Trusses: 4. Truss Method Of Joints Math

Now to finish this sheet, we've done all the easy members that did not need a lot of math. To finish this sheet, we're going to need a little bit of math. And to calculate all the diagonal members and the members, top chord, bottom chord, that we haven't calculated thus far because they were not priming into simple joints, We have to first review, this is the problem with structures. Everything is interrelated. Topics are not independent. So we have to go back and take a look at force resolution and remember what that means. If I have a slope of 1 to 1 versus a slope of run 2, rise 1 versus the rise is 2, the run is 1, versus a rise of 3 versus a run of 12, That should help me resolve a force, and I need this for solving the rest of the members of a truss. So in this first example, if the vertical component is 4 kips and the slope is 1 to 1, well then the horizontal is also head to tail and 4 kips. Now if the run is 2, the rise is 1, and I have already from another joint or something, If the horizontal is 4, that corresponds to the run. Well, then the rise is half as much because 1 is half of 2. Then this component must be half of 4. That's 2. And if the vertical is given as 4, then that corresponds to the 2 rise. Then the run is half of 4. So that would be 2. and if this were 4 kips, and the slope is rise of 3, run of 12, the run is 4 times the rise, then 4 times 4, and the arrowhead is head to tail. Very good. Let's see if we can tackle these problems now, but again, before moving forward, looks like I need the reactions for all these trusses to move forward. So let's calculate the reactions without too much math. Let's just remember that if this 3 kip here is above this reaction, then it's traveling straight down. This one got none of the 3. Likewise, the 9 at D is right on top of that reaction. You get 9 kips, you get 0. and the 6 skip looks like it's 3 and 3. So 3 goes to the left, 3 goes to the right, as long as these dimensions are equal, which I intend for them to be. Okay, so you're going to get from the 6 another 3. You're going to get the other 3 from the 6. Turns out the reaction on the left is 6 kips. The reactions on the right are 12 kips. Let's try again. Let's do it with cross number 3. Again, 3, 6, 9, the 3 goes straight into A, the 9 goes straight into E, the 6 splits 3 and 3. So this is 6, this is 12. Just like the previous problem, as long as the loads are 3, 6, 9, and the bottom panels are equal, the reactions are equal, 6 on the left, 12 on the right. So, looking at the next one, problem number 2, the 3 goes straight down, the 9 goes straight down, the 6 splits into 3 and 3. That gives me 6 and 12. And you guessed it by now, 6 and 12, 6 and 12. Because the loads are the same, 3, 6, 9. For this problem, number 6, loads are still 3, 6, 9. Reactions are still 3 and 3 and 3 and 9. Now, problem 7 looks like it involves math. I'm skipping it. I just want to talk about problem number 8. The reactions due to 3, 6, 9 would have been 6 and 12. But then this 4-kip messes with the vertical reactions. We did this with overturning moment, stabilizing moment, and the couple and all that stuff in a previous video. The point is, this reaction is a tie-down. This reaction is an upward, due to the 4 kip only. Maybe I should have explained it on problem 7. This is a tie-down. This one is a push-back. And this one has to be 4 kips because the sum of horizontals is 0. So due to the 4 kips, you have a tie-down on the right and a push-back on the left. So there is a tie-down on the left, but there was 12 kips to begin with. and there were 6 skips and now we added to it. So the reactions here are more than 6 on the left and a little bit less than 12 on the right without numbers. Okay, so let me take problem number 2 as an example and let me solve for the internal force in member AC, in member CE, in member EF, and in member FA. Those are the unknown forces in this thrust and they involve a little bit of math. I need the slope. What is the slope? One to one, meaning the rise equal the run, the vertical component equal to the horizontal component. So let me start at joint A. And let me say here's what joint A looks like. It's got three kips. It's got a reaction of six kips. And it has this green member that has two components that are equal. And then the bottom chord, I suspect, is in tension. So I'm going to color it blue. Now looking at this joint A, I have two horizontal components. One, two. And two unknowns is too many. I can't solve. But I have only one vertical unknown component. And I need to have equilibrium. Right now, I have six up. I have three down. Looks like I need to be three. and down. So down 3 plus another 3 is equal to 6 up. I'm in vertical equilibrium. Good to go. Now this arrowhead, oops, this arrowhead is pointing towards the joint, which means this one is pointing towards the joint, which means that component is compression. It's pointing towards the joint A. I don't see any other joint. So if it's towards the joint, so is its horizontal component. What was the slope? One to one. The vertical was three components. Well, then, sorry, three kips. Well, then, this one is three kips. And the force in the diagonal is this much. It's not three kips. It's a little bit more. Three square plus three square square root gives you 4.2 or something like that. I'm not interested in that number now. Now, let's check something quickly. Vertically, I have 3 and 3 equals 6. Very good. That one is taken care of. Horizontally, what do I have? 3. This has to balance that 3. So this one has to be 3 kips. And it's running away from joint A. So this one is 3 kips of tension. That's it. We solved a difficult joint. So the 3-kip blue transfers into tension. It was pulling at A, then it pulls at the joint F. It was 3-kips. It's 3-kips here. Let's zoom into joint F and say you have 3-kips pull on the left. You're gonna need 3-kips pull on the right. And now the 3-kip blue has reached here as tension. It's still pulling. Very good. Let's transfer the green load. This one. This one turned out to be compression. I should color it red, but I'm going to leave it green for now. It was pointing towards the joint A, then it points, pushes into the joint C. And it had two components that were pointing towards the joint C. The vertical was 3. The horizontal was 3. This one is 0 from a previous analysis. Then I need a 3 to balance the 3 horizontal. I need a 3, oops, this is 6 skips down. I have 3 up. I need another 3 up. Very good. Now this joint is in equilibrium. Let's check on it. horizontally, 3 and 3 cancel. Vertically, what's going on with vertically? 