SME
Mechanics of Materials (Solid Mechanics I) · Problem type
Principal stresses and Mohr’s circle
Use this for “A point carries σₓ = 80 MPa, σᵧ = −20 MPa and τₓᵧ = 40 MPa. Find the principal stresses, θₚ and τₘₐₓ.”
Diagram
σavg
=
σₓ + σᵧ
2
=
80 + −20
2
σavg = 30 MPa
R
=
√((
σₓ − σᵧ
2
)²
+
τₓᵧ²)
=
64
MPa.
σ₁
=
σavg
+
R
=
94
MPa;
σ₂
=
σavg
−
R
=
−34
MPa;
τₘₐₓ
=
R.
tan
2θₚ
=
2τₓᵧ
÷
(σₓ
−
σᵧ),
so
θₚ
=
19.3°:
the
turned
element
carries
σ₁
on
the
faces
turned
θₚ
and
no
shear.
τ
is
plotted
positive
down,
so
X
turns
to
σ₁
by
2θₚ
the
way
the
element
turns.
This is an example. Type your own values to start a new problem.
σavg (MPa)
=
σₓ (MPa)
+
(
σᵧ (MPa)
)
2
R (MPa)
=
−
(
)
2
2
+
(
τₓᵧ (MPa)
)
2
σ₁ (MPa)
=
σavg (MPa)
+
R (MPa)
σ₂
Smaller principal stress
calculated
θₚ
Principal angle
calculated · °
τₘₐₓ
Largest in-plane shear
calculated
Formulas
σavg
=
σₓ + σᵧ
2
+30
=
80 + (−20)
2
R
=
√((
σₓ − σᵧ
2
)²
+
τₓᵧ²)
64.0312
=
√((
80 − (−20)
2
)²
+
40
2
)
σ₁
=
σavg
+
R
+94.0312 = 30 + 64.0312
σ₂
=
σavg
−
R
−34.0312 = 30 − 64.0312
θₚ
=
0.5
×
tan⁻¹(2
×
τₓᵧ
σₓ − σᵧ
)
+19.3299
=
0.5
×
tan⁻¹(2
×
40
80 − (−20)
)
τₘₐₓ
=
R
64.0312 = 64.0312
Assumptions
•
Plane stress: nothing acts on the z faces. Tension is +; τₓᵧ is + up on the right face.
•
θₚ (−45° to 45°, counterclockwise +) turns x to σ₁’s plane when σₓ > σᵧ, else to σ₂’s.
•
τₘₐₓ = R is the largest in-plane shear, on planes 45° from the principal ones.
•
Type τₓᵧ: R gives only its size, and its sign sets which way θₚ turns.
Step-by-step
WE KNOW
σₓ = +80 MPa, σᵧ = −20 MPa, τₓᵧ = +40 MPa
FIND
average stress (σavg), circle radius (R), larger principal stress (σ₁), smaller principal stress (σ₂), principal angle (θₚ), largest in-plane shear (τₘₐₓ)
Step 1 · Find average stress (σavg)
Use σavg = (σₓ + σᵧ) ÷ 2
The circle’s center is the mean of the two normal stresses.
σavg
=
80 + (−20)
2
σavg
=
60
2
σavg = +30 MPa
Step 2 · Find circle radius (R)
Use R = √(((σₓ − σᵧ) ÷ 2)² + τₓᵧ²)
The radius reaches from the center to the point (σₓ, τₓᵧ): half the difference across, τₓᵧ down.
R
=
√((
80 − (−20)
2
)²
+
40
2
)
R
=
√(
(
100
2
)
2
+
40
2
)
R
=
√
50
2
+
40
2
R
=
√
2,500 + 1,600
R
=
√
4
,100
R
=
64.0312
MPa
Step 3 · Find larger principal stress (σ₁)
Use σ₁ = σavg + R
The circle’s right end on the σ axis.
σ₁
=
30
+
64.0312
σ₁
=
+94.0312
MPa
Step 4 · Find smaller principal stress (σ₂)
Use σ₂ = σavg − R
The circle’s left end on the σ axis.
σ₂
=
30
−
64.0312
σ₂
=
−34.0312
MPa
Step 5 · Find principal angle (θₚ)
Use θₚ = 0.5 × tan⁻¹(2 × τₓᵧ ÷ (σₓ − σᵧ))
On the circle the turn from the x point to the σ axis is 2θₚ; halve it for the element.
θₚ
=
0.5
×
tan⁻¹(2
×
40
80 − (−20)
)
θₚ = +19.3299°
Step 6 · Find largest in-plane shear (τₘₐₓ)
Use τₘₐₓ = R
The top of the circle: the largest in-plane shear is the radius.
τₘₐₓ = 64.0312 MPa
Check
Put the numbers back in. Both sides should match.
+30
=
80 + (−20)
2
✓
64.0312
=
√((
80 − (−20)
2
)²
+
40
2
)
✓
+94.0312 = 30 + 64.0312 ✓
−34.0312 = 30 − 64.0312 ✓
+19.3299
=
0.5
×
tan⁻¹(2
×
40
80 − (−20)
)
✓
64.0312 = 64.0312 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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