Understanding the concepts on Degrees of Freedom

Answer

Mass matrix (m) =

Modified mass matrix (M) = ɸ_n^T. m . ɸ_n

For the first mode (n=1), =

M₁ = 3.861 m

Modal property (L) = ɸ_n^T. m . l

Where, l = Influence Vector (Assuming 100% utilization of seismic force) =

For the first mode (n=1), =

L₁ = 4.121 m

Modal participation factor (⎾) = L/M

For the first mode (n=1), ⎾₁ = L₁/M₁ = 4.121/3.861 = 1.067

⎾₁ = 1.067

Similarly,

⎾₂ = -0.335

⎾₃ = 0.177

⎾₄ = -0.098

⎾₅ = 0.045

Effective modal mass (M*) = L_n²/M_n

For the first mode (n=1), M₁* = L₁²/M₁ = (4.121)²/3.861 = 4.398 m

M₁* = 4.398 m

Similarly,

M₂* = 0.436 m

M₃* = 0.121 m

M₄* = 0.038 m

M₅* = 0.008 m

Modal mass participation ratio = M_n*/ m_j

Where, m_j = Total mass of the building = m+m+m+m+m = 5 m

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

(n=1) = 0.88

Similarly,

Second mode, (n=2) = 0.09

Third mode, (n=3) = 0.02

Fourth mode, (n=4) = 0.01

Fifth mode, (n=5) = 0.002

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 (V_B) = M_n*. A_n

A₁ = 0.27 g (From the graph)

For the first mode (n=1), V_b1 = 4.398 m x 0.27 g = 1.187 mg

V_b1 = 1.187 mg

Similarly,

V_b2 = 0.327 mg

V_b3 = 0.126 mg

V_b4 = 0.039 mg

V_b5 = 0.008 mg

Average base shear = √( (V_b1²)+(V_b2²)+(V_b3²)+(V_b4²)+(V_b5²) ) = 1.15 mg

V_b = 1.15 mg