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# Stress vector and section plane

This exercise is about the determination of the stress vector and section plane for a given stress state and normal vector.

The following stress tensor is given at a point in space in the cartesian coordinate system:

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

$S = \myvec{1 & 2 & -2\\2 & 1 & 2\\-2 & 2 & 1} \cdot 100 \, Nmm^{-2}$

Sought are

a) the stress vector to the plane described through the normal unit vector

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

$\vec{n} = \frac{1}{3} \cdot \myvec{-2\\1\\2}$

b) the absolute value of the stress vector calculated in a) and

c) the equation that determines the section plane given by the normal unit vector.

## Solution

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

a) Calculation of the stress vector

The stress vector is calculated with the Cauchy stress equation:

$\tag{1} \vec{S} = S \cdot \vec{n}$

$\tag{2} \vec{S} = \myvec{1 & 2 & -2\\2 & 1 & 2\\-2 & 2 & 1} \cdot \myvec{-2\\1\\2} \cdot \frac{1}{3} \cdot 100 \, Nmm^{-2}$

$\tag{3} \vec{S} = \myvec{-4\\1\\0} \cdot \frac{100}{3} Nmm^{-2}$

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

b) Absolute value of the stress vector

The absolute value of the stress vector follows from

$\tag{4} | \vec{S} | = \sqrt{S_x^2+S_y^2+S_z^2}$

$\tag{5} | \vec{S} | = \sqrt{-4^2+1^2+0^2} \cdot \frac{100}{3} Nmm^{-2}$

$\tag{6} | \vec{S} | = 137,4 \, Nmm^{-2}$

$\newcommand{\myvec}{{\begin{bmatrix}#1\end{bmatrix}}}$ $\require{cancel}$

c) Equation for the plane from the normal vector

The general equation for a plane derived from a normal vector is

$\tag{7} a\cdot x + b\cdot y +c \cdot z = 0$

with the components of the normal vector (the given 1/3 can be neglected for the plane equation )

$\tag{8} \vec{n} = \myvec{a\\b\\c} \cdot \bcancel{\frac{1}{3}}$

Thus the equation for the section plane is

$\tag{9} -2x + y + 2z = 0$