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Cheese Press
#28081
02/08/12 09:09 PM
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Joined: May 2010
Posts: 946
D L Bahler
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I am right now working on building a cheese press. But there is one important matter I do not know how to calculate.
This is a beam style press, relying on the leverage of a large wooden beam to achieve pressing pressure. The beam is fixed at one end, and the other end moves freely up and down. You press the cheese by placing a block under the beam and bringing it down to put pressure on the cheese in the mold.
What I need to know is how to calculate the actual poundage placed on the cheese based on the leverage of the beam. For this particular cheese I need to attain 1500 pounds of pressure.
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Re: Cheese Press
[Re: D L Bahler]
#28085
02/09/12 02:23 PM
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Joined: Apr 2002
Posts: 895
daiku
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Ignoring the weight of the mechanisms, the mechanical advantage is simply the ratio of the distances from the fulcrum to the load and from the fulcrum to the force. So if the press (the load) is 3 feet from the fixed end (the fulcrum), and the lever (the force) is 15 feet long, you'll get a 5:1 advantage. So you'll need to apply 300 pounds of force to get 1500 pounds of pressure.
-- Clark Bremer Minneapolis Proud Member of the TFG
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Re: Cheese Press
[Re: daiku]
#28086
02/09/12 02:44 PM
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Joined: May 2010
Posts: 946
D L Bahler
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Yes, the simple mechanical advantage I understand. But what I need to know is if yoou have a large wooden beam, how do you factor the weight of that beam.
I have a hickory beam that will have 9 feet of working length, and the weight of that working portion is about 220 pounds
considering it a class 2 lever, how do you factor that 20 pounds? because it cannot all accurately be considered a force applied on the far end.
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Re: Cheese Press
[Re: D L Bahler]
#28087
02/09/12 03:49 PM
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Joined: Nov 2003
Posts: 687
Gabel
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Do you have to be that precise? I suppose you are looking at cantilevered beam with evenly distributed load (2.037 lbs/in) formulas or something similar.
I don't really like math that is that involved at this point in my life, so here's what I would do: stick a scale where the cheese goes and let the beam rest on it and see what the dead load is.
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Re: Cheese Press
[Re: Gabel]
#28088
02/09/12 05:14 PM
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Joined: May 2010
Posts: 946
D L Bahler
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It is not really a cantilevered beam.
Here is the design as I have it figured:
At one end there is a support structure with an H shape, one end of the beam is placed under the cross member of this H, which is about 4 feet above the pressing table. A block can be placed under the beam to hold this end up to make positioning the cheese mold easier. The other end of the beam is completely free to move up and down. Again, a block is placed under it to hold it up until you are ready to let it come down.
A wooden block is placed under the beam about 1 1/2 feet from the secured end (a little less) that is long enough to press against the cheese in the form.
Now the beam itself has 220 pounds of dead weight bearing down with some degree of mechanical advantage. I want to know how that transfers into the pressing block so I know how much additional weight I need to add to the end.
I would try the scales approach, but I do not have one that would not be destroyed by the weight of the press.
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Re: Cheese Press
[Re: D L Bahler]
#28089
02/09/12 05:16 PM
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Joined: May 2010
Posts: 946
D L Bahler
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Also, I am not asking for you to figure out how much weight, I just need to know how to figure it out.
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Re: Cheese Press
[Re: D L Bahler]
#28102
02/10/12 11:14 AM
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Joined: Jan 2008
Posts: 918
bmike
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Re: Cheese Press
[Re: bmike]
#28104
02/10/12 02:35 PM
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Joined: Apr 2002
Posts: 895
daiku
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I think for a lever calculation, you can just treat the weight as a point load at the midpoint. 133lbs?
-- Clark Bremer Minneapolis Proud Member of the TFG
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Re: Cheese Press
[Re: daiku]
#28105
02/10/12 03:46 PM
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Joined: Jan 2008
Posts: 918
bmike
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not mine, but a small one. a bit bigger:
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Re: Cheese Press
[Re: D L Bahler]
#29535
08/14/12 05:58 AM
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Joined: Jan 2012
Posts: 92
Stuart
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The trickiest piece of information you need is the load on the beam. Once you have this the process is easy.
You now the load on the beam, you will know or can decide/make up the dimensions of the beam as a lever. From that you can calculate the force transmitted to the top of the cheese. Then you just divide the force by the area that it is being transmitted to.
Does that make sense? If not I'll explain better at length.
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