Metal Sheet/Plate needed for Cider Press - Which type of metal/which thickness?

[bfbf]

Hi All,

I have recently built a cider press (pictured and videoed below). It has had one year of operation and was successful however whatever we put between the jack and the apples snapped/broke/bent (Including Oak, Break disks, plastic and a combination of the aforementioned). Which is not great. This year we used a 10 ton jack, I do want to upgrade to 20 tons next year though and I need something to go between the jack and the apples!

The base of the jack the new jack would be a bit bigger around 14cm by 15cm (that is the smallest it would be). The plate of metal would need to be 47cm by 47cm and have the jack on top of it and not bend too much.

I have no idea about which metal to use - which is strongest for the type of use I am using it for and how thick it should be or even where to get it from - this is where I thought you guys might be able to help? I am based in West Sussex.





Thanks for any help!

[PinkertonD]

Wow, nice job so far. The press looks good and the attention to detail on the woodwork looks first class. I wish all first time posters provided as much detail at the get go.

Doubling the jack pressure is going to take you into a new realm of pain I suspect. Not insurmountable, but at a guess after watching the video, a lot of stuff is going to need to be made thicker and stronger. That will make a lot of it heavier and harder to handle.

Fortunately by it's clever design, the press has great availability for adjustment to accommodate an increase in height of the "load." At this point, just looking at the press, I am guessing that the beams (RSJ's) will handle the 20-tons, but it would pay to check the bending load using the calculators here on this site by plugging in the real figures.

The "plates" beneath the catch tray will need to be as strong as the "plates" at the top with the jack. Since you are isolating the foodstuff from the metalwork with what looks like Polyethylene or Polypropylene (plastic) sheets the choice of metal is not so critical. In that case, common old Mild Steel hot-rolled plate will do fine.

Once again, using the Calculators on this site, (start here... https://www.engineersedge.com/calculators.htm )you can determine the workable sizing for the metal. I would start with 1/2" thickness given that the cost difference between 3/8" and 1/2" steel plate will be minimal. You will need one plate at the top and another under the catch tray. The disk brake rotor is a good idea as it distributes the load quite well during the pressing.

On another issue, I am wondering why you do not use the screw portion of the jack when you first run out of hydraulic movement. Releasing the pressure then unscrewing the top and hydraulically press again. Screw the top back in, place another packing piece (disk brake etc) and go again. Every second "push" will be using just the screw portion rather than another packing piece.

[bfbf]

Thanks for your detailed reply.

I know the RSJs will cope with the pressure upgrade, theoretically so should the uprights, not sure if it is something I want to risk though. The base is very unlikely to break as the pressure is even and there is mainly RSJ support underneath. It did not seem to bend at all with 10 tons on it.

I have worked out that the apples in theory should push up 9kg per square cm, then in the center there will be 94kg per square cm downforce. It seems very hard to put that into any of those formulas as I guess it is not something traditional to calculate. Are there any you could suggest I use.

I was thinking of using 12mm mild steel maybe with some cross supports some how? Again it would be hard to calculate the cross supports but I guess if it was possible to calculate that the 12mm steel plate would bend by 20mm, the assume the supports would make it better.

On another issue, I am wondering why you do not use the screw portion of the jack when you first run out of hydraulic movement. Releasing the pressure then unscrewing the top and hydraulically press again. Screw the top back in, place another packing piece (disk brake etc) and go again. Every second "push" will be using just the screw portion rather than another packing piece.

The reason for that is that without the break disk the wood broke and without the bit of metal on top of the break disc the break disc broke! Hence we had to get things in to spread the load as quickly as possible! When we have this metal plate and a better Jack the screw will be used far more.

[PinkertonD]

You (we) always design for the maximum pressure that can be applied rather than what it takes to squish the apples. So, 20-tons is the figure you need for the load. The apple load can vary dramatically because of many factors. Not worth worrying about, just use the 20-ton figure since you know 10-tons works.

Since it is a simple issue to calculate the beam strength using the website calculators, I would strongly suggest you plug in the appropriate figures for your beam and the threaded rod. Although I doubt a failure will be all that dangerous to human life, but may take off a finger or two. The noise will probably do more long term damage. Please do not guess since the real information is at your finger-tips.

The beams are no longer called RSJs but instead new profiles (shapes) and are now called UC, Universal Column or UB, Universal Beam. Yours looks to be UC where the flange width is almost the same as the beam height. You will need the Moment of Inertia for a UC but it is being used in a Beam mode, supported at each end and a load (20-tons) in the middle.

