Weak MS180 Construction

[blsnelling]

Can you elaborate on this a bit more?

If the compression is placed on the tie rod while it is vertical (top of stroke or bottom of stroke), there would seem to be less bending force placed on it than when it is at an angle.

Philbert

This has been discussed here in the past, but I'm not sure I agree. The crank is always pushing linearly along the axis of the rod during the compression stroke. So you're never putting "bending force" on the rod, but rather pushing the piston skirt into the cylinder wall. No? It doesn't seem to me that the crank could put any torsional strain on the rod with the roller bearings on both ends, even as it swings past TDC. I think we're creating a "wives tale" here.

 

[brages]

This has been discussed here in the past, but I'm not sure I agree. The crank is always pushing linearly along the axis of the rod during the compression stroke. So you're never putting "bending force" on the rod, but rather pushing the piston skirt into the cylinder wall. No? It doesn't seem to me that the crank could put any torsional strain on the rod with the roller bearings on both ends, even as it swings past TDC. I think we're creating a "wives tale" here.

Well, that's correct - the rod can only be loaded in compression or tension. No bending, due to the bearings on the ends.

But the force on the rod multiplied by the perpendicular distance from the centerline of the rod to the axis of the crank will be equal to the torque. (physics: T = F x r)

You want the perpendicular distance ("r") to be as large as possible, so you want the "arm" of the crank as horizontal as possible when you loosen the nut.

(Personally, I wouldn't use anything but an impact wrench on these cheap Stihls; I think it has been demonstrated that the connecting rods are just too weak.)

 

[stipes]

Brad...Just wonder...

When your TDC what kinda squish would you have?? Wonder if a smaller guage of rope would have locked everything up and not bend a crank??I'm just wondering and tryin to learn....

 

[Philbert]

This has been discussed here in the past, but I'm not sure I agree. The crank is always pushing linearly along the axis of the rod during the compression stroke. So you're never putting "bending force" on the rod, but rather pushing the piston skirt into the cylinder wall. No? It doesn't seem to me that the crank could put any torsional strain on the rod with the roller bearings on both ends, even as it swings past TDC. I think we're creating a "wives tale" here.

Thanks for your reply.

Think of the connecting rod being made of rubber, not metal, with the crank at 90 degrees (rod is at it's steepest angle) and the piston stopped (rope or whatever). If the crank keeps turning upward, and the piston does not move, the force would cause the rod to bend or buckle, just like in the photos.

If the rod was straight up and down, the force would cause the rod to compress along its axis before it eventually buckled.

Just thinking out loud.

Philbert

 

[Philbert]

You want the perpendicular distance ("r") to be as large as possible, so you want the "arm" of the crank as horizontal as possible when you loosen the nut.

This makes sense when you are looking at the force applied to the crank arm, but I think that we are looking at the force applied to the connecting rod (both the magnitude and direction) when we are looking at bending the rod.

The bearings reduce friction, but they do not change the direction of the force applied.

Philbert

 

[Taxmantoo]

Dynamically, you get all sorts of side loads on a conrod at speed.
Statically, you can't put a side load on it unless a bearing is frozen, all you can do is push and pull. A wooden rod would break under excessive compression, a steel rod bends, but there's no way to predict if it will bend forward or backward if its construction is symmetrical.

 

[brages]

This makes sense when you are looking at the force applied to the crank arm, but I think that we are looking at the force applied to the connecting rod (both the magnitude and direction) when we are looking at bending the rod.

The bearings reduce friction, but they do not change the direction of the force applied.

Philbert

The bearings allow rotation, so there's no moment (bending) in the rod.

And the force the crank applies to the con rod is equal and opposite to the force the con rod applies to the crank.

Because there's no moment applied to the con rod at its ends, the only direction that force can be applied to the con rod is axial. Otherwise, it would have an unbalanced set of forces on it.

 

[brages]

Dynamically, you get all sorts of side loads on a conrod at speed.

Yes, this is correct. The stuff I've written applies in a static situation.

 

[Philbert]

The bearings allow rotation, so there's no moment (bending) in the rod.

Again, that is not accurate. In a theoretical situation, where the rod was rigid and could not bend, the resultant forces would only be transmitted along the axis of the rod. Clearly, this is not happening here, because the rod is bending.

I'm not talking about torque/moment, I am talking about compressive forces (including horizontal and vertical components) transferred through bearings at each end of the rod which exceed the rod's ability to resist deformation.

And the force the crank applies to the con rod is equal and opposite to the force the con rod applies to the crank.

Yes, the resultant forces balance, but the component forces (horizontal and vertical) on each part vary with the position of the crank. When the crank is horizontal, a vertical force would provide the greatest resistance to rotation, however, at this point, the top of the connecting rod (at the piston) is farthest horizontally from the crank bearing (the bottom of the rod), which mean that the internal forces inside the rod (those which would affect its risk of bending) are very different than when the rod is aligned more vertically over the crank.

Philbert

 

[brages]

Again, that is not accurate. In a theoretical situation, where the rod was rigid and could not bend, the resultant forces would only be transmitted along the axis of the rod. Clearly, this is not happening here, because the rod is bending.

