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Mechanics of materials 6th edition solutions chapter 4Embed Size (px) 344 x 292429 x 357514 x 422599 x 487DESCRIPTIONSolutions to mechanics of material by ferdinand beer, for the 5th edition chapter 4 You're Reading a Free Preview Pages 8 to 13 are not shown in this preview. You can read a free preview pages 17 through 19 that don't appear in this preview. You can read a free preview pages 23 through 26 won't appear inthis preview. You can read a free preview pages 31 through 45 won't appear in this preview. You can read a free preview pages 53 through 57 won't appear in this preview. You can read a free preview pages 61 through 65 won't appear in this preview. You're reading a free sample page 76 that doesn't appear in this example. You're reading a free sample page 83 that doesn'tappear in this example. You can read a free preview pages 90 through 104 are not displayed in this preview. You can read a free preview pages 111 through 147 are not displayed in this preview. You can read a free sample pages 164 through 176 that are not displayed in this preview. You can read a free sample pages 180 to 183 that are not displayed in this preview. You canread a free sample page 187 not shown in this example. You can read a free preview pages from 191 through 198 are not displayed in this preview. Slideshare uses cookies to improve functionality and performance and provide you with relevant advertisements. If you continue to browse the site, you agree to the use of cookies on this website. See our User Agreement and PrivacyPolicy. Slideshare uses cookies to improve functionality and performance and provide you with relevant advertisements. If you continue to browse the site, you agree to the use of cookies on this website. For more information, see our privacy policy and user agreement. This survival guide textbook has been created for the textbook: Mechanics of Materials, edition: 6. Thiscomprehensive survival guide textbook covers the following chapters and their solutions. Mechanics of Materials is written by and affiliated with the ISBN: 9780073380285. Since 203 issues were answered in Chapter 4, more than 37857 students have looked at complete step-by-step solutions from this chapter. Chapter 4 contains 203 complete step-by-step solutions. CHAPTER 420 40 PROBLEM 4.1 20 20 A Knowing that the pair shown is acting in a vertical plane, determining the voltage at (a) point A, (b) point B.M 5 15 kN · m 80 20 B Dimensions in mm SOLUTION For rectangle: I 1 3 bra 12 Out-of-court angle: I1 1 (80)(120)3 12 I1 11.52 110 6 mm 4 I2 1 (40)(80)3 12 I2 1.70667 106 mm 4 Recess: Section: a) yA 40 mm I I I 2 11.52 10 6 m4 1.70667 109.81333 10 0.040 m 6 A 6 m4 m4 My A I (0.040 m15 103 ) (0.040) 9.81333 10 6 61.6 106 Pa 61.6 MPa A (b) yB 60 mm 0.060 m B MyB I (15 103 )( 0.060) 9.81333 10 6 106 Pa B 91.7 MPa OWN MATERIAL. Image copyright 2015 McGraw-Hill Education. This is proprietary material for authorized instructor use only. Not allowed for sale or distribution in any way. This documentmay not be copied, scanned, duplicated, forwarded, distributed or posted on a website in whole or in part. 447 447 In. 2 in. PROBLEM 4.2 M ! 