SME
Machine Design · Problem type
Brittle failure: Coulomb–Mohr
Use this for “Cast iron with Sᵤₜ = 200 MPa and Suc = 750 MPa carries σ₁ = 60 MPa and σ₃ = −90 MPa. Find n by Coulomb–Mohr.”
Diagram
The load: σA = σ₁ = 60, σB = σ₃ = −90 MPa.
Grow the load along its ray by n and it reaches each locus (the open marks); n > 1 means the load is inside.
This is an example. Type your own values to start a new problem.
σ₁ (MPa)
Sᵤₜ (MPa)
−
σ₃ (MPa)
Suc (MPa)
=
1
n
Formulas
σ₁
Sᵤₜ
−
σ₃
Suc
=
1
n
60
200
−
−90
750
=
1
2.381
Assumptions
•
Plane stress: sort the two in-plane principal stresses and 0, so σ₁ ≥ 0 ≥ σ₃.
•
A brittle material such as cast iron: weaker in tension (Sᵤₜ) than in compression (Suc).
•
Both stresses tensile: σ₃ = 0 and n = Sᵤₜ ÷ σ₁. Both compressive: σ₁ = 0.
Step-by-step
WE KNOW
Sᵤₜ = 200 MPa, Suc = 750 MPa, σ₁ = 60 MPa, σ₃ = −90 MPa
FIND
safety factor (n)
Step 1 · Find safety factor (n)
Use σ₁ ÷ Sᵤₜ − σ₃ ÷ Suc = 1 ÷ n
Each
stress
over
the
strength
on
its
own
side;
minus
a
negative
σ₃
adds,
then
turn
the
sum
over.
n
=
1
σ₁
÷ Sᵤₜ −
σ₃
÷ Suc
n
=
1
60 ÷ 200 − (−90) ÷ 750
n
=
1
0.3 − (−0.12)
n
=
1
0.42
n
=
2.381
Check
Put the numbers back in. Both sides should match.
60
200
−
−90
750
=
1
2.381
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.