tegangan elastisitas (elasticity stress)

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  • 8/11/2019 Tegangan Elastisitas (Elasticity Stress)

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    CE172 Introduction to Rock Mechanics, Spring, 2004 Prof. Steven Glaser University of California

    1

    Stress, and Strain, and Elasticity

    As engineers, we are usually vitally interested in forces and displacements This is because as

    engineers designing or evaluating a structure we are concerned with deformation orcompetence of the structure - will it fail at design load? Will there be so much deformation at

    design load that the structure will not function as desired?

    Importance of Forces for Rock Mechanics:

    1) There are pre-existing forces in the earth, from both gravity loading and active tectonic

    forces. Since rock engineers design structures in the earth, we have to know the forces we

    are designing for.

    2) When we construct a structure in the ground, we are disturbing the existing equilibrium of

    forces. We can dramatically alter the state of loading, whether by unloading (excavation)

    or extra loading (foundation or dam abutment).

    3) In order to design, we need computational tools that describe mechanical behavior of the

    materials using easily measurable variables. Engineering design criteria are all in

    terms of load and deformation (stress and strain).

    Loads are the product of forces distributed over the body in question (the dreaded

    engineering mechanics potato). In order to normalize the physical size of a given problem out

    of our calculations, we devise the concepts of stress and strain. In actuality, these are not straight-forward concepts! Stress and strain

    only exist as intellectual abstractions.

    Strictly speaking, stress is a point property, reducing the effects of a multitude of external

    forces to a andacting on a point on an arbitrary plane within. It is an imaginary

    mathematically defined quantity:

    (1)

    We never measure stress directly! We measure the displacement of a material with known

    properties, i.e. E; we calculate stress: = E. We can define other elastic constants:

    E= Youngs Modulus = . (2)

    F A loaded= extent extent =

    normalst re ss N, F

    A------- sh earst re ss S

    A-------

    A 0( )lim,;

    A 0( )lim=

    NN------

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    CE172 Introduction to Rock Mechanics, Spring, 2004 Prof. Steven Glaser University of California

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    For ease of later calculations, and understanding the physicalness of elasticity, let us define

    another elastic constant - Poissons Ratio ()- which is the ratio of lateral extension to axial

    compression. From the figure, we see that ranges between 0 (cork) and 0.5 (rubber).

    (3)

    For a 3-dimensional state of stress (prism of rock is confined by neighboring prisms of rock)

    we can define a bulk modulus (K):

    (4)

    and the compressional seismic velocity in the solid Vp= (5)

    Similar constants can be defined for shear:

    (6)

    shear modulus = G = (7)

    shear wave velocity = Vs= (8)

    For convenience, define one last constant, Lame`'s constant (9)

    Lets look at STRESS in detail.

    Stress is a tensor quantity! What the hell are tensors? Lets start with what we already know:

    Scalar quantities have magnitude only - temperature, speed, # of home runs, age, etc. They

    are characterized by a single numerical value

    la taxial------------ Ela t

    axial------------- l El

    a Ea------------------ ElEa

    la----------- E

    2

    2------= = = = =

    axial

    lat 2

    K E

    3 1 2( )------------------------=

    K

    ---- where is density

    shear stress = transverse force

    area of action-------------------------------------- shear strain = ; transverse displacement

    height of body---------------------------------------------------------= =

    --

    G

    ----

    E1 +( ) 1 2( )

    --------------------------------------=

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    CE172 Introduction to Rock Mechanics, Spring, 2004 Prof. Steven Glaser University of California

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    A vector is quantity that needs 2 or more numerical descriptors to characterize the quantity

    - velocity (speed and direction), temperature change (temp at many given times),

    distribution of discontinuities in a given direction, etc.

    for velocity we need 3 values - the Cartesian coordinates (really 4 - magnitude and 3

    directional cosines)

    for room temperature we might only need 2 descriptors - temperature readings and

    the time when the temperature readings were taken.

    (Note - characterizing locations in 2-coordinate systems is a common use of complex numbers.)

