MINISTRY OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF
TECHNICAL AND VOCATIONAL EDUCATION
Sample Questions & Worked Out Examples
For
Min 05037
GEOTECHNICAL ENGINEERING
B.E.
Mining Engineering
YANGON TECHNOLOGICAL UNIVERSITY
DEPARTMENT OF MINING ENGINEERING
Geotechnical Engineering
(Part I)
MIN. 05037
BE (Mining)
QUESTIONS AND ANSWERS
16 SEPTEMBER, 2003
YANGON
1
(QUESTIONS)
CHAPTER 1
STRESS AND STRAIN ANALYSIS
1.* Derive the mathematical formulations of the stresses in three dimensional stress
condition for the followings. (10 marks each)
(i) Stress invariants
(ii) Deviatoric stress invariants
2.* Explain the components of stress for the moments of the forces acting on a cube
element. (10 marks)
3.* Explain the components of a strain for small displacement of particles of a
deformed body and derive the relationship of strain and displacement for an
equilibrium condition for the analyses of engineering structure. (20 marks each)
4.* According to Hooke’s law, derive linear relationship of the components of stress
and strain for three dimensional stress-strain analyses. (10 marks each)
5. ** Explain as much as you know the mathematical formulations for the stress-strain
relationships of plain strain and plain stress analyses. Derive the D-matrices for
plain strain and plain stress analyses. (10 marks each)
6. ** Derive the linear elastic constitutive equation of isotropic material for the
followings. (10 marks each)
(i) Plane strain modeling
(ii) 3-D modeling
7.*** Derive and discuss as much as you know the mathematical formulation of the
condition for the compatibility of stress distribution to determine the state of stress
in a body submitted to the action of given forces. (10 marks each)
8. *** From the following diagrams, determine the principal stresses (σ1, σ2) and
shear, τmax, max stress by graphical method. (20 marks)
2
Given: The stress components σx, σy, τxy are known for any two perpendicular
planes. The points D and D1 represent the stress conditions on the two
coordinate planes.
τ
σy
D1
C
B
O
F
F1
σ
A
(a)
2α
b
a
D
σx
d
c
τ
x
σy
D1
P
σ2
τXY
σY
B
τXY
O
C
B
σ
σx
σX
D
C
τ
σ
α
N
σ1
(b)
y
(c)
CHAPTER 2
THEORY OF PLASTICITY
9.* Explain the basic assumptions of mechanical behaviors of materials in
geotechnical engineering purposes. (20 marks)
10.* Explain in detail the post failure behaviors of rock material with necessary
diagrams. (20 marks)
3
11.* Describe the ingredients in the determination of rock mass properties beyond
inelastic range of rock. Discuss briefly the basic concept of plastic property of
rocks. (10 marks each)
12.* Explain as much as you know the ingredients of plastic theory, which determines
the properties of material in beyond the elastic range. (20 marks)
13.* Derive the mathematical expressions of current values of yield function and
current values of yield stress of Hoek and Brown yield criterion and MohrCoulomb yield criterion for the numerical nonlinear elastic-plastic solution
processes. (20 marks)
14.** Describe the hardening function of a material, which determines the yield
function changes due to plastic straining and explain the different types of strain
hardening models. (20 marks)
15. ** Determine the hardening parameter, H, for an elasto-plastic material from
uniaxial test result. (10 marks)
16. *** Discuss in detail, the importance of most suitable yield criteria to obtain the
realistic results when using in numerical computation. (20 marks)
17. *** Discuss, with adequate diagrams, the effects of dilatancy on strength of rock
mass during the plastic flow. (20 marks)
18. *** Discuss as much as you know with necessary diagrams the connection
between the flow rule and dilatancy during plastic flow. (20 marks)
19.*** Discuss in detail the basic concept of Hoek and Brown yield criteria and give
the reasons for using this formula to obtain the realistic results for rock failure
process. (20 marks)
CHAPTER 3
NUMERICAL ANALYSIS
20.* Describe the basic concept of continuum mechanics and explain how to
understand the behavior of rock mass if it is applied as the continuum nature. (10
marks)
4
21.* Explain the solution procedure for solving the simultaneous equation with matrix
equation. (10 marks)
22.* The springs are applied at both ends by compression and tension forces as case 1
and case2 shown in following Figures. If axial loads of F1, 25 KN and F2 25 KN
are applied at both ends for case 1 and case 2 respectively as shown, determine the
displacements u1 and u2 at nodes A and B. (20 marks)
Given: k1=4200 KN/m and k2=7500 KN/m.
F1
u1
u2
k1
A A'
F2
B
B'
(a)
F1
u1
A'
u2
k2
A
B
F2
B′
Case 1
Equal Compression
Forces applied at both
ends A and B.
Case 2
Equal Tension Forces
applied at both ends A
and B.
(b)
23.* Explain or define the following finite element basics. (10 marks each)
(i) Elements
(ii) Shape function
24.* Derive the formulation for the coordinate transformation of an element from
local coordinates to global coordinates. (20 marks)
25.* Explain briefly the followings used in numerical computation. (10 marks each)
(i) Strain-displacement relationship (To define the strain matrix B)
(ii) Stress-strain relationship (To define the elastic D matrix)
26.* Derive the nodal force-displacement relationship for an element by using the
principle of virtual work done. (10 marks)
5
27. ** Explain the followings. (10 marks each)
(i) Body force
(ii) Surface tractions
28. ** Explain in detail the procedure for transformation from local stiffness matrix to
overall (global) stiffness matrix for simultaneous equations. (20 marks)
29. *** In most of the geotechnical engineering problems, the initial stress is prime
important to be considered. As the analyst for solving a geotechnical problem,
discuss how to specify the initial stress prior to analysis for a problem. (20 marks)
30.*** Explain and define equivalent nodal forces of the following loadings for
solving the geotechnical problems. (10 marks each)
(i) Gravity loading
(ii) Distributed loading
6
SAMPLE QUESTIONS AND ANSWER
1. Question
No (1)* Derive the mathematical formulations of the stresses in three dimensional
stress condition for the followings. (Chapter 1)
(i) Stress invariants,
(ii) Deviatoric stress invariants
Answer
(i) Stress invariants
The comments of stress depend on the chosen direction of the coordinate axes.
The principle stresses on the hand are invariants. Their values are affected by the
choice of reference axes. In general, three independents stress invariants may be
defined. For isotropic materials, the stress can usually be described in terms of
invariants.
n
Y
σn
σz
τx
σx
τzx
τzy
τy
σn
σn
X
τyz
τyx
X
σy
Z
Figure 1: Three Dimensional Stress System
When stress components are specified at a point (σij), the stress components
on any plane passing through this may be determined. Using static principles alone,
σij = the stresses in a plane whose unit normal is ‘n’ or components of ‘n’.
For equilibrium, resolve the forces in the 3-directions of coordinates giving
three equations:
σnx = l1σx
+ l2 τxy +
l3 τxz
σny = l1τyx + l2 σy + l3 τyz
7
σnz = l1τzx
+ l2 τzy + l3 σz
or matrix notation,
σnx
σny
=
σnz
σx
τxy
τxz
l1
τyx
σy
τyz
l2
τzx
τzy
