rod loading
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From: Bradenton, FL, USA
rod loading
I have a few questions about rod loading:
Is a connecting rod mostly subjected to compressive forces, or is tension just as much of a factor? I assume that the compressive force from combustion would be the greatest factor in how strong a connecting rod should be for a particular application. The only tension force the rod sees that I can think of is after the piston reaches TDC and starts the intake stroke. Is this correct?
Also, is there a way to find about how much force the rod is subjected to for a V8 engine with about 1000 fwhp?
Jon
Is a connecting rod mostly subjected to compressive forces, or is tension just as much of a factor? I assume that the compressive force from combustion would be the greatest factor in how strong a connecting rod should be for a particular application. The only tension force the rod sees that I can think of is after the piston reaches TDC and starts the intake stroke. Is this correct?
Also, is there a way to find about how much force the rod is subjected to for a V8 engine with about 1000 fwhp?
Jon
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Joined: Oct 2002
Posts: 2,931
From: Upstate NY
Re: rod loading
Originally posted by Jon 97TA
I have a few questions about rod loading:
Is a connecting rod mostly subjected to compressive forces, or is tension just as much of a factor? I assume that the compressive force from combustion would be the greatest factor in how strong a connecting rod should be for a particular application. The only tension force the rod sees that I can think of is after the piston reaches TDC and starts the intake stroke. Is this correct?
Tension is probably more of a factor in an NA applicatioin and is due to the g-loads of the piston, pin, rings and small end of the rod trying to pull themselves out of the top of the block, and is rpm dependant. The bad thing about tension is that the rod bolts and the rod cap take the loads. In compression the beam of the rod, which is much stronger, is taking all the load; the bolts and cap are along for the ride.
Also, is there a way to find about how much force the rod is subjected to for a V8 engine with about 1000 fwhp?
Sure. Some simulators give piston g's vs. rpm. Using F = ma you can calculate the tension force on the rod. For a 3.50 inch stroke engine at 6500 figure about 2700 g's on the piston at TDC trying to stretch the rod. Worst case is on the exhaust stroke with no compreession or combustion load to counteract it.
Maximum NA compressive forces sorta depend on how much detonation there is. At 1000 fwhp, the max pressure during combustion on a SBC might be 3000 psi. @ 20 or so lbs. of boost around torque peak with no detonation. This may be enough to fail the rods in compression.
On an NA engine I'd worry about tension failure, but on a 1000 hp blown engine both tension and compression are a concern.
Ask the rod manufacturer for a recommendation. I'd start with Carillo and Arrow.
I have a few questions about rod loading:
Is a connecting rod mostly subjected to compressive forces, or is tension just as much of a factor? I assume that the compressive force from combustion would be the greatest factor in how strong a connecting rod should be for a particular application. The only tension force the rod sees that I can think of is after the piston reaches TDC and starts the intake stroke. Is this correct?
Tension is probably more of a factor in an NA applicatioin and is due to the g-loads of the piston, pin, rings and small end of the rod trying to pull themselves out of the top of the block, and is rpm dependant. The bad thing about tension is that the rod bolts and the rod cap take the loads. In compression the beam of the rod, which is much stronger, is taking all the load; the bolts and cap are along for the ride.
Also, is there a way to find about how much force the rod is subjected to for a V8 engine with about 1000 fwhp?
Sure. Some simulators give piston g's vs. rpm. Using F = ma you can calculate the tension force on the rod. For a 3.50 inch stroke engine at 6500 figure about 2700 g's on the piston at TDC trying to stretch the rod. Worst case is on the exhaust stroke with no compreession or combustion load to counteract it.
Maximum NA compressive forces sorta depend on how much detonation there is. At 1000 fwhp, the max pressure during combustion on a SBC might be 3000 psi. @ 20 or so lbs. of boost around torque peak with no detonation. This may be enough to fail the rods in compression.
On an NA engine I'd worry about tension failure, but on a 1000 hp blown engine both tension and compression are a concern.
Ask the rod manufacturer for a recommendation. I'd start with Carillo and Arrow.
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Joined: Dec 1969
Posts: 1,147
From: Phila., PA
Of all the motors I've blown up quite a few were from "stretchy" or just plain snapped rod bolts. Never blew one up from a snapped connecting rod (as the primary CAUSE of the failure, not saying they didn't sometimes get snapped when the motor grenaded).
