# How to calc force required to steer an articulated vehicle?

• mekyy
In summary, you need to calculate the turning force and the friction required to steer an articulated heavy vehicle. The turning force is calculated using the equation F=Crr*N/R. The friction is calculated using the equation R=F/N.
mekyy
Ok so for a project of mine I need to calc. the force required to steer an articulated heavy on road vehicle and to overcome friction (so I need to calc frictional forces too). I understand I need to draw FBDs but it has been a while since i have done that and I have never worked with an articulated vehicle and it is very confusing for me.
I have been told that this can be worked by using only one side (left or right) of the vehicle as it is a dual-cylinder setup and calculating moment about center pivot. Is this moment the turning force? Actually, on second thought it would not be as it is a torque but how do I use that to get the turning force?

So what I know:
- vehicle's front axle and rear axle weights (both are single-axle but with 4 tyres front and rear)
-I know the hydraulic working pressure and area of the pistons and hence I have calculated the force required to extend and retract the cylinders (so the max. force that can be provided through the hydraulics to steer the vehicle)
-max articulation angle either side of center line (unsure if important to know)
-angle and length from center pivot to front cyl mount
-length mount to mount of cylinder
-length at 90 degrees from center pivot to cylinder
-length from center pivot to rear cyl mount
-offset angle as to which cylinder is mounted from the 90 degree position when steering is centered (value is 0.31 rad to the left of 90 degrees when looking at right hand side..so cylinders point inwards at front)

What do I do now, how do I go about calulting the turning force and the friction? Any help would be appreciated sorry for the length of this..

What will happen is the following: The midpoints of the front and rear axles will stay in place, the axles rotating about their respective midpoint as your articulate the vehicle.

This means that the wheels are in pure rolling. Hence the only resistive force is rolling resistance. You can get some rolling resistance coefficients values here as well.

Because - on the same axle - one wheel turn one way and the other, the other way, this will create a torque on your FBD. The other axle will create another torque, but in the other direction. So the input torque needs to be as high as the sum of these torques to initiate/maintain motion. Higher torque will create a greater rotational acceleration. So you will also have to define how fast you want this motion to happen to determine the turning force needed.

jack action said:
What will happen is the following: The midpoints of the front and rear axles will stay in place, the axles rotating about their respective midpoint as your articulate the vehicle.

This means that the wheels are in pure rolling. Hence the only resistive force is rolling resistance. You can get some rolling resistance coefficients values here as well.

Because - on the same axle - one wheel turn one way and the other, the other way, this will create a torque on your FBD. The other axle will create another torque, but in the other direction. So the input torque needs to be as high as the sum of these torques to initiate/maintain motion. Higher torque will create a greater rotational acceleration. So you will also have to define how fast you want this motion to happen to determine the turning force needed.

OK thanks for that starting to understand it a bit better. Using that equation for rolling resistance I am getting a force of 32 N for front right and 48 N for rear right tyres...for a 20 tonne vehicle seems too little? I also found another equation which yields closer to 100 N... F=Crr * N/R where R is wheel radius...unsure which to use

and with calculating the moment do i disregard the left side (as they would cancel)? and then only use half the axle lengths so moment about the centre of axle = roll resist. force *half axle length?? And then I just add this to that found for the other axle and this should give me my required torque? Or am i supposed to use these moments to then do another fbd to calculate the torque at the centre pivot and then add these together?

Because I have been given a max torque provided by cylinders which gives around 40 000 Nm at centre pivot...seem like my torques from my calcs are too small

actually come to think of it..since both moments on the same axle act at the center (thererfore spinning the axle in same direction) they would cause the moment to double??so calculate the moment on that axle for one side and double the answer to get total moment of axle??

also another thing...my "axle length" is actually length from centre of tyre to centre of opposite tyre on the axle...is that ok?

*Except I am using half that*

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lol hold up i was using kg not N

changed it and only gives me ~430 Nm for rear axle moment about centre of the axle...

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After looking at it some more, there will be some lateral displacement of the axles too. This introduces lateral friction which can be defined with the coefficient of friction of the tires.

After that, you need to do the FBD for both parts of the vehicle, each having its own mass and moment of inertia. For example:

RED: rolling resistance and lateral friction;
GREEN: Reaction at pin (same for both);
PURPLE: Input force from hydraulic cylinder (same for both; I've put only 1 cylinder);
BLUE: acceleration (linear & rotational) at CG.

mekyy
ok thanks for the help

So what I did was use the lateral force with mu=0.7

i then calculated the moments of both axle centres (this uses the small values of rolling resistance forces)...turns out you use the whole axle as the opposing directions of the rolling resistance forces actually double the moment not cancel it.

I then noted down these moment values...one was negative the other positive.

then I 'removed' the axles and treated the rest as a single beam with pivot at centre..i then calculated the moment about the centre summing the axle moments with the frictional forces*dist to pivot.

This gave me a reasonable answer of 30 000 Nm

The hydraulics provide 48 000 Nm at the pivot which is more then enough to overcome friction.

But...The 30 000 Nm is the torque required to only just get the vehicle turning right? But then this should be the max value since it is overcoming static friction and once turning at a stand still (and also while moving at speed) the torque required should be less

So, the extra torque supplied by hydraulics could be some sort of safety factor, having more than what is needed? Or would the extra value be there to get a certain turning velocity??

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mekyy said:
So, the extra torque supplied by hydraulics could be some sort of safety factor, having more than what is needed? Or would the extra value be there to get a certain turning velocity??
Yes, whatever force/torque that will exceed what is needed will translate into an acceleration of the motion; F=ma, T=Iα.

mekyy

## 1. How does weight distribution affect the force required to steer an articulated vehicle?

The weight distribution of an articulated vehicle can greatly impact the force required to steer it. If the weight is unevenly distributed, the vehicle may become imbalanced and require more force to steer in order to maintain stability. Additionally, the location of the center of gravity can also affect the force needed to steer.

## 2. What is the relationship between speed and the force required to steer an articulated vehicle?

The speed of an articulated vehicle can impact the force needed to steer it. As the speed increases, the force required to steer also increases. This is because at higher speeds, the momentum and inertia of the vehicle make it more difficult to turn or change direction.

## 3. How does the wheelbase of an articulated vehicle affect the force required to steer?

The wheelbase of an articulated vehicle refers to the distance between the front and rear axles. A longer wheelbase can make steering more difficult as it increases the turning radius and requires more force to maneuver the vehicle. On the other hand, a shorter wheelbase can make steering easier as it decreases the turning radius and requires less force to turn.

## 4. What role do the tires play in the force required to steer an articulated vehicle?

The type and condition of tires can greatly impact the force needed to steer an articulated vehicle. Worn or under-inflated tires can increase the friction between the tires and the road, making it more difficult to turn the vehicle. Additionally, different types of tires, such as all-season or winter tires, can also affect the force needed to steer in different weather and road conditions.

## 5. Can the design of the steering system affect the force required to steer an articulated vehicle?

Yes, the design of the steering system can play a significant role in the force required to steer an articulated vehicle. A well-designed steering system, with proper alignment and steering geometry, can make it easier to steer the vehicle with less force. On the other hand, a poorly designed steering system can make steering more difficult and require more force.

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