## I'm back!!

### I'm back!!

Hi everyone... Especially Walker!!... I'm back and finally ready to build my new trebuchet!!... Remember the 'endless bridge' theory??

### Re: I'm back!!

Now there's a name from the way-back past!

Time for the 'endless trebuchet'?

Time for the 'endless trebuchet'?

### Re: I'm back!!

... Still having trouble finding a good engineer to help with calculations for the arm so I've decided I will go ahead and do it manually... I'm using 25x25x1.6 (grade 350) square tubing for the core and will place it in a press to determine the length between supports... I'm guessing around 130mm (length will be determined by finding the point where the tube no longer bends)... That's the easy part!

### Re: I'm back!!

Dave, if you just need to know the load on a single piece of the tubing before it permanently bends, that's pretty simple. Just increase the load in the calculator until you reach the yield stress of the material you are using (which you'll need to look up for the grade of steel you are using).

The following link will be of help: http://www.engineersedge.com/beam_bendi ... ding2e.htm

That is for a beam simply supported on the ends, not with ends locked as if they would resist rotation at the end (which at certain points in a throwing arm would in fact be the case). There are other scenarios at this link, which may be of help too: http://www.engineersedge.com/beam_calc_menu.shtml

You will need to calculate the area moment of inertia for these calculators, which for a box beam is relatively simple. For a solid rectangle where b=base, h=height. You have a hollow rectangle (square actually), so you subtract the area moment of inertia of the "missing" part. So, , where b1 and h1 are the outer base and height, and and are the inner base and height. Since you are using a square tube, b and h are the same, so . Using your 25x25x1.6mm box tube (I assumed those were metric dimensions, so a roughly 1"x1"x1/16" tube), . Watch your units and make sure you use common ones (m vs mm, etc).

Hope that helps!

The following link will be of help: http://www.engineersedge.com/beam_bendi ... ding2e.htm

That is for a beam simply supported on the ends, not with ends locked as if they would resist rotation at the end (which at certain points in a throwing arm would in fact be the case). There are other scenarios at this link, which may be of help too: http://www.engineersedge.com/beam_calc_menu.shtml

You will need to calculate the area moment of inertia for these calculators, which for a box beam is relatively simple. For a solid rectangle where b=base, h=height. You have a hollow rectangle (square actually), so you subtract the area moment of inertia of the "missing" part. So, , where b1 and h1 are the outer base and height, and and are the inner base and height. Since you are using a square tube, b and h are the same, so . Using your 25x25x1.6mm box tube (I assumed those were metric dimensions, so a roughly 1"x1"x1/16" tube), . Watch your units and make sure you use common ones (m vs mm, etc).

Hope that helps!

Last edited by The Admin on Wed Jul 15, 2015 7:45 pm, edited 1 time in total.

**Reason:***made equations readable*### Re: I'm back!!

Thanks Madmattd... Sorry but the tubing would be standing vertically in the press and pressure applied downwards (axial compression I believe) on the tube to determine the overall length of the tube before it buckles. Once that length is determined (let's say it is 150mm for example) I would then add outrigger/struts at 300mm intervals and the diagonals (at 45 deg) would meet at the 150mm mark... So then I could build it up to maybe a metre of more in length and test again for buckling... Length of the built up truss would be determined (let's say at 1.5 metres for example) so that another layer of outer webbing can be added outside the original... Same way as the single tube.

So looking at the truss now there would be the central tube (25x25) plus struts, diagonals and two outer chords on both sides of the tube... The struts would be extended at every 3 metres for the outer webbing.

Hope that makes sense?... Sorry I don't have a drawing to show you for easier understanding as I'm only using iPhone

Thanks for your input,

Dave

So looking at the truss now there would be the central tube (25x25) plus struts, diagonals and two outer chords on both sides of the tube... The struts would be extended at every 3 metres for the outer webbing.

Hope that makes sense?... Sorry I don't have a drawing to show you for easier understanding as I'm only using iPhone

Thanks for your input,

Dave

### Re: I'm back!!

Oh, okay, you're talking column loading instead of beam loading. Still a fairly easy calculation (though involved) for one member. Once you get the full assembly and need to run the math on that, well, it's more complicated and that's where I let the computer do the crunching.

Check the following link out: http://www.engineersedge.com/column_buc ... _ideal.htm. Once welded up into a structure, you will effectively have fixed-fixed end conditions, so your "effective length", Le, will be 0.5*L.

