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Life Science Journal 2012;9(2)                     http://www.lifesciencesite.com



         Dynamic Stresses and Displacements around Cylindrical Cavities in an Infinite Elastic Medium under
                                    Moving Step Loads on the Cavity's Surface

                                          Hamid Mohsenimonfared 1       M.Nikkhahbahrami 2
1.
      Department of mechanical Engineering , Science & Research Branch , Islamic Azad University(IAU), Tehran, Iran
                       2.
                           Department of mechanical Engineering ,Tehran University , Tehran , Iran
                                         h-mohsenimonfared@ iau-arak.ac.ir

     Abstract: Potential functions and Fourier series method in the cylindrical coordinate system are employed to solve
     the problem of moving loads on the surface of a cylindrical bore in an infinite elastic medium. The steady-state
     dynamic equations of medium are uncoupled into Helmholtz equations, via given potentials. It is used that because
     of the superseismic nature of the problem, two mach cones are formed and opened toward the rear of the front in the
     medium. The stresses and displacements are obtained by using integral equations with certain boundary conditions.
     Finally, the dynamic stresses and displacements for step loads with axisymmetric and nonaxisymmetric cases are
     obtained and discussed in details via a numerical example. Moreover, effects of Mach numbers and poisson's ratio
     of medium on the values of stresses are discussed.
     [Hamid Mohsenimonfared, M. Nikkhahbahrami. Dynamic Stresses and Displacements around Cylindrical
     Cavities in an Infinite Elastic Medium under Moving Step Loads on the Cavity's Surface. Life Sci J
     2012;9(2):812-822] (ISSN:1097-8135). http://www.lifesciencesite.com. 121

     Keywords: Wave propagation, Fourier series method, cylindrical cavities, Helmholtz equations, moving load.

     1. Introduction                                                terms of Hankel functions, and finally the stresses
     1.1 General Remarks                                            and displacements are found by inversion of the
          Moving loads on the surfaces have been                    transformed quantities. In this paper the coefficients
     investigated by many researchers. Investigation on             of the stresses and displacements are found by
     dynamic stresses in solids is very significant in the          solving sets of coupled integral equations.
     study of dynamic strength of materials and in the                   The waves are expanded into Fourier series in
     design of underground structures subject to ground             terms of the angle,  , around the opening. The
     blasting waves. A related but considerably simpler             stress field of the wave is written in terms of potential
     problem has been treated by Biot (1952), who                   functions which satisfy the equations of motion.
     considered space- harmonic axisymmetric standing               These equations decoupled via introducing the
     waves and obtained a closed form solution. Another             potential functions and reduced to Helmholtz
     related problem was treated by Cole and Huth (1958),           equations that the potentials satisfy.
     who considered a line load progressing with a                       These potential functions are in integral form
     velocity V on the surface of an elastic half- space.           with unknown functions in the integrands. Therefore
     Because of the simpler geometry, they were able to             the Fourier series coefficients of the stresses and
     obtain a solution in closed form. Adrianus (2002)              displacements are also in integral form with unknown
     investigated the moving point load problem in soil             integrands. The applied boundary tractions (the step
     dynamics with a view to determine the ground                   loads) are expanded into a Fourier series in  and
     motion generated by a high-speed train traveling on            expressions for the stress and displacement
     a poorly consolidated soil with low shear wave speed.          components at points in the medium are derived for
     M.C.M. Bakker (1999) revisited the nonaxisymmet-               each term of the Fourier series as functions of the
     rical boundary value problem of a point load of                radial distance r from the cavity axis and the distance
     normal traction traveling over an elastic half-space.          z behind the wave front.
     M.Rahman (2001) considered the problem of a line                    The following three cases of step loads are
     load moving at a constant transonic speed across the           considered: normal to the surface, tangential to the
     surface of an elastic half-space and derived solution          surface in the direction of the axis of the bore, and
     of the problem by using the method of Fourier                  tangential to the circle of load application. These
     transform. Iavorskaia (1964) also studied diffraction          results can be used, by superposition, to determine
     of a plane longitudinal wave on circular cylinder.             the effects of other load patterns moving with the
     One basic method has been used for the solution of             velocity V in the direction of the axis of the bore.
     these problems, the solution is obtained by using an               Numerical solution of these equations gives the
     integral transform of the displacement potentials. The           values of the unknown functions. These values can
     resulting transformed equations are then solved in

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then be used to find the stresses and displacements on          2.    Governing equations and general solutions
the boundary and also anywhere in the medium.                        Consider a cylindrical cavity of radius r = a in a
                                                                linearly elastic, homogeneous, and isotropic medium
1.2 Problem Description                                         referred to a fixed coordinate system r, θ , z  whose
    The object of this work is to obtain stresses and           origin lies on the axis of the cavity.
displacements in an elastic medium in the vicinity of           A step load along the circle at z = -vt progresses
a cylindrical cavity which is engulfed by a plane               along the interior of the cavity with a velocity V such
stress wave of dilatational travelling parallel to the          that the stresses on the boundary r=a are:
axis of the cylinder, as shown in Figure 1.
     The step load has an arbitrary distribution P (θ)
along the circumference of the circle and moves with
                                                                σrr     σ (θ)U(z  vt)
                                                                    ra 1
                                                                                                                           (1)

a velocity V  C1  C 2 ; therefore, the speed is               σrθ 
                                                                    ra
                                                                         σ 2 (θ ) U(z  vt)                               (2)

                                                                σrz 
superseismic with respect to both the dilatational and
shear waves in the medium. Consequently, the                               σ3 (θ ) U(z  vt)                              (3)
disturbances which were initiated far behind the front                ra
on the boundary of the cavity cannot reach the                      Where the functions σ k (θθ)          define the
vicinity of the wave front for some time after the              distribution of the applied load. To determine the
incident wave passes.                                           steady state solution, a moving coordinate system (r ,
                                                                θ ,z) is introduced such that:
                                                                r  r , θ  θ , z  z  Vt                      (4)

                                                                The following treatment is restricted to the case
                                                                where the velocity V is greater than C1 and C2 , the
                                                                respective propagation velocities of dilatational and
                                                                equivoluminal waves in the medium. Hence
                                                                      V            V                             (5)
                                                                 M  1   1 , M      1 2
                                                                         C1                    C2
Figure 1. Moving step load                                      Where
                                                                          λ  2μ        μ                                  (6)
     Moreover, because of the super seismic nature of           C1              , C2 
                                                                             ρ          ρ
the problem, it should be expected that two mach
cones will be formed in the medium, as shown in                 The equations of motion in cylindrical coordinates, r ,
Figure 2. These cones should open toward the rear of             , z, for an elastic medium, may be expressed in the
the front. Furthermore, there can be no stresses or             following form:
displacements ahead of the leading front.                                                      u θ          Δ ρ  u r
                                                                                                                    2
                                                                              1                        λ
     If a coordinate system is assumed to move along             2u r             (u r  2      )  (  1)    
the cylinder with the wave front, it is seen that the                         r   2            θ      μ     r   μ t 2
state of stress at points close behind the wave front
                                                                                                              1 Δ ρ  u θ
                                                                                                                      2
                                                                              1                u r     λ
depends only on relative position of them with                   2uθ             (u  2
                                                                                  2 θ
                                                                                                   )  (  1)     
respect to the front. Thus, in the vicinity of the wave                       r                 θ      μ     r θ μ t 2
                                                                                         Δ ρ  u z
front, provided that the end of the cavity is far away,                                          2
                                                                                  λ
the problem may be treated as a steady-state case. In            2u z  (           1)   
                                                                                  μ      z μ t 2
other words, in the moving coordinate system, the
state of stress and displacement is independent of                                           (7)
time.
                                                                Where the dilatation , Δ , and the laplacian operator,
                                                                  2 , are given by:
                                                                       u    u 1 u θ u z
                                                                 Δ r  r             
                                                                        r    r r θ      z                     (8)
                                                                         2 1  1 2          2
                                                                   2
                                                                   2
                                                                                          
                                                                        r    r r r 2 θ 2 z 2
                                                                As mentioned earlier, the assumption of the existence
                                                                of a steady-state case and trans-formation form r , 
                                                                , z coordinates to r ,  , z results in elimination of
Figure 2. Geometry of the problem and the coordinate            the time variable, t , from the equations of motion.
systems


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This transformation is performed by the following                              φ n        2ψn
                                                                                           n
relations, as given in relations (4):                               u n, r                     
 z  z  Vt                                                                   r   rz r n
                                                                               n     n ψ n  n
                                                                u n, θ   φ n                              (13)
          ,   V                                                              r     r z       r
 z z t          z                                                        φ n   2 1  n 2 
Therefore equations (7) may be expressed as follows:                u n, z                    ψ
                                                                              z  r 2 r r r 2  n
                                                                                                  
