CHAPTER 4 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 (b) yB = 0.75 in. σB = − MyB (25)(0.75) =− I 28.854 σ A = −2.38 ksi 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 4.2 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 Outside rectangle: I1 = 1 (80)(120)3 12 I1 = 11.52 × 106 mm 4 = 11.52 × 10−6 m 4 Cutout: I2 = 1 (40)(80)3 12 I 2 = 1.70667 × 106 mm 4 = 1.70667 × 10−6 m 4 Section: (a) I = I1 − I 2 = 9.81333 × 10−6 m 4 y A = 40 mm = 0.040 m σA = − My A (15 × 103 )(0.040) =− = −61.6 × 106 Pa I 9.81333 × 10−6 σ A = −61.6 MPa (b) yB = −60 mm = −0.060 m σB = − MyB (15 × 103 )(−0.060) =− = 91.7 × 106 Pa −6 I 9.81333 × 10 σ B = 91.7 MPa 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 4.3 Using an allowable stress of 16 ksi, determine the largest couple that can be applied to each pipe. SOLUTION (a) I = π (r 4 4 o ) − ri4 = π 4 (0.64 − 0.54 ) = 52.7 × 10−3 in 4 c = 0.6 in. σ = Mc : I M = σI c = (16)(52.7 × 10−3 ) 0.6 M = 1.405 kip ⋅ in (b) π (0.7 4 − 0.54 ) = 139.49 × 10−3 in 4 4 c = 0.7 in. I = σ = Mc : I M = σI c = (16)(139.49 × 10−3 ) 0.7 M = 3.19 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 4.4 A nylon spacing bar has the cross section shown. Knowing that the allowable stress for the grade of nylon used is 24 MPa, determine the largest couple Mz that can be applied to the bar. SOLUTION I = I rect − I circle = = 1 3 π 4 bh − r 12 4 1 π (100)(80)3 − (25) 4 = 3.9599 × 106 mm 4 12 4 = 3.9599 × 10−6 m c= 80 = 40 mm = 0.040 m 2 σ = Mc : I Mz = σI c = (24 × 106 )(3.9599 × 10−6 ) = 2.38 × 103 N ⋅ m 0.040 M z = 2.38 kN ⋅ 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 4.5 A beam of the cross section shown is extruded from an aluminum alloy for which σ Y = 250 MPa and σ U = 450 MPa. Using a factor of safety of 3.00, determine the largest couple that can be applied to the beam when it is bent about the z-axis. SOLUTION Allowable stress. = σU F.S . = 450 = 150 MPa 3 = 150 × 106 Pa Moment of inertia about z-axis. 1 (16)(80)3 = 682.67 × 103 mm 4 12 1 I 2 = (16)(32)3 = 43.69 × 103 mm 4 12 I 3 = I1 = 682.67 × 103 mm 4 I1 = I = I1 + I 2 + I 3 = 1.40902 × 106 mm 4 = 1.40902 × 10−6 m 4 1 Mc with c = (80) = 40 mm = 0.040 m 2 I I σ (1.40902 × 10−6 )(150 × 106 ) = = 5.28 × 103 N ⋅ m M= 0.040 c σ= M = 5.28 kN ⋅ 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 4.6 Solve Prob. 4.5, assuming that the beam is bent about the y-axis. PROBLEM 4.5 A beam of the cross section shown is extruded from an aluminum alloy for which σ Y = 250 MPa and σ U = 450 MPa. Using a factor of safety of 3.00, determine the largest couple that can be applied to the beam when it is bent about the z-axis. SOLUTION = Allowable stress: σU F.S . = 450 = 150 MPa 3.00 = 150 × 106 Pa Moment of inertia about y-axis. 1 (80)(16)3 + (80)(16)(16) 2 = 354.987 × 103 mm 4 12 1 I 2 = (32)(16)3 = 10.923 × 103 mm 4 12 I 3 = I1 = 354.987 × 103 mm 4 I1 = I = I1 + I 2 + I 3 = 720.9 × 103 mm 4 = 720.9 × 10−9 m 4 1 Mc with c = (48) = 24 mm = 0.024 m 2 I I σ (720.9 × 10−9 )(150 × 106 ) = = 4.51 × 103 N ⋅ m M= 0.024 c σ= M = 4.51 kN ⋅ 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 4.7 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 4.8 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 Width = 4.060 in. I y = 3.86 in 4 For one rolled section, moment of inertia about axis b-b is I b = I y + Ad 2 = 3.86 + (3.83)(2.030) 2 = 19.643 in 4 For both sections, I z = 2 I b = 39.286 in 4 c = width = 4.060 in. σU 58 = = 19.333 ksi F .S . 3.0 σ I (19.333) (39.286) = all = 4.060 c σ all = M all σ= Mc I M all = 187.1 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 4.10 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 A, mm 2 y0 , mm A y0 , mm3 600 30 18 × 103 600 30 18 × 103 300 5 1.5 × 103 1500 Y0 = 37.5 × 103 37.5 × 103 = 25 mm 1500 Neutral axis lies 25 mm above the base. 1 (10)(60)3 + (600)(5)2 = 195 × 103 mm 4 I 2 = I1 = 195 mm 4 12 1 I 3 = (30)(10)3 + (300)(20) 2 = 122.5 × 103 mm 4 12 I = I1 + I 2 + I 3 = 512.5 × 103 mm 4 = 512.5 × 10−9 m 4 I1 = ytop = 35 mm = 0.035 m ybot = −25 mm = −0.025 m a = 150 mm = 0.150 m P = 10 × 103 N M = Pa = (10 × 103 )(0.150) = 1.5 ×103 N ⋅ m σ top = − σ bot = − M ytop I =− (1.5 × 103 )(0.035) = −102.4 × 106 Pa 512.5 × 10−9 M ybot (1.5 × 103 )(−0.025) =− = 73.2 × 106 Pa I 512.5 × 10−9 σ top = −102.4 MPa (compression) σ bot = 73.2 MPa (tension) 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 4.11 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 A y0 A y0 8 7.5 60 6 4 24 4 0.5 Σ 18 Yo = 2 86 86 = 4.778 in. 18 Neutral axis lies 4.778 in. above the base. 1 1 b1h13 + A1d12 = (8)(1)3 + (8)(2.772)2 = 59.94 in 4 12 12 1 1 I 2 = b2 h23 + A2 d 22 = (1)(6)3 + (6)(0.778)2 = 21.63 in 4 12 12 1 1 3 2 I 3 = b3 h3 + A3 d3 = (4)(1)3 + (4)(4.278) 2 = 73.54 in 4 12 12 I = I1 + I 2 + I 3 = 59.94 + 21.63 + 73.57 = 155.16 in 4 ytop = 3.222 in. ybot = −4.778 in. I1 = M − Pa = 0 M = Pa = (25)(20) = 500 kip ⋅ in σ top = − σ bot = − Mytop I =− (500)(3.222) 155.16 Mybot (500)(−4.778) =− I 155.16 σ top = −10.38 ksi (compression) σ bot = 15.40 ksi (tension) 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 4.12 Knowing that a beam of the cross section shown is bent about a horizontal axis and that the bending moment is 6 kN ⋅ m, determine the total force acting on the top flange. 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. d1 = 54 − 18 = 36 mm d 2 = 54 + 36 − 54 = 36 mm 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 4.12 (Continued) Moment of inertia of entire cross section: 1 1 b1h13 + A1d12 = (216)(36)3 + (216)(36)(36)2 = 10.9175 × 106 mm4 12 12 1 1 3 2 I 2 = b2 h2 + A2 d 2 = (72)(108)3 + (72)(108)(36)2 = 17.6360 × 106 mm 4 12 12 I = I1 + I 2 = 28.5535 × 106 mm4 = 28.5535 × 10−6 m 4 I1 = For the shaded area, A* = (216)(36) = 7776 mm 2 y * = 36 mm A* y * = 279.936 × 103 mm3 = 279.936 × 10−6 m3 F =− MA* y * (6 × 103 )(279.936 × 10−6 ) = I 28.5535 × 10−6 = 58.8 × 103 N F = 58.8 kN 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 4.13 Knowing that a beam of the cross section shown is bent about a horizontal axis and that the bending moment is 6 kN ⋅ m, determine the total force acting 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. d1 = 54 − 18 = 36 mm d 2 = 54 + 36 − 54 = 36 mm 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 4.13 (Continued) Moment of inertia of entire cross section: 1 1 b1h13 + A1d12 = (216)(36)3 + (216)(36)(36)2 = 10.9175 × 106 mm 4 12 12 1 1 I 2 = b2 h23 + A2 d 22 = (72)(108)3 + (72)(108)(36) 2 = 17.6360 × 106 mm 4 12 12 I = I1 + I 2 = 28.5535 × 106 mm 4 = 28.5535 × 10−6 m 4 I1 = For the shaded area, A* = (72)(90) = 6480 mm 2 y * = 45 mm A* y * = 291.6 × 103 mm3 = 291.6 × 10−6 m F= MA* y * (6 × 103 )(291.6 × 10−6 ) = I 28.5535 × 10−6 = 61.3 × 103 N F = 61.3 kN 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 4.15 The beam shown is made of a nylon for which the allowable stress is 24 MPa in tension and 30 MPa in compression. Determine the largest couple M that can be applied to the beam. SOLUTION A, mm 2 y0 , mm A y0 , mm3 600 22.5 13.5 × 103 300 7.5 2.25 × 103 Σ 900 Y0 = 15.75 × 103 15.5 × 103 = 17.5 mm 900 The neutral axis lies 17.5 mm above the bottom. ytop = 30 − 17.5 = 12.5 mm = 0.0125 m ybot = −17.5 mm = −0.0175 m 1 1 b1h13 + A1d12 = (40)(15)3 + (600)(5)2 = 26.25 × 103 mm 4 12 12 1 1 I 2 = b2 h23 + A2 d 22 = (20)(15)3 + (300)(10)2 = 35.625 × 103 mm 4 12 12 I = I1 + I 2 = 61.875 × 103 mm 4 = 61.875 × 10−9 m 4 I1 = |σ | = My I M= σI y Top: (tension side) M= (24 × 106 )(61.875 × 10−9 ) = 118.8 N ⋅ m 0.0125 Bottom: (compression) M= (30 × 106 )(61.875 × 10−9 ) = 106.1 N ⋅ m 0.0175 Choose smaller value. M = 106.1 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 4.16 Solve Prob. 4.15, assuming that d = 40 mm. PROBLEM 4.15 The beam shown is made of a nylon for which the allowable stress is 24 MPa in tension and 30 MPa in compression. Determine the largest couple M that can be applied to the beam. SOLUTION A, mm 2 y0 , mm A y0 , mm3 600 32.5 19.5 × 103 500 12.5 6.25 × 103 Σ 1100 Y0 = 25.75 × 103 25.75 × 103 = 23.41 mm 1100 The neutral axis lies 23.41 mm above the bottom. ytop = 40 − 23.41 = 16.59 mm = 0.01659 m ybot = −23.41 mm = −0.02341 m 1 1 b1h13 + A1d12 = (40)(15)3 + (600)(9.09)2 = 60.827 × 103 mm 4 12 12 1 1 I 2 = b2 h22 + A2 d 22 = (20)(25)3 + (500)(10.91)2 = 85.556 × 103 mm 4 12 12 I = I1 + I 2 = 146.383 × 103 mm 4 = 146.383 × 10−9 m 4 I1 = |σ | = My I M= σI y Top: (tension side) M= (24 × 106 )(146.383 × 10−9 ) = 212 N ⋅ m 0.01659 Bottom: (compression) M= (30 × 106 )(146.383 × 10−9 ) = 187.6 N ⋅ m 0.02341 Choose smaller value. M = 187.6 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 4.17 Knowing that for the extruded beam shown the allowable stress is 12 ksi in tension and 16 ksi in compression, determine the largest couple M that can be applied. SOLUTION A y0 2.25 1.25 2.8125 2.25 0.25 0.5625 4.50 Y= A y0 3.375 3.375 = 0.75 in. 4.50 The neutral axis lies 0.75 in. above bottom. ytop = 2.0 − 0.75 = 1.25 in., ybot = −0.75 in. 1 1 b1h13 + A1d12 = (1.5)(1.5)3 + (2.25)(0.5) 2 = 0.984375 in 4 12 12 1 1 I 2 = b2 h22 + A2 d 22 = (4.5)(0.5)3 + (2.25)(0.5)2 = 0.609375 in 4 12 12 I = I1 + I 2 = 1.59375 in 4 I1 = σ = My I M = σI y Top: (compression) M = (16)(1.59375) = 20.4 kip ⋅ in 1.25 Bottom: (tension) M = (12)(1.59375) = 25.5 kip ⋅ in 0.75 Choose the smaller as Mall. M all = 20.4 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 4.18 Knowing that for the casting shown the allowable stress is 5 ksi in tension and 18 ksi in compression, determine the largest couple M that can be applied. SOLUTION Locate the neutral axis and compute the moment of inertia. Y = Ai yi Ai I = 1 bi hi3 for rectangle 12 I = ( Ai di2 + I ) di = yi − Y I , in 4 Part A, in 2 yi , in. Ai yi , in 3 1 1.5 1.25 1.875 0.3333 0.1667 0.03125 2 1.0 0.75 0.75 0.1667 0.0277 0.02083 3 0.5 0.25 0.125 0.6667 0.2222 0.01042 Σ 3.0 0.4166 0.0625 Y = Ay 2.75 = = 0.9167 in. A 3.0 Ai di2 , in 4 di , in. 2.75 I = ( I + Ad 2 ) = 0.4166 + 0.0625 = 0.479 in 4 Allowable bending moment. σ = Tension at A: Mc I or M = σI c σ A ≤ 5 ksi c A = 0.583 in. M ≤ Compression at B: (5)(0.479) = 4.11 kip ⋅ in 0.583 σ B ≤ 18 ksi M ≤ The smaller value is the allowable value of M. cB = 0.917 in. (18)(0.479) = 9.40 kip ⋅ in 0.917 M = 4.11 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 4.19 Knowing that for the extruded beam shown the allowable stress is 120 MPa in tension and 150 MPa in compression, determine the largest couple M that can be applied. SOLUTION rectangle circular cutout A1 = (150)(250) = 37.5 × 103 mm 2 A2 = −π (50) 2 = − 7.85398 × 103 mm 2 A = A1 + A2 = 29.64602 × 103 mm 2 y1 = 0 mm y2 = −50 mm Y = ΣA y ΣA (37.5 × 103 )(0) + (−7.85393 × 103 )(−50) 29.64602 × 103 = 13.2463 mm Y = I X ′ = Σ( I + Ad 2 ) = I1 − I 2 1 = (150)(250)3 + (37.5 × 103 )(13.2463) 2 12 π − (50) 4 + (7.85398 × 103 )(50 + 13.2463) 2 4 = 201.892 × 106 − 36.3254 × 106 = 165.567 × 106 mm 4 = 165.567 × 10−6 m 4 Top: (tension side) c = 125 − 13.2463 = 111.7537 mm = 0.11175 m σ= Bottom: (compression side) M= I σ (165.567 × 10−6 )(120 × 106 ) = c 0.11175 = 177.79 × 103 N ⋅ m c = 125 + 13.2463 = 138.2463 mm = 0.13825 m σ= Choose the smaller. Mc I Mc I M= I σ (165.567 × 10−6 )(150 × 106 ) = c 0.13825 = 179.64 × 103 N ⋅ m M = 177.8 × 103 N ⋅ m M = 177.8 kN ⋅ 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 4.20 Knowing that for the extruded beam shown the allowable stress is 120 MPa in tension and 150 MPa in compression, determine the largest couple M that can be applied. SOLUTION A, mm 2 y0 , mm A y0 , mm3 d, mm 2160 27 58320 3 1080 36 38880 3 Σ 3240 Y = 97200 = 30 mm 3240 97200 The neutral axis lies 30 mm above the bottom. ytop = 54 − 30 = 24 mm = 0.024 m ybot = −30 mm = −0.030 m 1 1 b1h13 + A1d12 = (40)(54)3 + (40)(54)(3)2 = 544.32 × 103 mm 4 12 12 1 1 1 I 2 = b2 h22 + A2 d 22 = (40)(54)3 + (40)(54)(6)2 = 213.84 × 103 mm 4 36 36 2 3 4 I = I1 + I 2 = 758.16 × 10 mm = 758.16 × 10−9 m 4 I1 = |σ | = My I |M | = σI y Top: (tension side) M= (120 × 106 )(758.16 × 10−9 ) = 3.7908 × 103 N ⋅ m 0.024 Bottom: (compression) M= (150 × 106 )(758.16 × 10−9 ) = 3.7908 × 103 N ⋅ m 0.030 Choose the smaller as Mall. M all = 3.7908 × 103 N ⋅ m M all = 3.79 kN ⋅ 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 4.21 A steel band saw blade, that was originally straight, passes over 8-in.-diameter pulleys when mounted on a band saw. Determine the maximum stress in the blade, knowing that it is 0.018 in. thick and 0.625 in. wide. Use E = 29 × 106 psi. SOLUTION Band blade thickness: t = 0.018 in. Radius of pulley: r= 1 d = 4.000 in. 2 Radius of curvature of centerline of blade: 1 2 ρ = r + t = 4.009 in. c= 1 t = 0.009 in. 2 c 0.009 = 0.002245 4.009 Maximum strain: εm = Maximum stress: σ m = Eε m = (29 × 106 )(0.002245) ρ = σ m = 65.1 × 103 psi σ m = 65.1 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 4.22 Straight rods of 0.30-in. diameter and 200-ft length are sometimes used to clear underground conduits of obstructions or to thread wires through a new conduit. The rods are made of high-strength steel and, for storage and transportation, are wrapped on spools of 5-ft diameter. Assuming that the yield strength is not exceeded, determine (a) the maximum stress in a rod, when the rod, which is initially straight, is wrapped on a spool, (b) the corresponding bending moment in the rod. Use E = 29 × 106 psi . SOLUTION Radius of cross section: r = Moment of inertia: I = D = 5 ft = 60 in. ρ = 1 1 d = (0.30) = 0.15 in. 2 2 π 4 r4 = π 4 (0.15) 4 = 397.61 × 10−6 in 4 1 D = 30 in. 2 c = r = 0.15 in. (a) σ max = (b) M = EI ρ Ec ρ = = (29 × 106 )(0.15) = 145 × 103 psi 30 (29 × 106 )(397.61 × 10−6 ) 30 σ max = 145 ksi M = 384 lb ⋅ 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 4.23 A 900-mm strip of steel is bent into a full circle by two couples applied as shown. Determine (a) the maximum thickness t of the strip if the allowable stress of the steel is 420 MPa, (b) the corresponding moment M of the couples. Use E = 200 GPa . SOLUTION When the rod is bent into a full circle, the circumference is 900 mm. Since the circumference is equal to 2π times ρ, the radius of curvature, we get ρ = 900 mm = 143.24 mm = 0.14324 m 2π σ = Eε = Stress: Ec or ρ c= ρσ E For σ = 420 MPa and E = 200 GPa, c= (a) (0.14324)(420 × 106 ) = 0.3008 × 10−3 m 200 × 109 Maximum thickness: t = 2c = 0.6016 × 10−3 m t = 0.602 mm Moment of inertia for a rectangular section. I = (b) bt 3 (8 × 10−3 )(0.6016 × 10−3 )3 = = 145.16 × 10−15 m 4 12 12 Bending moment: M= M = EI ρ (200 × 109 )(145.16 × 10−15 ) = 0.203 N ⋅ m 0.14324 M = 0.203 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 4.24 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= 1 ρ (b) = Mc (60)(0.010) = = 75.0 × 106 Pa I 8 × 10−9 M 60 = = 37.5 × 10−3 m −1 EI (200 × 109 )(8 × 10 −9 ) 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 4.25 A couple of magnitude M is applied to a square bar of side a. For each of the orientations shown, determine the maximum stress and the curvature of the bar. SOLUTION 1 3 1 3 a4 bh = aa = 12 12 12 a c= 2 I= σ max 1 ρ a Mc M 2 = = 4 I a 12 = M M = EI E a 4 12 σ max = 1 ρ = 6M a3 12M Ea 4 For one triangle, the moment of inertia about its base is I1 = 1 3 1 bh = 12 12 I 2 = I1 = a 2 ) a4 24 I = I1 + I 2 = c= ( 3 a a4 2a = 24 2 a4 12 σ max = 1 ρ = Mc Ma/ 2 6 2 M = 4 = I a /12 a3 M M = EI E a 4 12 σ max = 1 ρ 8.49M a3 = 12M Ea 4 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 4.26 A portion of a square bar is removed by milling so that its cross section is as shown. The bar is then bent about its horizontal axis by a couple M. Considering the case where h = 0.9h0, express the maximum stress in the bar in the form σ m = kσ 0 , where σ 0 is the maximum stress that would have occurred if the original square bar had been bent by the same couple M, and determine the value of k. SOLUTION I = 4 I1 + 2 I 2 1 1 = (4) h h3 + (2) (2h0 − 2h) (h3 ) 12 3 1 4 4 4 = h 4 + h 0 h3 − h h3 = h 0 h3 − h 4 3 3 3 3 c=h σ= For the original square, 3M Mc Mh = = 3 4 4 I (4h 0 − 3h) h 2 h h −h 3 0 h = h0 , c = h0 . σ0 = 3M 3M = 3 2 (4h0 − 3h0 )h0 h0 h03 h03 σ = = = 0.950 σ 0 (4h0 − 3h) h 2 (4h0 − (3)(0.9)h0 )(0.9 h02 ) σ = 0.950 σ 0 k = 0.950 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 4.27 In Prob. 4.26, determine (a) the value of h for which the maximum stress σ m is as small as possible, (b) the corresponding value of k. PROBLEM 4.26 A portion of a square bar is removed by milling so that its cross section is as shown. The bar is then bent about its horizontal axis by a couple M. Considering the case where h = 0.9h0, express the maximum stress in the bar in the form σ m = kσ 0 , where σ 0 is the maximum stress that would have occurred if the original square bar had been bent by the same couple M, and determine the value of k. SOLUTION I = 4 I1 + 2 I 2 1 1 = (4) hh3 + (2) (2h0 − 2h) h3 12 3 1 4 4 4 4 = h − h 0 h3 − h3 = h 0 h3 − h 4 3 3 3 3 I 4 c=h = h 0 h 2 − h3 c 3 I d 4 is maximum at h 0 h 2 − h3 = 0. c dh 3 8 h 0 h − 3h 2 = 0 3 2 3 I 4 8 8 256 3 = h0 h0 − h0 = h0 729 c 3 9 9 For the original square, h = h0 c = h0 σ0 = h= σ= 8 h 0 9 Mc 729M = I 256h03 I0 1 3 = h0 c0 3 Mc0 3M = 2 I0 h0 σ 729 1 729 = ⋅ = = 0.949 σ 0 256 3 768 k = 0.949 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 4.28 A couple M will be applied to a beam of rectangular cross section that is to be sawed from a log of circular cross section. Determine the ratio d/b, for which (a) the maximum stress σm will be as small as possible, (b) the radius of curvature of the beam will be maximum. SOLUTION Let D be the diameter of the log. D2 = b2 + d 2 I = (a) 1 3 bd 12 d 2 = D 2 − b2 c= σm is the minimum when 1 d 2 I is maximum. c I 1 = b( D 2 − b 2 ) c 6 d I 1 2 = D − db c 6 d = (b) ρ = I 1 = bd 2 c 6 D2 − = 1 2 1 D b − b3 6 6 3 2 b =0 6 1 2 D = 3 b= 1 D 3 2 D 3 d = b 2 d = b 3 EI M ρ is maximum when I is maximum, 1 3 bd is maximum, or b 2d 6 is maximum. 12 ( D 2 − d 2 )d 6 is maximum. 6D 2d 5 − 8d 7 = 0 b= D2 − 3 2 1 D = D 4 2 d = 3 D 2 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 4.29 For the aluminum bar and loading of Sample Prob. 4.1, determine (a) the radius of curvature ρ ′ of a transverse cross section, (b) the angle between the sides of the bar that were originally vertical. Use E = 10.6 × 106 psi and v = 0.33. SOLUTION From Sample Prob. 4.1, I = 12.97 in 4 1 ρ (a) = M 103.8 × 103 = = 755 × 10−6 in −1 EI (10.6 × 106 )(12.97) 1 1 = v = (0.33) (755 × 10−6 ) = 249 × 10−6 in −1 ρ′ ρ ρ ′ = 4010 in. (b) M = 103.8 kip ⋅ in θ= length of arc b 3.25 = = = 810 × 10−6 rad ρ ′ 4010 radius ρ ′ = 334 ft θ = 0.0464° 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 4.30 For the bar and loading of Example 4.01, determine (a) the radius of curvature ρ , (b) the radius of curvature ρ ′ of a transverse cross section, (c) the angle between the sides of the bar that were originally vertical. Use E = 29 × 106 psi and v = 0.29. SOLUTION M = 30 kip ⋅ in, From Example 4.01, (a) (b) (c) 1 ρ = I = 1.042 in 4 M (30 × 103 ) = = 993 × 10−6 in −1 EI (29 × 106 )(1.042) ε 1 = vε = vc ρ =v ρ = 1007 in. c ρ′ 1 1 = v = (0.29)(993 × 10−6 )in.−1 = 288 × 10−6 in.