Properties of Rock
Examination 2020
Questionnaire
Lecture 1
1. Define the geological classification of rock.
Solution:
Geologists classify rocks according to how they are formed. The geological classification is
therefore a genetic one. Rocks are usually separated into three groups:
i.
ii.
iii.
Sedimentary rocks: rocks formed by the accumulation, compaction and cementation of
pieces of other rocks and possible organic debris;
Igneous rocks: rocks formed by the solidification of a hot molten rock called magma.
They consist of a complex crystalline assemblage of minerals such as quartz, feldspars,
micas, pyroxenes, amphibole and olivine;
Metamorphic rocks: rocks that are formed by transformation of existing rocks by the
action of temperature and pressure.
2. What is the objective of rock mechanical engineering.
Solution
For most engineering projects involving rocks, the objectives of rock mechanics are
essentially of three fold:
(i) Determine the properties of the rock and the rock mass associated with the project of
interest.
These properties may be physical, mechanical, hydraulic or thermal. Not all properties need
to be determined but only those that are deemed necessary. In addition to these properties,
the in situ stress field needs to be measured as well. The intact rock and rock mass
properties are usually determined in the laboratory and in the field, respectively.
(ii) Model and predict the behavior of the rock mass when subjected to the new loads
associated with the engineering structure to be built.
(iii) Finally, once the engineering structure is built and upon its completion, the third
objective is to observe and monitor its response and behavior with adequate
instrumentation.
Lecture 2
1. Please describe the mechanical weathering of rock
Solution:
Mechanical weathering causes disintegration of rocks into smaller pieces by exfoliation or
decrepitation (slaking). The chemical composition of the parent rock is not or is only slightly
altered. Mechanical weathering can result from the action of agents such as frost action, salt
crystallization, temperature changes(freezing and thawing), moisture changes(cycles of
wetting and drying), wind, glaciers, streams, unloading of rock masses (sheet jointing), and
biogenic processes (plants, animals, etc.). For instance, mechanical weathering is very active
in high mountains with cold climates (see Figure 1). The 9% increase in volume associated
with the transformation of water into ice as the temperature drops below 0°C can create
pressures large enough to crack rocks. A good example of this type of process can be found
in the Niagara Falls area where large blocks of dolomite detach from the rest of the rock mass
in the Spring and Summer seasons.
2. Please describe the chemical weathering of Rock
Solution:
This type of weathering creates new mineralsin place of the onesit destroysin the parent rock.
As rocks are exposed to atmospheric conditions at or near the ground surface, they react with
components of the atmosphere to form newminerals.Themost important atmospheric
reactants are oxygen, carbon dioxide, and water. In polluted air, other reactants are available
(acid rain problems associated with the release of sulfuric acid from coal-fired power plants,
sulfur dioxide and smoke emissions, nitrogen oxides from vehicle exhaust). Table 1 gives a
list of weathering reactions that have been recognized. In general, chemical weathering
reaction are exothermic and cause volume increases.
Solution is a reaction whereby a mineral completely dissolves during weathering. This type
of reaction depends on the solubility of the rock minerals. For instance, evaporite minerals
(salt, gypsum) dissolve quickly in water, whereas carbonate minerals are somewhat less
soluble.
Hydrolysisis the reaction between acidic weathering solutions and many of the silicate
minerals. Feldspars are transformed by hydrolysis as they react with hydrogen ions to form
various products including clay minerals. This phenomenon is responsible for the degradation
of granite and other plutonic rocks to a material that resembles more of a dense soil than a
rock. The disintegrated granite called grus, saprolite, or spheroidal granite consists of rounded
blocks surrounded by a mixture of detrital clays and resistant grains of quartz.
Hydration corresponds to the penetration of water into the lattice structure of minerals. A
good example is the hydration of anhydrite into gypsum which is often accompanied with large
volume increases and substantial swelling pressures.
Oxidation corresponds to the reaction of free oxygen with metallic elements. This reaction is
familiar to everyone as rust. In an oxidation reaction, the iron atoms contained in the minerals
lose one or more electrons each and then precipitate as different minerals or amorphous
substances. An example of oxidation reaction is the transformation of pyrite (FeS2) into iron
hydroxides that liberates sulfuric acid.
Lecture 4
1. What is the transection grade in Rock
Solution
The transection grade indicates the number of joints (n) related to a measuring distance, a
measuring area or a measuring volume therefore linear, areal and spatial transection grade:
(n / m, n / m2, n / m3)
Important is the formation of joints




