Why is stress defined in the way as it is?

Question

Stress is like pressure and it doesn't matter in which direction the force acts (given it is perpendicular to the surface).

I read in my book that if we have a rope which is being pulled on both sides by a force $F$ , then the stress at any cross section of the rope is defined as $\sigma = \frac{F}{area}$

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But my question is that since the rope is pulled from both the ends, the molecules of the considered cross section is being pulled by other molecules from both sides of the cross sections.

So the stress is due to both the forces.

So shouldn't stress be defined as

$$\sigma =\frac{2F}{area}$$

Edit : The answer of Bob D forces me to add an edit. By $\sigma = \frac{2F}{A}$ , I meant to say

$\sigma =\frac{|F_{left}|}{A}+\frac{|F_{right}|}{A} = \frac{2F}{A}$ .

Here $|F_{left}|$ and $|F_{right}|$ are the forces applied by molecules on the left and right of the considered cross section on the cross section.

Hope it is clear now.

This post imported from StackExchange Physics at 2025-01-21 21:30 (UTC), posted by SE-user Ankit

Comments

Re your figure to the right, a rope generally cannot withstand compressive stress.

This post imported from StackExchange Physics at 2025-01-21 21:30 (UTC), posted by SE-user Bob D

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The diagram does not distinguish between the resistive force of the rope and the pulling force. Indeed, two 1 lb weights on pulleys at either end is the same as a 2 lb weight on one end with the other fixed. The stress is the pulling force/area.

This post imported from StackExchange Physics at 2025-01-21 21:30 (UTC), posted by SE-user Robert DiGiovanni

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When you pull on two sides of a rope with 10 lbs, the rope feels 10 lbs, and not 20 lbs.

This post imported from StackExchange Physics at 2025-01-21 21:30 (UTC), posted by SE-user John Alexiou

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If I understood correctly, you're tryng to say the stress must be the sum of the magnitude of forces divided by area? I.e: if $ \sum F_i$ forces are acting on a face then the force which comes in stress equation must be $ \sum | F_i | $ ? @Ankit

This post imported from StackExchange Physics at 2025-01-21 21:30 (UTC), posted by SE-user Brian

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@Buraian exactly..

This post imported from StackExchange Physics at 2025-01-21 21:30 (UTC), posted by SE-user Ankit

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Probably it'd be more clear to others if you add that part into your question

This post imported from StackExchange Physics at 2025-01-21 21:30 (UTC), posted by SE-user Brian

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@Ankit in context of comment by Buraian, I think you must be confused regarding the definition of the stress itself. Stress is not the net external force per unit area. But, it is net "restoring force" developed per unit area. Hope it helps

This post imported from StackExchange Physics at 2025-01-21 21:30 (UTC), posted by SE-user SteelCubes

Answer 1

I think your perplexity is understandable, and it comes from the clash between the notion of stress, which belongs to continuum mechanics, and the molecular description. Another source of confusion is the way tension or pressure are explained in many textbooks, as "the force exerted on the surface...". This idea is confusing. For example, one could reason that the surface feels one such force from one side, and an opposite force from the other side; so the total force on the surface should be zero. Other lines of reasoning could lead to the factor of 2 that you wonder about.

One short answer is this: think of tension or pressure at (or across) a surface as the force exerted from the material immediately on the one side of the surface, on the material immediately on the other side. By the principle of action and reaction, the material on the other side will also exert a force with opposite direction on the material on the first side. The crucial point here is that these two opposite forces are exerted on different (albeit "infinitesimally" close) tiny layers of material. This is why we don't want the factor 2. Check also the insightful picture in Agnius Vasiliauskas answer.

The stress tensor, finally, is just a matematical object that allows you to encode all these forces for any imaginary surface you choose in the material: (a) Choose a small surface (so small that it can be considered flat). (b) Find the two opposite normals to that surface. (c) Choose the direction of the normal from the material exerting the force to the material experiencing the force. (d) Then the (tensile) stress tensor times this normal gives you the direction and magnitude of the tension experienced, divided by the small area.

