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 $\sigma_{\mathrm{n}}(\mathrm{AA})=2500 \mathrm{psf}$. Compute $\sigma_{\mathrm{xy}}$ and the shear stress on plane AA.

Solution: Sum forces normal to and tangential to AA in the element freebody above, with $\sigma_{\mathrm{n}}(\mathrm{AA})$ known and $\sigma_{\mathrm{xy}}$ unknown:
\begin{aligned} \sum \mathrm{F}{\mathrm{n}, \mathrm{AA}}= & 2500 \mathrm{~L}-\left(\sigma{\mathrm{xy}} \cos 30^{\circ}+2000 \sin 30^{\circ}\right) \mathrm{L} \sin 30^{\circ} \ & -\left(\sigma_{\mathrm{xy}} \sin 30^{\circ}+3000 \cos 30^{\circ}\right) \mathrm{L} \cos 30^{\circ}=0 \end{aligned}
Solve for $\sigma_{x y}=(2500-500-2250) / 0.866 \approx-\mathbf{2 8 9} \mathbf{l b f} / \mathbf{f t}^2 \quad$ Ans. (a)

In like manner, solve for the shear stress on plane $\mathrm{AA}$, using our result for $\sigma_{\mathrm{x}}$
\begin{aligned} \sum \mathrm{F}{\mathrm{t}, \mathrm{AA}}= & \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.

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
$$h_{c a p}=\frac{2 \mathrm{Y} \cos \theta}{\gamma R}=\frac{2(0.073 \mathrm{~N} / \mathrm{m}) \cos \left(0^{\circ}\right)}{\left(9790 \mathrm{~N} / \mathrm{m}^3\right)(0.0005 \mathrm{~m})}=0.030 \mathrm{~m}$$
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 $p=\gamma h_{\text {press }}=\left(9790 \mathrm{~N} / \mathrm{m}^3\right)(0.22 \mathrm{~m}) \approx 2160 \mathrm{~Pa}$ Ans.

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