SME
Materials Science & Material Properties · Problem type
Fatigue crack growth (Paris law)
Use this for “A plate with a 1 mm edge crack (Y = 1.12, KIc = 50 MPa√m) is cycled from 0 to 150 MPa. With C = 10⁻¹² and m = 3, how many cycles until it breaks?”
Chart
Level af = 28.19 meets the curve at Nf = 1,944,000
This is an example. Type your own values to start a new problem.
af (mm)
=
1
π
×
KIc (MPa√m)
×
Δσ (MPa)
2
Y is the geometry factor for the crack’s shape and place (1.12 for an edge crack); af is the length at which K reaches KIc and the plate breaks.
C
Paris constant
example
m
Paris exponent
example
a₀
Starting crack
example · mm
Nf
Cycles to failure
calculated · cycles
Formulas
af
=
1,000
×
(KIc ÷ (
Y
× Δσ))²
π
28.195
=
1,000
×
(50 ÷ (1.12 × 150))²
π
Nf
=
(
(
af
1,000
)
1 −
m
÷ 2
−
(
a₀
1,000
)
1 −
m
÷ 2
)
÷
((1
−
m
2
)
×
C
×
(
Y
× Δσ ×
√
π
)
m
)
1.9443 ×
10
6
=
(
(
28.195
1,000
)
1 − 3 ÷ 2
−
(
1
1,000
)
1 − 3 ÷ 2
)
÷
((1
−
3
2
)
×
(1
×
10
-12
)
×
(1.12
×
150
×
√
π
)³)
Assumptions
•
Y stays the same as the crack grows.
•
The load cycles from 0 to σₘₐₓ (R = 0), so Δσ = σₘₐₓ and ΔK = Kₘₐₓ.
•
m ≠ 2, so the growth law integrates to the power form shown.
•
C goes with crack lengths in m and ΔK in MPa√m; the steps turn mm into m.
Step-by-step
WE KNOW
a₀ = 1 mm, Δσ = 150 MPa, C = 1 × 10⁻¹², m = 3, Y = 1.12, KIc = 50 MPa√m
FIND
final crack (af), cycles to failure (Nf)
Step 1 · Find final crack (af)
Use af = 1,000 × (KIc ÷ (Y × Δσ))² ÷ π
The
crack
breaks
the
part
when
YΔσ
√
πa
reaches
KIc;
×
1000
gives
mm.
af
=
1,000
×
(50 ÷ (1.12 × 150))²
π
af
=
1,000
×
(
50
168
)
2
÷
π
af
=
1,000
×
0.2976
2
π
af
=
1,000
×
0.08857
π
af
=
88.57
π
af = 28.195 mm
Step 2 · Find cycles to failure (Nf)
Use Nf = ((af ÷ 1,000)1 − m ÷ 2 − (a₀ ÷ 1,000)1 − m ÷ 2) ÷ ((1 − m ÷ 2) × C × (Y × Δσ × √(π))ᵐ)
Separate
da/dN
=
C
(YΔσ
√
πa
)
m
and
add
up
the
cycles
from
a₀
to
af.
Nf
=
(
(
28.195
1,000
)
1 − 3 ÷ 2
−
(
1
1,000
)
1 − 3 ÷ 2
)
÷
((1
−
3
2
)
×
10
-12
×
(1.12
×
150
×
√
π
)³)
Nf
=
(
0.028195
−0.5
−
0.001
−0.5
)
÷
(−0.5
×
10
-12
×
297.77
3
)
Nf
=
5.9554 − 31.623
−0.5 ×
10
-12
× (2.6403 ×
10
7
)
Nf
=
−25.667
−1.3201 ×
10
-5
Nf
=
1.9443 ×
10
6
cycles
Check
Put the numbers back in. Both sides should match.
28.195
=
1,000
×
(50 ÷ (1.12 × 150))²
π
✓
1.9443 ×
10
6
=
(
(
28.195
1,000
)
1 − 3 ÷ 2
−
(
1
1,000
)
1 − 3 ÷ 2
)
÷
((1
−
3
2
)
×
10
-12
×
(1.12
×
150
×
√
π
)³)
✓
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