Building a truss

On my opinion it's too abundance in "heavy metal". And I would replace a big triangle on opposite (left) side to make a structure more stable.

Not my field at all,
I'm a Cat you see, but all the zig zigs below the top one don't seem to add much as far as I can see, so I'll agree with Caramba .
Maybe one big triangle like he suggests with a little cross bracing inside the triangle to stop it buckling.

Del _{(Please remember I'm just a Cat..so this is just gut feel<groan>)}

Try one of these programs to get some ideas. Any member that has a very low load is essentially useless.

http://www.jhu.edu/~virtlab/bridge/truss.htm

http://3d2f.com/programs/12-010-framemaster-download.shtml

http://3d2f.com/programs/33-283-framesolver-2d-download.shtml

The simplest structure is a sloping tension diagonal and a compression vertical. The supports should both be hinges. If the object of the game is to minimize the weight/load ratio, this is probably the most efficient structure, but it is not a truss.

If you must have a truss, then make the left support a hinge, the right support a roller and connect them with a horizontal member (which will carry zero stress).

Tension members are always more efficient than compression members because slenderness effects are not a factor in their design.

Many of the diagonal elements in compression are redundent since the diagonals under tensile loads will, when properly sized, be able to hold the loads without too much concern about buckling.

You have a good diagram...pick an arbitrary applied load and solve the system for all the loads in each element. Solving these kinds of systems is straigth out of 1st year engineernig statics text books so you should be able to find many similar examples if you need some details on these calculations.

Hello Guest,

I looked at your drawing and it does seem to have an unnecessary number of braces on the left. Or is this a roof section with the truss on the right?

I am not sure if you are working in wood or steel, but look at any house with a 'hip' coming down from the main vertical truss. That is what I think you want. It all depends on the load of course, at to the structure and the sizes of the wood or steel used.

Good luck!

I would suggest bending a piece into a long arc with the supporting angles applying pressure to the arc.

After all the arc is supposed to be the strongest structure.

Your diagarm has too many unnecessary braces. In building a truss, you also need to consider the span or the height it will cover. A truss is made up of a series of triangles, the longer the span or height it needs to cover the more series of triangles you need. Now, if the span or height is not an issue an ordinary triangle will do.

I'm in "Guest's" group and would like to thank you guys for many nice answers and would like to be more precise in this matter.

We are building in aluminium and the construction is supposed to have a height of 1meter. And it has to have a bottom plate that's a 400x400x4mm plate

and a bottom plate that's a 300x300x4mm plate

It has to withstand atleast a force of 500N in vertical forces. In horizontal forces there will be applied more and more pressure from the top of the construction(pushing on the top plate), so it is from these forces we have to withstand as much as possible.

The whole point of this is to get the best score from this formula Points= Force it can hadle/weight of the construction

Any other applies will be truly helpful=)
Thanks from a Marine Technologies student.

Here is a "3D" picture of how we might want it. We are abit curious if the middle pole in our triangle will help us with anything:P
http://img258.imageshack.us/my.php?image=trussbs4.jpg

The drawing is not practical. The long diagonal braces on the right side done not do anything. I would suggest to build your tower using a symmetrical digram, meaning you start with your base wide and tapers as it goes up until you meet your design criteria for dimension.

Hello Guest,

why do you not join?

The explanation you give here is a little confusing.

Is your bottom plate 30cm, or 40cm ? And what size is the 'angle' or flat you are build it with? For the size I think there is probably still to many struts on the left. The height you are building is about the same size as you might find in a dolls house. I have seen these kinds of things built from spaghetti also.

You are building a single truss which could go with other trusses to make a bridge or similar?

Try drawing the strut, with no drawing along-side please.

And it has to have a bottom plate that's a 400x400x4mm plate

and a bottom plate that's a 300x300x4mm plate

Hello babybear,

I think that Guest #2 has just changed the concept substantially. Looking at his sketch, it seems clear that the 300 x 300 plate is on the top, not the bottom.

What we are dealing with is a tower, not a truss. It could be built in the shape of a truncated pyramid with just four sloping columns, one located at each corner. The advantage would be that all four sides could be identical.

