土样的干密度rd=1.54g/cm3,含水量W=19.3%,土粒比重Gs=2.71.计算土的e,n和Sr.

Question

土样的干密度rd=1.54g/cm3,含水量W=19.3%,土粒比重Gs=2.71.计算土的e,n和Sr.

Answer 1

令vs=1;V=1+e;Vv=e
ws=rw×Vs×Gs=1×1×2.71=2.71;
v=Ws/rd=2.71/1.54=1.76;
e=Vv=v-vs=1.76-1=0.76;
n=e/1+e=0.76/1+0.76=0.43;
Ww=Ws×w=2.71×19.3%=0.52;
Vw=Ww/rw=0.52/1=0.52;
Sr=Vw/Vv=0.52/0.76=0.68;