SME
Advanced Solid Mechanics (Solid Mechanics II)
Stress and strain tensors
Use this for “σₓ = 60, σᵧ = 20, σz = −30 and τₓᵧ = 15 MPa, with τyz = τzx = 0. Find σ₁, σ₂, σ₃ and τₘₐₓ.”
Diagram
Sorted: σ₁ = 65, σ₂ = 15, σ₃ = −30 MPa.
The
biggest
circle,
from
σ₃
to
σ₁,
gives
τₘₐₓ
=
σ₁ − σ₃
2
=
47.5
MPa.
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
σz
Normal stress on z
example
τₓᵧ
Shear stress in xy
example
σ₁
Greatest principal stress
worked out (not typed)
σ₂
Middle principal stress
worked out (not typed)
σ₃
Least principal stress
worked out (not typed)
τₘₐₓ
Largest shear stress
worked out (not typed)
Formulas
σ₁
=
greatest
root
of
(σ
−
σz)(
σ
2
−
(σₓ
+
σᵧ)σ
+
σₓ
×
σᵧ
−
τₓᵧ²)
=
0
+65
=
greatest
root
of
(σ
−
(−30))(
σ
2
−
(60
+
20)σ
+
60
×
20
−
15
2
)
=
0
σ₂
=
middle
root
of
(σ
−
σz)(
σ
2
−
(σₓ
+
σᵧ)σ
+
σₓ
×
σᵧ
−
τₓᵧ²)
=
0
+15
=
middle
root
of
(σ
−
(−30))(
σ
2
−
(60
+
20)σ
+
60
×
20
−
15
2
)
=
0
σ₃
=
least
root
of
(σ
−
σz)(
σ
2
−
(σₓ
+
σᵧ)σ
+
σₓ
×
σᵧ
−
τₓᵧ²)
=
0
−30
=
least
root
of
(σ
−
(−30))(
σ
2
−
(60
+
20)σ
+
60
×
20
−
15
2
)
=
0
τₘₐₓ
=
σ₁
−
σ₃
2
47.5
=
65 − (−30)
2
Assumptions
•
τyz = τzx = 0, so σz is a principal stress; the xy circle gives the other two.
•
The principal stresses are the roots of det(σ − σI) = 0, ordered σ₁ ≥ σ₂ ≥ σ₃.
•
τₘₐₓ = (σ₁ − σ₃) ÷ 2 is the largest shear on any plane, not just in the xy plane.
•
Tension is +; τₓᵧ is + up on the right face.
Step-by-step
WE KNOW
σₓ = +60 MPa, σᵧ = +20 MPa, σz = −30 MPa, τₓᵧ = +15 MPa
FIND
greatest principal stress (σ₁), middle principal stress (σ₂), least principal stress (σ₃), largest shear stress (τₘₐₓ)
Step 1 · Find greatest principal stress (σ₁)
Use σ₁ = greatest root of (σ − σz)(σ² − (σₓ + σᵧ)σ + σₓ × σᵧ − τₓᵧ²) = 0
With τyz = τzx = 0 the determinant factors: σz is one root, and the xy circle σavg ± R gives the other two. Expand, find the roots and order them.
σ₁
=
greatest
of
−30,
15,
65
The
roots
of
σ
3
−
50σ²
−
1,425σ
+
29,250
=
0,
in
order:
−30,
15,
65.
Rejected: −30 and 15, since +65 is the greatest of the 3 roots.
σ₁
=
+65
MPa
Step 2 · Find middle principal stress (σ₂)
Use σ₂ = middle root of (σ − σz)(σ² − (σₓ + σᵧ)σ + σₓ × σᵧ − τₓᵧ²) = 0
With τyz = τzx = 0 the determinant factors: σz is one root, and the xy circle σavg ± R gives the other two. Expand, find the roots and order them.
σ₂
=
median
of
−30,
15,
65
Rejected: −30 and 65, since +15 is the second greatest of the 3 roots.
σ₂
=
+15
MPa
Step 3 · Find least principal stress (σ₃)
Use σ₃ = least root of (σ − σz)(σ² − (σₓ + σᵧ)σ + σₓ × σᵧ − τₓᵧ²) = 0
With τyz = τzx = 0 the determinant factors: σz is one root, and the xy circle σavg ± R gives the other two. Expand, find the roots and order them.
σ₃
=
least
of
−30,
15,
65
Rejected: 15 and 65, since −30 is the least of the 3 roots.
σ₃
=
−30
MPa
Step 4 · Find largest shear stress (τₘₐₓ)
Use τₘₐₓ = (σ₁ − σ₃) ÷ 2
The
biggest
of
the
three
Mohr
circles
runs
from
σ₃
to
σ₁
;
its
radius
is
the
largest
shear.
τₘₐₓ
=
65 − (−30)
2
τₘₐₓ
=
95
2
τₘₐₓ = 47.5 MPa
Check
Put the numbers back in. Both sides should match.
+65
=
greatest
root
of
(σ
−
(−30))(
σ
2
−
(60
+
20)σ
+
60
×
20
−
15
2
)
=
0
✓
+15
=
middle
root
of
(σ
−
(−30))(
σ
2
−
(60
+
20)σ
+
60
×
20
−
15
2
)
=
0
✓
−30
=
least
root
of
(σ
−
(−30))(
σ
2
−
(60
+
20)σ
+
60
×
20
−
15
2
)
=
0
✓
47.5
=
65 − (−30)
2
✓
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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Octahedral stresses
Use this for “The principal stresses are 65, 15 and −30 MPa. Find the octahedral normal and shear stresses.”
Strain rosette
Use this for “A 45° rosette reads εₐ = 600 με, εb = 300 με and εc = −200 με. Find γₓᵧ and the principal strains.”