Projetos e Elementos de Maquinas - Juvinall -exercicios Resolvidos - Materials

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. The stress is the load divided by the cross-sectional area. 2. The strength is the maximum value of stress a material will carry before failure. 3. The yield strength, Sy, is the value of stress at which significant plastic yielding first occurs. 4. The ultimate strength is the maximum value of stress a material will carry before fracture for nondynamic loading. 5. The elastic limit is the highest stress the material can withstand and still return exactly to its original length when unloaded. 6. 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. 7. 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). 8. 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 point 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 standard laboratory determination, whereas elastic limit and proportional limit are not. 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-1, 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. (b) From the textbook appendices, we select the following non-steel materials with Su values greater than 1378 MPa. Material | Su (MPa) Steel 1378 MPa Leaded beryllium copper C17300 (Appendix C-13) 1379 MPa (max) Duranickel 301 1448 MPa (max) CD aged bar (Appendix C-15) Rene 95 1620 MPa Superalloy Comment: Inspection of the materials listed in the appendix of the textbook reveals that steel is a relatively strong and stiff material. SOLUTION (3.2D) -- alternate 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 www.matweb.com we select the following materials with higher E values than steel: Modulus of Elasticity, ksi, greater than 30 x 106 psi Material | E (ksi) Steel 30000 ksi Beryllium, Be 40020 ksi Chromium, Cr 35960 ksi Recrystallized 75980 ksi Molybdenum, Mo Annealed 47850 ksi Osmium, Os Annealed 81200 ksi Rhenium, Re Annealed 68005 ksi Rhodium, Rh Annealed 52055 ksi Ruthenium, Ru Annealed 60300 ksi Technetium, Tc Annealed 46690 ksi Tungsten, W 58000 ksi (b) We select the following non-steel materials with ultimate tensile strength Su greater than 1378 MPa (200,000 psi) Material | Su (psi) AISI Grade 18Ni (300) Maraging Steel 293625 psi Aged, round bar Tested longitudinal, 75 mm Carpenter AerMet®-for-Tooling Tool Steel 300000 psi Double Aged 468°C BioDur™ 316LS Stainless Medical Implant Alloy 223445 psi 90% Cold Worked VascoMax® C-300 Specialty Steel 285070 psi Heat Treatment: 927°C (1700°F) + Age W-25 Re Tungsten Rhenium Alloy Deformed AISI A6, Type Tool Steel Austenzitized 830-870°C (1525-1600°F) AISI A9, Type Tool Steel Tempered at 500°C Titanium Ti-15Mo-5Zr ST 730°C, Aged 400°C AISI Type W2 Tool Steel Water quenched at 775°C (1425°F), and tempered AISI Type S5 Tool Steel Austenzitized 855-870°C (1575-1600°F) Oil quenched to 55 HRC Mo-47.5 Re Molybdenum Rhenium Alloy Deformed Iridium, Ir 290000 psi Rhenium, Re 304500 psi Rhodium, Rh 299860 psi Technetium, Tc 218950 psi As-Rolled Pt-20% Ni Alloy Hard 79Pt-15Rh-6Ru Alloy 851 Pt-8% W Alloy Hard Comment: Matweb is an award winning web site. 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: S_o = 66 ksi 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 A_o. On the basis of the new area, the yield strength is Sy = 62(1.1) = 68.2 ksi. The ultimate strength is Su = 66(1.1) = 72.6 ksi. At fracture, R increases to 2.5 on the graph. R = 2.5/1.1 = 2.27 Using Eq. (3.3) and Eq. (3.2) A_r = 1 - 1/R = 1 - 1/2.27 = 0.56 e = R - 1 = 2.27 - 1 = 1.27 or 127% 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: Su = 66 ksi Assumption: After cold working the stress-strain curve for the critical location starts at point J. Analysis: The area ratio at J is R = A_o/A_f = 1.2. The initial area is thus 1/1.2. The yield strength is Sy = 65(1.2) = 78 ksi. The ultimate strength is Su = 66(1.2) = 79.2 ksi. At fracture, R increases to 2.5 on the graph. R = 2.5/1.2 = 2.08 Using Eq. (3.3) and Eq. (3.2) A_r = 1 - 1/R = 1 - 1/2.08 = 0.519 e = R - 1 = 2.08 - 1 = 1.08 or 108% SOLUTION (3.5)\nKnown: A tensile specimen of a known material is loaded to the ultimate stress, then unloaded and reloaded to the ultimate stress point.