I suspect 12mm flat plate should be OK without stiffening strips but you could add (arc-weld) some 10mm x 25mm strips welded on edge, radiating out from the center in a modified "X" fashion (pic coming). Both top and bottom plates would need to be the same as the 20-tons is transferred equally through the entire system. Top plate, mushy stuff, tray, bottom plate.

[zeke]

To give you an idea, I conservatively estimated for plain 12mm steel plate, the maximum deflection under 20T would be about 16 mm. If you double the thickness you would get 2 mm (probably overkill)
Don,t know whether you would bother reinforcing (notwithstanding a good pattern suggested by Dave)

[bfbf]

Thanks for your replies - I think what I will do is go for a 12mm sheet then if need be weld some additional supports on. I will let you know how much it bends (if any!).

I will also ask for offcuts while at the metal place which will be a bit bigger than the base of the Jack being that it should spread the load a bit more.

Do you know how much I should expect to pay for a 50cm*50cm*12mm Mild Steel plate?


Regarding the 20mm threaded rods - it seems each has a tensile strength of 5 tons - so 20tons would be too much (assuming that it won't go perfectly evenly. I may look for some higher quality 20mm threaded rods, and perhaps some slightly longer ones. I suspect it may be cheaper to do that than have the holes expanded (my drill does not work through RSJ) and buy all new bolts of 25mm standard threaded rod.

[PinkertonD]

Can't help with prices in UK but metal suppliers here generally have off-cut bins where steel is usually around $0.40/lb.

I would seriously consider going to 25mm threaded rod straight off. But first use the calculators here to make sure that will do. You can open the holes up with a round file. Build a thirst for the Cider.

If Zeke says the plate will bow 16mm, you can rest assured it will bow 16mm. He is not as lazy as I am and will have checked the figures for you. However, a 16mm bow is probably not a bad thing and it should spring back from that. As you suggest waiting to see if you need to add stiffeners is a good idea.

[bfbf]

Cheers

I would seriously consider going to 25mm threaded rod straight off. But first use the calculators here to make sure that will do. You can open the holes up with a round file. Build a thirst for the Cider.

A round file, filing 5mm off 16 holes - not sure there would be any cider left by the time I had finished! Surely that would take days?

[PinkertonD]

A round file, filing 5mm off 16 holes - not sure there would be any cider left by the time I had finished! Surely that would take days?

All in a worthy pursuit. The other option, opening up large holes with a conventional two-flute drill is a recipe for lost digits and maybe even limbs unless you can have them done in a machine shop on something like a radial drill. The drill will grab the metal incredibly fast and dangerously and swing the drill-meat (you) around to break a wrist or two. On a radial drill, the metal can be firmly clamped to the machine-base and the drilling head locked in place before attempting metal removal.

Some night-time reading while you still have limbs attached and digits to tickle the keyboard.


You may get lucky and be able to borrow a multi-flute (maybe 4, 5 or 6) 26mm drill that will mitigate the problems.

Let me just mention, that many, many years back, I began my Engineering career in a machine shop as an apprentice Toolmaker. I saw my fair share of body parts removed, moved, reshaped and/or lost to a finely ground bloody mist. I fortunately, still have all my first-issued limbs, extremities and digits, but I saw quite a few people lose parts when not so wise as to listen to sound advice.

[bfbf]

I am trying to calculate how many tons each bar would be able to take if I brought some high tensile m20.

m20 (8.8) 1.5m span tensile strength 870 N/mm2

m20 (10.0) 1.5m span tensile strenth 1000 N/mm2

Originally I tried converting the Netwtons to Tons, then times it by the area but it came out at about 32 tons per bar - which i think is wrong

I then tried this calculator, but had no idea what I would put for a v or t. http://www.engineersedge.com/materia...calc-round.htm

Any pointers would be appreciated

[Kelly_Bramble]

I then tried this calculator, but had no idea what I would put for a v or t. https://www.engineersedge.com/materi...calc-round.htm

Any pointers would be appreciated

The calculator you used calculates torsional stress..

For the volume of a cylinder use: https://www.engineersedge.com/calcul...calculator.htm

All you need then is the weight per cubic foot - normally obtainable from the supplier or a generic material specification.

Do you know the material specification or other?

[RWOLFEJR]

A cabide burr would wallow the holes out quickly. They can be had pretty cheap and your drill would do the trick. Might have to stop and let your drill cool now and then.

