MSE 230 Assignment 4 Solutions Fall 2013 1 Find all 110 and 211 in 111 Confirm your answers using the dot product 110 11 0 211 11 2 1 1 0 0 1 1 0 11 1 0 1 1 0 1 1 1 2 2 11 2 1 1 1 2 1 1 2 1 z z 111 showing 110 111 showing 211 y x y x 2 a The key to doing this problem is to recognize that the lattice parameter is always the same even though the interplanar spacing changes for different sets of planes Use the 200 peak found in both FCC and BCC to calculate the lattice parameter 2dsin d 0 154nm 0 157nm 2sin 58 7 2 Since the crystal structure is cubic a d 2 2 0 2 0 2 0 314 nm If the metal has a FCC structure the peaks at 40 5 and 73 8 two theta will have Miller s indicies of 111 and 220 respectively h k l all odd or all even If the metal has a BCC structure the Miller s indicies will be 110 and 211 h k l even For the peak at 40 5 two theta d 0 154nm 0 222nm 2sin 40 5 2 For FCC a d 12 12 12 0 385 nm for BCC a d 12 12 0 2 0 315 nm Since the lattice parameter matches for a 110 Miller s index the metal has a BCC structure and the peaks at 40 5 and 73 8 two theta have Miller s indicies of 110 and 211 respectively b From above a 0 314 nm For a BCC crystal structure the relationship between the 3a lattice parameter and the atomic radius is r 0 136 nm This means that the 4 unknown metal is molybdenum 2a 0 111 nm Since this does not 4 match any of the metals it should be a tip that there was a problem with your conclusion above If you chose a FCC crystal structure above r 3 14 5 a From the tabulated data we are asked to compute M the number average n molecular weight This is carried out below Molecular wt Range Mean M 8 000 16 000 i xM i i 12 000 0 05 600 16 000 24 000 20 000 0 16 3200 24 000 32 000 28 000 0 24 6720 32 000 40 000 36 000 0 28 10 080 40 000 48 000 44 000 0 20 8800 48 000 56 000 52 000 0 07 3640 i x M n x M 33 040 g mol i i b From the tabulated data we are asked to compute M w the weight average molecular weight Molecular wt Range Mean M 8 000 16 000 12 000 0 02 240 16 000 24 000 20 000 0 10 2000 24 000 32 000 28 000 0 20 5600 32 000 40 000 36 000 0 30 10 800 40 000 48 000 44 000 0 27 11 880 48 000 56 000 52 000 0 11 5720 w i wM i i i M w w M 36 240 g mol i i 4 14 22 The tendency of a polymer to crystallize decreases with increasing molecular weight because as the chains become longer it is more difficult for all regions along adjacent chains to align so as to produce the ordered atomic array 5 a We are asked to compute the densities of totally crystalline and totally amorphous polyethylene and from Equation 14 8 From Equation 14 8 let C c a crystallinity 100 such that c s a s c a C Rearrangement of this expression leads to c C s s c a C s a 0 in which and are the variables for which solutions are to be found Since two values c a of and C are specified in the problem two equations may be constructed as follows s c C1 s1 s1 c a C 1 s1 a 0 c C2 s2 s2 c a C 2 s2 a 0 In which s1 3 0 965 g cm s2 0 925 g cm 3 C 0 768 and C 0 464 Solving the 1 2 above two equations leads to s1 s2 C1 C 2 C1 s1 C 2 s2 a 0 965 g cm3 0 925 g cm3 0 768 0 464 0 768 0 965 g cm3 0 464 0 925 g cm3 3 0 870 g cm And c 1 s1 C1 1 s1 s2 C2 C1 s2 C2 0 965 g cm3 0 925 g cm3 0 464 0 768 0 925 g cm3 0 464 1 0 0 965 g cm3 0 768 1 0 b Now we are to determine the crystallinity for s 3 0 950 g cm Again using Equation 14 8 crystallinity c s a x 100 s c a 0 998 g cm3 0 950 g cm3 0 950 g cm3 0 998 g cm3 65 7 3 0 998 g cm 0 870 g cm3 0 870 g cm3 x 100 x 85 MPa y 140 MPa z 0 0 33 E Al 70 GPa 70 x 103 MPa 6 y x x z E E E 140 MPa 85 MPa 0 33 0 70 000 MPa 70 000 MPa x 0 0005543 y y x z E E E y 0 0016 z y z x E E E z 0 001061 a AB y 25 cm 0 04 cm b CD x 25 cm 0 0139 cm c thickness z 2 cm 0 0021 cm the plate gets thinner during elastic deformation d Vo 40 cm x 40 cm x 2 cm 3200 cm3 Vf 40 cm x 40 cm x 40 cm y 40 cm x 2 cm z 2 cm 40 0 0005543 40 x 40 0 0016 40 x 2 0 001061 2 Vf 40 0222 cm x 40 064 cm x 1 99788 cm Vf 3203 5 cm3 V 3 5 cm3 Note that volume is not conserved during elastic deformation for 0 33
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