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EN3: Introduction to Engineering and Statics

Division of Engineering

Brown University

7. Equilibrium

Engineers need to manage forces in structures and machines. There most common problems that engineers face are:

1. To design a system that will apply a set of forces to an object. For example, many manufacturing processes (milling, cutting,injection molding, etc) need to apply large forces to the material being processed; and engineers have to design machines

capable of applying these forces. Vehicle design is a second example. An aircraft in flight must be designed carefully to

balance gravitational and aerodynamic forces, to ensure that the aircraft remains controllable. Similarly, a cars engine and

transmission must be designed to exert whatever forces are necessary to overcome air resistance, rolling resistance, etc

while the car travels.

2. To design a structure or machine that can support a given set of external forces for an extended period of time without

failure. Examples range from structural problems, to the design of artificial joints to design of micromachines;

To manage forces, engineers need to be able to answer two questions: (1) What forces are required to make a structure move or

deform in some prescribed way?; and (2) What internal forcesare induced in a component by external loading? A structure or

machine fails when these internal forces become larger than the materials strength.

Surprisingly, forces in most engineering systems can be understood using the principle ofstatic equilibrium. There are really only

three main situations where statics doesnt work, (because large accelerations occur)

1. Anything involving impact (crashing vehicles, exploding bombs, etc)
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2. Anything that travels along a curved path (an airplane in a turn, cornering vehicles)

3. Systems that contain rapidly rotating parts;

4. Forces induced by vibrations.

A static analysis always solves the same problem: given a set of known forces acting on a structure or machine, calculate a set of

unknown forces. The unknown forces could be internal forces, or some subset of unknown external forces.

The unknown forces are always calculated by solving equilibrium equationsfor the system.

7.1 Definition

Four and a half simple concepts are required to understand the concept of static equilibrium

1. If a structure is stationary it is in static equilibrium

2. If a machine moves at constant speed along a straight line without rotation it is in static equilibrium

3. The resultant force acting on a structure or machine that is in static equilibrium is zero

4. The resultant moment acting on a structure or machine that is in static equilibrium is zero.

4.5 It doesnt matter what point you choose to take moments about in a statics problem its zero about any point.

But beware! When you start solving dynamics problems, you will need to take moments about special points to applyequations of motion (usually the center of mass).

Thats it! Theres nothing more to statics. We can all go home now

But wait! How about a few examples and applications before you go? We have to dosomethingto earn your tuition dollars


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7.2 Examples of the use of static equilibrium to calculate unknown forces

When solving statics problems, well always follow the steps below:

1. Draw a clear picture showing the forces and moments acting on the object(s) of interest. Its important to show the

positions of the forces correctly;

2. Introduce an appropriate basis to be used for all vector calculations

3. Write down the forces acting on the system (introduce variables to describe unknown forces)

4. Write down the position of the forces

5. Find the moment of the all forces about any convenient point (you must use the same point for each force).

6. Write down any pure moments or torques acting on the system

7. Find the resultant force F

8. Find the resultant moment M

9. Set F=0 and M=0.

10. If you can, then solve the equations for unknown quantities of interest (often forces, but the unknowns could be other thingstoo, as we shall see).

11. If you have too many unknown forces and not enough equations at this point, you need to look for more equations. These

may come from (a) Force balance for other components; (b) Force laws (eg spring law, buoyancy force law, gravity law,

etc); (c) geometry.

12. When you have enough equations, solve them.

So lets try some problems and see how this works.

Example 1: A beam with weight Wis suspended by two spring-scales, as shown in the picture below.

(i) Find an expression for the force reading on each spring scale in terms of W,Land d.

(ii) If the spring scales both have stiffness k andun-stretched length a, find an expression for the difference in length bof the

scales, in terms of k, W,Land d.


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This is a standard equilibrium problem. We know the weight force acting on the beam. We dont know the forces exerted on the

beam by the springs. However, we doknow that the beam isnt moving and therefore must be in static equilibrium. We can try to

calculate the unknown spring forces by writing down equilibrium equations for the beam.

To solve the problem, we follow the steps in our recipe.

