SD Lecture05 Elastic Stress Waves

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    Soil Dynamics

    Lecture 05

    Elastic Stress Waves in Bars

    Luis A. Prieto-Portar, August 2006.

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    The propagation of stress in an elastic medium.

    When stress is applied to a body, that stress will propagate away from the point of

    application via stress waves. Different materials will propagate the stress at different

    speeds.

    For example, in sands the stress will propagate at about 1,000 feet/sec. In sandstones,the stress will propagate at 14,000 feet/sec.

    This lecture deals with the propagation of stress waves in elastic media in the form of

    rods, bars or beams.

    An example of this type of stress is that induced by a pile driving hammer striking

    the head of a pre-cast concrete pile (shown on the next three slides).

    The propagation of stress is central to the understanding of how dynamic loads

    propagate in soils, whether the loads come from gravity, wind, explosions, etc., orfrom earthquakes.

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    Stress and strain in elastic media.

    The notation for the normal and shear stresses in an idealized very small element of amuch larger elastic body is shown below.

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    1

    21

    2

    1

    2

    x

    y

    z

    xy yx

    yz zy

    zx xz

    x

    y

    z

    xy x

    yz y

    zx z

    Normal stresses ,

    Shear stresses

    uNormal strains

    x

    v

    yw

    z

    v u w vShear strains and

    x y y zw v u w

    y z z x

    u w v u

    z x x y

    =

    =

    =

    =

    =

    =

    = + =

    = + =

    = + =

    where is the components

    of rotation about the x, yand the z axes.

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    Constitutive Relations (Hookes law).

    In elastic and isotropic media, the stresses and the strains are related to each otherthrough relationships called constitutive relationships. These are:

    ( )

    ( )

    ( )

    1

    2

    12

    1

    2

    2 1

    x x y z

    y y z x

    z z x y

    xy xy

    yz yz

    zx zx

    where E is Young' s elastic mod ulus and is Poisson' s ratio.

    Shear stresses and shear strains are related via the shear mod ulus G,

    EG where G

    ( )G

    G

    = +

    = +

    = +

    = =+

    =

    =

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    ( ) ( )

    ( )

    22

    2

    1 1 2

    2

    x x

    y y

    z z

    x y z

    Normal stresses can also be exp ressed in t erms of strains,

    GG

    G

    where,E

    and called the volumetric strain

    Notice that ,

    G

    = += +

    = +

    = = + ++

    =+

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    Longitudinal Elastic Stress Waves in a Bar.

    The rod shown below is experiencing the propagation of a stress wave from left to

    right. Using Newtons second law of motion (F = ma) we can find the differential

    equation of motion.

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    ( )

    ( )

    2

    2

    2

    2

    2

    2

    2 22

    2 2

    1

    C

    A xW V uA x A ma a a

    g g g t

    Since g ,

    ux A x Assume A , and

    x t

    usince E E , replace in the above equation,

    x

    u uEx x t

    u uor v whe

    t x

    + + = = = =

    =

    = =

    = =

    =

    =

    C

    Cv is the longitudinal stress wav

    Ere v

    w e velohe ce yr .it

    =

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    ( ) ( )( )

    ( )

    1

    2

    C C

    P

    t C

    solution to find the displacement due to the stress wave could be ,

    u F v t x G v t x The function F v t x is the propagation going along x ( Block # ) at time t ,

    u F v t x

    At time t t the function is represented by Block # shown below.

    = + + +

    = +

    +

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    ( ) ( )

    ( ) ( ) ( )

    1 2

    t t C

    t t t

    C P

    C

    C

    Therefore,at time t t,

    u F v t t x x If the block moves left without changing its shape from position# to position# ,

    u u

    orF v t x F v t t x x

    or

    v t x

    Therefore, the longitudinal stress wave velocity v

    +

    +

    +

    = + +

    =

    + = + +

    =

    =

    ( )C

    x

    t

    Similarly, the functionG v t x represents a wave traveling to the right.

