Elastic analysis of pressurized thick hollow cylindrical shells with clamped-clamped ends/Plonasieniu hermetiniu cilindru su uzsandarintais galais tamprioji analize.
Ghannad, M. ; Nejad, M. Zamani
1. Introduction
There are many engineering applications of commonly used
structures, such as rods, annular disks, cylindrical and spherical
shells when subjected to different loading and boundary conditions.
Deformation and stress analysis of thick-walled cylinders subjected to
either internal or external pressure is an important topic in
engineering because of their rigorous applications in industry as well
as in daily life. For this reason, the classical problem of a
pressurized thick hollow cylinder has been the topic of a variety of
theoretical investigations.
Naghdi and Cooper [1], assuming the cross shear effect, formulated
the shear deformation theory (SDT). Mirsky and Hermann [2], derived the
solution of thick cylindrical shells of homogenous and isotropic
materials, using the first shear deformation theory (FSDT). Greenspon
[3], opted to make a comparison between the findings regarding the
different solutions obtained for cylindrical shells. Making use of
Mirsky-Hermann theory and the finite difference method (FDM), Ziv and
Perl [4] obtained the vibration response for semi-long cylindrical
shells. Using SDT and Frobenius series, Suzuki et. al. [5], obtained the
solution of the free vibration of cylindrical shells with variable
thickness, and Takashaki et. al. [6] obtained the same solution for
conical shells. A paper was published by Kang and Leissa [7] where
equations of motion and energy functional were derived for a
three-dimensional coordinate system. The field equations are utilized to
express such energy functional in terms of displacement components. The
stress state of two-layer hollow bars in which they are exposed to axial
load is analyzed [8]. The layers are made of isotropic, homogeneous,
linearly elastic material, and they are considered as concentric
cylinders.
Assuming that the material properties vary nonlinearly in the
radial direction and the Poisson's ratio is constant, Zamani Nejad
and Rahimi [9] obtained closed form solutions for one-dimensional
steady-state thermal stresses in a rotating functionally graded
pressurized thick-walled hollow circular cylinder. A complete and
consistent 3D set of field equations has been developed by tensor
analysis to characterize the behavior of FGM thick shells of revolution
with arbitrary curvature and variable thickness along the meridional
direction [10]. Using the analytical method for stress strain state of
two-layer mechanically inhomogeneous pipe subjected to internal pressure
at elastic plastic loading are analyzed by Brazenas and Vaiciulis [11].
This article presents the general method of derivation and the
analysis of an internally pressurized thick-walled cylinder shell with
clamped-clamped ends, taking into account the effect of shear stresses
and strains.
2. Classical theory
The plane elasticity theory (PET) or classical theory is based on
the assumption that the straight lines perpendicular to the central axis
of the cylinder remain unchanged after loading and deformation.
According to this theory, the deformations are axisymmetric and do not
change along the longitudinal cylinder. In other words, the elements do
not have any rotation, and the shear strain is assumed to be zero. Thus,
equilibrium equations are independent of one another, and the coupling
of the equations is deleted. Therefore,
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1)
and
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (2)
The differential equation based on the Navier Solution is
[d.sup.2][u.sub.r]/d[r.sup.2] + 1/r d[u.sub.r]/dr - 1/[r.sup.2]
[u.sub.r] = 0 (3)
The solution of the Eq. (3) is
[u.sub.r](r) = [C.sub.1]r + [C.sub.2]/r (4)
This method is applicable in problems in which shear stresses and
strains are considered zero. However, to solve the problems such as the
following it is not possible to use the PET
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (5)
3. Shear deformation theory (SDT)
In SDT, the straight lines perpendicular to the central axis of the
cylinder do not necessarily remain unchanged after loading and
deformation, suggesting that the deformations are axial axisymmetric and
change along the longitudinal cylinder. In other words, the elements
have rotation, and the shear strain is not zero.
