# Calculation for the location of support of vertical circular gate

**Problem**

A vertical circular gate in a tunnel 8 m in diameter has oil (sp. gr. 0.80) on one side and air on the other side as shown in Figure HD-73. If oil is 12 m above the invert and the air pressure is 40 kPa, where will a single support be located to hold the gate in position (above the invert of the gate)?

A. 1.36 m | C. 3.24 m |

B. 1.84 m | D. 2.62 m |

**Answer Key**

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**Solution**

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*F*

_{oil}and

*F*

_{air}are in equilibrium.

$F_{oil} = p_{cg}A$

$F_{oil} = 9.81(0.8)(8) \times \pi (4^2)$

$F_{oil} = 3155.87 ~ \text{kN}$

$F_{air} = p_{air}A$

$F_{air} = 40 \times \pi (4^2)$

$F_{air} = 2010.62 ~ \text{kN}$

Since *F*_{oil} > *F*_{air}, the hinge must be located below the point where *F*_{oil} is acting. The hinge is indicated by letter *H* in the figure shown.

Eccentricity

$e = \dfrac{I_g}{A\bar{Y}}$

$e = \dfrac{\frac{1}{4}\pi (4^4)}{\pi (4^2) \times 8}$

$e = 0.5 ~ \text{m}$

For equilibrium, sum of moment at *H* must be zero

$\Sigma M_H = 0$

$F_{air}(4 - y) = F_{oil}(4 - e - y)$

$2010.62(4 - y) = 3155.87(4 - 0.5 - y)$

$8042.48 - 2010.62y = 11\,045.545 - 3155.87y$

$1145.25y = 3003.065$

$y = 2.62 \, \text{m}$ ← *answer*

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