SME
Aerospace Structures · Problem type
Basquin’s law: cycles to failure
Use this for “An aluminum part has σ′f = 1000 MPa and b = −0.1. How many cycles does it last at a stress amplitude of 300 MPa?”
Diagram
σₐ
=
σ′f
(2N)
b
:
2N
=
(
300
1,000
)
1 ÷ −0.1
=
169,000
reversals,
N
=
84,700
cycles.
On log–log axes the power law is a straight line, of slope b.
This is an example. Type your own values to start a new problem.
σₐ (MPa)
=
σ′f (MPa)
×
(
)
N
=
2N
2
Formulas
σₐ
=
σ′f
×
(2
N
)
b
300
=
1,000
×
(169,400)
−0.1
N
=
2
N
2
84,680
=
169,400
2
Assumptions
•
Fully reversed stress of amplitude σₐ, the same every cycle, below yield.
•
Basquin’s law for long lives; aluminum has no endurance limit to stop the line.
•
One cycle is two reversals, so N is half of 2N.
Step-by-step
WE KNOW
σ′f = 1,000 MPa, b = −0.1, σₐ = 300 MPa
FIND
reversals to failure (2N), cycles to failure (N)
Step 1 · Find reversals to failure (2N)
Use σₐ = σ′f × (2N)ᵇ
Divide
by
σ′f,
then
raise
to
the
power
1
b
to
undo
the
exponent.
2
N
=
(
σₐ
σ′f
)
1 ÷
b
2
N
=
(
300
1,000
)
1 ÷ (−0.1)
2
N
=
0.3
-10
2
N
=
169,400
Step 2 · Find cycles to failure (N)
Use N = (2N) ÷ 2
Each cycle is two reversals.
N
=
169,350.8781
2
N
=
84,680
Check
Put the numbers back in. Both sides should match.
300
=
1,000
×
(169,400)
−0.1
✓
84,680
=
169,350
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.