JD 8340 engineering calculating offset torque

I'm working on an old John Deere 8340 4X4 articulated tractor. I have to replace the cratered front axle with a better one from a donor machine. I got the newer axle and decided to replace the rear seal at the pinion shaft. The keeper nut at that location requires being torqued to 440 ftlbs and then retorqued to 350 ftlbs. The size of the nut is 3-1/2 inches. I cut a wrench out of steel plate and welded a 3/4 inch socket into the plate 12 inches away from the center of the 3-1/2 hexagonal hole. I only have 1/2 inch drive torque wrenches. I did purchase a 4x1 torque multiplier off Ebay which works great. It has a 1/2 inch input socket and a 3/4 inch output lug. You attach a reaction bar to the head of the multiplier which is what does the hard work of withstanding the applied torque.
Here's my quandary. In order to achieve the correct torque values at the big nut, I must calculate the forces and moments applied to the home made wrench. I insert the torque multiplier 3/4 inch output lug into the socket I welded into the plate. The torque multiplier has a reaction arm which is attached to something solid. I then insert my 1/2 drive torque wrench which is 17 inches long into the input side of the torque multiplier and apply how much force to the torque wrench to get the required torque value at the big nut?
I think this is one of those dang indeterminate statics problems I always had trouble with in school. Anybody have an idea how to solve this problem? It is not as simple as it might seem and has me baffled.
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I'm just throwing some numbers together here...

But if you could put the torque multiplier directly over the nut, to get 440 ft lbs, at 4:1, you would go to 110 on the multiplier. Then being extended out 1 extra foot, go to 55 on the multiplier.

I think this would be in the ball park as long as the extension handle on the multiplier is held in a straight line with the 12" adapter.

This sounds too simple, I'm probably overlooking something...
 
There's a million websites out there that have calculators for offset adapters on a torque wrench, but the simplest way is to set up the adapter at 90? to the bar axis of the torque wrench and make NO correction.

You made that arm REALLY long, I'm not sure why, but still I don't think you'll be far off. This is not a "criticality one" application, simply divide the torque spec by the ratio of your multiplier (which I didn't see in your post) and set it all up with the offset adapter at 90? to the wrench and have at it. I've been there and done that with several 8630's. I'd use a bit of high strength Loctite, as well.

<img src = "http://www.cncexpo.com/Images/WrenchAdp.jpg">
 

The diagram in Bob's post is correct, the math that goes with it is as follows:

Tw = (Ta x L) / (L + E)

Where:

Tw = The setting on the torque wrench

Ta = Actual torque applied to nut

L = Length of torque wrench (17")

E = Length of adapter (12")

Case 1, The adapter is at a right angle (A = 90 degrees) to the torque wrench.

In this case E = 0 and there is no correction required for the adapter. However, since the torque multiplier is 4:1 the actual torque will be 4 times any setting on the torque wrench. Therefore, setting 110 on the torque wrench should apply 440 ft-lbs at the nut.

NOTE, the reaction arm of the torque multiplier must be maintained at a right angle (A = 90 degrees) to the adapter. As the nut rotates the reaction arm must also rotate to maintain, A = 90 degrees.

This is in agreement with the 110 torque wrench setting in Steve's post.

Case 2, Torque wrench and adapter are in a straight line.

In this case we have:

Tw = (440 x 17) / (17 + 12) = 7480 / 29 = 258 ft-lbs.

However, since you are using the torque multiplier any setting on the torque wrench gets multiplied by four. Therefore, we need to divide by four to use the above equation.

We have:

Tw = 258 / 4 = 64.5 ft-lbs

Setting 64.5 on the torque wrench should yield 440 ft-lbs on the nut.

NOTE, the torque multiplier reaction arm must be maintained at a right angle (A = 90 degrees) to the adapter. As the nut rotates the reaction arm must also rotate to maintain, A = 90 degrees.

