This set of Design of RC Structures Multiple Choice Questions & Answers (MCQs) focuses on “Foundations – Safe Bearing Capacity of Soil”.

1. Which of the following symbol is used to denote the safe bearing capacity of soil?

a) q_{u}

b) q_{0}

c) q_{nu}

d) q_{na}

View Answer

Explanation: The safe bearing capacity of soil is denoted by (q

_{0}). It is computed from the Terzaghi’s equations. The allowable bearing capacity should be taken smaller than the safe bearing capacity based on the ultimate capacity. It is also smaller of the allowable bearing pressure based on tolerable settlement.

2. Which of the following is the test for determining the safe bearing capacity of soil?

a) Plate load test

b) Direct shear test

c) Triaxial shear test

d) Unconfined compression test

View Answer

Explanation: The safe bearing capacity of soil is determined from plate load test. Other field tests that can be performed are drop weight method, standard penetration test. The ultimate bearing capacity is determined by direct shear test, triaxial shear test andunconfined compression test.

3. Which of the following is the equation of safe bearing capacity for strip footing?

a) q_{0} = 1 / F [cN_{c} + γD(N_{q} – 1)R_{w1} + 1.5 γBN_{γ}R_{w2}] + γD

b) q_{0} = 1 / F [cN_{c} + γD(N_{q} – 1)R_{w1} + 0.5 γBN_{γ}R_{w2}] + γD

c) q_{0} = 1 / F [1.3cN_{c} + γD(N_{q} – 1)R_{w1} + 0.4 γBN_{γ}R_{w2}] + γD

d) q_{0} = 1 / F [1.3cN_{c} + γD(N_{q} – 1)R_{w1} + 0.3 γBN_{γ}R_{w2}] + γD

View Answer

Explanation: The equation of safe bearing capacity for strip footing is given by:

q

_{0}= 1 / F [cN

_{c}+ γD(N

_{q}– 1)R

_{w1}+ 0.5 γBN

_{γ}R

_{w2}] + γD

Here(D) is depth of footing, (B) is width of footing.

(F) is the factor of safety (2 to 3).

Also N

_{c}, N

_{q}, N

_{γ}are the bearing capacity factors for general shear failures.

R

_{w1}and R

_{w2}are water reduction factors.

4. What is the equation of safe bearing capacity for a square footing?

a) q_{0} = 1 / F [cN_{c} + γD(N_{q} – 1)R_{w1} + 1.5 γBN_{γ}R_{w2}] + γD

b) q_{0} = 1 / F [cN_{c} + γD(N_{q} – 1)R_{w1} + 0.5 γBN_{γ}R_{w2}] + γD

c) q_{0} = 1 / F [1.3cN_{c} + γD(N_{q} – 1)R_{w1} + 0.4 γBN_{γ}R_{w2}] + γD

d) q_{0} = 1 / F [1.3cN_{c} + γD(N_{q} – 1)R_{w1} + 0.3 γBN_{γ}R_{w2}] + γD

View Answer

Explanation: The equation of safe bearing capacity for a square footing is:

q

_{0}= 1 / F [1.3cN

_{c}+ γD(N

_{q}– 1)R

_{w1}+ 0.4 γBN

_{γ}R

_{w2}] + γD.

Here (D) is depth of footing, (B) is width of footing.

(F) is the factor of safety (2 to 3).

The bearing capacity factors (N

_{c}, N

_{q}, N

_{γ}) are for general shear failure.

The water reduction factors are denoted by R

_{w1}and R

_{w2}.

5. Which expression is correct for safe bearing capacity of a circular footing?

a) q_{0} = 1 / F [cN_{c} + γD (N_{q} – 1)R_{w1} + 1.5 γBN_{γ}R_{w2}] + γD

b) q_{0} = 1 / F [cN_{c} + γD (N_{q} – 1)R_{w1} + 0.5 γBN_{γ}R_{w2}] + γD

c) q_{0} = 1 / F [1.3cN_{c} + γD (N_{q} – 1)R_{w1} + 0.4 γBN_{γ}R_{w2}] + γD

d) q_{0} = 1 / F [1.3cN_{c} + γD (N_{q} – 1)R_{w1} + 0.3 γBN_{γ}R_{w2}] + γD

View Answer

Explanation: The correct equation of safe bearing capacity of a circular footing is:

q

_{0}= 1 / F [1.3cN

_{c}+ γD (N

_{q}– 1)R

_{w1}+ 0.3 γBN

_{γ}R

_{w2}] + γD

Depth of footing is (D), diameter of the footing is (B).

(F) is the factor of safety (2 to 3).

For general shear failures (N

_{c}, N

_{q}, N

_{γ}) are the bearing capacity factors.

The water reduction factors are denoted by R

_{w1}and R

_{w2}.

For local shear failure N

_{c}‘, N

_{q}‘, N

_{γ}‘ is used.

6. What is the value of water reduction factor R_{w1}?

a) 0.5(1 + (Z_{w1} / 2D)

b) 0.5(2 + (Z_{w1} / D)

c) 1.5(1 + (Z_{w1} / D)

d) 0.5(1 + (Z_{w1} / D)

View Answer

Explanation: The value to be used in determining water reduction factor

(R

_{w1}) = 0.5(1 + (Z

_{w1}/ D).

Here (D) is the depth of footing, (Z

_{w1}) is the depth of water table from the ground.

Also the water reduction factor (R

_{w2}) is 0.5(1 + (Z

_{w2}/ B).

Here (B) is width of the footing.