σz
l3
----------
(a)
----------
(b)
Equation (a) may be simplified as follow:
σnj = σij lj
Where, lj are direction cosine of ‘n’ from equation (a). It is obvious that it is
possible to find a plane that consists only of normal stresses (maximum or minimum)
giving zero shear stresses. Such a plane is called principal plane and its corresponding
stresses are known as principal stresses.
Assuming that there is such a unique principal plane, it is true that
σnj = σ lj
----------
(c)
Where ‘σ’ is assumed to be the principal stress. Expending the notation given in
equation (c) in matrix form and substitute σnj from equation (a).
σnj = σ lj = 0 ⇒
σx
τxy
τxz
l1
τyx
σy
τyz
l2
τzx
τzy
σz
l3
l1
-σ
l2
=0
l3
or we can be further simplified,
(σx - σ) τxy
τxz
l1
τyx
(σy - σ)
τyz
l2
τzx
τzy
(σx - σ)
l3
=0
This matrix has non-trivial solution if its determinant is zero. Hence, if its
determinant is assigned to zero, the following characteristic is obtained.
σ3 – (σx + σy + σz) σ2 + (σxσy + σyσz + σxσz - τxy2 - τxz2 - τyz2) σ - (σxσyσz + 2τxy τxz
τyz + σzτxy2 - σyτxz2 - σxτyz2) = 0
Similarly,
Jl = σx + σy + σz = σx′ + σy′+ σz′
Jl = σxσy + σyσz + σxσz - τxy2 - τxz2 - τyz2
= σx′σy′ + σy′σz′ + σx′σz′ - τ′xy2 - τ′xz2 - τ′yz2
Jl = σxσyσz + 2τxy τxzτyz +
σzτxy2
-
σyτxz2
-
---------σxτyz2
(d)
8
= σx′ σy′σz′ + 2τ′xy τ′xz τ′yz + σz′τ′xy2 - σy′τ′xz2 - σx′τ′yz2
Jl, J2 and J3 are called invariants of the stress ‘σ’, which mean that their values
remain unchanged in any axis of reference and/or under any space transformation.
These invariants are further simplified if they are written in term of principal stresses,
shown below.
J1 = σ 1 + σ 2 + σ 3
J2 = σ1σ2 + σ2σ3 + σ3σ1
J3 = σ1σ2σ3
(ii) Deviatoric Stress
σ2
N′
σ2
N
σ′2
P′
P
σm
σ1 = σ2 = σ3
Q
O
σ3
σ1
Figure 2 Deviatoric Stress Components
Let σ1, σ2 and σ3 are axes of principal stresses thought a point ‘O’ of the material.
-
ON is the resultant stress at ‘O’ in some arbitrary direction.
-
Now ON may be represent by the sum of OP and PN.
The component OP is chosen to be equally inclined to all three axes. It direction
cosines will then be l1 = l2 = l3 = l/(3)½.
Consider the triangle ON’Q, the component of σ2 along OP is OQ (i.e., OQ =
σ2 cos β = σ2/(3)½), repeating this for σ1 and σ3 it can be seen that the length:
OP = (σ1 + σ2 + σ3)/(3)½
OP can be resolved along each of the coordinate axes, its component in the σ2
direction is 1/(3)½OP or = (σ1 + σ2 + σ3)/3 and similarly, its component along the
other two axes will be (σ1 + σ2 + σ3)/3 = σm is called the hydrostatic stress.
9
Similarly, PN may be resolved along principal axis, for the σ2 – axis the
distance σ′2
=
σ2 - σm being known as the deviatoric stress components. Thus the
deviatoric stress components are:
σ′1 = σ1 - (σ1 + σ2 + σ3)/3 = (2σ1 - σ2 - σ3)/3
σ′2 = σ2 - (σ1 + σ2 + σ3)/3 = (2σ2 - σ3 - σ1)/3
----------
(10)
σ′3 = σ3 - (σ1 + σ2 + σ3)/3 = (2σ3 - σ1 - σ2)/3
The distance PN is the vector sum of σ′1, σ′2 and σ′3 = (σ′12 + σ′12 + σ′32)½ and
is the resultant deviatoric stress.
J′1 = σ′1 + σ′2 + σ′3 = σii = 0
J′2 = ½ (σ′ii σ′ij)
= 1/6 (σ1 - σ2)2 + (σ2 - σ3)2 (σ3 - σ1)
----------
(11)
J′2 = 1/3 (σ′ij σ′ik σ′ki)
2(σ1 + σ2 + σ3)2 – 9(σ1σ2 + σ2σ3 = σ3σ1)
= σ1σ2σ3 + 1/27 (σ1 + σ2 + σ3)
.............................................................................................
2. Question
No (2)* Explain the components of stress for the moments of the forces acting on a
cube element. (Chapter 1)
σz
Y
σy
τzx
τxz
σx
τzy
τyz
σy
Z
τxy τyx
σx
σz
X
y
X
Z
Figure 2 Elementary Cube Element
The pair of parallel sides of a cube element can be seen in the above figure. In
this condition, the normal component of stress and two components of shearing
stresses.
10
To describe the stresses acting on the six sides of the element three symbols
σx, σy, σz are necessary for normal stresses; and also six symbols τxy, τyx, τxz,τzx, τyz,
τzy for six shearing stress.
In above figure, for a rectangular system of coordinates:
- The Y–axis is coincident with σN,
- The Z and X-axes are two shear stress components
At equilibrium condition, nine values of stress components are necessary for
the stresses in the body. For equilibrium condition of rotation of the element, τxz = τzx,
τxy = τyx, τyz = τzy. Stresses will be determined (i.e., σx, σy, σz, τxy, τxz, τyz).
In a 3-D elements volume of a continuum, it is easy to see from Figure that
can be written in a matrix from as:
σx
τxy
τxz
τyx
σy
τyz
τzx
τzy
σz
In tensor notation, it is convenient to write the above matrix as follows;
σx
τxy
τxz
τyx
σy
τyz
τzx
τzy
σz
=
σ11
σ12
σ13
σ21
σ22
σ23
σ31
σ32
σ33
It can be written as σij in short form. This σij is a tensor of degree 2 (i.e., has
components of 32 or i, j=1 - 3 in 3D space) when the indices of a tenser consist of a
repeated index such as σij, the summation is implicitly assumed, thus,
σij = σ31 + σ32 + σ33
The Kronecker delta (δ) is a special matrix denoted as δij. Such matrix consists
of a components but it has a property that δij will take a value of 1 when I=j or zero
otherwise, i.e.,
δij
1
for
i=j
0
for
i≠j
=
and thus, similarly δii means δ11 + δ22 + δ33 = 3
.............................................................................................
11
3. Question
No (6)** Derive the linear elastic constitutive equation of isotropic material for the
followings. (Chapter 1)
(i) Plane strain modeling
(ii) 3-D modeling
Answer:
When than one-D model is loaded as shown in figure below. The strain in the
z-direction may be obtained from the formula developed previously using:
εx = σz / E
However, if the volume (total volume) of the element is kept constant, there
will be other strains acting in the opposite sense in the other directions. The obvious
solutions would be to use proportional constants, such that,
εx = - υx εz = - υ σz / E
εy = - υY εz = - υ σz / E
σz
Z
X
Y
σz
Where η + η ≤ 1 to be physically possible and negative signs are used to
indicate that (εx, εy) and εz are acting in the opposite directions.
If the material is isotropic, it is reasonable to assume ηx = ηy = υ = Poisson’s
ratio. Combining all strains due to loadings of 3-different directions, the linear elastic
constitutive equation for an isotropic material can be obtained (i.e., the simultaneous
action of normal stresses σx, σy and σz uniformly distributed over 3-different sides of
material).
εx
εy
εz
σx
σx/E
-υσx/E
-υσx/E
σy
-υσy/E
σy/E
-υσy/E
σz
-υσz/E
-υσz/E
σz/E
12
This equation may be written in matrix form as follows:
ε = [D] σ
−1
or
σ = [D]ε
Where, [D] = linear elastic D – matrix
This equation is usually referred to as the constitutive relationship that
interconnects the σ and ε.
(i) Plane Strain Equation
For plane strain to exists γxy and γyz must be zero throughout the body and
variation of εx with respect to z must be zero (i.e., εz = 0). If εz is set to zero, the
results are as follows:
0 = [σz – υ (σx + σy]/E
σz = υ (σx + σy)
For plane strain with εz = 0 becomes
εx = 1/E [(1-υ2) σx – υ (1+υ)σy]
εx = 1/E [(1-υ2) σy – υ (1+υ)σx]
γxy = γxy/G = 2(1+υ) τxz/E
Therefore, constitutive equations become:
σ = [D]plane – strain ε
For linear elastic material the stress-strain or constitutive matrix D is given as
1 − ν
ν
0 


