ch03 - SOLUTION(3.213 Known We are to determine the modulus...

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Unformatted text preview: SOLUTION (3.213) Known: We are to determine the modulus of elasticity, ultimate tensile strength, elongation at break, and density for certain carbon and alloy steels. Find: Search the materials property database at http'.£fiufl}x,matweb.c0m and list the (1) modulus of elasticity, E, (2) ultimate tensile strength, S". (3) elongation at break in %, and (4) density in gI‘cc, for the following: (a) teels: [010 cold drawn, 1020 cold rolled, 1040 as roiled, 1050 as rolled, 1080 as rolled, and 1116 cold drawn; and (b) Alloy steels: 4140 annealed, 4340 annealed, and 4620 annealed. Analysis: (a) From www.matweb.com we find for W: AISI 1010, cold drawn Density 7.87 gfcc Tensile Strength, Ultimate 365 MPa 52900 psi Modulus of Elasticity 205 GPa 29700 ksi Elongation at Break 20 % AISI 1020 Steel, cold rolled Density 7.87 g/cc Tensile Strength, Ultimate 420 MPa 60900 psi Modulus of Elasticity 205 GPa 29700 ksi Elongation at Break 15 % AISI 1040 Steel, as rolled Density 7.845 gz’cc Tensile Strength, Ultimate 620 MP3 89900 psi Modulus of Elasticity 200 GPa 29000 ksi Elongation at Break 25 % A151 1050 Steel, as rolled Density 7.85 gfcc Tensile Strength, Ultimate 725 MPa 105000 psi Modulus of Elasticity 205 GPa 29700 ksi Elongation at Break 20 % A151 1080 Steel, as rolled Density 7.85 g/cc Tensile Strength, Ultimate 965 MPa 140000 psi Modulus of Elasticity 205 GPa 29700 ksi Elongation at Break 12 % AISI 1116 Steel, cold drawn Density 7.85 glee Tensile Strength, Ultimate 530 MP3 76900 psi - Modulus of Elasticity 205 GPa 29700 ksi Elongation at Break 17 % (b) From www.matweb.eom we find for Afloyggeels: AISI 4140 Steel, annealed at 815°C (1500°F) furnace cooled 11°C (20°F)lhour to 665°C (1230°F), air cooled, 25 mm (1 in.) round Density 7.85 glee Tensile Strength. Ultimate 655 MPa 95000 psi Elongation at Break 25.7 % Modulus of Elasticity 205 GPa 297’00 ksi AISI 4340 Steel, annealed, 25 mm round Density 7.85 glee Tensile Strength, Ultimate 745 Mle 108000 psi Elongation at Break 22 % Modulus of Elasticity 205 GPa 29700 ksi AISI 4620 Steel, annealed, 25 mm round Density 7.85 glee Tensile Strength, Ultimate 510 MPa 74000 psi Elongation at Break 31.3 % Modulus of Elasticity 205 GPa 29ml) ksi Comment: Matweb is an award winning web site. SOLUTION (3.3D) Known: We are to determine the modulus of elasticity, ultimate tensile strength. elongation at break, and density for certain carbon and alloy steels. Find: Search the materials property database at http',lfwww.matweb.com and list the (1) modulus of elasticity, E. (2) ultimate tensile strength, S", (3) elongation at break in %, and (4) density in gfcc, for the following: (a) Cast iron: ASTM class 20 and class 35; (0} W: 3003-H12, 3003-H18, 5052—H32, 5052-1138, 5052-0, 6061— T4, 6061-T9l. and 7075—0. Analysis: (a) From www.matwebaeom we find for Castings Standard gray iron test bars, as cast, ASTM class 20 Density 7.15 gfcc Tensile Strength, Ultimate 152 MPa 22000 psi Modulus of Elasticity 66 - 97 GPa 9570 - 14100 ksi Standard gray iron test bars, as cast, ASTM class 35 Density 7.15 