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Test - 3

Description: Test - 3
Number of Questions: 19
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Tags: Test - 3 Soil Mechanics

A footing 2 m x 1 m exerts a uniform pressure of 150 kN/mm2 on the soil.
Assuming a load dispersion of 2 vertical to 1 horizontal, the average vertical stress (kN/m2) at 1.0 m below the footing is

  1. 50

  2. 75

  3. 80

  4. 100

Answer: 1
Explanation:

 $\text{Average vertical stress}\\
= \frac{\text{total load}}{\text{total area at 1 m depth}}\\
= \frac{150 \times 2 \times 1}{( 2 + 0.5 + 0.5)(1 + 0.5 + 0.5)} = 50 kN/m^2$

A direct shear test was conducted on a cohesion-less soil (c = 0) specimen under a normal stress of 200 kN/m2. The specimen failed at a shear stress of 100 kN/m2.
The angle of internal friction of the soil (degrees) is

  1. 26.6

  2. 29.5

  3. 30.0

  4. 32.6

Answer: 1
Explanation:

 

A pile of 0.50 m diameter and length 10 m is embedded in a deposit of clay. The undrained strength parameters of the clay are cohesion = 60 kN/m2 and the angle in internal friction = 0. The skin friction capacity (kN) of the pile for an adhesion factor of 0.6, is

  1. 671

  2. 565

  3. 283

  4. 106

Answer: 2
Explanation:

 $\text{Skin friction capacity} \\
= \alpha . C \text{(perimeter $\times$ depth)} \\
= 0.6 \times 60 \times \pi \times 0.5 \times 10 \\
= 565.75 kN$

A saturated clay stratum draining both at the top and bottom undergoes 50 percent consolidation in 16 years under an applied load. If an additional drainage layer were present at the middle of the clay stratum, 50 percent consolidation would occur in

  1. 2 years

  2. 4 years

  3. 8 years

  4. 16 years

Answer: 2
Explanation:

 

A test plate 30 cm x 30 cm resting on a sand deposit settles by 10 mm under a certain loading intensity. A footing 150 cm x 200 cm resting on the same sand deposit and loaded to the same load intensity settles by

  1. 2.0 mm

  2. 27.8 mm

  3. 3.02 mm

  4. 50.0 mm

Answer: 2
Explanation:

 

A volume of 3.0 x 106 m3 of groundwater was pumped out from an unconfined aquifer, uniformly from an area of 5 km2. The pumping lowered the water table from initial level of 102 m to 99 m. The specific yield of the aquifer is

  1. 0.20

  2. 0.30

  3. 0.40

  4. 0.50

Answer: 1
Explanation:

 

The ground conditions at a site are shown in the figure below.

The saturated unit weight of the sand (kN/m3) is

  1. 15

  2. 18

  3. 21

  4. 24

Answer: 3
Explanation:

 

The ground conditions at a site are shown in the figure below.

The total stress, pore water pressure and effective stress (kN/m2) at the point P are, respectively

  1. 75, 50 and 25

  2. 90, 50 and 40

  3. 105, 50 and 55

  4. 120, 50 and 70

Answer: 3
Explanation:

 

A column is supported on a footing as shown in the figure below. The water table is at a depth of 10 m below the base of the footing.

The net ultimate bearing capacity (kN/m2) of the footing based on Terzaghi’s bearing capacity equation is

  1. 216

  2. 432

  3. 630

  4. 846

Answer: 3
Explanation:

 

The water content of a saturated soil and the specific gravity of soil solids were found to be 30% and 2.70 respectively. Assuming the unit weight of water to be 10 kN/m3, the saturated unit weight (kN/m3) and the void ratio of the soil are

  1. 19.4, 0.81

  2. 18.5, 0.30

  3. 19.4, 0.45

  4. 18.5, 0.45

Answer: 1
Explanation:

 

Sieve analysis on a dry soil sample of mass 1000 g showed that 980 g and 270 g of soil pass through 4.75 mm and 0.075 mm sieve respectively. The liquid limit and plastic limits of the soil fraction passing through 425$\mu$ sieves are 40% and 18% respectively. The soil may be classified as

  1. SC

  2. MI

  3. CI

  4. SM

Answer: 1
Explanation:

 

What is the ultimate capacity in kN of the pile group shown in the figure assuming the group to fail as a single block?

