Introduction to Structure and Properties Winter 2005 Final Exam March 17, 2005 TOTAL POINTS 37

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1 Materials 101 Introduction to Structure and Properties Winter 005 Final Exam March 17, 005 Solutions TOTAL POINTS 37 Problem 1: Tensile Test and Plastic Deformation (10 Points) A copper rod is deformed using a uniaxial tensile force of 16,000 N. Deformation continues until sufficient strain hardening has occurred such that the applied force is too small to allow further deformation. After deformation, the rod has a diameter of 0.01 m and a length of 1.5 m. Assume that copper follows the strain hardening law σ = Kε n with K = 310 MPa, and n = In the following, do not forget to state the units. (a) Briefly explain why many ductile materials exhibit an increase in yield stress with plastic deformation when deformed at room temperature. During plastic deformation at room temperature, many ductile materials exhibit and increase in dislocation density. As the dislocation density increases, interaction between dislocations inhibits dislocation motion, leading to an increase in yield strength. To get full credit the dislocation mechanisms had to mentioned. (b) Calculate the true stress that caused the deformation. σ T = F A = 16,000 π( 0.005m) = 03MPa (-1 point for wrong units) 1

2 (c) Calculate the true strain after deformation. lnσ T = ln K + n lnε T lnε T = lnσ T ln K = n ε T = 0.46 ln(03) ln(310) 0.54 (-1 point for wrong units) (d) Calculate the initial length of the rod before deformation began. ε T = ln l 0.46 = ln = exp(0.46) = 0.95m (-1 point for wrong units) (e) Suggest a way to further deform the rod using the same applied force. Heat the rod to a temperature at which recrystallization occurs. Note: it was stated that the same applied force (i.e. tensile) was to be used. Problem 3: Single Crystal Slip (8 Points) The slip systems in FCC are of type {111}<110> and the total number of slip systems is 1. (a) List the Miller indices of four unique slip planes in FCC. ( 111), ( 1 11),11 ( 1),111 ( )

3 (b) A single crystal of a FCC metal is loaded in tension along the [011] direction. Which slip systems do you expect to exhibit a non-zero resolved shear stress? In other words, for which slip systems is slip possible under the given loading conditions? Loaded in [011] direction. ( ) and ( 11 1 ) give zero resolved shear stress because cosφ=0 (the dot product is zero in => each case) This leaves the ( 111) and ( 1 11) planes. The ( 111) plane contains [ 1 10], [ 1 01],01 [ 1]. The ( 1 11) plane contains [ 110], [ 101],01 [ 1] but cos λ =0 for [ 0 1 1],01 [ 1] This leaves the following slip systems: ( 111)1 [ 10], ( 111)1 [ 01], ( 1 11)110 [ ], ( 1 11)101 [ ] 6 points Problem 5: Mechanisms of Slip (3 Points) Shown below is a schematic of a dislocation. (a) Indicate the Burgers vector and line direction of the dislocation. (b) A shear stress that is of sufficient magnitude to induce dislocation glide is applied as shown. Indicate the direction of dislocation motion. τ 3

4 motion b τ l 3 points Problem 6: Phase Diagrams and Strengthening of Metal Alloys (8 points) Shown below is the Ag-Cu phase diagram. 4

5 (a) State the main characteristics (structure, composition) of the phases that are labeled by α and β. Both are solid solutions and both have FCC structure. The α phase is Ag rich with a maximum solubility of 8.8 % for Cu at the eutectic temperature. The β phase is Cu-rich with a maximum solubility of 8% of Ag at the eutectic temperature. (b) State the eutectic temperature and concentration. 8.1 wt% Cu T e = 783 C (c) Sterling silver, an alloy containing approximately 9.5 wt% silver, is held for some time at a temperature in the α phase field, and then rapidly quenched to room temperature. What microstructure is likely? By what mechanisms is the quenched alloy strengthened? This will give single phase α phase that is supersaturated with Cu. Fast cooling does not allow precipitates to form. This alloy will be solid solution strengthened. Note that martensite is only ever found in Fe-C alloys. (d) Sterling silver can be heat treated to produce β particles in a matrix of α. However, little strengthening occurs, compared to, for example, the strengthening effect of Fe 3 C particles in BCC iron. Why is there a difference? Since both phases are FCC, there is not much effect in terms of hindering dislocation motion. The dislocations can easily cut through (same slip systems in precipitates and matrix). In contrast, Fe 3 C particles have a very different crystal structure from BCC and are also very hard (dislocations cannot shear through them). 5

6 Note that in both cases fine microstructure can be produced, so this was not the correct answer. Problem 7: Brittle Fracture (8 points) A thick plate of Al O 3 (alumina) contains an embedded elliptical crack (geometry factor = 1) and is broken in uniform tension (mode I loading). The fracture stress was 1 MPa and the critical crack size (c) was 60 mm. The Young s modulus of Al O 3 is 400 GPa. (a) Determine the fracture toughness. Do not forget to state the units. K I c = σ πc K I c =1MPa π m = 3.68MPa m (-1 point for incorrect units) (b) Determine the fracture surface energy. Do not forget to state the units. This problem could be solved with the equations on the formula sheet. K I c = γe γ = K I c E =17 J m (-1 point for incorrect units) (c) Explain, qualitatively, why is alumina brittle? Alumina has a very high yield strength and the cleavage surface energy is the only term providing resistance against fracture. (d) Explain why it is not good engineering practice to use the maximum tensile stress when designing mechanically reliable parts made of ceramics such as alumina. Flaw sizes vary and cause a large variation in the maximum tensile strength. 6

7 Potentially Useful Equations K I = σ πc σ γe πc cosθ = u 1 u + v 1 v + w 1 w u 1 + v 1 + w 1 u + v + w τ = σ cosθ cosφ lnε = l σ = F A A = πr 7