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
(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
RE: Plate Loading Test
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
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
RE: Plate Loading Test
This link has some scoop about plate loading tests.
http://www.theconstructioncivil.org/plate-load-tes...
I copied a section of it. Spelling not good.
RE: Plate Loading Test
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
RE: Plate Loading Test
RE: Plate Loading Test
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