[Jukis]

For hsc reduction after shim has bent maximum you can use lower viscosity oil. For lsc from mid valve it's a lot about faster response. That's why less float is better, also for providing more actual lsc. To reduce all float there would need to be more ifp pressure aka preloaded ifp spring. For inconsistency I mean that the damping isn't linear or digressive but becomes very progressive at parts of the curve, those translate into spikes for fast hits.

 

[chehlarov]

You make me wondering. Beofre last service, I ran 0.1mm mid valve shim with 0.4mm float.
Now I changed to 0.1mm mid valve shim with 0.6mm float. I am quite happy of the change - the fork feels very reactive and plush on rough terrain. It might be due to the service of lowers / airspring.

 

[Jukis]

When the damper reacts with delay it means it feels plush on single hit but it makes it go too deep on multiple hits to recover fast enough and it will reach the hard section of the spring too soon.

 

[Dano91]

For hsc reduction after shim has bent maximum you can use lower viscosity oil. For lsc from mid valve it's a lot about faster response. That's why less float is better, also for providing more actual lsc. To reduce all float there would need to be more ifp pressure aka preloaded ifp spring. For inconsistency I mean that the damping isn't linear or digressive but becomes very progressive at parts of the curve, those translate into spikes for fast hits.

Interesting, although it's a different damper, Ch2.1 RC2, when I tune one, I'm aiming for complete opposite - to increase the float, lower the damping on mid valve and do as much of damping on base valve as possible.

 

[Jukis]

Interesting, although it's a different damper, Ch2.1 RC2, when I tune one, I'm aiming for complete opposite - to increase the float, lower the damping on mid valve and do as much of damping on base valve as possible.

Mid valve is always faster and with float you aren't reducing hsc anyways. Lsc is always welcome for me, as long as it isn't harsh. Reason that base valve has delay for compression is that there is always air in the oil and air is compressible.

But you gotta remember that charger 2.1 does not have back pressure from ifp so if you have any bleed open in the basevalve you must always have enough float or the midvalve will cause cavitation for slow speed movements.

To be honest I didn't know about the damper reaction importance before Fox engineers started to talk about it and I had just happened to tune my damper in the same direction but for different reasons.. But it does work, I can feel the fork dampening even on this rocky section between streets, where there is rocks roughly 20cm from each other and I have never managed to get my fork to do more than one compression on them, now it compresses for each one and returns enough to be able to compress again. Overall the fork feels stiff(I have quite strong basevalve stack and only 1.4mm bleed hole for the adjuster) , but my hands aren't sore even after 4 hour mtb ride.

 

[chehlarov]

When the damper reacts with delay it means it feels plush on single hit but it makes it go too deep on multiple hits to recover fast enough and it will reach the hard section of the spring too soon.

I use a custom rebound stack with two stages. The LSR is only one shim 0.15mm, followed by 0.2mm gap and 2 * 0.15mm for HSR. The idea I had is to recover as fast as possible at small velocities, like in the case you described. I even think to reduce LSR further on next service (17.35x0.1 + 13.60x0.1).

 

[Dano91]

I use a custom rebound stack with two stages. The LSR is only one shim 0.15mm, followed by 0.2mm gap and 2 * 0.15mm for HSR. The idea I had is to recover as fast as possible at small velocities, like in the case you described. I even think to reduce LSR further on next service (17.35x0.1 + 13.60x0.1).

Although I never saw Grip damper IRL and I don't know it's geometry, 0.2mm thick crossover seems quite thick to me. Will it ever reach the 2nd stage?

 

[chehlarov]

I was wondering the same. To be more precise here is the change I recently made on the rebound of GRIP:

Before the change I rode with ~10clicks from open, now with~6clicks it has the same rebound. I test the rebound by compressing at ~40% and rapidly release the front wheel to check for wheel lift.
I actually tend to ride the new setup with ~10clicks as well and feels OK.
The addition of the 12mm shim clearly reduced rebound damping. Next change will be with two 8x0.1 shims.

p.s. Pike manual from 2014-2017 has light rebound tune, where 0.2mm cross over is set. Though it is different architecture.

p.s. What is "Grip damper IRL"?

