Geotechnical and Foundation Engineering - questions from AMIE exams (Winter 2018)

Answer the following questions (5 x 3)

Find a relation for porosity, n of soil with air content ac and per cent air void (na) of the soil.

Percentage air voids na = Va/V
The above relation can be written as,
na = (Va/Vv)/(Vv/V) = n x ac
where n is porosity.

Degree of saturation of the soil, s, and void ratio of the soil, e, is related to water content, W, of the soil by the following equation: s.e = G.W where G = specific gravity of soil solids. Prove the above statement.

Write down the basic equation
Manipulate the basic equation to get the desired equation.
You want to get e in the denominator, and you have Vv. You know that Vv = eVs and Vw is the weight of water divided by the unit weight of water. From the definition of water content, the weight of water is wWs.
Here is the algebra:
Vv = eVs
Vw = Www = wWsw
S = Vw/Vv = (wWsw)/eVs = Gsw/e

For normally consolidated clay having LL = 55% and PL = 20%, estimate the value of its compression index.

Cc = 0.007(wL - 7)
⇒ Cc = 0.007(55 - 7) = 0.33

“The allowable bearing capacity of purely cohesive soil, with a factor of safety = 3, is approximately equal to its unconfined compressive strength.” — Prove the statement.

Allowable bearing pressure
qa = qu/Fs 
where Fs is factor of safety generally taken as 3.

Find the value of the time factor, T, when the degree of consolidation is 50%.

Tv = (π/4)U2 = (π/4)(50/100)2 = 0.2

Choose the correct answers (5 x 1)

1. The phreatic line in an earthen dam is
(a) straight line
(b) parabolic
(c) circular
(d) elliptical

2. If the shearing stress is zero on two planes then the angle between these two planes, is
(a) 45⁰
(b) 90°
(c) 135°
(d) 225°

3. Contact pressure beneath a rigid fooling resting on cohesive soil is
(a) less at edge compared to meddle
(b) more at edges compared to meddle
(c) uniform throughout
(d) none of the above

4. Mechanical stabilization of soil is done with the help of
(a) cement
(b) lime
(c) bitumen
(d) proper grading

5. A soil having particles of nearby the same size is known as
(a) well graded
(b) gap graded
(c) uniformly graded
(d) poorly graded


1. (b) Hydrostatic pressure acts below the phreatic line, whereas atmospheric pressure exists above the phreatic line. This line separates a saturated soil mass from an unsaturated soil mass. It is not an equipotential line, but a flow line. For an earthen dam, the phreatic line approximately assumes the shape of a parabola.

2. (b) The planes on which shearing stress is zero are major and minor principal planes. These two planes are inclined at 90° to each other.

3. (c) The contact pressure of a rigid footing is assumed as uniform stress development. But in reality, if a rigid footing is placed on a cohesive soil, the stress level will be slightly higher at the edges of the footing and at the centre of the footing the stress level will be lower than the edges of the footing. The stress diagram will be in a parabolic curve from edge to centre, connecting the respective stress ordinates at edges and centre.

4. (d) Mechanical stabilization is the process of improving the properties of the soil by changing its gradation. Two or more types of natural soil are mixed to obtain a composite material which is superior to any of its components. To achieve the desired grading, sometimes the soil with coarse particles are added or the soils with fine particles are removed.

5. (c)
The coefficient of uniformity is given by:
Cu = D60/D10
For a soil having particles of nearly the same size, D10 = D60
Therefore, Cu = 1.
The value of Uniformity coefficient for different soils
For, Cu = 1 (Uniformly graded soil)
For, Cu > 6 (Well-graded sand)
For, Cu > 4 (Well graded gravel)

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