Stress Distribuion on Open-Endedcarbon Nanotubes - Draft-70251

8/3/2019 Stress Distribuion on Open-Endedcarbon Nanotubes - Draft-70251

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Proceedings of MicroNano08MicroNano2008

June 3-5, 2008, Clear Water Bay, Kowloon, Hong Kong

1 Copyright 2008 by ASME

DRAFT-70251

STRESS DISTRIBUION ON OPEN-ENDEDCARBON NANOTUBES

Kasra MomeniSharif University, Department of Mechanical Engineering of Technology, Azadi St., Tehran 1458889694, IRAN, Tel: +98

912 2004922, [email protected]

Aria Alasty

Sharif University, Department of Mechanical Engineering of Technology, Azadi St., Tehran 1458889694, IRAN, Tel: +9821 6616 5504, Fax: +98 21 6600 0021, [email protected]

ABSTRACT: The study of the stress distribution on open-ended Carbon Nanotubes (CNTs) placed in a composite

material is considered in this work and an analytical solution

for the stress distribution has been constructed. Finally some

parameters such as CNTs thickness and CNTs length are

considered and their effects on the distribution of stress have

been investigated. For finding the governing relations,

continuity equations for the axisymmetric problem in

cylindrical coordinate (r,,z) are considered. Then by

assuming some conditions and solving the governing

differential equations, using constitutive equations and

applying the boundary conditions, an equation relates the

stress applied to the representative volume element with the

stress distribution on the CNT has been found.

INTRODUCTIONAfter the discovery of CNTs by Iijima et al.[1] in 1991, many

researches have been done on this new carbon structure.

While most of the researches are on the production methods

[2-7], many of them are also on its mechanical properties.

These investigations show that CNTs have superior

mechanical properties compared to most of the otherstructures. This is due to the sp2 bounds in CNTs which have

a high bound strength [8].

Theoretical calculations and experimental measurements show

that the elastic module of CNTs in the range of 1-5 TPa [9-

11], which is significantly higher than that of a carbon fiber,

that is from 0.1 to 0.8 TPa [12]. The measured tensile

strength of CNTs is about 60 GPa [13;14]which is

significantly higher than carbon fibers which is about 4 GPa

[12]. Such superior mechanical properties make CNTs a

promising material to act as the reinforcement phase o

composite materials.

In recent years also a great effort was done in order to

implement CNTs as the reinforcement phase of

nanocomposites [15-19]. As in the case of conventiona

composite materials one of the key issues for determining the

strength of a composite material is the interfacial bonding

strength between the matrix and reinforcement phase

Different approaches were used for finding this bondingstrength from continuum models [20-23] to molecular

dynamics simulation [24] and experimental measurements

[25-28]. Many researches are also done in order to find the

stress distribution on CNTs placed in the matrix of composite

materials and mechanical behavior of CNTs embedded in the

matrix [29-31].

It has been noticed that despite the efforts for analyzing the

production process of open-ended CNTs [32-34], n

analytical work on the reinforcement effect of an open-ended

CNT embedded in a composite material was reported till now

Therefore, in this paper an open-ended CNT has been

considered as a reinforcement phase embedded in matrixwhere an axial load is applied at the ends of compound

medium.

GEOMETRY AND ASSUMPTIONSFig. 1 illustrates the geometry of the compound structure

under consideration for further analysis. It comprises of an

isotropic external cylinder with length of 2L and radius R

plus an open-ended CNT which is assumed isotropic with


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length of 2Lf and inner and outer radii of ri and ro,

respectively. As it is seen, the open-ended CNT is embedded

within the matrix and a perfect bounding is established

between them. For analysis of this problem a representative

volume element (RVE) of such structure will be considered.

In order to analyze this problem following assumptions are

made:

1. Both matrix and CNT are made of isotropic but

different materials.2. A perfect bonding between CNT and matrix exists.3. Radial strain is much smaller than the axial strain.

r

w

z

u

pp

.

4. The applied axial force will be transferred to theopen-ended CNT via surrounding matrix.

5. The cross section area of the CNT is smaller than theRVEs or in other words the volume fraction of the

fiber is much less than the volume fraction of the

matrix.

6. The induced radial and hoop stresses are much

smaller than the axial stress +rr


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( )

( )

( )

1

1

1

rr rr zz

zz rr

zz zz rr

rzrz

E

E

E

G

= +

= +

= +

=

(3-a)

(3-b)

(3-c)

(3-d)

in whichE, and G are the Youngs modulus, Poissons ratio

and shear modulus, respectively.