(1, 2) Heres the picture showing the forces; as well as a basis for our vectors. We know the weight acts at the center of mass, and

can look up the position of the center of mass in the table provided in section 2.1. We also know the spring scales will pull on thebeam along the line of the scale, and show the forces accordingly.


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Now we proceed to steps 3-9. Its convenient to assemble this information in aforce and moment balance tableas shownbelow.

Force and moment balance table for beam ABC. Origin at A.

Force/Moment Position Force Moment about origin

Weight

Force at A 0 0

Force at B

Sum (=0)

Note that you can use any of our short-cuts to compute the moments, if you wish. For a statics problem, it doesnt matter what point

you choose for the origin. But beware when you move on to dynamics, you will need to write down dynamical equations of motion

by taking moments about special points

Finally, step 10 - we collect and (try to) solve the governing equations. We get the equations by setting the i,jand kcomponents of


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resultant force and moment to zero individually. For the problem at hand, we have only two equations

We have two equations and two unknown forces, so we are ready to solve them. If, at this point, we had more unknowns than

equations, wed have to look for other principles to apply in order to provide additional equations for the unknowns.

The equations are easily solved for the unknown forces and

This answers part (i) of our problem.

To answer part (ii), we have to look for additional principles to apply we cant get any more information from statics. Elementary

geometry tells us that bis the difference in length of the two spring scales. Next, we recall that the length of each spring scale can be

computed from the force using the force-extension relation for a spring as

Therefore, the difference in length is given by

giving the solution to part (ii).

If youve taken courses in physics in high school, this procedure may look cumbersome and unnecessary to you. Bear with me. Its

true that simple problems like this one can be solved in your head. But the objective of this discussion is to develop a systematic

approach that can be used to solve 3D problems involving complex engineering systems, containing tens or even hundreds of

unknown forces. You can get completely lost if you try to solve these using an ad-hocprocedure.


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Example 2

The figure shows a force transducer mounted to the wheel of a car. Its purpose is to measure

the contact forces acting on the tires under realistic driving conditions. The contact forces act

where the wheel touches the ground, and in general, three components of moment and three

components of force may be present at the contact.

The transducer will measure three force components, and three moment components. The

transducer doesnt record contact forces directly. Instead, it records the forces and moments

exerted by the transducer on the wheels hub. These readings need to be corrected. Our

mission is to derive a formula that can be used to calculate the contact forces from the

transducer reading. Well neglect the wheels mass, to keep things simple.

This is again a standard statics problem. We know the forces applied on the wheel by the force

transducer (because the instrumentation tells us the values). We dont know the contact force,and will calculate it from the condition that the wheel is in static equilibrium

The picture shows forces acting on the wheel, and relevant dimensions. The force transducer records force components and

moments that are exerted by the transducer on the wheels hub. We have assumed that the transducer forces act on the outside

of the wheel hub at the center of the wheel, as shown in the picture. Our goal is to calculate a formula for the contact forces and

moments , in terms of the measured and , as well as geometrical variables.


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Well use static equilibrium to relate the two forces (actually this wont be correct if the car accelerates but the error will be small).

The force balance table for the wheel is shown below.

Force and moment balance table for wheel. Origin at force transducer.


Contact force

Force transducer 0


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Sum (=0)

This gives us 6 equations for the 6 unknown contact force and moment components

We could solve these by hand But were lazy. Instead, well use MAPLE


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So, the answer we are looking for is

Example 3:

The figure below shows a proposed design for a torque-wrench. The system is intended to measure the momentMthat is exerted


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by the wrench on a bolt. This will be accomplished by mounting three small force transducers at the locations shown. The user will

place the wrench over a bolt, and apply a force to the handle. As a result, the bolt will exert forces , and a moment on the

wrench at D. The transducers will record the forces acting on member (2) at points A, B, and C . The transducers are designed to

behave like springs, exerting only one component of force parallel to the transducer. The force readings must be used to calculate

.

The force transducer readings will be converted to a torque reading using a microprocessor built into the wrench. We have been

assigned the chore of working out the formulas that the microprocessor must use to deduce the torque.

Once again, this is a standard equilibrium problem. For a known set of forces and moment acting at D, we want to calculate the

force transducer readings. We can relate the force transducer readings to the forces and moments at D using the condition that

component ABCD is in static equilibrium.