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    The velocity of the particles in the stressed zone.

    There is a difference between the longitudinal stress wave propagation vc and thevelocity of the particles vc in the stressed zone.

    Consider a compression stress pulse of intensity xand duration tapplied to the rod.

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    The compressive stress pulse could be the result of a striking hammer upon a pre-cast

    concrete pile. When this stress pulse is initially applied, that zone of the rod or pile

    will be under compression. This compression wave will soon move to an adjacentzone, such that, during the time interval t the stress will have traveled a distance x,

    ( )

    C

    C

    x xC

    x v t

    A t an y time later on, that is , when t t' a segment of the pile of length x willconstitute the comressed zone. Hence,

    x v t'

    The elastic shortening u of the rod in t he wave packet x is thus,

    u x v t' where u is the disE E

    =

    >

    =

    = =

    x C

    placement of the end of the rod .

    The velocity of the particles u is therefore,

    vuu

    t' E

    = =

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    Therefore, theparticle velocity is a function of the intensity of stress x, whereas the

    longitudinal stress wave propagates at a velocity is a function of the material properties

    only, that is,

    The particle and the stress wave velocities are in the same direction when a

    compression stress is applied to the end of the pile (or rod).

    However, when a tensile stress is applied at the end of the rod, the particle and the

    stress wave velocities are in opposite directions.

    C

    Ev

    =

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    The reflection of elastic stress waves at the end of a rod or bar.

    What happens when the stress wave reaches the opposite end of the pile (or rod)?

    Notice the compression wave moving towards the right along the bar (or pile) in the

    figure (a) below, and the tensile wave moving left.

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    Notice that when the two stress waves meet at section a-a they cancel each others

    stress, but the particle velocity doubles!

    After the waves pass each other the stress and velocity go back to zero.

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    Notice that the section a-a is essentially like a free end. Therefore, in figure (d) below

    notice that a compression wave is reflected back as a tension wave. Both have the

    same magnitude and shape. Obviously, a tension wave approaching the pile (or rod)end will be reflected back as a compression wave.

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    In contrast to the compression versus tension example seen before, the figure below

    shows the effect of two identical compression waves traveling in opposite directions.

    When they cross each other (shown below) at section a-a, the stress is doubled but

    the particle velocity becomes zero.

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    After the two compression waves pass each other, the stress and particle velocity

    return to zero at section a-a.

    Notice again, that section a-a remains stationary and hence behaves as the fixed end

    of the rod or pile. Notice (below) that at the end, the compression wave is reflectedback also as a compression wave. At the end the stress is doubled.

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    Torsional waves in a rod (bar or pile).

    In the figure below, the rod is experiencing a torsional force or torque T; the bar willbe rotated an angle at a distancex.

    T

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    2

    2

    2 2 2 22

    2 2 2 2s

    s

    TT T x J

    x t

    where J is the polar moment of inertia of the bar cross sec tion.

    The torque T itself is given by,

    T JGx

    Re placing T in t he force equation yields,

    Gor v

    t x t x

    where,

    Gv Notice th

    + + =

    =

    = =

    = C

    sv is the velo

    Ee similarity t

    city of torsion

    o v

    wh al wa sere .ve

    =

    The differential equation for the torsional stress wave in a bar is found again,

    starting with Newtons second law (F = ma),

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    Former Secretary of State Henry Kissinger listens to a lecture on soil dynamics.

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    Longitudinal vibration of short bars.

