Solutions_for_HW2

Solutions_for_HW2 - Chapter 2 2-1 From Tables A-20 A-21...

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Unformatted text preview: Chapter 2 2-1 From Tables A-20, A-21, A—22, and A-24c, (a) UNS G10200 HR: Sm = 380 (55) MPa (kpsi), Syt = 210 (30) Mpa (kpsi) Ans. (b) SAE 1050 CD: SW = 690 (100) MPa (kpsi), Syt = 580 (84) Mpa (kpsi) Ans. (c) A181 1141 Q&T at 540°C (1000°F): Sm = 896 (130) MPa (kpsi), Sy, = 765 (111) Mpa (kpsi) Ans. (d) 2024-T4: Sm = 446 (64.8) MPa (kpsi), Sy, = 296 (43.0) Mpa (kpsi) Ans. (e) Ti-6Al—4V annealed: SW = 900 (130) MPa (kpsi), Syt = 830 (120) Mpa (kpsi) Ans. 2-2 (a) Maximize yield strength: Q&T at 425°C (800°F) Ans. (b)MaXimize elongation: Q&T at 650°C (1200°F) Ans. 2-3 1 Conversion of kN/m3 to kg/ m3 multiply by 1(103) / 9.81 = 102 A181 1018 CD steel: Tables A—20 and A—5 s 370(103) —y = ——~————— : 47.4 kN -m/kg Ans. ,0 76.5(102) 2011—T6 aluminum: Tables A—22 and A—5 s, 169(103) —)=——=62.3 ani/kg Ans. ,0 26.6(102) . Ti—6Al-4V titanium: Tables A—24c and A-5 S 830 103 —y:—()—=187 kN-m/kg Ans. ,0 43.4(102) ASTM No. 40 cast iron: Tables A-24a and A-5.Does not have a yield strength. Using the ultimate strength in tension 42.5 6.89 103 &=—(—)()=40.7 kN-m/kg Ans ,0 70.6(102) 2-4 A181 1018 CD steel: Table A—5 E 30000“) y _ 0.282 2011-T6 aluminum: Table A-5 10.4 106 52—h )=106(106) in Ans. 7 0.098 Ti—6Al-6V titanium: Table A—5 =106(106) in Ans. Chapter 2 — Rev. D, Page 1/19 2-5 V 2-6 6 Ezmflomofi) in Ans. 9/ 0.160 No. 40 cast iron: Table A-5 E 14.5(106) __————=55.8(106) in Ans. 9/ 0.260 2G(l+v)=E :> v: E“2G - 2G From Table A—5 30.0—2 11.5 Steel: v = ————(——l = 0.304 Ans. 2(11.5) 10.4 — 2 3.90 Aluminum: v = ~——-—(——) = 0.333 Ans. . 2(3.90) 18.0 — 2 7.0 Beryllium copper: v = ~—————(———)— 2 0.286 Ans. 2(70) , 14.5~2(6.0) Gray cast iron: v = _ = 0.208 Ans. A 2(6.0) (20.40 = 7r(0.503)2/4, a =P,- /AO For data in elastic range, 6 = A l / [0 = A Z / 2 For data in plastic range, 5 29—1-2140 —L-l =fi‘l—l lo 10 _ 10 A On the next two pages, the data and plots are presented. Figure (a) shows the linear part of the curve from data points 1—7. Figure (1)) shows data points l—lZ. Figure (c) shows the complete range. Note: The exact value of A0 is used Without rounding off. (b) From Fig. (a) the slope of the line from a linear regression is E = 30.5 Mpsiv Ans. From Fig. ([9) the equation for the dotted offset line is found to be 0 = 30.5(10‘55 — 61000 (1) The equation for the line between data points 8 and 9 is a = 7.60005); + 42 900 (2) Chapter 2 — Rev. D, Page 2/19 Solving Eqs. (1) and (2) simultaneously yields a- = 45.6 kpsi which is the 0.2 percent offset yield strength. Thus, Sy = 45.6 kpsi Ans. ' The ultimate strength from Figure (0) is St, = 85.6 kpsi Ans. The reduction in area is given by Eq. (2-12) is -A _ R: A0 f (100)=W(100):45.8 % Ans. A0 0.1987 Data Point 2000 0.0006 0.00030 10065 0.001 0.00050 4000 0.00115 0.0028 0.00140 0.00180 0.00445 0.00158 0.00461 0.01229 0.03281 0.05980 15 17000 0.1563 0.27136 16 16400 0.1307 0.52037 14800 0.1077 0.84506 74479 .0300 y, 3.05:,0075 a, “I 1.0012101 $1 1 E 20300 » +5131 5351 on f 10300 —Linear 15011031} 0 , a 0.000 0.001 0.001 0.002 Strain (a) Linear range Chapter 2 - Rev. D, Page 3/19 sum ‘3 asm 35ml x Stress (psi: 5 § 0.001 0.006 0.003 0.0.1.0 0.012 0.014 Straln (b) Offset yield Stress {955) 0.? 0.8 0.9 0.3 0.4 0.5 0.5 0.0 0.1 0.2 Strain (0) Complete range (c) The material is ductile since there is a large amount of deformation beyond yield. ((1) The closest material to the values of Sy, Sm, and R is SAE 1045 HR with Sy = 45 kpsi, SW = 82 kpsi, and R = 40 %. Ans. 2-7 To plot 0mm V8.8, the following equations are applied to the data. P O-true : 2 Eq. (2—4) Chapter 2 - Rev. D, Page 4/19 Eq.(2-l4), 1404;: 1 =1.25 A,. 1~W 1—0.20 Eq. (2-17), 6,. =1nfi10— = 1n1.25 = 0.223 <8“ A. Eq. (2—18), 5),, = 008;" =110(0.223)0’Z4 = 76.7 kpsi Ans. 174% increase Ans. Eq. (2—19), Su' 2 4g— = 61's = 76.9 kpsi Ans. 25% increase Ans. l— W l— 0.20 (b) Before: i = = 2.20 After: ~51"— : = 1.00 Ans. Sy 28 gy' 76.7 Lost most of its ductility 2—11 W: 0.20, (a) Before cold working: 2024-T4 aluminum alloy. Table A-22, Sy = 43.0 kpsi, St, = 64.8 kpsi, 0'0 = 100 kpsi, m = 0.15, «.7 = 0.18 After cold working: Eq. (2-16), Eu = m = 0.15 Eq. (2-14), £92;: 1 =1.25 Al. l—W 1—0.20 Eq. (2-17), 5,. :1n? : 1n1.25 = 0.223 >5f Material fractures. Ans. 2-12 For H3 = 275, Eq. (2—21), S” = 3.4(275) = 935 MPa Ans. 2-13 Gray cast iron, H B = 200. Eq. (2-22), S1, = 023(200) — 12.5 = 33.5 kpsi Ans. From Table A-24, this is probably ASTM No. 30 Gray cast iron Ans. 2-14 Eq. (2—21), 0.5HB = 100 :> H3 = 200 Ans. Chapter 2 - Rev. D, Page 8/19 ...
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Solutions_for_HW2 - Chapter 2 2-1 From Tables A-20 A-21...

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