Understanding the concepts on Degrees of Freedom
1)
Different mode shape values for different Time periods
Find :- Mn,Ln,Γn,Mn*,Vb
Formation of matrix :
Find ;
For finding Mn we have to find M1, M2, M3, M4, M5
Hence:
- M1 = 3.861m
- M2 = 3.861m
- M3 = 3.861m
- M4 = 3.861m
- M5 = 3.861m
Second we have to find Ln:
L1= 4.121m
Similarly
L2=−1.29m
L3=0.68m
L4=0.381m
L5=0.175m
Find 
Hence :
Γ1=1.067
Γ2=−0.336
Γ3=0.177
Γ4=0.099
Γ5=0.045
Find 
Participation Ratio = [M*n/Mi] Mi = Total mass, m1+m2+m3+m4+m5 = 5
For M*n We have to find (M1*, M2*, M3*, M4*, M5*)
M1* = [ L²1 / M1 ]
M*1 = 4.121² / 3.861
= 4.398 m P ratio = 0.8796
Similarly,
M*2 = 0.435m P ratio = 0.087
M*3 = 0.121m P ratio = 0.024
M*4 = 0.037m P ratio = 0.0074
M*5 = 0.008m P ratio = 0.0016
From the above participation factor ratio of model mass, we can understand that the sum of Model1 and Model2 are enough to obtain minimum 90% of the participation.
Hence; Modal-1 + Model-2 > 90%
0.8796 + 0.087 = 0.966 = 96%
Finally, obtain the base shear for each of the 5 modes of vibration and the combined base shear as per SRSS rule. The pseudo acceleration design spectrum to obtain the base shear for each mode is provided below.
T1=2, A1=0.27
T2=0.6852, A2= 0.76
T3=0.43446, A3= 1.03
T4=0.3383, A4=1.03
T5=0.2966, A5=1.03
For Vb ;
There are two Modes are to be considered such that a modal mass participating ratio of 90% is obtained
Modal-1 + Model-2 > 90%
0.8796 + 0.087 = 0.966 = 96%