In Fig. 1, the location of a typical point m (r), within the shell
element may be determined by R and z, as
r = R(x) + z (6)
where R represents the distance of middle surface from the axial
direction, and z is the distance of typical point from the middle
surface.
In Eq. (6), x and z must be as follows
- h/2 [less than or equal to] z [less than or equal to] h/2, 0
[less than or equal to] x [less than or equal to] L (7)
where h and L are the thickness and the length of the cylinder.
R(x) and inner and outer radii ([r.sub.i], [r.sub.o]) of the
cylinder are as follows
[r.sub.i] = R - h/2 = const., [r.sub.o] = R + h/2 = const. (8)
Based on PET, the radial displacement of the cylinder is
[u.sub.r](r) = [C.sub.1](R + z) + [C.sub.2]/R + z (9)
Using the Taylor's expansion for [absolute value of z/R] <
1,
[u.sub.r](r) = [C.sub.1] (R + z) + [C.sub.2]/R(1 - z/R +
[z.sup.2]/[R.sup.2] +...) (10)
Thus,
[u.sub.r](r) = [u.sub.0] + [u.sub.1]z + [u.sub.2][z.sup.2] +....
(11)
According to Eq. (11), the radial displacement is written in the
form of a polynomial function of z. When z = 0, it shows the
displacement of the mid-plane.
The general axisymmetric displacement field ([U.sub.x], [U.sub.z]),
in the first-order Mirsky-Hermann's theory could be expressed on
the basis of axial and radial displacements, as follows
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (12)
where u(x) and w(x) are the displacement components of the middle
surface. Also, [phi](x) and [psi](x) are the functions used to determine
the displacement field.
[FIGURE 1 OMITTED]
The strain-displacement relations in the cylindrical coordinates
system are
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (13)
In addition, the stresses on the basis of constitutive equations
for homogenous and isotropic materials are as follows
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (14)
where [[sigma]].sub.i] and [[epsilon].sub.i] are the stresses and
strains in the axial (x), circumferential ([theta]), and radial (z)
directions; [upsilon] and E are Poisson's ratio and Young's
modulus, respectively.
The normal forces ([N.sub.x],[N.sub.[theta]],[N.sub.z]),shear force
([Q.sub.x]), bending moments ([M.sub.x], [M.sub.[theta]], [M.sub.z]),
and the torsional moment ([M.sub.xz]) in terms of stress resultants are
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (15)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (16)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (17)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (18)
On the basis of the principle of virtual work, the variations of
strain energy are equal to the variations of the external work as
follows
[delta]U = [delta]W (19)
where U is the total strain energy of the elastic body and W is the
total external work due to internal pressure. The strain energy is
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (20)
and the external work is
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (21)
where P is internal pressure.
The variation of the strain energy is
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (22)
The resulting Eq. (22) will be
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (23)
and the variation of the external work is
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (24)
The resulting Eq. (24) will be
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (25)
Substituting Eqs. (13) and (14) into Eq. (19), and drawing upon
calculus of variation and the virtual work principle, we will have
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (26)
and the boundary conditions are
R[[[N.sub.x][delta]u + [M.sub.x][delta][phi] + [Q.sub.x][delta]w +
[M.sub.xz][delta][psi]].sup.L.sub.O] = 0 (27)
Eq. (27) states the boundary conditions which must exist at the two
ends of cylinder. In order to solve the set of differential equations
(26), forces and moments need to be expressed in terms of the components
of displacement field, using Eqs. (15) to (18). Thus, set of
differential equations (26) could be derived as follows
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (28)
The set of equations (28) is a set of linear nonhomogenous
equations with constant coefficients. The coefficients matrices
[[[[bar.A].sub.i]].sub.4x4], and force vector {[bar.F]} are
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (29)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (30)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (31)
{[bar.F]} = P/[lambda]E(R - h/2)[{0 0 -1 h/2}.sup.T] (32)
The parameters u and a are as follows
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (33)
where K is the shear correction factor that is embedded in the
shear stress term.
