Understanding the concepts on Degrees of Freedom

Q1): A five-story structure with 5 translational DOF is shown here. Each story has mass ‘m’. The eigenvalue problem has been solved and the 5 periods of vibration T_n and their corresponding mode shapes, φ_n has been derived.

`Q_1 = ((0.334 ), (0.641),(0.895),(1.078),(1.173))`

`Q_2 = ((-0.895),(-1.173),(-0.641),(0.334),(1.078))`

`Q_3= ((1.173),(0.334),(-1.078),(-0.641),(0.895))`

`Q_4=((-1.078),(0.895),(0.334),(-1.173),(0.641))`

`Q_5 = ((0.641),(-1.078),(1.173),(-0.893),(0.334))`

L_n , M_n and Γ_n for each of the five modes are to be computed from Equations below.

`M_n =Phi_n^Tm Phi_n`

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]]`

Solution:

`M_1 = Phi_1^TmPhi_1`

`M_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]]``((0.334 ), (0.641),(0.895),(1.078),(1.173))`

`M_1=3.8614 m`

Similarly

`M_2``=[[-0.895,-1.173,-0.641,0.334,1.078]]``[[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]]``((-0.895),(-1.173),(-0.641),(0.334),(1.078))`

`M_2`=3.861 m

`M_3`=3.861 m

`M_4`=3.861 m

`M_5`3.861 m

`L_n=Phi_n^Tml`

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]]`

l=`((1),(1),(1),(1),(1))`

Solution:

`L_1=Phi_1^Tml`

`L_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_1`= 4.121m

Similarly

`L_2=-1.29 m`

`L_3=0.68 m`

`L_4=0.381m`

`L_5= 0.175m`

`Gamma_n`=`L_n/M_n`

Solution :

`Gamma_1=L_1/M_1`

`4.121/3.8614`

`Gamma_1 =1.067`

Similarly :

`Gamma_2=-0.336`

`Gamma_3=0.177`

`Gamma_4=0.009`

`Gamma_5=0.045`

`M_n^**` =`L_n^2/M_n`

Participating ratio = `M_N^**/M_i`

`M_i=5m`

Solution :

`M_1^**=L_1^2/M_1`

`M_1^**=`4.398 m

Participating ratio = 0.8796

Similarly:

`M_2^**=0.435m`

Participating ratio = 0.087

`M_3^**=0.121m`

Participating ratio = 0.0242

`M_4^**`=0.037 m

Participating ratio = 7.4 x `10^-3`

`M_5^**=0.008 m`

Participating ratio = 1.6 x `10^-3`

Finally, obtain the base shear for each of the 5 modes of vibration and the combined base shear as per SRSS rule. The pseudo acceleration design spectrum to obtain the base shear for each mode is provided below.

`T_1=2` `A_1=0.27`

`T_2=0.6852` `A_2 =`0.76

`T_3 =0.43446` `A_3=`1.03

`T_4=0.3383` `A_4=1.03`

`T_5=0.2966` `A_5=1.03`

`V_(bn) = M_n^** A_n`

Solution :

`V_(b1) = M_1^** A_1`

`V_(b1) =`1.1874 m

similarly :

`V_(b2)=0.3306 m`

`V_(b3) = 0.12463 m`

`V_(b4)=0.3811m`

`V_(b5)=8.24 xx 10^(-3)`m

As per SRSS Method :

`V_b =sqrt ((1.1874)^2 + (0.3306)^2+(0.12463)^2+(0.3811)^2+(0.00824)^2)`

`V_b=1.457m`