1 Demonstrate that the minimum cation to anion radius ratio for a coordination number of 8 is 0 732 Solution This problem asks us to show that the minimum cation to anion radius ratio for a coordination number of 8 is 0 732 From the cubic unit cell shown below the unit cell edge length is 2rA and from the base of the unit cell x 2 2rA 2 2 rA 2 8rA2 Or x 2rA 2 Now from the triangle that involves x y and the unit cell edge x 2 2rA 2 y 2 2 rA 2rC 2 2rA 2 2 4rA2 2 rA 2rC 2 Which reduces to 2rA 3 1 2rC Or rC rA 3 1 0 732 2 Compute the atomic packing factor for the rock salt crystal structure in which rC rA 0 414 Solution This problem asks that we compute the atomic packing factor for the rock salt crystal structure when rC rA 0 414 From Equation 3 2 APF VS VC With regard to the sphere volume VS there are four cation and four anion spheres per unit cell Thus 4 4 VS 4 rA3 4 rC3 3 3 But since rC rA 0 414 VS 16 3 r 1 0 414 3 17 94 rA3 3 A Now for rC rA 0 414 the corner anions in Table 12 2 just touch one another along the cubic unit cell edges such that VC a 3 2 rA rC 3 3 2 rA 0 414rA 22 62 rA3 Thus APF VS VC 17 94 rA3 22 62 rA3 0 79 3 A hypothetical AX type of ceramic material is known to have a density of 2 65 g cm3 and a unit cell of cubic symmetry with a cell edge length of 0 43 nm The atomic weights of the A and X elements are 86 6 and 40 3 g mol respectively On the basis of this information which of the following crystal structures is are possible for this material rock salt cesium chloride or zinc blende Justify your choice s Solution We are asked to specify possible crystal structures for an AX type of ceramic material given its density 2 65 g cm3 that the unit cell has cubic symmetry with edge length of 0 43 nm 4 3 10 8 cm and the atomic weights of the A and X elements 86 6 and 40 3 g mol respectively Using Equation 12 1 and solving for n yields n 2 65 g cm3 4 30 VC N A AC 10 8 cm 3 unit cell AA 6 022 1023 formula units mol 86 6 40 3 g mol 1 00 formula units unit cell Of the three possible crystal structures only cesium chloride has one formula unit per unit cell and therefore is the only possibility 4 A three point bending test is performed on a glass specimen having a rectangular cross section of height d 5 mm 0 2 in and width b 10 mm 0 4 in the distance between support points is 45 mm 1 75 in a Compute the flexural strength if the load at fracture is 290 N 65 lbf b The point of maximum deflection y occurs at the center of the specimen and is described by y FL3 48EI where E is the modulus of elasticity and I is the cross sectional moment of inertia Compute y at a load of 266 N 60 lbf Solution a For this portion of the problem we are asked to compute the flexural strength for a glass specimen that is subjected to a three point bending test The flexural strength Equation 12 7a is just fs 3F f L 2bd 2 for a rectangular cross section Using the values given in the problem statement fs 3 290 N 45 10 3 m 7 83 10 7 N m2 78 3 MPa 10 660 psi 2 10 10 3 m 5 10 3 m 2 b We are now asked to compute the maximum deflection From Table 12 5 the elastic modulus E for glass is 69 GPa 10 106 psi Also the moment of inertia for a rectangular cross section Figure 12 32 is just I bd 3 12 Thus y 4 69 FL3 FL3 bd 3 4Ebd 3 48E 12 266 N 45 10 3 m 3 10 10 3 m 5 10 3 m 3 10 9 N m2 7 0 10 5 m 7 0 10 2 mm 2 5 10 3 in 5 A three point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 3 5 mm 0 14 in the specimen fractured at a load of 950 N 215 lbf when the distance between the support points was 50 mm 2 0 in Another test is to be performed on a specimen of this same material but one that has a square cross section of 12 mm 0 47 in length on each edge At what load would you expect this specimen to fracture if the support point separation is 40 mm 1 6 in Solution For this problem the load is given at which a circular specimen of aluminum oxide fractures when subjected to a three point bending test we are then are asked to determine the load at which a specimen of the same material having a square cross section fractures It is first necessary to compute the flexural strength of the aluminum oxide Equation 12 7b and then using this value we may calculate the value of Ff in Equation 12 7a From Equation 12 7b fs Ff L R 3 950 N 50 10 3 m 352 10 6 N m2 352 MPa 50 000 psi 3 5 10 3 m 3 Now solving for Ff from Equation 12 7a realizing that b d 12 mm yields Ff 2 fs d 3 3L 2 352 106 N m2 12 10 3 m 3 10 100 N 2165 lb f 3 40 10 3 m 6 Cite the two desirable characteristics of glasses Solution Two desirable characteristics of glasses are optical transparency and ease of fabrication 7 a Explain why residual thermal stresses are introduced into a glass piece when it is cooled b Are thermal stresses introduced upon heating Why or why not Solution a Residual thermal stresses are introduced into a glass piece when it is cooled because surface and interior regions cool at different rates and therefore contract different amounts since the material will experience very little if any deformation stresses are established b Yes thermal stresses will be introduced because of thermal expansion upon heating for the same reason as for thermal contraction upon cooling 8 Borosilicate glasses and fused silica are resistant to thermal shock Why is this so Solution Borosilicate glasses and fused silica are resistant to thermal shock because they have relatively low coefficients of thermal expansion therefore upon heating or cooling the difference in the degree of expansion or contraction across a cross section of a ware that is constructed from these materials will be relatively low 9 Compute repeat unit …
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