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Let $$\sigma=\left(\begin{array}{lcr} 1 & 1 & 0\\ 1&1&1\\0&1&1 \end{array}\right)$$ be the stress tensor. Find the stress vector acting on a plane through the point whose normal has components $\frac{1}{\sqrt{3}}(1,1,1)$ and deduce the component of this stress vector perpendicular to the plane. Find the angle between the stress vector and the normal to the plane.

To start off we obvious have that the plane has form $\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z=k$ where $k$ is a constant. Since this stress tensor does not depend on any $x,y,z$ am I right in assuming I can set this constant to be zero for simplicity? Then do I multiply the tensor and normal vector together to get the vector on the plane? Then just trigonometry it to get the angle?

Daniel Piggott
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The stress vector $\boldsymbol{t}$ acting on a surface at a point $\boldsymbol{x}$ is given by $$\boldsymbol{t} = \sigma\left(\boldsymbol{x}\right)\boldsymbol{n}\left(\boldsymbol{x}\right),$$ where $\sigma$ is the stress tensor and $\boldsymbol{n}$ is the outward unit normal to the surface at $\boldsymbol{x}$. As you have already noted, the stress tensor $\sigma$ and unit normal $\boldsymbol{n}$ is constant for all $\boldsymbol{x}$ on the plane. Hence the stress vector $\boldsymbol{t}$, is given by $$\boldsymbol{t}=\sigma\boldsymbol{n} = \left(\begin{array}{ccc} 1 & 1 & 0\\ 1 & 1 & 1\\ 0 & 1 & 1\end{array}\right) \left(\begin{array}{c} \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}\end{array}\right)= \left(\begin{array}{c} \frac{2}{\sqrt{3}}\\ \frac{3}{\sqrt{3}}\\ \frac{2}{\sqrt{3}}\end{array}\right).$$

As $\boldsymbol{n}$ has magnitude 1, then $\boldsymbol{c}$, the component of $\boldsymbol{t}$ perpendicular to the plane, is given by $$\boldsymbol{c} = (\boldsymbol{t}\cdot\boldsymbol{n})\boldsymbol{n}=\frac{7}{3}\boldsymbol{n}.$$

Finally $\theta$, the angle between the stress vector and the normal to the plane, is given by $$\theta=\cos^{-1}\left(\frac{\boldsymbol{t}\cdot\boldsymbol{n}}{|\boldsymbol{t}||\boldsymbol{n}|}\right)=\cos^{-1}\left(\frac{7/3}{\sqrt{17}/\sqrt{3}}\right).$$

Lythia
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