An Axisymmetric Bending and Shear Stress Analysis of of Functionally Graded Circular Plate Based on...

8/13/2019 An Axisymmetric Bending and Shear Stress Analysis of of Functionally Graded Circular Plate Based on Unconstrain

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IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308

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Volume: 02 Issue: 07 | Jul-2013, Available @ http://www.ijret.org 37

AN AXISYMMETRIC BENDING AND SHEAR STRESS ANALYSIS OF OF

FUNCTIONALLY GRADED CIRCULAR PLATE BASED ONUNCONSTRAINED THIRD ORDER SHEAR DEFORMATION THEORY

VIA DIFFERENTIAL QUADRATURE METHOD

HamadM.H 1., F. Tarlochan 2

Center for Innovation and Design, College of Engineering, UniversitiTenagaNasional, 43009 Selangor

Abstract In this study, based on the unconstrained third order shear deformation theory (UTSDT), numerical analysis of an axisymmetricbending and stresses of circular plate are investigated. The material properties are considerd to graded through the thickness of theverticlecoordinate, and follow a simple power of volume fraction of the constituents.governing equations are derived and DQM isused as an efficient numerical method for solving the differential equations.Two types of boundary conditions under the influence ofthe bending and body force are studied. The validation of the results is done by a comparison with another study ,which available inthe literature and found good agreement between two studies.

Index Terms :bending,shearstress,circularplate,UTSDT,GDQM .

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1. INTRODUCTION

Thick and thin Circular disk in structured componentsplays a

major role in engineering applications related to this area isthe static analysis thetypes of plates which arenotably crucialin their design ranging from automotive railway brake systemsto disks which constitute vital components particularly inturbo machines. Functionally,graded materials (FGMs) werefirst introduced in 1989 [1] whereby a number of researchers,because interested to study them .

In the past decade , many of the studies which carried out onthe FGMs disks concentrated on the conventional plate and thefirst order shear deformation theories . The conventional platetheory (CPT) furnishes accurate and reliableanalysis for thisplate . As the disk thickness increases CPT over predictsstresses response, because the transverse shear deformationand rotary inertia effects are neglected .So there a number ofshear deformation theories used to analyze moderately thickplate , first order theory and third order theory weredeveloped to incorporate the shear deformation effects , in thefirst order shear deformation theory (FSDT), the constantshear stress condition through thicknesses violates thestatically condition of zero shear stress at the free surface . Soits need for shear correction factor to modify the shear forces.The third order shear deformation theory (TSDT )predictsparabolic variation of shear stress through the thickness.Although the use of higher order plate theory leads to more anaccurate prediction of the global response quantities such asshear forces , deflections strain and stresses , it requires much

computation effort . Furthermore the use of the (TSDT) byReddy is constrained , because it considers the shear stressvanishes on the top and bottom surfaces of the plate , but this

limitation is solved by the unconstrainedthird order sheardeformation theory (TSDT) by Leuny [2].

In past decades several studies published on the static analysisof FGMs circular disks Reddy et al. [3] Study relates to axsymmetric bending of functionally graded circular and annularplates whereby the first order shear deformation plate theorywas used. Ma and Wang [4] analyzed further by discussing therelationship between axisymmetric bending and bucklingsolutions of FGM circular plates. Third-order plate theory andclassical plate theory were demonstrated and discussed indetail in their study. In addition, asymmetric flexural vibrationand an additional stability analysis of FGM circular plates was

included in thermal environment by using finite elementtechniquespresented by Prakash and Ganapathi [5]. Also,Three-dimensional free vibration of functionally gradedannular plates whereby boundary conditions were differentusing a Chebyshev-Ritz method was also studied by Dong [6].Malekzadeh et al. [7] Also showed how in thermalenvironment in-plane free vibration analysis of FGM thin-to-moderately thick deep circular arches. Third-order sheardeformation theory was used by Saidi et al. [8] To analyzeaxisymmetric bending and buckling of thick functionallygraded circular plates. Subsequently, fourth-order sheardeformation theory was researched by Sahraee and saidi [9] tostudy axisymmetric bending of thick functionally graded


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( ) ( )P z Pm Pc Vm Pc= + ( 1 )

Where:( )P z :material property at location z through thickness.m and

c denotes the metallic and ceramic constituents respictivlly .

