(Solution Manual) Applied Strength of Materials

Mott,Untener, 7

Engineering

Solution Manual

7th Edition

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Solution Manual for Applied Strength of Materials, The downloadable file consists of 376 pages, and the total size of the file is 8.38 MB. Purchase now only for highly discounted price.

Author(s): Mott, Untener

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Chapter 1 Basic Concepts in Strength of Materials 1.1 to 1.11 Answers in text. 1.12 𝑊 = 𝑚 ∙ 𝑔 = 1400 kg ∙ 9.81 m/s2 = 13 734 (kg ∙ m)/s2 = 14 × 103 N 𝑾 = 𝟏 3. 𝟕 𝐤 𝐍 1.13 Total Weight = 𝑚𝑔 = 3500 kg ∙ 9.81 m/s2 = 34.34 kN Each Front Wheel: 𝐹𝐹= (1 2) (0.40)(34.34 kN) = 6.87 𝐤𝐍 Each Rear Wheel: 𝐹𝑅= (1 2) (0.60)(34.34 kN) = 𝟏0.32 𝐤𝐍 1.14 Loading = Total Force / Area Total Force = 𝑚𝑔 = 5900 kg ∙ 9.81 m/s2 = 57.9 kN Area = (4.5 m)(3.5 m) = 15.8 m2 Loading = 57.9 kN⁄15.8 m2 = 3.66 kN⁄m2 = 𝟑.66 𝐤𝐏𝐚 1.15 For ce = 𝑚 𝑔 = 35 kg ∙ 9.81 m/s2 = 343 N 𝐾 4800 N/mK = Spring Scale = 4800 N⁄m = 𝐹/Δ𝐿 Δ𝐿 = 𝐹 = 343 N= 0.0715 m = 71.5 × 10−3 m = 71. 𝟓 𝐦 𝐦 1.16𝑚 =𝑤 𝑔=3250 lb 32.2 (ft/s2)= 101 lb∙s2 ft= 101 𝐬𝐥𝐮𝐠𝐬 1.17𝑚 =𝑤 𝑔=11 600 lb 32.2 (ft/s2)= 360 lb∙s2 ft= 𝟑60 𝐬𝐥𝐮𝐠𝐬 1.19𝑝 = 1700 psi ∙ 6.895 (kPa⁄psi) = 11 722 𝐤𝐏𝐚 1.20𝜎 = 24 300 psi ∙ 6. 895 (kPa psi⁄ ) =167 549 kPa = 𝟏68𝐌𝐏𝐚 Solutions Manual for Applied Strength of Materials, 7e by Robert Mott, Joseph Untener (All Chapters)1.21 𝑠𝑢=14 000 psi ∙ 6. 895 (kPa psi⁄ ) =96 500 kPa = 𝟗𝟔. 𝟓 𝐌𝐏𝐚 𝑠𝑢=76 000 psi ∙ 6. 895 (kPa psi⁄ ) =524 000 kPa = 𝟓𝟐𝟒 𝐌𝐏𝐚 1.22 𝑛 = 3600 rev min×2π rad rev×1 min 60s= 377 𝐫𝐚𝐝 𝐬 1.232 in𝐴 = 2 6.1 i n2 × (25.4 2mm)= 16 839 𝐦 𝐦𝟐 1.24 𝑦 = 0. 08 in∙25.4 (mm in⁄ ) = 𝟐. 𝟎𝟑 𝐦𝐦 1.25 Dimensions: 18 in × 25.4 (mm/ in) = 457 mm 12 in × 25.4 (mm /in) = 305 mm Area = (18 in )2=𝟑𝟐𝟒 𝐢𝐧𝟐 Area = (457 mm )2= 𝟐. 𝟎𝟗×𝟏𝟎𝟓 𝐦𝐦𝟐 Volume = 𝑉 = Area × Height 𝑉 = 324 in2×12 in =𝟑𝟖𝟖𝟖 𝐢𝐧𝟑 𝑉 = (1.5 ft)2× 1.0 ft= 𝟐. 𝟐𝟓 𝐟𝐭𝟑 𝑉 = (209 ×103 mm2)×305 mm = 𝟔. 𝟑𝟕×𝟏𝟎𝟕 𝐦𝐦𝟑 𝑉 = (0. 457 m)2 × 0.305 m = 0. 0637 m3= 𝟔. 𝟑𝟕×𝟏𝟎−𝟐 𝐦𝟑 1.26 𝐴 = 𝜋𝐷2⁄4 = 𝜋 (0.505 in)2⁄4 = 𝟎. 𝟐𝟎𝟎 𝐢 𝐧𝟐 𝐴 = 0. 200 in2×(25.4 mm )2 in2 =𝟏𝟐𝟗 𝐦𝐦𝟐 1.27 𝜎 =𝑃 𝐴 (𝜋𝐷= 2800 2 N 4⁄)=2800 N [𝜋(10 mm )2] 4⁄= 35.7 N mm2= 35. 𝟕 𝐌𝐏𝐚 1.28 𝜎 =𝑃 𝐴=18×1 03 N (12)(30) mm2= 50.7 N mm2= 50. 𝟕 𝐌𝐏𝐚 1.29 𝜎 =𝑃 𝐴=1150 lb (0.40 in)2= 7188 𝐩𝐬𝐢 1.30 𝜎 =𝑃 𝐴=1850 lb [𝜋(0.375 in)2]4⁄=𝟏𝟔 𝟕𝟓𝟎 𝐩𝐬𝐢 1.31 Load on Shelf = 𝑊 = 𝑚𝑔 = 1650 k g ∙ 9.8 1 m⁄s2 = 16 187 N 𝑊/2 = 8093 N O n each side ∑ 𝑀𝐴 = 0 = (8093 N )(600 m m) − 𝐶 𝑉(1200 m m) 𝐶𝑉 = 4047 N 𝐶 = 𝐶 𝑉/ sin 30° = 8093 N 𝐴= 𝐶 𝐴 [𝜋(12 mm )2]4⁄ 𝜎 = 𝑃 = 9025 N= 71.6 𝐌𝐏𝐚 1.32 𝜎 = 𝑃 𝐴=70000 lb [𝜋(10 in)2]/4= 891 𝐩𝐬𝐢 1.33 𝜎 = 𝑃 𝐴=(29500 lb)/3 (3.5 in)2=𝟖𝟎𝟑 𝐩𝐬𝐢 1.34 𝜎 = 𝑃 𝐴=3500 N (8.0 mm )2=𝟓𝟒. 𝟕 𝐌𝐏𝐚 1.35 𝑊 = 𝑚𝑔 = 4200 k g ∙ 9.8 1 m/s2 = 41.2 k N 𝐴𝐵 𝑋=𝐴𝐵sin35° 𝐴𝐵 𝑌=𝐴𝐵cos35° 𝐵𝐶 𝑋=𝐵𝐶sin55° 𝐵𝐶 𝑌=𝐵𝐶cos55° ∑𝐹𝑋= 0 = 𝐴𝐵 𝑋−𝐵𝐶 𝑋 0 = 𝐴𝐵sin35° −𝐵𝐶sin55° 𝐴𝐵=𝐵𝐶∙sin55° sin35°= 1. 428 𝐵𝐶 ∑𝐹𝑉= 0 = 𝐴𝐵 𝑌+𝐵𝐶 𝑌−41.2 kN=𝐴𝐵cos35° +𝐵𝐶cos55° −41.2 kN 0 = (1.428 𝐵𝐶)cos35° +𝐵𝐶cos55° −41.2 kN 41.2 kN=𝐵𝐶[1.170 + 0. 574 ]= 1. 743 𝐵𝐶 𝐵𝐶=41.2 kN 1.743=23.63 kN 𝐴𝐵= 1. 428 𝐵𝐶=33.75 kN Stress in Rod AB: 𝜎𝐴𝐵=𝐴𝐵 𝐴=33.75× 103 N [𝜋(20 mm )2]/4=𝟏𝟎𝟕. 𝟒 𝐌𝐏𝐚 Stress in Rod BC: 𝜎𝐵𝐶=𝐵𝐶 𝐴=23.63× 103 N [𝜋(20 mm )2]/4=𝟕𝟓. 𝟐 𝐌𝐏𝐚 Stress in Rod BD: 𝜎𝐵𝐷=𝐵𝐷 𝐴=41.2× 103 N [𝜋(20 mm )2]/4=𝟏𝟑𝟏. 𝟏 𝐌𝐏𝐚 1.36 𝐹=0.01097 𝑚𝑅𝑛2 =(0.01097 )(0.40)(0.60)(3000 )2 N 𝐹 = 23 695 N 𝐴 =𝜋(16 mm )2 4=201 mm2 𝜎 =𝐹 𝐴=23695 N 201 mm2=𝟏𝟏𝟖 𝐌𝐏𝐚 1.37 𝐴 = (30 m m)2 = 900 m m2 For AB: 𝐹𝐴𝐵=(110 −40+80) kN=150 kN 𝜎𝐴𝐵=𝐹𝐴𝐵 𝐴=150× 103 N 900 mm2=𝟏𝟔𝟕 𝐌𝐏𝐚 Tension For BC: 𝐹𝐵𝐶=110 −40=70 kN 𝜎𝐵𝐶=𝐹𝐵𝐶 𝐴=70× 103 N 900 mm2=𝟕𝟕. 𝟖 𝐌𝐏𝐚 Tension For CD: 𝐹𝐶𝐷=110 kN 𝜎𝐶𝐷=𝐹𝐶𝐷 𝐴=110× 103 N 900 mm2=𝟏𝟐𝟐 𝐌𝐏𝐚 Tension 1.38 Areas: A-C ; 𝐴1 = 𝜋(2 5)2/4 = 491 m m2 C-D; 𝐴2= 𝜋(16)2/4 = 201 mm2 For AB: 𝐹𝐴𝐵= −9. 