Question 2.10: A sector gate, of radius 4 m and length 5 m, controls the fl...
A sector gate, of radius 4 m and length 5 m, controls the flow of water in a horizontal channel. For the equilibrium condition shown in Fig. 2.33, determine the total thrust on the gate.
The horizontal component of the hydrostatic force is the thrust which would be exerted by the water on a projected plane surface in a vertical plane. The height of this projected surface is 4 sin 30° m = 2 m and its centroid is (1 + 2/2) m = 2 m below the free surface.
Therefore, the horizontal component of hydrostatic thrust
F_{H}=\rho g \bar{h} A=1000 \times 9.81 \times 2 \times(5 \times 2) N = 196.2 kN
The line of action of F_{H} passes through the centre of pressure which is at a distance z_{p} below the free surface, given by (see Eq. 2.44a)
y_{p}=\frac{I_{x^{\prime} x^{\prime}}}{A y_{c}}+y_{c} (2.44a)
z_{P}=2+\frac{5(2)^{3}}{12 \times(5 \times 2) \times 2}=2.167 m
The vertical component of the hydrostatic thrust F_{V}
= Weight of imaginary water contained in the volume ABDCEA
Now, Volume ABDCEA = Volume ABDEA + Volume OECO Volume ODCO
Volume A B D E A=5 \times A B \times B D=5 \times\left(4-4 \cos 30^{\circ}\right) \times 1
= 5 × 0.536
Volume O E C O=5 \times \pi \times(O C)^{2} \times 30 / 360
=5 \times \pi \times(4)^{2} \times(30 / 360)
Volume O D C O=5 \times \frac{1}{2} \times 4 \sin 30^{\circ} \times 4 \cos 30^{\circ}
=5 \times \frac{1}{2} \times 2 \times 4 \cos 30^{\circ}
Therefore,
F_{V}=1000 \times 9.81 \times 5\left[(0.536 \times 1)+\left(\pi \times 4^{2} \times \frac{30}{360}\right)\right.
\left.-\left(\frac{1}{2} \times 2 \times 4 \cos 30^{\circ}\right)\right] N =61.8 kN
The centre of gravity of the imaginary fluid volume ABDCEA is found by taking moments of the weights of all the elementary fluid volumes about BC. It is 0.237 m to the left of BC. The horizontal and vertical components, being co-planar, combine to give a single resultant force of magnitude F as
F=\left(F_{H}^{2}+F_{V}^{2}\right)^{1 / 2}=\left\{(196.2)^{2}+(61.8)^{2}\right\}^{1 / 2}=205.7 kN
at an angle \alpha=\tan ^{-1}(61.8 / 196.2) \approx 17.5^{\circ} to the horizontal.
Alternative method:
Consider an elemental area \delta A of the gate subtending a small angle d \theta at 0 (Fig. 2.33). Then the hydrostatic thrust dF on the area \delta A becomes d F=\rho g h \delta A. The horizontal and vertical components of dF are
d F_{H}=\rho g h \delta A \cos \theta
d F_{y}=\rho g h \delta A \sin \theta
where h is the vertical depth of area \delta A below the free surface.
Now h=(1+4 \sin \theta)
and \delta A=(4 d \theta \times 5)=20 d \theta
Therefore the total horizontal and vertical components are,
F_{H}=\int d F_{H}=1000 \times 9.81 \times 20 \int_{0}^{\pi / 6}(1+4 \sin \theta) \cos \theta d \theta N =196.2 kN
F_{V}=1000 \times 9.81 \times 20 \int_{0}^{\pi / 6}(1+4 \sin \theta) \sin \theta d \theta N =61.8 kN
Since all the elemental thrusts are perpendicular to the surface, their lines of action pass through O and that of the resultant force therefore also passes through O.