nonlinear bending response of functionally graded plates subjected to tránverse loads and in...

7/29/2019 Nonlinear bending response of functionally graded plates subjected to trnverse loads and in thermal environments

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562 H.-S. Shen / International Journal of Mechanical Sciences 44 (2002) 561 584

Many studies for thermal stress and linear thermal bending of FGM plates are available in the

literature, see, for example, Refs. [311]. However, investigations in nonlinear analysis of FGM

plates under thermal or mechanical loading are limited in number. Among those, Mizuguchi and

Ohnabe [12] employed the Poincare method to examine the large deection of simply supported,heated FGM thin plates with Youngs modulus varying symmetrically to the middle plane in thick-

ness direction. Based on the rst-order shear deformation plate theory, Praveen and Reddy [13]

analyzed nonlinear static and dynamic response of functionally graded ceramicmetal plates sub-

jected to transverse loads and temperature distribution by using nite element method. Reddy [14]

developed theoretical formulations for thick FGM plates according to the higher-order shear de-

formation plate theory. In Ref. [14], both Navier solutions for linear bending of simply supported

rectangular FGM plates and nite element models for nonlinear static and dynamic response were

presented.

Geometrically nonlinear bending solutions for homogeneous isotropic and composite laminated

plates subjected to mechanical and=or thermal loading and resting on an elastic foundation weregiven by Shen [1518]. In these studies the material properties were considered to be independent of

temperature. Recently, Yang and Shen [19] gave a large deection analysis of thin FGM rectangular

plates subjected to combined transverse and in-plane loads, but the numerical results were only for

a simple case of an FGM plate in a xed thermal environment.

The present paper extends the previous works to the case of FGM plates with two constituent

materials subjected to a transverse uniform or sinusoidal load and in thermal environments. The

material properties are assumed to be temperature-dependent, and graded in thickness direction ac-

cording to a volume fraction power-law distribution. The governing equations of the plate are based

on Reddys higher-order shear deformation plate theory that includes thermal eects [2023]. A

mixed Galerkin-perturbation technique is employed to determine the load-deection and load-bending

moment curves. The numerical illustrations show the nonlinear bending response of FGM plates un-der dierent sets of environmental conditions.

2. Theoretical development

Here we consider an FGM plate of length a, width b and thickness h, which is made from a

mixture of ceramics and metals. We assume that the composition is varied from the top to the

bottom surface, i.e. the top surface (Z = h=2) of the plate is ceramic-rich whereas the bottom

surface (Z = h=2) is metal-rich. In such a way, the eective material properties P, like Youngsmodulus E or thermal expansion coecient , can be expressed as

P=PtVc +PbVm (1)

in which Pt and Pb denote the temperature-dependent properties of the top and bottom surfaces

of the plate, respectively, and Vc and Vm are the ceramic and metal volume fractions and are

related by

Vc + Vm = 1 (2)

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H.-S. Shen / International Journal of Mechanical Sciences 44 (2002) 561 584 563

and we assume the volume fraction Vm follows a simple power law as

Vm =2Z + h2h

N

; (3)

where volume fraction index N dictates the material variation prole through the plate thickness and

may be varied to obtain the optimum distribution of component materials. It is noted that similar

denition can be found in Refs. [13,14], but is for Vc.

From Eqs. (1)(3), the eective Youngs modulus E and thermal expansion coecient of an

FGM plate can be written as

E= (Eb Et)

2Z + h

2h

N+Et; = (b t)

2Z + h

2h

N+ t: (4)

It is evident that when Z = h=2; E=Et and = t and when Z = h=2; E=Eb and = b. It is

assumed that Et; Eb; t and b are functions of temperature, but Poissons ratio depends weaklyon temperature change and is assumed to be a constant, as shown in Section 4, so that E and are

functions of temperature and position.

The plate is subjected to a transverse uniform load q = q0 or a sinusoidal load q = q0 sin(X=a)

sin(Y=b) combined with thermal loads. As usual, the reference coordinate system has its origin at

the corner of the plate on the middle plane. Let U ; V and W be the plate displacements parallel to

a right-hand set of axes (X; Y; Z), where X is longitudinal and Z is perpendicular to the plate. xand y are the mid-plane rotations of the normals about the Y and X axes, respectively. Let F(X; Y)

be the stress function for the stress resultants dened by Nx = F; yy; Ny = F; xx and Nxy = F; xy,where a comma denotes partial dierentiation with respect to the corresponding coordinates.

Based on Reddys higher-order shear deformation plate theory (HSDPT), including the thermal

eects, the governing dierential equations for an FGM plate undergoing moderately large rotations

in the von Karman sense can be derived as

L11( W) L12( x) L13( y) + L14( F) L15( NT

) L16( MT

) = L( W ; F) + q; (5)

L21( F) + L22( x) + L23( y) L24( W) + L25( NT

) = 12

L( W ; W); (6)

L31( W) + L32( x) L33( y) + L34( F) L35( NT

) L36( ST

) = 0; (7)

L41( W) L42( x) + L43( y) + L44( F) L45( NT

) L46( ST

) = 0; (8)

where linear operators Lij( ) and nonlinear operator L( ) are dened as in Appendix A.

Note that these plate equations show thermal coupling as well as the interaction of stretching and

bending.

The forces, moments and higher-order moments caused by elevated temperature are dened by

NT

xM

T

xP

T

x

NT

yM

T

yP

T

y

NT

xyM

T

xyP

T

xy

=

h=2h=2

(1; Z ; Z 3)

Ax

Ay

Axy

TdZ (9a)

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and

S

T

x

STy

ST

xy

=

MT

x

MTy

MT

xy

43h2

PT

x

PTy

PT

xy

; (9b)

where T is temperature rise from some reference temperature at which there are no thermal strains,

and

Ax

Ay

Axy

=

Q11 Q12 Q16

Q12 Q22 Q26

Q16 Q26 Q66

1 0

0 1

0 0

; (10)

where the thermal expansion coecient is given in detail in Eq. (4), and

Q11 = Q22 =E

1 2; Q12 =

E

1 2; Q16 = Q26 = 0; Q44 = Q55 = Q66 =

E

2(1 + )(11)

in which E is also given in detail in Eq. (4), and vary through the plate thickness.

