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PROBLEM 4.1 Knowing that the couple shown acts in a vertical plane, determine the stress at (a) point A, (b) point B. SOLUTION For rectangle: I= 1 3 bh 12 For cross sectional area: I = I1 + I 2 + I 3 = 1 1 1 (2)(1.5)3 + (2)(5.5)3 + (2)(1.5)3 = 28.854 in 4 12 12 12 (a) y A = 2.75 in. σA = − My A (25)(2.75) =− I 28.854 σ A = −2.38 ksi (b) yB = 0.75 in. σB = − MyB (25)(0.75) =− I 28.854 σ B = −0.650 ksi PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 4.7 Two W4 × 13 rolled sections are welded together as shown. Knowing that for the steel alloy used σ Y = 36 ksi and σ U = 58 ksi and using a factor of safety of 3.0, determine the largest couple that can be applied when the assembly is bent about the z axis. SOLUTION Properties of W4 × 13 rolled section. (See Appendix C.) Area = 3.83 in 2 Depth = 4.16 in. I x = 11.3 in 4 For one rolled section, moment of inertia about axis a-a is I a = I x + Ad 2 = 11.3 + (3.83)(2.08) 2 = 27.87 in 4 For both sections, I z = 2 I a = 55.74 in 4 c = depth = 4.16 in. M all σU 58 = 19.333 ksi F .S . 3.0 σ I (19.333) (55.74) = all = c 4.16 σ all = = σ= Mc I M all = 259 kip ⋅ in PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 4.9 Two vertical forces are applied to a beam of the cross section shown. Determine the maximum tensile and compressive stresses in portion BC of the beam. SOLUTION A1 = π 2 r2 = π 2 (25) 2 = 981.7 mm 2 A2 = bh = (50)(25) = 1250 mm 2 y = 4r (4)(25) = = 10.610 mm 3π 3π h 25 y2 = − = − = −12.5 mm 2 2 y1 = A1 y1 + A2 y2 (981.7)(10.610) + (1250)(−12.5) = = −2.334 mm A1 + A2 981.7 + 1250 I1 = I x1 − A1 y12 = π r 4 − A1 y12 = π (25) 4 − (981.7)(10.610) 2 = 42.886 × 106 mm 4 8 8 d1 = y1 − y = 10.610 − ( −2.334) = 12.944 mm I1 = I1 + A1d12 = 42.866 × 103 + (981.7)(12.944)2 = 207.35 × 103 mm 4 1 3 1 bh = (50)(25)3 = 65.104 × 103 mm 4 12 12 d 2 = y2 − y = −12.5 − (−2.334) = 10.166 mm I2 = I 2 = I 2 + A2 d 22 = 65.104 × 103 + (1250)(10.166)2 = 194.288 × 103 mm 4 I = I1 + I 2 = 401.16 × 103 mm 4 = 401.16 × 10−9 m 4 ytop = 25 + 2.334 = 27.334 mm = 0.027334 m ybot = −25 + 2.334 = −22.666 mm = −0.022666 m M − Pa = 0 : σ top = σ bot = −Mytop M = Pa = (4 × 103 )(300 × 10−3 ) = 1200 N ⋅ m (1200)(0.027334) = −81.76 × 106 Pa 401.16 × 10−9 σ top = −81.8 MPa −Mybot (1200)(−0.022666) =− = 67.80 × 106 Pa I 401.16 × 10−9 σ bot = 67.8 MPa I =− PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 4.14 Knowing that a beam of the cross section shown is bent about a horizontal axis and that the bending moment is 50 kip ⋅ in., determine the total force acting (a) on the top flange, (b) on the shaded portion of the web. SOLUTION The stress distribution over the entire cross-section is given by the bending stress formula: σx = − My I where y is a coordinate with its origin on the neutral axis and I is the moment of inertia of the entire cross sectional area. The force on the shaded portion is calculated from this stress distribution. Over an area element dA, the force is dF = σ x dA = − My dA I The total force on the shaded area is then F = dF = − My M dA = − I I ydA = − M * * y A I where y * is the centroidal coordinate of the shaded portion and A* is its area. Calculate the moment of inertia. 1 1 (6 in.)(7 in.)3 − (4 in.)(4 in.)3 = 150.17 in 4 12 12 M = 50 kip ⋅ in I = (a) Top flange: A* = (6 in.)