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# Invariants and principal stresses

This exercise is about the invariants and principal stresses of a stress tensor and addresses the following questions:

• How to calculate the invariants of a stress tensor?
• How to calculate the determinant of a stress tensor?
• How to calculate the principal stresses of a stress tensor?
• How to calculate the principal stress directions of a stress tensor?
• How to calculate a stress vector?

$\newcommand{\myvec}[1]{{\begin{bmatrix}#1\end{bmatrix}}}$

The following stress tensor was calculated for a point in space:

$S = \myvec{1 & 1 & 3\\1 & 5 & 1\\3 & 1 & 1} \, Nmm^{-2}$

Calculate:

a) the three invariants of S

b) the three principal stresses σ1, σ2 and σ3

c) the three main directions of stress (main axis system) and

d) the stress vector on the plane with

$\vec{n} = \frac{1}{13} \, \myvec{3\\4\\12}$

## Solution

$\newcommand{\myvec}[1]{{\begin{bmatrix}#1\end{bmatrix}}}$

The stress tensor is noted in the general form beforehand in order to capture the correct values ​​when calculating the invariants.

$S = \myvec{\sigma_x & \tau_{xy} & \tau_{xz}\\\tau_{yx} & \sigma_y & \tau_{yz}\\\tau_{zx} & \tau_{zy} & \sigma_z}$

reg. a)

### Invariant 1, Trace of the stress tensor

The general solution for the trace of the stress tensor is

$\tag{1} I_1 = \sigma_x + \sigma_y + \sigma_z$

$\tag{2} I_1 = (1 + 5 + 1) \, Nmm^{-2}$

$\tag{3} I_1 = 7 \, Nmm^{-2}$

### Invariant 2

The general solution for invariant 2 of the stress tensor is

$\tag{4} I_2 = \sigma_x \cdot \sigma_y + \sigma_y \cdot \sigma_z + \sigma_x \cdot \sigma_z -\tau_{xy}^2 - \tau_{xz}^2 - \tau_{yz}^2$

$\tag{5} I_2 = (1 \cdot 5 + 5 \cdot 1 + 1 \cdot 1 -1^2 - 3^2 - 1^2) \, N^2mm^{-4}$

$\tag{6} I_2 = 0$

### Invariant 3, Determinant of the stress tensor

The general solution for the determinant of the stress tensor is

$\tag{7} I_3 = det \,S = \sigma_x \cdot \sigma_y \cdot \sigma_z + \tau_{xy} \cdot \tau_{yz} \cdot \tau_{zx} + \tau_{xz} \cdot \tau_{yx} \cdot \tau_{zy} \, ...$

$... \, - \sigma_x \cdot \tau_{yz} \cdot \tau_{zy} - \tau_{xy} \cdot \tau_{yx} \cdot \sigma_z - \tau_{xz} \cdot \sigma_y \cdot \tau_{zx}$

$\tag{8} I_3 = (1 \cdot 5 \cdot 1 + 1 \cdot 1 \cdot 3 + 3 \cdot 1 \cdot 1 \, ...$

$... \, - 1 \cdot 1 \cdot 1 - 1 \cdot 1 \cdot 1 - 3 \cdot 5 \cdot 3)\, N^3mm^{-6}$

$\tag{9} I_3 = -36\, N^3mm^{-6}$

reg. b)

### Calculation of the 3 principal stresses

The characteristic equation of the stress tensor is solved to calculate the three principal stresses. This is:

$\tag{10} \sigma^3 - I_1 \cdot \sigma^2 - I_2 \cdot \sigma - I_3 = 0$

$\tag{11} \sigma^3 - 7\,Nmm^{-2} \cdot \sigma^2 + 36\,N^3mm^{-6} = 0$

The solution of this equation leads to the principal stresses

$\tag{12} \sigma_1 = -2\, Nmm^{-2}$

$\tag{13} \sigma_2 = 6\, Nmm^{-2}$

$\tag{14} \sigma_3 = 3\, Nmm^{-2}$

reg. c)

### Calculation of the main stress directions

The general system of equations for calculating a stress direction is, with n as the component of the normal vector and σ as the principal stress:

$\tag{15.I} (\sigma_x-\sigma) \cdot n_x + \tau_{xy} \cdot n_y + \tau_{xz} \cdot n_z = 0$

$\tag{15.II} \tau_{yx} \cdot n_x + (\sigma_y-\sigma) \cdot n_y + \tau_{yz} \cdot n_z = 0$

$\tag{15.III} \tau_{zx} \cdot n_x + \tau_{zy} \cdot n_y + (\sigma_z - \sigma) \cdot n_z = 0$

#### Establishing the systems of equations

The unit Nmm-2 is not included in the following equation systems in order to improve the readability of the equations.

