Oops, broke through top of piston on Stihl MS 180 CB-E

[SawTroll]

No you probably don't have to replace the cylinder.

 

[Aaron441]

You should just be able to look at it and tell the rod is bent. They usually don't just bend a little. If it looks ok, it probably is. After you put the saw back together, roll it over by hand. Make sure it is free and watch the piston thru the ports and see if it goes all the way up and down. If it does, you should be fine.

 

[willysmn]

Stihl bent rod discussion

I agree, Pioneerguy600. That is where the crank has the most leverage on the rod. The rod must bend sideways or not with the beam?

I believe that the rod is more likely to bend when the offset between the two connecting points are at their greatest, ..when the crank is at 90 deg, to TDC.

Pioneerguy600

 

[Saw Dr.]

I agree, Pioneerguy600. That is where the crank has the most leverage on the rod. The rod must bend sideways or not with the beam?

The crank has the most leverage on the rod at that point, but for a constant torque applied to the crankshaft, the force on the piston grows exponentially as it approaches TDC.

Try this sometime:

Take a saw with the top end off.
Put a breaker bar on the flywheel nut
Now position the crank with the throw 90 degrees and see how much force it takes to lift the breaker bar off the bench.
Take the crank and re-position it to 15 degrees to TDC and see how much force you have to apply to lift the breaker bar.

It will require several times the force to lift the breaker bar. Which one puts more strain on the rod? The breaker bar weighs the same and therefore applies the same amount of torque on the crank regardless of the throw position.

There some people I respect a whole lot posting up here, and I do not usually dig in like this. I am sorry to say it, but for a given torque applied to the crank (as you would do when trying to remove the FW or clutch) the stress on the rod is LOWEST at the 90 point.

 

[timberwolf]

The crank has the most leverage on the rod at that point, but for a constant torque applied to the crankshaft, the force on the piston grows exponentially as it approaches TDC.

....the stress on the rod is LOWEST at the 90 point.

Yes, and not exactly (I Think). You do need to considder rod angularity too, the point of lowest stress would be the same as the pont of highest piston speed. Which is actually a little towards TDC from 90 deg. This is somewhere about where there is a 90 deg angle between the rod and the crank.

 

[blsnelling]

If using an impact to remove the clutch on these, you'll want to remove the flywheel first. Otherswise you're likely to shear the cast in key off the flywheel like I did. I also had the clutch break from using one. Maybe I just need a smaller impact, lol

 

[Saw Dr.]

Yes, and not exactly (I Think). You do need to considder rod angularity too, the point of lowest stress would be the same as the pont of highest piston speed. Which is actually a little towards TDC from 90 deg. This is somewhere about where there is a 90 deg angle between the rod and the crank.

Well, I was making an approximation there. It should be the point where the centerline of the crank and pin make a 90 degree angle with the rod holes.

I think you and I are sying the same thing here. The point of highest piston speed is the lowest stress on the rod (approx 90 degrees) and therefore the point of lowest piston speed (Near TDC and BDC) is the HIGHEST rod stress for a given torque on the crankshaft.

This is not true as the engine runs, since the torque applied to the crankshaft by the piston varies through the entire cycle. The torque applied at TDC when the engine runs is ZERO, but at the approximate 90 degree point is at its highest.

 

[nmurph]

i always use an air impact set on max power. i usually do this with the plug in, but an impact will usually break the clutch loose with out the sparkplug to give compression. i would rather hit it once with enough force to break it loose than to have to let it bap, bap, bap to break it. i have never damaged a rod, crank or sheared a key, maybe i just haven't done enough.

 

[Saw Dr.]

I usually do what Brad does, and remove the FW first. Only takes a minute on most of them. I have not sheared a key that I know of, but the next one could be the first.

 

[MS260 Fan]

Brad, I have been trying to resist responding for the last 20 minutes or so. The rod sees the least load when the piston is slighty past halfway up the bore. (I.E. the centerline of the crankpin and crank axis is 90 degrees to the centerline of the rod holes) If it is only a few degrees from TDC, the compression force on the rod goes up exponentially for the same applied torque on the crank. This is the same principle (any physics) as a felling wedge applying a force to fell a tree rather than just hitting the tree with a hammer. If I were better at paintbrush, I'd draw a picture to illustrate.

I had never thought about this before. I agree with you Saw Dr. I think I can even put it into words.

For a given torque required to remove a flywheel nut etc. with a wrench, an equal but opposite torque has to be created by the crank and rod. If the two opposite torques are not equal, the piston will move.

Torque = force x distance. To reduce the force on the rod you want to maximize the length of the torque arm. The torque arm in this case is created by the length of crank between the centerline of the base of the rod and the crank shaft. That means increasing the horizontal distance shown in the attached. Like you said the horizontal distance between these centerlines is greatest when the centerlines are at 90 degrees to the centerline of the cylinder.

