Some thoughts on PSD connecting rod failure...
#1
Some thoughts on PSD connecting rod failure...
In this thread...
Idm stuff again...
https://www.ford-trucks.com/forums/743297-idm-stuff-again-2.html#post6182132
... I remarked that...
...Here I was thinking about the compressive force in the rod due to the build up of combustion pressure which typically reaches a maximum in the vicinity of 15* to 20* ATDC, and that got me to wondering just what are the various types and magnitudes of forces applied to PSD rods, and how closely do they approach the maximum force the rod can tolerate.
They're two basic mechanisms that generate forces on the connecting rod. The piston is accelerated up and down in a reciprocating motion, and this applies both tensional and compressive forces to the rod which are proportional to the piston weight and to (RPM)^2. During the power stroke a large compressive force is applied to the rod that's proportional to the cylinder pressure and the piston area, and during the compression and exhaust strokes smaller compressive forces are applied.
The stress in a rod is given by the applied force divided by the rod's cross-sectional area. There is a stress region of elastic deformation where steel can be stretched, but it returns to its original size when unloaded. The load stress at which it fails to return to its original size is called the yield stress. If the yield stress is exceeded the deformation is permanent, and for slightly higher stress an ultimate stress point is reached where failure is imminent, and continued operation even at lower stress can still cause fracture. This is apparently what happened to Jeremy whose rod broke on his way home after his dyno run at Smokin.
The yield stress for structural steel is 36,250 psi, for high strength alloy steel it's 100,000 psi, and for 4340 chromoly steel it's 125,000 psi. I'll assume that PSD rods have a yield stress of at least 50,000 psi, and that their cross-sectional area is at least 0.65 in2. Based on these assumptions, a PSD rod should be able to tolerate a force of at least 32,500 lb.
Consider the rod forces required to accelerate and de-accelerate the piston weight up and down as the crankshaft spins at a given RPM without the cylinder head installed. Starting at TDC, as the crankshaft rotates 90* the piston descends about 60% of the stroke, and over the next 90* of crankshaft rotation to BDC, the piston descends the remaining 40% of the stroke. The reason for this asymmetrical up and down motion is that as the crankshaft rotates the rod journal (crankpin) moves left and right by one half the stroke.
At 3,000 RPM the mean piston speed is 2,085 fpm or 24 mph, and the maximum piston speed is 3,440 fpm or 39 mph. Of course the speed is zero at both TDC and BDC, and due to the asymmetrical motion the maximum piston speed occurs at 74* ATDC. The piston acceleration is given by the rate of change of its speed, and the piston acceleration has a sharp maximum of 22,545 ft/sec2 = 705 g at TDC, and another rather broad maximum of 12,640 ft/sec2 = 395 g in the region from 120* ATDC to BDC.
I'm estimating that a PSD piston assembly including rings and wrist pin weighs no more than 4.5 lb, and using this weight and the 705 g acceleration gives a 3,173 lb peak tensional force on the rod as the piston is in the vicinity of TDC, and the 395 g acceleration gives a 1,778 lb peak compressive force on the rod as the piston is in the vicinity of BDC. These forces are only about 10% of the estimated 32,500 lb load limit. However, at 6,000 RPM the rod forces are increased by a factor of 4, and are now about 40% of the maximum allowed.
Now the story gets more interesting, and I've used several different ways to estimate the compressive force in the rod due to the combustion process, which depends on the boost pressure which fills the cylinder, the amount and type of fuel injected, and on the injection timing. A PSD piston has an area of 13.3 in2 so that a 1,000 psi cylinder pressure, for example, produces about a 13,300 lb compressive force in the rod.
I've seen data on a 400 FWHP 6 cylinder diesel that indicated maximum pressures in excess of 2,000 psi, and Jody mentioned that he's seen maximum pressures in excess of several thousand psi when he measures a PSD on a dyno using gauges hooked into the glow plug holes. At 2,000 psi there's a 26,600 lb compressive force on the rod, and that's about 82% of the estimated load limit.
One can also estimate rod forces by considering the repetitive TQ peaks that are applied through the rods to the crankshaft to generate an average TQ at a given RPM which in turn produces a given FWHP. For a single cylinder engine, the peak TQ is x15 greater than the mean TQ, for 6 cylinders it's x3.55, and for 8 cylinders it's x2.95. The stroke for a PSD is 4.18 in, and this gives a TQ moment arm of about 2.1 in at 90*, but only about 1.3 in the vicinity of 40* ATDC where peak TQ occurs. Therefore, a given average TQ needs to be multiplied by 12/1.3 = x9.2 to get the average rod force, and by an additional x2.95 to get the peak rod force, for a net factor of about x27. So to generate an average TQ of say 800 lb-ft, requires a peak rod force of about 21,600 lb, and this value is fairly consistent with a peak pressure of 2,000 psi which occurs at 20*ATDC vs the 40* ATDC for the TQ peak.