gfcc Tensile Strength, Ultimate 252 MPa 36500 psi Modulus of Elasticity 100 - 119 GPa 14500 — 17300 ksi (b) From www.matwebxom we find for Aluminum alloys: Aluminum 3003-H12 Density 2.73 gfcc Ultimate'l'ensile Strength 131 MPa 19000 psi Elongation at Break 10 % AA; Typical; 1(16 in. (1.6 mm) Thickness Elongation at Break 20 % AA; Typical; 112 in. (12.7 mm) Diameter Modulus of Elasticity 68.9 GPa 10000 ksi Aluminum 3003-1118 Density 2.?3 glcc Ultimate Tensile Strength 200 MPa 29000 psi Elongation at Break 10 % AA; Typical; 1K2 in. (12.7 mm) Diameter Elongation at Break 4 % AA; Typical; 1/16 in. (1.6 mm) Thickness Modulus of Elasticity 68.9 GPa Aluminum 5052-H32 Density 2.68 g/cc Ultimate Tensile Strength 228 MP3 33000 psi Elongation at Break 12 % AA;Typical; 1! 16 in. (1.6 mm) Thickness Elongation at Break 18 % AA; Typical; 1f2 in. (12.7 mm) Diameter Modulus of Elasticity 70.3 GPa Aluminum 5052-H38 Density 2.68 glcc Ultimate Tensile Strength 290 MP3 42000 psi Elongation at Break 7 % AA; Typical; 1116 in. (1.6 mm) Thickness Elongation at Break 8 % AA; Typical; 1/2 in. (12.7 mm) Diameter Modulus of Elasticity 70.3 GPa Aluminum 5052-0 Density 2.68 glcc Ultimate Tensile Strength 193 MP3 28000 psi Elongation at Break 25 % AA; Typical; 1:16 in. (1.6 mm) Thickness Elongation at Break 30 % AA; Typical; 1l2 in. (12.7 mm) Diameter Modulus of Elasticity 70.3 GPa Aluminum 6061-T4; 6061-T451 Density 2.? glee Ultimate Tensile Stren th 24] MP3 35000 psi Elongation at Break 2 % AA; Typical; 1:16 in. (1.6 mm) Thickness Elongation at Break 25 % AA; Typical; 1:2 in. (12.7 mm) Diameter Modulus of Elasticity 68.9 GPa Aluminum 6061-T91 Density 2.7 gi’cc Tensile Strength, Ultimate 405 MPa 58700 psi Elongation at Break 12 % Modulus of Elasticity 69 GPa Aluminum 7075-0 Density 2.31 g/cc Ultimate Tensile Strength 228 MPa 33000 psi Elongation at Break 16 % AA; Typical; 1:2 in. (12.7 mm) Diameter Elongation at Break 17 % AA; Typical; 1fl6 in. (1.6 mm) Thickness Modulus of Elasticity 71.7 GPa Comment: Matweb is an award winning web site. SOLUTION (3.?) Known: The critical location of a part made from known steel is cold worked during fabrication. Find: Estimate Su. Sy and the ductility. Schematic and Given Data: Assumption: After cold working the stress-strain curve for the critical location starts at p01 nt G. Analysis: 1. At point G in Fig. 3.2, the part has been permanently stretched to 1.1 times its initial length. Hence, its area is U 1.1 times its original area An. 0n the basis of the new area, the yield strength is Sy = 620.1) = 63.2 ksi. I The ultimate strength is Su = 660. I} = 72.6 ksi. I At fracture, R increases to 2.5 On the graph. R =2.5I1.1= 2.27 Using Eq. (3.3) and Eq. (3.2) PifierP — __.1_= _——1 = Ar‘l R l 2.27 0'56 I s=R—l=2.27-1=1.27or127% I SOLUTION {3.9) Known: A tensile specimen of a known material is loaded to the ultimate stress, then unloaded and reloaded to the ultimate stress point. Find: Estimate the values of o, 8, OT. ET for the first loading and the reloading. Schematic and Given Data: Hot Rolled 1020 Steel Assumption: After unloading the stress-strain curve starts at point H for the new specimen. Analysis: 1. For the initial sample, a = 66 ksi, E = 