Problem Set 4 Due: Tuesday, February 14, 11:00 AM 1. Tungsten in Tensile Test A single crystal of tungsten is oriented for a tensile test such that tensile stress is applied along the [001] direction. a) On which family of planes should slip most likely occur? Of the planes in this family, list which will feel a resolved shear stress. Which will not? Will any of the planes that do feel stress be more favored than the others? (10 pts) Tungsten is body-centered cubic. Slip in this crystal system usually occurs in the {110} family of planes. A plane will experience no shear stress if the Schmid factor is zero. This corresponds to the plane normal being either perpendicular or parallel to the applied stress. Of the {110} family: Perpendicular: (110) (1-10) Not perpendicular: (101) (011) (-101) (0-11) All of the non-perpendicular planes have the same angle with respect to their normal direction and the tensile axis, so none of them will be preferred over any of the other three. b) Choose one of the planes that experiences stress from part (a). If the critical resolved stress for this slip system of tungsten is 9.5 MPa, calculate the tensile stress needed to initiate slip along this plane in a close-packed direction. (8 pts) We can recognize visually that all of the planes from part (a) that feel shear stress have a normal 45 to the tensile axis, so φ =45. To find λ, we use the dot product. Also, note that by symmetry, all of the close-packed directions, <111> family, will form the same angle with the tensile axis, so we can choose any C.P. direction on the plane of choice and we ll get the same result. If we choose (011), the directions that lie on that plane are: [1-11], [-1-11], [-11-1] and [11-1]. Using relation from Callister, the stress required is 23.25 MPa. c) Briefly, why are HCP metals generally not as ductile as FCC or BCC metals? (2 pts) There are less slip systems that can be activated (3 as opposed to 12+ in FCC and BCC), so slip is more difficult to produce.
2. Mechanical Properties of Metals Consider a single crystal (FCC) metal cylinder 0.5m long with a radius of 0.02m. The length of the rod is in the <100> direction of the crystal. Featured below is the stress-strain behavior for the metal. a) Determine the Young s Modulus. (2 pts) The slope of the linear portion of the plot gives the elastic modulus. From the inset figure, E = Δσ/Δε ~ 150MPa/.001 = 150GPa b) Determine the yield strength for a strain offset of.002. (2 pts) The yield strength is given by the intersection of the curve with the parallel offset line featured in the figure above. From the figure, σ y = 240MPa c) Determine the maximum load that can be sustained by the cylinder. (4 pts) The tensile strength is given by the maximum of the curve. The figure gives a value of σ ~ 850MPa. The maximum load is then given by F= σa 0 = (850MPa)*π(.02m) 2 = 1.07 10 6 N
d) The rod experiences a tension along the <100> direction perpendicular to the circular end with a stress of 150MPa. Using the modulus of elasticity you found in part a and assuming entirely elastic deformation, calculate the elongation. (6 pts) σ = εe = (Δl/l 0 )*E Δl = σl 0 /E = (150MPa)(0.5m)/150GPa = 0.0005m e) Suppose that slip occurs on the (111) plane along the <1Ī0> direction. Calculate the critical resolved shear stress. (6 pts) 3. Mechanical Properties of a Gold Ring Still skeptical about the quality of the gold ring you received last week, you decide to test some of its mechanical properties. You know the Young s modulus for Gold is approximately 80 GPa (GPa = 10 9 Pa). a. The ring can be roughly approximated as a circular rod bent into a circle (imagine a donut with a large hole in the middle). The ring measures 2 cm in diameter from the center of the rod on one side to the other. You make a perfect cut through one side of the ring and unfold it such that it is a perfectly straight rod with a circular cross section with a radius of 1 mm. You apply a uniaxial force of 100 N by pulling on either side of the rod with your fingers. Calculate the cross sectional area of the ring, the length of the unfurled rod, the stress induced by this force on the rod, the strain and corresponding length change. (2 pts each for 10 pts total)
Cross sectional area of the ring = Pi*(.001) 2 = 3.1416*10-6 m 2. Length of the rod = circumference of ring = 2*pi*(.01) =.0628 meters. Stress = F/A = 100 N/(3.1416*10-6 m 2 ) = 3.18*10 7 Pa Strain = Stress/Young s Modulus = 3.18*10 7 /80*10 9 =.0003978 Multiply strain by length of rod to get a change in length of.000025 meters or.025 mm. b. You pull harder and harder until the gold ring/rod begins to plastically deform at 200 MPa. What is the strain at the onset of plastic deformation? (1 pts) Strain = 200 MPa / 80 GPa =.0025 c. You ve been working out so you continue to increase the load on the rod until you reach the tensile strength of gold (350 MPa), and the material begins to neck. The strain at this point is 0.2. Compute the true stress and true strain. (2 pts) True sress = 350 MPa(1+.2) = 420 MPa True strain = ln(1+.2) =.182 d. You remember the time your significant other forgot your anniversary, so you get angry and pull even harder until the ring/rod fractures at 1.4 times its original length. Quantify the ductility of gold by calculating its percent elongation %EL. (1 pt) %EL = 100*(1.4*l 0 -l 0 )/l 0 = 40% e. Use the above information to draw a reasonable quantitative sketch of the engineering stress-strain curve for this gold ring similar to figure 6.11. Specifically indicate: The coordinates of the proportional limit, the tensile strength, and strain at the fracture point (does not need to be drawn to scale). (6 pts, total, 1 pt for each coordinate on the drawing and 2 pts for a reasonable drawing).
