Weak MS180 Construction

[stihl only]

So has anyone ever bent a connecting rod trying to remove a clutch on anything ?????

There, I fixed it for you!

You're welcome

 

[Stihl 041S]

Crank position doesn't matter in this case.

Brad was trying to get the nut loose. If you have to tow a Cadilac and the chain breaks, it doesn't get moved. Yugo or Kenworth as a tow vehicle, the chain broke. The rod bent before the bolt loosened, no matter how you apply the force.

 

[Philbert]

As a mechanical engineer, you guys have peaked my interest as to the actual compressive forces in the piston rod during removal.

I drew some free body diagrams and came up with a formula for the compressive force in the piston rod . . .

This shows that the closer the piston is to TDC or BDC, the larger the force on the piston rod. It also shows that the least amount of force is exerted on the piston rod at ?=90. If blsnelling had too much or too little rope in the cylinder, he could have been far from the desired ?=90.

Airwolf, thanks for doing the heavy lifting on the math and physics. If you are willing to play with the diagrams a bit more, the applied questions are:

1. At what point (if any) in the cycle is the piston rod most effective at stopping the crank from turning;

2. At what point (if any) in the cycle is the piston rod most susceptible to buckling (this may not be the point at which the most force is applied).

Thanks.

Philbert

 

[timberwolf]

It also shows that the least amount of force is exerted on the piston rod at θ=90. If blsnelling had too much or too little rope in the cylinder, he could have been far from the desired θ=90.

Hmmm, see Tzed post. Max torque is at the point the rod is at 90 deg to the crank throw, not at 90 deg of crank rotation from TDC. That is that the piston if locked will have the maximum holding force on the crank when the rod is at 90 deg angle from big end center to the center line of the crank.

The exact ideal point will depend on the rod ratio, but in general it's more in the 75 to 80 deg from TDC area for rod ratios typical of a chainsaw. Edit to add thought (the longer the rod in relation to the stroke the closer the ideal point will move to 90 deg from TDC)

Here is a graph by my own calculations of piston velocity typical of a small stock chainsaw , peek velocity is about 77 or 78 deg, this is the point where the piston moves the most from a single degree of crank rotation. This then too is the point where the piston will have maximum mechanical advantage or holding force on the crank.

 

[timberwolf]

Here is a good article that explains why this is.

http://www.epi-eng.com/piston_engine_technology/piston_velocity_and_acceleration.htm

Food for thought, a wrench is going to be puting all the force required to move the nut through the rod, the closer to TDC the more this force will be amplified trough the vectors of the crank throw and rod angle. Impact gun will be putting some of the force on the rod but will be using the inertia of crank, flywheel to take some of this force off the rod.

 

[Tzed250]

A lot going on inside a recip engine. Also, the deltaV of the piston is not linear because of the changing angle of the connecting rod. You see many bent rods working on ATV's and PWC's. Operators who thumb the starter with water in the cylinder quickly find out about the incompressability of that liquid.

 

[windthrown]

See my post above WT. The crank has huge leverage on the rod near TDC. At this point you have very little movement of the rod little end for a great deal of movement at the crank. This is mechanical advantage. Camlocks work in the same way. The rod has the greatest force on the crank when the two form an "L". This is why you put a big cc engine at TDC to start it.

The way I see it, the crank at 90 degrees from TDC is applying all the force in the upward direction. Leverage or not, any movement on the crank with a frozen piston at 90 degrees is going to bend the rod. Whereas at TDC, all the movement of the crank is applying force perpendicular to the rod, and thus no energy is transfered to the rod to compress or bend it. It would seem from the photos in the posts here that the rods had massive bends in them. Thus the crank had to travel a good distance to do all that bending.


I took one survey course in statics at university.... a long time ago.

 

[Tzed250]

As a mechanical engineer, you guys have peaked my interest as to the actual compressive forces in the piston rod during removal.

I drew some free body diagrams and came up with a formula for the compressive force in the piston rod.

