Solution to two-dimensional elastic problems involving functionally graded material in radial co-ordinates

Abstract

A general solution is derived for two-dimensional elastic problems involving functionally graded material in radial co-ordinates, similar to Michell’s solution for homogeneous isotropic materials. Young’s modulus is graded along the radial direction (r) as a power-law function, while Poisson’s ratio is constant. The stresses are expanded as Fourier series in the tangential direction (θ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\theta $$\end{document}) with the coefficients as functions of r. Equilibrium and strain compatibility equations are reduced to ordinary differential equations in terms of these coefficients, which are solved to get the general forms of stresses and corresponding displacements. While analytical solutions exist for some of the standard problems with specific geometry and loading conditions, this paper presents a general solution for problems with arbitrary geometry and boundary/loading conditions. Solutions to the standard problems available in the literature are only special cases of the general solution derived here, as shown by validation in this paper.

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Noname manuscript No.
(will be inserted by the editor)
Solution to two-dimensional elastic problems involving1
functionally graded material in radial co-ordinates2
U. Arasan ·S. Venkatachalam ·H. Murthy3
4
Received: date / Accepted: date5
Abstract A general solution is derived for two-dimensional elastic problems in-6
volving functionally graded material in radial co-ordinates, similar to Michell’s7
solution for homogeneous isotropic materials. Young’s modulus is graded along8
the radial direction (r) as a power-law while the Poisson’s ratio is constant. The9
stresses are expanded as Fourier series in the tangential direction (θ) with the10
coefficients as functions of r. Equilibrium and strain compatibility equations are11
reduced to ordinary differential equations in terms of these coefficients, which are12
solved to get the general form of stresses and corresponding displacements. While13
analytical solutions exist for some of the standard problems with specific geometry14
and loading conditions, this paper presents a general solution for problems with15
arbitrary geometry and boundary/loading conditions. Solutions to the standard16
problems available in the literature are only special cases of the general solution17
derived here, as shown through validation in this paper.18
Keywords Functionally graded material ·Stress-based approach ·Power-law19
gradation ·Two-dimensional elasticity20
1 Introduction21
In mechanical applications, a single material may not be suitable for every function22
that needs to be performed i.e, a single material may not satisfy multiple opera-23
tional requirements of the components. For example, in gears, a harder peripheral24
surface is desirable for wear resistance while a softer core is desirable for better25
structural damping/ fracture toughness. If a hard coating is provided to the sur-26
face such that the core and the layer are of two different materials, discontinuous27
U. Arasan∗
Matelys - Research Lab, Vaulx-en-Velin F-69120, France
E-mail: arasan62su@gmail.com
S. Venkatachalam
Dept. of Aerospace Engg., Karunya Institute of Technology and Sciences, Coimbatore, India
H. Murthy
Dept. of Aerospace Engg., Indian Institute of Technology Madras, Chennai, India

2 U. Arasan et al.
stresses at the interface could lead to debonding of the surface from the core. To28
avoid this, it is desirable to have a continuous variation in material property from29
the surface to the core. Functionally graded material (FGM) is a class of materials30
having continuous variation of material properties from one point to another (typ-31
ically from the surface to core) to meet the operational requirements. Reddy [1]32
states that “functionally graded structures are those in which the volume fractions33
of two or more materials are varied continuously as a function of position along34
certain dimension(s) of the structure to achieve a required function”. In addition35
to applications in gears, FGM can be used to alleviate stress concentrations in36
mechanical components. A properly chosen gradation in properties could reduce37
thermal stresses, residual stresses and stress concentrations [2, 3].38
In FGM, since the properties are changing through the dimension (typically39
thickness) of a material, it can provide designers with the tailored material re-40
sponse and exceptional performance in thermal environments. Birman and Byrd41
[13] reviewed the applications of FGM in various engineering fields. Further appli-42
cations of FGM would require analysis tools to solve the problems involving more43
complex geometry and loading [14–17]. Such analysis tools provide confidence in44
using FGMs since their behaviour for a given application can be thoroughly char-45
acterized. While finite element analysis (FEA) is a good tool, analytical solutions46
are always desirable, at least for simpler geometries and boundary conditions, be-47
cause they would provide benchmark solutions and an idea about the effect of48
different design parameters.49
Some of the standard problems involving homogeneous, isotropic materials50
have been solved for FGMs as well. Solutions to the axisymmetric problem of a51
thick-walled cylinder made of FGM, subjected to internal and external pressure,52
are obtained by a few researchers [10, 18–23]. Sburlati [23] solved for both power-53
law and exponential law gradation along the radial direction and the results from54
the Navier’s equation (displacement approach) are validated with FEA. The ef-55
fect of FGM on stress concentration around a hole in an infinite plate subjected56
to remote stresses has also been studied [11, 24–27]. Sburlati et al. [11] obtained57
Airy’s stress function under plane stress condition by solving the ordinary differ-58
ential equations corresponding to strain compatibility, for an FGM ring around59
the hole with a radial power-law variation of Young’s modulus and a constant60
Poisson’s ratio. They validated the results with FEA and showed that a decrease61
in Young’s modulus of the inner ring with respect to the entire plate leads to an62
increase of hoop stress at the interface. They also indicated that by changing the63
gradient factor, a considerable reduction in stress concentration could be achieved.64
Each of these solutions available in the literature is for specific geometry and load-65
ing/boundary conditions. A general solution subsuming all these solutions could66
provide a powerful analytical tool.67
In elasticity approach, two-dimensional (2D) problems are typically solved us-68
ing Airy’s stress functions along with stress and displacement boundary condi-69
tions for homogeneous materials. Due to the unavailability of a general solution70
for FGM, solving each problem separately is time-consuming. Thus, a general71
solution is desirable to obtain the solutions for the stresses and displacements,72
as given by Michell [28, 29] for homogeneous materials. In this work, the gen-73
eral form of stresses and displacements are derived for the 2D elastic problems74
involving FGMs with a power-law variation for Young’s modulus in the radial75
direction and validated with existing analytical solutions and FEA. Although an-76

