Step-by-step solved example
emax = 0.85, emin = 0.35, in-place e = 0.50. Find Dr.
1. Numerator and span
emax − e = 0.35. emax − emin = 0.50.
2. Relative density
Dr = 0.35/0.50 = 0.70 = 70%.
Answer: 0.70 (70%)
emax = 0.85, emin = 0.35, in-place e = 0.50. Find Dr.
1. Numerator and span
emax − e = 0.35. emax − emin = 0.50.
2. Relative density
Dr = 0.35/0.50 = 0.70 = 70%.
Answer: 0.70 (70%)
1.e = emax means the sand is
2.e = emin corresponds to
3.emax = 0.90, emin = 0.40, e = 0.65. Dr =
4.Relative density is NOT used for
5.A dense sand typically has Dr about
6.As Dr increases, expected φ of a sand
7.γ_dmax corresponds to
8.emax = 0.80, emin = 0.40, Dr = 0.25. In-place e =
9.The unit-weight form of Dr includes an extra factor γ_dmax/γ_d because
10.SPT N-values in sand generally increase as Dr