The formula is
F=T/(Rsinθ)

where
F is the axial compressive force in the piston rod
T is the torque exerted by the wrench
R is the fixed distance from the center of the crank to the piston rod connection on the crank(not to be confused with "r" from previous posts)
θ is the angle of the crank from TDC

If we apply a torque at θ=90 degrees, the formula will give us a numerical value for F. Now try the same torque near TDC. Let's say θ=10 degrees. The formula shows that F will be 5.75 times larger at θ=10 than it was at θ=90. Now try the formula 10 degrees from BDC. At θ=170, F is also 5.75 times larger than it was at θ=90.

This shows that the closer the piston is to TDC or BDC, the larger the force on the piston rod. It also shows that the least amount of force is exerted on the piston rod at θ=90. If blsnelling had too much or too little rope in the cylinder, he could have been far from the desired θ=90.

The formula doesn't work at exactly TDC or BDC because you can't divide by zero. The rope wouldn't provide any resistance to turning at TDC anyway.

Hmmm, see Tzed post. Max torque is at the point the rod is at 90 deg to the crank throw, not at 90 deg of crank rotation from TDC. That is that the piston if locked will have the maximum holding force on the crank when the rod is at 90 deg angle from big end center to the center line of the crank.

The exact ideal point will depend on the rod ratio, but in general it's more in the 75 to 80 deg from TDC area for rod ratios typical of a chainsaw. Edit to add thought (the longer the rod in relation to the stroke the closer the ideal point will move to 90 deg from TDC)

Here is a graph by my own calculations of piston velocity typical of a small stock chainsaw , peek velocity is about 77 or 78 deg, this is the point where the piston moves the most from a single degree of crank rotation. This then too is the point where the piston will have maximum mechanical advantage or holding force on the crank.

Here is a good article that explains why this is.

http://www.epi-eng.com/piston_engine_technology/piston_velocity_and_acceleration.htm

Food for thought, a wrench is going to be puting all the force required to move the nut through the rod, the closer to TDC the more this force will be amplified trough the vectors of the crank throw and rod angle. Impact gun will be putting some of the force on the rod but will be using the inertia of crank, flywheel to take some of this force off the rod.

The way I see it, the crank at 90 degrees from TDC is applying all the force in the upward direction. Leverage or not, any movement on the crank with a frozen piston at 90 degrees is going to bend the rod. Whereas at TDC, all the movement of the crank is applying force perpendicular to the rod, and thus no energy is transfered to the rod to compress or bend it. It would seem from the photos in the posts here that the rods had massive bends in them. Thus the crank had to travel a good distance to do all that bending.


I took one survey course in statics at university.... a long time ago.

See above info...

 

[airwolf]

You are correct timberwolf. The point at which the rod and crank are 90 degrees from one another does not correspond to 90 degrees from TDC. I was mainly interested in how the magnitude of the force changes the farther the rod and crank get from perpendicular. I mistakenly took them to be the same. I will change my formula so that θ means the angle between rod and crank, instead of crank angle from TDC. It would be more complicated to rewrite the formula as a function of crank position since it depends on the rod to crank ratio for your particular saw.

Here are I will repost my formula and conclusions with the corrected meaning for θ.

The formula is
F=T/(Rsinθ)

where
F is the axial compressive force in the piston rod
T is the torque exerted by the wrench
R is the fixed distance from the center of the crank to the piston rod connection on the crank(not to be confused with "r" from previous posts)
θ is the difference in angle between the line of the rod and the line of the crank

So θ=0 is at TDC
θ=90 is when the rod and crank are perpendicular
θ=180 is at BDC

If we apply a torque at θ=90 degrees, the formula will give us a numerical value for F. Now try the same torque near TDC. Let's say θ=10 degrees. The formula shows that F will be 5.75 times larger at θ=10 than it was at θ=90. Now try the formula 10 degrees from BDC. At θ=170, F is also 5.75 times larger than it was at θ=90.

This shows that the closer the piston is to TDC or BDC, the larger the force on the piston rod. It also shows that the least amount of force is exerted on the piston rod at θ=90. If blsnelling had too much or too little rope in the cylinder, he could have been far from the desired θ=90.