    A tensor is devised quantity that has magnitude and direction that are themselves a

    function of the coordinate system chosen. Examples - stress, strain, permeability, physical

    fields (gravity, electromagnetic, waves, etc.), scary things from physics.

    Tensors are abstract objects which can be described by arrays of functions; each

    function of such an array is called a component. Components are functions of the

    selected coordinates....Tensor analysis offers at least the following

    advantages:...Physical concepts requiring many functions to express are formulated

    easily...Results are translated easily to forms appropriate to any coordinate

    system...Physical concepts can be expressed without reference to any particular

    coordinate system. - from Tensors of Geophysics for Mavericks and Mongrels, Hadsel,

    F., SEG Press.

    Normal Stress Components and Shear Stress Components

    From statics, remember that on any arbitrary plane cut through a material there are both

    normal andshear stresses. This combination gives us a tensor quantity. (This is only true for

    a material with shear strength, i.e. a solid.)

    The normal and shear stress components are the normal and shear forces per unit area, using

    the notation Fnfor normal force, Fsfor shear force, for normal stress and for shear stress.

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    CE172 Introduction to Rock Mechanics, Spring, 2004 Prof. Steven Glaser University of California

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    Fnand Fscan be easily found from the force magnitude Fand the directional cosine -

    . (10)

    For the case of stress (Fn/A), : (11)

    In the first case we only have to resolve 1 quantity - force. In the second case we have to

    resolve both normal force and the area the force acts upon, and the area changes depending

    on the arbitrary orientation of the plane through the object - the plane in question.

    Stress Components on a Small Cube Within the Rock

    For convenience and familiarity, lets use the Cartesian coordinate system, and enlarge our

    mathematical point to an engineers point - a cube. In this case, the normal forces will act

    perpendicular - normal- to the Cartesian planes. The shear stresses will, in general, not bealigned with the Cartesian planes. They need to be reduced to a pair of shear stresses acting

    parallel to the Cartesian plane - one each acting in the 2 remaining Cartesian directions:

    Each face of our unit cube - the x-face, the y-face, and the z-face - end up with 3 stress

    components acting on it. We now have the 9 stress components - 3 normal components and 6

    shear components.

    Fn F Fs;cos F sin= =

    N cos2=

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    CE172 Introduction to Rock Mechanics, Spring, 2004 Prof. Steven Glaser University of California

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    Notation convention: the first subscript to a stress value is the plane on which the component

    acts; the second subscript defines the direction in which the stress acts.

    Now, combine all the components into a matrix where

    rows = component on a given plane

    columns = stress components acting in a given direction

    (12)

    This is the world-famous Stress Tensor!!

    Symmetry of the Stress Matrix

    We have so for defined 9 stress components for a given point in a rock. Since this is statics, the

    rock is in equilibrium, so the forces and moments acting on the rock are in equilibrium.

    Remember from statics:

    xx xy xzyx yy yzzx zy zz

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    CE172 Introduction to Rock Mechanics, Spring, 2004 Prof. Steven Glaser University of California

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    e.g. (13)

    So, (14)

    and

    (15)

    giving us a symmetric matrix and only 6 independent stress components to define.

    Note

    - Stress is a Tensor. It takes 6 independent components to define the state of stress of a

    point on an arbitrary plane in a body.

    - Stress is very different from pressure! Pressure is hydrostatic, i.e. it acts in all directions

    uniformly, so is a scalar.

    - Whatever method is used to specify the stress state, you must have 6 independent pieces

    of information!!! (and we cannot directly measure all components)

    The Principal Stresses

    The actual values of the 3 normal and 3 shear components will vary according to the

    orientation of the reference cube with the rock. What if we pick an orientation such that the

    shear stresses go to zero, and the normal stresses go to their maximum (minimum)?

    The Principal stresses are the normal stresses on the 3 orthogonal cube faces when the

    orientation is such that the shear stresses are zero. We still need 6 independent pieces of

    information since the orientation (directional cosines) must be provided in addition to 1, 2,

    and 3(where 1> 2>3).

