SME
Machine Design
Fatigue
Use this for “A part’s stress cycles from 40 MPa to 200 MPa; Sᵤₜ = 600 MPa, Marin product 0.7. Find the Goodman factor of safety.”
Diagram
Goodman:
σₐ
Sₑ
+
σₘ
Sᵤₜ
=
80
210
+
120
600
=
1
n
,
so
n
=
1.72.
The load line from the origin meets the Goodman line 1.72 times as far out as the point: the part is safe.
Test bar endurance limit Sₑ′ = 300 MPa
Marin product k = 0.7
This is an example. Type your own values to start a new problem.
σₐ (MPa)
Sₑ (MPa)
+
σₘ (MPa)
Sᵤₜ (MPa)
=
1
n
Sₑ′
Test bar endurance limit
calculated
k
Marin product
example
σₘₐₓ
Largest stress in the cycle
example
σₘᵢₙ
Smallest stress in the cycle
example
Formulas
σₐ
=
σₘₐₓ − σₘᵢₙ
2
80
=
200 − 40
2
σₘ
=
σₘₐₓ + σₘᵢₙ
2
120
=
200 + 40
2
Sₑ′ = 0.5 × Sᵤₜ
300 = 0.5 × 600
Sₑ
=
k
×
Sₑ′
210 = 0.7 × 300
σₐ
Sₑ
+
σₘ
Sᵤₜ
=
1
n
80
210
+
120
600
=
1
1.7213
Assumptions
•
The same stress cycle repeats; stresses at the critical point stay under the yield strength.
•
Steel with Sᵤₜ up to 1400 MPa, so the test bar’s endurance limit is Sₑ′ = 0.5Sᵤₜ.
•
The Marin product k gathers surface, size, load, temperature and reliability factors.
•
Modified Goodman line, for infinite life; the steps work in MPa.
Step-by-step
WE KNOW
Sᵤₜ = 600 MPa, k = 0.7, σₘₐₓ = +200 MPa, σₘᵢₙ = +40 MPa
FIND
alternating stress (σₐ), mean stress (σₘ), test bar endurance limit (Sₑ′), part endurance limit (Sₑ), factor of safety (n)
Step 1 · Find alternating stress (σₐ)
Use σₐ = (σₘₐₓ − σₘᵢₙ) ÷ 2
Half the swing.
σₐ
=
200 − 40
2
σₐ
=
160
2
σₐ = 80 MPa
Step 2 · Find mean stress (σₘ)
Use σₘ = (σₘₐₓ + σₘᵢₙ) ÷ 2
The middle of the swing.
σₘ
=
200 + 40
2
σₘ
=
240
2
σₘ = 120 MPa
Step 3 · Find test bar endurance limit (Sₑ′)
Use Sₑ′ = 0.5 × Sᵤₜ
A polished steel test bar survives endlessly below about half its ultimate strength.
Sₑ′ = 0.5 × 600
Sₑ′ = 300 MPa
Step 4 · Find part endurance limit (Sₑ)
Use Sₑ = k × Sₑ′
The Marin factors take the real part below the polished test bar.
Sₑ = 0.7 × 300
Sₑ = 210 MPa
Step 5 · Find factor of safety (n)
Use σₐ ÷ Sₑ + σₘ ÷ Sᵤₜ = 1 ÷ n
Each
stress
over
its
strength
is
the
share
of
the
Goodman
line
used;
n
is
1
over
their
sum.
n
=
1
σₐ ÷ Sₑ + σₘ ÷ Sᵤₜ
n
=
1
80 ÷ 210 + 120 ÷ 600
n
=
1
0.381 + 0.2
n
=
1
0.581
n
=
1.7213
Check
Put the numbers back in. Both sides should match.
80
=
200 − 40
2
✓
120
=
200 + 40
2
✓
300 = 0.5 × 600 ✓
210 = 0.7 × 300 ✓
80
210
+
120
600
=
1
1.7213
✓
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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