Skip to main content
Loading Explore guides…
Truss Problem Analysis · Acedra
Home
Explore
Engineering
Truss Problem Analysis
Sign in to save
Chat
If YouTube asks you to sign in, open the video on YouTube.
Watch on YouTube
Reload player
Transcript
54 segments
Copy
Hide video
0:03
PROFESSOR: Good morning, guys. I'm sorry. Hello guys. I guess that's what my students want me to say. So hello, guys. How are you doing? I'm going to solve a problem with trusses today. It's a really nice problem. I'm going to show it here to you. I'm going to change to this view, so you can see it. I like drawing the problem with you, so you can do it with me at the same time and you start practicing as you draw it. So I have these diagonals here. I have this beam support, this beam support.
0:40
And this distance is 20 meters. This two are 15 and 15 meters. And then it has some forces applied, of course, at the joints. Remember, this is a truss. And because it's a truss, the forces should be applied to the joints. And all the joints should be pin. Otherwise, it's not a truss. Or it might look like a truss, but it doesn't behave like one.
1:12
OK, that's the problem. Originally, what the problem is asking you is calculate the internal force in this bar like that. I put it in red for you to identify. But I'm going to solve the whole problem from all the joints and all the bars, so you can see how the load transfer happens. The first thing that you do in these type of problems, assuming that you don't have labels, you have to put labels in the joints.
1:43
So you put here A, B, C, D, E, F, A, B, C, D, E F, G, H, I, K, L, M, N, O, P. So you start here. And of course, the bar that we are looking for is the bar CD, forces at the bar CD. Now, when you see this problem, your first instinct is to start with the reactions. And this is a pin, and this is a pin. So you're going to have B y, and then you're going to have Bx. And then you're going to have Ay and Ax.
2:24
And then you say, oh my god, look at that. 1 2 3 4-- 4 unknowns, 4 unknowns, 3 equations.
2:43
What does that mean? That means that the system is statically determinate. And we couldn't solve it in statics, but you know we can. We can because of a particular condition here, which is that we don't have this bar in this location. Now, let me show you that. If you say that you want to calculate this, then you're going to start from this joint and move it to this joint if you want to, right? No, you cannot move to this joint because you will have 1, 2, 3 unknowns.
3:15
So you start with this joint. And what do you do? You don't do anything. Or maybe you start at this joint. If you start at this joint, then you have 1, 2, 3, 4, unknowns. So you cannot start at this joint. Well, actually, yes, you can. You could start at this joint. I'm not telling you to do it like that, but this is what is going to happen to you. You're going to get stuck if you start like that. Or you can say summation of moments. OK, let's do summation of moments. Let's start from there.
3:43
Summation of the moments at A equal 0. And I'm going to do that. This is A. I'm going to do moment here. I'm going to start from this point. If I apply this force in this location, this is what is going to happen. And according to this convention, that is going to be negative. So you will have 200 times 15 and it's going to be negative. And then this one force is horizontal also. In this direction, this [INAUDIBLE] should be vertical from this point to that point, 100 times 30.
4:14
And then you go to this side, same situation. Horizontal force, vertical distance, you do that. 100 times 30 and 100-- and 200, I mean, times 15. And this one doesn't produce anything because the line of action passes through that point. The x doesn't produce anything for the same reason. And then you have the y that I assume going in this direction. So if the y goes in that direction, it would be plus because the rotation would happen like that when I apply the force like that, right-hand rule.
4:49
B y times vertical force horizontal distance, 20, equals 0. When you solve for that, you get that B y is equal to-- calculator again. 200 times 15, we're going to put it everything positive. Like if I pass it to the other side, plus 100 times 30 plus 100 times 30 plus 200 times 15 equals divided by 20 equals 600. That's the value for the reaction in y that I got, 600 newtons.
5:30
And then I can say summation of forces in y equals 0. And then I say that the only thing that I have in y is Ay and B y, so Ay plus B y. And B y is 600 newtons equals 0. So Ay is going to be negative 600 newtons, which actually means that I mess up when I assumed this. And Ay goes in that direction.
5:55
So I'm going to say that. I'm going to say that Ay is coming in this direction. This is the joint A. And this is 600 newtons here. Now, after this point, what should we do? I don't know. Summation of forces in x equals 0. Oh perfect. Yes, summation of forces in x equals 0. Ax minus Ax minus Bx plus all of these forces equals 0. I can't calculate the Ax and Bx unless I realize this joint. When it comes to this joint, you can see basically that this is a joint B. And then I'm
6:30
going to have this force here of 600 that we already calculated. And then I have this force of 100 newtons here. That means that the reaction at Bx should go in this direction, Bx. And this is the other force that I have, which is the force BD.