           1            u θ     λ     Δ ρ 2  u r
                                                2
 2u r      (u r  2       )  (  1)     V                       These equations may be obtained from the vector
           2
           r           θ        μ     r  μ   z 2
                                                                    equation
                                       1 Δ ρ 2  u θ
                                                  2
            1          u r      λ
 2uθ         (uθ  2      )  (  1)      V                       u  grad φ  curl 
           r2           θ       μ     r θ μ     z 2              Where  is the sum of two independent vectors as
                   Δ ρ 2  u z
                                   2
             λ                                                      follows:
 2u z    (  1)       V
             μ     z μ           z 2
                                                                        curl ψ
(9)
Stress components are given by
                                                                    The vectors  and ψ have only one non- zero
              u
 σrr  λΔ  2μ r                                                    component which is in the z- direction in both cases.
               r
               u    1 uθ                                           r  0            ψr  0
σθθ  λΔ  2μ ( r        )                                         θ  0            ψθ  0
                r r θ
                                                                    z              ψz  ψ
              u
σzz  λΔ  2μ z                                        (10)
               z                                                   By substitution of the values given in equations (13)
            1 u r u θ u θ                                         into the equations (9), it can be shown that the
σ rθ  μ(                     )
            r θ      r      r                                     potential functions satisfy the modified wave
            u     u                                               equations.
σ rz  μ( r  z )
             z     r                                                        V 2  φn
                                                                                     2
            u      1 u z                                           2φ n  (   )
σθz  μ( θ                )                                                  c1    z 2
             z     r θ
                                                                               V  ψn
                                                                                      2
                                                                                                               (14)
Displacement components u r , u θ and u z may be                     2ψn  ( )2
expressed in Fourier series:
                                                                              c2     z 2

                                                                              V 2  χn
                                                                                     2
                                                                    χn  ( )
                                                                      2
u r (r,θ, z)   u n, r (r, z)cosnθ                                           c2    z 2
               n0                                                  Stress components are expressed in Fourier series
                                             (11)
u θ (r,θ, z)   u n, θ (r, z)sinnθ                                 form as follows:
                n 1                                                                       
                                                                    σ rr (r,θ, z)   σ n,rr (r, z) cos nθ
u z (r,θ, z)   u n, z (r, z)cosnθ                                                       n 0
                n0                                                                         
Three potential functions are now introduced,                        σθθ (r,θ, z)   σn,θθ (r, z) cos nθ
                                                                                          n 0
                
φ(r, θ, z)   φ n (r, z)cosnθ                                                             
               n 0
                                                                     σzz (r,θ, z)   σn,zz (r, z) cos nθ
                                                                                         n 0
ψ(r, θ, z)   ψ n (r, z)cosnθ                         (12)                                                        (15)
                                                                                           
               n 0                                                  σrθ (r,θ, z)   σn,rθ (r, z) sinnθ
                                                                                         n 1
 (r,θ, z)    n (r, z)sinnθ
               n 1                                                                        
                                                                     σ rz (r,θ, z)   σ n,rz (r, z) cos nθ
The displacement components u n, r , u n, θ and u n, z                                    n 0
                                                                                           
are defined as follows:                                              σθz (r,θ, z)   σ n,θz (r, z) sin nθ
                                                                                          n 1
                                                                         Equations (11) and (15) may be substituted into
                                                                    equations (10), and as a result stress- displacement
                                                                    relations may be written for each term of the series:
                                                                                         u n,r             (16)
                                                                     σ n,rr  λΔ  2 μ
                                                                                           r

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                   u n,r    u n,θ                                                  β1  M1  1
                                                                                       2    2
                                                                                                       ,    β2  M2  1
σ n,θθ  λΔ  2 μ 
                   r n r        
                                                                                                             2    2
                                                                                   The differential equations (14) may be written in the
                                                                                     following form:
                         u n,z
σ n,zz  λΔ  2 μ                                                                     2φn       1 φn n2           2φn
                          z                                                                          2 φn  β12
                u n,r u θ  u n,θ                                                   R2        R R R            Z2
σ n,rθ  μ   n
                                
                  r    r    r                                                     2ψn      1 ψn n2                 2ψn
                                                                                                        2 ψn  β22                 (20)
             u n,r u n,z                                                           R2 R R R                        Z2
σ n,rz  μ 
             z  r                                                                2  n 1  n n 2             2  n
                                                                                                                        2
                                                                                                     2 n  β2
                                                                                      R 2 R R R                      Z 2
             u n,θ   u n,z 
σ n,θz  μ 
            z  n r        
                                                                                     It is seen that these equations have the same general
                                                                                   form as the differential equations of the cylindrical
where                                                                                waves obtained in reference (13). Therefore solutions
                                                                                     of equations (20) may be obtained in a manner
             u n,r           u n,r            u n,θ          u n,z                 similar to that in reference (13). These solutions are
  Δ                                  n                                           given in integral form as follows (see Appendix A for
          r       r           r       z
                                                                                     verification of the solutions):
Substitution of equations (13) into equations (16) and
                                                                                          ∞
application of the differential equations (14) result in                             φn  ∫ fn (Z - Rβ1coshu)coshnu du
                                                                                           0
the following equations for stress components:
                                                                                                                                     (21)
                                                                                     ψn  0 gn (Z  Rβ2coshu)coshnudu
a 2 σ n,rr
                 
                  M 2  2M1        
                  φ n,ZZ  2 φ n,RR  2ψ n,RRZ 
                    2
                          2
                                                                                            
  μ                                                                                   n  0 h n (Z  Rβ 2coshu)coshnu du
 2n       1    
   n,R   n                                                                            From consideration of the fact that the
 R        R    
                                                                                     disturbances are zero ahead of the wave front, it is
a 2 σ n,θθ
   μ
                              
          M 2  2 φ n,ZZ  2 φ n,RR 
             2
                                                                                     seen that the functions fn , gn and hn are zero for the
                                                                                     values      of      their    arguments      less    than
 2                   
               2                                                                     Rβ1 ,  Rβ2 and  Rβ2 ,respectively.
   ψ n,RZ  n ψ n,Z   2n   n,R  1  n 
                                                                                   Therefore the upper limits of the integrals may be
 R
            R         R 
                                      R                                            changed from  to the following values:
a 2 σ n,zz
      μ
                                          
               M 2  2M1  2 φ n,ZZ  2 M 2 1 ψ n,ZZZ
                  2
                        2
                                           2                                       u1  cosh-1(1 
                                                                                                          Z
                                                                                                        Rβ1
                                                                                                             )        for φn
                                                                                                                                       (22)
                                                                                                           Z
a 2 σ n,rθ   2n 
                                                                              (17)               -1
                                                                                     u2  cosh (1             )     for ψn & n
                          1    2n          1                                                         Rβ2
               φ n,R  φ n     ψ n,RZ  ψ n,Z  
  μ          R                R                  
             
                          R                  R
                                                                                     The integrals are then written with these limits:
                                                                                                                  
 M 2  1  n,ZZ  2  n,RR
    2
                                                                                              u
                                                                                     φn  01 fn ZRβ1coshu coshnudu
a 2 σ n,rz
      μ
               2φ n,RZ           
                                   M2  2
                                    2          ψ   n,RZZ
                                                              n
                                                               n,Z
                                                              R                               u2
                                                                                     ψn  0 g n ZRβ2 coshucoshnudu                 (23)


                                                                                    n  0 hn ZRβ2 coshucoshnudu
      2
  a σ n,θz    2n                n                                                           u
                 φ n,Z  M 2  2 ψ n,ZZ   n,RZ
                         2
                                                                                                 2
      μ       R                 R
     Where the second set of subscripts of
                                                                                     3. Expressions of stresses and displacements
 φn ,ψn and n represent the partial derivatives of
                                                                                       Substitution of equations (23) into equations (17)
these functions. R and Z are the dimensionless                                       gives the following expressions for stress
variables form:                                                                      components.
  r     z                                                              (18)
R , Z
  a     a                                                                              a2σn,rr
                                                                                               0 1 f (η1)[M22  2 
                                                                                                  u
The values M1 and M2 are defined as follows:                                             μ
                                                                                       2β12sinh 2u]cosh nu du                        (24a)
      V         V
M1      , M2                                                                         2β22 0 2 g(η2 )cosh2u coshnu du 
                                                                                              u
     C1         C2                                                                         2 u2 h(η )sinh 2u coshnu du
                                                                       (19)             β2 0           2
Let



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  a2σn,θθ                                                                       stress components must satisfy the                  following
                0 1 f (η1)[M22  2
                  u
                                                                                boundary conditions:
    μ
  2β12cosh2u] cosh nu du                                        (24b)               
                                                                                 σ n,rr r a  σ n1u(Z)
 2β22 0 2 g(η2 )sinh2u cosh nu du
       u                                                                        σ n,rz  r a  σ n2 u(Z)                               (26)
  β2 0 h(η2 )sinh 2ucosh nu du                                              σ n,rθ  r a  σ n3 u(Z)
     2 u2