The picture below shows the forces acting on component ABCD of the wrench. The forces acting at D are the forces and moment

exerted by the head of the bolt on the wrench they are therefore equal and opposite to the forces and moment exerted by the

wrench on the bolt.


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The transducer forces can be related to the forces acting at D using force and moment equilibrium. The force and moment balance

table is shown below

Force and moment balance table for wrench component. Origin at D.


Force and moment

exerted by bolt

0

Transducer A

Transducer B

Transducer C 0

Sum (=0)

The equilibrium equations are


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We have three equations, and three unknowns ( ). We can go ahead and solve them. MAPLE provides the formulas we are

looking for.

So the formulas are

A torque wrench usually only records the moment and does not measure the forces. Our calculations show that we couldcompute the moment using only two force transducers at A and B. We could cut cost by replacing the third force transducer with a

small stiff spring.


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Example 4:

The figure illustrates a proposed design for a floating oil platform. The platform has total weight W, and has center of gravity at the

location shown. The platform floats in seawater with mass density . The platform floats on three cylindrical legs, with radius ,

which must be extended or retracted appropriately to level the platform. Find a formula which predicts the length of each leg, as a

function of W, ,R, H,aand the acceleration due to gravityg.

This time it seems like the problem weve been asked to solve has nothing to do with forces we need to calculate the length of

each leg, not a bunch of forces! But of course, the length of each leg must be chosen to balance the forces acting on the oil platform

properly. Recall that the buoyancy force acting on an object is proportional to its submerged volume. The forces acting on each leg

are therefore controlled by lowering (which increases the force) or raising the leg.

S t l th bl ill fi t l l t th f th t d t t h l f th l tf t b l it Th ll


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So, to solve the problem, we will first calculate the force that needs to act on each leg of the platform to balance it. Then well use

the buoyancy force law to work out how much of the leg needs to be immersed.

We will begin by reviewing how buoyancy forces work.

Recall that

1. Buoyancy forces act perpendicular to the water surface

2. The force magnitude is equal to the weight of water displaced

3. The force acts at the center of mass of the displaced water.

The volume of water displaced by leg A is the submerged volume of the cylinder,

i.e.

.

The weight of water displaced by leg A is therefore

The force acts halfway up the submerged part of the leg, on the axis of the cylinder.

With this in mind, we can draw a free body diagram for the platform


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We can now balance forces acting on the platform

Force and moment balance table for platform. Origin at center of circular platform.


Weight

Force on leg A

Force on leg B

Force on leg C

Sum (=0)


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Sum (=0)


We have three unknown forces, and three equations, so we could go ahead and solve them. However, were not interested in the

forces were being paid to calculate the length of each leg.

We can get additional equations for the length of each leg using the buoyancy law.

We now have 6 unknowns ( and ) and six equations, so were all set. We can solve them by hand, or if lazy use

Maple. Here is the MAPLE solution (lazy? Me? No way!)


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(IMPORTANT NOTE: Weve used variables t60 to denote the angles (60 degrees). If you try to put in sin(60) into MAPLE, it will calculate sin(60radians) !! To get numbers, wed have to subst itute values for t60 in radians)

Example 5:

This is a goofy problem with no practical application whatsoever, for which we apologize But its designed to make an importantpoint.

In all the problems weve solved up to now, its been fairly obvious what part of the system we should isolate to balance forces. But

thats not always the case.

For example, suppose that two identical boxes, each of weight W, are suspended by the system of spring-balances as shown. (I said

this was a stupid problem). Calculate the forces shown by each spring balance.


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Obviously, we need to calculate the force readings by balancing forces for something. But what? The top box? The bottom one?

Both together?

Whenever youre stuck like this, the best thing to do is to write down equilibrium equations for everything you can think of, and then

look at the equations you end up with, and see if you can solve them.

Well work through all three possible combinations for the example. First, well get force balance equations for each box, then well

do the two together.

Heres what we get for the top box


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Here s what we get for the top box

Force and moment balance table for top box. Origin at CG.


Weight 0 0

Force A

Force B

Force E

Force F

Sum (=0)

Next, the bottom box


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Force and moment balance table for bottom box. Origin at CG.