I've pushed stock rods a LONG LONG way. Massive nitrous hits, vicious over-revving. They're strong little suckers. Can't say I've ever built 1000 HP, though.
Good quality rod bolts and standard high performance oriented rod prep is all I've ever done to them. I like ARP WaveLoc rod bolts myself. I think they're about the best rod bolts a weekend warrior or semi-serious racer can lay his hands on for reasonably money. Cheap insurance. They're the ONLY thing I'll put in a short-rod 400 and I've pushed some of them to 7000 RPMs for short periods (drag racing) without failure.
I've pushed stock rods a LONG LONG way. Massive nitrous hits, vicious over-revving. They're strong little suckers. Can't say I've ever built 1000 HP, though.
Good quality rod bolts and standard high performance oriented rod prep is all I've ever done to them. I like ARP WaveLoc rod bolts myself. I think they're about the best rod bolts a weekend warrior or semi-serious racer can lay his hands on for reasonably money. Cheap insurance. They're the ONLY thing I'll put in a short-rod 400 and I've pushed some of them to 7000 RPMs for short periods (drag racing) without failure.
Thread Starter
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Joined: Oct 2000
Posts: 49
From: Bradenton, FL, USA
I'm not sure how to word this, so if you don't understand it tell me what you don't understand, so I can clearify:
The reason I'm asking these questions is I have a project to do, where we are modeling the forces that a connecting rod will see. We just need a general idea of whether the force(s) we will have to worry about will be compression, tension, or both(can we ignore one or the other, or not?) and about how much force will actually be seen on the rod. We are using algor(an FEA program) and need to know what is the greatest force the rod will see to get an accurate representation.
oldsstroker:
Using a 500 gram piston, ls1 crank(3.62 stroke), and 6700 rpm(about 3000 g's), I found about 107,000 lbs of force on the rod. This seems like a lot, is it right?
here's what I used:
500g*(1 oz/28 g)*(1 lb/16 oz)* (32.2 ft/s^2)*3000=~107000 lb*ft/s^2
Also, after talking to a professor about this, he said that he would have thought that the compressive forces from combustion should be higher than the forces the rod experiences just due to the acceleration and deceleration of the components(piston, pin, rings, etc), I was thinking the same thing.
On a side note:
I think if I could find how much the cylinder pressure is, I could use pressure = force/(piston surface area), to find the combustion force.
Thanks for the responses so far, they have helped a lot
Jon
The reason I'm asking these questions is I have a project to do, where we are modeling the forces that a connecting rod will see. We just need a general idea of whether the force(s) we will have to worry about will be compression, tension, or both(can we ignore one or the other, or not?) and about how much force will actually be seen on the rod. We are using algor(an FEA program) and need to know what is the greatest force the rod will see to get an accurate representation.
oldsstroker:
Using a 500 gram piston, ls1 crank(3.62 stroke), and 6700 rpm(about 3000 g's), I found about 107,000 lbs of force on the rod. This seems like a lot, is it right?
here's what I used:
500g*(1 oz/28 g)*(1 lb/16 oz)* (32.2 ft/s^2)*3000=~107000 lb*ft/s^2
Also, after talking to a professor about this, he said that he would have thought that the compressive forces from combustion should be higher than the forces the rod experiences just due to the acceleration and deceleration of the components(piston, pin, rings, etc), I was thinking the same thing.
On a side note:
I think if I could find how much the cylinder pressure is, I could use pressure = force/(piston surface area), to find the combustion force.
Thanks for the responses so far, they have helped a lot
Jon
Registered User
Joined: Dec 1969
Posts: 1,147
From: Phila., PA
I'm not gonna jump in SStroker's area but just to maybe give a little frame of reference to talk around this topic.....
On the power stroke the force of combustion pushing down on the piston has to be greater than the tesnion genrated by sheer RPM form decelerating and then pulling the piston down the bore away from TDC again at high RPMs to generate ANY compression on the rod. In other words, you have to be pushing WICKED HARD on the top of the piston during combustion just to BREAK EVEN with the g-forces (tension) on the rod from decelerating and then accelerating the assembly away from TDC again.
However, on the EXHAUST stroke there is very little pressure above the piston so the rod (and rod bolts especially) see full effect of RPM-generated g-forces trying to "stretch out" the rod and rod bolts as the crank slows the piston down and then pulls it back from TDC.
