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Rock Dynamics
Lecture Topics: Concepts of
1-Strain energy
2-Wave Equations
3- Blast Waves
4- Wave Propagation in continuous and
Discontinuous media
Mid term presentation
Presented by: Abdolhakim Javid

STRAIN ENERGY
Introduction
• The concept of strain energy is particularly useful in the
determination of the effects of impact loadings on structures or
machine components.
• The strain-energy density of a material will be defined.
• strain energy of a member will be defined as the increase in energy
associated with the deformation of the member.
• strain energy is equal to the work done by a slowly increasing load
applied to the member.
Consider a rod BC of length L and uniform
cross-sectional area A, which is attached at
B to a fixed support, and subjected at C to a
slowly increasing axial load P .
(Fig. 1)

STRAIN ENERGY
Let us now consider the work dU done by
the load P as the rod elongates by a small
amount dx. This elementary work is equal
to the product of the magnitude P of the
load and of the small elongation dx. We
write:
and note that the expression obtained is equal to the element of area of
width dx located under the load-deformation diagram. The total work U
done by the load as the rod undergoes a deformation x1 is thus
and is equal to the area under the load-deformation diagram between x =0 and x =x1.
𝑑𝑈 = 𝑃𝑑𝑥 = 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑤𝑜𝑟𝑘
𝑈 = න
0
𝑥1
𝑃 𝑑𝑥 = 𝑡𝑜𝑡𝑎𝑙 𝑤𝑜𝑟𝑘
(Fig. 2)
Eq. (1)

STRAIN ENERGY
The work done by the load P as it is slowly applied to the rod must result in the
increase of some energy associated with the deformation of the rod. This energy
is referred to as the strain energy of the rod. We have, by definition,
In the case of a linear and elastic deformation,
the portion of the load-deformation diagram
involved can be represented by a straight line
of equation P=kx (Fig. 3). Substituting for P in
Eq.(1), we have
where P1 is the value of the load corresponding to the
deformation x1.
𝑈 = න
0
𝑥1
𝑃 𝑑𝑥 = 𝑡𝑜𝑡𝑎𝑙 𝑤𝑜𝑟𝑘 = 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑛𝑒𝑟𝑔𝑦
𝑈 = න
0
𝑥1
𝑘𝑥𝑑𝑥 =
1
2
𝑘𝑥1
2
=
1
2
𝑃1𝑥1
(Fig. 3)
Eq. (1)
Eq. (2)

STRAIN ENERGY Density
As the load-deformation diagram for a rod BC depends upon the length L and the
cross-sectional area A of the rod. The strain energy U defined by Eq. (1), therefore,
will also depend upon the dimensions of the rod.
In order to eliminate the effect of size from our discussion and direct our attention
to the properties of the material, the strain energy per unit volume will be
considered. Dividing the strain energy U by the volume V =AL of the rod (Fig.1),
and using Eq. (1), we have:
Recalling that P/A represents the normal stress σx
in the rod, and x/L the normal strain εx, we write:
The strain-energy density of a material will be defined as
“ Strain energy per unit volume”
𝑈
𝑉
= න
0
𝑥1
𝑃
𝐴
𝑑𝑥
𝐿
𝑢 = න
0
𝜀1
𝜎𝑥 𝑑𝜀𝑥 = strain energy density
where ε1 denotes the value of the strain corresponding
to the elongation x1. The strain energy per unit
volume, U/V, is referred to as the strain-energy
density and will be denoted by the letter u. We have,
therefore:
if SI metric units are used, the strain-energy density is expressed in J/m3 or its multiples kJ/m3 and MJ/m3
Eq. (3)

STRAIN ENERGY Density
Referring to Fig. 4, we note that the
strain-energy density u is equal to the
area under the stress-strain curve,
measured from εx =0 to εx= ε1.
If the material is unloaded, the stress
returns to zero, but there is a permanent
deformation represented by the strain εp,
and only the portion of the strain energy
per unit volume corresponding to the
triangular area is recovered.
The remainder of the energy spent in
deforming the material is dissipated in
the form of heat.
it will be seen that it is equal to
“area under the stress-strain
diagram of the material.”
(Fig. 4)

Modulus of Toughness
The value of the strain-energy density obtained by
setting εx = εR . where εR is the strain at rupture, is
known as the modulus of toughness of the
material .
It is equal to the area under the entire stress-
strain diagram (Fig. 5) and represents the energy
per unit volume required to cause the material to
rupture.
It is clear that the toughness of a material is
related to
• its ductility as well as
• to its ultimate strength
(Fig. 5)

