Problem 2 Transform the stresses in Flaman... [FREE SOLUTION]

Chapter 30: Problem 2

Transform the stresses in Flamant's solution into polar coordinates, or in other words, derive expressions for the stress components \(\sigma_{r r}, \sigma_{\theta \theta}\) and \(\sigma_{r \theta}\). Note that the result is somewhat peculiar.

Short Answer

Expert verified

Answer: In polar coordinates, the stress components in Flamant's solution are: - \(\sigma_{rr} = \frac{6P\sin^2\theta}{\pi d^3r^3}\) - \(\sigma_{\theta\theta} = \frac{6P\sin^2\theta}{\pi d^3r^3}\) - \(\sigma_{r\theta} = 0\) This result indicates that there's a uniform distribution of stresses with no shear stress component.

Step by step solution

Write down Flamant's solution in Cartesian coordinates

Flamant's solution for stresses in an infinite elastic medium is given in Cartesian coordinates as follows: - \(\sigma_{xx} = \frac{2P}{\pi d^3}\frac{3x^2y}{r^5}\) - \(\sigma_{yy} = \frac{2P}{\pi d^3}\frac{3y^3}{r^5}\) - \(\sigma_{xy} = \frac{2P}{\pi d^3}\frac{3xy^2}{r^5}\) Where \(P\) is the load applied at the origin, \(d\) is the distance between the origin and the point of interest, and \(r\) is the radial distance from the origin to the point of interest.

Convert the coordinates from Cartesian to polar

To transform the stress components, we first need to convert Cartesian coordinates \((x,y)\) to polar coordinates \((r,\theta)\), using the following relationships: - \(x = r \cos\theta\) - \(y = r \sin\theta\)

Transform the stress components into polar coordinates

We will now substitute the polar coordinates for the Cartesian coordinates in Flamant's solution to obtain the stress components in polar coordinates. 1. For \(\sigma_{rr}\): \(\sigma_{rr} = \sigma_{xx} \cos^2 \theta + 2\sigma_{xy} \cos \theta \sin \theta + \sigma_{yy} \sin^2 \theta\) 2. For \(\sigma_{\theta\theta}\): \(\sigma_{\theta\theta} = \sigma_{xx} \sin^2 \theta - 2\sigma_{xy} \cos \theta \sin \theta + \sigma_{yy} \cos^2 \theta\) 3. For \(\sigma_{r\theta}\): \(\sigma_{r\theta} = (\sigma_{yy} - \sigma_{xx}) \cos\theta \sin\theta + \sigma_{xy}(\cos^2\theta - \sin^2\theta)\)

Simplify the obtained stress components

Now, we will simplify the expressions for the stress components obtained in Step 3 by substituting the expressions for \(\sigma_{xx}\), \(\sigma_{yy}\), and \(\sigma_{xy}\) from Step 1. After simplifying, we get: - \(\sigma_{rr} = \frac{2P}{\pi d^3}\frac{3r^2\sin^2\theta}{r^5} = \frac{6P\sin^2\theta}{\pi d^3r^3}\) - \(\sigma_{\theta\theta} = \frac{2P}{\pi d^3}\frac{3r^2\sin^2\theta}{r^5} = \frac{6P\sin^2\theta}{\pi d^3r^3}\) - \(\sigma_{r\theta} = 0\)

Analyze the obtained results

The obtained results show that the radial and azimuthal stress components \(\sigma_{rr}\) and \(\sigma_{\theta\theta}\) are equal, and the shear stress component \(\sigma_{r\theta}\) is zero. This is a peculiar result, as it indicates that in Flamant's solution, there's a uniform distribution of stresses with no shear stress component.