Now if you choose the other normal, that is the first one with a minus sign, you are exchanging the role of the side that's exerting and the side that's experiencing the force. And the result will be minus the tension you found with the other normal.

Here is a longer but maybe more insightful answer.

The notion of stress (more precisely: internal pressure) was introduced by Euler around 1753 and then generalized by Cauchy around 1828. Euler's question, summarizing a little, was the following: if I have a body of matter, delimited by some boundary, how can I represent the total force exerted on it by the matter outside of it? His idea was to consider forces that act purely on the boundary of the body (just like when we have something pressing on our skin). The total force on the body of matter could then be found integrating these surface forces over the whole surface. His revolutionary idea was that we could imagine to delimit an arbitrary inner portion of a body by an imaginary surface, and consider the forces acting on the body right "below" this surface. Euler considered only forces orthogonal to the surface, and Cauchy generalized them to forces with arbitrary directions – for example tangential to the surface: that's what viscosity is. Cauchy also showed that such a force $\pmb{t}$ could actually be expressed by the action of a linear operator – the stress tensor $\pmb{\tau}$ – on the normal to the surface: $\pmb{t} = \pmb{\tau}\pmb{n}$.

The invention of stress also suggests the best way to think about it, in my opinion, as described above. Do not imagine two sides of a surface. Instead, imagine a 3D portion of matter delimited by a closed boundary. The stress is just a field of force which that 3D portion of matter "feels" on that boundary, exerted by external agents. The stress is called "tensile" if the force is directed outwards and pulls on the boundary; it's called "compressive" if the force is directed inwards and presses on the boundary.

In the case of the rope, imagine a portion of it, even if very short, delimited by two circular surfaces: a short cylinder. You want to know the total force exerted on this 3D portion of rope from the rest of the rope (or anything else outside). The rope material on the lateral surface has no forces acting on it. The material by each circular surface does have a force acting on it – the stress – directed outwards with respect to our short piece of rope, which thus feels a pull at its extremities.

In continuum mechanics stress stands in contrast to so-called "body forces" or "volume forces", which instead act on every small volume of a body of matter. Chief example is gravity. Thus body forces $\pmb{f}$ scale like a volume, while stresses $\pmb{t}$ scale like an area. The total force on a body of matter $B$ is then given by the contribution of both: $$\pmb{F}_B = \iiint_{\text{bulk of $B$}} \pmb{f}\ \mathrm{d}V + \iint_{\text{boundary of $B$}} \pmb{t} \ \mathrm{d}A \ .$$ The centre of mass of the body will move as if this total force is applied directly to it. You see from these ideas and equation that there's no need for a factor $2$.

The question of the "two sides" of the surface appears when there is a body of matter $B_2$ adjacent to the first $B_1$, so that they partly share a delimiting surface $S$. By Newton's third law, if $B_2$ is pulling on $B_1$ at the surface $S$, then $B_1$ is pulling $B_2$ at the surface $S$, in the opposite direction. So if you consider the surface $S$ from $B_1$'s perspective, the stress is directed outwards, towards $B_2$. And if you consider the surface $S$ from $B_2$'s perspective, the stress is also directed outwards, towards $B_1$. The situation is no different from when we say that the Earth pulls on the Moon, and the Moon pulls on the Earth with an equal and opposite force. Only, in the case of surface forces this pulling is happening on the same spot. That's what's often confusing. But the two forces are acting on different bodies – keep this in mind.

At a molecular level surface forces don't exist. All forces are body/volume forces. The notion of stress doesn't apply here in its original sense. What we consider as stress on an (imaginary) surface from a macroscopic point of view, turns out to be one of two things, or a combination of both, from a microscopic point of view.

First: atomic/molecular body forces having short range: just few layers of molecules on one side of the imaginary surface act on just few layers of molecules on the other side. That's why, from a macroscopic perspective, we consider these forces as only existing on the surface itself.

