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Calculating contact surface area of Cylindrical Contact

DFlemming

Plastic
Joined
Mar 27, 2015
Good Day fellow Machinists,

I hope you are all doing well...I have a quick question.
I intend to use 1" OD mild steel tube as a shaft in a 1" bore UCP type pillow block bearing unit. I am using SolidWorks Simulation to calculate the maximum stress on the steel tube shaft, BUT...I need to first determine what is the contact surface area between the steel tube shaft and the inside of the pillow block bearing unit. Knowing this I can calculate the pressure exerted on the steel tube.

Does anyone know of a way or a formula to find this? Shigley's Mechanical Eng. Design has a formula for Cylindrical Contact but if I use that formula with 100% contact it doesn't work out...or I'm just using it wrongly...

Also, does anyone know if SolidWorks can calculate this for me?

I look forward to your favourble responses,
Thank you,
David:D
 
Whats the goal ?

Usually .. tube =/= ie Does NOT Relate to calculations.
Also..
Pressure is not something related to shafts.
Twist, load, tension, pull, push, yes.

If you have a perfect surface, to a prefect bearing surface, it will have the area of the geometrical calculation.
BUT..
Its totally irrelevant if you are trying to calculate if the bearing is big enought, the shafts D is big enough, etc.

Critical speed in RPM (Q value in bearings) or twisting moment, or load moment will determine the lifetime or if it works.

Thats why my question.
What are you trying to figure out ?

Speed, load, and use are all critical parameters.
Impacts, twist, or rpm can all make it impossible for the given idea to work, depending on the parameters.
 
what hanermo said!

Whats the goal ?

Usually .. tube =/= ie Does NOT Relate to calculations.
Also..
Pressure is not something related to shafts.
Twist, load, tension, pull, push, yes.

If you have a perfect surface, to a prefect bearing surface, it will have the area of the geometrical calculation.
BUT..
Its totally irrelevant if you are trying to calculate if the bearing is big enought, the shafts D is big enough, etc.

Critical speed in RPM (Q value in bearings) or twisting moment, or load moment will determine the lifetime or if it works.

Thats why my question.
What are you trying to figure out ?

Speed, load, and use are all critical parameters.
Impacts, twist, or rpm can all make it impossible for the given idea to work, depending on the parameters.



Another aspect to consider. If the contact between two cylinders (ID/ OD) is taken to the extreme, the contact is a line of zero width. Running a quick calculation on loading, the load per unit area on zero area is infinate. At infinite load, the material will distort. Broadening the contact region, adding area and reducing the specific loading.

The contact area will always exactly equal the amount required to carry the load . A bit confusing, but...
That's how car tires work anyway.
 
Thanks for the quick replies...

I am actually designing a seesaw using 2" x 2" square hollow section (SHS) as the teeter beam. The two pieces of 1" OD steel tube will be welded on either side of the SHS at the midpoint of its length. The 1" tube pieces will go into two pillow block bearing units to allow the seesaw motion. As such, I am trying to determine the best thickness of tube to use.
Knowing the contact surface area would let me know how the reaction force is distributed on the steel tube, and I would be able to choose a suitable tube thickness (not to thin and at the same time not over-designed:typing:)

What do you think?
 
I think that you're overthinking the problem (and missing the point).

Any piece of tubing (not OD ground) is going to have significant variance in form and size affecting % contact with your bearings. Why deal with it? Use a solid bar rather than a tube. Additionally, using a solid will eliminate the need to plug the tube ends so tiny fingers won't probe.

Also, your stated plan to weld on two stubs is the hard way to go. Instead, drill through the 2 x 2, insert a solid bar and 100% weld around both sides. That approach will require less fixturing (fooling around) and result in better alignment of the two stubs.
 
Shear is the only performance criteria you need consider.

Thanks for the quick replies...

I am actually designing a seesaw using 2" x 2" square hollow section (SHS) as the teeter beam. The two pieces of 1" OD steel tube will be welded on either side of the SHS at the midpoint of its length. The 1" tube pieces will go into two pillow block bearing units to allow the seesaw motion. As such, I am trying to determine the best thickness of tube to use.
Knowing the contact surface area would let me know how the reaction force is distributed on the steel tube, and I would be able to choose a suitable tube thickness (not to thin and at the same time not over-designed:typing:)

What do you think?

Just give a 100:1 safety factor ;-) (Assuming it a play ground teeter totter ;-)
 
Thanks for the quick replies...

I am actually designing a seesaw using 2" x 2" square hollow section (SHS) as the teeter beam. The two pieces of 1" OD steel tube will be welded on either side of the SHS at the midpoint of its length. The 1" tube pieces will go into two pillow block bearing units to allow the seesaw motion. As such, I am trying to determine the best thickness of tube to use.
Knowing the contact surface area would let me know how the reaction force is distributed on the steel tube, and I would be able to choose a suitable tube thickness (not to thin and at the same time not over-designed:typing:)

What do you think?

