# Calculating Hydraulic Press Pressure on a Skateboard :)

Hi all,

I wonder if someone can help me with a calculation/estimation?
I have recently built a skateboard press to...press skateboards!
My question is, how do I calculate how much pressure is being applied to the skateboard?

Most presses are steel, however, not having any metal working skills I have made one out of wood.

General advice to press one deck, is applying 2 - 4 tonnes of pressure for x hours

I am using 2 x two tonne bottle jacks and a scissor jack, but as the pressure applied increases the wood starts to give.

Is there a very crude calculation or estimation which can be derived from this - the deck is sitting between the two molds?

Thanks!

berkeman

jrmichler
Mentor
When you say "the wood starts to give", do you mean that the top and bottom cross pieces bend? Does the wood crush under the bolts? What are sizes of the boards? We need some dimensions and a better description of what is happening.

Hi jrmichier,
The surrounding frame is constructed of 4" fence posts.
The frame starts to creak as their capacity to withstand the pressure is reached. I am sure there is a degree of bend in the top cross piece, but it is not instantly noticeable.
I have not noticed an issue with the bottom cross piece (but I admittedly have not looked).
With one's weight behind it (standing up and leaning while pressing down on the jack lever), it would break apart.
The effort to expand the jacks does noticeably increase throughout, but its hard to tell how much effort before it breaks - it did eventually split along the left side of the top cross beam coming from where the top bolt is inserted.

So, did you make any calculations before building your frame, or did you just eyeball it?

I eyeballed it - I knew it would withstand enough pressure to compress the deck to get a solid form (it's 7 x 1.5mm of maple veneer glued), but now wanting to remake the mold, and possibly the frame, I'm trying to determine how far off of the 2 - 4 tonnes I might be. If I'm nowhere near, I could get a fabricator to knock one up - at greater expense than just reinforcing this one.

Baluncore
That is a good start.

The problem appears to be that you are using a wooden outer frame that is in tension.
The bolts that hold it together are thin pins in shear, that are cutting into the softwood. Because the area of the bolt is small compared to the area of the skateboard, the wooden frame is failing at the bolts. The bolts may bend slightly but the compressive strength of the wood is very low.

As a quick fix I would wrap a loop of transport load chain about each jack, over the top rail and under the bottom rail, to make three tension loops about the frame that will oppose the jack pressure. Insert a wedge between the chain and the top rail, or use turnbuckles to pretension each chain.

You might consider gluing the wood, with the bolts only as clamps to hold the joints tight.
That way the area of wood carrying the shear will be much greater.

SCP and Lnewqban
Thanks.
I don't have chain, but guess I could use ratchet straps - if I need to invest more money I would put it into a steel one, so if fixing/reinforcing I need to do it with what I have.

Didn't think of gluing the wood first - would 4" glued faces add sufficient resistance?

Once done - how do I go about estimating (that's all I really need), the force in tonnage being applied?

Baluncore
I don't have chain, but guess I could use ratchet straps
Straps will not work if they stretch before the wood fails. You must use 2" wide straps, pretensioned, and rated to a couple of tons. What sort of tons are they, short, long, or tonne.

Didn't think of gluing the wood first - would 4" glued faces add sufficient resistance?
You have not answered post #2 so we don't know the diameter of the bolts you used. An engineer cannot answer a general question without numbers and detailed information.
If the bolts were 1/2" diameter there would be 0.5" * 4" = 2 sq inches of area.
If the 4" * 4" contact patch was glued, it would have 16 sq inches more, to carry shear forces.
But we don't yet know the bolt diameters.

Once done - how do I go about estimating (that's all I really need), the force in tonnage being applied?
The force applied by the jacks will be up to about 2 + 1 + 2 = 5 of your tons.
The framework will fail before that, depending on the timber and bolts used.

The sides of your frame have a useless 4" post joined with the 4 external plates and bolts. The sides of the bolts will push the end grain out of the vertical posts.
Consider replacing each pair of external plates with more than a full length of plate on each side. Then you can have one bolt in each corner, as is now, to clamp the glued faces.
The spare/other bolt at each corner, can go just outside/beyond the cross beam, with a steel plate or wedge, to bear the bolt force on a greater area of cross beam wood. It is then more than twice as strong as the first design, with the same number of bolts.
Using the longer side plates will keep the outer bolt holes away from the ends of the wood where it could easily split.

Baluncore
Double that.
The 4” * 4” glued face between a side plate and a cross beam makes 16 sq inches in shear, on either side, That is 32 square inches total at each corner. You might get away with one bolt only. If the top beam starts to bend it will tend to shear the glue joint.

Lnewqban
Gold Member
Here is my rough calculation:
You are basically trying to split that top wood beam along its fiber-line at both ends.
Only half of the height of the beam is resisting the force of the jacks.

Also, could you place three supports underneath the bottom shape, rather than two, in such way that those three supports align with the jacks?

Hi all, just to give a bit more information to the above responses before I respond in case it changes anything:

4" Posts
1.75" Plate Thickness
10mm Bolts

Clearer front view

Clearer side view

Clearer rear view

Clearer view of bottom beam - this sits about 5mm off the floor as the bottom of the legs aren't perfectly flush

You can just see where the split is on the right of the bolt

Clearer side view of the split

The wood isn't 'great' but good enough - this split is more through drying out than damage and is not in line with where the pressure is taking place

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The scab plates that connect sides to top and bottom beams appear very short. This puts the bolts quite close to the ends of the side members, increasing the likelihood of a tear-out.

