Answer :
Answer:
Gs = 2.647
e = 0.7986
Explanation:
We know that moist unit weight of soil is given as
[tex]\gamma_m \ or\ bulk\ density = \frac{(Gs+Se)\times \gamma_w}{(1+e)}[/tex]
where, [tex]\gamma_m[/tex] = moist unit weight of the soil
Gs = specific gravity of the soil
S = degree of saturation
e = void ratio
[tex]\gamma_w[/tex] = unit weight of water = 9.81 kN/m3
From data given we know that:
At 50% saturation,[tex] \gamma_m = 16.62 kN/m3[/tex]
puttng all value to get Gs value;
[tex]16.62= \frac{(Gs+0.5*e)\timees 9.81}{(1+e)}[/tex]
Gs - 1.194*e = 1.694 .........(1)
for saturaion 75%, unit weight = 17.71 KN/m3
[tex]17.71 = \frac{(Gs+0.75*e)\times 9.81}{(1+e)}[/tex]
Gs - 1.055*e = 1.805 .........(2)
solving both equations (1) and (2), we obtained;
Gs = 2.647
e = 0.7986
The void ratio and specific gravity of this question are;
Void ratio; e = 0.7999
Specific Gravity; G_s = 2.65
We are given;
Moist unit weight 1; γ₁ = 16.62 kN/m³
degree of saturation 1; S₁ = 50% = 0.5
Moist unit weight 2; γ₂ = 17.71 kN/m³
degree of saturation 1; S₂ = 75%
The formula for moist unit weight is;
γ = (G_s + Se)γ_w/(1 + e)
where;
G_s is specific gravity
e is void ratio
S is degree of saturation
γ_w is unit weight of water = 9.81 kN/m³
Thus, for γ₁ = 16.62 kN/m³ and S₁ = 0.5, we have;
16.62 = 9.81(G_s + 0.5e)/(1 + e)
⇒ 16.62(1 + e) = 9.81G_s + 4.905e
⇒ 16.62 + 16.62e = 9.81G_s + 4.905e
⇒ 11.715e + 16.62 = 9.81G_s
divide through by 9.81 to get;
1.1942e + 1.6942 = G_s ---eq 1
for γ₂ = 17.71 kN/m³ and S₂ = 0.75, we have;
17.71 = 9.81(G_s + 0.75e)/(1 + e)
⇒ 17.71(1 + e) = 9.81G_s + 7.3575e
⇒ 17.71 + 17.71e = 9.81G_s + 7.3575e
⇒ 10.3525e + 17.71 = 9.81G_s
divide through by 9.81 to get;
1.0553e + 1.8053 = G_s ---eq 2
Putting 1.0553e + 1.8053 for G_s in eq 1;
1.1942e + 1.6942 = 1.0553e + 1.8053
1.1942e - 1.0553e = 1.8053 - 1.6942
0.1389e = 0.1111
e = 0.1111/0.1389
e = 0.7999
Thus;
G_s = 1.0553(0.7999) + 1.8053
G_s = 2.65
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