ndjiji
Geotechnical
- Apr 9, 2014
- 25
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