 Test - 4

 Description: Test - 4 Number of Questions: 20 Created by: Yashbeer Singh Tags: Test - 4 Soil Mechanics Civil Engineering - CE

Water is flowing through the permeability apparatus as shown in the figure. The coefficient of permeability of the soil is k m/s and the porosity of the soil sample is 0.50. The total head, elevation head and pressure head in metres of water at the point R shown in the figure are respectively

1. 0.8, 0.4 and 0.4

2. 1.2, 0.4 and 0.8

3. 0.4, 0 and 0.4

4. 1.6, 0.4 and 1.2

Explanation: Water is flowing through the permeability apparatus as shown in the figure. The coefficient of permeability of the soil is km/s and the porosity of the soil sample is 0.50. What are the discharge velocity and seepage velocity through the soil sample?

1. k, 2k

2. 2/3k, 4/3k

3. 2k, k

4. 4/3k, 2/3k

Explanation:

$\text{Discharge velocity V = ki = k$\times$hydraulic gradient = k m/s} \\ \text{Seepage velocity} = \frac{v}{n} = \frac{k}{0.5} = 2k m/s$

A masonry dam is found on previous sand having porosity equal to 45% and specific gravity of sand particles is 2.65. For a desired factor of safety of 3 against sand boiling, the maximum permissible upward gradient will be

1. 0.225

2. 0.302

3. 1.0

4. None of these

Explanation: Water is pumped from a well tapping an unconfined aquifer at a certain discharge rate and the steady state drawdown (X) in an observation well is monitored. Subsequently, the pumping discharge is doubled and the steady state drawdown in the same observation well is found to be more than double (i.e. more than 2X). This disproportionate drawdown is caused by

1. well losses

2. decrease in the saturated thickness of the aquifer

3. nonlinear flow

4. delayed gravity yield

Explanation: The ground conditions at a site are as shown in the figure. The water table at the site which was initially at a depth of 5 m below the ground level got permanently lowered to a depth of 15 m below the ground level due to pumping of water over a few years. Assume the following data:
i. unit weight of water = 10 kN/m3
ii. unit weight of sand above water table = 8 kN/m3
iii. unit weight of sand and clay below the water table = 20 kN/m3
iv. coefficient of volume compressibility = 0.25 m2/MN What is the compression of the clay layer in mm due to the lowering of the water table?

1. 125

2. 100

3. 25

4. 0

Explanation: A double draining clay layer, 6 m thick, settles by 30 mm in three years under the influence of a certain loads. Its final consolidation settlement has been estimated to be 120 mm. If a thin layer of sand having negligible thickness is introduced at a depth of 1.5 m below the top surface, the final consolidation settlement of clay layer will be

1. 60 mm

2. 120 mm

3. 240 mm

4. None of these

Explanation:

Rate of settlement will not affect the final settlement and rate of settlement of clay layer will increase four times but total settlement of the clay layer will remain uneffected, hence 120mm.

The ground conditions at a site are as shown in the figure. The water table at the site which was initially at a depth of 5 m below the ground level got permanently lowered to a depth of 15 m below the ground level due to pumping of water over a few years. Assume the following data
i. unit weight of water = 10 kN/m3
ii. unit weight of sand above water table = 8 kN/m3
iii. unit weight of sand and clay below the water table = 20 kN/m3
iv. coefficient of volume compressibility = 0.25 m2/MN What is the change in the effective stress in kN/m2 at mid-depth of the clay layer due to the lowering of the water table?

1. 0

2. 20

3. 80

4. 100

Explanation: A 25 kN point load acts on the surface of an infinite elastic medium. The vertical pressure intensity in kN/m2 at a point 6.0 m below and 4.0 m away from the load will be

1. 132

2. 13.2

3. 1.32

4. 0.132

Explanation:

$\text{We have}, \sigma_z = \frac{3Q}{2 \pi z^2} \bigg[ \frac{1}{1+(r/z)^2}\bigg]^{5/2}\\ \hspace{2.3cm}=\frac{3 \times 25}{2 \pi \times 6^2} \times \frac{3Q}{2 \pi z^2} \bigg[ \frac{1}{1+(4/6)^2}\bigg]^{5/2}\\ \hspace{2.3cm} = 0.132 kN/m^2$

In a plate load test conducted on cohesionless soil, a 600 mm square test plate settles by 15 mm under a load intensity of 0.2 N/mm2. All conditions remaining the same, settlement of a 1 m square footing will be

