SME
Materials Science & Material Properties
Fracture and creep
Use this for “A creep test at 1000 K breaks after 1000 h. With C = 20, how long will the same stress last at 950 K?”
Chart
tᵣ(T) = 1023,000/T − 20
A log y-axis: each decade gets the same room.
This is an example. Type your own values to start a new problem.
LMP
=
T₁ (K)
(
C
+
log
10
(
t₁ (h)
))
LMP
=
T₂ (K)
(
C
+
log
10
(
t₂ (h)
))
Formulas
LMP
=
T₁
×
(
C
+
log₁₀(
t₁
))
23,000 = 1,000 × (20 + log₁₀(1,000))
LMP
=
T₂
×
(
C
+
log₁₀(
t₂
))
23,000 = 950 × (20 + log₁₀(16,237.7674))
Assumptions
•
The test and the service carry the same stress, so they share one LMP.
•
Temperatures are absolute, in kelvins (°C + 273); times are in hours.
•
C ≈ 20 fits most steels and superalloys.
Step-by-step
WE KNOW
T₂ = 950 K, T₁ = 1,000 K, t₁ = 1,000 h, C = 20
FIND
Larson–Miller parameter (LMP), service life (t₂)
Step 1 · Find Larson–Miller parameter (LMP)
Use LMP = T₁ × (C + log₁₀(t₁))
The
test
temperature
times
C
plus
the
log
of
its
life
in
hours.
LMP = 1,000 × (20 + log₁₀(1,000))
LMP = 1,000 × (20 + 3)
LMP = 1,000 × 23
LMP = 23,000
Step 2 · Find service life (t₂)
Use LMP = T₂ × (C + log₁₀(t₂))
Divide
LMP
by
T
to
get
C
+
log₁₀
t,
take
away
C
,
then
raise
10
to
that
power.
t₂
=
10
LMP ÷
T₂
−
C
t₂
=
10
23,000 ÷ 950 − 20
t₂
=
10
24.2105 − 20
t₂
=
10
4.2105
t₂
=
16,237.77
h
Check
Put the numbers back in. Both sides should match.
23,000 = 1,000 × (20 + log₁₀(1,000)) ✓
23,000 = 950 × (20 + log₁₀(16,237.77)) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
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