Riemann surface
matlab

P2.2 For the stress field of Fig. P2.1, change the known data to $\sigma_{\mathrm{xx}}=2000 \mathrm{psf}, \sigma_{\mathrm{yy}}=3000$ psf, and $\sigma_{\mathrm{n}}(\mathrm{AA})=2500$ psf. Compute $\sigma_{\mathrm{xy}}$ and the shear stress on plane AA.

In like manner, solve for the shear stress on plane $\mathrm{AA}$, using our result for $\sigma_{\mathrm{xy}}$ :
\begin{aligned} \sum \mathrm{F}{\mathrm{tAA}}= & \tau{\mathrm{AA}} \mathrm{L}-\left(2000 \cos 30^{\circ}+289 \sin 30^{\circ}\right) \mathrm{L} \sin 30^{\circ} \ & +\left(289 \cos 30^{\circ}+3000 \sin 30^{\circ}\right) \mathrm{L} \cos 30^{\circ}=0 \end{aligned}
Solve for $\tau_{\mathrm{AA}}=938-1515 \approx-\mathbf{5 7 7} \mathbf{l b f} / \mathbf{f t}^2 \quad$ Ans. (b)
This problem and Prob. P2.1 can also be solved using Mohr’s circle.

P2.3 A vertical clean glass piezometer tube has an inside diameter of $1 \mathrm{~mm}$. When a pressure is applied, water at $20^{\circ} \mathrm{C}$ rises into the tube to a height of $25 \mathrm{~cm}$. After correcting for surface tension, estimate the applied pressure in $\mathrm{Pa}$.

Solution: For water, let $Y=0.073 \mathrm{~N} / \mathrm{m}$, contact angle $\theta=0^{\circ}$, and $\gamma=9790 \mathrm{~N} / \mathrm{m}^3$. The capillary rise in the tube, from Example 1.9 of the text, is
$$h_{\text {cap }}=\frac{2 \mathrm{Y} \cos \theta}{\gamma R}=\frac{2(0.073 \mathrm{~N} / \mathrm{m}) \cos \left(0^{\circ}\right)}{\left(9790 \mathrm{~N} / \mathrm{m}^3\right)(0.0005 \mathrm{~m})}=0.030 \mathrm{~m}$$
Then the rise due to applied pressure is less by that amount: $h_{\text {press }}=0.25 \mathrm{~m}-0.03 \mathrm{~m}=0.22 \mathrm{~m}$. The applied pressure is estimated to be $p=\gamma h_{\text {press }}=\left(9790 \mathrm{~N} / \mathrm{m}^3\right)(0.22 \mathrm{~m}) \approx \mathbf{2 1 6 0} \mathbf{P a}$ Ans.

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