25 chicken · Inch. A Knowing that the pair shown is acting in a vertical plane, determine the voltage at (a) point A, (b) point B. 2. B 1.5 in. SOLUTION I For rectangle: 1 3 bra 12 For cross section area: I I1 I2 I3 1 (2)(1.5)3 12 1 (2)(5.5)3 12 1(2)(1.5)3 12 (a) yA 2.75 in. A My A I (25)(2.75) 28.854 (b) yB 0.75 in. B MyB I (25)(0.75) 28.854 28.854 in 4 A B 2.38 ksi 0.650 ksi OWN MATERIAL. Image copyright 2015 McGraw-Hill Education. This is proprietary material for authorized instructor use only. Not allowed for sale or distribution in any way. This document may not be copied, scanned, duplicated, forwarded,distributed or posted on a website in whole or in part. 448 200 mm PROBLEM 4.3 12 mm y C M x Using a permissible voltage of 155 MPa, determine the largest bending moment M that can be applied to the wide flange beam shown. Neglect the effect of fillets. 220 mm 8 mm 12 mm SOLUTION Moment of inertia around x-axis: I1 1 (200)(12)3 12 (200)(12)(104)2 2 25.9872 106 mm4 I2 1 (8)(196)3 12 I3 I1 25.9872 106 mm 4 I I1 I2 M Mx I 3 Mc with c I I with c 5.0197 106 mm 4 56.944 106 mm 4 1 (220) 2 110 mm 56.944 1 10 6 m4 0.110 m 155 106 Pa (56.944 10 6 )(155 106 ) 0.110 80.2 103 N m Mx 80.2 kN m OWN MATERIAL. Image copyright 2015 McGraw-Hill Education. This is proprietary material for authorized instructor use only. Not allowed for saleor distribution in any way. This document may not be copied, scanned, duplicated, forwarded, distributed or posted on a website, in whole or in part. 449 200 mm PROBLEM 4.4 12 mm y Solve Prob. 4.3, assuming that the broad-flanges beam is bent around the y-axis Mine for a few moments. C M x 220 mm PROBLEM 4.3. Using a permissible voltage of 155 MPa, determine thelargest bending torque M that can be applied to the wide flange beam shown. Neglect the effect of fillets. 8 mm 12 mm SOLUTION Moment of inertia around y-axis: I1 I2 1 (12)(200)3 8 106 mm 4 12 1 (196)(18)3 8.3627 103 mm 4 12 I3 I1 8 106 mm 4 I I1 I2 My I3 Mc with c I I with c 16.0084 106 mm 4 1 (200) 2 100 mm 16.0084 10 6 m4 0.100 m 155 106 Pa (16.0084 10 6 )(155 106) 0.100 24.8 103 N m My 24.8 kN m OWN Material. Image copyright 2015 McGraw-Hill Education. This is proprietary material for authorized instructor use only. Not allowed for sale or distribution in any way. This document may not be copied, scanned, duplicated, forwarded, distributed or posted on a website in whole or in part. 450 PROBLEM 4.5 0.1 in. 0.5 in. M1 Using apermissible voltage of 16 ksi, determine the largest pair that can be applied to each pipe. (a) 0,2 in. M2 (b) SOLUTION (a) I c ro4 ri4 4 0.6 in. Mc : I M 4 (0.64 I c 0.54 ) 52.7 10 3 in 4 (16)(52.7 10 3 ) 0.6 M (b) I c (0.74 4 0.7 in. Mc : I 0.54 ) M 139.49 10 I c 3 1.405 chicken in. in 4 10 3) 0.7 M 3.19 chicken in. OWN MATERIAL. Image copyright 2015 McGraw-Hill Education. This isproprietary material for authorized instructor use only. Not authorized for sale or in any way. This document may not be copied, scanned, duplicated, forwarded, distributed or posted on a website in whole or in part. 451 PROBLEM 4.6 r 5 20 mm A 30 mm B M 2.8 kN · m Knowing that the pair shown is acting in a vertical plane, determining the voltage at (a) point A, (b) point B. 30mm 120 mm SOLUTION I 1 (0,120 m)(0.06 m)3 12 2.1391 10 (a) A 6 2 12 4 (0.02 m) 4 mm 4 (2.8 103 N m)(0.03 m) 2.1391 10 6 mm 4 M yA I 39.3 MPa A (b) B M yB I (2.8 103 N m)(0.02 m) 2.1391 10 6 m 4 B 26.2 MPa OWN MATERIAL. Image copyright 2015 McGraw-Hill Education. This is proprietary material for authorized instructor use only. Not allowed for sale or distributionin any way. This document may not be copied, scanned, duplicated, forwarded, distributed or posted on a website in whole or in part. 452 PROBLEM 4.7 y Two W4 13 rolled sections are welded together as shown. Knowing