E


ν
[D]PLANE−STRAIN =
1− ν
0 

(1 + ν)(1 − 2ν )

1 − 2ν 
0
 0

2 

σ = [σx, σy, τxy]T
ε = [ εx, εy γxy]T
(ii) 3-D modeling
For linear material, the stress-strain matrix is given as:
13
(1 − ν)

 ν
E

[D] =
(1 − ν )(1 − 2ν)  0

 0

ν
0
(1 − ν)
ν
ν
(1 − ν)
0
0


0 

0 

1 − 2ν 
2 
0
----------
(21)
σ = [σx, σθ, σz, τxz ]T
ε = [ εx, εθ, εz γxz ]T
.............................................................................................
4. Question
No (8)*** From the following diagrams, determine the principal stresses (σ1, σ2) and
shear, τmax, max stress by graphical method. (Chapter 1)
Given: The stress components σx, σy, τxy are known for any two perpendicular
planes. The points D and D1 represent the stress conditions on the two coordinate
planes.
τ
σy
D1
C
B
O
F1
F
(a)
σ
A
2α
a
b
D
σx
d
c
τ
x
P
σy
τXY
σY
B
τXY
D1
σ2
O
A
C
B
σ
σ
σX
C
σx
τ
σ
D
y
σ1
(b)
(c)
α
N
14
Answer:
From diagram (a), for the planes perpendicular to the principal directions, the
points A and B is at the stresses of σx,σy. The stress components for any plane BC
with an angle α will be represented by coordinates of a point on the circle having AB
diameter (see diagram (c)). Let point D is the required point,
By using graphical method,
OF = OC + CF = ½ (σx + σy) + ½(σx - σy) sin 2α = σx cos2 α + σy sin2α
Or
σ = σx cos2 α + σy sin2α
(a)
DF = CD sin 2α = ½(σx - σy) sin 2α
τ = ½(σx - σy) sin 2α
(b)
The plane BC rotates about an axis perpendicular to the xy plane (see diagram
(c)). Angle, α varies from 0 to π/2.
The maximum shear stress, τmax is given in the diagram (a) by the maximum
ordinate of the circle or the radius of the circle.
τmax = ½(σx - σy)
(c)
If the stress components σx, σy, τxy are known for any two perpendicular
planes. From the diagram (a) and (b), the maximum and minimum principal stresses
(σ1, σ2) can be determined.
σ1 = OC + CD = ½ (σx + σy) + (½ (σx - σy))2 + τxy2) ½
(a’)
σ2 = OC – CD = ½ (σx + σy) - (½ (σx - σy))2 + τxy2) ½
(b’)
The maximum shearing stress is given by the radius of the circle;
τmax = ½(σx - σy) = (½ (σx - σy))2 + τxy2) ½
(c’)
.............................................................................................
5. Question
No (10)* Explain in detail the post failure behaviors of rock material with necessary
diagrams. (Chapter 2)
Answer:
When the rock material is the low strength compared with the high stress, the
rock material is deformed after initial yielding. Thus, the stability of material depends
on the deformational behavior of rock material.
15
In simplifying the post failure behavior of the rock for engineering viewpoint,
there are four main types of post failure behavior of rock as shown in following
Figure, namely, strain hardening, strain softening, perfectly plastic and perfectly
brittle behaviors.
The post failure behavior of rock is depended on the confining pressure, rate
of loading, rheological characteristics.
Strain softening: is one in which once the strength reaches its peak strength it
gradually decreases toward a residual value as total strain increases.
Strain hardening: the bearing capacity of rock increases with increasing total
strain.
Perfectly plastic: is a transitional mode between two but the post failure strength
of rock is not a function of strain.
Perfectly brittle: has no effectiveness on plastic behavior and can not be
Stress, σ
controlled after initial yielding.
Strain hardening
Initial
yielding
Perfectly plastic
σy
Strain softening
Perfectly
brittle
Strain, ε
εe
εvp
Figure 1 Post-failure Behavior of Rock
Stress
Y0
O
A
P
ε = ε-εy
Y′
εy
ε
Strain
Figure 2 Uniaxial stress-strain relationship of perfectly plastic material
16
Stress
B
Y′
A
Y0
P
ε = Permanent plastic strain
Recoverable elastic strain
O
εy
εy′
Strain
ε
P
Figure 3 Uniaxial stress-strain relationship of strain hardening material
..................................................................................................
6. Question
No (11)* Describe the ingredients in the determination of rock mass properties
beyond inelastic range of rock. Discuss briefly the basic concept of plastic property of
rocks. (Chapter 2)
Answer:
(a) There are three ingredients to the elasto-plastic stress-strain laws.
(i)
Yield function
(ii)
Hardening law
(iii)
Flow rule
The yield function which signals if the material is yielding plastically or not.
The hardening function indicates the manner in which the yield function changes (if at
all) due to plastic straining. Flow rule determines the direction of plastic straining.
(i) Yield Function
Yield function (F) is a scalar function of stress (stress components, principal
stress strain invariants) which indicates the onset of plastic strains and can be
symbolically written as
F (σ) = 0
………..
(a)
The behavior F (σ) < 0 implies elastic behavior and F(σ) > 0 is an impossible
situation.
17
σ3
σ3
F (σ, H) < 0;
Elastic (Yield)
domain
Yield surface, F (σ) = 0
F < 0;
Elastic
domain
F > 0; impossible stress
situation
σ2
σ2
F (σ, H) = 0;
Yield surface
Figure
(1)
σ1
σ1
Representation of the yield surface in stress space
Equation (a) represents a surface in one dimensional stress space. The space
enclosed by it is the elastic domain. For multi-axial behavior, the yield function can
be symbolically written as,
F (σ , H) = 0
………
(b)
Where, F = some function of stresses
H = Material parameter which is actually a hardening parameter
The basic assumptions of yield function are shown in figure (1). For multidimensional case,
When F (σ , H) = 0, the stress level lies on the yield surface and material
behavior is elasto-plastic. In this case, strain is partially recoverable, partially
permanent. When F (σ, H) > 0, the stresses are not allowed elasto-plastic; however, it
is allowed elasto-viscoplastic
(ii) Hardening law
The hardening law indicates how the threshold of yielding changes with the
plastic strain – the material hardens.
Strain components, principal strains, or plastic strain invariants. It is therefore more
generally written as
F (σ, εP) = 0
F (σ, H)
=0
or
(a)
(b)
Where H represents a scalar function of plastic strains called a hardening parameter.
Here again, F (σ, H) < 0, implies elastic response and a situation F (σ, H) >0 is
impossible.
18
P
Yield stress, Y
Y = f (ε )
Y0
P
Plastic strain, (ε )
Figure (2) Representation of hardening law in uniaxial condition
(iii) Flow rule
In the uniaxial examples, the direction of plastic strains is obvious that plastic
strains took place in the same direction as the stress. In the multi-axial case as the
stress and strains have in general six components and it is needed to specify the
direction of plastic straining at every stress state by what is called a flow rule. It can
be considered an equation of the following form:
∂F(σ, H)
d ε pij = dλ
∂σij
( )
………….
(a)
Where, d (εpij) = the components of the increments of plastic strain,
dλ = the plastic multiplier,
F (σ, H) = yield function
Equation (a) may be expressed graphically. This is done in figure (2) where
the vector normal to the yield surface has components which are the plastic strain
increment components. The stress axes become plastic strain increment axes for
representing these. For the purpose of graphical presentation the surface is shown in a
principal stress space.
The vector is normal to the ‘yield surface’ the normality condition is said it
applies. As the direction of plastic strains is associated with the current yield surface
the flow rule represented by equation (a) is also called an associated flow rule. The
vector representing plastic strain increments is referred to as flow rule vector.
∂F(σ, H)
d ε pij = dλ
………….
(b)
∂σij
Where Q (σ, H) ≠ F (σ,H)
( )
19
Here Q (σ, H) represents a ‘plastic potential’ function which is different from
yield function. The normality is not associated with the yield surface but with the
plastic potential surface Q, (See figure 2 b).
σ3
σ3
Current Yield stress
situation
Current Yield
stress situation
Yield surface and Plastic
potential surface
σ2
σ2
Yield surface
σ1
σ1
(a)
(b)
Figure (2) Associated (a), and non-associated (b), flow rule and normality condition
......................................................................................................
7. Question
No (13)* Derive the mathematical expressions of current values of yield function and
current values of yield stress of Hoek and Brown yield criterion and Mohr-Coulomb
yield criterion for the numerical nonlinear elastic-plastic solution processes. (Chapter
2)
Answer:
(a) Hoek and Brown yield criterion
Hoek and Brown (1980) proposed the empirical failure criterion and when
σ1>σ3, σ1 = σ3 + (m σc σ3 + s σc2)½
………….
(a)
Where, σc = uniaxial compressive strength of intact rock
Rewrite the equation (a) by invariants,
Equation (a) may be rewritten in terms of mean stress and stress invariant by
substituting the values of σ1 and σ3.
20
 mσ (σ ' ) 12

