Moments and Couples

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From the course by Georgia Institute of Technology

Fundamentals of Engineering Exam Review

203 ratings

Georgia Institute of Technology

Fundamentals of Engineering Exam Review

203 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

Statics

This module reviews the principles of statics: Forces and moments on rigid bodies that are in equilibrium. We first discuss Newton’s laws and basic concepts of what is a force, vectors, and the dimensions and units involved. Then we consider systems of forces and how to compute their resultants. We discuss the main characteristics of vectors and how to manipulate them. Then the meaning and computation of moments and couples. We discuss the concept of equilibrium of a rigid body and the categories of equilibrium in two dimensions. We show how to draw a meaningful free body diagram with different types of supports. Then how to analyze pulleys and compute static friction forces and solve problems involving friction. The concept and major characteristics of trusses are discussed, especially simple trusses, and we show how to analyze them by the method of joints and the method of sections. Finally, we analyze the geometrical properties of lines, areas, and volumes that are important in statics and mechanics of materials. These are moments of inertia, centroids, and polar moments of inertia of simple and composite objects. 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 3 hours | Difficulty Level: Medium

Basic Concepts10:47

Basic Concepts Continued13:25

Moments and Couples16:54

Equilibrium22:00

Equilibrium Examples30:10

Trusses Method of Sections12:52

Centroids and Moments of Inertia18:03

Centroids and Moments of Inertia Continued10:11

Meet the Instructors

Dr. Philip Roberts
Professor
School of Civil and Environmental Engineering

In this segment, I want a complete ad discussion of basic concepts of,

of statics by looking at moments and couples.

So, a moment of a force about a point we can think of

as the measure of a tendency of a force to turn or rotate a body.

And a moment can be either about a point or

about a line, but in this segment we'll only talk about moments about points.

So there are three definitions that are commonly used.

The first one is the so-called common sense one, and the second one,

the more general one, is the vector representation, and

finally, the use of Varignon's theorem.

And the appropriate section from the reference manual is shown down,

down at the bottom here.

So, the first one, the common sense,

so-called common sense definition, the way that we normally think about a moment

is simply the product of the force multiplied by the perpendicular

distance between the line of action of the force and the point we're taking about.

So, in this case, the perpendicular distance here, is d, so

the magnitude of the moment about the pipe is F times d.

And the dimensions of moment is a force times a distance.

So, it will either be units such as Newton meters, or pound force feet,

or pound inches for examples, will be typical units of the moment.

So, the common sense definition then,

is just the force times perpendicular distance.

The more general vector representation is shown here.

And I'll suppose that I have some force here and a line of action.

Then the more general definition, the vector definition is that the moment about

that point O is equal to the cross product of the radius vector and the force vector.

And we note that when you compute this, it's independent of what you take for

the radius vector.

So for example, I could take a radius vector here, or

I could take a radius vector here, it doesn't matter.

I will get the same answer.

It's independent of the actual vector so

long as it's in between the point and the line of action of the force.

The direction of the moment,

moment is, is a vector and therefore it has a magnitude and a direction.

And the ma,

the direction is given by the right hand rule, as shown over on the right here.

So, if we curl our fingers in the direction of the rotation of the moment,

then the direction of the moment vector is given by the,

the direction that our thumb is pointing in.

So in this example here the direction of the rotation is around here.

So from right hand rule, the moment is going to be in this direction.

So the moment is a vector which is perpendicular to

the plane passing through the point, and the line of action of the force.

And the direction is given by the right hand rule.

So generally then we have moment is equal to force times distance.

And another way to think of that is that the distance,

d, is equal to this distance, r sine alpha,

where alpha is this angle as shown here.

But that in turn I can write as f sine alpha times r.

But f sine alpha is the component of the force

which is perpendicular to that radius vector, r.

So, I can also think of the common sense definition as being the component

of the force, perpendicular to the radius vector, multiplied by the length of

the radius vector, is another way to think about it which is often useful.