    The solution to the differential equation for the longitudinal stress waves of short barsvibrating in their natural mode is given by,

    ( )

    ( )

    1 2

    1 2

    2 22

    2 2

    2

    n n

    n

    C

    u( x ,t ) U( x ) A sin t A cos t

    where,U( x ) is the amplitude of the displacement along the length of the rod ,

    A and A are cons tan ts and is the natural circular requency of vibration.

    u uBacksubstituting int o v

    t x

    u x , t

    = +

    =

    ( )

    ( )

    ( ) ( )( )

    ( )

    2

    2 2

    2

    2

    2

    1 2

    0

    0n

    n n

    c c

    u x , t x E t

    or

    U x , t U x

    x E

    The solution for the amplitude U x can be of the form,

    x xU x B sin B cos

    v v

    =

    + =

    = +

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    1) End conditions: free and free.

    At the ends, the stress (and strain) at the ends are zero. In other words, at x = 0

    dU(x)/dx = 0 and at x = L dU(x)/dx = 0. Differentiating our proposed solution U(x)(previous slide),

    1 2

    11

    22

    0 0

    0

    n n n n

    c c c c

    n

    c

    n n

    c c

    n

    c

    B x B xdU ( x )cos cos

    dx v v v v

    U sin g the first boundary condition yields ,

    Btherefore B

    v

    U sin g the sec ond boundary condition yields ,B x

    ( ) sin and sin ce B is not zero ,v v

    L

    n orv

    =

    = =

    =

    =

    21 2 3

    c n

    n c

    n v L

    therefore vL n

    Thu s , the equation for the am plitude U ( x ) is ,

    n xU ( x ) B cos where n , , ...

    L

    = =

    = =

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    The free-free end condition for the vibration of a longitudinal wave in a short bar.

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    2) End conditions: fixed and fixed.

    At x = 0 U(x) = 0 (that is, no displacement) and also at x = L U(x) = 0.

    2

    1 1

    1

    0

    0 0n n

    c c

    c nn c

    The first boundary condition yields,

    B

    and the sec ond boundary condition,

    L LB sin and since B yields n

    v v

    n v Lor therefore v

    L n

    Thus, the equation for the displacement amplitude U( x ) is,

    n xU( x ) B sin

    =

    = =

    = =

    = 1 2 3where n , , ...L

    =

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    The fixed-fixed end condition for the vibration of a longitudinal short bar.

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    3) End conditions: fixed and free.

    The boundary conditions for this case are,atx= 0 (fixed end) U(x) = 0

    atx=L (free end) dU(x)/dx= 0

    2

    1

    1

    0

    0

    2 1 1 2 32

    2

    n

    n n

    c c

    c

    The first boundary condition yields,

    U ( x ) B

    and the sec ond boundary condition ,

    B LdU ( x )cos

    dx v vL

    or ( n ) where n , , ...v

    Thus , the equation for the displacement am plitude U ( x ) is ,

    U ( x ) B sin

    = =

    = =

    = =

    =

    ( )2 1n x

    L

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    The fixed-free end condition for the vibration of a longitudinal short bar.

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    The torsional vibration of short bars.

    Torsional vibration is similar to longitudinal vibrations. The equation for the naturalmodes of vibration is given as,

    [ ]1 2

    2 22

    2 2

    n n

    s

    sn

    ( x,t ) ( x ) A sin t A cos t

    where is the amplitude of the angular distorsion.The solution of the differential equation,

    v with the solution above result s in,

    t xn v

    for the free free and the fixed fixed end conL

    = +

    =

    =

    ( )2 12 sn

    ditions, and

    n vfor the fixed free end condition.

    L

    =

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    References.

    Dowding C.H., Construction Vibrations, Prentice-Hall, New Jersey, 1996;

    Das, B., Principles of Soil Dynamics, PWS-Kent Publishing Co., Boston, 1993;

    Richart F.E., Hall J.R., Woods R.D., Vibrations of Soils and Foundations, Prentice-Hall Inc., New Jersey, 1970;

    Humar J.L., Dynamics of Structures, Prentice-Hall, New Jersey, 1990;

    Prakash S., Soil Dynamics, McGraw-Hill, New York, 1981;

    Timoshenko S.P., Goodier J.N., Theory of Elasticity, McGraw-Hill Book Company,

    New York, 1970;