Weight 0 0

Force A

Force B

Force E

Force F

Sum (=0)

Finally, the two boxes taken together as a single system


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Force and moment balance table for both boxes. Origin at CG for bottom box.


Weight (top) 0

Weight (bot) 0 0Force C

Force D

Force A

Force B

Sum (=0)


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Note in passing how we introduced the variable hto give the distance between the CGs of the two boxes. The problem didnt give

us this info and well find out later that the answer is actually independent of the height h but if you need a dimension in a problemas an intermediate step, its perfectly OK to go ahead and define one.

OK, lets check the equations we ended up with. The first set (eq1) were for the bottom box; the second (eq2) were for the top,

and the third (eq3) were for the two together

(1.1)

(1.2)

(1.3)

So now what? We have 9 equations, and only 6 unknowns. Can we solve them? Usually, no. But we could try throwing the whole

shebang into MAPLE and see what happens. It turns out that this actually works heres MAPLEs solution. (By the way - if

youve struggled to get solutions to some of the homework problems for this course, you are not alone... It took me over an hour to

fix all the sign errors in my calculations and typos in the MAPLE formulas before I could get this to work!)


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(IMPORTANT NOTE: Weve used variables t30 and t60 to denote the two angles (30 degrees and 60 degrees). If you try to put in sin(30) intoMAPLE it will calculate sin(30radians ) !! To get numbers wed have to subs titute values for t30and t60 in radians)


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MAPLE, it will calculate s in(30 radians ) !! To get numbers , we d have to subs titute values for t30 and t60 in radians)

Its surprising that this works. Usually, when you have more linear equations to solve than unknowns, you cant satisfy all the

equations. So how come it works here?

It works because our three sets of equations are not independent. If any two of them are satisfied, the third one is satisfied

automatically. We can easily check this with MAPLE. Heres what happens if we solve the first two and plug the solution into the

third.


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Although this is a goofy example, it illustrates a general feature of complex statics problems. Heres what we can conclude:

1. When you need to solve a statics problem that involves many different components connected together, you can write down

equilibrium equations for each component, or you can write down equilibrium equations for combinations of the components.

2. Generally the safest way to proceed is to get equations for each component, and solve them. For example, in the problem we

j t l d ld h itt d ilib i ti f h b b it lf d l d (1 1) d (1 2)


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just solved, we could have written down equilibrium equations for each box by itself, and solved (1.1) and (1.2).

3. However, its perfectly OK to get equilibrium equations for group(s) of components too. We see from our simple example that

we will get the same answer, and in many cases the equilibrium equations for a group of components may give us the answer we

are interested in much more quickly than if we were to try to analyze each component separately. For example, if you want to

calculate the forces acting on the wheels of the car, you would just write down equilibrium equations for the whole car, you

wouldnt do one for each nut, bolt, engine component, etc

When you solve a complex statics problem, its often worth getting equilibrium equations for lots of possible combinations, and then

selecting the set of equations that looks the simplest.

Example 6Another goofy example to make another point. Not all statics problems have a solution. Consider the following the

beam shown below is supported by a single spring balance. What force does the spring balance record?


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This problem is so simple we dont need to draw a force diagram heres the force balance table

Force and moment balance table for beam ABC. Origin at CG.


Weight 0 0

Force at A

Sum (=0)

So now we need to solve the equations

but theres no value of that can satisfy bothequations.

Thats because the system cannot be in static equilibrium in the configuration shown the beam would fall down. So, if you work

through a problem and find that you cant satisfy all the equilibrium equations, it means the system you are analyzing must move.


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Example 7: This is a final goofy problem to make another point. Some statics problems may not have a unique solution. Heres an

example. The beam shown below is supported by three spring balances. Calculate the force reading on each.

Heres the force and moment balance for this problem

Force and moment balance table for beam ABC. Origin at CG.

Force/Moment Position Force Moment about originWeight 0 0

Force at A

Force at B 0

Force at C

Sum (=0)

We now have two equations with three unknown forces


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We now have two equations with three unknown forces

There are lots of possible solutions to these equations eg

or many others.

So how can we get the force readings? We need more information. For example, if we were told the force reading on oneof the

balances we could get the other two, or if we were told that the springs all have the same stiffness and un-stretched length, we would

know all the forces had to be equal, and could again get a unique solution.

