§2.6 Total hydrostatic force on curved surface

In engineering practice, some action surfaces bearing water pressure are curved planes such as the surface of arch dams, arcuate gate and U-shaped aqueduct.The common curved surfaces are two-directional curved surfaces, i.e.cylinders with parallel generatrix.Thus, this book focuses on the total hydrostatic pressure exerted on the two-directional curved surface.

There is a two-directional curved surface MN (cylinder), with one side pressure-bearing.The generatrix is perpendicular to the map.The area of the curved plane is A. A coordinate system is built, in which Oxy plane and liquid surface overlap, and Oz axis downward, as illustrated in Fig.2-23a.In order to calculate the total hydrostatic pressure F on the two-directional curved plane, the components Fx and Fy of F need be calculated, respectively.

2.6.1 Horizontal component

The proj ection plane of the curved surface MN on the vertical plane is M＇N＇.The area of the proj ection plane is Ax.The liquid M＇N＇NM was taken as a detached body. The equilibrium of forces acting on the detached body was analyzed as shown in Fig. 2-23b.There are three forces exerted on the detached body, including F＇, F1and F2. The counterforce exerted by the liquid on the curved surface includes horizontal force F＇x, vertical force F＇y and the gravity W.Thus, the horizontal forces exerted on the detached body only include F＇and F＇x, which are equal in magnitude and opposite in direction.

According to the relationship between action and reaction, we can write

where Fx is the horizontal component of the force which the liquid exerts on the curved surface and pC is the pressure at the centroid of the proj ection plane Ax.

Equation(2-22)indicates the horizontal component Fx of the force exerted on the curved surface is equal to the force F'exerted on the vertical proj ection plane of the curved surface.The action line of Fx must be the action line of F＇.

2.6.2 Vertical component

The vertical component on the curved surface can be calculated on the basis of the volume of the liquid (called pressure prism)surrounded by the curved surface MN, the proj ection plane M0 N 0 of MN on the free surface and the vertical plane at the edge of the curved surface (see Fig.2-23c).This liquid is at hydrostatic equilibrium.The pressure on the free surface is zero.Vertical forces only include gravity W and the reaction force F＇z of the vertical force Fz exerted on the curved surface.Thus,

where Fz is the vertical force, which the liquid exerts on the curved plane; Az is the area of the proj ection of the curved plane on the horizontal planeis the volume of the pressure prism.

Equation(2-23)indicates the vertical component of the hydrostatic pressure exerted on the curved plane is equal to the weight of the liquid surrounded by MN, M0 N 0 and the vertical plane at the edge of MN.The location of the action line of the vertical component is the same as that of gravity W and the action line must pass through the center of the pressure prism.

2.6.3 Total hydrostatic pressure on the curved plane

In general, there is no single resultant force on the irregular plane.The horizontal component and the vertical component may not be on the same plane.For a two-directional plane, the horizontal and vertical components are on the same plane, which can be combined into a force F:

The included angle between F and horizontal line is:

The action line of the total hydrostatic pressure must pass through the intersection point of the action lines of Fx and Fz.The acting point of F locates at the intersection point of the curved plane and the action line of F.It is not necessary to calculate F in many practical engineering.Calculating the components of F, including the magnitude, direction and acting point, can meet the needs of many practical engineering.

2.6.4 Pressure prism

In equation(2-23), the integral represents geometric volume which is called pressure prism.In fact, the calculation of vertical component of fluid hydrostatic pressure on the curved plane is the calculation of pressure prism.So the concept of pressure prism is very important.

Pressure prism is a cylinder with the action surface as lower surface and the free surface as upper surface, which is cut out by a vertical line moving a circle along the edge of the action face.According to the definition, pressure prism is surrounded by three curved surfaces:(1)compressed surface; (2)vertical surface along the edge of the curved surface; (3)free surface or the extended surface of free surface.

1．Real pressure prism

When the liquid and pressure prism are on the same side of the curved surface as shown in Fig.2-24a, the direction of Fz is downward and Fz is equal to the water weight of the pressure prism in magnitude.This pressure prism is called real pressure prism. 2.Virtual pressure prism

When the liquid and pressure prism are on the different sides of the curved surface as shown in Fig.2-24b, the direction of Fz is upward and Fz is equal to the water weight of the pressure prism in magnitude.This imaginary pressure prism is called virtual pressure prism.

3．Superposition of pressure prism

For the curved surfaces whose horizontal proj ections overlap, separate the defined pressure prisms and superpose them.The pressure prism of half cylinder can be determined according the curved surfaces and (Fig.2-24c).For the virtual pressure prism derived from superposition ABC, the direction of Fz is upward.

2.6.5 Inference

(1)In a limited space, the variation of gas pressure with the depth is slight and the weight of gas is negligible.After analyzing the forces acting on the detached body, the pressure exerted by gas on the curved surface is:

Where Ax and Az is the proj ection area of the curved surface on the horizontal and vertical plane, respectively.