Title Suppressed Due to Excessive Length 3
alytical solutions exist for some of the standard problems with specific geometry77
and loading conditions, this paper presents a general solution (similar to Michell’s78
solution for homogeneous materials) for problems with arbitrary geometry and79
boundary/loading conditions. Further, all the standard problems solved earlier80
have been shown to be simple applications of this general solution. Moreover, the81
choice of gradation factor is unbounded for the proposed solution whereas the82
existing analytical solutions provided in the literature are valid only for certain83
ranges of gradation factor.84
2 General Elasticity Solutions for Problems involving FGM85
2.1 Governing Equations and Series Representation of Stresses86
In the absence of body forces, following are the equilibrium equations in the polar87
coordinate system.88
r∂σrr
∂r +∂σrθ
∂θ +σrr −σθθ = 0 (1)
89
r∂σrθ
∂r +∂σθθ
∂θ + 2σrθ = 0 (2)
where σrr , σrθ and σθθ are the radial, shear and tangential stress components90
respectively. The constitutive relations for plane stress/ strain are as given below.91
rr =σrr −α(σrr +σθ θ )
2G, θθ =σθθ −α(σrr +σθ θ )
2G, rθ =σrθ
2G(3)
where α=νfor plane strain and α=ν/(1 + ν) for plane stress. In this paper,92
Young’s modulus in FGM is assumed to vary as a power-law along the radial93
direction, with a constant Poisson’s ratio. Consequently, the shear modulus Galso94
varies as a power-law with a gradient factor mas given below.95
G(r) = G0rm⇒dG
dr =mG
r;d
dr
1
G=−m
rG (4)
Strain compatibility is given by the following equation.96
r∂rr
∂r −∂2rr
∂θ2−r2∂2θθ
∂r2−2r∂θ θ
∂r + 2r∂2rθ
∂r∂θ + 2 ∂rθ
∂θ = 0 (5)
Using the Eqs. (3) & (4), and further simplifying using the equilibrium equations,97
the compatibility equation is written in terms of stresses.98
r2∂2
∂r2+ (1 −2m)r∂
∂r +∂2
∂θ2(σrr +σθθ ) + β1σrr +β2σθ θ = 0 (6)
where β1=m2−m(m+ 1)
1−αand β2=m2+m
1−α.99
Any random stress distribution can be assumed to be a Fourier series in θwith100
coefficients as functions of r.101
σrr =L0(r) + ∞
X
n=1
[Ln(r) cos nθ +Mn(r) sin nθ] (7a)

4 U. Arasan et al.
102
σrθ =P0(r) + ∞
X
n=1
[Pn(r) cos nθ +Qn(r) sin nθ] (7b)
103
σθθ =R0(r) + ∞
X
n=1
[Rn(r) cos nθ +Sn(r) sin nθ] (7c)
By substituting the Eq. (7) into the Eq. (1) and comparing corresponding cos nθ,104
sin nθ terms and the pure functions of r, the following ordinary differential equa-105
tions are obtained.106
rL0
0+L0=R0(8a)
107
rL0
n+Ln+nQn−Rn= 0 (8b)
108
rM 0
n+Mn−nPn−Sn= 0 (8c)
where ()0is derivative with respect to r. Similarly, substituting Eq. (7) into second109
equilibrium (Eq. (2)) and compatibility (Eq. (6)) equations, the following equations110
are obtained.111
rP 0
0+ 2P0= 0 (9a)
112
rP 0
n+ 2Pn=−nSn(9b)
113
rQ0
n+ 2Qn=nRn(9c)
114
r2(L
00
0+R
00
0) + (1 −2m)r(L0
0+R0
0) + β1L0+β2R0= 0 (10a)
115
r2(L
00
n+R
00
n) + (1 −2m)r(L0
n+R0
n) + (β1−n2)Ln+ (β2−n2)Rn= 0 (10b)
116
r2(M
00
n+S
00
n) + (1 −2m)r(M0
n+S0
n) + (β1−n2)Mn+ (β2−n2)Sn= 0 (10c)
Eqs. (8c), (9b) & (10c) are similar to Eqs. (8b), (9c) & (10b) with Mn,−Pnand117
Snin place of Ln,Qnand Rnrespectively. Therefore, the solution will be similar118
to both sets of equations involving (M, P, S )&(L, Q, R). Therefore, the solution119
to only one of the two sets of equations is presented here.120
2.2 Solution to the Governing Equations121
There are effectively 3 sets of 3 equations each corresponding to stresses being a122
function of ronly (axisymmetric) and stresses corresponding to sin nθ & cos nθ.123
Each of these 3 sets has to be solved separately. As discussed earlier, the sets124
involving sin nθ & cos nθ are similar and by solving one set, the other can be125
written in a similar manner.126

Title Suppressed Due to Excessive Length 5
2.2.1 Terms involving only r127
L0and R0represent the axisymmetric components of the stresses. Substituting128
Eq. (8a) in Eq. (10a), the following equation is obtained.129
r3L
000
0+ (5 −2m)r2L
00
0+3−6m+m2+m
1−αrL0
0+2m2−m2
1−αL0= 0
(11)
Taking L0(r) = Araand dividing by L0, the following equation is obtained.130
(a−m)a2+ (2 −m)a−2m+m
1−α= 0; (12)
131
⇒a=m, m
2−1±sm
2+ 12
−m
1−α(or)
a=m, m
2−1±sm
2+ 2ρ2−12
+ 4ρ2(1 −ρ2)
(13)
where ρ=s(1 −2α)
2(1 −α)is always real. It can be shown that aiare always real for132
α > 0 and distinct except when α= 0.5. Substituting L0(r) = Arain Eq. (8a),133
R0= (1 + ai)L0. Eq. (9a) is solved to get P0(r) = a4/r2. The results for the134
stresses are presented in Table 1. Depending upon the values of mand α, the135
following special cases are observed.136
−For m= 0: a1=a2= 0 & a3=−2 for any α. This case corresponds to the137
homogeneous material as indicated in Mitchell’s solution.138
−For m=2&α= 0.5 (possible only for plane strain): a1= 2; a2=a3= 0.139
Since roots are repeated, for third root, L0(r) = ln r;R0(r) = 1+ln r. However,140
the solution is continuous and αcan never be greater than 0.5. Therefore, this141
solution could be obtained by substituting α= 0.499 in the standard solution142
(Table 1) and no explicit solution for this special case is required. However, for143
the sake of completeness, solution has been obtained for α= 0.5 and presented144
in Table A.1.145
2.2.2 Terms involving rand θ146
Eliminating Rnusing Eqs. (9c) & (8b), we get the following equation:
r(nL0
n−Q0
n) + (nLn−Qn) = −(n2−1)Qn⇒d
dr [r(nLn−Qn)] = −(n2−1)Qn
147
⇒nLn=Qn+−(n2−1) RQndr +k
r(14)
where kis the integration constant. Eq. (10b) is simplified by substituting Eqs. (14)148
& (9c).149
r3Q
000
n+ (6 −2m)r2Q
00
n+η1rQ0
n+η2Qn+η3−(n2−1) RQndr +k
r= 0 (15)

