Tool Cab: Selecting a strong enough piano hinge?

[Jim Neeley]

I'm envisioning a tool cabinet that could be up to about 5'x5'x16" deep + 3-4" deep doors. This will give me 30"x60" worth of tool mounting space and I'm trying decide how to select piano hinges that won't sag with tools hanging on the inside of each door, or if that size is even feasible. I suspect it'll be more of an issue of torque than weight of the tools and I know how to do the calculations but wonder if (1) is it possible and (2) is it necessary. I have no intuitive feel whether this is a no-problem or a no-go.

Any advice on where to start? :confused:

Jim

[Sean Richards]

Hi Jim

Try this

http://www.spep.com/public/Webpictures/iStore/images/HingeGuide.pdf

[Bruce Haugen]

I suspect that the info in the site that Sean provided will be more than enough to answer your question. My info is merely anedcotal.

My wall cabinet is entirely utilitarian, 40" X 40", poplar, skinned with 1/2" plywood glued into place, 1/2" box joints, with the door units 3" deep. The doors carry about half of my chisels (maybe 25 - 30 - I'm writing this away from home and going from memory), bow saws, Japanese saws, dovetail saws, framing and other squares. It's been like this for about 20 years and shows absolutely no signs of sagging, but as you can tell they don't carry the really heavy stuff like planes, etc.

One major consideration is where you intend to have this cabinet. Where will it hang? Is it above a bench, and if so, how will you get at the tools at the top? This is somewhat problematic for me, as I keep the things I don't use much way at the top, but they're relatively long things like a couple of braces, things that hang down a bit. Your proposed cabinet is 10" taller, so if it is above a bench you're going to have a very long reach. Plus, those doors are going to swing out 30", and you will constantly be moving them to either get at a tool or moving the door out of the way to get at your bench. I have a very small shop space and the wall cabinet is above my bench. Even at 20" wide, I'm always moving those doors. If I move the door on the right and hold it out of the way, then it's blocking a light.

[Jim Neeley]

Sean,

Thank you.. this is the kind of information I was looking for. The units of measure on the tables are a bit confusing, however, so I'm going to throw out a "straw man interpretation" of the table units for others to shoot at. Please, fire away!

The horizontal axis is "HINGE LENGTH IN FEET", whose description is obvious. FRom the table a 1/4" knuckle, a vertical load, with 0.040 cold-rolled steel a 24" hinge handles a "MAXIMIM BENDING MOMENT IN LBS" of 750 units. The confusion is in their labeling of the moment: read as the X-axis label it would be 750#, which makes no sense for moment. Therefore, the assumption would be that the units are in-lbs, even though their examples are in ft-lbs.

The text says "In vertical load, the strength increases with the square of the length". Therefore, with a 48" hinge on a 60" door, the load capacity would be (48/24)^2 * 750 in-lbs = 3000 in-lbs.

Assuming the load to be evenly distributed across the 30" wide door, the yield load would be 100#.

If a 1/2" knuckle, 0.075 material were used, the yield load would be (48/24)^2 * 1250 in-lbs = 5000 in-lbs.


Assuming the load to be evenly distributed across the 30" wide door, the yield load would be 160#.

Is this consistent with your interpretation?

Bruce,

My current thought is to place it on a rolling base 24-30" high and deep enough to hold my jointer plane. This would place the top at 7' to 7'-6" from the floor and along the top would be items such as my braces, cabinet gauge, winding sticks and other long items. It'll go behind me as I stand at my bench but even in my new (fairly large) shop, the 10' wingspan is the force encouraging me to make it more narrow.

Keep in mind that the numbers given are are the upper-end on it's size because those would be the most stressful on the hinge. In the end I'm expecting it'll be more like 4'x4'.

Jim

[Charlie Stanford]

Don't worry about the hinges, worry about the screws. Use good ones.

[Jessica Pierce-LaRose]

Don't worry about the hinges, worry about the screws. Use good ones.

That's been my experience - the weakest part of this equation isn't apt to be the metal in the hinges (although I suppose a very poorly made hinge with a loose fit might make things worse) but rather the attachment to the wood. I haven't much more than glanced at the link in question the other day, but remember piano hinges are used in a lot of other applications besides wooden cabinetry - we have things at work with various piano hinges attached to access panels, in many cases attached to heavy plastic or metal pieces in such a fashion that the hinge very well could be the weakest point in the equation. For a tool cabinet though, I'd imagine the screws are going to start pulling from the wood long before the hinges become an issue.

Besides just replacing the junky screws that came with my piano hinges, since I had the room, I replaced them with longer ones as well.

[Sean Richards]

Sean,

Thank you.. this is the kind of information I was looking for. The units of measure on the tables are a bit confusing, however, so I'm going to throw out a "straw man interpretation" of the table units for others to shoot at. Please, fire away!

The horizontal axis is "HINGE LENGTH IN FEET", whose description is obvious. FRom the table a 1/4" knuckle, a vertical load, with 0.040 cold-rolled steel a 24" hinge handles a "MAXIMIM BENDING MOMENT IN LBS" of 750 units. The confusion is in their labeling of the moment: read as the X-axis label it would be 750#, which makes no sense for moment. Therefore, the assumption would be that the units are in-lbs, even though their examples are in ft-lbs.

The text says "In vertical load, the strength increases with the square of the length". Therefore, with a 48" hinge on a 60" door, the load capacity would be (48/24)^2 * 750 in-lbs = 3000 in-lbs.

Assuming the load to be evenly distributed across the 30" wide door, the yield load would be 100#.

If a 1/2" knuckle, 0.075 material were used, the yield load would be (48/24)^2 * 1250 in-lbs = 5000 in-lbs.


Assuming the load to be evenly distributed across the 30" wide door, the yield load would be 160#.

Is this consistent with your interpretation?


Jim

Hi Jim,

My interpretation of the graph was that the Y axis units are lbf-ft. Ignoring that for a minute and using your first calc - if you have a moment of 3000 in-lbs on a 30" wide door and the load is evenly distributed then the yield should be 200lbs - centre of gravity being at 15".

For a more intuitive (at least for me) indication of the strength of a continuous hinge then using your 48" long 1/4" knuckle hinge as example you have ~192 load bearing points where the cross-sectional area of the hinge pin is 0.0050^2" (assuming pin diameter = 2 X 0.040" leaf thickness) giving a total cross-sectional area of 0.96^2" - pretty strong in other words.