Soil Mechanics Questions and Answers - Functional Relationships Part-1

1. The relationship between e, S and na is ______
a) \(n_a=e*\frac{1+S}{1+e}\)
b) \(n_a=e*\frac{1-S}{1-e}\)
c) \(n_a=\frac{1-S}{1+e}\)
d) \(n_a=e*\frac{1-S}{1+e}\)

  Discussion

Answer: d
Explanation: The percentage air voids is given by,
\(n_a=\frac{V_a}{V}\)
Va = Vv – Vw = e – ew
And V = Vs+Vv = 1+e
na=\(\frac{e-e_w}{1+e}\)
Since ew = eS
\(n_a=e*\frac{1-S}{1+e}\).

2. The relationship between na, ac, and n is ____
a) na=n*ac
b) na=n/ac
c) na=n+ac
d) na=n-ac

  Discussion

Answer: a
Explanation: The percentage air voids is given by,
ac = \(\frac{V_a}{V_v}\) and n = \(\frac{V_v}{V}\)
na = \(\frac{V_a}{V}\) = n * ac.

3. The relationship between γd, G and e is given by ______
a) γd = \(\frac{Gγ_w}{1-e}\)
b) γd = \(\frac{G+γ_w}{1+e}\)
c) γd = \(\frac{Gγ_w}{1+e}\)
d) γd = \(\frac{Gγ_w}{e}\)

  Discussion

Answer: c
Explanation: The dry unit weight is given by,
γd = \(\frac{W_d}{V} = \frac{γ_s*V_s}{V}\)
when Vs = 1, we have V = (1+e)
γd = γs*1/(1+e)
But γs = Gγw
γd = \(\frac{Gγ_w}{1+e}\).

4. The relationship between γsat, G and e is _______
a) \(γ_{sat}=\frac{(G*e) γ_w}{(1+e)}\)
b) \(γ_{sat}=\frac{(G-e) γ_w}{(1+e)}\)
c) \(γ_{sat}=\frac{(G+e) γ_w}{(1+e)}\)
d) \(γ_{sat}=\frac{(G+e)}{(1+e)}\)

  Discussion

Answer: c
Explanation: The saturated unit weight is given by,
\(γ_{sat}=\frac{W_{sat}}{V} = \frac{W_d+W_w}{V}\)
\(γ_{sat}=\frac{(γ_s*V_s + γ_w *V_w)}{V}\)
When Vs = 1, Vw = e and V = 1+e
\(γ_{sat}=\frac{(γ_s*1 + γ_w *e)}{(1+e)} = \frac{(Gγ_w+ γ_w e)}{(1+e)}\)
\(γ_{sat}=\frac{(G+e) γ_w}{(1+e)}\).

5. The relationship between γ, S, G and e is ______
a) γ=\(\frac{(GS+e) γ_w}{(1+e)}\)
b) γ=\(\frac{(G+eS) γ_w}{(1+e)}\)
c) γ=\(\frac{(G+e)S γ_w}{(1+e)}\)
d) γ=\(\frac{(G+e) γ_w}{(1+e)S}\)

  Discussion

Answer: b
Explanation: The bulk unit weight is given by,
\(γ=\frac{W}{V} = \frac{(γ_s*V_s+ γ_w *V_w)}{V}\)
When Vs = 1, Vw = ew and V = 1+e
γ = \(\frac{(γ_s*1 + γ_w * e_w)}{(1+e)}\)
since ew = e*S and γs = Gγw
γ = \(\frac{(Gγ_w*1 + γ_w * e*S)}{(1+e)}\)
γ=\(\frac{(G+eS) γ_w}{(1+e)}\).

6. The relationship between γ and γd can e derived when the degree of saturation is______
a) 0
b) 0.5
c) 0.8
d) 1

  Discussion

Answer: a
Explanation: The bulk unit weight is given by,
γ = \(\frac{(G+e)S γ_w}{(1+e)}\)
When the degree of saturation S=0,
the unit weight is \(\frac{(G+e) γ_w}{(1+e)}\) which is equal to γd
\(γ_d = \frac{Gγ_w}{(1+e)}\)
When the degree of saturation S=1,
the unit weight is \(\frac{(G+e) γ_w}{(1+e)}\) which is equal to γsat
\(γ_{sat}=\frac{(G+e) γ_w}{(1+e)}\).

7. The relationship between γ’, G and e is ______
a) \(γ’=\frac{(G+1) γ_w}{(1+e)}\)
b) \(γ’=\frac{(G-1) γ_w}{(1+e)}\)
c) \(γ’=\frac{(G-1) γ_w}{(1-e)}\)
d) \(γ’=\frac{G γ_w}{(1+e)}\)

  Discussion

Answer: b
Explanation: The submerged unit weight is given by,
γ’ = γsat – γw
γ’ = \(\frac{(G+e) γ_w}{(1+e)} – γ_w\)
\(γ’=\frac{(G-1) γ_w}{(1+e)}\).

8. The relationship between γd, γ and w is given by _______
a) γd = γ(1+w)
b) γd = γ *(1+w)
c) γd = \(\frac{γ}{(1+w)}\)
d) γd = γ+1+w

  Discussion

Answer: c
Explanation: The water content is given by,
\(w=\frac{W_w}{W_d}\)
hence \(1+w=\frac{(W_w + W_d)}{W_d} = \frac{W}{W_d}\)
\(W_d=\frac{W}{(1+w)}\)
\(γ_d=\frac{W_d}{V} = \frac{W}{(1+w)V}\)
γd = \(\frac{γ}{(1+w)}\).

9. The relationship between γsat, γd, γ and S is ______
a) γ = γd*S[γsatd]
b) γ = \(\frac{γ_d}{S[γ_{sat}-γ_d]}\)
c) γ = γd+S[γsatd]
d) γ = γd-S[γsatd]

  Discussion

Answer: c
Explanation: The bulk unit weight is given by,
\(γ = \frac{(G+eS) γ_w}{(1+e)} \,or\, γ = G\frac{γ_w}{(1+e)} + S\frac{eγ_w}{(1+e)}\)
\(γ = γ_d+S[\frac{(G+e) γ_w}{(1+e)} – \frac{G γ_w}{(1+e)}]\)
γ = γd+S[γsatd].

10. The relationship between γd, G, w and S is_______
a) γd = G γw*(1+(\(\frac{wG}{S}\)))
b) γd = G γw – (1+(\(\frac{wG}{S}\)))
c) γd = G γw/(1+(\(\frac{wG}{S}\)))
d) γd = G γw+(1+(\(\frac{wG}{S}\)))

  Discussion

Answer: c
Explanation: The dry unit weight is given by,
γd = \(\frac{Gγ_w}{(1+e)}\)
but e=\(\frac{wG}{S}\)
substituting e in γd
γd = G γw/(1+(\(\frac{wG}{S}\))).