6:51
Now, I could do it because I can say summation of forces in x equals 0. And I can solve Bx plus 100 equals 0 and Bx equals 100 newtons, of course. And summation of forces in y equals 0, and then I get that BD equals 600 newtons. And I like to put immediately here the direction. Remember, whenever the force enters the joint, it's in compression. So I'm going to put a compression there. But you see all the work that we just did here for calculating those reactions.
7:31
Once I have Bx, I know Ax because I can say that summation of forces in x equals 0. And then I have 100 plus 200 plus 100 plus 100 plus 200 plus 100 minus Bx equals 0-- no, I'm sorry-- minus Bx minus Ax.
7:57
Because it's here, so I can solve for Ax. This is 400 here, 200, 300, 400, 500, 700, 800 minus 600. That should be equal to 200 divided for that reaction, 200. This is not 600. This is 100 here, 100. I don't know. I got confused with this BD. No, it's Bx. And that happens to the student, also. So that means that, of course, this is live. I'm not editing anything. So Ax is going to be equal to 700. And 700 is going to be acting in that way. OK perfect. Now, it's good. 100 plus 700 equals--
8:44
but look all the work that I already did here and all the complications in order to do that. Instead of doing that, I could have started from the top. You just had to be a little bit smart, a little smart, and think and see what is happening. If you realize and you go to the joint E, which is the joint, you will see that you have directly only two unknown bars. And that's what I can do. I can use those two unknown bars. And I can use particle equilibrium
9:19
or use the method of the joints to do that. So if I go to this joint instead of starting from the bottom because you are used to start from the bottom-- no, from here. Now, you have that force of 100. And then you have this force EF. And then you have these other force EC. The direction that you place these forces, it doesn't mean anything. You can put it in any way. But if you do a little bit of analysis, mental analysis, before putting the direction, it's
9:50
going to help you out throughout the problem. The first thing that you should realize is, here, you have three elements because the force is considered like an element. Remember when I explained the 0-force member rule, I say two forces, three forces, or three bars or any combination of forces plus bars, you have three, which is one of the rules. And two of them are collinear. So that means that the one that is not collinear is 0. That means that this bar is 0.
10:18
If you didn't see that, it doesn't matter. You could do summation of forces in y equals 0, and then you will see it there. But now, if you see this force going in this direction, definitely EF has to go in the other direction for canceling each other. So then I do summation of forces in x equals 0 if you want to. And then you have 100, which is this one, minus EF equals 0 or EF equals 100 newton. And this is in compression because it's entering at the joint.
10:50
One of the things that I like to do when I do the truss, if it's a big truss like this one, I like to do a schematic of that truss with all the joints here, something like this.
11:09
And then I put all the joints with the forces, external forces that I have applied to those joints, 100, 200 100. I know it's newtons.
11:27
And I put a dotted line whatever the bars go because this is going to help me throughout the whole problem. I know this bar is 0. I already calculated that bar. And this is a joint EF, CD, and AB. This is what we have. And we have the reactions here. I don't know the direction. I'm not going to go through the reaction right now. I'm just going to start from top to bottom. So I already calculated this force EF, which is 100. Now, why do I do that?
12:11
Because now I have to go and analyze the joint F. The joint F you have two forces, FC and FD. So when I move to that joint, the joint-- I'm going to use this, continue here. When I move to the joint F, which is this joint here, I have this force, this force, and this force, 1, 2, 3, plus this force. This is EF. This is FC. And this is FD. Now, if you look at this drawing here, you can see that, if the force EF is entering at this joint, it's going to be entering also here.
12:57
That means that my force EF is going to be like that. Now, if you look at this drawing and you do a quick, quick analysis, you can see immediately that these two forces go to the right. And the only thing that can oppose to those two forces is this. So the component, the horizontal component of this force, should go to the left. And that only happens when this force points in this direction. And now, if you see that this is pointing down, the only thing that can cancel is this one pointing up.
13:31
You can do it in any way. You can put all of them in tension. And at the end, if you get a negative sign, you realize that it's in compression. And you just switch it. That's good. That's fine. But in this way, you lose less time, let's say. This is 15. This is 15. And this distance is 20. Now, you can calculate this angle. If you want to calculate this angle, you can, or this angle, you can. But you can also realize what you have there. You have a triangle form here, here, and here.
14:06
And this distance is 20, this distance is 15, and this is a diagonal here. If I use this triangle, 20 or 15, I can use also divide these by 5. And then I have 4, 3, 5. This is a 3, 4, 5 triangle. Or you can calculate the angle, whatever way. I like working with the triangles. But if you want to calculate that angle alpha there, it's fine. You can say that alpha is the inverse tangent of 15 divided by 20. And that is going to give you-- 15 divided by 20 inverse tangent-- 36.87 degrees.
