Since there are problems with 4G63 crankshafts fracturing at oil port at #4, I thought I'd share my thoughts on the stresses in our crankshafts in hopes that we can find a way to prolong its life.

The short version is the crank transfers all the engine power to the wheels through the last segment, which is the flywheel to last rod connection (#5 to #4). All diameters are thick and the rod connection is the smallest, which is where it cracks. The fracture is from oil port to end of rod connection, which tells me it is the flywheel causing more stress than that part can handle. My conclusion is that the combination of torque produced by engine with heavy launches exceed the crank strength at #4 to transfer between the two. My opinion is that #4 combustion is the straw that shocks the crank enough to crack the oil port over time. Taking a closer look, the oil port location is the cause because that compromises the strength of the crank at that connection. Since it is unlikely manufacturers will move the oil ports and waiting for changes can be years, I want to take a look at what we can do to minimize this problem.

To start, I want to set my views up so we can all follow along discuss easier. When I talk about rotation, I am viewing the crank from the accessory pulley looking from passenger side to driver's side (front is pulley assembly and rear is the flywheel). With this view, the crank rotates clockwise.

I also was to label each part of the crank starting with the front pulley area. We will label that part 0. Then where the rods connect, we will label that 1, 2, 3, and 4, corresponding to the cylinder number. And at the end where the flywheel is, let's label that 5. Here's a photo of what I'm talking about:

 

Engine basics:
A cylinder can be in suction, compression, power, or exhaust stroke. Only power stroke turns engine. Let's look at how power is delivered. Spark goes 1, 3, 4, 2. But let's reorder to get a better understanding: 3, 4, 2, 1. As you can see, power starts on inner cylinders and goes outward. Let's look at half, 3 & 4 (1 & 2 will be similar). 3 will make power for 1/2 revolution of crank, then 4 will make power next 1/2 revolution (then 2 & 1).

Let's look at what the crank sees. 3 will force crank to rotate clockwise from top to bottom. It will flex in that direction as it drags 2 and 4 to turn in same direction. Now imagine 4 switching frim compression stroke to power stroke. It will switch from being dragged to doing the drag as 3 switches from doing the dragging (power stroke) to being dragged (exhaust stroke). 0 & 5 will always be dragging (5 can do the dragging but we're not going to talk about misshifting here. ). Having a damper at pulley side is good to minimize the shock of when 1 & 2 switches from being to doing the dragging. Having a heavier flywheel is similar to a damper for 3 and 4. They are like stabilizers or shock absorbers to absorb the spikes and lows.

 

Crankshaft harmonics:
One cylinder will always be in power stroke. Each of the others will be in one of the other strokes. Let's take a closer look at a power stroke. At TDC, spark occurs and ignites the fuel mixture, which is at the highest compression. At this point, it is a violent explosion that releases massive heat energy, which expands the compressed air. As the piston moved down, the combustion chamber void increases where this burned mixed is allowed to expand, decreasing it's ability to push the piston down. So the biggest transfer of energy in the system (pistons to rods to crank) is at the beginning of ignition. So when an engine is producing 600 horsepower, the crankshaft is actually seeing MUCH more power at the rod connection because 600 hp is the average of all for cylinders continuously running. The second half of the power stroke produces no where near the first half.

Now imagine 5 when we use launch control. 4 see max drag because 5 is connected to 4. When the clutch slams shut on clutch engagement during launch control, all the energy spinning the crank hits the wall at 5. If you look at the thinkness of the crank (photo), the thinnest part closest to 5 is 4. More specifically, it is midpoint of 4 rod bearing (where the oil passage is located) to end. And that's where the fractures show up. The oil port is cut closer to the end near the flywheel. Why didn't they put it closer to other side or further back where it is stressed closer to bottom of rotation (which ia closer to end of crank [higher up] instead of middle)?

 

So why are the oil ports perpendicular to outward? At TDC during power stroke, all the combustion energy pushed downward through rod to crank and the rotational engergy (centripetal force) is outward. The holes being perpendicular doesn't see any of that power (good thing), but the bottom sees all the tension as that bearing bows toward the center of crank. It basically is stretching along the bottom (which is fine since it is solid the entire width). My question is, did manufacturers calculate the loads at different situations? Or was it just put at perpendicular to outter edges because it doesn't get affected by centripetal forces of rods swinging around?

For those that care (skip if you don't care), here's more info on centripetal force:




So how much force is rod caps pulling crank outward? Let's do a quick estimate: mass is about 1kg for rod and piston (during suction stroke).
Velocity is PI × diameter (100mm is 0.1m for 100mm cranks) × RPM (let's use 7800) × min/60 sec =
3.141592 × 0.1m × 130 rev/s = 40.84m/s
So Centripetal Force = 1kg × (40.84m/s)^2 ÷ 0.05m = 33,359 Kg-m/s^2 or 33,359 N = 7,500 pounds! (1 N = 0.22481 pounds) Each rod bolt has to hold 3,750 pounds! WHAT?!

