ch03 - SOLUTION (3.1D) Known: Definitions of the terms...

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Unformatted text preview: SOLUTION (3.1D) Known: Definitions of the terms stress strength, yield strength, ultimate strength, elastic limit, proportional limit, modulus of elasticity, and yield point are given in Section 3.2. Find: Write definitions of the above terms -- see Section 3.2. Analysis: 1. 2. 3. The stress is the load divided by the cross—sectional area. The strenth is the maximum value of stress a material will carry before failure. The yield strength, Sy, is the value of stress at which significant plastic yielding first occurs. The ultimate strength is the maximum value of stress a material will carry before fracture for nondynamic loading. The elastic limit is the highest stress the material can withstand and still return exactly to its original length when unloaded. The proportional limit is the stress at which the stress—strain curve first deviates (ever so slightly) from a straight line. Below the proportional limit, Hooke's law applies. The modulus of elasticity (Young's modulus) E, is the constant of proportionality between stress and strain (which is the slope of the curve between the origin and the proportional limit). The yield point of a material is a point for a material where appreciable yielding occurs suddenly at a clearly defined value of stress; for example, in soft steel. In other materials the onset of appreciable yielding occurs gradually, and the yield strength for these materials is determined by using the "offset method." This is illustrated in Fig. 3.1; it shows a line, offset an arbitrary amount of 0.2 percent of strain, drawn parallel to the straight-line portion of the original stress—strain diagram. Point B is the yield mint of the material at 0.2 percent offset. If the load is removed after yielding to point B, the specimen exhibits a 0.2 percent permanent elongation. Yield strength corresponding to a specified (very small) offset is a stande laboratory determination, whereas elastic limit and proportional limit are not. 3-1 SOLUTION (3 .2D) Known: The materials to be selected have E greater than 207 GPa and Su greater than 1378 MPa. Find: Identify five materials with (a) modulus of elasticity greater than steel, and (b) ultimate strength greater than 200 ksi. Analysis: (a) From Appendix C-l, we attempt to select the following materials with higher E values than steel, but inspection of the materials listed in the appendix of the textbook, reveals no materials with E values higher than that of steel. I (b) From the textbook appendices, we select the following non—steel materials with Su values greater than 1378 MPa. I Material Leaded beryllium copper 1379 MPa (max) C 17300 (A n ndix C—13) Duranickel 301 1448 MPa (max) CD aged bar (A - ndix C-15) Rene 95 1620 MPa Su - rallo Comment: Inspection of the materials listed in the appendix of the textbook reveals that steel is a relatively strong and stiff material. 3-2 SOLUTION (32D) —— alternate Known: The materials to be selected have E greater than 207 GPa and Su greater than 1378 NIPa. Find: Identify five materials with (a) modulus of elasticity greater than steel, and (b) ultimate strength greater than 200 ksi. Analysis: (a) From www.matweb.com we select the following materials with higher E values than steel: I Modulus of Elastici , ksi, ; eater than 30 x 106 E (ksi) 30000 ksi Be Ilium, Be 40020 ksi 35%" “1 Rec stallized Annealed Annealed Annealed Annealed Annealed Annealed Annealed 58000 ksi (b) We