where A = 2prt Hint: Choose a differential area element dA = Rt du Using dQ = y dA, formulate Q for a circular section from u to (p - u) and show that Q = 2R2t cos u, where cos u = 2R2 - y2>R ds du y u t dA = R t du dQ = y dA = yR t du Here y = R sin u Therefore dQ = R2 t sin u du p-u Q = p-u R2 t sin u du = R2 t(-cos u) | Lu u = R t [-cos (p - u) - (-cos u)] = 2R2 t cos u dI = y2 dA = y2 R t du = R3 t sin2 u du 2p I = L0 2p R3 t sin2 u du = R3 t 2p = t = sin 2u R3 t [u ]Η 2 R3 t [2p - 0] = pR3 t VQ V(2R2t cos u) V cos u = = It pR t pR t(2t) Here cos u = t = = L0 (1 - cos 2u) du 2R2 - y2 R V 2R2 - y2 pR2t Ans tmax occurs at y = 0; therefore tmax = V pR t A = 2pRt; therefore tmax = 2V A QED 519 R 07 Solutions 46060 5/26/10 2:04 PM Page 520 © 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 7–63 Determine the location e of the shear center, point O, for the thin-walled member having the cross section shown where b2 b1 The member segments have the same thickness t t h e b2 Section Properties: I = h t h2 t h3 + 2c(b1 + b2)ta b d = C h + 6(b1 + b2) D 12 12 Q1 = y¿A¿ = h ht (x )t = x 2 Q2 = y¿A¿ = h ht (x )t = x 2 2 Shear Flow Resultant: VQ1 q1 = = I q2 = VQ2 = I P A ht2 x1 B P A ht2 x2 B h C h + 6(b1 + b2) D h C h + 6(b1 + b2) D 6P t h2 12 C h + 6(b1 + b2) D = t h2 12 C h + 6(b1 + b2) D = 6P b1 (Ff)1 = L0 q1 dx1 = 6P x1 x2 b1 h C h + 6(b1 + b2) D L0 x1 dx1 3Pb21 = b2 (Ff)2 = L0 q2 dx2 = h C h + 6(b1 + b2) D 6P b2 h C h + 6(b1 + b2) D L0 x2 dx2 3Pb22 = h C h + 6(b1 + b2) D Shear Center: Summing moment about point A Pe = A Ff B h - A Ff B h Pe = e = 3Pb22 h C h + 6(b1 + b2) D 3(b22 - b21) h + 6(b1 + b2) (h) - 3Pb21 h C h + 6(b1 + b2) D (h) Ans Note that if b2 = b1, e = (I shape) 520 b1 O 07 Solutions 46060 5/26/10 2:04 PM Page 521 © 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 *7–64 Determine the location e of the shear center, point O, for the thin-walled member having the cross section shown The member segments have the same thickness t b d 45Њ O e Section Properties: I = = t a b(2d sin 45°)3 + C bt(d sin 45°)2 D 12 sin 45° td2 (d + 3b) Q = y¿A¿ = d sin 45° (xt) = (td sin 45°)x Shear Flow Resultant: qf = P(td sin 45°)x VQ 3P sin 45° = = x td2 I d(d + 3b) (d + 3b) b Ff = L0 b qfdx = 3P sin 45° 3b sin 45° P xdx = d(d + 3b) L0 2d(d + 3b) Shear Center: Summing moments about point A, Pe = Ff(2d sin 45°) Pe = c e = 3b2 sin 45° P d(2d sin 45°) 2d(d + 3b) 3b2 2(d + 3b) Ans 521 45Њ 07 Solutions 46060 5/26/10 2:04 PM Page 522 © 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 •7–65 Determine the location e of the shear center, point O, for the thin-walled member having a slit along its side Each element has a constant thickness t a e a t a Section Properties: I = 10 (2t)(2a)3 + C at A a2 B D = a t 12 Q1 = y1œ A¿ = y t (yt) = y2 2 Q2 = ©y¿A¿ = a at (at) + a(xt) = (a + 2x) 2 Shear Flow Resultant: q1 = P A 12 y2 B VQ1 3P = 10 = y I 20a a t P C at2 (a + 2x) D VQ2 3P = = (a + 2x) q2 = 10 I 20a a