−1 ρ′ ρ ρ ′ = 3470 in. length of arc b 0.8 = = = 230 × 10−6 rad ρ ′ 3470 radius θ = 0.01320° θ= 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 4.31 A W200 × 31.3 rolled-steel beam is subjected to a couple M of moment 45 kN ⋅ m. Knowing that E = 200 GPa and v = 0.29, determine (a) the radius of curvature ρ , (b) the radius of curvature ρ ′ of a transverse cross section. SOLUTION For W 200 × 31.3 rolled steel section, I = 31.4 × 106 mm4 = 31.4 × 10−6 m 4 (a) (b) 1 ρ = M 45 × 103 = = 7.17 × 10−3 m −1 EI (200 × 109 ) (31.4 × 10−6 ) 1 1 = v = (0.29) (7.17 × 10−3 ) = 2.07 × 10−3 m −1 ρ′ ρ ρ = 139.6 m ρ ′ = 481 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 4.32 It was assumed in Sec. 4.3 that the normal stresses σ y in a member in pure bending are negligible. For an initially straight elastic member of rectangular cross section, (a) derive an approximate expression for σ y as a function of y, (b) show that ( σ y ) max = −(c/2 ρ )(σ x )max and, thus, that σ y can be neglected in all practical situations. (Hint: Consider the free-body diagram of the portion of beam located below the surface of ordinate y and assume that the distribution of the stress σ x is still linear.) SOLUTION Denote the width of the beam by b and the length by L. θ= Using the free body diagram above, with Σ Fy = 0 : σy = − (a) σ y bL + 2 θ 2 sin L 2 σ x = −(σ x )max But, (σ ) σ y = x max ρc y −c cos (σ ) y2 ydy = x max ρc 2 y −c θ 2 y −c L ρ ≈1 σ x bdy sin σ x dy ≈ − θ L θ 2 y −c =0 σ x dy = − 1 ρ y −c σ x dy y c y σy = −c (σ x )max 2 ( y − c 2 ) 2ρ c The maximum value σ y occurs at y = 0 . (b) (σ y ) max = − (σ x )max c 2 (σ ) c = − x max 2ρ c 2ρ 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 4.33 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 Modulus of elasticity Allowable stress Brass 70 GPa 105 GPa 100 MPa 160 MPa 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 figure. For the transformed section, n1 1.0 b1 h13 = (8) (32)3 = 21.8453 × 103 mm 4 12 12 n 1.5 I 2 = 2 b2 h23 = (32)(32)3 = 131.072 × 103 mm 4 12 12 I 3 = I1 = 21.8453 × 103 mm 4 I1 = I = I1 + I 2 + I 3 = 174.7626 × 103 mm 4 = 174.7626 × 10−9 m 4 |σ | = Aluminum: σI ny σ = 100 × 106 Pa (100 × 106 )(174.7626 × 10−9 ) = 1.0923 × 103 N ⋅ m (1.0)(0.016) n = 1.5, | y | = 16 mm = 0.016 m, M= Choose the smaller value. M= n = 1.0, | y | = 16 mm = 0.016 m, M= Brass: nM y I σ = 160 × 106 Pa (160 × 106 )(174.7626 × 10−9 ) = 1.1651 × 103 N ⋅ m (1.5)(0.016) M = 1.092 × 103 N ⋅ m M = 1.092 kN ⋅ 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 4.35 For the composite bar indicated, determine the largest permissible bending moment when the bar is bent about a vertical axis. PROBLEM 4.35 Bar of Prob. 4.33. Modulus of elasticity Allowable stress Aluminum Brass 70 GPa 105 GPa 100 MPa 160 MPa 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 figure. For the transformed section, n1 1.0 (32) (8)3 + (1.0)[(32)(8)](20) 2 = 103.7653 × 103 mm 4 h1 b13 + n1 A1d12 = 12 12 n 1.5 (32)(32)3 = 131.072 × 103 mm 4 I 2 = 2 h2b23 = 12 12 I 3 = I1 = 103.7653 × 103 mm 4 I1 = I = I1 + I 2 + I 3 = 338.58 × 103 mm 4 = 338.58 × 10−9 m 4 |σ | = n My I Aluminum: M= σ = 100 × 106 Pa (100 × 106 )(338.58 × 10−9 ) = 1.411 × 103 N ⋅ m (1.0)(0.024) n = 1.5, | y | = 16 mm = 0.016 m, M= Choose the smaller value. ny n = 1.0, | y | = 24 mm = 0.024 m, M= Brass: σI σ = 160 × 106 Pa (160 × 106 )(338.58 × 10−9 ) = 2.257 × 103 N ⋅ m (1.5)(0.016) M = 1.411 × 103 N ⋅ m M = 1.411 kN ⋅ 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 4.36 For the composite bar indicated, determine the largest permissible bending moment when the bar is bent about a vertical axis. PROBLEM 4.36 Bar of Prob. 4.34. Modulus of elasticity Allowable stress Aluminum 70 GPa 100 MPa Brass 105 GPa 160 MPa 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 (32)(483 − 323 ) = 311.296 × 103 mm 4 h1 B13 − b13 = 12 12 n2 1.0 I 2 = h2b23 = (8)(32)3 = 21.8453 × 103 mm 4 12 12 I 3 = I 2 = 21.8453 × 103 mm 4 ( I1 = ) I = I1 + I 2 + I 3 = 354.99 × 103 mm 4 = 354.99 × 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 )(354.99 × 10−9 ) = 2.2187 × 103 N ⋅ m (1.0)(0.016) n = 1.5 | y | = 24 mm = 0.024 m σ = 160 × 106 Pa M= (160 × 106 )(354.99 × 10−9 ) = 1.57773 × 103 N ⋅ m (1.5)(0.024) Choose the smaller value. M = 1.57773 × 103 N ⋅ m M = 1.578 kN ⋅ 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 4.37 Wooden beams and steel plates are securely bolted together to form the composite member shown. Using the data given below, determine the largest permissible bending moment when the member is bent about a horizontal axis. Modulus of elasticity: Allowable stress: Wood Steel 2 × 106 psi 29 × 106 psi 2000 psi 22 ksi SOLUTION Use wood as the reference material. n = 1.0 in wood n = Es /Ew = 29/2 = 14.5 in steel For the transformed section, n1 1.0 b1h13 = (3)(10)3 = 250 in 4 12 12 n 14.5 1 3 4 I 2 = 2 b2 h23 = (10) = 604.17 in 12 12 2 I1 = I 3 = I1 = 250 in 4 I = I1 + I 2 + I 3 = 1104.2 in 4 σ = Wood: nMy I ∴ M = n = 1.0, M = Steel: ny y = 5 in., σ = 2000 psi (2000)(1104.2) = 441.7 × 103 lb ⋅ in (1.0)(5) n = 14.5, M = σI y = 5 in., σ = 22 ksi = 22 × 103 psi (22 × 103 )(1104.2) = 335.1 × 103 lb ⋅ in (14.5)(5) Choose the smaller value. M = 335 × 103 lb ⋅ in M = 335 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 4.38 Wooden beams and steel plates are securely bolted together to form the composite member shown. Using the data given below, determine the largest permissible bending moment when the member is bent about a horizontal axis. Modulus of elasticity: Allowable stress: Wood Steel 2 × 106 psi 29 × 106 psi 2000 psi 22 ksi SOLUTION Use wood as the reference material. n = 1.0 in wood n = Es /Ew = 29/2 = 14.5 in steel For the transformed section, I1 = n1 b1h13 + n1 A1d12 12 3 14.5 1 1 (5) + (14.5)(5) (5.25) 2 = 999.36 in 4 = 12 2 2 n 1.0 (6)(10)3 = 500 in 4 I 2 = 2 b2 h22 = 12 12 I 3 = I1 = 999.36 in 4 I = I1 + I 2 + I 3 = 2498.7 in 4 σ = Wood: nMy I ∴ M = n = 1.0, M = Steel: ny y = 5 in., σ = 2000 psi (2000)(2499) = 999.5 × 103 lb ⋅ in (1.0)(5) n = 14.5, M = σI y = 5.5 in., σ = 22 ksi = 22 × 103 psi (22 × 103 )(2499) = 689.3 × 103 lb ⋅ in (14.5)(5.5) Choose the smaller value. M = 689 × 103 lb ⋅ in M = 689 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 4.39 A steel bar and an aluminum bar are bonded together to form the composite beam shown. The modulus of elasticity for aluminum is 70 GPa and for steel is 200 GPa. Knowing that the beam is bent about a horizontal axis by a couple of moment M = 1500 N ⋅ m, determine the maximum stress in (a) the aluminum, (b) the steel. SOLUTION Use aluminum as the reference material. For aluminum, n =1 For steel, n = Es /Ea = 200/70 = 2.8571 Transformed section: Part 1 2 Σ Y0 = A, mm2 600 1200 nA, mm2 1714.3 1200 2914.3 109714 = 37.65 mm 2914.3 yo , mm 50 20 nAyo , mm3 85714 24000 109714 d, mm 12.35 17.65 d = y0 − Y0 n1 2.8571 (30)(20)3 + (1714.3)(12.35) 2 = 318.61 × 103 mm 4 b1 h13 + n1 A1 d12 = 12 12 n 1 I 2 = 2 b2 h23 + n2 A2 d 22 = (30)(40)3 + (1200)(17.65) 2 = 533.83 × 103 mm 4 12 12 3 I = I1 + I 2 = 852.44 × 10 mm 4 = 852.44 × 10−9 m 4 I1 = M = 1500 N ⋅ m Stress: (a) σ =− Aluminum: n My I n = 1, σa = − (b) Steel: y = −37.65 mm = −0.03765 m (1)(1500)(−0.03765) = 66.2 × 106 Pa 852.44 × 10−9 n = 2.8571, σs = − σ a = 66.2 MPa y = 60 − 37.65 = 22.35 mm = 0.02235 m n My (2.8571)(1500) (0.02235) =− = −112.4 × 106 Pa I 852.44 × 10−9 σ s = −112.4 MPa 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 4.40 A steel bar and an aluminum bar are bonded together to form the composite beam shown. The modulus of elasticity for aluminum is 70 GPa and for steel is 200 GPa. Knowing that the beam is bent about a horizontal axis by a couple of moment M = 1500 N ⋅ m, determine the maximum stress in (a) the aluminum, (b) the steel. SOLUTION Use aluminum as the reference material. For aluminum, n =1 For steel, n = Es /Ea = 200/70 = 2.8571 Transformed section: A, mm2 Part nA, mm2 yo , mm nAyo , mm3 d, mm 1 600 600 50 30000 25.53 2 1200 3428.5 20 68570 4.47 4028.5 Σ Y0 = 98570 = 24.47 mm 4028.5 98570 d = y0 − Y0 n1 1 b1 h13 + n1 A1 d12 = (30)(20)3 + (600)(25.53)2 = 411.07 × 103 mm 4 12 12 n 2.8571 I 2 = 2 b2 h23 + n2 A2 d 22 = (30)(40)3 + (3428.5)(4.47) 2 = 525.64 × 103 mm 4 12 12 I = I1 + I 2 = 936.71 × 103 mm 4 = 936.71 × 10−9 m 4 I1 = M = 1500 N ⋅ m Stress: (a) Aluminum: σ =− nM y I n = 1, σa = − (b) Steel: y = 60 − 24.47 = 35.53 mm = 0.03553 m (1)(1500)(0.03553) = −56.9 × 106 Pa −9 936.71 × 10 n = 2.8571, σs = − σ a = −56.9 MPa y = −24.47 mm = −0.02447 m (2.8571)(1500) (−0.02447) = 111.9 × 106 Pa −9 936.71 × 10 σ s = 111.9 MPa 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 Transformed section: 421.931 112.278 = 3.758 in. Yo = = wood = 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 y , in. 0.571 nAy , in 3 31.091 d, in. 3.216 72 6.22 447.84 2.433 478.93 126.456 Σ Y0 = nA, in2 54.456 478.93 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 4.43 For the composite beam indicated, determine the radius of curvature caused by the couple of moment 1500 N ⋅ m. Beam of Prob. 4.39. SOLUTION See solution to Prob. 4.39 for the calculation of I. I = 852.44 × 10−9 m 4 1 ρ = Ea = 70 × 109 Pa M 1500 = = 0.02513 m −1 9 EI (70 × 10 )(852.44 × 10−9 ) ρ = 39.8 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 4.44 For the composite bar indicated, determine the radius of curvature caused by the couple of moment 1500 N ⋅ m. Beam of Prob. 4.40. SOLUTION See solution to Prob. 4.40 for calculation of I. I = 936.71 × 10−9 m 4 1 ρ = Ea = 70 × 109 Pa M 1500 = = 0.02288 m −1 9 EI (70 × 10 )(936.71 × 10−9 ) ρ = 43.7 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 4.45 For the composite beam indicated, determine the radius of curvature caused by the couple of moment 450 kip ⋅ in. Beam of Prob. 4.41. SOLUTION See solution to Prob. 4.41 for calculation of I. I = 1873.77 in 4 Ew = 1.8 × 106 psi M = 450 kip ⋅ in = 450 × 103 lb ⋅ in 1 ρ = M 450 × 103 = = 133.42 × 10−6 in −1 6 EI (1.8 × 10 )(1873.77) ρ = 7495 in. = 625 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 4.46 For the composite beam indicated, determine the radius of curvature caused by the couple of moment 450 kip ⋅ in. Beam of Prob. 4.42. SOLUTION See solution to Prob. 4.42 for calculation of I. I = 1874.69 in 4 Ew = 1.8 × 106 psi M = 450 kip ⋅ in = 450 × 103 lb ⋅ in 1 ρ = M 450 × 103 = = 133.36 × 10−6 in −1 6 EI (1.8 × 10 )(1874.69) ρ = 7499 in. = 625 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 4.47 The reinforced concrete beam shown is subjected to a positive bending moment of 175 kN ⋅ m. Knowing that the modulus of elasticity is 25 GPa for the concrete and 200 GPa for the steel, determine (a) the stress in the steel, (b) the maximum stress in the concrete. SOLUTION n= Es 200 GPa = = 8.0 25 GPa Ec As = 4 ⋅ π π d 2 = (4) (25) 2 = 1.9635 × 103 mm 2 4 4 nAs = 15.708 × 103 mm 2 Locate the neutral axis: x − (15.708 × 103 )(480 − x) = 0 2 150 x 2 + 15.708 × 103 x − 7.5398 × 106 = 0 300 x Solve for x: −15.708 × 103 + (15.708 × 103 ) 2 + (4)(150)(7.5398 × 106 ) (2)(150) x = 177.87 mm, 480 − x = 302.13 mm x= 1 I = (300) x3 + (15.708 × 103 )(480 − x) 2 3 1 = (300)(177.87)3 + (15.708 × 103 )(302.13)2 3 = 1.9966 × 109 mm 4 = 1.9966 × 10−3 m 4 nMy σ =− I (a) Steel: y = −302.45 mm = −0.30245 m σ =− (b) Concrete: (8.0)(175 × 103 )(−0.30245) = 212 × 106 Pa −3 1.9966 × 10 σ = 212 MPa y = 177.87 mm = 0.17787 m σ =− (1.0)(175 × 103 )(0.17787) = −15.59 × 106 Pa 1.9966 × 10−3 σ = −15.59 MPa 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 4.48 Solve Prob. 4.47, assuming that the 300-mm width is increased to 350 mm. PROBLEM 4.47 The reinforced concrete beam shown is subjected to a positive bending moment of 175 kN ⋅ m. Knowing that the modulus of elasticity is 25 GPa for the concrete and 200 GPa for the steel, determine (a) the stress in the steel, (b) the maximum stress in the concrete. SOLUTION n= Es 200 GPa = = 8.0 Ec 25 GPa As = 4 π π d 2 = (4) (25)2 4 4 = 1.9635 × 103 mm 2 nAs = 15.708 × 103 mm 2 Locate the neutral axis: x − (15.708 × 103 )(480 − x) = 0 2 175 x 2 + 15.708 × 103 x − 7.5398 × 106 = 0 350 x Solve for x: −15.708 × 103 + (15.708 × 103 ) 2 + (4)(175)(7.5398 × 106 ) (2)(175) x = 167.48 mm, 480 − x = 312.52 mm x= 1 I = (350) x3 + (15.708 × 103 )(480 − x) 2 3 1 = (350)(167.48)3 + (15.708 × 103 )(312.52) 2 3 = 2.0823 × 109 mm 4 = 2.0823 × 10−3 m 4 nMy σ =− I (a) Steel: y = −312.52 mm = −0.31252 m σ =− (b) Concrete: (8.0)(175 × 103 )(−0.31252) = 210 × 106 Pa 2.0823 × 10−3 σ = 210 MPa y = 167.48 mm = 0.16748 m σ =− (1.0)(175 × 103 )(0.16748) = −14.08 × 106 Pa −3 2.0823 × 10 σ = −14.08 MPa 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 4.49 A concrete slab is reinforced by 16-mm-diameter steel rods placed on 180-mm centers as shown. The modulus of elasticity is 20 GPa for concrete and 200 GPa for steel. Using an allowable stress of 9 MPa for the concrete and of 120 MPa for the steel, determine the largest bending moment in a portion of slab 1 m wide. SOLUTION n= Es 200 GPa = = 10 20 GPa Ec Consider a section 180-mm wide with one steel rod. As = π d2 = π (16)2 = 201.06 mm 2 4 4 nAs = 2.0106 × 103 mm 2 Locate the neutral axis: x − (100 − x)(2.0106 × 103 ) = 0 2 90 x 2 + 2.0106 × 103 x − 201.06 × 103 = 0 180 x Solving for x: −2.0106 × 103 + (2.0106 × 103 ) 2 + (4)(90)(201.06 × 103 ) (2)(90) x = 37.397 mm 100 − x = 62.603 mm x= 1 I = (180) x3 + (2.0106 × 103 )(100 − x) 2 3 1 = (180)(37.397)3 + (2.0106 × 103 )(62.603) 2 3 = 11.018 × 106 mm 4 = 11.018 × 10−6 m 4 Concrete: σI σ =− nMy I n = 1, y = 37.397 mm = 0.037397 m, M= ∴ M= ny σ = 9 × 106 Pa (9 × 106 )(11.018 × 10−6 ) = 2.6516 × 103 N ⋅ m (1.0)(0.037397) 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 4.49 (Continued) n = 10, Steel: M= Choose the smaller value. y = 62.603 mm = 0.062603 m, σ = 120 × 106 Pa (120 × 106 )(11.018 × 10−6 ) = 2.1120 × 103 N ⋅ m (10)(0.062603) M = 2.1120 × 103 N ⋅ m The above is the allowable positive moment for a 180-mm wide section. For a 1-m = 1000-mm width, 1000 = 5.556 mutiply by 180 M = (5.556)(2.1120 × 103 ) = 11.73 × 103 N ⋅ m M = 11.73 kN ⋅ 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 4.50 A concrete slab is reinforced by 16-mm-diameter steel rods placed on 180-mm centers as shown. The modulus of elasticity is 20 GPa for concrete and 200 GPa for steel. Using an allowable stress of 9 MPa for the concrete and of 120 MPa for the steel, determine the largest allowable positive bending moment in a portion of slab 1 m wide. Solve Prob. 4.49, assuming that the spacing of the 16-mm-diameter rods is increased to 225 mm on centers. SOLUTION n= Es 200 GPa = = 10 20 GPa Ea Consider a section 225-mm wide with one steel rod. As = π d2 = π (16)2 = 201.06 mm 2 4 4 nAs = 2.0106 × 103 mm 2 Locate the neutral axis: x 225 x − (100 − x)(2.0106 × 103 ) = 0 2 112.5 x 2 + 2.0106 x − 201.06 × 103 = 0 Solving for x: −2.0106 × 103 + (2.0106 × 103 ) 2 + (4)(112.5)(201.06 × 103 ) (2)(112.5) x = 34.273 mm 100 − x = 65.727 1 I = (225) x3 + 2.0106 × 103 (100 − x) 2 3 1 = (225)(34.273)3 + (2.0106 × 103 )(65.727) 2 3 = 11.705 × 106 mm 4 = 11.705 × 10−6 m 4 x= |σ | = Concrete: nMy I n = 1, M= ∴ M = σI ny y = 34.273 mm = 0.034273 m, σ = 9 × 106 Pa (9 × 106 )(11.705 × 10−6 ) = 3.0738 × 103 N ⋅ m (1)(0.034273) 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 4.50 (Continued) n = 10, Steel: M= Choose the smaller value. y = 65.727 mm = 0.065727 m, σ = 120 × 106 Pa (120 × 106 )(11.705 × 10−6 ) = 2.1370 × 103 N ⋅ m (10)(0.065727) M = 2.1370 × 103 N ⋅ m The above is the allowable positive moment for a 225-mm-wide section. For a 1-m = 1000-mm section, 1000 = 4.4444 multiply by 225 M = (4.4444)(2.1370 × 103 ) = 9.50 × 103 N ⋅ m M = 9.50 kN ⋅ 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 4.51 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 2 π π 7 d 2 = (3) = 1.8040 in 2 4 4 8 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 π As = (4) (1) 2 = 3.1416 in 2 4 nAs = 25.133 in 2 Locate the neutral axis. x (24)(4)( x + 2) + (12 x) − (25.133)(17.5 − 4 − x) = 0 2 96 x + 192 + 6 x 2 − 339.3 + 25.133x = 0 x= Solve for x. 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 4.53 The design of a reinforced concrete beam is said to be balanced if the maximum stresses in the steel and concrete are equal, respectively, to the allowable stresses σ s and σ c . Show that to achieve a balanced design the distance x from the top of the beam to the neutral axis must be d x= 1+ σ s Ec σ c Es where Ec and Es are the moduli of elasticity of concrete and steel, respectively, and d is the distance from the top of the beam to the reinforcing steel. SOLUTION nM (d − x) Mx σc = I I σ s n( d − x ) d = =n −n σc x x σs = Eσ d 1 σs =1+ =1+ c s x n σc E sσ c x= d Ecσ s 1+ E sσ c 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 4.54 For the concrete beam shown, the modulus of elasticity is 3.5 × 106 psi for the concrete and 29 × 106 psi for the steel. Knowing that b = 8 in. and d = 22 in., and using an allowable stress of 1800 psi for the concrete and 20 ksi for the steel, determine (a) the required area As of the steel reinforcement if the beam is to be balanced, (b) the largest allowable bending moment. (See Prob. 4.53 for definition of a balanced beam.) The design of a reinforced concrete beam is said to be balanced if the maximum stresses in the steel and concrete are equal, respectively, to the allowable stresses σ s and σ c . SOLUTION Es 29 × 10 6 = = 8.2857 Ec 3.5 × 106 nM ( d − x) Mx σs = σc = I I σ s n( d − x ) d = =n −n σc x x n= 1 σs 1 20 × 103 d =1+ =1+ ⋅ = 2.3410 8.2857 1800 x n σc x = 0.42717 d = (0.42717)(22) = 9.398 in. Locate neutral axis: (a) As = bx d − x = 22 − 9.398 = 12.602 in . x − nAs ( d − x) = 0 2 bx 2 (8)(9.398) 2 = = 3.3835 in 2 2n(d − x) (2)(8.2857)(12.602) As = 3.38 in 2 1 1 I = bx3 + nAs ( d − x) 2 = (8)(9.398)3 + (8.2857)(3.3835)(12.602)2 = 6665.6 in 4 3 3 n My σI M= σ= I ny (b) Steel: Concrete: n = 1.0 y = 9.398 in. σ c = 1800 psi M= (1800)(6665.6) = 1.277 × 106 lb ⋅ in (1.0)(9.398) n = 8.2857 | y | = 12.602 in. σ s = 20 × 103 psi M= (20 × 103 )(6665.6) = 1.277 × 106 lb ⋅ in (8.2857)(12.602) Note that both values are the same for balanced design. M = 106.4 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 4.55 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 4.56 Five metal strips, each of 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 = 1.0 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 aluminum portion. For the transformed section, n1 1.5 b1h13 + n1 A1d12 = (40)(10)3 + (1.5)(40)(10)(25) 2 = 380 × 103 mm 4 12 12 n2 3.0 3 2 I 2 = b2 h2 + n2 A2 d 2 = (40)(10)3 + (3.0)(40)(10)(15)2 = 280 × 103 mm 4 12 12 n3 1.0 I 3 = b3 h33 = (40)(20)3 = 26.67 × 103 mm 4 12 12 I 4 = I 2 = 280 × 103 mm 4 I 5 = I1 = 380 × 103 mm 4 I1 = I= (a) σ =− nMy I Aluminum: σa = Steel: σs = (b) where 6 mm 4 = 1.3467 × 10−6 m 4 M = 1800 N ⋅ m n = 1, y = −10 mm = −0.010 m (1.0)(1800)(0.010) = 13.37 × 106 Pa 1.3467 × 10−6 Brass: σb = I = 1.3467 × 10 n = 1.5, y = −30 mm = −0.030 m (1.5)(1800)(0.030) = 60.1 × 106 Pa 1.3467 × 10−6 n = 3.0, σ b = 60.1 MPa y = −20 mm = −0.020 m (3.0)(1800)(0.020) = 80.1 × 106 Pa 1.3467 × 10−6 Radius of curvature. σ a = 13.37 MPa 1 ρ = M 1800 = = 0.01909 m −1 9 −6 Ea I (70 × 10 )(1.3467 × 10 ) σ s = 80.1 MPa ρ = 52.4 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 2.51327 Σ 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. PROBLEM 4.58 A steel pipe and an aluminum pipe are securely bonded together to form the composite beam shown. The modulus of elasticity is 200 GPa for the steel and 70 GPa for the aluminum. Knowing that the composite beam is bent by a couple of moment 500 N ⋅ m, determine the maximum stress (a) in the aluminum, (b) in the steel. SOLUTION Use aluminum as the reference material. n = 1.0 in aluminum n = Es / Ea = 200 / 70 = 2.857 in steel For the transformed section, Steel: I s = ns Aluminium: I a = na π 4 π (r 4 o (r 4 4 o π − ri4 = (2.857) (164 − 104 ) = 124.62 × 103 mm 4 4 ) π − ri4 = (1.0) (194 − 164 ) = 50.88 × 103 mm 4 4 ) I = I s + I a = 175.50 × 103 mm 4 = 175.5 × 10−9 m 4 (a) Aluminum: c = 19 mm = 0.019 m σa = na Mc (1.0)(500)(0.019) = = 54.1 × 106 Pa −9 I 175.5 × 10 σ a = 54.1 MPa (b) Steel: c = 16 mm = 0.016 m σs = ns Mc (2.857)(500)(0.016) = = 130.2 × 106 Pa −9 I 175.5 × 10 σ s = 130.2 MPa 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 4.59 The rectangular beam shown is made of a plastic for which the value of the modulus of elasticity in tension is one-half of its value in compression. For a bending moment M = 600 N ⋅ m, determine the maximum (a) tensile stress, (b) compressive stress. SOLUTION n= 1 on the tension side of neutral axis. 2 n = 1 on the compression side. Locate neutral axis. n1bx x h−x − n2 b(h − x) =0 2 2 1 2 1 bx − b(h − x) 2 = 0 2 4 1 1 (h − x) 2 x = ( h − x) 2 2 1 x= h = 0.41421 h = 41.421 mm 2 +1 h − x = 58.579 mm x2 = 1 1 I1 = n1 bx3 = (1) (50)(41.421)3 = 1.1844 × 106 mm 4 3 3 1 1 1 I 2 = n2 b(h − x)3 = (50)(58.579)3 = 1.6751 × 106 mm 4 3 2 3 I = I1 + I 2 = 2.8595 × 106 mm 4 = 2.8595 × 10−6 m 4 (a) Tensile stress: n= 1 , 2 σ =− (b) Compressive stress: n = 1, σ =− y = −58.579 mm = −0.058579 m nMy (0.5)(600)(−0.058579) =− = 6.15 × 106 Pa I 2.8595 × 10−6 σ t = 6.15 MPa y = 41.421 mm = 0.041421 m nMy (1.0)(600)(0.041421) =− = −8.69 × 106 Pa I 2.8595 × 10−6 σ c = −8.69 MPa 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 4.60* A rectangular beam is made of material for which the modulus of elasticity is Et in tension and Ec in compression. Show that the curvature of the beam in pure bending is 1 ρ = M Er I where Er = 4 Et Ec ( Et + Ec ) 2 SOLUTION Use Et as the reference modulus. Then Ec = nEt . Locate neutral axis. x h−x − b( h − x ) =0 2 2 nx 2 − ( h − x)2 = 0 nx = (h − x) nbx h x= I trans 1 ρ n +1 nh h−x= n +1 3 h 1 3 n 3 1 n 3 3 = bx + b(h − x) = bh + 3 n + 1 n + 1 3 3 ( ) = 1 n + n3/ 2 1 n 1+ n 1 3 bh = bh3 = × 3 3 3 n +1 3 n +1 3 = M M = Et I trans Er I ( ) ( ) where I = n ( ) n +1 2 bh3 1 3 bh 12 Er I = Et I trans Er = = Et I trans 12 = 3 × Et × I bh 3 ( 4 Et Ec /Et = ) ( Ec /Et + 1 2 n ( ) n +1 4 Et Ec Ec + Et ) 2 bh3 2 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 4.61 Semicircular grooves of radius r must be milled as shown in the sides of a steel member. Using an allowable stress of 60 MPa, determine the largest bending moment that can be applied to the member when (a) r = 9 mm, (b) r = 18 mm. SOLUTION (a) d = D − 2r = 108 − (2)(9) = 90 mm D 108 = = 1.20 d 90 From Fig. 4.32, r 9 = = 0.1 d 90 K = 2.07 1 (18)(90)3 12 = 1.0935 × 106 mm 4 I= = 1.0935 × 10−6 m 4 1 d = 45 mm = 0.045 m 2 KMc σ= I c= M= (b) σI Kc = (60 × 106 )(1.0935 × 10−6 ) = 704 (2.07)(0.045) M = 704 N ⋅ m d = 108 − (2)(18) = 72 mm D 108 r 18 = = 1.5 = = 0.25 d 72 d 72 1 c = d = 36 mm = 0.036 m 2 From Fig. 4.32, K = 1.61 1 I = (18)(72)3 = 559.87 × 103 mm 4 = 559.87 × 