                                                                                These equations are satisfied for each term n. The
  a 2σ n,zz                                                                     coefficients of the stress, σ n,rr , σ n,rθ and σ n,rz are
           (M 2      2
                           2M12     2)0 1 f (η1 )cosh nu du
                                        u
                                                                   (24c)
     μ
                                                                                expressed in equations (24) in integral form. These
   2β 2 0 g(η2 )cosh nu du
        2 u2
                                                                                integrals           include       the      unknown     functions
  a 2σ n,rθ                                                                      f (η1 ), g (η 2 ) and h (η 2 ) which are to be found by
              β12 0 1 f (η1 )sinh 2u sinh nu du 
                     u
                                                                                solving the set of three simultaneous integral
     μ                                                            (24d)
 β 2 2 0 2 g(η2 )sinh 2u sinh nu du
        u                                                                       equations. These values then may be substituted back
  β 2 0 h(η2 )cosh 2u cosh nu du
       2 u2                                                                     into the equations (24) and (25) to find the stress
                                                                                components and displacement components of the
  a2σ n,rz                                                                      waves at any point on the boundary or in the medium
             2β1 u1 f (η1)coshu cosh nu du 
     μ               0                                            (24e)         behind the move front.
         2       u 2 
 β2 (M2  2) 0 g (η2 )coshu coshnu du                                          5. Solution of the Boundary Equations
  β2 0 2 h(η2 )sinh u sinh nu du
       u                                                                            Numerical solution of the boundary equations
                                                                                requires finding numerical values of the functions
  a2σn,θz                                                                        f (η1 ), g (η 2 ) and h (η 2 ) .In the following para-
             2β10 1 f (η1)sinh u sinh nu du 
                    u
     μ                                                            (24f)         graph, the changes in variables are used. At the
 β2 (M22  2) 0 2 g(η2 )sinh u sinh nu du
                 u
                                                                                boundary, the radius R is fixed, R=1. Therefore the
  β2 0 2 h(η2 )coshu coshnu du
       u                                                                        arguments of the functions f , g  and h  are:
Substitution of equations (23) into equations (13)                              η1  Z  β1 coshu                                      (27)
gives the following expressions for displacement                                η2  Z  β 2 cos hu
components,                                                                                                                           (28)
                                                                                                   Z              Z
                                                                                We let :      ξ1         , ξ2      kξ1
                                                                                                   β1            β2
μu n,r      Rβ12 u1 f (η1 ) sinh u [n sinh nu cosh u
                   0  sinh u cosh nu] du              
  a        n2  1
             Rβ 2 2 u 2 g(η2 ) sinh u [n sinh nu cosh u                    Where K  β1 /β 2 . Equations (27) may be written
                     0 sinh u cosh nu] du                        (25a)
             n2  1                                                             as:
                   2
             Rβ 2 u 2 h(η2 ) sinh u [sinh nu cosh u                                  
                                                                                η1  β1 ξ1  coshu   
         
             n2  1
                       0 n sinh u cosh nu] du                                            
                                                                                η2  β 2 kξ1  cos hu    
μu n,θ         Rβ12    u f (η1) sinh u [n sinh u cosh nu
                                                                                A new variable  is now introduced by the following
                      0 1  sinhn u cosh u] du              
               2
  a           n 1                                                              relations:
                Rβ22    u g(η2 ) sinh u [n sinh u cosh nu      (25b)
                       0 2  sinh nu cosh u] du             
               n2  1                                                           coshu 1  ξ
              Rβ22 u 2 h(η2 ) sinh u[n sinh nu cosh u                                dξ      dξ
                    0                                                         du                                                    (29)
              n2  1  sinh u cosh nu] du                                            sin hu   2ξ  ξ 2
μu n,z      Rβ1 u 1
                 f (η1) sinhuSinhn u du 
  a          n 0                                                   (25c)        The limits of the integrals with this variable are as
            - Rβ 23 u 2
                   0 g(η2 ) sinh usinh nu du                               follows:
               n
                                                                                Low er limits, u  0 , ξ  0
              η1  Z  Rβ1 cosh u                                                             u  u1        , ξ  ξ1          (30)
Where                                                                           Upper limits 
              η2  Z  Rβ 2 cosh u
                                                                                              u  u2       , ξ  ξ 2  kξ1
And f(η1), g(η2 ),h(η2 )  fn (η1), gn (η2 ), hn (η2 ) and
                                                                                The upper limits are linear functions of Z and ξ 1 ;
primes represent the derivatives of the functions with
                                                                                therefore, in order to perform numerical integration,
respect to their arguments.
                                                                                the longitudinal axis ξ 1 is divided into small steps.
4. Boundary conditions
  In order to satisfy the condition of a traction                               At every point along this axis the numerical
boundary at the face of the cavity, r=a , three of the                          integration is performed and at each step only one
                                                                                new         value        of       the       functions


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 f (η1 ), g(η2 ) and h(η2 ) enters into     the                                                  σ rθ  r  a  σ n3 sin nθ U (Z)
computations.
                                                                                           K=3          σ rr  r  a  σ rz  r  a  0
As an example of the procedure of the numerical                                            The curves are shown for two sets of prameters:
integration, the component of the stress in the radial                                                     V
direction is given symbolically below:                                                     Case 1: M1   2 ; υ  0.25
                                                                                                           C1
 σn,rr    IF  I G  IH
           rr     rr         rr
                                                  (31)
                                                                                                             V
                                                                                            Case 2: M1         1.033 ; υ  0.25
Where
                                                 
                                                                                                             C1
IF  u1 f η1 M2  2  2 β1 sinh 2u cos hnudu
 rr                2
                              2
                                                                                                The values of M1 were chosen for application of
      0
Irr  2β2 u2 gη2 cos h2ucos hnudu
 G                                                                            (32)         the results to problems of some practical interest. The
         2 0                                                                               stress components in each case approach the static
Irr  β2  h η2 sin h2u sinhnududu
 H      2 u2 
                                                                                           plain strain solution as Z approaches infinity,
             0
                                                                                           indicating that mathematical model produces correct
The integrals IF ,Irr and IH at the pth step are
               rr
                   G
                           rr                                                              results for propagation of waves in the isotropic
expressed by using the convolution theorem in                                              medium. For those cases in which the static solutions
summation form as follows:                                                                 do not vanish, a typical overshoot above the value of

 rr       f    R  f   R rr  f 
IF  R  F
       rr    0    p
                           P 1

                           m 1
                                   F
                                   rr m    Pm
                                                  F
                                                       p       0
                                                                                           the static (long term) solutions is observed.
                                                                                           Moreover, a decrease in the Mach number M1
       R  g    R  g   R rr  g 
                            P 1                                             (33)          appears to compress the stress response curve into a
         G                                       G
IG
 rr      rr 0      p
                                   G
                                   rr m    Pm                     0
                                                           p                               smaller range of Z such that the asymptotic values of
                           m 1
       R  h    R  h   R rr  h 
                           P 1
         H                                        H                                      the lower value M1=1,033 are obtained for smaller
IH
 rr      rr 0      p
                                    H
                                    rr m    Pm                    0
                                                           p
                           m 1                                                            values of Z. Figures 12 and 13 show the stress
 Where (f) , (g) and (h) are the unknown functions to                                      components σ 0, θθ and σ 0, zz at the cavity boundary
be evaluated.
   At this stage of integration the values (f)m , (g)m ,                                   r=a for the axisymmetric loading case, n=0, for the
(h)m are known for m=0 to p-1.The only unknowns in                                         mach numbers M1=1.033, 1.5 and 2. As in the cases
these expressions are (f)p , (g)p , and (h)p . Similar                                     where n  0, the stress components in each case
expressions are written for the other components of                                        approach the static plain solutions as Z approaches
stress, at the pth step. The boundary conditions are                                       infinity. Figures 14 through 19 show the
now in the form of a set of three simultaneous linear                                      displacement components u n,r , u n,θ and u n,Z (n=1,
equations. Solution of this set results in the values of                                   2, 3, 4) for each of the three loading cases, k=1, 2, 3.
(f)p , (g)p , and (h)p . The procedure is then carried on                                  Figures 20 through 22 show the u 0,r and u 0, Z
to the (P+1)th step; Similar operations are performed                                      displacement components for the case n=0. These
to find the values of (f)p+1 , (g)p+1 , and (h)p+1 .                                       displacement results are shown for the           M1= 2,
 As mentioned previously, when the values f, g and h                                         1 / 4 case only.
are found at each step, these values are substituted                                            The only property of the material in the medium
into     the     expressions      for     ,u   ,u    and u                                 which enters into computations is its poisson's ratio.
                                                           n,r         n,θ    n, z
σ n, θθ , σ n, zz , σ n, θz to compute the numerical                                       Figures 23 through 26 represent the effect of this
                                                                                           parameter on the values of stress components for the
values of these stresses and displacements in the                                          axisymmetric loading case, n=0. The following
medium.                                                                                    values of poisson's ratio are used in this study:
6. Numerical Results and Conclusion                                                                     υ  0 , 0.15, 0.25 and 0.35
     For the non-axisymmetric loadings characterized                                            It is noticed that the change in poisson's ratio
by n > 0, numerical values of the stress components                                        does not have a large effect on the maximum value of
σ n,θθ , σ n,zz and σ n,zθ at the cavity boundary r=a                                      longitudinal stress for the load case k=2 (Figure 26),
are presented in this section. These stresses are given                                    while it affects considerably the value of longitudinal
for the cases n=1, 2 for each of the three step- traction                                  stress for the case load, k=1 (Figure 25) , and hoop
loading indicated below:                                                                   stress for the two cases, k=1,2 (Figures 23 and 24) for
Index                 Applied load                                                         smaller values of Z.

