# Why are 3 planes needed to define stress at a point?

My question is simple. Why do we need 9 different quantities, ie 1 normal stress and 2 shear stresses on 3 different planes, to define stress at a point?

example: http://www.geosci.usyd.edu.au/users/prey/Teaching/Geol-3101/Strain/stress.html

I think it should be enough to define the 3 stresses (normal and 2 shear) on a single plane through the point. For any other plane just use the projection of the 3 above to get the new values. Shouldn't this work?

TIA :-)

Orodruin
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Neither is correct. You need six independent quantities, the principal stresses (3 independent) and their directions (2 independent for the first direction, 1 for the second, 0 for the last). The stress tensor is a rank-2 symmetric tensor, which generally has 6 independent components.

Neither is correct. You need six independent quantities, the principal stresses (3 independent) and their directions (2 independent for the first direction, 1 for the second, 0 for the last). The stress tensor is a rank-2 symmetric tensor, which generally has 6 independent components.

Equilibrium renders the matrix symmetric reducing the need for 9 to 6 dependants right? But mathematically didn't we need 9 quantities to define stress and be able to express stresses in any plane/direction?

Can u please elaborate what you mean by "their directions"? I understand the principle stresses are the normal stresses. Thanks is advance!

Chestermiller
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The principal stresses are not the normal stresses (except in the principal directions).

One normal stress and two shear stresses on a specified plane is not sufficient to determine the stresses on a plane of arbitrary orientation. To do that, you need to specify the entire stress tensor. Even though, in terms of arbitrary orthogonal directions, there are nine components, the symmetry of the stress tensor reduces this to 6 independent components (3 normal stresses and 3 shear stresses). It is always possible to find 3 principal directions in which the stress tensor has only normal stress components (and no shear stress components).

One normal stress and two shear stresses on a specified plane is not sufficient to determine the stresses on a plane of arbitrary orientation. To do that, you need to specify the entire stress tensor.

Yes, this is exactly what I want to know. Why is this so, why is one normal and two shear not enough?

Thank you for the reply sir.

Chestermiller
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Yes, this is exactly what I want to know. Why is this so, why is one normal and two shear not enough?

Thank you for the reply sir.
To better idea of why, you need to review the derivation of the stress tensor.

To better idea of why, you need to review the derivation of the stress tensor.

Can you link me to a good source? I have tried many videos, websites but they all go like ' we need 3 planes to define stress at a point' or 'consider a infinitesimal cube at a point and its faces' assuming it as an axiom.

• Stephen Tashi
Chestermiller
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Can you link me to a good source? I have tried many videos, websites but they all go like ' we need 3 planes to define stress at a point' or 'consider a infinitesimal cube at a point and its faces' assuming it as an axiom.
I think it would be very worthwhile for you to learn about the stress tensor (and other 2nd order tensors) on the basis of Dyadic tensor notation. This is a very simple approach, and, for me and many many others, cleared up all my confusion about the essence of the stress tensor (as well as how to work with it). An excellent treatment of this is presented in Appendix A.3 of the famous textbook Transport Phenomena by Bird, Stewart, and Lightfoot (Wiley, New York).

I think it would be very worthwhile for you to learn about the stress tensor (and other 2nd order tensors) on the basis of Dyadic tensor notation. This is a very simple approach, and, for me and many many others, cleared up all my confusion about the essence of the stress tensor (as well as how to work with it). An excellent treatment of this is presented in Appendix A.3 of the famous textbook Transport Phenomena by Bird, Stewart, and Lightfoot (Wiley, New York).

Thank you Sir. I had a look at the reference and the closest thing I could find is this sentence "(In physical problems we often work with quantities that require the simultaneous specification of two directions. For example, the flux of x-momentum across a unit area of surface perpendicular to the у direction is a quantity of this type. Since this quantity is sometimes not the same as the flux of y-momentum perpendicular to the x direction, it is evident that specifying the two directions is not sufficient; we must also agree on the order in which the directions are given.)"

I suppose stress is like a flux quantity but I am still not sure why it requires knowing information of 3 planes (not 1,2 (too less) and not 4 (redundant)) to define it completely and in every plane possible

Chestermiller
Mentor
Thank you Sir. I had a look at the reference and the closest thing I could find is this sentence "(In physical problems we often work with quantities that require the simultaneous specification of two directions. For example, the flux of x-momentum across a unit area of surface perpendicular to the у direction is a quantity of this type. Since this quantity is sometimes not the same as the flux of y-momentum perpendicular to the x direction, it is evident that specifying the two directions is not sufficient; we must also agree on the order in which the directions are given.)"

I suppose stress is like a flux quantity but I am still not sure why it requires knowing information of 3 planes (not 1,2 (too less) and not 4 (redundant)) to define it completely and in every plane possible
In Bird et al, the treat stress as a momentum flux, which I don't really like.

In my judgment, you need to get some experience working some problems before you can better feel for how all this works. Have you seen any description of Mohr's Circle yet? This also can help you with your insight.

In Bird et al, the treat stress as a momentum flux, which I don't really like.

In my judgment, you need to get some experience working some problems before you can better feel for how all this works. Have you seen any description of Mohr's Circle yet? This also can help you with your insight.

I gave what you said a thought, starting with a simpler case of 2D. I realised that the same question I asked is valid for 2D too!!! Why 2 planes/square at a point is considered instead of one line.

I am, in some sense, confusing shear stress and normal stress. I will give it some more thinking a hopefully will reach a valid explanation for the same.

Thanks a lot sir.

• Chestermiller
Stephen Tashi