All the edges are assumed to be simply supported. Depending upon the in-plane behavior at the

edges, two cases, case 1 (referred to herein as movable edges) and case 2 (referred to herein as

immovable edges), will be considered.

Case 1: The edges are simply supported and freely movable in both the X- and Y-directions,

respectively.Case 2: All four edges are simply supported with no in-plane displacements, i.e. prevented from

moving in the X- and Y-directions.

For these two cases the associated boundary conditions are

X = 0; a:

W = y = 0; (12a)

Nxy = 0; (12b)

b

0

Nx dY = 0 (movable edges); (12c)

U = 0 (immovable edges); (12d)

Y = 0; b:

W = x = 0; (12e)

Nxy = 0; (12f)

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a0

Ny dX = 0 (movable edges); (12g)

V = 0 (immovable edges): (12h)

It is noted that Eqs. (5) (8) are identical to those of unsymmetric cross-ply laminated plates

under thermomechanical loading. The presence of stretching-bending coupling gives rise to bending

curvatures under the action of in-plane loading, no matter how small these loads may be. In this

situation the boundary condition of zero bending moment cannot be incorporated accurately. Because

of immovable edges are considered in the present analysis, Mx = 0 (at X = 0; a) and My = 0 (at

Y = 0; b) are not included in Eq. (12), as previously shown in Refs. [24,25].

The unit end-shortening relationships are

x

a= 1

ab

b

0

a

0

9 U9X

dX dY

= 1

ab

b0

a0

A11

92 F

9Y2+A12

92 F

9X2+

B11

4

3h2E11

9 x

9X+

B12

4

3h2E12

9 y

9Y

4

3h2

E11

92 W

9X2+E12

92 W

9Y2

1

2

9 W

9X

2 (A11 N

T

x +A

12N

T

y)

dX dY (13a)

y

b= 1

ab

a

0

b

0

9 V9Y

dY dX

= 1

ab

a0

b0

A22

92 F

9X2+A12

92 F

9Y2+

B21

4

3h2E21

9 x

9X

+

B22

4

3h2E22

9 y

9Y

4

3h2

E21

92 W

9X2+E22

92 W

9Y2

129

W9Y

2 (A12 NTx +A22 NTy)

dY dX; (13b)

where x and y are plate end-shortening displacements in the X- and Y-directions.

In Eqs. (13) and (16) below, the reduced stiness matrices [Aij]; [B

ij]; [D

ij]; [E

ij]; [F

ij] and

[Hij ] (i; j = 1; 2; 6) are functions of temperature and position, dened by

A = A1; B = A1B; D = D BA1B; E = A1E;

F = F EA1B; H = H EA1E; (14)

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where Aij ; Bij, etc., are the plate stinesses, dened by

(Aij; Bij; Dij ; Eij ; Fij ; Hij) = h=2

h=2

(Qij)(1; Z ; Z 2; Z3; Z4; Z6) dZ (i; j = 1; 2; 6); (15a)

(Aij; Dij; Fij) =

h=2h=2

(Qij)(1; Z2; Z4) dZ (i; j = 4; 5): (15b)

3. Analytical method and asymptotic solutions

Having developed the theory, we will try to solve Eqs. (5)(8) with boundary condition (12).

Before proceeding, it is convenient to rst dene the following dimensionless quantities for such

plates [with ijk in Eqs. (22) and (24) below are dened as in Appendix B]:

x = X=a; y = Y=b; = a=b; W = W =[D

11D

22A

11A

22]1=4;

F = F=[D11D

22]1=2; (x; y) = ( x; y)a=[D

11D

22A

11A

22]1=4;

14 = [D

22=D

11]1=2; 24 = [A

11=A

22]1=2; 5 = A

12=A

22;

(T1; T2) = (AT

x ; ATy)a

2=2[D11D

22]1=2;

(T3; T4; T6; T7) = (DT

x ; DTy ; F

Tx ; F

Ty )a

2=2h2D11;

(31; 41) = (a2=2)[A55 8D55=h

2 + 16F55=h4; A44 8D44=h

2 + 16F44=h4]=D11;

(Mx; My; Px; Py; M

T

x ; M

T

y ; P

T

x ; P

T

y )

= ( Mx; My; 4 Px=3h2; 4 Py=3h

2; MT

x ; MT

y ; 4 PT

x =3h2; 4 P

T

y=3h2)a2=2D11[D

11D

22A

11A

22]1=4;

q = q0a4=4D11[D

11D

22A

11A

22]1=4;

(x; y) = (x=a;y=b)b2=42[D11D

22A

11A

22]1=2 (16)

in which ATx (=ATy); D

Tx (=D

Ty ) and F

Tx (=F

Ty ) are dened by

ATx DT

x FT

x

ATy DTy F

Ty

=

h=2h=2

(1; Z ; Z 3)

Ax

Ay

dZ (17)

and the details of which can be found in Appendix C.The nonlinear governing Eqs. (5)(8) can then be written in dimensionless form as

L11(W) L12(x) L13(y) + 14L14(F) L16(MT) = 14

2L(W; F) + q; (18)

L21(F) + 24L22(x) + 24L23(y) 24L24(W) = 12

242L(W; W); (19)

L31(W) + L32(x) L33(y) + 14L34(F) L36(ST) = 0; (20)

L41(W) L42(x) + L43(y) + 14L44(F) L46(ST) = 0; (21)

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where

L11( ) = 11094

9x4+ 2112

2 94

9x29y2+ 114

4 94

9y4;

L12( ) = 1209

3

9x3+ 122

2 93

9x9y2;

L13( ) = 13193

9x29y+ 133

3 93

9y3;

L14( ) = 14094

9x4+ 142

2 94

9x29y2+ 144

4 94

9y4;

L16(MT) =

92

9x2(MTx ) + 2

92

9x9y

(MTxy) + 2 9

2

9y2(MTy );