                 σ rr  r a  σ n1 cos nθ U (Z)
K=1              σ rz  r  a  σ rθ  r a  0

                 σ rz  r  a  σ n2 cos nθ U (Z)
K=2              σ rr  r a  σ rθ  r  a  0



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                       σ n,
                                                                                                                                  0.8
                        1.
                        4
                       1.2
                           σ n1
                       1                                                                                                          0.6
                                                                           Static Value
                    0.8
                                                                                                                                                                                                                      Static Value
                    0.6                                   n=1(M1=2)                                                               0.4
                                                          n=1(M1=1.033)
                    0.4                                   n=2(M1=2)               0.25
                                                          n=2(M1=1.033)
                    0.2
                                                                                                                                                                                                                      Static Value
                                                                                                                                 0.2
                           0                                                                    Z=z/a
                               0       4         8     12        16       20      24       28
                    -0.2
                                                                                Static Value
                                                                                                                                       0
                                                                                                                                                                                                                               Z=z/a
                    -0.4
                                                                                                                                            0                 4           8           1          1        2               2              2
                   -0.6                                                                                                                                                               2          6
                                                                                                                                                                                                 n=1(M1=2)0               4              8
                                                                                                                                -0.2                                                             n=2(M1=2)
                                                                                                                                σ n,zz                                      0.25               n=1(M1=1.033)
                                                                                                                                                                                                 n=2(M1=1.033)
                                                                                                                                 σ n1
Figure 3. Stress σ at boundary due to step load ;
                                                                                                                                  -0.4

σ rr  σ n1 cos n u ( Z ); n  1,2

                   0                        4                    8                12                    16
                                                                                                                           Figure 6. Stress σ zz at boundary due to step load ;
            0.8
                                                                                                                           σ rr  σ n1 cos n u ( Z ); n  1,2

             0.4
                                                                                                                                                                                                                                Z=z/a
                                                                                                                                                  0               4           8       12         16          20         24          28
                                                                                       Static Value                                        0.5
              0                                                                                          Z=z/a
                                                                                                                                             0
                                                                                                                                                                                                                    Static Value
                                                                               n=1(M1=1.033)                                               -0.5
                                                                               n=1(M1=2)                                                                                                         0.25
            -0.4                                                                                                                            -1
                                                       0.25                  n=2(M1=1.033)
                                                                                                                                           -1.5                                                  n=1(M1=2)
                                                                               n=2(M1=2)
                                                                                                                                                                                                 n=2(M1=2)
                                                                                                                                            -2                                                   n=1(M1=1.033)
            -0.8
                       σ n,                                                                                                              -2.5
                                                                                                                                                                                                 n=2(M1=1.033)


                           σ n2                                                                                                             -3

            -1.2                                                                                                                           -3.5           σ n, zz
                                                                                                                                           -4
                                                                                                                                                              σ n2
Figure 4. Stress σ at boundary due to step load :                                                                                        -4.5



σ rz  σ n2 cos n u ( Z ); n  1,2
                                                                                                                           Figure 7. Stress σ zz at boundary due to step load ;
             0         4           8   12       16   20     24       28   32     36       40    44      48   Z=z/a
        0                                                                                                                  σ rz  σ n2 cos n u ( Z ); n  1,2


      -0.5
                                                                                                                                   σ n, zz 0.2
                                                                                       Static Value                                     σ n3 0.1

        -1                                                                                                                                            0                                                                              Z=z/a

                                                                                                                                                  -0.1
                                                                 n=1(M1=2)                                                                                                                                           Static Value

                                             0.25              n=2(M1=2)
                                                                 n=1(M1=1.033)
                                                                                                                                                  -0.2
      -1.5
                                                                 n=2(M1=1.033)                                                                    -0.3
                                                                                                                                                                                        0.25
                                                                                                                                                  -0.4
                                                                                                                                                                                                                     Static Value
                                                                                          Static Value                                            -0.5
       -2
                                                                                                                                                  -0.6
                                                                                                                                                                                                                  n =1(M1=2)
     σ n,θθ                                                                                                                                       -0.7                                                            n =2(M1=2)
                                                                                                                                                                                                                  n =1(M1=1.033)
      -σ                                                                                                                                          -0.8
         n3                                                                                                                                                                                                       n =2(M1=1.033)
      2.5
                                                                                                                                                  -0.9
                                                                                                                                                          0           4           8       12       16         20         24         28
Figure 5. Stress σ at boundary due to step load ;
σ r  σ n3 sin n u ( Z ); n  1,2
                                                                                                                           Figure 8. Stress σ zz at boundary due to step load ;
                                                                                                                           σ r  σ n3 sin n u ( Z ); n  1,2




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                                                                                              
     σ n,z                                                                                       ij , 0
                   0.4                                                                               1
       σ n1                                                                                       01

                                                                                                  0.8
         0.35
                                                                                                  0.6
          0.3
                                                                                               0.4
         0.25                                                                                  0.2
                                                                                                                                                                           Static Value
                                                                                                                                                                                                 Z=z/a
          0.2                                                                                     0        0
                                                                                                                            4                8                 1            16         20
                                       n=1(M1=2)                                              -0.2                                                             2M1=1.033
         0.15
                                       n=2(M1=2)                                                                                                                M1=1.033
                                       n=1(M1=1.033)                                          -0.4                                                              M1=1.5
          0.1                          n=2(M1=1.033)                                                                                                            M1=1.5
                                                                                              -0.6                                                              M1=2
         0.05                                                                                                                                                   M1=2
                                                                 Static Value                 -0.8                                                                       Static Value
          0                                                                                   -1
                   0     4    8 12 16 20 24 28 32 36 40 44 48
        -0.05                                                                                 -1.2
                                                                                              -1.4
          -0.1
                                                  0.25                                     Figure 12. Stresses at boundary due to
Figure 9. Stress σ z at boundary due to step load ;                                         axisymmetric step load
                                                                                             σ rr  σ n1u ( Z); n  0
σ rr  σ n1 cos n u ( Z ); n  1,2

         1.4                                                                                         ij , 0
                                                                                                     02
         1.2                                                                                                    1
                                                                 Static                                    0.5
          1                                                      Value                                          0                                                                               Z=z/a
                                                                                                                    0           4                8                            16           20
                                                     n=1(M1=2)                                             -0.5                                                  12
         0.8
                                                     n=2(M1=2)                                                 -1
                                                                                                                                                                             M1=1.033
                                                     n=1(M1=1.033)
                                                                                                                                                                             M1=1.5
                                                                                                           -1.5
         0.6
                                                     n=2(M1=1.033)                                                                                 0.25                    M1=2
                                                                                                           -2                                                                M1=1.033
         0.4                                                                                                                                                                 M1=1.5
                                                                                                       -2.5
                                                                                                                                                                             M1=2
                                                                                                               -3
         0.2
                                                                                                       -3.5

          0                                                                                                    -4
               0          4      8     12       16       20         24          28                         -4.5

                                       0.25                              Z=z/a



  Figure 10. Stress σ z at boundary due to step                                             Figure 13. Stresses at boundary due to
                                                                                             axisymmetric step load
          load ; σ rz  σ n2 cos n u(Z ); n  1,2
                                                                                             σrz  σn2 u ( Z ); n  0
        0.1
                                                                                                                            4            8        12           16     20          24   Z=z/a
                                                                                                                                                                                       2
                                                                                                                                                                                       8
                                                           Static Value                             0.00                0
         0                                                                     Z=z/a
               0          4     8     12    16        20       24         28                               -0.10
                                                                                                                                                        n =4
        -0.1                                                                                               -0.20
        -0.2                                                                                               -0.30
                                                                                                                                                 n =3

        -0.3
                                                                                                           -0.40
                                                                                                                                                                           STATIC
                                                                                                           -0.50                                                           VALUE
        -0.4                                                                                                                                              n =2
                                     n=1(M1=2)                                                             -0.60
        -0.5                         n=2(M1=2)
                                                                                                           -0.70
                                     n=1(M1=1.03
        -0.6                         3)
                                     n=2(M1=1.033)                                                         -0.80                                                           n =1

                                                                                                           -0.90
        -0.7                                                                                               -1.00
        -0.8                                                                                          -1.10
                   σ n,z
                                                                                                    un, r                                                             Static Value = -1.45
                       σ n3                       0.25
                                                                                                     a 01

Figure 11. Stress σz at boundary due to step load ;
                                                                                             Figure 14. Boundary displacement u n, r due to step
                       σ r  σ n3 sin n u( Z ); n  1,2
                                                                                             load               σ rr  σ n1 cos n u ( Z ); n  1,2,3,4 , M  2 ,   0.25                              ,
                                                                                                                                                           1
                                                                                              r   rz  0


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            un, r              un, z                                                                                                              0       4       8    12           16    20         24         28
                        
            a 02               a 01                                                                                                          0                                                                  Z=z/a
             0.05                                                                                                                          -0.1

                                                                                                                                           -0.2
                                                                                                                                                                                                 Static Value
             0.04
                                                                                                                                           -0.3

                                                                                                                                           -0.4
             0.03                                                                         n=1
                                                                                                                                           -0.5
                                                                                          n=2                                                                                                    Static Value

             0.02                                                                         n=3                                              -0.6
                                                                                          n=4                                              -0.7
             0.01                                                                                                                          -0.8
                                                                                                                                                                n=1
                                                                                       Static Value                                        -0.9                 n=2
                    0                                                                                 Z=z/a                                    -1               n=3
                                                                                                                                                                n=4
                                                                                                                                           -1.1
            -0.01
                                                                                                                                           -1.2

            -0.02                                                                                                                     un, 

                        0              4            8        12       16        20       24       28                                  a
                                                                                                                                               03



Figure 15. Boundary displacements due to step load                                                                       Figure 18. Boundary displacements due to step
;n=1,2,3,4 , M  2 ,   0.25
                            1
                                                                                                                         load ;n=1,2,3,4 , M1  2 ,   0.25
    u   n , r                  u   n ,   
                        
    a  03                      a  01
                                                                                 Static Value = 1.45
                                                                                                                                      0                 4       8       12        16        20             24      28
    0.9                                                                                                                                                                                                                  Z=z/a
                                                                                                                                 0
    0.8                               n=1                                                                                    -0.1
                                      n=2
    0.7
                                      n=3                                                                                    -0.2
                                      n=4                                                                                    -0.3