  1. 921.6

  2. 1177.6

  3. 2438.6

  4. 3481.6

Answer: 4
Explanation:

 

The factor of safety of an infinite soil slope shown in the figure having the properties c = 0, $\phi$ = 350 = $\gamma_{dry}$ = 16 kN/m3 and $\gamma_{sat}$ = 20 kN/m3 is approximately equal to

  1. 0.70

  2. 0.80

  3. 1.00

  4. 1.20

Answer: 1
Explanation:

 

The bearing capacity of a rectangular footing of plan dimensions 1.5 m x 3 m resting on the surface of a sand deposit was estimated as 600 kN/m2 when the water table is far below the base of the footing. The bearing capacities in kN/m2 when the water level rises to depths of 3 m, 1.5 m and 0.5 m below the base of the footing are

  1. 600, 600 and 400

  2. 600, 450 and 350

  3. 600, 500 and 250

  4. 600, 400 and 250

Answer: 1
Explanation:

 

To provide safety against piping failure, with a factor of safety of 5, what should be he maximum permissible exit gradient for soil with specific gravity of 2.5 and porosity of 0.35 ?

  1. 0.155

  2. 0.176

  3. 0.195

  4. 0.213

Answer: 3
Explanation:

 

For steady flow to a fully penetrating well in a confined acquifer, the drawdowns at radial distances of r1 and r2 from the well have been measured as s1 and s2 respectively, for a pumping rate of Q. The transmissivity of the aquifer is equal to

  1. $\frac{Q}{2\pi}\frac{In\frac{r_2}{r_1}}{(s_1-s_2)}$

  2. $\frac{Q}{2\pi} \frac{In(r_2-r_1)}{(s_1-s_2)}$

  3. $\frac{Q}{2\pi}In\bigg(\frac{r_2/r_1}{s_1-s_2}\bigg)$

  4. $2\pi Q \frac{r_2-r_1}{In\big(\frac{s_2}{s_1}\big)}$

Answer: 1
Explanation:

 

Figure given below shows a smooth vertical gravity retaining wall with cohesionless soil backfill having an angle of internal friction $\phi$ In the graphical representation of Rankine’s active earth pressure for the retaining wall shown in figure, length OP represents

  1. vertical stress at the base

  2. vertical stress at a height H/3 from the base

  3. lateral earth pressure at the base

  4. lateral earth pressure at a height H/3 from the base

Answer: 1
Explanation:

 Rankine's active earth pressure method is graphical method to determine the active earth pressure of the soil.OP dose not cut the mohr;s circle tangential, so OP represents the vertical stress at the base.

Match the following groups:

Group — I Group —II
P. Constant head permeability test 1. Pile foundations
Q. Consolidation test 2. Specific gravity
R. Pycnometer test 3. Clay soil
S. Negative skin friction 4. Sand
  1. P - 4, Q - 3, R - 2, 5 - 1

  2. P - 4, Q - 2, R - 3, S - 1

  3. P - 3, Q - 4, R - 2, 5 - 1

  4. P - 4, Q - 1, R - 2, S - 3

Answer: 1
Explanation:

 

The range of void ratio between which quick sand conditions occurs in cohesion less granular soil deposits is

  1. 0.4 - 0.5

  2. 0 - 6 - 0.7

  3. 0.8 - 0.9

  4. 1.0 - 1.1

Answer: 2
Explanation:

 $i_c=\frac{G-1}{(1+e)},\text{value of G=2.65 to 2.68}\\
and\hspace{0.1cm} i_c\approx 1 So,\text{e should vary from 0.66 to 0.7. }$

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