Vm :volume fraction of metal

22

p

mh z

V h

= ( 2 )

Where:

z: thickness coordinates( )2 2h z h

p : material constant.

As the material constant is equl to zero ( )0 p = , or equal toinfinity ( ) p = , the homogeneous isotropic material isobtained as a specializ case of functioaly graded material . infact , from equation (2) it possible to obtaian :

( )0 1, 0m c m p V V p z p= = = =

( )0, 1m c c p V V p z p= = = =

Fig.2 show that material profile through the FG plate forvarious of p.

According to relution (1), the elastic modulur E and density to be varied according to the above equation . andpoissonsratio is assumed to be constant.

( ) ( )m C m c E z E E V E = + + ( 3 )

( ) ( )m C m c z V = + ( 4 )

3. GOVERING EQUATIONSBased on the unconstrained thiredorder shear deformationtheory (UTSDT) ,displacement field in the cylindricalcoordinate system can be written as :

( ) ( ) ( ) ( ) ( ) ( )1 2,U r z u r f z r g z r = + + ( 5 )

( ) ( ),w r z w r = ( 6 )

Where : u ,w are the displacements of points on the middleplane (z=0)in the radial and vertical direction respictivly.

1 :smalltransrverse normal rotation about the -axises2 : smalltransrverse normal higher order rotation about -

axises( ) ( ), f z g z are shear functions.

From the previous[23], the displacement field has beenimproved by taking into consideration shear functions alongthe thinckess . indeed ,the model for the shear function in thisstudy has taken from previous work[24].

Strain- displacement relations:

( )3 31 2

rr

d d du z z z

dr dr dr

= +

(7a)

( )3 31 21 1 1 u z z zr r r = + (7b)

( )2 2 1 21 3 3rz dw z z dr = + (7c)0, 0 , 0r z z = = =

Stress-strain relations

r 11 12 r

12 22

rz 66 rz

0

0

0 0

Q Q

Q Q

Q

= ( 8 )

Where :( )

( )11 22 12 112,

2 1

E zQ Q Q Q

= = =

+

( )( )66E z

2 1Q

=

+

The total potential energy of circular plate:

f U V= + ( 9 )

a h / 2

r r rz rz

0 h / 2

U 2 rdzdr

= + + ( 10 )


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2 2 2 21 2 1

0

2 2 2 23 1 3 2

2 (

)

a

f V r u r

r r w rq w dr

= +

+

(11)

Where :, , f U V are the total potential energy , strain energy and thepotential energy of the body force and pressure loadrespectively.

By the principle of minimum total energy 0 =

( )

( ) ( )

( )

( )

( )

2 21

2 23 1 1

0

2 23 2

)

30

3

r

r r r

ar

r r

r r

d rN N r u

dr

d d rM rp M p r dr dr rR

dr r

d rp p rR r

dr

d r rq w

dr

+ + ( + + +

+ = + + +

(12)

Where:

r N , N :Stress resultantsr M , M :Stress couples,r p p :higher order stress couples

r :transverse shear resultantr R : higher order shear resultant1, 2 3, : constant proportional to the mean ,first and third

moment of the density along the thickness.

( ) /2

3

/2

( , , ) 1, ,h

r r r r

h

N M P z z dz

= (16a)

( ) ( ) /2

3

/2

, , 1, ,h

h

N M P z z dz

= (16b)

( ) ( ) /2

2

/2

, 1,h

r r rz

h

R z dz

= (16c)

( ) ( )( ) /2

31 2 3

/2

, , 1, ,h

h

z z z dz

= (16d)

From eqs. (7),(8) and (16), one can obtain the followingrelations:

( )

11

1 211 11 1 11 2

1

1 1

r du

N A udr r

d d B E E

dr r dr r

= + + + + (17a)

( )

11

1 211 11 1 11 2

1

1 1

du N A u

dr r

d d B E E dr r dr r

= +

+ + + (17b)

( )

11

1 211 11 1 11 2

1

1 1

r du

M B udr r

d d D F F

dr r dr r

= + + + + (17c)

( )

11

1 211 11 1 11 2

1

1 1

du M B u

dr r

d d

D F F dr r dr r

= +

+ + + (17d)

( )

