We were discussing the concept of Torsion or
twisting moment, Torque transmitted by a circular solid shaft and torque transmitted by a circular hollow shaft in our previous posts.

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*

*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

Ï„ = Shear stress at outer surface of
shaft

dA = Area of the small elementary of
circular ring

dA = 2ÐŸ x r x dr

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

q/r = Ï„ /R

q = Ï„ 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 = Ï„ x r/R x 2ÐŸ x r x dr

dF = Ï„/R x 2ÐŸ r

^{2}dr
Twisting moment at the circular elementary
ring could be determined as mentioned here

dT = Turning force x r

dT = Ï„/R x 2ÐŸ r

^{3}dr
dT = Ï„/R x r

^{2 }x (2ÐŸ x r x dr)
dT = Ï„/R x r

^{2 }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

Î¸ = 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

nice

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