7.3 Useful short-cuts to solve equilibrium problems

In the examples weve solved so far, weve always applied the same, formal, procedure to set up equilibrium equations. The

information was always presented in one or more force and moment balance tables.

Some simple problems can be solved using short-cuts, which you may have used extensively in introductory physics courses.

Below, we list a few of the more common short-cuts.

1If only two forces act on an object that is in static equilibrium, the forces must act along the same line, and must be equal

in magnitude and opposite in direction

You can show this using force and moment balance.

The shortcut is useful because for systems with only 2 forces acting on them we dont need to work through all that force balance


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The shortcut is useful, because for systems with only 2 forces acting on them, we don t need to work through all that force balance

table crud. We dont even have to mess with vectors. We can just use the fact that the two forces must have equal magnitudes and

opposite directions.

Example: Estimate the terminal velocity of a tennis ball, and human, falling through the air.

In both cases the objects are subject to only 2 forces: weight, and air drag.1. The weight force has magnitude ;

2. the air drag has magnitude , where is air density; Vis the velocity; is the drag coefficient andAis the projected

frontal area of the object.

The forces must be equal and opposite, so

Finally, we need to make an estimate of the numbers. This is a good chance to use our formal estimation procedures

Data/Assumptions:

1. Tennis ball

a. Mass: 0.057g (source http://physicsweb.org/article/world/12/6/3b. Area: diameter is 0.066m (source as above) so projected area is 0.0034

c. Drag coefficient 0.1-0.5 (smooth sphere result from section 2.7 of notes)

2. Human (assumed to be upright)

a. Mass 100kg

b. Area 0.125m (rough measurement of nearest human)

c. Drag coefficient 1.0-1.3 (source: http://aerodyn.org/Drag/tables.html#man
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Air density

Calculation

1. Tennis ball

2. Human

Error estimate

1. Tennis ball

a. Mass +/- 2%

b. Area +/-2%

c. Drag coeft +/- 50%

d. Air density +/- 5%

e. Total error = 59/2 % = 30% (the factor of 1/2 is because of the square root)

2. Human

a. Mass +/-20%

b. Area +/- 20%c. Drag coeft +/- 20%

d. Air density +/- 5%

e. Total error = 65/2% = 33% (again, the factor of 1/2 is the square root)

Example 2 we actually already made use of this principle when we described the forces exerted by a spring. Remember the

picture below?


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We said that the force exerted by the spring always had to act along the line of the spring. Why? Well, if we look at the spring,

there are only two forces acting on it. The forces act on the ends of the spring. Now we know that the forces have to be parallel;

equal in magnitude, and opposite in direction. This means that the forces mustact along the line of the spring otherwise the spring

wouldnt be in moment equilibrium, and would spin around.

2.If only three forces act on an object that is in static equilibrium, then (a) the forces must be coplanar and (b) the forces

must either act through the same point, or must be parallel.If the forces are not parallel, moment balance is satisfied

automatically. If the forces are parallel, two components of force balance and one component of moment balance are

satisfied automatically..

Example. The chair on a ski-lift has weight Wand is supported by two cables, which may be idealized as springs with stiffness kand

unstretched length a. Find a formula for the stretch of each cable in terms of W, , kand a.


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To calculate the extension of each cable, we clearly need to calculate the force thats stretching it, and then use the spring law. We

can calculate the cable forces by performing a force balance for the chair.

The picture below shows the forces acting on the chair. There are only three forces the weight; and the force exerted at O by the

two cables. Consequently, the three forces must be in the same plane, and must act through the same point. Both the forces exerted

by the cables clearly act through O, because thats where the cables are attached. For the weight force to act through O, the chair

must swing so that its CG is directly below O. The forces are not parallel, and so must be calculated using force balance moment

balance is satisfied automatically (if you dont see this, try taking moments about O!)


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We can do the force balance in a table

Force balance table for chair

Force exerted by cable OA

Force exerted by cable OB

Weight

Sum (=0)

This gives us, as expected, two equations for the two unknown forces in the cables.


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These can be solved (even without MAPLE!) to see that

where weve used the trig formula to get the simple formulas.

