Prop Shaft load parked on a grade

[JGM]

What would the torque on a driveshaft be for a 8 ton vehicle in park on a 16% grade? Tire radius is 19". Axle ratio it 7.1.

 

[berkeman]

What would the torque on a driveshaft be for a 8 ton vehicle in park on a 16% grade? Tire radius is 19". Axle ratio it 7.1.

Welcome to the PF.

Is this question for schoolwork? What is the context of the question?

 

[JGM]

Welcome to the PF.

Is this question for schoolwork? What is the context of the question?

Trying to determine the loads on an output shaft of a transmission when parked on a grade with different axle ratios. Not for school work. Just general knowledge.

 

[jack action]

If the rear wheels support the entire load caused by the slope, the force acting at the rear tires is $F_w = W\sin\theta$, where $W$ is the weight of the vehicle and $\theta$ is the angle of the slope (reference). $100\tan\theta = \%slope$ to find the angle of the slope (reference).

The wheel torque produced is $T_w = F_w r$, where $r$ is the tire radius.

The gear ratio reduces the torque seen by the driveshaft, so $T_d = \frac{T_w}{GR}$, where $GR$ is the axle gear ratio.

$\theta = \arctan\frac{16}{100} = 9.1°$

$F_w = (16000\ lb)\sin9.1° = 2530\ lb$

$T_w = (2530.5\ lb) * (1.583\ ft) = 4006\ lb.ft$

$T_d = \frac{4006\ lb.ft}{7} = 572\ lb.ft$

 

[JGM]

If the rear wheels support the entire load caused by the slope, the force acting at the rear tires is $F_w = W\sin\theta$, where $W$ is the weight of the vehicle and $\theta$ is the angle of the slope (reference). $100\tan\theta = \%slope$ to find the angle of the slope (reference).

The wheel torque produced is $T_w = F_w r$, where $r$ is the tire radius.

The gear ratio reduces the torque seen by the driveshaft, so $T_d = \frac{T_w}{GR}$, where $GR$ is the axle gear ratio.

$\theta = \arctan\frac{16}{100} = 9.1°$

$F_w = (16000\ lb)\sin9.1° = 2530\ lb$

$T_w = (2530.5\ lb) * (1.583\ ft) = 4006\ lb.ft$

$T_d = \frac{4006\ lb.ft}{7} = 572\ lb.ft$

Shouldn't this include the acceleration of gravity?
F_w = 16000 Lbs* 32.174 ft/s2

 

[bsheikho]

Shouldn't this include the acceleration of gravity?
F_w = 16000 Lbs* 32.174 ft/s2

pounds is a unit for weight not mass, already includes gravity.

 

[Baluncore]

I used SI units.
Mass = 8000 kg.
Force due to gravity = 8000 * 9.8 = 78400 Newton.
Wheel radius is 39” = conveniently 1 metre.
16% grade = 9.09 deg; Sin(9.09°) = 0.158
7:1 axle ratio.
Drive shaft torque = 1m * 78400N * 0.158 / 7 = 1769.5 Nm
1769.5 Nm = 1305.1 ft.lbs
This is quite different to jack action's 572. ft.lb

 

[jack action]

I used SI units.
Mass = 8000 kg.
Force due to gravity = 8000 * 9.8 = 78400 Newton.
Wheel radius is 39” = conveniently 1 metre.
16% grade = 9.09 deg; Sin(9.09°) = 0.158
7:1 axle ratio.
Drive shaft torque = 1m * 78400N * 0.158 / 7 = 1769.5 Nm
1769.5 Nm = 1305.1 ft.lbs
This is quite different to jack action's 572. ft.lb

That is because we don't use the same numbers:

• I use 1 ton = 2000 lb and you use 1 tonne = 1000 kg;
• I use a wheel radius of 19" and you use a wheel radius of 39".

Other than that, everything is the same!

 

[Baluncore]

Fixing the radius to 19” and using a mass of 8 Short Tons.
Mass 8 short ton = 7257.5 kg.
Force due to gravity = 7257.5 * 9.8 = 71123.5 Newton.
Wheel radius is 19” = inconveniently 0.4826 metre.
16% grade = 9.09 deg; Sin(9.09°) = 0.158
7:1 axle ratio.
Drive shaft torque = 0.4826 * 71123.5 * 0.158 / 7 = 774.75 Nm
774.75 Nm = 571.5 ft.lbs
We agree.

In Australia the standard Ton was a Long Ton = 1016.05 kg = 2240 lbs.
In 1966 that was replaced by the metric Tonne = 1000 kg = 2205 lbs.

Are vehicle weights in the USA always specified in Short Tons = 907.18 kg = 2000 lbs ?

 

[jack action]

Are vehicle weights in the USA always specified in Short Tons = 907.18 kg = 2000 lbs ?

https://en.wikipedia.org/wiki/Short_ton

I'm in Canada and a ton has always been 2000 lb around here. But we're with the SI system as well, so this is not a usual unit for us nowadays (unless we have to do business with our neighbor).

As for truck classification, it is done in pounds, so no problem there: