Question

A natural soil deposit has bulk unit weight of
18.44 kN/m and water content of 5%. Calculate
the amount of water required to be added to
1 mⓇ of soil to raise the water content to 15%.
Assume void ratio remains constant. If specific
gravity of solids is 2.67, then find the degree of
saturation of soil after addition of water.
[Take Yw = 9.81 kN/m]​

Answer 1

ANSWER:

ww=w×w5w×w5

w=5w=5=0.05

γ=18.44kn/m2γ=18.44kn/m2

G=2.67

Ww1=0.05wsWw1=0.05ws

Ww2=0.15WsWw2=0.15Ws

Quantity of water to be added =0.15ws−0.05ws0.15ws−0.05ws

γdγd=γ1+wγ1+w=18.441+0.05=17.56kn/m318.441+0.05=17.56kn/m3

Ws=γd×γWs=γd×γ

For 1m31m3 volume.

Ws=17.56×1=17.56kNWs=17.56×1=17.56kN

Ww=0.1×17.56=17.56kNWw=0.1×17.56=17.56kN

VwVw=Wwγw=1.7569.81=0.179m3Wwγw=1.7569.81=0.179m3 i.e 179 litres.

e=Gγmwγd−1=2.67∗9.8117.56−1Gγmwγd−1=2.67∗9.8117.56−1

e=0.492

weight of sand filling hole =1050-445=605g

unit weight of sand=1550100015501000=1.55g/cc

Volume of the hole =6051.556051.55=390.32cc

Insitu unit density ∫=761.25390.32∫=761.25390.32=1.95 g/cc

Insitu unit weight.γ=∫g=1.95∗9.81γ=∫g=1.95∗9.81= 19.13 kN\m^{3}

γ=19.13kN/m3γ=19.13kN/m3

The insitu unit weight of soil is 19.13 kN/m3kN/m3