3.22 Mechanical Behavior of materials PS8 Solution Due: April, 27, 2004 (Tuesday) before class (10:00am)

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1 3. Mechanical Behavior of materials PS8 Solution Due: April, 7, 004 (Tuesday before class (10:00am 8 1. Annealed copper have a dislocation density of approimately 10 cm. Calculate the total elastic strain energy for copper. Assume that the dislocations are screw dislocations for simplicity and dislocation core radius is r0 = 5b. Neglect the strain energy of the dislocation core. For copper G=40GPa and b=0.5nm. U el 9 Gb R Gb 1/ ρ (40 10 ( / L = ln ln = 4 r = π 0 4π 5b 4π = J / m Total strain energy is 1 ln 5 ( ( U el Total = = J / m.. An FCC single crystal of nickel is sheared by γ = Assuming that the dislocation density is equal to 10 8 cm - and that it remains constant, what is the average distance each dislocation will have to move? If the shear strain rate is 10-4 sec -1, what is the mean velocity of the dislocation? Assume that the Burgers vector of Ni is 0.5 nm. γ (i l = = = 4 10 m. ρ 1 9 b (10 ( (ii 4 γ 10 7 γ = ρbv. So v = = = 4 10 m / sec. ρ 1 b 10 ( The figure below shows the arrangement of atoms for a curved dislocation segment. The open circles represent the atomic planes just above the slip plane and the closed circles represent the atoms just below. (a What is the Burgers vector for the dislocation shown below? (b Indicate the regions of slipped material. (c Define the regions of the dislocation, whether screw, edge or mied. 1

2 y (d If the dislocation segment epands in its own plane under the influence of an applied stress, slip step will be formed when all components of the segment intersect the faces of the crystal. What will be the orientation of slip steps formed on the crystal faces assuming that shear stress ( τ = τ zy is applied along the y-direction?

3 Eplain in detail. I have labeled three corners P, Q, and R. With the application of a shear stress, τ, on the surface of the lattice, the dislocation will epand. The slip steps will be formed along PQ and parallel to QR. After time t T the slip step will encompass all of PQ with the top layer of atoms etending one bond length paralled to BC. 4. In rectangular coordinates, the stress field surrounding edge dislocation is given by y yy y = Gb π y(3 (1 v ( + y Gb = π y( (1 v ( + y Gb = π ( + y y (1 v ( + y zz = v + ( yy y For a single edge dislocations in a block of annealed pure copper, plot the stress field along the positive y-ais and eplain how your result relates to the need for a dislocation core. For =0, the stress becomes a function of y only. Gb 1 =, π (1 v y Gb 1 yy =, y = 0 π (1 v y 3

4 Normalizing by Gb / π (1 v For y=0, =0, stresses ehibit a singularity. Thus, we must analyze the stress away from the dislocation core. 5. (a Eplain why a second hardness measurement made adjacent to a previous measurement indicates a higher hardness. The previous hardness measurement leads to plastic deformation near the indented area. This plastically deformed region is thus work hardened. The work hardened area around the previous indentation results in a higher hardness when indented a second time. (b In class, we have shown how the motion of dislocations on the atomic scale results in macroscopic plastic deformation and ductility. How can we eplain brittle, crystalline solids such as ceramics that show essentially no plastic deformation? Can we conclude that ceramics have a perfect, dislocation free crystal structure? Brittle materials such as ceramics do have dislocations, but because of the stronger character of the ionic or covalent bonds in these materials, they are etremely difficult to move by glide through the lattice as we depicted in class. These materials typically fail by brittle fracture before stresses large enough to overcome the lattice resistance (opposing dislocation motion are attained. It should be noted that this is not true at high temperatures (where the lattice resistance goes down or if a large eternal pressure is imposed (fracture is suppressed by the high pressure. Under these conditions ceramics can actually show significant plasticity. 6. (a Maraging steels are relatively soft upon quenching from the austenitizing temperature range but strengthen greatly following eposure to a reheating treatment at intermediate temperature. Given that the carbon level of such steels is typically less than 4

5 0.03%, whereas the alloy contains additions of Ni, Mo, and Ti, speculate as to the probable strengthening mechanism that controls the strength of this class of alloys. Ferritic steels derive much of their strength from solid solution strengthening of carbon in the BCC lattice. In maraging steels, the carbon concentration is so low that the solid solution of carbon in the steel does not influence the strength of the steel significantly. If solid solution strengthening were present, then the steel would not be relatively soft after quenching from the austenitizing temperature. From the fact that strength increases following eposure to an intermediate temperature (aging after quenching, we can speculate that the strengthening comes from the precipitation of a second phase (NiMo and NiTi due to the presence of the alloying elements. Therefore, the probable strengthening mechanism in maraging steels is precipitation hardening. (b Eplain why metals show the ductile-brittle transition temperature behavior at low temperatures. The yield strength of metal decreases linearly with temperature and increases with ln γ. The plastic zone size, r p, of fracture sample depends on the yield strength. As temperature decreases, the yield strength of the metal increases and the size of plastic zone decreases. By suppressing the size of the plastic zone at low temperatures, the facture behavior of metal becomes more brittle. 7. At 5, an alloy yields at 170 MPa when strained at 0.0 s -1. The same alloy yields at 165 MPa when the strain rate decreased by a factor of ten and the temperature is kept the same. What test temperature would have to be used with a strain rate of 0.0sec -1 to reach a yield strength of 165 MPa? Assume that the energy per bond (Q b is 500 kj/mol. The yield strength of a material as a function of temperature and strain rate is kt γ 0 τ ( γ, T = τ ln. Q b γ Q b = 500 kj/mol = J/atom γ 0 τ (0.0, T = 98 = τ ln = 170MPa γ 0 τ (0.00, T = 98 = τ ln = 165MPa By solving the above equations, we have 5

6 51 1 γ = τ MPa 0 sec 0 = To find the temperature at which the yield strength is 165MPa for a strain rate of 0.0 s -1, T γ 0 τ (0.0, T = τ ln = 165MPa T=303.55K=

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