Fluid Statics-Forces on submerged plane surfaces

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From the course by Georgia Institute of Technology

Fundamentals of Engineering Exam Review

172 ratings

Georgia Institute of Technology

Fundamentals of Engineering Exam Review

172 ratings

The purpose of this course is to review the material covered in the Fundamentals of Engineering (FE) exam to enable the student to pass it. It will be presented in modules corresponding to the FE topics, particularly those in Civil and Mechanical Engineering. Each module will review main concepts, illustrate them with examples, and provide extensive practice problems.

From the lesson

Fluid Mechanics

This module reviews the basic principles of fluid mechanics particularly the topics covered in the FE Exam. It first discusses what a fluid is and how it is distinguished from a solid, basic characteristics of liquids and gases, and concepts of normal and shear forces and stresses. The major fluid properties are then discussed. Next fluid statics are addressed: pressure variation in homogeneous and stratified fluids and application to manometers; forces on submerged plane surfaces and buoyancy forces on fully and partially submerged objects.Flowing fluids are then covered. This includes the equations for conservation of mass (the continuity equation) and energy (the Bernoulli equation). These are then applied to velocity and flow measuring devices: the Pitot tube, and Venturi and orifice meters.The final topic is similitude and dimensional analysis. This includes concepts of fundamental dimensions and dimensional homogeneity, the Buckingham Pi theorem of dimensional analysis, and the conditions for complete similitude between a full-scale prototype flow situation and a small scale model.In all cases, basic ideas and equations are presented along with sample problems that illustrate the major ideas and provide practice on expected exam questions.Time: Approximately 6 hours | Difficulty Level: Medium

Fluid Properties- Introduction6:29

Fluid Properties-Density and Pressure10:37

Fluid Properties-Stresses Viscosity15:46

Fluid Properties-Surface tension10:21

Fluid Properties-Units and other properties4:16

Fluid Statics- Introduction and Pressure Variation12:13

Fluid Statics-Application to manometers and barometers14:55

Fluid Statics-Forces on submerged plane surfaces10:19

Fluid Statics-Forces on submerged plane surfaces continued9:28

Fluid Statics-Buoyancy and stability9:12

Continuity and Energy Equations: Continuity and mass conservation7:45

Continuity and Energy Equations: Energy equation9:12

Continuity and Energy Equations: Energy equation examples9:24

Flow Measurement-Pilot tubes10:51

Flow Measurement-Venturi meter4:27

Flow Measurement-Orifice meter9:17

Flow Measurement-Dimensions and units, Pi theorem12:24

Flow Measurement-Similitude9:26

Flow Measurement-Similitude examples13:06

Meet the Instructors

Dr. Philip Roberts
Professor
School of Civil and Environmental Engineering

Continuing our discussion of fluid statics.

The topic I want to address now is forces on submerged plane surfaces.

And, the question is this.

Suppose we have some surface which is submerged, say below water surface.

For example, the water in this tank here and for example the bottom surface here.

If this is horizontal, the pressure there is horizontal,

the pressure there is constant rather.

So the pressure, the force there is just the pressure multiplied by the area.

And this force I can replace with some resultant force which I'll call FR.

But, what about the vertical faces here?

In this case, the pressure is increasing linearly downwards

according to our hydrostatic pressure distribution.

And, again, I will have some resultant force here, which is

pushing on this plane, and in this case, it's going to be pushing to the left.

So the most general question is,

if I have some arbitrary plane surface of arbitrary shape, arbitrary inclination,

what is the magnitude of the resultant hydrostatic force acting?

Where does it act, and what is the direction of the resultant force?

And those are the issues we're going to address in this segment.

So to do this, I'll consider a completely arbitrary plane surface as shown here.

And this arbitrary plane surface,

if I look at this in edge view, then I see the view here.

And the inclination of this plane surface is

inclined at theta to the horizontal with the angle here.

And the resultant force I will assume is FR

at some location below the free surface.

Now, because the pressure forces varying

over this object I have to find the force on an element, and

then sum the forces over the whole area to get the results in force.

So, I'll take an element of the area here, which I'll call dA.

Which is submerged below the water surface by a height, h,

a depth y Where y is the depth measured down the sloping surface.

And the vertical depth of this element below the water surface is H.

So at any depth H the pressure is the hydrostatic

pressure which is specific weight times H, in other words gamma times H.

