lecture -3 fluid statics. force on a plane area no tangential force can exist within a fluid at...

Lecture -3

Fluid Statics

Force on a Plane Area

No tangential force can exist within a fluid at rest. All forces are normal to the surface.

If pressure is uniformly distributed over an area, the force is equal to pressure times the area and the point of application of force is at the centroid of the area.

For submerged horizontal areas, the pressure is uniform. In case of gases, the pressure variation with vertical distance

is very small due to low specific weight. Therefore when we compute the static fluid force exerted by a gas, we usually treat p as a constant. Thus for such cases:

(3.14) pAdAppdAF


In the case of liquids the distribution of pressure is generally not uniform, so further analysis is necessary.

Let us consider a vertical plane whose upper edge lies in the free surface of a liquid (Fig). Let this plane be perpendicular to the plane of the figure, so that MN is its edge.

Fig-3.15


The gage pressure will vary from zero at M to NK at N. The total force on one side of the plane is the sum of the products of the elementary areas and the pressure upon them.

From the pressure distribution, we can see that the resultant of this system of parallel forces must act at a point below the centroid of the area, since the centroid of an area is the point where the resultant of a system of uniform parallel forces would act.


If we lower the plane to position M'N', the change of pressure from M' to N' is less than it was from M to N. Hence the resultant pressure force will act nearer to the centroid of the plane surface.

The deeper we submerge the plane, the smaller the pressure variation becomes, and the closer the resultant moves to the centroid.

Fig-3.15

Force on a Plane Area (Sloping)

In Fig. let MN be the edge of a plane area making an angle with the horizontal. To the right we see the projection of this area onto a vertical plane.

Fig-3.16


The pressure distribution over the sloping area forms a pressure prism (MNKJ times width in Fig.), whose volume is equal to the total force F acting on the area.

If the width x is constant then we can easily compute the volume of the pressure prism, using a mean pressure = 0.5 (MJ+NK), and so obtain F.

If x varies, we must integrate to find F. Let h be the variable depth to any point and let y be the corresponding distance from OX, the intersection of the plane containing the area and the free surface.


Choose an element of area so that the pressure over it is uniform. Such an element is a horizontal strip, of width x, so dA = xdy. As p = h and h = y sin , the force dF on the horizontal strip is:

Where yc is, by definition, the distance from OX along the sloping plane to the centroid C of the area A.

(3.15) y sin ydAsin dFF

sin

c

AgIntegratin

dAyhdApdAdF


If hc is the vertical depth to the centroid, then,

And in general we have:

Thus we find the total force on any plane area submerged in a liquid by multiplying the specific weight of the liquid by the product of the area and the depth of its centroid. The value of F is independent of the angle of inclination of the plane so long as the depth of its centroid is unchanged.

Since hc is the pressure at the centroid, we can also say that the total force on any plane area submerged in a liquid is the product of the area and the pressure at its centroid.

sincc yh

(3.16) AhF c

Center of Pressure

The point of application of the resultant pressure force on a submerged area is called the center of pressure. We need to know its location whenever we wish to work with the moment of this force.

The most general way of looking at the problem of forces on a submerged plane area is through the use of the recently discussed pressure prism concept (Sec. 3.6 and Fig. 3.16). The line of action of the resultant pressure force must pass through the centroid of the pressure prism (volume). As noted earlier, this concept is very convenient to apply for simple areas such as rectangles. For example, if the submerged area in Fig. 3.15 is of constant width then we know that the centroid of the pressure prism on area MN is 2/3 MN below M.

Center of Pressure

Fig-3.16

Center of Pressure

If the shape of the area is not so regular, i.e., if the width x in Fig. 3.16 varies, then we must take moments and integrate. Taking OX in Fig. 3.16 as an axis of moments, the moment of an elementary force dF = y sin dA is:

and if yp denotes the distance to the center of pressure, using the basic concept that the moment of the resultant force equals the sum of the moments of the component forces,

Where we recognize that Io is the moment of inertia of the plane area about axis OX.

dAyydF sin 2

o2 I sin sin dAyFy p

Center of Pressure

If we divide this last expression by the value of F given by Eq. (3.15), we obtain

The product yc A is the static moment of area A about OX. Therefore Eq. (3.17) tells us that we can obtain the distance from the center of pressure to the axis where the plane (extended) intersects the liquid surface by dividing the moment of inertia of the area A about the surface axis by its static moment about the same axis.

)17.3( sin

sin

Ay

I

Ay

Iy

c

o

c

op

Center of Pressure

We may also express this in another form, by noting from the parallel axis theorem that:

Where Ic is the moment of Inertia of an area about its centroidal axis. By substituting for Io into Eq. 3.17.

