## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## WriteWise AI Model by Holooly Genius

Your Ultimate AI Essay Writer & Assistant.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

Products

## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## WriteWise AI Model by Holooly Genius

Your Ultimate AI Essay Writer & Assistant.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 10.6

A thin-walled hollow tube AB of conical geometry with constant wall thickness t and average diameters $d_a$ and $d_b$ as shown in Figure 10.23 is subjected to a pure torsion, T. Calculate the strain energy due to torsion of the tube. Assume material is Hookean.

## Verified Solution

For the thin-walled tube, let us calculate polar moment of inertia as follows:

$J=\left\lgroup \frac{\pi}{32} \right\rgroup\left[(d+t)^4-(d-t)^4\right]$

$=\frac{\pi}{32} d^4\left[\left\lgroup 1+\frac{t}{d} \right\rgroup^4-\left\lgroup 1-\frac{t}{d} \right\rgroup^4\right]$

$=\frac{\pi}{32} d^4\left\{\left[1+4\left\lgroup \frac{t}{d} \right\rgroup +\cdots\right]-\left[1-4\left\lgroup \frac{t}{d} \right\rgroup +\cdots\right]\right\}$

Neglecting higher powers of (t/d ) as $t<, we get

$J = \frac{\pi}{32} d^4\left\lgroup 8 \frac{t}{d} \right\rgroup$

$J=\frac{\pi}{4} d^3 t$                (1)

Let us place our coordinate x as shown in Figure 10.23. If d be the mean diameter at a distance x from end A, then

$d=d_{ a }+\left\lgroup \frac{d_{ b }-d_{ a }}{L} \right\rgroup x$             (2)

If we consider a differential portion of the hollow tube, we can consider its strain energy dU from Eq. (10.20) as

$U_{\text {torsion }}=\frac{T^2 L}{2 G J}$               (10.20)

$d U=\frac{T^2 d x}{2 G J(x)}$

The total strain energy of the tube is

$U=\int_0^L \frac{T^2 d x}{2 G J(x)}=\left\lgroup \frac{T^2}{2 G} \right\rgroup \int_0^L \frac{4 d x}{\pi d ^3 t}$

$=\left(\frac{2 T^2}{\pi G t}\right) \int_0^L \frac{ d x}{\left[d_{ a }+\left\lgroup \frac{d_{ b }-d_{ a }}{L} \right\rgroup x\right]^3}$

Therefore,          $U=\left\lgroup \frac{2 T^2}{\pi G t} \right\rgroup \frac{1}{2}\left\lgroup \frac{L}{d_{ b }-d_{ a }} \right\rgroup\left[-\frac{1}{\left\{d_{ a }+\left\lgroup \frac{d_{ b }-d_{ a }}{L} \right\rgroup x\right\}^2}\right]_0^L$

$=\frac{T^2}{\pi G t} \cdot \frac{L}{\left(d_{ b }-d_{ a }\right)}\left[-\frac{1}{d_{ b }^2}+\frac{1}{d_{ a }^2}\right]=\frac{T^2}{\pi G t} \frac{L}{\left(d_{ b }-d_{ a }\right)} \frac{d_{ b }^2-d_{ a }^2}{d_{ a }^2 d_{ b }^2}$

or            $U=\frac{T^2 L}{\pi G t} \frac{d_{ b }+d_{ a }}{d_{ a }^2 d_{ b }^2}$