6
$\begingroup$

I have a structural analysis question. I want to know if the hydrostatic pressure in a water tank is a live load or a dead load and why?

$\endgroup$
6
$\begingroup$

Eurocode looks into this with very much detail, see the "Loading Standards" list, attached. You can see that there is a document dedicated to actions in silos and tanks. This deals with hydrostatic forces.

But as a rule of thumb you have two types of loads:

Self Weight (Dead Load): Loads that are fixed and do not change during the lifespan of the structure. The load factors only accommodate variances in quality of workmanship and materials.

Imposed Loads (Live Loads): loads which may vary during the lifespan of the structure. This means anything which may change. This would include your hydrostatic pressure, since the water level may change.

Eurocode "Loading Standards"

| improve this answer | |
$\endgroup$
4
$\begingroup$

This will likely depend on your jurisdiction, but here is my interpretation based on US codes.

Dead loads are typically considered well-known and well-defined. For example, if you have a W8x13 beam, you know that that piece of steel is 13 pounds per linear foot. This is why the LRFD load factor for dead load is lower than other loads - there is less inherent variability in the loading. This is unlike a live load, where the occupancy of a building isn't well-known and is really just an educated guess based on the space's function.

Hydrostatic loads are very easy to calculate and, more importantly, very easy to calculate to a high degree of accuracy. For anything that acts like a fluid, the pressure at the bottom is simply the density of the fluid times the height of the fluid.

ASCE 7-05, Chapter 3 is titled Dead Loads, Soil Loads, and Hydrostatic Loads. Though it doesn't explicitly include hydrostatic pressures on tanks, it does discuss lateral pressures due to soil and groundwater.

In my opinion, the hydrostatic pressure on a tank wall is therefore a Dead Load. However, check with the authority having jurisdiction or your code of record to be certain.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ You forgot the $g$ in your formula for pressure ;) $\endgroup$ – Chris Mueller Oct 1 '15 at 21:26
3
$\begingroup$

@grfrazee says it's a dead load, @NamSandStorm and @hazzey say it's a live load, and now I'm going to stop half way between them.

It depends on what your tank is doing:

  • Is it a frequently flushed tank, which empties and fills frequently? Then that's a live load.
  • Is it a storage tank, whose fluid levels rarely ever change? Then that's a dead load.
  • Is it something in between, where, for example, there's a minimal level that will never (or rarely) be crossed, but the level itself may vary above that level? Then consider the minimal level a dead load and the variance a live load. The inverse case is also possible: if you have the storage tank above but you want to take into consideration the possible effects of the rare moments where it's flushed? Then consider its standard level as a dead load and apply a negative load as a live load.
| improve this answer | |
$\endgroup$
2
$\begingroup$

It may depend on your location, but in the USA most codes eventually reference back to ASCE 07 "Minimum design loads for buildings and other structures." In there, one load type is:

H = load due to lateral earth pressure, ground water pressure, or pressure of bulk material.

The load is then included the load combinations in the same place as live load.

It could be argued that in your case, the load is easy to calculate accurately. This could make it seem like it should be factored as a dead load.

| improve this answer | |
$\endgroup$
0
$\begingroup$

From a practical perspective, when in doubt (and if it does not lead to an overly conservative design) look at both cases, and after analysis choose the worst case scenario.

Whereas the codes are the guide (and there appears to be some contradiction between US and European codes from the information provided), in any case the final result must be a structure that can serve it's design purpose.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.