Short Answer:

The torsion equation links the applied torque , the shaft geometry, material stiffness and the resulting shear stress and twist:

It means torque per polar inertia equals shear stress divided by radius and equals shear modulus times angle of twist per unit length. From this we get the useful forms  and .

In simple words: when a circular shaft is twisted by a torque, the shear stress increases linearly from center to outer radius and the shaft rotates by an angle proportional to torque and length, and inversely proportional to stiffness and cross-sectional polar inertia.

Derive torsion equation

Detailed Explanation :

Purpose and assumptions
We derive the torsion equation for a straight, prismatic, circular shaft under elastic torsion. Main assumptions (simple and essential) are: the material is homogeneous and linear elastic (Hooke’s law), cross-sections remain plane and circular after twisting, deformations are small, and the shaft is subjected to pure torque (no bending or axial load). These assumptions let us relate geometry, material, and stresses simply.

Geometry and variables
Consider a circular shaft length , outer radius  (or  as a generic radial coordinate), subjected to a torque  at one end while the other end is fixed. A cross-section at a distance  from the fixed end rotates relative to a neighboring cross-section by a small angle. Let  be the total angle of twist over length . Let  be the shear modulus of the material,  the polar moment of inertia of the cross-section about its axis, and  the shear stress at radius .

Kinematics — strain distribution
Take a circular fiber at radius  from the center. After twisting, a cross section at  is rotated by angle  and the cross section at  by . The relative rotation over  is . A point on the circular fiber originally at arc length  experiences longitudinal displacement around the circumference equal to . The shear strain  in the material at radius  is the relative tangential displacement per unit length along the shaft:

If the total twist over length  is , then  for uniform twist, so .

Constitutive law — stress from strain
Using Hooke’s law in shear, shear stress  is proportional to shear strain:

Thus shear stress varies linearly with radius, zero at center  and maximum at outer surface :

Equilibrium — internal moment from stress distribution
The internal resisting torque produced by shear stresses in the cross-section must balance the applied torque . Consider an elemental ring at radius  of thickness . The shear stress  acting on the elemental area  — for a solid circular section we integrate over area — produces an elemental torque  (stress × area gives force, times radius gives moment). Integrating over the whole area gives total internal torque:

Substitute :

The integral  is the polar moment of inertia  of the cross-section about the shaft axis. So

Rearrange:

Relating stress and torque
From earlier  and the expression for  from the equilibrium equation , eliminate  to express shear stress in terms of torque:

Hence the compact torsion relation:

or equivalently

Polar moment of inertia 
For circular sections  has closed forms:

• Solid circular shaft of diameter  (radius ):

• Hollow circular shaft with outer radius  and inner radius :

These formulas let you compute numerical values of  and .

Interpretation and use

•  shows shear stress increases linearly with radius; maximum at the outer fiber.
•  shows twist grows with torque and length, and reduces with larger  (stiffer cross-section) or larger  (stiffer material).
• For hollow shafts, material is removed from the center where it contributes little to  but little to strength; hollow shafts are therefore efficient for torsion.

Limits and applicability
The derivation is valid for circular shafts under elastic torsion. Noncircular sections produce warping and a different stress state; for those, torsion analysis is more complex and involves warping functions. Also, if shear strains are large or material yields, linear relations no longer hold and plastic torsion formulas apply.

Quick example
A solid shaft  carrying : . Compute  and  using  for chosen material to check stresses and twist.

Conclusion

The torsion equation  follows directly from shear strain kinematics, Hooke’s law in shear, and equilibrium of internal torque. It gives the useful results  for stress distribution and  for angle of twist. These relations are central to designing shafts and other members subjected to torque, enabling engineers to check both shear capacity and acceptable twist.