select the following non—steel materials with ultimate tensile strength Su greater than 1378 MPa (200,000 psi). I Tensile Stren th, Ultimate, si, ; eater than 200 ksi AISI Grade 18Ni (300) Maraging 293625 psi Steel Aged, round bar Tested lon itudinal, 75 mm Carpenter AerMet®-for-Tooling Tool Steel Double A ed 468°C BioDurTM 316LS Stainless Medical 223445 psi 90% Cold Worked 3 3- VascoMax® C—300 Specialty Steel 285070 psi Heat Treatment: 927°C (1700°F) + A e W-25 Re Tungsten Rhenium Alloy 580000 psi Deformed AISI A6, Type Tool Steel 345100 psi Austenitized 830-870°C (1525- 1600°F) “900° F“ Temered at 500°C Titanium Ti-15Mo-52r 246500 psi ST 730°C, Aed 400°C AISI Type W2 Tool Steel 261000 psi Water quenched at 775°C (1425°F), and temered AISI Type SS Tool Steel Austenitized 855-870°C (1575— 1600°F) Oil uenched to 55 HRC Mo-47.5 Re Molybdenum Rhenium Alloy Deformed 290000?“ Cold—Drawn 30450” mi Deformed 29986" “i Hardened Technetium, Tc 218950 psi As—Rolled Pt-20% Ni Alloy 250125 psi Hard 79Pt-15Rh—6Ru 300150 psi A110 851 Pt-8% W Alloy 300150 psi Hard 323350 psi 464000 psi Comment: Matweb is an award winning web site. 3-4 SOLUTION (3.3) Known: The critical location of a part made from a 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 point 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 1/ 1.1 times its original area A0. 2. On the basis of the new area, the yield strength is Sy = 62(1.l) = 68.2 ksi. I 3. The ultimate strength is Su = 66(1.l) = 72.6 ksi. I 4. At fracture, R increases to 2.5 on the graph. 5. R=2.5/l.l =2.27 - 6. Using Eq. (3.3) and Eq. (3.2) — _ L: _ ; —_- Ar —1 R l 227 0.56 I e=R-l=2.27-l=l.27or127% I 3—5 SOLUTION (3.4) Known: The critical location of a part made from a known steel is cold worked during fabrication. Find: Estimate Su, Sy and the ductility. Schematic and Given Data: Hot Rolled 1020 Steel Assumption: After cold working the stress—strain curve for the critical location starts at point I. Analysis: 1. The area ratio at J is R = Ao/Af = 1.2. The initial area is thus 1/1.2. 2. The yield strength is Sy = 65(1.2) = 78 ksi. I 3. The ultimate strength is Su = 66(1.2) = 79.2 ksi. 4. At fracture, R increases to 2.5 on the graph. 5. R = 2.5/1.2 = 2.08 6. Using Eq. (3.3) and Eq. (3.2) _ 1 _ 1 __ -— _ = - = Ar 1 R 1 2.08 0.519 8 R 1 2.08 1.08 or 108% I 3—6 SOLUTION (3.5) 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 0', 8, 6T, 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, 0' = 66 ksi, 8 = 30%. I 2. For Figure 3.2, R = 1.3 at point H. 3. From Eq. (3.4), or: (SR = (66)(l.3) = 85.8 ksi. I 4. From Eq. (3.5), 8T=ln(1 + e) = In 1.30 = 0.26 = 26%. I 5. For the new specimen; 0' = 66(l.3) = 85.8 ksi. I 6. The new specimen behaves elastically, so a = 6/13 = 85.8/30,000 = .00286. I 7. Within the elastic range, CT = G and 8T 2 8. Therefore CT = 85.8 ksi and 81‘ = 0.29%. I Comment: Note also that 81‘ = ln(l + e) = ln(1.0029) = 0.29%. 3—7 SOLUTION (3.6) Known: A tensile specimen of a known material is loaded to a known stress, then unloaded and reloaded to the same stress point. Find: Estimate the values of o, 8, 6T, 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 G for the new specimen. Analysis: 1. For the initial sample, 6 = 62 ksi, e = 10%. I 2 For Figure 3.2, R = 1.1 at point G. 3. From Eq. (3.4), O'T= OR = (62)(1.1) = 68.2 ksi. I 4. From Eq. (3.5), 8T: ln(l + e) = In 1.10 = 0.095 = 9.5%. I 5 For the new specimen; 6 = 62(1.1) = 68.2 ksi. I 6 The new specimen behaves elastically, so 8 = o/E = 68.2/30,000 = .00227. I 7 Within the elastic range, O'T z 6 and ET z 