Some form of bracing is required on each of the four sides to provide stability. The details of the bracing will depend on the selection of members to which the students have access for their project.

One thing is clear. The latest sketch illustrates an unstable structure.

I think we should also be clear that, while we may offer some suggestions, it is their project and CR4 members cannot do the project for them.

Hello ba/ael:

just to say I fully agree with you on your summarisation. But, they have yet to explain or send drawings for any member or otherwise, to discuss? One says a truss, but, as you say the drawing is not a truss. Me thinks they should talk a little more and just see what is possible. The structure is not that big, after all! One thing I will say is, for a structure that small, the lower and upper plate thickness seems to thick?

I have yet to get used to the 'new' metric of 'Newtons', as mentioned for the force, is it? I am just talking from memory here. Sorry if I used the wrong term.

Take care............

A kilo Newton (kN) is 225 pounds (think of it as a football player). So a Newton is 0.225 pounds. And yes they are both units of force.

Hello ba/ael:

thank you! It is simple when you know how?

So in metrec terms a KN is 102.27 Kilos.

Hello babybear,

You could say that, but the SI metric buffs don't like us to equate units of force with units of mass. A mass of one kilo is one kilo whether it is on the earth, the moon or the sun but a mass of one kilo will weigh less on the moon than it does on earth.

Newton's famous equation F = ma says force = mass times acceleration. One Newton is defined as the force required to give a mass of one kilogram an acceleration of one meter per second squared, or

1 N = 1 kg.m/s²

Gravitational acceleration g = 9.81 m/s² approximately (it varies slightly from place to place because of flattening of the poles and the difference in density between land and ocean).

So on earth, a mass of 1 kg exerts a force on the surface of 9.81 N or 0.00981kN. To exert a force of 1 kN on earth, the mass would need to be approximately 102 kilograms.

For a planet with gravitational acceleration of 5, the same 1 kilogram mass would exert a force of only 5 N and a mass of 200 kg would be required to exert a force of 1 kN.

For years, we have grown used to the fact that the gravitational constant is included in the units, i.e. 1 pound force = 1 pound mass or 1 poundal as some have labeled it. Not so with the SI system.

Hello ba/ael

really appreciate the explanation, thanks. I worked this out once.............when I had nothing to do. But, I was not aware that 9.81 m/s² could be used or written as .00981KN. I should have figured that out really, and may well have but it slipped my mind? I know I worked out how much 'weight' was pushing down on a sheet of newspaper once and it was amazing. At the time I knew it as psi. But converted it to the metric system. I used up a lot of paper trying to verify my outcome, but I got there in the end.

Thank you my friend.

An example

Hello nick name:

Your example may be a little OTT, but, a good one none the less. As there is several types of truss on here.

Hi,

What means OTT?

Now being serious for the one who put the question:

Basic element in ALL designs is the triangle since it theoretically undeformable. This basic ides combined with the other comments lead to the optimal result: least weight for maximal load and highest mark!

You should analyse how many side you need for an optimal design and how weight and load are related to eachother.

My example was choosen because it is a VERY light structure mostly loaded by horizontal loads as wind and which was so optimized that if one would reduce it the dimensions of a sheet of paper it will weight less than the sheet!

It is interesting to think that it was build only for 1 year and it is still here after over a century and more than 1E6 visitors enjoy it every year. This an example of engineering!

Hello nick name:

OTT......."Over The Top".

And yes, the Eiffel Tower is beautiful!

Thanks I am glad every time I can get a new expression.

Hello nick nameI think acronyms are pretty cool as well. But, they are often used very local to you. In a profession, or just one Town? I usually Google the ones I am not sure off and that I might want to use,, as they can shorten phrases a lot. Particularly important when texting! Take care

If this is homework I would highly recommend reading up in a few good books and hammering through a few examples before your exam comes along. There is NO replacement for first hand knowledge.

Hello kkjensen:

Well said Sir! Of the two people whom have mentioned it here as their project, neither seems to be talking of the same shape, and niether seem to know what a truss is.

It is going to come to the test and they are not going get far as each other thinks they are building a different shape!

Hello kkjensen:

Thay should read my signature, especially the last bit.......Do and you understand!

Hear and you forget; see and you remember; do and you understand.