 

[Dano91]

If you want, I can check your rebound tune behavior in ReStackor, I just need the piston geometry or at least some high quality photo of it. Send me PM for more info if you re interested.

p.s. What is "Grip damper IRL"?

Sorry, I just meant that I never saw Grip damper In Real Life

 

[Jukis]

Although I never saw Grip damper IRL and I don't know it's geometry, 0.2mm thick crossover seems quite thick to me. Will it ever reach the 2nd stage?

There is lots of flow going through midvalve in both directions on this damper, tons of shim lift. It's due the large piston diameter and relatively small shaft diameter (20/10)

 

[chehlarov]

I could but I won't. I have already told you too much. I have simulator data on how it collides, but it's not my data. It collides because there is a huge flow and the shim bends a lot.

It's completely different damping on basevalve when you are still using perimeter piston instead of 4 port when piston is sanded flat. Yes my adjuster is now lsc bleed adjuster.

I was thinking on the risk of midvalve shim colliding the piston. I try to make rough estimation of the shim bending. Assuming:

• my weight is 650N (65kg);
• half of the weight is transfered to fork;
• the peak spring force is about 3 times the sage spring force;
• peak damping force is same as peak spring force;
• midvalvel shim produces half of the damping force (other half beeing basevalve and midvalve orifice damping)

650 / 2 * 3 / 2 = 487.5N should be produced from the midvalve shim bending.
I made an FEA model of a shim 16.6 x 7 x 0.1 (I use a little smaller OD than stock). The shim is supported on the inner edge and normal pressure is applied to produce the desired reaction of 487.5N.

The shim bends about 0.93mm at the OD. I don't see any collision risk. Here is an overlay:

Doubling the weight and hence the displacement might create a risk of collision with the flange. However, rider with such weight (65*2=130kg) will choose much stiffer midvalve, that will reduce shim displacement.
Looking forward your comments.

 

[Jukis]

^ There are so many things wrong in this, or I just dont understand what you are calculating. Where in your calculations you are using the 10mm shaft displacing oil for the basevalve and 20mm piston and its ports flowing through 3 ports (not a perimeter piston like you seem to have calculated, force being applied to whole perimeter of the shim, it bends in 3 spots and you cant really calculate it that simple..
Here is the actual simulation for the default grip 1 midvalve shim lift(its with a 0.15mm shim which is over 3 times as stiff as 0.1mm shim):

 

[Jukis]

double -remove

 

[upsha]

I think chehlarov was trying first to estimate the peak damping force needed for a hypothetical rider, assuming a particular BV/MV split, then back calculate what will be the corresponding mid-valve bend for a 0.1 shim. No shaft speed is involved here (or assuming this happens at the max shaft speed this rider will experience). He is not necessarily modeling what GRIP1 is doing.

He used “peak force matching” as his tuning philosophy and also chose to keep positive pressure difference from BV to MV, since he uses a split lower than the “pressure balance point”, which will be 75% force from MV using the sweep/damper cross section ratio.

I think his approximation that the whole shim face experiences the pressure difference would be fair for a lipped BV when the face is mostly flat. For MV, the port is where matters as you pointed out.

 

[Jukis]

Yeah, shims are very tough to bend if you place them ontop of a flat round tube and try to press from the middle.. Now if you take it from one side they bend easily with your fingers, its like magic.

 

[chehlarov]

upsha, you are correct - I match the expected shim force to its deformation. The Von Misses stresses from the same simulation are as follows:

Spring steels, like ASTM A 228 (one of the strongest), has design stress level of 1240MPa. From the plot above, stresses as high as 1900MPa are seen. There is no material with suitable properties for such deformation. The shim cannot withstand such extreme deformations --> the actual deformations are less.

