Question

A soil has a bulk unit density of 1.91 Mg/m3 and a water content of 9.5%. The value of Gs = 2.7. Calculate the void ratio and degree of saturation of the soil. What would be the values of density and water content if the soil were fully saturated at the same void ratio.

Answer 1

2.0978 g/cm^3

Step-by-step explanation:

Given data:

`p=1.91 g/cm^(3)`

w = 9.5 %

`G_(s)`= 2.70

e =?

`S_(r)` =?

solution:

`w=(mass of water)/(mass of solid) =(m_(w) )/(m_(s) )`

`e=(volume of solid)/(volume of solid) =(V_(s) )/(V_(s) )`

assume total volume
`V_(total)`=1
`cm^(3)`=
`V_(w) +V_(s) +V_(air)`

`p=(m_(total) )/(V_(total) )`

`m_(total) =1.91 g\\m_(total) =m_(w) +m_(s)\\w=(m_(w) )/(m_(s))\\ 0.095.m_(s)=m_(w)\\1.91=m_(s)+0.095.m_(s)\\m_(s)=1.744 g\\m_(w)=0.166 g\\`

`p_(w) =(m_(w) )/(V_(w) ) ==> V\\G_(S)=(m_(s) )/(V_(s).p_(w))\\2.70=(1.744)/(V_(s).1 ) \\\V_(s)=0.646 cm^(3) \\V_(v)=V_(T)-V_(S)=0.354 cm^(3) \\e=(V_(v))/(V_(S)) =(0.354)/(0.646)=0.55=55 percent\\`

`S_(r)=(w*Gs)/(e) =0.4666=46.6 percent\\if S_(r)=1 \\ w=??\\S_(r)=(w*Gs)/(e) \\\\w=0.204=20.4 percent\\`

now find
`p`

`p=(m_(t)(=m_(w) +m_(s) ))/(V_(t) )`

`p`=2.0978 g/cm^3