7. Which of the following is the empirical relation for allowable bearing pressure (q_{ρ}) based on maximum settlement?

a) 34.3 (N – 3)[(B + 0.3)^{2} / 2B].R_{w2}.R_{d}

b) 34.3 (N – 3)[(B + 0.3)^{2} / B].R_{w2}.R_{d}

c) 34.3 (N – 3)[(2B + 0.3)^{2} / 2B].R_{w2}.R_{d}

d) 34.3 (N – 3)[(B + 0.5)^{2} / 2B].R_{w2}.R_{d}

View Answer

Explanation: The allowable bearing pressure (q

_{ρ}) based on the maximum settlement of individual footing to 25 mm is given by empirical formula

(q

_{ρ}) = {34.3 (N – 3)[(B + 0.3)

^{2}/ 2B].R

_{w2}.R

_{d}}.

Here (N) is standard penetration number with applicable corrections, R

_{d}is depth factor.

8. Which of the following are the three components of total settlement (S)?

a) (S_{w}), (S_{c}), (S_{s})

b) (S_{i}), (S_{w}), (S_{s})

c) (S_{i}), (S_{c}), (S_{s})

d) (S_{i}), (S_{c}), (S_{w})

View Answer

Explanation: The total settlement of a footing in clay consists of three components (S

_{i}), (S

_{c}), and (S

_{s}). It is given by:

S = (S

_{i}+ S

_{c}+ S

_{s})

Here (S

_{i}) is immediate elastic settlement, (S

_{c}) is consolidation settlement, (S

_{s}) is settlement due to secondary consolidation of clay and (S) is total settlement.

9. What is the value of elastic settlement (S_{i})?

a) qB ((1 – μ^{2}) / 2E_{s}) I_{w}

b) qB ((1 – μ^{2}) / E_{s}) I_{w}

c) qB ((1 – μ) / E_{s}) I_{w}

d) qB ((1 – μ^{2}) / E_{s})

View Answer

Explanation: The immediate settlement (S

_{i}) is the elastic settlement. It is given by:

(S

_{i}) = [qB ((1 – μ

^{2}) / E

_{s}) I

_{w}]

Here (q) is intensity of contact pressure, (B) is the least lateral dimension of footing, (E

_{s}) is modulus of elasticity for soil. (I

_{w}) is influence factor, its value depends on the shape of footing.

10. It is difficult to determine (μ) and (E_{s}) for soils. However the entire [(1 – μ^{2}) / E_{s}) I_{w}] is determined by plate load tests.

a) True

b) False

View Answer

Explanation: The values of factor (μ) and modulus of elasticity (E

_{s}) are difficult to determine for soils. However the entire term [(1 – μ

^{2}) / E

_{s}) I

_{w}] is determined from plate load tests by using two or three different size plates of the same shape. The plot between (S

_{i}) and (qB) gives straight line. The slope of the plot is equal to [(1 – μ

^{2}) / E

_{s}) I

_{w}].

11. What is the expression for consolidation settlement (S_{c})?

a) (C / 1 + e_{0}) . H log_{10}((σ_{0} + Δσ) / σ_{0})

b) C (C_{c} / e_{0}) . H log_{10} ((σ_{0} + Δσ) / σ_{0})

c) C (C_{c} / 1 + e_{0}) . 2H log_{10} ((σ_{0} + Δσ) / σ_{0})

d) C (C_{c} / 1 + e_{0}) . H log_{10} ((σ_{0} + Δσ) / σ_{0})

View Answer

Explanation: The consolidation settlement is

(S

_{c}) = C (C

_{c}/ 1 + e

_{0}) . H log

_{10}((σ

_{0}+ Δσ) / σ

_{0})

Here (σ

_{0}) is effective overburden pressure due to soil over burden, (Δσ) vertical stress on footing, (e

_{0}) is initial void ratio, (H) thickness of compressible layer. The compression index is

(C

_{c}) = (0.009 (w

_{L}– 10))

C is correction factor, (w

_{L}) is liquid limit.

12. Which of the following is the Rankine formula for determining depth of foundation (h)?

a) p / γ[(1 – sinΦ) / (1 + sinΦ)]

b) p / γ[(1 – sinΦ) / (1 + sinΦ)]^{2}

c) p / γ[(1 + sinΦ) / (1 – sinΦ)]^{2}

d) p / γ[(sinΦ) / (1 + sinΦ)]^{2}

View Answer

Explanation: The Rankine formula for determining depth of foundation is given by:

(h) = {p / γ[(1 – sinΦ) / (1 + sinΦ)]

^{2}}

Here (p) is gross bearing capacity, (γ) is density of soil and (Φ) is angle of repose of soil.

IS: 1080 – 1985 requires that in all soils the minimum depth of 50 cm is necessary.

13. A reinforced concrete wall of 250 mm thickness carries a load of 500 kN/m inclusive of its own weight. What is the value of net upward pressure if safe bearing capacity of soil is 160 kN/m^{2}?

a) 142 kN/m^{2}

b) 142.86 kN/m^{2}

c) 140 kN/m^{2}

d) 145 kN/m^{2}

View Answer

Explanation: Size of footing

W = 500 kN/m, q

_{0}= 160 kN/m

^{2}

Assume self – weight of footing equal to 10% of the super imposed load.

W

^{‘}= 50 kN/m,

B = (W + W

^{‘}) / q

_{0}= (500 + 50) / 160 = 3.43 m

Adopt B = 3.5 m = 3500 mm

Net upward pressure p

_{0}= 500 / 3.5 = 143.86 kN/m

^{2}

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