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Plate Loading Test

Plate Loading Test

Plate Loading Test

(OP)
Hi,

(im not a native-english speaker,my english level is s0-s0, but i try to explain my best to you guys)

i just got a method statement of plate loading test for a foundation that i need to look at.I know that the result of plate loading test will not eventually reflecting the larger actual footing scenario as the depth of influence is not the same because of the size effect. But i still need to check the method of statement. could any one help me?

ok from s.i report, the calculated allowable bearing capacity for the foundation that will seat at 1.5m SAND layer b.g.l is 100kN/m2, so this contractor proposed that adjusted 25kN working load of 0.25m2 plate should be used for the test.

They gave us their adjustment;
100 x 1/4 = 25kN
1m2 x 1/4 = 0.25m2
so size plate is 0.5m x 0.5m = (0.25m2)

So my question are;
1.can the test be conducted using this working load (25kN) instead of 100 kN?
so cycle one: loading to 25kN then unloading to zero,
2nd cycle: loading to 25kN then unloading to zero.

i know for sand, q(footing) = q(plate) x (Width footing/width plate) ........... Braja Das formula

2. in bs1377-9, it didnt stated very clearly what settlement is ok. or do can we use 25mm (common one) as the limit of allowable settlement?

Thank you

RE: Plate Loading Test

The settlement that is allowable would be what is tolerable for the structure. Commonly it is taken as 25 mm in North America whereas in India, it is commonly as per IS taken as 40 mm. As for the load to apply to your plate, wouldn't you want to apply several loads including up past the "allowable" load that the soils report says would be the allowable bearing pressure? I would think so. I might suggest that you load to 50% of the working load (100 kPa), cycle back down to zero (or 10%) then head up to 100% of the working load, cycle back down to 50% and then at least to 125% of working load, and then back down. Might give some interesting data. Normally I've seen round plates used - 300 mm diameter rather than square plates.

RE: Plate Loading Test

(OP)
BigH thank you for the insight.

I was thinking of the same thing about the cycle because the plate laoding test cyles are more and less similar with what we do with the pile maintained load test.

The contractor proposed smaller 25kN loading as they make the plate smaller 0.25m2. they implying because the plate getting smaller so they should use fraction of the laod from the actual loading(of large 1m2 footing) as well. (i gave the calculation on my original question)

is it correct that 25kN is the 100% working load that should be imposed on the plate to get the effect of 100 kN/m2.

I have a question, how do you decide on how many kN or working load you will imposed on your plate from this 100kN/m2.

Thank you

RE: Plate Loading Test

(OP)
additional info:

proposed cycle
cycle one: loading to 25kN then unloading to zero,
2nd cycle: loading to 50kN then unloading to zero.
3rd cycle : loading to 75kN then nloading to zero

RE: Plate Loading Test

Remember that load is a force (or mass) and the pressure (100 kN/m2 or 100 kPa) is load/area. Bearing pressures are based on "pressure" not loads. L/(0.25x0.25) = 100 kN/m2 Therefore L = 0.25x0.25x100 or 0.0625x100 = 6.25 kN.

RE: Plate Loading Test

(OP)
BigH,

Are you implying that the contractor proposed calculation is incorrect? so actually to get that 100kPa or 100 kN/m2 at least need the working load (100%) is 6.25 kN for 0.25m x 0.25m plate. while if they wanna used 25kN then the area should be 0.5m x 0.5m? Or we can use any load for any size of plate and do the correlation during analysis using Braja Das formula:

for clay, q(footing) = q(plate)
for sand, q(footing) = q(plate) x (Width footing/width plate)


Oldestguy,

i quoted the paragraph from ur first link
"A test pit is dug at site up to the depth at which the foundation is proposed to be laid. The width of the pit should be at least 5 times the width of the test plate. At the centre of the pit a small square depression or hole is made whose size is equal to the size of the test plate and bottom level of which corresponds to the level of actual foundation. The depth of the hole should be such that the ratio of depth to width of the loaded area is approximately the same as the ratio of the actual depth to width of the foundation."

correct me if im wrong.
i) means that the depth of test from existing ground floor must be the depth proposed foundation level isn't it?
ii) the the ratio of size (thickness of plate/width of plate)= (thickness of actual footing/width of actual footing)?
iii) how about the loading?

however, i cant access ur 2nd link.

Thank you both of you. i really learn a lot!

RE: Plate Loading Test

All I did was detemine what 100 kPa would be for a plate of 0.25 by 0.25 m. Straight forward math. You could reference Bowles Foundation book (5th edition). it explains the plate load test (check the index page for corresponding page numbers).

RE: Plate Loading Test

(OP)
Bigh sorry the plate is 0.5m × 0.5m so area is 0.25m2. Not 0.25m x 0.25m. I didnt check back my original post for d size.

RE: Plate Loading Test

As to the loading now I don't have the wole articl copies, but here is what they say about loading.

The load is applied in regular increment of about 2KN or 1/5th of the expected ultimate bearing capacity, whichever is less. Settlement should be observed for each increment of load after an interval of 1, 4, 10, 20,
40 and 60 minutes and thereafter at hourly intervals until the rate of settlement becomes less than 002 mm per hour. The maximum load to be applied for the test should be about 15 times the expected ultimate bearing capacity of the soil.
In case of clayey soils the, time settlement curve should be plotted at each load stage and load should be increased to next stage either when the curve indicates that the settlement has exceeded 70 to 80% of the probable ultimate settlement at that stage or at the end of 24 hour period.

RE: Plate Loading Test

(OP)
Thank u oldestguy

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