Solution to Hydrostatic Bearing Integration Task

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SUMMARY

The forum discussion centers on solving a hydrostatic bearing integration task, specifically calculating the total load capacity using the formula W=\int_0^{R_o} p_r(2πr dr) + \int_{R_o}^R p(2πr dr). The radial pressure is expressed in terms of recess pressure as p=p_r{\frac{ln(R/r)}{ln(R/R_o)}}, and participants confirm that substituting this into the equation is necessary. A participant also points out a potential error in the original equation, suggesting it may be underestimated by a factor of 2.

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mathi85
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Hi everyone!
I would like to ask you for help with one of the tasks from my assignment. The rest of the assignment is done including some simple integration but I struggle with this one:

Task
"The total load capacity for a circular hydrostatic bearing is given as

##W=\int_0^{R_o} p_r(2πr dr) + \int_{R_o}^R p(2πr dr) ##

By expressing the radial pressure in terms of the recess pressure, and by step by step argument, show that:

##W={\frac{π}{2}}{\frac{R^2-R_o^2}{2ln(R/R_o)}}p_r ## "

I think that radial pressure in terms of recess pressure is:

##p=p_r{\frac{ln(R/r)}{ln(R/R_o)}} ##

I really cannot get my head around it. Shall I just substitute above equation for 'p'? Then I would get:

##W=\int_0^{R_o} p_r(2πr dr) + \int_{R_o}^R{\frac{p_r2πrdrln(R/r)}{ln(R/R_o)}} ##

Do I have to then sort both integrals and just add them up together?
 
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mathi85 said:
Hi everyone!
I would like to ask you for help with one of the tasks from my assignment. The rest of the assignment is done including some simple integration but I struggle with this one:

Task
"The total load capacity for a circular hydrostatic bearing is given as

##W=\int_0^{R_o} p_r(2πr dr) + \int_{R_o}^R p(2πr dr) ##

By expressing the radial pressure in terms of the recess pressure, and by step by step argument, show that:

##W={\frac{π}{2}}{\frac{R^2-R_o^2}{2ln(R/R_o)}}p_r ## "

I think that radial pressure in terms of recess pressure is:

##p=p_r{\frac{ln(R/r)}{ln(R/R_o)}} ##

I really cannot get my head around it. Shall I just substitute above equation for 'p'? Then I would get:

##W=\int_0^{R_o} p_r(2πr dr) + \int_{R_o}^R{\frac{p_r2πrdrln(R/r)}{ln(R/R_o)}} ##

Do I have to then sort both integrals and just add them up together?

Yes. That is exactly what the formula says.

BTW: I think the given answer is too small by a factor of 2.
 
Here is first part:

##\int_R_o^0 ##
 
mathi85 said:
Here is first part:

##\int_R_o^0 ##

Are you saying that your equation in the original post is wrong?
 
mathi85 said:
Here is first part:

##\int_R_o^0 ##
Is this what you meant to write?
$$\int_{R_0}^0$$
The LaTeX script for the above is \int_{R_0}^0. If a limit of integration is more than one character, you need to put it in braces - { }.
 

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