    0.6                                                                                                                      -0.4                                                                  Static Value
                                                                                                                             -0.5
    0.5                                                                                                                      -0.6
                                                                                                                                                                                           n=1
                                                                                                                             -0.7
    0.4                                                                              Static Value                                                                                          n=2
                                                                                                                             -0.8                                                          n=3
                                                                                                                             -0.9                                                          n=4
    0.3
                                                                                                                                                                                                     Static Value
                                                                                                                                 -1
    0.2
                                                                                     Static Value                            -1.1

    0.1                                                                                                                      -1.2
                                                                                                                             -1.3
                                                                                                           Z=z/a
        0                                                                                                                    u n, z 
            0               4                  8            12        16        20         24          28                    a  02



                                                                                                                         Figure 19. Boundary displacements due to step
Figure 16. Boundary displacements due to step load
                                                                                                                         load ;n=1,2,3,4 , M1  2 ,   0.25
;n=1,2,3,4 , M1                        2 ,   0.25
                                                                                                                             u
                                                                                                                                 0 , r     μ
        u n,              u n, z                                                                                           a σ
                                                                                                                                      01
                     
        a  02              a  03                                                                                            -0.7
             0.028
                                                                                                                             -0.6
             0.024
                                                                                                                             -0.5                                            STATIC
                0.02
                                                                                                                                                                             VALUE
             0.016                                                               n=1                                         -0.4
                                                                                 n=2

             0.012                                                               n=3
                                                                                 n=4
                                                                                                                             -0.3
             0.008
                                                                                                                             -0.2
             0.004
                                                                                       Static Value
                                                                                                                             -0.1
                        0
                                                                                                           Z=z/a
                            0              4            8        12        16    20        24         28
            -0.004                                                                                                                0
                                                                                                                                           0            4       8       12            16     20            24           28
            -0.008                                                                                                                                                                                                      Z=z/a

            -0.012
                                                                                                                         Figure 20. Boundary displacement                                             u          due to step
            -0.016                                                                                                                                                                                        0, r

                                                                                                                         load σ01 ; n=0 , M1  2 ,   0.25
Figure 17. Boundary displacements due to step
load ;n=1,2,3,4 , M 1                                    2 ,   0.25




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      u0, r         u0, z 
                                                                                                                        0 , 
       a 02         a 01
                                                                                                                           02
       0.08
                                                     0 . 25 , M  2
                                                         U ( Z )
                                                                                                                          0.3
                                                       rr        01
                                                                 on r  a
       0.07                                                0  
                                                       rz                                                                0.2
                                                       0      
                                                     rr          on r  a
       0.06
                                                      U ( Z )
                                                     rz    02                                                            0.1
       0.05
                                                                                                                              0                                                         z/a
                                                                                                                                  0       4       8    12   16      20       24        28
       0.04                                                                                                               -0.1
                                                                                                                                                                 0
       0.03                                                                                                               -0.2
                                                                                                                                                                 
       0.02                                                                                                               -0.3                                   0.15
       0.01                                                                                                               -0.4

       0.00
                0       4          8    12     16           20        24   28    32         36    40
                                                                                                 Z=z/a           Figure 24. Comparision of Hoop stress for different values of
                                                                                                                 poisson's ratio at boundary r=a due to axisymmetric step load

Figure 21. Boundary displacement                                                 u           due to step         σ02     ; n= 0,M1=2
                                                                                      0, r
                                                                                                                    0 , zz
load σ02 ; n= 0                         u          due to step load σ01 ; n= 0                                       01
                                            0, z
                                        , M  2 ,   0.25
                                                                                                                  0.6
                                           1
                                                                                                                  0.5
       u 0,z μ                                                                                                    0.4
       a σ 02                                                                                                                                                      0
                                                                                                                  0.3                                              0.15
         -3.2                                                                                                                                                      0.25
                                                                                                                  0.2                                              0.35
         -2.4                                                                                                     0.1
                                                                                                                                                                                   z/a
                                                                                                                     0
         -1.6                                                                                                              0          4       8        12    16         20        24      28
                                                                                                                 -0.1
         -0.8                                                                                                    -0.2

                0                                                                                                Figure 25. Comparision of Longitudinal stress for different
                         4 8 12 16 20 24 28 32 36 40                                                             values of poisson's ratio at boundary r=a due to axisymmetric
                                                 Z=z/a
                                                                                                                 step load σ 01 ;         n= 0, M1=2
Figure 22. Boundary displacement                                                 u           due to step            0 , zz
                                                                                     0, z
load σ 02 ; n= 0                                                                                                     02
                                                                                                                         0            4       8       12    16      20       24         28
      0 , 
                                                                                                                     0
       01                                                                                                                                                                              z/a
       0.8                                                                                                         -0.1
       0.6                                                                 0                                     -0.2
       0.4                                                                   0.15
       0.2                                                                   0.25                                -0.3
                                                                             0.35              z/a
         0                                                                                                         -0.4                                      0
      -0.2 0                   4         8             12             16    20         24          28              -0.5
      -0.4                                                                                                                                                   0.15
      -0.6                                                                                                         -0.6
      -0.8                                                                                                         -0.7
                                                                                                                                                             0.25
        -1
                                                                                                                   -0.8                                      0.35
      -1.2
      -1.4                                                                                                         -0.9
      -1.6                                                                                                           -1


 Figure 23. Comparision of Hoop stress for different values of                                                   Figure 26. Comparision of Longitudinal stress for different
 poisson's ratio at boundary r=a due to axisymmetric step load                                                   values of poisson's ratio at boundary r=a due to axisymmetric
                                        σ01        ; n= 0,M1=2                                                   step load    σ 02    ; n= 0, M1=2




                                                                                                           821
121 9168life0902 812_822[1]

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121 9168life0902 812_822[1]

