Brake Disc Design- Torque Acting On Wheel Stud Bolts

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Brake Disc Design- Torque Acting On Wheel Stud Bolts


I have currently been working on designing a brake disc assembly. The current problem i am having is trying to workout the force acting on each wheel stud.

The frictional force of the brake pad is 3.2 KN with an efective radius of 105MM from the center of the disc
The 4 wheel studs concentric to the disc have a pcd of 65MM

What would be the force acting on each stud? i need to know this so i can further calculate the minimum bolt diameter. I will attach the work that i have done so far however i doubt the initial calcualtions to find the force acting on each bolt is correct.




the link to my current workings are as follows

http://i175.photobucket.com/albums/w...pqscan0001.jpg [Broken]
 
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  • #2
jack action
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The torque at the brake system will be the same as the torque at the bolt circle. The torque is the force times the radius. In a perfect world, each of the 4 bolts will support ¼ of the force calculated with the known torque and known bolt circle radius.

Although you don't specify, if this for a car, you should used the torque found with the maximum friction force at the tire contact patch as the upper limit for maximum torque application. Don't forget there is weight transfer that increases normal tire load. One can assume that the ultimate vertical load on a single tire would be the total mass of the vehicle.
 
  • #3
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Thanks for your reply jack that has made it much clearer. Yes it is for a car, i have worked out that their is a weight distribution of 70 percent on the front wheels.

As you have sugested i will use the torque found with the normal friction force at the tire contact patch, for the shear force upperlimit.

Thanks for your sugestion it has been most helpful
 
  • #4
Ranger Mike
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70 percent front wheel weight seems way high to me. even under max braking with weight transfered to the front the figure seems very high
 
  • #5
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The vehicle parameters are:
V1=100 M/s
V2=50 M/s
T=10 s
M=100 Kgs
Therfore the vehicle is traveling at 100 m/s and slows down to 50 m/s in 10 seconds. As a results this means that the Total Required braking force is 5000N. From taking moments about vehicle FBD i found that the Normal Reaction Force In the front wheels was 7000N and 3000 N in the rear. Thefore this illustrates there is a 70/30 weight distribution under braking. As a result total front braking force required is 3500N (1750 per wheel). And a rear braking force a total of 1500N (750N per wheel).

However my aim is only to design the front Disc brakes in the situation
 
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  • #6
Ranger Mike
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yeah you may be right,,I went back to my track notes. Did you figure in CG??
.Let’s assume we have a 2500 pound car with a 50/50 static weight distribution. If we are only concerned with the vehicle at rest, it’s easy to determine the weight on each wheel. We just need to find some scales and weigh it. The sum of the front corner weights is equal to the front axle weight (1250 pounds), and the sum of the rear corner weights is equal to the rear axle weight (also 1250 pounds). The weight of the vehicle is of course equal to the sum of the two axle weights (our original 2500 pounds), and this weight can be thought of as acting through the vehicle’s center of gravity, or CG. Figure 1 sums it up nicely.

Note that when at rest, there are no horizontal (left or right) forces acting on the vehicle. All of the forces are acting in a vertical (up and down) direction. But what happens to the vehicle when we start to apply forces at the tire contact patch to try to stop it?

During braking, weight is transferred from the rear axle to the front axle. As in cornering where weight is transferred from the inside tires to the outside tires, we can feel this effect on our bodies as we are thrown against the seat belts. Consequently, we now need to add several more arrows to our illustration, but the most important factor is that our CG now has an deceleration acting on it.

Because the deceleration force acts at the CG of the vehicle, and because the CG of the vehicle is located somewhere above the ground, weight will transfer from the rear axle to the front axle in direct proportion to the rate of deceleration. In so many words, this is the effect of weight transfer under braking in living color.

This deceleration force is a function of a mechanical engineer’s most revered equation, F=ma, where F represents the forces acting at the contact patches, m represents the mass of the vehicle, and a represents the acceleration (or in our case, deceleration) of the vehicle. But enough of the engineering mumbo-jumbo – just have a look at these additional factors in Figure 2.



