Torsion equation or torsion constant is defined as the geometrical property of a bar’s cross-section that is involved in the axis of the bar that has a relationship between the angle of twist and applied torque whose SI unit is m4. The torsion equation is given as follows:

Derivation : -

Following are the assumptions made for the derivation of torsion equation:

• The material is homogeneous (elastic property throughout)
• The material should follow Hook’s law
• The material should have shear stress proportional to shear strain
• The cross-sectional area should be plane
• The circular section should be circular
• Every diameter of the material should rotate through the same angle
• The stress of the material should not exceed the elastic limit

$Angle\ in\ radius\ = \frac{arc}{radius}$

Let,

R = Radius of the circular shaft

D = Diameter of the circular shaft

dr = Thickness of small elementary circular ring

r = Radius of the small elementary of circular ring

q = Shear stress at a radius r from the centre of the circular shaft

τ = Shear stress at outer surface of shaft

dA = Area of the small elementary of circular ring

dA = 2П x r x dr

Shear stress, at a radius r from the centre, could be determined as

$\frac{q}{r} = \frac{τ}{R}$

$q = τ * \frac{r}{R}$

Turning force due to shear stress at a radius r from the centre could be determined as

dF = q x dA

$dF = τ * \frac{r}{R} * 2\pi * r * dr$

$dF = \frac{τ}{R} * 2\pi r^2dr$

Twisting moment at the circular elementary ring could be determined as

dT = Turning force x r

$dT = \frac{τ}{R} * 2П r^3dr$

$dT = \frac{τ}{R} * r^2 x (2\pi * r * dr)$

$dT = \frac{τ}{R} * r^2 * dA$

Total torque could be easily determined by integrating the above equation between limits 0 and R Therefore total torque transmitted by a circular solid shaft could be given in following way as displayed here in following figure.

Let us recall here the basic concept of Polar moment of inertia and we can write here the formula for polar moment inertia. Further, we will use this formula of polar moment of inertia in above equation. Polar moment of inertia

Therefore total torque transmitted by a circular solid shaft could be given by following equation as mentioned here.

Therefore, Finally we can write here the expression for torsion equation for circular shaft as -