σ mσ
'
2
F = 2(σ J2 ) cosθ −  c J2
sinθ − 3cosθ − J1 c + sσ c2 
3
3


(
1
)
1
2
 mσ (σ ' ) 12

'
2
c
J2
2
sinθ − 3cosθ − mσ c σ m + sσ c 
F = 2(σ J2 ) cosθ − 
3


(
1
)
=0
1
2
=0
The current value of yield function can be written as
4σ 'J2 cos 2 θ m(σ 'J2 )
F=
−
σc
3
1
2
(sinθ −
)
3cosθ + mσ m
……. (c)
The current value of yield stress or equivalent yield stress is
σ ey = sσ c
(b) Mohr–Coulomb Yield Criterion
The form of this yield criterion in terms of shear stress (τ) and effective
normal stress (σn) on the failure plane is given by:
τ| = σn tan φ + c
……. (a)
Where, c and φ are effective cohesion and angle of internal shearing friction
respectively
Alternative form of Mohr–Coulomb yield criterion can be expressed in terms
of effective principal stresses as
σ1 (1 – sinφ ) - σ3 (1 + sinφ ) – 2 c cosφ = 0
……..
(b)
Using the relationship between σ1, σ3 and the stress-components from the
geometry of the Mohr circle can be written for plane strain conditions as
F=
(σ x − σ y ) 2 + 4τ xy2 − (σx + σy) sin φ – 2 c cos φ = 0
(c)
In terms of σs , σd invariants, equation (2-12) becomes particular simple. It is
F = σd – 2 σs sinφ – 2 c cosφ = 0
(d)
This relationship follows immediately from equation (c) by introducing the definitions
of Sin σs and σd , or from equation (b). The current value of yield function is:
F = - σm sin φ +
Sin θ Sin φ 
9 
Cos θ −
 − c Cos φ
3 
3

(e)
The current value of yield stress or equivalent yield stress is:
σ ey = ccosφ
................................................................................
21
8. Question
No (25)* Explain briefly the followings used in numerical computation. (Chapter 3)
(i) Strain-displacement relationship for plane and axi-symmetric analysis (To
define the strain matrix B)
(ii) Stress-strain relationship (To define the elastic D matrix)
Answer:
(i) Strain-displacement relations
Let u and v be the x and y displacements which cause small strains εx, εy, and
γ. Then, restricting attention to two dimensions, we have:
Plane stress and strain
= - ∂u/∂x
(a)
εy = - ∂v/∂y
(b)
γ
(c)
εx
= - (∂u/∂y + ∂v/∂x)
(εz = 0 for plane strain)
Axi-symmetric
The relations are the same as equations (a-c) with x, u standing for radial and
y,v for axial. The hoop tangential strain εz is given by:
εz = -u/x
The minus sign is a consequence of our compression positive notation. The
strain equations are conveniently represented by a vector or column matrix), ε, having
3, 4, or 6 components accordingly as the problem is plane, axisymmetric, or threedimensional.
For finite element applications it is necessary to relate strain to the
displacements at element nodes. Let these be represented by a vector δe which in 2-D
is [u1, v1, u2, v2……un, vn]T. Using equation (a-c) to express u, v in terms of nodal
displacements one obtains:
ε = B δe
The matrix, B comprises a row of n (the number of nodes in the element) sub
matrices Bi which for plane problems take the form:
22
 ∂N i

 ∂x
Bi =  0

 ∂N
 i
 ∂y

0 

∂N i 
∂y 
∂N i 

∂x 
…………
(3.20a)

0 

∂N i 
∂y 
0 
∂N i 

∂x 
…………
(3.20b)
For axi-symmetric analysis,
 ∂N i

 ∂x
0
Bi =  N i
 x
 ∂N i

 ∂y
Note that for the axi-symmetric case the hoop strain, εz, is in the third position in ε.
(ii) Stress-Strain Relations
Generally these can be expressed:
∆σ = D ∆ε
Where σ contains the 3, 4, or 6 stress components corresponding to ε
‘∆’ indicates ‘change in’ which may or may not be small.
D is the corresponding square modulus matrix.
Its components are constant for linear (elastic) applications. For nonlinear they are deformation or stress dependent. D is usually symmetric.
.....................................................................................................
9. Question
No (26)* Derive the nodal force-displacement relationship for an element by using
the principle of virtual work done. (Chapter 3)
Answer:
For two-dimensional plane strain problems, the generalized displacement, δe
and nodal forces, Fe for an eight noded element can be expressed as:
δ(e) = [u1, v1, u2, v2 … u8, v8]T
F(e) = [Fx1, Fy1, Fx2, Fy2 ….Fx8, Fy8]T
23
Where, Fe = any external forces such as gravity loading, distributed edge loading,
temperature loading etc.
Stiffness
e
K
Nodal forces
e
F
Nodal displacements
e
δ
Figure (1) An Element
The principle of virtual work done will be used to calculate the stiffness of an
element. The external loads, F(e) are applied to the nodes and result the displacements,
δe. Let the internal stress at a point be σ and the internal strain corresponding be ε(e).
According to energy conservation, the external work done is equal to the internal
work done.
(δ (e) )T F(e) = ∫ (ε (e) )T σdv
(a)
v
From Equation ε(e) = [B] δ(e),
(ε(e))T = ([B] δ(e))T = BT (δ(e))T
Substituting the value of ε(e))T in Equation (a) becomes
F(e) = ∫ [B] σdv
T
v
........................................................................................................
10. Question
No (30)*** Explain and define equivalent nodal forces of the following loadings for
solving the geotechnical problems. (Chapter 3)
(i) Gravity loading
(ii) Distributed loading
Answer:
(i) Gravity loading
These forces, which are distributed over volume of elements, must be
converted to equivalent nodal forces. They will then form part or all right-hand side
vector R. Let Rb denote the body force contribution to R.
24
Denoting the body force intensity (force per unit volume) at x,y by the vector
p = (px,py)T, and using the virtual work principle with an asterisk to denote the virtual
displacements, we have:
[δ *]T R be = ∫ (u * p x + v * p y )dv
Using the virtual work principle to express u* and v* in terms of the nodal
values, and expanding the left-hand side:
n
n
n


