We were discussing the concept of Moment of resistance of beam section, Area moment of inertia, Direct and bending stresses
and Shear force and bending moment diagram in our previous posts.

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

Ï„ = 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’

Î¸ = Angle of twist i.e. Angle DOD’

Ï• = Shear strain i.e. Angle DCD’

####
*From
above diagram, we can write *

Tan Ï• = DD’/CD = DD’/L

Ï• = DD’/L

As Ï• is quite small and therefore we can say Tan Ï• =
Ï•

####
*From
above figure, we can also say that *

DD’ = OD x Î¸

DD’ = R x Î¸

Hence, we have two equations as mentioned here

Ï• = DD’/L

DD’ = R x Î¸

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

Ï• = R x Î¸ /L

Shear strain, Ï• = R x Î¸ /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 = Ï„ /(R x Î¸ /L)

C = Ï„ x L/R x Î¸

(C x Î¸) / L= Ï„ /R

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

####
**Ï„ = (C x Î¸ x R)/L**

For a given shaft and given applied torque, value of
C, Î¸ 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 (Ï„) Î± 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

Derivation of mass moment of inertia

Shear force and bending moment diagrams for a simply supported beam with a point load acting at midpoint of the loaded beam

Shear force and bending moment diagrams for a simply supported beam with a point load acting at midpoint of the loaded beam