So,

cco IAyI 2

Ay

IAyy

c

ccp

2

(3.18) Ay

Iyy

c

ccp

Center of Pressure

From this equation we see that the location of center of pressure P is independent of the angel that is, we can rotate the plane area about axis OX without affecting the location of P. Also, we see that P is always below the centroid C and that as the depth of immersion is increased, yc increases and therefore P approaches C.

We can find the lateral position of the centre of pressure P by considering that the area is made up of a series of elemental horizontal strips. The centre of pressure for each strip is at the midpoint of the strip. Since the moment of the resultant force F must be equal to the moment of the distributed force system about an axis, say, the y-axis.

Where Xp is the lateral distance from the selected y axis to the center of pressure P of the resultant force F, and xp is the lateral distance to the center of any elemental horizontal strip of area dA on which the pressure is p.

(3.19) pdAxFX pP

ProblemThe cubic tank shown in Fig. is half full of water. Find (a) the pressure on the bottom of the tank, (b) the force exerted by the fluids on a tank wall, and (c) the location of the center of pressure on a wall.

Solution:

kN

kNxxAh

mxy

kNxAp

b

kPa

xhpa

watercwater

c

airair

waterwaterair

8.4181.932F F Force Total

81.925.081.9F

surface, water top thebelow 5.015.0h

at isn applicatio of

area theof C centroid the, tank wall theof halflower on the

ondistributi pressure water varying thei.e., Bcomponent For

3248F

: watert the throughouacts which air, confined

theof pressure the todue on,distributi pressure uniform a has

AComponet sketch.on distributi pressure on the B andA labeled

,components twointo divided is end tank on theaction force The )(

81.1717.81kN/m

181.98p )(

BA

B

c

A

2

bott

surface. water top thebelow 1565.0y

)(yF)(yF)F(y :moments Taking

667.0y

2mbhA 1667.012/1212

I ,5.0y

:equation thisusingby find alsocan We

area. wallr wettedrectangula

on the pressure varyingfor the surface water top thebelow

667.0)1(3

2

3

2)(y

pressure.air uniform for the area

square 2m theof centroid the tosurface, water top thebelow

0)(y

aresurface, water top thebelowy distance as

forces,component theof pressure of centers oflocation The )(

p

pBpAp

p

2433

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mh

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water

ProblemWater and oil in an open storage tank are in contact with the end wall as shown in Fig. S3.7. (a) Find the pressure at the bottom (lowest point) of the tank caused by the liquids. Also find (b) the total force exerted on the end wall by the liquids, and (c) the depth of its center of pressure.

(a) P bott = oil h oil + water h water

= (0.8x62.4Ib/ft3) (1.5ft) + (62.4Ib/ft3) (1.0 ft)

= 137.3 lb/ft3 = 0.953psi

(b) The force acting on the end consists of three components, labeled A, B and D, on the pressure distribution sketch. Note that component B has a uniform pressure distribution, due to the oil (A) above, which acts throughout the liquid below.

As a preliminary, we note for the semicircular end area (r = 1 ft) that(i) A = r2/2 = 12/2 = 1.571 ft2;(ii) from Appendix A, Table A.7, the centroid is 4r/3= 0.424 ft

from the center of the circle, i.e., below the water top surface.

Solution:

For component A, i.e., the varying oil pressure distribution on the l.5-ft height of the end wall, the centroid C of the area of application is at:

hc= yc = 0.5(1.5 ft) = 0.75 ft below the free oil surface,

so, from Eq. (3.16),

FA = oil hc Aoil = (0.8 x 62.4)0.75(1.5 x 2) = 112.3Ib

For component B, the force FB on the water-wetted area of the end wall due to the uniform pressure produced by the 1.5-ft depth of oil above is

FB = pA = h A= (0.8 x 62.4 )1.5(12/2) = 117.6 lb

For component D, i.e., the varying pressure distribution due to the water (only) on the water-wetted area of the end wall, the centroid C is at

hc = yc = 0.424 ft below the water top surface,

so: FD= hc A = 62.4(0.424) 12/2 = 41.61b

Solution:

The total force F on the end of the tank is therefore

F = FA + FB + FD = 272 lb

(c) As a preliminary to locating the center of pressure, we note that for the semicircular end area with D = 2 ft,

(i) from Table A.7: I about the center of the circle, is I = D4/128 = 4/128 = 0.393 ft4, and

(ii) by the parallel axis theorem: Ic about the centroid, distance 0.424 ft below the center of the circle, is Ic = I + Ad2 = 0.393 + (12/2) (0.424) 2 = 0.1098 ft4.

