L7 Truss Analysis

Hinzugefügt: 2026-05-31

[00:00:00]so in the previous video we talked about the behavior of trusses and in this video we will be talking about how to actually analyze and calculate the magnitude of the forces in these members so let's recall the general principles that we learned for predicting direction of internal forces in trust members first we have zero Force members and we had three cases that we learned for when zero Force members occur first is when we have a load applied at a joint that has diagonal members attached to it as well as a vertical piece that leads to a joint that does not have diagonal members in this case the vertical piece will be a zero Force member also if we have a vertical piece with no loads applied to it at all from either side again that vertical piece will be a zero Force member next with tension members if we have a load applied downward at a joint with no diagonal pieces attached to it but diagonal pieces located up above that vertical member will be in tension also any diagonal member that follows the deflected shape of the truss will be in tension for compression members when we have a load applied at the top of a truss at a joint that does not have diagonal pieces framing into it but it does have a vertical piece that leads to a joint that does have diagonal pieces that vertical piece will be in compression and similarly to what we saw before with tension but this time the opposite anything that goes against the shape of this deflected Curve will be in compression also as we had with beams when we have simply supported end conditions and vertical downward loads the bottom cord will be in tension and the top chord will be in compression now let's use these guiding principles to help us analyze a truss say we have a truss that's configured like this with panels that are six feet wide typical and it is four feet deep and we have two Kip forces applied at each panel point this six foot horizontal by four foot vertical allows us to create a similar triangle and if similar triangle relationships don't sound familiar to you please go back and watch the videos in week one we had a lot of questions around the similar triangle relationship mathematical principle in our week one assignment and I had a lot of people trying to use trig functions when given a similar triangle relationship which actually led to a lot more confusion so when we're given defined geometries like this and we're able to set up similar triangles if you're someone who struggles with trig functions please use the similar triangle relationship it will make your life so much easier now let's start with our analysis let's say we want to find the force in this vertical member fbe as we discussed earlier this vertical piece has diagonals at both ends and so we can't say exactly what direction or magnitude the force in this member will be without some further analysis so how are we going to calculate the force in this member well the first tool that we have for analyzing trusses is called the method of joints and the basic principle for the method of joints is that if this truss is in static equilibrium then all joints must also be in equilibrium so this joint a this joint D any joint that we pick in this truss must be in static equilibrium and so we can draw a separate diagram of any of these joints and use static equilibrium equations to come up with internal forces so let's try to pick a joint that will be most advantageous for us to calculate the force in fbe we have two choices we could either pick joint B here or joint e now joint B has four members framing into it and therefore we would have four unknowns if we were to pick joint B joint e on the other hand only has three members framing into it and so we would only have three unknowns making joint e the much better pick so let's draw a free body diagram of joint e and we can see that this is relatively simple we have the joint e here the vertical piece as well as the bottom chord and this diagonal piece now let's go back to our Truss and we know that the bottom chord will be in tension we know that this diagonal will also be in tension so f d e will be in tension and so we can draw a force going in this direction f d e and note that any force that is in tension will be pulling away from The Joint anything that would be in compression would be pushing toward the joint next we have the diagonal piece fce will be pulling in this direction and we're solving for fbe which we do not know whether it will be in tension or compression yet but just for fun let's draw it in compression f b e now we also have one more force that we haven't drawn yet and that is this externally applied load and so let's draw that as well now with our method of joints we have two equations we have the sum of forces in the Y and the sum of forces in the X so let's start with the sum of forces in the Y we have our two Kip Force pointing downward we have our f b e Force also pointing downward next we have the vertical component of fce we'll call it fce y now for the sum of forces in the X we have our f d e which is completely horizontal and pointing to the left and we have our horizontal component of fce to the right and both of these equations must equal zero so we have a lot of unknowns here in total we have four unknowns but actually these two unknowns can be tied together very easily since they are part of one total Force vector and so really we have three unknowns so we have three unknowns but only two equations and so we cannot solve for the one unknown that we want this fbe so we need to ask ourselves if there is any additional information we know that may help us to solve for our unknown so let's scroll back up to the top here and try to look for any other way to give us more information about the forces at this joint e and let's concentrate on this joint C at joint C we know that there is going to be some vertical reaction force call it f c y that vertical reaction force fcy has to be equal to the vertical component of our f c e and that's because the only other member at this joint is FBC which is perfectly horizontal so if we can solve for fcy we can easily find f c e y which will allow us to calculate fbe so let's look at that vertical reaction at joint C now for solving the reactions of a truss it's actually very similar to solving for reactions for a beam in fact we can neglect everything that's happening down below and just think of this as a beam and as we've seen earlier with our beam analysis when we have a beam with loads applied symmetrically on both sides we know that half the load will go to this support on the right and the other half will go to this support on the left and so we can sum up all of our total applied forces divide by two and get our vertical reaction fcy and this vertical reaction fcy is equal to fce y the vertical component of the force in fce so this force is equal to 5 kips now let's go back to our method of joints so this f c e y is equal to five kips and so now we have everything we need to solve this equation f b e will be equal to five kips minus two kips which gives us three kips and since this force is positive we know that we have drawn the arrow in the correct direction and it is in fact a compressive Force so fbe is equal to three Kips of compressive Force now let's take this one step further and say we now want to solve for f d e the force in the bottom chord now we do have to look at this sum of forces in the X equation and as we discussed earlier we have a relationship between