Well, this is my best effort for know, and I've got to get ready to travel tomorrow. I'm sure that rod failure is a lot more complicated than this, and probably has a lot to do with the stress being concentrated at flaws and defects that were pre-existing or develop over time. Maybe the next time someone goes to a sled pulling event, they can use an old PSD rod to hook the sled to the truck, and see just how much force those rods can take!
Idm stuff again...
https://www.ford-trucks.com/forums/743297-idm-stuff-again-2.html#post6182132
... I remarked that...
They're two basic mechanisms that generate forces on the connecting rod. The piston is accelerated up and down in a reciprocating motion, and this applies both tensional and compressive forces to the rod which are proportional to the piston weight and to (RPM)^2. During the power stroke a large compressive force is applied to the rod that's proportional to the cylinder pressure and the piston area, and during the compression and exhaust strokes smaller compressive forces are applied.
The stress in a rod is given by the applied force divided by the rod's cross-sectional area. There is a stress region of elastic deformation where steel can be stretched, but it returns to its original size when unloaded. The load stress at which it fails to return to its original size is called the yield stress. If the yield stress is exceeded the deformation is permanent, and for slightly higher stress an ultimate stress point is reached where failure is imminent, and continued operation even at lower stress can still cause fracture. This is apparently what happened to Jeremy whose rod broke on his way home after his dyno run at Smokin.
The yield stress for structural steel is 36,250 psi, for high strength alloy steel it's 100,000 psi, and for 4340 chromoly steel it's 125,000 psi. I'll assume that PSD rods have a yield stress of at least 50,000 psi, and that their cross-sectional area is at least 0.65 in2. Based on these assumptions, a PSD rod should be able to tolerate a force of at least 32,500 lb.
Consider the rod forces required to accelerate and de-accelerate the piston weight up and down as the crankshaft spins at a given RPM without the cylinder head installed. Starting at TDC, as the crankshaft rotates 90* the piston descends about 60% of the stroke, and over the next 90* of crankshaft rotation to BDC, the piston descends the remaining 40% of the stroke. The reason for this asymmetrical up and down motion is that as the crankshaft rotates the rod journal (crankpin) moves left and right by one half the stroke.
At 3,000 RPM the mean piston speed is 2,085 fpm or 24 mph, and the maximum piston speed is 3,440 fpm or 39 mph. Of course the speed is zero at both TDC and BDC, and due to the asymmetrical motion the maximum piston speed occurs at 74* ATDC. The piston acceleration is given by the rate of change of its speed, and the piston acceleration has a sharp maximum of 22,545 ft/sec2 = 705 g at TDC, and another rather broad maximum of 12,640 ft/sec2 = 395 g in the region from 120* ATDC to BDC.
I'm estimating that a PSD piston assembly including rings and wrist pin weighs no more than 4.5 lb, and using this weight and the 705 g acceleration gives a 3,173 lb peak tensional force on the rod as the piston is in the vicinity of TDC, and the 395 g acceleration gives a 1,778 lb peak compressive force on the rod as the piston is in the vicinity of BDC. These forces are only about 10% of the estimated 32,500 lb load limit. However, at 6,000 RPM the rod forces are increased by a factor of 4, and are now about 40% of the maximum allowed.
Now the story gets more interesting, and I've used several different ways to estimate the compressive force in the rod due to the combustion process, which depends on the boost pressure which fills the cylinder, the amount and type of fuel injected, and on the injection timing. A PSD piston has an area of 13.3 in2 so that a 1,000 psi cylinder pressure, for example, produces about a 13,300 lb compressive force in the rod.
I've seen data on a 400 FWHP 6 cylinder diesel that indicated maximum pressures in excess of 2,000 psi, and Jody mentioned that he's seen maximum pressures in excess of several thousand psi when he measures a PSD on a dyno using gauges hooked into the glow plug holes. At 2,000 psi there's a 26,600 lb compressive force on the rod, and that's about 82% of the estimated load limit.