14:53
I'm going to work, though, with the triangle. So once you have this joint, then you're starting to do what? You have two possibilities only, summation of forces in x or summation of forces in y. By the way, I put my reactions here and my forces here in this drawing so I don't get lost later, assuming these are going to be correct. But I know these are correct. So what would you do first? If you do summation of forces in y, you still have two unknowns because it's
15:25
the vertical component of FC plus FD. So we're going to do a summation of forces in x. If I do summation of forces in x here, then I'm going to have EF. But EF I know how much is that because it's 100. So I have 100 plus 100, 100 plus 100, minus-- either I say FC cosine alpha, or I say FC. If this is alpha, the angle that I'm working is for 4/5 equals 0. And from here, you can solve for FC. And FC is going to be equal to 200 4/5 is 0.8 divided by 0.8. That is 250 newtons.
16:23
And remember, once you calculate that 250 newton and you see that the force is like that, please put a little T because it's leaving the joint. That means it's in tension. And that's going to help you a lot of troubles later when you do the summary for your problem. Now, we have FC. What do we have to do now? Summation of forces in y equals 0. So what do I have in y? I have negative FC.
16:51
Because it's this component that I'm talking about, either I use sine of FC, or I use 3/5, which is exactly that, sine of another C, sine of alpha. 3/5 plus FD equals 0. And from here I can solve for FD. Remember, FC is 250. So basically, what we have is FD equal 3/5 multiplied by 250 times-- 3/5 is-- 0.6. that's 150 newton. What direction? FD is entering at the joint. So that means that it's in compression. Now, I come back to my drawing here. And I put those values that I calculated, 250.
17:40
And this is FC. And this is 150, and it's FD. Now, we could move to whatever joint we want to because here I have two unknowns. No, here, I have 1, 2, 3 unknowns. So I'm not able to move to the joint B still. So I'm going to move to the joint C. When you move to the joint C-- and then I do it right here-- you put all the forces that you have in that joint. What are all the forces? Well, you have it here. You don't have anything here because it has zero force member.
18:17
So I'm not going to put it there, but you have this 200 force here. And then you have FC. Now, notice what is the advantage of using this. You see these FC there. I'm going to put it like that now, because if he's leaving the joint, he's going to be leaving the joint also. And this is 250. So I'm going to put that 250 in that direction here. And then I have these other two forces, which I had no idea still or yet in what direction they go. So I'm going to put them in the drawing like this.
18:53
And I'm going to do a little bit of analysis again. It's not that much. It's not that hard. I say, oh, look at these two forces. These two forces go to the right. The only thing that can oppose these two forces going to the right is if this force goes to the left. And this force is called CD. Now, do the analysis for this one again. This force is going up. The only possibility is if I have these force coming down. In that way, I don't have to change any sign. So I go back here, and I change.
19:23
I put my signs. I say this is going these way. And this is going in that way. So now, I do the analysis. Remember we have this triangle, but this triangle actually refers to this triangle here. So the angle alpha is this, or the angle alpha is this, whatever you want to say. Now, you have this one. This is called CA.
19:52
Now, if I want to calculate these forces, what way we're going to start? Actually, it doesn't make any difference. It's a little bit easier working with this, but just that a little bit easier. So I'm going to do a summation of forces in y equals 0.
20:13
Then I'm going to have 250. Remember, this is the angle alpha. I'm looking for y direction, so I'm looking for this component. This is going to be a sine of alpha, or it's going to be 3/5. 3/5 positive minus CA equals 0, which tells you that CA is going to be equal to 250 multiplied by 0.6 is 150 newton. And CA is leaving the joint, meaning it's going to be in tension-- again tension.
20:53
If you want to put it here, you can put it here, 150. And now, we do summation of forces in x equals 0. So you have 200 plus the horizontal component of this one, 250. Remember, this is alpha. If I use this angle, the horizontal component is going to be cosine alpha or is going to be 4/5 minus CD equals 0. And from here, you calculate CD. And CD is equal to 250 multiplied by 0.8. That's 200. And 200 plus 200 is equal 400. These pass to the other side, 400 newtons.
21:47
And I assume that force entering at the joint and this is positive. That means that that is the correct direction. And because it's entering at the joint, it's in cooperation. So now, we are done with that. OK, so we are here. We go back to our drawing. And I put the correct forces here. This CD is 400. And this CA is 150. Now, if that was the original question, we already calculated that bar. The bar CD was 400 in compression. But I wanted to continue and finish the problem.
22:32
Now, look at the work that we have to do only for calculating that bar. In the next video, I'm going to show you one solution with one step, only one step to calculate that bar.
22:46
Now, let's keep going. This bar goes here. That means this is 400 also. And this one was entering at this joint. That means that it's going to be entering also here, 150. This force is 200, and I have these two unknowns here.