6 U. Arasan et al.
Table 1: Stress components for the elasticity solution for FGM with G=G0rm
(functions of rand θfor all values of n).
σrr σrθσθθ Terms
rai0 (ai+ 1) raii=1to3
01/r20
rb1icos θ
sin θrb1isin θ
−cos θ(b1i+ 2) rb1icos θ
sin θi=1to3
mρ26=−2
1
rcos θ
sin θ∆
rsin θ
−cos θ∆
rcos θ
sin θmρ26=−2
m6= 2ρ2−4
1
r−bni cos nθ
sin nθn∆ni
r−bni sin nθ
−cos nθ(bni + 2)∆ni
r−bni cos nθ
sin nθi=1to4
bni ∈R
cos (φIln r)
r−φRcos nθ
sin nθnψ1(r)
r−φRsin nθ
−cos nθψ2(r)
r−φRcos nθ
sin nθbni ∈C
sin (φIln r)
r−φRcos nθ
sin nθnψ3(r)
r−φRsin nθ
−cos nθψ4(r)
r−φRcos nθ
sin nθi=1to4
a1=m;a2,3=m
2−1±sm
2+ 2ρ2−12
+ 4ρ2(1 −ρ2)
b11 =m−1; b12,13 =m
2−1±rm
2+ 2ρ22+ 4(1 −ρ4)
ρ=s1−2α
2(1 −α)>0; ∆=(1 −2ρ2)(m+ 1) −1
m+ 4 −2ρ2
bn1,n2=m
2−1±ql1+pl2;bn3,n4=m
2−1±ql1−pl2
l1=n2+m
2+ρ22+ 1 −ρ4;l2=n2(mρ2+ 2)2−(n2−1)m2(1 −ρ2)2
∆ni =bni + 1
bni −n2+ 2
ψ1(r) = ξ1cos(φIln r) + ξ2sin(φIln r); ψ2(r) = ξ3cos(φIln r) + ξ4sin(φIln r)
ψ3(r) = ξ1sin(φIln r)−ξ2cos(φIln r); ψ4(r) = ξ3sin(φIln r)−ξ4cos(φIln r)
φR= Re(bni); φI=|Im(bni )|
ξ1= [φ2
I+φR−n2+ 2(φR+ 1)]/ξ;ξ2= (n2−1)φI/ξ
ξ3= (φR+ 2)ξ1+φIξ2;ξ4= (φR+ 2)ξ2−φIξ1;ξ=φ2
I+φR−n2+ 22
η1= 5 −8m−2n2+β2;η2= 2n2(m−1) −2m−1 + β1+ 2β2;150
η3= 2m+ 1 −n2+β1
151
The above equation has general solution and a particular solution. The integral
term in Eqs. (14) & (15) is zero when n= 1. Therefore, when n= 1, it is a third
order differential equation in Q1. Substituting the general solution of the form
Qn(r) = Bnrbnin the Eq. (15) gives Eq. (16) for n= 1.
b3
1+ (3 −2m)b2
1+m2−6m−1 + m
1−αb1+ 3(m2−1) + m(1 −m)
1−α= 0;
152
⇒b11 =m−1; b12,13 ="m
2−1±rm
2+ 2ρ22+ 4(1 −ρ4)#(16)
It can be shown that b1i(solutions for n= 1) are real for all values of α&m
and the roots b12 &b13 would never be equal. For distinct values of roots b1i, the

Title Suppressed Due to Excessive Length 7
general and particular solutions from Eq. (15) result in (from Eqs. (8b) & (9c)),
L1(r) = Q1(r) = rb1i&R1(r) = (b1i+ 2)rb1iand L1(r) = 1/r &Q1(r) = R1(r) =
∆/r respectively, where
∆=(1 −2ρ2)(m+ 1) −1
m+ 4 −2ρ2
The stresses are presented in Table 1 for this case. Depending upon the values of153
mand α, the following special cases are observed.154
−For m= 0 (constant modulus, Mitchell’s solution), b1i=±1,−3.155
−For mρ2=−2 (⇒∆= 1) and 0 ≤α < 0.5, b11 =b12 =m−1 and b13 =156
−1. For one of the repeated roots, L1(r) = Q1(r) = rm−1ln rand R1(r) =157
rm−1[1+(m+1) ln r]. Additionally, since the stresses corresponding to b13 =−1158
and particular solution (∆= 1) are the same, the particular solution would159
be, L1(r)=(µ+ ln r)/r , Q1(r) = (ln r)/r and R1(r) = (1 + ln r)/r where160
µ= [m/(m+ 2)]2.161
−For m= 2ρ2−4, ∆becomes infinite, L1(r) = 0; Q1(r) = R1(r) = 1/r.162
These solutions could be obtained by approaching the corresponding values in the163
standard solution (Table 1) and no explicit solutions for these special cases are164
required. However, for the sake of completeness, solutions have been obtained for165
these special cases and presented in Table A.1.166
When n≥2, Eq. (15) is a fourth order equation in ¯
Qn=RQndr with four
roots since Qn=¯
Q0
nand so on. Particular solution for the Eq. (15) is obtained as
given below.
¯
Qn=ZQndr =k
n2−1⇒Qn= 0
Hence, only general solution is considered for n≥2. Substituting the general167
solution of the form Qn(r) = Bnrbnin the Eq. (15) gives Eq. (17).168
bn−m
2+ 12−l12
=l2;⇒bn=m
2−1±ql1±√l2∀n > 1 (17)
l1=m
2+ 12+n2−m
2(1 −α)=n2+m
2+ρ22+ 1 −ρ4;169
170
l2=n2(m+ 2)2−mn2(m+ 2)
1−α+m2
4(1 −α)2=n2(mρ2+ 2)2−(n2−1)m2(1 −ρ2)2
171
The term l2in bni represents a parabola in mwith curvature d2l2
dm2= 2(1 −172
4n2α+4n2α2). It can be observed that this curvature represents another parabola173
in αwith positive curvature 4n2(as n≥2) and zeros at α1,2=1
21∓r1−1
n2.174
Considering both plane strain and plane stress cases, for positive Poisson’s ratio175
(α∈[0,1
2]), only α1would be physically possible. Therefore, l2would be positive176
for the following conditions:177
m∈


(−∞, m1]||[m2,∞),if α∈[0, α1]
[m1, m2],if α∈α1,1
2(18)