Joint filling
Wall condition
Grade of joints opening or healing
Spatial position of joints relating to working direction
2. What kind of methods used in determination of the hardness in non-jointed rock?
Solution
Methods for non-jointed rock
 Single Axial Compressive Strength Test (UCS)
 Point load test
 Tensile Load Test (Tensile Strength)
3. Please describe the point load test and mention its advantages.
Solution:





Simple and fast test procedure, high number of samples possible
Relatively lightweight measuring device (25 kg) and therefore applicable on site
Easy consideration of stratifications, foliation and anisotropy
Indirect method for the determination of uniaxial compressive strengths
Meanwhile a large number of modifications to adapt to special cases
Lecture 5
1. Please explain about the Rock Mass Rating (RMR) Bienawski 1973.
Solution:
Six important parameter for determination of RMR
A.1 Uniaxial strength (UCS)
A.2 RQD (Rock Quality Designation)
A.3 Distance between joints
A.4 Joint situation
A.5 Ground water state
F Orientation of joints
RMR=A1-5+F
2. Please explain Rock Classification belonging to Q-System
Solution:
RQD - value
Jn - Joint set index
Jr - Roughness index of unfavourable divison surfaces
Ja - Weathering grade of joint
Jw - Joint water
SRF - Stress reduction factor
3. Please explain how to obtain RQD value?
Solution
RQD = [Ʃ Length of core pieces > 10 cm length / Total Length of core run] x100%
Lecture 6
1. Please explain about the Unified Soil Classification System (USCS)?
Solution:
The Unified Soil Classification System (USCS) is a soil classification system used in engineering
and geology to describe the texture and grain size of a soil. The classification system can be
applied to most unconsolidated materials, and is represented by a two-letter symbol. Each letter
is described below (with the exception of Pt):
2. What is the grading characteristics?
Solution:
3. Please explain about the consistency of the soil.
Solution
The consistency of a fine-grained soil refers to its firmness, and it varies with the water content
of the soil. A gradual increase in water content causes the soil to change from solid to semisolid to plastic to liquid states. The water contents at which the consistency changes from one
state to the other are called consistency limits (or Atterberg limits).
The three limits are known as the shrinkage limit (WS), plastic limit (WP), and liquid limit (WL)
as shown. The values of these limits can be obtained from laboratory tests.
The difference between the liquid limit and the plastic limit is known as the plasticity
index (IP), and it is in this range of water content that the soil has a plastic consistency. The
consistency of most soils in the field will be plastic or semi-solid.
Lecture 7
1. Please explain about the density determination of Soil.
Solution:
cohesive soils  Immersion Weighing generally possible because undisturbed samples can be
taken
for non-cohesive soils  density determination on site according to DIN 18125 (next slides) 
a. Sand replacement method: Excavation of a pit  Filling the pit with test sand (bulk
density known)  Weighing the mass before and after filling
b. Balloon method: Determination of the volume of the pit by means of a rubber
balloon, which is filled with water


in both processes, the excavated material is filled into airtight buckets
with excavation mass + water content + volume  density can be determined
2. Please describe how lime determination conducted?


