Trusses – Snow Load on Howe, Fink, and Pratt trusses

Intoduction

I want to work another snow load problem… using three different trusses. I found a pair of these, for a Fink truss and a Howe truss, on a professor’s university website – his name is Zig Herzog and his main statics page is this. Individual links for the two problems will follow.

Each of these trusses is 12 meters across the bottom, and 4 meters high. Each has a total snow load of 2400 Newtons. As ever, I have used Mathematica® for the computations and graphics.

One of the things I liked was that Herzog asked us to find the maximum values of stress (both compression and tension), and the total length of the beams used in each truss. In addition to using Fink and Howe trusses, I will do the Pratt truss again, with the parameters of this problem.

The purpose of this post was to see how different the trusses are, under the same load. In one respect they are the same: the maximum values of both compression and tension are the same for all three trusses. And that’s the summary!

Here are screenshots of the author’s assigned problem… for which he does provide solutions, so I’m not giving away any secrets.

For comparison, here’s the Pratt truss I used in the previous post.



Let me point out that after I drafted the running commentary for the Howe truss, I used it for the other two problems. You may find this boring – the same text repeated three times – but you shouldn’t have to worry about which problem contained a particular comment, because everything is in all three. (Except, of course, comments pertaining to specific problems. If anything, I might have forgotten to change something that I should have. And if it’s any consolation, if I have to make corrections, I may have to make them in three places. One so far, before publication.)

Howe truss

Here is a link specifically to Herzog’s Howe truss problem.

Note that this Howe truss is smaller than the one we worked (I’ll show it to you in a moment) – which is why I once remarked that I might have been working with a double Howe truss.

Anyway, we have numbered points.

A quick count: we have 13 beams (members, m = 13) and 8 joints (j = 8), and we want

2j = 16 = 13 + 3

so we want 3 possible reaction forces. But we’ve got 2. What the heck? All the loads are downward, so there are no horizontal reactions. But, as we’ve seen before, it will work out. Whether we assume the truss is rolling at one end – which is what Herzog did on his drawing, and what I will do this time – or just go ahead and solve 16 equations in 15 unknowns – as I have done once – it will work out.

I don’t need to use labels – after all, I have numbers instead of letters – but I want to keep this collection of code more powerful than less.

Herzog declared that it is fixed at 1, rolling at 5. For comparison, here’s Meriam’s drawing of the (double)Howe truss I used in a previous post:

I’m going to omit a lot of the computational details, but let me show you the starting point: get the vectors between every pair of joints:

This time I’m not going to show all the intermediate results. If you need to see them, go look at the two previous roof truss posts.

I make unit vectors u_ij out of the vectors between joints…. I clear my force arrays and then set F_ij = f_ij u_ij (but in such a way that only half the magnitudes f_ij are used, since f_ij = f_ji).

I need to outright declare that there are no forces in the following beams – because there are no such beams! I’m setting these magnitudes to zero:

Here are the external forces: P=600, and we have P/2 at the end, just as we did for my previous post, a snow load on a Pratt truss… first we divvy up the total load uniformly on the beams, and then we assign half of every beam load to its two joints. Since the joints at the ends of the top chord have only one loaded beam apiece, those two joints only see P/2.

A little trickery gets me from forces in beams… to equations… the left column is x-components and the right column is y-components:

(As I said, if you need to see what the intermediate steps look like, look at the previous two truss posts.)

Now solve! (The interior “Flatten” converts two sets of eight equations into one set of 16.)

Note that R1 = 0 as expected: there is no horizontal component of reaction force at joint 1.

Let me check the global balance. The external forces are… and their vectors sums are… zero:

Change the signs for the drawings.

Are 2-6 and 4-7 really unstressed? Yes.

Note: their lengths are 2 each, so if we were to remove those beams from the truss, we would subtract 4 from total length. I would not, however, actually recommend removing those two beams; in the real world, all these beams deform – our assumption that they do not deform gets us only a first approximation, and I’m not at all sure that it’s really safe to remove any beams as a result of the undeformed analysis.

