PSI - Issue 45

Oscar Zi Shao Ong et al. / Procedia Structural Integrity 45 (2023) 140–147 O.Z.S. Ong et al. / Structural Integrity Procedia 00 (2019) 000 – 000

142

3

  

 

,

11 E z

1 m CNT E E V z E     CNT

(4)

m

  V z V z E E   V z V z G G   2 22 CNT m CNT m   3 CNT m CNT m 12

 

,

22 E z

 

  

  

(5)

 

,

12 G z

 

  

  

(6)

   

22 E z E z 11

  v z v z    21 12

,

(7)

where  is the CNT efficiency parameter, which is dependent on the volume fraction of CNT. For * 1 2 0.11, 0.149, 0.934 CNT V      , * 1 2 0.14, 0.150, 0.941 CNT V      , * 1 2 0.17, 0.149, 1.381 CNT V      , and 2 3    (Zhong et al. 2018). The linear non-zero stresses of a FG CNT reinforced Kirchhoff plate are (Fu et al. 2016)

 

  

  

     

2

2

E

E

3 

3 

1 

2 

,

z

v

z

11

11

21

xx

2

2

1

1

12 21 v v x  

v v

y

x

y

12 21

(8)

 

  

  

  

  

2

2

2

E

E

3 

3 

3 

2 

1 

1 

2 

,

2

2

,

z

v

z

G

z

22

11

21

12

yy

xy

2

2

1

1

12 v v y   21

v v

x

y

x y x

y

x

  

12 21

where ε 1 , ε 2 and ε 3 are the axial, lateral and transverse displacements, respectively. The kinetic energy of the mass imperfection is then formulated

2

2

2

1 2

3     t                t t         1  2 

  x x y y dA     .

A 

K

M

(9)

0

0

mass

Utilising the Hamilton method, the equation of motions (axial, lateral and transverse) are obtained,

  

  

3

2

2

1          2 2 t

3 

3 

3 

2         y 

1        x  

2 2

2

3  

I

I

A

B

A

B

 





1

2

1

1

2

2

2

2

2

x

x t

x

y

 

(10)

  

2   3 

2    1 

  

  

1 

2 

2

2

2

) x x y y   ( ( 

)  

0,

D

D

M





 

 

 

1

2

0

0

2

y

y

x

x y

t

 

  

  

3

2

2         2 2  t

1

3 

3 

  

  

1 

2 

2

2 2

2  

2

I

I

D

D



 

1

2

1

2

2

x

y 

x

x y

y t

 

 

(11)

  

  

2

2

2        2  2 t 

3 

3 

2         y 

1        x  

2

2

2  

) x x y y    ( ( 

) 0 

,

M

C

B

C

B



0

0

1

1

2

2

2

2

y

y

x 1

3          2 2 t

4

4

 

  

  

 

3

3

2

1

3 

3 

1 

2 

I

I

I

1

2

3

2

2

2 2 2 x t   

2 2 2 y t   

x t

y t

 

 

 

     

  

3     2

2

2

3 

3 

2         y 

1         x 

x   y

3  

2  

4

2

) x x y   ( ( 

)

A

B

A

B

M

y

(12)

 

2

2

3

3

0

0

2

2

2

2

x

y

t

 

  

  

  

2

2

2

2

3 

3 

  

  

2   

1         x 

1 

2 

2

2  

2

2

4

0,

C

B

3 C B 

D

D





 

 

2

2

3

3

2

3

2

2

2

y

x y

y

x

x y

y

x

 

 

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