8. Therefore CT = 68.2 ksi and ET = 0.23%. I Comment: Note also that 81‘ = ln(l + 8) = ln(1.0023) = 0.23%. SOLUTION (3.7) Known: A tensile specimen of a known material is loaded to a known stress, then unloaded and reloaded to the same stress point. Find: Estimate the values of 0', 8, GT, 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 J for the new specimen. Analysis: 1. For the initial sample, (5 = 65 ksi, e = 20%. I 2 For Figure 3.2, R = 1.2 at point J. 3. From Eq. (3.4), CT = OR = (65)(1.2) = 78.0 ksi. I 4. From Eq. (3.5), 8T: ln(l + s)=ln1.20 = 0.18 = 18%. I 5 For the new specimen; 0' = 65( 1.2) = 78 ksi. I 6 The new specimen behaves elastically, so 8 = 6/13 = 78/30,000 = .0026. I 7 Within the elastic range, O'T z 6 and ET z 8. Therefore 61‘ = 78 ksi and 81‘ = 0.26%. I Comment: Note also that 81‘ = ln(l + e) = ln(1.0026) = 0.26%. 3-9 SOLUTION (3.8D) Known: A steel is to be selected from Appendix C-4a. Find: Estimate Su and Sy 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 = 57.3, Sy = 42.8, and Bhn =111. 2. Using Eq. (3.11), we can estimate Su from the Brinell hardness using: Su 2 KBHB where KB z 500 for most steels. Su = 500(111) = 55,500 psi I 3. Sy can be estimated by using Eq. (3.12). Sy = 1.05 Su - 30,000 psi = 1.05(55,500) - 30,000 = 28,275 psi. I Comments: 1. Equation (3.12) is a good estimate of the tensile yield strength of stress-relieved (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 z 57.3 ksi. 2. Experimental data would be helpful to refine the above equations for specific steels. 3-10 SOLUTION (3.9D) Known: A steel is to be selected from Appendix C-4a. Find: Estimate Su and Sy from the given values of Brinell hardness for the selected steel. Schematic and Given Data: Decision: Select ANSI 1040 annealed 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 ANSI 1040 annealed steel has Su = 75.3 ksi, Sy = 51.3 ksi, and Bhn = 149. 2. Using Eq. (3.11), we can estimate Su from the Brinell hardness value using: Su = KBHB where KB —~— 500 for most steels. SUB = 500(149) = 74,500 psi 3. Sy can be estimated by using Eq. (3.12). SyB = 1.05 Su - 30,000 psi = 1.05(74,500) - 30,000 = 48,225 psi. 4. Ratio of strength (Appendix C—4a values to Brinell hardness based values): 3“ — 75-3 — 1.011 s“B ‘ 74.5 Sy _ 51.3 _ 8y; 8.2 _ 1.064 Comments: 1. Equation (3.12) is a good estimate of the tensile yield strength of stress—relieved (not cold-worked) steels. 2. The Brinell hardness based strength values are slightly less than the Appendix C- 4a strength values. 3—11 SOLUTION (3.10D) Known: A steel is to be selected from Appendix C-4a. Find: Estimate Su and Sy from the given values of Brinell hardness for the selected steel. Schematic and Given Data: A181 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, Sy = 50.0 psi, Bhn = 179; (2) normalized Su = 75.5 psi, Sy = 32.0 psi, Bhn = 149; (3) annealed Su = 67.3 psi, Sy = 31.2 psi, Bhn = 126. 2. Using Eq. (3.11), we can estimate 8“. Su = KBHB where KB z 500 for most steels. Sy can be estimated by using Eq. (3.12). In as-rolled condition: Su = 500(179) = 89,500 psi Sy = 1.05 811 — 30,000 psi 2 1.05(89,500) - 30,000 = 63,975 psi. 5. In normalized condition: 11 = 500(149) = 74,500 psi. Sy = 1.05811 — 30,000 psi = 48,225 psi. 6. In annealed condition: Su = 500(126) = 63,000 psi. Sy = 1.058“ — 30,000 psi = 36,150 psi. 7. Ratio of strength (Appendix C—4a values to Brinell hardness based values): J>LM 3-12 as-rolled: 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. 