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Plate Load Test question

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04kebuts

Geotechnical
Joined
Dec 3, 2007
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Location
GB
I have been given the task by my technical director to confirm that the spreadsheet we have been using to calculate the loading that we use for the plate load test is correct.
We have both a 455mm plate and a 600mm plate.
The dial gauge on the jack is in bar.

I have tried - and failed - to achieve the same "answer" as the spreadsheet. I think my method is correct, though it must not be.
So my question is, how do you calculate the loading on a plate if a client requests we load to X kPa.

Thanks!

 
You need to find out what the pressure reading for the jack means in terms of total load the jack is applying....so that you can then divide this load by the plate area and convert into whatever units you want. The same pressure reading of hydraulic fluid in different models of jacks will result in different total loads.

There are typically manufacturer conversion tables that relate the pressure reading for a particular jack model to a load. But you really should also use a load cell in the test setup to measure the actual load......jack pressure readings and conversions can often be inaccurate.

So in summary, spreadsheet formulas relating jack pressure and plate pressure would depend on the specifics of the jack being used an how it's hydraulic pressure reading translates into load output.
 
So does this mean that the reading off the dial gauge on the jack is not the actual load that is being applied?

So my first port of call will be to get the jack gauge calibrated. We have recently purchased a new jack (still using the old gauge) so I am glad I asked this question before it was brought out to site.

Thanks for your reply!
 
Bar is a unit of pressure (i.e., about 1 tsf). If the hydraulic pressure is 1 tsf there is no reference to what the load (i.e., tons) will be on the piston that is applying pressure to your plate. Your plate has an area of 3 sf. If a hydraulic pressure of 1 tsf converts to a piston pressure of 300 pounds then your incremental stress ON THE GROUND would be 100 psf.

You need to calibrate your jack!

f-d

¡papá gordo ain’t no madre flaca!
 
Proper conversion from hydraulic pressure to jack load is important, and as noted may be done by direct computation, calibration curve or load cell. How you do this is an internal descion. However, as long as you use the same method as the spread sheet, the hand computation and the spreadsheet should agree. If they do not, copy the spreadsheet and work through each step and find out where you and the spreadsheet diverge. It is a relatively straight forward process except for computing the jack force. Someone may have pasted some calibration values into a formula that may no longer be accurate. Or a formula may be wrong. Rember the you may be correct and the spreadsheet may be wrong (Why would some one want to check it if they didn't see a possibility for error?)
 
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