But pressure is force per unit area, so the force on that element,

dF, is gamma h times the area, which is dA.

And the total force on that plane's surface is

the summation of the forces on these elemental areas, or in the limit,

the integral over the area of gamma h dA.

Now, we can evaluate that because from simple geometry, the vertical depth, h,

is just y times sin theta, so substituting that in, we find

that the result in force is the integral over the area of gamma y sin theta dA,

and taking out the constants, Gamma sine Theta, we get this expression.

But, this integral here, integral over the area of ydA,

is by definition, yC times A where yC is one over a,

integral ydA, which is the distance of the centroid of the area,

below the surface, measured down the sloping incline.

Therefore, the resultant force is gamma ayc sine theta,

but yc sine theta, this term here It's just

the vertical depth of the centroid below the water surface hc.

So the resulting force is gamma hc x A,

but this term gamma times hc is just the pressure at the centroid.

So therefore we arrive at a very simple equation that the result in force

is equal to the pressure at the centroid of the plane

multiplied by the area of that plane surface.

That is the magnitude of the result in force.

What is the direction?

Well I think intuitively you can see that the direction of

this resultant force must be perpendicular to the surface.

Why?

Because if it wasn't, let's suppose that the resultant force was at some arbitrary

angle, then there would be a component of force which is tangential to that surface.

Therefore we would have a tangential or a sheer stress and,

because of our definition of a fluid, the fluid would stop moving.

Therefore, the only way for this to be in static equilibrium is for

the resultant force to be perpendicular to the plane.

Now the resultant force is the pressure at the centroid multiplied by the area,

however, the resultant force does not act at the centroid.

It acts at a location which I'll call the center of pressure.

So here's the resultant force, FR, and

it's acting at the center or pressure here CP.

And the center pressure's defined as the point of application of the result and

force in other words, where a single equal and

opposite force must be added for that plane surface to be in equilibrium.

And we find this location by taking moments about this point O,

on the free surface.

So, the moment of the result in force is Fr times yr,

where yr is the location of the center of pressure,

measured down the slope from the free surface, and

that is equal to the summation of the moments on this elemental area.

The force on the element is ydF,

the moment arm times force, and the sum of the moments

on all of those elements is equal to the integral over the area of ydF.

But dF, we've already shown, is gamma y sin theta times dA.

So this in turn becomes taking the constants out of the equation,

gamma sine theta, integral over the area of Y squared dA.

But this quantity, integral over the area of Y squared dA,

is equal to the moment of inertia of the service about O which we'll denote by IXO.

So therefore, combining those equations we find that the location of

the center of pressure of the resultant force is IXO/ycA.

But that formula is not very convenient because

that gives the moment of inertia about this point.

It's much more convenient if we can express the equation in terms of

the moment of inertia about the centroid.

And to find that, we use the parallel axis theorem,

which states that the moment of inertia, about any arbitrary location

Ix0 is equal to the moment of inertia about the centroid,

plus the area multiplied by the distance squared

between the centroid and the point about which the moment of inertia is taken.

So combining that, we get this final expression.

That yR the location of the central pressure is equal to yC

the distance of the centroid plus IXC divided by yC times

A where IXC is now the moment of inertia about the centroid.

So here's that formula again, yR is yC plus ixc over yc times A and

what you notice about this formula is that ixc is always a positive quantity,

yC is a positive.

A is a positive a quantity.

Therefore, this term here is always a positive quantity.

In other words,

the distance of the center pressure below the centroid is always positive.

In other words, the center pressure is always below the centroid, it's deeper.

However, if I take this plane and push it downwards.

So I take this plane and move it downwards, submerge it more deeply,

IxC remains constant, A remains constant, but yC increases.

In other words, as I go deeper down, this quantity decreases.

It becomes smaller, in other words,

the center pressure approaches the centroid as the plate moves down.

And the distance here,

this distance is the distance of the resulting force below the centroid.

Now, this raises another question of course.

What is the moment of inertia of a plane's surface?

And this typically will look up in tables.

This is an extract of the table from

FE Reference Handbook and it gives moments of inertia and other properties for

various shapes, such as rectangles, triangles, etc.

And, this is a detail or

a blow up of that table, for triangles and also rectangles.

And in the next section we'll do some examples and

applications of these surfaces to different types of problems.

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