Friday, 8 September 2017

SHEAR STRESS PRODUCED IN A CIRCULAR SHAFT SUBJECTED TO TORSION

SHEAR STRESS PRODUCED IN A CIRCULAR SHAFT SUBJECTED TO TORSION


Now we are going further to start a new topic i.e. Torsion or twisting moment with the help of this post. We will also derive here the expression of shear stress developed in a circular shaft subjected with torsion.

So, what is torsion or twisting moment?

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 discuss here one case of circular shaft which will be subjected to torsion and we will also secure the expression for shear stress and shear strain developed in the material of the shaft.

Expression for shear stress developed in the circular shaft subjected to torsion

Let us consider a shaft of circular cross-section fixed at one end AA and free at other end BB as displayed here in following figure. Let us consider one line CD at the outer surface of the circular shaft.

Let us think that torque T is applied at the end of shaft BB in clock wise direction as displayed in following figure. Hence, shaft at end BB will be rotated in clockwise direction and therefore each cross-section of the circular shaft will be subjected to shear stresses.
We have following information from above figure
R = Radius of the circular shaft
L = Length of the circular shaft
T= Torque applied at the free end BB in clockwise direction
C = Modulus of rigidity of the shaft material
τ = Shear stress induced in the material of the shaft due to the application of torque T

CD will be shifted to CD’ and
D will be shifted to D’
OD will be shifted to OD’
θ = Angle of twist i.e. Angle DOD’
ϕ = Shear strain i.e. Angle DCD’

From above diagram, we can write

Tan ϕ = DD’/CD = DD’/L
ϕ = DD’/L
As ϕ is quite small and therefore we can say Tan ϕ = ϕ

From above figure, we can also say that

DD’ = OD x θ
DD’ = R x θ
 
Hence, we have two equations as mentioned here
ϕ = DD’/L
DD’ = R x θ

Let us use the value of DD’ = R x θ in above equation and we will have following equation as mentioned here

ϕ = R x θ /L
Shear strain, ϕ = R x θ /L

Let us remind the concept of modulus of rigidity
Modulus of rigidity is also termed as shear modulus and we can define it as ratio of shear stress to shear strain. Modulus of rigidity will be indicated by C.

Modulus of rigidity (C) = Shear stress /shear strain
C = τ /(R x θ /L)
C = τ x L/R x θ
(C x θ) / L= τ /R

Shear stress developed in the shaft material could be expressed as mentioned here

τ = (C x θ x R)/L

For a given shaft and given applied torque, value of C, θ and L will be constant and therefore shear stress developed in the shaft material will be directionally proportional to the radius of the shaft.

Shear stress (τ) α Radius of the shaft (R)

 
Shear stress at the outer surface of the shaft will be maximum and shear stress will be zero at the axis of the shaft.

Do you have suggestions? Please write in comment box.

We will now discuss Torque 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

Also read

Thermal properties of material

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