Simply put if this distance decreases by half, the force on the rod will have to double in order to maintain the constant torque. The force on the rod will approach infinity as the centerline distance approaches zero.

I apologize for the image of the torque arm. Found the first 2-stroke crank image on Google!

 

[timberwolf]

The torque applied at TDC when the engine runs is ZERO, but at the approximate 90 degree point is at its highest.

We are saying the same thing on the ~90 deg point, I was just adding a precision to it for historic significance I suppose. Not to detract from your point. it does depend a bit on the rod length to stroke ratio but on saws somewhere about 75-80 from TDC is where piston speeds are highest and the piston has the greatest mechanical advantage on the crank. If the rod were a mile long then the max advantage would be almost 90 deg, but as the rod gets shorter in relation to the stroke the maximum mechanical advantage point will move from 90 deg and get a bit closer to TDC.

Actually BDC would be the worst place to lock the piston as the piston velocity is lowest and the piston has even less mechanical advantage on the crank than at TDC. Anyway that is not really too importaint.

As far as torque being imparted from piston to crank if you considder the the enging in a static condition then it would stand to reason that the maximum torque would be applied to the crank when the piston had maximum mechanical advantage. There are two problems with this, one is that cylinder pressure must be looked at, peek cylinder pressure comes closer to 15-20 deg ATDC and falls drammaticaly by 90 deg ATDC, esp with an exhaust durration over 180. The second problem is that the forces in an engine are dynamic and actually a huge percentage of the energy from burn period is used in accelerating the piston downward, given this, it is the decelleration of the piston coming to a stop at BDC and again to TDC where much of the energy is actually transfered from the piston to the crank.

Took me a bit to get my head the idea that max torque to the crank did not need to happen somewhere between TDC and the exhaust port opening, but went through enough math and graphs on it to convince myself.

I know really of no importance to spinning off a nut on the end of the crank, but it is none the less interesting and has some value when looking at loading on bearings, the effect of reducing weight on pistons, the effect of rod length on the engine ect.

 

[Yoopermike]

One thing to remember when you go from 50:1 to 40:1 your running a leaner mix.
Especially going to 25:1.More oil less gas.
Sounds like you were on your way to a lean seize.

I always thought that a lean sieze was more gas to not enough oil. As in straight gassing, a saw it runs lean and fries the piston. A 25:1 mix is a richer oil content mix than say 50:1

 

[mowoodchopper]

I think the rope or the stop or the impact is fine, the problem is the piece of #### ms180.

 

[pioneerguy600]

The crank has the most leverage on the rod at that point, but for a constant torque applied to the crankshaft, the force on the piston grows exponentially as it approaches TDC.

Try this sometime:

Take a saw with the top end off.
Put a breaker bar on the flywheel nut
Now position the crank with the throw 90 degrees and see how much force it takes to lift the breaker bar off the bench.
Take the crank and re-position it to 15 degrees to TDC and see how much force you have to apply to lift the breaker bar.

It will require several times the force to lift the breaker bar. Which one puts more strain on the rod? The breaker bar weighs the same and therefore applies the same amount of torque on the crank regardless of the throw position.

There some people I respect a whole lot posting up here, and I do not usually dig in like this. I am sorry to say it, but for a given torque applied to the crank (as you would do when trying to remove the FW or clutch) the stress on the rod is LOWEST at the 90 point.

What you are saying is all true but bending the rod will happen much more when the piston is half down the bore, crank at 90 deg to TDC. A structural member will carry a heavier load when stood vertical between the two opposing forces, place that member at a 45 deg angle and it will fail under a lesser load, seen it many times so just my observations.

Pioneerguy600

 

[Scooterbum]

I always thought that a lean sieze was more gas to not enough oil. As in straight gassing, a saw it runs lean and fries the piston. A 25:1 mix is a richer oil content mix than say 50:1

Straight gassing a saw means no lube at all.
Lean seize is an air to fuel ratio.

 

[Tzed250]

And sheared the flywheel key, and broke a clutch. If procedure is followed carefully, they can be worked on successfully, but they are built VERY lightly, and will break easily.

Do not use rope. That's what caused my bent rod. The piston needs to be close to TDC with the least rod angle possible.

I used rope, and plenty of it, and that's what caused the bent rod. Thall proved that you can't bend the rod if using the proper stop, which leaves the piston much closer to TDC. Remember the video he did? He even replicated bending the rod when not using the proper stop, IIRC. In the case of a 170-180-210-230, I will never use rope again.

:hmm3grin2orange::hmm3grin2orange::hmm3grin2orange:


Those who cannot remember the past are doomed to repeat it!!!


It is amazing that you didn't learn the lesson from the first go around!!!!!

The crank cannot be near BDC or TDC. Those are the points at which the crank has the most leverage over the rod. I explained it back then. TW explained it today!!!!


