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Mechanical

Second stress invariant

Combination of the stress components that is unchanged by rotating the axes.

Ref: MEC-61
Rev: v1
Field: Mechanical
Result: I₂ [MPa²]

I_{2} = σ_{x} σ_{y} + σ_{y} σ_{z} + σ_{z} σ_{x} - {τ_{xy}}^{2} - {τ_{yz}}^{2} - {τ_{xz}}^{2}

I₂ = σ_xσ_y + σ_yσ_z + σ_zσ_x − τ_xy² − τ_yz² − τ_xz² takes the same value in every set of axes. It equals σ₁σ₂ + σ₂σ₃ + σ₃σ₁ and is the middle coefficient of the characteristic equation λ³ − I₁λ² + I₂λ − I₃ = 0, whose three roots are the principal stresses. It also gives von Mises directly, because σ_VM² = I₁² − 3I₂.

Combination of the stress components that is unchanged by rotating the axes. It is the middle coefficient of the characteristic equation whose roots are the principal stresses.

Calculator

Solve for

I₂Second invariantComputed

MPa²

σ_xNormal stress in x

MPa

−200300

σ_yNormal stress in y

MPa

−200300

σ_zNormal stress in z

MPa

−200300

τ_xyShear stress xy

MPa

−150150

τ_yzShear stress yz

MPa

−150150

τ_xzShear stress xz

MPa

−150150

First invariant I₁180 MPa
Sum of pairwise products σ₁σ₂ + σ₂σ₃ + σ₃σ₁ (should equal I₂)8,700 MPa²
Von Mises from invariants, √(I₁² − 3I₂)79.37 MPa

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Second invariant (MPa²) current value 8,700 MPa²

σ_zNormal stress in zKey driver

−26,50053,500

σ_xNormal stress in x

−19,30030,700

σ_yNormal stress in y

−19,30030,700

τ_xyShear stress xy

−12,9009,600

τ_yzShear stress yz

−13,800

τ_xzShear stress xz

−13,800

The lecture state gives I₂ = 8,700 MPa². With I₁ = 180 MPa, I₁² − 3I₂ = 6,300 and its square root is 79.373 MPa, the von Mises stress for this state.
The principal stresses 110, 50 and 20 MPa give 110×50 + 50×20 + 20×110 = 8,700 MPa², the same number found from the components, which is how the invariants are used to check a calculation.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing (MPa²)	Local step	Change in result
σ_z Normal stress in z	−200 to 300 MPa	−26,500	53,500	80,000	+10 MPa	+18.4 %
σ_x Normal stress in x	−200 to 300 MPa	−19,300	30,700	50,000	+10 MPa	+11.5 %
σ_y Normal stress in y	−200 to 300 MPa	−19,300	30,700	50,000	+10 MPa	+11.5 %
τ_xy Shear stress xy	−150 to 150 MPa	−12,900	9,600	22,500	+6 MPa	−4.55 %
τ_yz Shear stress yz	−150 to 150 MPa	−13,800	8,700	22,500	+6 MPa	−0.414 %
τ_xz Shear stress xz	−150 to 150 MPa	−13,800	8,700	22,500	+6 MPa	−0.414 %

Sweep of Shear stress xy

Shear stress xy (MPa)	Second invariant (MPa²)
−150	−12,900
−112.5	−3,056
−75	3,975
−37.5	8,194
0	9,600
37.5	8,194
75	3,975
112.5	−3,056
150	−12,900

Sweep one variable

Vary

Second invariant (MPa²)Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

I₂

Second invariant

Result · MPa²

What the equation gives you. Everything else feeds it.

σ_x

Normal stress in x

MPa · typical −200 to 300

Normal stress on the x face, in MPa. Tension is positive and compression negative, so the slider runs from −200 to +300 MPa. Raises I₂ by the change times (σ_y + σ_z). Raising it by 10 MPa (to 90 MPa) moves the result by +11.5 %, from 8,700 to 9,700 MPa².

σ_y

Normal stress in y

MPa · typical −200 to 300

Normal stress on the y face, in MPa. Tension is positive and compression negative, so the slider runs from −200 to +300 MPa. Raises I₂ by the change times (σ_x + σ_z). Raising it by 10 MPa (to 90 MPa) moves the result by +11.5 %, from 8,700 to 9,700 MPa².

σ_z

Normal stress in z

MPa · typical −200 to 300

Normal stress on the z face, in MPa. Tension is positive and compression negative, so the slider runs from −200 to +300 MPa. Raises I₂ by the change times (σ_x + σ_y). Raising it by 10 MPa (to 30 MPa) moves the result by +18.4 %, from 8,700 to 10,300 MPa².

τ_xy

Shear stress xy

MPa · typical −150 to 150

Shear stress τ_xy, in MPa, acting on the x face in the +y direction. Shear is signed here, so the slider runs from −150 to +150 MPa. Lowers I₂, through −τ_xy², whichever way the shear acts. Raising it by 10 MPa (to 40 MPa) moves the result by −8.05 %, from 8,700 to 8,000 MPa².

τ_yz

Shear stress yz

MPa · typical −150 to 150

Shear stress τ_yz, in MPa, acting on the y face in the +z direction. Shear is signed here, so the slider runs from −150 to +150 MPa. Lowers I₂, through −τ_yz², whichever way the shear acts. Raising it by 10 MPa (to 10 MPa) moves the result by −1.15 %, from 8,700 to 8,600 MPa².

τ_xz

Shear stress xz

MPa · typical −150 to 150

Shear stress τ_xz, in MPa, acting on the x face in the +z direction. Shear is signed here, so the slider runs from −150 to +150 MPa. Lowers I₂, through −τ_xz², whichever way the shear acts. Raising it by 10 MPa (to 10 MPa) moves the result by −1.15 %, from 8,700 to 8,600 MPa².

Worked example with the values above

Products of the normal stresses: σ_xσ_y + σ_yσ_z + σ_zσ_x = 9,600 MPa².
Squares of the shears: τ_xy² + τ_yz² + τ_xz² = 900 MPa².
I₂ = 9,600 − 900 = 8,700 MPa².

Assumptions and limits

A symmetric stress tensor with six independent components.
All stresses in one set of axes, the same units, tension positive.

Common mistakes

Leaving out the shear terms, which each subtract the square of the shear.
Mixing MPa and N/mm² with Pa in the same expression.

Statistics and uncertainty

I₂ is a quadratic quantity, so errors are weighted by the other stresses: an error of ε in σ_x changes I₂ by about ε(σ_y + σ_z) = 100 MPa² per MPa at the defaults. Stress states with large mean stress therefore amplify small errors in I₂, which is one reason to work with the deviatoric stress for yield checks.

Where it is used

Building the characteristic equation to find the principal stresses.
Checking a rotated stress state and finding von Mises from invariants.

Maths you need

Matrix determinant: sum of products of terms
Invariants: values that do not change with the axes

Related formulas

Practise this formula