65−12.32+ 4. 45= − 17.52 kN 𝜎𝐴𝐵=𝐹𝐴𝐵 𝐴1=−17.52× 103 N 491 mm2= − 𝟑𝟓. 𝟕 𝐌𝐏𝐚 Compression For BC: 𝐹𝐵𝐶= −9. 65−12.32= − 21.97 kN 𝜎𝐵𝐶=𝐹𝐵𝐶 𝐴1=−21.97× 103 N 491 mm2= − 𝟒𝟒. 𝟕 𝐌𝐏𝐚 Compression For CD: 𝐹𝐶𝐷= −9. 65 kN 𝜎𝐶𝐷=𝐹𝐶𝐷 𝐴2=−9.65× 103 N 201 mm2= − 𝟒𝟖. 𝟎 𝐌𝐏𝐚 Compression 1.39 𝐴 = 𝜋[(1.90)2−(1.61)2] 4= 0. 799 in2 [11 2 in Pipe-Appendix A-9(a)] For BC : 𝜎𝐵𝐶=𝐹𝐵𝐶 𝐴=2500 lb 0.799 in2=𝟑𝟏𝟐𝟗 𝐩𝐬𝐢 Tension For AB: 𝐹𝐴𝐵=2500 + 2(8000 cos30°)=16 356 lb 𝜎𝐴𝐵=𝐹𝐴𝐵 𝐴=16 356 lb 0.799 in2=𝟐𝟎 𝟒𝟕𝟏 𝐩𝐬𝐢 Tension 1.40 ∑ 𝑀𝐶 = 0 = 2800 (45) − 𝐹 𝐵𝐷(30) 𝐹𝐵𝐷=4200 lb 𝜎𝐵𝐷=𝐹𝐵𝐷 𝐴=4200 lb (2.0)(0.65) in2=𝟑𝟐𝟑𝟏 𝐩𝐬𝐢 Tension 1.41 𝐴𝐷 sin 30° = 5.2 5 kN 𝐴𝐷=10.5 kN=𝐶𝐷 𝐴𝐵=𝐴𝐷cos30° = 9. 09 kN =𝐵𝐶 Stresses: 𝐴𝐵,𝐵𝐶: 𝜎𝐴𝐵= 𝜎 𝐵𝐶=9.09× 103 N (12)( 30) mm2=𝟐𝟓. 𝟑 𝐌𝐏𝐚 Tension 𝐵𝐷: 𝜎𝐵𝐷=10.5×103 N (2)( 10)(30) mm2=𝟏𝟕. 𝟓 𝐌𝐏𝐚 Tension 𝐴𝐷,𝐶𝐷: 𝐴 = ( 30)2−(20)2=500 mm2 𝜎𝐴𝐷= 𝜎 𝐶𝐷=−10.5× 103 N 500 mm2= − 𝟐𝟏. 𝟎 𝐌𝐏𝐚 Compression 1.42 ∑𝑀𝐴 =0= 6000 (6)+12 0 00(12)−𝑅𝐹(18) 𝑅𝐹=10 000 lb ∑𝑀𝐹= 0 = 12 000 (6)+6000 (12)− 𝑅 𝐴(18) 𝑅𝐴=8000 lb 𝑅𝐴=𝐴𝐵sin 𝜃 = 𝐴𝐵(0.8) 𝐴𝐵=𝑅𝐴 0.8=8000 0.8=10 000 lb Compression 𝐴𝐷=𝐴𝐵cos𝜃 = 10 000 (0.6)=6000 lb Tension 𝐵𝐸sin 𝜃 + 6000 −𝐴𝐵sin 𝜃 = 0 𝐵𝐸=𝐴𝐵sin 𝜃− 6000 sin 𝜃=10 000(0.8)− 6000 0.8=2500 lb Tension 𝐵𝐶=𝐴𝐵cos𝜃 + 𝐵𝐸cos𝜃 = 10 000 (0.6)+2500 (0.6) [Continued on next page] 𝐵𝐶=7500 lb Compression 𝐵𝐶=𝐶𝐹cos𝜃 𝐶𝐹=𝐵𝐶 cos 𝜃=7500 0.6=12 500 lb Compression 𝐶𝐸=12 000 −𝐶𝐹sin 𝜃 = 12 000 − 12 500 (0.8) CE = 2000 lb Compression EF = CF cos  = 12 500 lb(0.6) = 7500 lb Tension Areas of members: Appendixes A-5(a) and A-6(a) AD, DE, EF – 2(0.484 in2) = 0.968 in2 BD, BE, CE – 0.484 in2 AB, BC, CF – 2(1.21 in2) = 2.42 in2 Stresses: AD = DE = 6000/0.968 = +6198 psi EF = 7500/0.968 = +7748 psi BD = 0 BE = 2500/0.484 = +5165 psi CE = -2000/0.484 = -4132 psi [NOTE: Compression members must be AB = -10 000/2.42 = -4132 psi checked for column buckling.] BC = -7500/2.42 = -3099 psi CF = -12 500/2.42 = -5165 psi 1.43 ∑𝑀𝐶 =0=(12.5 )(4.0) −𝐴𝐵(2.5) 𝐴𝐵=20 kN 𝜎 =20× 103 N (20)2 mm2=𝟓𝟎 𝐌𝐏𝐚 1.44 𝐴 = 𝜋(0.505)2 4= 0. 200 in2 𝜎 =𝐹 𝐴=12 600 lb 0.200 in2=𝟔𝟑 𝟎𝟎𝟎 𝐩𝐬𝐢 1.45 𝐴=(2.65)(1.40) + 2[(1.4 0)(0.5)(𝑡) ] =4.41 in2 𝜎 =𝐹 𝐴=52 000 lb 4.41 in2=𝟏𝟏 𝟕𝟗𝟏 𝐩𝐬𝐢 1.46 𝐴 = (80)(40) − (60)(15) + 𝜋(40)2 4=3557 mm2 𝜎 =𝐹 𝐴=640× 103 N 3557 mm2=𝟏𝟖𝟎 𝐌𝐏𝐚 1.47 Direct Shear – Single Shear 𝐴𝑆= [𝜋(12.0)2 4] mm2=113 mm2 𝜏 =𝐹 𝐴𝑆=16.5× 103 N 113 mm2=𝟏𝟒𝟔 𝐌𝐏𝐚 1.48 ∑ 𝐹𝐽 = 0 = 55 (145) − 𝐹 𝑃(45) 𝐹𝑃=177 N 𝐴𝑆=𝜋(3.0)2 4= 7. 07 mm2 𝜏 =𝐹𝑃 𝐴𝑆=177 N 7.07 mm2=𝟐𝟓. 𝟏 𝐌𝐏𝐚 1.49 From Problem 1-46: 𝐹 = 23 6 95 N 𝐴𝑆= 2 [𝜋(10)2 4] = 157 mm2 Double Shear 𝜏 =𝐹 𝐴𝑆=23 695 N 157 mm2=𝟏𝟓𝟏 𝐌𝐏𝐚 1.50 𝐴𝑆 = (3.0)(3.5) = 10.5 i n2 𝜏 =𝐹 𝐴𝑆 10.5 in= 1650 lb 2= 157 𝐩𝐬𝐢 1.51 𝐴𝑆 = [2(35) + 𝜋(8)](50) = 475.7 m m2 𝜏 =𝐹 𝐴𝑆=38.6× 103 N 475.7 mm2=𝟖𝟏. 𝟏 𝐌𝐏𝐚 1.52 𝐿 = √0.42 + 0.62 = 0.721 in 𝐴𝑆= [2 (1.60)+𝜋(0.8) 2+ 2(0. 721 )] 0. 194 𝐴𝑆= 1. 144 in2 𝜏 =𝐹 𝐴𝑆=45 000 lb 1.144 in2=𝟑𝟗 𝟑𝟐𝟒 𝐩𝐬𝐢 1.53 𝑇 = 𝐹 𝑆 ∙ 𝑅 𝐹𝑆=𝑇 𝑅=95 N∙m 35 mm /2∙103 mm m=5429 N 𝐴𝑆= 𝑏 ∙ 𝐿 = (10)(22)=220 mm2 𝜏 =𝐹𝑆 𝐴𝑆=5429 N 220 mm2=𝟐𝟒. 𝟕 𝐌𝐏𝐚 Pin is in single shear 𝑅=8000 lb∙in 1.0 in1.54 𝐹 𝑆 = 𝑇 = 8000 lb 𝐴𝑆= 𝑏 ∙ 𝐿 = (0.50)(2.25)= 1. 125 in2 𝜏 =𝐹𝑆 𝐴𝑆=8000 lb 1.125 in2=𝟕𝟏𝟏𝟏 𝐩𝐬𝐢 1.55 Pin: Double Shear; 𝐴𝑆 = 2[𝜋(0.5)2/4] = 0.3 93 in2 𝜏 =𝐹 𝐴𝑆=20 000 lb 0.393 in2=𝟓𝟎 𝟗𝟑𝟎 𝐩𝐬𝐢 Collar: Shear Collar from Connector Body 𝐴𝑆= 𝜋𝑑𝑡 = 𝜋(0. 875 )(0.1875) = 0. 5154 in2 𝜏 =𝐹 𝐴𝑆=20 000 lb 0.5154 in2=𝟑𝟖 𝟖𝟎𝟎 𝐩𝐬𝐢 1.56 ∑ 𝑀𝐴 = 0 = 800 (80) − 𝐵 𝑉(8) 𝐵𝑉=8000 lb 𝐵 =𝐵𝑉 cos 20°=8513 lb 𝐴𝑆= 2 [𝜋(0.375 )2 4] = 0. 221 in2 𝜏 =𝐵 𝐴𝑆=8513 lb 0.221 in2=𝟑𝟖 𝟓𝟒𝟎 𝐩𝐬𝐢 1.57 𝐴 𝑆 = (40)(12) = 480 m m2 𝜏 =𝐹 𝐴𝑆=88× 103 N 480 mm2=𝟏𝟖𝟑 𝐌𝐏𝐚 1.58 𝐴 𝑆 = (40)(120) = 4800 m m2 𝜏 =𝐹 𝐴𝑆=88.2× 103 N 4800 mm2=𝟏𝟖. 