Understanding the concepts on Degrees of Freedom

Answer:- 

First of all we should build a mass matrix.

Mass matrix (m) = [[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) = phi n^T m l 

where:

l = influence vector (Assuming 100% utilization of seismic force) 

[[1],[1],[1],[1],[1]]

For the first mode = (n=1)

[0.334,0.641,0.895,1.078,1.173] [[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]] [[1],[1],[1],[1],[1]]

L1= 4.121m

Modal participation factor  (r) = L/M

For the first mode (n=1), r = L1/M1

r1 = 4.121m/3.861m

r1 = 1.067

similarly:

r2 = -0.335

r3 = 0.177

r4 = -0.098

r5 = 0.045

Effective modal mass (M*) = Ln^2/Mn

For the first mode, M1* = L1^2/M1 (n=1)

M1* = (4.121 m)62/3.861m

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 building m+m+m+m+m = 5m

Modal mass participation ratio of first mode (n=1)

= 4.398m/5m = 0.88

88% of probability may happen in first mode

Similarly

Second mode (n=2) = 0.09 = 9%

Third mode (n=3) = 0.02 = 2%

Fourth mode (n=4) = 0.01 = 1%

Fifth mode = (n=5) = 0.002 = 0.2%

Adding first two mode mass participation ratios,We get 0.88+0.09 = 0.97 to achieve minimum 90% of modal mass participation ratio,first two modes are enough.

Base shear(VB) = Mn*.An

For the first mode (n=1) ,VB1 = 4.398m x 0.27g(0.27 is came from graph)

VB1 = 1.187mg

Similarly:

VB2 = 0.327 mg 

VB3 = 0.126mg

VB4 = 0.039mg

VB5 = 0.008mg

Average base shear is calculated by the method of square root of sum of square (SRSS)

sqrt((Vb1^2)+(Vb2^2)+(Vb3^2)+(Vb4^2)+(Vb5^2))

VB= 1.15mg