One can also estimate rod forces by considering the repetitive TQ peaks that are applied through the rods to the crankshaft to generate an average TQ at a given RPM which in turn produces a given FWHP. For a single cylinder engine, the peak TQ is x15 greater than the mean TQ, for 6 cylinders it's x3.55, and for 8 cylinders it's x2.95. The stroke for a PSD is 4.18 in, and this gives a TQ moment arm of about 2.1 in at 90*, but only about 1.3 in the vicinity of 40* ATDC where peak TQ occurs. Therefore, a given average TQ needs to be multiplied by 12/1.3 = x9.2 to get the average rod force, and by an additional x2.95 to get the peak rod force, for a net factor of about x27. So to generate an average TQ of say 800 lb-ft, requires a peak rod force of about 21,600 lb, and this value is fairly consistent with a peak pressure of 2,000 psi which occurs at 20*ATDC vs the 40* ATDC for the TQ peak.
Well, this is my best effort for know, and I've got to get ready to travel tomorrow. I'm sure that rod failure is a lot more complicated than this, and probably has a lot to do with the stress being concentrated at flaws and defects that were pre-existing or develop over time. Maybe the next time someone goes to a sled pulling event, they can use an old PSD rod to hook the sled to the truck, and see just how much force those rods can take!
#5
Actually, as explained below they only fail by sucking in high RPM gassers. In diesels, they get the crap pounded out of them and bend. However, you're correct in that the I-Beam rod is the most popular type of connecting rod because it's cheap to make but still fairly reliable for stock applications.
On the other hand, the more expensive H-Beam rods can better handle very large compressive loads. For high RPM gassers, rods typically fail by being torn apart in tension due to the large force needed to accelerate the pistons during their intake strokes, which is where the sucking takes place!
As shown in my analysis, for relatively low RPM high HP diesels, rods encounter much higher loads in compression than they do in tension. The tensile strength of a rod is almost entirely independent of its shape, and depends only on its cross-sectional area and material. Therefore, for high RPM gassers the rod shape (I-Beam or H-beam) has little bearing on it's likelihood of failing during sucking, as long as they're made from the same material and have equal cross sectional areas, but for high HP diesels H-beam rods are preferred.
On the other hand, the more expensive H-Beam rods can better handle very large compressive loads. For high RPM gassers, rods typically fail by being torn apart in tension due to the large force needed to accelerate the pistons during their intake strokes, which is where the sucking takes place!
As shown in my analysis, for relatively low RPM high HP diesels, rods encounter much higher loads in compression than they do in tension. The tensile strength of a rod is almost entirely independent of its shape, and depends only on its cross-sectional area and material. Therefore, for high RPM gassers the rod shape (I-Beam or H-beam) has little bearing on it's likelihood of failing during sucking, as long as they're made from the same material and have equal cross sectional areas, but for high HP diesels H-beam rods are preferred.
#7
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#9
On the other hand, the more expensive H-Beam rods can better handle very large compressive loads. For high RPM gassers, rods typically fail by being torn apart in tension due to the large force needed to accelerate the pistons during their intake strokes, which is where the sucking takes place!
#14
Great post, BTW!!! Makes me all the more likely to "settle" for something around 375RWHP instead of pushing the 400HP mark. Basically, where Kris is now.
#15
Check this link...
H-Beam Vs I Beam
Performance Connecting Rods
"But when you add the fact that some components of the combustion event attempt to screw the piston down the cylinder, the greater distance from the centerline of the pin to the edge of the beam gives the H-beam an advantage in resisting such twisting forces."
For equal material, cross-sectional area, etc... H beam is always the better choice for a high HP diesel because a diesel see's very high compressive loads, but because the RPM is low it see's only modest tension loads.
Remember, the tension load increases as (RPM)^2 so that it's a factor x4 higher for each doubling of the RPM. At a high enough RPM, gassers see more tension than compression, but diesels always see much more compression, and that's where H beam is always a superior geometry to handle the compression! The H beam resists the buckling and twisting forces of a compressive load better than does an I beam, everything else being equal.
H-Beam Vs I Beam
Performance Connecting Rods
"But when you add the fact that some components of the combustion event attempt to screw the piston down the cylinder, the greater distance from the centerline of the pin to the edge of the beam gives the H-beam an advantage in resisting such twisting forces."
For equal material, cross-sectional area, etc... H beam is always the better choice for a high HP diesel because a diesel see's very high compressive loads, but because the RPM is low it see's only modest tension loads.
Remember, the tension load increases as (RPM)^2 so that it's a factor x4 higher for each doubling of the RPM. At a high enough RPM, gassers see more tension than compression, but diesels always see much more compression, and that's where H beam is always a superior geometry to handle the compression! The H beam resists the buckling and twisting forces of a compressive load better than does an I beam, everything else being equal.