Step-by-step solved example
q=200 kPa, B=2 m, ν=0.5, Es=20 MPa, Is Id=1.0. Si?
1. (1−ν²)/Es
(1−0.25)/20000 kPa=0.75/20000=3.75e-5 /kPa.
2. Si
200×2×3.75e-5=0.015 m=15 mm.
Answer: Si = 15 mm
q=200 kPa, B=2 m, ν=0.5, Es=20 MPa, Is Id=1.0. Si?
1. (1−ν²)/Es
(1−0.25)/20000 kPa=0.75/20000=3.75e-5 /kPa.
2. Si
200×2×3.75e-5=0.015 m=15 mm.
Answer: Si = 15 mm
1.Consolidation settlement is additional and
2.ν=0.5 for
3.If Es doubles, Si
4.B is
5.Secondary compression Cα comes
6.q=100 kPa, B=3 m, Es=15 MPa, ν=0, I=1: Si=
7.A deeper embedment Id<1 typically
8.Sand settlements are mostly
9.Influence Is is larger for
10.Net q is