It is assumed that in the static state, for cylindrical shells K =
5/6 [12].
[[[bar.A].sub.3]] is irreversible and its reverse is needed in the
next calculations. In order to make [[[[bar.A].sub.3]].sup.-1], the
first equation in the set of Eqs. (26) is integrated.
R[N.sub.x] = [C.sub.0] (34)
In Eq. (28), it is apparent that does not exist, but du/dx does.
Taking du/dx as v,
u = [integral] vdx + [C.sub.7] (35)
Thus, set of differential Eqs. (28) could be deived as follows
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (36)
where, the coefficients matrices [[[A.sub.i]].sub.4x4], and force
vector {F} are
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (37)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (38)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (39)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (40)
The equations (36) are the set of nonhomogenous linear differential
equations with constant coefficients.
4. Analytical solution
Defining the differential operator P(D), Eq. (36) is written as
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (41)
Thus
P(D){y} = {F} (42)
The differential Eq. (42) has the general solution including
general solution for homogeneous case [{y}.sub.g] and particular
solution [{y}.sub.p], as follows
{y} = [{y}.sub.g] + [{y}.sub.p] (43)
For the general solution for homogeneous case, [{y}.sub.g] =
{V}[e.sup.mx] is substituted in P(D){y} = 0.
[e.sup.mx] [[m.sup.2][[A.sub.1]] + m[[A.sub.2]] + [[A.sub.3]]]{V} =
{0} (44)
Given that [e.sup.mx] [not equal to] 0, the following eigenvalue
problem is created.
[[m.sup.2][[A.sub.1]] + m[[A.sub.2]] + [[A.sub.3]]]{V} = {0} (45)
To obtain the eigenvalues, the determinant of coefficients must be
considered zero.
[absolute value of [[m.sup.2[[A.sub.1]] + m[[A.sub.2]] +
[[A.sub.3]]] = 0 (46)
Thus,
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (47)
where
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (48)
The result of the determinant above is a six-order polynomial which
is a function of m, the solution of which is a 6 eigenvalues [m.sub.i].
The eigenvalues are 3 pairs of conjugated root. Substituting the
calculated eigenvalues in Eq. (45), the corresponding eigenvectors
[{V}.sub.i] are obtained. Therefore, the general solution for
homogeneous Eq. (42) is
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (49)
The constants [C.sub.1] to [C.sub.6] are obtained by applying
boundary conditions. Given that {F} is comprised of constant parameters,
the particular solution is obtained as follows.
[{y}.sub.P] = [[[A.sub.3]].sup.-1] {F} (50)
Therefore, the general solution for Eq. (42) is
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (51)
In general, the problem consists of 8 unknown values of [C.sub.i],
including [C.sub.0] (Eq. 34), [C.sub.1] to [C.sub.6] (Eq. 51), and
[C.sub.7] (Eq. 35). By applying boundary conditions, one can obtain the
constants of [C.sub.i].
Given that the two ends of the cylinder are clamped-clamped, then
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (52)
Based on PET in the plane strain state, radial stress,
circumferential stress and radial displacement are as follows [13]:
[[sigma].sub.r] = P/[k.sup.2] - 1 [1 - [k.sup.2]/[([bar.r]).sup.2]]
(53)
[[sigma].sub.[theta]] = P/[k.sup.2] - 1 [1 -
[k.sup.2]/[([bar.r]).sup.2]] (54)
[[sigma].sub.x] = 2[upsilon]P/[k.sup.2] - 1 (55)
[u.sub.r] = P[r.sub.i][bar.r](1 + [upsilon])/E([k.sup.2] - 1)[1 -
2[upsilon] + [k.sup.2]/[([bar.r]).sup.2]] (56)
where [bar.r] = r/[r.sub.i] and k = [r.sub.o]/[r.sub.i].
