# What is the actual stress in a certain point in a body?

I'm reading my book on strength of materials and it says that the stress in a point depends on the orientation of the section that we are observing, and since there are infinite number of section that can go through a point, there can be infinite different stresses in that point.

So my question is what is the real stress in that point, that we use in out calculations? Is it the largest one?

Because obviously the "real" stress (if there is one) can't depend on the section we decide to look at, so I'm confused.

Let's start getting rid of the confusion by talking about something simpler than "stress", i.e. force, and keep it simple by talking about two dimensions, not three.

A "real" force on an object acts in some definite direction - for example the weight of an object acts vertically downwards. But to do any calculations with forces, we usually want to describe them using two components at right angles to each other.

For example, if you have an object is on an inclined plane and you want to consider how it moves, or whether it slips if there is friction, you often want to describe its weight using components parallel and perpendicular to the plane, not horizontal and vertical. You probably already learned how to "resolve" forces in this way.

The "real force" (the weight) doesn't depend on how we choose to describe it, but the description (the two components) does depend on the direction we choose for the components.

The same idea applies to other quantities, such as velocity, acceleration, etc. There is a name for the general concept - we call the "real thing" a vector and its "description" is the components of the vector in some coordinate system.

The same idea applies to stress. The "real stress" at a point in a body is more complicated than a vector. The name for the general concept is a "second-order tensor" ("vector" is just another name for a "first-order tensor") but in a strength of materials course you probably won't learn much about the general concept of a tensor, only some specific results about stresses and strains. In 3-dimensional space, a vector has 3 components, and a 2nd order tensor has 9 (9 = 3$$\times$$3). However, stress and strain are both "symmetric" tensors, which means there are only 6 independent components, not 9.

The engineering way to describe those 6 components is to consider a small rectangular block of material around a point, and think about the tension and shear forces on each face of the block. If we divide those forces by the area of the faces, we get a description of the three direct stress and three shear stress components at that point. (In two dimensions, there are two direct stresses and one shear stress, not three of each - other components are irrelevant).

Just like the with vectors, the size of the numbers representing the same "real stress" depends on the orientation of the little block.

The "rules" about how the numbers change when you change the orientation of the block are a bit more complicated than finding the components of a vector, and I guess that is what you are currently learning.

There is a basic problem in teaching all this. If you already know linear algebra at undergraduate level, the general idea of how to work with tensors should be straightforward, but you want to start doing practical engineering calculation with "stresses and strains" long before that point, so students get introduced to complicated-looking formulas and ideas like Mohr's circle as some sort of "magic," without the general understanding of what is going on and "why all this stuff works the way it does".

There is a second complication, in that engineers sometimes use the same words that mathematicians use, but with different meanings. For example it is common to write the six components of stress or strain in a row or column, which looks like a "vector with 6 components", even though it doesn't behave like a vector mathematically. And you can then write an equation linking stress and strain where the material properties data looks like a 6$$\times$$6 symmetric matrix which (for deep reasons, not just good luck) happens to have 21 independent entries, corresponding to the 21 independent components of the "real" fourth-order tensor. (In two dimensions the 6's become 3's).

That notation is very convenient in practice, but it completely obscures what is "really going on" in the underlying physics.

• I appreciate the effort you put into the answer, thank you very much. If I understand correctly, just like a vector can be described by 2 components, stress can be described with a tensor (6 independent components). But if we know the angle between the vector components and their magnitudes, we can calculate the magnitude of the vector (the "real" value). Can the same be done for stress? Can I calculate the "real" stress if I'm given a tensor? – user1477107 Oct 25 '19 at 16:30
• There's something called the "equivalent stress" or "von Mises stress" which is analogous to the magnitude of a vector. But the correspondence is not exact. The magnitude of a vector $\mathbf{v}$ is $\sqrt{\mathbf{v}\cdot\mathbf{v}}$. The analogous magnitude of a tensor $\boldsymbol{\sigma}$ is $\sqrt{\boldsymbol{\sigma}:\boldsymbol{\sigma}}$. If you think of the tensor as a nine-dimensional vector, the analogy is exact. – Biswajit Banerjee Oct 25 '19 at 21:08
• So is that stress the one we compare with the allowed stress at that point? – user1477107 Oct 26 '19 at 19:28
• If the "allowed" stress is computed using the same formula, then yes. Often, people just consider the deviatoric part of the stress (what you get after removing the part of stress that only causes volume change) and then equate the allowable stress to the magnitude of the deviatoric stress. – Biswajit Banerjee Oct 26 '19 at 21:30

Let's imagine you have a vertical rod with a mass m hanging from it. Let's call the vertical axis Z.

And we are interested in stresses on a point Q on the rod.

We are going to assume the rod is an isotropic material like steel, which is the case for the majority of cases, meaning it has only one uniform young modulus and a unique Poisson ratio along all the 3 axis. If no we have to handle it differently by continuum mechanics.

Then the stresses on point Q depend on the orientation of the plane we are interested in.

If we are interested in the stress on Q on a plane perpendicular to Z it is obviously, sigma= mg/A, area of the rod.

If we are interested in stress in any plane parallel to the Z the stress is zero. And if we are interested in stresses on Q along other angles we use Mohr circle.

That is why we can use a cable with no lateral resistance to do the lifting more efficiently. Because the lateral stress is zero.

In situations where we have at a given section interaction between several stresses, we find the principal stresses using the Mohr circle or the equations of Mohr.