Dripping with hydrochloric acid (semi-quantitative is generally sufficient)

<1%

1 to 2%
less calcareous soil weak effervescence

2 to 5%
calcareous soils significant effervescence

> 5% strongly calcareous soils strong effervescence
lime-free soils, no effervescence
quantitative results by experimental apparatus according to Scheibler (DIN 18129)

the amount of CO2 produced is measured

Calculation of the amount of total carbonate
3. Please explain about the Grain Size Distribution?
Solution:
Screening by test sieves with defined standard sieve set,
4 to 63 mm  square hole
0.063 to 2 mm  mesh
Standard sieve set (mesh sizes)
63 – 31,5 – 16 – 8 – 4 – 2 – 1 – 0,5 – 0,25 – 0,125 – 0,063 mm
Grain Sizes < 0,063 mm  Sedimentation analysis
•
Slurry to a suspension
•
Coarse fractions sink faster than the fines
•
Suspension changes the density over the time
Measurement via a areometer (hydrometer)
Lecture 8
1. What is the source of stress in the ground?
Solution
•
•
Geostatic stress

Due to the weight of soil above and around

Naturally occurring

Affected by geologic and human activities
Induced stresses

Caused by external loads and the surface ground
o
Building and any construction
o
Road
o
Dam
2. What is the principle stresses in Mohr’s Circle?
Solution:
additional explanation is required to construct Mohr’s circle and computing the
principle stresses from Mohr’s circle.
Lecture 9
1. Please explain about the Boussinesq formula for the point loads
Lecture 10
1. What is the hydraulic gradient of soil?
Solution:
When water flows through a saturated soil mass there is certain resistance for the flow because
of the presence of solid matter. However, the laws of fluid mechanics which are applicable for the
flow of fluids through pipes are also applicable to flow of water through soils. As per Bernoulli's
fluids through pipes are also applicable to flow of water through soils. As per Bernoulli's equation,
the total head at any point in water under steady flow condition may be expressed as:
Total head = pressure head + velocity head + elevation head
2. Please explain about Darcy’s law
Solution:
Darcy in 1856 derived an empirical formula for the behavior of flow through saturated soils. He
found that the quantity of water q per sec flowing through a cross-sectional area of soil under
hydraulic gradient i can be expressed by the formula
3. Please explain about the Constant head permeability test
Solution: Additional explanation is required other than written!
The constant head permeameter test is more suited for coarse grained soils such as gravelly
sand and coarse and medium sand.
4. Please describe direct determination of hydraulic conductivity of soils in place by
pumping test and explain one of the methods.
A well sunk into a water bearing stratum, termed an aquifer, and tapping free flowing ground
water having a free ground water table under atmospheric pressure, is termed a gravity or
unconfined well. A well sunk into an aquifer where the ground water flow is confined between
two impermeable soil layers, and is under pressure greater than atmospheric, is termed as
artesian or confined well.
Lecture 11
1. What is the seepage force or pressure?
The interaction between soils and percolating water has an important influence on:
1. The design of foundations and earth slopes,
2. The quantity of water that will be lost by percolation through a dam or its subsoil.
Foundation failures due to 'piping' are quite common. Piping is a phenomenon by which the soil
on the downstream sides of some hydraulic structures get lifted up due to excess pressure of
water. The pressure that is exerted on the soil due to the seepage of water is called the seepage
force or pressure. In the stability of slopes, the seepage force is a very important factor. Shear
strengths of soils are reduced due to the development of neutral stress or pore pressures. A
detailed understanding of the hydraulic conditions is therefore essential for a satisfactory design
of structures.
2. What is the Flownet and how to construct them?
A flow net for an isometric medium is a network of flow lines and equipotential lines intersecting
at right angles to each other.
The path which a particle of water follows in its course of seepage through a saturated soil mass
is called a flow line.
Equipotential lines are lines that intersect the flow lines at right angles. At all points along an
equipotential line, the water would rise in piezometric tubes to the same elevation known as the
piezometric head . Fig. 4.16 gives a typical example of a flow net for the flow below a sheet pile
wall. The head of water on the upstream side of the sheet pile is ht and on the downstream side
hd. The head lost as the water flows from the upstream to the downstream side
The properties of a flow net can be expressed as given below:
1. Flow and equipotential lines are smooth curves.
2. Flow lines and equipotential lines meet at right angles to each other
3. No two flow lines cross each other.