Modulus of Resilience
The modulus of resilience is equal to the area
under the straight-line portion OY of the
stress-strain diagram (Fig. 6) and represents
the energy per unit volume that the material
can absorb without yielding.
The capacity of a structure to withstand an
impact load without being permanently
deformed clearly depends upon the resilience
of the material used.
Since the modulus of toughness and the
modulus of resilience represent characteristic
values of the strain-energy density of the
material considered, they are both expressed
in J/m3
(Fig. 6)

Modulus of Resilience
If the stress σx remains within the proportional limit of the material, Hooke’s law
applies and we write
The value uY of the strain-energy density obtained by setting σx =σY where
is σY the yield strength, is called the modulus of resilience of the material.
We have
𝑢 = න
0
𝜀1
𝐸𝜀𝑥 𝑑𝜀𝑥 =
𝐸𝜀1
2
2
=
𝜎1
2
2𝐸
𝑢𝑌 =
𝜎𝑌
2
2𝐸
= modulus of resilience
Eq. (4)
𝜎𝑥 = 𝐸𝜀𝑥

2-Wave Equations
Theory of Elasticity
Rock Dynamics

Rock dynamics-presentation -javid.pdf

▫ Primary wave
▫ compressional wave
▫ Longitudinal wave
▫ Dilatation wave
▫ Irrotational wave
Longitudinal Waves

Transverse Waves
▫ Secondary wave
▫ Shear wave
▫ Rotational wave
▫ Transvers wave

A seismograph, or seismometer, is an instrument used to detect and record
seismic waves. Seismic waves are propagating vibrations that carry energy
from the source of an earthquake outward in all directions.

Theory of Elasticity
• Seismic waves are stress (mechanical) waves that are
generated as a response to acting on a material by a force.
• The force that generates this stress comes from a source of
seismic energy such artificial (dynamite, ... etc) or natural
earthquakes.
• The stress will produce strain (deformation) in the material
relating to elasticity theory.
• Therefore, we need to study a little bit of elasticity theory in
order to better understand the theory of seismic waves.

Stress
• There should be a maximum of 9 stress components associated
with every possible combination of the coordinate system axes
(xx, xy, xz, yx, yy, yz, zx, zy, zz).
• According to equilibrium (body is not moving but only
deformed as a result of stress application): ij = ji, meaning
that xy = yx, yz = zy, and zx = xz.
• If the force is perpendicular to the surface, we have a normal
stress (xx, yy, zz); while if it’s tangential to the surface, we
have a shearing stress (xy, yz, xz).

Stress
• The stress matrix composed of nine components of the stress:
𝜎 =
𝜎𝑥𝑥 𝜎𝑥𝑦 𝜎𝑥𝑧
𝜎𝑦𝑥 𝜎𝑦𝑦 𝜎𝑦𝑧
𝜎𝑧𝑥 𝜎𝑧𝑦 𝜎𝑧𝑧

Components of stress and strain
• If a stretching force is acting in the
x-y plane and the corresponding
motion is only occurred in the
direction of x- axis, we will have the
situation depicted in the
corresponded figure.
• The point P moves a distance u to
point P’ after stretching while point
Q moves a distance ux+ux to point
Q’.
y
x
P Q
P’ Q’
ux
x
x
x
u
u x
x

+ 


Normal Strain
 As we know that normal strain in x-
direction is know as the ratio
between the change of length of QP
to the original length of QP




y
x
P Q
P’ Q’
ux
x
Coordinates
P(x,y)
Q(x+x,y)
P’(x+u,y)
)
,
(
' y
x
x
x
u
u
x
Q
x
x 
+



+
+
QP
QP
P
Q
QP
of
length
original
QP
of
length
in
change
xx
−
=
=
'
'

u
x
x
x
x
u
u
x
P
Q x
+
−

+



+
+
=
'
'
x
x
x
QP −

+
=
x
ux
xx


=

Do similar processing for yy and zz.
x
x
u
u x
x

+ 


Shear Strain
• If a stretching force is acting in the x-y
plane and the corresponding motion is
induced either in the direction of x-
axis and y-axis, we will have the
situation depicted in the corresponded
figure.
• The infinitesimal rectangular PQRS
will have displaced and deformed into
the diamond P’Q’R’S’.
• After stretching, points P, Q, S and R
move to P’, Q’, S’, and R’ with
coordinates.
y
x
P Q
P’ Q’
ux
x
x
x
u
u
x
x 


