Jtam-A4.dvi


JOURNAL OF THEORETICAL

AND APPLIED MECHANICS

51, 4, pp. 1067-1079, Warsaw 2013

ELASTIC SOLUTION OF PRESSURIZED CLAMPED-CLAMPED

THICK CYLINDRICAL SHELLS MADE OF FUNCTIONALLY

GRADED MATERIALS

Mehdi Ghannad

Shahrood University of Technology, Mechanical Engineering Faculty, Shahrood, Iran

e-mail: mghannadk@shahroodut.ac.ir

Mohammad Zamani Nejad

Yasouj University, Mechanical Engineering Department, Yasouj, Iran

e-mail: m.zamani.n@gmail.com; m zamani@mail.yu.ac.ir

The effect of material inhomogeneity on displacements and stresses of an internally pres-
surized clamped-clamped thick hollow cylinder made of functionally graded materials is
investigated. Themodulus of elasticity is graded along the radial direction according to po-
wer functions of the radial direction. It is assumed that Poisson’s ratio is constant across the
cylinder thickness.Thegoverningdifferential equationsweregenerallyderived,makinguseof
the first-order shear deformation theory (FSDT). Following that, the set of non-homogenous
linear differential equations for the cylinderwith clamped-clamped endswas solved, and the
effect of loading and supports on the stresses and displacements was investigated. The pro-
blem was also solved, using the finite element method (FEM), the results of which were
compared with those of the analytical method.

Key words: clamped-clamped thick cylinder, functionally graded material (FGM), shear
deformation theory (SDT)

1. Introduction

Fromviewpoints of solidmechanics, functionally gradedmaterials (FGMs) arenon-homogeneous
elastic mediums. Using this property in the FGMs, an engineer can design composite materials
such that any portion of the materials reaches the same safety level. In the last two deca-
des, FGMs have been widely used in engineering applications, particularly in high-temperature
environment, microelectronic, power transmission equipment, etc. On the other hand, the thick
cylinder is a pressure vessel commonly used in industry. From the strength analysis, it is seen
that the most dangerous point is located at the inner portion of the cylinder, if homogeneous
materials are used. However, if one uses the FMGs for a thick cylinder, this situation can be
changed. Due to the importance of structural integrity, the safe design of such hollow cylin-
drical structures, in particular for functionally graded hollow annuli or tubes have attracted
considerable attention in recent years. Naghdi and Cooper (1956), assuming the cross shear
effect, formulated the shear deformation theory (SDT). Mirsky and Hermann (1958), derived
the solution of thick cylindrical shells of homogenous and isotropic materials, using the first
shear deformation theory (FSDT). Using the shear deformation theory and Frobenius series,
Suzuki et al. (1981), obtained the solution of the free vibration of cylindrical shells with variable
thickness. Kang (2007) derived the field equations for homogenous thick shells of revolution.
Eipakchi et al. (2003) obtained the solution of the homogenous and isotropic thick-walled cy-
lindrical shells with variable thickness, using the first-order shear deformation theory (FSDT)
and the perturbation technique. Hongjun et al. (2006), and Zhifei et al. (2007), provided elastic
analysis and the exact solution for stresses in FGM hollow cylinders in the state of plane strain



1068 M. Ghannad,M.Z. Nejad

with isotropic multi-layers based on Lamé’s solution. Nejad and Rahimi (2010), obtained stres-
ses in isotropic rotating thick-walled cylindrical pressure vessels made of a functionally graded
material as a function of the radial direction by using the theory of elasticity. Pan and Roy
(2006) derived exact solutions for multilayered FGM cylinders under static deformation. They
obtained these solutions by making use of the method of separation of variables and expressed
it in terms of the summation of the Fourier series in the circumferential direction. A complete
and consistent 3D set of field equations was developed by tensor analysis to characterize the
behavior of FGM thick shells of revolutionwith arbitrary curvature and variable thickness along
the meridional direction by Nejad et al. (2009). Using plane elasticity theory and Complemen-
tary Functions method, Tutuncu and Temel (2009) obtained axisymmetric displacements and
stresses in functionally-graded hollow cylinders, disks and spheres subjected to uniform internal
pressure. Assuming that the shear modulus varies in the radial direction either according to a
power law relation or an exponential function, Batra andNie (2010) obtained analytically plane
strain infinitesimal deformations of a non-axisymmetrically loaded hollow cylinder and of an
eccentric cylinder composed of a linear elastic isotropic and incompressible functionally graded
material. Using the Airy stress function, Nie and Batra (2010) presented exact solutions for
plane strain deformations of a functionally graded hollow cylinder with the inner and the outer
surfaces subjected to different boundary conditions. Making use of FSDT and the virtual work
principle, Ghannad and Nejad (2010) generally derived the differential equations governing the
homogenous and isotropic axisymmetric thick-walled cylinders with the same boundary condi-
tions at the two ends. Following that, the set of nonhomogenous linear differential equations for
the cylinderwith clamped-clamped endswas solved.Ghannad et al. (2012) also obtained the ge-
neral solution of the clamped-clamped thick cylindrical shells with variable thickness subjected
to constant internal pressure using the first shear deformation theory. Ostrowski and Michalak
(2011) investigated the effect of geometry andmaterial properties on the temperature field in a
two-phase hollow cylinder.

