- #1
DanielSu
- 5
- 1
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 :)
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 :)
