Lecture 10: Stress tensor Mohr circles states of stress strain

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Lecture 10 Stress tensor Mohr circles states
of stress strain
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stress acts on every surface that passes through
the point we can use three mutually
perpendicular planes to describe the
stress state at the point, which we approximate
as a cube
each of the three planes has one normal
component two shear components
therefore, 9 components necessary to
define stress at a point
3 normal 6 shear
convention for describing sij
i 1, 2, 3 j1, 2, 3
i plane on which component acts (defined by
perpendicular) j axis to which component is
parallel
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characteristics of the nine components on the
three surfaces
those with same first subscript act on same
plane
act on plane normal to x1
?11, ?12, ?13
act on plane normal to x2
?21, ?22, ?23
act on plane normal to x3
?31, ?32, ?33
those with the same second subscript act in same
direction
?11, ?21, ?31
parallel to x1
?12, ?22, ?32
parallel to x2
?13, ?23, ?33
parallel to x3
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more characteristics
those with same two subscripts
are normal stresses
?11, ?22, ?33
those with different two subscripts are shear
stresses, ?
?12, ?13, ?23 ?21, ?31, ?32
for cube to be in equilibrium (at rest not
moving, not spinning)
if they dont offset, block spins
?21?12
?23?32
therefore, only six are independent
?31?13
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the nine components (six of which are
independent) can be written in matrix
form
this is the stress tensor
remember, a vector is represented by 3
coordinates in a Cartesian reference frame
(x, y, z) this is a matrix with 3 components
stress at a point has nine components this is a
matrix with 9 components (second-rank tensor)
components on diagonal are normal stresses off
are shear stresses
describing stress as tensor allows relative ease
of manipulation can change reference frame
(rotation translation)
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principal stresses
when examining stress at a point, it is possible
to choose three mutually perpendicular planes on
which no shear stresses exist
think in two dimensions first
stress applied to plane
same stress
different orientation of plane
normal and shear components
only normal component shear stresses are zero
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in three dimensions, one combination of
orientations for the three mutually
perpendicular planes will cause the shear
stresses on all three planes to go to zero
this is the state defined by the principal
stresses
principal stresses are normal stresses that
are orthogonal to each other
principal planes are the planes across which
principal stresses act
(faces of the cube)
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for principal stresses (shear stresses are zero),
matrix simplifies
only one subscript kept
?1???2???????3?are the principal stresses where
?1 ???2 ???3
(remember, diagonal components of matrix are
normal stresses)
principal stresses define axes of stress
ellipsoid
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we use the principal stresses to define some
other terms
mean stress average of 3 principal stresses
deviatoric stress
using mean stress, we can separate stress tensor
into 2 components


mean stress hydrostatic/lithostatic pressure
e.g. weight of overburden same in every
direction isotropic
deviatoric stress produces distortion s1
always compression s3 always tension
anisotropic
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how will we determine stress in a rock?
compress a block of clay between two wooden
planks
what are normal and shear stresses on
an arbitrarily oriented plane in the block
of clay?
we will express the stresses on the arbitrary
plane (AB) (the plane AB makes the trace AB in
our diagram) in terms of the principal
stresses ?1, ?2, and ?3
?1 is from squeezing the clay ?2 and ?3 are equal
as we are doing our experiment in the room (both
are atmospheric pressure) we will ignore ?2
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take arbitrary plane, AB, that is at angle ? to
?3 (?3 is parallel to BC)
?1
assume plane has unit area, e.g. 1 unit2 thus AB
has unit length, e.g. 1 unit
?s
A
we find area of plane AB (length of trace AB)
D
?n
?3
..along AC (parallel to ?1)
?
area of AC 1 x sin ? sin q
B
C
..along BC (parallel to ?3)
area of BC 1 x cos ? cos q
next we consider forces acting on each plane
represented by AB, BC, and AC
force stress x area
force on side BC ?1 x cos ? ?1 cos q
force on side AC ?3 x sin ? ?3 sin q
force on AB is a normal force, ?n x 1 and a
shear force, ?s x 1
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for equilibrium, forces must be in balance (block
does not spin or move) balance parallel to AB
and balance perpendicular to AB
(direction CD)
?1
?s
A
D
perpendicular to AB
?n
force AB force BC resolved on CD
force AC resolved on CD
?3
?
B
C
remember force stress x area
1 x ?n ?1 cos ? cos ? ?3 sin ? sin ?
?n ?1 cos2 ? ?3 sin2 ?
if we substitute the following trigonometric
relationships
cos2 ? 1/2 (1 cos 2?? sin2 ? 1/2 (1 - cos
2??
into the above equation, we get..
?n 1/2 (?1 ?3) 1/2 (?1 - ?3 ) cos 2?
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1 x ?s ?1 cos ? sin ? - ?3 sin ? cos ?
?s (?1 - ?3) sin ? cos ?
?1
parallel to AB
force parallel AB force BC resolved on
AB force AC resolved on AB
?s
A
D
?n
?3
these 2 forces act in opposite direction so one
has to be negative
?
