Thursday, 21 September 2017

DERIVATION OF TORSIONAL EQUATION

DERIVATION OF TORSIONAL EQUATION


Now we are going further to start a new topic i.e. Derivation of torsional equation with the help of this post.

Before going ahead, let us recall the basic definition of twisting moment or torsion.

 
A shaft will said to be in torsion, if it will be subjected with two equal and opposite torques applied at its two ends.

When a shaft will be subjected to torsion or twisting moment, there will be developed shear stress and shear strain in the shaft material.

We will consider here one case of circular shaft which will be subjected to torsion and we will derive here the torsion equation for circular shaft.

We have following information from above figure

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 mentioned here
q/r = τ /R
q = τ x r/R

Turning force due to shear stress at a radius r from the centre could be determined as mentioned here
dF = q x dA
dF = τ x r/R x 2П x r x dr
dF = τ/R x 2П r2dr

Twisting moment at the circular elementary ring could be determined as mentioned here
dT = Turning force x r
dT = τ/R x 2П r3dr
dT = τ/R x r2 x (2П x r x dr)
dT = τ/R x r2 x 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.
We have already derived the expression for shear stress produced in a circular shaft subjected to torsion and therefore we have following result from that expression
Considering above two equations, we can write here the expression for torsion equation for circular shaft as displayed here.
Where, 
C = Modulus of rigidity
L = Length of shaft
θ = Angle of twist (Radian)

 
Do you have suggestions? Please write in comment box.

We will now discuss another topic i.e. Power transmitted by a circular solid shaft, in the category of strength of material, in our next post.

Reference:

Strength of material, By R. K. Bansal
Image Courtesy: Google

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