Test: Pressure Force on Curved Surfaces - Civil Engineering (CE) MCQ

Concept:
In the case of the curved surface, the direction of pressure varies from point to point. To solve this problem the total hydrostatic force F is to be resolved into the horizontal and vertical component F_h and F_v respectively.

• Thus, F_h is the total hydrostatic force on the projected area of the curved surface on the vertical plane and will act at the centre of pressure of the plane surface.
• F_y is the weight of the liquid contained in the portion extending vertically above the curved surface upto the free surface of the liquid and will act through the centre of gravity of the volume of the liquid contained of the portion
  

The vertical component of force per unit length of the curved surface,
F_y = Weight of water supported by AB

F_y = 9.81π kN

Concept:
For a plane surface of arbitrary shape immersed in a liquid in such a way that the plane of the surface makes an angle θ with the free surface of the liquid:
where A = Total area of an inclined surface, h̅ = Depth of centre of gravity of inclined area from a free surface, h* =
Distance of centre of pressure from the free surface of a liquid.
Centre of pressure:

For vertical body θ = 90° 

where I_G = Moment of Inertia of the body from its centroidal axis.
Calculation:
Given:
Width b = 2 m, Height h = 3 m
h̅ = 2.5 + h/2 = 2.5 + 3/2 = 4 m

Horizontal component

• The Horizontal component of the total pressure force is equal to the total pressure force on the projected area of the curved surface in the vertical plane.

Vertical component

• The vertical component is equal to the weight of liquid supported by the curved surface upto the free liquid surface.

Concept:
Net hydrostatic force exerted by fluid on flat surface, F = ρ × g × h̅ × A
Where, ρ = density of fluid, g = acceleration due to gravity, A = Area of the flat surface, h̅ = depth of centroid from free surface
This force is assumed to be acting at a point.
Centre of the pressure of rectangular lamina from free surface,
CP = 2/3h (h = height of rectangular lamina)

Calculation:
For left side of the barrier
Area = 4 m × 1 m = 4 m², h¯= 4 / 2 = 2m, 
Center of pressure from top surface of the water, CP₁ = 2 × 4 / 3 = 8 / 3m 
F₁= ρ × g × h̅ × A = 1000 × 10 × 2 × 4  = 80 kN
For right side of the barrier
Area = 1 m × 1 m, h¯= 1 / 2 = 0.5m,
Center of pressure from top surface of the water CP₂ = 2 × 1 / 3 = 2 / 3m
F₂ = ρ × g × h̅ × A = 1000 × 10 × (1 × 1) × 0.5 = 5 kN             
Distance of centre of pressure from bottom end:
CP₁ = h/3 = 4/3
CP₂ = h/3 = 1/3
These forces will act as point load on the barrier. If the barrier is treated as a cantilever and the fixed end of this cantilever beam is on the bottom side of the tank, then the loading conditions will look like the figure drawn below


F₁ = 80 kN, F₂ = 5 kN

From the above-simplified diagram:
A moment about the fixed end of barrier,

Concept:
Hydrostatic Forces:

• Fluid statics or hydrostatics is the branch of fluid mechanics in which stresses generated in a fluid system are determined when it is at rest or static condition.
• The force exerted by a static fluid on a surface either plane or curved, when the fluid comes in the contact with the surface is called Total pressure.
• The point of application of total hydrostatic force on the surface is known as the center of pressure.

Total Hydrostatic Force on a Horizontal Plane and inclined surface:
Consider a plane surface immersed in a static mass of liquid of specific weight γ, such that it is held in a horizontal position at a depth h below the free surface of the liquid and if the surface is in a vertical position such that the centroid of the surface is at depth of h̅ below the free surface. A is the area of the total surface, then the total hydrostatic force on the horizontal surface,
F = γAh̅ 
γh̅ is the pressure intensity at the centroid of the surface. Therefore, it can be stated that the total hydrostatic force on a plane surface is equal to the product of the surface area and the intensity of pressure at the centroid of the area.

Concept:
Whenever a static mass of fluid comes into contact with a surface the fluid exerts force upon that surface. The magnitude of this force is known as the hydrostatic force or total pressure force.
The magnitude of hydrostatic force is given by F = ρ g h̅̅  A N
Where ρ = Density of the fluid, h̅ = Depth of center of gravity of the surface from free liquid surface, A = Area of the surface
Calcualtion:
Given tank has the height twice as the width of the tank. 

Pressure force acting on the vertical surface:
The center of gravity of the vertical surface from the free surface will be
h¯= 2x/2 = xm
Area of the vertical surface = 2x × x = 2x²
Now the Pressure force ρ × g × x × 2x² = 2ρgx³
Now, the force on the bottom surface:
The center of gravity of the bottom surface from the free liquid surface = hieht of the tank = 2x
Area of the bottom surface = x × x = x²
Now the pressure force = ρ × g × 2x × x² = 2ρgx3
Therefore the ratio of the forces will be one as the forces on the bottom side and the vertical sides are equal. 

Forces over aircraft

• At any given time, there are four forces acting upon an aircraft. These forces are lift, weight (or gravity), drag and thrust.
• Lift is the key aerodynamic force that keeps objects in the air. It is the force that opposes weight; thus, lift helps to keep an aircraft in the air.
• Weight is the force that works vertically by pulling all objects, including aircraft, toward the centre of the Earth. In order to fly an aircraft, something (lift) needs to press it in the opposite direction of gravity. The weight of an object controls how strong the pressure (lift) will need to be. Lift is that pressure.
• Drag is a mechanical force generated by the interaction and contract of a solid body, such as an aeroplane, with a fluid (liquid or gas).
• Finally, the thrust is the force that is generated by the engines of an aircraft in order for the aircraft to move forward.

Bernoulli's principle in fluid dynamics

• We are able to explain how lift is generated for an aeroplane by gaining an understanding of the forces at work on an aeroplane and what principles guide those forces. First, it takes thrust to get the aeroplane moving - Newton’s first law at work. This law states that an object at rest remains at rest while an object in motion remains in motion unless acted upon by an external force.
• Bernoulli's principle can be used to calculate the lift force on an aerofoil if the behaviour of the fluid flow in the vicinity of the foil is known.
• For example, if the air flowing past the top surface of an aircraft wing is moving faster than the air flowing past the bottom surface, then Bernoulli's principle implies that the pressure on the surfaces of the wing will be lower above than below. This pressure difference results in an upwards lifting force. 

Concept:
Whenever a static mass of fluid comes into contact with a surface the fluid exerts force upon that surface. The magnitude of this force is known as the hydrostatic force or total pressure force.
The magnitude of hydrostatic force is given by F = ρgh̅A 
Where ρ = Density of the fluid, h̅ = Depth of center of gravity of the surface from free liquid surface, A = Area of the surface
Calculation:

Given, h = 10 m
h̅ = 10/2 = 5 m
Area per metre length = 10 × 1 = 10 m²                  
Force acting on vertical wall = ρ × g × h̅ × A
= 1000 × 9.81 × 5 × 10
= 490500 N
= 490.5 kN

Concept:

The force applied from the bottom should be such that it must balance the force applied due to water above the cap.
Force F here is vertical hydrostatic force
F = ρgV
Where V = volume of fluid above the curved surface
So, F = ρgV
V = [Volume of a cuboid  – Volume of hemisphere]