that for the steel alloy used Y 36 ksi and U 58 ksi and with a safety factor of 3.0, determine the largest pair that can be applied when the assembly is bentaround the z-axis. z C SOLUTION Properties of W4 13 rolled part. (See Appendix C.) Area 3.83 in 2 Width 4.060 in. Iy 3.86 in 4 For one rolled part, the moment of inertia around axis b-b For both sections, Ad 2 Ib Iy Iz 2 Ib c is all M all width U F . S. all I c 3.86 (3.83)(2.030)2 19.643 in 4 39.286 in 4 4.060 in. 58 19,333 ksi 3.0 (19,333)(39,286) 4,060 Mc I M all 187.1 chicken in. OWNMATERIAL. Image copyright 2015 McGraw-Hill Education. This is proprietary material for authorized instructor use only. Not allowed for sale or distribution in any way. This document may not be copied, scanned, duplicated, forwarded, distributed or posted on a website in whole or in part. 453 PROBLEM 4.8 y C Two W4 13 rolled sections are welded together as shown. Knowingthat for the steel alloy U 58 ksi used and with a safety factor of 3.0, determine the largest pair that can be applied when the assembly is bent around the z-axis. z SOLUTION Properties of W4 13 rolled part. (See Appendix C.) Field 3.83 in 2 Depth 4.16 in. Ix 11.3 in 4 For one rolled part, the moment of inertia around axis a-a For both sections, Ad 2 Ia Ix Iz 2Ia c all M is all depth U F .S. all I c 11.3 (3.83)(2.08)2 27.87 in 4 55.74 in 4 4.16 in. 58 19,333 ksi 3.0 (19,333)(55.74) 4.16 Mc I M all 259 chicken in. OWN MATERIAL. Image copyright 2015 McGraw-Hill Education. This is proprietary material for authorized instructor use only. Not allowed for sale or distribution in any way. This document may not be copied, scanned, duplicated, forwarded, distributed orposted on a website, in whole or in part. 454 PROBLEM 4.9 3 in. 3 in. 3 in. 6 in. Two vertical applied to a bar of the cross section shown. Determine the maximum tensile and pressure voltages in part BC of the beam. 2 in. A 15 chicken 15 chicken B C 40 in. 60 in. D40 in. SOLUTION A y0 A y0 18 5 90 18 1 18 36 Y0 108 36 108 3 in. Neutral axis is 3 in. above the base. I1 I2 I I M M1 1 b1h13 A1d12 (3)(6)3 (18)(2) 2 126 in 4 12 12 1 1 3 2 b2 h2 A 2 d 2 (9)(2))3 (18)(2) 2 78 in 4 12 12 I1 I 2 126 78 204 in 4 5 in. ybot Pa 0 Pa (15)(40) 600 chicken in. M ytop top bot 3 in. I (600)(5) 204 M ybot I (600)( 3) 204 top 14.71 ksi (compression) bone 8.82 ksi (voltage) OWN MATERIAL. Image copyright 2015 McGraw-Hill Education. This is proprietary material forauthorized instructor use only. Not allowed for sale or distribution in any way. This document may not be copied, scanned, duplicated, forwarded, distributed or posted on a website in whole or in part. 455 PROBLEM 4.10 8 in. 1 in. Two vertical forces are applied to a bar of the transverse section shown. Determine the maximum tensile and pressure voltages in part BC of the beam.6 in. 1 in. 1 in. 4 in. A 25 chicken 25 chicken B C 20 in. 60 in. D20 in. SOLUTION A y0 A y0 8 7.5 60 6 4 24 4 0.5 18 86 86 18 4.778 in. (8) (2,772) 2 59.94 in 4 (6)(0.778) 2 21.63 in 4 (4)(4,278) 2 73.54 in 4 Yo 2 Neutral axis is 4,778 in. above the base. I1 I2 I3 I ytop 1 1 (8)(1)3 b1h13 A1d12 12 12 1 1 (1)(6)3 b2 h23 A2 d 22 1 2 12 1 1 (4)(1)3 b3 h33 A3 d32 12 12 I1 I 2 I 3 59.94 21.634.778 in. 3,222 in. ybot M M Pa 0 Pa (25)(20) 500 chicken in. Mytop top bot 73.57 155.16 in 4 I Mybot I (500)(3.222) 155.16 (500)( 4.778) 155.16 top 10.38 ksi (compression) bone 15.40 ksi (voltage) OWN MATERIAL. Image copyright 2015 McGraw-Hill Education. This is proprietary material for