SME
Materials Science & Material Properties · Problem type
Pearlite and ferrite in a steel
Use this for “A 0.40 wt% C steel is cooled slowly through 727 °C. What shares of pearlite and proeutectoid ferrite form?”
Diagram
Lever
rule:
WP
=
C₀ − Cα
CE − Cα
=
0.378
0.738
=
0.512.
Wα′ = 1 − WP = 0.488.
Each phase’s share is the arm on the far side of C₀.
Below 727 °C the austenite becomes pearlite (α + Fe₃C).
This is an example. Type your own values to start a new problem.
WP
=
C₀ (wt%)
−
0.022
wt%
0.76
wt%
−
0.022
wt%
Wα′
Proeutectoid ferrite share
calculated
Formulas
WP
=
C₀
− 0.022
0.76 − 0.022
0.5122
=
0.4 − 0.022
0.76 − 0.022
Wα′ = 1 − WP
0.4878 = 1 − 0.5122
Assumptions
•
The eutectoid is at 0.76 wt% C and 727 °C (some texts write 0.77 or 0.8).
•
Ferrite holds 0.022 wt% C at 727 °C; the austenite there all becomes pearlite.
•
The steel is hypoeutectoid and cooled slowly, so the phases stay near equilibrium.
Step-by-step
WE KNOW
C₀ = 0.4 wt%
FIND
pearlite share (WP), proeutectoid ferrite share (Wα′)
Step 1 · Find pearlite share (WP)
Use WP = (C₀ − 0.022) ÷ (0.76 − 0.022)
The
lever
rule:
the
pearlite
share
is
the
arm
from
the
far
end
to
C₀
,
over
the
whole
tie
line.
WP
=
0.4 − 0.022
0.76 − 0.022
WP
=
0.378
0.738
WP = 0.5122
Step 2 · Find proeutectoid ferrite share (Wα′)
Use Wα′ = 1 − WP
The two shares add to 1, so the ferrite is the rest.
Wα′ = 1 − 0.5122
Wα′ = 0.4878
Check
Put the numbers back in. Both sides should match.
0.5122
=
0.4 − 0.022
0.76 − 0.022
✓
0.4878 = 1 − 0.5122 ✓
Topic check
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