10−9 m 4 12 M= σI Kc = (60 × 106 )(559.87 × 10−9 ) = 580 (1.61)(0.036) M = 580 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 4.62 Semicircular grooves of radius r must be milled as shown in the sides of a steel member. Knowing that M = 450 N ⋅ m, determine the maximum stress in the member when the radius r of the semicircular grooves is (a) r = 9 mm, (b) r = 18 mm. SOLUTION (a) d = D − 2r = 108 − (2)(9) = 90 mm D 108 = = 1.20 d 90 From Fig. 4.32, r 9 = = 0.1 d 90 K = 2.07 1 3 1 bh = (18)(90)3 12 12 = 1.0935 × 106 mm 4 I= = 1.0935 × 10−6 m 4 1 c = d = 45 mm = 0.045 m 2 KMc (2.07)(450)(0.045) = I 1.0935 × 10−6 = 38.3 × 106 Pa σ max = (b) σ max = 38.3 MPa d = D − 2r = 108 − (2)(18) = 72 mm D 108 = = 1.5 d 72 From Fig. 4.32, r 18 = = 0.25 d 72 K = 1.61 1 d = 72 mm = 0.036 m 2 1 I = (18)(72)3 = 559.87 × 103 mm 4 = 559.87 × 10−9 m 4 12 c= σ max = KMc (1.61)(450)(0.036) = = 46.6 × 106 Pa I 559.87 × 10−9 σ max = 46.6 MPa 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 4.63 Knowing that the allowable stress for the beam shown is 90 MPa, determine the allowable bending moment M when the radius r of the fillets is (a) 8 mm, (b) 12 mm. SOLUTION 1 3 1 bh = (8)(40)3 = 42.667 × 103 mm 4 = 42.667 × 10−9 m 4 12 12 c = 20 mm = 0.020 m I = D 80 mm = = 2.00 40 mm d (a) r 8 mm = = 0.2 d 40 mm σ max = K Mc I From Fig. 4.31, M = σ max I Kc = K = 1.50 (90 × 106 )(42.667 × 10−9 ) (1.50)(0.020) M = 128 N ⋅ m (b) r 12 mm = = 0.3 d 40 mm From Fig. 4.31, M = K = 1.35 (90 × 106 )(42.667 × 10−9 ) (1.35)(0.020) M = 142 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 4.64 Knowing that M = 250 N ⋅ m, determine the maximum stress in the beam shown when the radius r of the fillets is (a) 4 mm, (b) 8 mm. SOLUTION 1 3 1 bh = (8)(40)3 = 42.667 × 103 mm 4 = 42.667 × 10−9 m 4 12 12 c = 20 mm = 0.020 m I = D 80 mm = = 2.00 40 mm d (a) r 4 mm = = 0.10 d 40 mm σ max = K (b) K = 1.87 Mc (1.87)(250)(0.020) = = 219 × 106 Pa I 42.667 × 10−9 r 8 mm = = 0.20 d 40 mm σ max = K From Fig. 4.31, From Fig. 4.31, σ max = 219 MPa K = 1.50 Mc (1.50)(250)(0.020) = = 176 × 106 Pa I 42.667 × 10−9 σ max = 176 MPa 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 4.65 The allowable stress used in the design of a steel bar is 12 ksi. Determine the largest couple M that can be applied to the bar (a) if the bar is designed with grooves having semicircular portions of radius r = 34 in. as shown in Fig. a, (b) if the bar is redesigned by removing the material above and below the dashed lines as shown in Fig. b. SOLUTION 7 in. = 0.875 in. 8 D = 7.5 in. 3 in. = 0.75 in. 4 d = 5 in. r 0.75 = = 0.15 d 5 D 7.5 = = 1.5 d 5 t = Dimensions: Stress concentration factors: Figs. 4.32 and 4.31 Configuration (a): K = 1.92 Configuration (b): K = 1.58 I = Moment of inertia: c= 1 d = 2.5 in. 2 σm = KMc I r = 1 3 1 td = (0.875)(5)3 = 9.115 in 4 12 12 σ m = 12 ksi M = Iσ m Kc (a) M = (9.115)(12) (1.92)(2.5) M = 22.8 kip ⋅ in (b) M = (9.115)(12) (1.58)(2.5) M = 27.7 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 4.66 A couple of moment M = 20 kip ⋅ in. is to be applied to the end of a steel bar. Determine the maximum stress in the bar (a) if the bar is designed with grooves having semicircular portions of radius r = 12 in., as shown in Fig. a, (b) if the bar is redesigned by removing the material above and below the dashed line as shown in Fig. b. SOLUTION 7 in. = 0.875 in. 8 D = 7.5 in. 1 in. = 0.5 in. 2 d = 5 in. r 0.5 = = 0.10 d 5 D 7.5 = = 1.5 d 5 t = Dimensions: Stress concentration factors: Figs. 4.32 and 4.31 Configuration (a): K = 2.22 Configuration (b): K = 1.80 I = Moment of inertia: c= 1 d = 2.5 in. 2 Maximum stress: r = 1 3 1 td = (0.875)(5)3 = 9.115 in 4 12 12 M = 20 kip ⋅ in σm = KMc I (a) σm = (2.22)(20)(2.5) 9.115 σ m = 12.2 ksi (b) σm = (1.80)(20)(2.5) 9.115 σ m = 9.9 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 4.67 The prismatic bar shown is made of a steel that is assumed to be elastoplastic with σ Y = 300 MPa and is subjected to a couple M parallel to the x axis. Determine the moment M of the couple for which (a) yield first occurs, (b) the elastic core of the bar is 4 mm thick. SOLUTION (a) I= 1 3 1 bh = (12)(8)3 = 512 mm 4 12 12 = 512 × 10−12 m 4 c= 1 h = 4 mm = 0.004 m 2 (300 × 106 )(512 × 10−12 ) 0.004 c = 38.4 N ⋅ m MY = (b) yY = 1 (4) = 2 mm 2 σY I = yY 2 = = 0.5 c 4 1 y 2 3 M = M Y 1 − Y 2 3 c 3 1 = (38.4) 1 − (0.5) 2 2 3 = 52.8 N ⋅ m M Y = 38.4 N ⋅ m M = 52.8 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 4.68 Solve Prob. 4.67, assuming that the couple M is parallel to the z axis. PROBLEM 4.67 The prismatic bar shown is made of a steel that is assumed to be elastoplastic with σ Y = 300 MPa and is subjected to a couple M parallel to the x axis. Determine the moment M of the couple for which (a) yield first occurs, (b) the elastic core of the bar is 4 mm thick. SOLUTION (a) I= 1 3 1 bh = (8)(12)3 = 1.152 × 103 mm 4 12 12 = 1.152 × 10−9 m 4 1 c = h = 6 mm = 0.006 m 2 (300 × 106 )(1.152 × 10−9 ) c 0.006 = 57.6 N ⋅ m MY = (b) yY = 1 (4) = 2 mm 2 σY I = M Y = 57.6 N ⋅ m yY 2 1 = = c 6 3 M= 1 y 2 3 M Y 1 − Y 2 3 c 1 1 2 3 = (57.6) 1 − 2 3 3 = 83.2 N ⋅ m M = 83.2 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 4.69 The prismatic bar shown, made of a steel that is assumed to be elastoplastic with E = 29 × 106 psi and σ Y = 36 ksi, is subjected to a couple of 1350 lb ⋅ in. parallel to the z axis. Determine (a) the thickness of the elastic core, (b) the radius of curvature of the bar. SOLUTION 3 (a) I= 1 1 5 4 −3 = 10.1725 × 10 in 12 2 8 c= 15 = 0.3125 in. 28 (36 × 103 )(10.1725 × 10−3 ) 0.3125 c = 1171.872 lb ⋅ in MY = σY I = 3 M = MY 2 1 y 3 1 − Y 3 c yY (2)(1350) M = 3−2 = 3− = 0.83426 1171.872 c MY yY = (0.83426)(0.3125) = 0.26071 in. Thickness of elastic core: (b) yY = ε Y ρ = σY 2 yY = 0.521 in. ρ E y E (0.26071)(29 × 106 ) = 210.02 in. ρ= Y = σY 36 × 103 ρ = 17.50 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 4.70 Solve Prob. 4.69, assuming that the 1350-lb ⋅ in. couple is parallel to the y axis. PROBLEM 4.69 The prismatic bar shown, made of a steel that is assumed to be elastoplastic with E = 29 × 106 psi and σ Y = 36 ksi, is subjected to a couple of 1350 lb ⋅ in parallel to the z axis. Determine (a) the thickness of the elastic core, (b) the radius of curvature of the bar. SOLUTION 3 (a) I= 1 5 1 −3 4 = 6.5104 × 10 in 12 8 2 1 1 c = = 0.25 in. 2 2 (36 × 103 )(6.5104 × 10−3 ) 0.25 c = 937.5 lb ⋅ in MY = σY I = 2 3 1 yY M = 1 − 2 3 c yY (2)(1350) M = 3−2 = 3− = 0.34641 937.5 c MY yY = (0.34641)(0.25) = 0.086603 in. Thickness of elastic core: (b) yY = ε Y ρ = σY 2 yY = 0.1732 in. ρ E y E (0.086603)(29 × 106 ) = 69.763 in. ρ= Y = σY 36 × 103 ρ = 5.81 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 4.71 A bar of rectangular cross section shown is made of a steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 300 MPa. Determine the bending moment M for which (a) yield first occurs, (b) the plastic zones at the top and bottom of the bar are 12 mm thick. SOLUTION 1 (30)(40)3 = 160 × 103 mm 4 = 160 × 10−9 m 4 12 1 c = (40) = 20 mm = 0.020 m 2 I = (a) First yielding: MY = σ = Mc = σY I Iσ Y (160)(10−9 )(300 × 106 ) = = 2400 N ⋅ m c 0.020 M Y = 2.40 kN ⋅ m (b) Plastic zones are 12 mm thick: yY = 20 mm − 12 mm = 8 mm yY 8 mm = = 0.4 c 20 mm M = = 2 3 1 y M Y 1 − Y 2 3 c 3 1 (2400) 1 − (0.4)2 = 3408 N ⋅ m 2 3 M = 3.41 kN ⋅ 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 4.72 Bar AB is made of a steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 240 MPa. Determine the bending moment M for which the radius of curvature of the bar will be (a) 18 m, (b) 9 m. SOLUTION 1 (30)(40)3 = 160 × 103 mm 4 = 160 × 10−9 m 4 12 1 c = (40) = 20 mm = 0.020 m 2 I = Iσ Y (160 × 10−9 )(240 × 106 ) = = 1920 N ⋅ m c 0.020 M 1 1920 = Y = = 0.0600 m −1 EI ρY (200 × 109 )(160 × 10−9 ) MY = (a) 1 ρ = 18 m: ρ = 0.05556 m −1 < 0.0600 m −1 The bar is fully elastic. M = EI ρ = (200 × 109 )(160 × 10−9 ) = 1778 N ⋅ m 18 M = 1.778 kN ⋅ m (b) 1 ρ = 9 m: ρ = 0.11111 m −1 > 0.0600 m −1 The bar is partially plastic. σY = EyY ρ yY = ρσ Y E (9)(240 × 106 ) = 0.0108 m = 10.8 mm 200 × 109 yY 10.8 mm = = 0.54 c 20 mm yY = M = 3 MY 2 2 1 y 3 1 1 − Y = (1920) 1 − (0.54)2 = 2600 N ⋅ m c 3 2 3 M = 2.60 kN ⋅ 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 4.73 A beam of the cross section shown is made of a steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 240 MPa. For bending about the zaxis, determine the bending moment at which (a) yield first occurs, (b) the plastic zones at the top and bottom of the bar are 30-mm thick. SOLUTION (a) 1 3 1 bh = (60)(90)3 = 3.645 × 106 mm 4 = 3.645 × 10−6 m 4 12 12 1 c = h = 45 mm = 0.045 m 2 σ I (240 × 106 )(3.645 × 10−6 ) MY = Y = = 19.44 × 10 N ⋅ m c 0.045 I= M Y = 19.44 kN ⋅ m R1 = σ Y A1 = (240 × 106 )(0.060)(0.030) = 432 × 103 N y1 = 15 mm + 15 mm = 0.030 m 1 1 R2 = σ Y A2 = (240 × 106 )(0.060)(0.015) 2 2 = 108 × 103 N 2 y2 = (15 mm) = 10 mm = 0.010 m 3 (b) M = 2( R1 y1 + R2 y2 ) = 2[(432 × 103 )(0.030) + (108 × 103 )(0.010)] = 28.08 × 103 N ⋅ m M = 28.1 kN ⋅ 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 4.74 A beam of the cross section shown is made of a steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 240 MPa. For bending about the z axis, determine the bending moment at which (a) yield first occurs, (b) the plastic zones at the top and bottom of the bar are 30-mm thick. SOLUTION (a) 1 3 1 bh = (60)(90)3 = 3.645 × 106 mm 4 12 12 1 1 I cutout = bh3 = (30)(30)3 = 67.5 × 103 mm 4 12 12 I = 3.645 × 106 − 67.5 × 103 = 3.5775 × 106 mm 4 I rect = = 3.5775 × 10−6 mm 4 1 h = 45 mm = 0.045 m 2 σ I (240 × 106 )(3.5775 × 10−6 ) MY = Y = 0.045 c 3 = 19.08 × 10 N ⋅ m c= M Y = 19.08 kN ⋅ m R1 = σ Y A1 = (240 × 106 )(0.060)(0.030) = 432 × 103 N y1 = 15 mm + 15 mm = 30 mm = 0.030 m 1 1 σ Y A2 = (240 × 106 )(0.030)(0.015) = 54 × 103 N 2 2 2 y2 = (15 mm) = 10 mm = 0.010 m 3 R2 = (b) M = 2( R1 y1 + R2 y2 ) = 2[(432 × 103 )(0.030) + (54 × 103 )(0.010)] = 27.00 × 103 N ⋅ m M = 27.0 kN ⋅ 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 4.75 A beam of the cross section shown is made of a steel that is assumed to be elastoplastic with E = 29 × 106 psi and σ Y = 42 ksi. For bending about the z axis, determine the bending moment at which (a) yield first occurs, (b) the plastic zones at the top and bottom of the bar are 3 in. thick. SOLUTION (a) 1 1 b1h13 + A1d12 = (6)(3)3 + (6)(3)(3)2 = 175.5 in 4 12 12 1 1 3 I 2 = b2 h2 = (3)(3)3 = 6.75 in 4 12 12 I 3 = I1 = 175.5 in 4 I1 = I = I1 + I 2 + I 3 = 357.75 in 4 c = 4.5 in. MY = σY I c = (42)(357.75) 4.5 M Y = 3339 kip ⋅ in R1 = σ Y A1 = (42)(6)(3) = 756 kip y1 = 1.5 + 1.5 = 3 in. 1 1 R2 = σ Y A2 = (42)(3)(1.5) 2 2 = 94.5 kip 2 y2 = (1.5) = 1.0 in. 3 (b) M = 2( R1 y1 + R2 y2 ) = 2[(756)(3) + (94.5)(1.0)] M = 4725 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 4.76 A beam of the cross section shown is made of a steel that is assumed to be elastoplastic with E = 29 × 106 psi and σ Y = 42 ksi. For bending about the z axis, determine the bending moment at which (a) yield first occurs, (b) the plastic zones at the top and bottom of the bar are 3 in. thick. SOLUTION (a) 1 1 b1h13 + A1d12 = (3)(3)3 + (3)(3)(3) 2 = 87.75 in 4 12 12 1 1 I 2 = b2 h23 = (6)(3)3 = 13.5 in 4 12 12 I 3 = I1 = 87.75 in 4 I1 = I = I1 + I 2 + I 3 = 188.5 in 4 c = 4.5 in. σ I (42)(188.5) MY = Y = c 4.5 M Y = 1759 kip ⋅ in R1 = σ Y A1 = (42)(3)(3) = 378 kip y1 = 1.5 + 1.5 = 3.0 in. 1 1 R2 = σ Y A2 = (42)(6)(1.5) 2 2 = 189 kip 2 y2 = (1.5) = 1.0 in. 3 (b) M = 2( R1 y1 + R2 y2 ) = 2[(378)(3.0) + (189)(1.0)] M = 2646 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 4.77 For the beam indicated (of Prob. 4.73), determine (a) the fully plastic moment M p , (b) the shape factor of the cross section. SOLUTION From Problem 4.73, E = 200 GPa and σ Y = 240 MPa. A1 = (60)(45) = 2700 mm 2 = 2700 × 10−6 m 2 R = σ Y A1 = (240 × 106 )(2700 × 10−6 ) = 648 × 103 N d = 45 mm = 0.045 m (a) (b) M p = Rd = (648 × 103 )(0.045) = 29.16 × 103 N ⋅ m M p = 29.2 kN ⋅ m 1 3 1 bh = (60)(90)3 = 3.645 × 106 mm 4 = 3.645 × 10−6 m 4 12 12 c = 45 mm = 0.045 m I= MY = σY I c = (240 × 106 )(3.645 × 10−6 ) = 19.44 × 103 N ⋅ m 0.045 k= Mp MY = 29.16 19.44 k = 1.500 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 4.78 For the beam indicated (of Prob. 4.74), determine (a) the fully plastic moment M p , (b) the shape factor of the cross section. SOLUTION From Problem 4.74, (a) E = 200 GPa and σ Y = 240 MPa. R1 = σ Y A1 = (240 × 106 )(0.060)(0.030) = 432 × 103 N y1 = 15 mm + 15 mm = 30 mm = 0.030 m R2 = σ Y A2 = (240 × 106 )(0.030)(0.015) = 108 × 103 N 1 y2 = (15) = 7.5 mm = 0.0075 m 2 M p = 2( R1 y1 + R2 y2 ) = 2[( 432× 103 )(0.030) + (108× 103 )(0.0075)] = 27.54 × 103 N ⋅ m (b) M p = 27.5 kN ⋅ m 1 3 1 bh = (60)(90)3 = 3.645 × 106 mm 4 12 12 1 3 1 I cutout = bh = (30)(30)3 = 67.5 × 103 mm 4 12 12 I = I rect − I cutout = 3.645 × 106 − 67.5 × 103 = 3.5775 × 103 mm 4 I rect = = 3.5775 × 10−9 m 4 1 h = 45 mm = 0.045 m 2 σ I (240 × 106 )(3.5775 × 10−9 ) = 19.08 × 103 N ⋅ m MY = Y = 0.045 c M p 27.54 k= = MY 19.08 c= k = 1.443 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 4.79 For the beam indicated (of Prob. 4.75), determine (a) the fully plastic moment M p , (b) the shape factor of the cross section. SOLUTION From Problem 4.75, (a) E = 29 × 106 and σ Y = 42 ksi. R1 = σ Y A1 = (42)(6)(3) = 756 kip y1 = 1.5 + 1.5 = 3.0 in. R2 = σ Y A2 = (42)(3)(1.5) = 189 kip y2 = 1 (1.5) = 0.75 in. 2 M p = 2( R1 y1 + R2 y2 ) = 2[( 756)( 3.0) + (189)(0. 75)] (b) M p = 4819.5 kip ⋅ in 1 1 b1h13 + A1d12 = (6)(3)3 + (6)(3)(3)2 = 175.5 in 4 12 12 1 1 I 2 = b2 h23 = (3)(3)3 = 6.75 in 4 12 12 I 3 = I1 = 175.5 in 4 I1 = I = I1 + I 2 + I 3 = 357.75 in 4 c = 4.5 in. σY I (42)(357.75) = 3339 kip ⋅ in 4.5 4819.5 = k= MY 3339 MY = c Mp = k = 1.443 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 4.80 For the beam indicated (of Prob. 4.76), determine (a) the fully plastic moment M p , (b) the shape factor of the cross section. SOLUTION From Problem 4.76, (a) E = 29 × 106 psi and σ Y = 42 ksi. R1 = σ Y A1 = (42)(3)(3) = 378 kip y1 = 1.5 + 1.5 = 3.0 in. R2 = σ Y A2 = (42)(6)(1.5) = 378 kip y2 = 1 (1.5) = 0.75 in. 2 M p = 2( R1 y1 + R2 y2 ) = 2[( 378)( 3.0) + ( 378)(0. 75)] (b) M p = 2835 kip ⋅ in 1 1 b1h13 + A1d12 = (3)(3)3 + (3)(3)(3)2 = 87.75 in 4 12 12 1 1 3 I 2 = b2 h2 = (6)(3)3 = 13.5 in 4 12 12 I 3 = I1 = 87.75 in 4 I1 = I = I1 + I 2 + I 3 = 188.5 in 4 c = 4.5 in. σY I (42)(188.5) = 1759.3 kip ⋅ in c 4.5 Mp 2835 k= = MY 1759.3 MY = = k = 1.611 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 4.81 Determine the plastic moment M p of a steel beam of the cross section shown, assuming the steel to be elastoplastic with a yield strength of 240 MPa. SOLUTION A= For a semicircle: π 2 r 2; r = 4r 3π R = σY A Resultant force on semicircular section: Resultant moment on entire cross section: M p = 2 Rr = Data: 4 σY r3 3 σ Y = 240 MPa = 240 × 106 Pa, Mp = r = 18 mm = 0.018 m 4 (240 × 106 )(0.018)3 = 1866 N ⋅ m 3 M p = 1.866 kN ⋅ 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 4.82 Determine the plastic moment M p of a steel beam of the cross section shown, assuming the steel to be elastoplastic with a yield strength of 240 MPa. SOLUTION Total area: A = (50)(90) − (30)(30) = 3600 mm 2 1 A = 1800 mm 2 2 1 A 1800 x= 2 = = 36 mm b 50 A1 = (50)(36) = 1800 mm 2 , y1 = 18 mm, A1 y1 = 32.4 × 103 mm3 A2 = (50)(14) = 700 mm 2 , y2 = 7 mm, A2 y2 = 4.9 × 103 mm3 A3 = (20)(30) = 600 mm 2 , y3 = 29 mm, A3 y3 = 17.4 × 103 mm3 A4 = (50)(10) = 500 mm 2 , y4 = 49 mm, A4 y4 = 24.5 × 103 mm3 A1 y1 + A2 y2 + A3 y3 + A4 y4 = 79.2 × 103 mm3 = 79.2 × 10−6 m3 M p = σ Y ΣAi yi = (240 × 106 )(79.2 × 10−6 ) = 19.008 × 103 N ⋅ m M p = 19.01 kN ⋅ 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 4.83 Determine the plastic moment M p of a steel beam of the cross section shown, assuming the steel to be elastoplastic with a yield strength of 240 MPa. SOLUTION Total area: A= 1 (30)(36) = 540 mm 2 2 By similar triangles, b 30 = y 36 Since A1 = b= Half area: 1 A = 270 mm 2 = A1 2 5 y 6 1 5 2 by = y , 2 12 y2 = 12 A1 5 12 (270) = 25.4558 mm 5 b = 21.2132 mm 1 A1 = (21.2132)(25.4558) = 270 mm 2 = 270 × 10−6 m 2 2 A2 = (21.2132)(36 − 25.4558) = 223.676 mm 2 = 223.676 × 10−6 m 2 y= A3 = A − A1 − A2 = 46.324 mm 2 = 46.324 × 10−6 m 2 Ri = σ Y Ai = 240 × 106 Ai R1 = 64.8 × 103 N, R2 = 53.6822 × 103 N, R3 = 11.1178 × 103 N 1 y1 = y = 8.4853 mm = 8.4853 × 10−3 m 3 1 y2 = (36 − 25.4558) = 5.2721 mm = 5.2721 × 10−3 m 2 2 y3 = (36 − 25.4558) = 7.0295 mm = 7.0295 × 10−3 m 3 M p = R1 y1 + R2 y2 + R3 y3 = 911 N ⋅ m M p = 911 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 4.84 Determine the plastic moment M p of a steel beam of the cross section shown, assuming the steel to be elastoplastic with a yield strength of 240 MPa. SOLUTION Let c1 be the outer radius and c2 the inner radius. A1 y1 = Aa ya − Ab yb π 4c π 4c = c12 1 − c22 2 2 3π 2 3π 3 2 3 = c1 − c2 3 ( A2 y2 = A1 y1 = ) ( 2 3 c1 − c 22 3 ) M p = σ Y ( A1 y1 + A2 y2 ) = Data: ( 4 σ Y c13 − c32 3 ) σ Y = 240 MPa = 240 × 106 Pa c1 = 60 mm = 0.060 m c2 = 40 mm = 0.040 m 4 (240 × 106 )(0.0603 − 0.0403 ) 3 = 48.64 × 103 N ⋅ m Mp = M p = 48.6 kN ⋅ 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 4.85 Determine the plastic moment M p of the cross section shown, assuming the steel to be elastoplastic with a yield strength of 36 ksi. SOLUTION A = (1.8)(0.4) + (0.6)(1.2) = 1.44 in 2 Total area: 1 A = 0.72 in 2 2 1 A 0.72 x= 2 = = 1.2 in. b 0.6 Neutral axis lies 1.2 in. below the top. 1 1 A = 0.72 in 2 , y1 = (1.2) = 0.6 in. 2 2 1 1 2 A2 = A = 0.72 in , y2 = (0.4) = 0.2 in. 2 2 M p = σ Y ( A1 y1 + A2 y2 ) A1 = = (36)[(0.72)(0.6) + (0.72)(0.2)] = 20.7 kip ⋅ in M p = 20.7 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 4.86 Determine the plastic moment M p of the cross section shown, assuming the steel to be elastoplastic with a yield strength of 36 ksi. SOLUTION Total area: 1 1 1 A = (4) + (3) + (2) = 4.5 in 2 2 2 2 1 A = 2.25 in 2 2 A1 = 2.00 in 2 , y1 = 0.75, A1 y1 = 1.50 in 3 A2 = 0.25 in 2 , y2 = 0.25, A2 y2 = 0.0625 in 3 A3 = 1.25 in 2 , y3 = 1.25, A3 y3 = 1.5625 in 3 A4 = 1.00 in 2 , y4 = 2.75, A4 y4 = 2.75 in 3 M p = σ Y ( A1 y1 + A2 y2 + A3 y3 + A4 y4 ) = (36)(1.50 + 0.0625 + 1.5625 + 2.75) M p = 212 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 4.87 For the beam indicated (of Prob. 4.73), a couple of moment equal to the full plastic moment M p is applied and then removed. Using a yield strength of 240 MPa, determine the residual stress at y = 45 mm . SOLUTION M p = 29.16 × 103 N ⋅ m See solutions to Problems 4.73 and 4.77. I = 3.645 × 10−6 m 4 c = 0.045 m σ′ = Mp c M max y = I I UNLOADING LOADING σ′ = at y = c = 45 mm RESIDUAL STRESSES (29.16 × 103 )(0.045) = 360 × 106 Pa −6 3.645 × 10 σ res = σ ′ − σ Y = 360 × 106 − 240 × 106 = 120 × 106 Pa σ res = 120.0 MPa 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 4.88 For the beam indicated (of Prob. 4.74), a couple of moment equal to the full plastic moment M p is applied and then removed. Using a yield strength of 240 MPa, determine the residual stress at y = 45 mm . SOLUTION M p = 27.54 × 103 N ⋅ m (See solutions to Problems 4.74 and 4.78.) I = 3.5775 × 10−6 m 4 , LOADING c = 0.045 m σ′ = Mp c M max y = I I σ′ = (27.54 × 103 )(0.045) = 346.4 × 106 Pa −6 3.5775 × 10 UNLOADING at y =c RESIDUAL STRESSES σ res = σ ′ − σ Y = 346.4 × 106 − 240 × 106 = 106.4 × 106 Pa σ res = 106.4 MPa 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 4.89 A bending couple is applied to the bar indicated, causing plastic zones 3 in. thick to develop at the top and bottom of the bar. After the couple has been removed, determine (a) the residual stress at y = 4.5in. , (b) the points where the residual stress is zero, (c) the radius of curvature corresponding to the permanent deformation of the bar. SOLUTION See solution to Problem 4.75 for bending couple and stress distribution. M = 4725 kip ⋅ in σ Y = 42 ksi (a) yY = 1.5 in. I = 357.75 in 4 c = 4.5 in. Mc (4725)(4.5) = = 59.43 ksi I 357.75 MyY (4725)(1.5) σ ′′ = = = 19.81 ksi I 357.75 At y = c, σ res = σ ′ − σ Y = 59.43 − 42 = 17.43 ksi σ′ = At σ res = 0 ∴ y0 = (c) At y = yY , σ res = σ ′′ − σ Y = 19.81 − 42 = −22.19 ksi UNLOADING LOADING (b) E = 29 × 106 psi = 29 × 103 ksi RESIDUAL STRESSES My0 − σY = 0 I Iσ Y (357.75)(42) = = 3.18 in. M 4725 Answer: y0 = −3.18in., 0, 3.18 in. y = yY , σ res = −22.19 ksi σ =− Ey ρ ∴ ρ =− Ey σ = (29 × 103 )(1.5) = 1960 in. 22.19 ρ = 163.4 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 4.90 A bending couple is applied to the bar indicated, causing plastic zones 3 in. thick to develop at the top and bottom of the bar. After the couple has been removed, determine (a) the residual stress at y = 4.5in., (b) the points where the residual stress is zero, (c) the radius of curvature corresponding to the permanent deformation of the bar. SOLUTION See solution to Problem 4.76 for bending couple and stress distribution during loading. M = 