Determining radial force on shaft

[MechEng2023]

TL;DR Summary

In a mechanical system consisting of an air cylinder, linkage, and shaft, i'm trying to understand if changing the cylinder applied force will impact the radial force that exists between linkage and shaft.

Hello,

For the mechanical system below, i'm trying to understand the relationship between cylinder applied force and radial force.

The system consists of a shaft which is held in-place by a bushing. The shaft is connected to an air cylinder via a linkage. The air cylinder is on a fixed pivot and has an internal hard stop in both extended and retracted positions. Actuating the air cylinder between its extended and retracted positions rotates the shaft between 2 positions that are 90deg apart. The shaft doesn't rotate freely within the bushing - significant friction between the shaft and bushing exists.

I'm in the process of analyzing the radial force that the linkage will transfer to the shaft, using force triangles drawn at various positions throughout the shaft's 90deg rotation. What i'm struggling to understand is whether or not increasing the cylinder's applied force will result in an increase in the radial force that the linkage transfers to the shaft (The system's geometry and friction between shaft and bushing remain constant. The only variable changing is the cylinder's applied force)

For example, assume that 10 lbf and 10,000 lbf applied cylinder force are both enough to overcome friction and cause the shaft to rotate. Will the 10,000 lbf cylinder force result in a higher radial force being transferred from linkage to shaft? OR, does the system only generate enough force to overcome friction, and since friction is constant in this scenario, a 10lbf and 10,000 lbf cylinder applied force will result in identical radial forces being transferred from linkage to shaft?

Any insight is appreciated.
Thank you!

 

[berkeman]

Welcome to PF.

Can you upload a diagram of the system please? Use the "Attach files" link below the Edit window to upload a PDF or JPEG image of your mechanism. Thanks.

 

[MechEng2023]

Welcome to PF.

Can you upload a diagram of the system please? Use the "Attach files" link below the Edit window to upload a PDF or JPEG image of your mechanism. Thanks.

Thank you - 2 images of the mechanism are now attached

 

[Lnewqban]

Welcome, @MechEng2023 !

The cylinder can’t apply a force of greater magnitude than the resistance to its stroke, everything else remaining the same (as per Newton’s third law).

 

[MechEng2023]

Thanks for the quick reply!

2 follow up questions:

1) Does this mean that the max force value shown on the cylinder spec sheet represents what the cylinder is capable of applying to the system, and not what the cylinder actually applies to the system? (i.e. the actual force that the cylinder applies to the system is dependent on how much resistance exists in the system)

2) If my mechanical system were assumed to be frictionless, does this mean the applied cylinder force (F_CYL on my image) would be ~0 lbf?

 

[berkeman]

2) If my mechanical system were assumed to be frictionless, does this mean the applied cylinder force (F_CYL on my image) would be ~0 lbf?

What do you mean by "frictionless"? What is the counter-torque on the shaft that is being turned by the cylinder/piston? That should be part of your FBD...

 

[Baluncore]

The cylinder force is countered by an equal and opposite sideways reaction force from the shaft.
For the same torque applied to the shaft, but through a longer radius arm, will require less linear cylinder force, so will apply proportionally less side force to the shaft.

 

[Lnewqban]

Thanks for the quick reply!

2 follow up questions:

1) Does this mean that the max force value shown on the cylinder spec sheet represents what the cylinder is capable of applying to the system, and not what the cylinder actually applies to the system? (i.e. the actual force that the cylinder applies to the system is dependent on how much resistance exists in the system)

2) If my mechanical system were assumed to be frictionless, does this mean the applied cylinder force (F_CYL on my image) would be ~0 lbf?

1) Being a pneumatic cylinder, that force (and delivered work) is directly proportional to the internal pressure, which is also proportional to how quickly the internal volume can be filled.
For zero resistance, it will deliver a quick stroke until reaching an stopping limit.
For high resistance, the stroke will tend to be slower.
For resistance value higher than the cylinder specs, nothing will move, even if the internal pressure reaches a maximum value.

2) As the cylinder has a pivot, the forces represented in your diagram can’t exist. It can only push in line with its rod, which will automatically get aligned with the tangent to the perpendicular and variable distance to the center of the shaft being rotated.

 

[Tom.G]

2) As the cylinder has a pivot, the forces represented in your diagram can’t exist. It can only push in line with its rod, which will automatically get aligned with the tangent to the perpendicular and variable distance to the center of the shaft being rotated.

Huh??

There are an awful lot of steam locomotives that seem a rather successful implementation.

Cheers,
Tom

 

[Baluncore]

Huh??

Then how does an internal combustion engine manage its compression stroke to operate many millions of cars?

I think you have misinterpreted something.
The Air Cylinder Pivot is the equivalent of a piston gudgeon pin, (small end), in an ICE.

The effective length of the radius arm varies with crank angle. That length is measured from the shaft centre, to the nearest point on the cylinder rod axis, not from the crank pin on the radius arm, to the shaft centre.

 

[Tom.G]

@Baluncore...
You caught me between edits.
Current post is:

Huh??

There are an awful lot of steam locomotives that seem a rather successful implementation.

Cheers,
Tom

 

[Lnewqban]

Huh??

There are an awful lot of steam locomotives that seem a rather successful implementation.

Cheers,
Tom

Sorry Tom, I am unable to see the connection between your quotation and steam locomotives.
Would you mind explaining it further?

Perhaps my post should be rephrased?
As you may have noted, my command of the English language is poor.
Will the following editing be an improvement?:

“As the cylinder has a pivot, the forces represented in your diagram can’t exist as represented.”

 

[Tom.G]

“As the cylinder has a pivot, the forces represented in your diagram can’t exist as represented.”

I agree. As shown in the image one force seems to be a resultant and the other...?

My reference to a steam locomotive was to point out that a single drive link with a pivot joint in it (as usually seen), is conceptially identical to a pivoting steam cylinder and a solid drive link to the wheel.