In Figure 3 (the beginning of what we call a “fishbone diagram” – more on this later), we see how our 2500 pound vehicle with 50/50 weight distribution at rest transfers weight based upon deceleration. Under 1.0g of deceleration (and using some typical values for our vehicle geometry) we have removed 600 pounds from the rear axle and added it to the front axle. That means we have transferred almost 50% of the vehicle’s initial rear axle weight to the front axle!

At this point, the brake system with a 50/50 weight distribution is going to apply too much force to the rear brakes, causing them to lock before we’re getting as much work as we could out of the front brakes. Consequently, our hero is going to get that white-knuckled ride because he creates more tire slip in the rear than the front, and it’s going to take longer for him to stop because the front tires are not applying as much force as they could be.


If we look at the equations we have developed, we see that all of the following factors will affect the weight on an axle for any given moment in time:

· Weight distribution of the vehicle at rest
· CG height – the higher it is, the more weight gets transferred during a stop
· Wheelbase – the shorter it is, the more weight gets transferred during a stop

We also know from fundamental brake design that the following factors will affect how much brake torque is developed at each corner of the vehicle, and how much of that torque is transferred to the tire contact patch and reacted against the ground:

· Rotor effective diameter
· Caliper piston diameter
· Lining friction coefficients
· Tire traction coefficient properties

It is the combination of these two functions – braking force at the tire versus weight on that tire – that determine braking bias. Changing the CG height, wheelbase, or deceleration level will dictate a different force distribution, or bias, requirement for our brake system. Conversely, changing the effectiveness of the front brake components without changing the rear brake effectiveness can also cause our brake bias to change.

hope this helps
 

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  • #7
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once again ranger mike that is very helpful.However assuming that

The co-efficent of friction is 0.5 between the wheel and road
The wheel radius is 190.5 MM
The efective radius is 105 MM
The force of the vehicle acting directly downwards on the wheel is 6867N (700*9.81)
The required braking force acting on the outside wheel radius is 1750 newtnons.
The frictional braking force acting on the efective radius by the brake pad is 3200kn

If there were four bolts at a pcd of 65 MM what would be the tourqe acting on each bolt ? i understand its very tricky to visualize it without a diargam i will try to upload a quick sketch
 
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  • #8
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updated diagram the downward force at the top is the wieght transfered by the car
 

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  • #9
Ranger Mike
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The co-efficent of friction you ar using is too low..

The maximum braking force that a particular tire can generate is theoretically equal to the coefficient of friction of the tire-road interface multiplied by the amount of weight being supported by that corner of the car. For example, a tire supporting 500 pounds of vehicle weight with a peak tire-road coefficient of 0.8 (a typical street tire value) could generate, in theory, 400 pounds of braking force. Throw on a good race tire with a peak coefficient of 1.5, and the maximum rises to 750 pounds of braking force. More braking force means higher deceleration, so we again see the mathematical benefits of a sticky race tire.

On the other hand, if our race tire was now only supporting 300 pounds, the maximum force would drop from 750 pounds of braking force to 450 pounds of braking force – a reduction of 40%.

Since the amount of braking force generated by the tire is directionally proportional to the torque generated by the calipers, pads, and rotors, one could also say that reducing the weight on the tire reduces the maximum brake torque sustainable by that corner before lock-up occurs. In the example above, if an assumed 700 ft-lb. of brake torque is required to lock up a wheel supporting 500 pounds, then only 420 ft-lb. (a 40% reduction) would be required to lock up a wheel supporting 300 pounds of vehicle weight.

At first glance, one could surmise that in order to achieve perfect brake bias you could just:

1. Weigh the four corners of the car
2. Design the front and rear brake components to deliver torque in the same ratio as the front-to-rear weight distribution
3. Win races

In other words, for a rear-wheel-drive race car with 50/50 front/rear weight distribution it would appear that the front and rear brakes would need to generate the same amount of torque. At the same time, it would look like a production-based front-wheel-drive car with a 60/40 front/rear weight distribution would need front brakes with 50% more output (torque capability) than the rears because of the extra weight being supported by the nose of the car.

Like most things in life though, calculating brake bias is not as simple as it may appear at first glance. Designing a braking system to these static conditions would neglect the second most important factor in the brake bias equation – the effect of dynamic weight transfer during braking.