t a (Fw)1 = L0 a q1 dy = a Ff = L0 3P P y2 dy = 20 20a3 L0 a q2 dx = 3P (a + 2x)dx = P 10 20a L0 Shear Center: Summing moments about point A Pe = 2(Fw)1 (a) + Ff(2a) Pe = a e = P b a + a Pb 2a 20 10 a 10 Ans 522 O 07 Solutions 46060 5/26/10 2:04 PM Page 523 © 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 7–66 Determine the location e of the shear center, point O, for the thin-walled member having the cross section shown a 60Њ O a 60Њ a e Summing moments about A Pe = F2 a I = 13 ab t 1 (t)(a)3 + a b(a)3 = t a3 12 12 sin 30° q1 = V(a)(t)(a>4) q2 = q1 + F2 = = ta V a V(a>2)(t)(a>4) ta = q1 + V 2a V 4V V (a) + a b (a) = a 2a e = 223 a Ans 523 07 Solutions 46060 5/26/10 2:04 PM Page 524 © 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 7–67 Determine the location e of the shear center, point O, for the thin-walled member having the cross section shown The member segments have the same thickness t b t h O e h b Shear Flow Resultant: The shear force flows through as Indicated by F1, F2, and F3 on FBD (b) Hence, The horizontal force equilibrium is not satisfied (©Fx Z 0) In order to satisfy this equilibrium requirement F1 and F2 must be equal to zero Shear Center: Summing moments about point A Pe = F2(0) e = Ans Also, The shear flows through the section as indicated by F1, F2, F3 + ©F Z However, : x To satisfy this equation, the section must tip so that the resultant of : : : : F1 + F2 + F3 = P Also, due to the geometry, for calculating F1 and F3, we require F1 = F3 Hence, e = Ans 524 07 Solutions 46060 5/26/10 2:04 PM Page 525 © 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 *7–68 Determine the location e of the shear center, point O, for the beam having the cross section shown The thickness is t — r e r O I = (2)c r (t)(r>2)3 + (r>2)(t)ar + b d + Isemi-circle 12 = 1.583333t r3 + Isemi-circle p>2 Isemi-circle = p>2 L-p>2 (r sin u) t r du = t r3 L-p>2 sin2 u du p Isemi-circle = t r3 a b Thus, p I = 1.583333t r3 + t r3 a b = 3.15413t r3 r r Q = a b t a + rb + Lu p>2 r sin u (t r du) Q = 0.625 t r2 + t r2 cos u q = VQ P(0.625 + cos u)t r2 = I 3.15413 t r3 Summing moments about A: p>2 Pe = L-p>2 (q r du)r p>2 Pe = e = Pr (0.625 + cos u)du 3.15413 L-p>2 r (1.9634 + 2) 3.15413 e = 1.26 r Ans 525 — r 07 Solutions 46060 5/26/10 2:04 PM Page 526 © 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 •7–69 Determine the location e of the shear center, point O, for the thin-walled member having the cross section shown The member segments have the same thickness t h1 h O e h1 b Summing moments about A Pe = F(h) + 2V(b) h 1 (t)(h3) + 2b(t)a b + (t)[h3 - (h - 2h1)3] 12 12 I = = (1) t(h - 2h1)3 bth2 th3 + 12 Q1 = y¿A¿ = t(hy - 2h1 y + y2) (h - 2h1 + y)yt = 2 VQ Pt(hy - 2h1 y + y2) = I 2I q1 = V = L h1 Pt Pt hh1 2 (hy - 2h1 y + y2)dy = c - h31 d 2I L0 2I q1 dy = Q2 = ©y¿A¿ = 1 h (h - h1)h1 t + (x)(t) = t[h1 (h - h1) + hx] 2 VQ2 Pt = (h (h - h1) + hx) I 2I q2 = b F = L q2 dx = Pt Pt hb2 [h1 (h - h1) + hx]dx = ah1 hb - h21 b + b 2I L0 2I From Eq, (1) Pe = h2b2 Pt [h1 h2b - h21 hb + + hh21 b - h31 b] 2I I = t (2h3 + 6bh2 - (h - 2h1)3) 12 e = b(6h1 h2 + 3h2b - 8h31) t (6h1 h2b + 3h2b2 - 8h1 3b) = 12I 2h3 + 6bh2 - (h - 2h1)3 526 Ans 07 Solutions 46060 5/26/10 2:04 PM Page 527 © 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 7–70 Determine the location e of the shear center, point O, for the thin-walled member having the cross section shown t r a O a e Summing moments about A Pe = r dF L dA = t ds = t r du (1) y = r sin u dI = y2 dA = r2 sin2 u(t r du) = r3 t sin2 udu p+a I = r3 t L sin2 u du = r3 t Lp - a - cos 2u du = sin 2u p + a r3 t (u ) 2 p -Η a = sin 2(p + a) sin 2(p - a) r3 t c ap + a b - ap - a bd 2 = r3 t r3 t (2a - sin a cos a) = (2a - sin 2a) 2 dQ = y dA = r sin u(t r du) = r2 t sin u du u Q = r2 t q = L u sin u du = r2 t (-cos u)| Lp-a = r2 t(-cos u - cos a) = -r2 t(cos u + cos a) p-a P(-r2t)(cos u + cos a) -2P(cos u + cos a) VQ = = r3t I r(2a - sin 2a) (2a - sin 2a) dF = L q ds = L q r du p+p L = dF = 2P r -2P (cos u + cos a) du = (2a cos a - sin a) r(2a - sin 2a) Lp - a 2a - sin 2a 4P (sin a - a cos a) 2a - sin 2a 4P (sin a - a cos a) d 2a - sin 2a 4r (sin a - a cos a) e = 2a - sin 2a From Eq (1); P e = r c Ans 527 07 Solutions 46060 5/26/10 2:04 PM Page 528 © 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 7–71 Sketch the intensity of the shear-stress distribution acting over the beam’s cross-sectional area, and determine the resultant shear force acting on the segment AB The shear acting at the section is V = 35 kip Show that INA = 872.49 in4 C V in B A in Section Properties: y = 4(8)(8) + 11(6)(2) ©yA = = 5.1053 in ©A 8(8) + 6(2) INA = in (8) A 83 B + 8(8)(5.1053 - 4)2 12 + (2) A 63 B + 2(6)(11 - 5.1053)2 12 = 872.49 in4 (Q.E.D) Q1 = y1œ A¿ = (2.55265 + 0.5y1)(5.1053 - y1)(8) = 104.25 - 4y21 Q2 = y2œ A¿ = (4.44735 + 0.5y2)(8.8947 - y2)(2) = 79.12 - y22 Shear Stress: Applying the shear formula t = tCB = VQ , It VQ1 35(103)(104.25 - 4y21) = It 872.49(8) = {522.77 - 20.06y21} psi At y1 = 0, tCB = 523 psi At y1 = -2.8947 in tCB = 355 psi tAB = VQ2 35(103)(79.12 - y22) = It 872.49(2) = {1586.88 - 20.06y22} psi At y2 = 2.8947 in tAB = 1419 psi Resultant Shear Force: For segment AB VAB = L tAB dA 0.8947 in = L2.8947 in 0.8947 in = L2.8947 in in in A 1586.88 - 20.06y22 B (2dy) A 3173.76 - 40.12y22 B dy = 9957 lb = 9.96 kip Ans 528 07 Solutions 46060 5/26/10 2:04 PM Page 529 © 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 *7–72 The beam is fabricated from four boards nailed together as shown Determine the shear force each nail along the sides C and the top D must resist if the nails are uniformly spaced at s = in The beam is subjected to a shear of V = 4.5 kip in in in 10 in A in 12 in V B Section Properties: y = 0.5(10)(1) + 2(4)(2) + 7(12)(1) © yA = = 3.50 in ©A 10(1) + 4(2) + 12(1) INA = (10) A 13 B + (10)(1)(3.50 - 0.5)2 12 + (2) A 43 B + 2(4)(3.50 - 2)2 12 + (1) A 123 B + 1(12)(7 - 3.50)2 12 = 410.5 in4 QC = y1œ A¿ = 1.5(4)(1) = 6.00 in2 QD = y2œ A¿ = 3.50(12)(1) = 42.0 in2 Shear Flow: qC = VQC 4.5(103)(6.00) = = 65.773 lb>in I 410.5 qD = VQD 4.5(103)(42.0) = = 460.41 lb>in I 410.5 Hence, the shear force resisted by each nail is FC = qC s = (65.773 lb>in.)