Issue 60

C. O. Bulut et al., Frattura ed Integrità Strutturale, 60 (2022) 114-133; DOI: 10.3221/IGF-ESIS.60.09

Once the values from the Eqn. (10) are applied into Eqn. (8), Eqn. (11) is obtained

Mg A

Mg B

Mg C

Mg D

*

*

*

*

i

i

i

1 2 Ω i

1

1

1

A

B

C

D

,

,

,

(11)

i

i

i

i

1

1

1

1

 2 i

2

 2 i

2

 2 i

2

 2 i

m

m

m

m

i

i

i

i

The sum of both solution expresses the general solution of Eqn. (6) for 1 L V

. It can be provided as:

         sin cos t t

 

 ) ( N N 1 i h

N t

N

(

)

(

)

(12)

i

i

p

i

i

i

p

1

1

2

1

  0 i i N N   1

 

To be able to calculate  1 and  2 , initial states (

 0 0)

are used, and constant terms are calculated as

1

 

* A C

  *

i

i

i

1

1

* B D

*

 1

 2

 

,

(13)

i

i

1

1

 i

*

*

*

*

Since   , ,

1 2 1 1 i i i i A B C D have been found, Eqn. (12) can be stated as 1 1 , , ,

  t      sin[ ] cos t

 

 

 

*

*

*

*

  

 



 

  ]

1 N t i

N t

sin[ A t

B

t C

t D

t

cos

sinh[

cosh[

(14)

j

i

i

i

i

i

i

i

i

i

i

1

1

2

1

1

1

1

¨ i N can be found.

 , i i N N and

Using Eqn. (14),

  i N t in Eqn. (14) is applied into Eqn. (6) for beam subpart  1 t L V

1 i R

Once

of Eqn. (8), we get

   

   

  

           

   

  

  

   

İ

İ

İ

  M N N S vt g    2 ¨ 2 i

¨

  v N S vt   1 1 j ' j

 

 

2

''

1 1 N S vt j

v

1 1 j N S vt j

R

2

(15)

i

i

i

j

i

1

mL

  

j

j

j

1

1

1

Eqn. (14) is inserted into the right side of Eqn. (15), Eqn. (16) is found for  1 t L V ,   ¨ 2 i i i N N = 1 i R Eqn. (16) can be defined by taking the initial conditions into consideration as follows For the first beam segment,  1 t L v

(16)

L v

1

1

 

  

        t d i

 0

N t

R

sin

(17)

 

i

i

1

 i

For the second and third beam segments, the same is embodied and Eqn. (18) and Eqn. (19) are determined as follows. It should be noted that initial conditons are not zero now. For the second beam part,   1 2 L t L v v

L

L

1

2

v

v

1

1

 

 

  

        t d i

 0

      sin[ ] t d

N t

R

R

sin

 

(18)

i

i

i

i

1

2

 i

i L v

1

  2 L t L v v

For the third beam part,

119

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