How does stress in solids work?

Question

How do you figure out what the stress equations are? How do you find a solution to the stress equation?

Answer 1

In mechanics of deformable solids there are basically three major types of mechanical stress: (1) Normal Stress, (2) Shear Stress, and (3) Torsional Stress. Normal stress is due to the component of a force normal (perpendicular to) and areas. Shear stress is due to the component of a force parallel to an area. Torsional stress is due to the application of a torque (twisting force) typically times the radius of a shaft divided by the polar moment of inertia. If the loading of the material is in the linear elastic region, each type of stress is associated with specific equation that provides information on the amount of strain (deformation) that can occur for a given load. Normal strain, shear strain, and torsional strain.

In addition to the above types of stresses there are other special cases including but not limited to thermal stresses that result in thermal deformation due to the expansion or contraction of materials with temperature.

Three examples of equations relating are:

1. For uniaxial loading and deformation

$$ε=\frac{δ}{L}=\frac{P}{AE}=\frac{σ}{E}$$

Where $δ$ = the elastic longitudinal deformation

$L$ = original length

$P$ = the normal load (force)

$A$ = original cross sectional area

$ε$ = engineering strain (change in length per unit length).

$E$ = Young's Modulus, or Modulus of Elasticity.

$σ$ = the normal Engineering Stress

2. For shear stress and strain

$$Υ=\frac{τ}{G}$$

Where $Υ$ = shear strain

$τ$ = shear stress = $\frac{P}{A}$ where (P) is the load parallel to a surface $A$ (shear force)

$G$ = Shear modulus = $\frac{E}{2(1+ν)}$ where $ν$ is poisson's ratio

3. For torsional stress/strain for a solid or thick-walled shaft

$$τ=\frac{Tr}{J}$$

Where $τ$ = torsional stress

$T$ = torque

$r$ = shaft radius

$J$ = polar moment of inertia.

The angle of twist $Φ$ at the free end of a shaft of length $L$ at the free end with the other end fixed is

$$Φ=\frac{TL}{GJ}$$

The above are only special cases where only normal, shear, and torsional stresses are involved. Solids subjected to a combination of different loads involve more complex equations and tools. For example, Mohr's Circle is a tool for analyzing stresses due to combinations of normal and shearing loads on solids, called principle stresses, in the case of a two dimensional stress state.

Hope this helps.

Answer 2

I have no engineering knowledge. All this is a speculative guess and can be figured out using the math.

First of all, not all solid objects are even trying to form into a shape that exists in Euclidean geometry. That can either be from part of the object being heated. Or it can be from it inherently trying to form into a non-Euclidean shape. For example, Prince Rupert's drops are made by dripping molten glass into water and they are under strong compression on the surface and strong tension in the interior and if you snap the tail, all the tension energy is released and the whole thing explodes into many tiny pieces. I'll just talk about objects that don't have an internal molecular structure trying to form them into a non-Euclidean shape and would be free of stress if they were an a zero gravity environment and weren't rotating.

All stressed objects locally in any spot are just simply undergoing a linear transformation. Lets take for example any perfect crystal diamond that was etched into any shape with forces acting on it from the outside and putting it under stress in a gravity free environment. We have a function that assigns to each atom position in the lattice its position in space. Each part of it can locally be modelled by a liner transformation and let's just call one point in the lattice and its position in space the origin. We know that there are 9 degrees of freedom of linear transformation. Furthermore, suppose the amount of transformation from the untransformed state is infinitesimal, even the pure rotation part is infinitesimal.

I found a simple choice of basis where 3 of the components are pure rotation ones. Let γ be the infinitesimal number. Then here are the 9 component linear transformations.