Some other terminology which we'll use, sometimes we talk about moment,

or torque is the same thing.

And also a couple,

which is a special kind of a moment which we'll look at in a minute.

So the vector representation is the cross-product of the radius vector and

the force vector.

But what physically is a cross-product?

How can we think about that simply?

Well, let's suppose I have two vectors, A and

B, with some angle theta in between them.

Then the cross product C,

we know is going to be perpendicular to the plane containing those two.

And because I've taken A first here, going from A to B, it's going to be

in this direction as shown, the positive direction here from the right hand rule.

But physically what this is, if I complete the parallelogram here,

then the magnitude of the vector of the cross-product of A and

B is just AB sine theta.

But AB sine theta is simply the area of this parallelogram here,

is the simple way to think of that.

And from that you can see, of course, that the cross-product

of two parallel vectors is 0.

We note that B cross A is not equal to A cross B, in fact, it's the negative of it.

So in other words, if I go in the other direction, from B to A,

going this way, then my cross-product is going to be the same magnitude, but

in the opposite direction.

So B cross A is negative of A cross B.

The most general representation of the cross-product is by means of

the determinant formed by ijk, and then the xyz components of the radius vector,

and the xyz components of the force vector.

And to evaluate that we just cross out,

firstly to get the i component cross out i and j.

And that is equal to i times

the product of ry Fz minus rz Fy.

Then to to get the k component we cross out the j and

we have, and we alternate the sign so it's negative.

So it's negative rx Fz minus that.

And finally the z component crossed through the k row and column and that.

And this is our general expression for the cross-product.

Now, the last method,

Varignon's theorem, states that the moment of a force about the point is

the sum of the moments of the components of that force about the same point.

So, in other words, if I have a situation like this, and I want to get the moment of

this force about the point O, and let's suppose I have some resultant force here,

which can be, expressed in terms of its comp, components P and Q.

Then the moment of that resultant r about that point is,

by definition, the cross-product of r with the resultant r.

But R is equal to P plus Q.

Therefore the moment about O is r cross the sum of those two, P plus Q.

Expanding that out we get r cross P plus r cross Q.

In other words, it's just equal to the sum of the moments of this force and

this force, the components of the force.

And that is Varignon's theorem, which will often simplify problems.

So let's illustrate this by means of some examples.

And the first one we want to compute the moment of the force

about this point as shown.

So the moment of the point O is most nearly which of these alternatives?

So, I'll compute this by all three methods to show how we do it.

Firstly, the common sense method is shown here.

So, in that case,

the moment of the force is just force multiplied by the perpendicular distance,

where the perpendicular distance is this distance here.

And to get that distance we use some simple geometry.

So this distance is just equal to this distance which is d 4 cosine 40 degrees,

plus this distance which is 2 sin 40 degrees,

which is equal 4.35 meters.

And therefore the moment is 600,

the magnitude of the force times the moment arm, 4.35 is equal to

2610 Newton meters, so the answer is D.

And to express that in terms of vectors, if I put a coordinate system on here,

x, y, and z, is a right handed coordinate system,

and the moment is rotating about the z axis in this direction.

So in this case, from the right hand rule, we can see that the direction of

the vector in this case is actually going to be in the negative z direction.

In other words, the negative k direction.

So, the proper representation of the, of the moment as a vector

is equal to negative 2610 times k Newton meters.

Now, secondly, we'll do this by Varignon's theorem where we've now

decomposed the force into its horizontal component, Fx or

F1, which is 600 cosine 40 degrees, and

its y component F2, which is 600 sine 40 degrees in the negative y direction.

So now, the moment of our force is equal to the moment of this component,

which is 600 cosine 40 degrees multiplied by its moment arm which is this distance,

which is 4 meters, plus the moment of the y force here,

this force, multiplied by its moment arm which is this distance, which is 2 meters.