Lnewqban
Lnewqban
Gold Member
Hi all, just to give a bit more information to the above responses before I respond in case it changes anything:
...
You can just see where the split is on the right of the bolt
View attachment 270700

Clearer side view of the split
View attachment 270704
Compare pictures #5 and #6 with pic of post #10: same type of failure.
The connections of the bottom beam to the columns sems to be even weaker than the connections of the top beam.

I would built two inverted U shape out of metal, and would install it in such a way that it runs among the bolt holes and the top of the horizontal beam.
If you don't have metal, screwing some wood among the two plates and the top ends of the horizontal beam, would help.

Even better, extend the four top plates and use two additional bolts to hold another horizontal beam on top of existing.
Perhaps you will need to do the same for the bottom beam.

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Lnewqban - I didn't understand what you meant before, but now you have edited it it makes sense, thanks.
I will probably attempt something like this.
Once this is done, will this allow pressure calculation? I was initially thinking some kind of pressure registering material between two of the compressed surfaces, but maybe an equation will do it?

Lnewqban
Actually, there is no need for the bolts to pass through either horizontal member. If the side members (or the scab plates) extend far enough above and below the horizontal members, the bolts can simply pass across above the top member and below the bottom member. This will have the advantage of putting the shear into the top and bottom members by direct bearing on the full section.

With the bolts passing through the horizontal members, the bolt hole becomes a stress raiser and only allows the lower part of the top member (top part of the lower member) to be loaded.

The talk of pressure seems to be misleading. The actual pressure distribution will be far from uniform in this situation. About all we can hope to describe is the total force.

Lnewqban
With all this in mind - no need to replace the split cross beam?

With all this in mind - no need to replace the split cross beam?
I'd suggest replacing it for best results. Now that it is split, it can no longer support shear on that surface and an intact beam would.

Lnewqban
Hi all,

So, I took it apart, took out the cross beam bolt in each corner and used joist hangers instead.

(I didn't replace the split post due to not being able to source anything other than a full 8' length).

The deck is sitting in between the blue foam sheets.

So, back to the original question - how can I calculate how much pressure is being applied to the deck?

Thanks!

jack action
Gold Member
This guy surrounded the wood pieces with metal straps:

But if it was me - stuck with using wood - I would use screws instead of jacks. This one, from this video, is my favorite:

These guys, from this video, use clamps:

Finally, the simplest of them all (but seems to work):

The advantage of screws is that it is possible to mathematically convert the torque to tighten the screws to the pressure applied. A lot harder (or more expensive?) to evaluate with jacks. Even if it can be difficult to have a precise value, at least you know that the same torque will give the same pressure. Thus, once you have your recipe, you can reproduce it.

Lnewqban
Thanks jack action - I've seen everyone of those videos and many more when I was trying to decide how to do mine

All seemed to have pros and cons, but I think I just had more of the materials to hand to make the one I eventually chose.

The guys from the third image, their skateboard broke after a couple of hours - admittedly, I think they were using 1/32" veneers, but maybe lack of compression played a part?

With the screw ones, the glue only has about a 15 minute working time, so by the time you have thinly coated 12 faces with glue, it doesn't leave a lot of time (if any), to start screwing things down - bottle jacks are much quicker.

Lnewqban and jack action
jack action
I don't see how you can do that cheaply. Available 2-4 tons scale will cost you easily $1000. The idea is to get something of a known stiffness (i.e. a spring) and calculate its deformation while being squeeze between the jack and the frame. But something that deform enough to easily measure it AND that can resist 2-4 ton, that's though to find. I did made my own tool to weight a car once with four human weight scales and lever arms to de-multiply the weight measured ("reverse" mechanical advantage), but I don't see how to apply this for internal pressure with a system like yours. Lnewqban Lnewqban Gold Member Hi all, So, I took it apart, took out the cross beam bolt in each corner and used joist hangers instead. ... So, back to the original question - how can I calculate how much pressure is being applied to the deck? Thanks! That is a very good structural improvement. I believe that we can assume that by the time your hand feels an appreciable resistance from the levers of the jacks, they are delivering a force close to their ratings (2 tons for each of the hydraulic jacks). Let's say the three jacks are able of delivering around 5 tons; however, we don't know whether the structure can stand that combined force or not. Perhaps you could use some deformable gauge located between the two forms and pressured until the structure starts making funny noises. Then, you could measure the plastic deformation of that gauge and try to reproduce that result in a hydraulic press that is equipped with a manometer that tells you the force used to achieve such deformation. If you have the resources, you could also install a manometer to one of your hydraulic jacks and calculate the force via converting pressure to ram force. jack action ^Interesting vids, not sure I want to start down that road though :) I am able to apply more pressure (ergo receive more resistance feedback), with less creaking coming from the press. I've added 120kg to the top beam which is probably helping. I'm sure I could crank the jacks easily enough to break the press, but finding the 'biting' point and determining what percentage is left is obviously tricky. I was thinking there would be some kind of digital pressure plate on the market (although I guess if in existence this would be expensive). I've seen pressure indicating film, but I assume I would need two virtually flat surfaces to get an accurate measurement, and its not really an 'off-the-shelf' product. At this stage I'd be happy to get with a 250kg estimation! jrmichler Mentor I was thinking there would be some kind of digital pressure plate on the market There is, and it's called a load cell. They are very accurate, with typical accuracy specification on the order of +/-0.1%. My experience is that they really are that accurate. Here is a link to a good source: https://assets.omega.com/manuals/LC402.pdf. This is a pancake style load cell, another style is the button style. Typical cost for a good load cell is about$1000, and you need a Wheatstone bridge signal conditioner to use it. They are not protected against overload, one overload can destroy it. Proper mounting is essential.