(Solution Manual) Applied Strength of Materials SI Units Version

Mott,Untener, 6

Engineering

Solution Manual

6th Edition

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Solution Manual for Applied Strength of Materials SI Units Version, The downloadable file consists of 315 pages, and the total size of the file is 9.28 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 1.15 Answers in text. 1.16 𝑊=𝑚∙𝑔=1800 kg∙9.81 m/s2=17 658 (kg∙m)/s2=17×103 N 𝑾=𝟏𝟕.𝟕 𝐤𝐍 1.17 Total Weight =𝑚𝑔=4000 kg∙9.81 m/s2=39.24 kN Each Front Wheel: 𝐹𝐹=(1 2)(0.40)(39.24 kN)=𝟕.𝟖𝟓 𝐤𝐍 Each Rear Wheel: 𝐹𝑅=(1 2)(0.60)(39.24 kN)=𝟏𝟏.𝟕𝟕 𝐤𝐍 1.18 Loading = Total Force / Area Total Force =𝑚𝑔= 6800 kg∙9.81 m/s2=66.7 kN Area =(5.0 m)(3.5 m)=17.5 m2 Loading =66.7 kN 17.5 m2⁄ =3.81 kN m2⁄ =𝟑.𝟖𝟏 𝐤𝐏𝐚 1.19 Force = Weight = 𝑚𝑔=25 kg∙9.81 m/s2=245 N K = Spring Scale = 4500 Nm⁄ =𝐹/Δ𝐿 Δ𝐿=𝐹 𝐾=245 N 4500 N/m=0.0545 m=54.5×10−3 m=𝟓𝟒.𝟓 𝐦𝐦 1.22 𝜎=𝑃 𝐴=3200 N (𝜋𝐷24⁄)=3200 N [𝜋(10 mm )2]4⁄=40.7 N mm2=𝟒𝟎.𝟕 𝐌𝐏𝐚 1.23 𝜎=𝑃 𝐴=20×103 N (10)(30) mm2=66.7 N mm2=𝟔𝟔.𝟕 𝐌𝐏𝐚 1.24 𝜎=𝑃 𝐴=3500 N (0.010 m)2=𝟑𝟓.𝟎 𝐌𝐏𝐚 1.25 𝜎=𝑃 𝐴=8300 N [𝜋(9.0 mm )2]4⁄=𝟏𝟑𝟎 .𝟓 𝐌𝐏𝐚 1.26 Load on Shelf =𝑊=𝑚𝑔 =1840 kg∙9.81 ms2⁄ =18 050 N 𝑊/2=9025 N On each side ∑𝑀𝐴=0=(9025 N)(600 mm )−𝐶𝑉(1200 mm ) 𝐶𝑉=4512 N 𝐶=𝐶𝑉/sin30° =9025 N 𝜎=𝑃 𝐴=𝐶 𝐴=9025 N [𝜋(12 mm )2]4⁄=𝟕𝟗.𝟖 𝐌𝐏𝐚 1.27 𝜎=𝑃 𝐴=310 ×103 N [𝜋(0.2 m)2]/4=𝟗.𝟖𝟕 𝐌𝐏𝐚 1.28 𝜎=𝑃 𝐴=(132 000 N)/3 (85 mm )2=𝟔.𝟏 𝐌𝐏𝐚 1.29 𝜎=𝑃 𝐴=3500 N (8.0 mm )2=𝟓𝟒.𝟕 𝐌𝐏𝐚 1.30 𝑊=𝑚𝑔=4200 kg∙9.81 m/s2=41.2 kN 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 1.31 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 1.32 𝐴=(30 mm )2=900 mm2 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 1.33 Areas: A -C; 𝐴1=𝜋(25)2/4=491 mm2 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 Compre ssion For CD: 𝐹𝐶𝐷=−9.65 kN 𝜎𝐶𝐷=𝐹𝐶𝐷 𝐴2=−9.65×103 N 201 mm2=−𝟒𝟖.𝟎 𝐌𝐏𝐚 Compression 1.34 𝐴=515 .8 mm2 [𝐷𝑁 40 Pipe -Appendix A -9(b)] For BC : 𝜎𝐵𝐶=𝐹𝐵𝐶 𝐴=11 000 N 515 .8 mm2=𝟐𝟏.𝟑 𝐌𝐏𝐚 Tension For AB: 𝐹𝐴𝐵=11 000 +2(36 000 cos30°)=73 354 N 𝜎𝐴𝐵=𝐹𝐴𝐵 𝐴=73 354 N 515 .8 mm=𝟏𝟒𝟐 .𝟐 𝐌𝐏𝐚 Tension 1.35 ∑𝑀𝐶=0=13 000 N(1.2 m)−𝐹𝐵𝐷(0.8) 𝐹𝐵𝐷=19 500 N 𝜎𝐵𝐷=𝐹𝐵𝐷 𝐴=19 500 N (25)(16) mm2=𝟒𝟖.𝟖 𝐌𝐏𝐚 Tension 1.36 𝐴𝐷sin30° =5.25 kN 𝐴𝐷=10.5 kN=𝐶𝐷 𝐴𝐵=𝐴𝐷cos30° =9.09 kN=𝐵𝐶 Stresses: 𝐴𝐵,𝐵𝐶: 𝜎𝐴𝐵=𝜎𝐵𝐶=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=𝜎𝐶𝐷=−10.5×103 N 500 mm2=−𝟐𝟏.