[amit_purohit20]

Originally Posted by Samurai

Let's consider a modified pipe.

Attachment 1186027

This is a soild pipe, with a much strengthened center. Center is where the Superman is holding it. The thinner parts of the pipe will start twisting at 500Nm of torque, but the center won't twist until 2000Nm of torque.

If the human grip can be overcome with 50Nm and Superkid's grip requires 1000Nm to overcome, what happens when superman starts turning the center? I would say the Human side will start turning since the effective torque on that side is 50Nm (too less to twist the pipe). The superkid side of the pipe will be twisting because it can't handle above 500Nm of twisting force. This twist won't leak to the other side because the much bulkier center can easily withstand 500Nm of torque. Visualise, don't apply formulas.

If you still can't believe one side can rotate and other can twist... let's change the condition without changing the applied torque.

Let superman hold the center stationery. Let the human try to turn the pipe applying upto 50Nm. Let Superkid also try to turn the pipe, but he can go up to 1000Nm. As you can see effective torque everywhere is same as before. Now, the human side remains stationery, and the superkid side will twist once he exceeds 500Nm.

In the both the above cases, the human side will follow the superman (rotate or stationery), the superkid side will twist.

@Samurai - I prefer visualisation rather than applying formulas too.
Let us make clear that when you use the word "rotate". It means that the shaft should reach a certain rpm say 50 rpm and go on rotating for a period of time say half hour.

There is difference between rotate and twist. Twisting of a shaft is only possible in some degrees say not more than 20 degrees. Beyond which any reasonable shaft will break.

Case I
Superman first applies 25 Nm to the centre. Nothing happens except a very small negligible twist (We can neglect this twist). The shaft as a whole is not rotating.

Super man now applies 51 Nm of torque, will the shaft rotate? No because for the shaft to rotate superman needs to apply 50 +1000= 1050 Nm.
Both the sides of the shaft will twist by a very small degree equivalent to the torque applied by superman.

The statement that twist in thinner portions of shaft will start only after 500 Nm is wrong. Twist is not like a diode which will start functioning once a threshold of 500 Nm is reached. Twist has a linear relation to the shaft.

The twist starts with even 0.5 Nm and it goes on increasing as you increase the applied torque.

But let us assume that a major chunk of the twist will happen only after 500 Nm as per your case. Consider it like a threshold as you have stated. So no twist before 500 Nm, to simplify things.

Coming back to the point

Super man now applies 500 Nm of torque. The end of pipe towards super kid starts twisting.
Say after 500 Nm of torque the shaft thin ends will twist 1 degree for 1 Nm.
501 Nm - Twist 1 degree.
502 Nm - Twist 2 degree.
.
.
.
This will go on till we reach 20 degree where the shaft will break.
The centre portion of the shaft has turned by 20 degree.

Same will be the turn in the shaft on the human end as it it is attached firmly to the centre portion of the shaft which has only turned by 20 degree.
The centre portion of the shaft cannot rotate now because we have applied only 520 Nm which is less than the total torque required to rotate the shaft ie. 1050 Nm.

So the shaft on the human side has rotated only 20 degree till now.
Now even if you even assume that the shaft can twist max to 720 degree (which is next to impossible as the shaft will break before that) then

500 Nm + 720 Nm (used for the 720 degree twist) = 1220 Nm which is higher than the required 1050 Nm the shaft should be already rotating by now.

Infact the centre portion of the shaft will rotate at 1050 Nm so shaft will start rotating at 1050 Nm where the super end kid of the shaft will be twisted by 1050-500= 550 degrees.

In Case II

Yes here the superkids side will twist but not rotate.
Are you getting confused between twist and rotate?

[Sutripta]

Originally Posted by DKG

The least expensive way to replicate a locked diff effect in your 4x4 is to install a separate handbrake lever for each of the driving wheels. As you brake the slipping wheel more torque gets applied to the wheel which still has contact with the ground.

(Dual handle) fiddle brakes. Can be mimicked by the ABS (hardware) module running under a proprietary controller

Originally Posted by ex670c

Hi Sutripta & Amit,
3) However when the Front Axle is Engaged 50% of the Torque is sent to the Front Axle?

