SME
Advanced Solid Mechanics (Solid Mechanics II) · Problem type
Stress invariants
Use this for “The stress tensor is [[50, 20, 10], [20, −10, 10], [10, 10, 30]] MPa. Find I₁, I₂ and I₃.”
Diagram
I₃
I₃ = +50 × (−400) − +20 × 500 + +10 × 300
I₃ = −27,000 MPa³
Each first-row entry times the 2 × 2 left when its row and column are struck, signs +, −, +
This is an example. Type your own values to start a new problem.
σ
=
σₓ
τₓᵧ
τzx
τₓᵧ
σᵧ
τyz
τzx
τyz
σz
MPa
I₁
First invariant
worked out (not typed) · MPa
I₂
Second invariant
worked out (not typed) · MPa²
I₃
Third invariant
worked out (not typed) · MPa³
Formulas
I₁
=
σₓ
+
σᵧ
+
σz
+70 = 50 + (−10) + 30
I₂
=
σₓ
×
σᵧ
+
σᵧ
×
σz
+
σz
×
σₓ
−
τₓᵧ²
−
τyz
2
−
τzx
2
+100
=
50
×
(−10)
+
(−10)
×
30
+
30
×
50
−
20
2
−
10
2
−
10
2
I₃
=
σₓ
×
(σᵧ
×
σz
−
τyz
2
)
−
τₓᵧ
×
(τₓᵧ
×
σz
−
τyz
×
τzx)
+
τzx
×
(τₓᵧ
×
τyz
−
σᵧ
×
τzx)
−27,000
=
50
×
(−10
×
30
−
10
2
)
−
20
×
(20
×
30
−
10
×
10)
+
10
×
(20
×
10
−
(−10)
×
10)
Assumptions
•
The tensor is symmetric: τᵧₓ = τₓᵧ, τzy = τyz, τₓz = τzx.
•
I₁, I₂ and I₃ are the same in every axis set, since they are the coefficients of det(σ − σI) = 0.
•
Stresses are in MPa, so I₂ is in MPa² and I₃ = det σ in MPa³.
Step-by-step
WE KNOW
σₓ = +50 MPa, σᵧ = −10 MPa, σz = +30 MPa, τₓᵧ = +20 MPa, τyz = +10 MPa, τzx = +10 MPa
FIND
first invariant (I₁), second invariant (I₂), third invariant (I₃)
Step 1 · Find first invariant (I₁)
Use I₁ = σₓ + σᵧ + σz
The first invariant is the trace: add the three normal stresses on the diagonal.
I₁
=
50
+
(−10)
+
30
I₁
=
+70
MPa
Step 2 · Find second invariant (I₂)
Use I₂ = σₓ × σᵧ + σᵧ × σz + σz × σₓ − τₓᵧ² − τyz² − τzx²
The second invariant is the sum of the three 2 × 2 minors on the diagonal: each pair of normal stresses less its shear squared.
I₂
=
50
×
(−10)
+
(−10)
×
30
+
30
×
50
−
20
2
−
10
2
−
10
2
I₂
=
50
×
(−10)
+
(−10)
×
30
+
30
×
50
−
400
−
100
−
100
I₂
=
−500
+
(−300)
+
1,500
−
400
−
100
−
100
I₂
=
+100
MPa²
Step 3 · Find third invariant (I₃)
Use I₃ = σₓ × (σᵧ × σz − τyz²) − τₓᵧ × (τₓᵧ × σz − τyz × τzx) + τzx × (τₓᵧ × τyz − σᵧ × τzx)
The third invariant is the determinant: expand along the first row, each entry times its 2 × 2 minor, signs +, −, +.
I₃
=
50
×
(−10
×
30
−
10
2
)
−
20
×
(20
×
30
−
10
×
10)
+
10
×
(20
×
10
−
(−10)
×
10)
I₃
=
50
×
((−10)
×
30
−
100)
−
20
×
(600
−
100)
+
10
×
(200
−
(−100))
I₃
=
50
×
((−300)
−
100)
−
20
×
500
+
10
×
300
I₃
=
50
×
(−400)
−
20
×
500
+
10
×
300
I₃
=
−20,000
−
10,000
+
3,000
I₃
=
−27,000
MPa³
Check
Put the numbers back in. Both sides should match.
+70 = 50 + (−10) + 30 ✓
+100
=
50
×
(−10)
+
(−10)
×
30
+
30
×
50
−
20
2
−
10
2
−
10
2
✓
−27,000
=
50
×
(−10
×
30
−
10
2
)
−
20
×
(20
×
30
−
10
×
10)
+
10
×
(20
×
10
−
(−10)
×
10)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.