Tension kills and it happens almost purely from weight and RPM, regardless of power level produced and it kills on the exhaust stroke when there's no pressure above the piston. Compressive forces on the rod from combustion are somewhat or entirely offset by these same tension-generating forces during the power stroke at least at low-modest power levels.
You want the math on this? Well, that's not my area. I'm sure someone could help out with that. I think I did a not too terrible job of explaining it from the 30,000 foot level.
On the power stroke the force of combustion pushing down on the piston has to be greater than the tesnion genrated by sheer RPM form decelerating and then pulling the piston down the bore away from TDC again at high RPMs to generate ANY compression on the rod. In other words, you have to be pushing WICKED HARD on the top of the piston during combustion just to BREAK EVEN with the g-forces (tension) on the rod from decelerating and then accelerating the assembly away from TDC again.
However, on the EXHAUST stroke there is very little pressure above the piston so the rod (and rod bolts especially) see full effect of RPM-generated g-forces trying to "stretch out" the rod and rod bolts as the crank slows the piston down and then pulls it back from TDC.
Tension kills and it happens almost purely from weight and RPM, regardless of power level produced and it kills on the exhaust stroke when there's no pressure above the piston. Compressive forces on the rod from combustion are somewhat or entirely offset by these same tension-generating forces during the power stroke at least at low-modest power levels.
You want the math on this? Well, that's not my area. I'm sure someone could help out with that. I think I did a not too terrible job of explaining it from the 30,000 foot level.
Last edited by Damon; Nov 4, 2003 at 10:01 PM.
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Joined: Dec 1969
Posts: 1,147
From: Phila., PA
BTW- the force from combustion doesn't have to be greater than the RPM-generated tension forces on the rod to generate power.
It took my brain a LONG time to wrap around that idea, but it's true. All those forces going on accelerating and decelerating stuff inside the engine. Easy to get lost in it. Just remember that they all net out at ZERO (in the perfect, friction-free theoretical world). So you don't have to overcome the tension-generating g-forces, to make power. You just have to put SOME force pushing down on the piston during the combustion stroke and you'll be making power when you net it all out. All the accleration and deceleration will net out at zero and what you'll be left with is the force pushing down on the piston that you generate during the combustion stroke.
HELP!!! I'm once again officially over my head. I don't think I'm doing such a hot job explaining this.
It took my brain a LONG time to wrap around that idea, but it's true. All those forces going on accelerating and decelerating stuff inside the engine. Easy to get lost in it. Just remember that they all net out at ZERO (in the perfect, friction-free theoretical world). So you don't have to overcome the tension-generating g-forces, to make power. You just have to put SOME force pushing down on the piston during the combustion stroke and you'll be making power when you net it all out. All the accleration and deceleration will net out at zero and what you'll be left with is the force pushing down on the piston that you generate during the combustion stroke.
HELP!!! I'm once again officially over my head. I don't think I'm doing such a hot job explaining this.
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Joined: Oct 2002
Posts: 2,931
From: Upstate NY
Originally posted by Jon 97TA
The reason I'm asking these questions is I have a project to do, where we are modeling the forces that a connecting rod will see. We just need a general idea of whether the force(s) we will have to worry about will be compression, tension, or both(can we ignore one or the other, or not?) and about how much force will actually be seen on the rod. We are using algor(an FEA program) and need to know what is the greatest force the rod will see to get an accurate representation.
oldsstroker:
Using a 500 gram piston, ls1 crank(3.62 stroke), and 6700 rpm(about 3000 g's), I found about 107,000 lbs of force on the rod. This seems like a lot, is it right?
I think 500 gm is low for the piston, pin (140 gm for a strong one), rings and locks if you are talking about a 350 engine of 600++ blown hp. I'd use 600 gm or more.
Maybe you are modelling a small 4 cyl with 6700 max rpm. If you have your choice of engines, choose something you'd like to have, or something that exists, and see what you come up with.
On the other hand, Cup engines turn 9500+, have 525 gm min. rod weight (I think), and the lightest pistons and pins money can buy for their 800 hp application and NASCAR rules. This might make the piston/pin/ring weight around 450 gm, so 500 isn't that far off.
I'd certainly design for a overspeed, like one gets on a 4-1 downshift when a 4-3 was intended. The C5 M6 driveshaft has a bumper just for this reason. Bent pushrods are more easily fixed than broken rods!
