## Introduction

I want to close the recent examples of trusses by providing a sampler of truss designs. This is far from encyclopedic. In fact, this post is limited to planar trusses.

First, however, let me give you a link to an online calculator. I checked it out on the Howe truss with a snow load. As you can see, I scaled the loads by a factor of 10. I had to, for the program – not a big deal.

There are plenty of websites with information. You can search for yourself… you could start with the usual wiki article – the merit of which is that it has a lot of external links.

Personally, I have kept a link to bridge trusses in western PA and a link to roof trusses by an Australian contractor.

Okay, what do we have?

Oh, I relied on J.L. Meriam, “Statics”, John Wiley & Sons, 1966 – the book from which I took Examples 3 (Howe) and 4 (Pratt). I constructed the last five drawings in this post (“sign and cantilever trusses”) from his examples 107, 108, 110, 112, and 114.

In addition, he had a convenient one-page display (p. 128) of commonly used trusses in another book, “Engineering Mechanics: Statics and Dynamics”, Wiley, 1978, ISBN 0 471 01979 8. I made my own drawings based on his.

Finally, there is a very recent book, “Engineering Mechanics: Statics, Student Value Edition by J. L. Meriam and L. G. Kraige”, ISBN-13: 978-0470499771 (paperback). I haven’t held a copy in my hands, but based on two of the three reviews, this latest edition is like the earlier ones: it has tons of examples.

My examples were roof trusses… there are, as I’m sure you know, bridge trusses… there are also cantilever trusses… and there are trusses for signs… and trusses need not span two points at the same elevation….

## Roof trusses

We have seen two forms of a Howe truss: We have seen a simple Pratt truss, and there is a more complicated one: I’m sure that both of these designs can be extended to more beams in the obvious way.

Notice the distinction between the Howe and the Pratt: whether the diagonals which touch the center do so at the top or at the bottom. We have seen a simple Fink truss… and it, too, has a generalization: The joints at L and N… and the horizontal beams there… this is not a straight-forward generalization.

Then there is a Warren truss – which I want to compare to a Howe: ## Bridge Trusses

There are Howe, Pratt, and Warren bridge trusses. Let me compare them to the corresponding roof trusses. Here are the Howe roof and bridge trusses: Isn’t that odd? The diagonals go the other way.

Here are the Pratt roof and bridge trusses: Again, the diagonals go the other way.

Here are the Warren roof and bridge trusses… of course, the diagonals go the other way. I would, naively, have expected the pair to have the same name, but they don’t: that’s a Howe roof and a Pratt bridge.

What’s going on?

The roof trusses are designed for continuous loads on the top chord… the bridge trusses are designed for continuous loads on the bottom chord. Imagine that we take a Howe roof truss and turn it upside down to start on the Howe bridge truss. Turning it upside down flips the diagonals – and then the diagonals have the same relationship to the expected load.

At least, that’s how I figure it. (No pun intended, I swear.)

Then there are some bridge trusses for which I have not seen roof trusses.

Here’s a K truss. Here’s a Baltimore: Here is a Wichert truss. It is apparently intended as a statically determinate alternative to two trusses: Maybe I’ll work that out someday, the Wichert versus a pair of trusses.

## Sign and Cantilever Trusses

Here is a truss for a sign… It expects a continuous wind load from the right on beams BC and CD. Oh, it’s anchored at A and G.

Here is a truss which I think could be used for a sign: If so, it expects a continuous wind load on beam FG. Oh, of course it’s anchored at A and B.

Here is a cantilever truss: It’s anchored at A and J.

Here is an interesting truss: it is fixed at A and E. I imagine it supports a load at F. The key is that the anchor points are at different elevations. Here is another interesting truss: it is fixed at A and H. Again, the key is that the anchor points are at different elevations – and, in fact, one is on a vertical wall. I expect that this one supports a load at E. I also suspect that the last two trusses are actually 3D. They seem rather complicated for 2D supports. But I’m just guessing.

And that’s probably it for a while on trusses.

