# Interpreting stress at the ends of a bar

Consider a bar loaded in tension by distributed loads applied on its ends as shown in the figure. 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 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: 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?

• Does this answer your question? What is the difference between stress and pressure? Apr 27 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. Apr 27 at 12:46
• @BobD I have made some edits, about what I think. Apr 27 at 12:53
• 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! Apr 27 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). Apr 27 at 13:49