Modeling Material Microstructure - Continuum Mechanics of Anisotropic Materials - page 185

8 G m f c
1

2 G eff 2 G eff

13 ¼ 23 ¼

2 G m

þ n m ;

then from 2 G m ¼ E m /(1 þ n m ) it follows that

E m ð 1 þ n m 4 f c Þ

2 G eff

13 ¼ 2 G eff

23 ¼

:

2

ð 1 þ n m Þ

E eff
1

From the second formula of ( 7.26 ) one can see that E m ¼

1 3 f c , thus,

E eff
1

ð 1 þ n m 4 f c Þ

2 G eff
13

2 G eff
23

¼ ¼

2 ;

ð 1 3 f c Þð 1 þ n m Þ

or to the neglect of squared terms in f c ,

E eff
1

2 G eff
13

2 G eff
23

¼ ¼

2 ð

þ n m ð n m Þf c Þ:

1 3 1

ð 1 þ n m Þ

eff
12

eff
21

From the fourth formula of ( 7.26 ) one can write n ¼ n ¼ n m ð 3 n m 1 Þf c ,

thus obtaining the desired result

:

2 G eff 2 G eff E eff

1

eff
12

13 ¼ 23 ¼

1 þ n

In the special case when n m is 1/3, the expressions for the effective constants

( 7.26 ) simplify to

G eff
13

G eff
23

E eff
1

E eff
2

G m ¼ G m ¼ E m ¼ E m ¼

f c ;

1 3

(7.27)

E eff
3

eff
12

eff
21

eff
31

eff
32

eff
13

eff
23

n

n m ¼ n

n

n m ¼ n

n

n m ¼ n E m ¼

1 f c ;

n m ¼

1 ;

n m ¼

1 ;

n m ¼

1 2 f c :

From these results it is apparent that the Young's modulus and the shear modulus
in the transverse plane are more severely reduced than the axial Young's modulus
as the porosity increases. In the example below, the second of ( 7.27 ), E eff
3

= E m ¼ 1

f c , is constructed from a mechanics of materials argument.
Example Exercise 7.4.3

Problem : Recall the mechanics of materials formula for the deflection PL / AE

of a bar of cross-sectional area A , length L , and modulus E subjected to an axial
force P . Apply this formula to the bar with axially aligned cylindrical cavities
illustrated in Fig. 7.3 to show that E eff
3

d ¼

= E m ¼ 1 f c . The bar and the cylindrical

cavities are aligned in the x 3 direction.