8 U. Arasan et al.
where m1,2=−4n(1 −α)[n(1 −2α)±√n2−1]
1−4n2α(1 −α)and α1=1
21−r1−1
n2.178
For the above values of mand α, the corresponding value of l1is always greater179
than √l2. Therefore, bni(n > 1) would be real for the above conditions and
Fig. 1: Imaginary values of roots bni versus mfor n= 2 and α= 0.23 (for plane
stress condition with ν= 0.3).
180
complex otherwise. As an example, imaginary part of bni is shown in Fig. 1, for181
n= 2 and α= 0.23. Real values of roots bni lie between the lines m=m1&182
m=m2and complex values of roots bni lie outside.183
For real and distinct values of roots bni, using Eqs. (8b) & (9c), Ln(r) =184
rbni , Qn(r) = n∆nirbni and Rn(r) = (bni +2)∆nirbni where ∆ni = (bni +1)/(bni −185
n2+ 2). Depending upon the values of mand α, the following special cases are186
observed.187
−For m= 0 (constant modulus, Mitchell’s solution), bni =±n, (±n−2).188
−For m=m1,2,l2= 0 and bni are a pair of repeated roots, as shown in Figs. 1 &189
2a. For this case, Ln(r) = rbni ln r, Qn(r) = nΩ1(r)rbni and Rn(r) = Ω2(r)rbni
190
where Ω1(r) = −(n2−1)
(bni −n2+ 2)2+∆ni ln r&Ω2(r) = ∆ni + (bni + 2)Ω1(r).191
−For bn1(or bn3) = n2−2 (intersections of roots bn1and bn3with bni =n2−2
as shown in Fig. 2a), ∆n1(or ∆n3) becomes infinite which requires mto take
either of the following values.
m3,4=n2−3 + ρ2∓p(1 −ρ2)(5 −ρ2)−n2(1 −2ρ2)
It is observed that m3,4respect the condition m2< m3≤m4< m1. For
n= 2, m3,4= 0,2(1 + ρ2) which is always real as ρ > 0. This is indicated as

Title Suppressed Due to Excessive Length 9
(a) n= 2, α = 0.23. (b) n= 3, α = 0.23 < α3.
(c) n= 3, α =α3= 0.26. (d) n= 3, α = 0.3> α3.
Fig. 2: Real values of roots bni versus mfor different values of nand α.
points (0,1) and (2.70,1) in Fig. 2a. Additionally, to have the real values of m
for n > 2, αmust be in the range [0, α3] where
α3=1
n2+1
2"1−s1−1
n21−4
n2#.
Therefore, for n > 2, bni =n2−2 would be possible only when 0 ≤α≤192
α3. These conditions are explained for n= 3 (α3= 0.26) in Fig. 2. When193
α(= 0.23) < α3, two values of m(m3= 5.78 & m4= 6.93) are possible for194
bni =n2−2 as shown in Fig. 2b. When α=α3= 0.26, bni =n2−2 would195
occur at only one value of m(m3=m4= 6.32) as shown in Fig. 2c. When196
α(= 0.3) > α3,bni =n2−2 would never be possible as it leads to complex197
values of roots bni as shown in Fig. 2d. For this case, Ln(r) = 0, Qn(r) = rn2−2
198
and Rn(r) = nrn2−2.199

10 U. Arasan et al.
These solutions could be obtained by approaching the corresponding values in the200
standard solution (Table 1) and no explicit solutions for these special cases are201
required. However, for the sake of completeness, solutions have been obtained for202
these special cases and presented in Table A.1.203
For m > m1or m < m2, the roots bni become complex and would occur in204
two pairs as shown in Fig. 2a (only real parts are shown). For one of the two205
pairs, Ln(r) = rφRcos(φIln r), Qn(r) = nrφRψ1(r) and Rn(r) = rφRψ2(r). For206
another pair, Ln(r) = rφRsin(φIln r), Qn(r) = nrφRψ3(r) and Rn(r) = rφRψ4(r).207
The stresses corresponding to these cases are presented in Table 1 along with the208
functions ψ(r) and the constants φR&φI.209
Eqs. (8c), (9b) & (10c) are solved similarly to obtain the same roots and similar210
functions.211
2.3 Displacement Field212
The displacements ur&uθare obtained from the strain-displacement relations213
(Eq. (3)), substituting power-law, G=G0rm.214
rr =∂ur
∂r ⇒ur=Zσrr −α(σrr +σθθ )
2G0rmdr +f(θ) (19)
215
θθ =ur
r+1
r
∂uθ
∂θ ⇒uθ=Zσθθ −α(σrr +σθθ )
2G0rm−ur
rrdθ +g(r) (20)
where f(θ) and g(r) are arbitrary functions obtained during integration. They are216
evaluated by substituting Eqs. (19) & (20) into the shear strain.217
rθ =1
21
r
∂ur
∂θ +∂uθ
∂r −uθ
r=σrθ
2G0rm(21)
The process is described below for different terms in Tables 1. The corresponding218
displacements are presented together in Tables 2 & 3.219
2.3.1 Row 1 of Table 1: axisymmetric stress220
Displacements are evaluated from Eqs. (19) & (20).221
ur=1−(ai+ 2)α
2(ai−m+ 1)G0
rai−m+1 +f(θ) (22a)
222
uθ=
(1 −α)a2
i−(m−2)ai−2m+m
1−α
2(ai−m+ 1)G0
rai−m+1θ+g(r)−F(θ) (22b)
where F(θ) = Rf(θ)dθ. Note that the numerator in the above fraction is the same223
as the quadratic equation in brackets in Eq. (12). Consequently, the numerator is224
zero for a2and a3. It is non-zero only for a1=m. Hence,225
uθ=m
2G0
rθ +g(r)−F(θ) for a1=m;uθ=g(r)−F(θ) for a2, a3(23)

Title Suppressed Due to Excessive Length 11
g(r), F(θ) are obtained by substituting the displacements into Eq. (21).226
⇒F
00
(θ) + F(θ) = g(r)−rg0(r) = constant (24)
The solutions of the above equations have only rigid body displacements and hence,227
the solution is neglected whenever this equation arises.228
⇒ur=1−(ai+ 2)α
2(ai−m+ 1)Grai+1 i= 1,2,3 (25a)
229
uθ=m
2Grm+1θfor i= 1; uθ= 0 for i= 2,3 (25b)
It must be noted that when a2=m−1 or m= (1 −α)/α,ur= (m−1)/2G0.230
Corresponding displacements are given in Table 2.231
2.3.2 Row 2 of Table 1: σrr =σθθ = 0, σrθ = 1/r2
232
Displacements from Eqs. (19) & (20) are substituted in Eq. (21).
ur=f(θ); uθ=g(r)−F(θ);
rθ =r−m−2
2G0⇒F
00
(θ) + F(θ) = g(r)−rg0(r) + r−m−1
G0
= constant
General solution to the above equation represents rigid body displacements and233
only particular solution for g(r) is considered (⇒ur= 0, uθ=g(r)).234
g(r) = −r−m−1
(m+ 2)G0
for m6=−2; g(r) = rln r
G0
for m=−2 (26)
2.3.3 Row 3 of Table 1: rb1icos θterm in σrr and σθθ
235
Case 1: b1i6=m−1236
ur=1−(b1i+ 3)α
2(b1i−m+ 1)G0
rb1i−m+1 cos θ+f(θ) (27a)
237
uθ=b1i+ (b1i−m)[(1 −α)(2 + b1i)−α] + 1
2(b1i−m+ 1)G0
rb1i−m+1 sin θ+g(r)−F(θ) (27b)
Substituting shear stress and displacements in Eq. (21), only rigid body terms238
are obtained (Eq. (24)) for f(θ) and g(r), which are neglected. Whenever such239
conditions arise, f(θ) and g(r) are neglected, as in: sections 2.3.4, 2.3.5 (case 1),240
2.3.6 and 2.3.7.241