  • 1. Life Science Journal 2012;9(2) http://www.lifesciencesite.com Dynamic Stresses and Displacements around Cylindrical Cavities in an Infinite Elastic Medium under Moving Step Loads on the Cavity's Surface Hamid Mohsenimonfared 1 M.Nikkhahbahrami 2 1. Department of mechanical Engineering , Science & Research Branch , Islamic Azad University(IAU), Tehran, Iran 2. Department of mechanical Engineering ,Tehran University , Tehran , Iran h-mohsenimonfared@ iau-arak.ac.ir Abstract: Potential functions and Fourier series method in the cylindrical coordinate system are employed to solve the problem of moving loads on the surface of a cylindrical bore in an infinite elastic medium. The steady-state dynamic equations of medium are uncoupled into Helmholtz equations, via given potentials. It is used that because of the superseismic nature of the problem, two mach cones are formed and opened toward the rear of the front in the medium. The stresses and displacements are obtained by using integral equations with certain boundary conditions. Finally, the dynamic stresses and displacements for step loads with axisymmetric and nonaxisymmetric cases are obtained and discussed in details via a numerical example. Moreover, effects of Mach numbers and poisson's ratio of medium on the values of stresses are discussed. [Hamid Mohsenimonfared, M. Nikkhahbahrami. Dynamic Stresses and Displacements around Cylindrical Cavities in an Infinite Elastic Medium under Moving Step Loads on the Cavity's Surface. Life Sci J 2012;9(2):812-822] (ISSN:1097-8135). http://www.lifesciencesite.com. 121 Keywords: Wave propagation, Fourier series method, cylindrical cavities, Helmholtz equations, moving load. 1. Introduction terms of Hankel functions, and finally the stresses 1.1 General Remarks and displacements are found by inversion of the Moving loads on the surfaces have been transformed quantities. In this paper the coefficients investigated by many researchers. Investigation on of the stresses and displacements are found by dynamic stresses in solids is very significant in the solving sets of coupled integral equations. study of dynamic strength of materials and in the The waves are expanded into Fourier series in design of underground structures subject to ground terms of the angle,  , around the opening. The blasting waves. A related but considerably simpler stress field of the wave is written in terms of potential problem has been treated by Biot (1952), who functions which satisfy the equations of motion. considered space- harmonic axisymmetric standing These equations decoupled via introducing the waves and obtained a closed form solution. Another potential functions and reduced to Helmholtz related problem was treated by Cole and Huth (1958), equations that the potentials satisfy. who considered a line load progressing with a These potential functions are in integral form velocity V on the surface of an elastic half- space. with unknown functions in the integrands. Therefore Because of the simpler geometry, they were able to the Fourier series coefficients of the stresses and obtain a solution in closed form. Adrianus (2002) displacements are also in integral form with unknown investigated the moving point load problem in soil integrands. The applied boundary tractions (the step dynamics with a view to determine the ground loads) are expanded into a Fourier series in  and motion generated by a high-speed train traveling on expressions for the stress and displacement a poorly consolidated soil with low shear wave speed. components at points in the medium are derived for M.C.M. Bakker (1999) revisited the nonaxisymmet- each term of the Fourier series as functions of the rical boundary value problem of a point load of radial distance r from the cavity axis and the distance normal traction traveling over an elastic half-space. z behind the wave front. M.Rahman (2001) considered the problem of a line The following three cases of step loads are load moving at a constant transonic speed across the considered: normal to the surface, tangential to the surface of an elastic half-space and derived solution surface in the direction of the axis of the bore, and of the problem by using the method of Fourier tangential to the circle of load application. These transform. Iavorskaia (1964) also studied diffraction results can be used, by superposition, to determine of a plane longitudinal wave on circular cylinder. the effects of other load patterns moving with the One basic method has been used for the solution of velocity V in the direction of the axis of the bore. these problems, the solution is obtained by using an Numerical solution of these equations gives the integral transform of the displacement potentials. The values of the unknown functions. These values can resulting transformed equations are then solved in 812
  • 2. Life Science Journal 2012;9(2) http://www.lifesciencesite.com then be used to find the stresses and displacements on 2. Governing equations and general solutions the boundary and also anywhere in the medium. Consider a cylindrical cavity of radius r = a in a linearly elastic, homogeneous, and isotropic medium 1.2 Problem Description referred to a fixed coordinate system r, θ , z  whose The object of this work is to obtain stresses and origin lies on the axis of the cavity. displacements in an elastic medium in the vicinity of A step load along the circle at z = -vt progresses a cylindrical cavity which is engulfed by a plane along the interior of the cavity with a velocity V such stress wave of dilatational travelling parallel to the that the stresses on the boundary r=a are: axis of the cylinder, as shown in Figure 1. The step load has an arbitrary distribution P (θ) along the circumference of the circle and moves with σrr  σ (θ)U(z  vt) ra 1 (1) a velocity V  C1  C 2 ; therefore, the speed is σrθ  ra  σ 2 (θ ) U(z  vt) (2) σrz  superseismic with respect to both the dilatational and shear waves in the medium. Consequently, the  σ3 (θ ) U(z  vt) (3) disturbances which were initiated far behind the front ra on the boundary of the cavity cannot reach the Where the functions σ k (θθ) define the vicinity of the wave front for some time after the distribution of the applied load. To determine the incident wave passes. steady state solution, a moving coordinate system (r , θ ,z) is introduced such that: r  r , θ  θ , z  z  Vt (4) The following treatment is restricted to the case where the velocity V is greater than C1 and C2 , the respective propagation velocities of dilatational and equivoluminal waves in the medium. Hence V V (5) M  1 1 , M  1 2 C1 C2 Figure 1. Moving step load Where λ  2μ μ (6) Moreover, because of the super seismic nature of C1  , C2  ρ ρ the problem, it should be expected that two mach cones will be formed in the medium, as shown in The equations of motion in cylindrical coordinates, r , Figure 2. These cones should open toward the rear of  , z, for an elastic medium, may be expressed in the the front. Furthermore, there can be no stresses or following form: displacements ahead of the leading front. u θ Δ ρ  u r 2 1 λ If a coordinate system is assumed to move along  2u r  (u r  2 )  (  1)  the cylinder with the wave front, it is seen that the r 2 θ μ r μ t 2 state of stress at points close behind the wave front 1 Δ ρ  u θ 2 1 u r λ depends only on relative position of them with  2uθ  (u  2 2 θ )  (  1)  respect to the front. Thus, in the vicinity of the wave r θ μ r θ μ t 2 Δ ρ  u z front, provided that the end of the cavity is far away, 2 λ the problem may be treated as a steady-state case. In  2u z  (  1)  μ z μ t 2 other words, in the moving coordinate system, the state of stress and displacement is independent of (7) time. Where the dilatation , Δ , and the laplacian operator,  2 , are given by: u u 1 u θ u z Δ r  r   r r r θ z (8) 2 1  1 2 2   2 2   r r r r 2 θ 2 z 2 As mentioned earlier, the assumption of the existence of a steady-state case and trans-formation form r ,  , z coordinates to r ,  , z results in elimination of Figure 2. Geometry of the problem and the coordinate the time variable, t , from the equations of motion. systems 813
