Understanding the concepts on Degrees of Freedom

mass matrix M =`[[m,0,0],[0,m,0],[0,0,m]]`

modification of mass matrix M =`phin^t`.m.`phin`

for the first mode =[0.334 0.641 0.895 1.078 1.173] `[[m,0,0],[0,m,0],[0,0,m]]``[[0.334 ],[0.641],[0.895],[1.078],[1.173]]`

n=1

m1=3.861m

model property of L=`phin^t`.m.`iota`

where `iota`= influencing vector assuming 100% utilization of sesmic force=`[[1],[1],[1],[1],[1]]`

for the first mode =[0.334 0.641 0.895 1.078 1.173] `[[m,0,0],[0,m,0],[0,0,m]]``[[1],[1],[1],[1],[1]]`

n=1

L1=4.121m

model participation factor (`Gamma`)=L/m

for the first mode (`Gamma`)=L1/m1

n=1

`Gamma1`=4.121m/3.861m

`Gamma1`=1.067

similarly

`Gamma2`=-0.335`

`Gamma3`=0.177`

`Gamma4`=-0.098`

`Gamma5`=0.045`

effective model mass (M*)=`Ln^2/(mn)`

`or the first mode m1*= `(L1^2)/(m1)`

n=1

m1*`(4.121m)^2/(3.861m)`

m1*4.398m

similarly

m2*=0.436m

m3*=0.121m

m4*=0.038m

m5*=0.008m

model mass participation ratio =mn*/mj

where mj=total mass of the building

add the mass of each floor 'm'for the five floor building it is 5m

model mass participation ratio of first mode (n=1)=4.398m/5m

=0.88

similarly

2nd mode (n=2)=0.09

3rdd mode (n=3)=0.02

4th mode (n=4)=0.01

5th mode (n=5)=0.002

add first two modes mass participation ratios we get 0.88+0.09=0.97 to achive minimum 90% of model mass participation ratio two modes are enough

base shear Vb=Mn*.An

for the first mode Vb1=4.398m*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 Vb =`sqrt((Vb1^2)+(Vb2^2)+(Vb3^2)+(Vb4^2)+(Vb5^2))`

`sqrt((1.187mg)^2+(0.327mg)^2+(0.126mg)^2+(0.039mg)^2+(0.008mg)^2`

Vb=1.15mg