(3 in.) = 197 lb Ans FD = qD s = (460.41 lb>in.)(3 in.) = 1.38 kip Ans 529 in 07 Solutions 46060 5/26/10 2:04 PM Page 530 © 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 •7–73 The member is subjected to a shear force of V = kN Determine the shear flow at points A, B, and C The thickness of each thin-walled segment is 15 mm 200 mm B 100 mm A C V ϭ kN Section Properties: y = = © yA ©A 0.0075(0.2)(0.015) + 0.0575(0.115)(0.03) + 0.165(0.3)(0.015) 0.2(0.015) + 0.115(0.03) + 0.3(0.015) = 0.08798 m (0.2) A 0.0153 B + 0.2(0.015)(0.08798 - 0.0075)2 12 + (0.03) A 0.1153 B + 0.03(0.115)(0.08798 - 0.0575)2 12 + (0.015) A 0.33 B + 0.015(0.3)(0.165 - 0.08798)2 12 INA = = 86.93913 A 10 - B m4 QA = ' QB = y 1œ A¿ = 0.03048(0.115)(0.015) = 52.57705 A 10 - B m3 Ans QC = ©y¿A¿ = 0.03048(0.115)(0.015) + 0.08048(0.0925)(0.015) = 0.16424 A 10 - B m3 Shear Flow: qA = VQA = I Ans qB = VQB 2(103)(52.57705)(10 - 6) = 1.21 kN>m = I 86.93913(10 - 6) Ans qC = VQC 2(103)(0.16424)(10 - 3) = 3.78 kN>m = I 86.93913(10 - 6) Ans 530 300 mm 07 Solutions 46060 5/26/10 2:04 PM Page 531 © 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 7–74 The beam is constructed from four boards glued together at their seams If the glue can withstand 75 lb>in., what is the maximum vertical shear V that the beam can support? in 0.5 in Section Properties: INA = 1 (1) A 103 B + 2c (4) A 0.53 B + 4(0.5) A 1.752 B d 12 12 in 0.5 in = 95.667 in4 V Q = y¿A¿ = 1.75(4)(0.5) = 3.50 in3 in Shear Flow: There are two glue joints in this case, hence the allowable shear flow is 2(75) = 150 lb>in q = 150 = in 0.5 in 0.5 in VQ I V(3.50) 95.667 V = 4100 lb = 4.10 kip Ans 7–75 Solve Prob 7–74 if the beam is rotated 90° from the position shown in 0.5 in in 0.5 in V in in 0.5 in Section Properties: INA = 1 (10) A 53 B (9) A 43 B = 56.167 in4 12 12 Q = y¿A¿ = 2.25(10)(0.5) = 11.25 in3 Shear Flow: There are two glue joints in this case, hence the allowable shear flow is 2(75) = 150 lb>in q = 150 = VQ I V(11.25) 56.167 V = 749 lb Ans 531 0.5 in ... (0.01)(0.05)(0.02) + (0.055)(0 .07) (0.02) y = = 0.03625 m (0.05)(0.02) + (0 .07) (0.02) I = + B 20 mm 50 mm (0.05)(0.023) + (0.05)(0.02)(0.03625 - 0.01)2 12 (0.02)(0 .073 ) + (0.02)(0 .07) (0.055 - 0.03625)2... (0.01)(0.05)(0.02) + (0.055)(0 .07) (0.02) = 0.03625 m (0.05)(0.02) + (0 .07) (0.02) I = (0.05)(0.023) + (0.05)(0.02)(0.03625 - 0.01)2 12 + 20 mm 50 mm (0.02)(0 .073 ) + (0.02)(0 .07) (0.055 - 0.03625)2 = 1.78625(10... m, Fig b, Q as a function of y is Q = ©y¿A¿ = 0.09 (0.03)(0.3) + (0 .075 + y)(0 .075 - y)(0.1) = 1.09125(10 - 3) - 0.05 y2 For 0 .075 m y … 0.105 m, t = 0.3 m Thus, t = 600 (103) C 1.65375(10 - 3)