2646 kip ⋅ in I = 188.5 in 4 σ Y = 42 ksi (a) (c) E = 29 × 106 psi = 29 × 103 ksi c = 4.5 in. Mc (2646)(4.5) = = 63.17 ksi I 188.5 MyY (2646)(1.5) σ ′′ = = = 21.06 ksi I 188.5 σ′ = At y = c, σ res = σ ′ − σ Y = 63.17 − 42 = 21.17 ksi At y = yY , σ res = σ ′′ − σ Y = 21.06 − 42 σ res = −20.94 ksi UNLOADING LOADING (b) yY = 1.5in. My0 = σY I Iσ Y (188.5)(42) y0 = = = 2.992 in. M 2646 RESIDUAL STRESSES σ res = 0 ∴ At y = yY , σ =− Ey ρ Answer: y0 = −2.992 in., 0, 2.992 in. σ res = −20.94 ksi ∴ ρ =− Ey σ = (29 × 103 )(1.5) = 2077 in. 20.94 ρ = 173.1 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 4.91 A bending couple is applied to the beam of Prob. 4.73, causing plastic zones 30 mm thick to develop at the top and bottom of the beam. After the couple has been removed, determine (a) the residual stress at y = 45 mm, (b) the points where the residual stress is zero, (c) the radius of curvature corresponding to the permanent deformation of the beam. SOLUTION See solution to Problem 4.73 for bending couple and stress distribution during loading: M = 28.08 × 103 N ⋅ m I = 3.645 × 10−6 m 4 σ Y = 240 MPa (a) E = 200 GPa c = 0.045 m σ′ = Mc (28.08 × 103 )(0.045) = = 346.7 × 106 Pa = 346.7 MPa I 3.645 × 10−6 σ ′′ = M yY (28.08 × 103 )(0.015) = = 115.6 × 106 Pa = 115.6 MPa −6 I 3.645 × 10 At y = c, σ res = σ ′ − σ Y = 346.7 − 240 σ res = 106.7 MPa At y = yY , σ res = σ ′′ − σ Y = 115.6 − 240 σ res = −124.4 MPa UNLOADING LOADING (b) yY = 15 mm = 0.015 m σ res = 0 ∴ y0 = RESIDUAL STRESSES M y0 − σY = 0 I Iσ Y (3.645 × 10−6 )(240 × 106 ) = = 31.15 × 10−3 m = 31.15 mm M 28.08 × 103 Answer: y0 = −31.15 mm, 0, 31.15 mm (c) At y = yY , σ res = −124.4 × 106 Pa σ =− Ey ρ ∴ ρ =− Ey σ = (200 × 109 )(0.015) −124.4 × 106 ρ = 24.1 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 4.92 A beam of the cross section shown is made of a steel that is assumed to be elastoplastic with E = 29 × 106 psi and σ Y = 42 ksi . A bending couple is applied to the beam about z axis, causing plastic zones 2 in. thick to develop at the top and bottom of the beam. After the couple has been removed, determine (a) the residual stress at y = 2 in., (b) the points where the residual stress is zero, (c) the radius of curvature corresponding to the permanent deformation of the beam. SOLUTION See solution to Problem 4.76 for bending couple and stress distribution during loading. M = 406 kip ⋅ in σ Y = 42 ksi (a) yY = 1.0 in. I = 14.6667 in 4 E = 29 × 106 psi = 29 × 103 ksi c = 2 in. Mc (406)(2) = = 55.36 ksi I 14.6667 MyY (406)(1.0) σ ′′ = = = 27.68 ksi I 14.6667 σ′ = σ res = σ ′ − σ Y = 55.36 − 42 At y = c, At y = yY , σ res = σ ′′ − σ Y = 27.68 − 42 UNLOADING LOADING (b) σ res = 0 ∴ y0 = (c) σ =− Ey ρ σ res = −14.32 ksi RESIDUAL STRESSES M y0 − σY = 0 I Iσ Y (14.6667)(42) = = 1.517 in. M 406 At y = yY , σ res = 13.36 ksi Answer: y0 = −1.517 in., 0, 1.517 in. σ res = −14.32 ksi ∴ ρ =− Ey σ = (29 × 103 )(1.0) = 2025 in. 14.32 ρ = 168.8 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 4.93* A rectangular bar that is straight and unstressed is bent into an arc of circle of radius by two couples of moment M. After the couples are removed, it is observed that the radius of curvature of the bar is ρ R . Denoting by ρY the radius of curvature of the bar at the onset of yield, show that the radii of curvature satisfy the following relation: 1 ρR 2 1 3 ρ 1 ρ 1 − = 1 − ρ 2 ρY 3 ρY SOLUTION 1 ρY Let m denote = MY , EI M = 3 1 ρ2 M Y 1 − , 2 3 ρY2 M : MY M 3 ρ2 ρ2 = 1 − 2 ∴ = 3−2m MY 2 ρY ρY2 1 1 M 1 mM Y 1 m = − = − = − ρR ρ EI ρ EI ρ ρY m= = 1 ρ 1 3 ρ 1 ρ 2 m − = − − 1 1 1 ρ ρY ρ 2 ρY 3 ρY2 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 4.94 A solid bar of rectangular cross section is made of a material that is assumed to be elastoplastic. Denoting by M Y and ρY , respectively, the bending moment and radius of curvature at the onset of yield, determine (a) the radius of curvature when a couple of moment M = 1.25 M Y is applied to the bar, (b) the radius of curvature after the couple is removed. Check the results obtained by using the relation derived in Prob. 4.93. SOLUTION (a) 1 ρY = m= MY , EI M = 3 1 ρ2 M Y 1 − 2 3 ρY2 M 3 1 ρ2 = 1 − MY 2 3 ρY2 Let m = M = 1.25 MY ρ = 3 − 2m = 0.70711 ρY ρ = 0.70711ρY (b) 1 ρR = = 1 ρ − M 1 mM Y 1 m 1 1.25 = − = − = − EI EI 0.70711ρY ρ ρ ρY ρY 0.16421 ρY ρ R = 6.09ρY 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 4.95 The prismatic bar AB is made of a steel that is assumed to be elastoplastic and for which E = 200 GPa . Knowing that the radius of curvature of the bar is 2.4 m when a couple of moment M = 350 N ⋅ m is applied as shown, determine (a) the yield strength of the steel, (b) the thickness of the elastic core of the bar. SOLUTION M = 3 1 ρ2 M Y 1 − 2 3 ρY2 1 ρ 2σ Y2 1 − 3 E 2c 2 = 3 σY I 2 c = 3 σ Y b(2c)3 1 ρ 2σ Y2 1 − 2 12c 3 E 2c 2 1 ρ 2σ Y2 = σ Y bc 2 1 − 3 E 2c 2 (a) ρ 2σ Y2 bc 2σ y 1 − =M 3E 2c 2 Cubic equation for σ Y Data: E = 200 × 109 Pa M = 420 N ⋅ m ρ = 2.4 m b = 20 mm = 0.020 m c= 1 h = 8 mm = 0.008 m 2 (1.28 × 10−6 ) σ Y 1 − 750 × 10−21σ Y2 = 350 σ Y 1 − 750 × 10−21σ Y2 = 273.44 × 106 Solving by trial, (b) yY = σY ρ E = σ Y = 292 × 106 Pa σ Y = 292 MPa (292 × 106 )(2.4) = 3.504 × 10−3 m = 3.504 mm 200 × 109 thickness of elastic core = 2 yY = 7.01 mm 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 4.96 The prismatic bar AB is made of an aluminum alloy for which the tensile stress-strain diagram is as shown. Assuming that the σ -ε diagram is the same in compression as in tension, determine (a) the radius of curvature of the bar when the maximum stress is 250 MPa, (b) the corresponding value of the bending moment. (Hint: For part b, plot σ versus y and use an approximate method of integration.) SOLUTION (a) σ m = 250 MPa = 250 × 106 Pa ε m = 0.0064 from curve 1 h = 30 mm = 0.030 m 2 b = 40 mm = 0.040 m c= 1 ρ (b) = ε m 0.0064 = = 0.21333 m −1 c 0.030 ρ = 4.69 m ε = −ε m Strain distribution: M = − Bending couple: y = −ε mu c c −c u = where c 0 y ε 1 yσ bdy = 2b y |σ | d y = 2bc 2 0 u |σ | du = 2bc 2 J 1 where the integral J is given by 0 u |σ | du Evaluate J using a method of numerial integration. If Simpson’s rule is used, the integration formula is J = Δu Σwu |σ | 3 where w is a weighting factor. Using Δu = 0.25, we get the values given in the table below: u 0 0.25 0.5 0.75 1.00 |ε | 0 0.0016 0.0032 0.0048 0.0064 |σ |, (MPa) 0 110 180 225 250 u |σ |, (MPa) 0 27.5 90 168.75 250 w 1 4 2 4 1 wu |σ |, (MPa) 0 110 180 675 250 1215 ← Σwu |σ | (0.25)(1215) = 101.25 MPa = 101.25 × 106 Pa 3 M = (2)(0.040)(0.030)2 (101.25 × 106 ) = 7.29 × 103 N ⋅ m J = M = 7.29 kN ⋅ 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 4.97 The prismatic bar AB is made of a bronze alloy for which the tensile stress-strain diagram is as shown. Assuming that the σ − ε diagram is the same in compression as in tension, determine (a) the maximum stress in the bar when the radius of curvature of the bar is 100 in., (b) the corresponding value of the bending moment. (See hint given in Prob. 4.96.) SOLUTION (a) ρ = 100 in., b = 0.8 in., c = 0.6 in. εm = (b) c ρ = 0.6 = 0.006 100 Strain distribution: Bending couple: From the curve, ε = −ε m M = − y = −ε mu c c −c where u = c 0 σ m = 43 ksi y ε 1 y σ bdy = 2b y |σ | dy = 2bc 2 0 u |σ | du = 2bc 2 J 1 where the integral J is given by 0 u |σ | du Evaluate J using a method of numerial integration. If Simpson’s rule is used, the integration formula is J = Δu Σwu|σ | 3 where w is a weighting factor. Using Δu = 0.25, we get the values given the table below: u |ε | 0 0.25 0.5 0.75 1.00 0 0.0015 0.003 0.0045 0.006 |σ |, ksi 0 25 36 40 43 u |σ |, ksi 0 6.25 18 30 43 J = w 1 4 2 4 1 wu |σ |, ksi 0 25 36 120 43 224 ← Σwu |σ | (0.25) (224) = 18.67 ksi 3 M = (2)(0.8)(0.6)2 (18.67) M = 10.75 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 4.98 A prismatic bar of rectangular cross section is made of an alloy for which the stress-strain diagram can be represented by the relation ε = kσ n for σ > 0 and ε = − kσ n for σ < 0 . If a couple M is applied to the bar, show that the maximum stress is σm = 1 + 2n Mc 3n I SOLUTION Strain distribution: ε = −ε m y = −ε mu where c u = y c Bending couple: c c c M = − − c y σ bdy = 2b 0 y |σ | dy = 2bc 2 0 y dy |σ | c c 1 = 2 bc 2 0 u |σ | du For ε = Kσ n , ε m = Kσ m σ ε =u = εm σm Then n ∴ 1 | σ | = σ mu n 1 1 1 1+ 1n du M = 2bc 2 0 uσ mu n du = 2bc 2σ m 0 u 2+ 1 u n = 2bc σ m 2 + 1n 2 σm = 1 2n 2 0 = 2n + 1 bc σ m 2n + 1 M 2 bc 2 Recall: that I 1 b(2c)3 2 1 2 c = = bc 2 ∴ = 2 3 3 I c 12 c bc Then σm = 2n + 1 Mc 3n 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 4.99 A short wooden post supports a 6-kip axial load as shown. Determine the stress at point A when (a) b = 0, (b) b = 1.5in., (c) b = 3in. SOLUTION A = π r 2 = π (3) 2 = 28.27 in 2 π π r 4 = (3) 4 = 63.62 in 4 4 4 I 63.62 S = = = 21.206 in 3 c 3 P = 6 kips M = Pb I = (a) b=0 σ =− (b) M =0 P 6 =− = −0.212 ksi A 28.27 σ = −212 psi b = 1.5 in. M = (6)(1.5) = 9 kip ⋅ in σ =− P M 6 9 − =− − = −0.637 ksi 28.27 21.206 A S σ = −637 psi (c) b = 3 in. σ =− M = (6)(3) = 18 kip ⋅ in P M 6 18 − =− − = −1.061 ksi 28.27 21.206 A S σ = −1061 psi 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 4.100 As many as three axial loads, each of magnitude P = 10 kips, can be applied to the end of a W8 × 21 rolled-steel shape. Determine the stress at point A, (a) for the loading shown, (b) if loads are applied at points 1 and 2 only. SOLUTION For W8 × 21 Appendix C gives A = 6.16 in 2 , d = 8.28 in., I x = 75.3 in 4 At point A, (a) Centric loading: 1 d = 4.14 in. 2 F My − σ = A I y = F = 30 kips, M = 0 σ = (b) Eccentric loading: 30 6.16 σ = 4.87 ksi F = 2 P = 20 kips M = −(10)(3.5) = −35 kip ⋅ in σ = 20 (−35)(4.14) − 6.16 75.3 σ = 5.17 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 4.101 Knowing that the magnitude of the horizontal force P is 8 kN, determine the stress at (a) point A, (b) point B. SOLUTION A = (30)(24) = 720 mm 2 = 720 × 10−6 m 2 e = 45 − 12 = 33mm = 0.033 m 1 3 1 (30)(24)3 = 34.56 × 103 mm 4 = 34.56 × 10−9 m 4 bh = 12 12 1 c = (24 mm) = 12 mm = 0.012 m P = 8 × 103 N 2 I = M = Pe = (8 × 103 )(0.033) = 264 N ⋅ m (a) σA = − P Mc 8 × 103 (264)(0.012) − =− = = −102.8 × 106 Pa A I 720 × 10−6 34.56 × 10−9 σ A = −102.8 MPa (b) σB = − P Mc 8 × 103 (264)(0.012) + =− + = 80.6 × 106 Pa −6 A I 720 × 10 34.56 × 10−9 σ B = 80.6 MPa 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 4.102 The vertical portion of the press shown consists of a rectangular tube of wall thickness t = 10 mm. Knowing that the press has been tightened on wooden planks being glued together until P = 20 kN, determine the stress at (a) point A, (b) point B. SOLUTION Rectangular cutout is 60 mm × 40 mm. A = (80) (60) − (60)(40) = 2.4 × 103 mm 2 = 2.4 × 10−3 m 2 I = 1 1 (60)(80)3 − (40)(60)3 = 1.84 × 106 mm 4 12 12 = 1.84 × 10−6 m 4 c = 40 mm = 0.040 m e = 200 + 40 = 240 mm = 0.240 m P = 20 × 103 N M = Pe = (20 × 103 )(0.240) = 4.8 × 103 N ⋅ m (a) σA = P Mc 20 × 103 (4.8 × 103 )(0.040) + = + = 112.7 × 106 Pa A I 2.4 × 10−3 1.84 × 10−6 σ A = 112.7 MPa (b) σB = P Mc 20 × 103 (4.8 × 103 ) (0.040) − = − = −96.0 × 106 Pa A I 2.4 × 10−3 1.84 × 10−6 σ B = −96.0 MPa 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 4.103 Solve Prob. 4.102, assuming that t = 8 mm. PROBLEM 4.102 The vertical portion of the press shown consists of a rectangular tube of wall thickness t = 10 mm. Knowing that the press has been tightened on wooden planks being glued together until P = 20 kN, determine the stress at (a) point A, (b) point B. SOLUTION Rectangular cutout is 64 mm × 44 mm. A = (80) (60) − (64)(44) = 1.984 × 103 mm 2 = 1.984 × 10−3 mm 2 I = 1 1 (60)(80)3 − (44)(64)3 = 1.59881 × 106 mm 2 12 12 = 1.59881 × 10−6 m 4 c = 40 mm = 0.004 e = 200 + 40 = 240 mm = 0.240 m 3 P = 20 × 10 N M = Pe = (20 × 103 )(0.240) = 4.8 × 103 N ⋅ m (a) σA = P Mc 20 × 103 (4.8 × 103 )(0.040) + = + = 130.2 × 106 Pa A I 1.984 × 10−3 1.59881 × 10−6 (b) σB = P Mc 20 × 103 (4.8 × 103 ) (0.040) − = − = −110.0 × 106 Pa A I 1.984 × 10−3 1.59881 × 10−6 σ A = 130.2 MPa σ B = −110.0 MPa 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 4.104 Determine the stress at points A and B, (a) for the loading shown, (b) if the 60-kN loads are applied at points 1 and 2 only. SOLUTION (a) Loading is centric. P = 180 kN = 180 × 103 N A = (90)(240) = 21.6 × 103 mm 2 = 21.6 × 10−6 m 2 σ =− At A and B: P 180 × 103 = = −8.33 × 106 Pa −3 A 21.6 × 10 σ A = σ B = −8.33 MPa (b) Eccentric loading. P = 120 kN = 120 × 103 N M = (60 × 103 )(150 × 10−3 ) = 9.0 × 103 N ⋅ m 1 3 1 bh = (90)(240)3 = 103.68 × 106 mm 4 = 103.68 × 10−6 m 4 12 12 c = 120 mm = 0.120 m I = At A: σA = − At B: σB − P Mc 120 × 103 (9.0 × 103 )(0.120) − =− − = −15.97 × 106 Pa A I 21.6 × 10−3 103.68 × 10−6 P Mc 120 × 103 (9.0 × 103 )(0.120) + =− + = 4.86 × 106 Pa A I 21.6 × 10−3 103.68 × 10−6 σ A = −15.97 MPa σ B = 4.86 MPa 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 4.105 Knowing that the allowable stress in section ABD is 10 ksi, determine the largest force P that can be applied to the bracket shown. SOLUTION Statics: M = 2.45 P Cross section: A = (0.9)(1.2) = 1.08 in 2 c= 1 (0.9) = 0.45 in. 2 I = 1 (1.2)(0.9)3 = 0.0729 in 4 12 At point B: σ = −10 ksi P Mc − A I P (2.45P)(0.45) −10 = − − = −16.049P 1.08 0.0729 σ =− P = 0.623 kips P = 623 lb 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 4.106 Portions of a 12 × 12 -in. square bar have been bent to form the two machine components shown. Knowing that the allowable stress is 15 ksi, determine the maximum load that can be applied to each component. SOLUTION The maximum stress occurs at point B. σ B = −15 ksi = −15 × 103 psi σB = − 1 ec + A I K = where P Mc P Pec − =− − = − KP A I A I e = 1.0 in. A = (0.5) (0.5) = 0.25 in 2 I = 1 (0.5)(0.5)3 = 5.2083 × 10−3 in 4 for all centroidal axes. 12 (a) (a) c = 0.25 in. K = 1 (1.0) (0.25) + = 52 in −2 −3 0.25 5.2083 × 10 P=− (b) (b) c= σB K =− (−15 × 103 ) 52 P = 288 lb 0.5 = 0.35355 in. 2 K = 1 (1.0)(0.35355) + = 71.882 in −2 0.25 5.2083 × 10−3 P=− σB K =− (−15 × 103 ) 71.882 P = 209 lb 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 4.107 The four forces shown are applied to a rigid plate supported by a solid steel post of radius a. Knowing that P = 100 kN and a = 40 mm, determine the maximum stress in the post when (a) the force at D is removed, (b) the forces at C and D are removed. SOLUTION For a solid circular section of radius a, A = π a2 σ =− a4 M x = − Pa, σ =− F 4P =− 2 A πa Mz = 0 F M xz 3P (− Pa)(−a) 7P − =− 2 − =− 2 2 π A I a πa πa 4 Forces at C and D are removed: F = 2 P, Resultant bending couple: σ =− M = M x = − Pa, M x2 + M z2 = F Mc 2P − =− 2 − A I πa M z = − Pa 2 Pa 2 Pa a 2+4 2 P =− = −2.437 P/a 2 2 π a2 π a 4 P = 100 × 103 N, Numerical data: 4 Force at D is removed: F = 3P, (b) π F = 4 P, M x = M z = 0 Centric force: (a) I = a = 0.040 m Answers: (a) σ =− (7) (100 × 103 ) = −139.3 × 106 Pa 2 π (0.040) σ = −139.3 MPa (b) σ =− (2.437)(100 × 103 ) = −152.3 × 106 Pa (0.040) 2 σ = −152.3 MPa 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 4.108 A milling operation was used to remove a portion of a solid bar of square cross section. Knowing that a = 30 mm, d = 20 mm, and σ all = 60 MPa, determine the magnitude P of the largest forces that can be safely applied at the centers of the ends of the bar. SOLUTION A = ad , I = 1 ad 3 , 12 c= a d − 2 2 P Mc P 6Ped σ = + = + A I ad ad 3 3 P (a − d ) P σ = + = KP ad ad 2 1 d 2 e= Data: a = 30 mm = 0.030 m K = P= where K = 1 3(a − d ) + ad ad 2 d = 20 mm = 0.020 m 1 (3) (0.010) + = 4.1667 × 103 m −2 (0.030) (0.020) (0.030) (0.020) 2 σ K = 60 × 106 = 14.40 × 103 N 3 4.1667 × 10 P = 14.40 kN 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 4.109 A milling operation was used to remove a portion of a solid bar of square cross section. Forces of magnitude P = 18 kN are applied at the centers of the ends of the bar. Knowing that a = 30 mm and σ all = 135 MPa, determine the smallest allowable depth d of the milled portion of the bar. SOLUTION A = ad , e= I = 1 ad 3 , 12 c= 1 d 2 a d − 2 2 P 1 (a − d ) 1 d P Mc P Pec P 2 = P + 3P ( a − d ) = + = + 2 σ = + 1 A I ad I ad ad ad 2 ad 3 12 3P 2P 2P d − 3P = 0 σ = 2 − or σ d 2 + ad a d 2 1 2P 2P d = + 12 Pσ − 2σ a a Solving for d, a = 0.030 m, Data: d = P = 18 × 103 N, σ = 135 × 106 Pa 2 1 (2) (18 × 103 ) (2) (18 × 103 ) 3 6 + × × − 12(18 10 ) (135 10 ) 0.030 0.030 (2) (135 × 106 ) = 16.04 × 10−3 m d = 16.04 mm 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 4.110 A short column is made by nailing two 1 × 4-in. planks to a 2 × 4-in. timber. Determine the largest compressive stress created in the column by a 12-kip load applied as shown in the center of the top section of the timber if (a) the column is as described, (b) plank 1 is removed, (c) both planks are removed. SOLUTION (a) Centric loading: 4 in. × 4 in. cross section σ =− (b) P 12 =− A 16 Eccentric loading: 4 in. × 3 in. cross section 1 c = (3) = 1.5 in. 2 I = σ = −0.75 ksi A = (4) (3) = 12 in 2 e = 1.5 − 1.0 = 0.5 in. 1 3 1 bh = (4) (3)3 = 9 in 4 12 12 σ =− (c) A = (4)(4) = 16 in 2 P Pec 12 (12) (0.5) (1.5) − =− − A I 12 9 Centric loading: 4 in. × 2 in. cross section σ =− P 12 =− A 8 σ = −2.00 ksi A = (4) (2) = 8 in 2 σ = −1.50 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 4.111 An offset h must be introduced into a solid circular rod of diameter d. Knowing that the maximum stress after the offset is introduced must not exceed 5 times the stress in the rod when it is straight, determine the largest offset that can be used. SOLUTION For centric loading, σc = P A For eccentric loading, σe = P Phc + A I Given σe = 5 σc P Phc P + =5 A I A Phc P =4 I A ∴ π (4) d 4 4I 64 = 1d h= = cA d π 2 2 d 2 4 h = 0.500 d 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 4.112 An offset h must be introduced into a metal tube of 0.75-in. outer diameter and 0.08-in. wall thickness. Knowing that the maximum stress after the offset is introduced must not exceed 4 times the stress in the tube when it is straight, determine the largest offset that can be used. SOLUTION 1 d = 0.375 in. 2 c1 = c − t = 0.375 − 0.08 = 0.295 in. c= ( ) A = π c 2 − c12 = π (0.3752 − 0.2952 ) = 0.168389 in 2 I = π (c 4 4 ) − c14 = π 4 (0.3754 − 0.2954 ) −3 = 9.5835 × 10 in 4 For centric loading, σ cen = P A For eccentric loading, σ ecc = P Phc + A I σ ecc = 4 σ cen hc 3 = I A or h= P Phc P + =4 A I A 3I (3)(9.5835 × 10−3 ) = Ac (0.168389)(0.375) h = 0.455 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 4.113 A steel rod is welded to a steel plate to form the machine element shown. Knowing that the allowable stress is 135 MPa, determine (a) the largest force P that can be applied to the element, (b) the corresponding location of the neutral axis. Given: The centroid of the cross section is at C and I z = 4195 mm 4 . SOLUTION (a) A = (3)(18) + π 4 (6)2 = 82.27 mm 2 = 82.27 × 10−6 m 2 I = 4195 mm 4 = 4195 × 10−12 m 4 e = 13.12 mm = 0.01312 m Based on tensile stress at y = −13.12 mm = −0.01312 m σ = P Pec 1 ec + = + P = KP A I I A K = 1 ec 1 (0.01312)(0.01312) + = + = 53.188 × 103 m −2 −6 −12 A I 82.27 × 10 4195 × 10 P= (b) σ K = 135 × 106 = 2.538 × 103 N 3 53.188 × 10 Location of neutral axis. P = 2.54 kN σ =0 σ = P My P Pey − = − =0 A I A I ey 1 = I A y = I 4195 × 10−12 = = 3.89 × 10−3 m −6 Ae (82.27 × 10 )(0.01312) y = 3.89 mm The neutral axis lies 3.89 mm to the right of the centroid or 17.01 mm to the right of the line of action of the loads. 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 4.114 A vertical rod is attached at point A to the cast iron hanger shown. Knowing that the allowable stresses in the hanger are σ all = +5 ksi and σ all = −12 ksi , determine the largest downward force and the largest upward force that can be exerted by the rod. SOLUTION Ay (1 × 3)(0.5) + 2(3 × 0.25)(2.5) X = = A (1 × 3) + 2(3 × 0.75) X = 12.75 in 3 = 1.700 in. 7.5 in 2 A = 7.5 in 2 σ all = +5 ksi σ all = −12 ksi 1 I c = bh3 + Ad 2 12 1 1 (3)(1)3 + (3 × 1)(1.70 − 0.5) 2 + (1.5)(3)3 + (1.5 × 3)(2.5 − 1.70) 2 = 12 12 I c = 10.825 in 4 Downward Force. M = P(1.5 in.