+
S R
S’
R’
x
x
u
u
y
y 


+
uy
y
y
y
x
u
x
u 


+

Shear Strain
• The deformation in y coordinates in relative
to x-axis is given by
Coordinates
P(x,y) P’(x+ux,y+uy)
Q(x+x,y)
S(x,y+y)
)
,
(
' x
x
u
u
y
x
x
x
u
u
x
Q
y
y
x
x 


+
+

+



+
+
x
y
P
y
P
y
Q
y
Q
length
-
x
original
length
-
x
to
relative
y
in
change
xy

−
−
−
=
=
)
'
(
)
'
(

Substitute the coordinates of points P, Q, P’, and Q’ to get the shear-
strain component in the x-y plane
)
,
(
' y
y
x
u
x
u
y
y
y
y
x
u
ux
x
S 


+
+

+



+
+
y
x
P Q
P’
Q’
ux
x
x
x
u
u
x
x 


+
S R
S’
R’
x
x
u
u
y
y 


+
uy
y

Strain
• There are generally 9 strain components corresponding to the
9 stress components (xx, xy, xz, yx, yy, yz, zx, zy, zz)
because of equilibrium: ij = ji, meaning that xy = yx, yz =
zy, and zx = xz.
• We can define the following strains:
▫ Normal strains ( )
▫ Shear strains ( )
𝜀𝑥𝑥 =
𝜕𝑢
𝜕𝑥
, 𝜀𝑦𝑦 =
𝜕𝑣
𝜕𝑦
, 𝜀𝑧𝑧 =
𝜕𝑤
𝜕𝑧
𝜀𝑥𝑦 =
𝜕𝑣
𝜕𝑥
+
𝜕𝑢
𝜕𝑦
, 𝜀𝑦𝑧 =
𝜕𝑤
𝜕𝑦
+
𝜕𝑣
𝜕𝑧
, 𝜀𝑧𝑥 =
𝜕𝑢
𝜕𝑧
+
𝜕𝑤
𝜕𝑥

Strain
• Dilatation () is known as the change in volume
(V) per unit volume (V):
• The strain matrix composed of the nine components
of strain:
Δ =
Δ𝑉
𝑉
= 𝜀𝑥𝑥 + 𝜀𝑦𝑦 + 𝜀𝑧𝑧 =
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑦
+
𝜕𝑤
𝜕𝑧
𝜀 =
𝜀𝑥𝑥 𝜀𝑥𝑦 𝜀𝑥𝑧
𝜀𝑦𝑥 𝜀𝑦𝑦 𝜀𝑦𝑧
𝜀𝑧𝑥 𝜀𝑧𝑦 𝜀𝑧𝑧

Hook’s law
• It states that the strain is directly proportional to the stress
producing it.
• An elastic object is one that
returns to its original size
and shape after the act
forces have been removed.
• The energy is released in
the form of seismic waves
in earth materials are such
that Hooke’s law is always
satisfied.

Hook’s law
• In isotropic media, Hooke’s law takes the following form:
𝜎11
𝜎22
𝜎33
𝜎12
𝜎13
𝜎23
=
𝜆 + 2𝜇 𝜆 𝜆 0 0 0
𝜆 𝜆 + 2𝜇 𝜆 0 0 0
𝜆 𝜆 𝜆 + 2𝜇 0 0 0
0 0 0 𝜇 0 0
0 0 0 0 𝜇 0
0 0 0 0 0 𝜇
𝜀11
𝜀22
𝜀33
𝜀12
𝜀13
𝜀23

Hook’s law
• Hooke’s law in an isotropic medium is given by the following
index equations:
• These equations are sometimes called the constitutive
equations.
𝜎𝑖𝑗 = 𝜆 Δ + 2 𝜇 𝜀𝑖𝑗 (𝑖 = 𝑥, 𝑦, 𝑧)
𝜎𝑖𝑗 = 2 𝜇 𝜀𝑖𝑗 (𝑖 ≠ 𝑗, 𝑖, 𝑗 = 𝑥, 𝑦, 𝑧)

One dimensional wave equation
• To get the wave equation, we will develop Newton’s second
law towards our goal of expressing an equation of motion.
• Newton’s second law simply states:
𝐹 = 𝑚 ⋅ 𝑎