In this paper, taking into account the effect of shear stresses and strains, the generalmethod
of derivation and the analysis of an internally pressurized thick-walled cylinder shell made of a
functionally gradedmaterial with clamped-clamped ends are presented.

2. Problem formulation

Consider a clamped-clamped thick-walled isotropic FGMcylinderwith an inner radius ri, thick-
ness h, and length L subjected to internal pressures P.

Given that the radial coordinate r is normalized as r = r/ri. The modulus of elasticity E
is assumed to be radially dependent and is assumed to vary as follows

E(r)= Ei(r)
n =

Ei
rni
rn (2.1)

Here Ei is themodulus of elasticity at the inner surface ri and n is the inhomogeneity constant
determined empirically. Since the analysis was done for a thick wall cylindrical pressure vessel
of an isotropic FGM, and given that the variation of Poisson’s ratio ν for engineeringmaterials
is small, it is assumed as constant.

The range −1 ¬ n ¬ +1 to be used in the present study covers all the values of coordi-
nate exponent encountered in the references cited earlier. However, these values for n do not
necessarily represent a certain material.

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
anddeformation.According to this theory, thedeformations are axisymmetric anddonot change



Elastic solution of pressurized clamped-clamped ... 1069

along the longitudinal cylinder. In other words, the elements do not have any rotation, and the
shear strain is assumed to be zero

ur = ur(r) ux = ux(x)

γrx =
∂ur
∂x
+
∂ux
∂r
=0 ⇒ τrx =0

(2.2)

Therefore, equilibrium equations are independent of one another, and the coupling of the equ-
ations is deleted

dσr
dr
+
1

r
(σr−σθ)= 0

dσx
dx
=0 (2.3)

The differential equation based on the Navier solution in terms of the radial displacement is
(Ghannad et al., 2010)

r2
d2ur
dr2
+(n+1)r

dur
dr
+(nν∗−1)ur =0 (2.4)

ν∗ is defined based on boundary conditions. The solution to the Eq. (2.4) based on real roots is
as follows

ur(r)= C1r
m1 +C2r

m2 m1,2 =
1

2

[

−n±
√

n2−4(nν∗−1)
]

(2.5)

This method is applicable to problems in which the shear stress τrx and shear strain γrx, are
considered zero. However, to solve the problems such as the following, it is not possible to use
the PET

ur = ur(x,z) ux = ux(x,z)

γrx =
∂ur
∂x
+
∂ux
∂r
6=0 ⇒ τrx 6=0

(2.6)

In the shear deformation theory (SDT), the straight lines perpendicular to the central axis of
the cylinder do not necessarily remain unchanged after loading anddeformation, 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+z (2.7)

where R represents the distance of the middle surface from the axial direction, and z is the
distance of a typical point from themiddle surface.
In Eq. (2.7), x and z must be as follows

−
h

2
¬ z ¬

h

2
0¬ x ¬ L (2.8)

where h and L are the thickness and the length of the cylinder.
The general axisymmetric displacement field (Ux,Uz) in the first-order shear deformation

theory could be expressed on the basis of axial and radial displacements, as follows (Mirsky and
Hermann, 1958)

{

Ux
Uz

}

=

{

u(x)
w(x)

}

+

{

φ(x)
ψ(x)

}

z (2.9)



