Skip to main content

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
S=Vw/Vv
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

Answers

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)

---
  • The study material for AMIE/B Tech/Junior Engineer exams is available at https://amiestudycircle.com
  • If you like the post please share your thoughts in the comment section 


Comments

Popular posts from this blog

Mechanics of Fluids (Solved Numerical Problems)

Numerical The surface Tension of water in contact with air at 20°C is 0.0725 N/m. The pressure inside a droplet of water is to be 0.02 N/cm² greater than the outside pressure. Calculate the diameter of the droplet of water. (7 marks) (AMIE Summer 2023) Solution Surface tension, σ = 0.0725 N/m Pressure intensity, P = 0.02 N/m 2 P = 4σ/d Hence, the Diameter of the dropd = 4 x 0.0725/200 = 1.45 mm Numerical Find the surface tension in a soap bubble of 40 mm diameter when the inside pressure is 2.5 N/m² above atmospheric pressure. (7 marks) (AMIE Summer 2023) Answer: 0.0125 N/m Numerical The pressure outside the droplet of water of diameter 0.04 mm is 10.32 N/cm² (atmospheric pressure). Calculate the pressure within the droplet if surface tension is given as 0.0725 N/m of water. (AMIE Summer 2023, 7 marks) Answer: 0.725 N/cm 2   Numerical An open lank contains water up to a depth of 2 m and above it an oil of specific gravity 0.9 for a depth of 1 m. Find the pressure intensity (i) at t...

Energy Systems (Solved Numerical Problems)

Wind at 1 standard atmospheric pressure and \({15^0}C\) has velocity of 15 m/s, calculate (i) the total power density in the wind stream (ii) the maximum obtainable power density (iii) a reasonably obtainable power density (iv) total power (v) torque and axial thrust Given: turbine diameter = 120 m, and turbine operating speed = 40 rpm at maximum efficiency. Propeller type wind turbine is considered. (AMIE Winter 2023) Solution For air, the value of gas constant is R = 0.287 kJ/kg.K 1 atm = 1.01325 x 105 Pa Air density \(\rho  = \frac{P}{{RT}} = \frac{{1.01325x{{10}^5}}}{{287}}(288) = 1.226\,kg/{m^3}\) Total Power \({P_{total}} = \rho A{V_1}^3/2\) Power density \(\begin{array}{l}\frac{{{P_{total}}}}{A} = \frac{1}{2}\rho {V_1}^3\\ = \frac{1}{2}(1.226){(15)^3}\\ = 2068.87{\mkern 1mu} W/{m^2}\end{array}\) Maximum power density \(\begin{array}{l}\frac{{{P_{\max }}}}{A} = \frac{8}{{27}}\rho A{V^3}_1\\ = \frac{8}{{27}}(1.226){(15)^3}\\ = 1226{\mkern 1mu} W/{m^2}\end{array}\) Assuming eff...

Design of Electrical Systems (Solved Numerical Problems)

Important note There is something wrong with this question paper. It seems that instead of "Design of Electrical Systems" the IEI has given problems from "Electrical Machines". You should raise a complaint to director_eea@ieindia.org in this regard. Numerical A 120 V DC shunt motor draws a current of 200A. The armature resistance is 0.02 ohms and the shunt field resistance is 30 ohms. Find back emf. If the lap wound armature has 90 slots with 4 conductors per slots, at what speed will the motor run when flux per pole is 0.04 Wb?​ (AMIE Summer 2023, 8 marks) Solution The back EMF (E b ) of a DC motor can be calculated using the formula: E b = V - I a R a   Given: V = 120 V I a = 200 A R a = 0.02 ohms Substituting the values into the formula: E b = 120 − 200 × 0.02 = 120 − 4​ = 116 V Now, let's calculate the speed (N) at which the motor will run using the given flux per pole (φ p ). The formula to calculate the speed of a DC motor is: N = 60×E b /(P×φ p ) Wh...