Understanding the concepts on Degrees of Freedom
1) MASS MATRIX =(m 0 0 0 0)
(0 m 0 0 0)
(0 0 m 0 0)
(0 0 0 m 0)
(0 0 0 0 m)
Modified mass matrix (M) = CHIn^T *m*CHIn
For the first mode = (0.334 0.641 0.895 0.078 1.173) ** Mass matrix * (0.334 0.641 0.895 0.078 1.173)^t
(n=1)
M1=3.861m
Modal property (L) = CHIn^T*m*L
Where L = influence vector
Assuming 100% utilisation of seismic force =[1]
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For the first mode=[.334 .641 .895 .078 1.173] *mass matrix *Influence vector (n=1)
L1=4.121m
Modal participation factor (F)=L/M
for the first mode F=L1/M1
(n=1)
F1 = 4.121m/3.861m
F1=1.067
similarly , F2=0.335
F3=0.177
F4=0.096
F5=0.045
Effective Modal mass (M*)=Ln^2/Mn
For the first mode,M1*=L1*/M1
(n=1)
M1*=(4.121m/3.861m)
M1*=4.398m
SIMILARLY M2*=0.436m
M3*=0.121m
M4*=0.038m
M5*=0.008m
Modal mass participation ratio=Mn*/mj
where ,mj=Total mass of the building =m+m+m+m+m=5m
Modal mass participation ratio of first mode (n=1) =4.398m/5m=.88
similarly , second mode (n=2)=0.09
Third mode (n=3) = 0.02
Fourth mode (n=4) = 0.01
Adding first two mode mass participation ratios,we get 0.88+0.09=0.97 To achieve minimum 90% of model mass participation ratio, first two modes are enough
Base shear (Vb)=Mn*An
For the first mode , Vb1 =4.398 * 0.27g
(n=1) Vb1=1.187mg
similarly , Vb2=0.327mg
Vb3=0.126mg
Vb4=0.039mg
Vb5=0.008mg
Average Base shear is calculated by the method of squareroot of sum of square
=`(Vb1)^2+(Vb2)^2+(Vb3)^2+(Vb4)^2+(Vb5)^2`
Vb=1.15mg
m=mass g=gravity due to acceleration