5. Results and discussion
In this section, we present the results for a homogeneous and
isotropic thick hollow cylindrical shell with [r.sub.i] = 40 mm, h = 20
mm and L = 800 mm. The Young's modulus and Poisson's ratio,
respectively, have the values of E = 200 GPa and [upsilon] = 0.3. The
applied internal pressure is 80 MPa.
The analytical solution is carried out by writing the program in
MAPLE 12. The numerical solution is obtained through finite element
method (FEM).
Table presents the results of the different solutions for the
middle of the cylinder (x = L/2) and mid-layer (z = 0). The results
suggest that in points further away from the boundary it is possible to
make use of PET.
Fig. 2 shows the distribution of axial displacement at different
layers. At points away from the boundaries, axial displacement does not
show significant differences in different layers, while at points near
the boundaries, the reverse holds true. The distribution of radial
displacement at different layers is plotted in Fig. 3. The radial
displacement at points away from the boundaries depends on radius and
length. According to Figs. 2 and 3, the greatest axial and radial
displacement occurs in the internal surface (z = - h/2). Distribution of
circumferential stress in different layers is shown in Fig. 4. The
circumferential stress at all points depends on radius and length. The
circumferential stress at layers close to the external surface at points
near boundary is negative, and at other layers positive. The greatest
circumferential stress occurs in the internal surface (z = - h/2).
[FIGURE 1 OMITTED]
[FIGURE 2 OMITTED]
[FIGURE 3 OMITTED]
[FIGURE 4 OMITTED]
Fig. 5 shows the distribution of shear stress at different layers.
The shear stress at points away from the boundaries at different layers
is the same and trivial. However, at points near the boundaries, the
stress is significant, especially in the internal surface, which is the
greatest.
[FIGURE 5 OMITTED]
In the Figs. 6-10, displacement and stress distributions are
obtained using FSDT are compared with the solutions of FEM and are
presented in the form of graphs.
[FIGURE 6 OMITTED]
[FIGURE 7 OMITTED]
[FIGURE 8 OMITTED]
[FIGURE 9 OMITTED]
[FIGURE 10 OMITTED]
6. Conclusions
In the present study, the advantages as well as the disadvantages
of the PET (Lame' solution) for hollow thick-walled cylindrical
shells with different boundary conditions at the two ends were
indicated. Regarding the problems which could not be solved through PET,
the solution based on the FSDT is suggested. At the boundary areas (20
percent of the length of the cylinder) of a thick-walled cylinder with
clamped-clamped ends, having constant thickness and uniform pressure,
given that displacements and stresses are dependent on radius and
length, use cannot be made of PET, and FSDT must be used. In the areas
further away from the boundaries (80% of the length of the cylinder), as
the displacements and stresses along the cylinder remain constant and
dependent on radius, PET ought to be used. The shear stress in boundary
areas cannot be ignored, but in areas further away from the boundaries,
it can be ignored. Therefore, the PET can be used, provided that the
shear strain is zero. The maximum displacements and stresses in all the
areas of the cylinder occur on the internal surface. The analytical
solutions and the solutions carried out through the FEM show good
agreement.
Received June 02, 2010 Accepted September 27, 2010
References
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Article in Press.
M. Ghannad *, M. Zamani Nejad **
* Mechanical Engineering Faculty, Shahrood University of
Technology, Shahrood, Iran, E-mail:
[email protected]
** Mechanical Engineering Department, Yasouj University, Yasouj P.
O. Box: 75914-353 Iran, E-mail:
[email protected],
[email protected]
Table
Numerical results of the different solutions
[[sigma].sub.r], MPa [[sigma].sub.[theta]], MPa
FSDT -27.56 155.62
FEM -28.16 156.11
PET -28.16 156.16
[[sigma].sub.x], MPa [[sigma].sub.r], mm
FSDT 36.24 0.03826
FEM 36.22 0.03843
PET 38.40 0.03827