Help me understand PSI (pounds per square inch)

May 14, 2013 1:46 PM

When this c-clamp is rated at 6900 psi, does it really provide more pressure on a part than a 3-ton car?

So I want to be able to apply pressure on some parts I am working with, similar to what a wood vise does, but on a larger scale - about 150 square inches.

Currently I am just using a bunch of heavy concrete blocks - 4 of them, at about 50 pounds each - and laying them directly on the parts. Works fine.

Then I read up on a "paper press", which seems to be basically two blocks of wood that can be squeezed together. The claim was 2000 psi could be provided. Seemed like a lot, but I decided to see if I could make my own, rather than shelling out the $400. So my idea was to use a bunch of c-clamps to press the wood together to provide the pressure.

But I'm having a hard time believing that squeezing the wood using the clamps will provide more pressure than the concrete blocks I am currently using, or a 3-ton car for that matter, since the c-clamp is rated so high.

Now I know that the 6900 psi is probably on the 1 square inch where the two tips of the clamp meet, that this number is a factory tested value and not reflective of real-world applications, and that the further away one gets from the c-clamp tip touch point, the less pressure there is.

But if I'm dealing with a 10 x 15 in surface, and have clamps equally dispersed throughout the surface, I'm still finding it hard to believe that the pressure applied by these clamps can be that much or even half of that.

So my question is 2-fold: What does this c-clamp's rating actually mean? Does applying clamps throughout the wood surface really provide more pressure than my concrete block method, or for that matter, parking my car on top of it?

Thanks!

posted by bitteroldman to Science & Nature (14 answers total)

Can't help on the specifics of the clamp rating, but it'll definitely be greater pressure than your car. The pressure imparted by the car from its weight gets distributed over the contact patches of the tires. When you do the math, the pressure you'd get from your car ends up being the same as the tire pressure, about 30psi.

Putting what you'd want to squish under jack stands holding up the car, though, would do a better job
posted by hwyengr at 1:55 PM on May 14, 2013

The compressive strength of wood varies of course between different types of wood, but 6900 PSI is more than a lot of wood can withstand. I have no problem believing you could crush wood using a clamp, but not using cinderblocks or a car.

So, you want to use enough clamps that you don't have to put enough force on any one of them to damage the wood.
posted by aubilenon at 2:04 PM on May 14, 2013

The clamp has a Capacity Clamping Pressure rating of 6900lbs and the pads look to be around 1 square inch in area so 6900 PSI seems reasonable. How much force you would have to apply to the screw to get that much clamping force will depend on the friction of the mechanism and the pitch of the screw. You might not be able to obtain the rated clamping pressure.

Keep in mind that clamping pressure is dependant on surface area normal to the clamping force modified by the stiffness of the clamping surfaces. So if you were to take a small ball bearing and place it between one of the faces of the c-clamp and the work surface you could obtain 10s of thousands of psi pressing on the work but concentrated in a very small area. If your book press is 100 square inches in size and it can apply 2000 psi to all those inches that is the same as having a c-clamp capable of 200,000 PSI.

PS: this is why vacuum clamping can be so effective. Even though air pressure is only around 15PSI if you have a 2'x2' surface pulling a vacuum between the surface and the object you are clamping to results in 8500 lbs of pressure holding the objects together.
posted by Mitheral at 2:39 PM on May 14, 2013

That 6900 pound value is the load rating of the clamp and should be in pounds force, not pounds per square inch, despite the text on the website.

This is how much force you would have to apply before the body of the clamp bent or broke, or the screw collapsed.

The real limiting factors will be how rigid the boards you are clamping are, how far apart they are, and how hard you can twist the handle on the C-clamp. I wouldn't be surprised if you could generate about 1000 pounds force with an 8-inch C-clamp. And don't get those copper screw ones; they are weaker and more expensive and meant for welding applications because molten steel spatter won't stick to the threads.

So as a practical matter: yes, you're going to be able to generate a lot more force with clamps than with dead weight.
posted by Kakkerlak at 2:41 PM on May 14, 2013 [3 favorites]

Scaling that picture in Photoshop, it looks like the clamp's contact pad is a 1" circle, and the screw is about 6 threads per inch. The handle, if slid all the way to one extreme, sticks out about 5" from the centerline of the screw.