What dictates the above Split 50:50.
i) Type of T-Case.
ii) Differential Ratio

If we are talking of our jeeps with simple transfer cases, the torque split will be determined by the outside world. - the traction available at each wheel. If all wheels are essentially identical, then yes 50:50.

And if this split is 50:50, and the Front Wheels lose traction, Will the Rear Wheels lose traction?

No. 100% will then go to the working axle.

If the Front Wheels lose traction, will most of the 90-100% Torque be sent to the Rear Wheels?

Yes. See above

Originally Posted by ex670c

What is the Maximum Torque ONE WHEEL can deliver in a 4x4 with 3 Locked Differentials, Traction being Optimum.

a) approximately 50% of the Geared Torque (Gearbox/T-Case/Differential)

b) approximately 25% of the Geared Torque (Gearbox/T-Case/Differential)

Already answered by TSK.
I would like to change the wording to 'Maximun Torque that can be delivered to one wheel'. 100% (correct for appropriate gearing) if we ignore parasitic losses.

Regards
Sutripta

[Sutripta]

Originally Posted by amit_purohit20

The instumentation (which uses shaft deflection for finding torque) on the side of the free wheel will read nearly zero torque and the other side instrumentation will read full torque as found on the Crown wheel gear.

By the way are you a mechanical engineer?

And we agree once again!

You have any doubts that the torque (as read out by embedded strain gauges on the shaft) is fundamentally in error?

On this forum we are all comrades (during shared activities) or laymen (when we think we are discussing technical points)!

Regards
Layman Sutripta

[Samurai]

Originally Posted by amit_purohit20

@Samurai - I prefer visualisation rather than applying formulas too.

Glad to know. That is the trait of a good engineer. I always run programs in my head during the design phase. I write the code on the editor only if it runs fine in my head without crashing.

Originally Posted by amit_purohit20

Let us make clear that when you use the word "rotate". It means that the shaft should reach a certain rpm say 50 rpm and go on rotating for a period of time say half hour.

Well, I guess we do differ in our definitions. So we shall define that too.

Originally Posted by amit_purohit20

The statement that twist in thinner portions of shaft will start only after 500 Nm is wrong. Twist is not like a diode which will start functioning once a threshold of 500 Nm is reached. Twist has a linear relation to the shaft.

That wasn't my assumption. I understand twist starts happening from the very beginning, only the extent depends on the elasticity of the material used.

Imagine a rubber pipe connected to a light load, like a 1Kg iron ball. When you turn the pipe, it will first twist. Once the torque required to further twist the pipe exceeds the torque required to overcome the inertia of the iron ball, the ball will start rolling. Same happens in steel pipes too, except the initial twist in pipe is limited to nano meter or micro meters along the circumference of the pipe. Thus is it invisible to plain eyes. This twist is actually stored energy, which gets transferred to the other side when it starts turning. If the other side requires lot more energy than the pipe can safely store, it ends up destroying the pipe. But I digress.

Originally Posted by amit_purohit20

There is difference between rotate and twist. Twisting of a shaft is only possible in some degrees say not more than 20 degrees. Beyond which any reasonable shaft will break.

Let me describe my definition of rotate and twist. In fact, let me rename this twist as fatal twist or failure twist.

Rotate: The shaft/pipe safely transfers the energy to the other side. At first the pipe will twist until the elasticity of the pipe is overcome, and then the opposing force at other side is overcome. If this happens even at 1 rpm or 1/10 rpm, I call it rotating.

Failure twist: The shaft/pipe cannot safely transfer the energy to the other side. At first the elasticity of the pipe is overcome, but the opposing force at the other end is still very large. Upon supplying further torque, the pipe is twisted beyond the point of no return. The pipe is now mangled into a spiral or broken, this behavior depends on the property of the pipe material. Here the stored energy is used to destroy the pipe.

With the above definitions, now look back at the superman example.

[Sutripta]

^^^
Guys,
Will we get any different answers if we just say that
a) It is an uniform shaft (easier to work with conceptually and mathematically) that can
b) handle any amount of torque (ie remains within elastic limits) our imaginary experiment can generate?

I fail to see any advantage to this discussion in bringing in failure modes.

Regarding rotation/ twist: Once the system has reached a 'equilibrium/ steady state' the shaft turning by even 1 radian/ year is a rotation, if we expect it to go on.