L21( ) =94

9x4+ 2212

2 94

9x29y2+ 214

4 94

9y4;

L22( ) = 22093

9x3+ 222

2 93

9x9y2;

L23( ) = 23193

9x29y+ 233

3 93

9y3;

L24( ) = 2409

4

9x4 + 2422 9

4

9x29y2 + 2444 9

4

9y4 ;

L31( ) = 319

9x+ 310

93

9x3+ 312

2 93

9x9y2;

L32( ) = 31 32092

9x2 322

2 92

9y2;

L33( ) = 33192

9x9y;

L34( ) = L22( );

L36(ST) =

9

9x(STx ) +

9

9y(STxy);

L41( ) = 419

9y+ 411

93

9x29y+ 413

3 93

9y3;

L42( ) = L33( );

L43( ) = 41 43092

9x2 432

2 92

9y2;

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L44( ) = L23( );

L46(ST) =

9

9x

(STxy) + 9

9y

(STy );

L( ) =92

9x292

9y2 2

92

9x9y

92

9x9y+

92

9y292

9x2: (22)

The boundary conditions of Eq. (12) become

x = 0; :

W = y = 0; (23a)

F;xy =0; (23b)

0

292F

9y2dy = 0 (movable edges); (23c)

x = 0 (immovable edges); (23d)

y = 0; :

W = x = 0; (23e)

F;xy =0; (23f)

0

92

F9x2

dx = 0 (movable edges); (23g)

y = 0 (immovable edges) (23h)

and the unit end-shortening relationships become

x = 1

42224

0

0

224

2 92F

9y2 5

92F

9x2+ 24

511

9x

9x+ 233

9y

9y

24 611 92W

9x2+ 244

2 92W

9y2 1

224 9W

9x

2

+ (224T1 5T2) T dx dy;(24a)

y = 1

42224

0

0

92F

9x2 5

2 92F

9y2+ 24

220

9x

9x+ 522

9y

9y

24

240

92W

9x2+ 622

2 92W

9y2

1

224

2

9W

9y

2+ (T2 5T1) T

dy dx:

(24b)

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Applying Eqs. (18)(24), the nonlinear bending response of a simply supported FGM plate sub-

jected to a transverse uniform or sinusoidal load and in thermal environments is now determined by

a mixed Galerkin-perturbation technique. The essence of this procedure, in the present case, is to

assume that

W(x; y; ) =j=1

jwj(x; y); F(x; y; ) =j=0

jfj(x; y);

x(x; y; ) =j=1

jxj(x; y); y(x; y; ) =j=1

jyj(x; y); q =j=1

jj ;(25)

where is a small perturbation parameter and the rst term of wj(x; y) is assumed to have the form

w1(x; y) = A(1)11 sin mx sin ny: (26)

Then we expand the thermal-bending moments in the double Fourier sine series asMTx S

Tx

MTy STy

=

M(1)x S

(1)x

M(1)y S(1)y

i=1;3;:::

j=1;3;:::

1

ijsin ix sinjy; (27)

where M(1)

x ; M(1)y ; S

(1)x and S

(1)y , and all coecients in Eqs. (35)(37) below are given in detail in

Appendix D.

Substituting Eq. (25) into Eqs. (18)(21), collecting the terms of the same order of , gives a set

of perturbation equations. By using Eqs. (26) and (27) to solve these perturbation equations of each

order, the amplitudes of the terms wj(x; y); fj(x; y); xj(x; y) and yj(x; y) are determined step by

step, and j can be determined by the Galerkin procedure. As a result, up to third-order asymptotic

solutions can be obtained as

W = [A(1)11 sin mx sin ny] +

3[A(3)13 sin mx sin3ny +A

(3)31 sin 3mx sin ny] + O(

4); (28)

x = [C(1)11 cos mx sin ny] +

2[C(2)20 sin2mx] +

3[C(3)13 cos mx sin 3ny

+ C(3)31 cos3mx sin ny] + O(

4); (29)

y = [D(1)11 sin mx cos ny] +

2[D(2)02 sin 2ny] +

3[D(3)13 sin mx cos3ny

+D(3)31 sin 3mx cos ny] + O(

4); (30)

F= B(0)

00

y2

2 b(0)

00

x2

2 + [B(1)

11 sin mx sin ny]

+ 2

B(2)00y2

2 b(2)00

x2

2+B

(2)20 cos2mx +B

(2)02 cos2ny

+ 3[B(3)13 sin mx sin 3ny +B

(3)31 sin 3mx sin ny] + O(

4): (31)

Note that for boundary condition case 1, it is just necessary to take B(i)00 = b

(i)00 = 0 (i = 0; 2) in Eq.

(31), so that the asymptotic solutions have a similar form, and

q = 1 + 22 +

33 + O(4): (32)

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All coecients in Eqs. (28)(31) are related and can be written as functions of A(1)11 , so that Eqs.

(28) and (32) can be rewritten as

W = W(1)

(x; y)(A(1)11 ) + W

(3)

(x; y)(A(1)11 )

3

+ (33)

and

q = (1)q (A

(1)11 ) +

(2)q (A

(1)11 )

2 + (3)q (A(1)11 )

3 + : (34)

From Eqs. (33) and (34) the load-central deection relationship can be written as

q0a4

D11h=A

(0)W +A

(1)W

W

h

+A

(2)W

W

h

2+A

(3)W

W

h

3+ : (35)

Similarly, the bending moment-central deection relationships can be written as

Mxa2

D11h=A

(0)MX +A

(1)MX

Wh

+A

(2)MX

Wh

2+A

(3)MX

Wh

3+ ; (36)

Mya2

D11h=A

(0)MY +A

(1)MY

W

h

+A

(2)MY

W

h

2+A

(3)MY

W

h

3+ : (37)

Eqs. (35)(37) can be employed to obtain numerical results for the load-deection and load-bending

moment curves of an FGM plate subjected to a transverse uniform or sinusoidal load and in thermal

environments.