𝟒 𝐌𝐏𝐚 1.59 𝐴 𝑆 = 𝜋𝑑𝑡 = 𝜋(1 2)(8) = 301.6 m m2 𝜏 =𝐹 𝐴𝑆=22.3× 103 N 301.6 mm2=𝟕𝟑. 𝟗 𝐌𝐏𝐚 1.60 𝐴 𝑆 = 2[𝜋(12)2/4] = 226.2 m m2 Two Rivets – Single Shear 𝜏 =𝐹 𝐴𝑆=10.2× 103 N 226.2 mm2=𝟒𝟓. 𝟏 𝐌𝐏𝐚 1.61 𝐴 𝑆 = 4[𝜋(12)2/4] = 452.4 m m2 Two Rivets – Double Shear 𝜏 =𝐹 𝐴𝑆=10.2× 103 N 452.4 mm2=𝟐𝟐.𝟓𝟓 𝐌𝐏𝐚 Structural Shapes 1-62 [Appendix A-4(a)] 2×4 wood board is 1.50 in×3.50 in cross section. A = 5.25 in2 1-63 [Appendix A-4(a)] 4×12 wood board is 11.25 in×3.50 in cross section. A = 39.38 in2 1-64 [Appendix A-4(a)] 1 in size; 0.75 in: 2 in; 1.50 in: 3 in; 2.50 in: 4 in; 3.50 in: 6 in; 5.50 in: 8 in; 7.25 in: 10 in; 9.25 in: 12 in; 11.25 in. 1-65 [Appendix A-4(b)] Metric size similar to U.S. 2×4: 45 mm×90 mm; A = 4050 mm2 1-66 [App. A-4(b)] Metric size similar to U.S. 2×10: 45 mm×240 mm; A = 10 800 mm2 1-67 U.S. size Actual width Conversion Closest SI size Difference 1 in 0.75 in 19.05 mm 19 mm -0.05 mm 2 in 1.50 in 38.1 mm 40 mm +1.90 mm 3 in 2.50 in 63.5 mm 65 mm +1.50 mm 4 in 3.50 in 88.9 mm 90 mm +1.10 mm 6 in 5.50 in 139.7 mm 140 mm +0.30 mm 8 in 7.25 in 184.15 mm 190 mm +5.85 mm 10 in 9.25 in 234.95 mm 240 mm +5.05 mm 12 in 11.25 in 285.75 mm 290 mm +4.25 mm 1-68 [Appendix A-5(a)] L 6×6×3/4 weighs 28.8 lb/ft 6.0 ft length weighs: (6 ft)(28.8 lb/ft) = 172.8 lb; A = 8.46 in2 1-69 [Appendix A-5(b)] Angle 1.50×1.25×0.188 weighs 1.633 lb/ft 53.5 in length weighs: (53.5 in)(1.0 ft/12 in)(1.633 lb/ft) = 7.28 lb; A = 0.480 in2 1-70 [Appendix A-5(c)] Angle 40×20×3 (mm) weighs 12.883 N/m 1.45 m length weighs: (1.45 m)(12.883 N/m) = 18.68 N; A = 171 mm2 1-71 [Appendix A-6(a)] C 7×14.75 weighs 14.75 lb/ft; Depth = 7.00 in; A = 4.33 in2 148 in long section weighs: (14.75 lb/ft)(148 in)(1 ft/12 in) = 181.92 lb 1-72 [App. A-6(b)] Aluminum C 9×4.983 weighs 4.983 lb/ft; Depth = 9.00 in; A = 4.237 in2 37.5 in long section weighs: (4.983 lb/ft)(37.5 in)(1 ft/12 in) = 15.57 lb 1-73 [App. A-6( d)] Aluminum C 80×40 mm weighs 16.51 N/m; Depth = 80 mm; A = 608 mm2 3.80 m long section weighs: (16.1 N/m)(3.80 m) = 61.18 N 1-74 [Appendix A-7(a)] W 10×60 beam weighs 60.0 lb/ft; Depth = 10.2 in; A = 17.60 in2 290 in length weighs: (290 in)(1 ft/12 in)(60.0 lb/ft) = 1450 lb 1-75 [Appendix A-7(b)] S 24×121 beam weighs 121 lb/ft; Depth = 24.5 in; A = 35.5 in2 240 in length weighs: (240 in)(1 ft/12 in)(121 lb/ft) = 2420 lb 1-76 [Appendix A-7(e)] IPE I 330×160 beam weighs 471.7 N/m; A = 6261 mm2 2.55 m length weighs: (2.55 m)(471.7 N/m) = 1203 N Flange width = 160 mm; Flange thickness = 11.5 mm; web thickness = 7.50 mm 1-77 [Appendix A-7(f)] Wood I-shapes; 4 depths: 9.50 in; 11.75 in; 14.00 in; 16.00 in [Note: Many other depths commercially available] 1-78 [Appendix A-8(a)] HSS6×4×1/4 tube: t = 0.233 in (design thickness); A = 4.30 in2 1-79 [Appendix A-8(c)] Steel 150×50×4 mm tube weighs 116 N/m; A = 1536 mm2 Inside dimensions: 142×42 mm 1.22 m length weighs: (1.42 m)(116 N/m) = 164.7 N 1-80 [Appendix A-9(a)] 6-in Schedule 40 steel pipe: OD = 6.625 in; ID = 6.065 in Wall thickness = 0.280 in; A = 5.581 in2; AISC PIPE6STD. 1-81 [Appendix A-9(b)] 6-in Schedule 40 steel pipe is called: DN 150 size; OD = 168.3 mm; ID = 154.1 mm; Wall thickness = 7.11 mm; A = 3601 mm2 1-82 [Appendix A-9(c) ] Steel mechanical tube: 3-1/2×10; OD = 3.50 in; ID = 3.232 in; Wall thickness = 10 gauge = 0.134 in; A = 1.417 in2 1-83 [Appendix A-9(d)] For tube in Problem 1-93, OD = 3.50 in = 88.9 mm; A = 1.417 in2 × (645.2 mm2/in2) = 914.2 mm2; Closest SI metric tube: 90×3 mm; OD = 90 mm; A = 820 mm2Chapter 2 Design Properties of Materials Only those problems requiring numerical data are shown. 