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Pier engineering

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#226 macdonjh

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Posted 30 June 2020 - 08:35 AM

 

After that - consultations with architecture and constructor of the building. May be I`ll change type of the building footer (from slab to strip)... 

 

But really complicated consultation will be with my wife smile.gif 1 square meter pier is a serious point. Probably she will have much stronger arguments about pier then climate, type of soil and resistance of materials all together smile.gif Then I`ll have to accept solution of domes developer and make a "floating" footer of pier on vibration damper, placed on the mansard floor. We will see. 

 

Thank you a lot again! 

Agreed.  Engineering problems are usually easy to solve in comparison to making the solution aesthetically pleasing.



#227 macdonjh

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Posted 30 June 2020 - 09:16 AM

Using the load conditions at the beginning of this thread, 12" steel pipe deflects about 6 arc-sec.  Too much.  So, you either go bigger or just don't bump it in the first place.  

This has been gnawing at me.  So I redid my numbers.  Hopefully this won't be too boring for the denizens of this thread.  I thought about sending speedster a PM, but then thought some might find this interesting.

 

For a cantilevered beam (or column) with a force applied to the end:

d = PL3 / 3EI  (sorry, no Greek font, that "d" should be delta)

 

With a bit of rearranging, since speedster calculated 6" arc (or 0.007", 240" * TAN(6" arc) ):

P = 3dEI / L3

P = 3(0.007 in)(30x106 lb/in2)(279 in4) / (240 in)3 (note, I= 279 is for 12" x 0.375" wall pipe)

P = 12.7 lb (the load at the top of the 20'-0" tall pier pulling sideways).

 

Since speedster's criterion is 0.5" arc (0.0006 inches) deflection, I reorganized the deflection equation to find I:

I = PL3 / 3dE

I = (12.7 lb)(240 in)3 / 3(0.0006 in)(30x106 lb/in2)

I = 3251 in4

 

The area moment of inertia for a "hollow" circle is:

I = PI(R4 - r4) / 4

 

Since, at least in the US, "standard weight" pipe is available, I'll assume the wall thickness will be 0.375": 

I = PI(R4 - (R - 0.375 in)4) / 4

 

Therefore (I wish I hadn't thrown my scratch paper away, I don't remember how I calculated this without expanding the polynomial above): 

3251 < PI[ (15 in)4 - (14.625 in)4 ] / 4

3251 < 3829; therefore, 30" x 0.375" wall thickness pipe will be suitable.  For reference, the moment of inertia of 24" x 0.375" pipe is 1942 in4, so it isn't stiff enough to meet the 0.5" arc deflection criterion.

 

Those were the calculations behind my recommendation of 30" x 0.375" pipe for a 20'-0" tall pier with a maximum deflection of 0.5" arc under a 12.7 lb side load.  Or, in metric: 762mm x 9.5mm pipe for a 6.1m tall pier with a maximum deflection of 0.5" arc under a 5.8kg side load.



#228 speedster

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Posted 30 June 2020 - 03:51 PM

macdon,  your calcs are spot on.  The difference is that you are figuring 12.7# force and I'm figuring 5#.  The 5# is an arbitrary number to sort of simulate a bump of the pier and to have some defined load for calculating deflection difference in various pipe and concrete shapes.




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