Soil Mechnaics t1 With Solutions 2nd Sem 2014-15

BIRLA INSTITUTE OF TECHNOLOGY AND SCIENCE, PILANI HYDERBAD CAMPUS

Second Semester 2014-‘15 TEST 1 (CLOSED BOOK) - SOLUTIONS

Course No: CE F243 Course Name: Soil Mechanics Date: 21.02.2015 Maximum Marks: 45 Max Duration: 60 Min. Component Weightage: 15%

PART – I

I. Answer all the following questions on the main answer script [20x0.5=10]

1. ___________________ is popularly known as Father of Soil Mechanics (Karl Terzaghi)

2. Intrusive igneous rocks will have _________________ structure (Crystalline)

3. Desilication can be classified as ___________________ type of weathering (Chemical)

4. Fill the following

Type of the Soil Genesis

Alluvial Soils Running water

Lacustrine Soils Deposited in quiet lakes

Marine Soils Deposited in sea water

Aeolian Soils Wind

Colluvial Soils Gravitational force, as in land slides

Glacial Soils Deposited by Glaciers

5. If maximum portion of shear strength of the soil is contributed by angle of internal

friction, then that soil could be classified as ____________ (Sandy soils or sand or

cohesion less soils)

6. Specific Gravity of Clay is always ____________ than the specific gravity of sands

(More)

7. The index, which is defined as a ratio (LL – NMC)/(LL-PL) where LL is Liquid Limit, NMC

is Natural Moisture Content of the soil, PL is Plastic Limit; is known as ______________

(Consistency Index)

8. The ratio between dry density achieved in the field and that obtained in the laboratory is

usually known as ______________________ (Relative Compaction Density)

9. Show pictorially the densest particle arrangement possible for spherical shaped particles

10. During the conduct of plate load test, what is the standard size of the plate used (75cm

dia)

11. Write the expansion for SPT (Standard Penetration Test)

12. The fourth group of clay miners apart from Illite, Kaolinite, Montmorrilonite is

______________ (playgorskite)

13. Clay particles suspended as colloids are controlled by _____________ energy(Surface

Energy)

14. Unsymmetrical distribution of _____________crystals in clays make them polar

(Silicate)

15. It was found that the CBR at 5mm penetration is greater than CBR at 2.5mm penetration

levels in all the three trials being attempted, CBR at _____mm shall be considered as

the design CBR. (5mm)

PART II

Answer the following questions briefly [5x4=20]

1. Compare the variation of dry density with water content for cohesion-less soil (no fines),

and a well graded soil sample. Justify the reasons for the variations in the curves

S.No Compaction Curve of Well graded soil

Compaction Curve of Cohesion less soil (no fines)

1 (0.5)

(0.5)

2 Dry density is a function of water

content (0.5)

Little influence of water content on dry

density(0.5)

3 The dry density is maximum at

Optimum water content and it

varies with soil type

The maximum dry density is achieved at

fully saturated condition. Max. Bulking at

4% - 5% water content

3 Dry density increases with

increase in water content dry of

Optimum because the air voids

get reduced and Attractive forces

due to surface charges of soil are

predominant (0.5)

Dry Density reduces initially due to Bulking

of soil particles (there is an increase in

volume of voids and corresponding

reduction of dry density) (0.5)

4 For water contents wet of

Optimum compaction does not

reduce the air voids appreciably,

but the volume of voids filled with

water increases due to which the

dry density reduces. Repulsive

forces due to surface tension of

water is predominant in the soil

structure (0.5)

As the water content is further increased,

the bulking effect is reduced and the

increase in compactive effort breaks the

capillary tension and hydrostatic forces

overcome the effective pressure between

the soil particles. (0.5)

2. Define Zero air voids line and state the expression relating dry unit weight with water

content and degree of saturation. Compare the 10% air content line and 90% saturation

line using a suitable illustration

Zero Air Voids Line:

The theoretical curve which represents the fully saturated condition (S = 100 %), gives

the relation between water content and dry density for no air voids. It cannot be reached

by compaction (1)

Derivation: (2)

Wע d = Gע

(1+e)

Wע d = Gע

(1+Gw/S) Where G – Specific Gravity

e- Voids ratio w- water content S – Degree of Saturation

3. Discuss various states of the soil (with respect to consistency and volumetric changes)

with varied levels of water contents using a suitable illustration

Zero Air voids line

(0.5)

80% Saturation line

or 20% Air content

line (0.5)

4. Draw a neat diagram of dispersed structure and the usual structure of Montmorrilonite

minerals

5. Define briefly and write formulae for bulk density, dry density, density of solids, saturated

density and submerged density of soils.

PART III

Solve the following Numerical

1. A core cutter, 12.6 cm in height and 10.2 cm in diameter weighs 1071g when empty. It is

used to determine the in-situ unit weight of an embankment. The weight of the core

cutter full of soil is 2970g. If the moisture content is 6%, what are the in-situ dry weight

and porosity? If the embankment gets fully saturated due to heavy rains, what will be the

increase in water content and the bulk unit weight, if no volume change occurs. The

specific gravity of the soil solids is 2.69. [7.5]

2. Classify the composite soil having soil fractions as follows using textural classification

system. The Chart is appended with the paper [4.5]

Soil fraction with sizes ranging from 2mm to 4.75mm = 200gm

Soil fraction passing 0.05mm sieve and retained on 0.002mm sieve = 400gm

Soil fraction collected in the pan placed after 0.002mm sieve = 400gm

3. In the plasticity chart given below, write down the classification of the soil grouped under

zones 1 through 6. [3]