4. Numerical results and discussions

To study the thermal eect on the nonlinear bending behavior of FGM plates subjected to a

transverse uniform or sinusoidal load, many examples were solved numerically. In the numerical

analysis, asymptotic solutions up to third order, as given by Eqs. (35)(37) with deected modes

(m; n)=(1; 1), were used. Zirconia and titanium alloy were selected for the two constituent materials

of the plate in the present examples, referred to as ZrO2=Ti6Al 4V. However, the analysis is

equally applicable to other types of FGMs as well. The material properties P, such as Youngs

modulus E and thermal expansion coecient , can be expressed as a function of temperature, see

Ref. [26], as

P=P0(P1T1 + 1 +P1T +P2T2 +P3T3) (38)

in which T=T0 +T and T0 =300 K (room temperature), P0; P1; P1; P2 and P3 are the coecients

of temperature T(K) and are unique to the constituent materials. Typical values for Youngs modulus

E (Pa) and thermal expansion coecient (K) of Zirconia and Ti6Al4V (from Ref. [9]) are listed

in Table 1. Poissons ratio is assumed to be a constant, and = 0:28.

The accuracy and eectiveness of the present method for the nonlinear bending analysis of thin

and moderately thick, isotropic plates with movable or immovable edges were examined by many

comparison studies given in Shen [1518]. In addition, the load-center deection curves for a zir-

conia/aluminum square plate with dierent values of the volume fraction index n, dened as in

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Table 1

Temperature-dependent coecients for ceramics and metals, from Ref. [9]

Materials P0 P

1 P1 P2 P3

Zirconia E 244:27e + 9 0 1:371e 3 1:214e 6 3:681e 10

12:766e 6 0 1:491e 3 1:006e 5 6:778e 11

Ti6Al4V E 122:56e + 9 0 4:586e 4 0 0

7:5788e 6 0 6:638e 4 3:147e 6 0

0 10 150.0

0.2

0.4

0.6

Present

Praveen & Reddy [13]

1: aluminum2: n = 2.0

3: n = 1.0

4: n = 0.5

5: zirconia5

4

3

2

1zirconia/aluminum

= 1.0, b/h = 20

uniform load

q0b4/E0h

4

W/h

5

Fig. 1. Comparisons of load-central deection curves for a zirconia/aluminum square plate.

Refs. [13,14], subjected to a uniform transverse load are compared in Fig. 1 with numerical results

of Praveen and Reddy [13], using their material properties, i.e. for aluminum E= 70 GPa; = 0:3;

= 23:0 106=

C, and for zirconia E = 151 GPa; = 0:3, and = 10:0 106=

C. Note that in

Fig. 1, E0 is a referenced value of Youngs modulus, and E0 = 70 GPa. Clearly, the comparison is

reasonably well.

A parametric study has been carried out and typical results are shown in Figs. 27. It should

be appreciated that in all of these gures W=h; Mxb2=E0h

4 and q0b4=E0h

4 denote the dimensionless

central deection of the plate, central-bending moment and lateral pressure, respectively, where

E0= Youngs modulus of Ti6Al4V at T = 300 K.The results presented herein are for movable in-plane boundary conditions, unless it is stated

otherwise. In Figs. 37, the volume fraction index N = 0:5 and 5.0, and in Figs. 2, 3 and 57 the

plate width-to-thickness ratio b=h = 20.

Fig. 2 gives the load-deection and load-bending moment curves of ZrO 2=Ti6Al4V square plate

with dierent values of volume fraction index N subjected to a uniform pressure and under thermal

environmental condition T = 0 K. The results show that a fully titanium alloy plate (N = 0) has

highest deection and lowest bending moment. It can also be seen that the plate has higher deection

and lower bending moment when it has lower volume fraction. This is expected because the metallic

plate is the one with the lower stiness than the ceramic plate.

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2.0

uniform load

0 50 100 1500.0

0.5

1.0

1.5

W/h

q0b4/E

0h4 q0b

4/E0h4

Zr O2

N= 5.0 N= 2.0 N= 1.0 N= 0.5

Ti-6Al-4V

Zr O2

N= 5.0

N= 2.0

N= 1.0N= 0.5

Ti-6Al-4V

Zr O2/Ti-6Al-4V

=1.0, b/h = 20

T0 = 300 K

uniform load

Zr O2/Ti-6Al-4V

=1.0, b/h = 20

T0 = 300 K

0

1

2

3

0 50 100 150

Mx

b2/E0

h4

(a) (b)

Fig. 2. Eect of volume fraction index N on the nonlinear bending behavior of ZrO2=Ti6Al4V square plates under

uniform pressure: (a) load-central deection; and (b) load-bending moment.

1500 50 100 1500.0

0.5

1.0

1.5

2.0

W/h

q0b4/E0h

4

I: T= 0 K

II: T= 200 K

III: T= 300 K

N= 5.0(III)N= 0.5(III)N= 5.0(II)N= 0.5(II)N= 5.0(I)N= 0.5(I)

uniform load

T0=300 K

ZrO2/Ti-6Al-4V

=1.0, b/h= 20

0 50 100

q0b4/E0h

4

0

1

2

3

N= 5.0(III)

N= 0.5(III)

N= 5.0(II)

N= 0.5(II)

N= 5.0(I)

N= 0.5(I)

Mx

b2/E0

h4

I: T= 0K

II: T= 200 K

III: T= 300 K

uniform load

T0=300 K

ZrO2/Ti-6Al-4V

=1.0, b/h=20

(a) (b)

Fig. 3. Eect of temperature rise on the nonlinear bending behavior of ZrO2=Ti6Al4V square plates under uniform

pressure: (a) load-central deection; and (b) load-bending moment.

