Understanding the concepts on Degrees of Freedom

1. for modal propert

L1 = 4.12m

L2= -1.29m

L3 = 0.683m

L 4 = -0.386m

L5 = 0.174m

mass matrix =

M1= 3.86m

M2= -1.31m

M3 = 0.71m

M4 = 1.21m

M5=-0.07m

For modal participation factor

inverse vectoe 1 = 1.067

2= -0.335

3= 0.177

4= -0.098

5= 0.045

for effective modal mass

M1 = 4.398 m

M2 = 0.436 m

M3 = 0.12m

M4 = 0.038m

M5 = 0.008m

Total mass of the building is 5M

n1 = 0.88

n2 = 0.09

n3 = 0.02

n4 = 0.01

n5 = 0.002

To achieve minimum 90% of modal mass first two node are enough 0.88 + 0.09 = 0.97