Fixed Gantry - Trying again

[harryn]

I started to experiment with the requirements of the nearly 7 ft fixed gantry portion last night. Obviously the first step is to make it stiff enough that it does not flex (much) during cutting.

Shown in the pictures are my test setup, a simple dial indicator mounted on a small drill press, to measure the deflection of the test piece. In this case, I used a 7 ft span between the 2 x 4s, and measured a piece of 2x6 in wood and a 2 x 10 pice of Al honeycomb. (that stuff is amazingly light)

The deflection force was me standing on the beam. (around 250 Lbs)

Results
- 2x6 - sagged all the way to the floor even before my full weight was on it, so in excess of 2 inches
- Honeycomb - deflected 0.35 inches

The honeycomb is definitely more stiff, but I still need to improve this by 5x to have any chance of getting reasonable results.

Also shown, is my work bench showing the typical box construction method I like to use. It is just some kiln dried 2x4s in a box shape, with cross members every 16 inches. The top and bottom are 3/4 inch ply screwed and glued to the 2x4s. I am pretty happy with the stiffness of it as a work bench, but it is pretty heavy.

Maybe I should consider to make something similar, but with Al.

[PaulRowntree]

I assume that in the gantry the beam would be vertical, not horizontal as in the test photo. THen the test force of 250 lbs is to simulate the lateral cutting forces? 250 lbs seems very high, I have seen estimates more of the order of 50 lbs being used. Still, I expect your are targeting displacements more like 0.010" or less, not 0.070" (almost 2 mm).

[harryn]

Hi Paul - yes, exactly right. The 250 lbs just happens to be a convenient mass (me).

[OCNC]

Hi Paul - yes, exactly right. The 250 lbs just happens to be a convenient mass (me).

Perhaps convenient but not at all realistic. Using a Bosch router as you currently propose you could never get a 250 lb load at the cutting bit. You'd be hard pressed to hit 100 lbs. (I suppose though that if the bit is stalled and you just crash into the work piece then maybe depending on your motors.)

Chris

[harryn]

Perhaps convenient but not at all realistic. Using a Bosch router as you currently propose you could never get a 250 lb load at the cutting bit. You'd be hard pressed to hit 100 lbs. (I suppose though that if the bit is stalled and you just crash into the work piece then maybe depending on your motors.)

Chris

Hi Chris, thanks for looking at the thread.

Between the cutting and acceleration, 100 lbs is probably realistic enough. 100 / 250 x 0.35 = 0.140 inches, which I am sure you would agree is still way too much deflection for a primary frame member.

That is just the contribution from one piece, not the sum of all of the contibutors of a total system. Of all of the parts of a cnc router that contribute to a good result, a stiff frame is probably the lowest cost area where I can actually make a difference, especially at this early stage.

This test was a good reality check for me, as it is easy for me to underestimate what is really needed to make a stiff frame.

[OCNC]

That is just the contribution from one piece, not the sum of all of the contibutors of a total system. Of all of the parts of a cnc router that contribute to a good result, a stiff frame is probably the lowest cost area where I can actually make a difference, especially at this early stage.

Hi Harry,

I've got to agree with that. I just hate to see something overbuilt if it isn't necessary for the use.

Chris

[harryn]

Hi, several years ago, someone posted a build where the entire gantry went up and for the Z motion, rather than just the Z plate area. I wondered - does anyone know who built it or have a link ?

To help with envisioning it - think of a fixed gantry / moving table machine, where the entire gantry goes up and down. The router went along the Y axis, but did not move up and down by itself on a local moving Z setup.

Thanks

Harry

[harryn]

I went back and re-learned how to calculate area moment of inertia and deflection. I also cheated by double checking my work using engineersedge.com, something we definitely didn't have in 1980. Also, some great vidoes by a prof at IU Purdue.

