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 τ = G γ .

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

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.

Because it is torque and not force that causes twisting and torque is force time radius. Therefore the same force causes more torque and more twisting effect (angle of twist) the larger the radius.

The torsional angle of twist at the free end of a shaft that is fixed at the opposite end is given by the equation

$$Φ=\frac{TL}{IG}$$

where $T$ is the applied torque at the free end, L is the distance where the torque is applied from the fixed end of the shaft, $I$ is the polar moment of inertia, and $G$ is the shear modulus. The torque is the force applied perpendicular to the radius times the radius, or $T=Fr$. Therefore

$$Φ=\frac{FrL}{IG}$$

For a given force $F$, the larger the radius the greater the twisting effect (angle of twist) and the greater the deformation of the material.

Hope this helps.