Load analysis on a motorcross bike - need some advice

AI Thread Summary
The discussion focuses on analyzing the strength of a motocross bike frame under various load conditions, particularly from vertical falls. The calculations involve modeling the bike's suspension using spring constants for the front and rear springs, leading to derived impact forces based on different fall heights. Participants emphasize the importance of considering damping in shock absorbers, as it significantly affects suspension performance during landings. There is also a debate about the accuracy of the spring constants used, with suggestions that they may be closer than initially thought. Overall, the project is seen as an engaging challenge, with participants providing resources for further research.
DanielSu
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My task is to analyze a frame's strength of a specific motorcross using different loads, and to do this i decided to focus on vertical fall from various heights, since this will probably give the largest forces on the bike. The one can also experiment with different types of landings, i.e. landing on both wheels, backwheel etc.

So i reduced the problem and replaced the motorcross with a weight mg which is being supported by two springs which are parallel, so i can replace them with one spring constant k = k_rear + k_front.

Now let's go into the mechanics of the problem.
Lets say i have an unloaded spring with stiffness k and a weight mg over it. The weight is barely touching the top of the spring, such that the spring doesn't support it. I then release the weight and i want to find out what the impact load will be. So i reasoned that the potential energy must equal the energy of the spring according to
\frac{1}{2}kh_0^2 = mgh_0
or:
h_0 = 2\frac{mg}{k}

If i instead drop the weight from an additional height h above the unloaded spring, i get:
\frac{1}{2}kx^2 = mg\left(h+h_0\right)
or:
x = \sqrt{2\frac{mg}{k}\left(h+h_0\right)} = \sqrt{2\frac{mg}{k}\left(h+2\frac{mg}{k}\right)}
And the impact force would be:
F_i = kx = \sqrt{2kmg\left(h+2\frac{mg}{k}\right)} = \sqrt{2kmgh+4(mg)^2}
From here one can see that the stiffer the spring is, the stronger impact force one will get which seems logical.
Example:
k = 226 kN/m
m = 140 kg (50kg motorcross and 90 kg person)
g = 9.81 m/s^2
h = 1m - this is the height of the fall
Then we get an impact force of F_i = 25066 N
if i use h = 2m, F_i = 35343N

I find that at approximately 2,95m vertical fall the front spring bottoms (it has the shortest length) with a force of F_i = 43242 N. Are my calculations soundproof? Or at least reasonable?
One issue is that the springs have different spring constants, the front has 18 kN/m and the rear 208 kN/m, but used the fact that they were parallel to simplify my calculations but one could make the argument that the front will hit the bottom much earlier.

When i want to use these values in Ansys (Finite element analysis program) i of course can't put the entire load at one place but a big chunk of it is the driver putting at least 75% of his weight at the footpegs which will produce high stresses at some areas I am sure (i can't validate right now).

Im grateful for any advice in this matter :)
 
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Fun project! :smile:
DanielSu said:
the front has 18 kN/m and the rear 208 kN/m
Where did you get these numbers? The front and back spring rates are much closer than that, in my experience.

And you should start including the damping of the shock absorbers in your calculations from the beginning, IMO. The damping is a big part of how the suspension works, especially in landings...
DanielSu said:
the driver rider putting at least 75% of his weight at the footpegs
Much closer to 100% :biggrin:

MBwheelie_Reversed.jpg
 

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berkeman said:
Fun project! :smile:

Where did you get these numbers? The front and back spring rates are much closer than that, in my experience.

And you should start including the damping of the shock absorbers in your calculations from the beginning, IMO. The damping is a big part of how the suspension works, especially in landings...

Much closer to 100% :biggrin:

View attachment 224979
The front is from a downhill fork (from a bicycle) that's why it so small :) And yes usually they are probably much closer than that.
I thought the damping only was used to dampen the return movement once the spring was first compressed, anyway i got no data of the damping constant and i found by earlier calculations (they might've wrong of course) that it played little or no part in the beginning.
 
DanielSu said:
I thought the damping only was used to dampen the return movement once the spring was first compressed
No, there are damping rates for both compression and "rebound" (usually different). I'll try to find a datasheet for a typical MX shock for you...
DanielSu said:
i found by earlier calculations (they might've wrong of course) that it played little or no part in the beginning.
No, they play a huge part (see the picture of my CRF450R above).
 
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