• (x, y, z) → (x - γy, y + γx, z)
• (x, y, z) → (x - γz, y, z + γx)
• (x, y, z) → (x, y - γz, z + γy)
• (x, y, z) → (x + γy, y + γx, z)
• (x, y, z) → (x + γz, y, z + γx)
• (x, y, z) → (x, y + γz, z + γy)
• (x, y, z) → (x + γx, y, z)
• (x, y, z) → (x, y + γy, z)
• (x, y, z) → (x, y, z + γz)

Also at each spot, we have a power series centered around that spot. It does not necessarily mean the function is equal to the power series. We have 18 degrees of freedom of the second power terms, that is the terms of the form of a function that either squares one of the coordinates or multiplies 2 coordinates and then multiplies the result by one of the variables. In an ideal material, for some other infinitesimal quantity γ, the acceleration is also added component wise as follows

all 18 of them are

• (x, y, z) → (γx^2 + x, y, z) has acceleration (1, 0, 0)

• (x, y, z) → (γy^2 + x, y, z) has acceleration (1, 0, 0)

• (x, y, z) → (γz^2 + x, y, z) has acceleration (1, 0, 0)

• (x, y, z) → (γxy + x, y, z) has acceleration (0, 0, 0)

• (x, y, z) → (γxz + x, y, z)has acceleration (0, 0, 0)

• (x, y, z) → (γyz + x, y, z) has acceleration (0, 0, 0)

• (x, y, z) → (x, γx^2 + y, z) has acceleration (0, 1, 0)

• (x, y, z) → (x, γy^2 + y, z) has acceleration (0, 1, 0)

• (x, y, z) → (x, γz^2 + y, z) has acceleration (0, 1, 0)

• (x, y, z) → (x, γxy + y, z) has acceleration (0, 0, 0)

• (x, y, z) → (x, γxz + y, z) has acceleration (0, 0, 0)

• (x, y, z) → (x, γyz + y, z) has acceleration (0, 0, 0)

• (x, y, z) → (x, y, γx^2 + z) has acceleration (0, 0, 1)

• (x, y, z) → (x, y, γy^2 + z) has acceleration (0, 0, 1)

• (x, y, z) → (x, y, γz^2 + z) has acceleration (0, 0, 1)

• (x, y, z) → (x, y, γxy + z) has acceleration (0, 0, 0)

• (x, y, z) → (x, y, γxz + z) has acceleration (0, 0, 0)

• (x, y, z) → (x, y, γyz + z) has acceleration (0, 0, 0)

Also, for an ideal material, we lose 3 degrees of freedom of stress at the surface. For a surface perpendicular to the z axis, the 3 components of linear transformation that contribute to stress and must be 0 are

• (x, y, z) → (x + γz, y, z + γx)
• (x, y, z) → (x, y + γz, z + γy)
• (x, y, z) → (x, y, z + γz)

unless there is an outside force acting on that part of the surface.

Now finally, let's consider any object of an ideal material in a gravitational field that starts of with 0 velocity in the unstressed state and part of its surface confined to where it started. The gravitational field can be treated like acceleration so you now add to each interior part of the object gravitational acceleration. We can also treat it like there is a thin skin layer in the unbound surface parts confined to the 3 degrees of freedom of stress following different equations. To simplify things, I think we can even treat it like force is linearly proportional to velocity rather than acceleration and that the quadratic components contribute to velocity rather than acceleration. By adding time into the equation, we can see where it's going to end up after continuing to follow those equations. You just need to know how to determine how for any rotation that's not necessarily infinitesimal and infinitesimal linear transformation, if you apply the linear transformation then the rotation, which linear transformation do you have to apply after the rotation to get the same result if you apply the rotation first. Then you can get the equations for stress on a surface of any orientation. My answer at https://math.stackexchange.com/questions/3497890/how-do-you-prove-that-rotation-can-be-well-defined-in-the-third-dimension/3497891#3497891 tells you how you can generate a 3 dimensional rotation from a quaternion of magnitude 1.

For a nonideal material like diamond, I think it's not quite the same. However, stress is still a linear transformation of strain and acceleration is still a linear transformation of the quadratic terms. The way stress varies with strain and acceleration varies with the quadratic components is probably not preserved under rotation.