(2)If the liquid surface is not free surface, but in a closed container and the surface pressure is p0, the total pressure is:

Example 2-9 Two hemispherical surfaces were riveted to form a spherical container.There are n rivets.The spherical container is filled with the liquid with density ofρas shown in Fig.2-25a.Calculate the pulling force exerted on each rivet.

Solution:The upper hemisphere of the spherical container was considered as compressed curve.The pressure prism acting on the curved surface is described in Fig. 2-25 b.Then,

Thus,

Example 2-10 A steel plate with allowable stress of [σ]=150 MPa, was used to make a water pipe with inner diameter of D as shown in Fig.2-26a.The pressure inside the water pipe is 500 mH2O.Calculate the wall thickness of the water pipe.(Note:The pressure difference can be ignored due to the height difference at different points inside the pipe.)

Solution:A pipe segment with length of 1 m is considered as the obj ect of study. The pressure at arbitrary point on the wall of the pipe is equal.

The pipe wall was cut open along the pipe diameter.The half pipe segment was taken as a detached body.The forces exerted on the detached body were analyzed (see Fig.2-26b).The horizontal pressure exerted on the inner surface of the semi-cycle is equal to the pressure on the vertical proj ection plane of semi-cycle, i.e.F=p· Az=p·D×1.The pressure keeps balance with the tensile stress on the wall of semi-cycle, i.e.2 FT=F=pD.FT is assumed to be uniformly distributed along the wall thickness of the pipe.Then,

FT≤ [σ]·e×1

Thus,

Example 2-11 There is a half cylinder with a unit width (b=1 m).Whether the joint effects of buoyant force Fz and horizontal water pressure Fx, can create rotational moment with respect to the axle center or not(see Fig.2-27)?

Solution:

(1)Concept analysis

Because the action line of water pressure acting on the half cylinder is always perpendicular to the action surface and passes through the center of circle, the half cylinder below the water surface does not generate rotational moment.

(2)Prove it by means of calculation

The distance from the acting point of vertical force to the axle center is

Then,

So, the water pressure exerted on the half cylinder does not generate rotational moment with respect to the axle center.

Example 2-12 The diameter of cylinder is 2 m, which is horizontally-positioned.The measurements of various parts are shown in Fig.2-28a.There is water on the left side and no water on the right side.Calculate the horizontal component Fx and vertical component Fz of the total hydrostatic pressure exerted on the cylinder per meter.

Solution:CDHAB is the compressed plane of the cylinder.The horizontal component on the two sides of H AB plane offset each other.The horizontal component on the compressed surface CD H is

Fx=ρghCA(9800N/m3)×(1.2m+0.7m)×(1.4m)×(1m)=26068N

The vertical component Fz can be calculated through plotting the pressure prism of curved surface CDHAB.The curved surface CDHAB was divided into two sections (CD and DHAB).The pressure prisms of the two sections were plotted as shown in Fig.2-28b and Fig.2-28c.For the pressure prism of CD, the direction of Fz1 is downward.For the pressure prism of DHAB, the direction of Fz2 is upward.The two can counteract to each other partly.The pressure prism finally obtained is the hatching part as shown in Fig.2-28d.The vertical component Fz of the total hydrostatic pressure acting on the cylinder per meter is equal to water weight of DHABJ FGCD in magnitude.In order to facilitate the calculation, DHABJ FGCD was divided into three simple geometric figures such as rectangle, triangle and semi-circle.Then,

Example 2-13 There is a rectangular orifice on a clapboard.The height of the orifice is a=1.0 m, and the width is b=3m.A cylinder with a diameter of d=2m was used to block the orifice.Water is filled in both sides of the clapboard.h=2m and z=0.6m.Calculate the total hydrostatic pressure exerted on the cylinder.

Solution:Calculation by plotting pressure distribution curve and pressure prism.

With hydrostatic pressure acting on the both sides of clapboard, the horizontal pressure distribution of clapboard is described in Fig.2-19d.The pressure distribution of cylinder is rectangular distribution after the left and right pressure counteract to each other, as shown in Fig.2-29.

The left part of pressure prism is equal to area B AA＇B＇multiplied by the width b of the cylinder.Thus, the sum of the pressure prisms of the compressed curve on the both sides of the clapboard is equal to the volume of the cylinder, which is virtual pressure prism.The direction of Fz is upward.

After pressure distribution figure and the pressure prism were plotted, the horizontal component of the total hydrostatic pressure is:

Fx=a bρg z=(1.0m)×(3m)×(9800N/m3)×(0.6m)=17650N=17.65kN

The direction of Fx is rightward.

The vertical component of the total hydrostatic pressure is:

The direction of Fz is upward.

Thus, the total hydrostatic pressure acting on the cylinder is:

The pressure at arbitrary point on the surface of the cylinder passes through the center axis of the cylinder.Thus, the resultant force must pass through the center axis of the cylinder.The included angle between action line and the horizontal plane is

The depth of the acting point D below the water surface is