Equation of motion
• Using constitutive equations
and Newton’s second law, to
derive the wave equation in
one dimension In order to
obtain the equations of
motion for an elastic medium
we consider the variation in
stresses across a small
parallelepiped.
𝜎𝑥𝑥 +
𝜕𝜎𝑥𝑥
𝜕𝑥
𝑑𝑥
z
y
x 𝜎𝑥𝑦 +
𝜕𝜎𝑥𝑦
𝜕𝑦
𝑑𝑦
𝑑𝑥
𝑑𝑧
𝑑𝑦
𝜎𝑥𝑧 +
𝜕𝜎𝑥𝑧
𝜕𝑧
𝑑𝑧
𝜎𝑥𝑧
𝜎𝑥𝑦
𝜎𝑥𝑥

Equation of Motion
• Stresses acting on the surface of a small
parallelepiped parallel to the x-axis.
• Stresses acting on the front face do not
balance those acting on the back face.
• The parallelepiped is not in equilibrium
and motion is possible.
• If we first consider the forces acting in
the x-direction, hence the forces will be
acting on:
▫ Normal to back- and front faces,
▫ Tangential to the left- and right-hand
faces, and
▫ Tangential to the bottom and top
faces.
37
𝜎𝑥𝑥 +
𝜕𝜎𝑥𝑥
𝜕𝑥
𝑑𝑥
z
y
x 𝜎𝑥𝑦 +
𝜕𝜎𝑥𝑦
𝜕𝑦
𝑑𝑦
𝑑𝑥
𝑑𝑧
𝑑𝑦
𝜎𝑥𝑧 +
𝜕𝜎𝑥𝑧
𝜕𝑧
𝑑𝑧
𝜎𝑥𝑧
𝜎𝑥𝑦
𝜎𝑥𝑥

Equation of Motion
• Normal force acting on the back face
force = stress x area
• Normal force acting on the front face
• The difference between two forces is given the final normal force acting
on the sample in the x-direction:
𝐹 = 𝐹2 − 𝐹1
38
𝜎𝑥𝑥𝑑𝑦 𝑑𝑧
(𝜎𝑥𝑥 +
𝜕𝜎𝑥𝑥
𝜕𝑥
𝑑𝑥) 𝑑𝑦 𝑑𝑧
(𝜎𝑥𝑥 +
𝜕𝜎𝑥𝑥
𝜕𝑥
𝑑𝑥) 𝑑𝑦 𝑑𝑧 − 𝜎𝑥𝑥 𝑑𝑦 𝑑𝑧 =
𝜕𝜎𝑥𝑥
𝜕𝑥
𝑑𝑥 𝑑𝑦 𝑑𝑧
𝐹 = 𝜎 × 𝐴

Equation of Motion
• Tangential force acting on left-hand face
• Tangential force acting on right-hand face
• The difference between two forces is given one of tangential forces acting
on the sample in the x-direction
39
𝜎𝑥𝑦 𝑑𝑥 𝑑𝑧
(𝜎𝑥𝑦 +
𝜕𝜎𝑥𝑦
𝜕𝑦
𝑑𝑦) 𝑑𝑥 𝑑𝑧
(𝜎𝑥𝑦 +
𝜕𝜎𝑥𝑦
𝜕𝑦
𝑑𝑦) 𝑑𝑥 𝑑𝑧 − 𝜎𝑥𝑦 𝑑𝑥 𝑑𝑧 =
𝜕𝜎𝑥𝑦
𝜕𝑦
𝑑𝑥 𝑑𝑦 𝑑𝑧
Get the other tangential force acting on the sample in the x-direction

Equation of Motion
• The normal force can be balanced by the mass times the
acceleration of the cube, as given by Newton's law:
• where  • dxdydz is the mass(m).
• Cancelling out the volume term on each side, the equation can
be written in the following form
𝜕𝜎𝑥𝑥
𝜕𝑥
𝑑𝑥 𝑑𝑦 𝑑𝑧 = 𝜌 ⋅ 𝑑𝑥 𝑑𝑦 𝑑𝑧 ⋅
𝜕2𝑢𝑥
𝜕𝑡2
𝜕𝜎𝑥𝑥
𝜕𝑥
= 𝜌 ⋅
𝜕2
𝑢𝑥
𝜕𝑡2
𝐹 = 𝑚 ⋅ 𝑎