You totally threw the saw under the bus, then it was shown that it was you that was in error, not the engineers at Stihl. You were wrong then, and you were wrong again today. The man above gave you two ears and two eyes and only one mouth for a reason.


.

We are saying the same thing on the ~90 deg point, I was just adding a precision to it for historic significance I suppose. Not to detract from your point. it does depend a bit on the rod length to stroke ratio but on saws somewhere about 75-80 from TDC is where piston speeds are highest and the piston has the greatest mechanical advantage on the crank. If the rod were a mile long then the max advantage would be almost 90 deg, but as the rod gets shorter in relation to the stroke the maximum mechanical advantage point will move from 90 deg and get a bit closer to TDC.

Actually BDC would be the worst place to lock the piston as the piston velocity is lowest and the piston has even less mechanical advantage on the crank than at TDC. Anyway that is not really too important.

As far as torque being imparted from piston to crank if you consider the the engine in a static condition then it would stand to reason that the maximum torque would be applied to the crank when the piston had maximum mechanical advantage. There are two problems with this, one is that cylinder pressure must be looked at, peek cylinder pressure comes closer to 15-20 deg ATDC and falls dramatically by 90 deg ATDC, esp with an exhaust duration over 180. The second problem is that the forces in an engine are dynamic and actually a huge percentage of the energy from burn period is used in accelerating the piston downward, given this, it is the deceleration of the piston coming to a stop at BDC and again to TDC where much of the energy is actually transfered from the piston to the crank.

Took me a bit to get my head the idea that max torque to the crank did not need to happen somewhere between TDC and the exhaust port opening, but went through enough math and graphs on it to convince myself.

I know really of no importance to spinning off a nut on the end of the crank, but it is none the less interesting and has some value when looking at loading on bearings, the effect of reducing weight on pistons, the effect of rod length on the engine etc.

Nice post TW. I believe that ignition timing is right when it allows the combustion pressure to be near max as the rod and crank pass through perpendicularity.


P.S. I have used an impact for years and never had a problem. The impact wrench was turned well down though.

.

 

[THALL10326]

I used rope, and plenty of it, and that's what caused the bent rod. Thall proved that you can't bend the rod if using the proper stop, which leaves the piston much closer to TDC. Remember the video he did? He even replicated bending the rod when not using the proper stop, IIRC. In the case of a 170-180-210-230, I will never use rope again.

LOL, I remember that. In fact I still got that rod assembly laying around here somewhere. I used a torque wrench set up far as it would go, 80lbs, and used a piston stop, didn't phase the piston or rod and the wrench was clicking away. I finally bent the rod by putting well over 100 pounds of torque on top the piston. However the piston wasn't damaged in the process, only the rod.

A simple plastic piston stop works fine on a 017/170/018/180. The key to apply pressure gently, thats all thats required to disassemble those saws and use a torque wrench to proper specs when tightening things back together. I use these same guidelines on all saws I disassemble, whether it be a 017 or a 880. To date from over 40 years fooling with saws, 20 years professionally, I've yet to break a piston or bend a rod.

Those that are leary of a piston stop need only to remove a fan wheel from a BR500/550/600 blower. You'll be amazed just how strong a piston is. Those fan wheels tighen on their own while being run. It usually takes about 90lbs of torgue to break one loose. That 90lbs is being applied right to the top of the piston to hold the crank so the fanwheel can be loosened up. Pistons are much stronger than most realize. Far as using rope as a piston stop, if it was the proper way to do it the manufacturer would recommend it, none do that I know of...

 

[Fish]

Is this cra p firing up again?????

 

[THALL10326]

Is this cra p firing up again?????

Not a slug fest Fish, just talking about those saws and how to open them up without damaging them. I'm edge ma cating man. You know how to do it but some don't. Get to work with the schooling Fish, its your place to tell these guys how to do things, says who, me, nuff said. Gittttttttttter done,LOL

 

[Saw Dr.]

What you are saying is all true but bending the rod will happen much more when the piston is half down the bore, crank at 90 deg to TDC. A structural member will carry a heavier load when stood vertical between the two opposing forces, place that member at a 45 deg angle and it will fail under a lesser load, seen it many times so just my observations.

Pioneerguy600

The load on the rod is always vertical. The side of the piston takes the horizontal portion of the load when the rod is at an angle. This is where piston slap comes from. So the load that the rod sees is always vertical with respect to it's longitudinal axis. If a rod bearing is rusted solid, there may be a bending force introduced to the rod, but otherwise the load will be compression. In a running engine there may also be an alternating bending force due to the mass of the rod itself and the fact that is swings back and forth in the engine. This will not apply in a static state when you are trying to remove the clutch.

I am just going for some educated discussion here, and not trying rain on anyone's parade. In the end, we should all do what works for us. I just removed the FW on a Dolmar 123 with an impact, but would have been just as happy using a rope if I wasn't too lazy to take the plug out.....