In this article we will discuss about:- 1. Total Pressure on a Lamina Immersed in a Liquid 2. Centre of Pressure – Vertical Lamina 3. Centre of Pressure for an Inclined Lamina 4. Pressure on Curved Surfaces 5. Pressure on Lock Gates. [with formulas and examples]
Contents:
- Total Pressure on a Lamina Immersed in a Liquid
- Centre of Pressure – Vertical Lamina
- Centre of Pressure for an Inclined Lamina
- Pressure on Curved Surfaces
- Pressure on Lock Gates
1. Total Pressure on a Lamina Immersed in a Liquid:
Consider a lamina of area A immersed in a liquid of specific weight w. Consider an elemental area da of the lamina at a depth y below the free liquid surface.
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Intensity of pressure on the elemental area = p = wy
Pressure force on the elemental area = p da = wy da
∴ Total Pressure force on the whole lamina = P = Σw day = w Σday = w A y̅.
Where y̅ = depth of the centroid of the lamina below the free surface.
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2. Centre of Pressure – Vertical Lamina:
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On any lamina in a liquid body, the intensity of pressure will not be uniform, and the pressure intensity at any point depends on the depth of the point below the free surface. The resultant pressure P on an immersed lamina will act at a point C called the centre of pressure.
If the pressure intensity on the lamina had been uniform the resultant pressure would act at G, the centroid of the lamina. The pressure intensity being not uniform (being greater at greater depth), the resultant pressure will act at a point C below the centroid of the lamina.
3. Centre of Pressure for an Inclined Lamina:
Fig. 3.66 shows a lamina of area A immersed in a liquid and making an angle θ with the horizontal water surface. Consider an elemental area da of the lamina at a depth y below the free liquid surface.
Let the distance of the elemental area from the line of intersection of the plane of the lamina and the free liquid surface be x.
Intensity of pressure on the elemental area = p = wy
Force on the elemental area = pda = wyda
Total force on the lamina-
4. Pressure on Curved Surfaces
:
Figs. 3.87 (i) and (ii) shows two identical curved surfaces subjected to water pressure. In Fig. 3.87 (i) the curved surface is subjected to water pressure on the concave side while in Fig. 3.87 (ii) the curved surface is subjected to water pressure on the convex side. The position of the surface AB with respect to the free water surface is the same in the two cases. Obviously the total pressure force P on the two surfaces will be the same.
In Fig. 3.87 (i) the vertical component of P acts downwards while in Fig. 3.87 (ii) the vertical component of P acts upwards. We know in Fig. 3.87 (i) the vertical component of P is equal to the weight of the block of water A’ABB’. In Fig. 3.87 (ii) also the vertical component of P is equal to the weight of the block of water A’ABB’, but in this case the vertical component of P acts upwards.
5. Pressure on Lock Gates
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Fig. 3.139 shows the elevation and plan of the lock gates. Let AB and BC be the two lock gates. Each gate is supported on two hinges. In the closed position the gates meet at B exerting thrust on one another.
Let us consider the gate AB.
Let H1 and H2 be the depth of water on the two sides of the gate. Let P1 and P2 be the pressure exerted by water on the two sides of the gate.
Let P be the resultant water pressure on the gate.
Obviously, P = P1 – P2
Let N be the reaction at the common contact surface of the two gates. Let R be the resultant reaction of the top and bottom hinges. We will assume that the three forces, namely P, N and R are in the same horizontal plane. Let the inclination of the gate with the normal to the side of the lock be θ.
