Understanding the concepts on Degrees of Freedom

1. Aim: To solve a five storey structure with 5 translational Degree of freedom is as shown in figure with given mass 'm'.

Introduction: To solve a five storey structure with 5 translational Degree of freedom is as shown in figure with given mass 'm'.

Procedure:

Modal Property (L) = `(Phi_n)^T' `,m , t '

Where t = Influence vector (Assuming 100% utlization of seismic force) = `[[1],[1],[1],[1],[1]]`

For the first mode,

where 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]]`

`L_n`= 4.121m

Now modal participation factor (`gamma`) = `L_1/M_1`

where n=1

`gamma_1` = 4.121m/3.861m`

`gamma_1` = 1.067

In the same way, similarly,

`gamma_2` = -0.335

`gamma_3` = 0.177

`gamma_4` = -0.098

`gamma_5` = 0.045

Effective modal mass (`M^ast`) = `L_n^2/M_n`

For the first mode, `M_1ast = L_1^2/M_1`

where n=1

`M_1^* = (4.121m)^2/3.861m`

`M_1^*` = 4.398m

Similarly,

`M_2^* = 0.436m`

`M_3^*` = 0.121m

`M_4^*` = 0.038m

`M_5^*` = 0.008m

Modal mass participation ratio = `M_n^*/m_j`

where,

`m_j`= Total mass of the building = m+m+m+m+m

m/ = 5m

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

= 0.88

similarly, second mode (n=2) = 0.09

Third node (n=3) = 0.02

Fourth node (n=4) = 0.01

Fifth node (n=5) = 0.002

Adding 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`

For the first mode , `V_B1` = 4.398m x 0.27g

where n=1

`V_B1 = 1.187 mg`

Similarly,

`V_B2` = 0.327 mg

`V_B3` = 0.126mg

`V_B4` = 0.039mg

`V_B5` = 0.008mg

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

`sqrt_(Vb1^2) + (Vb2^2) + (Vb3^2) + (Vb4^2) + (Vb5^2)`

`V_B` = 1.15mg