Quote:
Originally Posted by CMW
...The moment (or torque in our case) along the beam is given by the linear equation M(x) = P * (L-x) where M is the moment, P is the force applied at the handle, L is the overall length of the beam, and x is a distance from the cantilever fixed end (the bolt in our case)...
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Very good, let's follow up on that.
We start by recognizing that the torque wrench plus the extension, represented here by the line between M and P, is what we call a rigid body. Yes, it pivots a tiny bit at the head (represented here as a point X inches to the right of the nut at M) in the process of measuring torque. But not much, so for our purposes we can assume that it is completely rigid. If we were evaluating its dynamic behavior it would be different, but we're not so it doesn't. When I'm flying I care about wing flutter, when I'm rock climbing I care about gate flutter, but I don't hardly ever care about wrench flutter.
Now, let's say we are using a torque wrench that is 36" from handle to drive lug. When we add the 3" extension to it, the system length L is 39". In feet, that is 39/12 = 3.25 feet.
Let's further say that we are looking for a torque at the nut of 100 ft-lbs. So let's solve our equation for P and find out how much force is required at the handle:
We start with:
M(x) = P * (L-x)
Solving for P and plugging in the numbers:
P = M(x)/(L - X)
= 100 /(3.25 - 0)
= 100 / 3.25
= 30.8 lbs
Let's run that back through the original equation to validate it:
M(x) = P * (L - x)
= 30.8 * (3.25 - 0)
= 30.8 * 3.25
= 100
So, we know that if we take a 39" bar and apply 30.8 lbs at the handle, we'll get our target torque at the nut.
Now, let's use the original equation again to see what the torque wrench reads when we are applying that 100 ft-lbs at the nut. Using our equation to find M(x), the moment 0.25 feet to the right of the nut where the head of the torque wrench is:
M(x) = P * (L - x )
= 30.8 * (3.25 - 0.25)
= 30.8 * 3
= 92.4
That tells us that with the 3" extension on the 36" torque wrench, the wrench will read or click at 92.4 ft-lbs when the torque at the nut is 100 ft-lbs.
That also tells us that the correction factor we would apply to the wrench reading or setting to get the correct 100 ft-lbs of torque when using the 3" extension is 92.4/100 = 0.924.
That is right in line with the equation in AC43.13 that is basically the rule of law among A&P and IAs and disregarded at their peril:
Y=T*L/(L+E)
=100*3/(3+0.25)
=92.3
Now, suppose that instead of the AC43.13 correction factor, we used the constant 0.8 correction factor that Allan dictates.
100 * 0.8 = 80
That would tell us that if we wanted 100 ft-lbs of torque and we believed Allan, we would either watch the wrench for 80 ft-lbs, or set the click for 80 ft-lbs.
Let's work this backwards a bit and use our original equation to see how much force that requires at the handle:
M(x)=P*(L-x)
We know that M(x) is the moment applied at the head of the torque wrench, which is 3" or 0.25 feet to the right of the nut. Solving for P and then plugging in the numbers, we get:
P = M(x)/(L - X)
= 80 /(3.25 - 0.25)
= 80 / 3
= 26.7 lbs
Now, let's use the original equation again and see what torque we get at the nut when we apply 26.7 lbs at the handle 3.25 feet away from the nut. Remember, the nut is 0 feet to the right of the nut (it is where it is, right?) so x=0:
M(x) = P * (L - x)
= 26.7 * (3.25 - 0)
= 26.7 * 3.25
= 86.8 ft-lbs
What that means is that if we apply the mythical "0.8 correction factor," when using a 36" torque wrench with a 3" extension we would get an actual torque at the nut of 86.8 ft-lbs. That is 13.2 ft-lbs or 13.2% less than our 100 ft-lb target, and in some circumstances may constitute a dangerously loose joint.
Thanks, Bob K.