Solution:

The locations of the centers of pressure, below the free oil surface, of the component forces are:

(yp)A = 2/3(1.5 ft) = 1.00 ft for the varying oil pressure on the oil-wetted area, and

(yp)B = yc = 1.5 + 0.424 = 1.924 ft to the centroid of the water-wetted semi circular area, for the uniform pressure on this area due to 1.5 ft of oil above the water; and

Eq. (3.18):

below the water top surface.

ftAy

IyDy

c

ccp 589.0

)2/1(424.0

1098.0424.0)(

2

Solution:

for the varying water pressure on the water-wetted semicircular area, =1.5 + 0.589 = 2.09 ft below the free oil surface Finally,

F(yp) =FA(yp)A +FB (yp)B + FD(yp)D

yp = 1.567 ft

Solution:

Exercise Problems:

1. A circular area of diameter d is vertical and submerged in a liquid. Its upper edge is coincident with the liquid surface. Derive an expression for the depth to its center of pressure.

2. If a triangle of height d and base b is vertical and submerged in a liquid with its base at the liquid surface, derive an expression for the depth to its center of pressure.

3. If a triangle of height d and base b is vertical and submerged in liquid with its vertex at the liquid surface, derive an expression for the depth to its center of pressure.

Force on a Curved Surface

On any curved surface such as MN in Fig (a), the force on the various elementary areas that make up the curved surface are different in direction and magnitude, so algebraic summation is impossible.

Hence we can apply Eq 3.16 (in previous section) only to plane area. But for non-planer areas, we can find the component forces in certain directions.

a. Horizontal Force on a Curved Surface:

We may project (extend) any irregular curved area MN (Fig a) onto a vertical plane whose trace is M’N’ (Fig b).

The projecting elements (M’MNN’), which are all horizontal, enclose a volume whose ends are the vertical plane M’N’ and the irregular area MN. This volume of liquid is in static equilibrium.


A force F’ acts on the projected vertical area M’N’. The horizontal force component Fx’ acts on the irregular end area MN and is equal and opposite to the Fx of Fig (a).

Gravity force W’ is vertical, and the lateral forces on all the horizontal projection elements are normal to these elements and hence normal to F’.

Thus the only horizontal forces on MNM’N” are F’ and Fx’, and therefore,

''

''

0

FFFand

FF

xx

x


Hence the horizontal force in any given direction on any area is equal to the force on the projection of that area onto a vertical plane normal to the given direction.

The line of action of Fx must be the same as that of F’. Above equation is applicable to liquid and gases.

b. Vertical Force on a Curved Surface:

We can find the vertical force Fz on a curved area, such as MN in Fig a, by considering the volume of liquid enclosed by the area and vertical elements extending to the free surface (MNKJ) (Fig c). This volume of liquid is in static equilibrium.

K J


Vertical forces on this volume of liquid are;

a. The force due to any gas (at pressure ) above the liquid.

b. The gravity force W downward.

c. Fz’ the upward vertical force on the irregular area MN. The force Fz’ (Fig-c) is equal and opposite to the force Fz

(Fig-a). Any other forces on the vertical elements are normal to the elements, and so are horizontal. Therefore

Gzz

Gz

FWFF

FWF

'

' 0

ApF GG Gp


Therefore the vertical force acting on any area is equal to the weight of the volume of liquid above it, plus any superimposed gas pressure force.

The line of action of Fz is the resultant of FG and W. FG must pass through the centroid of the plan (surface

projection) area, and W must pass through the center of gravity of the liquid volume.

c. Resultant Force on a Curved Surface:

In general, there is no single resultant force on an irregular area, because the horizontal and vertical forces, as found in the above discussion, may not be in the same plane.

But in certain cases these two forces will lie in the same plane and then we can combine them into a single force.

ProblemFind the horizontal and vertical components of the force exerted by the fluids on the horizontal cylinder in Fig. if :

a) The fluid to the left of cylinder is a gas confined in a closed tank at a pressure of 35 kPa.

b) The fluid to the left of cylinder is water with a free surface at an elevation coincident with the uppermost part of the cylinder.

Assume in both cases that the atmospheric pressure occurs to the right of the cylinder.

Solution:

upward /35)1(/35F

Thus .12sin30af is

gas for the plane horizontal aon projectionnet theHence cd'.' surface on the

that toopposite and equal is ac'' surface on the gas theof force verticalThe

right the to/5.130)73.3(/35F

:gas For the (a)

73.32cos30 2 ef diagram,

handedright thefrom is, diagram)left (seeion consideratunder surface

lcylindrica theofportion theof plane verticalaon projectionnet The

2z

o

22x

o

mkNmmkNpA

m

mkNmmkNpA

m

x

upward /0.100

)21(30cos230sin22

12

360

2109.81kN/m

3)2(1 liquid of volumehatched-cross ofweight

abca volumeofweight -abcdefa volumeof weight

ca surfaceon force downward - cde surfaceon force upwardFNet

right the to68.3kN/m

73.373.32

181.9F

:fluid For the (b)

223

z

x

mkN

mxx

xxxAh

oo

c

ooo 30180210 12

3

lecture -3 fluid statics. force on a plane area no tangential force can exist within a fluid at...

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