Referred to Fig. 1, the governing boundary conditions for the

reinforced and pure matrix are as follows:

| 0,m r Rt = =rr

3|m z L et = =

r

(4-a)

(4-b)

where the interfacial traction continuity conditions are:

, ,| | f f o f f of m

L z L r r L z L r r t t < < = < < ==r r

, ,| | f i o f i of m

L r r r z L r r r t t= < < = <


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Now by integrating the Eq. (1-b) over the cross sectional area

with respect to rfrom ro toR and using Eq. (15-b) one would

get:

f

o

o

o

m

zz

rR

r

dz

d

22

2

= (17)

where,

( )( )

2 2( ) ( , )

12

o

Rm mzz zz

ro

z r z r drR r

=

(18)

A close inspection of Eq.(18) indicates that them

zz will be afunction of z, moreover we have;

)(zgz

m

zz =

(19)

Were g(z) is an unknown function that must be determined.

By substituting Eq. (19) in (1-b) and integrating over the

cross-section with respect to rform ro toR leads to:

2 2

2 2

( )( ) ( )

2

2

m m

rz rz

zz r

g R r g r

R r r

= =

(20)

After combining Eq. (20) and Eq. (5-b), we have:

2 2( )

2 foo

o

zr

gR r

=

(21)

After back substitution ofg(z) in Eq. (20),

2 2

2 2( )

m forz o

o

rr R r

R r r

=

(22)

In view of assumption III i.e.r

w

z

u

pp and Eqs. (2-d)

and (3-d),

,f

f f

rz

wG

r

=

mm m

rz

wG

r

=

(23-a)

(23-b)

Using Eqs. (23) and Eq. (22) we have,

2 2

2 2

mf m oo

o

R r r wG

R r r r

=

(24)

After integrating Eq. (24) over the cross-section with respec

to rfrom rotoR, one would get:

( ) ( )

2 2

2 2 2ln 1 2

o

m m

R rf m oo

o o o

w wR rG

r R R r R r

=

(25)

Now we substitute fo from Eq.(25) into Eq. (22) which

leads to,

( ) ( )

2 2

2 2 2( )

ln 1 2

o

m m

R rm m

rz

o o

rw wR r

Gr R R r R r

=

(26

By back substitution ofm

rz from Eq. (26) into Eq. (23-b) andcarrying out the integration, it results in:

( )( )

( )( )( ) ( )

2 2 2

2 2 2( , )

ln 1 2

ln 1 2o

o

m m

R r o om mr r

o o

r z

w w R r r r r w w

R R r R r

= +

(27

By recalling assumption VI, i.e. zzrr pp+ for bothmatrix and the fiber and using Eqs. (2-c) and (3-c) we have:

,z

wE

f

ff

zz

=

z

wE

m

mm

zz

=

(28-a

(28-b

Substituting Eq. (27) into Eqs. (28-b) results in:

( ) ( )( ) ( )

[ ]oo rr

m

zzRr

m

zz

oo

oo

rr

m

zz

m

zzrRrRR

rrrrR===

+= ||

21ln

21ln|

222

222

(29)

Now, consider the force balance over the cross-section of the

composite alongzaxis:

drrdrrRo

i o

r

r

R

r

m

zz

f

zz)2()2(2 += (30)

Using Eqs. (7), (29) and (30) we have:

( ) ( )( ) ( ) ( )

( ) ( )

++

+=

=

==

22222

22222

222

|

341ln

21ln||

RrRrr

rRrRrRR

rRrRR

orr

m

zzoi

f

zz

ooo

oorr

m

zzRr

m

zz

o

o

(31)

From Eqs. (9), (25), (28) and (31) it follows that,


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( ) ( )( ) ( )( )22224

22222

22

22

2

2

341ln

|

1

1

ooo

or

m

zzoi

f

zz

moi

o

f

zz

rRrRrRR

RrRrr

rr

rR

dz

d

o

++

+

=

(32)

In view of assumption II, i.e. perfect bounding between two

media:

oooo rr

f

zzf

m

rr

m

zrr

f

zzrr

m

zzE

E==== == |||| (33)

In view of assumption V, i.e. low volume fraction of the fiber,

f

zzf

m

rr

m

zz

f

zzf

zzE

Eo

= = | (34)

From Eq. (34) and Eq. (32) it follows that:

( ) ( )

( ) ( )( )22224

22222

22

22

2

2

341ln

1

1

ooo

of

m

oi

f

zz

moi

o

f

zz

rRrRrRR

RRrE

Err

rr

rR

dz

d

+

+

+

=

(35)

By considering as follows,

( ) ( )( )22224

22

22

341ln

1

1

1

ooo

moi

o

rRrRrRR

rr

rR

+

=

(36)

Eq. (35) becomes:

( ) ( ) 22222

2

2

RRrEErr

dzd fzzof

m

oi

f

zz=

+ (37)