4. Stress and Strain in Composites Two materials 1 and 2 are joined together (such that they cannot come detached) in two different configurations, in series and in parallel, both shown below. The materials are equal in size and shape, but have unequal moduli of elasticity. In both situations, compressive stress of force F is applied to the center of both the top and bottom faces, which each have an area A. In the case of the materials in parallel, a rigid loading element is placed on the top and bottom faces to ensure uniform loading. 1 2 1 2 a) Write an expression for the total strain (ε) and total stress (σ) in both scenarios. (5 pts) In Series (due to force balance) The definition of strain is And the lengths in our system are defined as In Parallel (because they are held rigidly and must deform by the same amount) (okay to stop here) (but working through to here will make solving part b) easier) b) Determine the total effective modulus of elasticity (E) in both scenarios. (5 pts) In series
In Parallel (Divide each term by 2) Thus, c) Which of the two scenarios will have the higher the yield strength (σ y )? (Hint: begin by thinking about the yield strength of each material on its own and then by which material the yield strength in each scenario will be dominated.) (5 pts) In series, the yield strength will be limited by the material with the lower yield strength. Thus the yield strength will be higher for the scenario in parallel. d) Now let s consider the real material combinations. Let s consider only the situation in the materials are in series. If we consider two material combinations, Al/Ti and Ni/Steel, determine what the effective modulus of elasticity and effective yield strength of these two material combinations will be. (5 pts) Sketch qualitatively the stress-strain curves for both scenarios indicating the regions of elastic deformation and region of plastic deformation with relative slopes and relative yield strengths indicated on your sketch. (Physical property values for these materials can be found in Table 6.1 and 6.2 in your book). Modulus of Elasticity (GPa) Yield Strength (MPa) Al 69 35 Ni 207 138 Steel 207 180 Ti 107 450 For Al/Ti In the elastic deformation region, the slopes of the stress-strain curves are linear and are equal to the modulus of elasticity. Plugging into the expressions from part b. For the material in series, E= 84GPa. At some stress, the material will begin to deform plastically, so the stress-strain curve becomes non-linear. The stress at which the transition from elastic to plastic deformation occurs is the yield strength. For the blocks in series, the yield strength is 35 MPa.
For Ni/Steel For the material in series, E= 207 GPa. For the blocks in series, the yield strength is 138 MPa. 138 MPa 35 MPa.0415%.065% The Al/Ti curve has a lower slope (lower modulus of elasticity), and a lower yield strength. The Ni/steel curve has a higher slope (higher modulus of elasticity), and a higher yield strength. 5. Dislocations in Ceramics a) In general, is it easy for a dislocation to move in a ceramic material (as compared to a metal)? (2 pts) It is not easier for a dislocation to move in a ceramic material compared to a metal. b) Why is it easier/harder for a dislocation to move in covalent ceramics? (3 pts) It is harder for a dislocation to move in covalent ceramics due to the directional bonding present in covalent bonds. There is a very large energy cost to change the angle of these bonds. This increases the magnitude of the critical resolved shear stress required to break the covalent bonds allowing dislocation motion to occur. c) Why is it easier/harder for a dislocation to move in ionic ceramics? (3 pts) It is harder for a dislocation to move in ionic ceramics due to the repulsion between anions and cations. All ions want to have nearest neighbors of the opposite charge. For
a dislocation to move it needs to be able to maintain this keeping no anions neighboring anions or cations neighboring cations. In the following figure an extra half plane of atoms is present in the top half of the crystal. Parts d, e and f refer to this figure. d) Is the dislocation shown an edge or screw dislocation? (1 pts) This is an edge dislocation. e) If another identical sense dislocation is present close to this dislocation, how will these dislocations interact (i.e. what effect do forces between the dislocations have on the pair of dislocations)? (3 pts) The identical sense dislocations will feel a repulsive force between each other. If the conditions are such that the dislocations may move, the dislocations will move apart. f) If instead of the identical dislocation added in part e, an opposite sense dislocation (extra half plane of atoms in the bottom half of the crystal) is present. How will the dislocations interact? (2 pts) The opposite sense dislocations will feel an attractive force between each other. If conditions are such that the dislocations may move, the dislocations will move together. g) What happens if two opposite sense dislocations encounter each other while gliding along the same slip plane? (3 pts) The two dislocations will annihilate each other. h) Imagine a semiconductor device which performs better when there are fewer dislocations present. The semiconductor crystal has a distribution of different sense dislocations. It has been found that in this semiconductor crystal, dislocation motion can occur when heated to a designated temperature. What processing could be done in order to improve performance of the semiconductor? (3 pts) Heating the semiconductor crystal to allow the dislocations to move will cause similar sense dislocations to spread apart while opposite sense dislocations will move together. If this heating is done for sufficient time then opposite sense dislocations may encounter
each other and annihilate and similar sense dislocations may move to the edge of the crystal. This would lower the number of dislocations present in the semiconductor crystal making it perform better.