1070 M. Ghannad,M.Z. Nejad

Fig. 1. Cross section of the thick cylinder with clamped-clamped ends

where u(x) and w(x) are the displacement components of the middle surface. Also, φ(x) and
ψ(x) are functions of the displacement field.
The strain-displacement relations in the cylindrical coordinates system are

εx =
∂Ux
∂x
=
du

dx
+
dφ

dx
z εθ =

Uz
r
=
1

R+z
(w+ψz)

εz =
∂Uz
∂z
= ψ γxz =

∂Ux
∂z
+
∂Uz
∂x
=
(

φ+
dw

dx

)

+
dψ

dx
z

(2.10)

The distribution of elasticity modulus basis on Eq. (2.1) and Eq. (2.7) is

E(r)= Ei
(R+z

ri

)n
=
Ei
rni
(R+z)n = E(z) (2.11)

The stresses on the basis of constitutive equations for homogenous and isotropic materials are
as follows

σi = λE(z)[(1−ν)εi+ν(εj +εk)] i 6= j 6= k

τxz = λE(z)
[

(1−2ν)
γxz
2

]

λ =
1

(1+ν)(1−2ν)

(2.12)

where σi and εi are the stresses and strains in the axial x, circumferential θ, and radial z
directions, respectively.
The normal forces Nx, Nθ, Nz, shear force Qx, bending moments Mx, Mθ, Mz, and the

torsional moment Mxz in terms of stress resultants are

{Nx,Nθ,Nz}=

h/2
∫

−h/2

[

σx
(

1+
z

R

)

,σθ,σz
(

1+
z

R

)]

dz

{Mx,Mθ,Mz}=

h/2
∫

−h/2

[

σx
(

1+
z

R

)

,σθ,σz
(

1+
z

R

)]

z dz

Qx =

h/2
∫

−h/2

τxz
(

1+
z

R

)

dz Mxz =

h/2
∫

−h/2

τxz
(

1+
z

R

)

z dz

(2.13)

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

δU = δW (2.14)



Elastic solution of pressurized clamped-clamped ... 1071

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

U =

∫∫∫

V

U∗ dV dV = r dr dθ dx =(R+z)dxdθ dz

U∗ =
1

2
(σxεx+σθεθ +σzεz + τxzγxz)

(2.15)

and the external work is

W =

∫∫

S

(f ·u) dS dS = ridθdx =
(

R−
h

2

)

dθ dx f ·u= PUz (2.16)

The variation of the strain energy is

δU = R

2π
∫

0

L
∫

0

h/2
∫

−h/2

δU∗
(

1+
z

R

)

) dz dxdθ (2.17)

Resulting Eq. (2.17) will be

δU

2π
= R

L
∫

0

h/2
∫

−h/2

(σxδεx+σθδεθ +σzδεz + τxzδγxz)
(

1+
z

R

)

dz dx (2.18)

and the variation of the external work is

δW =

2π
∫

0

L
∫

0

PδUz
(

R−
h

2

)

dxdθ (2.19)

Resulting Eq. (2.19) will be

δW

2π
= P

L
∫

0

δUz
(

R−
h

2

)

dx (2.20)

Substituting Eqs. (2.10) and (2.12) into Eq. (2.14), and drawing upon calculus of variation and
the virtual work principle, we will have

R
dNx
dx
=0 R

dMx
dx
−RQx =0

R
dQx
dx
−Nθ =−P

(

R− h
2

)

R
dMxz
dx
−Mθ−RNz = P

h

2

(

R−
h

2

)

(2.21)

and the boundary conditions at the two ends of cylinder are

R
[

Nxδu+Mxδφ+Qxδw+Mxzδψ
]L

0
=0 (2.22)

In order to solve the set of differential equations (2.21), forces andmoments need to be expressed
in terms of the components of the displacement field, using Eqs. (2.13). Thus, set of differential
equations (2.21) could be derived as follows

A1
d2

dx2
y+A2

d

dx
y+A3y=F y= [u,φ,w,ψ]

T (2.23)



1072 M. Ghannad,M.Z. Nejad

The set of equations (2.23) is a set of linearnon-homogenous equationswith constant coefficients.

