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Hw 5 Solutions.pdf

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MSE 2610 Mechanical Properties of Materials Fall 2009 Homework #5 solutions 1.) A single crystal has only one slip plane with a normal vector that makes an angle of 60° with the tensile axis. The three slip directions have angles of 38°, 45°, and 84° with respect to the tensile axis, and the critical resolved shear stress (CRSS) is 3.6 MPa. The shear modulus G for zinc is 42 GPa. a.) 6 points; 1 for equation, 2 for explanation 1 for units, 2 for correct answer. Determine the tensile stress at which plastic deformation commences. τCRSS =σAcos λ( )cos θ( ) Because the only choice is among the slip directions, the largest critical resolved shear stress will occur when cos(λ) is the largest. σA = τCRSScos λ( )cos θ( )= (3.6MPa)cos 38o( )cos 60o( )= 9.14MPa σA = τCRSScos λ( )cos θ( )= (3.6MPa)cos 45o( )cos 60o( )=10.2MPa σA = τCRSScos λ( )cos θ( )= (3.6MPa)cos 84o( )cos 60o( )= 68.9MPa b.) 4 points for correct answer. Some may have used b = a/3<1120>, in this case give 2 for the expression and 2 for the correct answer (which is the same as <1120> has a length of 3a). The lattice parameters for zinc are a = 2.66 Å and c = 4.94 Å. What is the magnitude of the Burgers vector b? Because the lattice must remain unchanged after a dislocation has slipped, the Burgers vector must be 1 lattice parameter in the basal plane. The Burgers vector must be 2.66 Å. c.) 5 points; 1 for equation, 1 for units, 1 for work, 2 for correct answer. What is the critical force per unit length (of dislocation) required to move a straight dislocation in zinc? The critical resolved shear stress is the stress when slip starts. The minimum amount this dislocation can travel is 1 Burgers vector. F l =τCRSSb= (3.6MPa)(2.66×10 −10 m)= 9.6x10−4 N/m d.) 5 points; 1 for equation, 1 for units, 1 for work, 2 for correct answer. What is the length of the shortest dislocation line segment that could act as a double-ended dislocation source (i.e., Frank-Read source), if the maximum shear stress acting on the dislocation segment is 10.0 MPa? b=a When the dislocation gets pinned at 2 ends, it can act as a Frank-Read source only if the shear stress is high enough to expand the loop past the critical point. τmax = 2Gbd d= 2Gbτ max = 2(42GPa)(2.66×10 −10 m) 10.0MPa d= 2.2 µm 2.) 15 points; 3 for each correct σo and kHP, and 3 for graph (-1 for unlabled axes, no units…) The following grain size d and yield stress σy data were obtained for a steel and an aluminum alloy. Steel Aluminum alloy d (µm) σy (MPa) d (µm) σy (MPa) 406 93 42 223 106 129 16 225 75 145 11 225 43 158 8.5 226 30 189 5.0 231 16 233 3.1 238 Show graphically that the behavior of the steel and aluminum alloy is consistent with the Hall-Petch relation and determine σo and kHP for each material. Hall-Petch Relation for a Steel and an Aluminum Alloy σ = 0.6895/√d + 60.466 R2 = 0.992 σ = 0.0364/√d + 215.48 R2 = 0.9118 50 100 150 200 250 300 0 100 200 300 400 500 600 1/√d (m-1/2) Yi eld S tre ss (M Pa ) Steel Aluminum Linear (Steel) Linear (Aluminum) d d/2 Steel Aluminum alloy kHP 0.690 0.0364 σo 60.5 215 3.) Consider a polycrystalline aluminum alloy that has been strengthened with a dispersion of hard, uncuttable second phase particles. The tensile yield strength of the alloy is measured to be σy= 700MPa. (a) 5 points; 1 for equation, 1 for units, 1 for work, 2 for correct answer. Given that the average particle spacing l ~ 0.05 µm, estimate the contribution to the tensile yield strength that is due only to this dispersion of hard, uncuttable particles within the alloy. Start by deriving an expression from the diagram (e.g. force balance) for the critical resolved shear stress for dislocation extrusion in terms of G, l, and b. State any assumptions. τc = 2Tbl ,T =Gb2 →τc = 2Gbl σy,particles 2τc →σy,particles 4Gbl We know that b is the atomic spacing on the slip plane in the slip direction: b= 12a 110 = a 22 ∴σy,particles 4Gl a 22 = 4(27E9)(0.405E−9) 22(0.05E−6) = 619MPa (b) 5 points; 1 for equation, 1 for units, 1 for work, 2 for correct answer. What is the portion of the tensile yield strength that is due to other mechanisms σo (i.e., the yield stress of the material in the absence of second phase particles)? σy,total =σy,particles +σy,o = 700MPa= 618MPa+σy,o σy,o = 81MPa (c) 5 points; 1 for equation, 1 for units, 1 for work, 2 for correct answer. This alloy is to be used for the