SME
Finite Element Analysis
Interpreting FEA results
Use this for “A plate with a hole has σₙₒₘ = 50 MPa and Kₜ = 2.5. FEA shows a 131 MPa peak. With SY = 250 MPa, does it agree, and what is n?”
Chart
131 MPa
Stress concentration factor Kₜ = 2.5
Difference e = +4.8%
Factor of safety n = 1.9084
This is an example. Type your own values to start a new problem.
σhand (MPa)
=
Kₜ
×
σₙₒₘ (MPa)
n
=
SY (MPa)
÷
σFEA (MPa)
e
Difference
calculated · %
Formulas
σhand = Kₜ × σₙₒₘ
125 = 2.5 × 50
e
=
σFEA − σhand
σhand
×
100
+4.8
=
131 − 125
125
×
100
n
=
SY
σFEA
1.9084
=
250
131
Assumptions
•
The hand check is σhand = Kₜσₙₒₘ, with Kₜ read from a chart for the same shape and load.
•
A few percent apart is good agreement; a large gap points to the mesh, loads or supports.
•
Linear elastic: n = SY ÷ σFEA at the peak, and n below 1 means local yielding.
•
The difference e is + when the FEA peak is above the hand value.
Step-by-step
WE KNOW
σₙₒₘ = 50 MPa, Kₜ = 2.5, σFEA = 131 MPa, SY = 250 MPa
FIND
hand peak stress (σhand), difference (e), factor of safety (n)
Step 1 · Find hand peak stress (σhand)
Use σhand = Kₜ × σₙₒₘ
The stress raiser multiplies the nominal stress by Kₜ: this is the peak a hand calculation expects.
σhand = 2.5 × 50
σhand = 125 MPa
Step 2 · Find difference (e)
Use e = (σFEA − σhand) ÷ σhand × 100
Compare the two peaks: the gap over the hand value, as a percent.
e
=
131 − 125
125
×
100
e
=
6
125
×
100
e
=
0.048
×
100
e
=
+4.8%
Step 3 · Find factor of safety (n)
Use n = SY ÷ σFEA
Compare the yield strength with the peak the model found: their quotient is the factor of safety.
n
=
250
131
n
=
1.9084
Check
Put the numbers back in. Both sides should match.
125 = 2.5 × 50 ✓
+4.8
=
131 − 125
125
×
100
✓
1.9084
=
250
131
✓
Topic check
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