A3 is irreversible and its reverse is needed in the next calculations. In order to make A
−1

3 , the
first equation in the set of Eqs. (2.21) is integrated

RNx = C0 (2.24)

In Eq. (2.23), it is apparent that u does not exist, but du/dx does. Taking du/dx as v

u =

∫

v dx+C7 (2.25)

Thus, set of differential equations (2.23) could be derived as follows:

A1
d2

dx2
y+A2

d

dx
y+A3y=F y= [v,φ,w,ψ]

T (2.26)

The set of equations (2.26) is a set of linearnon-homogenous equationswith constant coefficients.

3. Analytical solution

Defining the differential operator P(D), Eq. (2.26) is written as

P(D)=A1D
2+A2D+A3 D =

d

dx
D2 =

d2

dx2
(3.1)

Thus

P(D)y=F (3.2)

Differential Eq. (3.2) has a general solution yg and particular solution yp, as follows

y=yg +yp (3.3)

For the general solution, yg =Ve
mx is substituted in homogeneous Eq. (3.2)

emx(m2A1+mA2+A3)V=0 (3.4)

Given that emx 6=0, the following eigenvalue problem is created

(m2A1+mA2+A3)V=0 (3.5)

To obtain the eigenvalues, the determinant of coefficients must be considered zero
∣

∣

∣m2A1+mA2+A3
∣

∣

∣=0 (3.6)

The result of the determinant above is a six-order polynomial which is a function of m, the
solution of which is 6 eigenvalues mi. The eigenvalues are 3 pairs of conjugated roots. Substitu-
ting the calculated eigenvalues into Eq. (3.5), the corresponding eigenvectors Vi are obtained.
Therefore, the general solution for homogeneous Eq. (3.2) is

yg =
6
∑

i=1

CiVie
mix (3.7)

Given that F is comprised of constant parameters, the particular solution is obtained as follows

yp =A
−1
3 F (3.8)



Elastic solution of pressurized clamped-clamped ... 1073

Therefore, the total solution for Eq. (3.2) is

y=
6
∑

i=1

CiVie
mix+A−13 F (3.9)

In total, the problem consists of 8 unknown values of Ci, including C0 (Eq. (2.24)), C1 to C6
(Eq. (3.9)), and C7 (Eq. (2.25)). By applying boundary conditions, one can obtain the constants
of Ci. Given that the two ends of the cylinder are clamped-clamped, then



















u
φ
w
ψ



















x=0

=



















u
φ
w
ψ



















x=L

=



















0
0
0
0



















(3.10)

4. Solution of FGM cylinders

In order to analyzeFGMcylinders, substitutingEq. (2.11) intoEqs. (2.12) and thenbyusingEq.
(2.10), the stress resultants in terms of the displacement field are obtained. With substituting
stress resultants into Eqs (2.21), a set of nonhomogeneous differential equations with variable
coefficients is obtained. The constants used are as follows

k =
ro
ri

α = lnk β =
k−1

kri
µ = K

(1−2ν

2

)

(4.1)

where K is the shear correction factor.

4.1. Inhomogeneity constant of n =−1

Based on Eq. (2.11), the modulus of elasticity is

E(z) =
Eiri
R+z

(4.2)

The coefficients matrices [Ai]]4×4 and the force vector F in Eq. (2.26) are

A1 =











0 0 0 0

0 (1−ν)h
3

12
0 0

0 0 µh 0

0 0 0 µh
3

12











A2 =











0 0 0 0
0 0 ν(h−Rα)−µh ν(R2α−Rh)
0 −ν(h−Rα)+µh 0 0
0 −ν(R2α−Rh) 0 0











A3 =











(1−ν)h 0 να ν(2h−Rα)
0 −µh 0 0
−να 0 −(1−ν)β −α+(1+ν)Rβ

−ν(2h−Rα) 0 −α+(1−ν)Rβ 2(Rα−h)− (1−ν)R2β











F=
1

λEiri

[

C0,0,−P
(

R−
h

2

)

,P
h

2

(

R−
h

2

)]T

(4.3)



1074 M. Ghannad,M.Z. Nejad

To obtain the eigenvalues mi, the determinant of coefficients (Eq. (3.6)) must be considered
zero
∣

∣

∣

∣

∣

∣

∣

∣

∣

(1−ν)h 0 να ν(2h−Rα)

0 m2(1−ν)h
3

12
−µh mν(h−Rα)−mµh mν(R2α−Rh)

−να −mν(h−Rα)+mµh m2µh− (1−ν)β −α+(1+ν)Rβ
−ν(2h−Rα) −mν(R2α−Rh) −α+(1−ν)Rβ b44

∣

∣

∣

∣

∣

∣

∣

∣

∣

=0 (4.4)

where

b44 = m
2µ
h3

12
+2(Rα−h)− (1−ν)R2β

With solving of the obtained polynomial of the above determinant, the eigenvalues mi and
the following eigenvectors Vi are calculated. Finally, the total solution (Eq. (3.9)), with using
boundary conditions is obtained.

