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.
Trusses
Loading...
來自 Georgia Institute of Technology 的課程
工程考试复习基础
184 個評分
從本節課中
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
與講師見面
Dr. Philip RobertsProfessor
School of Civil and Environmental Engineering
In this section of our discussion of statics, I want to talk about trusses.
And these are the major topics that we'll be looking at in trusses.
And in particular, in this segment, we'll first look at what a truss is,
in particular a simple truss and how we define it.
And then how we analyze it by using the method of joints.
So firstly, a truss or
a simple truss consists of rigid members, which are pin jointed.
And, this is a simple diagram of a, of a truss which can come in various shapes.
And trusses, generally speaking, are ridged structures and
are consisting only of two force members.
So the top example here, right here.
This is a simple truss.
It's rigid.
If I apply a reasonable load to any of the nodes there,
A, B, or C, it won't deform in any major way.
The second one however, is not a simple rigid truss because in this case if I
apply a force, say at B the structure will deform, so it's not a simple truss.
Generally simple trusses can be constructed by starting off with
a triangle such as the top on there, A,
B, C, adding a node and adding two new members.
So for example, in this case I start off with A, B, C,
I add another node at D and two members, BC, BD and CD.
Then that is also a simple truss.
Then I can add another node at AF and
two new members, and that is still a simple truss.
And then, finally another node at E and two members CE and DE, etc.
And in this way I can build up to arbitrarily complicated truss systems.
Although, they are still what we call simple trusses.
Now, each member in a truss is a two-force member, which means that it is under the,
under the action of only two-forces, which are applied at its ends or the nodes.
And those forces are collinear.
In other words, they are linear with the member and
apply along the axis of the member, as shown here.
So for example, two-force members are shown here.
They can be either under tension in which the four case,
the forces are trying to elongate them, or under compression,
in which case the forces are trying to compress them or shorten them.
And this in turn gives rise to a sign convention that
we will denote a member which is under tension as being positive.
And a member which is under compression as being negative.
And these are other examples of trusses, more complicated, but
again, these, these all fall under the general heading of simple trusses.
We can have two dimensional trusses in which case all of the forces in
the moments are in a single plane along with the the planes of the members.
And this we call a plane truss.
Or we can have three dimensional trusses which are also called space trusses.
But in this case we will only be looking at two dimensional plane trusses.
There are two major methods of analysis that we'll look at.
The first one is the method of joints and the second one, the method of sections.
So let's look first at the method of joints.
So, in this method, if we have a truss we can imagine
that all of the members are detached from their pins or
their node points, and because the structure is a whole in equi,
is in equilibrium each member must also be in equilibrium.
Regardless of the actual force which is acting on it.
Therefore, the equilibrium of the entire truss
is reduced to simple equilibrium of the node points or the pin connections.
So, let's take a very simple truss here, A, F, B, a triangle.
And the approach here is that we will isolate the joints and
apply the equilibrium equations at each joint, or each node point.
And the equations that we have are summation of the forces in,
for example, the horizontal direction, the x direction.
And summation of the forces in the vertical,
say, the y direction, are equal to 0.
We have two equations.
We don't have a moment equation because there are no moments about each node,
because all of the forces pass through the node point, therefore,
they exert no moment about it.
So we just have two equilibrium equations at each node point.
So, if I isolate the node it would look something like this.
For example, the node A here, and if I look at the,
the pin that's connecting those two points and
look at the forces on it, for example, this member here AB.
If I as, I assume its tension,
then the forces acting on that are acting out towards that trying to elongate it.
Conversely, the force direction on the pin is away from the pin,
if the member is in tension.
Similarly, if the known member is in compression,
then the forces are pushing inwards.
And the direction of the force on the node point is pushing toward the node point.
And also, possibly, we have reaction forces.
For example, an upward reaction, R1, at the node A.
So, to analyze this, generally speaking,
we'll start at a joint with the fewest arms or the fewest members.
And then we'll proceed one pin at a time, drawing the free body diagram of each pin.
And proceed and progress our way through the trust until we've solved for
all the members of interest.
And there will always be at least pin,
one pin with only two unknowns that we can therefore solve.
And then we can progress.
And this will always be true for
simple plane trusses of the types that we're analyzing here.
Now, generally if you don't know the force directions,
you can just make some guess at the direction.
And at the end if it turns out, the force turns out to be negative,
that just means that your guess of the direction was, was incorrect.
I recommend however,
that you assume at the beginning that all of the members are in tension.
In other words, when you draw a free body diagram of a node point like this,
your unknown forces are all pointing away from the node point.
Here's an easy way to be consistent.
So let's shell that by means of an example.
So here we have a statically determinant simple truss and
we can see it's statically determinant because we have three unknowns.
A, we have a pin joint, so
generally speaking we have a horizontal and vertical component of reaction, but
D is a roller joint, so we only have a vertical component.
So therefore we have three unknowns here, but we have three equilibrium equations
for the truss as a whole, therefore it's statically determinant.
