Tall Tower Bending

See thread immediately below.

Somebody must have the bizarre opinion that tower deflection formulas have nothing to do with several of the falling towers in the video. Oh, well.

If the conical section is short, you should be able to figure the deflection of the bottom part and top part using your formula and add them together to get pretty close to the total deflection.

do you have a reference for what you'r sayin!!

I can't point you to a reference, just the way I see it.

But the top of lower part (less the conical section) is the base of the upper part. So deflection of the top of the lower section moves the base of the upper section.

If you can ignore deflection in the conical section, the deflection of the top should be the sum of the deflection of the lower section calculated with its parameters and the deflection of the upper section (by itself) calculated with its parameters.

I think you're going to need more formulas...

Formulae

may you suggest some;and thanx

may you suggest some references or formulas ;and thanx

E=mc²

What - no picture of a Saturn 5 rocket? There has to be a picture of a Saturn 5 rocket! <:-(

Is this for a blow up tower?

no dude!

Does this help?

got no idea ! i mean this conical part is a real problem , by the way in this handbook the writer mentioned that it's treated as a cantilever beam!

Easy!

thank you so mutch but a more practical formula would better useful

Would you like ketchup with that?

It is a cantilever beam. The beam is anchored at the ground and the wind drag creates a distributed force. Calculate the force profile by diameter x drag coefficient of a cylinder x wind speed squared. Treat the conical section as varying diameter cylinder.

You can calculate the deflection piecemeal and add the results using the principle of superposition.

https://www.google.com/search?q=how+to+solve+cantilever+beam+deflection+problem&espv=2&biw=1421&bih=748&tbm=isch&tbo=u&source=univ&sa=X&ved=0ahUKEwjAh9rzvp_MAhWDg4MKHZRaAooQsAQIPw&dpr=0.9#imgrc=foAJpCRvuZzr1M%3A

Was that a coincidence that the angle from vertical of the pasted formula is exactly the same as the Tower of Pisa (as of 2016). Wow!

It means don't hang out under a wind turbine when only the junior engineer participated in calculating stress and strain, and material strength, especially if the high winds are particularly gusty that day. Stay away, and play another day!

thanx blooper i ment that !

If you know enough about beam theory and calculus you can derive a formula for deflection of a cone subject to a constant pressure from the wind. If the complete tower were a cone (or a truncated cone) it might be worth going to the trouble, but in the case here I'm sure Rixter in #2 is right in saying the conical section can be ignored. It's not an exact science!

Ref your #6, what's wrong with treating it as a cantilever, what else would it be? I tried some typical figures comparing your formula with a cantilever in Roark and got the same result.

Another thing - if you work out wind pressure from 0.5*ρ*V², it's normal to use a factor around 0.5 (if memory serves) in calculating the force (pressure x projected area) but your formula doesn't include this.

Don't forget the wind force causes a tensile stress on the side facing the wind, which should be added to the stress due to internal pressure for comparison with allowable stress. May not be significant but worth checking.

Insert "then a miracle happens" between the "H" and "(". (;>)

I fail to see why parentheses are even needed, when 12PDH⁴/8EI will do quite well indeed.

No, the formula is correct as it stands, but could be simplified to 12³PDH⁴/8EI.

Whoever wrote it has confused things by putting some lengths in feet and others in inches, but if you sort out the factors of 12 that's how it comes out. Also I checked a typical case against Roark and got same answer.

My bad, I forgot to cube the 12.