SME
Machine Design
Failure theories
Use this for “A shaft point has σₓ = 120 MPa, σᵧ = −40 MPa and τₓᵧ = 50 MPa; Sᵧ = 350 MPa. Find n by von Mises and by Tresca.”
Diagram
The
load:
σA
=
134,
σB
=
−54.3
MPa
(σ₃
=
0
beside
them).
σ′
=
√
σₓ² − σₓσᵧ +
σᵧ
2
+ 3τₓᵧ²
=
168
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.
σₓ
Normal stress on x
example
σᵧ
Normal stress on y
example
τₓᵧ
Shear stress
example
σ₁
Larger principal stress
calculated
σ₂
Smaller principal stress
calculated
σ′
Von Mises stress
calculated
Sᵧ
Yield strength
example
n
Safety factor, von Mises
calculated
τₘₐₓ
Largest shear stress
calculated
nT
Safety factor, Tresca
calculated
Formulas
σ₁
=
σₓ + σᵧ
2
+
√((
σₓ − σᵧ
2
)²
+
τₓᵧ²)
134.3398
=
120 + (−40)
2
+
√((
120 − (−40)
2
)²
+
50
2
)
σ₂
=
σₓ + σᵧ
2
−
√((
σₓ − σᵧ
2
)²
+
τₓᵧ²)
−54.3398
=
120 + (−40)
2
−
√((
120 − (−40)
2
)²
+
50
2
)
σₓ
+
σᵧ
=
σ₁
+
σ₂
120 + (−40) = 134.3398 + (−54.3398)
σ′
=
√
σₓ² − σₓ × σᵧ +
σᵧ
2
+ 3 × τₓᵧ²
168.226
=
√(
120
2
−
120
×
(−40)
+
(−40)
2
+
3
×
50
2
)
σ′
=
√
σ₁
2
−
σ₁
×
σ₂
+
σ₂
2
168.226
=
√(
134.3398
2
−
134.3398
×
(−54.3398)
+
(−54.3398)
2
)
n
=
Sᵧ
σ′
2.0805
=
350
168.226
τₘₐₓ
=
max(
σ₁
−
σ₂
,
σ₁
, −
σ₂
)
2
94.3398 = max(134.3398 − (−54.3398), 134.3398, −(−54.3398)) ÷ 2
nT
=
Sᵧ
2 × τₘₐₓ
1.855
=
350
2 × 94.3398
Assumptions
•
Plane stress: σ₃ = 0 on the z faces. Tension is +; τₓᵧ is + up on the right face.
•
A ductile material: it yields at Sᵧ in tension and in compression alike.
•
With σ₃ = 0 there are three Mohr circles; τₘₐₓ is the radius of the largest.
•
Von Mises fits tests best; Tresca gives a smaller n, so it is the safer estimate.
Step-by-step
WE KNOW
σₓ = +120 MPa, σᵧ = −40 MPa, τₓᵧ = +50 MPa, Sᵧ = 350 MPa
FIND
larger principal stress (σ₁), smaller principal stress (σ₂), von Mises stress (σ′), safety factor, von Mises (n), largest shear stress (τₘₐₓ), safety factor, Tresca (nT)
Step 1 · Find larger principal stress (σ₁)
Use σ₁ = (σₓ + σᵧ) ÷ 2 + √(((σₓ − σᵧ) ÷ 2)² + τₓᵧ²)
Mohr’s
circle:
its
center
is
the
mean
of
σₓ
and
σᵧ,
and
σ₁
is
the
center
plus
the
radius.
σ₁
=
120 + (−40)
2
+
√((
120 − (−40)
2
)²
+
50
2
)
σ₁
=
120 + (−40)
2
+
√(
(
160
2
)
2
+
50
2
)
σ₁
=
120 + (−40)
2
+
√
80
2
+
50
2
σ₁
=
80
2
+
√
6,400 + 2,500
σ₁
=
80
2
+
√
8
,900
σ₁
=
80
2
+
94.34
σ₁
=
40
+
94.34
σ₁
=
134.3398
MPa
Step 2 · Find smaller principal stress (σ₂)
Use σ₂ = (σₓ + σᵧ) ÷ 2 − √(((σₓ − σᵧ) ÷ 2)² + τₓᵧ²)
The circle’s left end: the center less the radius.
σ₂
=
120 + (−40)
2
−
√((
120 − (−40)
2
)²
+
50
2
)
σ₂
=
120 + (−40)
2
−
√(
(
160
2
)
2
+
50
2
)
σ₂
=
120 + (−40)
2
−
√
80
2
+
50
2
σ₂
=
80
2
−
√
6,400 + 2,500
σ₂
=
80
2
−
√
8
,900
σ₂
=
80
2
−
94.34
σ₂
=
40
−
94.34
σ₂
=
−54.3398
MPa
Step 3 · Find von Mises stress (σ′)
Use σ′ = √(σₓ² − σₓ × σᵧ + σᵧ² + 3 × τₓᵧ²)
Von Mises (distortion energy) for plane stress, straight from σₓ, σᵧ and τₓᵧ.
σ′
=
√(
120
2
−
120
×
(−40)
+
(−40)
2
+
3
×
50
2
)
σ′ = √(14,400 − 120 × (−40) + 1,600 + 3 × 2,500)
σ′ = √(14,400 − (−4,800) + 1,600 + 7,500)
σ′
=
√
28
,300
σ′ = 168.226 MPa
Step 4 · Find safety factor, von Mises (n)
Use n = Sᵧ ÷ σ′
Von
Mises:
yield
when
σ′
reaches
Sᵧ,
so
n
is
how
many
times
σ′
fits
under
Sᵧ.
n
=
350
168.226
n
=
2.0805
Step 5 · Find largest shear stress (τₘₐₓ)
Use τₘₐₓ = max(σ₁ − σ₂, σ₁, −σ₂) ÷ 2
The
three
circles
span
σ₁
to
σ₂
,
σ₁
to
0
and
0
to
σ₂
;
the
widest
one’s
radius
is
τₘₐₓ.
τₘₐₓ = max(134.3398 − (−54.3398), 134.3398, −(−54.3398)) ÷ 2
τₘₐₓ = 94.3398 MPa
Step 6 · Find safety factor, Tresca (nT)
Use nT = Sᵧ ÷ (2 × τₘₐₓ)
Tresca:
yield
when
τₘₐₓ
reaches
Sᵧ
2,
the
shear
in
a
tensile
test
at
yield.
nT
=
350
2 × 94.3398
nT
=
350
188.6796
nT = 1.855
Check
Put the numbers back in. Both sides should match.
134.3398
=
120 + (−40)
2
+
√((
120 − (−40)
2
)²
+
50
2
)
✓
−54.3398
=
120 + (−40)
2
−
√((
120 − (−40)
2
)²
+
50
2
)
✓
120 + (−40) = 134.3398 + (−54.3398) ✓
168.226
=
√(
120
2
−
120
×
(−40)
+
(−40)
2
+
3
×
50
2
)
✓
168.226
=
√(
134.3398
2
−
134.3398
×
(−54.3398)
+
(−54.3398)
2
)
✓
2.0805
=
350
168.226
✓
94.3398 = max(134.3398 − (−54.3398), 134.3398, −(−54.3398)) ÷ 2 ✓
1.855
=
350
2 × 94.3398
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
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.”
Peak stress at a hole
Use this for “A 50 mm × 10 mm plate with a 10 mm hole carries 20 kN; Kₜ = 2.5 from the chart. Find the nominal and peak stresses.”