Hi,
It's been a long time since I've been to school, and never did specialize in physics. But I now have a series of many questions that will build on each other I am hoping someone can help me with. I can only ask so man questions in each session as my next question set will depend on previous answers. And I'm looking for general understandings more than actual formulas.
Ok, Here it starts:
Suppose I have a hydraulic jack with a one-inch square piston to keep things simple. And suspended above this jack is a fixed plate.
-1- If I pump up this jack to 100 psi then there will be 100 pounds pressing against this top plate, correct?
-2- And if I place a 1 inch cube on top of the jack's piston, then pump the jack (I will always pump it to 100 psi and keep pumping so it stays at 100 psi unless I state otherwise) even if the cube is crushed slightly, there will still be 100 pounds of pressure squishing the cube and 100 pounds of pressure pressing against the top plate, correct?
-3- And if I put two cubes, one on top of the other, on top of the jack and pump it, each cube will be being squished by 100 pounds of pressure, and 100 pounds of pressure will be pressing against the top plate, correct?
-4- And if I put a 1 in. by 3 in. plate on top of the jack, (and assume the plate will not bend or fall off) and place the 1 in cube on top of this 3 in plate, pump the jack, then regardless where I place the cube on this plate, there will be 100 pounds of pressure squishing this cube against the top plate, is this correct?
-5- But if this is correct, then every sq in of that 3 in plate has a "squishing" potential of 100 pounds, correct?
-6- But if that's true, then without the cube, will that 100 psi jack be exerting 300 pounds of pressure against that top plate simply because I put a 3 in plate directly on top of the jack's piston? If true, it seems like we're getting more pressure from no-where with no additional jack pressure "work".
-7- And if I put two 1 in cubes directly on top of the 3 in plate, next to each other, will each cube be squished by 100 pounds or will the pressure spread and each cube be squished by 50 pounds? And the same question if I put 3 one-inch cubes on top of the plate - 100 pounds squishing each cube or 33.3 pounds? And the same question with one cube 3 in long.
-8- And will evenly stacking more cubes on top of each other change the answers?
I have lots more variations of questions on this "pressing" situation but I need to see the answers before I know which way to go with my next set of questions.
Please keep your answers very simple.
And thank you much for your help and education.
steve
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Ah the engineer in me is pulling out the calculator...
6ton * 2000lbs/ton = 12kip force
12kip / 4 = 3 kip (double shear on 2 bolts = 4)
(3/4)^2*pi/4 = .442 in^2
3kip / .442 in^2 = 6.8ksi
36ksi * .4 = 14.4ksi allowable based on A36 steel rod
6.8 / 14.4 = .472 < 1 therefore you'd be fine.
The only problem I see is with bearing... if the bearing area is too small you may see some localized yielding... which might bugger up the rods beyond use after a while...
3 kip / 14.4 ksi = .208 in^2
.208 in^2 / .75in = .277in
This means you need to support the bolt for a minimum of .277" each place it is in shear on both sides of the shear plane.
That's what I'd do..I'm sure someone could argue something different. But my 12 ton only has 3/4" bar, although that is higher strength material than A36 which is pretty much any run of the mill steel. So go with some 3/4" bar and you should be fine...or upgrade to some nice alloy steel and never think twice about it.
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