3-6 and 3-7 in compression? Yes.

(Let me remind you that my convention is that a positive f points inward along the beam, but it is the internal beam force on the joint… the force on the beam, therefore, points outward – which is tension. So negative f_ij is compression, while positive f_ij is tension. Finally, since my intuition goes the other way, my drawing of the forces shows the forces on the beams.)

Is the entire bottom chord in tension? Yes.

Let’s get a table of them all, and get the minimum and maximum:

That is, the maximum compression and the maximum tension respectively are 1622.5 and 1350 Newtons. The maximum tension occurs in all of the bottom-chord beams; the maximum compression occurs in the two end-beams of the top chord.

What is the total lengths of beams? And if we were to subtract 4 meters, by removing the two unstressed beams…

That is, we need 41.6333 meters of wood for the truss as computed; 4 meters less if we remove two 2-meter beams.

Fink truss

Here is a link specifically to Herzog’s Fink truss problem.

Here is Herzog’s drawing:

Again, we have numbered points.

A quick count: we have 11 beams (members, m = 11) and 7 joints (j = 7), and we want

2j = 14 = 11 + 3

so we want 3 possible reaction forces. But we’ve got 2. What the heck? All the loads are downward, so there are no horizontal reactions. But, as we’ve seen before, it will work out. Whether we assume the truss is rolling at one end – which is what Herzog did on his drawing, and what I will do this time – or just go ahead and solve 14 equations in 13 unknowns – as I have done once – it will work out.

I don’t need to use labels – after all, I have numbers instead of letters – but I want to keep this collection of code more powerful than less.

Herzog declared that it is fixed at 1, rolling at 4.

I’m going to omit a lot of the computational details, but let me show you the starting point: get the vectors between every pair of joints:

This time I’m not going to show all the intermediate results. If you need to see them, go look at the two previous roof truss posts.

I make unit vectors u_ij out of the vectors between joints…. I clear my force arrays and then set F_ij = f_ij u_ij (but in such a way that only half the magnitudes f_ij are used, since f_ij = f_ji).

I need to outright declare that there are no forces in the following beams – because there are no such beams! I’m setting these magnitudes to zero:

Here are the external forces: P=600, and we have P/2 at the end, just as we did for my previous post, a snow load on a Pratt truss… first we divvy up the total load uniformly on the beams, and then we assign half of every beam load to its two joints. Since the joints at the ends of the top chord have only one loaded beam apiece, those two joints only see P/2.

A little trickery gets me from forces in beams… to equations… the left column is x-components and the right column is y-components:

(As I said, if you need to see what the intermediate steps look like, look at the previous two truss posts.)

Now solve! (The interior “Flatten” converts two sets of eight equations into one set of 16.)

Note that R1 = 0 as expected: there is no horizontal component of reaction force at joint 1.

Let me check the global balance. The external forces are… and their vectors sums are… zero:

Change the signs for the drawings.

3-6 and 2-5 in compression? Yes.

(Let me remind you that my convention is that a positive f points inward along the beam, but it is the internal beam force on the joint… the force on the beam, therefore, points outward – which is tension. So negative f_ij is compression, while positive f_ij is tension. Finally, since my intuition goes the other way, my drawing of the forces shows the forces on the beams.)

what about 2-7 and 3-7? Tension.

Is the entire bottom chord in tension? Yes.

Let’s get a table of them all, and get the minimum and maximum:

That is, the maximum compression and the maximum tension respectively are 1622.5 and 1350 Newtons – exactly as for the Howe truss. But this time the maximum tension occurs only in the two end-beams of the bottom-chord beams; the maximum compression again occurs in the two end-beams of the top chord.

What is the total lengths of beams?