  • 3. Life Science Journal 2012;9(2) http://www.lifesciencesite.com This transformation is performed by the following φ n  2ψn n relations, as given in relations (4): u n, r     z  z  Vt r rz r n n n ψ n  n     u n, θ   φ n   (13)  , V r r z r z z t z φ n   2 1  n 2  Therefore equations (7) may be expressed as follows: u n, z     ψ z  r 2 r r r 2  n   1 u θ λ Δ ρ 2  u r 2  2u r  (u r  2 )  (  1)  V These equations may be obtained from the vector 2 r θ μ r μ z 2 equation 1 Δ ρ 2  u θ 2 1 u r λ  2uθ  (uθ  2 )  (  1)  V u  grad φ  curl  r2 θ μ r θ μ z 2 Where  is the sum of two independent vectors as Δ ρ 2  u z 2 λ follows:  2u z  (  1)  V μ z μ z 2     curl ψ (9) Stress components are given by The vectors  and ψ have only one non- zero u σrr  λΔ  2μ r component which is in the z- direction in both cases. r u 1 uθ r  0 ψr  0 σθθ  λΔ  2μ ( r  ) θ  0 ψθ  0 r r θ z   ψz  ψ u σzz  λΔ  2μ z (10) z By substitution of the values given in equations (13) 1 u r u θ u θ into the equations (9), it can be shown that the σ rθ  μ(   ) r θ r r potential functions satisfy the modified wave u u equations. σ rz  μ( r  z ) z r V 2  φn 2 u 1 u z  2φ n  ( ) σθz  μ( θ  ) c1 z 2 z r θ V  ψn 2 (14) Displacement components u r , u θ and u z may be  2ψn  ( )2 expressed in Fourier series: c2 z 2 V 2  χn 2   χn  ( ) 2 u r (r,θ, z)   u n, r (r, z)cosnθ c2 z 2 n0 Stress components are expressed in Fourier series  (11) u θ (r,θ, z)   u n, θ (r, z)sinnθ form as follows: n 1   σ rr (r,θ, z)   σ n,rr (r, z) cos nθ u z (r,θ, z)   u n, z (r, z)cosnθ n 0 n0  Three potential functions are now introduced, σθθ (r,θ, z)   σn,θθ (r, z) cos nθ n 0  φ(r, θ, z)   φ n (r, z)cosnθ  n 0 σzz (r,θ, z)   σn,zz (r, z) cos nθ  n 0 ψ(r, θ, z)   ψ n (r, z)cosnθ (12) (15)  n 0 σrθ (r,θ, z)   σn,rθ (r, z) sinnθ  n 1  (r,θ, z)    n (r, z)sinnθ n 1  σ rz (r,θ, z)   σ n,rz (r, z) cos nθ The displacement components u n, r , u n, θ and u n, z n 0  are defined as follows: σθz (r,θ, z)   σ n,θz (r, z) sin nθ n 1 Equations (11) and (15) may be substituted into equations (10), and as a result stress- displacement relations may be written for each term of the series:  u n,r (16) σ n,rr  λΔ  2 μ r 814
  • 4. Life Science Journal 2012;9(2) http://www.lifesciencesite.com  u n,r u n,θ  β1  M1  1 2 2 , β2  M2  1 σ n,θθ  λΔ  2 μ   r n r   2 2   The differential equations (14) may be written in the following form:  u n,z σ n,zz  λΔ  2 μ  2φn 1 φn n2  2φn z   2 φn  β12  u n,r u θ  u n,θ  R2 R R R Z2 σ n,rθ  μ   n      r r r    2ψn 1 ψn n2  2ψn   2 ψn  β22 (20)   u n,r u n,z  R2 R R R Z2 σ n,rz  μ    z  r    2  n 1  n n 2 2  n 2     2 n  β2 R 2 R R R Z 2   u n,θ u n,z  σ n,θz  μ   z  n r   It is seen that these equations have the same general   form as the differential equations of the cylindrical where waves obtained in reference (13). Therefore solutions of equations (20) may be obtained in a manner u n,r u n,r u n,θ  u n,z similar to that in reference (13). These solutions are Δ  n  given in integral form as follows (see Appendix A for r r r z verification of the solutions): Substitution of equations (13) into equations (16) and ∞ application of the differential equations (14) result in φn  ∫ fn (Z - Rβ1coshu)coshnu du 0 the following equations for stress components:  (21) ψn  0 gn (Z  Rβ2coshu)coshnudu a 2 σ n,rr   M 2  2M1  φ n,ZZ  2 φ n,RR  2ψ n,RRZ  2 2  μ  n  0 h n (Z  Rβ 2coshu)coshnu du 2n  1     n,R   n  From consideration of the fact that the R  R  disturbances are zero ahead of the wave front, it is a 2 σ n,θθ μ    M 2  2 φ n,ZZ  2 φ n,RR  2 seen that the functions fn , gn and hn are zero for the values of their arguments less than 2  2 Rβ1 ,  Rβ2 and  Rβ2 ,respectively.  ψ n,RZ  n ψ n,Z   2n   n,R  1  n    Therefore the upper limits of the integrals may be R  R  R   R  changed from  to the following values: a 2 σ n,zz μ    M 2  2M1  2 φ n,ZZ  2 M 2 1 ψ n,ZZZ 2 2 2   u1  cosh-1(1  Z Rβ1 ) for φn (22) Z a 2 σ n,rθ 2n  (17) -1 u2  cosh (1  ) for ψn & n 1  2n  1  Rβ2   φ n,R  φ n    ψ n,RZ  ψ n,Z   μ R   R     R R The integrals are then written with these limits:    M 2  1  n,ZZ  2  n,RR 2 u φn  01 fn ZRβ1coshu coshnudu a 2 σ n,rz μ  2φ n,RZ   M2  2 2 ψ n,RZZ n   n,Z R u2 ψn  0 g n ZRβ2 coshucoshnudu (23)    n  0 hn ZRβ2 coshucoshnudu 2 a σ n,θz 2n n u φ n,Z  M 2  2 ψ n,ZZ   n,RZ  2 2 μ R R Where the second set of subscripts of 3. Expressions of stresses and displacements φn ,ψn and n represent the partial derivatives of Substitution of equations (23) into equations (17) these functions. R and Z are the dimensionless gives the following expressions for stress variables form: components. r z (18) R , Z a a a2σn,rr  0 1 f (η1)[M22  2  u The values M1 and M2 are defined as follows: μ  2β12sinh 2u]cosh nu du  (24a) V V M1  , M2   2β22 0 2 g(η2 )cosh2u coshnu du  u C1 C2 2 u2 h(η )sinh 2u coshnu du (19)  β2 0 2 Let 815
  • 5. Life Science Journal 2012;9(2) http://www.lifesciencesite.com a2σn,θθ stress components must satisfy the following  0 1 f (η1)[M22  2 u boundary conditions: μ  2β12cosh2u] cosh nu du  (24b)  σ n,rr r a  σ n1u(Z) 2β22 0 2 g(η2 )sinh2u cosh nu du u σ n,rz  r a  σ n2 u(Z) (26)  β2 0 h(η2 )sinh 2ucosh nu du σ n,rθ  r a  σ n3 u(Z) 2 u2 These equations are satisfied for each term n. The a 2σ n,zz coefficients of the stress, σ n,rr , σ n,rθ and σ n,rz are  (M 2  2 2M12 2)0 1 f (η1 )cosh nu du u (24c) μ expressed in equations (24) in integral form. These  2β 2 0 g(η2 )cosh nu du 2 u2 integrals include the unknown functions a 2σ n,rθ f (η1 ), g (η 2 ) and h (η 2 ) which are to be found by  β12 0 1 f (η1 )sinh 2u sinh nu du  u solving the set of three simultaneous integral μ (24d) β 2 2 0 2 g(η2 )sinh 2u sinh nu du u equations. These values then may be substituted back  β 2 0 h(η2 )cosh 2u cosh nu du 2 u2 into the equations (24) and (25) to find the stress components and displacement components of the a2σ n,rz waves at any point on the boundary or in the medium  2β1 u1 f (η1)coshu cosh nu du  μ 0 (24e) behind the move front. 2 u 2  β2 (M2  2) 0 g (η2 )coshu coshnu du 5. Solution of the Boundary Equations  β2 0 2 h(η2 )sinh u sinh nu du u Numerical solution of the boundary equations requires finding numerical values of the functions a2σn,θz f (η1 ), g (η 2 ) and h (η 2 ) .In the following para-  2β10 1 f (η1)sinh u sinh nu du  u μ (24f) graph, the changes in variables are used. At the β2 (M22  2) 0 2 g(η2 )sinh u sinh nu du u boundary, the radius R is fixed, R=1. Therefore the  β2 0 2 h(η2 )coshu coshnu du u arguments of the functions f , g  and h  are: Substitution of equations (23) into equations (13) η1  Z  β1 coshu (27) gives the following expressions for displacement η2  Z  β 2 cos hu components, (28) Z Z We let : ξ1  , ξ2   kξ1 β1 β2 μu n,r  Rβ12 u1 f (η1 ) sinh u [n sinh nu cosh u  0  sinh u cosh nu] du  a n2  1  Rβ 2 2 u 2 g(η2 ) sinh u [n sinh nu cosh u  Where K  β1 /β 2 . Equations (27) may be written  0 sinh u cosh nu] du (25a) n2  1 as: 2  Rβ 2 u 2 h(η2 ) sinh u [sinh nu cosh u   η1  β1 ξ1  coshu   n2  1 0 n sinh u cosh nu] du  η2  β 2 kξ1  cos hu  μu n,θ  Rβ12 u f (η1) sinh u [n sinh u cosh nu A new variable  is now introduced by the following  0 1  sinhn u cosh u] du  2 a n 1 relations:  Rβ22 u g(η2 ) sinh u [n sinh u cosh nu (25b)  0 2  sinh nu cosh u] du  n2  1 coshu 1  ξ Rβ22 u 2 h(η2 ) sinh u[n sinh nu cosh u dξ dξ  0 du   (29) n2  1  sinh u cosh nu] du sin hu 2ξ  ξ 2 μu n,z Rβ1 u 1   f (η1) sinhuSinhn u du  a n 0 (25c) The limits of the integrals with this variable are as - Rβ 23 u 2  0 g(η2 ) sinh usinh nu du follows: n Low er limits, u  0 , ξ  0 η1  Z  Rβ1 cosh u u  u1 , ξ  ξ1 (30) Where Upper limits  η2  Z  Rβ 2 cosh u u  u2 , ξ  ξ 2  kξ1 And f(η1), g(η2 ),h(η2 )  fn (η1), gn (η2 ), hn (η2 ) and The upper limits are linear functions of Z and ξ 1 ; primes represent the derivatives of the functions with therefore, in order to perform numerical integration, respect to their arguments. the longitudinal axis ξ 1 is divided into small steps. 4. Boundary conditions In order to satisfy the condition of a traction At every point along this axis the numerical boundary at the face of the cavity, r=a , three of the integration is performed and at each step only one new value of the functions 816