Finally, we can calculate the extension of each cable using the spring force law

, so that

Example: A bicycle is suspended from two bungee cords (think of them as springs), as shown below. Find the position of the

center of mass of the bicycle in terms of the length dand the two angles and , expressing your answer as a position vector relative

to the attachment point A.


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Only three forces act on the bike weight, and the force exerted on the bike by the two bungee cords. The forces must therefore

act through the same point and must be coplanar. The center of gravity must be in the vertical plane, and must be located at the

intersection of lines through the cords. We can use elementary geometry to calculate the components of the position vector, using thetriangle sketched below

Heres one way to calculatexandy. Geometry on the triangle shown gives


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We can solve these forxandy

where weve used the trig formula to simplify the result. The position vector is

2.If only four non-coplanar forces act on an object that is in static equilibrium, then the forces must either act through the

same point, or must be parallel.If the forces are not parallel, moment balance is satisfied automatically. If the forces are

parallel, two components of force balance and one component of moment balance are satisfied automatically..

This short-cut is not as useful as the others It would tell you, for example, that lines through the three cords lifting the box shown in

the picture below would all have to meet at a point above or below the CG of the box


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7.4 Case Study using Forces and Equilibrium: A simple aircraft control model

Aircraft designers need to know how design parameters affect aircraft performance. The aerodynamicists need to know how the

design of wings, fuselage and powerplant influence performance. The control systems engineers need to have models of aircraft

performance to design fly-by-wire systems and autopilots.

Weve covered enough theory in this course to be able to develop a quite sophisticated model of aircraft behavior. Well work

through a 2D example, which will illustrate a number of very strange aspects of aircraft control and performance. The analysis showsbeautifully the power of physics and mathematics to explain the behavior of exceedingly complex engineering systems. This case

study is presented for interest only we dont expect you to be able to do an analysis like this one (yet. But youll be doing much

more sophisticated calculations very soon!)


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The picture shows key geometrical parameters that affect aircraft performance. Some remarks:1. The aircraft travels parallel to the green line. If the aircraft is climbing, if the aircraft is descending

2. The horizontal `tail of the aircraft is known as a `stabilator. The pilot can rotate the stabilator, by pulling or pushing on the

control column. We use to denote the stabilator angle.

3. The `angle of attack of the airplane is the angle is the angle between its longitudinal axis and its direction of travel (actually this

is not quite right, but the definition is close enough)

The aircraft designer can set the following parameters

1. Wing area2. The distance of wing center of lift behind the center of gravity of the aircraft

3. The stabilator area

4. The distance of stabilators center of lift behind the center of gravity of the aircraft

5. The aircraft weight W

6. The maximum engine thrust (this is a distributed load acting on the propeller well model it as a statically equivalent forceP

acting on the spinner)

7. The distance from the CG to the propeller


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The following issues are of interest in aircraft design

1. What wing area is required to support a reasonable aircraft weight?

2. What engine power is required to give a reasonable cruise speed and rate of climb?

3. How is performance affected by the location of the centers of lift?

If you want to design an autopilot for the plane, or if you want to know how to fly it, you need to know

1. How do you control the planes speed?

2. How do you make the plane climb or descend?

(A lot of people think that the stabilator controls climb or descent, and engine power controls speed, but as we shall see, the reverse

turns out to be true!)

We will develop formulas that answer all these questions, by analyzing the forces that act on the aircraft during steady flight.

To begin, we need to review lift and drag forces acting on wings. Recall that a wing is subjected to lift and drag forces as shown in

the picture below.

The formulas for lift and drag force are


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Here, is the air density, Vis the speed of the air past the wing, and is the wing area.

The lift and drag coefficients and vary with the angle of attack of the airfoil as

where , and are more or less constant for any given airfoil shape, for practical ranges of Reynolds number.

We can now draw a force diagram for the plane, and perform a force and moment balance. We assume that the plane travels at

constant speed Vat a constant climb angle. The drag and lift forces act parallel and perpendicular to the direction of travel.

