Issue34

M. Ševčík et al, Frattura ed Integrità Strutturale, 34 (2015) 216-225; DOI: 10.3221/IGF-ESIS.34.23

dx EI M W o 2

2 1

 

(4)

where M o

is bending moment in the segment, E I

is bending stiffness of the segment. The MMB specimen can be virtually

divided into three segments, see Fig. 3.

Figure 4: Scheme of location of forces P u

and P d .

Figure 3: Virtual division of the MMB specimen into three segments.

Each of these three segments has specific bending stiffness that corresponds to the stacking sequence of the segment. For the MMB specimen the strain energy stored in the system can be calculated as follows.

2

2

2

  

  

    d 2

  

  

   IE xP 33 3

  

  

xP P

 

L xP xP

d

d

 xP 1

u

2

L

 L

a

a

2

2

d

4

4

2 1

2 1

2 1

2 1

2

2

 0

 0

 L

W

dx

dx

dx

dx

(5)

u

1

2

3

4

IE 11

IE 22

IE 33

a

where E 1,2,3

are bending moduli of each segment calculated by classical laminate theory, P u and P d

are components of the

loading force P , see Fig. 4, that can be calculated from equations [21]:

g L c

L c

P  

P

P

(6)

u

g

g L c

  

  

L c

 

     1

P  

P

P

1

(7)

d

g

where c , c g , L are dimensions shown in Fig. 2, Pg is gravity force induced by weight of the loading lever. The displacement of the loading lever below the force P is then:         3 3 3 3 2 Pc LL a P Pc La cPa  

u   

12 2



u

u

d

d

(8)

P

11 ILE

22 I LE

33 I LE

3

12

where E i I i ( i = 1,2,3) are bending stiffnesses of the i-th segment of the MMB specimen. As soon as the displacement of the lever below the force P is known it is possible to calculate the compliance of the MMB specimen using following relation:

P u C P  .

(9)

Finally, the total strain energy release rate can be calculated using Eq. (2) in the following form:

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