We were discussing the concept
of stress and strain and also we have discussed the different
types of stress and also different
types of strain as well as concept
of Poisson ratio in
our previous posts. We have also seen the concept of Hook’s
Law , types
of modulus of elasticity in our recent post.

Now we are going further to start our discussion to
understand “Elongation of uniformly tapering circular rod”, in subject of
strength of material, with the help of this post.

###
**Let us see here the elongation of
uniformly tapering circular rod**

First we
will understand here, what is a uniformly tapering circular rod?

A circular rod is basically taper uniformly from one end to another end throughout the length and therefore its one end will be of larger diameter and other end will be of smaller diameter.

Let us
consider the uniformly tapering circular rod as shown in figure, length of the uniformly
tapering circular rod is L and larger diameter of the rod is D

_{1}at one end and as we have discussed that circular rod will be uniformly tapered and hence other end diameter of the circular rod will be smaller and let us assume that diameter of other end is D_{2}.
Let us consider that uniformly
tapering circular rod is subjected with an axial tensile load P and it is
displayed in above figure.

Let us
consider one infinitesimal smaller element of length dx and its diameter will
be at a distance x from its larger diameter end as displayed in above figure.

Let us
consider that diameter of infinitesimal smaller element is D

_{x}
D

_{x}= D_{1}-[(D_{1}-D_{2})/L] X
D

_{x}= D_{1}- KX
Where we
have assumed that K= (D

_{1}-D_{2})/L
Let us
consider that area of cross section of circular bar at a distance x from its
larger diameter end is A

_{x}and we will determine area as mentioned here.
A

_{x}= (П/4) D_{x}^{2}
A

_{x}= (П/4) (D_{1}- KX)^{ 2}###
**Stress**

Let us
consider that stress induced in circular bar at a distance x from its larger
diameter end is σ

_{x}and we will determine stress as mentioned here.
σ

_{x}= P/ A_{x}
σ

_{x}= P/ [(П/4) (D_{1}- KX)^{ 2}]
σ

_{x}= 4P/ [П (D_{1}- KX)^{ 2}]###
**Strain**

Let us consider that strain induced in circular bar at a distance x from its larger diameter end is Ԑ

_{x}and we will determine strain as mentioned here.

Strain =
Stress / Young’s modulus of elasticity

Ԑ

_{x}= σ_{x}/E
Ԑ

_{x}= 4P/ [П E (D_{1}- KX)^{ 2}]###
**Change in length of infinitesimal
smaller element**

Change in
length of infinitesimal smaller element will be determined by recalling the
concept of strain.

Δ dx = Ԑ

_{x}. dx
Where, Ԑ

_{x}= 4P/ [П E (D_{1}- KX)^{ 2}]
Now we
will determine the total change in length of the uniformly
tapering circular rod
by integrating the above equation from 0 to L.

And we can say that elongation of uniformly tapering circular rod will be calculated with the help of following result.

Do you
have any suggestions? Please write in comment box

###
**Reference:**

Strength of material, By R. K. Bansal

Image Courtesy: Google

We will see another important topic i.e. Stress analysis of bars of varying sections in the category of strength of material.