The forces acting on the plane are shown below


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Heres the force balance table. The basis vectors are parallel and perpendicular to the direction of travel

Force and moment balance table for plane. Origin at CG.Force/Moment Position Force Moment about origin

Weight 0 0

Force on wing

Force on stabilator

Engine thrust

Sum (=0)



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The trig functions in these formulas are a pain. Its convenient to simplify things by noting that for any reasonable operating conditions

and are both small. Thus, approximately,

This simplifies the equilibrium equations to

We can get formulas for the lift and drag forces from the airfoil equations. For the wing, , so that

For the stabilator, the angle of attack is . Therefore

We can now eliminate the lift and drag forces from our equations. Well make one further simplification: if all the angles are small,

then and . With these simplifications, the equilibrium equations reduce to


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Bummer! These equations are still too complicated to see whats going on. Well make one last simplification. In most practical

airplane designs, the wing area is much greater than the stabilator area (well see why this is in a second). Well simplify the first twoequilibrium equations by assuming that . Then, our equations are

(1.4)

(1.5)

(1.6)

Finally, we have something we can work with. Lets see what these formulas tell us.

How to fly an airplane!

The equations tell us how to fly an airplane. Lets see how this works out

1. Airspeed is controlled completely by angle of attack

Specifically, from equation (1.5), we see that

(1.7)

So, if you want to speed up, you need to decrease , by lowering the nose of the airplane. If you want to slow down, raise the

nose. Changing engine power will not change airspeed at all.


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g g g p g p

2. Climb rate is determined (mostly) by engine power setting

We can show this by using the preceding formula (1.7) to eliminate the airspeed from the first equilibrium equation (1.4), and then

solving the equation for climb angle. This shows that

Increasing engine thrustPwill increase the climb angle . Decreasing it will decrease the climb angle. If theres no engine power,

is negative, so the airplane descends (Duh!!).

The angle of attack does have an influence on climb rate, but its influence is rather counter-

intuitive. The function

is sketched in the figure on the right. It has a minimum value at

Usually, one flies with an angle of attack to the left of the minimum. Under these conditions,

increasing will reduce and so cause the airplane to climb, as youd expect.

But during slow flight the angle of attack is large, and you can end up on the `wrong side ofthe minimum then, pitching the airplane up will actually increasethe descent rate (and will also slow you down). This can lead to

serious problems. During landing, if you get caught in an unexpected down-draft, the natural reflex is to haul back on the control

column to try to arrest the descent. If youre on the wrong side of the critical angle, this will increase descent rate causing you to

raise the nose even more until you stall. Big bummer. The correct response is to add power.

3. Best rate of climb speed, and best glide speeds

We learn a few more interesting things about aircraft behavior from the last couple of equations. During takeoff, you want to


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g g p q g , y

maximize your climb rate, so you apply max power to the engine, and set the pitch attitude to

How do you know when you get the correct pitch attitude? Manufacturers publish a `best rate of climb airspeed for every airplane.

Weve seen that airspeed is completely determined by . So by flying at this published airspeed you know youve set your pitchcorrectly. Advanced aircraft have sensors that measure the angle of attack directly.

Secondly, in the event of an engine failure, you want to minimize your descent rate, so as to give you more options in selecting a site

for a forced landing (or at least to give your passengers more time to get their affairs in order). Again, descent rate is minimized by

flying at the optimal pitch attitude

You find this pitch attitude by flying at the published `best glide speed for the airplane. This airspeed is usually close to the best rate

of climb speed (our simple theory predicts that it should be identical, but of course we made lots of approximations).

4. The role of the stabilator

So far, we dont know how to actually change the angle of attack . The third of our equilibrium equations shows the pilot how to

use the controls to vary .

We can simplify the equation by eliminating aircraft speed Vfrom equation (1.6), using equation (1.7). If we rearrange the result and

again make the approximation that , we find that

This shows that the angle of attack is directly proportional to the stabilator angle . The pilot can therefore set angle of attack by

rotating the stabilator.


