工程材料科学和设计答案.doc

Problems - Chapter 5 1. FIND: Calculate the stress on a tensioned fiber. GIVEN: The fiber diameter is 25 micrometers. The elongational load is 25 g. ASSUMPTIONS: The engineering stress is requested. DATA: Acceleration due to gravity is 9.8 m/sec2. A Newton is a kg-m/sec2. A Pascal is a N/m2. A MPa is 106 Pa. SOLUTION: Stress is force per unit area. The cross-sectional area is SYMBOL 112 \f SymbolR2 = 1963.5 square micrometers. The force is 25 g (kg/1000g)(9.8 m/sec2) = 0.245 N. Thus, the stress is SYMBOL 115 \f Symbol = F/A = COMMENTS: You must learn to do these sorts of problems, including the conversions. 2. GIVEN: FCC Cu with ao = 0.362nm REQUIRED: A) Lowest energy Burgers vector, B) Length in terms of radius of Cu atom, C) Family of planes SOLUTION: We note that the Burgers vector is the shortest vector that connects crystallographically equivalent positions. A diagram of the structure is shown below: FCC structure with (111) shown We note that atoms lying along face diagonals touch and are crystallographically equivalent. Therefore, the shortest vector connecting equivalent positions is ? face diagonal. For example, one such vector is as shown in (111). A. The length of this vector is B. By inspection, the size of the vector is 2 Cu atom radii. C. Slip occurs in the most densely packed plane which is of the type {111}. These are the smoothest planes and contain the smallest Burgers vector. This means that the dislocations move easily and the energy is low. 3. GIVEN: ?b? = 0.288nm in Ag REQUIRED: Find lattice parameter SOLUTION: Recall the Ag is FCC. For FCC structures the Burgers vector is ? a face diagonal as shown. We see that 4. A. FCC structure The (111) plane is shown in a unit cell with all atoms shown. Atoms touch along face diagonals. The (111) plane is the most closely packed, and the vectors shown connect equivalent atomic position. Thus etc. Then in general B. For