𝟎 𝐌𝐏𝐚 Compression 1.37 ∑𝑀𝐴=0=25(1.5)+50(3)−𝑅𝐹(4.5) 𝑅𝐹=41.7 kN ∑𝑀𝐹=0=25(3)+50(1.5)−𝑅𝐴(4.5) 𝑅𝐴=33.3 kN 𝑅𝐴=𝐴𝐵sin𝜃=𝐴𝐵(0.8) 𝐴𝐵=41.6 kN Compression 𝐴𝐷=𝐴𝐵cos𝜃=25 kN Tension 𝐵𝐷 =0 𝐵𝐸sin𝜃+25−𝐴𝐵sin𝜃=0 𝐵𝐸=𝐴𝐵sin𝜃−25 sin𝜃=41.63(0.8)−25 0.8=10.4 kN Tension 𝐵𝐶=𝐴𝐵cos𝜃+𝐵𝐸cos𝜃=41.63(0.6)+10.4(0.6) 𝐵𝐶=31.2 kN Compression 𝐵𝐶=𝐶𝐹cos𝜃 𝐶𝐹=𝐵𝐶 cos 𝜃=31 200 0.6= 52.0 kN Compression 𝐶𝐸=50−𝐶𝐹sin𝜃=50− 52.0(0.8) CE = 8.4 kN Compression EF = CF cos  = 52.0 N(0.6) = 31.2 kN Tension Areas of members : Appendixes A -5(c) and A -6(d) AD, DE, EF – 2(475 mm2) = 950 mm2 BD, BE, CE = 475 mm2 AB, BC, CF – 2(608 mm2) = 1216 mm2 Stresses : AD = DE = 25 000/950 = +26.3 MPa EF = 31 200/950 = +32.8 MPa BD = 0 BE = 10 400 /475 = +21.9 MPa CE = -8400/475 = -17.7 MPa [NOTE: Compression members must be AB = -41 600/1216 = -34.2 MPa checked for column buckling.] BC = -31 200/1216 = -25.7 MPa CF = -52 000/1216 = -42.8 MPa 1.38 ∑𝑀𝐶=0=(12.5)(4.0)−𝐴𝐵(2.5) 𝐴𝐵=20 kN 𝜎=20×103 N (20)2 mm2=𝟓𝟎 𝐌𝐏𝐚 1.39 𝐴=𝜋(13)2 4=132 .7 mm2 𝜎=𝐹 𝐴=56 000 N 132 .7 mm2=𝟒𝟐𝟐 𝐌𝐏𝐚 1.40 𝐴=(68)(36)+2[(36)(12)(1/2)]=2880 mm2 𝜎=𝐹 𝐴=230 000 N 2880 mm2=𝟕𝟗.𝟗 𝐌𝐏𝐚 1.41 𝐴=(80)(40)−(60)(15)+𝜋(40)2 4=3557 mm2 𝜎=𝐹 𝐴=640 ×103 N 3557 mm2=𝟏𝟖𝟎 𝐌𝐏𝐚 1.42 Direct Shear – Single Shear 𝐴𝑆=[𝜋(12.0)2 4] mm2=113 mm2 𝜏=𝐹 𝐴𝑆=16.5×103 N 113 mm2=𝟏𝟒𝟔 𝐌𝐏𝐚 1.43 ∑𝐹𝐽=0=55(145 )−𝐹𝑃(45) 𝐹𝑃=177 N 𝐴𝑆=𝜋(3.0)2 4=7.07 mm2 𝜏=𝐹𝑃 𝐴𝑆=177 N 7.07 mm2=𝟐𝟓.𝟏 𝐌𝐏𝐚 1.44 From Problem 1 -46: 𝐹=23 695 N 𝐴𝑆=2[𝜋(10)2 4]=157 mm2 Double Shear 𝜏=𝐹 𝐴𝑆=23 695 N 157 mm2=𝟏𝟓𝟏 𝐌𝐏𝐚 1.45 𝐴𝑆=(75)(90)=6750 mm2 𝜏=𝐹 𝐴𝑆=1800 N 6750 mm2=𝟏.𝟐 𝐌𝐏𝐚 1.46 𝐴𝑆=[2(35)+𝜋(8)](50)=475 .7 mm2 𝜏=𝐹 𝐴𝑆=38.6×103 N 475 .7 mm2=𝟖𝟏.𝟏 𝐌𝐏𝐚 1.47 𝐿=√152+102=18.03 mm 𝐴𝑆=[2(40)+𝜋(20) 2+2(18)]5 𝐴𝑆=737 mm2 𝜏=𝐹 𝐴𝑆=200 000 N 737 mm2=𝟐𝟕𝟏 𝐌𝐏𝐚 1.48 𝑇=𝐹𝑆∙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 1.49 𝐹𝑆=𝑇 𝑅=900 N∙m 0.025 m=36 000 N 𝐴𝑆=𝑏∙𝐿=(12)(60)=720 mm2 𝜏=𝐹𝑆 𝐴𝑆=36 000 N 720 mm2=𝟓𝟎 𝐌𝐏𝐚 1.50 Pin: Double Shear; 𝐴𝑆=2[𝜋(13)2/4]=265 .5 mm2 𝜏=𝐹 𝐴𝑆=90 000 N 265 .5 mm2=𝟑𝟑𝟗 𝐌𝐏𝐚 Collar: Shear Collar from Connector Body 𝐴𝑆=𝜋𝑑𝑡 =𝜋(22)(5)=345 .6 mm2 𝜏=𝐹 𝐴𝑆=90 000 N 345 .6 mm2=𝟐𝟔𝟎 .𝟒 𝐌𝐏𝐚 1.51 ∑𝑀𝐴=0=3600 (2)−𝐵𝑉(0.2) 𝐵𝑉=36 000 N 𝐵=𝐵𝑉 cos 20°=38 310 N 𝐴𝑆=2[𝜋(10)2 4]=157 mm2 𝜏=𝐵 𝐴𝑆=38 310 N 157 mm2=𝟐𝟒𝟒 𝐌𝐏𝐚 1.52 𝐴𝑆=(40)(12)=480 mm2 𝜏=𝐹 𝐴𝑆=88×103 N 480 mm2=𝟏𝟖𝟑 𝐌𝐏𝐚 1.53 𝐴𝑆=(40)(120 )=4800 mm2 𝜏=𝐹 𝐴𝑆=88.2×103 N 4800 mm2=𝟏𝟖.𝟒 𝐌𝐏𝐚 1.54 𝐴𝑆=𝜋𝑑𝑡 =𝜋(12)(8)=301 .6 mm2 𝜏=𝐹 𝐴𝑆=22.3×103 N 301 .6 mm2=𝟕𝟑.𝟗 𝐌𝐏𝐚 1.55 𝐴𝑆=2[𝜋(12)2/4]=226 .2 mm2 Two Rivets – Single Shear 𝜏=𝐹 𝐴𝑆=10.2×103 N 226 .2 mm2=𝟒𝟓.𝟏 𝐌𝐏𝐚 1.56 𝐴𝑆=4[𝜋(12)2/4]=452 .4 mm2 Two Rivets – Double Shear 𝜏=𝐹 𝐴𝑆=10.2×103 N 452 .4 mm2=𝟐𝟐.𝟓𝟓 𝐌𝐏𝐚 Structural Shapes 1-57 [Appendix A -4(b)] A = 4050 mm2, 6.075 ×104 mm3 with long side vertical 1-58 [Appe ndix A -5(c)] 20.719 x 2 = 41.438 N, 275 mm2 1-59 [Appendix A -6(d)] 16.51 x 3.8 = 62.74 N, A = 608 mm2 1-60 [Appendix A -7(e)] 471.7 x 2.55= 1203 N, A = 6261 mm2 1-61 [Appendix A -8(c)] 33.5 N/m, A = 444 mm2 1-62 [App. A -8(c)] A = 9344 mm2, 704 N/m 1-63 [App. A -9(b)] 73.03 mm, A = 1099 mm2 1-64 [App. A -9(b)] 219.1 mm, A = 5419 mm2 1-65 [Appendix A -9(d)] A = 1335 mm2, 100.6 N/m 1-66 [Appendix A -9(d)] 30 mm x 4 mm Chapter 2 Design Properties of Materials Only those problems requiring numerical data are shown. 