## What is Torsion Equation?

When a shaft is subjected to a couple, shear stress will be induced in the shaft. If the axis of rotation of the couple coincides with the longitudinal axis of the shaft, then the shaft is said to be subjected to pure torsion. The theory of pure torsion is the study of the behavior of the shaft in torsion without accounting for the bending moment caused by one's weight or other longitudinal forces.

The torsion equation is derived from the theory of pure torsion that relates applied torque, torsional shear stress generated in the shaft, angle of twist, and geometrical properties of the shaft’s cross-section.

## Torsion Equation Formula

The torsion equation formula for a shaft of uniform cross-section along the length is given as

T/J = τ_{max}/R = Gθ/L

Where

- T = torque applied to the shaft
- J = polar moment of inertia
- τ
_{max}= maximum shear stress produced by the shaft - R = radius of the shaft
- G = modulus of rigidity of the material of the shaft
- θ= angle of twist in radians
- L = length of the shaft

## Assumptions in Torsion Equation

There are some assumptions involved in the torsion equation. These assumptions need to be studied before performing torsion equation derivation. These assumptions are listed below:

- Material is isotropic and homogeneous
- Material is linearly elastic and obeys Hooke’s law.
- The shaft should have a uniform cross-section and be perfectly straight.
- The cross-section of the shaft should remain plane before and after twisting.
- All the diameters of the cross-section should rotate at the same angle.

## Torsion Equation Derivation

Consider a solid cylindrical shaft of radius R and length L as shown in the figure. A twisting moment T is applied at one end of the shaft and a balancing couple of equal magnitude on the other end. This torque will result in angular deformation in the shaft.

Assume a line PM on the surface of the shaft parallel to the longitudinal axis before the deformation of the shaft and after the deformation due to torsion, the line becomes PN. The angle MPN = represents the shear strain at the surface of the shaft. The arc MN will be,

MN=LΦ (∵ Φ is small)

Φ=MN/L…(i)

Let angle MON =θ which is the angular movement of the radius OM due to twisting. It is also called the angle of twist. The shear stress will vary from zero at the centre of the shaft to maximum at the surface. Let τ_{max} be the maximum shear stress at the surface of the shaft

We know that the modulus of rigidity,

G=shear stress/Shear strain=τ_{max}/Φ…(ii)

τ_{max}=G

From eq. (i)

τ_{max}=G×MN/L

But, MN=Rθ

τ_{max}=GRθ/L

τ_{max}R=Gθ/L…(iii)

Let be the shear stress at an interior section of the shaft at a radius of r.

τ/r=Gθ/L

τ_{max}/R=τ/r=Gθ/L…(iv)

### Moment of Resistance

Consider an elemental area dA at any radius r on the cross-section of the shaft. The shear stress developed by this elemental area will be

τ=τ_{max}r/R [from eq. (iv)]

Shear force produced by the elemental area,

τdA=τ_{max}/R)rdA

Therefore, the moment of resistance produced by the elemental area

τdA=τ_{max}/R)rdA.r

τdA=(τ_{max}/R)dA.r^{2}

Total moment of resistance produced by the entire cross-section of the shaft

T=(τ_{max}/R)ΣdA.r^{2}

We know that ΣdA.r^{2} is the polar moment of inertia of the shaft (moment of inertia about the longitudinal axis of the shaft).

T=(τ_{max}/R). J

Where J = polar moment of inertia of the shaft

T/J=τ_{max}/R…(v)

From equations (iii) and (v) we get the torsion equation as

T/J=τ_{max}/R=Gθ/L

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