Basic Concepts Continued

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Do curso por Georgia Institute of Technology

Fundamentos da Engenharia - Revisão

184 classificações

Georgia Institute of Technology

Fundamentos da Engenharia - Revisão

184 classificações

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.

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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

Conheça os instrutores

Dr. Philip Roberts
Professor
School of Civil and Environmental Engineering

Now I want to continue our discussion of some basic concepts and statics and,

in this segment, we'll look at some basic ideas of vectors and

manipulations of vectors.

So, the reason that vectors are important, of course,

is that force is equal to, is a vector.

And the relevant sections from the manual are shown over on the right-hand side.

But some of the elementary concepts are, let's suppose that we have a vector

here v, which is composed of its rectangular components in the x and

y directions, Vx and Vy as shown.

So we can write this in vector notation as v as a vector is equal to Vx,

where Vx is the x component.

Multiplied by I, where I is the unit vector in the x direction.

Plus Vy, the component in the y direction.

Times j, where j is the unit vector in the y direction.

So Vx and Vy are the rectangular components of V,

where Vx is equal to V cosine theta.

Where theta is the angle between the x axis and

the vector as shown here and Vy is equal to v sine theta where v is equal to

the magnitude of the vector is the square root squared of Vx squared

plus Vy squared, and sometimes we write the magnitude like this.

Within two straight lines but normally I'll omit the two straight lines.

So v without the vector symbol on it stands for

the scale of magnitude of the vector.

The angle theta is equal to the arctangent of Vy over Vx

and the idea of a negative vector.

Is simply, a negative vector is just a vector with the same magnitude,

the same line of action, but simply in the opposite direction.

The generalization of this in three dimensions is shown here where we have x,

y, and z.

Coordinates as defined by the right hand rule for x, y, and z.

In this case, the generalization of the expression for the vector is Vx times i.

Vy times j plus Vc times k.

Where k is the unit vector in the z direction.

And the magnitude of the vector is as shown there,

the square root of the sum of the squares of the individual components.

Another concept which is very useful is the idea of a unit vector.

And the unit vector is a vector of un, length unity or one.

Which lies in the direction of the vector of interest.

So, for

example in this case if n is the unit vector in the direction of the vector v.

We can write the vector v as simply being equal to magnitude V

multiplied by the unit vector n.

And m in turn is given by this expression, ie, kj, ey,

ke, c where ex, ey and

ez are the components of the unit vector in the xy and z directions.

Another thing that is useful, is the idea of direction cosines, and

these are the cosines of the,

angles that the vector V makes with the x, y, and z directions.

And they are shown on the diagram here.

So, for example theta x is, this angle right here

sorry, theta x, is this angle right here, et cetera.

And the significance of these is that they are the scalar components

of the unit vector along that line.

So, in other words, n, is equal to i, cosine theta x, et cetera.

So cosine theta x is just equal to ex,

the component of the unit vector in the x direction, et cetera.

A simple example, we have a two dimensional vector here.

The vector is a magnitude 10 Newtons and is in the direction as shown.

So, in this case, the angles, theta x are, as shown,

is equal to a 120 degrees.

Theta y, the angle to the y axis, is 30 degrees.

And theta z, the angle to the z axis, is 90 degrees.

So, the z axis in this case is just coming out here in this direction.

The direction cosines are by definition ex is cosine theta x,

is cosine a 120 degrees, is minus one-half,

ey cosine theta y is 0.866, and ez cosine theta z is 0.

So we can write the individual components of this force this way,

Fx is equal to F, the magnitude of the force, times ex is therefore

equal to 10 times minus one-half is equal to minus 5 Newtons.

So that component is right here minus 5.

The y component, Fy, is F times ey is this component

is equal to 10 times 0.866 or

8.66 Newtons.

And Fz, we can obviously tell, is going to be 0,

10 times 0 is 0.

And we can also write the vector force in this way.

In general, which is therefor,

equal to minus 5i plus 8.66j, in this example.

We can add vectors by simple adding them in a force polygon, head to tail.

So in this case, we have two vectors, V1 and V2.

And the vector, the sum of those two, is just V1 plus V2,

which we can do in this form here.

But, we can also do this by adding the component in the different directions.