2-14 𝑠𝑢=90 ksi (621 MPa ); 𝑠𝑦=60 ksi (414 MPa );25% Elongation [Appendix A- 10] Because % Elongation >5%, it is ductile. 2-15 1020 HR: 36% Elongation – Greater ductility [Appendix A- 10] 1040 HR: 25% Elongation 2-16 SAE 1141 OQT 700: High sulfur alloy steel with 0.41% carbon, quenched in oil, tempered at 700 ℉. [Appendix A-10] 2-17 Yes, AISI 1141 steel has: 𝑠𝑦=172 ksi @ OQT 700 , 𝑠𝑦=129 ksi @ OQT 900 By interpolation, 𝑠𝑦≈150 ksi @ OQT 800 . [Appendix A-10] 2-18 𝐸 = 30×106 psi (207 GPa) For all carbon and alloy steels. [Appendix A- 10] 2-19 𝑊𝑡 = Density × Volume = (0. 283 lb /in3)(1.0)(4.0)( 14.5) in3=𝟏𝟔. 𝟒 𝐥𝐛 [Appendix A- 10] Value of lbm= Value of lb force ( Wt) 2-20 Volume = Area × Length =𝜋 4(50)2×250 = 4. 909 ×105 mm3 Steel Bar: Mass =7680 kg m3×4.909× 105 mm3 1×1 m3 (103 mm )3= 3. 77 kg [Appendix A- 10] 𝑊𝑡 = 𝑚 ∙ 𝑔 = 3. 77 kg∙ 9.81 m/s2=36.98 kg ∙ m/ s2=𝟑𝟔.𝟗𝟖 𝐍 2-21 Magnesium would stretch more because it has a lower 𝐸. 𝐸𝑀𝑎𝑔 =45 GPa; 𝐸 𝑇𝑖=114 GPa; Ti is stiffer. [Appendix A- 11] 2-23 Alloy of aluminum with silicon and magnesium. Heat treated to T6 temper. 2-24 [Appendix A- 14] 2-29 𝑠𝑢𝑡=40 ksi; 𝑠 𝑢𝑐= 140 ksi [Appendix A- 13] 2-31 Bending 𝜎𝑑=1450 psi; Tension 𝜎𝑑= 850 psi; Compression 1000 psi Parallel to grain, 385 psi Perpendicular to grain; Shear 𝜏𝑑=95 psi [Appendix A-15] 2-32 2000 to 7000 psi [Section 2- 11]𝑠𝑢 𝑠𝑦 𝐸 Density 6061 -0 18 ksi 8 ksi 10×106 psi 0.10 lb/ in3 6061 -T4 35 ksi 21 ksi 10×106 psi 0.10 lb/ in3 6061 -T6 45 ksi 40 ksi 10×106 psi 0.10 lb/ in3 2-44 Graphite fibers. 2-45 S-glass, quartz fibers, tungsten fibers coated with silicon carbide 2-51 Material Specific strength (in) Ratio to SAE 1020 Graphite/Epoxy (High Strength) 4.86×106 25.0 Aramid/Epoxy Composite 4.00×106 20.6 Boron/Epoxy Composite 3.60×106 18.5 Graphite/Epoxy (Ultra-high mod.) 2.76×106 14.2 Glass/Epoxy Composite 1.87×106 9.63 Titanium Ti-6Al- 4V 1.00×106 5.15 SAE 5160 OQT 700 Steel 0.929×106 4.78 Aluminum 7075- T6 0.822×106 4.23 Aluminum 6061- T6 0.459×106 2.36 SAE 1020 HR Steel 0.194×106 1.00 2-52 Material Specific modulus (in) Ratio to SAE 1020 Graphite/Epoxy (Ultra-high mod.) 8.28×108 7.81 Boron/Epoxy Composite 4.00×108 3.77 Graphite/Epoxy (High Strength) 3.45×108 3.25 Aramid/Epoxy Composite 2.20×108 2.07 SAE 1020 HR Steel 1.06×108 1.00 SAE 5160 OQT 700 Steel 1.06×108 1.00 Titanium Ti-6Al- 4V 1.03×108 0.97 Aluminum 6061- T6 1.02×108 0.96 Aluminum 7075- T6 0.99×108 0.93 Glass/Epoxy Composite 0.66×108 0.62 Solutions to Problems 2-58 to 2-69 : Some data are approximated from Figure P2-66. The most accurate values are for ultimate strength (b.) and % elongation (f.). The elastic limit (d.) is estimated between the proportional limit (c.) and yield strength (a.). The modulus of elasticity (e.) is computed from [  stress /  strain], the slope of the straight- line part of each curve. Listed materials are found in Appendixes A- 10 – A-14, matching su, sy, E, and % elongation. 2-58 a. sy = 73 ksi; Offset method b.su = 83 ksi c.sp = 60 ksi d.sel = 67 ksi e.E = 10.0×106 psi f.11% elongation g.Ductile h.Aluminum i.7075-T 6 3-60 a. sy = 62 ksi; Offset method b.su = 75 ksi c.sp = 50 ksi d.sel = 56 ksi e.E = 16.7×106 psi f.15% elongation g.Ductile h.Copper alloy i.C54400 Bronze-Hard 4-62 a. No yield strength-Brittle b.su = 55 ksi c.sp = 50 ksi d.sel = 53 ksi e.E = 20.0×106 psi f.0.5% elongation g.Brittle h.Cast iron i.ASTM A48 Grade 60 5-64 a. sy = 35 ksi; Yield point b.su = 57 ksi c.sp = 30 ksi d.sel = 27 ksi e.E = 26.0×106 psi f.21% elongation g.Ductile h.Structural Steel i.ASTM A362-59 a. sy = 173 ksi; Yield point b.su = 187 ksi c.sp = 162 ksi d.sel = 168 ksi e.E = 29.0×106 psi f.15% elongation g.Ductile h.Steel i.SAE 4140 OQT 900 3-61 a. sy = 49 ksi; Yield point b.su = 65 ksi c.sp = 46 ksi d.sel = 48 ksi e.E = 26.5×106 psi f.36% elongation g.Ductile h.Steel i.SAE 1020 CD 4-63 a. sy = 53 ksi; Offset Method b.su = 59 ksi c.sp = 31 ksi d.sel = 42 ksi e.E = 12.0×106 psi f.5.0% elongation g.Borderline Ductile/ Brittle h.Zinc i.Cast ZA-1 2 5-65 a. sy = 19 ksi; Offset Method b.su = 40 ksi c.sp = 14 ksi d.sel = 17 ksi e.E = 6.0×106 psi f.5.0% elongation g.Borderline Ductile/ Brittle h.Magnesium i.ASTM AZ 63A-T 62-66 a. sy = 155 ksi; Offset Method b.su = 170 ksi c.sp = 142 ksi d.sel = 149 ksi e.E = 16.5×106 psi f.8% elongation g.Ductile h.Titanium i.6Al-4 V 3-68 a. sy = 80 ksi; Offset Method b.su = 90 ksi c.sp = 62 ksi d.sel = 71 ksi e.E = 26.0×106 psi f.15% elongation g.Ductile h.Stainless Steel i.SAE 430 – Full hard2-67 a. sy = 40 ksi; Offset Method b.su = 45 ksi c.sp = 30 ksi d.sel = 35 ksi e.E = 10.0×106 psi f.17% elongation g.Ductile h.Aluminum i.6061-T 6 3-69 a. sy = 80 ksi; Offset Method b.su = 95 ksi c.sp = 55 ksi d.sel = 68 ksi e.E = 26.0×106 psi f.2.0% elongation g.Brittle – Does not yield h.Malleable Iron i.ASTM A220 Grade 80002Chapter 3 Design of Members Under Direct Stresses 3-1 𝜎 =𝑃 𝐴=8.75×103 N π(12 m m)2⁄4= 77.4 MPa = 𝜎 𝑑 = 𝑠𝑦 2 Required 𝑠𝑦 = 2(𝜎𝑑) = 2(77.4 MPa ) = 155 𝐌𝐏𝐚 Aluminum 2014-T4 has 𝑠𝑦 = 290 MPa (App. A-14) 3-2 𝜎 =𝑃 𝐴=18.0×1 03 N (10)(30)mm2= 60.0 MPa = 𝜎 𝑑 = 𝑠𝑢 8 Required 𝑠𝑢 = 8(𝜎𝑑) = 8(60.0 MPa ) = 480 𝐌𝐏𝐚 Plus good ductility SAE 1141 Annealed has 𝑠𝑢 = 600 MPa; 26% Elongation (A-1 0) 3-3 𝜎 =𝑃 𝐴=1650 lb (0.38 in)2= 11 427 psi = 𝜎 𝑑 = 𝑠𝑢 8 Required 𝑠𝑢 = 8(𝜎𝑑) = 8(11 427 ) = 91 416 𝐩𝐬𝐢 Plus good ductility SAE 1141 WQT 1300 has 𝑠 𝑢 = 94 ksi; 2 8% Elongation (A-1 0) 3-4 𝜎 =𝑃 𝐴=1850 lb π(0.375 in)24⁄=16 750 psi = 𝜎 𝑑= 0. 60𝑠 𝑦 (AISC) Required 𝑠𝑦=𝜎𝑑 0.60= 16 750 psi 0.60=𝟐𝟕 𝟗𝟎𝟎 𝐩𝐬𝐢 ASTM A36 Structural steel has 𝑠𝑦=36 000 psi (A-12) 3-5 𝜎 =𝑃 𝐴=5200 lb 5.25 in2=𝟗𝟗𝟎 𝐩𝐬𝐢 Too High; 𝜎𝑎𝑙𝑙𝑜𝑤 =575 psi (A-15) [Area from Appendix A-4(a)] 3-6 a) No. 1 grade Douglas Fir has 𝜎𝑎𝑙𝑙𝑜𝑤 =1050 psi (A-15) b) To use No. 2 grade Southern Pine: 𝜎𝑎𝑙𝑙𝑜𝑤 =575 psi Required area =𝑃 𝜎=5200 lb 575 lb/in2= 9. 04 in2; [From Appendix A-4(a)] Use 2 × 8 or 4 × 4 , or two 2 × 4 wood boards 3-7 𝜎 =𝑃 𝐴; 𝐴 =𝑃 𝜎=6400 lb 12 000 lb/in2= 0. 533 in2= 𝜋𝐷2/4 Required 𝐷 = √4𝐴/ 𝜋=√4(0. 533 in2)/𝜋= 𝟎. 𝟖𝟐𝟒 𝐢𝐧 Specif y 7 8in (0.875 in) or 1.00 in diameter. 3-8 Total mass =1150 +6350 =7500 kg; 1875 kg on each strap. 𝐹 = 𝑚 ∙ 𝑔 = 1875 kg∙ 9.81m s2=18 394 N Required 𝐴 =𝑃 𝜎=18 394 N 70 N/ mm2=263 mm2= (𝑤)(8 mm ) Required 𝑤 = 263 mm2/8 mm =𝟑𝟐. 𝟖 𝐦𝐦 ; Specify w = 35.0 mm 3-9 Force on shelf =1840 kg∙ 9.81m s2=18 050 N: 9025 N/side 𝐶𝑣= 𝐴 𝑦= 9025 N/2 = 4513 N 𝐶 =𝐶𝑣 sin20°=13 194 N C = Force in rod Required 𝐴 =𝐶 𝜎𝑑=13 194 N 110 N/ mm2=120 mm2= 𝜋𝐷2/4 Required 𝐷 = √4𝐴/ 𝜋=√4(120 mm2)/π=𝟏𝟐. 𝟒 𝐦𝐦 Specify 𝑫 = 𝟏𝟒. 𝟎 𝐦𝐦 3-10 𝜎 =𝑃 𝐴=70 000 lb π(8.0 in)24⁄= 1393 psi =1 4× Rated strength (Sec. 2- 11) Required rated strength = 4(1393 )=5570 psi Specify 6000 psi Rated strength 3-11 Load on each block =29 500 lb/3 = 9833 lb 𝜎 =𝑃 𝐴=9833 lb 12.25 in2=803 psi [Area from Appendix A-4(a)] If compression is perpendicular to grain, no suitable wood listed. If parallel to grain: No. 1 Southern Pine - 𝜎𝑎𝑙𝑙.=850 psi (A-15) No. 2 Hemlock - 𝜎𝑎𝑙𝑙.=800 psi No. 2 Douglas Fir - 𝜎𝑎𝑙𝑙.=1000 psi 3-12 𝜎𝑎𝑙𝑙𝑜𝑤 =Rated Strength 4=20.7 MPa 4= 5. 18 MPa (See Section 2 − 11) Required 𝐴 =𝑃 𝜎=1.50× 106 N 5.18 N/ mm2= 2. 90×105 mm2= 𝜋𝐷2/4 Required 𝐷 = √4𝐴/ 𝜋=607 mm ; Specify 700 mm Diameter 3-13 𝑠𝑢=483 MPa = 𝜎 =𝑃 𝐴; 𝑃 = 𝜎 ∙ 𝐴 𝐴 = 𝜋( 122−102)/4 = 34.56 mm2 𝑃 = 𝜎 ∙ 𝐴 =483 N mm2∙34.56 mm2=16.7 × 103 N = 𝟏𝟔. 