Since you seem to have an idea of how much gap you're happy with in this press... and it really isn't critical... Why not bolt or weld solid uprights and lose the all-thread? Eight good 20mm bolts and nuts will be less expensive than high strength all-thread.

Another thing picked up on watching another fellows press was that with a 20 ton jack he couldn't and didn't need to utilize all of the tonnage. When he approached full power he said the cheese cloth started to rip.

If you're going to buy a new bigger jack maybe consider an air over hydraulic to save yourself all the work? Looks like 20 ton can be had for under $100... and saw a new one for auction at less than 50. If you have an air compressor it'd save you a lot of work. Maybe consider fixing the rod end of the jack to your top beam to save lifting it up there every batch.

Far as the plate goes Dave's suggestion of the "modified X" or honeycomb stiffening could allow you to get away with lighter material to put on top of your stack of apples too.

[bfbf]

Hi,

I have been trying (and failing) to work out how much a threaded rod can take load wise.

It is m20 8.8 High Tensile Strength and has a tensile strength of 870 Newton Metres (870 N/mm2). The distance between the RSJs would be 1meter with 50cm of threaded rod to go through the RSJs.

I have been trying this formula https://www.engineersedge.com/materi...calc-round.htm (Cylinder Stress and Deflection with Applied Torsion)

But I am missing a, t,e and w

Any pointers you could give?

[PinkertonD]

Yup, look up the tension and shear strength calculator for threads. Your threaded rod is all in tension so tensile failure is the thing to work on.

The distance between nuts and beams is of no importance as the thread will fail in tension at the root (smallest diameter) of the thread.

[bfbf]

I could not find any threads about that.

I did the following maths.

Area in mm^2 is 314.15mm^2 per bar (10^2xpie)

= 0.486933474 square inches inches

870 Nmm2 = 126,200psi

126200psi*0.486933474 = 61,451 pounds per bar which is 30.73 tons per bar?

Seems wrong

[Kelly_Bramble]

I could not find any threads about that.

I did the following maths.

Area in mm^2 is 314.15mm^2 per bar (10^2xpie)

= 0.486933474 square inches inches

870 Nmm2 = 126,200psi

126200psi*0.486933474 = 61,451 pounds per bar which is 30.73 tons per bar?

Seems wrong

Well - no...

Assuming the material is Carbon Steel, like ASTM 1020 or equivalent, the materials should weigh approximately .28 lbs/in³

So, assuming that each bar is 3 ft long (36 inches).

Weight = [0.486933474 in² x 36 in.] x .28 lbs/in³

Weight = 4.9 lbs per rod.

That "126,200psi" number sounds like the the material strength specification - not the density...



For material weight numbers (generic and not exact), see:

https://www.engineersedge.com/manufa..._materials.htm

[bfbf]

Thanks - I was trying to work out the materials strength - how strong it is if it was being pulled at each end type thing. 32 tons sounded more than expected

Is the weight important to calculating that?

[Kelly_Bramble]

Thanks - I was trying to work out the materials strength - how strong it is if it was being pulled at each end type thing. 32 tons sounded more than expected

Is the weight important to calculating that?

Oops - sorry I thought for whatever reason you where calculating weight per rod.


The math when pure tension is applied is simply the thread tensile area (or the smallest cross sectional diameter) multiplied by the modulus of elasticity to get the theoretical yield strength or force.

tensile-stress.png

Thread tensile area can be calculated with:

https://www.engineersedge.com/thread...tress_area.htm

So, your calculation of:

126200psi*0.486933474 = 61,451 pounds per bar which is 30.73 tons per bar

Is correct for yield....

You should apply a FOS - of about three (3)

See: https://www.engineersedge.com/analys...ety-review.htm

So, your working load should be around 20K lbs per rod - if the material spec is right.

[PinkertonD]

I meant you to use the Tensile area Calculators and from there use the Tensile Strength for the steel the threaded rod is made of.

Material Tensile Strength x Tensile Area = ...

[bfbf]

Thanks Kelly :-D

I meant you to use the Tensile area Calculators and from there use the Tensile Strength for the steel the threaded rod is made of.

Material Tensile Strength x Tensile Area = ...

Ye I worked that out, was looking at fourm threads originally. But the good news is I worked it out. Got a slightly lower number than Kelly, but was still over 30 tons per bar.

I got this number 0.486933474 as 0.47559 (Which I guess is because that is right after taking into account the thread).