Direct Acting Force Analysis

[Ahmad Rezq]

Hello,
I'am trying to analyze the forces on direct acting suspension which the damper locates on the lower arm.
I used a method which was explained by Erik in this topic
http://www.fsae.com/forums/showthread.php?11179-analysis-of-wishbones

I divided the problem into two force analysis problems
First solving the upright forces 6 unknown 6 equations and the results are ( two forces on upper wishbone , force on tie rod , 3 components Fx,Fy,Fz on Lower wishbone).
Second solving the lower wishbones with these three force components in 7 unknowns the force goes to damper and the 3 forces on each inner point of the wishbone fore and aft. so it's an overrefined problem which i need to do assumptions i assumed that the fore point won't take a force in x direction i solved the problem and got the unknowns.

i built a matlab code i hope i didn't make some mistakes in calculations and i got some results which are as follow
simulation at 1.7g deceleration (just trial) Force Are In Newton

Upper arm forces : -3456 and 3226.2
Tie rod force : -1092.2
Damper force : -1106.1
lower arm forces
outer force components (-3391.9,-591.5,291.89)
Inner forces components
fore point (0,-6085,-1182.9)
aft point (3130,5871,1676.7)

the results don't seem acceptable , looking at the force on the aft point the direction of this force will bend the arm catastrophically. i did some changes in my assumptions but i got similar results.
435
(Force Direction Is Shown In The Image)
can't figure out what is wrong ?
any suggestions

[Ahmad Rezq]

Seems like i found my mistake and i can now have acceptable results.

[Z]

Ahmad,

It would be good if you could post an image of all the forces shown as arrows acting on the various parts. So something like a perspective CAD view of the whole suspension, with scaled arrows added to the picture (eg. 3,000 N force = 30 cm arrow?).

This visualisation of the problem might help other students better understand it all.

Z

[Ahmad Rezq]

Z
I didn't check the topic for a long time due to my semester exam so accept my apologize.
Here you are:-
1 – The data from the Matlab code I built.

Left wheel suspension coordinates(mm):

Wheel patch (-228.6,590,0)

Lower wishbone:
Fore (-223,250,115)
Aft (-393,250,87)
Upright (-225,530,152)

Upper wishbone:
Fore (-228,280,218)
Aft (-398,280,151)
Upright (-233,519,306])

Steering
Rack (-175,215,100)
Upright (-165,520,160)

Damper
Mount (Lower Wishbone) (-250,440,170)
Chassis ((-300,300,320)

Analysis is done on braking 1.7g
Force Components (-1646,0,968) N

Results
F_upper_fore = +3082 N
F_upper_Aft = -3303 N
F_tie= -1283 N
Fxyz_(Lower arm outer ball joint)=(3403,765,-230) N
F_damper = -1500 N
F_lower_fore = (0,-6302,-746)N
F_lower_aft = (3048,6073,1600)N

Also i draw two 3D sketches using SW one is the suspension links + the other is scaled force results from my code.
https://www.mediafire.com/?glu1lj6xo8g8iqu

Notes
Upper arm + Tie Rod + Damper
Solid line = Tension
Center line = Compression

The Lower arms force direction is started from the ball joints

[Ahmad Rezq]

I would like to raise the topic again, since my calculations leaded me to about 700 mpa stresses on lower wishbones size ( 18X8 mm tube )
how did you guys handle the direct acting or push - pull rod mounted on the wishbone ?

[Z]

Ahmad,

I am unable to view your image at the "mediafile" site (I just get lots of ads, eg. "Engineered"-electric-toothbrushes, scantily-clad-girls-selling???, etc.).

Can you try to post the image direct on the thread?

Z

[Ahmad Rezq]

Z
I will prepare the image for you.
but i have a question
why not to solve the problem of direct acting suspension system by the simple method 6 forces in 6 links ( two force members ).
assuming that the damper mounting point ( regardless my first image ) is very close to lower wishbone end.
if my calculations are right the other method ( considering the bending on the lower arm , ie the diviation of the force line of action from wishbone tube axis ) may lead me to oversize my lower arm, also the assumptions in the second problem ( solving for the the lower a-arm forces ) it will be different depending on the analysis case ( cornering , braking . . . etc ) .