4.2. Inhomogeneity constant of n =+1

Based on Eq. (2.11), the modulus of elasticity is

E(z) =
Ei
ri
(R+z) (4.5)

The coefficients matrices [Ai]4×4, and force vector F in Eq. (2.26) are

A1 =

















0 0 0 0

0 (1−ν)
(

R2h3

12
+ h

5

80

)

0 0

0 0 µ
(

R2h+ h
3

12

)

µRh
3

6

0 0 µRh
3

6
µ
(

R2h3

12
+ h

5

80

)

















A2 =















0 (1−ν)Rh
3

6
0 0

(1−ν)Rh
3

6
0 −µ

(

R2h+ h
3

12

)

+νh
3

12
−(2µ−3ν)Rh

3

12

0 µ
(

R2h+ h
3

12

)

−νh
3

12
0 0

0 (2µ−3ν)Rh
3

12
0 0















A3 =

















(1−ν)
(

R2h+ h
3

12

)

0 νRh ν
(

R2h+ h
3

6

)

0 −µ
(

R2h+ h
3

12

)

0 0

−νRh 0 −(1−ν)h −νRh

−ν
(

R2h+ h
3

6

)

0 −νRh −(1−ν)R2h− h
3

6

















F=
ri
λEi

[

C0,0,−P
(

R−
h

2

)

,P
h

2

(

R−
h

2

)]T

(4.6)

To obtain the eigenvalues mi, the determinant of coefficients (Eq. (3.6)) must be considered
zero

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

c11 m(1−ν)
Rh3

6
νRh ν

(

R2h+ h
3

6

)

m(1−ν)Rh
3

6
c22 c23 −m(2µ−3ν)

Rh3

12

−νRh c32 c33 m
2µRh

3

12
−νRh

−ν
(

R2h+ h
3

6

)

−m(2µ−3ν)Rh
3

12
m2µRh

3

12
−νRh c44

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

=0 (4.7)



Elastic solution of pressurized clamped-clamped ... 1075

where

c11 =(1−ν)
(

R2h+
h3

12

)

c22 = m
2(1−ν)

(R2h3

12
+
h5

80

)

−µ
(

R2h+
h3

12

)

c23 =−c32 =−mµ
(

R2h+
h3

12

)

+mν
h3

12

c33 = m
2µ
(

R2h+
h3

12

)

− (1−ν)h

c44 = m
2µ
(R2h3

12
+
h5

80

)

− (1−ν)R2h−
h3

6

Similarly, with solving of thr obtained polynomials of Eq. (4.7), the eigenvalues mi and the
following eigenvectors Vi are calculated, and then, rhe total solution (Eq. (3.9)), with using
boundary conditions is obtained.

Based onPET in the plane strain state, the radial stress, circumferential stress, axial stress,
and radial displacement are as follows (Ghannad et al., 2010)

σr =
P(r)n−1

km1 −km2
(km2 rm1 −km1 rm2)

σθ =
P(r)n−1

km1 −km2

[ 1−ν +νm1
(1−ν)m1+ν

km2 rm1 −
1−ν +νm2
(1−ν)m2+ν

km1 rm2
]

σx = ν(σr+σθ)

(4.8)

and

ur =
P(ri)(1+ν)(1−2ν)

Ei(km1 −km2)

[ 1

(1−ν)m1+ν
km2 rm1 −

1

(1−ν)m2+ν
km1 rm2

]

(4.9)

where r = r/ri and k = ro/ri.

5. Results and discussion

In this section, we present the results for a nonhomogeneous and isotropic clamped-clamped
thick hollow cylindrical shell with ri = 40mm, R = 50mm, h = 20mm and L = 800mm
for n = +1, n = 0 (Ghannad and Nejad, 2010), and n = −1. The modulus of elasticity at
the internal radius and Poisson’s ratio have values of Ei =200GPa and ν = 0.3, respectively.
The applied internal pressure is 80MPa. The shear correction factor K = 5/6 is considered
(Vlachoutsis, 1992; Jemielita, 2002). The analytical solution is carried out by writing a pro-
gram in MAPLE 13. The numerical solution is obtained through the finite element method
(FEM).