  • 6. Life Science Journal 2012;9(2) http://www.lifesciencesite.com f (η1 ), g(η2 ) and h(η2 ) enters into the σ rθ  r  a  σ n3 sin nθ U (Z) computations. K=3 σ rr  r  a  σ rz  r  a  0 As an example of the procedure of the numerical The curves are shown for two sets of prameters: integration, the component of the stress in the radial V direction is given symbolically below: Case 1: M1   2 ; υ  0.25 C1 σn,rr  IF  I G  IH rr rr rr (31) V Case 2: M1  1.033 ; υ  0.25 Where   C1 IF  u1 f η1 M2  2  2 β1 sinh 2u cos hnudu rr 2 2 The values of M1 were chosen for application of 0 Irr  2β2 u2 gη2 cos h2ucos hnudu G (32) the results to problems of some practical interest. The 2 0 stress components in each case approach the static Irr  β2  h η2 sin h2u sinhnududu H 2 u2  plain strain solution as Z approaches infinity, 0 indicating that mathematical model produces correct The integrals IF ,Irr and IH at the pth step are rr G rr results for propagation of waves in the isotropic expressed by using the convolution theorem in medium. For those cases in which the static solutions summation form as follows: do not vanish, a typical overshoot above the value of rr   f    R  f   R rr  f  IF  R  F rr 0 p P 1 m 1 F rr m Pm F p 0 the static (long term) solutions is observed. Moreover, a decrease in the Mach number M1  R  g    R  g   R rr  g  P 1 (33) appears to compress the stress response curve into a G G IG rr rr 0 p G rr m Pm 0 p smaller range of Z such that the asymptotic values of m 1  R  h    R  h   R rr  h  P 1 H H the lower value M1=1,033 are obtained for smaller IH rr rr 0 p H rr m Pm 0 p m 1 values of Z. Figures 12 and 13 show the stress Where (f) , (g) and (h) are the unknown functions to components σ 0, θθ and σ 0, zz at the cavity boundary be evaluated. At this stage of integration the values (f)m , (g)m , r=a for the axisymmetric loading case, n=0, for the (h)m are known for m=0 to p-1.The only unknowns in mach numbers M1=1.033, 1.5 and 2. As in the cases these expressions are (f)p , (g)p , and (h)p . Similar where n  0, the stress components in each case expressions are written for the other components of approach the static plain solutions as Z approaches stress, at the pth step. The boundary conditions are infinity. Figures 14 through 19 show the now in the form of a set of three simultaneous linear displacement components u n,r , u n,θ and u n,Z (n=1, equations. Solution of this set results in the values of 2, 3, 4) for each of the three loading cases, k=1, 2, 3. (f)p , (g)p , and (h)p . The procedure is then carried on Figures 20 through 22 show the u 0,r and u 0, Z to the (P+1)th step; Similar operations are performed displacement components for the case n=0. These to find the values of (f)p+1 , (g)p+1 , and (h)p+1 . displacement results are shown for the M1= 2, As mentioned previously, when the values f, g and h   1 / 4 case only. are found at each step, these values are substituted The only property of the material in the medium into the expressions for ,u ,u and u which enters into computations is its poisson's ratio. n,r n,θ n, z σ n, θθ , σ n, zz , σ n, θz to compute the numerical Figures 23 through 26 represent the effect of this parameter on the values of stress components for the values of these stresses and displacements in the axisymmetric loading case, n=0. The following medium. values of poisson's ratio are used in this study: 6. Numerical Results and Conclusion υ  0 , 0.15, 0.25 and 0.35 For the non-axisymmetric loadings characterized It is noticed that the change in poisson's ratio by n > 0, numerical values of the stress components does not have a large effect on the maximum value of σ n,θθ , σ n,zz and σ n,zθ at the cavity boundary r=a longitudinal stress for the load case k=2 (Figure 26), are presented in this section. These stresses are given while it affects considerably the value of longitudinal for the cases n=1, 2 for each of the three step- traction stress for the case load, k=1 (Figure 25) , and hoop loading indicated below: stress for the two cases, k=1,2 (Figures 23 and 24) for Index Applied load smaller values of Z. σ rr  r a  σ n1 cos nθ U (Z) K=1 σ rz  r  a  σ rθ  r a  0 σ rz  r  a  σ n2 cos nθ U (Z) K=2 σ rr  r a  σ rθ  r  a  0 817
  • 7. Life Science Journal 2012;9(2) http://www.lifesciencesite.com σ n, 0.8 1. 4 1.2 σ n1 1 0.6 Static Value 0.8 Static Value 0.6 n=1(M1=2) 0.4 n=1(M1=1.033) 0.4 n=2(M1=2)   0.25 n=2(M1=1.033) 0.2 Static Value 0.2 0 Z=z/a 0 4 8 12 16 20 24 28 -0.2 Static Value 0 Z=z/a -0.4 0 4 8 1 1 2 2 2 -0.6 2 6 n=1(M1=2)0 4 8 -0.2 n=2(M1=2) σ n,zz   0.25 n=1(M1=1.033) n=2(M1=1.033) σ n1 Figure 3. Stress σ at boundary due to step load ; -0.4 σ rr  σ n1 cos n u ( Z ); n  1,2 0 4 8 12 16 Figure 6. Stress σ zz at boundary due to step load ; 0.8 σ rr  σ n1 cos n u ( Z ); n  1,2 0.4 Z=z/a 0 4 8 12 16 20 24 28 Static Value 0.5 0 Z=z/a 0 Static Value n=1(M1=1.033) -0.5 n=1(M1=2)   0.25 -0.4 -1   0.25 n=2(M1=1.033) -1.5 n=1(M1=2) n=2(M1=2) n=2(M1=2) -2 n=1(M1=1.033) -0.8 σ n, -2.5 n=2(M1=1.033) σ n2 -3 -1.2 -3.5 σ n, zz -4 σ n2 Figure 4. Stress σ at boundary due to step load : -4.5 σ rz  σ n2 cos n u ( Z ); n  1,2 Figure 7. Stress σ zz at boundary due to step load ; 0 4 8 12 16 20 24 28 32 36 40 44 48 Z=z/a 0 σ rz  σ n2 cos n u ( Z ); n  1,2 -0.5 σ n, zz 0.2 Static Value σ n3 0.1 -1 0 Z=z/a -0.1 n=1(M1=2) Static Value   0.25 n=2(M1=2) n=1(M1=1.033) -0.2 -1.5 n=2(M1=1.033) -0.3   0.25 -0.4 Static Value Static Value -0.5 -2 -0.6 n =1(M1=2) σ n,θθ -0.7 n =2(M1=2) n =1(M1=1.033) -σ -0.8 n3 n =2(M1=1.033) 2.5 -0.9 0 4 8 12 16 20 24 28 Figure 5. Stress σ at boundary due to step load ; σ r  σ n3 sin n u ( Z ); n  1,2 Figure 8. Stress σ zz at boundary due to step load ; σ r  σ n3 sin n u ( Z ); n  1,2 818
  • 8. Life Science Journal 2012;9(2) http://www.lifesciencesite.com  σ n,z ij , 0 0.4  1 σ n1 01 0.8 0.35 0.6 0.3 0.4 0.25 0.2 Static Value Z=z/a 0.2 0 0 4 8 1 16 20 n=1(M1=2) -0.2 2M1=1.033 0.15 n=2(M1=2) M1=1.033 n=1(M1=1.033) -0.4 M1=1.5 0.1 n=2(M1=1.033) M1=1.5 -0.6 M1=2 0.05 M1=2 Static Value -0.8   Static Value 0 -1 0 4 8 12 16 20 24 28 32 36 40 44 48 -0.05 -1.2 -1.4 -0.1   0.25 Figure 12. Stresses at boundary due to Figure 9. Stress σ z at boundary due to step load ; axisymmetric step load σ rr  σ n1u ( Z); n  0 σ rr  σ n1 cos n u ( Z ); n  1,2 1.4  ij , 0  02 1.2 1 Static 0.5 1 Value 0 Z=z/a 0 4 8 16 20 n=1(M1=2) -0.5 12 0.8 n=2(M1=2) -1 M1=1.033 n=1(M1=1.033) M1=1.5 -1.5 0.6 n=2(M1=1.033)   0.25 M1=2 -2 M1=1.033 0.4 M1=1.5 -2.5 M1=2 -3 0.2 -3.5 0 -4 0 4 8 12 16 20 24 28 -4.5   0.25 Z=z/a Figure 10. Stress σ z at boundary due to step Figure 13. Stresses at boundary due to axisymmetric step load load ; σ rz  σ n2 cos n u(Z ); n  1,2 σrz  σn2 u ( Z ); n  0 0.1 4 8 12 16 20 24 Z=z/a 2 8 Static Value 0.00 0 0 Z=z/a 0 4 8 12 16 20 24 28 -0.10 n =4 -0.1 -0.20 -0.2 -0.30 n =3 -0.3 -0.40 STATIC -0.50 VALUE -0.4 n =2 n=1(M1=2) -0.60 -0.5 n=2(M1=2) -0.70 n=1(M1=1.03 -0.6 3) n=2(M1=1.033) -0.80 n =1 -0.90 -0.7 -1.00 -0.8 -1.10 σ n,z un, r  Static Value = -1.45 σ n3   0.25 a 01 Figure 11. Stress σz at boundary due to step load ; Figure 14. Boundary displacement u n, r due to step σ r  σ n3 sin n u( Z ); n  1,2 load σ rr  σ n1 cos n u ( Z ); n  1,2,3,4 , M  2 ,   0.25 , 1  r   rz  0 819
  • 9. Life Science Journal 2012;9(2) http://www.lifesciencesite.com un, r  un, z  0 4 8 12 16 20 24 28  a 02 a 01 0 Z=z/a 0.05 -0.1 -0.2 Static Value 0.04 -0.3 -0.4 0.03 n=1 -0.5 n=2 Static Value 0.02 n=3 -0.6 n=4 -0.7 0.01 -0.8 n=1 Static Value -0.9 n=2 0 Z=z/a -1 n=3 n=4 -1.1 -0.01 -1.2 -0.02 un,  0 4 8 12 16 20 24 28 a 03 Figure 15. Boundary displacements due to step load Figure 18. Boundary displacements due to step ;n=1,2,3,4 , M  2 ,   0.25 1 load ;n=1,2,3,4 , M1  2 ,   0.25 u n , r  u n ,   a  03 a  01 Static Value = 1.45 0 4 8 12 16 20 24 28 0.9 Z=z/a 0 0.8 n=1 -0.1 n=2 0.7 n=3 -0.2 n=4 -0.3 0.6 -0.4 Static Value -0.5 0.5 -0.6 n=1 -0.7 0.4 Static Value n=2 -0.8 n=3 -0.9 n=4 0.3 Static Value -1 0.2 Static Value -1.1 0.1 -1.2 -1.3 Z=z/a 0 u n, z  0 4 8 12 16 20 24 28 a  02 Figure 19. Boundary displacements due to step Figure 16. Boundary displacements due to step load load ;n=1,2,3,4 , M1  2 ,   0.25 ;n=1,2,3,4 , M1  2 ,   0.25 u 0 , r μ u n,   u n, z  a σ 01   a  02 a  03 -0.7 0.028 -0.6 0.024 -0.5 STATIC 0.02 VALUE 0.016 n=1 -0.4 n=2 0.012 n=3 n=4 -0.3 0.008 -0.2 0.004 Static Value -0.1 0 Z=z/a 0 4 8 12 16 20 24 28 -0.004 0 0 4 8 12 16 20 24 28 -0.008 Z=z/a -0.012 Figure 20. Boundary displacement u due to step -0.016 0, r load σ01 ; n=0 , M1  2 ,   0.25 Figure 17. Boundary displacements due to step load ;n=1,2,3,4 , M 1  2 ,   0.25 820
  • 10. Life Science Journal 2012;9(2) http://www.lifesciencesite.com u0, r  u0, z    0 ,  a 02 a 01  02 0.08   0 . 25 , M  2    U ( Z ) 0.3 rr 01  on r  a 0.07  0  rz  0.2  0  rr  on r  a 0.06    U ( Z ) rz 02  0.1 0.05 0 z/a 0 4 8 12 16 20 24 28 0.04 -0.1 0 0.03 -0.2  0.02 -0.3 0.15 0.01 -0.4 0.00 0 4 8 12 16 20 24 28 32 36 40 Z=z/a Figure 24. Comparision of Hoop stress for different values of poisson's ratio at boundary r=a due to axisymmetric step load Figure 21. Boundary displacement u due to step σ02 ; n= 0,M1=2 0, r  0 , zz load σ02 ; n= 0 u due to step load σ01 ; n= 0  01 0, z , M  2 ,   0.25 0.6 1 0.5 u 0,z μ 0.4 a σ 02 0 0.3 0.15 -3.2 0.25 0.2 0.35 -2.4 0.1 z/a 0 -1.6 0 4 8 12 16 20 24 28 -0.1 -0.8 -0.2 0 Figure 25. Comparision of Longitudinal stress for different 4 8 12 16 20 24 28 32 36 40 values of poisson's ratio at boundary r=a due to axisymmetric Z=z/a step load σ 01 ; n= 0, M1=2 Figure 22. Boundary displacement u due to step  0 , zz 0, z load σ 02 ; n= 0  02 0 4 8 12 16 20 24 28  0 ,  0  01 z/a 0.8 -0.1 0.6 0 -0.2 0.4   0.15 0.2   0.25 -0.3   0.35 z/a 0 -0.4 0 -0.2 0 4 8 12 16 20 24 28 -0.5 -0.4 0.15 -0.6 -0.6 -0.8 -0.7 0.25 -1 -0.8 0.35 -1.2 -1.4 -0.9 -1.6 -1 Figure 23. Comparision of Hoop stress for different values of Figure 26. Comparision of Longitudinal stress for different poisson's ratio at boundary r=a due to axisymmetric step load values of poisson's ratio at boundary r=a due to axisymmetric σ01 ; n= 0,M1=2 step load σ 02 ; n= 0, M1=2 821