# RTG crane -- hydraulic motor and linkage design to turn these wheels

Hi all
I would like to ask your professional guide in designing a hydraulic system. I attached a file, showing sketch of the RTG crane wheel structure.

Here are the conditions:

1. The red-line is the hydraulic cylinder with "X" as the fix point.
2. The hydraulic cylinder will stroke out from point "a" to point "b", resulting the rubber tyre to turn to 90o from the original position.
3. The yellow-line is a connecting rod that will move the adjacent rubber tyre accordingly.
4. The load is calculated to be 17 ton per tyre.
5. Both hydraulic cylinders are powered from the same single power pack.
6. The rubber tyres are sitting on a steel plate.

My question is, how much is the force required in order to move the rubber tyre to 90o position?

Regards

#### Attachments

• wheel turn.pdf
96 KB · Views: 135

berkeman
Mentor
Hi all
I would like to ask your professional guide in designing a hydraulic system. I attached a file, showing sketch of the RTG crane wheel structure.

Here are the conditions:

1. The red-line is the hydraulic cylinder with "X" as the fix point.
2. The hydraulic cylinder will stroke out from point "a" to point "b", resulting the rubber tyre to turn to 90o from the original position.
3. The yellow-line is a connecting rod that will move the adjacent rubber tyre accordingly.
4. The load is calculated to be 17 ton per tyre.
5. Both hydraulic cylinders are powered from the same single power pack.
6. The rubber tyres are sitting on a steel plate.

My question is, how much is the force required in order to move the rubber tyre to 90o position?

Regards
17 tons on each rubber tire? Yikes. You need to figure out the rolling resistance of those tires when under load. I'm assuming they are solid?

17 tons on each rubber tire? Yikes. You need to figure out the rolling resistance of those tires when under load. I'm assuming they are solid?

Yes. 17 ton per tyre.
The tires are air-inflated tyre.

Baluncore
My question is, how much is the force required in order to move the rubber tyre to 90o position?
This is not a rolling resistance problem. It appears to actually be an estimate of the force required to steer the wheels to point sideways. The size of the contact patch might be ignored and the rubber steel contact friction overcome without translation.

The key figure required is the coefficient of friction between the steel plate and the tread. That could be reduced significantly by water as a lubricant. Steering while translating would also reduce the force required.

Area is not important, nor is the speed of steering.
Force is = 17 ton * gravity * coefficient of friction.
But what sort of ton is that? Short US, Long UK or Tonne = 1000kg ?

berkeman
anorlunda
Staff Emeritus
The most difficult case turning the wheel will be when it is in mud, or in a rut. That is very difficult to calculate. Experience from other cranes would be a better source.

You will also need protection against the wheel getting stuck and unable to turn.

Baluncore
The most difficult case turning the wheel will be when it is in mud, or in a rut. That is very difficult to calculate.
6. The rubber tyres are sitting on a steel plate.
The mechanical parts must handle the maximum possible hydraulic forces. There will be a limit to those forces set by the hydraulic bypass pressure limiter.
The worst case coefficient of static friction of dry rubber on a solid material is unusual. It can range from 1.0 to 4.0; The type of rubber used will make a big difference.

You can probably measure the static friction force by parking one steering wheel on a steel plate, in dry weather. Then use a long arm or a winch truck to rotate the plate on the ground, under the wheel. Measure the hydraulic pressure peak in the closed hydraulic steering cylinder to get an estimate of the minimum pressure that will be needed to turn the wheel later. Obviously, the friction between the plate and the ground will not influence the hydraulic pressure estimate.

Tom.G
Since th OP is asking for
guide in designing a hydraulic system.
It would seem the worst case would be use the 4.0 coefficient of friction that @Baluncore supplied, times the 17 ton loading, at the equivalent point of the tyre footprint patch. That will give a torque, which will allow calculation of the force needed at the lever arm of the hydraulic cylinder attachment.

The 'equivalent point of the tyre footprint patch' could probably be best answered by the tyre manufacturer, although someone here could perhaps come up with an approximation.

Cheers,
Tom

Coefficient of friction for tyre on steel is 0.3

Baluncore
Coefficient of friction for tyre on steel is 0.3
Can you give a reference for that value of μs = 0.3 ?

berkeman
Mentor
Coefficient of friction for tyre on steel is 0.3
Can you give a reference for that value of μs = 0.3 ?
Yeah, it looks to be a bit higher than 0.3, unless he is referring to some special rubber...

https://www.engineersedge.com/coeffients_of_friction.htm

#### Attachments

4.9 KB · Views: 276
Baluncore
Yeah, it looks to be a bit higher than 0.3, unless he is referring to some special rubber...
Those “60 A belt” coefficients may be for Vee belts pulled into a pulley groove. Tool steel and stainless steel may have quite different values to mild steel plate due to their surface chemistry. Tyre on asphalt and on grass do not help.

The confidence of malemdk's μs = 0.3 with the the “tyre on steel” specificity, must have come from some source. It is sufficiently questionable that I should not really believe it until I see the reference. Maybe it allowed for a wet track, hard rubber, or just a guess by an examiner when setting an exam question.

It is easier to turn the steering when the wheel is rolling because the tread and tyre distortion can walk on the surface. If the linkage is able to handle the peak hydraulic pressures, and some form of cross-flow pressure relief is fitted, then an exceptionally sticky tyre, on a hot day, will not be a disaster. It will require a slight travel, or a bucket of water to prevent scuffing damage to the tyre. But that does not have any importance in the hydraulic actuator design.

What if; each pair of wheels were mounted on a common rocking axle, close on either side of a vertical steering post, then when the crane was stationary, any steering direction change would generate a travel. Indeed, independent wheel drive motors could be controlled in a way that assists the steering. Maybe the steering could then be done entirely without a hydraulic cylinder?

berkeman
JBA
Gold Member
Even given an accurate friction factor the amount of torque required to rotate the tire is going to depend on the tire patch dimensions because the amount of total torque will be affected by the distance of the friction from the rotating center of the tire.

Baluncore
Even given an accurate friction factor the amount of torque required to rotate the tire is going to depend on the tire patch dimensions because the amount of total torque will be affected by the distance of the friction from the rotating center of the tire.
Yes, but in this case there is a rubbery difference between static and sliding friction. I believe you are considering sliding friction.

Why do you suggest that the static frictional force is a function of the distance that the surfaces will move once sliding friction comes into play?

The prediction is complicated by the flexibility of the rubber and the independent movements that will be taken by each sub-patch, da, of the tyre contact patch area. But in principle, to break the static glue should not be a function related to radius. I see it as the force or energy needed to break a fixed number of chemical bonds.

The distance that each da moves will be determined by the angle and the radius from the fixed point of rotation. That will decide the energy required to steer the wheel. The power will be determined by how quickly it is being done.

This is not a rolling resistance problem. It appears to actually be an estimate of the force required to steer the wheels to point sideways. The size of the contact patch might be ignored and the rubber steel contact friction overcome without translation.

The key figure required is the coefficient of friction between the steel plate and the tread. That could be reduced significantly by water as a lubricant. Steering while translating would also reduce the force required.

Area is not important, nor is the speed of steering.
Force is = 17 ton * gravity * coefficient of friction.
But what sort of ton is that? Short US, Long UK or Tonne = 1000kg ?
it is 17 x 1,000 kg