Southern Methodist University

Bobby B. Lyle School of Engineering

ENCE/ME 2342 Fluid Mechanics

Roger O. Dickey, Ph.D., P.E.

II. HYDROSTATICS

C. Hydrostatic Forces on Plane Surfaces

Reading Assignment:

Chapter 2 Fluid Statics, Sections 2.8 and 2.9

C. Hydrostatic Forces on Plane Surfaces

When a surface is submerged in a fluid, forces develop on the surface due to fluid pressure, i.e., normal stress or normal force per unit area. For static fluids where there are no shear stresses, the magnitude of the resultant force,

FR, applied to any arbitrary surface is simply

the product of the average pressure, p, acting on the surface times the area of the surface, A,

i.e., FR = pA.

Determination of pressure forces is important for the analysis and design of:

• Pressure vessels and conduits

• Ships and submarines

• Storage tanks and impoundments

• Dams and other hydraulic structures

• Grout curtains and retaining walls for groundwater

Notation for Integrating (Summing) Physical

Quantities over Areas and Volumes

Planar Areas

Let (x,y) be a physical quantity per unit area.

Then, the total amount, T, contained within an

arbitrary fixed Region in the xy-plane is:

R

TR

T dxdydA or

3-Dimensional Surfaces in Space

Let (x,y,z) be a physical quantity per unit area.

Then, the total amount, T, contained on an

arbitrary fixed 3-dimensional Surface in xyz-

space is:

S

T dA dAor

3-Dimensional Volumes in Space

Let (x,y,z) be a physical quantity per unit

volume. Then, the total amount, T, contained

within an arbitrary fixed Volume in xyz-space is:

V

TV

T dxdydzdV or

Engineering Short-hand for Integrals

Engineering textbooks and references often use a short-hand notation for area, surface, and volume integrals:

V V

TT ρdVρdV

S CS

TT dAdA

A

TR

T dAdA Standard Math Engineering Short-hand

Horizontal Planar Surfaces

Consider an arbitrary horizontal surface submerged in a liquid:

z

y

x

dF = pdA

dA

Arbitrary Region of total area A in the xy-plane

Then, the pressure force acting on any differential area element, dA, within the planar Region is:

Integrating over the entire area yields the resultant pressure force,

pdAdF

AA

pdAdF

Planar surface integrals

Pressure is constant along any horizontal surface immersed in a static fluid, thus p may be factored from the integral on the right-hand side of the equation:

Completing the integration yields the appropriate expression for the resultant force,

FR, on the planar surface of total area, A:

AA

dApdF

pAFR

Consider an open-top tank of total planar area A, filled to depth h with a liquid having specific weight γ as shown in Figure 2.16 (a), p. 60:

If the tank bottom lies in a horizontal plane, the

fundamental principles of hydrostatics dictate

that: (i) pressure, p, is uniform everywhere over

the bottom, and (ii) p = γh. Then, the resultant

pressure force acting on the tank bottom of area,

A, is given by:

hAFR

The y-coordinate, yR, of the point of application

of the resultant pressure force can be determined by summation of moments around the x-axis. In other words, the moment created by the resultant pressure force times its distance from the x-axis,

FRyR, must exactly equal the sum of the

moments created by all differential pressure forces times their distances from the x-axis, ydF:

A

RR ydFyF

Substituting (γh)A for FR on the left-hand side

of the equation, and (γh)dA for dF on the right-

hand side yields,

A

R dAhyAyh

Factor the constant (γh) from the integral on the

right-hand side, divide through by (γh), and solve

for yR:

By definition, the expression on the right-hand

side of the equation is the y-coordinate of the

centroid, yC, of a planar Region of area A lying in

the xy-plane, hence, yR = yC.

A

ydAy A

R

The x-coordinate, xR, for the resultant pressure

force is determined in a similar manner using

summation of moments about the y-axis:

Once again, by definition, the expression on the

right-hand side of the equation is the x-coordinate

of the centroid, xC, of a planar Region of area A

lying in the xy-plane, hence, xR = xC.

