Calculation of normal and shear stress on a plane

Frequently it is necessary to calculate the normal and the shear stress on an arbitrary plane (with unit normal vector \( n \)) that crosses a rigid body in equilibrium. Faults or cracks that cross the rock mass and may lead to rock failure are cases that require the estimation of the stress components on a plane. Another example may be the calculation of the normal and shear stress on the failure surface of a specimen during a typical rock mechanics experiment.

The components of the stress vector \( T \) acting on any plane crossing an arbitrary point inside a rigid body may be calculated as follows:

\[ T_{i}^{(n)}=\sigma_{ij}n_{j} \]

where \( \sigma_{ij} \) is the stress tensor describing the stress state at that point and \( n_{j} \) are the components of the unit normal vector of the plane. In the above Eq. (1), the summation convention has been used.

The stress vector can be broken down into two components, the normal stress and the shear stress as shown in Fig. 1.

Normal and shear component of the stress vector on a plane
Figure 1: Normal and shear component of the stress vector on a plane

The magnitude of the normal component of the stress vector is calculated by:

\[ \sigma_{n}=T_{i}^{(n)}n_{i}=\sigma_{ij}n_{i}n_{j} \]

and the magnitude of the shear stress is calculated by:

\[ \tau_{n}^{2}=(T^{(n)})^{2}-\sigma_{n}^{2} \]


The stress state at a point is given by the following stress tensor:

\[ \sigma_{ij}=\left[\begin{array}{ccc}5 & 2 & 6 \\ 2 & 3& 4 \\ 6 & 4 & 1\end{array}\right] \]

For a plane with unit normal:

\[ n=\left(\frac{1}{3},\frac{1}{2},\frac{\sqrt{23}}{6}\right) \]

calculate the normal and shear components of the stress vector.

Show solution...

Firstly we will calculate the magnitude of the normal stress by using equation (2):

\[ \begin{array}{rl}\sigma_{n}=&\sigma_{11}n_{1}^{2}+\sigma_{22}n_{2}^{2}+\sigma_{33}n_{3}^{2}\\ &+2(\sigma_{12}n_{1}n_{2}+\sigma_{23}n_{2}n_{3}+\sigma_{13}n_{1}n_{3})\\=&9\end{array} \]

In order to calculate the shear component, we must calculate the components of the stress vector by virtue of equation (1):

\[ \begin{array}{l}T_{1}=\sigma_{11}n_{1}+\sigma_{12}n_{2}+\sigma_{13}n_{3}=7.5\\T_{2}=\sigma_{21}n_{1}+\sigma_{22}n_{2}+\sigma_{23}n_{3}=5.4\\T_{3}=\sigma_{31}n_{1}+\sigma_{32}n_{2}+\sigma_{33}n_{3}=4.8\end{array} \]

Hence, the magnitude of the stress vector squared is given by:

\[ (T^{(n)})^{2}=T_{1}^{2}+T_{2}^{2}+T_{3}^{2}=108.45 \]

Finally, the magnitude of the shear stress component is:

\[ |\tau_{n}|=\sqrt{(T^{(n)})^2-\sigma_{n}^{2}}=5.2 \]

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  • antonie

    Extremely helpful! Thanks a lot for the example.

  • Pantelis Liolios

    You are welcome antonie.

  • Tommy

    it is very clear and straight! thanks a lot! helpful!


    Hi, Incredible response time from you, professor Liolios. really interesting to say the least.


    Thanks for your help! I was stucked with Mohr circle but now it is clearer for me. I appreciate the examples.

  • Pantelis Liolios

    I'm glad that I helped you Edgar. Thank you for your comment.

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