written 6.6 years ago by
teamques10
★ 69k
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modified 6.6 years ago
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Given:-
ww=w×w5
w=5=0.05
γ=18.44kn/m2
G=2.67
Ww1=0.05ws
Ww2=0.15Ws
Quantity of water to be added =0.15ws−0.05ws
γd=γ1+w=18.441+0.05=17.56kn/m3
Ws=γd×γ
For 1m3 volume.
Ws=17.56×1=17.56kN
Ww=0.1×17.56=17.56kN
Vw=Wwγw=1.7569.81=0.179m3 i.e 179 litres.
e=Gγmwγd−1=2.67∗9.8117.56−1
e=0.492
weight of sand filling hole =1050-445=605g
unit weight of sand=15501000=1.55g/cc
Volume of the hole =6051.55=390.32cc
Insitu unit density ∫=761.25390.32=1.95 g/cc
Insitu unit weight.γ=∫g=1.95∗9.81= 19.13 kN\m^{3}
γ=19.13kN/m3
The insitu unit weight of soil is 19.13 kN/m3