Bore, Stroke, and block strength

[Circlotron]

x2 on what Madbill said. Same amount of torque equals same amount of stress on mains.

That is the opposite of what I thought he said...

Tom is correctly interpreting my words.

Let's try this: Torque results from a force applied to one end of a moment arm (in this case the crank throw) but that force must be balanced by an equal reaction torque at the other end, in this case the force being provided by the main journal and bolts.

Example: imagine trying to torque a bolt with a swivel socket and extension on your ratchet. Can you do this properly without supporting the ratchet head with your other hand? (Hint: NO.) Does the requires support vary with the applied torque? (Yes) Does it diminish if the torque remains the same, but the ratchet handle is lengthened? (If it does, do you no longer need to support the ratchet head at all if the handle is infinitely long?)

Now I'm not as sure as I was before, but I'm still leaning toward the support point needing proportionally reduced support. I don't care who's right, me or someone else, as long as we find the truth of the matter. Thank you for illustrating this *excellent* subject to think about.

[Ken_Parkman]

On a simplistic basis a force diagram will tell you at a given cylinder pressure the torque will be the same for the same cubic inches. But if this is achieved small bore/large stroke combination the force exerted on the piston and therefore the crankpin is smaller. Because of the longer stroke the torque is the same, but the force is smaller. As the vertical component is smaller, the vertical load on the main bearing from cylinder pressure is smaller.

Of course as stroke is longer it also means the acceleration/inertia loads are higher as those are linear with stroke.

So main bearing loads due to cylinder pressure go down with a small bore/long stroke, but inertial loads go up. Assuming other things constant which are usually not.

[digger]

as above so it comes down to where the dominant loads are, normally loads due to gas pressure are a fair bit higher than inertia loads depending on specifics however typically north of 8000 rpm give or take it can cross over on a NA such that inertia loads are higher

[Circlotron]

Example: imagine trying to torque a bolt with a swivel socket and extension on your ratchet. Can you do this properly without supporting the ratchet head with your other hand? (Hint: NO.) Does the requires support vary with the applied torque? (Yes) Does it diminish if the torque remains the same, but the ratchet handle is lengthened? (If it does, do you no longer need to support the ratchet head at all if the handle is infinitely long?)

Let's change the situation slightly. Suppose you put a flat ring spanner on the head of a bolt that is screwed down flush with a surface. Put 100 lb force on a 1 foot arm or 10 lb force on a 10 foot arm. The torque applied to the bolt head is the same in both cases of course, but is the lateral force applied to bolt head the same or different? Is there in fact any lateral force?

[Circlotron]

^^ Further to this, suppose we had a ring spanner that extended both sides of the bolt head, sort of like a box spanner or a wheel brace for undoing wheel nuts. Lets say one end is 1 foot long with 100 lb force applied and the other end is 10 foot long with 10 lb force applied. Total torque is 100+100 lb/ft. Seeing they are trying to rotate the same direction around the axis of the bolt there will presumably be no lateral force on the bolt despite 10 lb vs 100 lb force applied to opposite ends of the ring spanner. Comments please.

[Juho_]

Yes the cranks are very strong, but the blocks are known to crack at the water pump area. That is on the smaller bore engines that don't have the cylinder wall splitting problem.

All the same it is interesting, to me, that when they went for 300hp/300ft-lb they upped the stroke from 90 to 93.something. And the next series of engines at that power level kept the long stroke but reverted the bore to 81mm and these blocks don't crack. If mine ever cracks, it will get one of those 81mm bore blocks, but given that the engine has done nearly the equivalent of driving to the Moon, it's probably not going to.

I haven't heard about 81mm bore block cracking issues before.
They have been run up to ~150hp per cylinder without issues. Of course any knock etc. condition will cause damage -> crack very easily.

Volvo probably wanted to have more power at low rpms.
6 cylinder variant in the early form got replaced in 1995 I think with 2.5l and 3l (2.9) versions which had smaller cams and egr stuff also.. the ecu changed too at some point ('96?)

And the smaller cams because customers did not really like the lack of bottom end torque.
That's what I've read - perhaps the same goes in 5cyl variant but just longer stroke?

