Understanding the concepts on Degrees of Freedom

AIM

To compute the values of the seismic parameters based on the given structural data.

PROCEDURE

We know that

`phi_1= [[0.334],[0.641],[0.895],[1.078],[1.175]]`

`phi_2= [[-0.895],[-1.173],[-0.641],[0.334],[1.078]]`

`phi_3= [[1.173],[0.334],[-1.078],[-.0641],[0.895]]`

`phi_4= [[-1.078],[0.895],[0.334],[-1.173],[0.641]]`

`phi_5= [[0.641],[-1.078],[1.173],[-0.895],[0.334]]`

Formation of the 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]]`

Influence vector:

Since the degrees of freedom move along the direction of the lateral force, the influence vector will be a unit

`iota = [[1],[1],[1],[1],[1]]`

To calculate `M_n, L_n, Gamma_n and M_n^ast` we use the following formulas

1. Modified mass matrix (M): `M_n= phi_n^T*m*phi_n`
2. Modal property(L): `L_n= phi_n^T*m*iota`
3. Modal participation factor: `Gamma_n= frac {L_n}{M_n}`
4. Effective modal mass: `M_n^ast = frac {L_n^2}{M_n}`

Substituting the values of `phi` in the equation 1, we get

`M_1= [[0.334, 0.641, 0.895, 1.078,1 .175]]*m*[[0.334],[0.641],[0.895],[1.078],[1.175]]`

`M_1=3.861m`

`M_2= [[-0.895, -1.173, -0.641, 0.334 ,1.078]]*m*[[-0.895],[-1.173],[-0.641],[0.334],[1.078]]`

`M_2=3.85m`

`M_3= [[1.173, 0.334, -1.078, -0.641, 0.895]]*m*[[1.173],[0.334],[-1.078],[-.0641],[0.895]]`

`M_3=3.861m`

`M_4= [[-1.078, 0.895, 0.334, -1.173, 0.641]]*m*[[-1.078],[0.895],[0.334],[-1.173],[0.641]]`

`M_4=3.861m`

`M_5= [[0.641, -1.078, 1.173, -0.895, 0.334]]*m*[[0.641],[-1.078],[1.173],[-0.895],[0.334]]`

`M_5=3.861m`

Substituting the values of `phi` in the equation 2, we get

`L_1= [[0.334, 0.641, 0.895, 1.078,1 .175]]*m*[[1],[1],[1],[1],[1]]`

`L_1=4.123m`

`L_2= [[-0.895, -1.173, -0.641, 0.334 ,1.078]]*m*[[1],[1],[1],[1],[1]]`

`L_2=-1.29m`

`L_3= [[1.173, 0.334, -1.078, -0.641, 0.895]]*m*[[1],[1],[1],[1],[1]]`

`L_3=0.68m`

`L_4= [[-1.078, 0.895, 0.334, -1.173, 0.641]]*m*[[1],[1],[1],[1],[1]]`

`L_4=0.381m`

`L_5= [[0.641, -1.078, 1.173, -0.895, 0.334]]*m*[[1],[1],[1],[1],[1]]`

`L_5=0.175m`

Substituting the values of `L_n andM_n`in the equation 3, we get

`Gamma_1=frac{4.123m}{3.861m} = 1.068`

`Gamma_2=frac{-1.29m}{3.85m} = -0.336`

`Gamma_3=frac{0.68m}{3.861m} = 0.1761`

`Gamma_4=frac{0.381m}{3.861m} = 0.099`

`Gamma_5=frac{0.175m}{3.861m} = 0.045`

Substituting the values of `L_n andM_n`in the equation 4, we get

`M_1^ast= frac{(4.123m)^2}{3.861} =4.398m`

`M_2^ast= frac{(-1.29m)^2}{3.85} =0.432m`

`M_3^ast= frac{(0.68m)^2}{3.861} =0.119m`

`M_4^ast= frac{(0.381m)^2}{3.861} =0.0357m`

`M_5^ast= frac{(0.175m)^2}{3.861} =0.0078m`

We can verify the total mass of the building, m+m+m+m+m=5m

Adding all effective modal mass, `M_1^ast+M_2^ast+M_3^ast+M_4^ast+M_5^ast = 5m`

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

For first mode `frac{4.398}{5} = 0.879`

For second mode `frac{0.432}{5} = 0.086`

For third mode `frac{0.121}{5} = 0.0242`

For fourth mode `frac{0.037}{5} = 0.0074`

For fifth mode `frac{0.008}{5} = 0.0016`

From the above participation ratios of modal mass, we can understand that the sum of the first two modes i.e. Mode-1 and Mode-2 is enough to obtain a minimum of 90% of the participation.

0.879 + 0.086` = 0.965 = 96.5%

To calculate the base shear

`V_B = M_N^ast * A_n`

Pseudo acceleration obtained from the graph are as follows:

For `T_1, A_1=0.27g`

`T_2, A_2=0.76g`

`T_3, A_3=1.03g`

`T_4, A_4=1.03g`

`T_5, A_5=1.03g`

`V_(b1) = 4.398m * 0.27g = 1.187mg`

`V_(b2) = 0.432m * 0.76g = 0.328mg`

`V_(b3) = 0.119m * 1.03g = 0.123mg`

`V_(b4) = 0.0357m * 1.03g = 0.037mg`

`V_(b5) = 0.0078m * 1.03g = 0.008mg`

Average base shear is calculated by the SRSS method

`sqrt(V_(b1)^2+V_(b2)^2+V_(b3)^2+V_(b4)^2+V_(b5)^2)`

`=1.15mg`