4. No two flow or equipotential lines start from the same point
3. Please explain about the Laplace’s equation
Lecture 12
1. What is the effective stress and pore water pressure?
Solution:
The pressure transmitted through grain to grain at the contact points through a soil mass is
termed as intergranular or effective pressure. It is known as effective pressure since this
pressure is responsible for the decrease in the void ratio or increase in the frictional resistance
of a soil mass.
If the pores of a soil mass are filled with water and if a pressure induced into the pore water,
tries to separate the grains, this pressure is termed as pore water pressure or neutral stress.
2. Explain about the capillary rise of water in soil.
Solution:
In contrast to capillary tubes the continuous voids in soils have a variable width. They
communicate with each other in all directions and constitute an intricate network of voids.
When water rises into the network from below, the lower part of the network becomes
completely saturated. In the upper part, however, the water occupies only the narrowest
voids and the wider areas remain filled with air.
3. What is the surface tension of water?
Solution:
Surface tension could be defined as the property of the surface of a liquid that allows it to
resist an external force, due to the cohesive nature of the water molecules.
Surface tension is a force that exists at the surface of the meniscus. Along the line of
contact between the meniscus in a tube and the walls of the tube itself, the surface tension,
Ts, is expressed as the force per unit length acting in the direction of the tangent as shown in
Fig. 5.7(a). The components of this force along the wall and perpendicular to the wall
Lecture 13
1. Please describe the total compression of the saturated clay.
Solution:
The total compression of a saturated clay strata under excess effective pressure may be
considered as the sum of
1. Immediate compression, This portion of the settlement of a structure which occurs more
or less simultaneously with the applied loads is referred to as the initial or immediate
settlement. This settlement is due to the immediate compression of the soil layer under
undrained condition and is calculated by assuming the soil mass to behave as an elastic
soil.
2. Primary consolidation, This portion of the settlement that is due to the primary
consolidation is called primary consolidation settlement or compression. At the present
time the only theory of practical value for estimating time-dependent settlement due to
volume changes, that is under primary consolidation is the one-dimensional theory.
3. Secondary compression. This compression is supposed to start after the primary
consolidation ceases, that is after the excess pore water pressure approaches zero. It is
often assumed that secondary compression proceeds linearly with the logarithm of time.
However, a satisfactory treatment of this phenomenon has not been formulated for
computing settlement under this category
2. What is the normally consolidated and over consolidated clay
Solution:
A clay is said to be normally consolidated if the present effective overburden pressure po is the
maximum pressure to which the layer has ever been subjected at any time in its history, whereas
a clay layer is said to be overconsolidated if the layer was subjected at one time in its history to
a greater effective overburden pressure, pc, than the present pressure, po. The ratio pc/po is called
the overconsolidation ratio (OCR).
3. Please explain about one-dimensional consolidation test The Standard Onedimensional Consolidation Test
Solution:
The main purpose of the consolidation test on soil samples is to obtain the necessary
information about the compressibility properties of a saturated soil for use in determining the
magnitude and rate of settlement of structures.
Loads are applied in steps in such a way that the successive load intensity, p, is twice the
preceding one. The load intensities commonly used being 1/4, 1/2,1, 2,4, 8, and 16 tons/ft2 (25,
50, 100,200,400, 800 and 1600 kN/m2). Each load is allowed to stand until compression has
practically ceased (no longer than 24 hours). The dial readings are taken at elapsed times of
1/4, 1/2, 1,2,4, 8, 15, 30, 60, 120, 240, 480 and 1440 minutes from the time the new increment
of load is put on the sample (or at elpased times as per requirements).
4. Please explain graphical method to determine preconsolidation pressure?
The earliest and the most widely used method was the one proposed by Casagrande (1936).
The method involves locating the point of maximum curvature, 5, on the laboratory e-log p curve
of an undisturbed sample as shown in Fig. 7.8. From B, a tangent is drawn to the curve and a
horizontal line is also constructed. The angle between these two lines is then bisected. The
abscissa of the point of intersection of this bisector with the upward extension of the inclined
straight part corresponds to the preconsolidation pressure pc,