3-13 SOLUTION (3.11) Known: An A181 4340 steel part is heat treated to 217 Bhn. A second A181 4340 steel part is heat treated to 363 Bhn. Find: Estimate values of Sn and Sy for both parts. Schematic and Given Data: Part 1, 217 Bhn Part 2, 363 Bhn Suz? Sy=? Su=? Sy=? 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. Using Eq. (3.11), we can estimate Su. Su = KBHB where KB 2 500 for most steels. For Part 1, Su 2 500(217) = 108,500 psi; for Part 2, Su = 500(363) = 181,500 psi I 2. Sy can be estimated by using Eq. (3.12). Sy = 1.05 Su — 30,000 psi. For Part 1, Sy = 1.05(108,500) — 30,000 = 83,925 psi. For Part 2, Sy = 1.05(181,500) - 30,000 = 160,575 psi. I 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 helpful to refine the above equations for specific steels. 3~14 SOLUTION (3 . 1 2) Known: An A181 1020 steel part is heat treated to 111 Bhn. Find: Estimate Sn and Sy. Schematic and Given Data: 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. Using Eq. (3.11), we can estimate Su. Su 2 KBHB where K}; = 500 for most steels. u = 500(111) = 55,500 psi I 2. Sy can be estimated by using Eq. (3.12). Sy = 1.05 Su - 30,000 psi = 1.05(55,500) - 30,000 = 28,275 psi. I 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 helpful to refine the above equations for specific steels. 3-15 SOLUTION (3.13) Known: An AISI 3140 steel part is heat treated to 210 Bhn. Find: Estimate Su and Sy. Schematic and Given Data: A181 3140 Steel Heat Treated 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. Using Eq. (3.11), we can estimate Su. Su = KBHB where KB z 500 for most steels. Su = 500(210) = 105,000 psi I 2. Sy can be estimated by using Eq. (3.12). Sy = 105 Sn — 30,000 psi = 1.05(105,000) - 30,000 = 80,250 psi. I 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 helpful to refine the above equations for specific steels. 3-16 SOLUTION (3.14) Known: The hardness versus distance curve for a Jominy end-quench test of AISI 4340 steel is given. Find: Sketch similar curves for low-alloy steel and for plain carbon steel. Schematic and Given Data: . 4340 . Low Alloy . Plain Carbon 58 Rc 58 Rc 58 Rc / 4340 /_ Low Alloy O\\l CO ——l——1 kl] O Hardness (Rockwell C) U.) 4:- O O \ Plain Carbon rd no 000 10 20 30 40 Distance From Quenched End (mm) Assumption: We assume that the hardness for the non heat treated end of the J ominy test specimen properties are similar to 4340 steel. Analysis: From Appendix C—4a, for 1040 Normalized, HB = 170 and for 4340 Normalized, H}; = 363. We would expect at the end of the Jominy bar that the hardness of the 1040 would be approximately 50% of the 4340 hardness. Comment: The J orniny curves show the effectiveness of the alloying elements in imparting hardenability to steel. 3-17 SOLUTION (3.15D) Known: Four given applications require steel. Find: Choose between (a) 0.1% C and 0.4% C, and between (b) plain carbon and alloy steel. Schematic and Given Data: (a) Machine Frame (b) Round Rod 0: (c) Irregular Shaped Part (d) Rail CaI Wheel Eflaflr Assumption: Material is of usual quality. Analysis: (a) Machine frame. All steel materials provide approximately the same rigidity (30 X 106 psi). In this application, since stresses are low and we also want the cost to be low, we would select plain carbon because of low price and 0.1% C because of low strength. (b) Small round rod. For high bending and torsional stresses we select 0.4% C and to keep cost down we select plain carbon steel. (c) Large irregular shaped part. For high stress we select 0.4% C and for ease in heat treating we select alloy steel. (d) Rail car wheel. For low cost we select carbon steel and for low interior strength we choose 0.1% C. If we carburized the 0.1% C steel we can obtain a high strength wear surface. Comment: In summary, (1) alloy steel is more expensive but it is more easily heat treated than plain carbon steel, and (2) 0.4% C steel has greater strength than 0.1% C steel. 