# Circular Gate with Water on One Side and Air on the Other Side

**Situation**

The figure below shows a vertical circular gate in a 3-m diameter tunnel with water on one side and air on the other side.

- Find the horizontal reaction at the hinge.

A. 412 kN

B. 408 kN

C. 410 kN

D. 414 kN - How far from the invert of the tunnel is the hydrostatic force acting on the gate?

A. 1.45 m

B. 1.43 m

C. 1.47 m

D. 1.41 m - Where will the hinge support be located (measured from the invert) to hold the gate in position?

A. 1.42 m

B. 1.46 m

C. 1.44 m

D. 1.40 m

**Solution**

Force on the gate due to air pressure
$F_{air} = p_{air}A = 45\left[ \frac{1}{4}\pi (3^2) \right]$
$F_w = \gamma_w \bar{h} A = 9.81(1.5)\left[ \frac{1}{4}\pi (3^2) \right]$
$\Sigma F_x = 0$
$e = \dfrac{I_g}{A\bar{y}} = \dfrac{\frac{1}{64}\pi (3^4)}{\frac{1}{4}\pi (3^2) \times 10.5}$
$\Sigma M_O = 0$

$F_{air} = 318.09 ~ \text{kN}$

Force on the gate due to water

$F_w = 728.10 ~ \text{kN}$

Horizontal force at the hinge support

$R_O = 728.10 - 318.09$

$R_O = 410.01 ~ \text{kN}$ ← [ C ] *answer for part 1*

Location of *F _{w}* from the invert

$e = 0.0536 ~ \text{m}$

$y = 1.5 - e = 1.5 - 0.0536$

$y = 1.4464 ~ \text{m}$ ← [ A ] *answer for part 2*

Location of hinge support

$(y - z)F_w = (1.5 - z)F_{air}$

$(1.4464 - z)728.10 = (1.5 - z)318.09$

$z = 1.4048 ~ \text{m}$ ← [ D ] *answer for part 3*

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