Jukis, the deformed shape you are showng has about 3 times larger displacement and 1.5^3=3.375 time stiffer. This means the stresses would be 3 * 3.375 = 10.1 times higher, compared to the once shown above. These are extremely high stresses that no spring material can withstand. I believe there is some hard error in the calcaultions or assumptions. (p.s. the restactor seems to assumed fixed inner edge, while it is pinned for the midvalve)

Uniform loading vs 3 port loading
The simulation I made is with uniform loading and it creates compression stresses around the periphery. Those stresses are seen from the vector plot of the third pricipal stresses:

These tangential compression stresses could/will quite easily create buckling of the outer edge of the shim. In reality, the 3 port position breaks the symetry and for sure will create the edge buckling. I tried to evaluate the effect, by appling a load pattern in 3 zones while having the same reaction force.The simulation became numerically unstable, but managed to get the following results.

As expected, the shape is 3rd order. The maximum vertical dispalcementhas increased from 0.9mm in the uniform model to 4.87mm (wow) - becoming similar to what Jukis show. However, the von Mises stresses are extrmely high:

The last scenario (3rd mode) is impossible - if this was happening, shim will always crack/yield. I suspect during the start of load increase the buckles in much higher mode, making it stiffer and preventing developing 3rd mode. I created minor load disturbance that produced the following results:

Peak displacement 1.06mm - little higher compared to no buckling.

Yield stresses - similar to no buckling model.

Principal compression stresses increased on the periphery.

Final thoughts
The assumed reaction force of 487.5N is high for 0.1mm shim. The real force is lower.
The high mode buckling might be yielding the periphery of the shim, permanently changing the shape. Next service, I will measure the overall thickness for any increase.
Using 0.15mm shim (instead of 0.1mm) will reduce buckling effects as the slenderness ratio will be decreased.
Buckling might not be repeatable with each cycle - damping might not be consistent.
The buckling analysis is challenging and it is better to be handled with explicit simulation, for example with LS Dyna.

 

[Jukis]

Force is actually higher, not the force that leads to most bottom out events but force when you hit a rock that you didnt notice at high speed. It produces a short stroke very high speed shaft movement (over 5m/s, possibly over 10m/s) and that is where you dont want excessive damping coming into play, many people describe dampers that produce high damping force in very high speeds as "spiking". Force that leads to bottom out events is likely around same as the spring force like you have assumed, those hits dont produce so high shaft speed unless you happen to hit something huge completely undetected by yourself. Bottom outs usually happen on landings and they produce less than 2m/s speed, the mid valve shim doesnt bend so much at those speeds yet.

 

[chehlarov]

Let's simply the question and reduce the assumptions to "What is the maximum acceptable deformation of shim 17.35x7x0.15, while keeping the stresses below 1240MPa?"
Uniform pressure is applied till Von Misses stresses reach 1240MPa:

Maximum deformation is 0.561mm, compared to 3mm as provided from ReStackor.

(FYI: The resultant reaction force is 99.97N. The applied uniform pressure is 0.5MPa.)
Any deflection higher than 0.561mm will create cracks in the shim.

I made a non-axisymetrical loading as well:

The achieved maximum displacement is 1.05mm. The resultant reaction force is 56N. Pressure pattern 0.3MPa and 0.25MPa.
I doubt ReStackor is using nonuniform loading.
The 3mm deflection, indicated by ReStackor, will crack the shim.

 

[T3mppu]

upsha, you are correct - I match the expected shim force to its deformation. The Von Misses stresses from the same simulation are as follows:
View attachment 2096254
Spring steels, like ASTM A 228 (one of the strongest), has design stress level of 1240MPa. From the plot above, stresses as high as 1900MPa are seen. There is no material with suitable properties for such deformation. The shim cannot withstand such extreme deformations --> the actual deformations are less.

