SME
Materials Science & Material Properties · Problem type
Steady creep rate at a new stress and temperature
Use this for “A steel creeps at 2 × 10⁻⁸ s⁻¹ under 100 MPa at 900 K (n = 5, Q = 250 kJ/mol). Find its rate under 120 MPa at 950 K.”
Chart
ε̇₂(σ) = 2 × 10⁻⁸·5.804·(σ/100)⁵
Log–log axes: a power law draws as a straight line.
This is an example. Type your own values to start a new problem.
ε̇₁
Known rate
example · s⁻¹
σ₁
Known stress
example · MPa
T₁
Known temperature
example · K
σ₂
New stress
example · MPa
T₂
New temperature
example · K
n
Stress exponent
example
Q
Activation energy
example · kJ/mol
ε̇₂
New rate
calculated · s⁻¹
Formulas
ε̇₂
=
ε̇₁
×
(
σ₂
σ₁
)
n
×
e^((
Q
×
1,000
8.314
)
×
(
1
T₁
−
1
T₂
))
2.8882 ×
10
-7
=
2 ×
10
-8
×
(
120
100
)
5
×
e^((250
×
1,000
8.314
)
×
(
1
900
−
1
950
))
Assumptions
•
Steady (secondary) creep by one mechanism, so A, n and Q stay the same.
•
Temperatures are in kelvins; R = 8.314 J/(mol·K), so Q in kJ/mol is multiplied by 1000.
•
Stresses share one unit; only their ratio enters.
Step-by-step
WE KNOW
σ₂ = 120 MPa, T₂ = 950 K, ε̇₁ = 2 × 10⁻⁸ s⁻¹, σ₁ = 100 MPa, T₁ = 900 K, n = 5, Q = 250 kJ/mol
FIND
new rate (ε̇₂)
Step 1 · Find new rate (ε̇₂)
Use ε̇₂ = ε̇₁ × (σ₂ ÷ σ₁)ⁿ × e(Q × 1,000 ÷ 8.314) × (1 ÷ T₁ − 1 ÷ T₂)
Dividing
the
two
rates
cancels
A:
scale
ε̇₁
by
the
stress
ratio
to
the
n
and
the
Arrhenius
factor.
ε̇₂
=
2 ×
10
-8
×
(
120
100
)
5
×
e^((250
×
1,000
8.314
)
×
(
1
900
−
1
950
))
ε̇₂
=
(2
×
10
-8
)
×
(
120
100
)
5
×
e
250 × 1,000 ÷ 8.314 × (0.00111111 − 0.00105263)
ε̇₂
=
(2
×
10
-8
)
×
(
120
100
)
5
×
e
250 × 1,000 ÷ 8.314 × (5.848 ×
10
-5
)
ε̇₂
=
(2
×
10
-8
)
×
1.2
5
×
e
250,000 ÷ 8.314 × (5.848 ×
10
-5
)
ε̇₂
=
(2
×
10
-8
)
×
1.2
5
×
e
30,069.8 × (5.848 ×
10
-5
)
ε̇₂
=
(2
×
10
-8
)
×
1.2
5
×
e
1.75848
ε̇₂
=
(2
×
10
-8
)
×
2.48832
×
5.80361
ε̇₂
=
(4.97664
×
10
-8
)
×
5.80361
ε̇₂
=
2.8882 ×
10
-7
s
-1
Check
Put the numbers back in. Both sides should match.
2.8882 ×
10
-7
=
2 ×
10
-8
×
(
120
100
)
5
×
e^((250
×
1,000
8.314
)
×
(
1
900
−
1
950
))
✓
Topic check
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