2-14 𝑠𝑢=621 MPa ; 𝑠𝑦=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=1186 MPa @ OQT 700 [Appendix A -10] 2-18 𝐸=207 GPa for al l carbon and alloy steels. [Appendix A -10] 2-19 Mass =Density ×Volume =(7680kg m3)(0.03)(0.1)(0.35)m3=𝟖.𝟏 𝐤𝐠 [Appendix A -10] 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 ; 𝐸𝑇𝑖=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=276 MPa ; 𝑠𝑢𝑐=965 MPa [Appendix A -13] 2-31 Bending 𝜎𝑑=10.0 MPa ; Tension 𝜎𝑑=5.9 MPa ; Compression 6.9 MPa Parallel to grain, 2.65 MPa Perpendicular to grain; Shear 𝜏𝑑=0.66 MPa [Appendix A -15] 2-32 14 - 48 MPa [Section 2 -11] 𝑠𝑢 𝑠𝑦 𝐸 Density 6061 -0 124 MPa 55 MPa 69 𝐺𝑃𝑎 2770 km /m3 6061 -T4 241 MPa 145 MPa 69 𝐺𝑃𝑎 2770 km/ m3 6061 -T6 310 MPa 276 MPa 69 GPa 2770 km/ m3 2-44 Graphite fibers. 2-45 S-glass, quartz fibers, tungsten fibers coated with silicon carbide 2-51 Material Specific strength (m ) Ratio to SAE 1020 Graphite/Epoxy (High Strength) 1.24 ×105 25.0 Aramid/Epoxy Composite 1.02 ×105 20.6 Boron/Epoxy Composite 9.14 ×104 18.5 Graphite/Ep oxy (Ultra -high mod.) 7.01 ×104 14.2 Glass/Epoxy Composite 4.75 ×104 9.63 Titanium Ti -6Al-4V 2.54 ×104 5.15 SAE 5160 OQT 700 Steel 2.36 ×104 4.78 Aluminum 7075 -T6 2.09 ×104 4.23 Aluminum 6061 -T6 1.17 ×104 2.36 SAE 1020 HR Steel 4.94 ×103 1.00 2-52 Material Specific modulus (m ) Ratio to SAE 1020 Graphite/Epoxy (Ultra -high mod .) 2.10 ×107 7.81 Boron/Epoxy Composite 1.01 ×107 3.77 Graphite/Epoxy (High Strength) 8.76 ×106 3.25 Aramid/Epoxy Composite 5.59 ×106 2.07 SAE 1020 HR Steel 2.69 ×106 1.00 SAE 5160 OQT 700 Steel 2.69 ×106 1.00 Titanium Ti -6Al-4V 2.61 ×106 0.97 Aluminum 6061 -T6 2.59 ×106 0.96 Aluminum 7075 -T6 2.51 ×106 0.93 Glass/Epoxy Composite 1.68 ×106 0.62 2-60 Vm = 1.0 – Vf = 1.0 – 0.60 = 0.40 2-61 See Equation (2 -10) 2-62 See Equations (2 -11), (2 -12), (2 -13), (2 -14) See Table 2-15 for data. Use Equation (2 -10): See Table 2 -15 for data. Use Equation (2 -10): Use Equation (2 -10): See Table 2 -15 for data. Solutions to Problems 2 -66 to 2 -77: 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 Append ixes A -10 – A-14, matching su, sy, E, and % elongation. 