Anyone know how much shock force a shock tower needs to hold?

breakdown

New Member
Oct 8, 2009
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making a new shock tower and trying to figure out how heavy duty I need without going overboard. Using Fox position sensitive shocks and no dampeners or limit straps.
 

badassmav

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Jun 11, 2013
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Jamul

  • A good baseline number which represents the pounds of force that either of the front corners of your car would see at full bump is roughly the total weight of your car. This is based on a worst case scenario where you hit a g-out, at speed, unexpectedly. The math is pretty basic, and is based on an impact force of 4-5 g's (hence, the total load being around the total weight of your car. A 1,600 lb. car, assuming GVW is perfectly distributed, weighs 400 lbs. at each wheel/corner, times 4-5 g's, equals 1,600 or 2,000 lbs. respectively at the center of the wheel). An impact like this will cause shaft speeds of the shock to exceed 100 feet per second:eek:!
    Your question poses a great example of why I like to design my suspensions with as low a shock motion ratio as possible (stroking a longer travel shock, and using lighter, lower rate springs). The mounting "brackets" or tabs for the shocks can be much lighter, and therefore easier to fit in a given space when the shock is not over worked.
    Let me share some hard numbers with you. Awhile back, I was talking to Mark (Queen) about his front suspension, and learned that he runs a 350# over 450# spring set up, for a combined primary spring rate of 400 lbs./inch, and a secondary rate of 450 lbs./inch. He is running the +3 1/2" lonestar kit if I'm not mistaken, and it utilizes a stock length shock, and stock location upper shock mounts. Assuming 18" of wheel travel stroking a stock length shock at 8.375" for a .46 motion ratio, and 1/3 of bump is realized at ride ht., each shock mount will be preloaded with 1,350 lbs. static load at ride ht. and a whopping 3,600lbs. at full bump!
    Compare that to our 2015 RZR Xp4 1000 Monster Energy class 1900 UTV weighing in at just under 2,000 lbs., with 17.5" of front wheel travel stroking a 12" shock at a .69 motion ratio, and a spring set up of 150# over 250# for a primary rate of 200 lbs./inch, and a secondary rate of 250lbs./inch, and you get much lower figures. At ride height, the static load on the shock mount is 800 lbs., and the total load at full bump is 2,800 lbs. That is 800 pounds less at each corner that the shock mounts have to support, on a car weighing approx. 200 pounds more than its competitor!
    So, some simple math says that car "A" (Queen Racing) weighs 20% less than car "B" (The Monster Energy XP4), yet sees 23% more loading on the shock mounts at full bump! God, I love math! It doesn't lie or manipulate. Only tells things exactly as they are!
    The 2 valuable lessons here are:
    1) Spring rate and wheel rate are two entirely different things, and:
    2) As leverages upon springs/shocks are increased, the end results (loading upon a specific member or structure) are not directly proportional to the change in leverage, but rather inversely proportional to it. Kind of analogous to that compounding interest loan I took out when I bought my first new truck in 1982!
    So, the question of the day is: Given the loading at each corner, how will you determine the thickness, design, and bracing the shock mount will need to repeatedly sustain impacts as described above with out failing, or pre maturely fatiguing? Simplest solution would be to input your data into an interactive application such as Solid Works, then multiply the results by 1.25 "Overkill is the key to success" (a phrase I coined in 1989 after building a heavier class 8 than most).
    Ah yes! I love taking my brain for a walk!
 
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mearsman

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Nov 2, 2011
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Always thought provoking and educational, thanks for the session professor Johnson
 

badassmav

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Jamul
I appreciate the opportunity you afforded me. It's a good change of pace to all of the whining that's been going on around here! But I guess if it provokes needed changes in the pro 1900 class, it's all good. Please note that if you follow my model of thinking when building your suspension, be prepared to spend more time to dial it in. There are a ton of cars with stock shock geometry and/or lengths, but very few with modified layouts as outlined above. Therefore, there are a ton of good shocks to bolt into a stock configuration, while not many, if any at all, that will work out of the box when making significant changes to its mounting and working ratios. The reward is, that although other fab guys are reading and learning from my sharing here in the Underground (c'mon guys, you know it's your guilty pleasure! Ha ha!), most still decide to go with what is safe for them and their clients, therefore, giving you the advantage, hands down, when it comes to suspension performance and reliability. Don't be like me and over think it though. Get your project done within your means, Then drive, learn, and most of all, share! Ain't many secrets left in race car design. The indy and F-1 geniuses assured us of that in their awesomeness of the late 50's to early 70's!
 
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UTVUGJake

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Feb 11, 2013
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Love your knowledge of vehicle dynamics and particularly your willingness to make the effort to type it out and share Badassmav. Good info here.
 
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meathooker

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Mar 4, 2015
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.
  • Let me share some hard numbers with you. Awhile back, I was talking to Mark (Queen) about his front suspension, and learned that he runs a 350# over 450# spring set up, for a combined primary spring rate of 400 lbs./inch, and a secondary rate of 450 lbs./inch.

