Why are parts of a cylinder closer to centre elongated less during torsion?

Question

Say we have a fixed cylinder with a radius of $R$. Looking at any cross-section, the shear stress at a radius ($r$) is given by $\tau = G \gamma $.

I understand that the particles at the outer radius ($R$) twist (or elongate) more than those at any inner radius ($r$) so it makes sense that they would have more stress. However, I was wondering why it is that the outer radius particles twist more than the inner in the first place if the same force is applied at all radiuses.

To make it more clear:

Picture a large circle being twisted due to a shear force $F$. Why is it that a circle with a smaller radius will twist less due to the same shear force $F$? What is the reason behind this?

Answer 1

First, picture two disks, placed parallel, and close together. Draw a line through the center of each disk. Line up the lines. Are all the points on the lines the same distance apart? Now rotate one disk by a few degrees. Are all points on the lines the same distance apart? Have the lines drawn further apart at the center of the disks, or at the edges?

Second, consider your statement "why it is that the outer radius particles twist more than the inner in the first place if the same force is applied at all radiuses". Why do you believe this to be true? Can it be true in an isotropic material?

Answer 2

In fact the twist angle is assumed to be constant from the center of the cylinder out to its circumference. If we call the twist caused strain at any distance r from the center $ds, \ then\ \theta= \frac{ds}{r} \quad and \ ds=\theta*r$

Shear is $t=G*\epsilon= G*\theta*r \quad \text{or y in your equation}= G*yr$

Answer 3

Looking at your final question, it's important to make sure one thing is clear: a bigger cross-section (i.e. cylinder) will always twist less than a smaller one.

Yes, the outermost fiber of the cross-section will be rotated more than internal fibers, but the outermost fiber of a large cross-section will rotate less than the outermost fiber of a small cross-section. And therefore obviously the inner fiber of the large cross-section at a distance from the center equal to the smaller cross-section's radius will rotate even less.

That's because the rotation is proportional to the stress on each fiber of the cross-section. The stress is a reaction to the torsion, so Newton's Third Law tells us that the resultant "inner torsion" created by the stress must be equal to the external torsion. This inner torsion is equal to the product of the stress at a given point by its distance from the center, integrated over the entire cross-section. A larger cross-section can distribute that "inner torsion" over a larger area, so the resultant stress (and therefore deformation, by Hooke's Law) is lower than for a smaller cross-section.

As for why the outermost fibers deform more than the inner ones, there are a few ways to think about it:

1. We just assume it does. This is similar to the assumptions of Euler-Bernoulli Beam Theory, one of which is that sections that are parallel remain parallel under bending. Likewise, we assume that cross-sections under torsion behave as if suffering a solid-body rotation, with the relationships between all points within the cross-section being maintained. In a cylinder, this naturally implies that the more distant fibers must suffer more deformations: a point near the center barely has to move to rotate 1°, while one at the edge must move a lot more.

   Obviously, this isn't a blind assumption: experimental data shows this is very close to the truth.

2. If you think of rotations in terms of angles, it just kind of makes sense. As described above, if you are rotating the whole cross-section by a certain angle, of course the outer fibers will be more deformed. The only other option would be to say that the angle of rotation isn't equal across the cross-section: i.e. that instead of all fibers rotating 1° and therefore having different deformations, that they have equal deformations and therefore different rotations.

   Again, a good reason to say this isn't the case is because experimental data tells us it isn't. It would also lead to weird questions like "how to interpret the deformation at the center?"

Answer 4

If I understood correctly the question is "why it is that the outer radius particles twist more than the inner in the first place if the same force is applied at all radiuses."

(As kamran said), for a given crosssection they do not twist more. The twist angle is the same. What is different is how much the layer travels given a radius $r$

Imagine if the the solid beam is constructed by thin walled cylinders. One inside the other and they are bonded. If you hold the inner cylinder and try to rotate the outer then what will happen is that the shear forces between each layer will cause the layers in between to move. Each layer will sweep along the adjacent layer.

So a way to think about it is the following:

• Every layer will try to have the same twist angle
• However, the outer layers need to translate more to maintain that twist angle
• The greater translation is equivalent to more strains.
• More strains results in higher stresses.