1. less than 15 mm

2. greater than 25 mm

3. 15.60 mm

4. 20.50 mm

Explanation: The unit volume of a mass of saturated soil is subjected to horizontal seepage. The saturated unit weight is 22 kN/m3 and the hydraulic gradient is 0.3. The resultant body force on the soil mass is

1. 1.98 kN

2. 6.6 kN

3. 11.49 kN

4. 22.97 kN

Explanation: A granular soil possesses saturated density of 20 kN/m3. Its effective angle friction is 35 degrees. If the desired factor of safety is 1.5, the safe angle of slope for this soil, when seepage occurs at and parallel to the surface, will be

1. 25o

2. 23o

3. 20o

4. 13o

Explanation: In an undrained triaxial test on a saturated clay, the Poisson’s ratio is

1. $\frac{\sigma_3}{(\sigma_1 + \sigma_3)}$

2. $\frac{\sigma_3}{(\sigma_1 - \sigma_3)}$

3. $\frac{\sigma_1 - \sigma_3}{\sigma_3}$

4. $\frac{\sigma_1 + \sigma_3}{\sigma_3}$

Explanation: The ratio of saturated unit weight to dry unit weight of dry unit weight is 1.25. If the specific gravity of solids (Gs) is 2.56, the void ratio of the soil is

1. 0.625

2. 0.663

3. 0.944

4. 1.325

Explanation: Two circular footings of diameters D1 and D2 are resting on the surface of the same purely cohesive soil. The ratio of their gross ultimate bearing capacities is

1. D1/D2

2. 1.0

3. $D_1^1/ D_2^2$

4. D2/D1

Explanation:

For two circular footings, the ratio of gross ultimate bearing capacities does not depend upon the size or the circular footing but depends on the shape. So, it will remain same, hence ratio = 1.0.

A 10 m thick clay layer is underlain by a sand layer of 20 m depth (see figure below). The water table is 5 m below the surface of the clay layer. The soil above the water table is capillary saturated. The value of gsat is 19 kN/m3. The unit weight of water is gw. If now the water table rises to the surface, the effective stress at a point P on the interface will 1. increase by 5$\gamma w$

2. remain unchanged

3. decrease by 5 gw

4. decrease by 10 gw

Explanation:

$\text{Effective stress = total stress - pore water pressure}\\ \text{We know that rise of water due to capillary action acts as s surcharge for the calculation of effective pressure. So, when the water table rises upto ground level surcharged will be zero and reduction of the effective stress will be equal to$5 \gamma_w$so reduced by$5 \gamma _w.$}$

A clayey soil has a maximum dry density of 16 kN/m3 and optimum moisture content of 12%. A contractor during the construction of core of an earth dam obtained the dry density 15.2 kN/m3 and water content 11%. This construction is acceptable because.

1. the density is less than the maximum dry density and water content is on dry side of optimum.

2. the compaction density is very low and water content is less than 12%.

3. the compaction is done on the dry side of the optimum.

4. both the dry density and water content of the compacted soil are within the desirable limits

Explanation: Which of the following statement is NOT true in the context of capillary pressure in soils ?

1. Water is under tension in capillary zone

2. Pore water pressure is negative in capillary zone

3. Effective stress increases due to capillary pressure

4. Capillary pressure is more in coarse grained soils

Explanation:

Void ratio and size of voids is less in fine grained soil than coarse grained soil hence capillary size will be more in fine grained soil and less in coarse grained soil.

Root time method is used to determine

1. T, time factor

2. cv' coefficient of consolidation

3. av' coefficient of compressibility

4. mv' coefficient of volume compressibility

Explanation: There are two footings resting on the ground surface. One footing is square of dimension 'B'. The other is strip footing of width 'B'. Both of them are subjected to a loading intensity of q. The pressure intensity at any depth below the base of the footing along the centerline would be

1. equal in both footings

2. large for square footing and small for strip footing

3. large for strip footing and small or square footing

4. more for strip footing at shallow depth ($\leq$ B) and more for square footing at large depth (>B)

Explanation:

$\text{Large for strip footing and small for square footing}\\ L/b \text{ for square footing = 1}\\ L/b\text{ for strip footing = }\infty \\ \therefore \hspace{2.3cm}I_{strip}>I_{square}$

Negative skin friction in a soil is considered when the pile is constructed through a

1. fill material

2. dense coarse sand

3. over consolidated stiff clay

4. dense fine sand