A

xdAx A

R

This analysis proves that for horizontal surfaces

submerged to depth h in a static, incompressible

fluid having a free surface, the following apply:

• The resultant pressure force is,

• The resultant pressure force acts through the

centroid of the planar area; centroidal xy-

coordinates, (xC, yC), for common geometric

shapes are shown in Figure 2.18, p. 62:

hAFR

Figure 2.18 (a), p. 62Geometric Properties of Plane Shapes

Figure 2.18 (b), p. 62Geometric Properties of Plane Shapes (continued)

Figure 2.18 (c), p. 62Geometric Properties of Plane Shapes (continued)

Figure 2.18 (d), p. 62Geometric Properties of Plane Shapes (continued)

Figure 2.18 (e), p. 62Geometric Properties of Plane Shapes (concluded)

Inclined Planar Surfaces

Now consider the more general case of a planar

surface of arbitrary shape, having its y-axis

inclined at an arbitrary angle from the

horizontal, and immersed in a liquid having a

free surface as illustrated in Figure 2.17, p. 60:

Fig. 2.17,p. 60InclinedPlanarSurface

Pressure on the surface is not constant but, instead, varies with depth according to the hydrostatic distribution. Once again, the pressure force acting normal to any differential area element, dA, within the planar Region is

For a hydrostatic pressure distribution, p = γh, thus,

pdAdF

dAhdF

Now consider the geometric relationship between the depth, h, and the y-coordinate of the differential area element, dA:

Free Surface (pressure = p0)

y

hy

h

dA sin sin yhyh

Substituting ysin() for h in the equation for dF,

Integrating over the entire planar surface area,

Completing the integration on the left-hand side of

the equation yields the resultant pressure force, FR,

and factoring the constant γsin() from the surface integral on the right-hand side results in,

dAθydF sin

AA

dAθydF sin

A

R ydAF sin

By definition, the y-coordinate of the centroid, yC,

of a planar Region of area A lying in the xy-plane,

is,

Substituting AyC for the integral on the right-hand

side of the equation for FR yields,

A

CA

C ydAAyA

ydAy

sinCR AyF

From geometric considerations, as shown above,

Finally, substituting hC for yCsin() in the equation

for FR,ApFAhF CRCR or

sinCC yh

[Equation (2.18), p. 61 – Modified]

*Important Points –

1. Derived result FR = pCA proves that the

magnitude of the resultant pressure force is the product of the total area, A, and the pressure at

the centroid of the planar area, pC

2. FR = pCA holds whether a free surface is present

or not. For example, when there is an

overpressure, p0, applied inside a pressure vessel

Equation (2.8), p. 45 may be applied at the centroid of the planar area yielding

0php CC

3. Derived result FR = γhCA proves that the

resultant pressure force is independent of the angle , depending only upon the depth to the centroid of the planar area. Indeed, this general result also applies to horizontal

surfaces ( = 0) where h = hC everywhere on

the surface.

4. Because hydrostatic pressure increases with depth, pressures acting below the centroid of an inclined planar surface are greater than pressures acting above it. Consequently, the resultant pressure force does not act through the centroid of inclined surfaces, instead it acts through a point called the pressure center located at a depth somewhere below the centroid. Consider the hydrostatic pressure on a vertical ( = 90) tank wall:

2b 2

b

C

C = Centroid of the Planar AreaCP = Center of Pressure

Figure 2.19, p. 65 Modified

2h

2h

5. γgases << γliquids resulting in negligible variation

in gas pressure with depth over inclined planar

surfaces, hence the resultant gas pressure force

acts through the centroid of the area to a close

approximation.

6. In most cases, atmospheric pressure acts

equally on all sides of planar surfaces, thus

atmospheric pressure forces tend to counteract

making no contribution to the resultant

pressure force, as illustrated for a vertical tank

wall in Figure 2.22, p. 67:

Figure 2.22, p. 67 Atmospheric pressure forces act equally on both sides of most surfaces, producing no net contribution to the resultant pressure force.

Engineered systems relying on counteracting atmospheric pressure forces may fail when inadvertently exposed to overpressure or under-pressure: (1) doors are difficult to push open from the low pressure side of some rooms and buildings, and (2) under-pressurized tanks can be crushed:

Under-pressurized Chemical Storage Tank crushed “like a beer can” by atmospheric pressure

The y-coordinate, yR, of the pressure center for

the resultant pressure force can be determined

by summation of moments around the x-axis:

Since,

Similarly,

A

RR ydFyF

AyFAhF CRCR sin

dAydF sin

Substituting for F and dF yields,

Factoring the constant γsin() from the integral

on the right-hand side, and regrouping terms on

both sides of the equation,

A

RC dAyyyAy sin)sin(

A

RC dAyAyy 2sin)sin(

Divide through by γsin() and solve for the y-

coordinate of the resultant pressure force, yR,

By definition, the numerator on the right-hand

side of the equation is the second moment of the

area or moment of inertia, Ix, about the x-axis.