We can solve it by statics alone.
So generally the procedure would be to first determine the external
loads by applying a free body diagram to the entire truss and then proceed joint by
joint through the structure until you've determined all your members in interest.
So, in this case then the free body diagram for
the entire truss looks like this, and generally of course we would expect
a horizontal component of reaction forces A, at A.
However, there aren't any other external forces acting, so we see that,
that force is going to be zero.
So we could then go ahead and
determine the reaction forces R1 and R2 by normal statics.
Now, if we apply the method of joints,
let's suppose we start first at the node or node or joint A.
Our free body diagram looks like this.
The upward reaction force is R1, and the forces in the members AF and
AB, I denote by AF and AB, and
the angle of the member AF, I'll assume is equal to theta.
And I've drawn these arrows pointing away from the node point.
In other words,
I'm assuming that the forces in each of those two members are intention.
So, the equilibrium equations,
first I'll apply summation of the forces in the horizontal or the x direction.
Assuming that the rightward direction is positive.
So, here I have AB plus,
because I'm assuming it's to the right, AF cosine theta is equal to 0.
Next, I have my vertical equilibrium equation.
Sum Fy is equal to 0.
And here we have R1, and the vertical component of AF,
in other words, R1 plus AF sine theta is equal to 0.
So here you see we have two equations in two unknowns, AF and AB.
And we can solve those two equations for those two unknowns.
Next, we turn our attention to node F, this one right here.
And here is our free body diagram for
the node F with the three forces, which again, I'm assuming are in tension.
The arrows are pointing away from the node point.
And a similar procedure.
First, I sum the forces in the horizontal, the x direction, so
I have FE minus, because it's pointing to the left,
I have cosine theta is equal to 0, and in the vertical direction,
sum Fy, we have FB and the vertical component of AF.
So we have minus AF sine theta minus FB is equal to 0.
However, we've already solved for the force in the member AF.
We already know this.
So again, we have two equations, and we can solve for the two unknowns.
So we just progress in this way progressing to the next member,
the next node, until we've solved for all of them.
And that's the basic method of joints.
So, let's do an example on that.
I have a simple truss, which is loaded as, as shown, so
we have a pin joint here at A, and a roller at C, and a vertical weight
of mass 75 kilograms hanging, and the triangle here is an equilateral triangle.
Each length of each side is two meters.
So, the first question is what is the force in the member AB?
In other words, this member right here.
It's most nearly which of these alternatives?
And is it in compression or tension?
So for analysis we'll start at the node B and
draw a free body diagram and this is what the free body diagram looks like.
The tension in the cable here is W, which is equal to the mass of the weight
times acceleration due to gravity and the force is in AB and
CB are as shown where again, I'm assuming that they are in tension.
So, applying that equilibrium conditions, we sum the force in the horizontal first.
Sum Fx is equal to 0, and here all we have are the horizontal forces,
horizontal components of the forces in AB and CB.
In other words, minus AB cosine theta minus CB cosine theta
is equal to 0, where I'm assuming that the positive direction is to the right.
So from this then, it follows that CB is equal to minus AB.
In other words, the tension for the, the force in those two members is the same,
but one of them is in compression and the other one is in tension.
Next, I send the forces in the vertical direction sum Fy is equal to 0.
And here we have the weight force and
the vertical component of the forces in AB and CB.
Which gives me this equation, and substituting for
CB from this equation back into here, we see that AB is equal to W over
2 sine 30 degrees, the weight is equal to mass times acceleration due to gravity.
And substituting it in the numbers, we find that the answer is 736 Newtons.
And because this turned out to be a positive number from our sine convention,
we know that, that member is in tension.
So the correct answer is B.
The next part of this,
the same diagram, but now we want to compute the, the force in the member AC.
In other words, this one right here, and these are the possible answers.
So, the same procedure.
Now, I draw a free body diagram for the node C here, and this is a roller joint,
so the only component of the reaction force there is a horizontal force which
I'll call RC and the forces in the two members I denote by AC and AB.
And the angle of those sloping rods,
because this is an equilateral triangle is 30 degrees.
So, again we apply our equilibrium conditions and in this case
the simplest way is, is just to sum the forces in the vertical direction.
Sum Fy is equal to 0, and here then all we have is AC and
the vertical component of CB.
In other words, AC plus CB sine 30 degrees is equal to 0,
from which we see that AC is equal to minus CB sine 30 degrees,
but we already know from here that CB is equal to minus AB,
therefore, substituting in, we see that AC is equal
to minus minus 736 sine 30 degrees,
which is equal to positive 368 Newtons.
And again, because this is a positive number,
the force in that member must be in tension.
So, the correct answer is B.
So, out of these members here.
We can see that these two members are in tension, and
this one is going to be in compression.
But, I recommend that you don't try to second guess or
guess ahead of time, whether they're in tension or compression.
Just assume that they're all in tension and let the signs work themselves out.
So this completes our discussion of the method of joints applied to trusses.