Now recalculate using 9,000 RPM. I get 41.124 m/s. Force jumps to 9,985 pounds, so 4,993 pounds per rod bolt. There is a factor to adjust this. (Again, this is assuming rod, piston, and wrist pin weighs 1,000 g)
Is this why rod bolts stretch?

 

Let's assume rod is 150mm 25% of this is 37.5mm, which is the longest length of triangle. The bottom length of triangle is 12.5mm, which becomes 0.25. So reduce the above final forces by 25% and that's what force the crank sees. So the 3,555 pounds is reduced to 2,666 pounds per rod bolt connection. The bolt needs to be able to keep the rod and rod cap together to withstand this force (and not allow the 2 pieces to separate). You will need to ask manufacturer to check what the rod & rod cap strength is to compare.

But getting back to finding a potential cause for the fracture, we need to think about how the crank rotates. At TDC, combustion is most violent. As crank rotates energy drop. #4 is already twisting from 1-3, but add in #4 power stroke and lots of centripetal force from high revs and we see failures. This surge + powers from 1-3 is getting slammed with extreme resistance from launch control. It gets worse for those with slicks launching at redline.

 

You're very close. "Rod Journal #4 is also responsible for absorbing and transferring the power generated from piston rod assembly #4"
The difference at #4 from #1 - #3 is that #1 - #3 all have counterweights spinning with them. #4 during launch doesn't get the aid of counterweight because that energy doesn't get a chance to transfer the energy to counterweight since it is directly sent to flywheel (#5).

There are discussions about the different rods. H and I beams are the different designs. I beams are known to be stronger to resist longitudinal bending (perpendicular) forces within the rod. One side of the I sees tension and the other side sees compression. I beams have the thicker part at these stress areas. So it's the perfect rod right?

Not so fast.
Power stroke at TDC compresses the rod straight down and slightly at an angle as rod rotates. So at TDC, entire rod is compressed from wrist pin to crank. I beam doesn't have much cross sectional area at the rod side (wrist pin to bid end side) because the thickest (top and bottom parts of the I beam). Those thick parts are not in the same plane. If you look at the H beams, it has more usable cross sectional area so it can handle more power with less compression deflection. Next, let's examine as it rotates. I beam will start transferring the power through the inner I of the beam. Problem is power is getting there through the inner I so it won't make as much power. Imagine opening a door by pushing in middle of door vs pushing at knob. It takes twice as much power to open right? Same idea here. So even if we use 100mm crank, power delivery is similar to smaller stroke. So where whould I beams shine? It will stretch less at high RPMs. The design is better suited for high centripetal forces. The catch is it is more prone to twisting along the rod.
I personally will go with H beams because the cross sectional area is better to transfer the power from wrist pin to rod at any rotation position. It isn't as strong as I beams at high RPMs, but it doesn't have the mass at the edges (so reduces the need to reinforce the ends).

So, in short, that are the differences?
I beams will transfer less power to crank because it compresses more and transfers power on inner thicker cross section of rod longitudinally. This is actually better for the crank with the fracturing problem (how do you like that?), but owners will crank up the boost to compensate for the less power. [Can someone verify if I beam rods make slightly less power than H beams? My guess is it is negligible].
I expect the H beams to rev slightly faster since the cross sectional area is in the right places. Just be aware that the rod can flex more than I beams during high RPMs.

 

Don

This is very good info & an excellent "White Paper" on the theory behind crankshaft dynamics. Let me re-read it again & see if this sinks in

Well done! May have to consider to Sticky this

Joe

 

Sticky? It's only my thoughts on it. At least wait for the 100th version (this is my second draft). Others could have their own ideas. I plan to build my engine with more thoughts on how to protect #4 oil port until manufacturers fix this weak point. I have other ideas that I'll keep to myself since there are lots of people on the other side and there is no proof either side is right.

Please go through the calculations. I'm too brain dead to go through it again

 

And thinking further about piston rings, I don't see how we can prevent cylinder walls from ovaling at the bottom. Stroker crankshafts makes the piston pivoting pressures worse if wrist pins are not moved closer to the piston oil rings.

 

angular velocity = 2*pi*f

 

What's f?

 

F is frequency

easier to calculate since f is just 1/T

where T is period

 

Good point, we can simplify the formula using angular velocity × radius.

 

Frequency in revolutions per second

 

once you have w, then w = v/r

easy way to get your vector value (V)....

now that you have the V, you have to calculate your acceleration and deceleration of the rod/piston assembly as it goes up and down.
ill have to think some more on how to do that...
i think its 200 level statics tho.
haha