11

1 211 11 1 11 2

1

1 1

r du

P E udr r

d d F H H

dr r dr r

= + + + + (17e)

( )44 1 44 1 23r dw

Q A Ddr

+ +

= (17f )

( )44 1 44 1 23r dw R D F dr + +

=(17h)

Where:11 11, 11 11 11 11, , , , A B D E F H : are the circulaer disk stiffness

coefficients


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( )( )

( ) ( )

11 11, 11 11 11

/22 3 4 6

2 /2

, , ,

1, , , , ,1

h

h

A B D E H

E z

z z z z z dz

=

(18a)

( ) ( )

( ) ( ) /2

2 444 44 44

/2

, , 1, , ,2 1

h

h

E z A D F z z z dz

=

(18b)

From equation (12) , the equilibrium equations areu :

( ) 2 21 0r d

rN N r dr

+ = (19a)

1 :

( ) ( )

( )

2 22 3

3

0

r r r r d d

rM rP M P rQ rRdr dr

r

+ + +

= (19b)

2 :

( )2 2

33 0r r d

rP P rR r dr + = (19c)

:w

( ) 0r d

rQ rqdr

+ = (19d)

Equilibrium equations in terms of displacements:

( ) ( )

( ) ( )

221

11 11 11 11 112 2

111 11 11 11

22 22 2

1 11 11 11 2 12

1

1 10

d d u du A A A u B E r

dr r dr dr

d B E B E dr

d d E E E r

r dr r

r dr

r

+ + +

+

+ + =

(20a)

( ) ( ) ( )

( )( ) ( )

( )

( ) ( ) ( )

( ) ( )

2

11 11 11 11 11 112

2 111 11 11

2 211 11 11 44 44 44 1

22 2

11 11 2

2 2 2211 11 11 11 44 44 2

2 244 44 2 3

1

2

12 6 9

13 9

3

d u du B E r B E B E u

dr r dr

d D F H dr

D F H A D F r r

d F H r

dr d

F H F H D F rdr r

dw A D r

dr

+ +

+ + + +

+

+ + + + +

+ 0=

(20b)

( )

( ) ( ) ( )

221

11 11 11 11 112 2

111 11 11 11 44 44 1

21 1

11 11 11 442

2 22 44 3

1

1 3 9

1 9

3 0

d d u du E E E u F H r

dr r dr dr d

F H F H D F r dr r

d d H r H H E r

dr r dr dw

D r r dr

r

+ +

+ + + + + +

+ =

(20c)

( ) ( )

( ) ( )

144 44 44 44 1

22

44 44 2 44 2

3 3

3 3

0

d A D r A D

dr

d d w D r D A r

dr dr rq

+

+ +

= (20d)

Using the following dimensionless parameters for simplicity .

r R

a=

,

wW

h=

,2

uhU

a=

, = ,ha

= ,

611

111

A h H

= ,

511

211

B h

H =

,

411

311

D h

H =

,

311

411

E h H

= ,

211

511

F h H

= ,

644

611

A h H

= ,

444

711

D h H

= ,

244

811

F h H

=


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:Q pressure parameter

6

11

*h qQ H

=

: bodyforec parameters

7 21

111

h a H

=

8 22 3

211

( )h H

=

4 23

311

43

haH

=

the equations of motion in dimensionless form are:

2 21 1

1 1 1 2 4 2 42 2

222 2

2 4 1 4 4 4 2 12

1 4 43 3

4 1 4 4 4 13 3 3 3 12

d U du d d R U R

dr r dRdr dR

d d R R

R dRdR

+ + + + =

(21a)

( )

2

2 4 2 4 2 42

3 521 1

3 5 3 5 1 526 7 8 2

22

52

4 4 4 1

3 3 3

8 16 13 98 16 8 16 8 16

3 9 3 9 3 918 16

8 163 9

d U dU R U

dR r dR

Rd d R

dRdR R

d d R

dR

+ +

+ + + + + +

( ) ( )2 5 7 8 2 6 72

22

8 16 1 14 16 4

3 9

1

RdR R

dw R R

dR

+ + + +

=

(21b)

( )