[TiJei]

Hi Ahmad,

1. In case of an outboard brake you should apply the forces at the tire contact patch to the road. To me it looks like you applying loads at the wheel centre.

2. If your "6 forces 6 links" approach is something like a framework from rods this is basically a suitable approach as long as the force lines in each a-arm meet in a single point. It is thereby not so important that the mounting point of the damper is close to the joint of the tubes. Only the force lines need to meet in one location. The problem is that this should also be the case with the suspension in compression. Therefore it might be worth analysing different geometrical states of the suspension. If the force lines meet in one point in the a-arm the bending is reduced to the region outside of the damper joint and the out of plane component of the frame joints is reduced. If you can not reduce the bending in the a-arm for packaging reasons you should consider a-arms made from something different than just tubes.

Regards

[Ahmad Rezq]

Tijei,

1. In case of an outboard brake you should apply the forces at the tire contact patch to the road. To me it looks like you applying loads at the wheel centre.

I realized the reason you thought that i apply the force on tire center. but no the forces are on tire contact patch.


If your "6 forces 6 links" approach is something like a framework from rods this is basically a suitable approach as long as the force lines in each a-arm meet in a single point. It is thereby not so important that the mounting point of the damper is close to the joint of the tubes. Only the force lines need to meet in one location.

Correct me if i'am wrong , in tow force member for the wishbone and the damper mounting on it they are 3 links so how will the damper force line of action meet the 2 forces on wishbone two links.

My main goal regardless the suspension package which has been frozen is to learn, and analyze this problem the suitable way it should be treated.

[Z]

Ahmad,


...why not to solve the problem of direct acting suspension system by the simple method 6 forces in 6 links ( two force members ) assuming that the damper mounting point ... is very close to lower wishbone end.

Yes, most of the "Analysis of Wishbones" thread is aimed at just that.

I would say that as long as your Spring-Damper centreline passes LESS than ~30 mm from the wishbone's outer BJ you can do this simplified analysis. Many FSAE cars have the SD centreline passing MORE than 30 mm from the outer BJ, but many of these cars also end up with a bent or buckled wishbone when they hit a big bump on that rough test track "out back".

If your SD centreline passes about 100 mm or more from the outer BJ, then you definitely should do the analysis the way described in the later section of "AoW" thread.

More importantly, you should design the lower wishbone to take the resulting bending loads. This is NOT HARD. Mainly you want the wishbone "thicker in the middle", perhaps like a wide and low "A"-shaped triangle in front-view. For typical FSAE loads, an "A" about 50 mm high at its apex where it connects to the SD, tapering to ~20 mm high at the inner and outer BJs, should be OK.

TiJei's point 2 makes sense.

Z

[Ahmad Rezq]

Z
- I realize that the closer the damper mounting point from wishbone outer ball joint the simple analysis works. but why 30 mm or why less than 100 d do these values from experience or what ?
My damper mounting point is 70 mm far away from wishbone end point.

- Also for the analysis considering the bending on lower arm. as earlier written when solving the force analysis for the lower wishbone we have one force on the damper and 3 components (Fx,Fy,Fz) on each end of the wishbone. on what base the assumption to eliminate one unknown in order to solve the problem should go. i think it will be different depending on each case you analysis your wishbone ie cornering , acceleration would be different assumptions. combined case also would be different ( correct me if i'am wrong ) . also one must consider in this case the force acting on the ball joint axis which will affect the ball joint selection process.

[Z]

Ahmad,

The "less than 30 mm" is just based on experience. I have seen Teams with about this much offset who have completed a competition (ie. car went OK on smooth tracks), but have then buckled the wishbone when hitting a biggish bump in driver training the next year.

With your 70 mm offset of SD-centreline from outer-BJ, you MUST design the wishbone to be strong enough in bending.

I have posted more details on the "AoW" thread where more people are likely to see it.

Z