(1.5 in.) = 9 in 2 F = (b) Half web: 50 kip ⋅ in (9 in 2 )(2.75 in.) = 8.24 kips 150.17 in 4 A* = (2 in.)(2 in.) = 4 in 2 F = y * = 2 in. + 0.75 in. = 2.75 in. F = 8.24 kips y * = 1 in. 50 kip ⋅ in (4 in 2 )(1 in.) = 1.332 kips 150.17 in 4 F = 1.332 kips PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 4.24 A 60 N ⋅ m couple is applied to the steel bar shown. (a) Assuming that the couple is applied about the z axis as shown, determine the maximum stress and the radius of curvature of the bar. (b) Solve part a, assuming that the couple is applied about the y axis. Use E = 200 GPa. SOLUTION (a) Bending about z-axis. 1 3 1 bh = (12)(20)3 = 8 × 103 mm4 = 8 × 10−9 m 4 12 12 20 c= = 10 mm = 0.010 m 2 I= σ= Mc (60)(0.010) = = 75.0 × 106 Pa I 8 × 10−9 1 M 60 = = 37.5 × 10−3 m −1 EI (200 × 109 )(8 × 10 −9 ) ρ (b) = σ = 75.0 MPa ρ = 26.7 m Bending about y-axis. 1 3 1 bh = (20)(12)3 = 2.88 × 103 mm4 = 2.88 × 10−9 m4 12 12 12 c= = 6 mm = 0.006 m 2 Mc (60)(0.006) = = 125.0 × 106 Pa σ= I 2.88 × 10−9 I= 1 ρ = M 60 = = 104.17 × 10−3 m −1 −9 9 EI (200 × 10 )(2.88 × 10 ) σ = 125.0 MPa ρ = 9.60 m PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 4.34 A bar having the cross section shown has been formed by securely bonding brass and aluminum stock. Using the data given below, determine the largest permissible bending moment when the composite bar is bent about a horizontal axis. Aluminum Brass 70 GPa 105 GPa 100 MPa 160 MPa Modulus of elasticity Allowable stress SOLUTION Use aluminum as the reference material. For aluminum, n = 1.0 For brass, n = Eb /Ea = 105/70 = 1.5 Values of n are shown on the sketch. For the transformed section, n1 1.5 (8)(32)3 = 32.768 × 103 mm 4 b1h13 = 12 12 n 1.0 I 2 = 2 b2 H 23 − h23 = (32)(323 − 163 ) = 76.459 × 103 mm 4 12 12 I 3 = I1 = 32.768 × 103 mm 4 I1 = ( ) I = I1 + I 2 + I 3 = 141.995 × 103 mm 4 = 141.995 × 10−9 m 4 |σ | = Aluminum: M= σI ny n = 1.0, | y | = 16 mm = 0.016 m, σ = 100 × 106 Pa M= Brass: nMy I (100 × 106 )(141.995 × 10−9 ) = 887.47 N ⋅ m (1.0)(0.016) n = 1.5, | y | = 16 mm = 0.016 m, σ = 160 × 106 Pa M= Choose the smaller value. (160 × 106 )(141.995 × 10−9 ) = 946.63 N ⋅ m (1.5)(0.016) M = 887 N ⋅ m PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 4.41 The 6 × 12-in. timber beam has been strengthened by bolting to it the steel reinforcement shown. The modulus of elasticity for wood is 1.8 × 106 psi and for steel, 29 × 106 psi. Knowing that the beam is bent about a horizontal axis by a couple of moment M = 450 kip ⋅ in., determine the maximum stress in (a) the wood, (b) the steel. SOLUTION Use wood as the reference material. For wood, n =1 For steel, n = Es / Ew = 29 /1.8 = 16.1111 = wood Transformed section: 421.931 112.278 = 3.758 in. Yo = = steel A, in 2 nA, in 2 72 72 2.5 40.278 yo 6 −0.25 112.278 nA yo , in 3 432 −10.069 421.931 The neutral axis lies 3.758 in. above the wood-steel interface. n1 1 b1h13 + n1 A1d12 = (6)(12)3 + (72)(6 − 3.758)2 = 1225.91 in 4 12 12 n2 16.1111 I 2 = b2 h23 + n2 A2 d 22 = (5) (0.5)3 + (40.278)(3.578 + 0.25)2 = 647.87 in 4 12 12 I = I1 + I 2 = 1873.77 in 4 I1 = σ =− M = 450 kip ⋅ in (a) Wood: n = 1, y = 12 − 3.758 = 8.242 in σw = − (b) Steel: (1) (450) (8.242) = −1.979 ksi 1873.77 n = 16.1111, σs = − nMy I σ w = −1.979 ksi y = −3.758 − 0.5 = −4.258 in (16.1111) (450) (−4.258) = 16.48 ksi 1873.77 σ s = 16.48 ksi PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 4.42 The 6 × 12-in. timber beam has been strengthened by bolting to it the steel reinforcement shown. The modulus of elasticity for wood is 1.8 × 106 psi and for steel, 29 × 106 psi. Knowing that the beam is bent about a horizontal axis by a couple of moment M = 450 kip ⋅ in., determine the maximum stress in (a) the wood, (b) the steel. SOLUTION Use wood as the reference material. For wood, n =1 For steel, n= Es 29 × 106 = = 16.1111 Ew 1.8 × 106 For C8 × 11.5 channel section, A = 3.38 in 2 , t w = 0.220 in., x = 0.571 in., I y = 1.32 in 4 For the composite section, the centroid of the channel (part 1) lies 0.571 in. above the bottom of the section. The centroid of the wood (part 2) lies 0.220 + 6.00 = 6.22 in. above the bottom. Transformed section: A, in2 3.38 Part 1 72 2 Σ Y0 = nA, in2 54.456 y , in. 0.571 nAy , in 3 31.091 d, in. 3.216 72 6.22 447.84 2.433 478.93 126.456 478.93 in 3 = 3.787 in. 126.456 in 2 d = y0 − Y0 The neutral axis lies 3.787 in. above the bottom of the section. I1 = n1 I1 + n1 A1d12 = (16.1111)(1.32) + (54.456)(3.216)2 = 584.49 in 4 n2 1 b2 h23 + n2 A2 d 22 = (6)(12)3 + (72)(2.433)2 = 1290.20 in 4 12 12 4 I = I1 + I 2 = 1874.69 in I2 = M = 450 kip ⋅ in (a) Wood: n = 1, σw = − (b) Steel: n My I y = 12 + 0.220 − 3.787 = 8.433 in. σ =− (1)(450)(8.433) = −2.02 ksi 1874.69 n = 16.1111, σs = − σ w = −2.02 ksi y = −3.787 in. (16.1111) (450) (−3.787) = 14.65 ksi 1874.67 σ s = 14.65 ksi PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 4.51 A concrete beam is reinforced by three steel rods placed as shown. The modulus of elasticity is 3 × 106 psi for the concrete and 29 × 106 psi for the steel. Using an allowable stress of 1350 psi for the concrete and 20 ksi for the steel, determine the largest allowable positive bending moment in the beam. SOLUTION n= Es 29 × 10 6 = = 9.67 Ec 3 × 106 As = 3 π 4 d 2 = (3) π 4 7 8 2 = 1.8040 in 2 nAs = 17.438 in 2 x − (17.438)(14 − x) = 0 2 4 x 2 + 17.438 x − 244.14 = 0 Locate the neutral axis: 8x Solve for x. x= −17.438 + 17.4382 + (4)(4)(244.14) = 5.6326 in. (2)(4) 14 − x = 8.3674 in. I = 1 3 1 8x + nAs (14 − x)2 = (8)(5.6326)3 + (17.438)(8.3674) 2 = 1697.45 in 4 3 3 σ = nMy I ∴ M= σI ny n = 1.0, Concrete: M = M = Choose the smaller value. σ = 1350 psi (1350)(1697.45) = 406.835 × 103 lb ⋅ in = 407 kip ⋅ in (1.0)(5.6326) n = 9.67, Steel: y = 5.6326 in., y = 8.3674 in., σ = 20 × 103 psi (20 × 103 )(1697.45) = 419.72 lb ⋅ in = 420 kip ⋅ in (9.67)(8.3674) M = 407 kip ⋅ in M = 33.9 kip ⋅ ft PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 4.52 Knowing that the bending moment in the reinforced concrete beam is +100 kip ⋅ ft and that the modulus of elasticity is 3.625 × 106 psi for the concrete and 29 × 106 psi for the steel, determine (a) the stress in the steel, (b) the maximum stress in the concrete. SOLUTION n= Es 29 × 106 = = 8.0 Ec 3.625 × 106 As = (4) π 4 (1) 2 = 3.1416 in 2 nAs = 25.133 in 2 Locate the neutral axis. (24)(4)( x + 2) + (12 x) 96 x + 192 + 6 x 2 − 339.3 + 25.133x = 0 x= Solve for x. x − (25.133)(17.5 − 4 − x) = 0 2 or 6 x 2 + 121.133x − 147.3 = 0 −121.133 + (121.133) 2 + (4)(6)(147.3) = 1.150 in. (2)(6) d3 = 17.5 − 4 − x = 12.350 in. 1 b1h13 + 12 1 I 2 = b2 x3 = 3 I1 = A1d12 = 1 (24)(4)3 + (24)(4)(3.150) 2 = 1080.6 in 4 12 1 (12)(1.150)3 = 6.1 in 4 3 I 3 = nA3d32 = (25.133)(12.350) 2 = 