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MSE 230 Solutions for Assignment 11 Fall 2014 Problem 1* Gray iron is weak and brittle in tension because the tips of the graphite flakes act as points of stress concentration. Problem 2* a. Hardness testing or tensile testing. The yield strength (and hardness) of 5052 is typically greater than 6061. However, this is unreliable because you do not know if the material has undergone the standard strain hardening or heat treating condition. If nonstandard conditions were used 5052 may have a lower yield stress than 6061. b. X-ray diffraction. 5052 is a single phase alloy and will have one set of x-ray diffraction peaks. 6061 is a two-phase alloy and one set of peaks will correspond to the lattice parameter associated with the matrix while a second set of peaks will correspond to the second phase particles. Problem 3 This problem asks that we specify and compare the microstructures and mechanical properties in the heat-affected weld zones for 1080 and 4340 alloys assuming that the average cooling rate is 10°C/s. Figure 10.27 shows the continuous cooling transformation diagram for an iron-carbon alloy of eutectoid composition (1080), and, in addition, cooling curves that delineate changes in microstructure. For a cooling rate of 10°C/s (which is less than 35°C/s) the resulting microstructure will be totally pearlite--probably a reasonably fine pearlite. Figure 10.28 shows the CCT diagram for 4340 steel. From this diagram note that a cooling rate of 10°C/s produces a totally martensitic structure. Pearlite is softer and more ductile than martensite, and, therefore, is most likely more desirable in the case of a weld. Problem 4* This problem is concerned with the precipitation-hardening of copper-rich Cu-Be alloys.

(a) The range of compositions over which these alloys may be precipitation hardened is between approximately 0.2 wt% Be (the maximum solubility of Be in Cu at about 300°C) and 2.7 wt% Be (the maximum solubility of Be in Cu at 866°C).

(b) The heat treatment procedure, of course, will depend on the composition chosen. First of all, the solution heat treatment must be carried out at a temperature within the α phase region, after which, the specimen is quenched to room temperature. Finally, the precipitation heat treatment is conducted at a temperature within the α + γ2 phase region.

For example, for a 1.5 wt% Be-98.5 wt% Cu alloy, the solution heat treating temperature must be between about 600°C and 900°C, while the precipitation heat treatment

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would be below 600°C, and probably above 300°C. Below 300°C, diffusion rates are low, and heat treatment times would be relatively long.

Problem 5 From the class lecture notes

σc =E ⋅ α ⋅ ΔT(1 − ν)

=70,000MPa ⋅ 3 ⋅10−6 / °C( ) ⋅100°C

(1 − 0.2)= 26.3 MPa

Recall in class we assumed that glass had a strength (σc) of 70 MPa. The actual strength of glass is greatly affected by surface flaws (i.e. scratching glass lowers the strength making it easy to cut glass). Thus, the apparent decrease in strength (70 MPa to 26.3 MPa) with exuberant scrubbing was most likely due to scratches introduced during scrubbing. Critical flaw size: K = σ πa

ac =1π

KIC

σ# $ % &

' (

2

=1π

0.7 MPa m26.3 MPa

#

$ %

&

' (

2

= 2.25x10−4 m

or 225 µm to prevent this decrease in strength: • physically polish away the larger surface flaws • chemically etch the glass to smooth out the scratches • fire polishing: heat the glass, causing localized flow to smooth out the scratches Problem 6* There are a couple of ways to approach this problem. Approach 1: As the interior layer cools, it wants to shrink but is prevented from doing so by the already rigid outer layer. The amount of thermal contraction that the interior layer would undergo if unconstrained is given by: ε = αΔT For pyrex, ε = (3x10-6/°C) (680°C) = 2.04 x 10-3 Since the interior is constrained, the residual tensile stress is: σ = εE = (2.04 x 10-3) (70 x 103 MPa) σ = 143 MPa This residual tensile stress on the interior causes a larger surface compressive stress since the

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surface layer is thinner than the interior layer. The surface compressive stress can be estimated by: (thickness of exterior layer) σext. = (thickness of interior layer) σint. σext. =

(6 mm)(143 MPa)4 mm

σext. = 215 MPa Repeating this calculation for soda-lime glass gives: σint. = 476 MPa tensile stress σext. = 714 MPa compressive stress much larger due to the larger coefficient of thermal expansion. Approach 2: You should come up with a similar estimate (in fact somewhat larger) applying the equation for thermal stresses (from Prob. 2) to this situation.

.

Temperature Distribution

Position (mm)

Tem

pera

ture

(°

C)

20°C

700°C

2 8 10

Stress Distribution

10

Position (mm)

Stre

ss

(MPa

)

-714 MPa

-215 MPa

143 MPa

476 MPa

PyrexSoda-Lime Glass

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Problem 7 a) 60% of theoretical density → volume fraction of porosity = 0.40 From Chap. 12.11, equation 12.9, E = Eo (1-1.9P + 0.9P2) = 393 GPa (1-1.9(0.4) + 0.9(0.4)2) E = 151 GPa b) When the powder compact is placed in the furnace at 0.95Tm it will undergo densification via sintering (see Chap. 13.12). Therefore, the volume fraction of porosity will decrease and the modulus of the powder compact will increase (see Fig. 12.34). The plot of modulus of elasticity vs. time should be asymptotic to E0 because the modulus of elasticity of the powder compact will not exceed the Young's modulus of Al2O3.

Problem 8* Refer to the SiO2-Al2O3 phase diagram, Fig. 12.27 on page 480 of Callister. a) 70 wt.%Al2O3 - 30 wt.%SiO2 Minimum temperature to form liquid → 1588°C Weight fraction of liquid: First let's get the phases and phase compositions from the mullite and liquid two phase region: Mullite: 71 wt.%Al2O3 Liquid: 8 wt.%Al2O3

WL =CMUL − COCMUL − CL

=71 − 7071− 8

= 0.016

.

Mod

ulus

of E

last

icity

(GPa

) E0

Time

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or ≈1.6 wt.% liquid….sounds good for liquid phase sintering. b) 80 wt.%Al2O3 - 20 wt.% SiO2 Minimum temperature to form liquid → 1891°C. Phases and compositions: Alumina (Al2O3): 100 wt.% Al2O3 Liquid: ≈75 wt.% Al2O3

WL =CAl2 O3

− CO

CAl 2O3 − CL

=100 − 80100 − 75

= 0.80

or ≈ 80 wt.% liquid….sounds lousy for liquid phase sintering because most of the material present is liquid and very little is solid.