+1.70 in.) = (3.20 in.) P At D: σ D = + P Mc + A I P (3.20) P(1.70) + 7.5 10.825 + 5 = P(+0.6359) + 5 ksi = At E: σ E = + P = 7.86 kips ↓ P Mc − A I P (3.20) P(2.30) − 7.5 10.825 − 12 = P(−0.5466) − 12 ksi = We choose the smaller value. P = 21.95 kips ↓ P = 7.96 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 4.114 (Continued) Upward Force. M = P(1.5 in. + 1.70 in.) = (3.20 in.) P At D: σ D = + P Mc − A I P (3.20) P(1.70) − 7.5 10.825 −12 = P(−0.6359) −12 ksi = − At E: σ E = + P = 18.87 kips ↑ P Mc + A I P (3.20) P(2.30) + 7.5 10.825 +5 = P(+0.5466) +5 ksi = We choose the smaller value. P = 9.15 kips ↑ P = 9.15 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 4.115 Solve Prob. 4.114, assuming that the vertical rod is attached at point B instead of point A. PROBLEM 4.114 A vertical rod is attached at point A to the cast iron hanger shown. Knowing that the allowable stresses in the hanger are σ all = +5 ksi and σ all = −12 ksi , determine the largest downward force and the largest upward force that can be exerted by the rod. SOLUTION Ay (1 × 3)(0.5) + 2(3 × 0.25)(2.5) = X = A (1 × 3) + 2(3 × 0.75) X = 12.75 in 3 = 1.700 in. 7.5 in 2 A = 7.5 in 2 σ all = +5 ksi σ all = −12 ksi 1 I c = bh3 + Ad 2 12 1 1 = (3)(1)3 + (3 × 1)(1.70 − 0.5) 2 + (1.5)(3)3 + (1.5 × 3)(2.5 − 1.70) 2 12 12 I c = 10.825 in 4 Downward Force. σ all = +5 ksi σ all = −12 ksi M = (2.30 in. + 1.5 in.) = (3.80 in.) P At D: σ D = + P Mc − A I P (3.80) P(1.70) − 7.5 10.825 − 12 = P(−0.4634) −12 ksi = + At E: σ E = + P = 25.9 kips ↓ P Mc + A I P (3.80) P(2.30) + 7.5 10.825 +5 = P(+0.9407) +5 ksi = + We choose the smaller value. P = 5.32 kips ↓ P = 5.32 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 4.115 (Continued) Upward Force. σ all = +5 ksi M = (2.30 in. + 1.5 in.)P = (3.80 in.)P σ all = −12 ksi At D: σ D = − P Mc + A I P (3.80) P(1.70) + 7.5 10.825 5 = P(+0.4634) 5 ksi = − At E: σ E = − P = 10.79 kips ↑ P Mc − A I P (3.80) P(2.30) − 7.5 10.825 −12 = P(−0.9407) −12 ksi = − We choose the smaller value. P = 12.76 kips ↑ P = 10.79 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 4.116 Three steel plates, each of 25 × 150-mm cross section, are welded together to form a short H-shaped column. Later, for architectural reasons, a 25-mm strip is removed from each side of one of the flanges. Knowing that the load remains centric with respect to the original cross section, and that the allowable stress is 100 MPa, determine the largest force P (a) that could be applied to the original column, (b) that can be applied to the modified column. SOLUTION (a) Centric loading: σ =− P A A = (3)(150)(25) = 11.25 × 103 mm2 = 11.25 × 10−3 m 2 P = −σ A = −(−100 × 106 )(11.25 × 10−3 ) = 1.125 × 106 N (b) P = 1125 kN Eccentric loading (reduced cross section): A, 103 mm 2 y , mm A y (103 mm3 ) 3.75 87.5 328.125 76.5625 3.75 0 0 10.9375 2.50 −87.5 −218.75 98.4375 Σ 10.00 d , mm 109.375 Y = ΣAy 109.375 × 103 = = 10.9375 mm ΣA 10.00 × 103 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 4.116 (Continued) The centroid lies 10.9375 mm from the midpoint of the web. 1 1 b1h13 + A1d12 = (150)(25)3 + (3.75 × 103 )(76.5625) 2 = 22.177 × 106 mm 4 12 12 1 1 I2 = b2h23 + A2d 22 = (25) (150)3 + (3.75 × 103 )(10.9375) 2 = 7.480 × 106 mm 4 12 12 1 1 I3 = b3h33 + A3d32 = (100)(25)3 + (2.50 × 103 ) (98.4375)2 = 24.355 × 106 mm 4 12 12 I1 = I = I1 + I 2 + I 3 = 54.012 × 106 mm 4 = 54.012 × 10−6 m 4 c = 10.9375 + 75 + 25 = 110.9375 mm = 0.1109375 m M = Pe σ =− K = where e = 10.4375 mm = 10.4375 × 10−3 m P Mc P Pec − =− − = − KP A I A I A = 10.00 × 10−3 m 2 1 ec 1 (101.9375 × 10−3 )(0.1109375) = 122.465 m −2 + = + −6 −3 A I 10.00 × 10 54.012 × 10 P=− σ K =− (−100 × 106 ) = 817 × 103 N 122.465 P = 817 kN 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 4.117 A vertical force P of magnitude 20 kips is applied at point C located on the axis of symmetry of the cross section of a short column. Knowing that y = 5 in., determine (a) the stress at point A, (b) the stress at point B, (c) the location of the neutral axis. SOLUTION Locate centroid. Part A, in 2 y , in. 12 5 8 2 Σ 20 Eccentricity of load: e = 5 − 3.8 = 1.2 in. I1 = Ay , in 3 60 16 76 1 (6)(2)3 + (12)(1.2)2 = 21.28 in 4 12 y = I2 = ΣAy 76 = = 3.8 in. ΣA 20 1 (2)(4)3 + (8)(1.8) 2 = 36.587 in 4 12 I = I1 + I 2 = 57.867 in 4 (a) Stress at A: c A = 3.8 in. σA = − (b) σ A = 0.576 ksi Stress at B: cB = 6 − 3.8 = 2.2 in. σB = − (c) P Pec A 20 20(1.2)(3.8) + =− + 20 57.867 A I P PecB 20 20(1.2)(2.2) + =− − 20 57.867 A I σ B = −1.912 ksi P Pea ea 1 + =0 ∴ = A I I A I 57.867 a = = = 2.411 in. Ae (20)(1.2) Location of neutral axis: σ = 0 σ =− Neutral axis lies 2.411 in. below centroid or 3.8 − 2.411 = 1.389 in. above point A. Answer: 1.389 in. from point A. 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 4.118 A vertical force P is applied at point C located on the axis of symmetry of the cross section of a short column. Determine the range of values of y for which tensile stresses do not occur in the column. SOLUTION Locate centroid. A, in 2 Σ Eccentricity of load: 12 8 20 e = y − 3.8 in. I1 = y , in. Ay , in 3 5 2 60 16 76 y = ΣAi yi 76 = = 3.8 in. ΣAi 20 y = e + 3.8 in. 1 (6)(2)3 + (12)(1.2)2 = 21.28 in 4 12 1 (2)(4)3 + (8)(1.8)2 = 36.587 in 4 12 I2 = I = I1 + I 2 = 57.867 in 4 If stress at A equa ls zero, c A = 3.8 in. σA = − e= P Pec A + =0 A I ∴ ec A 1 = I A I 57.867 = = 0.761 in. Ac A (20)(3.8) y = 0.761 + 3.8 = 4.561 in. If stress at B equa ls zero. cB = 6 − 3.8 = 2.2 in. P PecB ecB 1 − =0 ∴ =− A I I A I 57.867 e=− =− = −1.315 in. AcB (20)(2.2) σB = − y = −1.315 + 3.8 = 2.485 in. Answer: 2.485 in. < y < 4.561 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 4.119 Knowing that the clamp shown has been tightened until P = 400 N, determine (a) the stress at point A, (b) the stress at point B, (c) the location of the neutral axis of section a − a. SOLUTION Cross section: Rectangle + Circle A1 = (20 mm)(4 mm) = 80 mm 2 y1 = 1 (20 mm) = 10 mm 2 A2 = π (2 mm)2 = 4π mm 2 y2 = 20 − 2 = 18 mm Ay (80)(10) + (4π )(18) cB = y = = = 11.086 mm 80 + 4π A c A = 20 − y = 8.914 mm d1 = 11.086 − 10 = 1.086 mm d 2 = 18 − 11.086 = 6.914 mm I1 = I1 + A1d12 = I 2 = I 2 + A2d 22 = 1 (4)(20)3 + (80)(1.086) 2 = 2.761 × 103 mm 4 12 π 4 (2) 4 + (4π )(6.914)2 = 0.613 × 103 mm 4 I = I1 + I 2 = 3.374 × 103 mm 4 = 3.374 × 10−9 m 4 A = A1 + A2 = 92.566 mm 2 = 92.566 × 10−6 m 2 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 4.119 (Continued) e = 32 + 8.914 = 40.914 mm = 0.040914 m M = Pe = (400 N)(0.040914 m) = 16.3656 N ⋅ m (a) Point A: σA = P Mc 400 (16.3656)(8.914 × 10−3 ) + = + −6 A I 92.566 × 10 3.374 × 10−9 = 4.321 × 106 + 43.23 × 106 = 47.55 × 106 Pa (b) Point B: σB = P Mc 400 (16.3656)(11.086) − = − A I 92.566 × 10−6 3.374 × 10−9 = 4.321 × 106 − 53.72 × 106 = −49.45 × 106 Pa (c) σ A = 47.6 MPa Neutral axis: σ B = −49.4 MPa By proportions, a 20 = 47.55 47.55 + 49.45 a = 9.80 mm 9.80 mm below top of section 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 4.120 The four bars shown have the same cross-sectional area. For the given loadings, show that (a) the maximum compressive stresses are in the ratio 4:5:7:9, (b) the maximum tensile stresses are in the ratio 2:3:5:3. (Note: the cross section of the triangular bar is an equilateral triangle.) SOLUTION Stresses: At A, σA = − Aec A P Pec A P − = − 1 + A I A I At B, σB = − P PecB P AecB + = − 1 A I A I 1 4 1 1 2 A1 = a , I1 = 12 a , c A = cB = 2 a, e = 2 a 1 1 (a 2 ) a a P 2 2 σ A = − 1 + 1 2 A a 12 2 1 1 (a ) a a 2 2 − 1 σ = P B A 1 2 a 12 a π 2 2 , I 2 = c4 , e = c A2 = π c = a ∴ c = 4 π P (π c 2 )(c)(c) σ A = − 1 + π 4 A2 c 4 P (π c 2 )(c)(c) σ B = − 1 π 4 A2 c 4 σ A = −4 P A1 σB = 2 P A1 σ A = −5 P A2 σB = 3 P A2 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 4.120 (Continued) 2 1 2 a I3 = a4 e = c A3 = a c = 2 12 2 2 a a ( a 2 ) 2 2 P σ A = − 1 + 1 4 A3 a 12 2 2 a ( a 2 ) 2 a 2 P −1 σ B = 1 4 A3 a 12 σ A = −7 P A3 σB = 5 P A3 3 2 s s s P 4 3 3 σ A = − 1 + A4 3 4 s 96 σ A = −9 P A4 3 2 s s s P 4 3 2 3 σB = − 1 A4 3 4 s 96 σB = 3 P A4 A4 = 1 3 3 2 (s) s = s 2 2 4 3 1 3 3 4 I4 = s s = s 36 2 96 cA = 2 3 s =e s= 3 2 3 cB = s 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 4.121 The C-shaped steel bar is used as a dynamometer to determine the magnitude P of the forces shown. Knowing that the cross section of the bar is a square of side 40 mm and that strain on the inner edge was measured and found to be 450 μ, determine the magnitude P of the forces. Use E = 200 GPa. SOLUTION At the strain gage location, σ = Eε = (200 × 109 )(450 × 10−6 ) = 90 × 106 Pa A = (40)(40) = 1600 mm 2 = 1600 × 10−6 m 2 1 (40)(40)3 = 213.33 × 103 mm 4 = 213.33 × 10−9 m 4 12 e = 80 + 20 = 100 mm = 0.100 m I = c = 20 mm = 0.020 m σ = P Mc P Pec + = + = KP A I A I K = 1 ec 1 (0.100)(0.020) + = + = 10.00 × 103 m −2 −6 A I 1600 × 10 213.33 × 10−9 P= σ K = 90 × 106 = 9.00 × 103 N 10.00 × 103 P = 9.00 kN 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 4.122 An eccentric force P is applied as shown to a steel bar of 25 × 90-mm cross section. The strains at A and B have been measured and found to be ε A = +350 μ ε B = −70 μ Knowing that E = 200 GPa, determine (a) the distance d, (b) the magnitude of the force P. SOLUTION 1 h = 45 mm = 0.045 m 2 A = bh − (25)(90) = 2.25 × 103 mm 2 = 2.25 × 10−3 m 2 h = 15 + 45 + 30 = 90 mm b = 25 mm c= 1 3 1 bh = (25)(90)3 = 1.51875 × 106 mm 4 = 1.51875 × 10−6 m 4 12 12 y A = 60 − 45 = 15 mm = 0.015 m yB = 15 − 45 = −30 mm = −0.030 m I= Stresses from strain gages at A and B: σ A = Eε A = (200 × 109 )(350 × 10−6 ) = 70 × 106 Pa σ B = Eε B = (200 × 109 )( −70 × 10−6 ) = −14 × 106 Pa Subtracting, σA −σB = − M =− σA = P MyA − A I (1) σB = P MyB − A I (2) M ( y A − yB ) I I (σ A − σ B ) (1.51875 × 10−6 )(84 × 106 ) =− = −2835 N ⋅ m 0.045 y A − yB Multiplying (2) by y A and (1) by yB and subtracting, P= (a) (b) y Aσ B − yBσ A = ( y A − yB ) P A A( y Aσ B − yBσ A ) (2.25 × 10−3 )[(0.015)(−14 × 106 ) − (−0.030)(70 × 106 )] = = 94.5 × 103 N 0.045 y A − yB M = − Pd ∴ d = − −2835 M =− = 0.030 m P 94.5 × 103 d = 30.0 mm P = 94.5 kN 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 4.123 Solve Prob. 4.122, assuming that the measured strains are ε A = +600 μ ε B = +420 μ PROBLEM 4.122 An eccentric force P is applied as shown to a steel bar of 25 × 90-mm cross section. The strains at A and B have been measured and found to be ε A = +350μ ε B = −70 μ Knowing that E = 200 GPa, determine (a) the distance d, (b) the magnitude of the force P. SOLUTION 1 h = 45 mm = 0.045 m 2 A = bh = (25)(90) = 2.25 × 103 mm 2 = 2.25 × 10−3 m 2 h = 15 + 45 + 30 = 90 mm b = 25 mm c= 1 3 1 bh = (25)(90)3 = 1.51875 × 106 mm 4 = 1.51875 × 10−6 m 4 12 12 y A = 60 − 45 = 15 mm = 0.015 m yB = 15 − 45 = −30 mm = −0.030 m I= Stresses from strain gages at A and B: σ A = Eε A = (200 × 109 )(600 × 10−6 ) = 120 × 106 Pa σ B = Eε B = (200 × 109 )(420 × 10−6 ) = 84 × 106 Pa Subtracting, σA = P My A − A I (1) σB = P MyB − A I (2) σA −σB = − M =− M ( y A − yB ) I I (σ A − σ B ) (1.51875 × 10−6 )(36 × 106 ) =− = −1215 N ⋅ m 0.045 y A − yB Multiplying (2) by y A and (1) by yB and subtracting, P= y Aσ B − yBσ A = ( y A − yB ) P A A( y Aσ B − yBσ A ) (2.25 × 10−3 )[(0.015)(84 × 106 ) − ( −0.030)(120 × 106 )] = = 243 × 103 N y A − yB 0.045 M = − Pd (a) (b) ∴ d =− −1215 M =− = 5 × 10−3 m 3 P 243 × 10 d = 5.00 mm P = 243 kN 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 4.124 A short length of a W8 × 31 rolled-steel shape supports a rigid plate on which two loads P and Q are applied as shown. The strains at two points A and B on the centerline of the outer faces of the flanges have been measured and found to be ε A = −550 × 10−6 in./in. ε B = −680 × 10−6 in./in. Knowing that E = 29 × 106 psi, determine the magnitude of each load. SOLUTION Strains: ε A = −550 × 10−6 in./in. εC = Stresses: ε B = −680 × 10−6 in./in. 1 1 (ε A + ε B ) = (−550 − 680)10−6 = −615 × 10−6 in./in. 2 2 σ A = Eε A = (29 × 106 psi)(−550 × 10−6 in./in.) = −15.95 ksi σ C = Eε C = (29 × 106 psi)(−615 × 10−6 in./in.) = −17.935 ksi A = 9.13 in 2 W8 × 31 S = 27.5 in 3 M = (4.5 in.)( P − Q) At point C: σC = − P+Q P+Q ; − 17.835 ksi = − A 9.13 in 2 P + Q = 162.83 kips (1) At point A: σA = − P+Q M − A S −15.95 ksi = −17.835 ksi − Solve simultaneously, (4.5 in.)( P − Q) ; 27.5 in 3 P − Q = −11.52 kips (2) P = 25.7 kips Q = 87.2 kips P = 25.7 kips ↓ Q = 87.2 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 4.125 Solve Prob. 4.124, assuming that the measured strains are ε A = +35 × 10−6 in./in. ε B = −450 × 10−6 in./in. PROBLEM 4.124 A short length of a W8 × 31 rolled-steel shape supports a rigid plate on which two loads P and Q are applied as shown. The strains at two points A and B on the centerline of the outer faces of the flanges have been measured and found to be ε A = −550 × 10−6 in./in. ε B = −680 × 10−6 in./in. Knowing that E = 29 × 106 psi, determine the magnitude of each load. SOLUTION See solution and figures of Prob. 4.124. ε A = +35 × 10−6 in./in.; εC = ε B = −450 × 10−6 in./in. 1 1 (ε A + ε B ) = (35 − 450)10−6 in./in. = −207.5 × 10−6 in./in. 2 2 Stresses: σ A = Eε A = (29 × 106 psi)(+35 × 10−6 in./in.) = +1.015 ksi σ C = Eε C = (29 × 106 psi)(−207.5 × 10−6 in./in.) = −6.0175 ksi At point C: σC = − P+Q P+Q ; − 6.0175 ksi = − A 9.13 in 2 P + Q = 54.94 kips (1) At point A: σA = − P+Q M − A S +1.015 ksi = −6.0175 − (4.5 in.)( P − Q) 27.5 in 3 P − Q = −42.98 kips (2) Solve simultaneously, P = 5.98 kips Q = 49.0 kips P = 5.98 kips ↓ Q = 49.0 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 4.126 The eccentric axial force P acts at point D, which must be located 25 mm below the top surface of the steel bar shown. For P = 60 kN, determine (a) the depth d of the bar for which the tensile stress at point A is maximum, (b) the corresponding stress at point A. SOLUTION 1 bd 3 12 1 1 c= d e= d −a 2 2 P Pec σA = + A I 1 1 P 1 12 2 d − a 2 d P 4 6a σA = + = − 2 b d d3 b d d A = bd I= ( (a) )( ) Depth d for maximum σ A : Differentiate with respect to d. dσ A P 4 12a = − 2 + 3 = 0 dd b d d (b) σA = 60 × 103 40 × 10−3 4 (6)(25 × 10−3 ) 6 − = 40 × 10 Pa −3 (75 × 10−3 ) 2 75 × 10 d = 3a d = 75 mm σ A = 40 MPa 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 4.127 The couple M is applied to a beam of the cross section shown in a plane forming an angle β with the vertical. Determine the stress at (a) point A, (b) point B, (c) point D. SOLUTION 1 (80)(100)3 = 6.6667 × 106 mm 4 = 6.6667 × 10−6 m 4 12 1 I y = (100)(80)3 = 4.2667 × 106 mm 4 = 4.2667 × 10−6 m 4 12 y A = − yB = − yD = 50 mm Iz = z A = z B = − z D = 40 mm M y = −250 sin 30° = −125 N ⋅ m M z = 250 cos 30° = 216.51 N ⋅ m (a) σA = − M z yA M y zA (216.51)(0.050) (−125)(0.040) + =− + Iz Iy 6.6667 × 10−6 4.2667 × 10−6 = −2.80 × 106 Pa (b) σB = − M z yB M y z B (216.51)(−0.050) ( −125)(0.040) + =− + Iz Iy 6.6667 × 10−6 4.2667 × 10−6 = 0.452 × 103 Pa (c) σD = − σ A = −2.80 MPa σ B = 0.452 MPa M z yD M y z D (216.51)(−0.050) (−125)(−0.040) + =− + Iz Iy 6.6667 × 10−6 4.2667 × 10−6 = 2.80 × 106 Pa σ D = 2.80 MPa 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 4.128 The couple M is applied to a beam of the cross section shown in a plane forming an angle β with the vertical. Determine the stress at (a) point A, (b) point B, (c) point D. SOLUTION 1 (80)(32)3 = 218.45 × 103 mm 4 = 218.45 × 10−9 m 4 12 1 I y = (32)(80)3 = 1.36533 × 106 mm 4 = 1.36533 × 10−6 m 4 12 y A = yB = − yD = 16 mm Iz = z A = − z B = − z D = 40 mm M y = 300 cos 30° = 259.81 N ⋅ m (a) σA = − M z = 300sin 30° = 150 N ⋅ m M z yA M y zA (150)(16 × 10−3 ) (259.81)(40 × 10−3 ) + =− + Iz Iy 218.45 × 10−9 1.36533 × 10−6 = −3.37 × 106 Pa (b) σB = − M z yB M y z B (150)(16 × 10−3 ) (259.81)(−40 × 10−3 ) + =− + Iz Iy 218.45 × 10−9 1.36533 × 10−6 = −18.60 × 106 Pa (c) σD = − σ A = −3.37 MPa σ B = −18.60 MPa M z yD M y z D (150)(−16 × 10−3 ) (259.81)(−40 × 10−3 ) + =− + Iz Iy 218.45 × 10−9 1.36533 × 10−6 = 3.37 × 106 Pa σ D = 3.37 MPa 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 4.129 The couple M is applied to a beam of the cross section shown in a plane forming an angle β with the vertical. Determine the stress at (a) point A, (b) point B, (c) point D. SOLUTION M z = −60 sin 40° = −38.567 kip ⋅ in M y = 60 cos 40° = 45.963 kip ⋅ in y A = yB = − yD = 3 in. z A = − z B = − z D = 5 in. (a) σA = − Iz = 1 π (10)(6)3 − 2 (1)2 = 178.429 in 4 12 4 Iy = 1 π (6)(10)3 − 2 (1)4 + π (1)2 (2.5) 2 = 459.16 in 4 12 4 M z yA M y zA ( −38.567)(3) (45.963)(5) + =− + 178.429 459.16 Iz Iy = 1.149 ksi (b) σB = − M z yB M y z B (−38.567)(3) (45.963)(−5) + =− + 178.429 459.16 Iz Iy = 0.1479 (c) σD = − σ A = 1.149 ksi σ B = 0.1479 ksi M z yD M y z D (−38.567)(−3) (45.963)(−5) + =− + 178.429 459.16 Iz Iy = −1.149 ksi σ D = −1.149 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 4.130 The couple M is applied to a beam of the cross section shown in a plane forming an angle β with the vertical. Determine the stress at (a) point A, (b) point B, (c) point D. SOLUTION Locate centroid. A, in 2 z , in. Az , in 3 16 −1 −16 8 2 16 Σ 24 0 The centroid lies at point C. 1 1 (2)(8)3 + (4)(2)3 = 88 in 4 12 12 1 1 I y = (8)(2)3 + (2)(4)3 = 64 in 4 3 3 y A = − yB = 1 in., yD = −4 in. Iz = z A = z B = −4 in., zD = 0 M z = 10 cos 20° = 9.3969 kip ⋅ in M y = 10 sin 20° = 3.4202 kip ⋅ in (a) σA = − M z yA M y zA (9.3969)(1) (3.4202)(−4) + =− + 88 64 Iz Iy (b) σB = − M z yB M y z B (9.3969)( −1) (3.4202)(−4) + =− + 88 64 Iz Iy σ B = −0.107 ksi (c) σD = − M z yD M y z D (9.3969)(−4) (3.4202)(0) + =− + 88 64 Iz Iy σ D = 0.427 ksi σ A = 0.321 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 4.131 The couple M is applied to a beam of the cross section shown in a plane forming an angle β with the vertical. Determine the stress at (a) point A, (b) point B, (c) point D. SOLUTION M y = 25sin 15° = 6.4705 kN ⋅ m M z = 25cos15° = 24.148 kN ⋅ m 1 1 (80)(90)3 + (80)(30)3 = 5.04 × 106 mm 4 12 12 −6 4 I y = 5.04 × 10 m Iy = 1 1 1 I z = (90)(60)3 + (60)(20)3 + (30)(100)3 = 16.64 × 106 mm 4 = 16.64 × 10−6 m 4 3 3 3 Stress: (a) σA = σ = M yz Iy − Mzy Iz (6.4705 kN ⋅ m)(0.045 m) (24.148 kN ⋅ m)(0.060 m) − 5.04 × 10−6 m 4 16.64 × 10−6 m 4 = 57.772 MPa − 87.072 MPa (b) σB = (6.4705 kN ⋅ m)( −0.045 m) (24.148 kN ⋅ m)(0.060 m) − 5.04 × 10−6 m 4 16.64 × 10−6 m 4 = −57.772 MPa − 87.072 MPa (c) σD = σ A = −29.3 MPa σ B = −144.8 MPa (6.4705 kN ⋅ m)(−0.015 m) (24.148 kN ⋅ m)( −0.100 m) − 5.04 × 10−6 m 4 16.64 × 10−6 m 4 = −19.257 MPa + 145.12 MPa σ D = −125.9 MPa 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 4.132 The couple M is applied to a beam of the cross section shown in a plane forming an angle β with the vertical. Determine the stress at (a) point A, (b) point B, (c) point D. SOLUTION Flange: 1 (8)(0.5)3 + (8)(0.5)(4.75) 2 12 = 90.333 in 4 Iz = Iy = 1 (0.5)(8)3 = 21.333 in 4 12 1 (0.3)(9)3 = 18.225 in 4 12 1 I y = (9)(0.3)3 = 0.02025 in 4 12 Iz = Web: I z = (2)(90.333) + 18.225 = 198.89 in 4 Total: I y = (2)(21.333) + 0.02025 = 42.687 in 4 y A = yB = − yD = 5 in. z A = − z B = − zC = 4 in. M z = 250 cos 30° = 216.51 kip ⋅ in M y = −250 sin 30° = −125 kip ⋅ in (a) σA = − M z yA M y zA (216.51)(5) (−125)(4) + =− + 198.89 42.687 Iz Iy (b) σB = − M z yB M y z B (216.51)(5) (−125)(−4) + =− + 198.89 42.687 Iz Iy (c) σD = − M z yD M y z D (216.51)( −5) ( −125)(−4) + =− + 198.89 42.687 Iz Iy σ A = −17.16 ksi σ B = 6.27 ksi σ D = 17.16 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 4.133 The couple M is applied to a beam of the cross section shown in a plane forming an angle β with the vertical. Determine the stress at (a) point A, (b) point B, (c) point D. SOLUTION 1 1 (4.8)(2.4)3 − (4)(1.6)3 = 4.1643 in 4 12 12 1 1 I y = (2.4)(4.8)3 − (1.6)(4)3 = 13.5851 in 4 12 12 y A = yB = − yD = 1.2 in. Iz = z A = − z B = − z D = 2.4 in. M z = 75sin15° = 19.4114 kip ⋅ in M y = 75cos15° = 72.444 kip ⋅ in (a) σA = − M z yA M y zA (19.4114)(1.2) (72.444)(2.4) + =− + 4.1643 13.5851 Iz Iy σ A = 7.20 ksi (b) σB = − M z yB M y z B (19.4114)(1.2) (72.444)(−2.4) + =− + 4.1643 13.5851 Iz Iy σ B = −18.39 ksi (c) σD = − M z yD M y z D (19.4114)(−1.2) (72.444)(−2.4) + =− + 4.1643 13.5851 Iz Iy σ D = −7.20 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 4.134 The couple M is applied to a beam of the cross section shown in a plane forming an angle β with the vertical. Determine the stress at (a) point A, (b) point B, (c) point D. SOLUTION π 2 8 4 π 4r π I z = r − r 2 = − r 8 2 3π 8 9π 4 = (0.109757)(20) 4 = 17.5611 × 10−3 mm 4 = 17.5611 × 10−9 m 4 Iy = π 8 r4 = π (20)4 8 = 62.832 × 10−3 mm 4 = 62.832 × 10−9 m 4 4r (4)(20) =− = −8.4883 mm 3π 3π yB = 20 − 8.4883 = 11.5117 mm y A = yD = − z A = − z D = 20 mm zB = 0 M z = 100 cos 30° = 86.603 N ⋅ m M y = 100sin 30° = 50 N ⋅ m (a) σA = − (86.603)(−8.4883 × 10−3 ) (50)(20 × 10−3 ) M z yA M y z A + =− + Iz Iy 17.5611 × 10−9 62.832 × 10−9 = 57.8 × 106 Pa (b) σB = − σ A = 57.8 MPa −3 (86.603)(11.5117 × 10 ) (50)(0) M z yB M y z B + =− + −9 Iz Iy 17.5611 × 10 62.832 × 10−9 = −56.8 × 106 Pa (c) σD = − σ B = −56.8 MPa −3 3 (86.603)(−8.4883 × 10 ) (50)(−20 × 10 ) M z yD M y z D + =− + Iz Iy 17.5611 × 10−9 62.832 × 10−9 = 25.9 × 106 Pa σ D =25.9 MPa 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 4.135 The couple M acts in a vertical plane and is applied to a beam oriented as shown. Determine (a) the angle that the neutral axis forms with the horizontal, (b) the maximum tensile stress in the beam. SOLUTION For C 200 × 17.1 rolled steel shape, I z = 0.538 × 106 mm 4 = 0.538 × 10−6 m 4 I y = 13.4 × 106 mm 4 = 13.4 × 10−6 m 4 zE = zD = − z A = − zB = yD = yB = −14.4 mm 1 (203) = 101.5 mm 2 yE = y A = 57 − 14.4 = 42.6 mm M z = (2.8 × 103 ) cos 10° = 2.7575 × 103 N ⋅ m M y = (2.8 × 103 ) sin 10° = 486.21 N ⋅ m (a) Angle of neutral axis. tan ϕ = Iz 0.538 tan θ = tan10° = 0.007079 13.4 Iy ϕ = 0.4056° α = 10° − 0.4056° (b) α = 9.59° Maximum tensile stress occurs at