Issue 74

E. Sharaf et alii, Fracture and Structural Integrity, 74 (2025) 262-293; DOI: 10.3221/IGF-ESIS.74.17

To perform this task, return again to the two-storey frame shown in Fig. 3, but the mass of the second floor is set by m 2 , and the mass of the first floor is omitted; then the structure mass matrix takes the form:

0 0 0 m

 

  

 

M

(18)

2

Again, by substituting into the eigen equation:

2

2 eq K M   

0

(19)

Then the equivalent mass is expressed as:

K

eq



(20)

m

eq

2  eq

2

To get the mass contribution factor, apply the following equation:

m K

eq

eq

 

(21)

f

0

2

m m 

2

2 2 eq

By substituting into Eqn. (21) for α =I b /I c and β 2 =L/h 2 , obtaining:

2 4 h    4

2 3 h h 1 2

4

3 h h 2 1 2

2 2 2 1 2 h h

3 2 1 2 h h

2 3 2 1 2 h h

2 2 2 2 1 2 h h

2 2 2 1 2 h h

2 4 2 2 h



2 1 h    4







2 1 h 

















(4

3

6

8

3

3

)

1



f

0 (22) To simplify calculations, the formula can be modified using h 1 =n ⋅ h 2 in Eqn. (22), where h 2 is the floor height, .h 1 is the height at which the mass has accumulated from the previous floor, and n is the floor number to which the mass was moved to simplify the calculation. 3 ( (4 h 2 2   1 1 2 2 2 2 2      4 )) h h h

2 4 n n                    2 4 3 2 2 3 2 2 2 2 2 3 (n (4 2 2 4 3 4 6 8 4 2 n)) n n n n n n

2 2 n

2   2 n

2 

3  

(4

3

)

2

2



f

(23)

0

2

2

2 





The following equation is obtained by replacing

, in Eqn. (23):



4

4

3 2 3      n   n   2 4 6 8 n n

2 2

2   3 n 

2

(4 4 3* n n  







n

n

3

)



f

(24)

0

3

2

( (4 4 n 

n    

2 ) n 

A simplified form of Equation (22) can be obtained by setting the stiffness ratio α =1, which corresponds to the case where the beam and column sections are identical. This assumption leads to the following simplified expression:

2 3 2 1 2 h h

2 2 2 2 1 2 h h

2 3 2 1 2 h h

2 4 2 2 h    4 2 1 h 3  

3 h h 2 1 2

2 2 2 1 2 h h

3 4   2 1 2 h h h h h  1

3



















(

3

3

4

6 h

8

1 2 4 4 )



f

(25)

0

3 h h 1 1 ( (4 2

2

2 1 h    2 2 h

4   

))

2 2

By substituting in Eq. (24) with h 1 =n*h 2

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