What' is lateral force on tall, narrow water containment (say 5' wi x 50' hi x infinite length)? 1

[Beanbag1]

Hi everyone. I am calculating wall pressure of 3120 psf at bottom of tank, with the resultant force = 78 kips per ft of wall, but the wt of the water in the containment structure is only 16 kips per ft of wall. In designing the concrete walls, can I use lateral force of 16 kips or do I need to use 78 kips? How can the design lateral load exceed the wt of the water in the containment structure? I noticed that when the width of the containment structure exceeds half the ht of the containment structure, then the lateral force always is less than the wt of the water in the containment structure. But in this case where the width of the containment structure is less than the half the ht, then can I use a reduced lateral force (= to the wt of the water in the containment structure) in designing my walls? Thank you.

 

[1503-44]

Wall pressure varies with height of water level below its surface. Wall pressure at any depth in psf = 62.4 x D where D = depth of water below the water's surface in feet.

Weight of water in the tank is not a factor of pressure on the wall.
It is a function of volume of water in the tank.
Weight of water in the tank is 62.4 x Water height x tank area.

3120 psf means you have 50ft of water in the tank.

At a 0ft depth below the water surface, the water pressure on the wall is 0 x 62.4 = 0 psfAt a 10ft depth below the water surface, the water pressure on the wall is 10 x 62.4 = 624 psf
At a 50ft depth below the water surface, the water pressure on the wall is 50 x 62.4 = 3120 psf

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[Beanbag1]

Thank you 1503-44.
To be clear, what would the lateral load be, say for designing a retaining wall for a theoretically very deep, but say only 1-ft wide column of fluid? Would it be equal to the area under the lateral pressure diagram, or would it be limited to the lesser of that area and the weight of the fluid column derived from the volume calc? If the area under the pressure diagram exceeds the weight of the fluid column, then the lateral force would exceed the weight of the fluid column, which intuitively doesn't make sense. For example, we wish to construct a continuously long (plan view dimension) sheet pile retaining wall say 50-ft high (from low ground to high ground) but say only 1-ft away (horizontal dimension) from an existing limestone cliff (relatively straight cliff wall), with the intent being to fill the 1-ft wide x 50-ft tall space with (wet) concrete and turn it into a tall finished wall. To determine the lateral force acting on the wall and appropriately size the steel sheet pile section, should I use the area under the pressure diagram, or, should I assume it is equal to the weight of the wet concrete (as determined by the volume calc)? Civil engineers typically do not design tall narrow fluid containment structures, and I have not seen an example like this in any foundation engineering textbook.
Thank you,

 

[BridgeSmith]

The pressure on the wall is the result of the depth of the water, period. It doesn't matter how wide or long the tank/container is.

So, yes, it's a triangular distribution with zero psf at the top, 3120 psf at the bottom. It has a total force of 78 klf, but that's fairly irrelevant, since it will need to be designed structurally on a sectional basis, with each section designed for the force exerted on it. I would use 1 foot increments, and just start at the bottom designing for 3.12 klf, and work your way up.

 

[1503-44]

The key is whether you are either talking about pressure or force. They are related, but ...

Load, as in a load applied to a structure can either be a force, or a pressure.
Force can be applied as a point load, or it can be the result of applying a pressure to a certain area.

A force of 150 lbs can be applied as a point load of 150 lbs, involving no area, or it can be the result of applying a pressure of 150 lbs/ft2 on a 1 ft2 area. Both equal a force of 150 lbs, but 150lbs/ft2 pressure only equals 150 lbs, if the area = 1 ft2. If the area is 2 ft2, then that = 150 lbs/ft2 pressure x 2 ft2 = 300 lbs.

Water pressure increases linearly with depth, from 0 at the surface to 62.4 lbs/ft2 x depth ft.
So for a 10 ft depth, the pressure at the surface is 0, and the pressure at the bottom is 62.4 x 10 ft = 624 lbs/ft2.

If a box is 10ft tall, 1 ft wide and 1 ft long, the weight of the water is volume x 62.4.
Volume is 1x1x10 ft3 x 62.4 lbs/ft3 =624lbs Since only 1 ft2 is in contact with the floor, that's a weight force of 624 lbs, but it is also a weight pressure of 624 lbs/ft2, on the floor.

624 psf is also the lateral pressure in the water at a 10ft depth on the side of the box wall. It is a triangular load applied by water pressure on each vertical wall of the box. 0 at the top and 624 lbs /ft2 at the bottom. Since it is not a constant pressure, we need to find the average pressure x the wall area. P_avg = (0+624)/2 = 312 lbs/ft2. Now we can get the force as 312 lbs/ft2 x 10 ft height x 1ft length = 324 lbs/ft2 x 10ft2 = 3200 lbs. If the length of the box was 2 ft, the wall area would then be 10ft height x 2 ft length, or 20ft2 and the force on that wall would be 312 lbs/ft2 x 20 ft2 = 6240 lbs.

Note that the short width wall, the one with the 1ft width only has 3200 lbs of lateral load.
A 10ft high box, 1 ft wide and 2 ft long has a pressure on the bottom of 624 lbs/ft2, but it has 2 ft2 of area, so the force on the bottom of the box is 624 lbs/ft2 x 2 ft2 = 1248 lbs total force. And note that 10ft high x 1 ft wide x 2 ft long is a volume of 20ft3. 20ft3 x 62.4 lbs /ft3 = 1248 lbs of water in the box. That weight force also corresponds to the average pressure on the floor of (624 + 624)/2 = 624 lbs/ft2, but now x 2 ft2 of floor area = 1248 lbs of water.

Lateral pressure and floor pressure depend only on depth of the water. Forces depend on depth of the water and the area of the wall, or the area of the floor that the average pressure that is applied to each one.


--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[jayrod12]

Our hydraulics teacher always preached how a 1/8" wide column infinitely long applied the same pressure with depth as a 1 mile wide column infinitely long. Thickness of water column has no effect on the actual design pressures if there is an infinite length of column.

 

[1503-44]

Yeah. A 1000 ft high 1/8" plastic tube has the same static pressure as a 1000ft high 6ft diameter pipe. Pressure is wierd. You can have a million psi on a pin head and 0 force. 0.25 psi on a roof top and it gets blown away.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[geotechguy1]

This thread is giving me PTSD about a final exam question a 1st year engineering professor gave us about water pressure on the hoover dam assuming two of them were back to back and separated by a meter. I think the entire class got it wrong (i.e. everyone thought the pressure from a one meter wide column of water would be different than the pressure from the same height of water in a lake)

 

[1503-44]

Ah, but would the water levels between the dams ever be the same as the lake? What was the flow rate of the first dam's power turbines and the second dam's power turbines? I see some potential for a wide variation in water levels. I'm way past PTSD.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[Compositepro]

The way to keep the water level the same in both lakes and between the dams is to put holes in both dams. Thus, simple logic shows the there is no net force on the dams because there is no difference between the dams being there or not being there.