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Notice also that during flight (otherwise the wing generates negative lift not a good idea!). This means that the stabilator angle

is negative, and moreover

This means that for the stabilator, so that the lift force on the stabilator actually acts downwards. In retrospect, this is

obvious: the aircrafts weight exerts a clockwise moment about the center of lift of the wing, so the stabilator must exert a counter-

clockwise moment. You might think you could improve airplane design by moving the CG behind the wings center of lift: in this case

the tail would contribute to lift. But dont try this at home, kids, its been found (the hard way) that moving the CG in this way

makes the aircraft dangerous (it may not recover from a stall). A better fix is to put the stabilator aheadof the main wing. A few

aircraft designs actually do this (including the Wright brothers plane) the small wing ahead of the main wings is called a `canard.

Design an aircraft

Finally, we can use our equations to help design an aircraft. Well try to design a small general aviation single engine aircraft, meeting

the following specifications

1. Cruise speed 120 knots

2. Max rate of climb 1500 ft/min

Parameters that we dont have much control over are

1. Aircraft weight W(determined by structural considerations). Well assume 2500lb

2. Air density . Varies with temperature, air pressure and humidity theres a neat Java calculator that will compute air density

for any given temperature at http://rshelq.home.sprynet.com/density_altitude.htm. For a standard atmosphere, .

3. Airfoil parameters , and . Typical numbers for these were given in Section 2. We will use , , and

(the latter number was obtained by assuming that the aspect ratio of the wing is of order 0.1).

Our design variables are:
http://pdfcrowd.com/http://pdfcrowd.com/redirect/?url=http%3a%2f%2fwww.engin.brown.edu%2fcourses%2fen3%2fnotes%2fStatics%2fEquilibrium%2fEquilibrium.htm&id=ma-130826000602-acd65c71http://pdfcrowd.com/customize/http://pdfcrowd.com/html-to-pdf-api/?ref=pdfhttp://rshelq.home.sprynet.com/density_altitude.htm


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g

1. Wing area

2. The distance of wing center of lift behind the center of gravity of the aircraft


4. The distance of stabilator center of lift behind the center of gravity of the aircraft5. The maximum engine thrust

6. The distance from the CG to the propeller

We have several design constraints. Most importantly, we need to make sure that we keep and small (less than 10 degrees)

under normal operating conditions, partly to make sure our calculations are correct, but mostly to ensure that the airplane remains

well away from stall.

We can estimate the wing area using equation (1.5), which gives

Lets design to cruise at a 2 degree AOA. Then, plugging in numbers for all the variables and converting to SI units, we get

The position of the aircrafts wings lift center behind the CG , the distance to the stabilator, and the stabilator area are

determined by a number of factors that we cant address completely with our simple calculations. The position of the CG will move

as the plane is loaded we need to ensure that the CG remains ahead of the lift center to ensure that the aircraft remains stable. We

need to keep the stabilator area as small as we can, and we need to make sure that the stabilators angle of attack is always less than

that of the wing, to make sure the wing stalls before the stabilator (pilots are trained to deal with wing stalls, but not stabilator stalls)

The stabilators AOA can be calculated as

As a design constraint, well choose to make . This requires


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Substituting numbers for , and rearranging this equation shows that

We now have to trade off this constraint against the requirement that the stabilator area should be as small as we can manage.

Clearly, to make the stabilator area small, we need to make as small as we can, and make as large as we can manage. But if

is too small theres a risk the plane will be unstable; while if is too large we will end up with a ridiculous looking airplane and

probably make the structural engineers rip their hair out.

Choosing

is a possible compromise.

Finally we need to select the engine thrust to give us the target climb rate of 1500 ft/min. The climb rate is related to the climb angle

and the aircrafts speed V through the formula

To get the best climb rate, the aircraft must fly at the optimal angle of attack

so that the aircraft speed during climb will be

(The predicted best rate of climb speed is typical of small aircraft of this type, so our calculations seem to be working out nicely.


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Either that, or were extremely lucky!)

Therefore

Finally, we can calculate the required engine thrust from the formula

To summarize, our design variables are

1. Wing area

2. The distance of wing center of lift behind the center of gravity of the aircraft


4. The distance of stabilator center of lift behind the center of gravity of the aircraft

5. The maximum engine thrust6. The distance from the CG to the propeller We forgot this one Theres nothing in our calculations that tell us what

should be. It needs to be big enough to fit the engine in. Now that we know the engine thrust, we could calculate the size

of engine required but the courses that teach that stuff will cost you another $60000 in tuition and fees