be
+
=
+
R be
u
*
R
v
*
p
N
u
*
p
N i v *i dv

∑
∑
∑
IX
i
iy
i
x
i
i
y
∫
i =1
i =1
 i =1

(
)
Again, since the components of δ*, i.e. u1*, v1*, etc. are independent we can
equate their coefficients, i.e.
be
R ix = ∫ p x N i dv
…………… (a)
R iy = ∫ p y N i dv
be
…………… (b)
with i = 1,2,…n.( node numbers of an element)
Equations (a-b) are used to determine the equivalent nodal forces comprising Rbe.
(ii) Distributed loading
These can be divided into pressures (p) acting normal to an element side and
shear stresses (q) acting a side. The sign convention adopted here is that p is positive
when it acts on the element, and q is positive when it is acting anti-clock-wise around
the element.
P′
v*
B
θ
q
P u*
P
y
AB = element side
A
x
Figure (1) Tractions on Element Side
25
Applying the principle of virtual work, this to an element edge having k nodes,
using superposed s to indicate ‘surface’ (‘e’ can now stand for ‘edge’ rather than
‘element’)’ and referring to figure 3.20, we obtain
∑ (R
n
i =1
se
IX
)
u *i + R seiy v *i = ∫ [p(v * Cosθ − u * Sinθ) + q(v * Sinθ + u * Cosθ)]dA
Integration is over the side area, i.e. length times out-of-plane thickness. Again using
the principle of virtual work to relate u*, v* to nodal values, and using the
independence of the latter we obtain
se
R ix = ∫ (pSinθ − q cos θ)N i dA
R iy = ∫ (pCosθ − qSinθ)N i dA
se
…………… (1a)
…………… (1b)
Note that for a horizontal surface with θ = 1800 (i.e. level ground, θ = 0 would
represent a ceiling).
R ix = ∫ qN i dA
se
R iy = ∫ pN i dA
se
…………… (2a)
…………… (2b)
Rs is obtained by assembling the edge contributions Rse from all the loaded
edges in the F.E. mesh.
...............................................END..............................................
YANGON TECHNOLOGICAL UNIVERSITY
DEPARTMENT OF MINING ENGINEERING
Geotechnical Engineering
(Part II)
Min. 05037
BE (Mining)
QUESTIONS AND ANSWERS
16 SEPTEMBER, 2003
YANGON
(QUESTIONS)
CHAPTER 1
NUMERICAL PROGRAMMING
1.* Outline the path of calculation procedure of Finite Element Method (FEM)
analysis (or sequence of computer analysis) within the elastic range. (20 marks)
2.* Construct the simple FEM mesh and explain the calculation procedure for
numerical analysis. (20 marks)
3.* Explain, briefly, the basic program organization of FEM analysis for numerical
solution processes for elastic behavior of rock. (10 marks)
4.* Discuss briefly the basic steps of finite element process and the possible errors
involved in using the finite element method for particular condition. (10 marks)
5.* Discuss the importance of the most suitable rock failure criteria to use in
excavation system designs. Give reasons for your discussion. (10 marks each)
6.* Explain the detailed procedure for the assembly of overall structural stiffness
matrix for solving the simultaneous equations. (20 marks)
7.* For underground stability analysis, describe, briefly, the procedure for calculating
the stress field around any underground opening and give an example. (10 marks)
8.** Derive the mathematical expressions for the (i) current value of yield stress and
(ii) current value of yield function (Use Tresca and Von Mises failure criterions).
(10 marks each)
9.** Outline, with the aid of flow sheet, the computation sequence for the twodimensional elastic-plastic program which can be used in the analysis of rock
movement around mine excavation. (20 marks)
10. ** Write subroutines for the elastic matrix-D and strain matrix-B by using FOR77
language in the calculation of numerical program for plain strain analysis. (10
marks each)
11. ** Sketch the FEM mesh for numerical analysis of rock movement around the
tunnel.
(i) Enumerate the necessary data for the FEM analysis.
(ii) Finally discuss, with example, the principles of development for the flow
sheet of main computer program to carry out the successful implementation of
FEM analysis. (10 marks each)
12. ** Explain with the aid of mathematical expression elastic-plastic analysis of rock
movement paying particular attention on: (10 marks each)
(i) The current value of yield stress
(ii) the current value of yield function (effective stress value)
(Use Mohr-Coulomb and Hoek-Brown failure criteria)
13. *** Discuss with reference to the methods:
(i) The factors which could have influenced their selection of most suitable
method, which gives the realistic result for solving the geotechnical problems.
(ii) Results and their application. (10 marks each)
14. *** Derive the mathematical expressions of current values of yield function and
current values of yield stress of Hoek and Brown yield criterion and MohrCoulomb yield criterion for the numerical nonlinear elastic-plastic solution
processes. (10 marks)
CHAPTER 2
REVIEW OF UPDATING NEW ROCK MASS CLASSIFICATION
15.* Write down the short notes on the followings: (10 marks)
(i) Terzaghi’s rock mass classification
(ii) Rock Structure Rating
16.* Discuss in detail the influences of the joint frequency, orientation and strength
along the critical joint sets of discontinuities on reduction in the strength of jointed
rocks. (20 marks)
17.* What are the various types of elastic modulii? Give the essential features to
ensure the results that are applicable in design calculations. (10 marks each)
18.* For detailed site investigation of tunnel, it is required to measure the geological
data. As a design engineer, what data are required? Give the required engineering
geological data for implementation of the in situ condition of rock mass properties
and discuss the limitations of data obtained. (10 marks each)
19.* Discuss as much as you know the followings.
(i) Quantification of the geological parameters, which influence the engineering
properties of rock mass.
(ii) Influence of the characteristics of discontinuity on the strength of rock mass.
(10 marks each)
20.* Input data to enable rock mass classifications have been compiled for al
structural regions along the tunnel route. The following table (2-1) is the collected
data of one of these structural regions.
Table (2.1) Summarized Record of Engineering Geological Data
Project Name: ……………………..
Site of survey: ………………………
Conducted by: ……………………...
Date: 07-12 - 2002
Rock Type and Origin
Structural Region: Between………………………
Shale with interbedded sandstone
DRILL CORE QUALITY (RQD)
WALL ROCK OF DISCONTINUITY
Unweathered: ……………………………………
Excellent quality: 90 – 100 % ………...……
Good quality:
75 – 90 % ………….……
Fair quality:
50 – 75 % ………72……
Poor quality:
25 – 50 % ………………
Very poor quality:
Slightly weathered: ……………SL.……………
Moderately weathered: …………………………
Highly weathered: …………………….…………
Completely weathered: …………………………
< 25 % ……………..…
Residual Soil: ……………………………………
GROUND WATER
STRENGTH OF INTACT ROCK
INFLOW per 10 m
Designation
Of tunnel length
Uniaxial compressive or Point Load
Liter/min: …………….
Strength, MPa
Or
Very high:
WATER PRESSURE
KPa: ………………
Over 250
High:
Or
strength indix, MPa
………..
> 10
100 – 250 ………..
………..
4 – 10 ………..
Medium high:
50 – 100
…60....
2–4
GENERAL CONDITIONS (completely dry, damp, wet,
Moderate:
25 – 50
…26…..
1–2
dripppping or flowing under low/medium or high pressure):
Low:
5 – 25
………..
<1
………Dripping……………….
Very Low:
1–5
………..
………..
………..
SPACING OF DISCONTINUITIES
Set 1
Very wide:
Over 2 m
Set 2
Set 3
……………..
……………..
Set 4
……………..
Wide:
0.6 – 2 m
……………..
……0.65…....
Moderately:
200 – 600 mm
……300..…..
……………..
……………..
……………..
60 – 200 mm
……………..
……………..
……………..
……………..
< 60 mm
……………..
……………..
……………..
……………..
Close:
Very close:
……………..
……………..
……………..
Note: These values are obtained from a joint survey and not from bore-hole logs.
STRIKE AND DIP ORIENTATIONS
Set 1
Strike
N23E
(from N5E
Set 2
Strike
N23E
(from N40E
Set 3
Strike
…….
(from N5E
Set 4
Strike
…….
(from N5E
to
N35E)
Dip: 20 SE
N60E)
Dip: 20 SE
to
N35E)
Dip: ………
to
N35E)
Dip: ………
to
NOTE: Refer all directions to magnetic north
Table (2-1) cont.
CONDITION OF DISCONTINUITIES
PERSISTENCE (CONTINUITY)
Set 1
Set 2
Set 3
Set 4
Very low:
< 1m
………….
……………
…………….
………..
Low:
1–3m
………….
……………
…………….
………..
Medium:
3 – 10 m
………….
……………
…………….
………..
High:
10 – 20 m
……15….
……18……
…………….
………..
Very high:
> 20 m
………….
……………
…………….
………..
Very tight joints: < 0.1mm
………….
……………
…………….
………..
Tight:
………….
……………
…………….
………..
Moderately open joints: 0.5 – 2.5 mm
……1.0….
……1.5..…
…………….
………..
Open joints:
………….
…………….
………..
SEPARATION (APERTURE)
0.1 – 0.5 mm
2.5 – 10 mm
Very wide aperture:
> 10 mm
………….
……………
……………
…………….
………..
ROUGHNESS (state also if surfaces are stepped, undulating or planer)
Very rough:
………….
……………
…………….
………..
Rough:
……R….
………R…
…………….
………..
Slightly rough:
………….
……………
…………….
………..
Smooth:
………….
……………
…………….
………..
Slickened:
………….
……………
…………….
………..
FILLING (GOUGE)
Type:
Thickness:
………….
………….
……………
……………
…………….
…………….
………..
………..
Uniaxial compressive strength, MPa:
………….
……………
…………….
………..
Seepage:
………….
……………
…………….
………..
Classify the rock mass conditions in accordance with: (10 marks each)
(i) Terzaghi
(ii) RSR concept
(ii) Geomechanics Classification (RMR)
(ii) Q-system
21.* From table (2-1), calculate the rock loads by means of each of the mass
classification systems (Terzaghi, RSR concept, Geomechanics Classification
(RMR) and Q-system) for drill and blast tunnel. (5 marks each)