    MO

    l

    2-----

    l( )2xy

    l2-----

    l( )2yx 0= =

    xy yx yz zy ;= xz zx ;=;=

    xx xy xzyx yy yzzx zy zz

    xx xy xzxy yy yzxz yz zz

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    All Unsupported (free) Excavation Surfaces are principal Stress Planes

    An unsupported excavated surface has no shear stresses acting on it - They are principal stress

    planes. In addition, the normal stress acting on these faces are zero (by definition). For

    example:

    in this case, the Principal stress tensor would be:

    . (16)

    0 0 0

    0yy yz0 yz zz

    3 0=( ) 0 0

    0 1 0

    0 0 2

    =

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    Stress Invariants of the 3-D Stress Tensor

    There are three Tensor quantities (invariants) that are constant no matter what the orientationof the reference cube - I1, I2, and I3.

    (17)

    (18)

    (for principal stresses- ) (19)

    (for principal stresses - ) (20)

    Also define Octahedral Stress:

    (Can change material state, behavior)(21)

    (Causes material damage) (22)

    ** Nondeviatoric stress does not cause distortion (damage, within reason), only volumetric

    change.

    I1 x y z+ + 3mean hydrostatic= = =

    I2 xy yz zx+ +( ) 2xy

    2yz

    2zx+ + +=

    I2 12 23 31+ +( )=

    I3 xyz 2xyyzzx x2 yz y2 zx z2 xy+= I3 123=

    A

    C

    B

    P

    nT(n)

    T(-e1)

    T(-e3)

    T(-e2)

    x1

    x3

    x2

    h

    oc t1 2 3+ +( )

    3----------------------------------- meanI1

    3---- NondeviatoricStress= = = =

    oc t2

    3------- I1

    23I2+ DeviatoricStress= =

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    CE172 Introduction to Rock Mechanics, Spring, 2004 Prof. Steven Glaser University of California

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    ** Deviatoric stress causes shear distortion and material damage.

    The stress state can always be broken down into Deviatoric and Nondeviatiric components:

    (23)

    For the hydrostatic case:

    , and define the Bulk Modulus K: (24)

    (25)

    Example

    Say you measure the following stresses, and you want to decompose your stress matrix into

    the deviatoric and nondeviatoric components

    (26)

    (27)

    therefore (28)

    1 xx= xy xz

    xy 2 yy= yzxz yz 3 zz=

    m 0 0

    0 m 0

    0 0 m

    1 m( ) xy xz

    xy 2 m( ) yzxz yz 3 m( )

    +=

    TotalStressState

    =

    Hydrostatic Stress State (volume)

    + Stress Deviation (distortion)

    [ ]

    P0 0 0

    0 P0 0

    0 0 P0

    and xx, yy zzP

    0

    1 2( )

    E---------------------------= = = =

    KP0

    Vo l-------------

    E

    3 1 2( )------------------------= =

    10 5 4

    5 20 12

    4 12 30

    so xx=10MP a yy; =20MP a zz; =30MP a and mean;10 20 30+ +

    3------------------------------ 20MP a= =

    10 5 4

    5 20 12

    4 12 30

    20 0 0

    0 20 0

    0 0 20

    10 5 4

    5 0 12

    4 12 10

    +=

    TotalStressState

    =

    HydrostaticStress State(volume)

    +StressDeviation(distortion)

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    Strain

    Strain, , is a Tensor quantity just like stress. In fact, as long as the material is

    Elastic, there is an equivalence between the two. This allows us to estimate strains(displacements) if we know the stress (load) to be applied, or estimate the applied stress

    (loads) by measuring the strains (displacements).

    e.g. For design of a dam abutment, we need to estimate the amount of horizontal

    deformation we should expect from the design retained pool acting on the dam.

    e.g. We need to know the in situ stresses acting on a new tunnel bore so that we can design

    the proper support system. We can measure how the tunnel walls strain inward and

    calculate the design stresses.