1500 50 1000.0

0.5

1.0

1.5

2.0

1: b/h = 20

2: b/h = 10

I: T= 0 K

II: T= 200 K

W/h

N= 5.0 (II & 2)

N= 0.5 (II & 2)N= 5.0 (I & 2)

N= 0.5 (I & 2)N= 5.0 (II & 1)

N= 0.5 (II & 1)N= 5.0 (I & 1)

N= 0.5 (I & 1)

uniform load

T0=300 K

ZrO2/Ti-6Al-4V

=1.0

q0b4/E0h

4q0b

4/E0h4

0 50 100 1500.0

0.5

1.0

1.5

2.0

2.5

N= 5.0 (II & 2)

N= 0.5 (II & 2)N= 5.0 (I & 2)

N= 0.5 (I & 2)

N= 5.0 (II & 1)

N= 0.5 (II & 1)

N= 5.0 (I & 1)

N= 0.5 (I & 1)

I: T= 0 K

II: T= 200 K

Mx

b2/

E0

h4

1: b/h = 20

2: b/h = 10

uniform load

T0=300 K

ZrO2/Ti-6Al-4V

=1.0

(a) (b)

Fig. 4. Eect of plate thickness ratio b=h on the nonlinear bending behavior of ZrO2=Ti6Al4V square plates under


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13/24


0 50 100 1500.0

0.5

1.0

1.5

2.0

1: = 1.0

2: = 1.5

q0b4/E0h

4

I: T= 0 K

II: T= 200 K

uniform load

T0 = 300 K

ZrO2/Ti-6Al-4V

b/h =20

W/h

N= 5.0 (II & 2)

N= 0.5 (II & 2)

N= 5.0 (I & 2)

N= 0.5 (I & 2)

N= 5.0 (II & 1)

N= 0.5 (II & 1)

N= 5.0 (I & 1)

N= 0.5 (I & 1)

q0b4/E0h

4

0.0

0.5

1.0

1.5

2.0

0 20 40 60 80

1: = 1.0

2: = 1.5N= 5.0 (II & 2)

N= 0.5 (II & 2)

N= 5.0 (I & 2)

N= 0.5 (I & 2)

N= 5.0 (II & 1)

N= 0.5 (II & 1)

N= 5.0 (I & 1)

N= 0.5 (I & 1)

I: T= 0 K

II: T= 200 K

Mx

b2/E0

h4

uniform load

T0

= 300 K

ZrO2/Ti-6Al-4V

b/h = 20

(a) (b)

Fig. 5. Eect of plate aspect ratio on the nonlinear bending behavior of ZrO2=Ti6Al4V plates under uniform pressure:

(a) load-central deection; and (b) load-bending moment.

0 50 100 150

0

1

2

31: movable edges

2: immovable edges

q0b4/E0h

4

I: T= 0 K

II: T= 200 K

uniform load

T0

= 300 K

ZrO2/Ti-6Al-4V

= 1.0, b/h = 20

W/h

N= 5.0 (II & 2)

N= 0.5 (II & 2)

N= 5.0 (I & 2)

N= 0.5 (I & 2)

N= 5.0 (II & 1)

N= 0.5 (II & 1)

N= 5.0 (I & 1)

N= 0.5 (I & 1)

0

1

2

3

4

0 50 100 150

N= 5.0 (II & 2)

N= 0.5 (II & 2)

N= 5.0 (I & 2)

N= 0.5 (I & 2)

N= 5.0 (II & 1)

N= 0.5 (II & 1)

N= 5.0 (I & 1)

N= 0.5 (I & 1)

I: T= 0 K

II: T= 200 K

Mx

b2/E0

h4

1: movable edges

2: immovable edges

uniform load

T0

= 300 K

ZrO2/Ti-6Al-4V

= 1.0, b/h = 20

q0b4/E0h

4(a) (b)

Fig. 6. Eect of in-plane boundary conditions on the nonlinear bending behavior of ZrO 2=Ti6Al4V square plates under


0 50 100 1500.0

0.5

1.0

1.5

2.0

1: uniform load

2: sinusoidal load

ZrO2/Ti-6Al-4V

= 1.0, b/h = 20T

0= 300 K

q0b4/E0h

4

I: T= 0 KII: T= 200 K

W/h

N= 5.0 (II & 2)

N= 0.5 (II & 2)N= 5.0 (I & 2)

N= 0.5 (I & 2)

N= 5.0 (II & 1)

N= 0.5 (II & 1)

N= 5.0 (I & 1)

N= 0.5 (I & 1)

q0b4/E0h

4

0.0

0.5

1.0

1.5

2.0

2.5

0 50 100 150

ZrO2/Ti-6Al-4V

= 1.0, b/h = 20T0 = 300 K

1: uniform load

2: sinusoidal load

N= 5.0 (II & 2)

N= 0.5 (II & 2)

N= 5.0 (I & 2)

N= 0.5 (I & 2)

N= 5.0 (II & 1)

N= 0.5 (II & 1)

N= 5.0 (I & 1)

N= 0.5 (I & 1)

I: T= 0 K

II: T= 200 K

Mxb

2/E0

h4

(a) (b)

Fig. 7. Comparisons of nonlinear responses of ZrO2=Ti6Al4V square plates subjected to a uniform or sinusoidal load

and in thermal environments: (a) load-central deection; and (b) load-bending moment.

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14/24


Fig. 3 gives the load-deection and load-bending moment curves of ZrO2=Ti 6Al4V square

plates subjected to a uniform pressure and under three sets of thermal environmental conditions,

referred to as IIII. For environmental case I, T = 0 K, for environmental case II, T = 200 K

and for environmental case III, T = 300 K. Because the thermal expansion at the top surface ishigher than that at the bottom surface, this expansion results in an upward deection. It is seen that

the deections are reduced, but the bending moments are increased with increases in temperature.

Note that for environmental cases II and III the deections are close to each other when W=h 1:5.

Fig. 4 gives the load-deection and load-bending moment curves of ZrO2=Ti 6Al4V square

plates with dierent values of width-to-thickness ratio b=h (=10 and 20) under two environmental

conditions I and II. As expected, these results show that the dimensionless central deections are

decreased, but the bending moments are increased, by increasing plate width-to-thickness ratio b=h

for both cases I and II.