compressor blades at the low temperature end of a small gas turbine where the operating stresses are 300 MPa. During operation, the blade temperature is raised, causing the particles to slowly coarsen such that the average spacing increases with time according to l = 0.05 + 0.01t0.35, where l is in µm and t is in hours. If the elevated temperature only has a significant effect on the second phase particles (i.e., σo is independent of temperature), estimate the critical particle spacing (i.e., particle spacing at which blades begin to yield under operating conditions). Now, yield occurs at 300MPa, due to high operating temperature. σyield = 300MPa 300MPa=σy,particles +σy,o = 4Gl a 22 +81MPa 219MPa≥ 4Gl c a 2 2 ∴lc = 4(27E9)(0.405E−9) 22(219E6) = 0.14µm (d) 5 points; 1 for equation, 1 for units, 1 for work, 2 for correct answer. Using the result in (c), estimate the lifetime of the blades (i.e., time to yield). lc = 0.14µm= 0.05+0.01t0.35 t= 550hrs (e) 5 points; 1 for equation, 1 for units, 1 for work, 2 for correct answer. What is the yield stress of the particle if the particles can be “cut” by the dislocations at an angle of 5o(i.e., the particles are very weak). σy,cuttable,weak = 2Fcbl Fc2T,Fc = 2Tsinθ,T =Gb2 σy,cuttable,weak = 4Gbsinθl sinθ = 4(27E9)(a 22 )sin(5) 0.05E−6 sin(5) =16MPa Potentially useful information for aluminum: Al is FCC with a = 0.405 nm Young’s modulus E = 70 GPa Shear modulus G = 27 GPa Poisson’s ratio ν = 0.3 4.) Consider the Al-Mg phase diagram below: a) 4 points; 1 for each. List the solid solutions. A solid solution is defined as a single phase region with the same lattice type everywhere and random positioning of the element atoms, this occurs in the Al, Al2Mg2, Al12Mg17 and Mg phase fields. If I have a lump of aluminum with 15 wt% Mg at 500 ºC at equilibrium it will be part solid and part liquid. b)6 points; 2 for expression, 2 for using correct numbers (accept +1 either way) and 2 for correct answer (accept anything between 30 – 50 % if working is correct) How much of it will be liquid? Using the lever rule: xL = 15 – 9 = 40% 24 – 9 c)5 points; accept anything between 8 – 10 %. -2 if confused Al and Mg (eg “9 wt % Al”), -2 if state % rather than wt%. What will be the composition of the solid present? Read from the graph: cs = 9 wt% Mg 5) Fe-C Phase Diagram a.) 4 points; 1 for temp, 1 for composition (or 2 for indicated on graph), 2 for equilibrium phases. Locate the eutectic point. What phase transition is occurring as the steel is cooled through the eutectic point? - The eutectic point is at 1154 oC and composition of 4.35% C (95.65% Fe). As the alloy is cooled through the eutectic point, it starts as a liquid and forms a 2-phase austenite and cementite alloy. b.) 4 points; 2 for temperature, 2 for composition. What is the composition and temperature of the eutectoid in this diagram? - The eutectoid point is at 738 oC and a composition of 0.68% C (99.32% Fe) c.) 5 points; -1 for each temperature, composition or equilibrium phase missed in the description below. Describe the phase transitions as a 0.5%-C steel is heated up from 600 ˚C to 1400 ˚C. - At 600 oC, the alloy is a 2-phase ferrite and cementite microstructure. At 738 oC, the alloy transitions to a 2-phase ferrite and austenite microstructure. At approximately 770oC, the alloy changes to a single phase Austenite microstructure. The alloy remains a single-phase austenite to 1400oC. d.) 6 points; 1 for expression, 1 for putting in correct values, 1 for correct answer for austenite and for liquid. What equilibrium phases are present in a 2.0%-C steel at 1300 ˚C? What is composition and mass percent of each phase present? CL = 3.0 wt% C (2.9-3.1 acceptable) CA = 1.3 wt% C (1.2-1.4 acceptable) Eutectic Eutectoid %wt2.41%1003.10.3 3.10.2CC CCLiquid% %wt8.58%1003.10.3 0.20.3CC CCAustenite% AL So AL oL =×−−=−−= =×−−=−−= e.) 6 points; 1 for expression, 1 for putting in correct values, 1 for correct answer for austenite and for ferrite. What equilibrium phases are present in a 0.25%-C steel at 740 ˚C? What is composition and mass percent of each phase present? This is close enough to the eutectic temperature that the values on the eutectic line can be used. CA = 0.68 wt% C CF = 0.022 wt% C (The diagram makes it look like zero, so zero is acceptable, but the real value is 0.022 wt%) %wt7.34%100022.068.0 022.025.0CC CCAustenite% %wt3.65%100022.068.0 25.068.0CC CCFerrite% FA Fo FA oA =×−−=−−= =×−−=−−= Drew Microsoft Word - HW 5 solutions_2_.doc