221 1

4 4 4 5 52 2

22 2

5 7 8 12 2

28 2 7 32

1 4 4

3 3

4 1 1 4 43 12

3 3 3

4 1 1 112 3

3

d d d U dU R U R

dR R dRdR dR

d d R R

R dRdR

dw R R R

R dR

+ +

+

+ + =

(21c)

( ) ( )1 26 7 6 7 1 72

7 2 7 62

4 4 4

4

d d R R

dR dR

d w dw R QRdRdR

+

+ + = (21d)

Boundary ConditionsClamped circular plateAt 0 R =

0,U = 1 0 = , 2 0 = , 0 ,

dwdR

=0 RQ = (22a)

At 1 R =

0,U = 1 0 = , 2 0 = , 0W = (22b)

Roller support circular plateAt 0 R =

0,U = 1 0 = , 2 0 = , 0

dwdR

= (22c)

At R a=

0,W = 0r N = , 0r M = 0r P = (22d)

4. IMPLEMENTATION OF GDQ METHOD

The generalized differential quadrature (DQ) method is

adopted to solve the differential equations of the annularplate. The core of the DQ method is that the derivative of a

function in a domain ( )0 x L is approximated as aweighted linear summation of a function values at all discretepoints in that domain. Thus, DQ method changes thegoverning differential equations into a set of correspondingsimultaneous equations. To demonstrate the DQ method ,consider the rth derivative of a function f(x) can be estimatedas

( )( )

r nr

X ik k rk 1

f x D f x

x i =

=

i=1,2,.,n

Where xi are the discrete points in the variable domain ,rik D

,and f( xk ) are the weighting coefficient and the functionvalue at the discrete points.Thus, for the first-order derivatives, the weighting coefficients can be calculated as [25]

( ) ( )( ) ( )

1 iik

i k k

xD

x x x

=

i , k = 1 , 2 ,.n , i k


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Where( ) ( )

n

i i k i 1

x x x =

= i , k = 1 , 2 ,..n , i k

After that , the domain of the annular plate is divided into ngrade points in r direction. Chebyshev polynomial is the bestmethod to evaluate the grid points in the domain of theplate[25 ]:

ii 1

r 0.5* 1 cos n 1

= i = 1 , 2 ,..n

The governing eqs. (21) can be discretized according to theGDQ method as follows:

( ) ( ) ( )

( )

( ) ( )

12 1 1 1 2 4 1

1 1 1

1 12 4 1 2 4

1

2 2 2 24 2 4 2 4 1

1 1

1 4, , ,3

4 4 1,

3 3

4 4 4 1, ,

3 3 3

n n n

i j j i i ji j j j

n

j ji j

n n

i j j j ii j j

R a i j u a i j u u R a i j R

a i j R

R a i j a i j R R

= = =

=

= =

+ + +

+ =

(23a)

( ) ( )

( ) ( )

( )

2 4 2 2 4 1 2 4

1 1

1 13 5 2 3 5 1

1 1

13 5 6 7 8 52

4 4 4 1, ,

3 3 3

8 16 8 16, ,

3 9 3 9

8 16 1 1 4 168 16

3 9 3 9

n n

i j j ii j j

n n

i j j

j j

i ji

R a i j u a i j u u R

R a i j a i j

R R

= =

= =

+ +

+ + +

+ + +

( )

( ) ( )

( ) ( )

22

1

25 1 5 7 8 2

1

2 26 7 1 2

1

,

4 16 4 16 1 1, 4 16

3 9 3 9

14 ,

n

i j

j

n

j ii

j

n

j i j i

j

R a i j

a i j R R

R a i j w R

=

=

=

+ +

=

(23b)

( ) ( ) ( )

( ) ( ) ( ) ( )

14 2 4 1 4 5 2 5

1 1 1

1 1 2 21 5 7 8 2 12

1 1 1

28 72

1 4 4, , ,

3 3

4 1 1 4 4, 3 12 , ,

3 3 3

4 1 1 112 3

3

n n n

i j j i i ji

j j j

n n n

j i j i j ji

j j j

i ji

R a i j u a i j u u R a i j R

a i j R R a i j a i j R

R R

= = =

= = =

+ + +

+

+ +

( ) 21 31

,n

i j i

j

R a i j w R

==

(23c)

( ) ( ) ( )

( ) ( ) ( ) ( )

1 16 7 1 6 7

1

2 27 1 7 6 2 6 1

1 1 1

4 , 4

4 , 4 , ,

n

i j j j

n n n

i j j i j j j j j

R a i j

R a i j R a i j w a i j w QR

=

= = =

+

+ + =

(23d)

The boundry conditions can be disaretized by the DQ on:

Clampled circular plate .