12 U. Arasan et al.
Table 2: Displacements correspond to FGM elasticity stress components with G=G0rmfor n= 0,1.
σrr 2Gur2GuθCase
rai
k1rm+1 mrm+1θi= 1
1−(1 −2ρ2)(ai+ 1)
2(ai−m+ 1)(1 −ρ2)rai+1 0i= 2,3
a26=m−1
(m−1)rm0a2=m−1
00−2/[(m+ 2)r]m6=−2
0 2 ln r/r m =−2
rb1icos θ
sin θ
Γ1i
2(1 −ρ2)rb1i+1 cos θ
sin θΓ2i
2(1 −ρ2)rb1i+1 sin θ
−cos θmρ26=−2
i= 1 to 3
rm
2(1 −ρ2)k1ln rcos θ+k2cos θ−k2θsin θ
k1ln rsin θ+k2sin θ+k2θcos θrm
2(1 −ρ2)−k1ln rsin θ+k3sin θ−k2θcos θ
k1ln rcos θ−k3cos θ−k2θsin θmρ26=−2
b11 =m−1
1
rcos θ
sin θ−2
m1 + ρ2+mρ2
m+ 4 −2ρ2cos θ
sin θ2
m1 + ρ2−m(1 −2ρ2)
m+ 4 −2ρ2 sin θ
−cos θmρ26=−2
m6= 2ρ2−4
Γ1i=1−(b1i+ 2)(1 −2ρ2)
(b1i−m+ 1) ;Γ2i= (bni + 2) −(1 −2ρ2)−Γ1i;k1= 1 −(1 −2ρ2)(m+ 1); k2=−(mρ2+ 2); k3=m+ 2ρ2

Title Suppressed Due to Excessive Length 13
Table 3: Displacements correspond to FGM elasticity stress components with G=G0rmfor n≥2.
σrr 2Gur2GuθCase
rbni cos nθ
sin nθ
χ1ni
2(1 −ρ2)rbni+1 cos nθ
sin nθχ2ni
2(1 −ρ2)nrbni+1 sin nθ
−cos nθ
κ1
2(1 −ρ2)rmcos nθ
sin nθκ2
2(1 −ρ2)nrmsin nθ
−cos nθbni =m−1
cos (φIln r)
r−φRcos nθ
sin nθω1cos(φIln r) + ω2sin(φIln r)
r−(φR+1) cos nθ
sin nθω3cos(φIln r) + ω4sin(φIln r)
nr−(φR+1) sin nθ
−cos nθbni ∈C
sin (φIln r)
r−φRcos nθ
sin nθω1sin(φIln r)−ω2cos(φIln r)
r−(φR+1) cos nθ
sin nθω3sin(φIln r)−ω4cos(φIln r)
nr−(φR+1) sin nθ
−cos nθ
χ1ni =(bni −n2+ 2) −(bni + 2)(bni + 1)(1 −2ρ2)
(bni −m+ 1)(bni −n2+ 2) ;κ1=m2n2+m+ 1+(m+ 1) n2−1+mn21−2ρ2
(n2−1) (n2−m−1)
χ2ni =(bni + 2)(bni + 1)
(bni −n2+ 2) −(1 −2ρ2)−2(1 −ρ2)χ1ni;κ2=−n2m2+ 3m+n2−1 + m1−2ρ2
(n2−1) (n2−m−1)
ω1ω= (1 −2ρ2)[ξ4φI−ξ3(φR−m+ 1)] + (φR−m+ 1); ω2ω=φI−(1 −2ρ2) [(φR−m+ 1)ξ4+ξ3φI]
ω3ω=ξ3ω−(1 −2ρ2) [ω+ξ4φI−ξ3(φR−m+ 1)] −(φR−m+ 1)
ω4ω=ξ4h(φR−m+ 1)2+φ2
Ii+ (1 −2ρ2) [ξ4(φR−m+ 1) + ξ3φI] ; ω= 2(1 −ρ2)h(φR−m+ 1)2+φ2
Ii

14 U. Arasan et al.
Case 2: b1i=m−1242
ur=[1 −(m+ 2)α] ln r
2G0
cos θ+f(θ) (28a)
243
uθ=[(m+ 2)α−1](ln r−1) + m
2G0
sin θ+g(r)−F(θ) (28b)
The displacements and shear stress (Table 1) are substituted in Eq. (21).
F
00
(θ) + F(θ)−4(1 −α) + m(1 −2α)
2G0
sin θ=g(r)−rg0(r) = constant
⇒F(θ) = −4(1 −α) + m(1 −2α)
4G0
θcos θ(without rigid body terms)
2.3.4 Row 4 of Table 1: (cos θ/r)term in σrr and σθθ
244
ur=(∆+ 1)α−1
2mG0
r−mcos θ;uθ=m∆ −(m+ 1)(1 + ∆)α+ 1
2mG0
r−msin θ(29)
2.3.5 Row 5 of Table 1: rbni cos nθ term in σrr and σθθ
245
Case 1: bni 6=m−1246
ur=n(1 −α)−(bni + 2)α∆ni
2n(bni −m+ 1)G0
rbni−m+1 cos nθ (30a)
247
uθ=(bni + 2)[(bni −m)(1 −α) + 1]∆ni −n[(bni −m)α+ 1]
2n2(bni −m+ 1)G0
rbni−m+1 sin nθ
(30b)
Case 2: bni =m−1248
Here, m=1−2α±p1−4n2α(1 −α)
2α=ρ2±p(1 −ρ2)2−n2(1 −2ρ2)
1−2ρ2where249
α∈[0, α1] for mto be real.250
The displacements corresponding to this case are given below.251
ur=f(θ); uθ=−m(1 −2α) + m2(1 −α) + α(n2−1)
2n(n2−m−1)G0
sin nθ +g(r)−F(θ) (31)
The displacements and shear stress (Table 1) are substituted in Eq. (21).
F
00
(θ) + F(θ) + m[2(n2−α) + 1] + m2(1 −α) + α(n2−1)
2n(n2−m−1)G0
sin nθ =g(r)−rg0(r)
⇒F(θ) = m[2(n2−α) + 1] + m2(1 −α) + α(n2−1)
2n(n2−1)(n2−m−1)G0
sin nθ
2.3.6 Row 6 of Table 1: rφRcos (φIln r) cos nθ term in σrr and rφRψ2(r) cos nθ252
term σθθ
253
ur=ω1cos(φIln r) + ω2sin(φIln r)
2G0
rφR−m+1 cos nθ (32a)
254
uθ=ω3cos(φIln r) + ω4sin(φIln r)
2nG0
rφR−m+1 sin nθ (32b)

