Posted by *Curbie* on February 25, 2009, 2:14 pm

Jim,

Sounds like you're where I want to be, metaphorically, spend a little

time on Euler, spend a little time on logs. Bliss.

I found the kind of site I like (almost) yesterday, its has an example

with not only the explanations to the math, but answers to check the

equations.

https://ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=me&chap_sec .1&page=case_sol

I felt the whole thing was making perfect mathematical sense until the

last equation (the "almost" above), I was getting 7,955 for Pcr and

the example said I should get 1,593. I tore my spreadsheet expression

apart Pcr =PI() ^ 2 * E * I / L ^ 2 but couldn't find my problem.

Cylinder Outside Diameter (do) 1.5 Inches

Cylinder Inside Diameter (di) 1.4375 Inches

Cylinder Wall Thickness (Tc) 0.0625 Inches

Cylinder Radius (Radius) 0.75 Inches

Cylinder Outside Radius (ro) 0.75 Inches

Cylinder Inside Radius (ri) 0.6875 Inches

Moment of Inertia (I) 0.073043851

Length of Mast (Lm) 4 Feet

Length of Member in Feet (Lf) 4.272 Feet

Length of Member in Inches (L) 51.26402247 Inches

Young's modulus (E) 29,000,000 PSI

Factor of Safety (Fs) 2.00

Critical Buckling Load (Pcr) 7955 Lbs.

Will hit it again today.

Thanks for help and advise on "statics", I read the site and made

notes, haven't quite figured out where it fit in, but I'm sure to run

into "it's place" in the scheme of things.

Curbie

On Wed, 25 Feb 2009 04:41:48 -0800 (PST), Jim Wilkins

*>>*

*>> Thank you so much for your time and patients with me.*

*>> On to Critical Buckling Load of a columns.*

*>>*

*>> Curbie*

*>The math there is strange and non-intuitive (except to Euler) but it*

*>really works. I made some 2x4x12' shear legs to lift logs onto a*

*>sawmill and overloaded one side right to the critical point. As*

*>predicted the leg bent and was stable at any degree of curvature. I*

*>could bend or straighten the center of it with one finger even though*

*>it was supporting a 350 Lb log.*

*>Jim Wilkins*

*>keeping l/d<50*

Posted by *Jim Wilkins* on February 25, 2009, 6:20 pm

*> ...but couldn't find my problem.*

*> Cylinder Outside Diameter (do) 1.5 Inches *

*> Cylinder Inside Diameter (di) 1.4375 Inches *

*> Cylinder Wall Thickness (Tc) 0.0625 Inches *

Posted by *Curbie* on February 25, 2009, 9:56 pm

Jim,

Well, Cylinder Outside Diameter (do) & Cylinder Wall Thickness (Tc)

are just input to feed radius calculations, which in turn feed the

Moment of Inertia (I) calculation =PI() / 4 * ((ro ^ 4) - (ri ^ 4))

Since the results from the Moment of Inertia (I) calculation matches

the results given in the example I presume that's not my problem. The

only calculation which is not producing the same results as the

examples is for Critical Buckling Load Pcr=PI() ^ 2 * E * I / L ^ 2

I don't expect you find my problem, thanks for trying though, I

thought something might jump out at you (mathematically or

syntactically) from the Critical Buckling Load (Pcr) expression.

I'll just keep hammering away at it, it's what I do, learn the

properties of thing long enough to define them in computer equation

then move on and let the programs do their work.

Thank for your help.

Curbie

On Wed, 25 Feb 2009 10:20:40 -0800 (PST), Jim Wilkins

*>> ...but couldn't find my problem.*

*>>*

*>> Cylinder Outside Diameter (do) 1.5 Inches *

*>> Cylinder Inside Diameter (di) 1.4375 Inches *

*>> Cylinder Wall Thickness (Tc) 0.0625 Inches *

Posted by *Jim Wilkins* on February 25, 2009, 10:41 pm

*> Jim,*

*> Well, Cylinder Outside Diameter (do) & Cylinder Wall Thickness (Tc)*

*> are just input to feed radius calculations, which in turn feed the*

*> Moment of Inertia (I) calculation =PI() / 4 * ((ro ^ 4) - (ri ^ 4))*

If the wall thickness is 0.062, the radii are 0.750 and 0.6875. 1/2 of

1.437 is 0.71875.

Posted by *daestrom* on February 25, 2009, 9:45 pm

*> Jim,*

*> Sounds like you're where I want to be, metaphorically, spend a little*

*> time on Euler, spend a little time on logs. Bliss.*

*> I found the kind of site I like (almost) yesterday, its has an example*

*> with not only the explanations to the math, but answers to check the*

*> equations.*

*>*

*
https://ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=me&chap_sec .1&page=case_sol *
*> I felt the whole thing was making perfect mathematical sense until the*

*> last equation (the "almost" above), I was getting 7,955 for Pcr and*

*> the example said I should get 1,593. I tore my spreadsheet expression*

*> apart Pcr =PI() ^ 2 * E * I / L ^ 2 but couldn't find my problem.*

*> Cylinder Outside Diameter (do) 1.5 Inches*

*> Cylinder Inside Diameter (di) 1.4375 Inches*

*> Cylinder Wall Thickness (Tc) 0.0625 Inches*

Wouldn't that be .03125?? (1.5-1.4375)/2 There are two walls

I run into this a lot with heat exchanger fluid calcs. I get the ID and the

wall thickness and have to calculate the OD to figure out the volume of

water displaced in the shell side.

daestrom

>>>> Thank you so much for your time and patients with me.>> On to Critical Buckling Load of a columns.>>>> Curbie>The math there is strange and non-intuitive (except to Euler) but it>really works. I made some 2x4x12' shear legs to lift logs onto a>sawmill and overloaded one side right to the critical point. As>predicted the leg bent and was stable at any degree of curvature. I>could bend or straighten the center of it with one finger even though>it was supporting a 350 Lb log.>Jim Wilkins>keeping l/d<50