You'll need to work through that page, there are different formulas to use depending on the "slenderness ratio" versus a "critical slenderness ratio". When finding the critical slenderness ratio, σy is the yield stress of your column material. Note that I had an error in my formula for I, the area moment of inertia, in my last post, which I have gone back and fixed. You'll have to do a couple of iterative runs to get an idea of the load/length. My hunch is that you will end up as a "short column" setup, which uses Johnson's formula on that page, but you may be able to get away with Euler's after all.

I also found a decent-looking set of lecture notes on this topic that may be of help (it's been a while since I've looked at this stuff. Definitely a use it or lose it type of thing, though it's slowly coming back to me): http://www.colorado.edu/engineering/CAS ... Lect26.pdf

Or, yea, run the real-life test

Check the following link out: http://www.engineersedge.com/column_buc ... _ideal.htm. Once welded up into a structure, you will effectively have fixed-fixed end conditions, so your "effective length", Le, will be 0.5*L.

You'll need to work through that page, there are different formulas to use depending on the "slenderness ratio" versus a "critical slenderness ratio". When finding the critical slenderness ratio, σy is the yield stress of your column material. Note that I had an error in my formula for I, the area moment of inertia, in my last post, which I have gone back and fixed. You'll have to do a couple of iterative runs to get an idea of the load/length. My hunch is that you will end up as a "short column" setup, which uses Johnson's formula on that page, but you may be able to get away with Euler's after all.

I also found a decent-looking set of lecture notes on this topic that may be of help (it's been a while since I've looked at this stuff. Definitely a use it or lose it type of thing, though it's slowly coming back to me): http://www.colorado.edu/engineering/CAS ... Lect26.pdf

Or, yea, run the real-life test

### Re: I'm back!!

Thanks again Madmattd, Can you suggest a good link where you can choose any size tubing with a grid and I could build it up?

I'm hopeless with computers so I have employed a friend to help me who is very good on the keyboard as also very patient. I will be doing a manual test with the single tube initially so we can compare/confirm and possibly with the built up truss as well.

Thanks for your excellent help so far and I hope I can call on you again later on as I'm certain we will come to some problems along the way.

I'm hopeless with computers so I have employed a friend to help me who is very good on the keyboard as also very patient. I will be doing a manual test with the single tube initially so we can compare/confirm and possibly with the built up truss as well.

Thanks for your excellent help so far and I hope I can call on you again later on as I'm certain we will come to some problems along the way.

### Re: I'm back!!

Do you mean where you can build up the truss and run stress calculations on it? If so no, I think to get that you have to step up to full engineering CAD packages, such as Solidworks, ProEngineer, NX, etc. ANSYS would do the job too. None are cheap, and they have a pretty significant learning curve if you are new to CAD.Dave wrote:Can you suggest a good link where you can choose any size tubing with a grid and I could build it up?

There may be a program out there to at least help you build up a truss to get cut lengths, angles, etc, but I don't know of one to send you to.

Keep us posted, I remember hearing the stories of "endless bridge Dave" but I think you had mostly stopped posting on theHurl by the time I showed up.

### Re: I'm back!!

Might have found a suitable operater... Lives in Thailand and has access to everything I need by the sounds of it... She is a mechanical engineer with a co worker who is structural so should have it covered... I will also be manual testing and comparing to computer models... Starting very soon I hope!!

### Re: I'm back!!

I can't seem to find any information on the present world record for hurling... The following is a reply from Guinness World Records Facebook site but am hoping someone here might be able to give me some information... Surely there is something on trebuchets??... I recently met with a mechanical engineer who said that a FAT Treb holds the present record of 1,600 metres (1mile)...

Hi Dave,

We do have this from our archives - I'm not sure if it's currently recognised by our records team but it has been in the past.

The greatest recorded distance for a catapult shot is 415 m (1,362 ft) by James M. Pfotenhauer (USA), using a patented 5.22 m (17 ft 18 in) Monarch IV Supershot and a lead ball on Ski Hill Road, Escanaba, Michigan, USA on 10 September 1977.

Best wishes,

Dan

Community Manager

Hi Dave,

We do have this from our archives - I'm not sure if it's currently recognised by our records team but it has been in the past.

The greatest recorded distance for a catapult shot is 415 m (1,362 ft) by James M. Pfotenhauer (USA), using a patented 5.22 m (17 ft 18 in) Monarch IV Supershot and a lead ball on Ski Hill Road, Escanaba, Michigan, USA on 10 September 1977.

Best wishes,

Dan

Community Manager