pipe heated nonuniformly

logistic_guy

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

pipe_strain.png
 
Normal strain is given by:

ϵ=ΔLL\displaystyle \epsilon = \frac{\Delta L}{L}

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

ΔL=ϵL\displaystyle \Delta L = \epsilon L

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

ΔL=0Lϵ dL=0π/2ϵR dθ=20π/20.05cosθ dθ\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


=110sinθ0π/2=110(sinπ/2sin0)=110(10)=0.1 ft\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}}
 
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