If you really want to go nuts, look at a Top Fuel engine. Unimaginable compressive forces during power on: plugs melt about half way into the run, and the nitro is reacting in a couple of modes with burning the least violent one. Combustion continues into the exhaust stroke!
At the end of the run, or if the driver pedals, the engine goes from full power to almost no power while still at 8000 rpm.
Or how about Pro Stock. Probably 10,000 rpm with 4.50 diameter or bigger diameter pistons that will support 1300+ hp. Now there are some loads!
here's what I used:
500g*(1 oz/28 g)*(1 lb/16 oz)* (32.2 ft/s^2)*3000=~107000 lb*ft/s^2
Hmmm. lb*ft/s^2 isn't a pound of force is it?
Using the MKS system, with 500 gm piston mass, I get:
500gm * kg/1000 gm * 3000 g * 9.8 M/sec^2*g = 14700 kg*M/sec^2 (which is Newtons) * .2248 lbs (force) /Newton = 3304 lbf (pounds force)
That's your answer divided by acc. of gravity.
I may be a little fuzzy on this, but I think I'm correct.
FWIW:
If you had 107,000 lbf pulling on a rod with a minimum cross-sectional area of about .3 in^2 (which is generous), the stress would be over 350,000 psi. Oops! Boom! Additionally the stress on 2 rod bolts .375 dia. would be about 485,000 psi each which is about twice the failure point of EXCELLENT rods loaded once.
Also, after talking to a professor about this, he said that he would have thought that the compressive forces from combustion should be higher than the forces the rod experiences just due to the acceleration and deceleration of the components(piston, pin, rings, etc), I was thinking the same thing.
As others pointed out, It's the rod cap and bolts that are the weak points in tension loading. Theoretically you don't even need a rod cap installed during the compression loading. Rod beams are fairly easy to make strong enough in compression unless you are at the outer ends of the envelope like the 525 gm steel Cup rod.
On a side note:
I think if I could find how much the cylinder pressure is, I could use pressure = force/(piston surface area), to find the combustion force.
Thanks for the responses so far, they have helped a lot
Jon
The reason I'm asking these questions is I have a project to do, where we are modeling the forces that a connecting rod will see. We just need a general idea of whether the force(s) we will have to worry about will be compression, tension, or both(can we ignore one or the other, or not?) and about how much force will actually be seen on the rod. We are using algor(an FEA program) and need to know what is the greatest force the rod will see to get an accurate representation.
oldsstroker:
Using a 500 gram piston, ls1 crank(3.62 stroke), and 6700 rpm(about 3000 g's), I found about 107,000 lbs of force on the rod. This seems like a lot, is it right?
I think 500 gm is low for the piston, pin (140 gm for a strong one), rings and locks if you are talking about a 350 engine of 600++ blown hp. I'd use 600 gm or more.
Maybe you are modelling a small 4 cyl with 6700 max rpm. If you have your choice of engines, choose something you'd like to have, or something that exists, and see what you come up with.
On the other hand, Cup engines turn 9500+, have 525 gm min. rod weight (I think), and the lightest pistons and pins money can buy for their 800 hp application and NASCAR rules. This might make the piston/pin/ring weight around 450 gm, so 500 isn't that far off.
I'd certainly design for a overspeed, like one gets on a 4-1 downshift when a 4-3 was intended. The C5 M6 driveshaft has a bumper just for this reason. Bent pushrods are more easily fixed than broken rods!
If you really want to go nuts, look at a Top Fuel engine. Unimaginable compressive forces during power on: plugs melt about half way into the run, and the nitro is reacting in a couple of modes with burning the least violent one. Combustion continues into the exhaust stroke!
At the end of the run, or if the driver pedals, the engine goes from full power to almost no power while still at 8000 rpm.
Or how about Pro Stock. Probably 10,000 rpm with 4.50 diameter or bigger diameter pistons that will support 1300+ hp. Now there are some loads!
here's what I used:
500g*(1 oz/28 g)*(1 lb/16 oz)* (32.2 ft/s^2)*3000=~107000 lb*ft/s^2
Hmmm. lb*ft/s^2 isn't a pound of force is it?
Using the MKS system, with 500 gm piston mass, I get:
500gm * kg/1000 gm * 3000 g * 9.8 M/sec^2*g = 14700 kg*M/sec^2 (which is Newtons) * .2248 lbs (force) /Newton = 3304 lbf (pounds force)
That's your answer divided by acc. of gravity.
