Fluid Statics-Forces on submerged plane surfaces continued

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From the course by Georgia Institute of Technology

Fundamentals of Engineering Exam Review

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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

Now, I want to continue our discussion of fluid statics by giving

some examples of calculations of forces on submerged plane surfaces.

Again, the basic formula we had for the location of the center of pressure or

resultant force was yR is yc + i xc divided by yc times A.

And at the end of the last segment where I showed

a number of examples of different shapes and moments of inertia.

Most probably the most common you'll encounter though are simple rectangles and

circles, and your relevant formulas for those are given here.

For a rectangle of width b and height a, the area is obviously b times a.

The moment of inertia is one-twelfth times the width, times the height cubed.

And the centroid by symmetry is obviously in the middle.

For a circle the area is pi R squared.

The moment of inertia about any axis, the horizontal or

the vertical through the centroid is pi R to the 4th over 4.

And again by symmetry obviously the centroid is in the middle.

So let me do some examples on that.

Firstly, for a vertical wall, for example a dam holding back water.

The pressure distribution is hydrostatic which is linear.

As a gauge pressure it increases linearly as we go down from zero

at the free surface.

And for rectangular wall the centroid is in the middle by symmetry.

In other words, it's at a depth of h of 2, below the water surface.

So the resultant force is Fr,

which is the pressure at the centroid, multiplied by the area.

And because this is symmetrical,

the pressure of the centroid is gamma times h over 2.

And the area is the water depth age multiplied by the width which is b,

which is equal to gamma h squared over 2 times the width b.

It acts at

a depth yr = yc + I xc over yc times A, but

in this case the wall is vertical therefore h,

the vertical depths is the same as y.

So in this case we have hR.

The location of the resultant force is hc plus I xc over hc times A.

For rectangular face, Ixc is bh cubed over 12,

and the depth of the centroid is half the water depth, h over 2.

So putting all of those together,

we find that the location of the resultant force, hR is two-thirds h.

In other words, it acts two-thirds h, down from the water surface.

Or conversely, one-third h, up from the bottom.

So we note from this that the resultant force is the average pressure

times the area.

And that simple formula works also for

a wall which is partially submerged, like this.

So if I'm just interested in this portion of a gate or a plug, for example.

Which is below the water surface then again, the resultant force is the pressure

at the centroid of that gate multiplied by area which is equal to

gamma h1 + h2 over 2, where h1 is the depth of the top of that gate.

Below the water surface, and h2 is the depth of the bottom of the gate.

The pressure distribution is shown over on the right here, and

that is equal to p1 + p2 over 2,

where p1 is the pressure at the top of the gate, p2 is the pressure at the bottom.

The area is h2- h1 times b, and the same formula applies.

But note that that only applies for a gate of constant width,

in other words a rectangular wall.

Because in that case the force in both cases doesn't act at the average depth.

It also applies to an inclined plane, the same formulas.

And these are the corresponding pages or reference from the Reference Handbook,

which just give the same formulas that I just derived.

Now let's do some numerical examples.

The first one, we have a water tank here which is connected to a round pipe.

So this is a round pipe here, 6 feet in diameter.

And it's sealed off by a circular plug.

If the specific weight of the water is 62.4 pounds per cubic foot,

the resultant force on the plug is most nearly which of these alternatives?

So the resultant force is given by our usual formula.

That the resultant force is the pressure of the centroid multiplied by the area.

In this case the bisymmetry,

the centroid here is in the middle of the plug.

So if the depth of the centroid below the water surface is hc,

the pressure there is gamma times hc, so the force is gamma hc times A.

The area, this is a round plug, so the area is pi d squared over 4.

So plugging in the number, specific way to 62.4.

The depth of the centroid below the surface is 9 plus half the diameter is 3.

Which is 12 feet times pi d squared over 4, diameter is 6 feet.

So the answer is 21,200 pound force and the best answer is B, 21,000 pounds.

Next, the same problem, but

the question now is the depth of the location of the resultant force

on the plug below the water surface is most nearly which of these alternatives?

So here our solution, we apply because this is a vertical plug,

the face here is vertical.

We have hr, the location of the resultant force.

So let's suppose that the resultant force

here is hR below the water surface,

is hc the depth in centroid plus I xc, over hc times a.

The moment of inertia of a circle is pi R to the 4th over 4.

In this case the radius is 3.

So that's pi times 3 to the 4th over 4 is

equal to 63.6 units of feet to the 4th,

the area is pi R squared which is 28.3 square feet.

So therefore Ixc over hc times A is 63.6 over

12 times 28.3 which is 0.19 feet.

And the answer is hc the depth of the resultant force

is 12 + 0.19 is equal to 12.19 feet.

So the closest answer is D.

The final example is a sloping gate.

So here we have a sloping gate.

And it's 4 feet wide into the drawing, and

it's held in place by a cable here, as shown.

The resultant force,

due to the water on the gate is most nearly which of these alternatives?

So here we have, the solution is again,

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

which is equal to gamma times the depth of the centroid multiplied by area.

Which is, and in this case, we have a sloping incline.

So the depth of the centroid.

The centroid by symmetry is in the middle of the gate, right here.

Which is 3 feet right here, measured by sloping.

And that is equal to yc times sine theta, times the area,

plugging in the numbers that's 62.4 specific weight

times 6 over 2, 3 feet times sin 60 degrees,

multiplied by the area of the gate which is 6 feet by 4 feet.

So the answer is 3890 pounds.

And the closest answer is D which is 3900 pounds force.

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