Question 10.3: A conical rod of length L and circular cross-section as show...

We position our x-coordinate as shown in Figure 10.19 itself and assume the diameter at that location to be d(x). Clearly,

\frac{d(x)-d_1}{d_2-d_1}=\frac{x}{L}

or              d(x)=d_1+\left(d_2-d_1\right) \frac{x}{L}              (1)

Now let us take free-body diagram of the portion of the rod whose length is x as shown in Figure 10.19. Now

\sigma_{x x}=\frac{P}{A(x)}, \quad \text { where } A(x)=\frac{\pi d^2(x)}{4}                  (2)

But, d \tilde{u}=\sigma_{x x} d ∈_{x x} for uniaxial stress case and ∈_{x x}=\sigma_{x x} / E as the material is Hookean. So

d \tilde{u}=\frac{\sigma_{x x} d \sigma_{x x}}{E}

Assuming \tilde{u}=0 \text { when } \sigma_{x x}=0 and integrating above, we get

\tilde{u}=\frac{\sigma_{x x}^2}{2 E}=\left\lgroup\frac{P^2}{2 E} \right\rgroup\left\{\frac{1}{A(x)}\right\}^2                   (3)

Hence,

U=\underset{\sout{V}}{\iiint} \tilde{u} d \sout{V} =\int_0^L \tilde{u} A(x) d x

Putting \tilde{u} from Eq. (3), we get

U=\int_0^L\left\lgroup\frac{P^2}{2 E} \right\rgroup \frac{ d x}{A(x)}

=\int_0^L\left\lgroup \frac{P^2}{2 E} \right\rgroup\left\lgroup \frac{4}{\pi} \right\rgroup \frac{ d x}{\left[d_1+\left\lgroup \frac{d_2-d_1}{L} \right\rgroup x\right]^2}

=\left\lgroup\frac{2 P^2}{\pi E}\right\rgroup\left\lgroup\frac{L}{d_2-d_1}\right\rgroup\left[-\frac{1}{d_1+\left\lgroup\frac{d_2-d_1}{L}\right\rgroup x}\right]_0^L

=\frac{2 P^2}{\pi E} \cdot \frac{L}{d_2-d_1}\left\lgroup -\frac{1}{d_2}+\frac{1}{d_1}\right\rgroup

or          U=\frac{2 P^2}{\pi E} \frac{L}{d_1 d_2}