Issue 39

M. Shariati et alii, Frattura ed Integrità Strutturale, 39 (2017) 166-180; DOI: 10.3221/IGF-ESIS.39.17

We assume that the functionally graded cylinder is made of epoxy-phase and glass-phase. In this study, micromechanical models for conventional composites [13, 14] are employed to calculate the properties of functionally graded cylinder. According to these models, the shear modulus of rigidity ( P ) and bulk modulus ( K ) of FGM cylinder are obtained by the following equations.

i P

P

V

i

m

P P

(2-3)

m

P

P

V

m i m m m m K 9 8 P P

1

§ ¨ ¨ ©

· ¸ ¸ ¹

P

m

P

K

6

2

m m

V K K

i

i

m

K K

(2-4)

m

m i V K K K m m

m

1

4 / 3 P

In these equations subscript i and m refer to inclusion and matrix, respectively. Young’s modulus ( E ), Poisson’s ratio ( Q ) and density U of FGM cylinder are calculated as follows.

K 9 / 3 P P

E K

(2-5)

(2-6)

K K 3 2 / 2 3 Q P P

(2-7)

i i V V U U

U

m m

To incorporate these formulas into the extended finite element model, the value of each material property is computed at each individual node based on micromechanical models. Utilizing the generalized isoparametric graded finite elements, introduced by Kim and Paulino [15], material properties gradation is considered in an element. In this method, material properties such as elastic modulus ( E ), Poisson’s ratio ( Q ), and mass density ( U ) at Gauss points can be interpolated using shape functions from their nodal point values as follows [15]:

h

h

h

i i ¦ ¦ ¦ i i ,¬¬ Q U

E N E N ,¬¬ Q

i i N h U

ne ¬,¬¬¬¬¬¬ 1, 2, , ¬¬ }

(2-8)

i

i

i

1

1

1

G ENERAL PROBLEM FORMULATION

W

ith regard to the geometry of cracked cylinders and loading conditions, considered problems are axisymmetric. In this kind of problems, material properties and forces are constant in circumferential direction; also, displacement along ƨ-axis (u ƨ ) vanishes. The general governing equations are the equations of motion in

cylindrical coordinates, which are simplified as follows for axisymmetric problems [16].

w

z W

2

w

r V

r

w

u

rz

r

V

U

(3-1)

r

rf

r

T

r

2

w

w

r

w

t

2

w

W

r

w

z V

w

u

rz

z

U

r

rf

r

¬

(3-2)

z

2

w

w

r

z

w

t

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