Title Suppressed Due to Excessive Length 15
2.3.7 Row 7 of Table 1: rφRsin (φIln r) cos nθ term in σrr and rφRψ4(r) sin nθ255
term σθθ
256
ur=ω1sin(φIln r)−ω2cos(φIln r)
2G0
rφR−m+1 cos nθ (33a)
257
uθ=ω3sin(φIln r)−ω4cos(φIln r)
2nG0
rφR−m+1 sin nθ (33b)
Similar to the displacements corresponding to the cosine terms in σrr &σθθ ,258
the displacements corresponding to sine terms can also be derived. In Tables 2259
& 3, all these displacements are presented in a compact form. Additionally, the260
displacements are presented in Table A.2 for the special cases listed in Table A.1.261
In the following section, the above-developed solutions are applied to solve some262
of the standard two dimensional elasticity problems. Solutions to these standard263
problems for FGMs have been obtained separately by various researchers [11,264
23, 30–32] by assuming stress functions of a particular form. The current effort265
gives a general solution from which all of these solutions can be easily derived266
and hence, this general solution subsumes all the individual solutions presented267
by other researchers. Further, these solutions are restricted to only a range of268
gradient values (m) while the presented general solution is valid for any value269
of m. Comparison of results obtained from the general solution with the specific270
solutions provided by earlier researchers where available and with FEM where no271
earlier solutions are available, provides confidence in the methodology.272
3 Applications to 2D Axisymmetric Problems273
For 2D axisymmetric problems, only first row in Table 1 is relevant for stresses as274
given below (σrθ = 0, σrr and σθ θ are functions of ronly).275
σrr =A1ra1+A2ra2+A3ra3(34a)
276
σθθ =A1(1 + a1)ra1+A2(1 + a2)ra2+A3(1 + a3)ra3(34b)
The above stresses are valid only for distinct values of roots ai. Using these stresses,277
axisymmetric problems like thick cylindrical pressure vessel and curved beam with278
end moments can be solved.279
3.1 Thick-walled Cylinder Subjected to Pressure Loading280
Consider a thick-walled FGM cylinder (inner radius = ri, outer radius = ro)281
subjected to internal (Pi) and external pressure (Po), with power-law variation in G282
along the radial direction (Fig. 3). If we consider the displacements corresponding283
to the three terms as provided in Table 2, the first term (i= 1) has θ(which is284
non-periodic) in its uθdisplacement. This implies that uθat any given θand at285
θ+ 2πare different for any given radius within the domain. However, θand θ+ 2π286
represent the same location in the physical domain and hence the displacements287
should be same. The only way this can be satisfied is if the first term is identically288

16 U. Arasan et al.
zero, i.e., A1= 0.The stress boundary conditions with remaining terms are given289
as below.290
σrr (ri, θ) = A2ra2
i+A3ra3
i=−Pi;σrr (ro, θ) = A2ra2
o+A3ra3
o=−Po(35)
A2and A3are obtained by solving the above equations.291
A2=ra2
oPi−ra2
iPo
ra2
ira1
o−ra1
ira2
o
;A3=−ra1
oPi−ra1
iPo
ra2
ira1
o−ra1
ira2
o
(36)
For the values of ν= 0.3, r0/ri= 2 and m={−2,−0.5,0.5,2}, the variations292
of normalized tangential stress with respect to normalized radius are shown in293
Fig. 3 for plane strain condition. The above solution is compared with the existing294
analytical solution [23] and they are found to be identical. The above solution as
Fig. 3: Tangential stress variations for thick-walled FGM cylinder with external
pressure (Po) and without internal pressure (Pi= 0).
295
well as solution presented by Sburlati [23] are not valid for m= 2 & ρ= 0 (or296
ν= 0.5) as it leads to repeated roots (a2=a3= 0). For this special condition,297
the corresponding stresses are taken from Table A.1 as follows.298
σrr =A2+A3ln r;σθθ =A2+A3(1 + ln r) (37)
299
σrr (ri, θ) = A2+A3ln ri=−Pi;σrr (ro, θ ) = A2+A3ln ro=−Po(38)
300
⇒A2=Poln ri−Piln ro
ln(ro/ri);A3=Pi−Po
ln(ro/ri)(39)

Title Suppressed Due to Excessive Length 17
Fig. 4: Tangential stress variations of FGM curved beam subjected to end mo-
ments.
The variation of normalized tangential stress with respect to normalized radius301
for the above case (ν= 0.5) is plotted in Fig. 3 and compared with FEM solution302
(as there is no existing analytical solution in the literature for this case) with303
ν= 0.499. This special case need not be explicitly considered if a close enough304
solution can be obtained rather than an exact solution. Such a closer solution305
can be simply obtained by considering ν= 0.499 (close to 0.5) in Eq. (36) as306
well as in Sburlati [23] solution. Fig. 3 shows that the results for the special case307
(Eq. (39)) matches well with the FEM solution as well as the standard solution308
taking ν= 0.499. This shows that the special case (m= 2 & ρ= 0) need not be309
considered explicitly.310
3.2 Curved Beam with End Moments311
Consider a curved FGM beam (inner radius = ri, outer radius = ro) subjected to312
pure moment M, with power-law variation in Galong the radial direction (Fig. 4).313
The stress boundary conditions are given as follows:314
σrr (ri) = σrr (ro) = 0; Zro
ri
σθθ rdr =−M(40)
Substituting for stresses from Eq. (34) in Eq. (40), the constants A1, A2and A3
315
are obtained.316


A1
A2
A3
=M
N(2 + a1)(2 + a2)(2 + a3)

ra2
ira3
o−ra3
ira2
o
ra3
ira1
o−ra1
ira3
o
ra1
ira2
o−ra2
ira1
o
(41a)

18 U. Arasan et al.
N= (r2+a2+a3
ira1
o+ra1
ir2+a2+a3
o)(2 + a1)(a2−a3)
+(r2+a1+a3
ira2
o+ra2
ir2+a1+a3
o)(2 + a2)(a3−a1)
317
+(r2+a1+a2
ira3
o+ra3
ir2+a1+a2
o)(2 + a3)(a1−a2) (41b)
For the values of ν= 0.3, r0/ri= 4 and m={−1,−0.5,0.01,0.5,1}, the variations318
of normalized tangential stress with respect to normalized radius are shown in319
Fig. 4 for plane stress condition. From the plot, it can be observed that the solution320
approaches that of the homogeneous material [8] as mapproaches zero. Also, the321
above solution is compared with the existing analytical solution [31] and they are322
found to be identical. For plane stress condition, the two solutions can be compared323
by the following relations between the constants Ai(Eq. (41)) and Ci(defined by324
Arslan and Eraslan [31]): A1=C1
rm
om(1+ν), A2,3=C3,4
rm
o, except when m=−2. For325
m=−2, the denominator N= 0 in Eq. (41) and the constants are evaluated326
separately:327


