Determine the horizontal and vertical components of reaction at hinge A and the normal reaction at B caused by the water pressure. The gate has a width of 3 m.
Understand the Problem
The question is asking to determine the horizontal and vertical components of reaction forces at hinge A of the gate, as well as the normal reaction force at point B, due to water pressure acting on the gate. The given width of the gate is 3 meters, and there are dimensions indicating the vertical height of water. The approach will involve applying principles of fluid mechanics and static equilibrium.
Answer
The horizontal reaction at hinge A is $H = 264.3 \, kN$, the vertical reaction is $V = 38 \, kN$, and the normal reaction force at B is $N = 226.3 \, kN$.
Answer for screen readers
The horizontal and vertical components of reaction at hinge A are $H = 264.3 , kN$ and $V = 38 , kN$, respectively. The normal reaction force at B is $N = 226.3 , kN$.
Steps to Solve
- Determine the Hydrostatic Pressure Force
The hydrostatic pressure force ($F_p$) can be calculated using the formula:
$$ F_p = \rho g h A $$
Where:
- $\rho$ is the density of water (approximately $1000 , kg/m^3$)
- $g$ is the acceleration due to gravity (approximately $9.81 , m/s^2$)
- $h$ is the height of water (3 m in this case)
- $A$ is the area of the gate submerged in water
In this case:
$$ A = \text{width} \times \text{height} = 3 , m \times 3 , m = 9 , m^2 $$
So the pressure force is:
$$ F_p = 1000 \times 9.81 \times 3 \times 9 = 264.3 , kN $$
- Calculate the Center of Pressure
The center of pressure ($h_{cp}$) below the surface is given by:
$$ h_{cp} = \frac{I_G}{A \cdot \bar{y}} + \bar{y} $$
Where:
- $\bar{y}$ is the depth to the centroid of the area (1.5 m for a 3 m high gate)
- $I_G$ is the moment of inertia about the horizontal axis through the centroid, which for a rectangle is given by:
$$ I_G = \frac{b \cdot h^3}{12} $$
For our case with $b = 3 , m$ and $h = 3 , m$:
$$ I_G = \frac{3 \cdot 3^3}{12} = 6.75 , m^4 $$
Now calculate the center of pressure:
$$ h_{cp} = \frac{6.75}{9 \cdot 1.5} + 1.5 = 1.75 , m $$
- Determine the Horizontal and Vertical Components of the Forces
The horizontal component of the force ($H$) at hinge A is equal to the hydrostatic force since there's no horizontal movement:
$$ H = F_p = 264.3 , kN $$
The vertical component ($V$) can be analyzed by the equilibrium of forces. The vertical reaction at hinge A and the normal force at B must balance the resultant force of the hydrostatics acting downwards:
$$ V + N = F_p $$
- Compute the Normal Reaction Force at B
Calculate the moment about hinge A for equilibrium:
$$ N \cdot h_{cp} = F_p \cdot \bar{y} $$
Substituting the values:
$$ N \cdot 1.75 = 264.3 \cdot 1.5 $$
Thus,
$$ N = \frac{264.3 \cdot 1.5}{1.75} = 226.3 , kN $$
Finally substituting back to find the vertical force component:
$$ V = F_p - N = 264.3 - 226.3 = 38 , kN $$
The horizontal and vertical components of reaction at hinge A are $H = 264.3 , kN$ and $V = 38 , kN$, respectively. The normal reaction force at B is $N = 226.3 , kN$.
More Information
The calculations involved hydrostatic pressure, the center of pressure, and applying equilibrium conditions to find the forces acting on the gate submerged in water.
Tips
- Failing to calculate the correct area for the hydrostatic force.
- Forgetting to use the correct depth for centroid when calculating the center of pressure.
- Not considering equilibrium conditions correctly.