Question 3.P.20: The shape described as x = 0.2y² forms the wall of a gate. D......
The shape described as x = 0.2y² forms the wall of a gate. Derive expressions for the horizontal force, vertical force and the moment on the gate with respect to O, as shown in Fig. P. 3.20. Calculate the values for y = 3 m.
Consider unit width,
Horizontal force = γ A \bar{h} = γ × y × (y/2) = γ y²/2
This force acts at y/3 from the bottom. Vertical force: weight of volume above the surface. Assuming unit width, the volume = area × width
A=\int_{0}^{y}x.d y=0.2\,\int_{0}^{y}y^{2}\,d y \\ = 0.2 y³/3 = xy/3
Vertical force = (xy/3) × γ × width
= (0.2 y³/3) × γ × width
The position of line of action can be determined taking moment about the y axis. Let it be \bar{x} \mathrm{from} y axis.
\overline{{{x}}}\cdot x y/3=\left.\int_{0}^{y}x\cdot d y\cdot x/2=\frac{0.04}{2}\right.\int_{0}^{y}y^{4}\ d y\ =\frac{0.04\ y^{5}}{10}=\frac{x^{2}\ y}{10},
∴ \bar{x} = (3x/10) = 0.06 y^{2}Clockwise moment about
O = (γ y^{2}/2) × (y/3) + (0.2 y^{3}/3) γ × 0.06 y^{2}
For y = 3 m and unit width
Horizontal force = 9810 (3 × 3/2) = 44145 \mathrm{N}
Vertical force = 0.2 y^{3} γ/3 = 0.2 × 27 × 9810/3 = 17658 \mathrm{N}
Clockwise moment = (γ × 27/6) (1 + 0.024 × 9) = 53680 \mathrm{N m}
The direction of the force with the vertical can be found using
\tan^{– 1} (44145/17658) = 68.2°
Resultant = (44145^{2} + 17658^{2})^{0.5} = 47546 \mathrm{N}
To locate the actual line of action of the force, perpendicular distance from O × force = moment
∴ Distance = 53680/47456 = 1.129 m
It cuts the vertical from O at 1.129/sin 68.2 = 1.216 m