One other thing we need to define, because it is an artificial example, is the nature of the forces at the two ends of the shaft. Do they act like slipper clutches, or shear pins? We are assuming that these will behave like slipper clutches.

Regards
Sutripta

[Samurai]

Originally Posted by Sutripta

Will we get any different answers if we just say that a) It is an uniform shaft (easier to work with conceptually and mathematically) that can b) handle any amount of torque (ie remains within elastic limits) our imaginary experiment can generate?

Well, this is exactly the situation explained by my original human experiment.

Originally Posted by Samurai

Let's take a human example.

Hold a 6ft long metal pipe at the center, horizontally to the ground. Have a strong guy hold one end tightly. And there is a small child holding the other end. Let's say we need 50Nm to overcome the strong guy and only 1Nm to overcome the child.

Now, I don't know how torque measuring devices work. But if you have a way to measure the torque at each end, they ought to measure 50Nm and 1Nm once the shaft is rotating due to the 51Nm torque applied at the center. The 1Nm side will have less twist (elastic) than the 50Nm side too. Is that how torque is sensed or measured, by the extent of elastic twist?

[Sutripta]

Originally Posted by Samurai

Well, this is exactly the situation explained by my original human experiment.

So why change it?

Is that how torque is sensed or measured, by the extent of elastic twist?

Embedded strain gauges.
Same principle as used in load cells. Except that previously getting millivolt level signals out reliably/ repeatably' using brushes and sliprings, in a practical industrial environment is not a job for the fainthearted.
Much easier nowadays.
HBM used to have a comprehensive tutorial on strain gauge measurements (printed). Don't know if softcopy exists.

Regards
Sutripta

[Samurai]

Originally Posted by Sutripta

So why change it?

Well, I thought if I showed only one side fails, I could show torque is distributed, without having to measure it. If you can measure it without breaking, we can go back to normal human example.

Originally Posted by Sutripta

Embedded strain gauges.
Same principle as used in load cells. Except that previously getting millivolt level signals out reliably/ repeatably' using brushes and sliprings, in a practical industrial environment is not a job for the fainthearted.

Ah, strain gauges... I remember studying these as part of electric measurements subject in BE. The resistance changes altering the current flow when force is applied.

[amit_purohit20]

Originally Posted by Samurai

Let's consider a modified pipe.

Attachment 1186027

This is a soild pipe, with a much strengthened center. Center is where the Superman is holding it. The thinner parts of the pipe will start twisting at 500Nm of torque, but the center won't twist until 2000Nm of torque.

If the human grip can be overcome with 50Nm and Superkid's grip requires 1000Nm to overcome, what happens when superman starts turning the center? I would say the Human side will start turning since the effective torque on that side is 50Nm (too less to twist the pipe). The superkid side of the pipe will be twisting because it can't handle above 500Nm of twisting force. This twist won't leak to the other side because the much bulkier center can easily withstand 500Nm of torque. Visualise, don't apply formulas.

If you still can't believe one side can rotate and other can twist... let's change the condition without changing the applied torque.

Let superman hold the center stationery. Let the human try to turn the pipe applying upto 50Nm. Let Superkid also try to turn the pipe, but he can go up to 1000Nm. As you can see effective torque everywhere is same as before. Now, the human side remains stationery, and the superkid side will twist once he exceeds 500Nm.

In the both the above cases, the human side will follow the superman (rotate or stationery), the superkid side will twist.

Going with your new definitions,

Again in Case I the human side pipe will rotate and the superkid side will go on twisting. The rotation (your definition) in the human side is caused at the expense of the twisting of the otherside of shaft and is exactly equal.

In Case II
I agree Human side will be stationary and SuperKids side will twist.

But again you are not giving/using practical examples by comparing rubber and steel, also by bringing out twisting of shafts I really do not understand what point you want to put forth?

Also your definitions of turning of human side shaft end is based on the thought that the shaft on the other side will go on twisting for a substantial amount of time and angle.


Now consider the shaft as steel only and that the shaft can twist only 1 degree and not beyond that, what difference does it create?

As normally in design we do not wan't the shafts to twist much. We can safely assume in practical situations the twist as zero.

Now with a shaft which doesnot twist can you try to put forth the point which you want to make?