Jukis, the deformed shape you are showng has about 3 times larger displacement and 1.5^3=3.375 time stiffer. This means the stresses would be 3 * 3.375 = 10.1 times higher, compared to the once shown above. These are extremely high stresses that no spring material can withstand. I believe there is some hard error in the calcaultions or assumptions. (p.s. the restactor seems to assumed fixed inner edge, while it is pinned for the midvalve)

Uniform loading vs 3 port loading
The simulation I made is with uniform loading and it creates compression stresses around the periphery. Those stresses are seen from the vector plot of the third pricipal stresses:
View attachment 2096256

These tangential compression stresses could/will quite easily create buckling of the outer edge of the shim. In reality, the 3 port position breaks the symetry and for sure will create the edge buckling. I tried to evaluate the effect, by appling a load pattern in 3 zones while having the same reaction force.The simulation became numerically unstable, but managed to get the following results.
View attachment 2096285
As expected, the shape is 3rd order. The maximum vertical dispalcementhas increased from 0.9mm in the uniform model to 4.87mm (wow) - becoming similar to what Jukis show. However, the von Mises stresses are extrmely high:
View attachment 2096286
The last scenario (3rd mode) is impossible - if this was happening, shim will always crack/yield. I suspect during the start of load increase the buckles in much higher mode, making it stiffer and preventing developing 3rd mode. I created minor load disturbance that produced the following results:
View attachment 2096287

Peak displacement 1.06mm - little higher compared to no buckling. View attachment 2096289

Yield stresses - similar to no buckling model.
View attachment 2096290

Principal compression stresses increased on the periphery.

Final thoughts
The assumed reaction force of 487.5N is high for 0.1mm shim. The real force is lower.
The high mode buckling might be yielding the periphery of the shim, permanently changing the shape. Next service, I will measure the overall thickness for any increase.
Using 0.15mm shim (instead of 0.1mm) will reduce buckling effects as the slenderness ratio will be decreased.
Buckling might not be repeatable with each cycle - damping might not be consistent.
The buckling analysis is challenging and it is better to be handled with explicit simulation, for example with LS Dyna.

Nice simulations!

Many shimms are advertised being Sandvik 20C http://cdna.terasrenki.com/ds/1.1274_C100S_AISI-1095_Ck101_20C_Datasheet_1.pdf
Tensile Strength (is it the same as your yield strength? I am no expert on these) is mentioned to be over 2000MPa for 0.1-0.2 shimms.

How much could these shimms bend before damage?

 

[chehlarov]

Nice simulations!

Many shimms are advertised being Sandvik 20C http://cdna.terasrenki.com/ds/1.1274_C100S_AISI-1095_Ck101_20C_Datasheet_1.pdf
Tensile Strength (is it the same as your yield strength? I am no expert on these) is mentioned to be over 2000MPa for 0.1-0.2 shimms.
View attachment 2098378

How much could these shimms bend before damage?

Thank you for the question!
This is a long topic, I will try to make it short.
Yiled stress of metals is the stress above which plastification happens; the detail will not return to its initial shape.
(ultimate) tensile stress (UTS) is the stress at which the material will crack. Yield is lower than UTS, espetially in annealed. conditions. Yield can be increased by cold working (and tempering) to UTS level. Shim's final thickness is achieved by cold rolling. Cold rolling is work hardening process needed to achieve tigh tolerances. It also increases the UTS; the thinner the final thickness, the higher the UTS.
Design stress can be compared to UTS, only when the detail is loaded normally. In case of pure shear (like helical springs) the UTS should be reduced by diving to sqrt(3).
Shim bending produce mainly normal stresses and the designer should limit the stresses up to yield limit with some safety margin.
All of the above is valid if the detail is loaded once or few time at maximum. However, shims are bended millions (maybe billions) of times during their lifespan. Every suspension compression creates one fatigue cycle on the midvalve and basevalve shims.The shim detail must be designed for fatigue. The design is done using S-N curves, which are typically hard to find. The S-N curves basically show how many cycles can be endured at cirtain stress amplitude. There are many correction factors I will not cover here (like is it tension-tension, tension-compression). Luckily ferritic steels (like Sandvic 20C) have enduarance limit. If the stresses are kept below the enduarance limit, the detail has infinate fatgiue life. As a designer, I would design a shim with stress levels below the enduarance limit. Typically, the enduarance limit for ferritic steels is accepted as 0.5 * UTS. Considering the particular 0.1mm shim, 0.5 * 2100MPa = 1050MPa enduarance limit. If infinite fatgiue life is desired, the stresses should be kept below 1050MPa. (Above I was using 1240MPa for ASTM A 228 - music wire).

Hope this bring some clarity.