    NOTE WELL: Strain is an intellectual abstraction, a point property which we cannot

    measure directly. We measure l over a known length (gage length). A strain

    measurement without the gage length is a deceptive snare!

    For normal loading, this linear relationship between stress and strain is through Youngs

    Modulus (E) as first stated by Robert Hooke (1678) in an anagram in the Times of London - as

    the extension so is the force.

    (29)

    More generally, (30)

    where (31)

    [S] is called the compliance matrix, and is the inverse of the Stiffness matrix. The Compliance

    matrix is also symmetrical, so we actually have 21 independent elastic constants to measure to

    fully characterize an elastic material, rather than 36. In the generalized case we have a fully

    ll 0( )lim l=

    xx xxExx=

    S =

    xxyyzzxy

    yzzx

    and

    xxyyzzxy

    yzzx

    and S

    S11 S12 S13 S14 S15 S16

    S12 S22 S23 S24 S25 S26

    S13 S23 S33 S34 S35 S36

    S14 S24 S34 S44 S45 S46

    S15 S25 S35 S45 S55 S56

    S16 S26 S36 S46 S56 S66

    = = =

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    anisotropic material, but in practice we cannot measure 21 elastic constants, so simplify - in

    the pure elastic case (isotropic, homogeneous elastic solid) we only need 2 elastic constants,

    e.g. Eand G.

    What does the Stiffness Matrix mean?

    (32)

    For a first approximation, assume that there is no coupling between normal and shear

    components and no coupling of shear components acting in different directions. This makes

    components and disappear. By definition the direct relation between normal strain and

    normal stress is 1/E, so will be the reciprocals of Ei. By the same logic, the rest of the maindiagonal will be 1/Gij. From the definition of Poissons ratio, = lateral/axial, which links

    indirect coupling of normal components, we know is -ij/Ei.

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    CE172 Introduction to Rock Mechanics, Spring, 2004 Prof. Steven Glaser University of California

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    The result is the following compliance matrix (33)

    This is the orthotropic case, and we have 9 independent elastic constants - the 3 Youngs

    Moduli, the 3 shear Moduli, and the 3 Poissons ratios. This could happen if there are 3

    orthogonal different sets of discontinuities in a rock mass,

    Simplifying even further, we have the transverse isotopic case. An example is laminated or

    layered rocks and rock masses. If the plane of isotropy is parallel to the Cartesian axes 1 and

    2:

    E1= E2= E (34)

    E3= E' (35)

    G' = G (36)

    12= 21= (37)

    v13= v23= v' (38)

    1

    E1------

    21E2-------

    31E3------- 0 0 0

    21E2-------

    1

    E2------

    32E3------- 0 0 0

    31E3-------

    32E3-------

    1

    E3------ 0 0 0

    0 0 01

    G1------ 0 0

    0 0 0 01

    G2------ 0

    0 0 0 0 01

    G3------

    S=

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    G12 G23, and G23= G31= G' so, (39)

    We now need to measure 5 elastic constants.

    Finally, assume complete isotropy -

    E1= E2= E3= E (40)

    12= 23= 31= (41)

    G12= G23= G31= G (42)

    and , giving (42)

    (43)

    * Remember that we also have volumetric strains, , which will become important when

    we look at poroelastic behavior.

    1

    E---

    E---

    'E'---- 0 0 0

    E---

    1

    E--- '

    E'---- 0 0 0

    'E'---- 'E'

    ---- 1E'---- 0 0 0

    0 0 02 1 +( )

    E-------------------- 0 0

    0 0 0 01

    G'----- 0

    0 0 0 0 01

    G'-----

    S=

    G E2 1 +( )--------------------=

    1

    E---

    1 0 0 0 1 0 0 0 1 0 0 00 0 0 2 1 +( ) 0 0

    0 0 0 0 2 1 +( ) 00 0 0 0 0 2 1 +( )

    S=

    vo lvol

    ------------