1 21 1 1 0U = = = ,

( )11

, 0n

j j

a i j w=

=at 0 R = (24a)

1 2

0n n n nU W = = = =

1 R = (24b)Roller support circular plate .

1 21 1 1 0U = = = ,

( )11

, 0n

j j

a i j w=

= at 0 R = (25a)

0nW =at 1 R = (25b)

( ) ( ) ( )

( )

11 1 1 2 4 1

1 1

12 4 4

2 21 4

1

4, ,

3

4 43 3

4, 0 ..at R=1

3

n n

j n j j j

n

n

j n j

a n j u U a n j

a n j

= =

=

+ + +

=

(25c)

( ) ( ) ( )

( )

12 1 2 3 5 1

1 1

13 5 5

2 21 5

1

4, ,

3

4 43 3

4, 0...........at R=1

3

n n

j n j j j

n

n

j n j

a n j u U a n j

a n j

= =

=

+ + +

=

(25d)


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( ) ( ) ( )

( )

14 1 4 5 1

1 1

1 25 1

1

2

4, ,

3

4 4,

3 3

4 0.............at R=1

3

n n

j n j j j

n

n j j

n

a n j u U a n j

a n j

= =

=

+ + +

=

(25e)

The congregation of the governing equations and the relatedboundary conditions lead to a set of simultaneous linearalgebraic equations which can be write in matrix form as:

( )

( )

( )

( )

bb bd b

db dd d

k k 0

k k q

= (26)

Where

bbk

and

bdk

are the stiffness matrices of boundary ofthe boundary conditions and the size of it are 8 8 and 8 4n-8 respectively.

dbk and ddk are stiffness matrices of governing equations andhave size of 4n-8 8 and 4 n-8 4n-8 respectively.

The vector b

contains the displacements corresponds to theboundary points and is eliminated using the staticcondensation technique. The stiffness matrix eq.26 can bereduced into the form of

{ } { }1db bb bd dd dk k k k q + = (27)

From the above equation the vector of domain displacements

d can be evaluated.

5. VALIDATION OF THE RESULTS

In order to examine the accuracy and efficiency of the resultsof this paper a comparative study made with another study toimplement this , an axisymmetric bending of a clamped androller-support functionally graded circular plate underuniformly distributed load q . metallic volume fractionpower law distribution through plate thickness and all materialproperties are getting from Reddy et al. [3] As shown inthetable 1 . The evaluation of the comparison .Between thepresented numerical analysis and Reddys exact result areillustrated in table2 for dimensionless maximum deflection ,

4

64 cWD

qa With( )

3

212 1c

c E h

D

=

Where h and a are the

thickness and radius of the circular plate. The conclusion fromthese comparisons , an excellent agreement between theseresults.

Table 1. Mechanical properties of ceramic and metal ofcircular FGM plate [ 3 ]

material Youngsmodulus( Gpa)

Poisons ratio,v

ceramic 151 0.288metal 70 0.288

Table 2 .Comparisons of the result got in the present paper tothe result got by Reddy etal.[3] for maximum dimantion less

diflaction of FGM circular plate for different values of p.

MaterialconstantP

Reddy [3] PresentClampedplate

Rollersupportplate

Clampedplate

Rollersupportplate

0 2.979 10.822 2.979 10.8222 1.623 5.925 1.608 5.9214 1.473 5.414 1.467 5.4106 1.404 5.155 1.399 5.1508 1.362 4.993 1.357 4.98910 1.333 4.882 1.329 4.88015 1.289 4.714 1.287 4.71320 1.265 4.619 1.264 4.61325 1.250 4.559 1.248 4.558

30 1.239 4.517 1.238 4.516535 1.231 4.486 1.230 4.48640 1.225 4.463 1.229 4.46250 1.216 4.429 1.216 4.429100 1.199 4.359 1.1987 4.359

6. RESULTS AND DISCUSSION

In order to demonstrate the bending and stress analysis of FGcircular plate numerically by unconstrained third order sheardeformation theory via a generalized differential quadraturemethod, two cases are studied in this study , clamped circularplate and roller support circular disk .