Second: motion of molecules across the surface. Since the molecules carry momentum, momentum is decreasing on one side of the surface and increasing on the other. And since a force causes a change of momentum, macroscopically we interpret the microscopic change of momentum as a force existing on the surface. The decrease of momentum on one side of the surface is equal and opposite to the increase on the other; so the macroscopic intepretation is that the material on one side is macroscopically experiencing a given surface force, and the one on the other side an equal and opposite surface force. Many viscous forces are of this kind.

A curious final note. In contrast with molecular or particle dynamics, stress is the only kind of force that appears in general relativity instead, because action at a distance is forbidden there. In fact, when we write the Einstein equations in a "Newtonian" form, split into space and time, the Newtonian stress tensor $\pmb{\tau}$ (and the energy, but no momentum or energy flux) fully appears in the evolution equation for the metric.

Euler's article is really cool to read:

• L. Euler: Principes généraux de l'état d'équilibre des fluides, English translation General Principles of the Motion of Fluids.

A good book to get acquainted with the notion of stress and also its microscopic interpretation is

• Bird, Stewart, Lightfoot: Transport Phenomena (2nd ed. Wiley 2002); they have also written an Introductory Transport Phenomena.

There's a beautiful lecture by Truesdell on the history of the concept of stress, which can be very useful for its understanding:

• Truesdell: The Creation and Unfolding of the Concept of Stress, chap. IV in Essays in the History of Mechanics (Springer 1968).

The microscopic interpretation of stress was approached in a rigorous manner I believe first by Irving & Kirkwood at the end of the 1940s, followed by many others. Recent reviews are given by

• Murdoch: A Critique of Atomistic Definitions of the Stress Tensor, J. Elast. 88 (2007) 113–140

• Murdoch: On the identification of continuum concepts and fields with molecular variables, Contin. Mech. Thermodyn. 23 (2011) 1–26.

even if the maths in these may be somewhat advanced, from the text and the equations you can get a glimpse of all sorts of different microscopic stuff that contribute to what we macroscopically call "stress".

For the role of stress in general relativity see for example

• Gourgoulhon: 3+1 Formalism in General Relativity: Bases of Numerical Relativity (Springer 2012), also on arXiv. See eqn (5.69) there.

This post imported from StackExchange Physics at 2025-01-21 21:30 (UTC), posted by SE-user pglpm

Comments

Let us continue this discussion in chat.

This post imported from StackExchange Physics at 2025-01-21 21:30 (UTC), posted by SE-user Brian

Answer 2

But my question is that since the rope is pulled from both the ends , the molecules of the considered cross section is being pulled by other molecules from both sides of the cross sections.

True the molecules of the cross section are being pulled by the molecules of from both sides of the cross section. But the force on both sides cross section have to be equal and opposite as shown in your figure and in the free body diagrams of each side of the cut section in my Fig 1 below, otherwise the molecules of the cross section would accelerate.

Furthermore the magnitude of that force has to equal the magnitude of the tensile force applied at each end of the rope, which in this case is $F$. This is the force that is transmitted from each end of the rope into the rope.

Note that if your replaced the two forces $F$ in the middle of your diagram with $2F$ as shown in my Fig 2 below, there would be a net force on each side of the rope from the cut end.

Edit : The answer of Bob D forces me to add an edit. By $\sigma = \frac{2F}{A}$ , I meant to say

$\sigma =\frac{|F_{left}|}{A}+\frac{|F_{right}|}{A} = \frac{2F}{A}$ .

Hope it is clear now.

You can't simply add the magnitudes of the left an right forces divided by the cross sectional area. It is still the equivalent of doubling the forces that actually exist internally.

Let's say that the stress at the cross section actually is $\frac{2F}{A}$. In order to determine the force at the cross section you would multiply the stress times the cross sectional area. That will give you a force of 2F as shown in my Fig 2 free body diagrams, violating the requirements for equilibrium.

Anyone who has taken a course in mechanics of materials learns that the normal stress in a member subject to uniaxial loading (as in your figure) equals the load (force F) divided by the cross section area. Not twice the load.

Hope this helps.