A1
A2
A3
=Mγ
2(rγ
o−rγ
i) + γ(rγ
i+rγ
o) ln(ri/ro)

−(rγ
i+rγ
o)
1
(riro)γ
(42)
where γ=p2(1 + ν). Again, the results for this special case (Eq. (42)) matches328
well with the standard solution (Eq. 41) taking m=−1.99. This shows that the329
special case need not be considered explicitly.330
4 Curved Beam with End Shear331
Consider a curved FGM beam (inner radius = ri, outer radius = ro) subjected to332
shear load P, with power-law variation in Galong the radial direction (Fig. 5).333
The stress boundary conditions are given as follows:334
σrr (ri, θ) = σrr (ro, θ ) = 0; σr θ (ri, θ) = σrθ (ro, θ) = 0 (43a)
335
Zro
ri
σθθ r, θdr =−Psin θ;Zro
ri
σrθ (r, θ)dr =Pcos θ(43b)
From the Table 1, the following stress terms are taken so that σrθ has cos θterms336
and σθθ has sin θterms (for m6=−2/ρ2⇒∆6= 1):337
σrr =B1rb11 +B2rb12 +B3rb13 +B4
rsin θ(44a)
338
σrθ =−B1rb11 +B2rb12 +B3rb13 +B4∆
rcos θ(44b)
339
σθθ =B1(2 + b11)rb11 +B2(2 + b12 )rb12 +B3(2 + b13 )rb13 +B4∆
rsin θ(44c)

Title Suppressed Due to Excessive Length 19
Fig. 5: Tangential stress variations of FGM curved beam subjected to end shear.
By solving Eq. (43), the constants B1, B2&B3are obtained and B4= 0.340


B1
B2
B3
=P
S(1 + b11)(1 + b12 )(1 + b13)

rb13
irb12
o−rb12
irb13
o
rb11
irb13
o−rb13
irb11
o
rb12
irb11
o−rb11
irb12
o

(45a)
S= (r1+b12+b13
irb11
o+rb11
ir1+b12+b13
o)(1 + b11)(b12 −b13 )
+(r1+b11+b13
irb12
o+rb12
ir1+b11+b13
o)(1 + b12)(b13 −b11 )
341
+(r1+b11+b12
irb13
o+rb13
ir1+b11+b12
o)(1 + b13)(b11 −b12 ) (45b)
For the values of ν= 0.3, r0/ri= 4 and m={−1,−0.5,0.01,0.5,1}, the varia-342
tions of normalized tangential stress with respect to normalized radius are shown343
in Fig. 5 for plane stress condition. From the plot, it can be observed that the344
solution approaches that of the homogeneous material [8] as mapproaches zero.345
The above solution is compared with the existing analytical solution [32] and they346
are found to be identical.347
The above solutions are valid except for m=−2/ρ2as it leads to same value for348
b11 &b12 (corresponding stress solutions (m=−2/ρ2) are given in Table A.1) and349
the stresses obtained for this case are verified with Zhang et al. [32]. Nevertheless,350
a near solution can also be achieved by letting mvalue close to −2/ρ2(m=−6.01351
for ν= 0.3) in Eq. (44) as shown in the Fig. 6. Consequently, this special case352
need not be considered explicitly.353

20 U. Arasan et al.
Fig. 6: Tangential stress variations of FGM curved beam subjected to end shear:
comparison for the case m=−2/ρ2.
5 Infinite Plate with FGM Ring Around the Hole354
Consider an infinite plate with FGM ring around the hole subjected to a uniaxial355
tensile load at its outer boundary as shown in Fig. 7. Let riand robe the inner356
and outer radii of the ring respectively. The power-law variation of shear modulus357
(G) is considered from the inner radius rito the outer radius ro(G=G0rm, ri≤358
r≤ro). Beyond the outer radius, the material is considered to be homogeneous359
and the properties are same as that of the material at the outer layer of the FGM360
ring (G0rm
o=Gplate). This is similar to the assumption made by Sburlati et al.361
[11]. Since the problem is symmetric, all the symmetric terms are taken from362
Table 1 for real values as well as complex values of roots bni. Their corresponding363
displacements are taken from Tables 2 and 3 respectively. Following are the stress364
and displacement boundary and interface conditions.365
σrr (ri, θ) = 0; σrθ(ri, θ) = 0,(46a)
366
σrr (∞, θ) = σ0
2(1 + cos 2θ); σrθ(∞, θ ) = −σ0
2sin 2θ, (46b)
367
σrr (r−
o, θ) = σrr (r+
o, θ); σrθ (r−
o, θ) = σrθ (r+
o, θ),(46c)
368
ur(r−
o, θ) = ur(r+
o, θ); uθ(r−
o, θ) = uθ(r+
o, θ) (46d)
Since the constant coefficients of the stresses found using the above conditions are369
not in a simple form, the solutions are given for specific cases (plane stress) as370
shown in Table 4. It must be noted that for plane stress condition, the value of m371
has to be between -2.19 and 9.39 (Eq. (18)) for ν= 0.3 to have the roots (bni) to372

Title Suppressed Due to Excessive Length 21
Fig. 7: Variation of tangential stress at the surface of the hole. For complex values
of roots bni (m=−3 & −5), analytical solutions are compared with FEM solutions.
be real (Fig. 1). For the same problem, the limits for mhas been indicated to be373
from 0 to 8(2−√3)
ν−7+4√3by Sburlati [30]. The upper limits obtained from Eq. (18) and374
by Sburlati [30] are same. However, the lower limit can be extended up to -2.19.
Table 4: Normalized tangential stress at inner radius for various values of m(in-
finite plate with FGM ring around the hole for ν= 0.3 and ro/ri= 2).
m σθθ /σ0
-5 5.6228–13.2888 cos 2θ
bni ∈C-3 3.3271–7.0065 cos 2θ
-2.191 2.5345 −5.1801 cos 2θ
-2.19 2.5335 −5.1781 cos 2θ
bni ∈R
-1 1.5825 −3.1637 cos 2θ
0.0001 0.9999 −1.9999 cos 2θ
1 0.5998 −1.2165 cos 2θ
375
Variations of the normalized tangential stress along the inner boundary of the376
hole (at r=ri) are plotted in Fig. 7. From this plot, it can be inferred that377
the solution approaches that of the homogeneous as gradient factor tends to zero.378
The solution is also verified with the existing analytical solution [11, 30] for real379
roots when n= 2. Further, for the complex roots, a value of m=−2.191 is380
considered to determine the stresses. The result obtained is compared with the381
stresses obtained for m=−2.19 (lower limit for real roots). The variations of the382
normalized tangential stress for both of these cases are reported in Table 4 and383

