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Solutions for Practice Questions # 1 MECH 321 Winter 2014 Question 1: (a) Do all metals have the same slip systems? Why or why not? (b) FCC metals are often recommended for use at low temperatures, particularly when any sudden loading of the part is expected. Explain. (c) A steel part can be made by powder metallurgy (compacting iron powder particles and sintering to produce a solid) or by machining from a solid steel block. Which part is expected to have the higher toughness? Explain. (d) Upon what three criteria are factors of safety based? Solution: (a) All metals do not have the same slip system. The reason for this is that for most metals, the slip system will consist of the most densely packed crystallographic plane, and within that plane the most closely packed direction. This plane and direction will vary from crystal structure to crystal structure. (b) FCC metals do not normally display a transition temperature; instead the impact energies decrease slowly with decreasing temperature and, in at least some cases (such as some aluminum alloys), the energies even increase at low temperatures. The FCC metals can obtain large ductilities, giving large areas beneath the true stress-strain curve (c) Parts produced by powder metallurgy often contain considerable amounts of porosity due to incomplete sintering; the porosity provides sites at which cracks might easily nucleate. Parts machined from solid steel are less likely to contain flaws that would nucleate cracks, therefore improving toughness. (d) The criteria upon which factors of safety are based are (1) consequences of failure, (2) previous experience, (3) accuracy of measurement of mechanical forces and/or material properties, and (4) economics. Question 2: The following data were collected from a 12-mm-diameter test specimen of magnesium (lo = 30.00 mm): After fracture, the gage length is 32.61 mm and the diameter is 11.74 mm. Plot the data and calculate (a) the 0.2% offset yield strength (b) the tensile strength (c) the modulus of elasticity (d) the %Elongation (e) the %Reduction in area (f) the engineering stress at fracture (g) the true stress at fracture (h) the modulus of resilience. Solution: σ = F / (π/4)(12 mm)2 = F/113.1 Ɛ = (l - 30)/30 (a) 0.2% offset yield strength = 186 MPa (b) tensile strength = 238.7 MPa (c) E = (132.6 − 0) / (0.00296 − 0) = 44,800 MPa = 44.8 GPa (d) %Elongation =(32.61 − 30)/30× 100 = 8.7% (e) %Reduction in area = ((π/4)(12)2 − (π/4)(11.74)2)/ (π/4)(12)2 × 100 = 4.3% (f) engineering stress at fracture = 221 MPa (g) true stress at fracture = 25,000 N / (π/4)(11.74)2 = 231 MPa (h) From Equation (6.14), the modulus of resilience is just Ur = σy2/2E which, using data computed in the problem yields a value of Ur = (186×106)2/(2×44.8×109)= 3.861×105 J/m3 Question 3: For a brass alloy, the following engineering stresses produce the corresponding plastic engineering strains, prior to necking: Engineering Stress(MPa) Engineering Strain 235 0.194 250 0.296 On the basis of this information, compute the engineering stress necessary to produce an engineering strain of 0.25. Solution: For this problem we first need to convert engineering stresses and strains to true stresses and strains so that the constants K and n in Equation (6.19) may be determined. Since σ = σ(1 + Ɛ) then T σ = (235 MPa)(1 + 0.194) = 280 MPa T1 σ = (250 MPa)(1 + 0.296) = 324 MPa T2 Similarly for strains, since ƐT = ln(1 + Ɛ) then Ɛ = ln(1 + 0.194) = 0.177 T 1 Ɛ = ln(1 + 0.296) = 0.259 T 2 Taking the logarithm of Equation (6.19), we get: log σ = log K + n log ƐT T which allows us to set up two simultaneous equations for the above pairs of true stresses and true strains, with K and n as unknowns. Thus log(280) = log K + n log(0.177) log(324) = log K + n log(0.259) Solving for these two expressions yields K = 543 MPa and n = 0.383. Now, converting Ɛ= 0.25 to true strain, ƐT = ln(1 + 0.25) = 0.223 The corresponding σ to give this value of ƐT is just T 0.383 σ = K ƐTn = (543 MPa)(0.223) T Now converting this σ to an engineering stress: T = 306 MPa 306 MPa σ = σ /1+ Ɛ = 1 + 0.25 = 245 MPa T Question 4: A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 20.000 and 20.025 mm, respectively, and its final length is 74.96 mm, compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are 105 GPa and 39.7 GPa, respectively. Solution: This problem asks that we compute the original length of a cylindrical specimen that is stressed in compression. It is first convenient to compute the lateral strain εx as In order to determine the longitudinal strain εz we need Poisson's ratio, which may be computed using Equation 6.9; solving for ν yields Now εz may be computed from Equation 6.8 as Now solving for l0 using Equation 6.2 Question 5: A three-point bending test is performed on a glass specimen having a rectangular cross section of height d = 5 mm (0.2 in.) and width b = 10 mm (0.4 in.); the distance between support points is 45 mm (1.75 in.). (a) Compute the flexural strength if the load at fracture is 290 N (65 lbf). (b) The point of maximum deflection Δy occurs at the center of the specimen and is described by where E is the modulus of elasticity and I is the cross-sectional moment of inertia. Compute Δy at a load of 266 N (60 lbf). Solution: (a) For this portion of the problem we are asked to compute the flexural strength for a glass specimen that is subjected to a three-point bending test. The flexural strength (Equation 12.7a) is just for a rectangular cross-section. Using the values given in the problem statement, (b) We are now asked to compute the maximum deflection. From Table 12.5, the elastic modulus (E) for glass is 69 GPa (10 × 106 psi). Also, the moment of inertia for a rectangular cross section (Figure 12.32) is just Thus