Eq. (37) represents an ordinary differential equation with

constant coefficients and has a solution as follows:

( ) ( )

( ) ( )

( ) ( )2222

2

2222

2222

RrE

E

rr

R

zRrE

ErrchB

zRrE

ErrshA

of

m

oi

of

m

oi

of

m

oi

f

zz

+

+

+

+

+=

(38)

Putting Eq. (38) into Eq. (9) results in,

( ) ( )

( ) ( )

( ) ( )

( ) ( )( )

2 2 2 2

2 2 2 2

2 2 2 2

2 2 2 2

2 2

2

mf

zz i o of

m

i o of

m

i o of

m f

o oi o of

i o

E A r r r R

E

Ech r r r R z

E

E B r r r R

E

E r sh r r r R z

E r r

= +

+ +

+

+ =

(39)

Eq. (39) can be put in the following form as:

( ) ( )2 2

2 2

2 1

2

f

o o i o

i o o

r r r A ch z B sh z

r r r

= +

(40)

in which is,

( ) ( ) +=2222 Rr

EErr of

m

oi (41)

Using Eqs. (31), (34) and (38) in (29) result in,

( ) ( )( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( )

2 2 2

2

2 2 2 2 2

2 2 2 2 2 2 2

2 2 2 2 2

2

2 2 2 2

ln 1 2

ln 1 4 3

ln 1 2

ln 1 4 3

o of

zz

o o o

m

o o o i ofm

f

o o o

m

o i of

R r r r r R

R R r R r R r

ER R r R r r r R r

EE

E R R r R r R r

R A sh z B ch z E

r r R r E

= +

+

+ +

+

(42)

Putting Eq. (40) in Eq. (14-b) results in,

( ) ( ){ }2

21

2

f irz

rrch z B sh z

r

= +

(43


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Finally, the use of Eq. (40) in (22) gives

( )( ) ( )

2 2 2

2 22

m i orz

o

r r Rr A ch z B sh z

rR r

= +

(44)

Eqs. (38), (40) and (42) to (44) represent some expressions for

unknowns stresses of , ,f f mzz rz zz and

m

rz . The only

remaining thing is specifying the constants A and B in theserelations. To calculate the value of these two constant

coefficients the pure matrix region must be considered. That is

a matrix with an embedded solid virtual CNT with zero inner

radius and outer radius equal to the outer radius of the actual

open-ended CNT in the reinforced region.

Solution in Pure Matrix Region

Equation (38) for this region will be as follows:

( ) ( )( )

( ) ( ) ( )

2 2 2

2 4 2 2 2 2

2 2 2

2 4 2 2 2 2

1

1 ln 1 4 3

1

1 ln 1 4 3

mo

zz

o m o o o

o

o m o o o

R r R A sh z

r R R r R r R r

R r R B ch z r R R r R r R r

= + +

+ +

(45)

A'andB'are constants, the same asA andB in Eq. (38) but in

the pure matrix region.

By substituting Eq. (45) into Eq. (9) and the consequent result

into Eq.(14-b) one would get:

( ) ( )( )2

fm

rz

r A ch z B sh z = + (46)

in which:

( ) ( )( )222242

2

22

341ln1

1

ooomo

o

rRrRrRR

R

r

rR

+

=

(47)

In order to calculate the unknown coefficients A and B we

will apply the boundary conditions which is listed in Eqs. (4-

b) and (5-b). After doing so and some further calculations,A

andBwill be obtained as followings:

( ) ( ) ( )

2

2 2 2 2

' 0, ' 1m

fo i of

RA B

Ech Lr r R r

E

= = +

(48)

Substituting the values ofAandBin equations (38), (40) and

(42-44), the following relations will result in:

( ) ( )

( )

( ) ( ) ( )

( ) ( ) ( )( )

( ) ( )( ) ( ) ( )

2 2

2 2 2 2 2 2 2 2

2 2 2

2 2 2 2

2 2 2

2

2 2 2 2 2

1

12

ln 1 2

ln 1 4 3

f

zz m m

fo i o o i of f

f i oo m

o fo i of

o om

zz

o o o

m

ch zR R

E Ech Lr r R r r r R r

E E

r r R sh z

Er ch Lr r R r

E

R r r r r R

R R r R r R r

E

E

= + + +

= +

= +

( ) ( ) ( ) ( )

( ) ( ) ( )

( )

( ) ( ) ( ) ( ) ( )

2 2 2 2 2 2 2

2 2 2 2 2

2 2

2 2 2 2 2 2 2 2

ln 1 2

ln 1 4 3

1

m

o o o i of

f

o o o

m m

fo i of o i of

ER R r R r r r R r

E

R R r R r R r

ch z R R

Ech L Er r R r r r R r

E E

+

+ + +

( )

( ) ( ) ( )

2 2

22 2 2 2

1 12

f irz m

fo i of

sh z rr R

Er ch Lr r R r

E

= +

( ) ( )( ) ( ) ( )

2 2 2 2

2 22 2 2 2

12

m i orz m

foo i of

sh z r r R Rr

Er ch LR rr r R r

E

= +

(49)

In order to verify the validity of the above relations if the

inner radius of the open-ended CNT, ri, approach zero, which

corresponds to a rod shape fiber, the formulas will be the same

as the ones for the case of rod shape cylinder obtained by Gao

and Li [39].

