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By newton's third law, all internal forces in a portion of volume must be zero. Thus the total force is due to other volume portions exerting force onto the surface of said portion. However, if one just expands the portion to be the entire volume, then all the forces inside must sum to zero.

Stress equilibrium (as described in Landau and Lifshitz) is defined as the internal stresses in every volume element must balance and thus $\Sigma F_i = 0$.

I am clearly missing something fundamental about stress. These two descriptions seem identical to me. Does this mean that an object must be in equilibrium to analyze the stress? What would an example be of an object experiencing stress but not in equilibrium?

Can someone in general please explain what exactly stress is and how the infinitesimal case applies to macroscopic volumes? I am at a loss :(

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    $\begingroup$ Take a look at residual stress. With real materials in their manufacture there can be stresses that are locked in and the solid can be in equilibrium. Sometimes the residual stresses can be time and temperature dependent and slowly be relieved over time. $\endgroup$
    – UVphoton
    Commented Aug 2, 2022 at 0:43

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If we make an imaginary cut in the body, the surface force of side A on the side B is equal to the surface force of side B on side A by the Newton's third law. But the forces on volumes are zero only on the equilibrium.

One example is a unstressed body in free fall. The only force acting on each elementary volume is its weight, and is clearly non zero.

Usually, for an object lying on the ground, each elementary volume has its weight being balanced by the surface forces, resulting in compressive stresses in the vertical direction and tensile or compressive stresses sideways, according to the particular configuration.

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As you say in your first paragraph, while the internal forces in a portion sum to zero, the portions may exert unbalanced forces on one another. These forces are internal to the object as a whole, but external to each portion of the object. (Lets call these 'external-internal' forces.) For a simple example, stretch a piece of elastic, and release both ends simultaneously. The middle will exert unbalanced forces on both ends.

The internal forces inside the piece of elastic as a whole sum to zero, but internal forces acting on portions of the elastic do not. The centre of mass of the piece of elastic as a whole does not accelerate, but the portions of the elastic at either end accelerate towards the middle.

For another example, if a body like a rubber ball rotates, the movement of the outer parts of the body is pulled into a circle by unbalanced stresses in the body. These forces all pull towards the centre, and the net internal force on the body as a whole is zero, but the net internal force on parts of the ball is non-zero.

If the external-internal forces on every individual portion of a body also sum to zero, this is called stress equillibrium. The main example of interest is a static, non-rotating body (like a suspension bridge or a tower block), but there are other cases that fit. Consider the special case of of a dust consisting of mutually non-interacting particles in which there are no internal forces at all, only external ones (dust collapsing under gravity, photons in a rotating mirrored box, etc.).

If the force applied to part $i$ by part $j$ is $F_{ij}$ then we have $F_{ij}+F_{ji}=0$ and thus $\sum_{i,j}F_{ij}=0$. Think of it as an anti-symmetric matrix (with a 3D force vector inserted in every slot, rather than a single number), all of its terms must add to zero. If the net force on part $i$ is $F_i=\sum_j F_{ij}$, then we can write the condition for stress equillibrium as $F_i=0$ for every $i$. Every row and every column of the matrix must individually add to zero.

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  • $\begingroup$ Thank you so much this is a lot more clear now. I just have one part that I don’t understand still. If the sum of internal forces in any arbitrary portion must be zero. Then because it is arbitrary it would seem that would imply that there can be no unbalanced forces exerted on each other. Since if you have two portions exerting an unbalanced force on each other, then you can draw a portion which is just the sum of those two portions and then the internal forces in this portion must sum to zero. Then this would imply the forces are balanced which is a contradiction so my logic must be flawed. $\endgroup$ Commented Aug 2, 2022 at 19:59

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