𝟕 𝐤𝐍 3-14 𝐴 = ( 40 mm)2=1600 mm2; 𝜎𝑎𝑙𝑙𝑜𝑤 = 1. 69 MPa Perpendicular to Grain (A-15) 𝜎𝑎𝑙𝑙𝑜𝑤 = 5. 52 MPa Parallel to Grain (#2 Hemlock) 𝑃 = 𝜎 ∙ 𝐴 = (1.69Nmm2⁄ )( 1600 mm2)= 𝟐. 𝟕𝟎 𝐤𝐍 - Perpendicular to Grain 𝑃 = 𝜎 ∙ 𝐴 = (5.52Nmm2⁄ )( 1600 mm2)= 𝟖. 𝟖𝟑 𝐤𝐍 - Parallel to Grain 3-15 𝜎𝑑= 0. 60 𝑠 𝑦= 0. 60(50 ksi )=30.0 ksi =𝑃 𝐴 [From AISC method] [From Appendix A- 12] 𝑠 𝑦=50 ksi if 𝐷 < 0. 75 in Required 𝐴 =𝑃 𝜎𝑑=4000 lb 30 000 lb/in2= 0. 133 in2= 𝜋𝐷2/4 Required 𝐷 = √4𝐴/ 𝜋=√4(0. 133 in2)/𝜋= 0. 412 in Specify 𝑫 =𝟕 𝟏𝟔 𝐢𝐧= 𝟎. 𝟒𝟑𝟕𝟓 𝐢𝐧 or 𝟎.𝟓𝟎𝟎 𝐢𝐧 (OK- < 0. 75 in ) 3-16 𝐴 = (2.65)(1.40)+ 2 [1 2(1.40)(0.5)] = 4. 41 in2 𝜎 =𝑃 𝐴=52 000 lb 4.41 in2=𝟏𝟏 𝟕𝟗𝟏 𝐩𝐬𝐢 = 𝜎 𝑑=𝑠𝑢𝑐 𝑁 𝑁 =𝑠𝑢𝑐 𝜎=80 000 psi 11 791 psi= 𝟔. 𝟕𝟖 3-17 For shock loading – ductile metal: 𝜎𝑑=𝑠𝑢𝑐 12 𝜎𝑑=1650 MPa 12=137 .5 MPa [From Appendix A- 14] Required 𝐴 =𝑃 𝜎𝑑=135× 103 N 137.5 N/ mm2=982 mm2=𝐵𝐻 = 𝐵 (2𝐵)= 2𝐵2 Required 𝐵 = √𝐴/2=√982 /2=22.2 mm ; 𝐻 = 44.4 mm Specify 𝑩 = 𝟐𝟓. 𝟎 𝐦𝐦 ; 𝑯 = 𝟓𝟎. 𝟎 𝐦𝐦 3-18 𝜎𝑑=𝑠𝑢 8=8000 psi 8=1000 psi =𝑃 𝐴 [From Appendix A- 16] Required 𝐴 =𝑃 𝜎𝑑=110 lb 1000 lb/in2= 0. 110 in2= (0. 20)(𝑤) Required 𝑤 =𝐴 0.20=0.110 in2 0.20 in= 0. 550 in ; Specify 𝒘 = 𝟎. 𝟔𝟎𝟎 𝐢𝐧 3-19 𝐴 = (80)(40)−(60)(15)+𝜋(40)2 4=3557 mm2 𝜎 =𝑃 𝐴=640× 103 N 3557 mm2=180 MPa = 𝜎 𝑑 For ductile metals: 𝜎𝑑=𝑠𝑦 2; Required 𝑠𝑦= 2(180 )=360 MPa Possible metals: SAE 1040 HR, 𝑠𝑦=414 MPa [Appendix A- 10)] SAE 4140 Annealed, 𝑠𝑦=414 MPa [Appendix A- 10] C54400 Bronze, 𝑠𝑦=393 MPa [Appendix A-11 )] Aluminum 2014-T6, 𝑠𝑦=414 MPa [Appendix A- 14 SAE 1020 CD, 𝑠𝑦=441 MPa [Appendix A-10] 3-20 See Problem 1-48: 𝜎𝑚𝑎𝑥 = − 48.0 MPa (Compression) 𝜎𝑑=(0.6)𝑠𝑦= 0.6 (248 MPa )=149 MPa (AISC Method) (A- 12) OK for compressive stress. Column bucking should be checked. 3-21 𝜎 = 50 MPa in member AB – See Problem 1- 53. 𝜎𝑑=𝑠𝑢 8; Required 𝑠𝑢= 8𝜎 = 8 (50)=400 MPa Aluminum 2014-T4, 𝑠𝑢=427 MPa; 20% Elongation. [Appendix A- 14] 3-22 𝐴 = 0. 944 in2 [App. A − 4 (a)]; 𝜎𝑑=(0.60)𝑠𝑦= 0. 60(36 000 psi)=21 600 psi 𝑃𝑎𝑙𝑙𝑜𝑤 = 𝜎 𝑑∙ 𝐴 = (21 600 lb/in2)(0.944 in2) = 𝟐𝟎 𝟑𝟗𝟎 𝐥𝐛 Elastic Deformation in Tension and Compression Members 3-23 𝜎𝑎=800 psi (A − 15); 𝐸 = 1. 40×106 psi; 𝐿 = (6.0 ft)× (12in ft) = 72 in 𝛿 =𝜎𝐿 𝐸=800 psi( 72 in) 1.40× 106 psi= 𝟎. 𝟎𝟒𝟏 𝐢𝐧 3-24 𝐴 = (0.75)(12)= 9 mm2; 𝜎 =𝐹 𝐴=90 N 9.0 mm2=10.0 MPa a)ABS; 𝑠𝑢=41 MPa (OK); 𝐸 = 2480 N/mm2 𝛿 =𝐹𝐿 𝐸𝐴=(90 N)( 375 mm) (2480 N/mm2)(9 mm2)= 𝟏. 𝟓𝟕 𝐦𝐦 b)Phenolic; 𝑠𝑢=45 MPa (OK); 𝐸 = 7580 MPa = 7580 N/mm2 𝛿 =𝐹𝐿 𝐸𝐴=(90)(375) (7580 )(9)= 𝟎. 𝟒𝟗𝟓 𝐦𝐦 3-25 𝐷𝑜= 2. 50 in; 𝐷 𝑖= 2. 50− 2(0.085 )= 2. 33 in ; 𝐸 = 10.6 × 106 psi 𝐴 =𝜋(𝐷𝑜2−𝐷 𝑖2) 4= 0. 645 in2 𝛿 =𝐹𝐿 𝐸𝐴; 𝐹 =𝛿𝐸𝐴 𝐿=(0.005 in)(10.6× 106 psi)(0. 645 in2) 14.5 in=𝟐𝟑𝟓𝟕 𝐥𝐛 𝜎 =𝐹 𝐴=2357 lb 0.645 in2=𝟑𝟔𝟓𝟓 𝐩𝐬𝐢 (OK); 𝑠𝑦=42 000 psi 3-26 1) Weigh it. Density of aluminum = 0.10 lb /in3 (Weight density) Density of magnesium = 0.066 lb/in3 2) Exert a known load and measure deformation. 𝐸 for aluminum = 10×106 psi 𝐸 for magnesium = 6.5 × 106 psi 3-27 a) SAE 1020 HR Steel: Strength : 𝜎𝑑=𝑠𝑢 8=448 8=56.0 MPa Required 𝐴 =𝐹 𝜎𝑑=3500 N 56.0 N/ mm2 =62.5 mm2=𝜋𝐷2 4 Required 𝐷 = √4𝐴 𝜋=√4(62.5 ) 𝜋= 8. 92 mm Deformation : 𝐸 = 207 GPa = 207 000 MPa = 207 000 N/mm2 Required 𝐴 =𝐹𝐿 𝐸𝛿=(3500 N)( 630 mm ) (207 000 N/ mm2)(0.12 mm)=88.8 mm2=𝜋𝐷2 4 Required 𝐷 = √4𝐴 𝜋=𝟏𝟎.𝟔𝟑 𝐦𝐦 - Governs b) SAE 4140 OQT 700 Steel is much stronger than SAE 1020 HR. Therefore: Deformation governs. 𝐸 is the same. Required 𝑫 = 𝟏𝟎.𝟔𝟑 𝐦𝐦 c) Aluminum 6061- T6: Strength : 𝜎𝑑=𝑠𝑢 8=310 MPa 8=38.75 MPa Required 𝐴 =𝐹 𝜎𝑑=3500 N 38.75 N/ mm2=90.32 mm2 Required 𝐷 = √4𝐴 𝜋=10.7 mm Deformation : 𝐸 = 69 GPa = 69 000 MPa = 69 000 N/mm2 Required 𝐴 =𝐹𝐿 𝐸𝛿=(3500)( 630) (69 000)(0. 