2-66 a. s y = 503 MPa ; Offset method 2-67 a. s y = 1190 MPa ; Yield point b. s u = 572 MPa b. s u = 1290 MPa c. s p = 413 MPa c. s p = 1117 MPa d. s el = 462 MPa d. s el = 1158 MPa e. E = 69×109 Pa e. E = 200 ×109 Pa f. 11% elongation f. 15% elongation g. Ductile g. Ductile h. Aluminum h. Steel i. 7075 -T6 i. SAE 4140 OQT 900 2-68 a. s y = 427 MPa ; Offset method 2-69 a. s y = 338 MPa ; Yield point b. s u = 517 MPa b. s u = 448 MPa c. s p = 345 MPa c. s p = 317 MPa d. s el = 386 MPa d. s el = 331 MPa e. E = 115 ×109 Pa e. E = 183 ×109 Pa f. 15% elongation f. 36% elongation g. Ductile g. Ductile h. Copper alloy h. Steel i. C54400 Bronze -Hard i. SAE 1020 CD 2-70 a. No yield strength -Brittle 2-71 a. s y = 365 MPa ; Offset Method b. s u = 379 MPa b. s u = 407 MPa c. s p = 345 MPa c. s p = 214 MPa d. s el = 365 MPa d. s el = 290 MPa e. E = 140 ×109 Pa e. E = 82.7 ×109 Pa f. 0.5% elongation f. 5.0% elongation g. Brittle g. Borderline Ductile/Brittle h. Cast iron h. Zinc i. ASTM A48 Grade 60 i. Cast ZA -12 2-72 a. s y = 241 MPa ; Yield point 2-73 a. s y = 131 MPa ; Offset Method b. s u = 393 MPa b. s u = 276 MPa c. s p = 207 MPa c. s p = 97 MPa d. s el = 186 MPa d. s el = 117 MPa e. E = 190 ×109 Pa e. E = 41 ×109 Pa f. 21% elongation f. 5.0% elongation g. Ductile g. Borderline Ductile/Brittle h. Structural Steel h. Magnesium i. ASTM A36 i. ASTM AZ 63A -T6 2-74 a. s y = 1069 MPa ; Offset Method 2-75 a. s y = 276 MPa ; Offset Method b. s u = 1173 MPa b. s u = 310 MPa c. s p = 979 MPa c. s p = 207 MPa d. s el = 1027 MPa d. s el = 241 MPa e. E = 114 ×109 Pa e. E = 69 ×109 Pa f. 8% elongation f. 17% elongation g. Ductile g. Ductile h. Titanium h. Aluminum i. 6Al -4V i. 6061 -T6 2-76 a. s y = 552 MPa ; Offset Method 2-77 a. s y = 552 MPa ; Offset Method b. s u = 627 MPa b. s u = 655 MPa c. s p = 428 MPa c. s p = 379 MPa d. s el = 490 MPa d. s el = 469 MPa e. E = 180 ×109 Pa e. E = 179 ×109 Pa f. 15% elongation f. 2.0% elongation g. Ductile g. Brittle – Does not yield h. Stainless Steel h. Malleable Iron i. SAE 430 – Full hard i. ASTM A220 Grade 80002 Chapter 3 Design of Members Under Direct Stresses 3-1 𝜎=𝑃 𝐴=8.50×103 N π(10 mm )24⁄=108 MPa =𝜎𝑑=𝑠𝑦 2 Required 𝑠𝑦=2(𝜎𝑑)=2(108 MPa )=𝟐𝟏𝟔 𝐌𝐏𝐚 Aluminum 2014 -T4 has 𝑠𝑦=290 MPa (App. A -14) 3-2 𝜎=𝑃 𝐴=20.0×103 N (10)(30)mm2=66.7 MPa =𝜎𝑑=𝑠𝑢 8 Required 𝑠𝑢=8(𝜎𝑑)=8(66.7 MPa )=𝟓𝟑𝟑 𝐌𝐏𝐚 Plus good ductility SAE 1141 Annealed has 𝑠𝑢=600 MPa ;26% Elongation (A -10) 3-3 𝜎=𝑃 𝐴=7650 N (10 mm )2=76.5 MPa = =𝜎𝑑=𝑠𝑢 8 Required 𝑠𝑢=8(𝜎𝑑)=8(76.5)=𝟔𝟏𝟐 𝐌𝐏𝐚 Plus good ductility SAE 1040 Hot Rolled has 𝑠𝑢=621 MPa ;25% Elongation (A -10) 3-4 𝜎=𝑃 𝐴=8250 N π(9.5 mm )24⁄=116 .4 MPa =𝜎𝑑=0.60𝑠𝑦 (AISC) Required 𝑠𝑦=𝜎𝑑 0.60=1i6.4 MPa 0.60=𝟏𝟗𝟒 .𝟎 𝐌𝐏𝐚 ASTM A36 Structural steel has 𝑠𝑦=248 MPa (A-12) 3-5 𝜎=𝑃 𝐴=23 100 N 4050 mm2=𝟓.