SME
Aerospace Structures
Fatigue
Use this for “A wing fitting cycles between 200 and 40 MPa. With Sₑ = 300 MPa and Sᵤ = 600 MPa, what is its Goodman safety factor?”
Diagram
Goodman:
σₐ
Sₑ
+
σₘ
Sᵤₜ
=
80
300
+
120
600
=
1
n
,
so
n
=
2.14.
The load line from the origin meets the Goodman line 2.14 times as far out as the point: the part is safe.
This is an example. Type your own values to start a new problem.
σₘ (MPa)
=
σₘₐₓ (MPa)
+
σₘᵢₙ (MPa)
2
σₐ (MPa)
=
σₘₐₓ (MPa)
−
σₘᵢₙ (MPa)
2
1
n
=
σₐ (MPa)
Sₑ (MPa)
+
σₘ (MPa)
Sᵤ (MPa)
Formulas
σₘ
=
σₘₐₓ + σₘᵢₙ
2
120
=
200 + 40
2
σₐ
=
σₘₐₓ − σₘᵢₙ
2
80
=
200 − 40
2
1
n
=
σₐ
Sₑ
+
σₘ
Sᵤ
1
2.143
=
80
300
+
120
600
Assumptions
•
Sₑ is the fatigue strength at the design life; aluminum has no endurance limit.
•
The modified Goodman line runs from Sₑ on the σₐ axis to Sᵤ on the σₘ axis.
•
Every cycle is the same, with a tensile or zero mean stress, below yield.
Step-by-step
WE KNOW
σₘₐₓ = 200 MPa, σₘᵢₙ = 40 MPa, Sₑ = 300 MPa, Sᵤ = 600 MPa
FIND
mean stress (σₘ), alternating stress (σₐ), safety factor (n)
Step 1 · Find mean stress (σₘ)
Use σₘ = (σₘₐₓ + σₘᵢₙ) ÷ 2
The mean stress is halfway between the largest and smallest stress.
σₘ
=
200 + 40
2
σₘ
=
240
2
σₘ = 120 MPa
Step 2 · Find alternating stress (σₐ)
Use σₐ = (σₘₐₓ − σₘᵢₙ) ÷ 2
The alternating stress is half the range from σₘᵢₙ to σₘₐₓ.
σₐ
=
200 − 40
2
σₐ
=
160
2
σₐ = 80 MPa
Step 3 · Find safety factor (n)
Use 1 ÷ n = σₐ ÷ Sₑ + σₘ ÷ Sᵤ
Each
stress
over
its
strength
is
the
share
of
the
Goodman
line
used;
n
is
one
over
their
sum.
n
=
1
σₐ ÷ Sₑ + σₘ ÷ Sᵤ
n
=
1
80 ÷ 300 + 120 ÷ 600
n
=
1
0.2667 + 0.2
n
=
1
0.4667
n
=
2.143
Check
Put the numbers back in. Both sides should match.
120
=
200 + 40
2
✓
80
=
200 − 40
2
✓
1
2.143
=
80
300
+
120
600
✓
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Miner’s rule: damage from two load blocks
Topic check
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More problem types
Miner’s rule: damage from two load blocks
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