Mohr's Circle and Principal Stresses

In case of the general 3-dimensional state stress in structures is represented by 6 stress values.
For this situation principal stresses and principal stress directions are calculated.
The principal stresses σ₁, σ₂, σ₃ are the eigenvalues of the stress tensor S:

σ_x	τ_xy	τ_xz
τ_yx	σ_y	τ_yz
τ_zx	τ_zy	σ_z

The principal stresses and the Mohr circles are visualized graphically.
In the shaded areas between the circles possible pairs of stress values (σ, τ) may be found.
The 3 red points represent stress values according to the given values with repespect to a xyz-coordinate system:
(σ_x, (τ_xy²+τ_xz²)^1/2), (σ_y, (τ_yx²+τ_yz²)^1/2) , (σ_z, (τ_zx²+τ_zy²)^1/2)
The yellow points mark the principal stresses. Under the associated directions there will be no shear stresses.

A special case is the 2-dimensional stress state given by σ_x, σ_y, τ_xy whereas σ_z=τ_yz=τ_zx=0.
In this case we can determine 2 points on Mohr's circle:
(σ_x, τ_xy) and (σ_y, -τ_xy).
The center of the circle is ((σ_x + σ_y)/2, 0).
Two principal normal stresses can be easily determined directly from the given stress values:
σ_1,2 = (σ_x+σ_y)/2 + ± (τ_xy²+(σ_x-σ_y)²/4)^1/2
One of the three principal normal stresses is always 0, and the corresponding principal normal stress direction is the z-direction.
This yields a third yellow dot at (0, 0).

For a normalized direction vector n the normal stress σ_n and shear stress τ_n will be calculated:
σ_n = n^T S n
|τ_n| = (n^TS^T S n - σ_n²)^1/2.

Example Calculation

The following stress state is given:
σ_x = 1
σ_y = 2
σ_z = 1
τ_xy = 1
τ_yz = 1
τ_xz = 0

This leads to the stress tensor S

1	1	0
1	2	1
0	1	1

The characteristic polynomial of the corresponding matrix is determined as solutions to the equation
det(S - λ I) = 0.

The determinant here is specific

| 1-λ  1   0  |
| 1  2-λ  1  |
| 0   1  1-λ |

The expansion of the determinant yields the characteristic polynomial:
λ³ - 4 λ² + 3 λ.

The zeros of this characteristic polynomial are the three principal normal stresses:
σ₁ = 3
σ₂ = 1
σ₃ = 0

Now sought is the normal stress and tangential stress for a surface with normal (1,1,1)^T.

The normalized and transposed vector is then n^T = (1,1,1)/√3
Furthermore you get:
n^T S = (2, 4, 2)/√3
and thus
σ_n = n^T S n = 8/3
and
|τ_n| = (n^TS^T S n - σ_n²)^1/2 = (24/3 - 64/9)^1/2 = (8/9)^1/2 = 0.9428

Significance of Principal Normal Stresses

One problem with stress calculations, e.g., using FEM, is that in the 3-dimensional case, 6 stresses occur at each point.
These 6 stress values also depend on the choice of the respective global coordinate system.

If one now calculates the principal normal stresses, one only has to deal with 3 numerical values, and
these are independent of the choice of the respective global coordinate system.
Three stress values are still too many for comparison with permissible maximum values for the respective material.

To clarify the question of permissibility, failure hypotheses are needed that characterize the respective material.

For example, the hypothesis of maximum distortion energy states that the following equivalent stress can be derived from the 3 principal normal stresses:

σ_v = ( ( (σ₁-σ₂)² + (σ₂-σ₃)² + (σ₃-σ₁)²)/2 )^1/2.

That is, only the failure hypothesis condenses the stress information into a single-number parameter that is suitable for comparison with permissible values.

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