PSI - Issue 18

Yaroslav Dubyk et al. / Procedia Structural Integrity 18 (2019) 622–629 Yaroslav Dubyk and Iryna Seliverstova / Structural Integrity Procedia 00 (2019) 000–000

625

4



  



  

1

1

mR

n





  

  

2

2

1 2

1 2 1

1 2

l n

l n



  

 1      

u C n  mn

v

C









 

mn

3

R m 

l

m Rl 

R

(13)



  

n



  

  

2 l n l n 

3

1 2

mn



w mn

0

C











3 R m m Rl R  

Rl



  





1

1     n

m

  



  

3

1 2

1 2

l n

n l





u C C   mn

v

n m

 











mn

3

Rl

Rl

m

Rl

m R

R









      

      

2

1    

mn

 

  

2 R n m 

(14)

3 3 m

2 





  

3

Rl

Rl

l

2 n Rl

2 n R m 

w mn

C

N

N











xx



m

l



4 2

  

  

1

n l

l



 1  

  

2 n m

2

mn

 







2 R m 

Rl

R mn

Thus, the problem is reduced to the solution of the algebraic system of equations (12)-(14) with respect to the unknown coefficients , , u v w mn mn mn C C C . Finding these unknowns allows us to determine all the force resultants, stresses and deformations. The solutions (12)-(14) are quite simple and allow us to obtain a number of practical results, in particular, we can analyze the influence of nonlinearity on force parameters of dents. It should be noted, that the solution for a harmonic dent is the basic for constructing the solution for a single dent by means of decomposition into double Fourier series. 2.2. Single dent Finding the coefficients of the double row and using solutions for a harmonic dent, it is easy to find a solution for a single dent by adding. For a dent shape from Eq. (2), the Fourier coefficients can be found from the integrals:

2

2

2

   

       

exp        

   

x 

l



0   x     x

0        

0   x     x

1 1

1 2

1 2

 

, 2

m n

1 exp



p

N

dxd











(15)

xx

x

l



0

l

  

   

       

exp        

   

2

2

2



l



0        

0        

0   x     x

1 1

1 2

1 2

 

, 2  0 m n

1 exp



p

N

dxd











(16)





l



l

  

Finding the coefficients of the double Fourier series by Eqs. (15) and (16) and together with the solution for harmonic dent, it is easily to find a solution for a single dent, using coefficient summation:   cos cos u v w x mn mn mn n m H m N C Rm C l n C l n x lR l           (17)   cos cos u v w mn mn mn n m H m N C Rm C ln C l n x lR l           (18)





D

m



n m 





 

2

2

2 w C Rm C ln C l n n      w v

cos cos 

M

x

 

(19)

x

mn

mn

mn

2 2

l R

l