12)=266 .3 mm2 - Governs Required 𝐷 = √4𝐴 𝜋=𝟏𝟖. 𝟒 𝐦𝐦 Mass = Volume × Density = 𝐴 × 𝐿 × Density Steel: Mass =(88.8 mm2)(630 mm )(7680 kg m3⁄ ) ×1 m3 (103 mm )3= 𝟎. 𝟒𝟑𝟎 𝐤𝐠 Aluminum: Mass =(266.3 )(630 )(2770 ) 109 = 𝟎. 𝟒𝟔𝟓 𝐤𝐠 3-28 𝐴 =𝜋𝐷2 4=𝜋(12)2 4= 113.1 m m2; 𝜎 = 𝐹 𝐴 = 16 0 00 N 113 .1 mm2= 141 MPa - Low for steel - OK 𝛿 =𝐹𝐿 𝐸𝐴=(16.0×1 03 N)(240 mm) (207 000 N/mm2)(113 .1 mm2) = 𝟎. 𝟏𝟔4 𝐦 𝐦 3-29 a) Ti -6AL- 4V: 𝐸 = 114 GPa = 114 000 MPa = 114 000 N/mm2, 𝑠𝑦=1070 MPa 𝛿 =𝐹𝐿 𝐸𝐴=(5000 N)( 1250 mm ) (114 000 N/mm2)(64 mm2)= 𝟎. 𝟖𝟓𝟕 𝐦𝐦 ; 𝜎 =𝐹 𝐴=78.1 MPa - Low – OK b) SAE 501 OQT 1000 Steel: 𝐸 = 200 GPa = 200 000 N/mm2, 𝑠𝑦=931 MPa 𝛿 =𝐹𝐿 𝐸𝐴=(5000)( 1250) (200 000 )(64)= 𝟎. 𝟒𝟖𝟖 𝐦𝐦3-30 Required 𝐴 =𝐹 𝜎𝑑=35 000 lb 21 600 lb/in2= 1. 620 in2: Use L 𝟒 × 𝟒 ×𝟏 𝟒; 𝐴 = 1. 94 in2 𝛿 =𝐹𝐿 𝐸𝐴=(35 000 lb)(13 ft)(12 in/ft) (1.94 in2)(29× 106 lb/in2)= 𝟎. 𝟎𝟗𝟕 𝐢𝐧 3-31 Elongation : 𝛿 =𝐹𝐿 𝐸𝐴=(450 lb)(8.40 in) (30× 106 lb/in2)(.25)(. 125) in2= 𝟎. 𝟎𝟎𝟒𝟎 𝐢𝐧 Compression : 𝛿 =𝐹𝐿 𝐸𝐴=(50)(8. 40) (30× 106)(.25)(. 125 )= 𝟎. 𝟎𝟎𝟎𝟒𝟓 𝐢𝐧 Stress 𝜎𝑡=𝐹𝑡 𝐴=450 0.03125=14 400 psi < 𝑠 𝑦 for any steel. Repeated load: 𝜎𝑑=𝑠𝑢 8; 𝑠𝑢𝑚𝑖𝑛= 8(14.4 ksi )=115 ksi; SAE 1141 OQT 1100, 𝑠𝑢=116 ksi 3-32 Lower Section : 𝛿1=𝐹1𝐿1 𝐸𝐴=(5000 )(10) (30× 106)(.50)= 0. 0033 in Mid Section : 𝛿2=𝐹2𝐿2 𝐸𝐴=(7000 )(15) (30×106)(.50)= 0. 0070 in Top Section : 𝛿3=𝐹3𝐿3 𝐸𝐴=(10 500 )(25) (30× 106)(.50)= 0. 0175 in Total : 𝛿𝑇= 𝛿 1+ 𝛿 2+ 𝛿 3= 𝟎. 𝟎𝟐𝟕𝟖 𝐢𝐧 Max stress in top section – assume static load. 𝜎 =𝐹 𝐴=10 500 lb 0.50 in2=21 000 psi 𝜎𝑑=𝑠𝑦 2; 𝑠𝑦𝑚𝑖𝑛= 2(21 ksi )=42 ksi OK for any steel in Appendix A- 10. 3-33 𝐴 =𝜋(𝐷𝑜2−𝐷 𝑖2) 4=𝜋(1.252−1.1262) 4= 0. 2314 in2 𝐹 =𝛿𝐸𝐴 𝐿=(0.050 in)(10× 106lb in2)(0.2314 in2) 36.0 in=3214 lb 𝜎 =𝐹 𝐴=3214 lb 0.2314 in2=𝟏𝟑 𝟗𝟎𝟎 𝐩𝐬𝐢 - Too High 𝜎𝑑=𝑠𝑢 8=45 000 psi 8=5625 psi 3-34 𝐴 =𝜋(𝐷𝑜2−𝐷 𝑖2) 4=𝜋(0.752−0.5632) 4= 0. 1928 in2 𝜎 =𝐹 𝐴=2500 lb 0.1928 in2=12 964 psi = 𝜎𝑑=𝑠𝑦 2; Required 𝑠𝑦=25 930 psi Aluminum 3003-H18 has 𝑠𝑦=27 000 psi 𝛿 =𝐹𝐿 𝐸𝐴=(2500 lb)(8.75 ft)(12 in/ft) (10× 106 lb/in2)(0.1928 in2)= 𝟎. 𝟏𝟑𝟔 𝐢𝐧3-35 𝐴 =𝜋 4(𝐷𝑜2− 𝐷 𝑖2) =𝜋 4(562−482)=653 mm2 𝛿 =𝐹𝐿 𝐸𝐴=(18.2×103 N)( 40 mm ) (69 000 N/mm2)(653 mm2)= 𝟎. 𝟎𝟏𝟔 𝐦𝐦 𝜎 =𝐹 𝐴= 18.2× 103 N 653 mm2=𝟐𝟕. 𝟗 𝐌𝐏𝐚 3-36 𝐴 =𝜋𝐷2 4=𝜋(0. 375 )2 4= 0. 1104 in2 𝜎 =𝐹 𝐴=1600 lb 0.1104 in2=𝟏𝟒 𝟓𝟎𝟎 𝐩𝐬𝐢 𝛿 =𝐹𝐿 𝐸𝐴=(1600 lb)(1 75 ft)(1 2 in/f t) (30× 106 lb/in2)(0.1104 in2)= 1.01 𝐢𝐧 3-37 𝐴 = ( 30 mm)2=900 mm2; 𝐸 = 114 GPa = 114 000 N/mm2 𝐹𝐴𝐵=110 +80−40=150 kN Tension – maximum tensile force 𝐹𝐵𝐶=110 −40=70 kN Tension 𝐹𝐶𝐷=110 kN Tension 𝜎𝑚𝑎𝑥 =150 000 N 900 mm2=167 MPa – in section A-B 𝑠𝑦=1070 MPa - OK 𝛿𝑇= 𝛿 𝐴𝐵+ 𝛿 𝐵𝐶+ 𝛿 𝐶𝐷=𝐹𝐴𝐵𝐿 𝐸𝐴+𝐹𝐵𝐶𝐿 𝐸𝐴+𝐹𝐶𝐷𝐿 𝐸𝐴 𝛿𝑇=(150× 103 N)(250 mm ) (114 000 N mm2 ⁄ )(900 mm2)+(70× 103)(250 ) (114 000 )(900 )+(110× 103)(250 ) (114 000 )(900 ) 𝛿𝑇= 0. 365 mm + 0. 171 mm + 0. 268 mm = 𝟎. 𝟖𝟎𝟒 𝐦𝐦 3-38 𝐸 =𝐹𝐿 𝛿𝐴=(10 000 lb)(10.0 in) (0.023 in)(π(0.75)24⁄)in2= 𝟗. 𝟖𝟒×𝟏𝟎𝟔 𝐩𝐬𝐢 Probably aluminum : 𝐸 = 10.0 × 106 psi 𝜎 =𝐹 𝐴=10 000 lb 0.4418 in2=22 635 psi [See Appendix A-14] 𝜎 < 𝑠 𝑦 for most tempers of most alloys. Ensure no yielding occurred. 3-39 𝐴𝐴𝐵= 𝐴 𝐵𝐶=𝜋(25)2 4=491 mm2: 𝐴𝐶𝐷=𝜋(16)2 4 =201 mm2 𝐹𝐴𝐵= −9. 65−12.32+ 4. 45= − 17.52 kN (Compression); 𝐿𝐴𝐵=80 mm 𝐹𝐵𝐶= −9. 65−12.32= − 21.97 kN (Compression); 𝐿𝐵𝐶=100 mm 𝐹𝐶𝐷= −9. 65 kN (Compression); 𝐿𝐶𝐷=120 mm 𝜎𝐵𝐶=𝐹𝐵𝐶 𝐴𝐵𝐶=𝟒𝟒. 𝟕 𝐌𝐏𝐚 [Continued on next page] 𝜎𝐶𝐷=𝐹𝐶𝐷 𝐴𝐶𝐷=𝟒𝟖. 𝟎 𝐌𝐏𝐚 Max 𝑠𝑢=72.0 MPa ; N =𝑠𝑢 𝜎𝐶𝐷=72 48= 1. 50 OK for static load. 𝐸 = 2960 N/mm2 𝛿𝐴𝐷=𝐹𝐴𝐵𝐿𝐴𝐵 𝐸𝐴𝐴𝐵+𝐹𝐵𝐶𝐿𝐵𝐶 𝐸𝐴 𝐵𝐶+𝐹𝐶𝐷𝐿𝐶𝐷 𝐸𝐴𝐶𝐷 𝛿𝐴𝐷= (−17.52× 103 N)( 80 mm) (2960N mm2)(491 mm2)+(−21.97× 103)(100) (2960)( 491 )+(−9. 65× 103)(120) (2960)( 201) 𝛿𝐴𝐷= −0. 9644 mm − 1. 512 mm − 1. 946 mm = −𝟒. 