Like the corner carvers, the brake guys are always looking to achieve maximum accelerations, but of course these accelerations are now really decelerations. Stopping distance is everything and every single foot counts. Remember: outbraking your opponent by just two feet every lap for a twenty lap sprint race can result in a three to four car length advantage at the checkered flag. Attention to detail matters.

As braking force is continuously increased, one end of the car must eventually break traction. If the front wheels lock up and turn into little piles of molten rubber first we say that the car is “front biased”, as the front tires are the limiting factor for deceleration. In the not-so-desirable situation where the rear tires are the first to lock we say that the car is “rear biased”, but the driver would probably have a few more choice adjectives to add. In either case, however, one end of the car has given up before the other, limiting the ultimate deceleration capability of the car.

The car with perfectly balanced brake bias will, however, be the last one to hit the brakes going down the back straight. By distributing the braking forces so that all four tires are simultaneously generating their maximum deceleration, stopping distance will be minimized and our hero will quickly find his way to victory lane. Just like neutral handling, balanced brake bias is our ticket to lower lap times.

All that said, once the braking system has achieved its perfect balance, it is still up to the tires to generate the braking forces. It’s still the tires that are stopping the car, but a poorly designed braking system can lengthen stopping distances significantly, expensive sticky tires or not.


While we can do calculations to determine what the optimum front-to-rear brake bias should be under all conditions, the difficult part is creating a brake system that can actually keep up with all of this. Our hero racer has it a little easier than those of us building cars for the real world. If he knows what his maximum deceleration capability is due to the tires he’s using, he can tune his brake system for that specific deceleration level. The good part is, if he tunes his vehicle for this 1.5g decel condition, because of the way weight transfer works, his car will be more front-biased in lower traction conditions, such as rain.

Back to the “fishbone diagram” mentioned earlier. Figure 3 shows front and rear axle weight versus deceleration of the vehicle. Now let’s look at it now as a percentage of the total vehicle weight. We can add on top of this chart the front-to-rear balance of the brake system. For example, if we use the exact same brake components at the front and rear axles of the car, they will each perform 50% of the braking, and the chart will look like Figure 4.

Evaluating this chart, we see that the vehicle will always be rear-biased. That is, the rear brakes will always be applying more force at the tire contact patch than the weight of the rear axle can sustain. This vehicle will always lock the rear brakes before the front. Not so good.

Most cars, however, have brakes at the rear that are smaller than the front. There are a lot of reasons for doing this, and one of them is to help provide the correct brake bias. Also, most cars have a proportioning valve which limits the amount of brake pressure seen at the rear calipers. If we look at the same chart with a more realistic braking system (one that takes into account these effects) it might look like the chart in Figure 5.
 

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  • #10
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There should be NO torque acting on the wheel studs.

All these analyses are good and will come in handy, but first and foremost a properly designed brake-disk assembly should have the bolts only in tension. The tension should be large enough that the friction developed by the resulting clamp will never be overcome by the forces acting on the wheel.

So the diameter of the bolts should be dependant on how much tension you need to accomplish this, not shear capability.

This design will give you no relative motion between the wheel, disk, and bolts, no cyclic loading, and it will be overall less prone to failure.

Google "bolted joint" for more information.
 
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  • #11
Mech_Engineer
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There should be NO torque acting on the wheel studs.

All these analyses are good and will come in handy, but first and foremost a properly designed brake-disk assembly should have the bolts only in tension. The tension should large enough that the friction developed by the resulting clamp will never be overcome by the forces acting on the wheel.

X2!

If the bolted joint is working correctly (e.g. the wheel studs are properly torqued) there should be no force acting on the wheel studs other than tension from preloading them. Any exteral forces should be taken by friction in the joint in the forms of shear, tension/compression, torsion about the bolt pattern, and bending. All externally applied forces should be less than the preload applied to the joint by the torqued/tensioned wheel studs. You should design the bolted joint to have enough wheel studs with enough tension to never slip or move; if it does, your design fails.
 
  • #12
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X3. Started a reply to this and evidently didn't post it.
 

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