22.* From table (2-1), determine the self-supporting span and the maximum span
possible for the encountered rock mass condition by means of each of the mass
classification systems (Geomechanics Classification (RMR) and Q-system). (10
marks each)
23.* From table (2-1), estimate the stand-up time, rock mass deformability (Em), the
angle of internal friction and cohesion of rock mass. Recommend the support
requirements for primary support (10 marks each)
24.** In geotechnical problems, there are numbers of links or correlations to use the
laboratory measured data to in situ values. Discuss in detail, with formulation, the
correlations between the rock mass quality (Use RMR) and laboratory test results.
(20 marks)
25. ** As a Geotechnical design Engineer, you are responsible for site
characterization of a tunnel to determine the quantitative data. Discuss what the
factors should be collected for using CSIR rock mass classification system. Give
an example as you like. (20 marks)
26. *** A vertical cross section of the outline of the proposed underground working is
shown in figure (1), together with the location of the underground water table. As a
designer, give the hydrological, geological data and mechanical properties of ore
and host rock mass, which must be considered for providing the successful design
of underground mine working in view of geomechanics. Briefly describe the
integrated surveying to obtain these properties. (10 marks each)
Water Level
100 m
Ore
Figure (1) Orientation of ore body
27. *** As a mining engineer, you should do the bench blasting design to be taken
into account in an understanding of rock failure mechanics for the different
confined stresses. Discuss as much as you know the importance of suitable yield
criterion for the proper understanding of the real rock mass properties in designing.
(20 marks)
CHAPTER 3
ROCK MASS PROPERTIES
28.* Write down the following short note: (10 Marks each)
(i) GSI rock mass classification
(ii) Correlation between intact rock properties and rock mass.
29.* Discuss in detail the Hoek and Brown strength parameters of rock mass from the
results of GSI rock mass classification. (20 marks)
30.** Discuss as much as you know the post-failure behavior of rock mass for strain
softening material by reducing the GSI value of rock mass quality. (20 marks)
31.*** Discuss the correlation between the CSIR (RMR) rock mass classification and
GSI rock mass classification to verify the data with each other. (20 marks)
32. *** As a rock engineering, discuss briefly how to understand the effect of
geological discontinuities on rock mass behavior in tunneling and explain the
required technical data for successful tunnel excavation. (20 marks)
CHAPTER 4
DESIGN OF ROCK REINFORCEMENT
33.* Discuss the following. (10 marks each)
(i) Bond strength of rock bolt and cable
(ii) Effects of grouting quality on the stiffness of rock mass
34.* Write the short notes the followings. (10 marks each)
(i) Ground treatment by Grouting
(ii) Ground treating by rock bolt reinforcement.
35.** Discuss briefly the applications of rock bolts and dowels to match the
applicability of rock bolts and dowels and rock mass nature in underground
excavations and tunneling. Show the rock mass nature with necessary figures (20
marks)
36.*** Discuss the effects of length and spacing of rock bolt and cable on the strength
of surrounding rock mass in design consideration of tunnel or underground
excavation.
SAMPLE QUESTIONS AND ANSWERS
1. Question
No (1)* Outline the path of calculation procedure of Finite Element Method (FEM)
analysis (or sequence of computer analysis) within the elastic range of rock mass
around tunnel. In addition, sketch the FEM mesh for numerical analysis. (Chapter 1)
Answer:
The basic procedure of the computer program can be summarized in the
following figure (1). For the complete solution of a problem by the finite element
method, analysis is carried out using input data, which describe fully the idealized
structure and its loading, and produces output consisting of tabulated nodal
displacements and element stresses.
To specify the problem it is necessary to provide the computer with input data.
This material consists of data specifying the geometry of the idealized structure, its
material properties and the way it is loaded and supported in space. The data also
include certain control numbers that may help the generality and efficiency of the
program and should be supplied early in the input data, such as the total number of
nodes and elements or a control number which indicates whether the problem is plane
stress, plane strain or plate flexure.
For FEM analysis, the following details are needed specifically.
i. To be numbered the nodal points 1…i…n, the node number i and its coordinates (xi, yi) are supplied.
ii. To be numbered 1…N for each element its nodal numbers (connectivity), the
corresponding nodal co-ordinates of the element can be supplied.
iii. To be determined the dimensions of the element and its position in the overall
structure. It is also necessary to specify the material properties of the element,
i.e. Young’s Modulus E and Poisson’s ratio v.
iv. The applied loads.
v. The ‘fixed nodes’−corresponding to the various types of supports.
Input Data:
- Geometry
- Material properties
- Loading
- Support condition
Evaluate
Individual Element
Stiffness matrix
e
[K ]
Assemble
Overall Stiffness matrix for
structure
[K]
Apply Boundary Conditions
Apply Boundary Conditions
Solve
{F}= [K] {δ}
Evaluate Stresses
e
{σ(x,y)}= [H] {δ }
Print Results
Figure (1) Sequence of Computer Analysis
...........................................................................................
2. Question
No. (2)* Construct the simple FEM mesh and explain the calculation procedure for
numerical analysis. (Chapter 1)
Answer: (FEM mesh for tunnel, slope etc. for geotechnical problems should be
constructed)
A deep cantilever carrying a load of 10 KN can be shown in following Figure
(2) and Figure (3). The cantilever is assumed fully fixed at the left-hand edge. The
idealization gives the following information (taking the origin of the co-ordinate
system at node 1).
100
5 mm
50 mm
y, v, Fy
x, u, Fx
10KN
Figure 2 Deep Cantilever
4
2
6
2
4
1
1
3
50
3
50
5
Figure 3 Coarse finite element idealization
i. Number of nodes: 6.
ii. Number of element: 4.
iii. Type of element: triangular plane stress, uniform thickness.
iv. All elements have the same material properties throughout.
v. At each of the six nodes there are two degrees of freedom. For example, at node
3 there is a horizontal deflection component u3 and a vertical deflection
component v3. The boundary conditions may be specified as follows:
u1 = v1 = u2 = 0
vi. The nodal co-ordinates and applied loading may be written as
Node
1
2
3
4
5
6
x-co-ordinate
0
0
50
50
100
100
y-co-ordinate
0
50
0
50
0
50
Fx
−
−
0
0
0
0
Fy
−
−
0
0
- 10
0
If the loading at a joint is known, then the displacements of that joint are not
known and vice versa. The forces Fx and Fy refer to externally applied loads.
vii. Numbering the element nodes in an anti-clockwise manner, the element
connectivity and material properties may be written as
Element
Connectivity
E
v
t
1
1
3
2
200
0.3
5
2
2
3
4
200
0.3
5
3
3
5
4
200
0.3
5
4
4
5
6
200
0.3
5
...........................................................................................
3. Question
No. (10)** Write subroutines for the elastic matrix-D and strain matrix-B by using
FOR77 language in the calculation of numerical program for plain strain analysis.
(Chapter 1)
TO STUDY THE ALL SUBROUTINES OF LINEAR ELASTIC PROGRAMING
Answer:
Descriptions of Program
NSTR1 = Number of stress component
YOUNG, POISS = Material properties
The results are stored in array DMATX (3,3) and BMATX (3,16) for 8 node, 3
point rule analyses respectively.
C******************************************************************
SUBROUTINE MODPS(DMATX,LPROP,MMATS,NTYPE,PROPS)
C******************************************************************
C*** THIS SUBROUTINE EVALUATES THE D-MATRIX
C******************************************************************
DIMENSION DMATX(3,3),PROPS(1,8)
YOUNG=PROPS(LPROP,1)
POISS=PROPS(LPROP,2)
DO 10 ISTR1=1,NSTR1
DO 10 JSTR1=1,NSTR1
DMATX(ISTR1,JSTR1)=0.0
10 CONTINUE
C******************************************************************
C*** D-MATRIX FOR PLANE STRAIN CASE
C******************************************************************
CONST=YOUNG*(1.0-POISS)/((1.0+POISS)*(1.0-2.0*POISS))
DMATX(1,1)=CONST
DMATX(2,2)=CONST
DMATX(1,2)=CONST*POISS/(1.0-POISS)
DMATX(2,1)=CONST*POISS/(1.0-POISS)
DMATX(3,3)=CONST*(1.0-2.0*POISS)/(2.0*(1.0-POISS))
RETURN
END
C******************************************************************
C******************************************************************
SUBROUTINE BMATPS(BMATX,CARTD,NNODE,SHAPE,GPCOD,NTYPE,KGASP)
C******************************************************************
C*** THIS SUBROUTINE EVALUATES THE STRAIN-DISPLACEMENT MATRIX
C******************************************************************
DIMENSION BMATX(3,16),CARTD(2,8),SHAPE(8),GPCOD(2,9)
NGASH=0
DO 10 INODE=1,NNODE
MGASH=NGASH+1
NGASH=MGASH+1
BMATX(1,MGASH)=CARTD(1,INODE)
BMATX(1,NGASH)=0.0
BMATX(2,MGASH)=0.0