I may be a little fuzzy on this, but I think I'm correct.
FWIW:
If you had 107,000 lbf pulling on a rod with a minimum cross-sectional area of about .3 in^2 (which is generous), the stress would be over 350,000 psi. Oops! Boom! Additionally the stress on 2 rod bolts .375 dia. would be about 485,000 psi each which is about twice the failure point of EXCELLENT rods loaded once.
Also, after talking to a professor about this, he said that he would have thought that the compressive forces from combustion should be higher than the forces the rod experiences just due to the acceleration and deceleration of the components(piston, pin, rings, etc), I was thinking the same thing.
As others pointed out, It's the rod cap and bolts that are the weak points in tension loading. Theoretically you don't even need a rod cap installed during the compression loading. Rod beams are fairly easy to make strong enough in compression unless you are at the outer ends of the envelope like the 525 gm steel Cup rod.
On a side note:
I think if I could find how much the cylinder pressure is, I could use pressure = force/(piston surface area), to find the combustion force.
Thanks for the responses so far, they have helped a lot
Jon
Jon (aka Old SStroker)
Thread Starter
Registered User
Joined: Oct 2000
Posts: 49
From: Bradenton, FL, USA
Oops, I forgot 1 lb/16 oz was already a pound force. 3304 makes a lot more sense than 107000.
It all makes sense now.
Thanks for all the good info.
I'll get an A for sure now.
Jon
It all makes sense now.
Thanks for all the good info.
I'll get an A for sure now.Jon
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Joined: Jul 2001
Posts: 727
From: Wolverine Lake, MI
Jon, you're getting some good replies here, but for some quick numbers to throw at you to help your calculations:
1. For compression loading a good assumption for peak cylinder pressure, at around 15 deg. ATDC is 950 psi for a stock engine.
2. The peak loading on rod components will almost always be as OldSS says, at peak overrev of the engine and no throttle - fuel shutoff. That is a rod bolt killer. Cylinder pressure here will be at a minimum and rpm will be highest. When we do FEA runs for this kind of condition we assume motoring cylinder pressure to be the worst case. A guesstimate for that would be 400-500 psi if I remember right.
3. Don't forget that the maximum tension loading on the rod will focus on the cap/bolt area. That loading isn't just your piston/pin. It also includes the whole mass of the rod above the portion of the cap you're looking at. That can easily add 500g of mass to your calculations when you're looking at the loading on the bolt area.
4. I agree with OldSS's guess of 600g for the piston/pin for the LS1. I believe the piston is well over 400g, and I'm sure the piston pin is closer to 150g. Also, the rings should weigh 40g or so.
1. For compression loading a good assumption for peak cylinder pressure, at around 15 deg. ATDC is 950 psi for a stock engine.
2. The peak loading on rod components will almost always be as OldSS says, at peak overrev of the engine and no throttle - fuel shutoff. That is a rod bolt killer. Cylinder pressure here will be at a minimum and rpm will be highest. When we do FEA runs for this kind of condition we assume motoring cylinder pressure to be the worst case. A guesstimate for that would be 400-500 psi if I remember right.
3. Don't forget that the maximum tension loading on the rod will focus on the cap/bolt area. That loading isn't just your piston/pin. It also includes the whole mass of the rod above the portion of the cap you're looking at. That can easily add 500g of mass to your calculations when you're looking at the loading on the bolt area.
4. I agree with OldSS's guess of 600g for the piston/pin for the LS1. I believe the piston is well over 400g, and I'm sure the piston pin is closer to 150g. Also, the rings should weigh 40g or so.
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Joined: Oct 2002
Posts: 2,931
From: Upstate NY
Originally posted by 94bird
4. I agree with OldSS's guess of 600g for the piston/pin for the LS1. I believe the piston is well over 400g, and I'm sure the piston pin is closer to 150g. Also, the rings should weigh 40g or so.
4. I agree with OldSS's guess of 600g for the piston/pin for the LS1. I believe the piston is well over 400g, and I'm sure the piston pin is closer to 150g. Also, the rings should weigh 40g or so.
Sorry, I just had to do that.
Good points, 94bird. Especially the bit about everything above the rod split line trying to stretch the bolts and cap at TDC reversal.
Jon (the old one)
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