# Allan Keys

## Main Question or Discussion Point

Hi guys

I'm attempting to calculate the pressure you can apply with allan keys. But the last time I did any physics was about 12 years ago, so rusty would be an understatement!

I'd appreciate it if you could point me in the right direction equation wise.

Cheers

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Danger
Gold Member
Welcome to PF. There are a lot more variables here than you might realize at first peek. You have to know not only the length of the lever arm, but also the material that the wrench is made of and how large it is. Some will bend, break, or strip out long before the screw is torqued to your requirment. Others can apply so much force that the screw will shear off or strip out.

Assuming you are more interested in the physics perspective, than the (also interesting) material science perspective, the force on the point of contact between the allen wrench and the screwhead is proportional to the length of the wrench, while the screw-in-out component of force is proportional to the cosine of the pitch of the spiral threads on the screw.

Danger
Gold Member
Good response, Crosson. Your wording brought another thought to mind, though. If the wrench is not a perfect fit for the screw (ie: not all flats in proper contact), is there a way to determine the load on the 'corners'? That, in my experience, seems to be one of the main causes of hex key failure.

berkeman
Mentor
I went ahead and moved this to the ME and Aero forum. Seemed like a better match -- hope that's okay with you folks.

What? :surprised

Currently I'm not worried about the material side of things, lets just assume both the key and the screw are made of some infinitely strong material. I'll worry about breaking screws and stuff when I get my head round the torque bit!

Lets walk before I can run.

the force on the point of contact between the allen wrench and the screwhead is proportional to the length of the wrench
So what your saying is the force applied is related to the length of the key - leverage, ok that makes sense.

while the screw-in-out component of force is proportional to the cosine of the pitch of the spiral threads on the screw
Does this mean, the torque out from the key to the screw depends on how tightly threaded the screw is?

Can I look at the equation that explains this?

FredGarvin

Are you trying to calculate how much torque you can apply with an allen wrench or are you trying to calculate how much torque the wrench itself will be able to take? Perhaps you are trying to find out how much clamping an allen head screw can produce?

By your answers to the last question it seems like you just want to know about how much torque you can provide via the wrench itself. It really is unclear.

Yes your right, all I want to know is how much torque I can apply with an allen key.

FredGarvin
In that case, the torque you can apply to the head is the basic equation for a moment which is a M=f*d where

M - torque (moment) in in*Lbf
F - Force applied to the end of the wrench in Lbf.
d - perpendicular distance of force to center of rotation in inches.

Of course, knowing the nature of allen heads, the wrench will likely bend or give out before you apply an appreciable torque. Hope this helps.

Yes it does, coulple more questions now!

Is Lbf - pounds per foot?

How much force can a normal person apply to the end?

Maybe this one is more a round about figure than a value set in stone.

"Thou shalt not apply more than 100Lbf of force to an allen key", doesn't really cut it as the 11th commandment!

Let me ask this: Shouldn't a machinist's handbook be referred to to determine how much torque should be applied to a particular type screw?

If you assume an infinite strenth material for both wrench and screw, wouldn't the amount of torque possible be infinite?

berkeman
Mentor
Yes it does, coulple more questions now!

Is Lbf - pounds per foot?
No. Lbf is pounds * foot, or commonly called "foot pounds".

How much force can a normal person apply to the end?
Depends on how strong they are, and how they can brace themselves. On a little Allen key, you could probably apply a few pounds of force before things start to bend. On a 24" breaker bar, you can easily apply a steady 50 pounds or so to the end.

FredGarvin
Whoops. My bad. I always specify Lbf (pounds force) and Lbm (pounds mass).

Of course we would have to look at material constraints. I was simply answering the OP thinking that his limiting variable would be how much force a person could apply by hand.

berkeman
Mentor

FredGarvin said:
M - torque (moment) in in*Lbf
Doh! That'll teach me to jump into the middle of a thread without reading more carefully.