30%. I 2. For Figure 3.2, R = 1.3 at point H. 3. From Eq. (3.4), GT = UR = (66)(1_3) = 85.8 ksi. I 4. From Eq. (3.5), 81-2an + e) = in 1.30 = 0.26 = 26%. I S. For the new specimen; 0 = 660.3) = 85.8 ksi. I 6. The new specimen behaves elastically, so 5 = GEE = 85.880300 = .00286. I 7. Within the elastic range, or - o and flu 6. Therefore CT = 85.8 ksi and s1— = 0.29%. I Comment: Note also that 81‘ = ln(1 + e) = ln(l .0029) = 0.29%. SOLUTION (3.130) Known: A steel is to be selected from Appendix (3-43. Find: Estimate Su and S); from the given value of Brinell hardness for the steel selected. Schematic and Given Data: Decision: Select ANSI 1020 annealed. Assumptions: 1. The experimentally determined relationship of ultimate strength to hardness is sufficiently accurate. 2. The experimentally developed relationship of yield strength to ultimate strength is sufficiently accurate for our purposes. Analysis: 1. Appendix C—4a shows that ANSI 1020 annealed steel has Su a 52.3, S}; = 42.8, and Bhn =111. 2. Using Eq. (3.11), we can estimate Su from the Brinell hardness using: Su = KBHB where KB as 500 for most steels. Su = 500011) = 55,500 psi I 3. S): can be estimated by using Eq. (3.12). 55: = 1.05 Su - 30,000 psi = 1.0565500) - 30,000 = 28,275 psi. I Comments: 1. Equation (3.12;) is a good estimate of the tensile yield strength of Stresswrelieved (not cold-worked) steels. Note that the estimated value of Su from the Brinell hardness of 55.5 ksi is close to the value given in Appendix C—4a of Su 2 57.3 ksi. 2. Experimental data would be helpful to refine the above equations for Specific steels. SOLUTION (3.15D) Known: A steel is to be selected from Appendix C-4a. Find: Estimate Su and S), from the given values of Brinell hardness for the selected Steel. Schematic and Given Data: AISI 1030 Steel as—rolled normalized annealed Decision: Select ANSI 1030 steel. Assumptions: 1. The experimentally determined relationship of ultimate strength to hardness is sufficiently accurate. 2. The experimentally developed relationship of yield strength to ultimate strength is sufficiently accurate for our purposes. Analysis: 1. Appendix C-4a shows that for ANSI 1030 the strengths are as follows: (1) as— rolled Su = 80.0 psi, 33; = 50.0 psi, Bhn = 179; (2) normalized Su 2 75.5 psi, 85; = 32.0 psi, Bhn = 149; (3) annealed Sll = 67.3 psi, S), = 31.2 psi, Bhn = 126. 2. Using Eq. (3.11), we can estimate Su. Su = KBHB where KB =5 500 for most steels. 3. Sy can be estimated by using Eq. (3.12]. 4. In as-rolled condition: Su : 500(179) = 89,500 psi Sy = 1.05 Su - 30,000 psi = l.05(89,500) - 30,000 = 63,975 psi. 5. In normalized condition: Sn = 500(149) = 74,500 psi. S3,, = 1.058” - 30,000 psi = 48,225 psi. 6. In annealed condition: Su = 500(126) = 63,000 psi. 8., = 1.058u — 30,000 psi = 36.150 psi. 7. Ratio of strength (Appendix C—4a values to Brinell hardness based values): i- 80.0 _ 894 YB ' 31-2 =.862 Su; 63.0 ‘sy 36.2 B i—filfi= 1.068 Comments: 1. Equation (3.12) is a good estimate of the tensile yield strength of stress-relieved (not cold-worked) steels. 2. Experimental data would be necessary to refine the above equations. ...
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