Step-by-step solved example
Cα=0.02, H=4 m, ep=1.0, t1=1 yr, t2=10 yr. Ss?
1. log
log10(10/1)=1.
2. Ss
0.02×4/(2)×1=0.04 m=40 mm.
Answer: Ss = 40 mm
Cα=0.02, H=4 m, ep=1.0, t1=1 yr, t2=10 yr. Ss?
1. log
log10(10/1)=1.
2. Ss
0.02×4/(2)×1=0.04 m=40 mm.
Answer: Ss = 40 mm
1.t2/t1 = 100 → log=
2.Cα/Cc is often about
3.If you start the log at tp too early you
4.Peat settlements are dominated by
5.H=8 m, Cα=0.01, 1+ep=2, one log cycle: Ss=
6.Preloading longer than tp can
7.e vs log t after tp is
8.Units of Cα are
9.Primary Sc uses Cc or Cr and
10.A 10-fold time increase at constant Cα