Shear on fasteners.

[Necrobata]

Hello my people,

Ive been asigned to calculate the shear strengh over the fasteners that goes in the. ehh.. tire (not english speaker sry).

Ive been looking on the Milliken and the springer about and only thing i find is the transversal & longitudinal force transfers.

A faculty spoke to me about doing the following:

1- With the centriful aceleration ( a=v^2/r ) and the second Newton law ( F=m.a) i could find the force over our car, soo the speed is Skidpads (45km) and the radious from the same event, that gave me abour 4000N

2- Then im assuming that the weight distribution is 40/60, the 60% of those 4000N goes to the back. I assumed that the most stressed fastener would be the one in the rear out of the curve.

3- I made a free body diagram of the car, in a transversal plane that cuts rear tires and has the double A arms on it. (you can see the section of the chassis, that cuts the middle of the tires, the tires and four arms)

4- I calculated the transversal weight transfer, that gave me the vertical forces on the tires and added to the FBD the lateral force (60% of the 4000)

5_ then i separated the tires from the arm nodes, but here i have a DOUBT, im assuming that, being a push rod fron suspension, all of the vertical forces are in the lower ball joint, am i right on this? the upper ball joint will only take horizontal strenght?

6-then, after having all of those forces, i sum the moments centered in the upper left joint, that cuts off the force that goes at the opposite side (because M=F.d and distance is 0), leaving only the lower ones (now you have a 2 incognite ecuation), then moment in any lower joint and you have another ecuation with the other forces, that leaves you with 3 ecuations (vertical forces, horizontal forces, and moment) and 4 incognites, the 4 forces...

Then i got stuck, is there a way to calculate this in a simpler way? can someone point me in the right direction?

Thanks!

[Adambomb]

Sounds like you're on the right track. For #5, yes, you are correct, all normal loads take a path from the pushrod or pullrod to whatever ball joint they are connected to, through just half of the upright to the wheel bearings, to the tire and wheel, to the ground.

If you're looking at one plane at a time (front view or side view), you should only have 2 force and 1 moment equation. Sorry if I am not interpreting this right, but it sounds like you are trying to calculate the moment in the lower ball joint? With that I must ask, will the lower ball joint support a moment in that particular direction?

Also don't forget to look at max braking, braking forces and moments can be quite severe.

[Necrobata]

I actually made a mistake, i took moments in two places, but after that i realized each moment threw at me the same ecuation.

The thing that is holding me back is that i have 5 horizontal forces, in wich 4 are incog (the 4 forces acting in the ball joint) but only 3 ecuations, that cant be solved :S

That would help me picking the fasteners, then Rodend Inserts, then the housings for suspension, then the A arms and then ill finally find the Toad of truth

Aw, and ill also take a look on the braking situation, thanks!

[Adambomb]

Ahh, I see. So the problem is splitting up the 4 forces in the ball joint, as in the two from each member of each arm? Why not just take a purely simplified front view, where you only have a combined ball joint force. Then you just have an upper ball joint force, lower ball joint force, and tire reaction. Then once you have the ball joint forces figured out you can do another FBD of the top view and figure out the forces in each member of each a-arm.

[revolutionary]

Don't forget to figure out the forces on the lug nuts/wheel studs. The wheel will be trying to rotate as it is stopping...