# Plane stress concept with a fixed end

Under plane stress (z direction perpendicular to the plane), there shouldn’t be any z stress component. Then if one end of the 2D model is fixed, does that mean the displacement on that fixed boundary is completely zero (u=v=w=0), but that will generally violate the stress component along z being 0.

Sure.
is plane stress never applicable here under this BC?

Chestermiller
Mentor
is plane stress never applicable here under this BC?
How can it be? It requires the z principal stress and one other principal stress to be equal.

How can it be? It requires the z principal stress and one other principal stress to be equal.
i don't get you by 'equal'

Chestermiller
Mentor
i don't get you by 'equal'
If the stretching were in the x-direction (i.e., BC applied at x = 0), the strains in the y- and z-directions would be zero. That would mean that the stresses in the y and z directions would have to be equal.

If the stretching were in the x-direction (i.e., BC applied at x = 0), the strains in the y- and z-directions would be zero. That would mean that the stresses in the y and z directions would have to be equal.
'the strains in the y- and z-directions would be zero' for x=0 or for all x?

Chestermiller
Mentor
'the strains in the y- and z-directions would be zero' for x=0 or for all x?
At x = 0.

At x = 0.
at x=0, stress y=stress z not equal to 0, because strain x isn't 0. but plane stress requires stress z to be 0.

Chestermiller
Mentor
at x=0, stress y=stress z not equal to 0, because strain x isn't 0. but plane stress requires stress z to be 0.
Right

Right
but you seemed to disagree with plane stress being not applicable here.

Chestermiller
Mentor
but you seemed to disagree with plane stress being not applicable here.
Sure. So...?

Chestermiller
Mentor
Well, on 2nd thought, you can have a shear at the boundary like ##\partial v/\partial x## which would satisfy the requirement of plane strain.

I don’t really understand this discussion. You can apply fixed constraint (all available DOFs* = 0) in plane stress finite element analysis. For example take a rectangular cantilever beam. Fix it at one end and apply load. Your software will give you correct stresses.

* all available DOFs - translations in X and Y direction (for some elememts also rotation about the Z axis), Z translation is simply not considered in 2D analysis

Chestermiller
Mentor
In close proximity to the cantilever (say within a beam thickness), you don’t have plane strain. The beam equations are only an approximation to the full theory of elasticity equations.

Let's consider an example - cantilever beam 1.5 m long, rectangular section 0.1x0.2 m. Loaded with 16000 N at the free end. Here are the results (maximum stress):
- 3D model: 36,98 MPa
- 2D plane stress model: 36,88 MPa
- beam model: 36,1 MPa
- analytical: 36 MPa
The only problem is with stress singularity caused by the unrealistic assumption of fixed constraint itself (both in case of 3D and 2D model) but it can be ignored and the results read slightly away from the very end of the beam.

Let's consider an example - cantilever beam 1.5 m long, rectangular section 0.1x0.2 m. Loaded with 16000 N at the free end. Here are the results (maximum stress):
- 3D model: 36,98 MPa
- 2D plane stress model: 36,88 MPa
- beam model: 36,1 MPa
- analytical: 36 MPa
The only problem is with stress singularity caused by the unrealistic assumption of fixed constraint itself (both in case of 3D and 2D model) but it can be ignored and the results read slightly away from the very end of the beam.
we have been discussing the plane stress theory. of course these results differ little as you go away from the fixed end.

Sure. So...?
plane stress says stress z=0.
so is plane stress applicable here?

Chestermiller
Mentor
plane stress says stress z=0.
so is plane stress applicable here?
As I said, it certainly applies to shear at the boundary you described.

As I said, it certainly applies to shear at the boundary you described.
but principal stress z is already not 0, violating plane stress requirements, why do we mention shear anyway?

Chestermiller
Mentor
but principal stress z is already not 0, violating plane stress requirements, why do we mention shear anyway?
It is zero if we apply shear at your boundary.

Chestermiller
Mentor
If the boundary is at x = 0, and u = w = 0, and ##v=\gamma x##, then, at the boundary, u = v = w = 0 and $$\sigma_{xx}=\sigma_{yy}=\sigma_{zz}=0$$ and $$\sigma_{xy}=G\gamma$$where G is the shear modulus. So we have plane strain throughout for this homogeneous deformation even though the principal in-plane stresses are not zero, even at the boundary.

If the boundary is at x = 0, and u = w = 0, and ##v=\gamma x##, then, at the boundary, u = v = w = 0 and $$\sigma_{xx}=\sigma_{yy}=\sigma_{zz}=0$$ and $$\sigma_{xy}=G\gamma$$where G is the shear modulus. So we have plane strain throughout for this homogeneous deformation even though the principal in-plane stresses are not zero, even at the boundary.
to fix the end with shear is really a brilliant idea of rendering the shaky plane stress applicable. but on 'the principal in-plane stresses are not zero, even at the boundary.', aren't the principal in-plane stresses 0 as you wrote in your equation?

Chestermiller
Mentor
to fix the end with shear is really a brilliant idea of rendering the shaky plane stress applicable. but on 'the principal in-plane stresses are not zero, even at the boundary.', aren't the principal in-plane stresses 0 as you wrote in your equation?
Do you know how to determine the principal stresses for this specific state of stress? They are not zero. They are at two different angles to the boundary, offset from one another by 90 degrees.

Do you know how to determine the principal stresses for this specific state of stress? They are not zero. They are at two different angles to the boundary, offset from one another by 90 degrees.
sorry i got it wrong and the principal stresses aren't 0. then we can come to the conclusion that for plane stress to be applicable when 1 end is fixed, that fixed end has to be sheared.