6.1 Clamped Circular Plate

In the below , the results are presented in dimensionless form.

Fig (2) shows the dimensionless deflection

wh

along the

dimensionless radius (R ) under bending load with differentvalues of material constant P.As it expected , the deflectionof the metallic plate (p=0) higher than that of the FG plate(p>0) , because of the FG plate more stiffer than that of thepure metal plate .


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Figure 2: Bending distribution (w/h) for clamped circularplatewith (R )for different value of p.

Fig (3) present the variety of the dimensionless shear stress

*

c E

= through the thickness of the plate with the

dimensionless thickness coordinate variable

zh

for different

values of material constant P and plate geometry

ha

. It is

seen that the values of shear stress decrease with increasingin the thickness to the radius ratio ( / )h a As well as , it isobservable that , there is increasing in the shear stress with theincreasing of the material constant P , moreover , it can beseen that the maximum value of shear stress does not occur in

the mid plane0

zh

= ,the reason of this case belongs to thenature of non-homogenous of the mechanical properties of theFGMs.

Figure 3: Shear stress distribution through thikness forclamped circular plate (a) h a 0.1, p 0.5 , (b)

h a 0.1, p 2 , (c) h a 0.2,p 0.5 ,(d) h a 0.2,p 2


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6.2 Roller Support Circular Disk

Fig (4) shows the radial stress parameter

* r

r c E

=

with the

body force parameter( )

, for different values of the materialconstantp.it is clear that the redial stress parameter varyinglinearly with the body force parameter , moreover the radialstress in the FG disk higher than that of pure metallic disk( )0P = .

Fig (5) illustrates the variation of the radial stress parameter

with R At0.25

zh

= and0.25

zh

= .It is clear thatunder uniform distributed load and body force , the radialstresses for FG plate higher than thatofthe pure metal plate( )0P = , because of the FG plate has a higher density.

Fig (6) present the dimensionless deflection

wh

with ( ) R of

the disk under bending load with different values of materialconstant p .it is clear that,the deflection of the pure metal disk(p=0) higher than that of FG plate ,because of the rigidty ofFG plate.

Fig (7) shows the distribution of shear stress parameter (

* )through the thickness of the plate for different value of

thickness to radial variation ( ha ) and material constant P .It is

clear that the behaviors of the shear stress through thethickness of plate in this case of roller support conditionsimilar to the case of the clamped boundary condition .

Figure 4: Radial stress distribution body force ( ) for rollersupport circular disk for different value of p.

Figure 5: Radial stress configralion with (R) ath

0.25 for roller support circular disk for different value of p.

Figure 6: Radial stress configralion with (R) ath

0.25 for roller support circular disk for different value of p.


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Figure 7: Shear stress distribution through thikness for rollersupport circular plate (a) h a 0.1, p 0.5 , (b)

h a 0.1, p 2 , (c) h a 0.2,p 0.5 ,(d)h a 0.2, p 2

Figure 8: Bending distribution ( h ) for roller support circularplate with (R )fordifferent value of p.

CONCLUSIONS

An axismmatric bending and stress analysis of functionallygraded circular plate under uniform body force and uniformdistributed load by unconstrateed third order sheardeformation theory via generalized differential quadraturemethod (DQM) the numerical solution of the unconstrateedthird order shear deformation theory can be applied todifferent case , of boundary condition , as well as , it can beapplied to different loading condition , in contrast to theanalytical solution limited to bending load .

REFERENCES

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BIOGRAPHIESF. Tarlochan was born in Malaysia. Heobtained his Bachelors in MechanicalEngineering and Masters in BiomedicalEngineering from Purdue University,USA. His PhD was from Universiti PutraMalaysia. He is currently an AssociateProfessor at UNITEN and heads the

Center for Innovation and Design.Email:[email protected]

Hamad M. H was born in Iraq Heobtained his Bachelors in Mechanical

Engineering and Masters in Engineeringfrom IRAQ. He is currently pursuing hisPhD at UNITEN in the field of appliedmechanics.

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An Axisymmetric Bending and Shear Stress Analysis of of Functionally Graded Circular Plate Based on...

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