This is not in agreement with the 55 torque wrench setting in Steve's post.

Steve, how did you arrive at the 55 setting? Maybe I made an error.

To test; the two methods, Case 1 and Case 2 should give the same results. To test, pick a lower torque value for the nut that can be applied using both Case 1 and Case 2. Using Case 1, torque the nut and mark the location it stops. Now, back it off and torque to the same value using Case 2 - the marks should line up. This does not verify the actual torque on the nut is correct, however is does verify the correction for the adapter.
 
The calculations above ignore one thing:

Since you have the torque wrench and the torque multiplier twelve inches away from the center of you big hex cutout; the torque from both will add together.

Why? Because your super duper largest crow's foot wrench in the world puts your torque wrench and the torque multiplier twelve inches away from the center of the nut and there will be two forces trying to turn the nut:

Force 1: The force that you, yourself put on the torque wrench...multiplied by the length of the new moment arm that you have created; which is 12" + 17" (imagine that the torque multipler wasn't there)

The torque on the nut from the torque wrench alone is equal to the setting on the wrench...multiplied by the new moment arm (29") divided by the original moment arm of the torque wrench (17").... that comes out to a multiplier of 1.7

Force 2: The force that the torque mulitplier's reaction arm puts on (in a perfect world) an immovable object that it is resting against. The multiplier there is 4

By superposition, these forces and their corresponding torques would add together.

So...

Using Tw as above for the wrench setting and Ta for the actual torque felt at the nut:

Ta = Tw X 1.7 + Tw X 4

Ta = Tw X 5.7

Tw = Ta/5.7

Tw = 77.19

This assumes that your torque wrench is in line with your super-duper largest crow's foot adapter in the world, and that the reaction bar of the torque multipler is also generally in line with the torque wrench...and that the reaction bar is against something that will be able to hold around 800 pounds of force without moving.
 
Sorry...skipped a number there.

My calculations assumed the 440ft/lb target... that's how I got

440/5.7 = 77.19

When you do the 350 ft/lb target...you would have

350/5.7 = 61.4
 

And in reality, the reaction arm of the torque multiplier doesn't have to be "in-line" with anything. The internal gears will perform the proper angular translation of force. But...it will still have something like 800 lbs of force at the end of that 6" stub.
 

I really regret that last statement, as soon as I hit submit.

Under normal circumstances, when the torque multiplier is on top of the fastener being torqued...yes...you don't care if the arm of the multiplier is in line with the torque wrench.

In this case, though...since you have translated everything over a foot...it's really important to keep the torque multipler arm, your homemade crow's foot, and the torque wrench all in line to get proper numbers. Otherwise, you have to start multiplying by the cosine of a bunch of unknown angles :)
 

Lastcowboy,

I agree you are correct the two torques add and the principle of superposition is the proper method to use. You summarized it very well in your post, thanks for the correction.

I do believe the reaction arm needs to be at 90 degrees to the adapter. The reaction force needs to act through the center of the nut, otherwise another torque is created.

Ken
 



In the end, though, if I didn't want the mystery of the torque multiplier...and if I didn't have something that could hold the force of its reaction arm...I would try to extend the distance that the socket is welded from the center of the hex cutout.

I would start with a torque that is easily attainable on a 1/2" torque wrench...say 90 ft-lbs...for Tw

That's a much more simple calculation:

Ta = Tw X ((L +17)/17)... when L is in inches

That you can manipulate to separate out the unknown L...

17 X Ta = Tw (L +17)

(17 X Ta) - (Tw X 17) = Tw X L

L = ((17 X Ta) - (Tw X 17))/Tw

If I put in 440 for Ta and 90 for Tw... I get...

L = 66 inches

So, if you move the socket hole out to 5' 6" from the center of the hex cut; you could put a torque wrench on there and get 440 ft-lbs at the nut with a 90 ft-lb reading on the wrench...without any multiplier at all.
 

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