6, 3, and 3. Good to go. So let me send this last 3 here. It was 3 on the horizontal, 3 on the vertical, and pointing towards the joint C. So it points towards the joint E. Now let's check this joint. It looks like, vertically, I have 9 and 3. I have a 12 here. I have a 12. Good to go. Horizontally, blue, horizontally, it looks like 3 to the right, 3 to the left. Now I know that all this work is correct. Very good. So we can go on and solve all these trusses in the same way we just did this one. I would like to, I'll give you a key to this one, and I would like to just solve number 1 and number 6 because they have different slopes. So, solving number 1, I've already found quite a few of the members, and there is the member AC. It has two components, and the two components are in the ratio of run of 2, rise of 1. So let me zoom into joint A. There's the joint A. There's a vertical AB. It had three kips of compression at B. Well, then it transfers down to here as three kips into the joint at B, therefore into the joint at A. The reaction here is equal to six kips. Let's zoom in here and see what we've got. Excuse me. I think this one is tension because it's the bottom chord of a simply supported truss. Excellent. So, in the vertical direction, what color was vertical? Whatever. This color. So, in the vertical direction, I have three. I have six. I have this one. In the horizontal direction, I have horizontal and horizontal. There's two horizontal unknowns. That's too many. Let's look at verticals. It looks like in the vertical direction, we have 6 up, 6 up, and 3 down, and we need another 3 down. Very good. Now, the run is double the rise for the member AC. So, the run has to be double the rise of 3. It's pointing towards the joint A, then it has to point towards the joint A because this is a compression member. So, how much is it? I don't know. It's that much. but its components are 3 on the vertical, 6 on the horizontal, which means that this one, to cancel the horizontals, has to be 6. Let's go to the next joint. I'll take the 6 with me, 6 in tension, pulling on the joint A, it pulls that F, and I'll take the force in member AC with me, and the member AC looks like it's doing that with two components, pointing towards the joint C. If a member is in compression, it's two components point towards the joint or compression. So the rise is 3, the run is 6. And this member here, BC, was 0 at B, then it's 0 at C. Likewise, the member CD, sorry, the member CD was 0 at D, then it's 0 at C. So, zooming in and let's solve joint C. There is a diagonal here. It has two components, and the run is double the rise. It looks like we had six skips out at external load. We have three up. We need another three up. If the three up for member CE, I think, Then the horizontal component is double and pointing towards a joint. And this one cancels this one. And this one, 6 is equal to 3 plus 3. Good to go. This was a good joint. Let me take that information and bring it down to joint E. The final joint should be a check. There's a run of 6, a rise of 3. 3 and 6. The arrows were pointing towards the joint at C. Well, then they point towards the joint at E. Let me solve joint F very quickly. There are 6 kips. There needs to be an opposite 6. That's tension. And it pulls on E with 6 kips of tension. The reaction is 12 kips. We found it earlier. And I have 9 kips coming from D down to E. This is 9 kips. Its arrowhead is towards the joint, then the arrowhead here is towards the joint. And let's see if we have equilibrium for this joint E. We have horizontally or vertically first, 9 plus 3 equals 12. Good to go. horizontally, we have 6 and 6, good to go. This checks. Let me solve the last problem down here. This one has a rise of 2, a run of 1, and we found quite a few of the members. So let me start again at joint A, and this time, oops, the rise is double the run. And I still think that bottom chord is in tension. And let me see what I have. I got 3 that I inherited from B up there. I have a reaction of 6 skips. And therefore, I'm still going to need a 3. Down. 3 down to help the other 3 down make the 6 up. Down means it's pointing towards the joint A, means compression. In this truss, the rise is double the run. Well, then the run is half of the rise, and therefore this is 1.5. The diagonal AC is 1.5 square plus 3 square square root. I don't want that answer right now. I'm just resolving the truss and its components. So, let's see what's going on horizontally. I have a 1.5 and a blue. Must be the blue is 1.5 in the opposite direction. It's running away from joint A, so that's tension. And it goes to joint F as tension, and it comes out of joint F as tension, because there's a zero member there. So this one here is 1 1⁄2, and also pulling or tension. And there's one joint remaining, sorry, two joints, C and E. Let's go to C. And let's bring this information with us up there, So let me pull out my green pen and let me say the components of member AC are 1.5 horizontal and larger component double on the vertical. The arrows were pointing towards the joint or compression. Well, then they must point towards the joint C or compression. This one has two components. This one is zero. This is six skips external load. So again, Then, I'm going to need a 3 plus a 3 on the left to make the 6 down, and a 1.5 because of slope. The rise is double the run. The run is half the rise. The rise is 3 kips. The run is half of 3 kips. Good. Let's bring this information with us down here. The vertical is 3 towards the joint. The horizontal is 1.5 towards the joint. Let's check that this one works horizontally. 1 1⁄2, 1 1⁄2, good to go. Vertically, 3 plus 9 equals 12. This is good to go. So I'd like to make a conclusion here, other than the math, a word conclusion. When you squish a truss, you squished it here. It used to be 1 to 1, you squished it, then you relaxed it, you made it deeper. When you take away the depth from a truss, or when you squish it, this 3 did not change. And the run was 6. Sorry, the run here was 3. Here it was double of 3 because you reduced the depth. Here it became half of 3. So, in essence, when you change the depth of the truss, the vertical components all remain the same. the horizontal components pay the price. When you squish the truss the horizontal component doubles. When you double the depth the horizontal component became 1.5 versus 3. Okay? Good.