Could you expand on this? It's been a decade since I took diff eq but I recall this being a more complicated relationship between the main and primary spring.



And I love reading your posts! It's like being back in school, but much more interesting.
 

meathooker

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Mar 4, 2015
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is this the formula you would use??

Primary Spring Rate X Secondary Spring Rate / Primary Spring Rate + Secondary Spring Rate. = Act. Spring Rate
 

COGNITO

Cognito Motorsports - Official UTVUnderground Spon
Apr 30, 2009
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  • A good baseline number which represents the pounds of force that either of the front corners of your car would see at full bump is roughly the total weight of your car. This is based on a worst case scenario where you hit a g-out, at speed, unexpectedly. The math is pretty basic, and is based on an impact force of 4-5 g's (hence, the total load being around the total weight of your car. A 1,600 lb. car, assuming GVW is perfectly distributed, weighs 400 lbs. at each wheel/corner, times 4-5 g's, equals 1,600 or 2,000 lbs. respectively at the center of the wheel). An impact like this will cause shaft speeds of the shock to exceed 100 feet per second:eek:!
    Your question poses a great example of why I like to design my suspensions with as low a shock motion ratio as possible (stroking a longer travel shock, and using lighter, lower rate springs). The mounting "brackets" or tabs for the shocks can be much lighter, and therefore easier to fit in a given space when the shock is not over worked.
    Let me share some hard numbers with you. Awhile back, I was talking to Mark (Queen) about his front suspension, and learned that he runs a 350# over 450# spring set up, for a combined primary spring rate of 400 lbs./inch, and a secondary rate of 450 lbs./inch. He is running the +3 1/2" lonestar kit if I'm not mistaken, and it utilizes a stock length shock, and stock location upper shock mounts. Assuming 18" of wheel travel stroking a stock length shock at 8.375" for a .46 motion ratio, and 1/3 of bump is realized at ride ht., each shock mount will be preloaded with 1,350 lbs. static load at ride ht. and a whopping 3,600lbs. at full bump!
    Compare that to our 2015 RZR Xp4 1000 Monster Energy class 1900 UTV weighing in at just under 2,000 lbs., with 17.5" of front wheel travel stroking a 12" shock at a .69 motion ratio, and a spring set up of 150# over 250# for a primary rate of 200 lbs./inch, and a secondary rate of 250lbs./inch, and you get much lower figures. At ride height, the static load on the shock mount is 800 lbs., and the total load at full bump is 2,800 lbs. That is 800 pounds less at each corner that the shock mounts have to support, on a car weighing approx. 200 pounds more than its competitor!
    So, some simple math says that car "A" (Queen Racing) weighs 20% less than car "B" (The Monster Energy XP4), yet sees 23% more loading on the shock mounts at full bump! God, I love math! It doesn't lie or manipulate. Only tells things exactly as they are!
    The 2 valuable lessons here are:
    1) Spring rate and wheel rate are two entirely different things, and:
    2) As leverages upon springs/shocks are increased, the end results (loading upon a specific member or structure) are not directly proportional to the change in leverage, but rather inversely proportional to it. Kind of analogous to that compounding interest loan I took out when I bought my first new truck in 1982!
    So, the question of the day is: Given the loading at each corner, how will you determine the thickness, design, and bracing the shock mount will need to repeatedly sustain impacts as described above with out failing, or pre maturely fatiguing? Simplest solution would be to input your data into an interactive application such as Solid Works, then multiply the results by 1.25 "Overkill is the key to success" (a phrase I coined in 1989 after building a heavier class 8 than most).
    Ah yes! I love taking my brain for a walk!
Reid you aren't calculating the combined rate properly
 

rico

New Member
Oct 16, 2014
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Lehi, UT
I typically don't comment much on this forum, but this thread has peaked my interest.

It is important to note that the shock towers will absorb the same amount of load no matter what the spring set up is, the only difference will be at what amount of force you bottom the shock out at. Once the shock reaches full bump it acts as a rigid member and transmits the remaining un-absorbed/dissipated load into the chassis (in addition to the compressed spring load). Since force is a function of mass and acceleration, no matter what the spring set up is, given the same acceleration the more massive car will experience a higher level of force acting on the system.

It seems like wishful thinking to assume that the vehicle will behave in an evenly distributed way, especially when we are considering worst case scenario. What happens when you land with simultaneous pitch, yaw, and roll, bringing everything down on one wheel for milliseconds? Simplistically this could quadruple the load from the "evenly distributed" example.

Perhaps as a starting point one might consider what they would rather fail, the shock mount or the bolt pinning the shock in place. I would say the bolt since that seems much easier to fix. Calculate the force required to shear the bolt, and that tells you the minimum load that the shock mount must resist. Then just remember not to upgrade your bolts :)
 
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