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wytbishop wrote:Read what I wrote again. The Strut bar will not be in compression because the inside tower is more stationary than the outer.

This is a quick sketch I did last night. The strut bar is definitely in tension.

I'm not a physics expert. Can you expand on your abbreviations?

It actually makes sense to me after thinking about it, with only the parts in the picture at play, but does the distribution of the mass of the car have an affect on the forces experienced by the bar?

Ryan wrote:Now I completely agree with you
When it comes to pivots, because they are mounted solid to the towers(cannot rotate on any axis), one or two pivots would not change the strength, but I believe three pivot points would cause it to fold, defeating the purpose.

Yet the lower portion of the suspension is essentially this. Interesting how everything works together.

I agree that the rear strut bar is in tension. My rear Corksport is adjustable, so what I did for autocross is that i tightened up the rear as much as I could while still being able to put it on. It was pretty tricky to put on, but with those quick-releases, it would be tricky to take out, but the hard part would be to put it back on...i'd be a sumo on each side of the mx-3 pushing against each side so the holes would line up.

tehbrookzorz wrote:
Yet the lower portion of the suspension is essentially this. Interesting how everything works together.

This is because the sway bar is mounted solid to the frame.

Clip two paperclips together. Hold the bottoms tightly and push them together. Should be rigid (for a paperclip). Put three together, and push them together. Still works, right? As long as you hold the bottoms tight. Take four.... epic fail. Take five and imagine the middle one is secured. Works again.

Fw = force due to the mass of the vehicle. This force tends to act directly along the line of the control arm. There is some weight acting on the top of the strut tower, but this is so complex to analyze that it has to be neglected for sanity reasons. The vast majority of the weight of the car is acting on the steering knuckle (or spindle in the rear wheel). The bar i nthe middle between the control arms is there to represent the frame of the car. It is rigid making each of the control arms, either front or rear, act as a "2 force member".

A "2 force member" is one which is usually pinned at both ends and must, by definition, be in pure commpression or pure tension. A 2 force member cannot generate any other reaction. The strut bar is also a 2 force member.

N = the "Normal" force. This is the force due to gravity. Gravity always acts straight down, however more of the weight of the vehicle is acting on the outside wheel when cornering so there are 2 values for N.

Ff = the force due to Friction. The force due to friction is directly proportionate to the Normal force and the coefficient of friction between the rubber and road surface. Ff = (Mu) * N, where Mu = ~0.7 for dry road.

Ma = the "Moment" about the point A. Moment is the fancy name for torque. The total Moment about A is equal to the sum of the forces multipied by their perpendicular distance to that point. Because "Fw" acts directly through the points A and B, it has a perpendicular distance of zero and so creates no moment. But the friction force does and because N1 is greater than N2, Ff1 is greater than Ff2 and the subsequent Moment or torque about A is greater than at B. That means that the deflection of the top of the strut at A is greater than at B as well. This is implied by the size of the arrows at the top of each strut.

Because strut A experiences greater torque and the top of the strut deflects more than the top of strut B, the bar is in tension.