iic forcesonplanesurfaces
TRANSCRIPT
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.