𝟒𝟐 𝐦𝐦 (Shorter) 3-40 𝜎𝐶=𝐹 𝐴𝐶=64 000 lb 50.27 in2=1273 psi ; Let 𝜎𝐷> 4𝜎 𝐶=5092 psi Use 𝑓𝐶′=6000 psi Rated: Then 𝐸 = 4.7 × 106 psi (Sec. 2- 11) 𝐴𝐶=𝜋(8)2 4=50.27 in2 ; 𝐴𝑇= 6. 02 in2 [App. A -8(a)] 𝐸𝑇=29×106 psi Structural Steel 𝛿 = 𝛿 𝐶+ 𝛿 𝑇=(64 000 lb)(3.0 ft)(12 in/ft) (4.7× 106 lb/in2)(50.27 in2)+(64 000)(8.6)( 12) (29× 106)(6.02) 𝛿 = 0. 0098 in+ 0. 0378 in= 𝟎. 𝟎𝟒𝟕𝟔 𝐢𝐧 𝜎𝑇=64 000 6.02=10 631 psi < 𝑠 𝑦 for ASTM A36 structural steel. 3-41 a) 𝛿 =𝐹𝐿 𝐸𝐴=(120 lb)(10.5 ft)(12 in/ft) (17.6×106 lb/in2)(0.00322 in2)= 𝟎. 𝟐𝟕𝟔 𝐢𝐧 𝐴 =𝜋𝐷2 4=𝜋(0. 064)2 4= 0. 00322 in2 𝜎 =𝐹 𝐴=120 lb 0.00322 in2=37 300 psi ; Close to 𝑠𝑦=44 000 psi b)=𝐹 𝐴=200 lb 0.00322 in2=62 200 psi ; This exceeds the ultimate strength of the wire. It should break. 3-42 𝜎 =𝐹 𝐴=25 lb (0.006 )(0.75) in2=𝟓𝟓𝟓𝟔 𝐩𝐬𝐢 𝛿 =𝐹𝐿 𝐸𝐴=(25 lb)(25 ft)(12 in/ft) (30× 106 lb/in2)(0.0045 in2)= 𝟎. 𝟎𝟓𝟓𝟔 𝐢𝐧 3-43 𝐹 = 𝜎 𝑎𝑙𝑙𝑜𝑤 × 𝐴 = (550 lb/in2 )(12.25 in2)=𝟔𝟕𝟑𝟕 𝐥𝐛 𝛿 =−𝐹𝐿 𝐸𝐴=−(6737 lb)(10.75 ft)(12 in/ft) (1 300 000 lb/in2)(12.25 in2)= −𝟎. 𝟎𝟓𝟓 𝐢𝐧 3-44 𝐴 = (2002−1502)=17 500 mm2 𝐹 = 𝜎 ∙ 𝐴 = (− 200 N/mm2)(17 500 mm2) = −3.5 × 106 N = −𝟑. 𝟓𝟎 𝐌𝐍 𝛿 =𝜎𝐿 𝐸=(−200 N/ mm2)(1800 mm ) (165 000 N/ mm2)= −𝟐. 𝟏𝟖 𝐦𝐦 3-45 𝐴 =𝜋𝐷2 4=𝜋(3.0)2 4= 7. 07 mm2 ; Assume mass of plate is small. 𝐹 =𝛿𝐸𝐴 𝐿=(6.0 mm )(110 000 N/ mm2)(7.07 mm2) 3600 mm=𝟏𝟐𝟗𝟔 𝐍 𝜎 =𝐹 𝐴=1296 N 7.07 mm2=𝟏𝟖𝟑 𝐌𝐏𝐚 (Less than 𝑠𝑦) 𝑚 = 𝑤/ 𝑔 = 1296 N/9.81 m/s2=132 N ∙ s2/m = 𝟏𝟑𝟐 𝐤𝐠 3-46 𝜎 = 50 MPa (Prob. 1-53); 𝛿 =𝜎𝐿 𝐸=(50 N/mm2)(1250 mm) 69 000 N/ mm2 = 𝟎. 𝟗𝟎𝟔 𝐦𝐦 3-47 No. 2 Southern Pine: 𝜎𝐶 𝐴𝐿𝐿𝑂𝑊𝐴𝐵𝐿𝐸 = 3.7 9 MPa: 𝐸 = 9.0 GPa 90 mm x 90 m m Post. 𝐴 = 8100 m m2, 𝐿 = 4.2 5 m 𝐴; 𝐹 = 𝜎 × 𝐴 = 3. 79 𝜎 =𝐹 𝑀𝐴𝑋 𝐶 𝐴𝐿𝐿N mm2× 8100 mm2=30 699 N 𝛿=𝐹𝐿 𝐸𝐴=(30 699 N )(4.25 m ) (9.0 × 109 mN 2)(8100 mm2)×(103 2mm )2 m 𝛿 = 1. 790 x 10−3 m = 𝟏. 𝟕𝟗𝟎 𝐦𝐦3-48 𝛿 𝑀𝐴𝑋 = 0.5 0 mm ; Repeated axial tensile load. Specify maximum permissible load. SAE 4140 OQT 1300. 𝐸 = 207 GPa a) Deformation: 𝛿 =𝐹𝐿 𝐸𝐴 ; then 𝐹𝑀𝐴𝑋 =𝛿𝐸𝐴 𝐿 𝐴 = 20 x 30=600 mm2 ; 𝐿 = 700 mm 𝐹𝑀𝐴𝑋 =(0.50 mm )(207 x 109 N/m2)(600 mm2) 700 mm x1 m2 (103 mm )2 𝐹𝑀𝐴𝑋 =88.7 x 103 N = 𝟖𝟖. 𝟕 𝐤𝐍 b)Stress: 𝜎𝑑=𝑠𝑢 8=814 MPa 8=101 .8 MPa =F A 𝐹𝑀𝐴𝑋 = 𝜎 𝑑𝐴 = (101 .8 N mm2)(600 mm2)=61 000 N = 61.0 kN Stress governs the design. 𝑭𝑴𝑨𝑿 =𝟔𝟏. 𝟎 𝐤𝐍 30 3-49 Figure P3-79. Compute total elongation. 2014 −T6 Aluminum; 𝐸 = 73 GPa =73 x 109 Pa 𝐸 = 73 x 103 MPa = 73 x 103 N/mm2; 𝑠𝑦=414 MPa 𝛿1=𝐹1𝐿1 𝐸𝐴1=(40 x 103N)(30 mm ) (73 x106N mm2)(100 mm2) = 0.164 mm A1 = (10 mm)2 = 100 mm2 𝛿2=𝐹2𝐿2 𝐸𝐴2=(40 x 103)(40) (7.3 x 103)(400 )= 0. 0548 mm A2 = (20 mm)2 = 400 mm2 𝛿3=𝐹3𝐿3 𝐸𝐴2=(40 x 103)(50) (73 x 103)(225 )= 0. 1218 mm A3 = (15 mm)2 = 225 mm2 𝛿𝑇𝑂𝑇𝐴𝐿 = 𝛿 1+ 𝛿 2+ 𝛿 3= 0. 341 mm Check stress in 1): 𝜎1=𝐹1 𝐴1=40 x 103N 100 mm2=𝟒𝟎𝟎 𝐌𝐏𝐚 Close to 𝒔𝒚 – Not Safe 3-50 Force Analysis: 𝑚 = 680 k g 𝑊 = 𝑚𝑔 =(680 kg)(9.81 m s) = 6671 N = Downward force at C. Sine Law: 6671𝑁 sin75°=𝐴𝐶 sin35°=𝐵𝐶 sin70° 𝐴𝐶=6671 N(sin 35°)/ sin 75° =3961 N 𝐵𝐶=6671 N(sin 70°)/ sin 75° =6490 N Stress: Rod Area = 𝐴 =𝜋(8.00 mm )2 4=50.27 mm2 𝜎𝐵𝐶=𝐵𝐶 𝐴=6490 N 50.27 mm2=129 MPa 𝜎𝐴𝐶=𝐴𝐶 𝐴=3961 N 50.27 mm2=78.8 MPa Assume static load: Safe for any steel Deflection: 𝛿𝐴𝐿=𝐴𝐶(𝐿1) 𝐸𝐴; 𝛿𝐵𝐶=𝐵𝐶(𝐿2) 𝐸𝐴 𝐿1=14.000 m cos 20°=14.898 m; 𝐿2=7.000 m cos 55°=12.204 m Original vertical distance from ceiling to C: tan55° =𝐵𝐷 𝐶𝐷; 𝐵𝐷 =𝐶𝐷tan 55° = 7. 000 tan55°; 𝐵𝐷 = 9. 997 m 𝛿𝐴𝐶=𝐴𝑐𝐿1 𝐸𝐴=(3961 N )(14.898 m ) (207 x109N m2)(50.27 mm2)x(103 mm )2 𝑚2 𝛿𝐴𝐶= 5. 67 x 10−3 m = 𝟓. 𝟔𝟕 𝐦𝐦 𝛿𝐵𝐶=𝐵𝐶(𝐿2) 𝐸𝐴=(6490 N )(12.204 m ) (207 x109N m2)(50.27 mm2)x (103 mm )2 𝑚2 = 7. 61 x 10−3 m = 7.61 mm 10 mm 20 mm 15 mm Square cross sections 30 40 50 40 kN 40 kN AC 7.00 in W = 6671 N W = 6671 N AC BC Vector Triangle 14.00 in A C E L1 L B C D FBD of joint C BC