Issue 54

T. I. J. Brito et alii, Frattura ed Integrità Strutturale, 54 (2020) 1-20; DOI: 10.3221/IGF-ESIS.54.01

being [ F 0 ] b the elastic flexibility matrix, expressed as follows [38]:

         

          

2

2

2

b U M M M N       b U

U

b

2   M

i

j

i

i

i

2

2

2

U

U

U

  0 F

b

b

b

2

b

 

 

M M

M N

M

i

j

j

i

j

2

2

2

b U M N M N N         b U U

b

2

i

i

j

i

i

2

4 sin 3 2 cos      

b 

3sin cos 

b 

R U

2

sin cos 

b 

b 

1 2

1 2

b

b

b

b

b

b

b

2

2

2

b 

b 

M

EI

R AE

sin

sin

i

b

b 

b   

sin cos 

b 

2 U M M b

R

2sin

sin cos 

b 

b 

  

1 2

1 2

b

b

b

2

2

b 

b 

EI

R AE

sin

sin

i

j

b

2

2

5sin 5 cos    

b 

b 

b 

 

3 2 cos  

b 

R U

cos

2

1 2

b

b

b

b

b

b

2

 

M N

b 

EI

sin

i

i

  

b 

sin sin cos  

b 

b 

b 

cos

1 2

b

b

2

b 

AE

sin

b b R U  

sin cos 

b 

2

in cos 

b 

b 

s

1 2

1 2

b

b j

b

2

2

2

b 

b 

M

EI

R AE

sin

sin

b

2

sin R U 

 

b 

b 

b   

sin cos 

b 

2

cos

1 2

b

b

b

b

2

 

M N

b 

sin sin sin cos EI  

j

i

  

b 

b 

b 

b 

cos

1 2

b

b

2

b 

AE

sin

(16)

3 R U 

2

b 

b 

b 

b 

b 

 

2 3sin cos  

b 

3sin

cos

cos

2

b

b

b

b

2

2

b   b

N

sin sin sin cos EI  

i

b   R

  

b 

b 

b 

c

os

b

b

2

b 

AE

sin

Deformation equivalence hypothesis For the sake of simplicity, consider initially a uniaxial problem where a damaged straight bar is axially loaded. By definition, the damage variable ω represents the micro-cracks density [11]. Then, such bar resists to the axial force by means of the effective stress i.e. the Cauchy stress σ divided by (1 – ω ). The following expression, named strain equivalence hypothesis [11], is given by substituting the effective stress in the Hooke’s law:   1 p E           (17) where E is the Young’s modulus, ε is the total strain and ε p is the plastic strain. The previous relation can be rewritten as follows:



p

p e        d

p

   

(18)

E E

E

1

1

6

Made with FlippingBook Ebook Creator