So that is equal to that quantity, which is, again,

equal to 2610 Newton meters, which of course is the same answer as it must be.

Now, finally, I'll do this problem again, but now by vector methods, so

in this case, our vector diagram looks like this.

Here is our resultant force, which is passing along this line of action here, A,

B, C, passing through this point here.

And as usual, I can represent that by its horizontal and

vertical components, F1 and F2.

The radius vector I will take

as being the vector from O to A, although again, it wouldn't make any difference,

I could have taken this vector, or this vector, whatever is, is convenient.

But in this case I'll take it as here.

And here is our general expression for the moment, the cross-product of r and F.

So the radius vector is 2 units horizontal and

4 units vertical, in other words, 2i plus 4j.

The force vector in terms of its horizontal and

vertical components from the previous slide is 600 times

i cosine 40 degrees is the horizontal component.

And 600 times minus j because it's in the negative y direction, sine 40 degrees.

So, the cross-product of those two looks like this.

And again, expanding this out in all it's full glory, is equal to that.

But we realize right away that when we expand this out,

we're only going to have a component of the vector in the z direction.

In other words, out of the picture.

So the first two terms here, the i and

j terms, are going to drop out, which we can see intuitively,

but which will also follow when you actually plug in the numbers.

So all we'll have left is the k component, the z component

which is rx Fy minus ry r.

Fx is equal to k, and here's the arithmetic so the components are as shown.

And evaluating that, we end up with negative 2610k Newton meters.

Again, the same answer as it must be.

But in this case, it's given us the, the vector solution directly.

So again, the answer is 2610 Newton meters is the magnitude.

Now another special case of a moment is a couple.

And a couple is defined as the moment of two equal in magnitude, but

opposite direction, non-collinear forces.

In other words, they're parallel to each other but

they're not along the same line, they have some distance between them, as shown here.

So, we have two forces our F and minus F, equal magnitude,

parallel to each other but different directions, separated by a distance d.

And the moment of this couple is equal to one of the forces

multiplied by the perpendicular distance between them.

Or more generally, it's the cross-product of any radius vector between them,

where any radius vector here is here.

And again, it doesn't matter which radius vector I take when I compute that couple.

And the direction again, is given by the right hand rule.

So for this example, the direction of the moment or

the couple would be perpendicular to the plane containing the forces,

and in the direction shown by the right hand rule.

The mag, the vector here is a free vector.

In other words, it doesn't matter, when you're drawing a free body diagram,

it doesn't matter where you draw this about, because it doesn't matter where

you take the moment about, you'll get the same answer.

To illustrate that we have an example with a lug wrench here.

So in this example we have a lug wrench with two arms, of each 14 inches.

And we're applying an upward force of 50 pounds on the leng, left hand arm, and

a downward force of 50 pounds on the right hand arm.

So this is a couple.

There's no net force in any direction.

It satisfies all of our definitions.

And the moment which is exerted is most nearly which of these alternatives?

So the moment is equal to, or

the couple, is equal to the force times the distance between them.

So the force is 50 pounds, the distance between them is 28 inches, so

the answer is 1400 pound inches, which is A.

And if we want to express that as a vector,

if we put in our coordinate system on here again x, y, and

z, we see that, applying our right hand rule, that the direction of the couple

is going to be in the negative z direction this way.

In other words, expressing that moment or couple as a vector,

it's equal to minus 1400 times k pound inches.

Just to demonstrate that the magnitude of the couple

doesn't depend on the location where you take it about,

let's compute it about this point here, the center of the lug wrench.

So in that case, the couple is equal to the moment of this force about that point

plus the moment of this force about that point.

In other words, that's equal to 50 times its moment arm, 14 inches,

plus 50 times 14, again, is equal to 1400 pound inches.

In other words, it doesn't matter where you take the the point about when

you're computing the moment due to a couple.

You get the same answer at any point.

And this concludes our discussion of moments and

couples, and our discussion of basic concepts of statics.

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