So V1 is equal to iVx1 plus jy1,

where Vx1 and Vy1 are the x and

y components of the vector V1.

Similarly, V2 is iVx2 et cetera, so adding these 2 together,

we have V1 plus V2 is equal to iVx1 plus Vx2, et cetera.

In other words we can add the vectors by just adding their

components in the x and y and z directions separately.

An example, we have two vectors here

vector V1 has magnitude 4 units and V2, 3units.

At angles of 45 and 30 degrees, as shown to the x axis.

And the question is, the magnitude of the vec,

the vector sum of these two vectors is most nearly which of these alternatives.

So, here, this is what the answer's going to be if I add these vectorially.

So the resultant is shown here and it's at some angle alpha to the horizontal.

And we can do this graphically by drawing these to scale and adding them up.

But to do it arithmetically, here's our equation, V the resultant

is V1 plus V2 and if I add the components separately.

The x components of V1 and V2 and the y components of V1 and

V2, I get this, so expanding this out,

I get i times the x component of V1, is this quantity,

is 4 cosine 40 degrees, plus the x component of V2,

is this quantity, which is 3 cosine 30 degrees, et cetera.

And the answer is 5.43i plus 1.33j is the resultant vector right here.

But we're asked for the magnitude.

So the magnitude is the square root of the sum of the coefficients,

or the called, the squares of the components, which is equal to that.

Which is equal to 5.59 units.

So the closest answer, the best answer, is B.

If we want to find the angle, alpha,

then alpha is given by the arctangent of Vy over Vx.

Which is equal to that, which is equal to 13.8 degrees is alpha.

Another example, we have two forces acting through this bracket,

three kilonewtons at an angle of and this is a three, four, five triangle here.

And another force, two kilonewtons here at an angle of 30 degrees.

And the first question is the magnitude of the resultant force is most nearly,

which of these?

So in this case the resultant force which I'll call R is just

F1 plus F2 and I'll do this by again,

by adding the components in the x direction and the y direction.

So the x component of the resulting force is the sum of the x component of

the individual forces, F1x plus F2x.

And that is equal to, so the horizontal component of the 3 kilonewton force,

that is a three, four, five triangle.

Is three 3 times three-fifths, plus 2,

times cosine 30 degrees, is equal to 3.53 kilonewtons.

So that is 3.53 horizontal.

The y component y is the summation of the y components

F1y plus F2y is equal to negative 3, because this

force is in the downward direction, negative direction times four-fifths plus,

because this component is in the positive y direction 2 times sine 30 degrees

is equal to minus 1.40 kilonewtons, which is in this direction.

So, the resultant force here, is right here.

The resultant force, the square root of the sum of the squares of

the coefficients, is equal to 3.8 kiloNewtons.

And the best answer is B.

Continuing with the same example, the next question is the angle of the resultant

force to the horizontal is most nearly which of these alternatives?

So the angle is this angle right here.

I'll call it theta.

So that angle is arctangent of Ry over Rx is

up tangent negative 1.4 over 3.53 is equal to negative or

minus 21.6 degrees so the closest answer is C.

Now, some other ideas about vectors.

We've already shown that if we have a vector f then the components in the x and

y direction are f sine theta and f cosine theta.

So in this case, we have the,

if we want to get the component of a force in any arbitrary direction.

Let's suppose I have two vectors here, F and Q, with an angle theta between them.

Then the component of F in the Q direction is F cosine theta.

But an easier way to get that is by means of the Dot product.

And the Dot product of two vectors is a scalar.

Which is the Orthogonal component of the vector in the direction of another.

In other words, it's the projection in the direction of that second vector.

So, the Dot product in this case of F dot Q, is FQ cosine theta,

which is of course a scalar quantity.

And we note that this is reversible, F dot Q is equal to Q dot F.

And if we want to find the component of the force

in the direction of any other vector, for example, Q we can fo,

form that by taking the dot product of F dot with n,

where n is the unit vector in the direction of the vector Q.

Which is given by Q as a vector divided by the magnitude of Q.

So the generalization of that, of the dot product of two vectors,

F dot Q is FxQx, in other words,

the product of the x components of the two vectors plus FyQy plus FzQz.

And the value of this is that it makes it very easy to get

the component of a force in any other direction by vector methods.

So this completes our discussion of vectors.

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