statically indeterminant problem 2

[Oakley45]

Hello

I wonder if anyone can help/guide me with a problem i have, the figure below shows a free body diagram of a brake caliper, the force on the body acts at 30 degrees from the vertical at point C, I resolved this to get the two forces shown in the x and y direction.

I have determined the values for v1 and v2 but i cannot decide how to get the values for h1 and h2, they cannot be the same due to the direction of the force. The problem appears to be statically indeterminant so is there another way i could solve this problem?
Any suggestions appreciated

 

[rb1957]

your problem is actually a "three force body"

this means that the forces intercept at a point

this should remove your redundancy

 

[hydtools]

Set up and solve summation of moments equations.

e.g.: sum moments about B = 0 = braking force * its moment arm - resultant force at B * its moment arm

Ted

 

[rb1957]

sum moments won't distinguish H1 from H2

 

[eromlignod]

Your diagram doesn't adequately show your constraints. Are joints A and B fixed (pivoted about an anchored eye)? If this is the case then, yes, it is indeterminate. There is no way to determine whether member AB is in compression or tension simply from the geometry.

If one of the joints is a roller/slider, then it is determinate. Assuming your pivoting point is A and slide point is B, there should be no H2 at all. If H2 is a frictional force then it is an externally applied one and is dependent on motion of the rubbing member, so it has to be calculated from the normal force V2 (easy to find) and the coefficient of friction.

Don
Kansas City

 

[Oakley45]

points a and b are both fixed, they cannot move in any direction. Below is the caliper that this problem is based on, the caliper is the one at the top left hand side. my diagram is a simplified version of this situation

 

[rb1957]

i think you can apply the "three force body" approach to this.

better would be an FEM, particularly if "A"and "B" are those groups of 3 fasteners (capable of taking some moment).

 

[BobM3]

Keeping in mind what Don says, if it is statically indeterminant then you would need to know something about relative stiffnesses of the two joints.

 

[Oakley45]

sorry i should have said A and B are the mounting points where the caliper is mounted to the frame.

I am just looking into the use of the three force body method

 

[eromlignod]

If points A and B are fixed, then it is statically indeterminate, even if they are pivoted.

Here's why. If both joints are perfectly fit, with zero tolerance (and they won't be), and the holes in both members are spaced perfectly the same (and they won't be) then there is no net force between them. But then if the temperature changes by a half a degree, there is, because the disk and link will expand or contract by different amounts. There is no way to simply measure this linkage and determine the forces on A or B since forces are invisible.

So, what if one hole is snug and the other is loose? Well, first of all, now they are technically not "fixed". If joint A is snug and B is loose then the angle at which the journal B acts on hole B depends on the clearance and the accuracy of the hole/journal centers. If the holes are .001" farther apart than the journals are, and the clearance at joint B is .002", then the bearing point of contact can be geometrically calculated...and it *won't* be tangential to the arc about A...depending on tolerances, it might not even be close. And, again, when the temperature changes, all bets are off and the angle could change radically.

As you can see, there can be no definite answer for this situation, since it is statically indeterminate. If B were designed as a free roller on a flat surface then the forces at B would not change with tolerances or temperatures and could be determined easily with statics.

Don
Kansas City

 

[floattuber]

I think your most practical solution is to 1) assume the forces are split 50/50 2) strain gauge your parts and measure the forces 3) or FEA.

 

[cbrn]

Hi,
as regards FEA, yes, it may be a good option, but:
1- if you schematize it as "two-beams" (for an extremely simple system like that you can even do it by hand), you will have to make assumptions about the stiffnesses of each beam
2- if you analyze the whole "thing" as a solid model, then you will need first of all a FEA package, then a little time and lastly a lot of attention about boundary-conditions' application.

Regards

 

[zekeman]

If A and B are bolted, then the fricion forces generated by the bolting forces times the coefficient of static friction might account for all of the loading at A and B; but their distribution aould still make it indeterminate. On the other hand if they are insufficient to sustain the lateral load, then the system would slide to the right and be restrained by by the first bolt to touch the edge of the hole in the frame and therefore, ( as a practical matter) one of the bolts would take the full lateral force.
So, my contention is that if the friction handles the load, then it is patently indeterminate, but if not, the geometry determines which bolt wins the total load battle. And if this is a design problem, then you designe each member to take the full lateral load.

 

[cbrn]

Good point, Zekeman !!!

 

[Oakley45]

the three force approach will as far as i can see give both forces to be equal on each bolt, I have also looked at the bolt group loading method but this also gives the horizontal load for each bolt as the same.

 

[BobM3]

Even with FEA you will have to know the stiffness of the mating piece. The stiffer load path reacts more of the load.

 

[eromlignod]

If it's bolted tightly at both points, Then the whole system (caliper + bicycle frame) could be considered as one large, contiguous part and analyzed via FEA. There might still be residual internal stresses, but this is usually neglected by FEA anyway. Of course, this is probably a lot more sophisticated a solution than Oakley45 was counting on.

As I stated before, the problem with any mathematical solution for a loose connection is that the direction of bearing contact can vary widely depending on variances in hole diameter, bolt diameter, hole spacing (both parts) and even hole/bolt runout. And the closer the fit, the more radical the changes in free-body component forces and direction for even small manufacturing variations and temperatures. This could vary a lot from assembly to assembly and under differing field conditions.

If one bolted connection is *really* loose, then you could probably approximate a solution by assuming that the loose journal acts in a tangential direction along the arc about the fixed joint. Perhaps this is sufficient for your purposes.

Don
Kansas City

 

[rb1957]

i'd've thought that if the disc acts like a lump (has reasonably the same stiffness in all directions), then the three force solution should work. sure that's brushing over a bunch of real world issues with the assembly, but that's what you get if you want to approximate things.

if the disc has a directional stiffness (stiff in one direction, weak in another), if the connections are more fixed than pinned, then FE would solve these issues.

 

[eromlignod]

This is a disk brake. The caliper wouldn't be mounted to the disk. It would ostensibly be mounted to the bicycle (from the outside) and pinch the disk, which would be mounted to a wheel.

Don
Kansas City