That is, we need 39.8286 meters of wood for the truss. Recall that the total length of beams for the Howe truss was

Pratt truss

I am using Meriam’s drawing for the Pratt truss, but with Herzog’s parameters (4 m tall, 12 m wide, total snow load 2400 N).

We have letter labels instead of numbers.

A quick count: we have 13 beams (members, m = 13) and 8 joints (j = 8), and we want

2j = 16 = 13 + 3

so we want 3 possible reaction forces. But we’ve got 2. What the heck? All the loads are downward, so there are no horizontal reactions. But, as we’ve seen before, it will work out. Whether we assume the truss is rolling at one end – which is what Herzog did on his drawing, and what I will do this time – or just go ahead and solve 16 equations in 15 unknowns – as I have done once – it will work out.

In keeping with Herzog’s conventions, I declare that the truss is fixed at A, rolling at E.

I’m going to omit a lot of the computational details, but let me show you the starting point: get the vectors between every pair of joints:

This time I’m not going to show all the intermediate results. If you need to see them, go look at the two previous roof truss posts.

I make unit vectors u_ij out of the vectors between joints…. I clear my force arrays and then set F_ij = f_ij u_ij (but in such a way that only half the magnitudes f_ij are used, since f_ij = f_ji).

I need to outright declare that there are no forces in the following beams – because there are no such beams! I’m setting these magnitudes to zero:

Here are the external forces: P=600, and we have P/2 at the end, just as we did for my previous post, a snow load on a Pratt truss… first we divvy up the total load uniformly on the beams, and then we assign half of every beam load to its two joints. Since the joints at the ends of the top chord have only one loaded beam apiece, those two joints only see P/2.

A little trickery gets me from forces in beams… to equations… the left column is x-components and the right column is y-components:

(As I said, if you need to see what the intermediate steps look like, look at the previous two truss posts.)

Now solve! (The interior “Flatten” converts two sets of eight equations into one set of 16.)

Note that R1 = 0 as expected: there is no horizontal component of reaction force at joint A.

Let me check the global balance. The external forces are… and their vectors sums are… zero:

Change the signs for the drawings.

CG unstressed? Yes:

Note: its length is 4, so if we were to remove that beam from the truss, we would subtract 4 from total length. I would not, however, actually recommend removing that beam; in the real world, all these beams deform – our assumption that they do not deform gets us only a first approximation, and I’m not at all sure that it’s really safe to remove any beams as a result of the undeformed analysis.

BH and DF compression? Yes.

(Let me remind you that my convention is that a positive f points inward along the beam, but it is the internal beam force on the joint… the force on the beam, therefore, points outward – which is tension. So negative f_ij is compression, while positive f_ij is tension. Finally, since my intuition goes the other way, my drawing of the forces shows the forces on the beams.)

CH and CF in tension? Yes.

Let’s get a table of them all, and get the minimum and maximum:

That is, the maximum compression and the maximum tension respectively are 1622.5 and 1350 Newtons, just as in the Howe and Fink trusses. The maximum tension occurs in the two end-beams of the bottom-chord beams; the maximum compression occurs in the two end-beams of the top chord.

What is the total lengths of beams? And if we were to subtract 4 meters, by removing the unstressed beam…

That is, we need 44.4222 meters of wood for the truss as computed; 4 meters less if we remove one 4-meter beam. Recall the design values for the Howe and the Fink:

As designed, the Fink requires less wood… even with a beam removed, the Pratt requires more than the Fink… but if we remove two beams from the Howe, it would require less wood than the Fink. But I would not remove any beams.

I will point out that the maximum tension and compression are the same for all three trusses.

It is, apparently, also a matter of interest: how long are the beams in which the maxima occur? But I don’t really know much about that. Yet. (Okay, Okay: I would guess that the buckling – collapse – of a beam in compression depends on the stress and the length.)

One Response to “Trusses – Snow Load on Howe, Fink, and Pratt trusses”

  1. sola Says:

    cool. thanks


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