B
C
if we substitute the following trigonometric
relationship
sin ? cos ????????(???????
into the above equation, we get..
?s 1/2 (?1 - ?3) sin 2?
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from these equations, which are the fundamental
stress equations
?n 1/2 (?1 ?3) 1/2 (?1 - ?3 ) cos 2?
?s 1/2 (?1 - ?3) sin 2?
we can see
planes of maximum normal stress are at
angles of 0 with respect to ?3 cos
2? reaches maximum value at that angle (cos
2????????????????????)
(remember ? was defined as angle relative to ?3)
planes of maximum shear stress are at angles
of 45 with respect to ?3 sin 2?
reaches maximum value at that angle (cos
2????????????????45 and 2? is 90)
so faults should form at 45 from ?3 ..
(or from ?1 as 45 is equidistant between the
two)
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the equations we derived
?n 1/2 (?1 ?3) 1/2 (?1 - ?3 ) cos 2?
?s 1/2 (?1 - ?3) sin 2?
can be rearranged and squared (why, you
are wondering, would we do that?) (well, Otto
Mohr did that about 150 years ago so we will, too)
??n - 1/2 (?1 ?3)2 1/2 (?1 - ?3 )2 cos 2 2?
?s2 1/2 (?1 - ?3 )2 sin 2 2?
now we add the two equations together (Otto did
that ,too)
??n - 1/2 (?1 ?3)2 ?s2 1/2 (?1 - ?3 )2
(cos 2 2?? sin 2 2??
using trigonometric relationship (cos 2 2?? sin
2 2??? 1 gives
??n - 1/2 (?1 ?3)2 ?s2 1/2 (?1 - ?3 )2
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the resulting equation
??n - 1/2 (?1 ?3)2 ?s2 1/2 (?1 - ?3 )2
has the form (x-a) 2 y 2 r 2
which is the equation for a circle with radius,
r, centered on the x-axis at a distance, a,
from origin
in this case radius, r 1/2 (?1 - ?3 )
distance, a, from origin 1/2 (?1
?3) x-axis ?n
(?1 - ?3 ) is the differential stress 2r
diameter of circle
1/2 (?1 ?3) is the mean stress center of
circle
this graphical representation is the Mohr Circle
(for Otto)
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2??
(?1 - ?3 )
Mohr circle
?s
P
?sP
?n
?3
?1
?nP
?1
P plane to left
(?1 ?3 )
(?1 - ?3 )
?
?3
2
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remember that the expression for the circle was
done in terms of 2?
point P represents a plane that is ??degrees from
?3 plots 2q counterclockwise from ?1 (note
original diagram in upper right)
normal and shear stresses on point P are ?nP and
?sP, respectively
?nP 1/2(?1 - ?3 ) 1/2(?1 - ?3 )cos 2? ?sP
1/2(?1 - ?3 ) sin 2?
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let us look at Mohr circle more closely
two planes oriented at angle ? and its
complement, 90-? which have equal shear
stresses but different normal stresses two
planes with equal normal stresses but shear
stresses of opposite sign
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Mohr circle also tells us along which planes
shear stress is greatest
planes with greatest shear stress make an angle
of 2? 90 or angle of ?45 with respect to
principal stresses
magnitude of ?s is equal to radius of circle,
1/2(?1 - ?3 ), which is twice (?1 - ?3 ) which
is the differential stress
thus ?d 2?s
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some other things to note about the Mohr circle
for this example, both ?1 and ?3 are positive
(compressive)
for ?3 as tensile stress, ?3 must be
negative
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let us look at an example.(from the lab textbook)
suppose ?????1 is oriented east-west,
horizontal, and equal to 40 MPa ?3 is
vertical and equal to 20 MPa determine normal
and shear stresses on a fault plane that strikes
north-south and dips 55 west
angle, ?? between ?3 and fault plane is 35
2? 70
but which direction on Mohr circle?
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convention for shear stresses for Mohr circle
diagram sinistral is considered positive
() dextral considered negative (-)
positive (sinistral)
negative (dextral)
our example is dextral and thus, negative
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to construct Mohr circle, we know
?????s
?????1 is equal to 40 MPa ?3 is equal to 20
MPa
2? 70 and is negative
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?????n
20
40
70
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?????n 33.4 ????s -9.4
P
?????1 ?3
distance from origin of center of circle
(4020)/230
2
?????1 - ?3
diameter of circle 40-2020
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some special states of stress
uniaxial stress only one non-zero principal
stress, i.e. ?1
biaxial stress one principal stress equals zero,
two do not, i.e. ?1 gt ?3 ?2 0
triaxial stress three non-zero principal
stresses, i.e. ?1 gt ?2 gt ?3
isotropic stress three principal stresses are
equal, i.e. ?1 ?2 ?3
axial stress two of three principal stresses are
equal, i.e. ?1 ?2 or ?2 ?3
lithostatic pressure weight of overlying column
of rock Plithostatic r????gz (density,
gravity, depth)
hydrostatic pressure weight of column of fluid
in interconnected pore spaces imagine being
in a swimming pool Phydrostatic
rfluid gz (density, gravity, depth)