point D. σD = − M z yD M y z D (2.7575 × 103 )(−14.4 × 10−3 ) (486.21)(0.1015) + =− + Iz Iy 0.538 × 10−6 13.4 × 10−6 = 73.807 × 106 + 3.682 × 106 = 77.5 × 106 Pa σ D = 77.5 MPa 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 4.136 The couple M acts in a vertical plane and is applied to a beam oriented as shown. Determine (a) the angle that the neutral axis forms with the horizontal, (b) the maximum tensile stress in the beam. SOLUTION For W 310 × 38.7 rolled steel shape, I z = 85.1 × 106 mm 4 = 85.1 × 10−6 m 4 I y = 7.27 × 106 mm 4 = 7.27 × 10−6 m 4 1 y A = yB = − yD = − yE = (310) = 155 mm 2 1 z A = z E = − z B = − z D = (165) = 82.5 mm 2 M z = (16 × 103 ) cos 15° = 15.455 × 103 N ⋅ m M y = (16 × 103 ) sin 15° = 4.1411 × 103 N ⋅ m (a) Angle of neutral axis. tan ϕ = Iz 85.1 × 10−6 tan θ = tan15° = 3.1365 Iy 7.27 × 10−6 ϕ = 72.3° α = 72.3° − 15° (b) α = 57.3° Maximum tensile stress occurs at point E. σE = − =− M z yE M y z E + Iz Iy (15.455 × 103 )(−155 × 10−3 ) (4.1411 × 103 )(82.5 × 10−3 ) + 85.1 × 10−6 7.27 × 10−6 = 75.1 × 106 Pa σ E = 75.1 MPa 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 4.137 The couple M acts in a vertical plane and is applied to a beam oriented as shown. Determine (a) the angle that the neutral axis forms with the horizontal, (b) the maximum tensile stress in the beam. SOLUTION I z′ = 21.4 in 4 I y′ = 6.74 in 4 z ′A = z B′ = 0.859 in. y A = −4 in. z D′ = −4 + 0.859 in. = −3.141 in. yB′ = 4 in. yD′ = −0.25 in. M y′ = −15 sin 45° = −10.6066 kip ⋅ in M z′ = 15 cos 45° = 10.6066 kip ⋅ in (a) Angle of neutral axis. tan ϕ = I z′ 21.4 tan θ = tan (−45°) = 3.1751 6.74 I y′ ϕ = −72.5° α = 72.5° − 45° α = 27.5° (b) The maximum tensile stress occurs at point D. σD = − M z yD M y z D (10.6066)(−0.25) (−10.6066)(−3.141) + =− + 21.4 6.74 Iz Iy = 0.12391 + 4.9429 σ D = 5.07 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 4.138 The couple M acts in a vertical plane and is applied to a beam oriented as shown. Determine (a) the angle that the neutral axis forms with the horizontal, (b) the maximum tensile stress in the beam. SOLUTION I z′ = 176.9 × 103 mm 4 = 176.9 × 10−9 m 4 I y′ = 281 × 103 mm 4 = 281 × 10−9 m 4 y′E = −18.57 mm, z E = 25 mm M z ′ = 400 cos 30° = 346.41 N ⋅ m M y′ = 400sin 30° = 200 N ⋅ m (a) tan ϕ = I z′ 176.9 × 10−9 tan θ = ⋅ tan 30° = 0.36346 I y′ 281 × 10−9 ϕ = 19.97° α = 30° − 19.97° (b) α = 10.03° Maximum tensile stress occurs at point E. σE = − M y ′ z′E (346.41)(−18.57 × 10−3 ) (200)(25 × 10−3 ) M z′ y′E + =− + I z′ I y′ 176.9 × 10−9 281 × 10−9 = 36.36 × 106 + 17.79 × 106 = 54.2 × 106 Pa σ E = 54.2 MPa 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 4.139 The couple M acts in a vertical plane and is applied to a beam oriented as shown. Determine (a) the angle that the neutral axis forms with the horizontal, (b) the maximum tensile stress in the beam. SOLUTION Bending moments: M y′ = −35sin15° = −9.059 kip ⋅ in M z ′ = −35cos15° = 33.807 kip ⋅ in Moments of inertia: 1 1 (2.4)(4)3 − (2 × 0.4)(2)3 = 12.267 in 4 12 12 1 1 (2)(2.4)3 + (2)(1.6)3 = 2.987 in 4 = 12 12 I y′ = I z′ (a) Neutral axis: tan φ = I z′ 2.987 in 4 tan θ = tan(−15°) = −0.06525 I y′ 12.267 in 4 φ = 3.73° α = 15° − φ = 15° − 3.73° = 11.27° α = 11.3° (b) Maximum tensile stress at D: y′D = −1.2 in. σD = M y′ z D I y′ z′D = −2 in. − (−9.059 kip ⋅ in)(−2 in.) (33.807 kip ⋅ in)(−1.2 in.) M z ′ y′D = − I z′ 12.267 in 4 2.987 in 4 = 1.477 ksi + 13.582 ksi = 15.059 ksi σ D = 15.06 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 4.140 The couple M acts in a vertical plane and is applied to a beam oriented as shown. Determine (a) the angle that the neutral axis forms with the horizontal, (b) the maximum tensile stress in the beam. SOLUTION I z′ = 53.6 × 103 mm 4 = 53.6 × 10−9 m 4 I y′ = 14.77 × 103 mm 4 = 14.77 × 10−9 m 4 M z′ = 120 sin 70° = 112.763 N ⋅ m M y′ = 120 cos 70° = 41.042 N ⋅ m (a) Angle of neutral axis. θ = 20°. tan ϕ = I z′ I y′ tan θ = 53.6 × 10−9 tan 20° = 1.32084 14.77 × 10−9 ϕ = 52.871° α = 52.871° − 20° (b) α = 32.9° The maximum tensile stress occurs at point E. yE′ = −16 mm = −0.016 m z E = 10 mm = 0.010 m M y ′ M y′ z E′ σ E = − z′ E + I z′ I y′ =− (112.763)(−0.016) (41.042)(0.010) + 53.6 × 10−9 14.77 × 10−9 = 61.448 × 106 Pa σ E = 61.4 MPa 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 4.141 The couple M acts in a vertical plane and is applied to a beam oriented as shown. Determine the stress at point A. SOLUTION 1 I y = 2 (7.2)(2.4)3 = 66.355 in 4 3 1 I z = 2 (2.4)(7.2)3 + (2.4)(7.2)(1.2) 2 = 199.066 in 4 12 I yz = 2 {(2.4)(7.2)(1.2)(1.2)} = 49.766 in 4 Using Mohr’s circle determine the principal axes and principal moments of inertia. Y : (66.355 in 4 , 49.766 in 4 ) Z : (199.066 in 4 , − 49.766 in 4 ) E : (132.710 in 4 , 0) 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 4.141 (Continued) DY 49.766 = DE 66.355 2θ m = 36.87° θ m = 18.435° tan 2θ m = 2 2 R = DE + DY = 82.944 in 4 I u = 132.710 − 82.944 = 49.766 in 4 I v = 132.710 + 82.944 = 215.654 in 4 M u = 125sin18.435° = 39.529 kip ⋅ in M v = 125cos18.435° = 118.585 kip ⋅ in u A = 4.8 cos 18.435° + 2.4 sin 18.435° = 5.3126 in. ν A = −4.8 sin 18.435° + 2.4 cos 18.435° = 0.7589 in. M u Mν σA = − ν A + u A Iν =− Iu (118.585)(5.3126) (39.529)(0.7589) + 215.654 49.766 = −2.32 ksi σ A = −2.32 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 4.142 The couple M acts in a vertical plane and is applied to a beam oriented as shown. Determine the stress at point A. SOLUTION Using Mohr’s circle determine the principal axes and principal moments of inertia. Y : (8.7, 8.3) in 4 Z : (24.5, − 8.3) in 4 E : (16.6, 0) in 4 EF = 7.9 in 4 FZ = 8.3 in 4 R = 7.92 + 8.32 = 11.46 in 4 FZ 8.3 = = 1.0506 EF 7.9 I v = 16.6 + 11.46 = 28.06 in 4 tan 2θ m = θ m = 23.2° I u = 16.6 − 11.46 = 5.14 in 4 M u = M sin θ m = (60) sin 23.2° = 23.64 kip ⋅ in M v = M cos θ m = (60) cos 23.2° = 55.15 kip ⋅ in u A = y A cos θ m + z A sin θ m = −3.92cos 23.2° − 1.08 sin 23.2° = −4.03 in. v A = z A cos θ m − y A sin θ m = −1.08cos 23.2° + 3.92 sin 23.2° = 0.552 in. σA = − M vu A M u vA (55.15)( −4.03) (23.64)(0.552) + + =− 28.06 5.14 Iv Iu σ A = 10.46 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 4.143 The couple M acts in a vertical plane and is applied to a beam oriented as shown. Determine the stress at point A. SOLUTION Using Mohr’s circle determine the principal axes and principal moments of inertia. Y : (1.894, 0.800) × 106 mm 4 Z : (0.614, 0.800) × 106 mm 4 E : (1.254, 0) × 106 mm 4 2 2 R = EF + FZ = 0.6402 + 0.8002 × 10−6 = 1.0245 × 106 mm 4 I v = (1.254 − 1.0245) × 106 mm 4 = 0.2295 × 106 mm 4 = 0.2295 × 10−6 m 4 I u = (1.254 + 1.0245) × 106 mm 4 = 2.2785 × 106 mm 4 = 2.2785 × 10−6 m 4 FZ 0.800 × 106 θ m = 25.67° = = 1.25 FE 0.640 × 106 M v = M cos θ m = (1.2 × 103 ) cos 25.67° = 1.0816 × 103 N ⋅ m tan 2θ m = M u = − M sin θ m = −(1.2 × 103 ) sin 25.67° = −0.5198 × 103 N ⋅ m u A = y A cos θ m − z A sin θ m = 45 cos 25.67° − 45 sin 25.67° = 21.07 mm v A = z A cos θ m + y A sin θ m = 45 cos 25.67° + 45 sin 25.67° = 60.05 mm σA = − M vu A M u vA (1.0816 × 103 )(21.07 × 10−3 ) (−0.5198 × 103 )(60.05 × 10−3 ) =− + + Iu Iv 0.2295 × 10−6 2.2785 × 10−6 = 113.0 × 106 Pa σ A = 113.0 MPa 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 4.144 The tube shown has a uniform wall thickness of 12 mm. For the loading given, determine (a) the stress at points A and B, (b) the point where the neutral axis intersects line ABD. SOLUTION Add y- and z-axes as shown. Cross section is a 75 mm × 125 mm rectangle with a 51 mm × 101 mm rectangular cutout. 1 1 (75)(125)3 − (51)(101)3 = 7.8283 × 106 mm 4 = 7.8283 × 10−6 m 4 12 12 1 1 I y = (125)(75)3 − (101)(51)3 = 3.2781 × 103 mm 4 = 3.2781 × 10−6 m 4 12 12 A = (75)(125) − (51)(101) = 4.224 × 103 mm 2 = 4.224 × 10−3 m 2 Iz = Resultant force and bending couples: P = 14 + 28 + 28 = 70 kN = 70 × 103 N M z = −(62.5 mm)(14 kN) + (62.5 mm)(28kN) + (62.5 mm)(28 kN) = 2625 N ⋅ m M y = −(37.5 mm)(14 kN) + (37.5 mm)(28 kN) + (37.5 mm)(28 kN) = −525 N ⋅ m (a) σA = 70 × 103 (2625)(−0.0625) (−525)(0.0375) P M z yA M y zA − + = − + −3 A Iz Iy 4.224 × 10 7.8283 × 10−6 3.2781 × 10−6 = 31.524 × 106 Pa σB = σ A = 31.5 MPa 70 × 103 (2625)(0.0625) (−525)(0.0375) P M z yB M y z B − + = − + −3 A Iz Iy 4.224 × 10 7.8283 × 10−6 3.2781 × 10−6 = −10.39 × 106 Pa (b) σ B = −10.39 MPa Let point H be the point where the neutral axis intersects AB. z H = 0.0375 m, 0= yH = ?, σ H = 0 P M z yH M y z H − + A Iz Iy P Mz H 7.8283 × 10−6 70 × 103 ( −525)(0.0375) + + = − 3 A 2625 I y 3.2781 × 10−6 4.224 × 10 = 0.03151 m = 31.51 mm yH = Iz Mz 31.51 + 62.5 = 94.0 mm Answer: 94.0 mm above point A. 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 4.145 Solve Prob. 4.144, assuming that the 28-kN force at point E is removed. PROBLEM 4.144 The tube shown has a uniform wall thickness of 12 mm. For the loading given, determine (a) the stress at points A and B, (b) the point where the neutral axis intersects line ABD. SOLUTION Add y- and z-axes as shown. Cross section is a 75 mm × 125 mm rectangle with a 51 mm × 101 mm rectangular cutout. 1 1 (75)(125)3 − (51)(101)3 = 7.8283 × 106 mm 4 = 7.8283 × 10−6 m 4 12 12 1 1 3 I y = (125)(75) − (101)(51)3 = 3.2781 × 106 mm 4 = 3.2781 × 10−6 m 4 12 12 A = (75)(125) − (51)(101) = 4.224 × 103 mm 2 = 4.224 × 10−3 m 2 Iz = Resultant force and bending couples: P = 14 + 28 = 42 kN = 42 × 103 N M z = −(62.5 mm)(14 kN) + (62.5 mm)(28 kN) = 875 N ⋅ m M y = −(37.5 mm)(14 kN) + (37.5 mm)(28 kN) = 525 N ⋅ m (a) σA = 42 × 103 (875)(−0.0625) (525)(0.0375) P M z yA M y zA − + = − + −3 A Iz Iy 4.224 × 10 7.8283 × 10−6 3.2781 × 10−6 = 22.935 × 106 Pa σB = σ A = 22.9 MPa P M z yB M y z B 42 × 103 (875)(0.0625) (525)(0.0375) − + = − + −3 A Iz Iy 4.224 × 10 7.8283 × 10−6 3.2781 × 10−6 = 8.9631 × 106 Pa (b) σ B = 8.96 MPa Let point K be the point where the neutral axis intersects BD. z K = ?, 0= yK = 0.0625 m, σ H = 0 P M z yH M y z H − + A Iz Iy I y M z yH P 3.2781 × 10−6 (875)(0.0625) 42 × 103 − = − −6 M y Iz A 525 4.224 × 10−3 7.8283 × 10 = −0.018465 m = −18.465 mm zH = 37.5 + 18.465 = 56.0 mm Answer: 56.0 mm to the right of point B. 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 4.146 A rigid circular plate of 125-mm radius is attached to a solid 150 × 200-mm rectangular post, with the center of the plate directly above the center of the post. If a 4-kN force P is applied at E with θ = 30°, determine (a) the stress at point A, (b) the stress at point B, (c) the point where the neutral axis intersects line ABD. SOLUTION P = 4 × 103 N (compression) M x = − PR sin 30° = −(4 × 103 )(125 × 10−3 ) sin 30° = −250 N ⋅ m M z = − PR cos 30° = −(4 × 103 )(125 × 10−3 ) cos 30° = −433 N ⋅ m 1 (200)(150)3 = 56.25 × 106 mm 4 = 56.25 × 10−6 m 4 12 1 (150)(200)3 = 100 × 106 mm 4 = 100 × 10−6 m 4 Iz = 12 − x A = xB = 100 mm z A = z B = 75 mm Ix = A = (200)(150) = 30 × 103 mm 2 = 30 × 10−3 m 2 (a) σA = − P M x zA M z xA 4 × 103 (−250)(75 × 10−3 ) (−433)(−100 × 10−3 ) − + =− − + −3 A Ix Iz 30 × 10 56.25 × 10−6 100 × 10−6 σ A = 633 × 103 Pa = 633 kPa (b) σB = − P M x z B M z xB 4 × 103 (−250)(75 × 10−3 ) (−433)(100 × 10−3 ) − + =− − + A Ix Iz 30 × 10−3 56.25 × 10−6 100 × 10−6 σ B = −233 × 103 Pa = −233 kPa (c) Let G be the point on AB where the neutral axis intersects. σG = 0 zG = 75 mm σG = − P M x zG M z xG − + =0 A Ix Iz xG = xG = ? I z P M x Z G 100 × 10−6 + = Mz A Ix −433 = 46.2 × 10−3 m = 46.2 mm (−250)(75 × 10−3 ) 4 × 103 + −3 56.25 × 10−6 30 × 10 Point G lies 146.2 mm from point A 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 4.147 4.147 In Prob. 4.146, determine (a) the value of θ for which the stress at D reaches it largest value, (b) the corresponding values of the stress, at A, B, C, and D. 4.146 A rigid circular plate of 125-mm radius is attached to a solid 150 × 200-mm rectangular post, with the center of the plate directly above the center of the post. If a 4-kN force P is applied at E with θ = 30°, determine (a) the stress at point A, (b) the stress at point B, (c) the point where the neutral axis intersects line ABD. SOLUTION P = 4 × 103 N (a) PR = (4 × 103 )(125 × 10−3 ) = 500 N ⋅ m M x = − PR sin θ = −500sin θ M x = − PR cos θ = −500cos θ 1 (200)(150)3 = 56.25 × 106 mm 4 = 56.25 × 10−6 m 4 2 1 I z = (150)(200)3 = 100 × 106 mm 4 = 100 × 10−6 m 4 2 xD = 100 mm z D = −75 mm Ix = A = (200)(150) = 30 × 103 mm 2 = 30 × 10−3 m 2 σ =− 1 Rz sin θ P M xz M z x Rx cos θ − + = −P − + A Ix Iz Ix Iz A For σ to be a maximum, dσ =0 dθ with z = z D , x = xD dσ D Rz cos θ Rx sin θ = − P 0 + D + D =0 dθ Ix IZ sin θ I z (100 × 10−6 )(−75 × 10−3 ) 4 = tan θ = − z D = − = −6 −3 cos θ I x xD (56.25 × 10 )(100 × 10 ) 3 sin θ = 0.8, (b) σA = − cos θ = 0.6, θ = 53.1° P M x z A M z xA 4 × 103 (500)(0.8)(75 × 10−3 ) (500)(0.6)(−100 × 10−3 ) − + =− + − A Ix Iz 30 × 10−3 56.25 × 10−6 100 × 10−6 = (−0.13333 + 0.53333 + 0.300) × 106 Pa = 0.700 × 106 Pa σ A = 700 kPa σ B = (−0.13333 + 0.53333 − 0.300) × 106 Pa = 0.100 × 106 Pa σ B = 100 kPa σ C = (−0.13333 + 0 + 0) × 106 Pa σ D = (−0.13333 − 0.53333 − 0.300) × 106 Pa = − 0.967 × 106 Pa σ C = −133.3kPa σ D = −967 kPa 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 4.148 Knowing that P = 90 kips, determine the largest distance a for which the maximum compressive stress dose not exceed 18 ksi. SOLUTION A = (5 in.)(6 in.) − 2(2 in.)(4 in.) = 14 in 2 1 1 (5 in.)(6 in.)3 − 2 (2 in.)(4 in.)3 = 68.67 in 4 12 12 1 1 I z = 2 (1 in.)(5 in.)3 + (4 in.)(1 in.)3 = 21.17 in 4 12 12 Ix = Force-couple system at C: P= P For P = 90 kips: P = 90 kips M x = (90 kips)(2.5 in.) = 225 kip ⋅ in Maximum compressive stress at B: σB = − −18 ksi = − M x = P(2.5 in.) M z = Pa M z = (90 kips) a σ B = −18 ksi P M x (3 in.) M z (2.5 in.) − − A Ix Iz 90 kips (225 kip ⋅ in)(3 in.) (90 kips) a (2.5 in.) − − 14 in 2 68.67 in 4 21.17 in 4 −18 = −6.429 − 9.830 − 10.628a −1.741 = −10.628 a a = 0.1638 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 4.149 Knowing that a = 1.25 in., determine the largest value of P that can be applied without exceeding either of the following allowable stresses: σ ten = 10 ksi σ comp = 18 ksi SOLUTION A = (5 in.)(6 in.) − (2)(2 in.)(4 in.) = 14 in 2 1 1 (5 in.)(6 in.)3 − 2 (2 in.)(4 in.)3 = 68.67 in 4 12 12 1 1 I z = 2 (1 in.)(5 in.)3 + (4 in.)(1 in.)3 = 21.17 in 4 12 12 Ix = For a = 1.25 in., Force-couple system at C: P = P M x = P(2.5 in.) M y = Pa = (1.25in.) Maximum compressive stress at B: σB = − −18 ksi = − σ B = −18 ksi P M x (3 in.) M z (2.5 in.) − − A Ix Iz P P(2.5 in.)(3 in.) P(1.25in.)(2.5in.) − − 2 14 in 68.67 in 4 21.17 in 4 −18 = − 0.0714P − 0.1092P − 0.1476 P −18 = 0.3282 P P = 54.8 kips Maximum tensile stress at D: σD = − σ D = +10 ksi P M x (3 in.) M z (2.5 in.) + + A Ix Iz +10 ksi = − 0.0714P + 0.1092P + 0.1476P 10 = 0.1854 P P = 53.9 kips The smaller value of P is the largest allowable value. P = 53.9 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 4.150 The Z section shown is subjected to a couple M 0 acting in a vertical plane. Determine the largest permissible value of the moment M 0 of the couple if the maximum stress is not to exceed 80 MPa. Given: I max = 2.28 × 10−6 mm 4 , I min = 0.23 × 10−6 mm 4 , principal axes 25.7° and 64.3° . SOLUTION I v = I max = 2.28 × 106 mm 4 = 2.28 × 10−6 m 4 I u = I min = 0.23 × 106 mm 4 = 0.23 × 10−6 m 4 M v = M 0 cos 64.3° M u = M 0 sin 64.3° θ = 64.3° I tan ϕ = v tan θ Iu 2.28 × 10−6 tan 64.3° 0.23 × 10−6 = 20.597 = ϕ = 87.22° Points A and B are farthest from the neutral axis. uB = yB cos 64.3° + z B sin 64.3° = (−45) cos 64.3° + (−35) sin 64.3° = −51.05 mm vB = z B cos 64.3° − yB sin 64.3° = (−35) cos 64.3° − (−45) sin 64.3° = +25.37 mm σB = − 80 × 106 = − M vu B M u v B + Iv Iu (M 0 cos 64.3°)(−51.05 × 10−3 ) (M 0 sin 64.3°)(25.37 × 10−3 ) + 0.23 × 10−6 2.28 × 10−6 = 109.1 × 103 M 0 M0 = 80 × 106 109.1 × 103 M 0 = 733 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 4.151 Solve Prob. 4.150 assuming that the couple M 0 acts in a horizontal plane. PROBLEM 4.150 The Z section shown is subjected to a couple M 0 acting in a vertical plane. Determine the largest permissible value of the moment M 0 of the couple if the maximum stress is not to exceed 80 MPa. Given: I max = 2.28 × 10−6 mm 4 , and 64.3° . principal axes 25.7° I min = 0.23 × 10−6 mm 4 , SOLUTION I v = I min = 0.23 × 106 mm 4 = 0.23 × 106 m 4 I u = I max = 2.28 × 106 mm 4 = 2.28 × 106 m 4 M v = M 0 cos 64.3° M u = M 0 sin 64.3° θ = 64.3° tan ϕ = Iv tan θ Iu 0.23 × 10−6 tan 64.3° 2.28 × 10−6 = 0.20961 = ϕ = 11.84° Points D and E are farthest from the neutral axis. uD = yD cos 25.7° − z D sin 25.7° = (−5) cos 25.7° − 45 sin 25.7° = −24.02 mm vD = z D cos 25.7° + yD sin 25.7° = 45cos 25.7° + (−5) sin 25.7° = 38.38 mm σD = − M v u D M u vD (M D cos 64.3°)(−24.02 × 10−3 ) + =− Iv Iu 0.23 × 10−6 + (M 0 sin 64.3°)(38.38 × 10−3 ) 2.28 × 10−6 80 × 106 = 60.48 × 103 M 0 M 0 = 1.323 × 103 N ⋅ m M 0 = 1.323 kN ⋅ 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 4.152 A beam having the cross section shown is subjected to a couple M 0 that acts in a vertical plane. Determine the largest permissible value of the moment M 0 of the couple if the maximum stress in the beam is not |to exceed 12 ksi. Given: I y = I z = 11.3 in.4 , A = 4.75 in.2 , kmin = 0.983 in. (Hint: By reason of symmetry, the principal axes form 2 an angle of 45° with the coordinate axes. Use the relations I min = Akmin and I min + I max = I y + I z ) SOLUTION M u = M 0 sin 45° = 0.70711 M 0 M v = M 0 cos 45° = 0.7071 M 0 2 = (4.75)(0.983) 2 = 4.59 in 4 I min = Akmin I max = I y + I z − I min = 11.3 + 11.3 − 4.59 = 18.01 in 4 u B = yB cos 45° + z B sin 45° = −3.57 cos 45° + 0.93 sin 45° = −1.866 in. vB = z B cos 45° − yB sin 45° = 0.93 cos 45° − (−3.57) sin 45° = 3.182 in. σB = − u M v u B M u vB v + = −0.70711 M 0 − B + B Iv Iu I max I min (−1.866) 3.182 = 0.70711 M 0 − + = 0.4124 M 0 4.59 18.01 M0 = σB 0.4124 = 12 0.4124 M 0 = 29.1 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 4.153 Solve Prob. 4.152, assuming that the couple M 0 acts in a horizontal plane. PROBLEM 4.152 A beam having the cross section shown is subjected to a couple M 0 that acts in a vertical plane. Determine the largest permissible value of the moment M 0 of the couple if the maximum stress in the beam is not to exceed 12 ksi. Given: I y = I z = 11.3 in.4, A = 4.75 in.2, kmin = 0.983 in. (Hint: By reason of symmetry, the principal axes form an angle of 45° with the coordinate 2 and I min + I max = I y + I z ) axes. Use the relations I min = Akmin SOLUTION M u = M 0 cos 45° = 0.70711 M 0 M v = − M 0 sin 45° = − 0.70711 M 0 2 = (4.75)(0.983) 2 = 4.59 in 4 I min = Akmin I max = I y + I z − I min = 11.3 + 11.3 − 4.59 = 18.01 in 4 u D = yD cos 45° + z D sin 45° = −0.93 cos 45° + (−3.57 sin 45°) = −1.866 in. vD = z D cos 45° − yD sin 45° = (−3.57) cos 45° − (0.93) sin 45° = 3.182 in. σD = − u M vu D M u v D v + = −0.70711 M 0 − D + D Iv Iu I max I min (−1.866) 3.182 = 0.70711 M 0 − + = 0.4124 M 0 4.59 18.01 M0 = σD 0.4124 = 12 0.4124 M 0 = 29.1 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 4.154 An extruded aluminum member having the cross section shown is subjected to a couple acting in a vertical plane. Determine the largest permissible value of the moment M 0 of the couple if the maximum stress is not to exceed 12 ksi. Given: I max = 0.957 in 4 , I min = 0.427 in 4 , principal axes 29.4° and 60.6° . SOLUTION I u = I max = 0.957 in 4 I v = I min = 0.427 in 4 M u = M 0 sin 29.4°, M v = M 0 cos 29.4° θ = 29.4° tan ϕ = Iv 0.427 tan θ = tan 29.4° Iu 0.957 = 0.2514 ϕ = 14.11° Point A is farthest from the neutral axis. y A = −0.75 in., z A = −0.75 in. u A = y A cos 29.4° + z A sin 29.4° = −1.0216 in. v A = z A cos 29.4° − y A sin 29.4° = −0.2852 in. σA = − M vu A MV (M cos 29.4°)(−1.0216) ( M 0 sin 29.4°)(−0.2852) + u A =− 0 + Iv Iu 0.427 0.957 = 1.9381 M 0 M0 = σA 1.9381 = 12 1.9381 M 0 = 6.19 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 4.155 A couple M 0 acting in a vertical plane is applied to a W12 × 16 rolled-steel beam, whose web forms an angle θ with the vertical. Denoting by σ 0 the maximum stress in the beam when θ = 0, determine the angle of inclination θ of the beam for which the maximum stress is 2σ 0. SOLUTION For W 12 × 16 rolled steel section, I z = 103 in 4 I y = 2.82 in 4 d = 11.99 in. b f = 3.990 in. yA = − tan ϕ = d 2 zA = bf 2 Iz 103 tan θ = tan θ = 36.52 tan θ Iy 2.82 Point A is farthest from the neutral axis. M y = M 0 sin θ σA = − For θ = 0, M z = M 0 cos θ M 0b f I zb f M z yA M y zA M d M d + = 0 cos θ + sin θ = 0 1 + tan θ cos θ Iz Iy I yd 2I z 2I y 2 I z σ0 = M 0d 2I z σ A = σ 0 1 + I 2b f I yd tan θ = tan θ cos θ = 2σ 0 I yd I zb f (103)(3.990) 2 − 1 (2.82)(11.99) cos θ 2 − 1 tan θ = 0.082273 cos θ Assuming cos θ ≈ 1, we get θ = 4.70° 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 4.156 Show that, if a solid rectangular beam is bent by a couple applied in a plane containing one diagonal of a rectangular cross section, the neutral axis will lie along the other diagonal. SOLUTION b h M z = M cos θ , M z = M sin θ tan θ = Iz = 1 3 bh 12 Iy = 1 3 hb 12 1 3 bh Iz b h tan ϕ = tan θ = 12 ⋅ = 1 Iy b hb3 h 12 The neutral axis passes through corner A of the diagonal AD. 