Crank bolt torque for timing belt change (10v) - why an exten sion mutiplies torque

KUNZ,BOB (HP-Boise,ex1) bob.kunz at hp.com
Mon Dec 1 10:54:11 EST 2003


OK. I see the issue here. Let's get some of the torque stuff straight. This
will be fun in ascii.

Take this example of a three point straight line with forces and torques.

1---------2---------------------3
     d1              d2

Call the torque at 1 = t1
Torque at 2 = t2
Distance between 1 and 2 = d1
Distance between 2 and 3 = d2
The force exerted at 3 = f (the force is you pushing down the page in our 2D
example)

We all agree that t2 = d2 * f

Since the joint at position 2 cannot rotate, then we can also say that

t1 = (d1+d2) * f

Substitute f = t2/d2 (from the first equation) into the second equation and
you get:

t1 = ( (d1+d2) * t2 ) / d2

which you can also rewrite as:

t1 = t2 * (d1+d2)/d2

Therefore the torque delivered to point 1 is the torque delivered at point 2
multiplied by the ratio of the overall length to the length of the torque
wrench. This is how the torque gets multiplied. It does in fact matter how
long the torque wrench is and how long the extension is.

Similarly, if you put an extension on the handle end of a torque wrench, you
have a similar equation with the ratios of forces:

1------------2----------------3
      d1            d2

Force at 3 = f3 (this is you standing on the extension)
Force at 2 = f2
Torque at 1 = t

Then t = d1 * f2
And t = (d1+d2) * f3 (again because joint 2 cannot rotate)

Equating these two gives:

d1 * f2 = (d1+d2) * f3

solving for f2:

f2 = f3 * (d1+d2)/d1

So, hanging out at the end of an extension bar changes the force applied to
the torque wrench by the ratio of (d1+d2)/d1, not d2/d1 as some would
suspect. The wrench will still "click" at the correct setting. You get more
leverage that you thought you would.

This one you can try real easily on a lug nut. Torque set to 80 ft lbs,
wrench is 2 ft long. The force you need at the end of the wrench is 40 lbs.
Now put a 4 ft extension on it. What's the force you need now? It is:

40 * 2/(2+4) = 13 1/3 lbs

Did you expect 20 lbs?

--bob

'86 5ks Avant (original owner, that's why my equations look like FORTRAN)
'02 TTQR



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