EN380 Homework #7 Solution
1. What are the criteria for two metals to form a perfect substitutional alloy (that is, these two will form
the same crystal phase at all potential concentrations of each species)?
• Atomic radii must not differ by more than 15%.
• Crystalline Structure must be the same (BCC and BCC, FCC and FCC, etc.).
• No appreciable difference in electronegativity.
• Elements should have the same valence state.
2. Sketch a qualitative plot of σY vs % elongation for the following steel microstructures:
(a) Ferrite (α)
(b) Bainite
(c) Martinsite
(d) Coarse Pearlite
(e) Fine Pearlite
1
3. The phase diagram for iron-iron carbide (F e - F e3 C) is shown below. For a 0.95% C steel (i.e. 1095
series) at a temperature just below the eutectoid temperature (727◦ C) determine:
(a) % C present in the ferrite (α). 0.02%
(b) % C present in the cementite (F e3 C). 6.67%
orange CF e3 C −C0 6.67%−0.95%
(c) total % of the steel that is ferrite (α). Wα = purple+orange = CF e3 C −Cα = 6.67%−0.02% = 86.02%
(d) total % of the steel that is cementite (F e3 C).
purple C0 −Cα 0.95%−0.02%
WF e3 C = purple+orange CF e3 C −Cα = 6.67%−0.02% = 13.98%
2
(e) % of steel that is pearlite.
This is the same as finding how much austenite, γ, was present just above the eutectoid temper-
ature. Draw the tie line just above 727◦ C.
Cα = 0.02% Cγ = 0.8%
green CF e3 C − C0 6.67% − 0.95%
Wγ = = = = 97.44%
red + green CF e3 C − Cγ 6.67% − 0.8%
(f) % of steel that is proeutectoid ferrite.
As C0 is greater than the eutectoid composition of 0.8%, there cannot be any α present above/before
the eutectoid transformation temperature. → 0%
(g) % of steel that is eutectoid ferrite.
As C0 is greater than the eutectoid composition of 0.8%, all of the α present must have formed
in the pearlite as eutectoid α ⇒ Wα, Eutectoid = Wα, T otal = 86.02% .
(h) % of steel that is proeutectoid cementite.
This is the same as finding how much cementite, F e3 C, was present just above the eutectoid
temperature
Draw the tie line just above 727◦ C.
Cα = 0.02% Cγ = 0.8%
red C0 − Cγ 0.95% − 0.8%
Wpro−eutectoid F e3 C = = = = 2.56%
red + green CF e3 C − Cγ 6.67% − 0.8%
(i) % of steel that is eutectoid cementite.
We already know how much cementite is present from part (d) and how much of this F e3 C was
proeutectoid from (h). Thus:
Weutectoid F e3 C = Wtotal F e3 C − Wproeutectoid F e3 C = 13.98% − 2.56% = 11.43%
(j) Compare this material qualitatively with steel with the eutectoid composition of 0.8%C (assuming
similar thermal history) in terms of:
i. Cementite content
This material has more Cementite than 1080 steel
3
ii. Strength (σY and σU T S )
With a higher cementite content (including in the grain boundaries) it would be expected to
be stronger (higher σY and σU T S ) than 1080 steel.
iii. Ductility
With a higher cementite content (including in the grain boundaries) it would be expected to
be less ductile/more brittle than 1080 steel.
iv. Toughness
Similarly it would be expected to be less tough (similar to ductility) than 1080 steel.
v. Hardness
Hardness should follow strength → harder than 1080 steel.
4
15
4. A steel plate is passed through rollers which reduce the thickness from 2 in to 1 16 in in a single pass.
What is the % C.W.? Qualitatively describe what happens to the yield strength of the plate as a result
and show this using a sketch of the σ − curve before and after rolling.
15 1
t0 − tf 2 in − 1 16 in in
% C.W. = · 100% = · 100% = 16 · 100% = 3.125% C.W.
t0 2 in 2 in
5. An aluminum rod of Diameter D = 2 in is extruded into the T-slotted below. What is the % C.W. if
the final profile area is A = 1.216 in2 .
π
A0 − Af 4 · D02 − Af
% C.W. = · 100% = π 2 · 100%
A0 4 · D0
π
4 · (2 in)2 − 1.216 in2
% C.W. = π 2
· 100%
4 · (2 in)
= 61.29% C.W.
5
6. Label the Eutectoid steel T T T diagram below to show:
(a) Regions where the stable phase is: (b) Cooling curves that result in:
i. Austenite i. 100% Coarse Pearlite
ii. Coarse Pearlite (α + F e3 C) ii. 50% Fine Pearlite + 50% Martinsite
iii. Fine Pearlite (α + F e3 C)
iii. 100% Martinsite
iv. Upper Bainite
v. Lower Bainite (For each case, label the quencing medium (or me-
vi. Martensite dia) used to obtain the final microstructure)