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 4.157 A beam of unsymmetric cross section is subjected to a couple M 0 acting in the horizontal plane xz. Show that the stress at point A, of coordinates y and z, is σA = zI z − yI yz 2 I y I z − I yz My where Iy, Iz, and Iyz denote the moments and product of inertia of the cross section with respect to the coordinate axes, and My the moment of the couple. SOLUTION The stress σ A varies linearly with the coordinates y and z. Since the axial force is zero, the y- and z-axes are centroidal axes: σ A = C1 y + C2 z where C1 and C2 are constants. M z = − yσ AdA = −C1 y 2dA − C2 yzdA = − I zC1 − I yzC2 = 0 C1 = − I yz Iz C2 M y = zσ AdA = C1 yz dA + C2 z 2dA = I yzC1 + I yC2 − I yz ( I yz Iz C2 + I yC2 ) 2 I z M y = I y I z − I yz C2 C2 = IzM y 2 I y I z − I yz C1 = − σA = I yz M y 2 I y I z − I yz I z z − I yz y 2 I y I z − I yz My 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 4.158 A beam of unsymmetric cross section is subjected to a couple M 0 acting in the vertical plane xy. Show that the stress at point A, of coordinates y and z, is σA = yI y − zI yz 2 I y I z − I yz Mz where I y , I z , and I yz denote the moments and product of inertia of the cross section with respect to the coordinate axes, and M z the moment of the couple. SOLUTION The stress σ A varies linearly with the coordinates y and z. Since the axial force is zero, the y- and z-axes are centroidal axes: σ A = C1 y + C2 z where C1 and C2 are constants. M y = zσ AdA = C1 yzdA + C2 z 2dA = I yzC1 + I yC2 = 0 C2 = − I yz Iy C1 M z = − yσ Adz = −C1 y 2dA + C2 yzdA I yz C1 = − I Z C1 − I yz Iy ( ) 2 I y M z = − I y I z − I yz C1 C1 = − C2 = + σA = − I yM z 2 I y I z − I yz I yz M z 2 I y I z − I yz I y y − I yz 2 2 I y I z − I yz Mz 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 4.159 (a) Show that, if a vertical force P is applied at point A of the section shown, the equation of the neutral axis BD is zA xA 2 x + 2 z = −1 rz rx where rz and rx denote the radius of gyration of the cross section with respect to the z axis and the x axis, respectively. (b) Further show that, if a vertical force Q is applied at any point located on line BD, the stress at point A will be zero. SOLUTION Definitions: rx2 = (a) M x = Pz A σE = − =− Ix I , rz2 = z A A M z = − Px A P M z xE M x z E P Px A xE Pz z + − =− − − A 2E 2 A Iz Ix A Arz Arx z P xA 1 + 2 xE + A2 z E = 0 A rz rx if E lies on neutral axis. z x 1 + A2 x + A2 z = 0, rz rx (b) M x = PzE σA = − zA xA 2 x + 2 z = −1 rz rx M z = − PxE P M z xA M x z A P Px x Pz z + − = − − E 2 A − E 2A A Iz Iy A Arz Arx = 0 by equation from part (a). 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 4.160 (a) Show that the stress at corner A of the prismatic member shown in part a of the figure will be zero if the vertical force P is applied at a point located on the line x z + =1 b /6 h /6 (b) Further show that, if no tensile stress is to occur in the member, the force P must be applied at a point located within the area bounded by the line found in part a and three similar lines corresponding to the condition of zero stress at B, C, and D, respectively. This area, shown in part b of the figure, is known as the kern of the cross section. SOLUTION 1 3 1 3 hb Ix = bh 12 12 h b zA = − xA = − 2 2 Iz = A = bh Let P be the load point. M z = − PxP σA = − σ A = 0, 1− M x = Pz P P M z xA M x z A + − A Iz Ix ( ) =− (− PxP ) − b2 ( Pz P − 2h ) P + − 3 1 hb 1 bh3 bh 12 12 =− P x z 1− P − P bh b/6 h/6 x z − =0 b/6 h/6 (a) For (b) At point E: z = 0 ∴ xE = b/6 At point F: x = 0 ∴ z F = h /6 x z + =1 b/6 h/6 If the line of action ( xP , z P ) lies within the portion marked TA , a tensile stress will occur at corner A. By considering σ B = 0, σ C = 0, and σ D = 0, the other portions producing tensile stresses are identified. 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 4.161 For the machine component and loading shown, determine the stress at point A when (a) h = 2 in., (b) h = 2.6 in. SOLUTION M = −4 kip ⋅ in Rectangular cross section: r = (a) A = bh r2 = 3 in. r1 = r2 − h 1 h (r1 + r2 ), R = , e=r −R r 2 ln 2 r1 h = 2 in. A = (0.75)(2) = 1.5 in 2 r1 = 3 − 2 = 1 in. R= 2 = 1.8205 in. ln 13 At point A: σA = r = 1 (3 + 1) = 2 in. 2 e = 2 − 1.8205 = 0.1795 in. r = r1 = 1 in. M (r − R) (−4)(1 − 1.8205) = = 12.19 ksi Aer (1.5)(0.1795)(1) σ A = 12.19 ksi (b) h = 2.6 in. A = (0.75)(2.6) = 1.95 in 2 r = r1 = 3 − 2.6 = 0.4 in. R= 2.6 = 1.2904 in. 3 ln 0.4 At point A: σA = 1 (3 + 0.4) = 1.7 in. 2 e = 1.7 − 1.2904 = 0.4906 in. r = r1 = 0.4 in. M (r − R ) (−4)(0.4 − 1.2904) = = 11.15 ksi Aer (1.95)(0.4096)(0.4) σ A = 11.15 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 4.162 For the machine component and loading shown, determine the stress at points A and B when h = 2.5 in. SOLUTION M = −4 kip ⋅ in Rectangular cross section: h = 2.5 in. b = 0.75 in. A = 1.875 in.2 r2 = 3 in. r1 = r2 − h = 0.5 in. 1 1 (r1 + r2 ) = (0.5 + 3.0) = 1.75 in. 2 2 2.5 h R = r = 3.0 = 1.3953 in. 2 ln 0.5 ln r = r1 e = r − R = 1.75 − 1.3953 = 0.3547 in. At point A: r = r1 = 0.5 in. σA = At point B: M (r − R ) (−4 kip ⋅ in)(0.5 in. − 1.3953 in.) = Aer (0.75 in.)(2.5 in.)(0.3547 in.)(0.5 in.) σ A = 10.77 ksi r = r2 = 3 in. σB = M (r − R ) (−4 kip ⋅ in)(3 in. − 1.3953 in.) = Aer (0.75 in. × 2.5 in.)(0.3547 in.)(3 in.) σ B = −3.22 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 4.163 The curved portion of the bar shown has an inner radius of 20 mm. Knowing that the allowable stress in the bar is 150 MPa, determine the largest permissible distance a from the line of action of the 3-kN force to the vertical plane containing the center of curvature of the bar. SOLUTION Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the section. The bending couple is M = P (a + r ) For the rectangular section, the neutral axis for bending couple only lies at R= h ln r2 r1 Also e = r − R The maximum compressive stress occurs at point A. It is given by σA = − P My A P P (a + r ) y A P − =− − = −K A Aer1 A Aer1 A with y A = R − r1 Thus, K =1+ Data: (a + r )( R − r1) er1 h = 25 mm, r1 = 20 mm, r2 = 45 mm, r = 32.5 mm R= 25 = 30.8288 mm, e = 32.5 − 30.8288 = 1.6712 mm 45 ln 20 b = 25 mm, A = bh = (25)(25) = 625 mm 2 = 625 × 10−6 m 2 R − r1 = 10.8288 mm P = 3 × 103 N ⋅ m, σ A = −150 × 106 Pa (−150 × 106 )(625 × 10−6 ) = 31.25 P 3 × 103 ( K − 1)er1 (30.25)(1.6712)(20) a+r = = = 93.37 mm R − r1 10.8288 K =− σ AA =− a = 93.37 − 32.5 a = 60.9 mm 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 4.164 The curved portion of the bar shown has an inner radius of 20 mm. Knowing that the line of action of the 3-kN force is located at a distance a = 60 mm from the vertical plane containing the center of curvature of the bar, determine the largest compressive stress in the bar. SOLUTION Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the section. The bending couple is M = P (a + r ) For the rectangular section, the neutral axis for bending couple only lies at R= h ln r2 r1 . Also e = r − R The maximum compressive stress occurs at point A. It is given by σA = − P My A P P (a + r ) y A P − =− − = −K A Aer1 A Aer1 A with y A = R − r1 Thus, K =1+ Data: h = 25 mm, r1 = 20 mm, r2 = 45 mm, r = 32.5 mm R= (a + r )( R − r1) er1 25 = 30.8288 mm, e = 32.5 − 30.8288 = 1.6712 mm 45 ln 20 b = 25 mm, A = bh = (25)(25) = 625 mm 2 = 625 × 10−6 m 2 a = 60 mm, a + r = 92.5 mm, R − r1 = 10.8288 mm (92.5)(10.8288) K = 1+ = 30.968 P = 30 × 103 N (1.6712)(20) σA = KP (30.968)(3 × 103 ) =− = −148.6 × 106 Pa A 625 × 10−6 σ A = −148.6 MPa 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 4.165 The curved bar shown has a cross section of 40 × 60 mm and an inner radius r1 = 15 mm. For the loading shown, determine the largest tensile and compressive stresses. SOLUTION h = 40 mm, r1 = 15 mm, r2 = 55mm A = (60)(40) = 2400 mm 2 = 2400 × 10−6 m 2 R= r = h 40 = = 30.786 mm r2 55 ln ln r1 40 1 (r1 + r2 ) = 35 mm 2 e = r − R = 4.214 mm At r = 15mm, σ =− σ =− My Aer y = 30.786 − 15 = 15.756 mm (120)(15.786 × 10−3 ) = −12.49 × 10−6 Pa (2400 × 10−6 )(4.214 × 10−3 )(15 × 10−3 ) σ = −12.49MPa At r = 55mm, σ =− y = 30.786 − 55 = −24.214 mm (120)(−24.214 × 10−3 ) = 5.22 × 106 Pa (2400 × 10−6 )(4.214 × 10−3 )(55 × 10−3 ) σ = 5.22 MPa 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 4.166 For the curved bar and loading shown, determine the percent error introduced in the computation of the maximum stress by assuming that the bar is straight. Consider the case when (a) r1 = 20 mm, (b) r1 = 200 mm, (c) r1 = 2 m. SOLUTION h = 40 mm, A = (60)(40) = 2400 mm2 = 2400 × 10−6 m 2 , M = 120 N ⋅ m 1 3 1 bh = (60)(40)3 = 0.32 × 106 mm 4 = 0.32 × 10 −6 mm 4 , 12 12 I = c= 1 h = 20 mm 2 Assuming that the bar is straight σs = − (a) Mc (120)(20 × 10−8 ) =− = 7.5 × 106 Pa = 7.5 MPa I (0.32 × 10−6 ) r1 = 20 mm r2 = 60 mm h 40 = = 36.4096 mm 60 r2 ln ln r1 20 R= r = 1 (r1 + r2 ) = 40 mm 2 σa = r1 − R = −16.4096 mm e = r − R = 3.5904 mm M (r1 − R) (120)(−16.4096 × 10−3 ) = = −11.426 × 106 Pa = −11.426 MPa Aer (2400 × 10−6 )(3.5904 × 10−3 )(20 × 10−3 ) % error = −11.426 − (−7.5) × 100% = −34.4% −11.426 For parts (b) and (c), we get the values in the table below: (a) (b) (c) r1, mm r2 , mm 20 200 2000 60 240 2040 R, mm 36.4096 219.3926 2019.9340 r , mm 40 220 2020 e, mm 3.5904 0.6074 0.0660 σ , MPa −11.426 −7.982 −7.546 % error −34.4 % 6.0 % 0.6 % 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 4.167 The curved bar shown has a cross section of 30 × 30 mm. Knowing that a = 60 mm, determine the stress at (a) point A, (b) point B. SOLUTION Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the section. The bending couple is M = P (a + r ) For the rectangular section, the neutral axis for bending couple only lies at R= Also h . r ln 2 r1 e=r −R The maximum compressive stress occurs at point A. It is given by σA = − P My A P P (a + r ) y A − =− − A Aer1 A Aer1 = −K Thus, P A with K =1+ y A = R − r1 (a + r )( R − r1) er1 h = 30 mm, r1 = 20 mm, r2 = 50 mm, r = 35 mm Data: R= 30 = 32.7407 mm, e = 35 − 32.7407 = 2.2593 mm 50 ln 20 b = 30 mm, A = bh = (30)(30) = 900 mm 2 = 900 × 10−6 m 2 a = 60 mm, a + r = 95 mm, R − r1 = 12.7407 mm K =1+ (95)(12.7407) = 27.786 (2.2593)(20) P = 5 × 103 N 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 4.167 (Continued) (a) σA = − KP (27.786)(5 × 103 ) =− = −154.4 × 106 Pa A 900 × 10−6 σ A = −154.4 MPa (b) At point B, yB = r2 − R = 50 − 32.7407 = 17.2953 mm P MyB P (a + r ) y B + = −1 + A Aer2 A er2 K ′P (a + r ) y B where K ′ = = −1 A er2 σB = − K′ = (95)(17.2953) − 1 = 13.545 (2.2593)(50) σB = (13.545)(5 × 103 ) = 75.2 × 106 Pa −6 900 × 10 σ B = 75.2 MPa 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 4.168 The curved bar shown has a cross section of 30 × 30 mm. Knowing that the allowable compressive stress is 175 MPa, determine the largest allowable distance a. SOLUTION Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the section. The bending couple is M = P (a + r ) For the rectangular section, the neutral axis for bending couple only lies at R= h . r ln 2 r1 Also e = r − R The maximum compressive stress occurs at point A . It is given by σA = − P My A P P (a + r ) y A − =− − A Aer1 A Aer1 P with y A = R − r1 A (a + r )( R − r1) Thus, K = 1 + er1 = −K h = 30 mm, r1 = 20 mm, r2 = 50 mm, r = 35 mm, R = Data: (1) 30 = 32.7407 mm ln 50 20 e = 35 − 32.7407 = 2.2593 mm, b = 30 mm, R − r1 = 12.7407 mm, a = ? σ A = −175 MPa = −175 × 106 Pa, P = 5 kN = 5 × 103 N K =− Solving (1) for a+r = Aσ A (900 × 10−6 )(−175 × 106 ) =− = 31.5 P 5 × 103 a + r, a+r = ( K − 1)er1 R − r1 (30.5)(2.2593)(20) = 108.17 mm 12.7407 a = 108.17 mm − 35 mm a = 73.2 mm 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 4.169 Steel links having the cross section shown are available with different central angles β . Knowing that the allowable stress is 12 ksi, determine the largest force P that can be applied to a link for which β = 90°. SOLUTION Reduce section force to a force-couple system at G; the centroid of the cross section AB. β a = r 1 − cos 2 The bending couple is M = − Pa. For the rectangular section, the neutral axis for bending couple only lies at R= h ln r2 r1 . Also e = r − R At point A the tensile stress is K =1+ where P= Data: σA = P My A P Pay A P ay P − = + = 1 + A = K A Aer1 A Aer1 A er1 A ay A er1 and y A = R − r1 Aσ A K r = 1.2 in., r1 = 0.8 in., r2 = 1.6 in., h = 0.8 in., b = 0.3 in. 0.8 A = (0.3)(0.8) = 0.24 in 2 R = 1.6 = 1.154156 in. ln 0.8 e = 1.2 − 1.154156 = 0.045844 in., y A = 1.154156 − 0.8 = 0.35416 in. a = 1.2(1 − cos 45°) = 0.35147 in. K =1+ P= (0.35147)(0.35416) = 4.3940 (0.045844)(0.8) (0.24)(12) = 0.65544 kips 4.3940 P = 655 lb 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 4.170 Solve Prob. 4.169, assuming that β = 60°. PROBLEM 4.169 Steel links having the cross section shown are available with different central angles β. Knowing that the allowable stress is 12 ksi, determine the largest force P that can be applied to a link for which β = 90°. SOLUTION Reduce section force to a force-couple system at G, the centroid of the cross section AB. β a = r 1 − cos 2 The bending couple is M = − Pa. For the rectangular section, the neutral axis for bending couple only lies at R= h r2 r1 ln . Also e = r − R At point A, the tensile stress is σA = K =1+ where P= Data: P My A P Pay A P ay P − = + = 1 + A = K A Aer1 A Aer1 A er1 A ay A er1 and y A = R − r1 Aσ A K r = 1.2 in., r1 = 0.8 in., r2 = 1.6 in., h = 0.8 in., b = 0.3 in. 0.8 A = (0.3)(0.8) = 0.24 in 2 R = 1.6 = 1.154156 in. ln 0.8 e = 1.2 − 1.154156 = 0.045844 in. y A = 1.154156 − 0.8 = 0.35416 in. a = (1.2)(1 − cos 30°) = 0.160770 in. K =1+ P= (0.160770)(0.35416) = 2.5525 (0.045844)(0.8) (0.24)(12) = 1.128 kips 2.5525 P = 1128 lb 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 4.171 A machine component has a T-shaped cross section that is orientated as shown. Knowing that M = 2500 N ⋅ m, determine the stress at (a) point A, (b) point B. SOLUTION Properties of the cross section. r , mm Part b, mm R= A bi hi Ai = = 1 ri +1 r dA bi ln bi ln i +1 r ri ri h, mm 40 90 110 1 20 2 60 2200 R= = 77.852 mm, 28.2588 A, mm 2 bi ln 50 20 ri +1 , mm ri r = Ai ri Ai ri , mm 1000 16.2186 65 1200 12.0402 100 2200 28.2588 185000 r = = 84.091 mm, e = r − R = 6.239 mm 2200 Ai ri , mm3 65,000 120,000 185,000 Stresses. (a) Point A: σA = (b) M (r − R ) (2.5 kN ⋅ m)(0.040 m − 0.0778517 m) = Aer (2.2 × 10−3 m 2 )(6.2392 × 10−3 m)(0.040 m) Point B: σB = r = rA = 40 mm σ A = −172.4 MPa r = rB = 110 mm M (r − R ) (2.5 kN ⋅ m)(0.110 m − 0.0778517 m) = Aer (2.2 × 10−3 m 2 )(6.2392 × 10−3 m)(0.110 m) σ B = 53.2 MPa 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 4.172 Assuming that the couple shown is replaced by a vertical 10-kN force attached at point D and acting downward, determine the stress at (a) point A, (b) point B. SOLUTION Properties of the cross section. R = r , mm 40 90 110 Part b, mm 1 20 2 60 2200 R= = 77.852 mm, 28.2588 A bi hi Ai = = 1 r r dA bi ln i +1 bi ln i +1 r ri ri h, mm 50 20 A, mm 2 ri +1 , mm ri 16.2186 12.0402 28.2588 bi ln 1000 1200 2200 185000 r = = 84.091 mm, 2200 r = Ai ri Ai ri , mm Ai ri , mm3 65 100 65,000 120,000 185,000 e = r − R = 6.239 mm Force-couple system at the centroid E. P = 10 kN M = (10 kN)(100 mm + 84.0909 mm) = 1840.9 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 4.172 (Continued) Stresses. (a) Point A: σA = − r = rA = 40 mm P M ( r − R) 10 kN (1840.9 N ⋅ m)(0.040 m − 0.0778517 m) + =− + −3 2 A Aer 2.2 × 10 m (2.2 × 10−3 m 2 )(6.2392 × 10−3 m)(0.040 m) = −4.545 MPa − 126.913 MPa σ A = −131.5 MPa (b) Point B: σB = − r = rB = 110 mm P M ( r − R) 10 kN (1840.9 N ⋅ m)(0.110 m − 0.0778517 m) + =− + −3 2 A Aer 2.2 × 10 m (2.2 × 10−3 m 2 )(6.2392 × 10−3 m)(0.110 m) = −4.545 MPa + 39.196 MPa σ B = 34.7 MPa 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 4.173 Three plates are welded together to form the curved beam shown. For the given loading, determine the distance e between the neutral axis and the centroid of the cross section. SOLUTION R= r = r 3 3.5 5.5 6 Part ΣA Σbi hi ΣA = = 1 r Σ r dA Σb ln ri +1 Σbi ln i +1 i ri ri ΣAri ΣA b h A b ln ri +1 ri r Ar 3 0.5 1.5 0.462452 3.25 4.875 0.5 2 1.0 0.225993 4.5 4.5 2 0.5 1.0 0.174023 5.75 5.75 3.5 0.862468 Σ 15.125 3.5 15.125 = 4.05812 in., r = = 4.32143 in. 0.862468 3.5 e = r − R = 0.26331 in. R= e = 0.263 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 4.174 Three plates are welded together to form the curved beam shown. For M = 8 kip ⋅ in., determine the stress at (a) point A, (b) point B, (c) the centroid of the cross section. SOLUTION R= r = r 3 3.5 5.5 6 Part b ΣA Σbi hi ΣA = = 1 r r Σ r dA Σb ln i +1 Σbi ln i +1 i ri ri ΣAri ΣA h A b ln r Ar 3 0.5 1.5 0.462452 3.25 4.875 0.5 2 1.0 0.225993 4.5 4.5 2 0.5 1.0 0.174023 5.75 5.75 3.5 0.862468 Σ 3.5 = 4.05812 in., 0.862468 e = r − R = 0.26331 in. R= (a) ri +1 ri 15.125 15.125 = 4.32143 in. 3.5 M = −8 kip ⋅ in r = y A = R − r1 = 4.05812 − 3 = 1.05812 in. σA = − My A (−8)(1.05812) =− Aer1 (3.5)(0.26331)(3) (b) yB = R − r2 = 4.05812 − 6 = −1.94188 in. σB = − (c) yC = R − r = −e σC = − MyB (−8)(−1.94188) =− Aer2 (3.5)(0.26331)(6) MyC Me M −8 =− =− =− Aer Aer Ar (3.5)(4.32143) σ A = 3.06 ksi σ B = −2.81 ksi σ C = 0.529 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 4.175 The split ring shown has an inner radius r1 = 20 mm and a circular cross section of diameter d = 32 mm. For the loading shown, determine the stress at (a) point A, (b) point B. SOLUTION 1 d = 16 mm, r1 = 20 mm, r2 = r1 + d = 52 mm 2 r = r1 + c = 36 mm c= 1 1 r + r 2 − c 2 = 36 + 362 − 162 2 2 = 34.1245 mm R= e = r − R = 1.8755 mm A= π 4 d2 = π 4 (32)2 = 804.25 mm 2 = 804.25 × 10−6 m 2 P = 2.5 × 103 N (a) Point A: M = Pr = (2.5 × 103 )(36 × 10−3 ) = 90 N ⋅ m y A = R − r1 = 34.1245 − 20 = 14.125 mm σA = − P My A 2.5 × 103 (90)(14.1245 × 10−3 ) − =− − A Aer1 804.25 × 10−6 (804.25 × 10−6 )(1.8755 × 10−3 )(20 × 10−3 ) = −45.2 × 106 Pa (b) Point B: σ A = −45.2 MPa yB = R − r2 = 34.1245 − 52 = −17.8755 mm σB = − P MyB 2.5 × 103 (90)(−17.8755 × 10−3 ) − =− − A Aer2 804.25 × 10−6 (804.25 × 10−6 )(1.8755 × 10−3 )(52 × 10−3 ) = 17.40 × 106 Pa σ B = 17.40 MPa 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 4.176 The split ring shown has an inner radius r1 = 16 mm and a circular cross section of diameter d = 32 mm. For the loading shown, determine the stress at (a) point A, (b) point B. SOLUTION 1 d = 16 mm, r1 = 16 mm, r2 = r1 + d = 48 mm 2 r = r1 + c = 32 mm c= 1 1 r + r 2 − c 2 = 32 + 322 − 162 2 2 = 29.8564 mm R= e = r − R = 2.1436 mm A= π 4 d2 = π 4 (32)2 = 804.25 mm 2 = 804.25 × 10−6 m 2 P = 2.5 × 103 N (a) Point A: M = Pr = (2.5 × 103 )(32 × 10−3 ) = 80 N ⋅ m y A = R − r1 = 29.8564 − 16 = 13.8564 mm σA = − P My A 2.5 × 103 (80)(13.8564 × 10−3 ) − =− − A Aer1 804.25 × 10−6 (804.25 × 10−6 )(2.1436 × 10−3 )(16 × 10−3 ) = −43.3 × 106 Pa (b) Point B: σ A = −43.3 MPa yB = R − r2 = 29.8564 − 48 = −18.1436 mm σB = − P MyB 2.5 × 106 (80)(−18.1436 × 10−3 ) − =− − A Aer2 804.25 × 10−6 (804.25 × 10−6 )(2.1436 × 10−3 )(48 × 10−3 ) = 14.43 × 106 Pa σ B = 14.43 MPa 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 4.177 The curved bar shown has a circular cross section of 32-mm diameter. Determine the largest couple M that can be applied to the bar about a horizontal axis if the maximum stress is not to exceed 60 MPa. SOLUTION c = 16 mm r = 12 + 16 = 28 mm 1 r + r 2 − c2 2 1 = 28 + 282 − 162 = 25.4891 mm 2 R= e = r − R = 28 − 25.4891 = 2.5109 mm σ max occurs at A, which lies at the inner radius. It is given by σ max = My A from which Aer1 M = Aer1 σ max Also, A = π c 2 = π (16) 2 = 804.25 mm 2 Data: y A = R − r1 = 25.4891 − 12 = 13.4891 mm M = yA . (804.25 × 10−6 )(2.5109 × 10−3 )(12 × 10−3 )(60 × 106 ) 13.4891 × 10−3 M = 107.8 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 4.178 The bar shown has a circular cross section of 0.6 in. diameter. Knowing that a = 1.2 in., determine the stress at (a) point A, (b) point B. SOLUTION 1 d = 0.3 in. r = 0.5 + 0.3 = 0.8 in. 2 1 1 R = r + r 2 − c 2 = 0.8 + 0.82 − 0.32 2 2 = 0.77081 in. c= e = r − R = 0.02919 in. A = π c 2 = π (0.3)2 = 0.28274 in 2 P = 50 lb M = − P(a + r ) = −50(1.2 + 0.8) = −100 lb ⋅ in y A = R − r1 = 0.77081 − 0.5 = 0.27081 in. yB = R − r2 = 0.77081 − 1.1 = −0.32919 in. (a) σA = P My A 50 (−100)(0.27081) − = − = 6.74 × 103 psi A Aer1 0.28274 (0.28274)(0.02919)(0.5) σ A = 6.74 ksi (b) σB = P MyB 50 (−100)(−0.32919) − = − = −3.45 × 103 psi A Aer2 0.28274 (0.28274)(0.02919)(1.1) σ B = −3.45 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 4.179 The bar shown has a circular cross section of 0.6 in. diameter. Knowing that the allowable stress is 8 ksi, determine the largest permissible distance a from the line of action of the 50-lb forces to the plane containing the center of curvature of the bar. SOLUTION 1 d = 0.3 in., r = 0.5 + 0.3 = 0.8 in. 2 1 1 R = r + r 2 − c 2 = 0.8 + 0.82 − 0.32 2 2 = 0.77081 in. e = r − R = 0.02919 in. c= A = π c 2 = π (0.3)2 = 0.28274 in 2 M = − P (a + r ) y A = R − r1 = 0.77081 − 0.5 = 0.27081 in. σA = = P My A P P( a + r ) y A P (a + r ) y A − = + = 1 + A Aer1 A Aer1 A er1 KP A where K =1+ (a + r ) y A er1 (8 × 103 )(0.28274) = 45.238 P 50 ( K − 1)er1 (44.238)(0.02919)(0.5) a+r = = = 2.384 in. yA 0.27081 K = σ AA = a = 2.384 − 0.8 a = 1.584 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 4.180 Knowing that P = 10 kN , determine the stress at (a) point A, (b) point B. SOLUTION Locate the centroid D of the cross section. r = 100 mm + 90 mm = 130 mm 3 Force-couple system at D. P = 10 kN M = Pr = (10 kN)(130 mm) = 1300 N ⋅ m Triangular cross section. A= 1 1 bh = (90 mm)(80 mm) 2 2 = 3600 mm 2 = 3600 × 10−6 m 2 1 1 h (90) 45 mm 2 2 R= = = r2 r2 190 190 ln − 1 ln − 1 0.355025 h r1 90 100 R = 126.752 mm e = r − R = 130 mm − 126.752 mm = 3.248 mm (a) Point A: σA = − rA = 100 mm = 0.100 m P M (rA − R) 10 kN (1300 N ⋅ m)(0.100 m − 0.126752 m) + =− + −6 2 A AerA 3600 × 10 m (3600 × 10 −6 m 2 )(3248 × 10−3 m)(0.100 m) = −2.778 MPa − 29.743 MPa (b) Point B: σB = − σ A = −32.5 MPa rB = 190 mm = 0.190 m P M (rB − R) 10 kN (1300 N ⋅ m)(0.190 m − 0.126752 m) + =− + −6 2 A AerB 3600 × 10 m (3600 × 10−6 m 2 )(3.248 × 10−3 m)(0.190 m) = −2.778 MPa + 37.01 MPa σ B = +34.2 MPa 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 4.181 Knowing that M = 5 kip ⋅ in., determine the stress at (a) point A, (b) point B. SOLUTION 1 1 bh = (2.5)(3) = 3.75 in 2 2 2 r = 2 + 1 = 3.00000 in. A= b1 = 2.5 in., r1 = 2 in., b2 = 0, r2 = 5 in. Use formula for trapezoid with b2 = 0. 