3833.3 in 4 I = I1 + I 2 + I 3 = 4920 in 4 σ =− (a) Steel: nMy I n = 8.0 y = −12.350 in. σs = − (b) Concrete: n = 1.0, σc = − where M = 100 kip ⋅ ft = 1200 kip ⋅ in. (8.0)(1200)(−12.350) 4920 σ s = 24.1 ksi y = 4 + 1.150 = 5.150 in. (1.0)(1200)(5.150) 4920 σ c = −1.256 ksi PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 4.55 Five metal strips, each 40 mm wide, are bonded together to form the composite beam shown. The modulus of elasticity is 210 GPa for the steel, 105 GPa for the brass, and 70 GPa for the aluminum. Knowing that the beam is bent about a horizontal axis by a couple of moment 1800 N ⋅ m, determine (a) the maximum stress in each of the three metals, (b) the radius of curvature of the composite beam. SOLUTION Use aluminum as the reference material. n = 1 in aluminum. n = Es / Ea = 210 / 70 = 3 in steel. n = Eb / Ea = 105 / 70 = 1.5 in brass. Due to symmetry of both the material arrangement and the geometry, the neutral axis passes through the center of the steel portion. For the transformed section, n1 1 b1h13 + n1 A1d12 = (40)(10)3 + (40)(10)(25) 2 = 253.33 × 103 mm 4 12 12 n2 1.5 3 2 (40)(10)3 + (1.5)(40)(10)(15) 2 = 140 × 103 mm 4 I 2 = b2 h2 + n2 A2 d 2 = 12 12 n 3.0 (40)(20)3 = 80 × 103 mm 4 I 3 = 3 b3 h33 = 12 12 I 4 = I 2 = 140 × 103 mm 4 I 5 = I1 = 253.33 × 103 mm 4 I1 = I= (a) = 866.66 × 103 mm 4 = 866.66 × 10−9 m 4 nMy where M = 1800 N ⋅ m I n = 1.0 y = −30 mm = 0.030 m Aluminum: σ =− σa = (1.0)(1800)(0.030) = 62.3 × 106 Pa −9 866.66 × 10 Brass: σb = Steel: σs = (b) I n = 1.5 y = −20 mm = −0.020 m (1.5)(1800)(0.020) = 62.3 × 106 Pa −9 866.66 × 10 n = 3.0 σ b = 62.3 MPa y = −10 mm = −0.010 m (3.0)(1800)(0.010) = 62.3 × 106 Pa −9 866.66 × 10 Radius of curvature. σ a = 62.3 MPa 1 ρ = M 1800 = = 0.02967 m −1 9 Ea I (70 × 10 )(866.66 × 10−9 ) σ s = 62.3 MPa ρ = 33.7 m PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 4.57 The composite beam shown is formed by bonding together a brass rod and an aluminum rod of semicircular cross sections. The modulus of elasticity is 15 × 106 psi for the brass and 10 × 106 psi for the aluminum. Knowing that the composite beam is bent about a horizontal axis by couples of moment 8 kip ⋅ in., determine the maximum stress (a) in the brass, (b) in the aluminum. SOLUTION For each semicircle, yo = r = 0.8 in. π A= 2 4r (4)(0.8) = = 0.33953 in. 3π 3π r 2 = 1.00531 in 2 , I base = π 8 r 4 = 0.160850 in 4 I = I base − Ayo2 = 0.160850 − (1.00531)(0.33953)2 = 0.044953 in 4 Use aluminum as the reference material. n = 1.0 in aluminum n= Eb 15 × 106 = = 1.5 in brass Ea 10 × 106 Locate the neutral axis. A, in2 nA, in2 yo , in. 1.00531 1.50796 0.33953 0.51200 1.00531 1.00531 −0.33953 −0.34133 Σ 2.51327 nAyo , in 3 Yo = 0.17067 = 0.06791 in. 2.51327 The neutral axis lies 0.06791 in. above the material interface. 0.17067 d1 = 0.33953 − 0.06791 = 0.27162 in., d 2 = 0.33953 + 0.06791 = 0.40744 in. I1 = n1I + n1 Ad12 = (1.5)(0.044957) + (1.5)(1.00531)(0.27162)2 = 0.17869 in 4 I 2 = n2 I + n2 Ad 22 = (1.0)(0.044957) + (1.0)(1.00531)(0.40744)2 = 0.21185 in 4 I = I1 + I 2 = 0.39054 in 4 (a) Brass: n = 1.5, y = 0.8 − 0.06791 = 0.73209 in. σ =− (b) Aluminium: n = 1.0, nMy (1.5)(8)(0.73209) =− I 0.39054 σ = −22.5 ksi y = −0.8 − 0.06791 = −0.86791 in. σ =− nMy (1.0)(8)(−0.86791) =− I 0.39054 σ = 17.78 ksi PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.