\nFind: Estimate the values of σ, e, σT, eT for the first loading and the reloading.\nSchematic and Given Data:\n\nStress\nS_u = 66 ksi\n\nHot Rolled 1020 Steel\n\nStrain\n\nAssumption: After unloading the stress-strain curve starts at point H for the new specimen.\n\nAnalysis:\n1. For the initial sample, σ = 66 ksi, e = 30%.\n2. For Figure 3.2, R = 1.3 at point H.\n3. From Eq. (3.4), σT = σR = (66)(1.3) = 85.8 ksi.\n4. From Eq. (3.5), eT = ln(1 + e) = ln 1.30 = 0.26 = 26%.\n5. For the new specimen; σ = 66(1.3) = 85.8 ksi.\n6. The new specimen behaves elastically, so e = σ/E = 85.8/30,000 = .00286.\n7. Within the elastic range, σT = σ and eT = e. Therefore σT = 85.8 ksi and eT = 0.29%.\n\nComment: Note also that eT = ln(1 + e) = ln(1.0029) = 0.29%. SOLUTION (3.6)\nKnown: A tensile specimen of a known material is loaded to a known stress, then unloaded and reloaded to the same stress point.\n\nFind: Estimate the values of σ, e, σT, eT for the first loading and the reloading.\nSchematic and Given Data:\n\nStress\nS_u = 66 ksi\n\nHot Rolled 1020 Steel\n\nStrain\n\nAssumption: After unloading, the stress-strain curve starts at point G for the new specimen.\n\nAnalysis:\n1. For the initial sample, σ = 62 ksi, e = 10%.\n2. For Figure 3.2, R = 1.1 at point G.\n3. From Eq. (3.4), σT = σR = (62)(1.1) = 68.2 ksi.\n4. From Eq. (3.5), eT = ln(1 + e) = ln 1.10 = 0.095 = 9.5%.\n5. For the new specimen; σ = 62(1.1) = 68.2 ksi.\n6. The new specimen behaves elastically, so e = σ/E = 68.2/30,000 = .00227.\n7. Within the elastic range, σT = σ and eT = e. Therefore σT = 68.2 ksi and eT = 0.23%.\n\nComment: Note also that eT = ln(1 + e) = ln(1.0023) = 0.23%. SOLUTION (3.7)\nKnown: A tensile specimen of a known material is loaded to a known stress, then unloaded and reloaded to the same stress point.\n\nFind: Estimate the values of σ, e, σT, eT for the first loading and the reloading.\nSchematic and Given Data:\n\nStress\nS_u = 66 ksi\n\nHot Rolled 1020 Steel\n\nStrain\n\nAssumption: After unloading the stress-strain curve starts at point J for the new specimen.\n\nAnalysis:\n1. For the initial sample, σ = 65 ksi, e = 20%.\n2. For Figure 3.2, R = 1.2 at point J.\n3. From Eq. (3.4), σT = σR = (65)(1.2) = 78.0 ksi.\n4. From Eq. (3.5), eT = ln(1 + e) = ln 1.20 = 0.18 = 18%.\n5. For the new specimen; σ = 65(1.2) = 78 ksi.\n6. The new specimen behaves elastically, so e = σ/E = 78/30,000 = .0026.\n7. Within the elastic range, σT = σ and eT = e. Therefore σT = 78 ksi and eT = 0.26%.\n\nComment: Note also that eT = ln(1 + e) = ln(1.0026) = 0.26%. SOLUTION (3.8D)\nKnown: A steel is to be selected from Appendix C-4a.\nFind: Estimate Su and Sy from the given value of Brinell hardness for the steel selected.\nSchematic and Given Data:\n\n AISI\n Steel\n Su = ?\n Sy = ?\n\nDecision: Select ANSI 1020 annealed.\n\nAssumptions:\n1. The experimentally determined relationship of ultimate strength to hardness is sufficiently accurate.\n2. The experimentally developed relationship of yield strength to ultimate strength is sufficiently accurate for our purposes.\n\nAnalysis:\n1. Appendix C-4a shows that ANSI 1020 annealed steel has Su = 57.3, Sy = 42.8, and Bhn = 111.\n2. Using Eq. (3.11), we can estimate Su from the Brinell hardness using:\n Su = KBH\n where KB = 500 for most steels.\n Su = 500(111) = 55,500 psi\n3. Sy can be estimated by using Eq. (3.12):\n Sy = 1.05 Su - 30,000 psi = 1.05(55,500) - 30,000 = 28,275 psi.\n\nComments:\n1. 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 = 57.3 ksi.\n2. Experimental data would be helpful to refine the above equations for specific steels.\n\n3-10