1) The document analyzes stresses and displacements around cylindrical cavities in an elastic medium caused by moving step loads on the cavity surface.
2) Governing equations are derived using potential functions and Fourier series in cylindrical coordinates. The equations are reduced to Helmholtz equations for the potentials.
3) Solutions are obtained for axisymmetric and non-axisymmetric step loads moving parallel to the cavity axis. Numerical solutions provide stress and displacement values.
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
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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)
ra 1
(1)
a velocity V C1 C 2 ; therefore, the speed is σrθ
ra
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 ra
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 rz 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
n0 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
n0
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 ZRβ1coshu 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 ZRβ2 coshucoshnudu (23)
n 0 hn ZRβ2 coshucoshnudu
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-
2β10 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 gη2 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 Pm
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 Pm 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 Pm 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 σz 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