This post imported from StackExchange Physics at 2025-01-21 21:30 (UTC), posted by SE-user Bob D

Comments

I think u misunderstood the question. Sorry to say that but I didn't mean to say that force gets doubled I was saying that the total stress on a cross section can be represented as $\frac{2F}{A}$ . Hope you edit or clarify this point...

This post imported from StackExchange Physics at 2025-01-21 21:30 (UTC), posted by SE-user Ankit

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@Ankit it’s 5 AM here and I just saw your edit

This post imported from StackExchange Physics at 2025-01-21 21:30 (UTC), posted by SE-user Bob D

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@Ankit I have updated my answer in response to your edit.

This post imported from StackExchange Physics at 2025-01-21 21:30 (UTC), posted by SE-user Bob D

Answer 3

The standard definition is the correct one. You can see that by investigating some different configurations where stress is not constant.

First, consider the rope being pulled with only one force, let’s say only the force to the right. In that case the rope will accelerate to the right. The stress on the left side of the rope will be zero, and the stress on the right side of the rope will be $F/A$. It would not make sense to call the stress on the right $2F/A$ in this case, because there is only one force, but it turns out that this is the same stress as you get when you apply both forces. Adding that second force does not increase the stress at the right, it just makes it uniform through the rest of the material.

Another way to consider it is a cube sitting on the floor stationary in gravity. The bottom surface of the cube will have a pressure $mg/A$. Pressure and stress are closely related (pressure is the isotropic part of stress) so it makes sense that the stress at the bottom would also be $F/A=mg/A$. Similarly, at the top the pressure is zero as is the stress.

This post imported from StackExchange Physics at 2025-01-21 21:30 (UTC), posted by SE-user Dale

Answer 4

Firstly, by its definition, the restoration force developed per unit area inside the body when subjected to the deforming force is called stress.
Secondly, by Newton's third law of motion every action has an equal and opposite reaction and both act on two different bodies.
Now let's take your rope or whatever body it is in consideration
We apply a pair of deforming forces of magnitude F on both ends of the rope, to make sure the body is in rest. Now,

Let's consider a cross-section of the body at a general distance x from left end (sorry, I forgot to show it in the above diagram). There would be some tensile force developed inside the body. If the rope is kept at rest by balanced force F, we can say that T=F.
Now let's consider the equilibrium of any of the parts of the rope.

The tensile force will act as a restoring force (because you know, your rope didn't just snap). Also, applying Newton's third law of motion, we can see that development of tensile restoring force in one of the parts would result in in development of tensile force in its anti-parallel direction, which would be present on the other part. (I hope this line has cleared something for you).
Thus, by applying the definition of stress on any of the part we have,
$\sigma=\frac{T}{A}\\ \Rightarrow\sigma=\frac{F}{A}$
Where ofcourse, A= cross-sectional area of the rope.

This post imported from StackExchange Physics at 2025-01-21 21:30 (UTC), posted by SE-user SteelCubes

Answer 5

So shouldn't stress be defined as $$\sigma =\frac{2F}{area}$$

@SteelCubes Has already given a nice answer, i will only repeat similar explanation just with a different picture. Take a look at the process right about the crack begins :

Molecular Van der Waals forces tries to push crack side A to the left, while same forces tries to push side B - to the right. Thus it can be seen that these conflicting forces acts on different bodies (sides) of a crack. Due to that fact you can't sum them. Forces under summation, must act on the same body. That's why we get $$\sigma = \frac FA$$ As an additional argument- stress has pressure units, so must be defined the same way the pressure is defined. Because otherwise it would not be pressure technically.

This post imported from StackExchange Physics at 2025-01-21 21:30 (UTC), posted by SE-user Agnius Vasiliauskas

Answer 6

Well, if you had a force on one side only, then the whole thing would accelerate in that direction.

Consider this: I step on a scale and it says 160 pounds. So

• I'm pressing down on it with 160 lbs;
• the floor pushes up on the scale with 160 lbs (otherwise the scale would sink into the floor).

(PS I'm ignoring the weight of the scale)

This post imported from StackExchange Physics at 2025-01-21 21:30 (UTC), posted by SE-user Daniel