I think you don't have a mentor to gently slap you upside the head.
Life is never as it is drawn in the CAD.
Under load, your tube will move and distort, hence the contact area and loading will change.
The math will still work out right, your starting number will move around continuously changing.
Bob
 
I think that you're overthinking the problem (and missing the point).

Any piece of tubing (not OD ground) is going to have significant variance in form and size affecting % contact with your bearings. Why deal with it? Use a solid bar rather than a tube. Additionally, using a solid will eliminate the need to plug the tube ends so tiny fingers won't probe.

Also, your stated plan to weld on two stubs is the hard way to go. Instead, drill through the 2 x 2, insert a solid bar and 100% weld around both sides. That approach will require less fixturing (fooling around) and result in better alignment of the two stubs.

Yes, running the tube through the SHS does made more sense...I understand what you are saying about the tube vs. the solid bar...but solid bar will definitely be overkill for the intended part, hence the reason I opted for tube instead and want to fine tune it by analyzing the stresses to best choose a thickness for it...
 
Another aspect to consider. If the contact between two cylinders (ID/ OD) is taken to the extreme, the contact is a line of zero width. Running a quick calculation on loading, the load per unit area on zero area is infinate. At infinite load, the material will distort. Broadening the contact region, adding area and reducing the specific loading.

The contact area will always exactly equal the amount required to carry the load . A bit confusing, but...
That's how car tires work anyway.

Not sure if I follow your explanation correctly...but if the contact is taken to the extreme, wouldn't the contact area be a rectangle measuring the circumference of the shaft x the thru-length of the bearing surface?
 
Good Day fellow Machinists,

I hope you are all doing well...I have a quick question.
I intend to use 1" OD mild steel tube as a shaft in a 1" bore UCP type pillow block bearing unit. I am using SolidWorks Simulation to calculate the maximum stress on the steel tube shaft, BUT...I need to first determine what is the contact surface area between the steel tube shaft and the inside of the pillow block bearing unit. Knowing this I can calculate the pressure exerted on the steel tube.

Does anyone know of a way or a formula to find this? Shigley's Mechanical Eng. Design has a formula for Cylindrical Contact but if I use that formula with 100% contact it doesn't work out...or I'm just using it wrongly...

Also, does anyone know if SolidWorks can calculate this for me?

I look forward to your favourble responses,
Thank you,
David:D

...seems to me it would be pi (3.14159) x 1 (the diameter) x the width of the inner bearing race...

...for example...if the bearing race is 1 inch wide...it would be...3.14159 x 1 x 1 = 3.14159 square inches...
 
...seems to me it would be pi (3.14159) x 1 (the diameter) x the width of the inner bearing race...

...for example...if the bearing race is 1 inch wide...it would be...3.14159 x 1 x 1 = 3.14159 square inches...

hmmm...this formula appears quite concise...do you mind telling me how you derived it?
 
hmmm...this formula appears quite concise...do you mind telling me how you derived it?

...first you have to accept the fact that pi x the diameter is the circumference...in this case 1 (the diameter) x pi (3.14159)= 3.14159...

...then picture (if it was possible) to split that bearing race on the circumference and lay it out flat...the length of the rectangle generated is 3.14159 inches...the surface area of a rectangle is length x width...

...if you use my example of the bearing race being 1 inch wide...length x width would be 3.14159 x 1 = 3.14159 square inches...
 
Not sure if I follow your explanation correctly...but if the contact is taken to the extreme, wouldn't the contact area be a rectangle measuring the circumference of the shaft x the thru-length of the bearing surface?

Magnify your perspective.
The support "bearing must be of larger ID than the shaft (tube in your case) ID. If both are perfect cylinders, the contact can only be a line, infinitely narrow. No matter what the length, if only a line contact, the loading is infinite, regardless of weight etc.

So, the two elements, shaft and support MUST distort from perfect roundness. ...Then we return to the car tire analogy.

I suggest you investigate "hoop stress", or revert to shear values. For your intended purposes, just over build. Kids will thank you for generations.
 
...first you have to accept the fact that pi x the diameter is the circumference...in this case 1 (the diameter) x pi (3.14159)= 3.14159...

...then picture (if it was possible) to split that bearing race on the circumference and lay it out flat...the length of the rectangle generated is 3.14159 inches...the surface area of a rectangle is length x width...

...if you use my example of the bearing race being 1 inch wide...length x width would be 3.14159 x 1 = 3.14159 square inches...

Oh ok...I see, so this is assuming 100% contact between the OD of the shaft and the ID of the bearing race...
 
Magnify your perspective.
The support "bearing must be of larger ID than the shaft (tube in your case) ID. If both are perfect cylinders, the contact can only be a line, infinitely narrow. No matter what the length, if only a line contact, the loading is infinite, regardless of weight etc.

...that's why bearings are press fit on shafts...
 
You have lost me.. A "light press fit" is NOT a bearing fit.

How about a "tight or loose running fit"
Oil or grease lubricated?
pressurized or open...
 








 
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