Originally Posted by Sutripta

^^^
Guys,
Will we get any different answers if we just say that
a) It is an uniform shaft (easier to work with conceptually and mathematically) that can
b) handle any amount of torque (ie remains within elastic limits) our imaginary experiment can generate?

I fail to see any advantage to this discussion in bringing in failure modes.

Regarding rotation/ twist: Once the system has reached a 'equilibrium/ steady state' the shaft turning by even 1 radian/ year is a rotation, if we expect it to go on.

One other thing we need to define, because it is an artificial example, is the nature of the forces at the two ends of the shaft. Do they act like slipper clutches, or shear pins? We are assuming that these will behave like slipper clutches.

Regards
Sutripta

I agree and please elaborate your point of Slipper clutches and shear pins....

@Samurai I agree with you for the 51 Nm torque thing.

[Samurai]

Originally Posted by amit_purohit20

@Samurai I agree with you for the 51 Nm torque thing.

You mean you agree that in the human example, one side will read 50Nm and other side will read 1Nm on the respective strain gauge?

[amit_purohit20]

Originally Posted by Samurai

You mean you agree that in the human example, one side will read 50Nm and other side will read 1Nm on the respective strain gauge?

I think I already agreed to the point in the above posts!

Just that I dont use the word "distribution" because its not the shaft or the locked differential which distributes the shaft in the torque but its the different resistances acting up at different ends of the shaft. This causes the different torque conditions in the same shaft at different places of the shaft.

Ultimately the torque required is 51 Nm to rotate the shaft.

[DKG]

Let's discuss open differentials and whether a torque split occurs there. Is torque at either end always equal in an open diff? And if so why and how does that happen?

[Samurai]

Originally Posted by amit_purohit20

I think I already agreed to the point in the above posts!

Perfect! I just wanted to make sure.

Originally Posted by amit_purohit20

but its the different resistances acting up at different ends of the shaft. This causes the different torque conditions in the same shaft at different places of the shaft.

I can live with that.

Originally Posted by amit_purohit20

Just that I dont use the word "distribution" because its not the shaft or the locked differential which distributes the shaft in the torque

This you have to discuss with Sutripta, I am nowhere near qualified to argue about this one.

Originally Posted by DKG

Let's discuss open differentials and whether a torque split occurs there. Is torque at either end always equal in an open diff? And if so why and how does that happen?

Yes, it is always equal and the torque is determined by the wheel with least traction/resistance.

It can be easily explained, just see this video. See what happens when one wheel is stopped, the propeller will not be stressed, it will continue turning the free wheel without putting any extra effort demanded by the stopped wheel. That won't happen in a locked differential.

http://www.youtube.com/watch?v=yYAw79386WI

[amit_purohit20]

Originally Posted by Samurai

Perfect! I just wanted to make sure.

Ah! You were driving me crazy with your rotation and twisting theory. I got your point from the last post but to that 50 + 1Nm theory I had agreed to it from my post which talked about half shafts and strain guages.
Thanks to you and Sutripta Sir I could brush up and also learn some fundamentals which I had not thought in detail. Hats off to you guys.
I learnt one thing from you, that if one has the passion one can really learn/do anything.

[Samurai]

Originally Posted by amit_purohit20

Ah! You were driving me crazy with your rotation and twisting theory.

Well, this isn't my field. So I had to use all kind of funny unorthodox examples like superman or safe energy transfer ideas.

Anyway, I am glad we have reached the conclusion of a discussion that was kick-started by my statement here two weeks back, also quoting it below.

Originally Posted by Samurai

I think Eaton MLD turns both wheels at the same speed irrespective of their traction. But torque would be different, if the traction at each wheel is different.

In order to prove this, we had to first establish engine generates torque based on load. I didn't think that would be contested, but it was. Once that was firmly established, we focused on establishing how in a locked differential, torque at the propeller is the sum of torque required at each wheel. Notice how I avoided the word distribution.

I hope we have proven that to the satisfaction of most. I'll also add a poll to this thread to see how many people agree or disagree with the conclusions.

PS: This has been the most enjoyable and most intellectually challenging TBHP thread of all time for me. Learned a boatload of stuff on this thread. My sincere thanks to all for the lively discussion, and especially to Sutripta and Amit.