The rod was loaded "axially" (along its length) and bent because it failed in "column buckling".

http://en.wikipedia.org/wiki/Buckling#Buckling_in_columns

 

[Philbert]

The rod was loaded "axially" (along its length) and bent because it failed in "column buckling". http://en.wikipedia.org/wiki/Buckling#Buckling_in_columns

I love Wikipedia. Great article. Thanks. It uses more technical language to say the same thing,
". . . analysis of buckling makes use of an axial load eccentricity that introduces a moment, which does not form part of the primary forces to which the member is subjected."

'Buckling' may be the appropriate technical term to use, as opposed to 'bending' - we tend to be less formal in this forum. Unless you have another definition of 'axial load eccentricity', I think that it means that not all of the force is directed directly down the middle of the rod.

Not trying to be a smart a**, just trying to get a practical, mechanical understanding of where a connecting rod would be most susceptible to getting bent ('buckled').

Philbert

 

[blsnelling]

The rod was loaded "axially" (along its length) and bent because it failed in "column buckling".

http://en.wikipedia.org/wiki/Buckling#Buckling_in_columns

Which is not from a side load. That's my whole point. I don't know the physics to argue this. I'm just trying to look at it logically. I don't believe the rod bent because the angle of the crank. I believe it was "compressed" until it could stand no more load.

If there are side force to consider, then I'd think you would want the crank in a position such that the rod is at a right angle to the throw of the crank.

 

[Tzed250]

Which is not from a side load. That's my whole point. I don't know the physics to argue this. I'm just trying to look at it logically. I don't believe the rod bent because the angle of the crank. I believe it was "compressed" until it could stand no more load.

If there are side force to consider, then I'd think you would want the crank in a position such that the rod is at a right angle to the throw of the crank.

Right...the conrod has the greatest leverage on the crank when the bores of the conrod form a line that is perpendicular to the centers of the crank journals, not when the crank is at 90 deg. before or after TDC. Coincidentally, this is also the point at which you would like to see max pressure ATDC for max torque.

 

[Stihl 041S]

Tzed;well said. Short and to the point.

 

[ironballs]

Did you get the oiler fixed? My saw does the same thing spit n sputter.

 

[blsnelling]

I haven't looked at it yet.

 

[windthrown]

You want the perpendicular distance ("r") to be as large as possible, so you want the "arm" of the crank as horizontal as possible when you loosen the nut.

I think that you mean to say that you want r to be as small as possible (if I read it right), in relation to the perpendicular distance of the crank rod bearing relative to the piston movement. In which case, a smaller amount of rope and the piston frozen near TDC would allow for more torque on the bolt to remove the clutch and flywheel without bending the piston rod in the process.

In other words, the position of the crank does make a huge differnce. The leverage of the arm of the crank will be far less with the piston at or near TDC (or BDC), compared to when the crank is 90 degrees from TDC. At a full right angle, the force from the crank on the connecting rod will be at its potential maximum due to the leverage of the crank relative to the piston rod. All of the force from the crank at 90 degrees is parallel to the piston movement. Nearer to TDC, less and less force is going 'up' to compress the rod, as compared to when it is at 90 degrees. All the force from the crank at TDC is perpendicular to the piston travel (meaning at TDC, there is no force being exerted on the rod to compress it if you wrench on the crankshaft).

And that would seem to verify why I have never bent a rod in a 1123 type saw, even using my foot and full leg strength on a scrench to remove a clutch. I had one beer too many on that one, and pushed it the wrong way with all my weight. That one had a rag strip stuffed in through the exhaust port, and just enough to keep the piston from clearing TDC. Once I realized my mistake (after having another beer) I reversed the scrench and the clutch popped right off. That 210 is still running fine.

 

[Tzed250]

I think that you mean to say that you want r to be as small as possible (if I read it right), in relation to the perpendicular distance of the crank rod bearing relative to the piston movement. In which case, a smaller amount of rope and the piston frozen near TDC would allow for more torque on the bolt to remove the clutch and flywheel without bending the piston rod in the process.

In other words, the position of the crank does make a huge differnce. The leverage of the arm of the crank will be far less with the piston at or near TDC (or BDC), compared to when the crank is 90 degrees from TDC. At a full right angle, the force from the crank on the connecting rod will be at its potential maximum due to the leverage of the crank relative to the piston rod. All of the force from the crank at 90 degrees is parallel to the piston movement. Nearer to TDC, less and less force is going 'up' to compress the rod, as compared to when it is at 90 degrees. All the force from the crank at TDC is perpendicular to the piston travel (meaning at TDC, there is no force being exerted on the rod to compress it if you wrench on the crankshaft).

And that would seem to verify why I have never bent a rod in a 1123 type saw, even using my foot and full leg strength on a scrench to remove a clutch. I had one beer too many on that one, and pushed it the wrong way with all my weight. That one had a rag strip stuffed in through the exhaust port, and just enough to keep the piston from clearing TDC. Once I realized my mistake (after having another beer) I reversed the scrench and the clutch popped right off. That 210 is still running fine.

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.

 

[airwolf]

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.

 

[brages]

So has anyone ever bent a connecting rod trying to remove a clutch on anything other than a cheap Stihl?