𝟕 𝐌𝐏𝐚 Too High; 𝜎𝑎𝑙𝑙𝑜𝑤 =4.0 MPa (A-15) [Area from Appendix A -4(b)] 3-6 a) No. 1 grade Douglas Fir has 𝜎𝑎𝑙𝑙𝑜𝑤 =7.2 MPa (A-15) b) To use No. 2 grade Southern Pine: 𝜎𝑎𝑙𝑙𝑜𝑤 =4.0 MPa Required area = 𝑃 𝜎=23 100 N 4.0 MPa=5775 mm2; [From Appendix A -4(b)] Use 2×8 or 90×90, or two 90×45 wood boards 3-7 𝜎=𝑃 𝐴;𝐴=𝑃 𝜎=28 500 N 82 MPa=348 mm2=𝜋𝐷2/4 Required 𝐷=√4𝐴/𝜋=√4(348 mm2)/𝜋=𝟐𝟏.𝟎𝟑 𝐦𝐦 Specif y 22 or 25 mm 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=𝐴𝑦=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=300 000 N π(200 mm )24⁄=9.6 MPa =1 4× Rated s trength (Sec. 2 -11) Requir ed rated s trength =4(9.6 MPa )=38.2 MPa Specify 41.4 MPa (6000 psi ) Rated s trength 3-11 Load on each block =131 000 N 3=43 700 N 𝜎=𝑃 𝐴=43 700 N 8100 mm2=5.4 MPa [Area from Appendix A -4(b)] If compression is perpendicular to grain, no suitable wood listed. If parallel to grain: No. 1 Southern Pine - 𝜎𝑎𝑙𝑙.=5.86 MPa (A-15) No. 2 Hemlock - 𝜎𝑎𝑙𝑙.=5.52 MPa No. 2 Douglas Fir - 𝜎𝑎𝑙𝑙.=6.9 MPa 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 𝑠𝑦=0.60(345 MPa )=207 MPa =𝑃 𝐴 [From AISC method] [From Appendix A -12] 𝑠𝑦=345 MPa if 𝐷<0.75 in Required 𝐴=𝑃 𝜎𝑑=17 800 N 207 N/mm2=86.0 mm2=𝜋𝐷2/4 Required 𝐷=√4𝐴/𝜋=√4(0.133 in2)/𝜋=10.5 mm Specify 𝑫=𝟏𝟎.𝟓 𝐦𝐦 or 𝟏𝟏 𝐦𝐦 (OK - <0.75 in) 3-16 𝐴=(80)(36)+2[1 2(36)(15)]=3420 mm2 𝜎=𝑃 𝐴=232 000 N 3420 mm2=𝟔𝟕.𝟖 𝐌𝐏𝐚 =𝜎𝑑=𝑠𝑢𝑐 𝑁 𝑁=𝑠𝑢𝑐 𝜎=552 MPa 67.8 MPa=𝟖.𝟏 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=55 MPa 8=6.9 MPa =𝑃 𝐴 [From Appendix A -16] Required 𝐴=𝑃 𝜎𝑑=500 N 6.9 MPa= 72.5 mm2=(5.0)(𝑤) Required 𝑤=𝐴 5 mm=72.5 mm2 5 mm=14.5 mm ; Specify 𝒘=𝟏𝟒.𝟓 𝐦𝐦 3-19 𝐴=(80)(40)−(60)(15)+𝜋(40)2 4=3557 mm2 𝜎=𝑃 𝐴=640 ×103 N 3557 mm2=180 MPa =𝜎𝑑 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 P roblem 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=475 mm2 [App .A−5(c)]; 𝜎𝑑=(0.60)𝑠𝑦=0.60(248 MPa )=148 .8 MPa 𝑃𝑎𝑙𝑙𝑜𝑤 =𝜎𝑑∙𝐴= (148 .8 MPa )(475 mm2)=𝟕𝟎 𝟔𝟖𝟎 𝐍 Elastic Deformation in Tension and Compression Members 3-23 𝜎𝑎=5.52 MPa (A−15); 𝐸=9.7×103N mm2; 𝐿=2 m 𝛿=𝜎𝐿 𝐸=5.52 MPa (2 m) 9.7×103 MPa=𝟏.𝟏 𝐦𝐦 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=64 mm ; 𝐷𝑖=64−2(2 mm )=60 mm ; 𝐸=73.1×109 Pa 𝐴=𝜋(𝐷𝑜2−𝐷𝑖2) 4=389 .6 mm2 𝛿=𝐹𝐿 𝐸𝐴;𝐹=𝛿𝐸𝐴 𝐿=(0.13 mm )(73.1×109 N/m2)(389 .6 mm2) 370 mm=𝟏𝟎 𝟎𝟎𝟓 𝐍 𝜎=𝐹 𝐴=10 005 N 389 .6 mm2=𝟐𝟓.𝟕 𝐌𝐏𝐚 (OK); 𝑠𝑦=290 MPa 3-26 1) Weigh it. Density of aluminum = 2770 kg/m3 Density of magnesium = 1830 kg/m3 2) Exert a known load and measure deformation. 