Question

Derive the expression for analysis of uniformly tapering circular rod subjected to an axial load p

Answer 1

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 D1 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 D2.

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 Dx
Dx = D1-[(D1-D2)/L] X
Dx = D1- KX
Where we have assumed that K= (D1-D2)/L

Let us consider that area of cross section of circular bar at a distance x from its larger diameter end is Ax and we will determine area as mentioned here.
Ax = (П/4) Dx2
Ax = (П/4) (D1- KX) 2
Stress

Let us consider that stress induced in circular bar at a distance x from its larger diameter end is σx and we will determine stress as mentioned here.
σx = P/ Ax
σx = P/ [(П/4) (D1- KX) 2]
σx = 4P/ [П (D1- KX) 2]
Strain



Let us consider that strain induced in circular bar at a distance x from its larger diameter end is Ԑx and we will determine strain as mentioned here.

Strain = Stress / Young’s modulus of elasticity
Ԑx = σx /E
Ԑx = 4P/ [П E (D1- 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 = Ԑx. dx
Where, Ԑx = 4P/ [П E (D1- 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.

Answer 2

Answer:

The expression of uniformly tapering circular rod subjected to an axial load $P$ for analysis is $\delta L=\frac{4PL}{\pi E d_{1}d_{2}}$.

Step-by-step explanation:

The stress at any cross-section will be found by dividing the load by the world of the cross-section and extension may be found by integrating extensions of a little length over the full length of the bar.

We shall consider the subsequent cases of the variable cross-section.

Let us consider a circular bar of length$L$ tapering uniformly from the diameter $d_{1}$ at the larger end to the diameter $d_{2}$ at the littleend, and subjected to axial tensile load $P$ as shown within the figure.

Considering a little strip of length $dx$at a distance $x$ from the larger end.

The diameter of the elementary strip is

$\begin{aligned}D&=d_{1}- \frac{[(d_{1} - d_{2})x]}{L}\\&= d_{1} - kx\end$

where$k=\frac{(d_{1}-d_{2})}{L}$

The cross-sectional area of the strip,

$\begin{aligned}A_{x}&=\frac{\pi}{4}D^{2}\\ &=\frac{\pi}{4}(d_{1}-kx)^2\end$

Stress within the strip,

$\begin{aligned}\sigma_{x}&=\frac{P}{A_{x}}\\ &=\frac{P}{\frac{\pi}{4}(d_{1}-kx)^2}\\ &=\frac{4P}{\pi(d_{1}-kx)^2}\end$

Strain within the strip,

$\begin{aligned}\epsilon_{x}&=\frac{\sigma_{x}}{E}\\ &=\frac{4P}{\pi(d_{1}-kx)^2E}\end$

Elongation of the strip,

$\begin{aligned}\delta L_{x}&=\epsilon_{x}dx\\ &=\frac{4Pdx}{\pi(d_{1}-kx)^2E}\end$

The total elongation of this tapering bar will be found out by integrating the above expression between the boundaries $x=0$ to $x=L$.

$\begin{aligned}\delta L&=\int_{0}^{L}\frac{4Pdx}{\pi(d_{1}-kx)^2E}\\ &=\frac{4P}{\pi E}\int_{0}^{L}\frac{dx}{(d_{1}-kx)^{2}}\\ &=\frac{4P}{\pi E}\left[\frac{(d_{1}-kx)^{-1}}{(-1)\times (-k)}\right]^{L}_{0}\\ &=\frac{4P}{\pi E K}\left[\frac{1}{d_{1}-kx}\right]_{0}^{L}\end$

Putting the worth of$k=\frac{d_{1}-d_{2}}{L}$ within the above expression, we obtain

$\begin{aligned}\delta L&=\frac{4PL}{\pi E(d_{1}-d_{2})}\left[\frac{1}{d_{1}-\frac{(d_{1}-d_{2})L}{L}}-\frac{1}{d_1}\right]\\ &=\frac{4PL}{\pi E(d_1-d_2)}\left[\frac{1}{d_{2}}-\frac{1}{d_{1}}\right]\\ &=\frac{4PL}{\pi Ed_{1}d_{2}}\end$

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