2.4l diesel variant also uses 93.2mm stroke and has always used if I remember correctly.
2.5l is a VW 10 valve diesel and it's a different thing.

[Tom Walker]

2 engines of same displacement but one small bore/long stroke, other engine big bore/short stroke.
What the small bore engine lacks in piston surface area is made up by the leverage of the long stroke.
What the short stroke engine lacks in leverage is made up by the big bores' greater surface area of the piston.
All else being equal for this simple analogy, both will have identical torque output, identical stress on the mains.
Stress at the rod journals with the 2 different stroke cranks putting out identical torque will be different, stress at the mains I believe to be the same.
Thats my story and I am sticking to it

[digger]

free body diagram of static condition for single cylinder (cirlce on LHS is main brg on RHS is crank pin)

small bore long stroke has less force
large bore short stroke has more force

but the torque is the same as the proportional increase in force due to the larger bore is exactly offset by the reduced radius



this is a simplified example to illustrate basics

[midnightbluS10]

Example: imagine trying to torque a bolt with a swivel socket and extension on your ratchet. Can you do this properly without supporting the ratchet head with your other hand? (Hint: NO.) Does the requires support vary with the applied torque? (Yes) Does it diminish if the torque remains the same, but the ratchet handle is lengthened? (If it does, do you no longer need to support the ratchet head at all if the handle is infinitely long?)

Let's change the situation slightly. Suppose you put a flat ring spanner on the head of a bolt that is screwed down flush with a surface. Put 100 lb force on a 1 foot arm or 10 lb force on a 10 foot arm. The torque applied to the bolt head is the same in both cases of course, but is the lateral force applied to bolt head the same or different? Is there in fact any lateral force?

What's a flat ring spanner? We don't call any of our tools spanners in the US. I assume it's some sort of wrench.

[Circlotron]

Example: imagine trying to torque a bolt with a swivel socket and extension on your ratchet. Can you do this properly without supporting the ratchet head with your other hand? (Hint: NO.) Does the requires support vary with the applied torque? (Yes) Does it diminish if the torque remains the same, but the ratchet handle is lengthened? (If it does, do you no longer need to support the ratchet head at all if the handle is infinitely long?)

Let's change the situation slightly. Suppose you put a flat ring spanner on the head of a bolt that is screwed down flush with a surface. Put 100 lb force on a 1 foot arm or 10 lb force on a 10 foot arm. The torque applied to the bolt head is the same in both cases of course, but is the lateral force applied to bolt head the same or different? Is there in fact any lateral force?

What's a flat ring spanner? We don't call any of our tools spanners in the US. I assume it's some sort of wrench.

One that doesn't have an offset at each end.
Like this:


Not like this:

[strokersix]

free body diagram of static condition for single cylinder (cirlce on LHS is main brg on RHS is crank pin)

small bore long stroke has less force
large bore short stroke has more force

but the torque is the same as the proportional increase in force due to the larger bore is exactly offset by the reduced radius



this is a simplified example to illustrate basics

Thanks for the FBD, digger. This is what I've been trying to say several times with words. Hope the illustration will help others.

[Tom Walker]

The old saying"A picture is worth a thousand words" is so true. Digger just saved many words!

[peejay]

x2 on what Madbill said. Same amount of torque equals same amount of stress on mains.

That is the opposite of what I thought he said...

Tom is correctly interpreting my words.

Let's try this: Torque results from a force applied to one end of a moment arm (in this case the crank throw) but that force must be balanced by an equal reaction torque at the other end, in this case the force being provided by the main journal and bolts.

Example: imagine trying to torque a bolt with a swivel socket and extension on your ratchet. Can you do this properly without supporting the ratchet head with your other hand? (Hint: NO.) Does the requires support vary with the applied torque? (Yes) Does it diminish if the torque remains the same, but the ratchet handle is lengthened? (If it does, do you no longer need to support the ratchet head at all if the handle is infinitely long?)

The answer to your final question is yes. Equal and opposite reaction, and all that. If the handle was infinitely long, the amount of force on the handle would be infinitely small, and that same infinitely small reaction force would be needed to stabilize the other end.