unbalance of 4 cylinder engines 1

[gio1]

Hello

It is known that in-line 4 cylinder engines have fully balanced 1st order forces and moments and also 2nd order moments, but second order forces are unbalanced.

If you sketch the crankshaft and draw the second order forces you will see that they will point all in the same direction, and therefore are unbalanced as it should be.

However if you try to calculate the moment generated by these forces about any point on the crank axis you will end up with a non-zero value (unless you calculate the moments about the median point of the crank) while moments should be zero...

We are calculating FREE moments so reaction (bearing) forces on the crank should not be considered

where is the catch?

Thanks

 

[EHudson]

The second order forces generate a force in the ‘vertical’ direction, vertical meaning parallel to the axis of the cylinder bores in this case.

In-line 4 cylinder cranks are normally arranged symmetrically about the center bearing. The forces of the two end piston/rod assemblies produce moments about the center bearing that cancel each other. The inner cylinders cancel each other in a similar fashion. The forces then are usually balanced about the center (bearing) of the crank so there is no rocking (fore-aft) moment just forces in the vertical direction.

The typical 4 stroke in-line 4 produces secondary forces that add to each other. A ‘flat’ or opposed 4 cylinder 4 stroke, on the order of an old VW, would have secondary forces that cancel.

If you use the formula for reciprocating forces developed by binomial expansion or similar series expansion it will contain a term having: [R/L x cosine(2?)] . It is this term that is ‘responsible’ for the secondary forces. (The cosine(?) term generates the primary forces.)

Your posting suggests that you have calculated these forces and understand that they are aligned and produce no external moment. The aligned forces that you have calculated will all work together in an effort to move the engine up and down in the mounts.

 

[gio1]

Thanks for your reply

I definitely agree with your explanation. My point is: from a formal point of view, if no free moments are present then the total sum of moments should be zero *regardless* of the point along the crank about which these are calculated.

The reason for my question is that I am creating a spreadsheet for calculating 1st and 2nd order forces&moments for different types of engine configurations. Moments are always calculated with respect to one of the cranks ends and this works for all cases but not for in-line 4, where I end up with a free moment of
M=6.F.a (F being the secondary force and a the inter-bore distance) this despite the fact that the position of the point should not affect the calculations.

How can I make things work out from a mathematical point of view?

 

[GregLocock]

Well, if I take moments about cyinder one I get +aF from cyl 2, +2*a*F from cyl 3 and -3* a*F from cyl 4. Result zero.


Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[EHudson]

You are correct, the position that you choose to calculate the moments should not matter. In an in-line 4 the center bearing is the easy choice. Other engines have different points that make the calculation easier by hand.

In an in-line 3 for examplle, I like to take moments about the center of crankpins 1 & 3 in order to get the c'wt values.

You should not have a moment from secondaries in an in-line 4 with symmetry about the center bearing. Just forces. Are you summing the moments from both ends of the crank? That is, calculating moments from one end then the other and adding them? I think that is the correct approach in a general spreadsheet. Could your problem be that you are only calculating the moments from one end?

 

[gio1]

GregLocock

The contribution from cyl4 is +3.a.F because all forces are alingned in the vertical direction, so the sum is not 0


EHudson

I think my problem is what you said: I am not summing up the moments from two different points along the crank! This would automatically leave out moments generated by forces without rocking effect

Thanks

gio1

 

[GregLocock]

No it isn't. Go and look at a crankshaft. See which way which piston moves. Big clue, the end ones go up as the middle pair goes down.



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[gio1]

Here we are talking about SECONDARY forces, not primary forces!
Secondary forces will all point in the same direction because they contain the term COS(2.t) with t being the crank angle, so

for cyl 1: t=0 cos(2t)=1
for cyl 2: t=180 cos(2t)=1
for cyl 3: t=180 cos(2t)=1
for cyl 4: t=0 cos(2t)=1

so they all point vertically

 

[GregLocock]

Hang on. You know you are wrong, there are no external moments. I'm not going to do your homework for you. You have made a fundamental error somewhere. Quite where is your problem, not mine.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.