authorized instructor use only. Not allowed for sale or distribution in any way. Thisdocument may not be copied, scanned, duplicated, forwarded, distributed or posted on a website in whole or in part. 456 10 mm PROBLEM 4.11 10 mm 10 kN 10 kN B 50 mm C A D Two vertical forces are applied to a beam of the cross section shown. Determine the maximum tensile and pressure voltages in part BC of the beam. 10 mm 50 mm 250 mm 150 mm 150 mmSOLUTION A, mm 2 y0 , mm A y0 , mm3 600 30 18 103 600 30 18 103 300 5 1.5 103 37.5 103 1500 Y0 37.5 103 1500 25 mm Neutral axis is 25 mm above the base. I1 I3 I ytop 35 mm 0.035 m a 150 mm M bone ybot 25 mm 0.025 m 0.150 m P 10 10 3 N (10 103 )(0.150) 1.5 103 N m Pa Mytop top 1 (10)(60)3 (600)(6 (00) 5)2 195 103 mm 4 I 2 I1 195 mm 4 12 1 (30)(10)3 (300)(20 ) 2 122.5 103 mm 4 12 I1 I 2 I 3 512.5 103 mm 4 512.5 10 9 m 4 I M ybot I (1.5 103 ) (0.035) 512.5 10 9 102.4 106 Pa (1.5 103 )(100.025) 512.5 10 9 73.2 106 Pa top 102.4 MPa (compression) bot 73.2 MPa (voltage) OWN MATERIAL. Image copyright 2015 McGraw-Hill Education. This is proprietary material for authorized instructor use only. Not allowed for sale ordistribution in any way. This document may not be copied, scanned, duplicated, forwarded, distributed or posted on a website in whole or in part. 457 PROBLEM 4.12 216 mm y z Knowing that a bar of the displayed curved around a horizontal axis and that the bending moment is 6 kN m determines the total force working on the shaded part of the web. 36 mm 54 mm C 108 mm 72mm SOLUTION SOLUTION voltage distribution over the entire cross section is given by the bending voltage formula: My I x where y is a coordinate with its origin on the neutral axis and I the moment of inertia of the entire cross section area. The force on the shaded part is calculated on the basis of this voltage distribution. Over an area element dA is the force dF My dA I x dA Thetotal force on the shaded area is then F My dA I dF M I y dA M * * y A I where y * is the centeroidal coordinate of the shaded part and A* is the area. d1 54 18 36 mm d2 54 36 54 36 mm OWN MATERIAL. Image copyright 2015 McGraw-Hill Education. This is proprietary material for authorized instructor use only. Not allowed for sale or distribution in any way. This document maynot be copied, scanned, duplicated, forwarded, posted in whole or in part on a website. 458 PROBLEM 4.12 (Continued) Moment of inertia of whole cross section: I1 I2 I 1 1 (216)(36)3 (216)(36)(36)2 10.9175 106 mm4 b1h13 A1d12 12 12 1 1 b2 h23 A2 d 22 (72)(108)3 (72)(108)(36)2 17.6360 106 mm 4 12 12 I1 I 2 28.5535 106 mm 4 28.5535 10 6 m 4 For the shady area , A* (72)(90) y* 45 mm A* y * F 6480 mm 2 291,6 103 mm3 MA* y * I 291,6 10 6 m (6 103 )(291.6 10 6 ) 28.5535 10 6 61.3 103 N F 61.3 kN OWN MATERIAL. Image copyright 2015 McGraw-Hill Education. This is proprietary material for authorized instructor use only. Not allowed for sale or distribution in any way. This document may not be copied, scanned, duplicated, forwarded,distributed or posted on a website in whole or in part. 459 20 mm 12 mm PROBLEM 4.13 20 mm y 12 mm 24 mm z Knowing that a beam of the cross section shown is curved around a horizontal axis and that the bending moment is 4 kN m, determine the total force working on the shaded part of the beam. 20 mm C 20 mm 24 mm SOLUTION Dimensions in mm: Iz 1 1 (12 12)(88)3(40)(40)3 12 12 1 3629 106 0.213 106 1.5763 106 mm 4 1.5763 10 6 m 4 For use in Prob. 4.14, Iy 1 1 (88)(64)3 (24 24)(40)3 12 12 1.9224 10