23:06
Let's do that. The joint D-- 150 200 400. And then I have this bar and this bar. This is the bar DA or AD. And this is the bar DB-- DA, this one, and DB, this one. Mental analysis, again-- look, these two forces go to the right. The only thing opposing or possible opposing is this one. And now, you have these two. Of the forces going down, the only one able to oppose that force is that one. Now, I can use this triangle if I want to, also. Or I can use this triangle.
23:58
Which one were we using before? We were using this triangle. So I'm going to keep using that triangle. Remember, this is alpha. This is 4. This is 3. And this is 5. Now, this triangle is this triangle. What should we do here first, summation of forces in y? No, because I have two unknowns. So I do summation of forces x equals 0. And if I do that, then I have positive 400 plus 200 minus DA-- this is alpha-- multiplied by cosine. I'm looking for the horizontal. Or multiply by 4/5 equals 0.
24:40
And from here, I can solve for DA. And DA is going to be equal 600. 600 divided by 0.8 because 4/5 is 0.8. This is 750 newton. And DA is leaving the joint, meaning is going to be in tension. And then I do summation of forces in y equals 0. And then you get negative 150 plus DB, which is my unknown, minus the vertical component of DA. It's DA. The vertical component is this one, meaning we are using sine of the angle or using 3/5.
25:24
That's it, right? 1, 2, 3, 1, 2, 3 equals 0. And DA, remember, is this one here. So if I do that, then we have negative 150. Pass it to the side positive. This is 750 times 0.6 plus 150. That is 600. So that means that DB is equal to 600. Look, it was in compression, so compression. DB is in compression. We go back to our drawing here. And then we say, OK, we calculated DB, which is going in this direction, and DA, which is going in this direction. And that force DA was 750.
26:09
And this one DB was 600. Good. Remember, this bar is not there in the original problem. I don't know where I drew that there. The original problem was this one here.
26:27
The only thing that we have to calculate is nothing. If we want to calculate the reactions, we can solve for the reactions. Because, now, I can say, OK, no problem. I go to the joint B if I wanted to. I go to the joint B. And I have all my other forces. This force here should be 600 in this way, meaning that this force here is 100. This force here is 600. And at this pin, I'm going to have two reaction because, remember, this is a pin that we have here.
27:00
The two reactions that I'm going to have is B y and Bx. And you can see perfectly that B y is going to be equal to 600 newton, and Bx is going to be equal to 100 newton.
27:16
And we could move to this other joint now, joint A. Either you do summation of forces in the overall beam if you want to or overall truss. Or you can go to the joint A and say, OK, joint A, I have this force here, 100. I have this 150 leaving the joint, so it's going to be leaving the joint here also, 150. And this one, 750 was leaving the joint. So it's going to be leaving this joint also, 750. And the only missing part are my reactions. Where are my reactions?
27:50
Well, you can do the same analysis. If these two point to the right, remember this was a pin here. If this is a pin, I'm going to have a reaction going-- those two go to the right. These reactions should go to the left. And it's Ax. And if this goes up and this goes up, Ay should go down, Ay. And then I do summation of forces in x and in y. And I go on the [? bar. ?] Let's see if I do summation of forces in x equals 0, once again, notice our triangle. Our triangle is this triangle here.
28:29
So this is alpha all the time. Remember, our triangle is this triangle here, alpha 3, 4, 5. No-- 4, 3, 5. If this is our triangle here and I do summation of forces in x, then I'm going to have positive 100 minus Ax plus 750. And this 750 should be multiplied by cosine alpha or multiplied by 4/5, which is Ax is going to be equal to 750 times 0.8 plus 100, 700. Ax equals 700 newton. And if you do summation of forces in y, you're going to get that B y is going to be--
29:28
Ay, I'm sorry, is going to be equal to 600 newtons. You can do that. You can do that, then you can solve it in your own if you want. Or if you have these reactions here, you can say, OK, summation of forces in x in the overall structure external forces only. Then I have 100, 200, 100. Those are 400 and 400 are 800. And I have 700 now from this Bx going in this direction, 100. So that means Ax that should be 700 also in that direction. You can do it like that if you want to.
30:03
And remember, we calculated this B y. We could have calculated from the beginning, as I did before. And once you realize that B y is 600 in this direction, you could do summation of forces in the overall structure external. When it's external, you don't count the bars, only the external forces. And you would realize that this force should be 600 in the opposite direction because there's nothing else to oppose the movement. Look everything that we have to do-- only for calculating that bar.
30:30
Now, watch the next video. It's going to be just 3 minutes or 3 minutes. And you're going to see how easy it is to solve this problem if the only thing that we are asked to calculate is the bar CB. See you next time, guys. Keep watching.
Here to help you learn
Sign in to keep a private conversation with your course tutor.
Ask me anything about the material…
Sign in to chat