Trebuchet – Sizing the Axle (Part 2)

7 11 2008

Ok, so in Sizing the Axle (Part 1) we found out how to determine the forces that act on the main axle.  We want to determine the appropriate size for the main axle so that it doesn’t deform the main axle.

The first step is to draw a free body diagram of the forces acting on the main axle.  Figure 1shows the two forces from the A-Frame acting on the axle (pointing up) and the force from the throwing arm acting on the main axle (point down).  The Force the Beam exerts on the axle is the 2048 lbs force that we calculated in Sizing the Axle (Part 1).  For a quick reminder

W = 2*W*(1-cos(θ)) = 2048lbs.


Figure 1

So the axle will fail if it gets too much stress in it.  So we need a way to determine the stress inside the axle.  From Machinery’s Handbook V27 page 261 we can see that the stress in a beam under this type of loading can be found by

stress = -WL/4Z     (I)


W is the load on the beam.

L is the length of the beam (in our case

Z is the section modulus of the cross-section of the beam

Wait ….what?  What does that mean?  That doesn’t mean much to me.  However it’s easy to find out.

Z is determined by

Z= I/c    (2)


I is the moment of Inertia.

c is the distance from the neutral axis to the extreme fiber

The moment of inertia can be found using integrals, but we are just going to find it in a book.  See figure 2.


Figure 2

So we need to know the deminsions of the axle, but that’s what we are trying to find out.  Well, it can get pretty difficult to find out what those values are directly, so I suggest picking a size and them following this guide.  We were lucky enough to find some pipe dumpster diving.  We found some schedule 40 3″ pipe. Now 3″ pipe is just a name. The actual dimensions are as follows

So now knowing the inner and outter dimension we can determine the moment of inertia for our scavenged pipe.

I = π(D^4-d^4)/64 = π(3.5^4 – 3.068^4)/64 = 3.129in^4

and in this case the distance from the neutral axis to the outer most edge is

c = D/2 = 1.75 in


Z = I/c = 3.129/1.75 = 1.788 in^3

We know have all of the information to calculate the stress in our axle.

Stress = -WL/4Z = -2100*32/4*1.788 =~ 9400 lbs/in^2 or 9400 psi.

Ok, so we found out what the stress is in the pipe. The next step depends on the material that your pipe is made of.  We know that it is some type of steel, but there are a lot of different types of steel out there each with there own mechanical properties.  So we are going to use a very low grade steel.  We are going to assume that the allowable bending stress is 30,000 psi.

So the stress in the pipe is much less than 30,000 psi so we should be good, but we need to calculate what the safety factor is.

Safety Factor = Allowable bending stress/ calculated bending stress = 30,000/9400 = ~3

I must say that the force on the beam needs to be readjusted to include the mass from the beam.  However, we don’t have the exact number but we do have an estimate.   I went ahead and recalculated the force and it comes out to be about 2400 lbs (rounded up).

So the stress comes out to be

Stress = -WL/4Z = -2100*32/4*1.788 =~ 9400 lbs/in^2 or 10800 psi.

And our new Safety Factor is

S.F. = 30000/10800 = ~ 2.7

That’s still is pretty strong.  It’s important to have a large safety factor because you can’t account for everything!  So be conservative in your calcuations.  If your calculations don’t have a large enough S.F.  pick a larger pipe size and start again, or use a very strong type of steel.

Other Resources

Trebuchet Design

Cement Blocks


Sizing the Axle (Part 1)

Sizing the Axle (Part 2)


Car Throwing Trebuchet

Cool Trebuchet Pictures

The bucket


Compressed Air Rockets (Part2)

13 09 2008

So what this project is missing is two things. The first thing and most obvious is paper rockets. We’ll cover that later. Right now I wanted to describe how to make a detonation box or the ignition switch.

Most of the parts can be found at Radio Shack.  The part they may not have is the illuminate momentary push button, but they have plenty of buttons that aren’t illuminated.  The parts that are needed are detailed below:


9 Volt Battery

Single Pull Single Throw Switch

Momentary Push Button (LED illuminate if possible)

Binding Posts

Solderless Banana plugs

Altoids tin (or any other brand or some sort of enclosure)

This is what the final design will look like.

Note:  The binding posts are not shown in this picture.  That was thought up after the picture.  When I get the chance I will update the picture and take some pictures of the guts.

Compressed Air Rocketry (Part 1)

12 09 2008

Compressed Air Rocketry is well….just that rocketry using compressed air.  I got this idea from my friend Blakestradamus.  He has build these before and is quite the wizard.  He was going to build one for some cub scouts that were coming over to his house to investigate engineering careers.  I told him that I wanted to build one and that we could be two for twice the cost!  What a deal!

I’m going to try and detail everything that I did to assemble the Compressed Air Rocket Launch Pad.

All of the materials can be picked up at your local hardware store.  They should have all of these parts, but if not you can alter the design to the parts that are at your disposal.


2″ PVC Tubing  – about 5 feet.  This will be cut into 3 different pieces.

2″ PVC End Cap – 1

2″ 90° Elbow – 2

2″ coupler – 1

2″ to 1″ Reducer – 1

3/4″ slip / 1″ threaded joint – 2

1″ Sprinkler Valve – 1

1″ 90° Elbow / 1/2″ threaded  – 1

1/2″ thread/slip joint – 1

Compressor air hose connector – 1

Other materials not shown

PVC primer

PVC glue/cement


NASA Paper Rockets

Rocket Templates

Air Power Rockets