The distribution of radial displacement at different layers for n = +1 is plotted in Fig. 2.
The radial displacement at points away from the boundaries depends on the radius whereas at
points near boundary, the radial displacement depends on the radius and length. Figure 3 shows
the distribution of axial displacement for n =+1 at different layers. At points away from the
boundaries, the axial displacement does not show significant differences in different layers, while
at points near the boundaries, the reverse holds true. The distribution of circumferential stress
in different layers for n = +1 is shown in Fig. 4. In the same layer, at points away from the
boundaries, the circumferential stress depends on the radius whereas at points near boundary,



1076 M. Ghannad,M.Z. Nejad

the circumferential stress depends on the radius and length. The circumferential stress at layers
close to the external surface at points near the boundary is negative, and at other layers positive.

Fig. 2. Radial (a) and axial (b) displacement distribution in different layers (n =+1)

Fig. 3. Circumferential (a) and shear (b) stress distribution in different layers (n =+1)

According to Figs. 2 and 3, the greatest axial and radial displacements, circumferential and
shear stresses occur in the internal surface. The distribution of shear stress at different layers
is shown in Fig. 3b. The shear stress at points away from the boundaries at different layers
is the same and trivial. However, at points near the boundaries, the shear stress is significant,
especially in the internal surface, which is the greatest. The distribution of radial displacement
of the cylinder for different values of n in themiddle layer is shown inFig. 4a. It is observed that
the radial displacement depends on thematerial inhomogeneity constant and there is a decrease
in the value of the radial displacement as n increases. The distribution of axial displacement of
the cylinder for different values of n in the middle layer is shown in Fig. 4b. As the value of
n increases, the axial displacement decreases. The radial stress distribution for different values
of n in the middle layer of the cylinder is shown in Fig. 5a. According to this figure, changes in
the inhomogeneity constant do not have a significant effect on the radial stress, and the shear
stress at points away from the boundaries at different layers is the same. The distribution of
axial stress of the cylinder for different values of n in the middle layer is shown in Fig. 5b. It is
observed that the axial stress depends on thematerial inhomogeneity constant. The distribution
of circumferential stress of the cylinder for different values of n in the middle layer is shown
in Fig. 10. According to this figure, at points near boundaries, changes in the inhomogeneity
constant do not have a significant influence on the circumferential stress. It is seen fromFig. 6a
thatatpointsaway fromtheboundaries forhighervalues of n, the circumferential stress increase.
Figure 11 shows the distribution of shear stress for different values of n along themiddle layer.
The shear stress at points away from the boundaries for different values of n is the same and
trivial. However, at points near the boundaries, the stress is significant.



Elastic solution of pressurized clamped-clamped ... 1077

Fig. 4. Radial (a) and axial (b) displacement distribution in the middle surface

Fig. 5. Radial (a) and axial (b) stress distribution in the middle surface

Fig. 6. Circumferential (a) and shear (b) stress distribution in the middle surface

Table 1 presents the results of different solutions for the middle of the FGM cylinder
(x = L/2) and the middle surface (z = 0). The results suggest that in points further away
from the boundary it is possible to make use of PET.

Table 1. Numerical data from FSDT, FEM, and PET calculations for n =−1 and n =+1 in
x = L/2 and z =0

n =−1 n =+1
σr σθ σx ur σr σθ σx ur
[MPa] [MPa] [MPa] [mm] [MPa] [MPa] [MPa] [mm]

FSDT −27.34 149.3 34.49 0.04600 −27.52 160 37.57 0.03140

FEM −24.79 152.61 35.78 0.04570 −31.64 157.36 36.05 0.03188

PET −24.82 151.08 37.88 0.04600 −31.61 159.12 38.25 0.03143



1078 M. Ghannad,M.Z. Nejad

6. Conclusions

In the present study, the advantages as well as the disadvantages of the PET (Lamé solution)
for hollow thick-walled cylindrical shells were indicated. Regarding the problems which could
not be solved through PET, the solution based on the FSDT is suggested. At the boundary
areas of a thick-walled cylinder with clamped-clamped or clamped-free 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, 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.

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Manuscript received December 24, 2011; accepted for print June 25, 2013