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Calculating spring rates, force and length?
I require a car spring to take a corner weight of 571kg and compress the spring to 242mm at rest. I also need it to compress a further 35mm to 207mm by dropping the vehicle 100mm (ie driving off a kerb). Therefore I need the force as it hits the ground to calculate the spring rate to compress it 35mm.
From this I need to then work out spring free length and number of active coils; the 2 variables as I see it.
I do not know strength of steel normally used but do know the internal spring diameter is 75mm and the maximum wire diameter can be 15.5mm.
(My initial calcs show a spring rate of 536lb/in with a spring free length of 331mm and 8 active coils. But this may be completely wrong).
Had a good thought about your problem for almost a day now. Very interesting problem!
The hardest part is finding the spring rate. Then again, it depends how you model it. I don't see why you can't use work-energy principles. Anyhow, the most accurate way, in my opinion, is by using knowledge of free-underdamped oscillations. Number-crunching too intense - I don't have time (and rusty memory too) to do that sorry. But I'll outline what I would do.
When an equilibrium system is disturbed it must subsequently go through vibrations.
The instant when the car drops down from the kerb the spring is extended by 100m which in turn decreases the compression and this makes the car drop down into it as the forces (mg and kx) no longer equate to each other. The car does not free-fall since the spring would still be compressed.
If we isolate the spring the situation would be like as if the spring never moved but only the mass (car) is pulled upwards by 100mm from equilibrium point (when the spring was 242mm). This assumes the wheel instantly touches lower-ground when it rolls off kerb. If there were no damping the mass would compress 100mm pass the equilibrium point and oscilation would continue forever - the passengers won't like that.
The first "overshoot" peak will be 35mm as you've stated. You could calculate "backwards" from using the peak values (The first inital peak of 100m and next opposite peak of 35mm) to calculate the natural frequency which equates to sqrt(k/m). M is the mass and k the spring rate. You might also need to calculate the damping cooefficient.
After that I think it pretty much straght forward.
G = 78GPa (steel rigidity)
d = 15.5mm (wire diameter)
D = 75mm (coil diameter)
n = no. of coils
n=(Gd^4) / (8kD^3)
And from initial conditions of spring, x0 uncompressed spring, x1 equilibrium spring length.
ke=mg
k(x0-x1)=mg
k(x0-242mm)=571g
x0 = [571g+k(242mm)]/k
As you see the real problem is finding k! :[
----added----
I varified that stiffness of spring is dependant on the damping of the system. Which is obvious really. Stiffer means lesser the damping and vice versa. Your problem cannot (if you want the problem to be solved with a the more real-world model) be solved without information about damping of the suspension system.
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