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CAUCHY-GREEN strain tensor

In this exercise the CAUCHY-GREEN strain tensor for a given stress state is determined.

Task

For an ideally elastic material, the following stress state is given at a point S:

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

\[ S = \myvec{4 & 2 & 0\\2 & 1 & 0\\0 & 0 & 0} \cdot 100 \, Nmm^{-2} \]

Determine the CAUCHY-GREEN strain tensor!

Solution

It is a plain state of tension. Even for a plane stress state, deformation can occur in three directions. With the elongations εx, … and shearings γxy, … the general form of the CAUCHY-GREEN strain tensor reads:

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

\[\tag{1} C = \myvec{\epsilon_x & \frac{1}{2} \gamma_{xy} & \frac{1}{2} \gamma_{zx}\\ \ \frac{1}{2} \gamma_{xy} & \epsilon_y & \frac{1}{2} \gamma_{zy}\\\frac{1}{2} \gamma_{xz} & \frac{1}{2} \gamma_{zy} & \epsilon_z} \]

The following applies to the elongations

\[ \tag{2} \epsilon_x = \frac{1}{E} \left[ \sigma_x -- \nu \left( \sigma_y + \sigma_z \right) \right] \]

\[ \tag{3} \epsilon_y = \frac{1}{E} \left[ \sigma_y -- \nu \left( \sigma_x + \sigma_z \right) \right] \]

\[ \tag{4} \epsilon_z = \frac{1}{E} \left[ \sigma_z -- \nu \left( \sigma_x + \sigma_y \right) \right] \]

E is the modulus of elasticity and ν is the Poisson’s ratio.

The stresses can be read off directly from S, so that the following strains result. (The units for the stresses are omitted for better readability.)

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

\[ \tag{5} \epsilon_x = \frac{400-100 \nu}{E} \]

\[ \tag{6} \epsilon_y = \frac{100-400 \nu}{E} \]

\[ \tag{7} \epsilon_z = -500 \frac{\nu}{E} \]

The following applies to the shearings

\[ \tag{8} \tau_{xy} = G \gamma_{xy} \]

\[ \tag{9} \tau_{xz} = G \gamma_{xz} \]

\[ \tag{10} \tau_{yz} = G \gamma_{yz} \]

with G as the shear module

\[ \tag{11} G = \frac{E}{2(1+\nu)} \]

This gives the CAUCHY-GREEN strain tensor to

\[\tag{12} C = \myvec{\frac{400-100 \nu}{E} & \frac{200}{2G} & 0\\ \ \frac{200}{2G} & \frac{100-400 \nu}{E} & 0\\0 & 0 & -500 \frac{\nu}{E}} \]

or with (11)

\[\tag{13} C = \frac{1}{E} \myvec{400-100 \nu & 200+200\nu & 0\\ \ 200+200 \nu & 100-400 \nu & 0\\0 & 0 & -500 \nu} \]

To make the result more tangible, we use the material properties of steel, E = 210000N / mm² and ν = 0.3. The result is then

\[\tag{14} C_{Steel} = \myvec{1.76 & 1.24 & 0\\ \ 1.24 & -0.095 & 0\\0 & 0 & 0.71} \cdot 10^{-3} \]

The determined result appears plausible.

Further tasks from strength theory can be found in the category Engineering Mechanics II.