BMATX(2,NGASH)=CARTD(2,INODE)
BMATX(3,MGASH)=CARTD(2,INODE)
BMATX(3,NGASH)=CARTD(1,INODE)
IF(NTYPE.NE.3) GO TO 10
BMATX(4,MGASH)=SHAPE(INODE)/GPCOD(1,KGASP)
BMATX(4,NGASH)=0.0
10 CONTINUE
RETURN
END
C******************************************************************
...........................................................................................
4. Question
No (15) * Write down the short notes on the followings: (Chapter 2)
(i) Terzaghi’s Rock Mass Classification
(ii) Rock Structure Rating
Answer:
(i) Terzaghi’s Rock Mass Classification
The design of tunnel support can be estimated by Terzaghi’s rock loads. This
method estimates based on descriptive classification. The clear and concise
definitions and the practical comments included in these descriptions are good
examples of the type of engineering geology information, which is most useful for
engineering design.
Terzaghi’s descriptions are:
-Intact rock contains neither joints nor hair cracks. Hence, if it breaks, it
breaks across sound rock. On account of the injury to the rock due to blasting, spalls
may drop off the roof several hours or days after blasting. This is known as a spalling
condition. Hard, intact rock may also be encountered in the popping condition
involving the spontaneous and violent detachment of rock slabs from the sides or roof.
Stratified rock consists of individual strata with little or no resistance against
separation along the boundaries between the strata. The strata may or may not be
weakened by transverse joints. In such rock the spalling condition is quite common.
Moderately jointed rock contains joints and hair cracks, but the blocks
between joints are locally grown together or so intimately interlocked that vertical
walls do not require lateral support. In rocks of this type, both spalling and popping
conditions may be encountered.
Blocky and seamy rock consists of chemically intact or almost intact rock
fragments which are entirely separated from each other and imperfectly interlocked.
In such rock, vertical walls may require lateral support.
Crushed but chemically intact rock has the character of crusher run. If most
or all of the fragments are as small as fine sand grains and no re-cementation has
taken place, crushed rock below the water table exhibits the properties of a water
bearing sand. 
Squeezing rock slowly advances into the tunnel without perceptible volume
increase. A prerequisite for squeeze is a high percentage of microscopic and submicroscopic particles of micaceous minerals or clay minerals with a low swelling
capacity.
Swelling rock advances into the tunnel chiefly on account of expansion. The
capacity to swell seems to be limited to those rocks that contain clay minerals such as
montmorillonite, with a high swelling capacity.
(ii) Rock Structure Rating
Wickham et al (1972) described a quantitative method for describing the
quality of a rock mass and for selecting appropriate support on the basis of their Rock
Structure Rating.
(RSR) classification: Most of the case histories used in the development of this
system, were for relatively small tunnels supported by means of steel sets, although
historically this system was the first to make reference to shotcrete support. In spite of
this limitation, it is worth examining the RSR system in some detail since it
demonstrates the logic involved in developing a quasi-quantitative rock mass
classification system.
The significance of the RSR system, in the context of this discussion, is that it
introduced the concept of rating each of the components listed below to arrive at a
numerical value of RSR = A + B + C.
1. Parameter A, Geology: General appraisal of geological structure on the basis
of:
a. Rock type origin (igneous, metamorphic, sedimentary).
b. Rock hardness (hard, medium, soft, decomposed).
c.
Geologic
structure
faulted/folded,
(massive,
slightly
faulted/folded,
moderately
intensely faulted/folded).
2. Parameter B, Geometry: Effect of discontinuity pattern with respect to the
direction of the tunnel drive on the basis of:
a. Joint spacing.
b. Joint orientation (strike and dip).
c. Direction of tunnel drive.
3. Parameter C: Effect of groundwater inflow and joint condition on the basis of:
a. Overall rock mass quality on the basis of A and B combined.
b. Joint condition (good, fair, poor).
c. Amount of water inflow (in gallons per minute per 1000 feet of tunnel).
Note that the RSR classification used Imperial units and three tables to
evaluate the rating of each of these parameters to arrive at the RSR value (maximum
RSR = 100).
...........................................................................................
5. Question
No. (20)* Input data to enable rock mass classifications have been compiled for al
structural regions along the tunnel route. The following table (2-1) is the collected
data of one of these structural regions.
Classify the rock mass conditions in accordance with: (Chapter 2)
(i) Terzaghi
(ii) RSR concept
(ii) Geomechanics Classification (RMR)
(ii) Q-system
Solution:
Item 1: Classification of rock mass conditions
(a) Terzaghi: ‘Moderately blocky and seamy’ (RQD = approx. 72%)
(b) RSR concept:
-Rock type: soft sedimentary rock;
-Slightly faulted and folded;
Parameter A = 15;
-Spacing: moderate to blocky;
-Strike approx, perpendicular to tunnel axis, dip 0 – 20°;
Parameter: B = 30;
-Water inflow: moderate;
-Joint conditions:
- fair (moderately open, rough and weathered); For A + B = 45,
Parameter C = 16;
Therefore: RSR = 15 + 30 + 16 = 16.
(c) Geomechanics Classification (RMR):
-Intact rock strength, σc = 50 MPa
Rating = 4;
-Drill core quality, RQD = 55 – 85%; Ave. 72%
Rating = 13;
-Spacing of discontinuities, range: 50 mm to 0.9 meters
Rating 10;
-Condition of discontinuities: separation 0.8 mm to 1.1 mm, slightly
weathered, rough surfaces
Rating 25;
-Groundwater: dripping water, low pressure, flow 25 -125 liters/min.
Rating 4;
-Basic RMR: 4 + 13 + 10 + 25 + 4 = 56 without adjustment for orientation of
discontinuities;
Fair orientation: Adjustment: -5,
Adjusted RMR = 56 – 5 = 51;
(d) Q-System:
- RQD = 72 % (average); Jn = 6, two joint sets and random;
- Jr = 1.5, rough, planar joints; Ja = 1.0, unaltered joint walls, surface staining
only; Jw = 0.5, possible large water inflow;
- SRF = 1.0, medium stress, σc/σ1 = 50/0.91 = 55.
Summary:
Classification
Terzaghi
RSR
RMR
Q
Result
Moderately blocky & seamy
61
51
Fair rock mass
9.0
Fair rock mass
6. Question
No. (21)* From table (2-1), calculate the rock loads by means of each of the mass
classification systems (Terzaghi, RSR concept, Geomechanics Classification
(RMR) and Q-system) for drill and blast tunnel. (Chapter 2)
Solution:
Drill-and-blast diameter: 74 m + 0.6 m, over break = 8.0 m
Machine-bored diameter: 7.4 m
Shale density: 2660 kg/m3 (166/ft3)
Method
Terzaghi
Drill-and-blast
hp = 0.35C = 0.7B = 0.7 × 8.0 = 5.6 m
rock load P = γhp = 0.146 MPa
(1.52 t/ft2)
From Fig. 6.3, P = 0.067 MPa
(1.4 kip/ft2)
RSR = 61
RMR = 51
P = γht = 0.102 MPa
Q = 9.0
TBM
hp = 0.45B = 3.3 m
P = 0.09 MPa (0.9 t/ft2)
TBM adjustment, Fig. 6.2
RSR = 69.5, P = 0.034 MPa
(0.7 kip/ft2)
TBM adjustment via conversion to RSR
(eqn. 6.12) and Fig. 6.2 RMR = 74,
P = 0.049 MPa
TBM adjustment via conversion to RSR
(eqn. 6.13) and Fig. 6.2.
Q = 54
P = 0.0321 MPa
Summary of rock-loads in KPa (1MPa = 1000 KPa)
Method
Drill-and-blast
TBM
Terzaghi
RSR
RMR
Q
146
67
102
63
90
34
49
32
...........................................................................................
7. Question
No. (22)* From table (2-1), determine the self-supporting span and the maximum
span possible for the encountered rock mass condition by means of each of the
mass classification systems (Geomechanics Classification (RMR) and Q-system).
(Chapter 2)
Solution:
Span Vs stand-up time
Self supporting span
Maximum span
RMR = 51
204 m
10.5 m
Q = 9 (ESR = 1.6)
8m
80 m [D = 2 (1.6 × 9)0.4]
...........................................................................................................
8. Question
No (23)* From table (2-1), estimate the stand-up time, rock mass deformability (Em),
the angle of internal friction and cohesion of rock mass. Recommend the support
requirements for primary support (Chapter 2)
Solution:
-RMR = 51 and Span = 8 m
Stand-up time: approximately 100 hours or 4-5 days
Deformability: RMR = 56 (no adjustment for joint orientations)
E = 2 RMR – 100 = 12 GPa (1.74 × 106 psi)
c = 192 KPa
φ = 39° (from table)
Terzaghi
RSR
RMR
Q system
Drill and blast-light to medium steel sets spaced 1.5 m. Concrete lining
Drill and blast-6H25 ribs on 2 m centers plus final lining
Drill and blast-systematic bolts 3.5 m long – spaced 1.5 m, shotcrete 50 to 100 mm
in roof and 30 mm on walls, were mesh in crown
Drill and blast – 3 m long rock bolts spaced 1.5 m and 50 mm thick shotcrete
...........................................................................................
9. Question
No (29)* Explain in detail the Hoek and Brown strength parameters of rock mass
from the results of GSI rock mass classification. (Chapter 3)
Answer:
Once the Geological Strength Index has been estimated, the parameters that
describe the rock mass strength characteristics are calculated as follows:
 GSI − 100 
m b = mi exp

