Understanding the concepts on Degrees of Freedom

1.

For modal mass

Substitute the 5 number of stories to Modified mass matrix (M)

n= 1, n=2, n=3, n=4, n=5

Multiplying the equation of the mass matrix then we will get the M1, M2, M3, M4, M5 values which are all same values  

For model property (L)

Substitute the values from 1 to 5 in this equation 

The new symbol is present in this equation called influence vector

We are assuming 100% utilization of seismic force

For modal participation factor (Γ) 

modal participation factor (Γ) = L/M

For first mode Γ = L1/M1

Γ1 = 4.121/3.861 = 1.067

Similarly,

Γ2 = -1.297/3.861 = -0.335

Γ3 = 0.683/3.861 = 0.177

Γ4 = -0.381/3.861 = -0.098

Γ5 = 0.175/3.861 = 0.045

For effective modal mass (M*) = L_n² /M_n 

For first mode (M*) = L_n² /M_n 

M1* = (4.121)²/3.861 = 4.398m

Similarly,

M2* = (-1.297)²/3.861 = 0.436m

M3* = (0.683)²/3.861 = 0.121m

M4* = (-0.381)²/3.861 = 0.038m

M5* = (0.175)²/3.861 = 0.008m

For modal mass participation ratio = M_n*/ mj   

Where mj = Total mass of the building = m+m+m+m+m = 5m

Modal mass participation ratio of first mode (n=1) = 4.398/5 = 0.88

Similarly, 

 Modal mass participation ratio of second mode (n=2) = 0.436/5 = 0.09

Modal mass participation ratio of third mode (n=3) = 0.121/5 = 0.02

Modal mass participation ratio of fourth mode (n=4) = 0.038/5 = 0.01

Modal mass participation ratio of fifth mode (n=5) = 0.008/5 = 0.002

Adding first two mode of mass participation ratios we get = 0.88+0.09 = 0.97

To achive a minimum 90% of total mass participation ratio, first two modes are enough

For base shear Vb = M_n*x A_n 

For first mode of base shear Vb1 = 4.398x0.27 = 1.187mg 

Similarly,

For second mode of base shear Vb2 = 0.436x0.27 = 0.327mg 

For first mode of base shear Vb3 = 0.121x0.27 = 0.126mg 

For first mode of base shear Vb4 = 0.038x0.27 = 0.039mg 

 For first mode of base shear Vb5 = 0.008x0.27 = 0.008mg 

Average base shear is calculated by 

((V_b1² + V_b2² + V_b3²  + V_b4² + V_b5² )^1/2

Substituting all the values of base shear then we will get answer

Vb =1.23mg 

The psudo acceleration value = 0.27 is taken from this graph