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Consider a bar loaded in tension by distributed loads applied on its ends as shown in the figure. enter image description here

The stress at any cross section of this bar will be

$$\sigma = \frac{P}{A}$$

From what I know about stress it measures the intensity of internal resistive forces and is defined as internal resistive forces acting on a unit area. Thus stress is associated with internal forces (considering the bar as the system).

If I take an element somewhere in the middle of the bar (the blue element in the fig. below), both the left and right faces of this element will be subjected to internal forces and hence stress $\sigma$ as shown

enter image description here

Now let us take an element at the end of the bar (in green). One face of this element (the left one) is acted upon by internal forces and hence stress $\sigma$ but the other face (right one) is acted upon by external forces and hence I cannot show that a stress acts on the other surface too (since an external force acts on the right face and stress is based on internal force).

In the textbook they take an element at the ends and show that on both the faces of the green element a stress acts. But how is this consistent with the definition of stress, you cannot associate stress with an external force.

So my question is, why does the book show that on the right face too a stress acts even though clearly an external force is acting on it?

How do I interpret taking a stress element at the ends?

My take:

enter image description here

What if we remove a very thin slice of the material first (in red) and then take an element as shown in green? In this way both the surfaces of the green element will be subjected to internal forces and hence stress $\sigma$. Moreover since the red slice was very thin, we can still say that this element was taken at the end. Would that be a way to interpret the stresses at the ends on an element?

Harshit Rajput
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    Does this answer your question? What is the difference between stress and pressure? – Bob D Apr 27 '22 at 12:31
  • @BobD I did go through this answer before posting this question and I'm afraid it doesn't answer my question. If I follow the answers given in that question then (which I agree with), on the left face of the green element a stress will act and on the right face a pressure. Even though their magnitudes would be same, but on one face a stress and on the other face a pressure, two different quantities, still seems a little muddled to me. – Harshit Rajput Apr 27 '22 at 12:46
  • @BobD I have made some edits, about what I think. – Harshit Rajput Apr 27 '22 at 12:53
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    Stress and pressure are not "different quantities". They are both force per unit area. One is externally applied the other the internal reaction. On an element they need to be equal and opposite for the element to be in equilibrium. Sorry, but I don't understand why they seem "muddled" to you, so I can't help. Good luck! – Bob D Apr 27 '22 at 13:09
  • @BobD I call them different quantities because one measures the intensity of internal forces and the other external. The idea of one face acted upon by stress and the other by pressure seems muddled to me. I say this because, for instance to draw Mohr's Circle we need a stress element, which must be acted upon by stresses, but pressure is not a stress (because stress is based on internal forces). – Harshit Rajput Apr 27 '22 at 13:49
  • Sorry, but your response hasn’t altered my view – Bob D Apr 27 '22 at 13:51
  • @HarshitRajput “stress is based on internal force [only]” and “you cannot associate stress with an external force” are false. If you drop these misconceptions, the issue is resolved. – Chemomechanics Apr 27 '22 at 14:08
  • @Chemomechanics The very first class that I took on mechanics of materials it was pointed out to me that stress is associated with internal forces only. I found this being consistent over textbooks too and even Wikipedia. So I've build a strong conception about stress being associated with internal resistive forces. – Harshit Rajput Apr 27 '22 at 14:23
  • For example we can have a body which is acted upon by a single external force, and if the body moves, there will be no internal forces developed and hence no stress. The external force acting per unit area won't be considered as stress. – Harshit Rajput Apr 27 '22 at 14:27

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The statements "stress is based on internal force [only]" and "you cannot associate stress with an external force" are misconceptions. Elasticity theory, among other fields, doesn't distinguish between your blue element (with a right-side stress applied by the same material as the element material) and your green element (with a right-side stress applied by a distributed load).

You added that

we can have a body which is acted upon by a single external force, and if the body moves, there will be no internal forces developed and hence no stress. The external force acting per unit area won't be considered as stress.

This is untrue. A single external force applied to a body accelerates it while also deforming it due to the inertia of the rest of the body, creating nonzero stress states. You can demonstrate this by applying an axial force to the middle of a loose, compliant spring; along with acceleration, you'll obtain compression on one side and elongation on the other side, corresponding to compressive and tensile stresses, respectively, in the frame of the spring. (Alternatively, if you apply a lateral force, you'll see acceleration plus cantilever bending on either side.) If you haven't been taught this formally, it could be because introductory physics and engineering dynamics classes often idealize bodies as perfectly rigid. But no real material is perfectly rigid.

Chemomechanics
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