𝐸 for aluminum = 10×106 Pa 𝐸 for magnesium = 69×109 Pa 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 stronge r 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 kgm3⁄ )×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 mm2; 𝜎=𝐹 𝐴=17 000 N 113 .1 mm2=150 MPa - Low for steel - OK 𝛿=𝐹𝐿 𝐸𝐴=(17.0×103 N)(220 mm ) (207 000 N/mm2)(113 .1 mm2) =𝟎.𝟏𝟔𝟎 𝐦𝐦 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 𝐦𝐦 3-30 Required 𝐴=𝐹 𝜎𝑑=156 000 N 149 MPa=1047 mm2: Use L 𝟏𝟎𝟎 ×𝟏𝟎𝟎 ×𝟏𝟎; 𝐴=1900 mm2 𝛿=𝐹𝐿 𝐸𝐴=(156 000 N)(4 m) (1900mm2 10002)(200 ×109 Pa)=𝟏.𝟔 𝐦𝐦 3-31 Elongation : 𝛿=𝐹𝐿 𝐸𝐴=(2000 N)(0.215 m) (207 ×109 Pa)(0.06)(0.003 )m2=𝟎.𝟎𝟎𝟎𝟏𝟐 𝐦=𝟎.𝟏𝟐 𝐦𝐦 Compression : 𝛿=𝐹𝐿 𝐸𝐴=(220 N)(0.215 m) (207 ×109)(.06)(.003 )=𝟎.𝟎𝟏𝟑 𝐦𝐦 3-32 Lower Section : 𝛿1=𝐹1𝐿1 𝐸𝐴=(22 200 )(.25) (207 ×109)(0.00032 )=0.084 mm Mid Section : 𝛿2=𝐹2𝐿2 𝐸𝐴=(31 100 )(.40) (207 ×109)(0.00032 )=0.187 mm Top Section : 𝛿3=𝐹3𝐿3 𝐸𝐴=(47 600 )(.65) (207 ×109)(0.00032 )=0.467 mm Total : 𝛿𝑇=𝛿1+𝛿2+𝛿3=𝟎.𝟕𝟑𝟖 𝐦𝐦 Max stress in top section – assume static load. 𝜎=𝐹 𝐴=47 600 N 0.00032 m2=149 MPa 𝜎𝑑=𝑠𝑦 2; 𝑠𝑦𝑚𝑖𝑛=2(149 ksi)=298 MPa OK for any steel in A ppendix A -10. 3-33 𝐴=𝜋(𝐷𝑜2−𝐷𝑖2) 4=𝜋(0.0322−0.0282) 4=188 .5 mm2 𝐹=𝛿𝐸𝐴 𝐿=(0.0013 m)(69×109 Pa)(188 .5 mm2) 1.0 m=16 908 N 𝜎=𝐹 𝐴=16 908 N 188 .5 mm2=𝟖𝟗.𝟖 𝐌𝐏𝐚 - Too High 𝜎𝑑=𝑠𝑢 8=310 MPa 8=39 MPa 3-34 𝐴=𝜋(𝐷𝑜2−𝐷𝑖2) 4=𝜋(192−142) 4= 130 mm2 𝜎=𝐹 𝐴=11 000 N 130 mm2=84.6 MPa = 𝜎𝑑=𝑠𝑦 2; Required 𝑠𝑦=169 MPa Aluminum 3003 -H18 has 𝑠𝑦=186 MPa 𝛿=𝐹𝐿 𝐸𝐴=(11 000 N)(2.65 m) (69×109 Pa)(130 mm2)=0.0032 m=𝟑.𝟐 𝐦𝐦 3-35 𝐴=𝜋 4(𝐷𝑜2−𝐷𝑖2)=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(10)2 4=78.5 mm2 𝜎=𝐹 𝐴=7100 N 78.5 mm2=𝟗𝟎.𝟒 𝐌𝐏𝐚 𝛿=𝐹𝐿 𝐸𝐴=(7100 N)(48 m) (207 ×109 N/m2)( 78.5 mm2)= 𝟐𝟏.𝟎 𝐦𝐦 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=𝛿𝐴𝐵+𝛿𝐵𝐶+𝛿𝐶𝐷=𝐹𝐴𝐵𝐿 𝐸𝐴+𝐹𝐵𝐶𝐿 𝐸𝐴+𝐹𝐶𝐷𝐿 𝐸𝐴 𝛿𝑇=(150 ×103 N)(250 mm ) (114 000 Nmm2 ⁄ )(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=(45 000 N)(250 mm ) (0.58 mm )(π(20 mm )24⁄)=𝟔𝟏.𝟖×𝟏𝟎𝟑𝐍 𝐦𝐦𝟐=𝟔𝟏.𝟖 𝐆𝐏𝐚 Probably aluminum : 𝐸=69×109 Pa 𝜎=𝐹 𝐴=45 000 N 314 mm2= 143 MPa [See App endix A-14] 𝜎<𝑠𝑦 for most tempers of most alloys. Ensure no yielding occurred. 3-39 𝐴𝐴𝐵=𝐴𝐵𝐶=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 𝐸𝐴𝐵𝐶+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=285 000 N 31 416 mm2=9.1 MPa ; Let 𝜎𝐷>4𝜎𝐶=36.4 MPa Use 𝑓𝐶′=41.4 MPa =6000 psi, Then 𝐸=32.4×109 Pa (Sec. 2 -11) 𝐴𝐶=𝜋(200 )2 4=31 416 mm2 ; 𝐴𝑇=1536 mm2 [App. A -8(c)] 𝐸𝑇=200 ×109 Pa Structural Steel 𝛿=𝛿𝐶+𝛿𝑇=(285 000 N)(1.0 m) (32.4×109 Pa)(31 416 mm2)+(285 000 )(2.6) (200 ×109)(1536 ) 𝛿=0.28 mm +2.41 mm =𝟐.𝟔𝟗 𝐦𝐦 𝜎𝑇=285 000 1536= 185 MPa <𝑠𝑦 for ASTM A36 structural steel. 3-41 a) 𝛿=𝐹𝐿 𝐸𝐴=(530 N)(3.2 M) (117 ×109 Pa)(2.0 mm2)=𝟕.𝟐 𝐦𝐦 𝐴=𝜋𝐷2 4=𝜋(1.6)2 4=2.0 mm2 𝜎=𝐹 𝐴=530 N 2.0 mm2=265 MPa ; Close to 𝑠𝑦=303 MPa b) =𝐹 𝐴=890 N 2.0 mm2=445 MPa ; This exceeds the ultimate strength of the wire. It should break. 