Foundation Engineering Questions and Answers Part-7

1. For a dry backfill with no surcharge, the active earth pressure intensity is _________
a) pa=Ka γH
b) pa=γH
c) pa=Ka H
d) pa=Ka γ

Answer: a
Explanation: The active earth pressure intensity is given by,
pa=Ka γH,
where, \(\frac{σ_1}{σ_3} = \frac{σ_v}{σ_h} = K_a=\frac{1}{tan^2(45°+\frac{φ}{2})},\)
γ=unit weight of the back fill
H= height of the retaining wall.

2. The resultant active pressure per unit length of wall for dry backfill with no surcharge is _______
a) \(P_a=\frac{1}{2}K_aγH^2\)
b) Pa=γH2
c) Pa=Ka γH2
d) Pa=Ka H2

Answer: a
Explanation: The pressure intensity p0 is given by,
pa=Ka γz,
The total earth pressure P0 at rest per unit length is,
\(P_a=\int_0^HK_a γz.dz\)
∴ \(P_a=\frac{1}{2}K_aγH^2\)

3. The resultant active pressure per unit length of wall for dry backfill with no surcharge acting at _________ above the base of wall.
a) H/2
b) H
c) H/6
d) H/3

Answer: d
Explanation: The pressure distribution of the stresses in the retaining wall due to the backfill is triangular one. Since the pressure distribution is triangular, the resultant active pressure per unit length of wall will act at the centroid of the wall, which is at a distance of H/3 from the base of the wall.

4. For a submerged backfill, the active earth pressure is given by _________
a) pa=Kaγ’z
b) pa=Kaγ’z-γwz
c) pa=Kaγ’z+γwz
d) pa=Kaγ’z*γwz

Answer: c
Explanation: For a submerged backfill, the lateral pressure is made up of two components,
1) lateral pressure due to submerged weight of backfill
2) lateral pressure due to water
pa=Kaγ’z+γwz.

5. If free water stands on both side of a retaining wall, the lateral earth pressure is given by ____
a) pa=Ka γ’ z
b) pa=Ka γ’ z-γw z
c) pa=Ka γ’ z+γw z
d) pa=Ka γ’ z*γw z

Answer: a
Explanation: When the free water stands on both side of a retaining wall, the water pressure is not considered and the net pressure is given by,
pa=Ka γ’ z.

6. If the angle of internal friction decreases, then Ka ___________
a) decreases
b) increases
c) equal to zero
d) does not change

Answer: b
Explanation: Since the coefficient of earth pressure for active state of plastic equilibrium is given by,
\(K_a=\frac{1-sinφ}{1+sinφ}.\) Therefore, from the equation it is clear that as the angle of internal friction decreases, then Ka increases.

7. For the same value of φ, the backfill is partly submerged to height H2 and the backfill is moist to a depth H1. Find the lateral pressure intensity at the base of wall.
a) pa=Ka γH1-Ka γ’ H2w H2
b) pa=Ka γH1+Ka γ’ H2w H2
c) pa=Ka γH1w H2
d) pa=Ka γH1+Ka γ’ H2

Answer: b
Explanation: When the backfill is partly submerged to height H2 and the backfill is moist to a depth H1, lateral pressure intensity is due to:
1) lateral pressure due to moist weight of backfill
2) lateral pressure due to saturated weight of backfill
3) lateral pressure due to water
pa=Ka γH1+Ka γ’ H2w H2.

8. For the different value of φ, the backfill is partly submerged to height H2 and the backfill is moist to a depth H1. Find the lateral pressure intensity at the base of wall for φ12.
a) pa=Ka2 γH1-Ka2 γ’ H2w H2
b) pa=Ka2 γH1+Ka2 γ’ H2w H2
c) pa=Ka2 γH1w H2
d) pa=Ka2 γH1+Ka2 γ’ H2

Answer: b
Explanation: For the different value of φ, the coefficient of active earth pressure is different. Since, the lateral pressure intensity is due to:
1) lateral pressure due to moist weight of backfill
2) lateral pressure due to saturated weight of backfill
3) lateral pressure due to water
pa=Ka2 γH1+Ka2 γ’ H2w H2. We have to consider Ka2 over Ka1 as the φ12.

9. If the backfill carries a uniform surcharge q, then the lateral pressure at the depth of wall H is ____________
a) pa=Ka γz+Ka q
b) pa=Ka γz-Ka q
c) pa=Ka γz*Ka q
d) pa=Ka γz/Ka q

Answer: a
Explanation: When the backfill is horizontal and carries a surcharge q, then the vertical pressure increment will be by q. Due to this, the lateral pressure will increase by Ka q. Hence, lateral pressure at the depth of wall H is pa=Ka γz+Ka q.

10. The height of fill Ze, equivalent to uniform surcharge intensity is __________
a) q/γ
b) q-γ
c) q+γ
d) q*γ

Answer: a
Explanation: The height of fill Ze, equivalent to uniform surcharge intensity is given by,
Ka γze=Ka q,
∴ \(z_e=\frac{q}{γ}.\)