AWWA TANK DESIGN

[ENGJP22]

Hi guys, I hope you are okay, I have a question, Im designing a steel tank per AWWA, there is
an equation to calculate de compression stress in the shell, they are using the overturning moment
and the seismic vertical component, but I wonder why in the portion corresponding to the overturning moment they are dividing the moment by the Area, do you know why?

Thanks in advance

 

[TLHS]

The plastic section modulus of a thin walled circle is pi*r^2*thickness. So moment / area of the circle would be the compression / unit length of wall at with a plastic stress distribution.

 

[JStephen]

Actually, it's elastic stress distribution in a thin-wall cylinder, as in beams, sigma = Mc/I = MR/(pi*R^3*t) = M/(pi*R^2)/t. The relationship to area of the circle is just more or less accidental. The 1.273 is 4/pi.

 

[ENGJP22]

is it ok to take the Moment and divide it by the diameter of the thin wall cylinder and assume that a half o the cilynder is in compression and the other half in tension by dividing that couple force by one half of the perimeter of the circle?

 

[JStephen]

Just use the equation as written. The equation assumes a linear variation of bending stress across the diameter, not sure exactly what your process would give.

 

[HTURKAK]

The answer is No!!!! with your approach , you will calculate less than the stress developing and unsafe..

-This formula is used for calculation of maximum shell compression when the tank is self anchored .
- the first term in the parenthesis wt( 1+ 0.4Av ) is the weight of the tank shell and portion of the roof reacting on the shell, for 1.0 meter of perimeter

- the second term is essentially M/ W ( W is section modulus of thin cylindrical shell W= (π*D^2)/4

- the last term ( 1/1000ts) is to find stress for line load with 1.0 m length


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