3-42 𝜎=𝐹 𝐴=110 N (0.15)(19) mm2=𝟑𝟖.𝟔 𝐌𝐏𝐚 𝛿=𝐹𝐿 𝐸𝐴=(110 N)(8 m) (207 ×109 N/m2)(2.85 mm2)=𝟏.𝟒𝟗 𝐦𝐦 3-43 𝐹=𝜎𝑎𝑙𝑙𝑜𝑤 ×𝐴=(3.79 MPa )(8100 mm2)=𝟑𝟎 𝟔𝟗𝟗 𝐍 𝛿=−𝐹𝐿 𝐸𝐴=−(30 699 N)(3.5 m) (9 000 000 000 N/m2)(8100 mm2)=−𝟏.𝟒𝟕 𝐦𝐦 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 Thermal Deformation and Thermal Stress 3-47 𝛿=𝛼𝐿(∆𝑡)=(10.8×10−6 C−1)(25 m)(80 C)=𝟐𝟏.𝟔 𝐦𝐦 3-48 𝛿=𝛼𝐿(∆𝑡)=(11.3×10−6 ℃−1)(12 000 mm )(77 ℃)=𝟏𝟎.𝟒 𝐦𝐦 3-49 𝜎=𝐸𝛼(∆𝑡)=(207 ×109 Pa)(11.3×10−6 ℃−1)(77 ℃)=𝟏𝟖𝟎 𝐌𝐏𝐚 (High) 3-50 𝛿=𝛼𝐿(∆𝑡)=(11.3×10−6 ℃−1)(625 mm )(156 ℃)=𝟏.𝟏𝟎 𝐦𝐦 3-51 a) 𝛿=𝛼𝐿(∆𝑡)=(11.3×10−6 ℃−1)(625 mm )(65 ℃)=𝟎.𝟒𝟓𝟗 𝐦𝐦 b) 𝜎=𝐸𝛼(∆𝑡)=(207 ×109 Pa)(11.3×10−6 ℃−1)(91 ℃)=𝟐𝟏𝟑 𝐌𝐏𝐚 3-52 𝛿=𝛼𝐿(∆𝑡)=(10.8×10−6 C−1)(42 m)(43 C)=0.0195 m=𝟏𝟗.𝟓 𝐦𝐦 This is the total width. Each end could be 9.7 mm . 3-53 Deck could expand by a total of 12.8 mm without stress. Require d ∆𝑡=𝛿 𝛼𝐿=12.8 mm /1000 (10.8×10−6 C−1)(42 m)=28.2 C 𝑡2=𝑡1+∆𝑡=0+28.2=28.2 ˚C Remaining ∆𝑡=43 ˚C−28.2 ˚C=14.8 ˚C 𝜎=𝐸𝛼(∆𝑡)=(26.2×109 Pa)(10.8×10−6 ˚C−1)(14.8 ˚C)=𝟒.𝟐 𝐌𝐏𝐚 3-54 ∆𝑡=45 ˚C−15 ˚C=30˚C 𝛿=𝛼𝐿(∆𝑡)=(10.8×10−6 ˚C−1)(42 m)(30 ˚C)=𝟎.𝟎𝟏𝟑𝟔 𝐦=𝟏𝟑.𝟔 𝐦𝐦 3-55 𝛿=𝜋(55.300 )−𝜋(55.100 )=0.2𝜋 mm (Change in circumference) ∆𝑡=𝛿 𝛼𝐿=0.2𝜋 mm (16.9×10−6 ℃−1)(𝜋)(55.100 mm )=214 .8 ℃ 𝑡2=𝑡1+∆𝑡=20+214 .8=𝟐𝟑𝟒 .𝟖 ℃ 3-56 For first part of cooling, ring is unrestrained until its diameter gets to 55.200 mm ∙δ=55.300 −55.200 =0.100 mm ∆𝑡=𝛿 𝛼𝐿=0.100 mm (16.9×10−6 ℃−1)(55.200 mm )=107 .2 ℃ 𝑡2=𝑡1−∆𝑡=234 .8 ℃−107 .2 ℃=127 .6 ℃ Additional ∆𝑡=127 .6−20=107 .6 ℃ 𝜎=𝐸𝛼(∆𝑡)=(193 ×109 Pa)(16.9×10−6 ℃−1)(107 .6 ℃)=𝟑𝟓𝟏 𝐌𝐏𝐚 3-57 𝛿Brass =𝛼𝐿(∆𝑡)=(20.5×10−6 ℃−1)(4200 mm )(75 ℃)=𝟔.𝟒𝟔 𝐦𝐦 𝛿SS=𝛼𝐿(∆𝑡)=(10.4×10−6 ℃−1)(4500 mm )(75 ℃)=𝟑.𝟓𝟏 𝐦𝐦 3-58 𝛿=𝛼𝐿(∆𝑡)=(11.7×10−6 ˚C−1)(12 m)(105 ˚C)=0.0147 m=𝟏𝟒.𝟕 𝐦𝐦 3-59 Initial expansion of 0.10 mm is unrestrained. Require d ∆𝑡=𝛿 𝛼𝐿=0.10 mm (25.2×10−6 ℃−1)(250 mm )=15.9 ℃ 𝑡2=𝑡1+∆𝑡=20 ℃+15.9 ℃=35.9 ℃ Add itional ∆𝑡=70 ℃−35.9 ℃=34.1 ℃ while bar is r estrained 𝜎=𝐸𝛼(∆𝑡)=(45×109 Pa)(25.2×10−6 ℃−1)(34.1 ℃)=𝟑𝟖.𝟕 𝐌𝐏𝐚 3-60 𝜎=𝐸𝛼(∆𝑡)=(207 ×109 Pa)(11.3×10−6 ℃−1)(90 ℃)=𝟐𝟏𝟏 𝐌𝐏𝐚 3-61 𝛿1=266 .83−266 .70=0.13 mm Unrestrained ∆𝑡1=𝛿 𝛼𝐿=0.13 mm (23.4×10−6 ˚C−1)(266 .83 mm )=20.8 ˚C 𝑡2=𝑡1+∆𝑡1=24+20.8=44.8 ˚C Additional ∆𝑡=200 −44.8=155 .2 ˚C while r estrained 𝜎=𝐸𝛼(∆𝑡)=(69×109 Pa)(23.4×10−6 ˚C−1)(155 .2 ˚C)=𝟐𝟓𝟏 𝐌𝐏𝐚 For 6061 -T4, 𝑠𝑢=241 MPa ; Bar should fail. Also , because bar is in compression, it may buckle. 3-62 𝛿𝑝=𝛼𝐿(∆𝑡)=(95.4×10−6 ˚C−1)(760 mm )(100 −18)˚C=𝟓.𝟗𝟓 𝐦𝐦 𝛿𝑇𝑖=𝛼𝐿(∆𝑡)=(9.5×10−6 ˚C−1)(760 mm )(82 ˚C)=𝟎.𝟓𝟗 𝐦𝐦 𝛿𝑝−𝛿𝑇𝑖=5.36 mm tan𝜃=5.36 610 =0.00876 𝜽=𝟎.