28


For GSI > 25, i.e. rock masses of good to reasonable quality, the original
Hoek-Brown criterion is applicable with
 GSI − 100 
s = exp

9


a=0.5
For GSI < 25, i.e. rock masses of very poor quality, the modified Hoek-Brown
criterion applies with
s = 0 and
GSI
a = 0.65 −
200
The choice of GSI = 25 for the switch between the original and modified
criteria is purely arbitrary.
For better quality rock masses (GSI>25), the value of GSI can be estimated
directly from the 1976 version of Bieniawski’s Rock Mass Rating, with the
Groundwater rating set to 10 (dry) and the Adjustment for Joint Orientation set to 0
(very favorable).
For very poor quality rock masses the value of RMR is very difficult to
estimate and the balance between the ratings no longer gives a reliable basis for
estimating rock mass strength. Consequently, Bieniawski’s RMR classification should
not be used for estimating the GSI values for poor quality rock masses.
If the 1989 version of Bieniawski’s RMR classification (Bieniawski 1989) is
used, then GSI = RMR89’ - 5 where RMR89’ has the Groundwater rating set to 15 and
the Adjustment for Joint Orientation set to zero.
One of the practical problems which arise when assessing the value of GSI in
the field is related to blast damage. There is a considerable difference in the
appearance of a rock face which has been excavated by controlled blasting and a face
which has been damaged by bulk blasting. Wherever possible, the undamaged face
should be used to estimate the value of GSI since the overall aim is to determine the
properties of the undisturbed rock mass.
Where all the visible faces have been damaged by blasting, some attempt
should be made to compensate for the lower values of GSI. In recently blasted faces,
new discontinuity surfaces will have been created by the blast and these will give a
GSI value that may be as much as 10 points lower than that for the undisturbed rock
mass.
Borehole cores can be used to estimate the GSI value. For reasonable quality
rock masses (GSI > 25) the best approach is to evaluate the core in terms of
Bieniawski’s RMR classification to estimate the GSI value from RMR.
For poor quality rock masses (GSI < 25), relatively few intact core pieces
longer than 100 mm are recovered and it becomes difficult to determine a reliable
value for RMR. In these circumstances, the physical appearance of the material
recovered in the core should be used as a basis for estimating GSI.
GSI in the field is related to blast damage. There is a considerable difference
in the appearance of a rock face which has been excavated by controlled blasting and
a face which has been damaged by bulk blasting.
Wherever possible, the undamaged face should be used to estimate the value
of GSI since the overall aim is to determine the properties of the undisturbed rock
mass.
...................................................END.......................................................