The formula doesn't work at exactly TDC or BDC because you can't divide by zero. The rope wouldn't provide any resistance to turning at TDC anyway.

Philbert- in answer to your questions
1. It would be optimal to stop the crank from turning when the piston rod and crank are perpendicular, since this is when the piston rod is stressed the least.

2.The buckling conditions don't change at different crank angles, because the boundary conditions of the rod don't change. The rod only experiences axial forces. Buckling is more likely to occur closer to TDC or BDC, only because of the increased axial force in the rod.

 

[timberwolf]

In static terms there is no force that can be put on a rod but 100% straight line pull or 100% straight line push as each end is centered on a bearing. When the crank is somewhere anywhere away from TDC or BDC the rod will be on an angle but any force on it can only push down it's length, there is no other way for the force to go as both ends of the rod are free to pivot and piston can only move in one dimention.

LOL, the math on this is still pretty easy, move the piston pin off center in the piston and move the bore line off the centerline of the crank and it gets fun.

Airwolf, you sure have all the terms and mathmatical structure down, most of that part is lost on me, I go at math from a different spatial more physical approach.

 

[airwolf]

In static terms there is no force that can be put on a rod but 100% straight line pull or 100% straight line push as each end is centered on a bearing. When the crank is somewhere anywhere away from TDC or BDC the rod will be on an angle but any force on it can only push down it's length, there is no other way for the force to go as both ends of the rod are free to pivot and piston can only move in one dimention.

well said timberwolf!

 

[Tzed250]

The rod gives up like a beer can being crushed...


For those that need a practical demonstration, hook up some compressed air to the cylinder of a saw, put a wrench on the crank and then see if you can rotate to TDC. Then put the at TDC and see how much easier it is to move the crank back and forth through say 15 deg. +/-. BTW, don't let the wrench get away from you.

 

[scotclayshooter]

I had a MS180 presented to me for a bit of tweaking, The oiler wasnt working and the air cleaner was missing!

I whipped the clutch off with the 18V impact wrench and a bit of rope without any problems and all the oiler parts were fine

The oiler pipe turned out to be blocked at the 90° bend, I used my pressure testing bulb to pressurise the tank via the oiler pipe and quick removal poped the dirt out

I had the impending doom feeling before touching this saw but its fine now!

 

[blsnelling]

I had a MS180 presented to me for a bit of tweaking, The oiler wasnt working and the air cleaner was missing!

I whipped the clutch off with the 18V impact wrench and a bit of rope without any problems and all the oiler parts were fine

The oiler pipe turned out to be blocked at the 90° bend, I used my pressure testing bulb to pressurise the tank via the oiler pipe and quick removal poped the dirt out

I had the impending doom feeling before touching this saw but its fine now!

I would have removed the flywheel first to remove the possibility of stripping the cast in key.

 

[scotclayshooter]

I would have removed the flywheel first to remove the possibility of stripping the cast in key.

But if the piston is against a rope against the top of the head how can any force be applied to the flywheel?????

 

[blsnelling]

But if the piston is against a rope against the top of the head how can any force be applied to the flywheel?????

You would still have impact action going on on the crank. The rope's going to have some give in it. I just know that the little cast in keys are easy to shear. Just don't ask how I know, lol

 

[belgian]

You would still have impact action going on on the crank. The rope's going to have some give in it. I just know that the little cast in keys are easy to shear. Just don't ask how I know, lol

that's why i don't use an impact !!!! :agree2:

 

[scotclayshooter]

You would still have impact action going on on the crank. The rope's going to have some give in it. I just know that the little cast in keys are easy to shear. Just don't ask how I know, lol

Again as long as the nut is tight on the flywheel ( I presume its on a taper?)
I cant see how it should shear the key.

I beleve you though!

Hard to belive it still works!

 

[blsnelling]

Nice air filtration, ehh?

 

[Fastcast]

Hard to belive it still works!

:censored:......

Some would be better off not owning things.....lol