Understanding the concepts on Degrees of Freedom

Modal property (L)=Φn.m.l

where l=Influence vector (assuming 100% utilization of seismic force)

 For the 1st mode=

<!-- [if gte vml 1]> <![endif]--><!-- [if !vml]--><!--[endif]--><!-- [if gte vml 1]> <![endif]--><!-- [if !vml]--><!--[endif]-->                                                  M    0     0       0     0                1     

                                                   0     m    0       0      0              1

    [0.334  0.641  0.895             0     0     m     0       0              1

                     1.078  1.173]         0      0     0     m      0              1

                                                    0       0      0     0    m              1

<!-- [if gte vml 1]> <![endif]--><!-- [if !vml]--> <!--[endif]--><!-- [if gte vml 1]> <![endif]--><!-- [if !vml]-->
<!--[endif]-->                                                          

 n=1

L1=4.121M

Modal participation factor Γ =L/M

For the 1st mode Γ=L1/M1

n=1

Γ1=4.121m/3.861m

Γ1=1.067

Γ2=-0.335

Γ3=0.177

Γ4=-0.098

Γ5=0.045

Effective modal mass (M*)=Ln2/Mn

for the 1st mode M*1=L12/M1

n=1

M*1=(4.121)2/3.861

M*1=4.398M

Similarly, M*2=0.436M

M*3=0.121m

M*4=0.038m

M*5=0.008m

modal mass capacity ratio = M*n/Mj

where mj=total mass of the building = m+m+m+m+m=5m

modal mass participation ratio of 1st mode (n=1) = 4.398m/5m=0.88

similarly, 2nd mode (n=2)=0.09

3rd mode (n=3)=0.02

4th mode (n=4)=0.01

5th mode (n=5)=0.002

adding 1st two mode mass participation ratios, we get 0.88+0.09=0.97

To achieve minimum 90% of modal mass participation ratio 1st two modes are enough.

Base shear (Vb) = M*n.An

for 1st mode Vb1=4.398mx0.27g

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 square root of square.

√[(Vb1)2+(Vb2)2+(Vb3)2+(Vb4)2+(Vb5)2]

Vb = 1.15mg