Fig. 5 shows the eect of plate aspect ratio (=1:0 and 1.5) on the nonlinear bending behavior of

ZrO2=Ti6Al4V rectangular plates under two environmental conditions. It can be seen that centraldeections are increased but the bending moments are decreased as changes from 1.0 to 1.5.

Fig. 6 shows the eect of in-plane boundary conditions on the nonlinear bending behavior of

ZrO2=Ti6Al4V square plates under two environmental conditions. To this end, the load-deection

and load-bending moment curves of ZrO2=Ti6Al4V square plates under movable and immovable

in-plane boundary conditions are displayed. The results show that the plate with immovable edges

will undergo less deection with smaller bending moments.

Fig. 7 compares the load-deection curves and load-bending moment curves of ZrO2=Ti6Al

4V square plates under two cases of transverse loading conditions along with two environmental

conditions. It can be seen that both load-deection and load-bending moment curves of the plate

subjected to a sinusoidal load are lower than those of the plate subjected to a uniform load.

5. Concluding remarks

The nonlinear bending behavior of functionally graded rectangular plates subjected to a transverse

uniform or sinusoidal load and in thermal environments has been presented. Material properties

are assumed to be temperature-dependent, and graded in the thickness direction according to a

simple power-law distribution in terms of the volume fractions of the constituents. A mixed Galerkin-

perturbation technique is employed to determine the load-deection and load-bending moment curves.

Numerical results are for ZrO2=Ti6Al4V plates. In eect, the results provide information about

nonlinear bending responses of FGM plates for dierent proportions of the ceramic and metal. Theyalso conrm that the characteristics of nonlinear bending are signicantly inuenced by temperature

rise, the character of in-plane boundary conditions, transverse shear deformation, plate aspect ratio

as well as volume fraction distributions.

Acknowledgements

This work is supported in part by the National Natural Science Foundation of China under Grant

59975058. The author is grateful for this nancial support.

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15/24


Appendix A

In Eqs. (4)(7)

L11( ) =4

3h2

F11

94

9X4+ (F12 +F

21 + 4F

66)9

4

9X29Y2+F22

94

9Y4

;

L12( ) =

D11

4

3h2F11

93

9X3+

(D12 + 2D

66) 4

3h2(F12 + 2F

66)

93

9X9Y2;

L13( ) =

(D12 + 2D

66) 4

3h2(F21 + 2F

66)

93

9X29Y+

D22

4

3h2F22

93

9Y3;

L14( ) = B

21

94

9X4 + (B

11 +B

22 2B

66)

94

9X29Y2 +B

12

94

9Y4 ;

L15( NT

) =9

2

9X2(B11 N

T

x +B

21N

T

y) + 29

2

9X9Y(B66 N

T

xy) +9

2

9Y2(B12 N

T

x +B

22N

T

y);

L16( MT

) =92

9X2( M

T

x ) + 292

9X9Y( M

T

xy) +92

9Y2( M

T

y);

L21( ) = A

22

94

9X4+ (2A12 +A

66)94

9X29Y2+A11

94

9Y4;

L22

( ) = B21

4

3h2E

21 939X3

+ (B11

B66

) 4

3h2(E

11E

66) 93

9X9Y2;

L23( ) =

(B22 B

66) 4

3h2(E22 E

66)

93

9X29Y+

B12

4

3h2E12

93

9Y3;

L24( ) =4

3h2

E21

94

9X4+ (E11 +E

22 2E

66)94

9X29Y2+E12

94

9Y4

;

L25( NT

) =92

9X2(A12 N

T

x +A

22N

T

y) 92

9X9Y(A66 N

T

xy) +92

9Y2(A11 N

T

x +A

12N

T

y);

L31( ) =A55

8

h2 D55 +

16

h4 F55 99X +

4

3h2

(F

11

4

3h2 H

11)

93

9X3

+

(F21 + 2F

66) 4

3h2(H12 + 2H

66)

93

9X9Y2

;

L32( ) =

A55

8

h2D55 +

16

h4F55

D11

8

3h2F11 +

16

9h4H11

92

9X2

D66

8

3h2F66 +

16

9h4H66

92

9Y2;

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16/24


L33( ) =

(D12 +D

66) 4

3h2(F12 +F

21 + 2F

66) +16

9h4(H12 +H

66)

92

9X9Y;

L34( ) = L22( );

L35( NT

) =9

9X

B11

4

3h2E11

N

T

x +

B21

4

3h2E21

N

T

y

+9

9Y

B66

4

3h2E66

N

T

xy

;

L36( ST

) =9

9X( S

T

x ) +9

9Y( S

T

xy);

L41( ) =A44 8

h2D44 + 16

h4F44

99Y

+ 43h2

(F12 + 2F

66) 43h2(H12 + 2H

66)

93

9X29Y

+

F22

4

3h2H22

93

9Y3

;

L42( ) = L33( );

L43( ) =

A44

8

h2D44 +

16

h4F44

D66

8

3h2F66 +

16

9h4H66

92

9X2

D22 83h2

F22 + 169h4H22

92

9Y2;

L44( ) = L23( );

L45( NT

) =9

9X

B66

4

3h2E66

N

T

xy

+

9

9Y

B12

4

3h2E12

N

T

x +

B22

4

3h2E22

N

T

y

;

L46( ST

) =9

9X( S

T

xy) +9

9Y( S

T

y );

L( ) =92

9X292

9Y2 2

92

9X9Y

92

9X9Y+

92

9Y292

9X2: (A.1)

Appendix B

In Eqs. (22) and (24)

(110; 112; 114) = (4=3h2)[F11; (F

12 +F

21 + 4F

66)=2; F

22]=D

11;

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17/24


(120; 122) = [D

11 4F

11=3h2; (D12 + 2D

66) 4(F

12 + 2F

66)=3h2]=D11;

(131; 133) = [(D

12 + 2D

66) 4(F

21 + 2F

66)=3h2; D22 4F

22=3h2]=D11;

(140; 142; 144) = [B

21; (B

11 +B

22 2B

66); B

12]=[D

11D

22A

11A

22]1=4;

(212; 214) = (A

12 +A

66=2; A

11)=A

22;

