Hydrostatic Pressure on a curved surface 1

[dittowizard]

I need a formula to figure the pressure present on a curved form (tank)

the form is a circle 10ft in dia. and 40ft tall. the form is standing vertical on solid concrete. the bottom is flat. the fluid placed in the form has a density of 2.4. there are no additional forces acting on the fluid (only atmospheric, the form has an open top)

I need know the pressure on the form at any point.

Dittowizard@infoave.net

 

[LPPE]

Density times depth at point of interest.

 

[dittowizard]

Is this formula correct? Density*height=psf or 2.4*40=96psf or 2.4*480=1152psi

the weight of the (at max cap)column placed on the base is 471,238 lbs (150lbs/cuft)

the formula given looks to be low. Please verify.

Dittowizard

 

[abhio]

Pressure in t/m2 = depth of the point from the water surface in m

 

[Briansch]

dittowizard, the density of 2.4 you mentioned for this fluid doesn't seem right. You did not give any units. Are you sure this is not specific gravity in which case the density would be 2.4 * 62.4 pcf (water) = 150 pcf. The pressure then at any depth is 150 * h (measured from the surface). Keep in mind though that the pressure varies linearly with depth along the verical face. The resultant force is at the centroid of the pressure prism (not the centroid of the area - fluid on one side) according to my fluids book. If we roll out your form so it's pi * d long or 31.4' and contain this fluid against it, the resultant force would be 150 * 40 * 40/2 * 31.4 = 3.768 x 10^6 lbs acting at 2/3 * 40 = 26.8' below the surface.

Maybe someone could do some QA and check this.

Briansch

 

[krd]

Briansch - no you're wrong. The question looks like pressure in a pipe, not force on a flat plate. The pressure in a pipe 40 ft high full of fluid at 150 pcf is 6,000 psf at the bottom of the pipe and varies linearly as you go up to zero at the top.
Dottowizard - if you're placing a concrete column 40 ft high you can reduce the maximum internal pressure according to ACI formula which accounts for the concrete at the bottom starting to set before you get the full 40 ft placed. Depends on pour rate and temperature.

 

[Briansch]

Thanks krd. Maybe making it more complicated than it is.

Dittowizard, what are you building?

 

[Qshake]

By the way Dittowizard, it is doubtfull if you can get forms to allow a 40' concrete placement. In many cases you will have to allow for the lower lifts to set up somewhat prior to continued placement. Most forms (ready-made steel forms) are rated based on the framing (kicker) system that is employed. Other arrangements should be checked by a registered engineer.

 

[dittowizard]

Thank you for the reply. This information will help. We do mostly concrete work, however, this is going to be a concrete tank. Someone asked the original question of me and I was not sure of the answer due to the curved surface.

Brainsch was correct when he stated,"Are you sure this is not specific gravity in which case the density would be 2.4 * 62.4 pcf (water) = 150 pcf."

krd was correct when he stated,"The question looks like pressure in a pipe, not force on a flat plate."

It looks like Denisty(pcf)*Height(f)=Pressure(psf) is the formula.(?)
Example: a fluid colum with a density of 90(pcf) and 4(f) tall will have an outward pressure on a curved surface at the lowest point of 360(psf)

 

[butelja]

I think that tall (i.e. - grain elevators, 100+ ft tall) cylindrical concrete structures are often formed by slip forming. In this method, the forms are slowly (a few feet per day) jacked up as the additional concrete and steel is added. The idea is that the forms are only on the order of ~10 ft tall, and the concrete solidifies as you go. Think of it as a continuous, rather than a batch manufacturing process.

Having a mechanical background, this is just about all the level of detail I know, but I've watched a couple of these projects in construction.

 

[af246]

The hydrostatic pressure is equal to the density of the fluid times the height of the column, The pressure at any point does not depend on the volume but on the height and on the density. Example: height of vessel equals 30 ft. and the fluid is water (i.e. density = 62.4 lb/cu.ft.) thus pressure at bottom of vessel is 62,4 x 30 = 18,720 lb/sq.ft

 

[austim]

Hi, all.

This thread appears to take the biscuit for the number of confused responses, but maybe that is because Dittowizard initially referred to concrete as a fluid.