1 h 2 (b + b ) 1 2 2 R= r2 (b1r2 − b2r1) ln − h(b1 − b2 ) r1 = (0.5)(3) 2 (2.5 + 0) = 2.84548 in. [(2.5)(5) − (0)(2)] ln 52 − (3)(2.5 − 0) e = r − R = 0.15452 in. (a) y A = R − r1 = 0.84548 in. σA = − (b) M = 5 kip ⋅ in My A (5) (0.84548) =− Aer1 (3.75)(0.15452) (2) σ A = −3.65 ksi yB = R − r2 = −2.15452 in. σB = − MyB (5)(−2.15452) =− Aer2 (3.75)(0.15452)(5) σ B = 3.72 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 4.182 Knowing that M = 5 kip ⋅ in, determine the stress at (a) point A, (b) point B. SOLUTION A = 1 (2.5)(3) = 3.75 in 2 2 r = 2 + 2 = 4.00000 in. b1 = 0, r1 = 2 in., Use formula for trapezoid with b2 = 2.5 in., r2 = 5 in. b1 = 0. 1 h 2 (b + b ) 1 2 2 R= r2 (b1r2 − b2r1) ln − h (b1 − b2 ) r1 (0.5) (3) 2 (0 + 2.5) = 3.85466 in. [(0)(5) − (2.5)(2)] ln 5 − (3)(0 − 2.5) 2 e = r − R = 0.14534 in. M = 5 kip ⋅ in = (a) y A = R − r1 = 1.85466 in. σA = − (b) My A (5) (1.85466) =− Aer1 (3.75)(0.14534) (2) σ A = −8.51 ksi yB = R − r2 = −1.14534 in. σB = − MyB (5) (−1.14534) =− Aer2 (3.75)(0.14534) (5) σ B = 2.10 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 4.183 For the curved beam and loading shown, determine the stress at (a) point A, (b) point B. SOLUTION Locate centroid. A, mm 2 r , mm Ar , mm3 600 45 27 × 103 300 55 16.5 × 103 Σ 900 r = R= = 43.5 × 103 43.5 × 103 = 48.333 mm 900 1 2 h 2 (b1 + b2 ) (b1r2 − b2r1) ln r2 r1 − h(b2 − b1) (0.5)(30) 2 (40 + 20) = 46.8608 mm − (30)(40 − 20) [(40)(65) − (20)(35)]ln 65 35 e = r − R = 1.4725 mm (a) y A = R − r1 = 11.8608 mm σA = − (b) M = −250 N ⋅ m My A (−250)(11.8608 × 10−3 ) =− = 63.9 × 106 Pa Aer1 (900 × 10−6 )(1.4725 × 10−3 )(35 × 10−3 ) σ A = 63.9 MPa yB = R − r2 = −18.1392 mm σB = − MyB (−250)(−18.1392 × 10−3 ) =− = −52.6 × 106 Pa Aer2 (900 × 10−6 )(1.4725 × 10−3 )(65 × 10−3 ) σ B = −52.6 MPa 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 4.184 For the crane hook shown, determine the largest tensile stress in section a-a. SOLUTION Locate centroid. A, mm 2 r , mm Ar , mm3 1050 60 63 × 103 750 80 60 × 103 Σ 1800 r= Force-couple system at centroid: 103 × 103 103 × 103 = 63.333 mm 1800 P = 15 × 103 N M = − Pr = −(15 × 103 )(68.333 × 10−3 ) = −1.025 × 103 N ⋅ m 1 2 R= = h 2 (b1 + b2 ) (b1r2 − b2 r1 ) ln r2 r1 − h(b1 − b2 ) (0.5)(60) 2 (35 + 25) = 63.878 mm [(35)(100) − (25)(40)]ln 100 − (60)(35 + 25) 40 e = r − R = 4.452 mm Maximum tensile stress occurs at point A. y A = R − r1 = 23.878 mm σA = = P My A − A Aer1 15 × 103 −(1.025 × 103 )(23.878 × 10−3 ) − 1800 × 10−6 (1800 × 10 −6 )(4.452 × 10−3 )(40 × 10−3 ) = 84.7 × 106 Pa σ A = 84.7 MPa 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 4.185 Knowing that the machine component shown has a trapezoidal cross section with a = 3.5 in. and b = 2.5 in., determine the stress at (a) point A, (b) point B. SOLUTION Locate centroid. A, in 2 r , in. Ar , in 3 10.5 6 63 7.5 8 60 Σ 18 r= R= = 123 123 = 6.8333 in. 18 1 2 h (b1 + b2 ) 2 (b1r2 − b2 r1 ) ln r2 r1 − h(b1 − b2 ) (0.5)(6) 2 (3.5 + 2.5) = 6.3878 in. [(3.5)(10) − (2.5)(4)]ln 104 − (6)(3.5 − 2.5) e = r − R = 0.4452 in. (a) (b) y A = R − r1 = 6.3878 − 4 = 2.3878 in. My (80)(2.3878) σA = − A = − (18)(0.4452)(4) Aer1 yB = R − r2 = 6.3878 − 10 = −3.6122 in. My (80)( −3.6122) σB = − B = − (18)(0.4452)(10) Aer2 M = 80 kip ⋅ in σ A = −5.96 ksi σ B = 3.61 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 4.186 Knowing that the machine component shown has a trapezoidal cross section with a = 2.5 in. and b = 3.5 in., determine the stress at (a) point A, (b) point B. SOLUTION Locate centroid. A, in 2 r , in. Ar , in 3 7.5 6 45 10.5 8 84 Σ 18 r= R= = 129 129 = 7.1667 in. 18 1 2 h (b1 + b2 ) 2 (b1r2 − b2 r1 ) ln r2 r1 − h(b1 − b2 ) (0.5)(6)2 (2.5 + 3.5) = 6.7168 in. [(2.5)(10) − (3.5)(4)]ln 104 − (6)(2.5 − 3.5) e = r − R = 0.4499 in. (a) y A = R − r1 = 2.7168 in. σA = − (b) M = 80 kip ⋅ in My A (80)(2.7168) =− (18)(0.4499)(4) Aer1 σ A = −6.71 ksi yB = R − r2 = −3.2832 in. σB = − MyB (80)( −3.2832) =− (18)(0.4499)(10) Aer2 σ B = 3.24 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 4.187 Show that if the cross section of a curved beam consists of two or more rectangles, the radius R of the neutral surface can be expressed as R= A r r b2 r b3 ln 2 3 4 r1 r2 r3 b1 where A is the total area of the cross section. SOLUTION R= = Note that for each rectangle, ΣA A = 1 Σ dA Σ b ln ri +1 i r r i A r Σ ln i +1 ri bi = A r b1 r b2 r b3 ln 2 3 4 r1 r2 r3 ri + 1 dr 1 d A= bi ri r r ri + 1 dr ri +1 = bi = bi ln r2 r ri 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 4.188 Using Eq. (4.66), derive the expression for R given in Fig. 4.73 for a circular cross section. SOLUTION Use polar coordinate β as shown. Let w be the width as a function of β w = 2 c sin β r = r − c cos β dr = c sin β d β dA = w dr = 2c 2 sin 2 β d β dA = r dA = r =2 =2 2 c 2 sin β dβ r − c cos β π 0 c 2 (1 − cos 2 β ) dβ r − c cos β π 0 r 2 − c 2 cos 2 β − (r 2 − c 2 ) dβ r − c cos β π 0 π 0 (r + c cos β ) d β − 2(r 2 − c 2 ) = 2r β π 0 2 + 2c sin β 2 − 2( r − c ) π 0 dr r − c cos β π 0 2 r 2 − c2 tan −1 ( ) r 2 − c 2 tan 1 β 2 r +c π 0 π = 2r (π − 0) + 2c(0 − 0) − 4 r 2 − c 2 ⋅ − 0 2 = 2π r − 2π r 2 − c 2 A = π c2 R= 1 = 2 A c2 r + r 2 − c2 1 π c2 = = × dA 2π r − 2π r 2 − c 2 2 r − r 2 − c 2 r + r 2 − c 2 r ( c 2 r + r 2 − c2 2 2 2 r − (r − c ) ) = 1 c (r + 2 2 r 2 − c2 c 2 )=1 r+ 2( ) r 2 − c2 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 4.189 Using Eq. (4.66), derive the expression for R given in Fig. 4.73 for a trapezoidal cross section. SOLUTION The section width w varies linearly with r. w = c0 + c1r w = b1 at r = r1 and w = b2 at r = r2 b1 = c0 + c1r1 b2 = c0 + c1r2 b1 − b2 = c1 (r1 − r2 ) = −c1h b −b c1 = − 1 2 h r2 b1 − r1b2 = ( r2 − r1 )c0 = hc0 r b − rb c0 = 2 1 1 2 h r2 w r2 c + c r dA 0 1 dr = dr = r1 r r1 r r r2 r2 = c0 ln r + c1 r r1 r1 r = c0 ln 2 + c1 (r2 − r1 ) r1 r2b1 − r1b2 r2 b1 − b2 = ln − h h r1 h r b − rb r = 2 1 1 2 ln 2 − (b1 − b2 ) h r1 1 A = (b1 + b2 ) h 2 1 2 h (b1 + b2 ) A 2 R= = r dA (r2 b1 − r1b2 ) ln r2 − h(b1 − b2 ) r 1 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 4.190 Using Equation (4.66), derive the expression for R given in Fig. 4.73 for a triangular cross section. SOLUTION The section width w varies linearly with r. w = c0 + c1r w = b at r = r1 and w = 0 at r = r2 b = c0 + c1r1 0 = c0 + c1r2 b = c1 ( r1 − r2 ) = −c1h br b c1 = − and c0 = −c1r2 = 2 h h r r 2 w 2 c + c r dA 0 1 dr = dr = r r r r 1 r 1 r2 r2 = c0 ln r + c1 r r1 r1 r = c0 ln 2 + c1 (r2 − r1 ) r1 br r b = 2 ln 2 − h h r1 h r br r r = 2 ln 2 − b = b 2 ln 2 − 1 h r1 h r1 1 A = bh 2 1 1 bh h A 2 R= = = r 2r r r 2 dA ln r2 − 1 b h2 ln r2 − 1 h r 1 ( 1 ) 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 4.191 For a curved bar of rectangular cross section subjected to a bending couple M, show that the radial stress at the neutral surface is r1 M R σr = 1 − − ln Ae R r1 and compute the value of σ r for the curved bar of Examples 4.10 and 4.11. (Hint: consider the free-body diagram of the portion of the beam located above the neutral surface.) SOLUTION M (r − R) M MR = − Aer Ae Aer For portion above the neutral axis, the resultant force is σr = At radial distance r, H = σ r dA = R r1 σ r bdr Mb R MRb R dr dr − Ae r1 Ae r1 r r1 Mb MRb R MbR R = ( R − r1 ) − ln = 1 − − ln Ae Ae r1 Ae R r1 = Fr = σ r cos β dA Resultant of σ n : = θ /2 θ − /2 σ r cos β (bRd β ) = σ r bR θ /2 = σ r bR sin β For equilibrium: Fr − 2 H sin 2σ r bR sin θ 2 θ 2 −θ /2 θ /2 θ − /2 cos β d β = 2σ r bR sin θ 2 =0 −2 θ MbR r1 R 1 − − ln sin = 0 Ae R r1 2 σr = M r1 R 1 − − ln Ae R r1 Using results of Examples 4.10 and 4.11 as data, M = 8 kip ⋅ in, σr = A = 3.75 in 2 , R = 5.9686 in., e = 0.0314 in., r1 = 5.25 in. 8 5.25 5.9686 − ln 1− (3.75)(0.0314) 5.9686 5.25 σ r = −0.54 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 4.192 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 A y0 A y0 18 5 90 18 1 18 Σ 36 Y0 = 108 108 = 3 in. 36 Neutral axis lies 3 in. above the base. 1 1 b1h13 + A1d12 = (3)(6)3 + (18)(2) 2 = 126 in 4 12 12 1 1 I 2 = b2 h23 + A2 d 22 = (9)(2)3 + (18)(2) 2 = 78 in 4 12 12 I = I1 + I 2 = 126 + 78 = 204 in 4 I1 = ytop = 5 in. ybot = −3 in. M − Pa = 0 M = Pa = (15)(40) = 600 kip ⋅ in σ top = − σ bot = − M ytop I =− (600)(5) 204 M ybot (600)(−3) =− I 204 σ top = −14.71 ksi (compression) σ bot = 8.82 ksi (tension) 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 4.193 Straight rods of 6-mm diameter and 30-m length are stored by coiling the rods inside a drum of 1.25-m inside diameter. Assuming that the yield strength is not exceeded, determine (a) the maximum stress in a coiled rod, (b) the corresponding bending moment in the rod. Use E = 200 GPa. SOLUTION D = inside diameter of the drum. Let 1 d, 2 ρ = radius of curvature of centerline of rods when bent. d = diameter of rod, ρ= I= (a) σ max = (b) M= Ec EI ρ ρ = = c= 1 1 1 1 D − d = (1.25) − (6 × 10−3 ) = 0.622 m 2 2 2 2 π 4 c4 = π 4 (0.003) 4 = 63.617 × 10−12 m 4 (200 × 109 )(0.003) = 965 × 106 Pa 0.622 (200 × 109 )(63.617 × 10−12 ) = 20.5 N ⋅ m 0.622 σ = 965 MPa M = 20.5 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 4.194 Knowing that for the beam shown the allowable stress is 12 ksi in tension and 16 ksi in compression, determine the largest couple M that can be applied. SOLUTION = rectangle = semi-circular cutout A1 = (2.4)(1.2) = 2.88 in 2 A2 = π 2 (0.75) 2 = 0.8836 in 2 A = 2.88 − 0.8836 = 1.9964 in 2 y1 = 0.6 in. 4r (4)(0.75) = = 0.3183 in. 3π 3π ΣA y (2.88)(0.6) − (0.8836)(0.3183) Y = = = 0.7247 in. ΣA 1.9964 y2 = Neutral axis lies 0.7247 in. above the bottom. Moment of inertia about the base: 1 π 1 π I b = bh3 − r 4 = (2.4)(1.2)3 − (0.75)4 = 1.25815 in 4 3 8 3 8 Centroidal moment of inertia: I = I b − AY 2 = 1.25815 − (1.9964)(0.7247) 2 = 0.2097 in 4 ytop = 1.2 − 0.7247 = 0.4753 in., ybot = −0.7247 in. |σ | = My I M= σI y Top: (tension side) M= (12)(0.2097) = 5.29 kip ⋅ in 0.4753 Bottom: (compression) M= (16)(0.2097) = 4.63 kip ⋅ in 0.7247 Choose the smaller value. M = 4.63 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 4.195 In order to increase corrosion resistance, a 2-mm-thick cladding of aluminum has been added to a steel bar as shown. The modulus of elasticity is 200 GPa for steel and 70 GPa for aluminum. For a bending moment of 300 N ⋅ m, determine (a) the maximum stress in the steel, (b) the maximum stress in the aluminum, (c) the radius of curvature of the bar. SOLUTION Use aluminum as the reference material. n = 1 in aluminum. n= Es 200 = = 2.857 in steel. Ea 70 Cross section geometry: Steel: As = (46 mm)(26 mm) = 1196 mm 2 Is = 1 (46 mm)(26 mm)3 = 67,375 mm 4 12 Aluminum: Aa = (50 mm)(30 mm) − 1196 mm 2 = 304 mm 2 Ia = 1 (50 mm)(30 mm3 ) − 67,375mm 4 = 45,125mm 4 12 Transformed section. I = na I a + ns I s = (1)(45,125) + (2.857)(67,375) = 237, 615 mm 4 = 237.615 × 10−9 m 4 Bending moment. (a) ys = 13mm = 0.013m ns Mys (2.857)(300)(0.013) = = 46.9 × 106 Pa I 237.615 × 10−9 Maximum stress in aluminum: σa = (c) ns = 2.857 Maximum stress in steel: σs = (b) M = 300 N ⋅ m na = 1, ya = 15 mm = 0.015 m na Mya (1)(300)(0.015) = = 18.94 × 106 Pa I 237.615 × 10−9 Radius of curvative: ρ = EI M ρ = σ s = 46.9 MPa (70 × 109 )(237.615 × 10−9 ) 300 σ a = 18.94 MPa ρ = 55.4 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 4.196 A single vertical force P is applied to a short steel post as shown. Gages located at A, B, and C indicate the following strains: ∈A = −500μ ∈B = −1000μ ∈C = −200μ Knowing that E = 29 × 106 psi, determine (a) the magnitude of P, (b) the line of action of P, (c) the corresponding strain at the hidden edge of the post, where x = −2.5 in. and z = −1.5 in. SOLUTION 1 (5)(3)3 = 11.25 in 4 12 M x = Pz M z = − Px Ix = Iz = 1 (3)(5)3 = 31.25 in 4 12 A = (5)(3) = 15 in 2 x A = −2.5 in., xB = 2.5 in., xC = 2.5 in., xD = −2.5 in. z A = 1.5 in., z B = 1.5 in., zC = −1.5 in., z D = −1.5 in. σ A = Eε A = (29 × 106 )(−500 × 10−6 ) = −14,500 psi = −14.5 ksi σ B = Eε B = (29 × 106 )(−1000 × 10−6 ) = −29,000 psi = −29 ksi σ C = Eε C = (29 × 106 )(−200 × 10−6 ) = −5800 psi = −5.8 ksi σA = − P M x z A M z xA − + = −0.06667 P − 0.13333 M x − 0.08 M z A Ix Iz (1) σB = − P M x z B M z xB − + = −0.06667 P − 0.13333 M x + 0.08 M z A Ix Iz (2) σC = − P M x zC M x − + z C = −0.06667 P + 0.13333 M x + 0.08 M z A Ix Iz (3) Substituting the values for σ A , σ B , and σ C into (1), (2), and (3) and solving the simultaneous equations gives M x = 87 kip ⋅ in, M z = −90.625 kip ⋅ in, (a) P = 152.25 kips x=− z = Mz −90.625 =− P 152.25 Mx 87 = P 152.25 σD = − (c) (b) x = 0.595 in. z = 0.571 in. P M x z D M z xD − + = −0.06667 P + 0.13333 M x − 0.08 M z A Ix Iz = −(0.06667)(152.25) + (0.13333)(87) − (0.08)(−90.625) = 8.70 ksi Strain at hidden edge: ε = σD E = 8.70 × 103 29 × 106 ε = 300 u 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 4.197 For the split ring shown, determine the stress at (a) point A, (b) point B. SOLUTION r1 = 1 40 = 20 mm, 2 r2 = A = (14)(25) = 350 mm 2 1 (90) = 45 mm h = r2 − r1 = 25 mm 2 25 h R = r = 45 = 30.8288 mm 2 ln 20 ln r 1 1 r = (r1 + r2 ) = 32.5 mm 2 e = r − R = 1.6712 mm Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the cross section. The bending couple is M = Pa = P r = (2500)(32.5 × 10−3 ) = 81.25 N ⋅ m (a) Point A: rA = 20 mm y A = 30.8288 − 20 = 10.8288 mm P My A 2500 (81.25)(10.8288 × 10−3 ) − =− − A AeR 350 × 10−6 (350 × 10−6 )(1.6712 × 10−3 )(20 × 10−3 ) 6 = −82.4 × 10 Pa σA = − (b) Point B: rB = 45 mm σ A = −82.4 MPa yB = 30.8288 − 45 = −14.1712 mm 2500 (81.25)( −14.1712 × 10−3 ) P MyB − =− − A AerB 350 × 10−6 (350 × 10−6 )(1.6712 × 10−3 )(45 × 10−3 ) = 36.6 × 106 Pa σB = − σ B = 36.6 MPa 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 4.198 A couple M of moment 8 kN ⋅ m acting in a vertical plane is applied to a W200 × 19.3 rolled-steel beam as shown. Determine (a) the angle that the neutral axis forms with the horizontal plane, (b) the maximum stress in the beam. SOLUTION For W200 × 19.3 rolled steel sector, I z′ = 16.6 × 106 mm 4 = 16.6 × 10−6 m 4 I y′ = 1.15 × 106 mm 4 = 1.15 × 10−6 m 4 203 = 101.5 mm 2 102 = zE = = 51 mm 2 y A = yB = − yD = − yE = z A = − zB = − zD M z = (8 × 103 ) cos 5° = 7.9696 × 103 N ⋅ m M y = −(8 × 103 )sin 5° = −0.6972 × 103 N ⋅ m (a) tan ϕ = Iz 16.6 × 10−6 tan θ = tan(−5°) = −1.2629 Iy 1.15 × 10−6 ϕ = −51.6° α = 51.6° − 5° (b) α = 46.6° Maximum tensile stress occurs at point D. σD = − (7.9696 × 103 )(−101.5 × 10−3 ) (−0.6972 × 103 )(−51 × 10−3 ) M z yD M y z D + =− + Iz Iy 16.6 × 10−6 1.15 × 10−6 = 79.6 × 106 Pa σ D = 79.6 MPa 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 4.199 Determine the maximum stress in each of the two machine elements shown. SOLUTION For each case, M = (400)(2.5) = 1000 lb ⋅ in At the minimum section, 1 (0.5)(1.5)3 = 0.140625 in 4 12 c = 0.75 in. I = (a) D/d = 3/1.5 = 2 r/d = 0.3/1.5 = 0.2 From Fig 4.32, σ max = (b) KMc (1.75)(1000)(0.75) = = 9.33 × 103 psi I 0.140625 D/d = 3/1.5 = 2 From Fig. 4.31, σ max = K = 1.75 σ max = 9.33 ksi r/d = 0.3/1.5 = 0.2 K = 1.50 KMc (1.50)(1000)(0.75) = = 8.00 × 103 psi I 0.140625 σ max = 8.00 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 4.200 The shape shown was formed by bending a thin steel plate. Assuming that the thickness t is small compared to the length a of a side of the shape, determine the stress (a) at A, (b) at B, (c) at C. SOLUTION Moment of inertia about centroid: a 1 2 2t 12 2 1 = ta3 12 ( I= Area: ) 3 a a A = 2 2t = 2at , c = 2 2 2 ( ) (a) P P Pec P σA = − = − A I 2at (b) P P Pec P σB = + = + 2at A I (c) σC = σ A ( )( ) a 2 2 1 ta 12 a 2 2 3 ( )( ) a 2 1 ta 3 12 a 2 σA = − P 2at σB = − 2P at σC = − P 2at 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 4.201 Three 120 × 10-mm steel plates have been welded together to form the beam shown. Assuming that the steel is elastoplastic with E = 200 GPa and σ Y = 300 MPa, determine (a) the bending moment for which the plastic zones at the top and bottom of the beam are 40 mm thick, (b) the corresponding radius of curvature of the beam. SOLUTION A1 = (120)(10) = 1200 mm 2 R1 = σ Y A1 = (300 × 106 )(1200 × 10− 6 ) = 360 × 103 N A2 = (30)(10) = 300 mm 2 R2 = σ Y A2 = (300 × 106 )(300 × 10−6 ) = 90 × 103 N A3 = (30)(10) = 300 mm 2 1 1 R3 = σ Y A2 = (300 × 106 )(300 × 10−6 ) = 45 × 103 N 2 2 y1 = 65 mm = 65 × 10−3 m y2 = 45 mm = 45 × 10−3 m (a) M = 2( R1 y1 + R2 y2 + R3 y3 ) = 2{(360)(65) + (90)(45) + (45)(20)} = 56.7 × 103 N ⋅ m (b) y3 = 20 mm = 20 × 10−3 m yY ρ = σY E ρ= EyY σY M = 56.7 kN ⋅ m = (200 × 109 )(30 × 10−3 ) 300 × 106 ρ = 20 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 4.202 A short column is made by nailing four 1 × 4-in. planks to a 4 × 4-in. timber. Determine the largest compressive stress created in the column by a 16-kip load applied as shown in the center of the top section of the timber if (a) the column is as described, (b) plank 1 is removed, (c) planks 1 and 2 are removed, (d) planks 1, 2, and 3 are removed, (e) all planks are removed. SOLUTION (a) M =0 Centric loading: σ =− P A A = (4)(4) + (4)(1)(4) = 32 in 2 σ =− (b) Eccentric loading: M = Pe σ =− 16 × 103 32 σ = −500 psi P Pec − A I A = (4) (4) + (3)(1)(3) = 28 in 2 e= y ΣAy (1)(4)(2.5) y= = = 0.35714 in. 28 A I = Σ( I + Ad 2 ) = + 1 (4)(1)3 + (4)(1)(2.14286) 2 = 53.762 in 4 12 σ =− (c) Centric loading: 1 (6)(4)3 + (6) (4)(0.35714) 2 12 16 × 103 (16 × 103 ) (0.35714) (2.35714) − 28 53.762 M =0 σ =− σ = −822 psi P A A = (6) (4) = 24 in 2 σ =− 16 × 103 24 σ = −667 psi 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 4.202 (Continued) (d) Eccentric loading: M = Pe σ =− P Pec − A I A = (4) (1) = (1)(4)(1) = 20 in 2 x = 2.5 − 2 = 0.5 in. 1 I = (4)(5)3 = 41.667 in 4 12 σ =− (e) Centric loading: M =0 16 × 103 (16 × 103 ) (0.5) (2.5) − 20 41.667 σ =− e=x σ = −1280 psi P A A = (4)(4) = 16 in 2 σ =− 16 × 103 16 σ = −1000 psi 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 4.203 Two thin strips of the same material and same cross section are bent by couples of the same magnitude and glued together. After the two surfaces of contact have been securely bonded, the couples are removed. Denoting by σ1 the maximum stress and by ρ1 the radius of curvature of each strip while the couples were applied, determine (a) the final stresses at points A, B, C, and D, (b) the final radius of curvature. SOLUTION Let b = width and t = thickness of one strip. Loading one strip, M = M1 I1 = 1 3 1 bt , c = t 12 2 M 1c σ M1 = I bt 2 1 M 12 M1 = 1 = ρ1 EI1 Et 3 σ1 = After M1 is applied to each of the strips, the stresses are those given in the sketch above. They are σ A = −σ1, σ B = σ1, σ C = −σ1, σ D = σ 1 The total bending couple is 2M1. After the strips are glued together, this couple is removed. M ′ = 2 M 1, I ′ = 1 2 b(2t )3 = bt 3 12 3 c=t The stresses removed are σ′ = − M ′y 2M y 3M1 y = − 2 13 = − I bt bt 2 3 σ ′A = − 3M1 1 3M 1 1 = − σ1, σ ′B = σ C′ = 0, σ ′D = = σ1 2 2 2 bt bt 2 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 4.203 (Continued) (a) Final stresses: 1 2 σ A = −σ1 − (− σ 1) 1 2 σ A = − σ1 σ B = σ1 σ C = −σ1 1 2 σ D = σ1 − σ1 1 ρ′ (b) = σD = 1 σ1 2 M′ 2M 3M 1 1 1 = 2 13 = = EI ′ 4 ρ′ E 3 bt Et 3 Final radius: 1 ρ = 1 ρ1 − 1 1 1 1 3 1 = − = ρ ′ ρ1 4 ρ1 4 ρ1 ρ = 4 ρ1 3 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.