MSE 230 Assignment 4 Solutions Fall 2014 1.* Find all <110> and <211> in 111( ) . Confirm your answers using the dot product.

<110> <211> 1 1 0[ ] 11 2 [ ] 1 1 0[ ] 1 1 2[ ] 0 1 1[ ] 2 11[ ] 0 11 [ ] 2 1 1 [ ] 1 0 1[ ] 1 2 1[ ] 1 0 1 [ ] 1 2 1 [ ]

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.

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λ = 2dsinθ d =0.154nm2sin 58.7 2( )

= 0.157nm

Since the crystal structure is cubic:

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a = d 22 + 02 + 02 = 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:

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d =0.154nm

2sin 40.5 2( )= 0.222nm

z

x

y

(111) showing <110>z

x

y

(111) showing <211>

For FCC,

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a = d 12 +12 +12 = 0.385 nm , for BCC,

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a = d 12 +12 + 02 = 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 lattice

parameter and the atomic radius is:

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r =3a4

= 0.136 nm. This means that the unknown metal is

molybdenum.

If you chose a FCC crystal structure above,

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r =2a4

= 0.111nm. Since this does not match any of

the metals, it should be a tip that there was a problem with your conclusion above.

3)* (a) From the tabulated data, we are asked to compute M_

n, the number- average molecular weight.

This is carried out below.

Molecular wt Range Mean Mi xi xiMi

8,000-16,000 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

____________________________

M_

n = ΣxiMi = 33,040 g/mol

(b) From the tabulated data, we are asked to compute M_

w, the weight- average molecular weight.

Molecular wt. Range Mean Mi wi wiMi

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

___________________________

M_

w = ΣwiMi = 36,240 g/mol

4. Make a qualitatively correct of the molecular weight distributions for two batches of polyethylene that were polymerized under different conditions. Batch 1 was polymerized such that

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M w >> M n and Batch 2 was polymerized such that

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M w ≈ M n .

See Figure 14.4 in Callister. As we discussed in class, when shorter and longer polymer chains are mixed together, the longer polymer chains skew the weight average molecular weight Mw( ) to the right

compared to the number average molecular weight Mn( ) . Therefore, when there is a broad molecular weight distribution, as on the left,

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M w >> M n . When all of the chains are of similar length,

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M w ≈ M n , and the molecular weight distribution is relatively narrow.

Mn

Mw

5)* 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.

6)* (a) We are asked to compute the densities of totally crystalline and totally amorphous polyethylene [ρc and ρa from Equation (14.8)]. From Equation (14.8) let C = (% crystallinity)/100, such that

C = ρc ρs − ρa( )ρs ρc − ρa( )

Rearrangement of this expression leads to

ρc Cρs − ρs( ) + ρcρa − Cρsρa = 0

in which ρc and ρa are the variables for which solutions are to be found. Since two values of ρs and

C are specified in the problem, two equations may be constructed as follows:

ρc C1ρs1 − ρs1( ) + ρcρa − C1ρs1ρa = 0

ρc C2 ρs2 − ρs2( ) + ρcρa − C2 ρs2 ρa = 0

In which ρs1 = 0.965 g/cm3, ρs2 = 0.925 g/cm3, C1 = 0.768, and C2 = 0.464. Solving the above

two equations leads to

ρa = ρs1ρs2 C1 − C2( )C1ρs1 − C2 ρs2

= 0.965 g/cm3( ) 0.925 g/cm3( ) 0.768 − 0.464( )

(0.768) 0.965 g/cm3( ) − (0.464) 0.925 g/cm3( ) = 0.870 g/cm3

And

ρc = ρs1ρs2 C2 − C1( )

ρs2 C2 − 1( ) - ρs1 C1 − 1( )

= 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) = 0.998 g/cm3

(b) Now we are to determine the % crystallinity for ρs = 0.950 g/cm3. Again, using Equation (14.8)

% crystallinity = ρc ρs − ρa( )ρs ρc − ρa( )

x 100

= 0.998 g/cm3( ) 0.950 g/cm3 − 0.870 g/cm3( )0.950 g/cm3( ) 0.998 g/cm3 − 0.870 g/cm3( )

x 100

= 65.7% 7. σx = 85 MPa, σy = 140 MPa, σz = 0, ν = 0.33 E(Al) = 70 GPa = 70 x 103 MPa First we need to determine the elastic strain in the x, y and z directions. We can then take these strains and determine: a) the change in the circle diameter in the y-direction by multiplying the diameter by the strain in the y-direction, b) the change in the circle diameter in the x-direction by multiplying the diameter by the strain in the x-direction, c) the change in the plate thickness by multiplying the plate thickness by the strain in the z-direction, and d) determine the change in the plate volume by taking the original volume (40x40x2 cm3) and subtracting it from the volume of the plate under the applied stress.

εx =

σxE

− υσyE− υ

σzE⇒

=85 MPa

70,000 MPa− 0.33 140 MPa

70,000 MPa& '

( ) − 0

εx = 0.0005543

εy = σyE− υ

σxE

− υσzE

εy = 0.0016

εz = σzE− υ

σxE

− υσyE

εz = -0.001061 a) ΔAB = εy (25 cm) = 0.04 cm b) ΔCD = εx (25 cm) = 0.0139 cm c) Δ thickness = εz (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 + εx(40 cm)[ ]x 40 cm + εy (40 cm)[ ]x 2 cm + εz (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

ΔV = 3.5 cm3

Note that volume is not conserved during elastic deformation for ν = 0.33. If you redid the calculation for ν = 0.33 you should find that V0 and Vf are the same.