Ay

dAyy

C

AR

2

Substituting Ix for the integral in the numerator on

the right,

Apply the Parallel Axis Theorem from Statics — the moment of inertia about the x-axis equals the moment of inertia about a parallel axis passing through the centroid of the area plus the product of the total area times the square of the y-coordinate of the centroid:

AyIyC

xR

2CxCx AyII

Now, substituting for Ix in the numerator on the

right-hand side of the equation for yR:

Finally, simplifying the right-hand side,

AyAyIy

C

CxCR

2

CC

xCR y

AyIy

Equation (2.19), p. 61

Equations for computing moments of inertia, IxC

and IyC, with respect to orthogonal axes passing

through the centroid of common geometric

shapes are illustrated in Figure 2.18, p. 62, as

shown above.

The x-coordinate, xR, of the pressure center for

the resultant pressure force can be determined

by summation of moments around the y-axis:

Since,

Similarly,

A

RR xdFxF

AyFAhF CRCR sin

dAydF sin

Substituting for F and dF yields,

Factoring the constant γsin() from the integral

on the right-hand side, and regrouping terms on

both sides of the equation,

A

RC dAyxxAy sin)sin(

A

RC xydAAxy sin)sin(

Divide through by γsin() and solve for the x-

coordinate of the resultant pressure force, xR,

By definition, the numerator on the right-hand

side of the equation is the product of inertia, Ixy,

with respect to the x and y-axes.

Ay

xydAx

C

AR

Substituting Ixy for the integral in the numerator on

the right:

Apply the Parallel Axis Theorem from Statics —the product of inertia with respect to xy-axes equals the product of inertia about a parallel set of orthogonal axes passing through the centroid of the area plus the product of the total area times the x- and y-coordinates of the centroid:

AyI

xC

xyR

CCxyCxy yAxII

Now, substituting for Ixy in the numerator on the

right-hand side of the equation for xR,

Finally, simplifying the right-hand side:

CC

xyCR x

AyI

x

Equation (2.20), p. 62

AyyAxI

xC

CCxyCR

Equations for computing products of inertia with

respect to xy-axes passing through the centroid

of common geometric shapes are illustrated in

Figure 2.18, p. 62, as shown above.

Pressure Prism

A scaled, graphical representation of pressure, p, acting on submerged surfaces is sometimes useful, especially for rectangular shapes. Consider a rectangular planar surface of dimensions W × L having its y-axis inclined at an arbitrary angle from the horizontal, having an orthogonal p-axis normal to the surface, and immersed in a liquid having a free surface as illustrated in Figure 2.21, p. 66:

Figure 2.21, p. 66 Modifiedy

p

L

W = dimension into the page, parallel to the x-axis

A 3-dimensional pressure prism is created in xyp-coordinate space by scaling the height to a surface located above the xy-plane proportionally to the pressure. Pressure prisms have the following properties:

1. The magnitude of the resultant pressure force equals the volume, , of the prism. For rectangular shapes as in Figure 2.21, p. 66;

Rearranging,

V

LWhhV 1221

Since FR = γhCA, this proves that

2. Resultant pressure force acts through the centroid of the prism. For rectangular shapes, as

in Figure 2.21, p. 66; (i) yC is located a distance

L/3 up from the bottom of the surface, and (ii)

xC is centrally located a distance W/2 from either

side of the surface due to symmetry.

LWhhV2

12

hC A

RFV

2b 2

b

C

C = Centroid of the Planar AreaCP = Center of Pressure

These properties are further illustrated for a vertical, rectangular tank wall in Figure 2.19, p. 65Modified:

2h

2h

Pressure prisms are not generally recommended

for non-rectangular shapes because the

centroidal coordinates for other geometries may

be unknown and difficult to determine.

Refer to Handout II.C. Hydrostatic Forces on

Plane Surfaces Example Problem.

Homework No. 6 Hydrostatic pressure forces

on submerged planar surfaces.