(a) You can forget all reference to "density times height" for depths below the point where the fluid concrete has commenced to gain shear strength. Below that point, (which is dependent on the rate of concreting and the concrete mix involved) lateral pressure is less than vertical, and the concrete no longer behaves as a pure fluid.

(b) The curvature of the forms has no practical effect on the lateral pressure on them, but the distance between inner and outer forms may have an effect (due to "arching&quot.

(c) Just design the forms as if they were flat, and use the established equations for pressure on formwork, taking into account the type of cement, additives used, rate of pour (in ft/hour [or m/hr] rise in the forms).

and dittowizard - zero marks out of ten for mixing your units and physical concepts. 2.4 was never a density, it was the specific gravity. The density would be about 2.4*62.5 = 150 pcf, giving you 150*40 =6000 psf [as krd noted]

 

[brad]

austim,
Amen,brother
Brad

 

[Dinosaur]

DittoWizard(?!)

The pressure is unit weight times depth measured from the free surface.

The company I work for recently designed a bridge using single concrete column piers (we refer to these as hammerhead piers). The pier columns were seven feet in diameter for the shorter ones and ten feet in diameter for the taller ones. The forms were made of steel and included a number of stiffeners to help them hold their shape when they were being handled. They came in quarters and could be bolted together to permit pour depths of around forty feet. The reason the pours were limited to forty feet is because that is the stock length of a reinforcing bar used for the primary reinforcement. The Contractor set a cage forty feet tall (less the lap requirement) and the form and poured the next lift. I don't know how he sealed the forms at the base, but theoretically you could just clamp the forms tight to the previous hardened concrete.

What is your application?

 

[sven]

Ditto,
Further to Austim's post, the main factor governing lateral concrete pressure is your proposed rate of pouring and slump of the concrete. Most commercial formwork systems have a limit of for example 2-3m vertical per hour, for wide elements with limited concrete arching to the formwork.

If you are considering pressure on the formwork, I would suggest using available design tables for a square column of the same cross-sectional area. If you are designing bolted steel formwork, then the "pressures" from those tables can be applied to provide a hoop force to design your bolts.

 

[dittowizard]

I believe at some point this questions scope got away from the original. I did say the density (not specific gravity) was 2.4. I never said this was a concrete form, only that this would be a concrete tank. The design spec. for the tank is for a material with a specific gravity of 2.4.

I needed a formula to find the blowout pressure on the side wall of the tank.

Example: If the walls of the tank were steel and they had a bolted seam. How much pressure would there be trying to pull the seam open or how much (tensile) pressure would placed on the bolts?

Example and question: The tank is 10' Dia or has a Circumference of 31.4' or (considering the lower 1' ring of wall) 31.4 sqft of surface. If the pressure on the walls is [40'(tall)x150pcf(density)=6000psf] 6000(psf)x31.4(sf)=188,400pound of pressure on the seam in the lower 1' of the ring.

??Is this correct??

I hope this will clarify my question.

 

[alan54]

Not correct.

This IS "pressure in a pipe". Think of it as a vertical cylinder with a east half and a west half. Pressure acts in all directions, so the force pushing on the east half is:

Pressure*Diameter*Height, which is same as force pushing on west half (I hope, or tank would be moving!).

The tank wall area holding the two halves together is:

2*Height*Wall Thickness, so the wall stress is:

S=Press*Dia*Height/(2*Height*Thk) = PD/2t (this cannot be applied to your ENTIRE tank because you have a varying hydrostatic head (P); you could apply it if pressure was due to a gas; read on...).

(Note circumference, or area of the wall, is not a factor - i.e., on either of our halves, the components of the forces acting on our curved wall that are not in a direction of directly east or west, say the north force component, are canceled out by the south force components.)

For you, P varies with depth in tank so the force on the bottom 1 foot of the cylinder where the pressure is ~6000 psf is 6000psf*10'*1'=60,000 lbs, where as on the top 1 foot of the cylinder where the pressure is 0 psf at top and 150 psf at bottom (average =75) is 75psf*10'*1'=750 lbs. As you can see, the required wall thickness will vary with depth; this is why large API tanks are often built with varying wall thicknesses.

For resisting "blowout", you need to substitute in an acceptable "allowable stress" when calculating the required wall thickness. If bolts are involved, they also need to be sized to carry the load.

Remember, thick and strong at the bottom, no problem at the top. Hope this helps.