(220; 222) = [B

21 4E

21=3h2; (B11 B

66) 4(E

11 E

66)=3h2]=[D11D

22A

11A

22]1=4;

(231; 233) = [(B

22 B

66) 4(E

22 E

66)=3h2; B12 4E

12=3h2]=[D11D

22A

11A

22]1=4;

(240; 242; 244) = (4=3h2)[E21; (E

11 +E

22 2E

66); E

12]=[D

11D

22A

11A

22]1=4;

(310; 312) = (4=3h2)[F11 4H

11=3h2; (F21 + 2F

66) 4(H

12 + 2H

66)=3h2]=D11;

(320; 322) = (D

11 8F

11=3h2

+ 16H

11=9h4

; D

66 8F

66=3h2

+ 16H

66=9h4

)=D

11;

331 = [(D

12 +D

66) 4(F

12 +F

21 + 2F

66)=3h2 + 16(H12 +H

66)=9h4]=D11;

(411; 413) = (4=3h2)[(F12 + 2F

66) 4(H

12 + 2H

66)=3h2; F22 4H

22=3h2]=D11;

(430; 432) = (D

66 8F

66=3h2 + 16H66=9h

4; D22 8F

22=3h2 + 16H22=9h

4)=D11;

(511; 522) = (B

11 4E

11=3h2; B22 4E

22=3h2)=[D11D

22A

11A

22]1=4;

(611; 622) = (4=3h2)(E11; E

22)=[D

11D

22A

11A

22]1=4;

(711; 722) = (B

11; B

22)=[D

11D

22A

11A

22]1=4;

(812; 821) = (D

12 4F

12=3h2; D12 4F

21=3h2)=D11;

(912; 921) = (4=3h2)(F12; F

21)=D

1 : (B.1)

Appendix C

In Eq. (17)

ATx =h

1 (b t)(Eb Et)

1

2N + 1+ [t(Eb Et) + (b t)Et]

1

N + 1+ tEt

;

DTx =h2

1

(b t)(Eb Et)

N

2(N + 1)(2N + 1)

+ [t(Eb Et) + (b t)Et]N

2(N + 1)(N + 2)

;

FTx =h4

1

(b t)(Eb Et)

1

8(2N + 1)

3

8(N + 1)(2N + 1)

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18/24


+3

2(N + 1)(2N + 1)(2N + 3)

3

2(N + 1)(N + 2)(2N + 1)(2N + 3)

+ [t(Eb Et) + (b t)Et]

1

8(N + 1)

3

4(N + 1)(N + 2)

+3

(N + 1)(N + 2)(N + 3)

6

(N + 1)(N + 2)(N + 3)(N + 4)

: (C.1)

Appendix D

In Eq. (27)M(1)x S

(1)x

M(1)y S(1)y

=

16

2T

( W =h)

T3 T3 T6

T4 T4 T7

A

(1)11 (D.1)

and in Eqs. (35)(37)

A(0)W =

4 T[(T3m2 + T4n

22) 11];

A(1)W =

4C11[S11 CW1];

A(2)W =

4W22 h

[D11D

22A

11A

22]1=4

;

A(3)W =

4C11{(313 + 331)[S11 CW1] +1

1614242}

h2

[D11D

22A

11A

22]1=2

;

A(0)

MX = 92416T(T3

X11)=11025 + CX0;

A(1)

MX = 2X11;

A(2)

MX = 2X22

h

[D11D

22A

11A

22]1=4

;

A(3)

MX = 2[X11(313 + 331) X33]

h2

[D

11D

22A

11A

22]1=2;

A(0)

MY = 92416 T(T4

Y11)=11025 + CY0;

A(1)

MY = 2Y11;

A(2)

MY = 2Y22

h

[D11D

22A

11A

22]1=4

;

A(3)

MY = 2[Y11(313 + 331) Y33]

h2

[D11D

22A

11A

22]1=2

(D.2)

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19/24


in which

2 =m4

7+

n44

6+ C22; S11 = g08 + 1424m

2n22g05g07

g06;

S13 = g138 + 14249m2n22

g135g137

g136; S31 = g318 + 14249m

2n22g315g317

g316;

W22 =1

3C111424m

2n22

8

6+

9

7+ 4

g05

g06

; 11 = g

08 + 1424g05g07

g06;

6 = 1 + 14242230

4m2

41 + 3224m2; 7 =

224 + 1424

2223

4n22

31 + 3224n22;

8 = 140 1202204m2

31 + 3204m2

; 9 = 144 1332334n22

41 + 4324n2

2

;

18 = 722 8122204m2

31 + 3204m2; 19 = 711 821233

4n22

41 + 4324n22;

J13 = S13 C13; J31 = S31 C31;

313 =1

161424

m2

7J13; 331 =

1

161424

n22

6J31;

X11 = (110m2 + 921n

22) + 1424(140m2 + 711n

22)g05

g06

+ 120m2

g04g00

1424 g02g05g00g06

+ 821n22

g03g00

1424 g01g05g00g06

;

X22 =1

81424

m219

7+

n228

6+ CX2

; X33 =

116

1424(X13 + X31);

X13 = 1424m4

7J13

(140m

2 + 7119n22)

m2120g132 + 9n22821g131

g130

g135

g136

+m4

7J13

(110m

2 + 9219n22) +

m2120g134 + 9n22821g133

g130

;

X31 = 1424 n4

4

6J31

(1409m

2 + 711n22) 9m

2

120g312 + n2

2

821g311g310

g315g316

+n44

6J31

(1109m

2 + 921n22) +

9m2120g314 + n22821g313

g310

;

Y11 = (912m2 + 114n

22) + 1424(722m2 + 144n

22)g05

g06

+ 812m2

g04

g00 1424

g02g05

g00g06

+ 133n

22

g03

g00 1424

g01g05

g00g06

;

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Y22 =1

81424

m29

7+

n2218

6+ CY2

; Y33 =

116

1424(Y13 + Y31);

