Braking calculations while going down a slope pullling a load

In summary: I'll just offend them without remorse)In summary, the brake system on a 42,000 lb machine can stop the tractor and trailer without issue. However, the slope of the road must be greater than 11.1525 degrees for the brakes to have an effect. If the tires are smaller than 9.4 ft in diameter, the rears will lock up.
  • #1
Brian P
4
0
I'm trying to find out how to see if my brakes will lock up or my tires will continue to roll through the following scenario. I have a machine that weighs 42,000 lbs. I'm pulling a 110,00 lbs trailer with load (no brakes on the trailer). This is a rubber tired machine going down a 8 degree slope at 5 mph. I'm using a 0.7 coefficient of friction (rubber on good concrete) and rolling resistance of 15 lb/1000 lb. The weights on the axles of the machine are 31,500 lbs on the front axle and 10,500 lbs on the rear axle. I have 622,000 in-lb of braking in each axle. If I slam on the brakes, will the wheels lock up? What is the calculation I'm missing? I have found calulations on stopping distances and times if the wheels are locked up and the tires are slidding but I need to prove they will lock up. Sorry if this a bad question but I have been out of the engineering field for a while and have just started back into it so I'm trying to learn it over again.
Thanks.
 
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  • #2
Welcome to PF, Brian.
There's an obvious typo in your post, regarding the trailer weight. I certainly hope that you mean 11,000 lbs., not 110,000.
 
  • #3
No, it is not a typo. I work in the coal mining industry, we haul giant size loads.
 
  • #4
Just stick with the .7 coeffiecient of friction. With a 42000 lb tractor, the maximum braking force is .7 x 42000 lbs = 29400 lbs. Total weight of the tractor and trailer is 42000 lbs + 110000 lbs = 152000 lbs. On a 8 degree slope, the component of gravity in the direction of the slope is sin(8 degrees) x 152000 lbs = 21154 lbs, which is less than the maximum braking force of 29400 lbs, so there's enough braking force to slow down and stop the tractor+trailer.

If the slope is > 11.1525 degrees, then the component of gravity in the direction of the slope is greater than the braking force and there tractor+trailer would accelerate even with maximum braking. You only have about 3 degrees of safety margin here so the slopes need to be made accurately.

This also assumes that the brakes don't fade. The brakes are performing 21154 lb ft of work for every foot braked, at 5 mph, this is 21154 lb x 5 mph x (5280 ft/mile) x (hour / 3600 sec) x (sec/550 lb ft) = 282 hp worth of heat generation.

I have 622,000 in-lb of braking in each axle.
This information doesn't help unless you know the effective diameter (or radius) of the tires. At the front, there is 31500 lbs time the .7 coefficient of friction = 22050 lbs of force at the surface of the tires, or 11025 lbs per front tire. Braking torque is 622000/12 = 51833 ft lbs. This limits the tire radius to 51833 (ft lb)/11025(lb) = 4.7 ft for maximum braking, or 9.4 ft diameter. If only the braking force to control the tractor and trailer is considered, then the maximum allowable tire radius increases to (29040/21154) x 4.7 ft = 6.45 ft, or a diamter of 12.9 ft.
 
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  • #5
Brian, I meant no offense about there being a typo, but there is one. What I didn't know was whether you put the comma in the wrong place, or if you left off a trailing zero. :tongue:
Since I thought that this was something to be done on a public road, I assumed the former. Sorry for the misunderstanding.
 
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  • #6
Brian P said:
If I slam on the brakes, will the wheels lock up?
If the front tire diameter is less than 9.4 ft, then yes. The rears will lock up unless the braking bias is adjusted to balance with the load differential on the front and rear axles (31500lbs versus 10500lbs).

Brian P said:
I work in the coal mining industry, we haul giant size loads.
Picture of a coal hauling truck, or a truck for those that think that a Hummer just isn't big enough to impress their neighbors:

coaltrk1.jpg
 
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  • #7
Thanks for all the responses everyone. I have not used a site like this before and it has been extremely helpful. I did not take offense Danger, sorry if I came across that way. It has been few years since I've been using my engineering degree, now I'm back in the field instead of managing. My tires are 38" diameter. My machine is actually a diesel underground hauler. Not as big as the truck in the picture though!:)
 
  • #8
Brian P said:
I did not take offense Danger, sorry if I came across that way.

You didn't actually, but being a Canuck I always try to cover all of my don't-offend bases just in case. (Unless someone richly deserves it, in which case I'll offend him within an inch of his life. :biggrin:)
I'm not qualified to even begin helping you anyhow (never finished high-school). My post was more of just a greeting and to clarify something that I thought might confuse other readers.
 
  • #9
I've been out for a week at the mines but I have a another question. How do I find the distance it takes my machine to stop on going down the slope at the speed I have stated?
 
  • #10
The net (braking) force on the tractor + trailer is: 29400lbs - 21154 lbs = 8246lbs.

Mass of the system is 152000 lb mass.

Acceleration = - 8246lbs / 152000lbs = - .05425 g x (32.174 ft /(sec^2 x g) => - 1.7454 ft / sec^2

Initial speed is 5mph = 7.3333 ft / sec

Time to stop is (7.3333 ft / sec) / (1.7454 ft / sec ^2) = 4.2015 seconds.

Distance = V0 t + 1/2 A t^2 = (7.3333 ft / sec) (4.2014 sec) + 1/2 (-1.7454 ft / sec^2) (4.2015 sec)^2 = 15.4 feet
 
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1. How do I calculate the braking force needed while going down a slope pulling a load?

To calculate the braking force needed, you will need to know the mass of the load, the angle of the slope, and the coefficient of friction between the load and the ground. The formula for calculating braking force is: Fb = m*g*sinθ + μ*m*g*cosθ, where Fb is the braking force, m is the mass of the load, g is the acceleration due to gravity (9.8 m/s²), θ is the angle of the slope, and μ is the coefficient of friction.

2. How does the angle of the slope affect braking calculations?

The angle of the slope plays a significant role in braking calculations. The steeper the slope, the greater the force needed to slow down or stop the load. This is because the force of gravity pulls the load down the slope, increasing the overall force applied to the load.

3. What is the coefficient of friction and how does it impact braking calculations?

The coefficient of friction is a measure of the force required to move one surface over another. In the context of braking calculations, it represents the resistance between the load and the ground. A higher coefficient of friction means there is more resistance, requiring a greater braking force to slow down or stop the load.

4. Can the braking force be greater than the weight of the load?

Yes, the braking force can be greater than the weight of the load. This can happen if the slope is very steep, the load is heavy, or the coefficient of friction is high. In these cases, the braking force will need to be greater than the weight of the load to prevent it from rolling down the slope.

5. Is there a specific formula for calculating braking distance?

Yes, the formula for calculating braking distance is: d = (v²)/(2*µ*g*cosθ), where d is the braking distance, v is the initial velocity of the load, µ is the coefficient of friction, g is the acceleration due to gravity, and θ is the angle of the slope. This formula assumes a constant braking force and does not take into account other factors such as air resistance.

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