Equation of Motion
• Now we may use Hooke's law to replace stress with
displacement:
• Now, substituting for xx, and remembering that the medium is
uniform so that k, m, and r are constants, we have
𝜎𝑥𝑥 = (𝜆 + 2𝜇) ⋅ 𝜀𝑥𝑥 = (𝜆 + 2𝜇) ⋅
𝜕𝑢𝑥
𝜕𝑥
= (𝑘 +
4
3
𝜇) ⋅
𝜕𝑢𝑥
𝜕𝑥
𝜕
𝜕𝑥
(𝜆 + 2𝜇) ⋅
𝜕𝑢𝑥
𝜕𝑥
= 𝜌 ⋅
𝜕2
𝑢𝑥
𝜕𝑡2

Equation of Motion
• The final form of the last equation can be written in the form;
• This equation equates force per unit volume to mass per unit
volume times acceleration.
• The equation means that Pressure is given by the average of
the normal stress components the may cause a change in
volume per unit volume.
𝜕2𝑢𝑥
𝜕𝑡2
=
(𝜆 + 2𝜇)
𝜌
⋅
𝜕2𝑢𝑥
𝜕𝑥2

Equation of Motion
• For an applied pressure P producing a volume change V of a
volume V, substituting the k is the modulus of
incompressibility (bulk modulus) in the last equation, we will
find:
𝜕2
𝑢𝑥
𝜕𝑡2
=
(𝑘 +
4
3
𝜇)
𝜌
⋅
𝜕2
𝑢𝑥
𝜕𝑥2
𝜕2
𝑢𝑥
𝜕𝑡2
= 𝑉
𝑝
2
⋅
𝜕2
𝑢𝑥
𝜕𝑥2
𝑉
𝑝 =
(𝑘 +
4
3
𝜇)
𝜌
Giving P wave equation

Blast waves
• Shock wave, strong pressure wave in any elastic
medium such as air, water, or a solid substance,
produced by supersonic aircraft, explosions,
lightning, or other phenomena that create violent
changes in pressure
OA: elastic region
AB - plastic region
BC - shock region
𝝈𝟐 = 𝟏𝟎𝝈𝟏
𝜎2
𝜎1

Blast waves
• A blast wave is an area of pressure expanding supersonically
outward from an explosive core. It has a leading shock front of
compressed gases. The blast wave is followed by a blast wind of
negative pressure, which sucks items back in towards the center.
The extent of damage caused by the blast wave mainly depends on
five factors:
- Peak of the initial positive pressure wave
- Duration of overpressure
- Medium of explosion
- Distance from the incident blast wave
- Degree of focusing because of a confined
area or walls
Blast waves from explosions that occur near or within
hard solid surfaces can be amplified two to nine times
because of shock wave reflection, causing an increase
in their destructive potential (Stewart, 2004)

Blast Waves
Types of explosive charges and related shapes of
shock wave propagation
(Non-planar waves)

Blast Waves
specific impulse:
Represents the area beneath the pressure-
time curve from arrival time to the end of
the positive phase
𝑖𝑠 = න
𝑡𝑎
𝑡𝑎+𝑡d
𝑃𝑠 𝑡 𝑑𝑡

• Pressure-time history for a blast wave is commonly described
by the Friedlander equation:
Blast Waves
Where:
𝑃𝑠 represents the blast wave overpressure,
b is the waveform parameter,
𝑡0 is the positive phase duration,
and
t is considered time

4- Wave Propagation in Continuous
and Discontinuous media
Rock Dynamics

Introduction
• Wave propagation in rock masses and its influence on the
stability of geotechnical structures are some of the most
important topics in rock dynamics and earthquake
engineering. Rock joints discontinuities play an important
role on.
• wave propagation: when an elastic wave impinges a
joint, part of the energy is transmitted and part is
reflected. The amplitude of the transmitted and reflected
waves depends on the joint model assumed, on its
geometrical properties (spacing, length, thickness) and on
the frequency content.

The energy partitioning of seismic wave
• When an incoming P-wave strikes an interface between two
isotropic homogeneous elastic media at an angle other than the
vertical, a portion of the P-wave energy is converted into S-wave
energy, which gets reflected and transmitted in the same way as
the P-wave does. This is known as the energy partitioning of
seismic wave.

Laboratory tests for wave propagation studies
Intact specimen and specimens with smooth fractures

Laboratory tests for wave propagation studies
Specimens with tooth fractures

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