22 U. Arasan et al.
observed to be in close. Additionally, for m=−3 and −5 (bni becomes complex),384
the normalized tangential stresses obtained using analytical and FEM solutions385
are compared as shown in Fig. 7. It is observed that the FEM solution is in good386
agreement with the analytical solution. The solution when roots become complex387
have not been discussed in the literature.388
6 Conclusions389
The complete general solution for stresses and displacements for two-dimensional390
elastic problems involving functionally graded material (FGM) with power-law391
variation along the radial direction has been derived, similar to Michell’s solution392
for homogeneous isotropic materials. These functions have been obtained by as-393
suming Fourier series for stresses and using both equilibrium and compatibility394
equations. A rigorous analysis has been done to determine the limits within which395
each of the solutions are valid and special cases have been solved explicitly.396
To verify the approach, the following standard problems have been solved.397
–Thick-walled FGM cylinder subjected to internal and external pressure398
–Curved FGM beam subjected to end moments399
–Curved FGM beam subjected to end shear400
–Infinite plate with FGM ring around a hole, subjected to a uniaxial tensile load401
The solutions obtained for the above problems are compared with analytical so-402
lutions where they exist and with finite element analysis for the cases where no403
analytical solutions exist. The solutions are also compared with that of the homo-404
geneous material by letting the gradient factor closer to zero. Each of the existing405
analytical solutions is for a standard problem with specific geometry and loading406
conditions and are further valid only for certain ranges of gradient factor (m). The407
general solution provided in this paper is applicable to problems with arbitrary408
geometry and boundary/loading conditions and for any value of m.409

Title Suppressed Due to Excessive Length 23
A Stresses and displacements for special cases410
Table A.1: Stress components of special cases for the elasticity solution for FGM
with G=G0rm(functions of rand θfor all values of n).
σrr σrθσθθ Terms
ln r0 1 + ln rm= 2 & ρ= 0
⇒a2,3= 0
01
rsin θ
−cos θ1
rcos θ
sin θmρ26=−2
m= 2ρ2−4
1
rcos θ
sin θ1
rsin θ
−cos θ1
rcos θ
sin θmρ2=−2
µ+ ln r
rcos θ
sin θln r
rsin θ
−cos θ1 + ln r
rcos θ
sin θ⇒
1
r1−mcos θ
sin θ1
r1−msin θ
−cos θm+ 1
r1−mcos θ
sin θrepeated
roots
ln r
r1−mcos θ
sin θln r
r1−msin θ
−cos θ1+(m+ 1) ln r
r1−mcos θ
sin θ−1, m −1
If l2= 0, bn3=bn1and bn4=bn2, the solution corresponding to bn1&bn2
are from the 5th row of Table 1. Other two terms are as given below
ln r
r−bni cos nθ
sin nθnΩ1(r)
r−bni sin nθ
−cos nθΩ2(r)
r−bni cos nθ
sin nθi= 3,4
0rn2−2sin nθ
−cos nθnrn2−2cos nθ
sin nθbn1(or bn3) =
n2−2
µ=m2
(m+ 2)2
Ω1(r) = −(n2−1)
(bni −n2+ 2)2+∆ni ln r;Ω2(r) = ∆ni + (bni + 2)Ω1(r)

24 U. Arasan et al.
Table A.2: Displacements correspond to FGM elasticity stress components of special cases with G=G0rmfor all n.
σrr 2Gur2GuθCase
ln rrm−1/20m= 2 & ρ= 0
0m+ 3
m(m+ 2) cos θ
sin θ−2m+ 3
m(m+ 2) sin θ
−cos θmρ26=−2
m= 2ρ2−4
1
rcos θ
sin θ2
m(m+ 2) cos θ
sin θ−2
mm+ 1
m+ 2  sin θ
−cos θmρ2=−2
µ+ ln r
rcos θ
sin θ µ1+2 ln r
m(m+ 2) cos θ
sin θ−µ2+2(m+ 1) ln r
m(m+ 2)  sin θ
−cos θ⇒
1
r1−mcos θ
sin θ−(m+ 2) ln r
2rmcos θ
sin θm−2+(m+ 2) ln r
2rmsin θ
−cos θrepeated roots
ln r
r1−mcos θ
sin θrm
4m+ 8 Λ1(r) cos θ−m2θsin θ
Λ1(r) sin θ+m2θcos θrm
4m+ 8 Λ2(r) sin θ−m2θcos θ
−Λ2(r) cos θ−m2θsin θ−1, m −1
rbni ln rcos nθ
sin nθ λ1+λ2ln r
2(1 −ρ2)rbni+1 cos nθ
sin nθ λ3+λ4ln r
2(1 −ρ2)nrbni+1 sin nθ
−cos nθl2= 0
0
−n(1 −2ρ2)
2(1 −ρ2)(n2−m−1)r1−n2cos nθ
sin nθ−(n2−m−1) + (1 −2ρ2)
2(1 −ρ2)(n2−m−1)r1−n2sin nθ
−cos nθbn1(or bn3)
=n2−2
m46=n2−1
−3n
n2−1rmcos nθ
sin nθn2+ 2
n2−1rmsin nθ
−cos nθbn1(or bn3)
=n2−2
m4=n2−1
µ1=8(2m+ 3)
m2(m+ 2)3+4
m(m+ 2) ;µ2=4(m+ 1) −m2
m(m+ 2)3+µ1;Λ1(r) = m2/2−2(m+ 4) ln r−[(m+ 2) ln r]2
Λ2(r) = (m+ 2) [m+ 2 (m−2) ln r]−Λ1(r)
λ1=(1 −2ρ2)
(bni −m+ 1)(bni −n2+ 2) m(bni + 2)
bni −m+ 1 +n2(bni + 1)
bni −n2+ 2 −1
(bni −m+ 1)2
λ2=1
(bni −m+ 1) −(1 −2ρ2)(bni + 1)(bni + 2)
(bni −m+ 1)(bni −n2+ 2)
λ3=(2bni + 3)
(bni −n2+ 2) −(bni + 1)(bni + 2)
(bni −n2+ 2)2−2(1 −ρ2)λ1;λ4=(bni + 1)(bni + 2)
(bni −n2+ 2) −(1 −2ρ2)−2(1 −ρ2)λ2

Title Suppressed Due to Excessive Length 25
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