NUMERICAL SOLUTION AND CASE STUDYBased on derived relations, for an open-ended CNT placed in

the epoxy matrix and under an axial constant traction, the

shear stress distribution in the medium is calculated and theresults are represented in different graphs. The controlling

parameters considered are taken to be the same as the on

used in Gao and Li [40]. Further more we assumed that the

CNT of interest is single wall, therefore we considered the

following parameters, Ef = 1000 GPa, Em = 2.41 GPa, vm =

0.35, ro=0.471nm, ri=0, 0.131nm, R= 5ro, and aspect ratios

Lf/ro=12.8, 10.1 and 7.8.

When we considerri =0 the problem would become the same

as when the CNT is closed-end. The results for this case are

shown in Figs. 4-5, which are the same as the results reported

by [41]. Consideringri =0.131, the mean stress in the open

ended CNT and the shear stress at the interface have been

shown in Figs. 6-7. Comparing the maximum mean stress inthe fiber, which occurs at the middle of the open-ended CNT

in Figs. 4 and 6 we will find that the maximum mean stress is

greater for the open-ended CNT compared with closed-end

CNT. It is plausible due to the fact that open-ended CNT has

a lower cross-section area compared with a closed-end CNT

of same outer diameter, and the stress is inversely promotiona

to the cross-section area.


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As it has been shown in Figs. 4 and 6, the maximum mean

axial stress occurs in the middle of the CNT and its maximum

increases by increasing its length.

FIG. 4 DISTRIBUTION OF NON-DIMENSIONAL MEANSTRESS VS. AXIAL POSITION OF CNT

NONDIMENSIONALIZED BY ITS DIAMETER, RO=0.

FIG. 5 NON-DIMENSIONAL SHEAR STRESSDISTRIBUTION VS. FIBER LENGTH ON THE HOLLOW

CARBON FIBER (LF = 0.01M, RO = 0.03CM)





The results show that as in the case of closed-end CNT the

maximum shear stress occurs at the extremes of the open-

ended CNTs while it will vanish in the middle due to

symmetry. Also the longer fibers have higher stress

transmission capability compared with the shorter ones, whichmeans the larger the open-ended CNTs aspect ratio, the

higher is the shear stress that the CNT can reach. As it is seen

in these figures, increasing the length of fiber will induce the

shear stress distribution becomes a local phenomenon i.e. as

the fiber length increases the shear jump will represent itsel

more locally towards the end of fiber length.

Comparing the maximum mean stress in the open-ended CNT

and closed-end CNT we will find that in all the cases the


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maximum mean stress in the open-ended CNT is higher than

the one in the close-ended one by a factor of about 1.08. In

addition by comparing the maximum shear stress of the open-

ended CNTs with the closed-ended CNTs we will find that the

maximum shear stresses of the open-ended CNTs are higher

than the close-ended ones by a factor of about 1.0008. This

value drops for shorter nanotubes to the factor of 1.0004.

CONCLUSIONAs it can be seen in the presented results the ability of theopen-ended CNT to stress transfer will be improved by

increasing its aspect ratio which is defined as the ratio of its

outer diameter and length. The shear stress transfer will also

increase by increasing the nanotubes aspect ratio as it has been

represented in Figs 5 and 7. It is also expected that the stress

transfer increase by reducing the inner diameter of the hollow

fiber because the shear stress is proportional to

the ( ) 222 oio rrr . It is also expected that the reduction in theinner radius of the fiber to have higher effect compared with

increasing the outer radius due to the proportionality of the

transferred stress to the

2

ir compared with its proportionalityto or .

Due to the above statements, it is expected that a rod-shaped

fiber has higher stress transferability compared with a hollow

cylindrical shaped fiber. Although it might be true for the

case of which the matrix does not penetrate inside the hollow

part but if it does it can cause increasing in the stress

transferability of the fiber due to increasing in the area of the

contact between the fiber and matrix which in this case can be

nearly twice the case of no matrix penetration. Therefore to

have a rational comparison between a rod-shaped fiber and a

hollow cylindrical fiber which the matrix is penetrated inside

its hollow part, further investigations needs to be done.

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