# Equality of Normal Stresses on opposite faces of the element

Consider a body, which is arbitrarily loaded by some loads. I take an element at some point of the body as shown, and show the stresses acting on its faces.

To specify a face I will be using the axis which is perpendicular to it. For eg. the front face is the +ve x face, and opposite to it is the -ve x face.

All the sources that I'm following for studying mechanics of materials, state that the normal stresses at opposite faces should be equal.

However, I feel that might not necessarily be the case. In order to satisfy the equilibrium along x axis, for example, the normal stresses on the +x and -x can be different and the equilibrium could be established by shear stresses acting along the x axis, in the +y,-y, +z,-z faces.

So, in this fashion too, the element can be in equilibrium, even though the normal stresses in +x and -x are not the same. Same arguments can hold true for equilibrium along y and z axes.

Then why all the sources, state that they must be equal?

To further point out that normal stresses on opposite faces can be different consider a beam in which the bending moment varies along the length. If I take an element as shown

in this element the normal stresses will be different because the bending moment on the two sections are different.

## 2 Answers

If you included the shear stresses then you would have also to account for rotational equilibrium.

• I made an edit, NMech, to better keep my point ,please have a look Feb 21, 2022 at 20:57
• In the edit are you questioning about the different bending moments on the cross-section?
– NMech
Feb 22, 2022 at 5:52
• Yes, the BM varies along the length of the beam, and hence if we take any element of length dx (in green) , the normal stress on its opposite faces will be different (bcz BM is different). Which brings me back to my original ques, that the normal stresses on opposite faces can be different. But the general state of stress at a point (as I show in the black n white image) says normal stresses will be equal on opposite faces. Feb 22, 2022 at 6:01
• I think this may be able to address some of your concerns. IMHO, the penultimate sentence is the essence of it.
– NMech
Feb 22, 2022 at 6:51
• @NMech: Many of these questions arise because of Cauchy's approach, which is used almost universally to introduce students to mechanics. In addition, for structural elements, the fact that forces are integrated over the cross-section/thickness is never made explicit, causing further confusion. See youtube.com/watch?v=_Oa1MDDNEb4 for a discussion of the issue. Feb 23, 2022 at 0:43

The classic stress analysis was performed on an "infinitive" small element, or you may think it is performed on a "particle" of a solid element, with all other effects, such as weight and deformations but the direct stresses, are ignored for ease of formulation and understanding.

ADD:

The stresses on the element are the state of stresses derived from the special cases of "uniaxial stress" and "pure shear" on a "stress element" with the assumption that the dimensions of the stress element to be infinitesimally small (the dimensions approach zero).

The figure in your revision represents the "flexural/bending" stress which is a different type/state of stress resulting from the effect of internal stress change from point to point over a defined distance "dx".

• I made an edit, r13 to better keep my point, please have a look Feb 21, 2022 at 20:56
• Is it that since the element is infinitesimally small, the normal stresses will be very close to each other in value, so that I can assume them to be equal? Feb 22, 2022 at 5:40
• Yes, IMO, that's been the point behind the assumption. Structural engineering is a branch of science, but exact. We often make assumptions to simplify the problems based on the "significance" of a matter in the end result. Otherwise, we will never learn and get the job done.
– r13
Feb 22, 2022 at 16:36