FM_TOC 46060 6/22/10 11:26 AM Page iii CONTENTS To the Instructor iv Stress Strain 73 Mechanical Properties of Materials 92 Axial Load 122 Torsion 214 Bending 329 Transverse Shear 472 Combined Loadings 532 Stress Transformation 619 10 Strain Transformation 738 11 Design of Beams and Shafts 830 12 Deflection of Beams and Shafts 883 13 Buckling of Columns 1038 14 Energy Methods 1159 06 Solutions 46060_Part1 5/27/10 3:51 PM Page 329 © 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 6–1 Draw the shear and moment diagrams for the shaft The bearings at A and B exert only vertical reactions on the shaft B A 800 mm 250 mm 24 kN 6–2 Draw the shear and moment diagrams for the simply supported beam kN M kNm A B 2m 329 2m 2m 06 Solutions 46060_Part1 5/27/10 3:51 PM Page 330 © 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 6–3 The engine crane is used to support the engine, which has a weight of 1200 lb Draw the shear and moment diagrams of the boom ABC when it is in the horizontal position shown a + ©MA = 0; F (3) - 1200(8) = 0; A + c ©Fy = 0; - Ay + + ©F = 0; ; x Ax - (4000) - 1200 = 0; (4000) = 0; A ft ft B FA = 4000 lb ft Ay = 2000 lb Ax = 2400 lb *6–4 Draw the shear and moment diagrams for the cantilever beam kN/m A kNm 2m The free-body diagram of the beam’s right segment sectioned through an arbitrary point shown in Fig a will be used to write the shear and moment equations of the beam + c ©Fy = 0; C V - 2(2 - x) = V = {4 - 2x} kN‚ (1) a + ©M = 0; - M - 2(2 - x) c (2 - x) d - = M = {-x2 + 4x - 10}kN # m‚(2) The shear and moment diagrams shown in Figs b and c are plotted using Eqs (1) and (2), respectively The value of the shear and moment at x = is evaluated using Eqs (1) and (2) Vx = = - 2(0) = kN Mx = = C - + 4(0) - 10 D = - 10kN # m 330 06 Solutions 46060_Part1 5/27/10 3:51 PM Page 331 © 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 6–5 Draw the shear and moment diagrams for the beam 10 kN kN 15 kNm 2m 3m 6–6 Draw the shear and moment diagrams for the overhang beam kN/m C A B 2m 4m 6–7 Draw the shear and moment diagrams for the compound beam which is pin connected at B kip kip A C B ft 331 ft ft ft 06 Solutions 46060_Part1 5/27/10 3:51 PM Page 332 © 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 *6–8 Draw the shear and moment diagrams for the simply supported beam 150 lb/ft 300 lbft A B 12 ft The free-body diagram of the beam’s left segment sectioned through an arbitrary point shown in Fig b will be used to write the shear and moment equations The intensity of the triangular distributed load at the point of sectioning is w = 150 a x b = 12.5x 12 Referring to Fig b, + c ©Fy = 0; a + ©M = 0; M + 275 - (12.5x)(x) - V = V = {275 - 6.25x2}lb‚ (1) x (12.5x)(x)a b - 275x = M = {275x - 2.083x3}lb # ft‚(2) The shear and moment diagrams shown in Figs c and d are plotted using Eqs (1) and (2), respectively The location where the shear is equal to zero can be obtained by setting V = in Eq (1) = 275 - 6.25x2 x = 6.633 ft The value of the moment at x = 6.633 ft (V = 0) is evaluated using Eq (2) M x = 6.633 ft = 275(6.633) - 2.083(6.633)3 = 1216 lb # ft 332 06 Solutions 46060_Part1 5/27/10 3:51 PM Page 333 © 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 6–9 Draw the shear and moment diagrams for the beam Hint: The 20-kip load must be replaced by equivalent loadings at point C on the axis of the beam 15 kip ft A C ft 20 kip B ft ft 6–10 Members ABC and BD of the counter chair are rigidly connected at B and the smooth collar at D is allowed to move freely along the vertical slot Draw the shear and moment diagrams for member ABC Equations of Equilibrium: Referring to the free-body diagram of the frame shown in Fig a, + c ©Fy = 0; P 150 lb Ay - 150 = C A Ay = 150 lb a + ©MA = 0; B 1.5 ft 1.5 ft ND(1.5) - 150(3) = D ND = 300 lb Shear and Moment Diagram: The couple moment acting on B due to ND is MB = 300(1.5) = 450 lb # ft The loading acting on member ABC is shown in Fig b and the shear and moment diagrams are shown in Figs c and d 333 1.5 ft 06 Solutions 46060_Part1 5/27/10 3:51 PM Page 334 © 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 6–11 The