Principal stress σ1

The system of equations is

$\tag{16.I} (1+2) \cdot n_{x1} + 1 \cdot n_{y1} + 3 \cdot n_{z1} = 0$

$\tag{16.II} 1 \cdot n_{x1} + (5+2) \cdot n_{y1} + 1 \cdot n_{z1} = 0$

$\tag{16.III} 3 \cdot n_{x1} + 1 \cdot n_{y1} + (1 +2) \cdot n_{z1} = 0$

From this it follows immediately

$\tag{17} n_{y1} = 0$

and from equation (16.II)

$\tag{18} n_{x1} + n_{z1} = 0$

$\tag{19} n_{x1} = - n_{z1}$

The absolut value of the normal vector is 1, therefore applies:

$\tag{20} | \vec{n_1}| = 1$

$\tag{21} 1 = \sqrt{n_{x1}^2+n_{y1}^2+n_{z1}^2}$

or with the previously calculated ny1 and nz1

$\tag{21} 2 \cdot n_{x1}^2 = 1$

$\tag{22} n_{x1} = \frac{1}{\sqrt{2}}$

$\tag{23} n_{z1} = - \frac{1}{\sqrt{2}}$

The normal vector for the principal stress 1 thus is

$\tag{24} \vec{n_1} = \frac{1}{\sqrt{2}}\myvec{1\\0\\-1}$

Principal stress σ2

The system of equations is

$\tag{25.I} -5 \cdot n_{x2} + 1 \cdot n_{y2} + 3 \cdot n_{z2} = 0$

$\tag{25.II} 1 \cdot n_{x2} -1 \cdot n_{y2} + 1 \cdot n_{z2} = 0$

$\tag{25.III} 3 \cdot n_{x2} + 1 \cdot n_{y2} -5 \cdot n_{z2} = 0$

The procedure for solving the problem is the same as for calculating the first principal stress, so it is not listed again here. The second normal vector is

$\tag{26} \vec{n_2} = \frac{1}{\sqrt{6}}\myvec{1\\2\\1}$

Principal stress σ3

The system of equations is

$\tag{27.I} -2 \cdot n_{x3} + 1 \cdot n_{y3} + 3 \cdot n_{z3} = 0$

$\tag{27.II} 1 \cdot n_{x3} + 2 \cdot n_{y3} + 1 \cdot n_{z3} = 0$

$\tag{27.III} 3 \cdot n_{x3} + 1 \cdot n_{y3} -2 \cdot n_{z3} = 0$

Here, too, no further demonstration of the solution is required, since it is analogous to main stress direction 1. The normal vector for the third principal stress is

$\tag{28} \vec{n_3} = \frac{1}{\sqrt{3}}\myvec{1\\-1\\1}$

reg. d)

### Calculation of the stress vector

The stress vector is calculated using Cauchy's stress equation:

$\tag{29} \vec{S} = S \cdot \vec{n}$

$\tag{30} \vec{S} = \frac{1}{13} \cdot \myvec{1 & 1 & 3\\1 & 5 & 1\\3 & 1 & 1} \cdot \, \myvec{3\\4\\12} \, Nmm^{-2}$

$\tag{31} \vec{S} = \frac{1}{13} \cdot \myvec{1\cdot3+1\cdot4+3\cdot12\\1\cdot3+5\cdot4+1\cdot12\\3\cdot3+1\cdot4+1\cdot12} Nmm^{-2}$

$\tag{32} \vec{S} = \frac{1}{13} \cdot \myvec{43\\35\\25} Nmm^{-2}$

We have come to the end, all of the above questions have been answered.

We have some interesting links in German language about the calculation of tensors and especially invariants. Maybe the language isn't so important in mathematical expressions: the article about tensor algebra at Wikipedia and our compiled sources for exercises and solution regarding engineering mechanics.