So, for a decent, first order of magnitued calculation:
- assume 7 ft beam, just freely supported on the ends - not being tensioned at the ends
- 8 x 8 x 1/4 inch
- 6061 Al
- Just assumed 100 lbs for the beam mass
- Assumed 100 lbs for the load on the beam in the center

First order of magnitude calc - 0.003 inch deflection

This is close enough for my router frame, so now I need to start looking around to see what this would cost me, either in Al tube or to make up a tube like structure using honeycomb.

I saw one on-line metals shop quote indicating around $ 300 - 400 for such a tube.

[PaulRowntree]

Hi Harry;
Engineers Edge is a great site, although I haven't found sections and calculators for C-channels or I-beams.
8x8 seems like a HUGE beam. Would a smaller steel tube be an option? 6"x3"x0.25" or even 6"x2"x0.25" steel would have similar max deflections (but perhaps be less fun to drill and tap...). It would still weigh ~100 lbs, and cost about $100
EDIT : If this is for the gantry, then even 100 lbs seems to be high, and may be inflating your requirements. A router and carriage would weigh 20-30 lbs, and any cutting forces would act against this weight. Lateral forces are similar. So you could probably recalculate for say 50 lbs, and get by with a much smaller (cheaper) main beam.

[OCNC]

So, for a decent, first order of magnitued calculation:
- assume 7 ft beam, just freely supported on the ends - not being tensioned at the ends
- 8 x 8 x 1/4 inch
- 6061 Al
- Just assumed 100 lbs for the beam mass
- Assumed 100 lbs for the load on the beam in the center

First order of magnitude calc - 0.003 inch deflection

I just ran your beam through MDSolids. I used the actual weight of the beam based on a cross sectional area of 7.5 in^2 and an aluminum density of 0.098 lbs/in^3. This gives the beam a self-weight of about 65 lbs or 9.3 lbs per ft. The deflection from self weight alone is 0.0006 in. and with a 50 lb point load added to the middle of the beam (the router and carriage) the deflection increases to 0.0014 in. All calculations were done using the software.

Chris

[OCNC]

Hi Harry;
Engineers Edge is a great site, although I haven't found sections and calculators for C-channels or I-beams.
8x8 seems like a HUGE beam. Would a smaller steel tube be an option? 6"x3"x0.25" or even 6"x2"x0.25" steel would have similar max deflections (but perhaps be less fun to drill and tap...).

6"x2"x0.25" weighs 12.21 lbs/ft so the self weight is a little bit higher than the 8"x8"x0.25" square section. The deflection under the same loading conditions is 0.003" in the 6" direction. In the 2" direction assuming a 50 lb cutting load (no self weight) the deflection is 0.0015". Again the software is doing the calculations.

Chris

[harryn]

Paul and Chris,

Thank you very much for running those numbers for me and the suggestions. I agree, I was a little taken back at the 8 x 8 x 1/4 numbers and size.

That starts to push the weight of this thing well beyond a portable, load it in the back of the min van kind of project. If I am not careful, it could quickly spiral up to 400 - 500 lbs.

If it is not too much bother, would it be possible to run one more calc for me?
- 6 x 6 x 1/4 in Al tube
- Unsupported distance of 5 ft (so 2 ft shorter)

- An extra 1/4 in Al plate running the length of it on the router side (will be used to make a dead flat surface for the rail mount)
- One version of this is to have this Al plate on the bottom of the gantry
- The the other version is to have it on the side of the gantry

- 100 lb load vertical and horizontal load (just for internal consistency if nothing else, and 100 lbs is easy to scale from)

My hand calculation (without the extra 1/4 in plate) is less than 0.001 inches, so I am not sure if I messed up.

I even tried a 4 x 4 x 1/4 inch, and at a 5 ft length, the deflection still looks very low - less than 0.001 inch.

What I am thinking, is to change the gantry to allow use a 6 ft beam (since they are easier to get) and only 5 ft would really be hanging in space.

I guess that cubed length term in the deflection equation really plays hard.

I am using (force ) x (length cubed) / (48) x (modulus = 10,000,000) x (area moment of inertia)

Thanks

Harry