overhanging beam has been fabricated with a projected arm BD on it Draw the shear and moment diagrams for the beam ABC if it supports a load of 800 lb Hint: The loading in the supporting strut DE must be replaced by equivalent loads at point B on the axis of the beam E 800 lb B Support Reactions: a + ©MC = 0; ft D ft C A 800(10) - FDE(4) - FDE(2) = 5 ft ft FDE = 2000 lb + c ©Fy = 0; - 800 + + ©F = 0; : x - Cx + (2000) - Cy = (2000) = Cy = 400 lb Cx = 1600 lb Shear and Moment Diagram: *6–12 A reinforced concrete pier is used to support the stringers for a bridge deck Draw the shear and moment diagrams for the pier when it is subjected to the stringer loads shown Assume the columns at A and B exert only vertical reactions on the pier 60 kN 60 kN 35 kN 35 kN 35 kN m m 1.5 m 1.5 m m m A 334 B 06 Solutions 46060_Part1 5/27/10 3:51 PM Page 335 © 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 6–13 Draw the shear and moment diagrams for the compound beam It is supported by a smooth plate at A which slides within the groove and so it cannot support a vertical force, although it can support a moment and axial load P Support Reactions: P A D B C From the FBD of segment BD a + ©MC = 0; + c ©Fy = 0; + ©F = 0; : x By (a) - P(a) = Cy - P - P = By = P a a a a Cy = 2P Bx = From the FBD of segment AB a + ©MA = 0; + c ©Fy = 0; P(2a) - P(a) - MA = MA = Pa P - P = (equilibrium is statisfied!) 6–14 The industrial robot is held in the stationary position shown Draw the shear and moment diagrams of the arm ABC if it is pin connected at A and connected to a hydraulic cylinder (two-force member) BD Assume the arm and grip have a uniform weight of 1.5 lbin and support the load of 40 lb at C in A 10 in B 50 in 120 D 335 C 06 Solutions 46060_Part1 5/27/10 3:51 PM Page 336 © 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 *6–16 Draw the shear and moment diagrams for the shaft and determine the shear and moment throughout the shaft as a function of x The bearings at A and B exert only vertical reactions on the shaft 500 lb 800 lb A B x ft For x ft + c ©Fy = 220 - V = a + ©MNA = V = 220 lb‚ Ans M - 220x = M = (220x) lb ft‚ Ans For ft x ft + c ©Fy = 0; 220 - 800 - V = V = - 580 lb a + ©MNA = 0; Ans M + 800(x - 3) - 220x = M = {- 580x + 2400} lb ft‚ Ans For ft x … ft + c ©Fy = 0; a + ©MNA = 0; V - 500 = V = 500 lb‚ Ans - M - 500(5.5 - x) - 250 = M = (500x - 3000) lb ft Ans 336 ft 0.5 ft 0.5 ft 06 Solutions 46060_Part1 5/27/10 3:51 PM Page 337 © 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 •6–17 Draw the shear and moment diagrams for the cantilevered beam 300 lb 200 lb/ft A ft The free-body diagram of the beam’s left segment sectioned through an arbitrary point shown in Fig b will be used to write the shear and moment equations The intensity of the triangular distributed load at the point of sectioning is x w = 200 a b = 33.33x Referring to Fig b, + c ©Fy = 0; - 300 - a + ©M = 0; M + (33.33x)(x) - V = V = {- 300 - 16.67x2} lb (1) x (33.33x)(x)a b + 300x = M = {-300x - 5.556x3} lb # ft (2) The shear and moment diagrams shown in Figs c and d are plotted using Eqs (1) and (2), respectively 337 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 45⬚ 45⬚ 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] a 2.5 - 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) qmax = a b = 641 lb>in 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 ] 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 + c (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 45⬚ 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 45⬚ 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 Pb2a 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 60⬚ O a 60⬚ 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) a r + 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 bt 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 - 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 Lp-a sin u du = r2 t (- cos u)| = 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 35(103)(104.25 - 4y21) VQ1 = 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 ⫽ 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