Y13 = 1424m4

7J13

(722m

2 + 1449n22)

m2812g132 + 9n22133g131

g130

g135

g136

+m4

7J13

(912m

2 + 1149n22) +

m2812g134 + 9n22133g133

g130

;

Y31 = 1424n44

6J31

(7229m

2 + 144n22)

9m2812g312 + n22133g311

g310

g315

g316

+n44

6J31 (9129m2 + 114n

22) +9m2812g314 + n

22133g313

g310 ;X11 = 120m

2 g

04

g00+ 821n

22g03g00

; Y11 = 812m2 g

04

g00+ 133n

22g03g00

;

g00 = (31 + 320m2 + 322n

22)(41 + 430m2 + 432n

22) 2331m2n22;

g01 = (31 + 320m2 + 322n

22)(231m2 + 233n

22) 331n22(220m

2 + 222n22);

g02 = (41 + 430m2 + 432n

22)(220m2 + 222n

22) 331m2(231m

2 + 233n22);

g03 = (31 + 320m2 + 322n

22)(41 411m2 413n

22) 331m2(31 310m

2 312n22);

g

03 = (31 + 320m2

+ 322n2

2

)(T4 T7) 331m2

(T3 T6);

g04 = (41 + 430m2 + 432n

22)(31 310m2 312n

22) 331n22(41 411m

2 413n22);

g04 = (41 + 430m2 + 432n

22)(T3 T6) 331n22(T4 T7);

g05 = (240m4 + 242m

2n2 + 244n44)

+m2(220m

2 + 222n22)g04 + n

22(231m2 + 233n

22)g03

g00;

g05 =m2(220m

2 + 222n22)g04 + n

22(231m2 + 233n

22)g03

g00

;

g06 = (m4 + 2212m

2n22 + 214n44)

+ 1424m2(220m

2 + 222n22)g02 + n

22(231m2 + 233n

22)g01

g00;

g07 = (140m4 + 142m

2n2 + 144n44)

m2(120m

2 + 122n22)g02 + n

22(131m2 + 133n

22)g01

g00;

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g08 = (110m4 + 2112m

2n22 + 114n44)

+

m2(120m2 + 122n

22)g04 + n22(131m

2 + 133n22)g03

g00 ;

g08 =m2(120m

2 + 122n22)g04 + n

22(131m2 + 133n

22)g03g00

;

g130 = (31 + 320m2 + 3229n

22)(41 + 430m2 + 4329n

22) 23319m2n22;

g131 = (31 + 320m2 + 3229n

22)(231m2 + 2339n

22) 3319n22(220m

2 + 2229n22);

g132 = (41 + 430m2 + 4329n

22)(220m2 + 2229n

22) 331m2(231m

2 + 2339n22);

g133 = (31 + 320m2

+ 3229n2

2

)(41 411m2

4139n2

2

)

331m2(31 310m

2 3129n22);

g134 = (41 + 430m2 + 4329n

22)(31 310m2 3129n

22)

3319n22(41 411m

2 4139n22);

g135 = (240m4 + 9242m

2n2 + 81244n44)

+m2(220m

2 + 2229n22)g134 + 9n

22(231m2 + 2339n

22)g133

g130

;

g136 = (m4 + 18212m

2n22 + 81214n44)

+ 1424m2(220m

2 + 2229n22)g132 + 9n

22(231m2 + 2339n

22)g131

g130;

g137 = (140m4 + 9142m

2n2 + 81144n44)

m2(120m

2 + 1229n22)g132 + 9n

22(131m2 + 1339n

22)g131

g130;

g138 = (110m4 + 18112m2n22 + 11481n44)

+m2(120m

2 + 1229n22)g134 + 9n

22(131m2 + 1339n

22)g133

g130;

g310 = (31 + 3209m2 + 322n

22)(41 + 4309m2 + 432n

22) 23319m2n22;

g311 = (31 + 3209m2 + 322n

22)(2319m2 + 233n

22) 331n22(2209m

2 + 222n22);

g312 = (41 + 4309m2 + 432n

22)(2209m2 + 222n

22) 3319m2(2319m

2 + 233n22);

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g313 = (31 + 3209m2 + 322n

22)(41 4119m2 413n

22)

3319m2(31 3109m

2 312n22);

g314 = (41 + 4309m2 + 432n

22)(31 3109m2 312n

22)

331n22(41 4119m

2 413n22);

g315 = (81240m4 + 9242m

2n2 + 244n44)

+9m2(220m

2 + 222n22)g314 + n

22(2319m2 + 233n

22)g313

g310;

g316 = (81m4 + 18212m

2n22 + 214n44)

+ 14249m2(2209m

2 + 222n22)g312 + n

22(2319m2 + 233n

22)g311

g310;

g317 = (81140m4 + 9142m

2n2 + 144n44)

9m2(1209m

2 + 122n22)g312 + n

22(1319m2 + 133n

22)g311

g310;

g318 = (81110m4 + 18112m

2n22 + 114n44)

+ 9m2

(1209m2

+ 122n2

2

)g314 + n2

2

(1319m2

+ 133n2

2

)g313g310

(D.3)

and

C11 =2

16mn (for uniform load); (D.4a)

C11 = 1 (for sinusoidal load) (D.4b)

in the above equations, for the case of movable edges

CW1 = CX0 = CY0 = CX2 = CY2 = C22 = C13 = C31 = 0 (D.5a)

and for the case of immovable edges

CW1 = 14(T1m2 + T2n

22) T;

CX0 = 214(711T1 + 140T2) T

[D11D

22A

11A

22]1=4

h;

CY0 = 214(144T1 + 722T2) T

[D11D

22A

11A

22]1=4

h;

C22 = 2(m4 + 224n

44) + 25m2n22

224 25

;

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C13 = 14(T1m2 + 9T2n

22) T; C31 = 14(9T1m2 + T2n

22) T;

CX2 = 711m2 + 5n

22

224 25+ 140

5m2 + 224n

22

224 25;

CY2 = 144m2 + 5n

22

224 25

+ 7225m

2 + 224n22

224 25

: (D.5b)

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