The area of the contact patch is ~0.785 square inches. 0.785*6900=5416.

The 5" long handle will follow a 10" diameter circle as it goes around. The circumference of a 10" diameter circle is 10*pi=31.4"

So, for every 31.4" you move the end of the handle, the screw advances 1/6 of one inch (0.167").

31.4/0.167=188.02, meaning that if you ignore friction, the screw multiplies the force applied to the end of the handle about 188 times.

5416/188=28.8 pounds, which is how hard you have to press on the end of the handle, in our magical frictionless explanatory universe, to develop 6900 PSI at the clamp pad.

BUT, I agree with Kakkerlak that the rating is probably more about what it takes to break the clamp, not the force it can practically develop. Also, friction is real, and it matters quite a lot.
posted by jon1270 at 3:25 PM on May 14, 2013 [2 favorites]

Keep in mind that clamping pressure is dependant on surface area normal to the clamping force modified by the stiffness of the clamping surfaces.

In fact, I would expect that any good C-Clamp would be able to put a deep dent into a piece of wood. Once the wood starts denting, any extra force applied by the clamp isn't transmitted to what's under the piece of wood, but instead goes into deforming the wood fibers.
posted by radwolf76 at 3:27 PM on May 14, 2013 [1 favorite]

Once the wood starts denting, any extra force applied by the clamp isn't transmitted to what's under the piece of wood, but instead goes into deforming the wood fibers.

Softwoods start to compress at well under 1000 PSI perpendicular to the grain. Even very hard domestic hardwoods, like hickory, can't take much more than 2000 PSI. Parallel to the grain, both hardwoods and soft are much stronger.
posted by jon1270 at 3:35 PM on May 14, 2013

True, jon1270. I guess it'd depend if the 10 x 15 piece of wood the OP is using is actually a slice of log. :-)
posted by radwolf76 at 3:38 PM on May 14, 2013

Imagine you had two perfectly flat, perfectly ridged plates and you put whatever you want to clamp in between those plates and torqued that clamp down upon the whole assembly. Assuming the C-clamp pad is a about 1 square inch, 6900 divided by the area of the smaller of the two plates would be your clamping force.

In real life, you're not going to have that. Instead, you can either make clamping cauls to distribute the force of the clamps over a larger area, or, if you're going to be doing this sort of thing a lot, require careful alignment and don't have a friend to give you an extra set of hands, build something like a veneer press.

If you don't believe a c-clamp can generate that kind of force, think about how hard you have to pull on a 1/4 steel bar to rip it in half. Now go overtighten a 5/16 bolt and see what happens. (Do this in some sort of contrived system that is nowhere near an engine block. Don't ask how I know this.)
posted by Kid Charlemagne at 4:42 PM on May 14, 2013 [1 favorite]

Response by poster: lot of interesting information here, thanks!

i was actually going to use a piece of softwood; didn't consider that after a given amount of pressure, it would start to deform.

and i guess pressure !=weight, which is why parking a car on top of the wood wouldn't necessarily be effective.
posted by bitteroldman at 4:58 PM on May 14, 2013

Response by poster: actually, scratch that pressure !=weight comment - i was oversimplifying
posted by bitteroldman at 5:16 PM on May 14, 2013

and i guess pressure !=weight, which is why parking a car on top of the wood wouldn't necessarily be effective.

That's right -- the weight of a car would be spread out over a relatively large area (the contact area of the tire, divided by four) while the PSI of the clamp is only spread over the small contact area of the clamp itself -- so even though it's a comparatively small amount of force, it will have a higher PSI as the force is concentrated on a tiny area.

actually, scratch that pressure !=weight comment - i was oversimplifying

No, that's actually perfectly correct; pressure is not equal to weight. Particularly when measured in per-square-inch units. A million pounds spread over a million square inches is 1 PSI. A million pounds spread over a single square inch is a million PSI.
posted by ook at 5:29 PM on May 14, 2013

On my tablet so not much typing, but look up clamping cauls, they distribute clamp pressure evenly, unlike solid blocks.
posted by seanmpuckett at 6:45 PM on May 14, 2013

You can still use the softwood and C-clamps, just don't go crazy when you tighten them down.
posted by Kid Charlemagne at 9:10 PM on May 14, 2013