### 3 Responses to “An Overview of Truss Designs”

1. prof dr mircea orasanu Says:

in these situations must stated that appear many more aspects for mathematics mechanics as stated prof dr mircea orasanu and prof drd horia orasanu and followed that in same situations prof dr Constantin Udriste exposed and posed problem that CONSTRAINTS OPTIMIZATIONS and other chapters can be approached with these apparatus for obtained of a results as a classical field. Its currents (ja) can be seen as the Nother charges derived from the KG Lagrangian L=(1/2)∂aϕ∂aϕ+(1/2)m2ϕ2, so that real-valued ϕ for which ja=0 should describe a classical non-charged field, while complex-valued ϕ are to be identified with classical fields carrying charges. Quantum theory is obtained either by introducing the quantum commutation relations on the field variables, for example, the equal-time commutation relations on (the now field-operators) ϕ and π=∂L/∂(∂tϕ), or going to the path-integral formalism by working out the partition function Z0[J]=∫Dϕei∫L+Jϕ, whose functional derivatives generates the theory Green’s functions. In any case, one then obtains a consistent, Lorentz invariant, theory for spinless particles, as a quantum theory of the field ϕ
. (Note. For Fermi-Dirac fields, one replaces commutation relations by anticommutation ones, or in the path-integral approach, one introduce Grassman variables).
Remark. The Fourier decomposition of the field operator ϕ
in terms of the (positive and negative frequency) plane-wave solutions ϕ(+)k and ϕ(−)k, namely, ϕ∝∫(d3k/ωk−−√)(ϕ(+)kak+ϕ(−)ka+k), shows at once that: (i) only positive-energy solutions exists, but (ii) the theory must be contain many-particle states, where the coefficients ak and a+k
acts as annihilation and creation operators.
4. Finally, two further remarks. (i) About the the role of the term eimt
in the derivation of the Sch. eq. from the KG, the author of the question should give more details about what reference he is using to study this. Nevertheless, I guess it must came from the expression of the time-evolution operator using the relativistic Hamiltonian, U(t)=eip2+m2√t. In this case, the non-relativistic limit is ∝eimt+ip2t/2m
, and so the first term can be ignored since it only gives an overall phase to the quantum state. (ii) As a further inquire, suppose now you begin with the Dirac equation, and then decide to take its non-relativistic limit. How one obtains the spinless Schrodinger eq. from it? (Tip: the state space of nonrelativistic QM will be recovered from the nonrelativistic limit of the Dirac eq. as an eigenspace of the spin operator, namely, the nonrelativistic wave-functions are going to arise as spin eigenfunctions.)
5. I think that the best way to relate the KG current for a complex scalar field to electric charge is by discussing the gauge-invariance of the theory. Begin with the KG Lagrangian for a complex field:
L=∂aϕ∂aϕ∗+m2ϕϕ∗.
Under the transformation (ϕ,ϕ∗)↦(ϕe−iΛ,ϕ∗eiΛ)
for a constant Λ, the complex KG Lagrangian is invariant. Using Noether theorem, you can show that under δϕ=−iΛϕ, δϕ∗=iΛϕ∗

2. prof dr mircea orasanu Says:

thus that Roof truss are good consolidated as observed prof dr mircea orasanu and prof drd horia orasanu and followed that are imposed more many tip of LAGRANGIAN and CONSTRAINTS OPTIMIZATIONS with examples

3. prof dr mircea orasanu Says:

in many situations appear MATH MECHANICS as is known and admitted by prof dr mircea orasanu and followed with To complete the solution of the Poisson’s equation for the problem in Figure 3.9-1, we only need to treat Poisson’s equation with zero boundary condition shown in Figure 3.9-2.
Figure 3.9-2 A thin rectangular plate with all non-homogeneous boundary conditions.
= f(x,y) (3.9-2)
The Poisson’s equation for this case has the following boundary conditions
u(0,y) = 0, u(a,y) = 0, 0 < y < b
u(x,0) = 0, u(x,b) = 0, 0 < x < a
Since the solution in any direction x or y with homogenous solution is the sin function, we try the following function that satisfies the zero boundary conditions
u1(x,y) = Emnsin sin (3.9-3)
The constants Emn are to be determined by substituting (3.9-3) into the equation (3.9-2)
=  Emn sin sin
=  Emn sin sin
 Emn sin sin = f(x,y) (3.9-4)
Equation (3.9-4) is a double Fourier sine series expansion of f(x,y), therefore
Emn =  sin( x) sin( y)dxdy
In this equation mn =
Example 3.9-1. ———————————————————————————-
Solve the following equation in a 11 square (0 < x < 1, 0 < y < 1)
= u + 3
with the following boundary conditions
u(0,y) = 0, u(1,y) = 0, 0 < y < 1
u(x,0) = 0, u(x,1) = 0, 0 < x < 1