3-18 SOLUTION (3.16D) Known: The site http://www.matweb.com provides the properties of aluminum 7075—0. Find: Compare the material properties for aluminum 707 5-0 from the web to those given in Appendix C-10 of the textbook. Analysis: 1 . From www.matweb.c0m we obtain for aluminum 7075-0: Key Words: Aluminum 7075-0; UNS A97075; ISO AlZn5.5MgCu(A); Aluminium 7075-0; AA7075—O; PHYSICAL PROPERTIES Density, g/cc 2.81 Hardness, Brinell 60 500 kg load with 10 mm ball Hardness, Knoop 80 Converted from Brinell Hardness Value Hardness, Vickers 68 Converted from Brinell Hardness Value MECHANICAL PROPERTIES Tensile Strength, Ultimate, MPa 220, 31,900 psi Tensile Strength, Yield, MPa 95, 13,775 psi Elongation %; break 17, In 5 cm; Sample 1.6 mm thick 17 % Modulus of Elasticity, GPa 72, Average of Tension and Compression. In Aluminum alloys, the compressive modulus is typically 2% greater than the tensile modulus, 10,440 ksi Poissons Ratio 0.33 Shear Modulus, GPa 26.9, 3,901 ksi Shear Strength, MPa 150, 21,750 psi THERMAL PROPERTIES CTE, linear 20°C, um/m-OC, 23.6, 20-100°C, 13 pin/in~°F CTE, linear 250°C, um/m—°C, 25.2, Average over the range 20-300°C, 14 pin/in— °F Heat Capacity, J/g-°C, 0.96, 0.23 BTU/lb-°F Thermal Conductivity, W/m—K, 173, 1,201 BTU-in/hr-ft2-°F Melting Point, 0C, 477, Solidus, 891 °F Solidus, °C, 477, 891 °F Liquidus, 0C, 635, 1,175 °F ELECTRICAL PROPERTIES Electrical Resistivity, Ohm—cm 0.0000038 I From Appendix C—10 of the textbook for 7075—0 aluminum we find: Aluminum 7075-0 Brinell Hardness 60 Tensile Strength, Ultimate, 38 ksi, MPa 230 Tensile Strength, Yield, 15 ksi, MPa 105 Elongation in 2 in %; 16 I 3—19 Comments: 1. The web wins again, providing a vast amount of information. 2. The tensile strength and the yield strength values from Appendix 010 are higher than those from www.matweb.com. SOLUTION (3.17D) Known: A designers favorite materials are 1020 steel, 1040 steel, 4340 steel, 2024- T4 aluminum, nylon 6/6, and acetal. Find: Compare the specific material properties for each material. Assumptions: The material properties presented in Appendix C are sufficient for comparison purposes. Decision: The material properties compared will be tensile strength, yield strength, elongation, and hardness. Analysis: 1. From Appendix C-4a, we obtain properties for normalized steel in 1 inch round sections. 2. From Appendix C- 10, we obtain properties for wrought aluminum alloy for 1/2 inch sizes. ' 3. From Appendix C-18a, we obtain mechanical properties of plastics. MATERIAL Tensile Strength Yield Strength Elongation (%) Brinell (W3) (MP 3) Hardness 1020 steel 441.3 346.5 35.8 131 1040 steel 589.5 374.0 28.0 170 4340 steel 1279.0 861.8 12.2 363 2024-T4 470 325 19 120 4. Inspection of the table provides a comparison of the properties of the six materials. Comments: The site http://www.matweb.com will provide additional material property information. 3—20 ...
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This note was uploaded on 03/27/2009 for the course MECHENG EN.530.215 taught by Professor Wang during the Spring '09 term at Johns Hopkins.

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ch03 - SOLUTION (3.1D) Known: Definitions of the terms...

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