fcey and fcex and that relationship is through similar triangles so let's grab this relationship and replicate it down here so as we learned in week one if we have a vertical component and we want to solve for the horizontal component all we need to do is apply a similar triangle relationship which is similar to a unit conversion so we have fcey and we want fce X so to convert fcey we need to multiply by the relationship between the vertical and the horizontal component of our similar triangle so we want the horizontal six and we are dividing by the vertical four and so this is our similar triangle relationship and now we can plug this formula into this equation and so here we get that f d e is equal to f c e y times six over four which is equal to five kips times six divided by four equals seven and a half cups and again this Force came out positive so we drew this arrow in the correct direction and it is 7.5 Kips of tension now we can continue to now we can continue to work like this hopping from joint to joint and slowly working our way across the truss to solve for any of the forces in any of these members but let's say we wanted to solve directly for this Force FDB and let's say we didn't want to go from joint to joint in order to find it and instead we wanted to find it directly well for that we would use the method of sections and the method of section states that if a truss is in equilibrium all sections must also be in equilibrium so if I were to cut this Truss through the member that we are trying to solve for and separate this portion of the truss it must be in static equilibrium and the benefit of the method of sections over the method of joints is that now we have three equations we are introducing the sum of moments so let's use the method of sections to solve for the force that we want in the member DB and note that what I've drawn so far for this free body diagram is clearly not in static equilibrium so what is it that allows this piece to be in static equilibrium well it would be the internal forces in each of these members that we have cut and we know that the force here will be in compression the force here will be in tension and this diagonal force will also be in tension as before anything in tension pulls anything in compression pushes so now let's apply the proper labels to each of these forces that we've drawn looking at our original Truss so the top chord will be a b the bottom chord will be d e now let's set up our static equilibrium equations for the sum of forces in the Y we have five Kips up minus two Kips down minus the vertical component of f DB and that must be equal to zero and let's simplify this equation before we move further we should get that f d b y will be equal to three kips and here we can actually stop and consider the fact that we have already solved for one component of our force that we want and so all we need to do to find the total force is to recall our similar triangle relationship six four seven point two one so we have our fdby we want the total FDB so FDB will be equal to fdby we want the hypotenuse we have the vertical so the relationship will be 7.21 over 4 and now we plug in our values and we get 5.4 kips of tension force now let's say we wanted to solve for the other forces as well we have three unknowns here and we have three equations so we should be able to do that and let's suppose that we didn't already solve for fde so if we do our sum of forces in the X we will have f a b minus the horizontal component of f DB minus f d e equal to zero and here we have already found FDB so we know its horizontal component so we can simplify this equation to f a b minus f d e is equal to f d b x which is equal to f d b y times we want the horizontal we have the vertical so that will be our unit conversion that is equal to three kips times six over four and so the horizontal component of FDB is equal to four and a half kips so four and a half Kips is equal to f a b minus f d e so now we have an equation with two unknowns neither of which appear in our first equation that we established with our sum of forces in the Y and so we need another equation which will be the sum of moments and let's try to pick a point that will be most advantageous for us to solve for our final two unknowns so looking at this diagram recall that the sum of moments can be taken at any point and when we sum moments we will be multiplying the force times its perpendicular distance to that point so for example if we were to some moments at this joint e this two Kip Force here would generate zero moment because it acts through that joint also fde would generate no moment at this joint but we would have to calculate the moment due to FDB Fab and this five Kip reaction force that of course is doable because we know FDB we know the five kit force and that would leave us with one unknown here however solving for the moment due to this FDB force is a little bit tricky and so let's try to pick a better point let's pick the point where these two forces converge this point here which would be joint d now this joint is outside of where we've applied our cut but that doesn't matter theoretically we can sum moments out here or over here it doesn't matter it will still add up to zero so that's some moments about joint d the moments due to FDB and fde will both be zero because they act through the point that we are choosing the perpendicular distance from f a b to our joint is four feet the depth of the truss and let's recall that we are using a standard coordinate system with positive y in the vertical positive x to the right and positive moment and the counterclockwise Direction this f a b force will apply a clockwise moment about joint D and so it will be a negative moment similarly this two Kip Force will apply a negative moment and its perpendicular distance to Joint d is the width of a panel which is six feet lastly we have our five Kip reaction force which applies a moment in the counterclockwise Direction around joint d and so it will generate a positive moment and its perpendicular distance to Joint D is two panels or 12 feet and all of this will be equal to zero and now we have only one unknown are f a b and so we can rearrange this equation to give us f a b equal to minus 12 foot clips Plus 60 foot kips divided by four feet and we get that f a b is equal to 12 Kips of compressive Force now that we have f a b we can take this value plug it into this equation to get us f d e so we have 12 kips minus f d e is equal to four and a half tips which gives us f d e equal to seven and a half cups of tensile force and we can compare this to what we solved earlier using method of sections and in fact here we did get that fde is equal to seven and a half Kips tension so to summarize all the forces that we've solved for here I've drawn them on our original truss diagram and we can compare these forces to what we find in our analysis program so here we have the exact same Truss with panels six feet wide and a total depth of four feet and if I look at just those members that we solved for and turn on the axial forces here you can compare our results to the results in the program and I know that the text is kind of small sorry about that so in summary these are the two tools that we have for analyzing trusses the method of joints which is useful for solving for forces and vertical members though its limitations are that we can only use two equations and the method of sections which has an advantage over the method of joints in that we now have a third equation