𝟓𝟎𝟑 𝐝𝐞𝐠𝐫𝐞𝐞𝐬 3-63 𝛿=𝛼𝐿(∆𝑡)=(11.3 x 10−6 ˚C−1)(10 m)(22−(−26))˚C=𝟓.𝟒 𝐦𝐦 3-64 Total 𝛿𝑇=0.50 mm = 0.0005 m= 𝛿𝑆+ 𝛿𝐵 610 mm 5.36 mm 𝛿𝑇= 𝛼𝑆𝐿𝑆(∆𝑡)+ 𝛼𝐵𝐿𝐵(∆𝑡)= ∆𝑡 (𝛼𝑆𝐿𝑆+ 𝛼𝐵𝐿𝐵) ∆𝑡=𝛿𝑇 𝛼𝑆𝐿𝑆+𝛼𝐵𝐿𝐵=0.0005 m (10.4×10−6 ℃−1)(2.800 m)+(20.5 x 106℃−1)(1.900 m) ∆𝑡=7.35 ℃ ;𝑡2=𝑡1+ ∆𝑡=20+7.35=𝟐𝟕.𝟑𝟓 ℃ 3-65 Added 𝜎=𝐸𝛼∆𝑡=(193 ×109Pa)(16.9×10−6 ℃−1)(35 ℃)=114 MPa Final 𝜎=40+114 =𝟏𝟓𝟒 𝐌𝐏𝐚 3-66 When heated, wire would relax. ∆𝑡=𝜎 𝐸𝛼=40 x 106Pa (193 x 109Pa)(16.9 x 10−6 ℃−1)=12.3 ℃ 𝑡2=𝑡1+∆𝑡=20 ℃+12.3 ℃=𝟑𝟐.𝟑 ℃ Members Made from Two Materials 3-67 𝐴𝑆=1502−1302=5600 mm2;𝐴𝐿=1302=16900 mm2 𝜎𝐶=𝐹𝐸𝐶 𝐴𝑆𝐸𝑆+𝐴𝐶𝐴𝐶=(900 X 103 N)(32.4 GPa ) (5600 mm2)(207 GPa )+(16900 mm2)(32.4 GPa )=𝟏𝟕.𝟏 𝐌𝐏𝐚 𝜎𝑆=𝜎𝐶𝐸𝐶 𝐸𝐶=(17.1 MPa )(207 GPa ) 32.4 GPa=𝟏𝟎𝟗 𝐌𝐏𝐚 3-68 𝐴𝑆=2900 mm2(App A−8(c) );𝐴𝐶=(140 mm )2=19 600 mm2 𝐸𝑆=207 x 109 Pa ;𝐸𝐶=32.4 x 109 Pa (Section 2−11) Let 𝜎𝐶=𝜎𝑑𝐶=10 MPa ;Then 𝜎𝑆<𝜎𝑑𝑆=149 MPa 𝐹=𝜎𝑑𝐶(𝐴𝑆𝐸𝑆+𝐴𝐶𝐸𝐶) 𝐸𝐶=10 x 106 [(2900 )(207 x 109)+19 600 (32.4 x 109)] 32.4 x 109 𝑭=𝟔𝟎𝟎 𝟓𝟎𝟎 𝐍 ;Check 𝜎𝑆= 𝜎𝐶𝐸𝑆 𝐸𝐶=10 MPa (207 32.4)=𝟔𝟑.𝟗 𝐌𝐏𝐚 −OK 3-69 Let 𝜎𝐶=10 MPa ; 𝜎𝑆= 𝜎𝐶𝐸𝑆 𝐸𝐶=10200 32.4=61.7 MPa −OK 𝐹 𝜎𝐶=(𝐴𝑆𝐸𝑆+𝐴𝐶𝐸𝐶) 𝐸𝐶=𝐴𝑆𝐸𝑆 𝐸𝐶+𝐴𝐶=𝐴𝑆(200 32.4)+𝐴𝐶=6.17𝐴𝑆+𝐴𝐶 𝐴𝑆=𝑏2−(𝑏−2𝑡)2=𝑏2−[(𝑏−2(13)]2=𝑏2−(𝑏−26)2=𝑏2−(𝑏2−52𝑏+676 = 52𝑏+676 𝐴𝐶=(𝑏−26)2=𝑏2−52𝑏+676 𝐹 𝜎𝐶=6.17(52𝑏+676 )+𝑏2−2𝑏+1=𝑏2+10.77𝑏−5.38 𝐹 𝜎𝐶=22 000 10 000 000=220 = 𝑏2+10.77𝑏−5.38 𝑏2+10.77𝑏−227 .8=0 By quadratic e quation , 𝒃=𝟑𝟒𝟖 𝐦𝐦 3-70 𝐴𝑆=𝐴𝐴=2𝜋(6 mm )2/4=56.55 mm2 𝜎𝐴=𝑃𝐸𝐴 𝐴𝑆𝐸𝑆+𝐴𝐴𝐸𝐴=(11.3 x 103N)(69 GPa ) (50.55 mm2)(207 GPa )+(56.55 mm2)(69 𝐺𝑃𝑎 )=49.5 MPa 𝜎𝑆=𝜎𝐴𝐸𝑆 𝐸𝐴=(49.5 MPa )(207 GPa ) 69 GPa=𝟏𝟒𝟖 .𝟓 𝐌𝐏𝐚 3-71 𝑊=𝑚𝑔=2265 kg  9.81 m s2=22.22 x 103N:AS=2𝐴𝐶 If 𝜎𝑆=552 MPa 2=276 MPa ;then σC=𝜎𝑆𝐸𝐶 𝐸𝑆=276 MPa (131 200)=181 MPa 𝜎𝐶 𝐴𝐿𝐿𝑂𝑊 =𝑠𝑦 2=1000 2=500 MPa − OK 𝜎𝐶=𝑃𝐸𝐶 𝐴𝑆𝐸𝑆+𝐴𝐶𝐸𝐶=𝑃𝐸𝐶 2𝐴𝑆𝐸𝑆+𝐴𝐶𝐸𝐶=𝑃𝐸𝐶 𝐴𝑐(2𝐸𝑆+𝐸𝐶) 𝐴𝐶=𝑃𝐸𝐶 𝜎[2𝐸𝑆+𝐸𝐶]=(22.22 x 103N)(131 GPa ) (181N mm2)[2(200 )+131 ]GPa=30.3 mm2 𝐷=√4𝐴 𝜋=√4(30.3) 𝜋=𝟔.𝟐𝟎 𝐦𝐦 =𝐖𝐢𝐫𝐞 𝐃𝐢𝐚𝐦𝐞𝐭𝐞𝐫 3-72 1) Compute force required to deflect aluminum r od 0.12 mm . 𝑃1=𝛿𝐸𝐴 𝐿=(0.12 mm )(69 000N mm2)(15 394 mm2) 1650 .12 mm=𝟕𝟕.𝟐𝟒 𝐤𝐍 2) Additional load available: 𝑃2=350 kN−77.24 kN=𝟐𝟕𝟐 .𝟕𝟔 𝐤𝐍 3) Both members deflect the same amount under 𝑃2 𝛿𝐴=𝛿𝑆;𝑃𝐴𝐿𝐴 𝐸𝐴𝐴𝐴=𝑃𝑆𝐿𝑆 𝐸𝑆𝐴𝑆 [𝐿𝐴=𝐿𝑆] 𝑃𝐴=𝑃𝑆(𝐸𝐴𝐴𝐴 𝐸𝑆𝐴𝑆)=𝑃𝑆(69 GPa )(15 394 mm2) (207 GPa )(3601 mm2)=𝟏.𝟒𝟐𝟓 𝑷𝑺 DN 150 steel pipe: 𝐴𝑆=3601 mm2 [App endix A-9(b) ] 4) 𝑃𝑆+𝑃𝐴=𝑃2=272 .76 kN 𝑃𝑆+1.425 𝑃𝑆=2.425 𝑃𝑆=272 .76 kN 𝑷𝑺=𝟏𝟏𝟐 .𝟒𝟖 𝐤𝐍 𝑃𝐴=272 .76−112 .48=160 .28 kN 5) Total load on alum inum : 𝑃𝐴𝑇=𝑃1+𝑃𝐴=77.24+160 .28=𝟐𝟑𝟕 .𝟓𝟐 𝐤𝐍 6) Stresses: 𝜎𝑆=𝑃𝑆 𝐴𝑆=112 .48 x 103N 3601 mm2=𝟑𝟏.𝟐𝟒 𝐌𝐏𝐚 𝜎𝐴=𝑃𝐴𝑇 𝐴𝐴=237 .52 x 103N 15 394 mm2=𝟏𝟓.𝟒𝟑 𝐌𝐏𝐚 3-73 1) The nut moves 1.25 mm in one turn 2) The force created causes the tube to shorten and t he bolts to get longer.