pipe heated nonuniformly

[logistic_guy]

The curved pipe has an original radius of \(\displaystyle 2 \ \text{ft}\). If it is heated nonuniformly, so that the normal strain along its length is \(\displaystyle \epsilon = 0.05\cos\theta\), determine the increase in length of the pipe.

 

[logistic_guy]

Normal strain is given by:

\(\displaystyle \epsilon = \frac{\Delta L}{L}\)

The increase in length is \(\displaystyle \Delta L\), then the formula becomes:

\(\displaystyle \Delta L = \epsilon L\)

Since the pipe is heated nonuniformly, we have to integrate over the length.

\(\displaystyle \Delta L = \int_{0}^{L}\epsilon \ dL = \int_{0}^{\pi/2}\epsilon R \ d\theta = 2\int_{0}^{\pi/2}0.05\cos\theta \ d\theta\)


\(\displaystyle =\frac{1}{10}\sin\theta \bigg|_{0}^{\pi/2} = \frac{1}{10}(\sin \pi/2 - \sin 0) = \frac{1}{10}(1 - 0) = \textcolor{blue}{0.1 \ \text{ft}}\)