Understanding the concepts on Degrees of Freedom
Ans.1.
As Given in the question ,
The structure has mass m on all its storeys with 5 DOF.
Let us first calculate the Mass Matrix,
So,
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]]`
And the formula for modified mass matrix will be :-
M =`Phiunderset(n) overset(T)``*m* phiunderset(n)`
Now ,
For Mode 1 i.e. n = 1
`Munderset1` =`[[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]]`
After Calculating the value and solving the equation ,
We get the value of `Munderset1` =3.861 m
For Calculation of Modal Property (L) ,
we know,
Modal Property (L) = `Phiunderset(n) overset(T)``*m* iota`
where, `iota` =influence vector (which shows how much load has been transfered in that particular direction where the load has been applied) [Assuming 100% utilization of the seismic forces]
Similiarly, For Mode 1 where n=1,
Value of Modal property ,
`Lunderset1` =`[[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]]`
`Lunderset1` = 4.1213 m
Similarly Calculating the values for `Munderset2` ,`Munderset3`,`Munderset4`,`Munderset5`, `Lunderset2` , `Lunderset3`, `Lunderset4`, `Lunderset5`
We get,
`Munderset2` = 3.861 m
`Munderset3` = 3.861 m
`Munderset4` = 3.861 m
`Munderset5` = 3.861 m
and
`Lunderset2` = -1.297 m
`Lunderset3` = 0.683 m
`Lunderset4` = -0.381 m
`Lunderset5` = 0.175 m
Therefore,
Calculating Modal participitation factor
`Lambda = L/M`
So,
For First Mode (n=1)
`Lambda = frac(L1)(M1)`
`Lambda1 = frac(4.121m)(3.861m)`
`Lambda1 =1.067`
`For Mode 2(n=1)`
`Lambda = frac(L2)(M2)`
`Lambda2 = frac(-1.297m)(3.861m)`
`Lambda2 =-0.3359`
`For Mode 3 (n=1)`
`Lambda = frac(L3)(M3)`
`Lambda3= frac(0.683m)(3.861m)`
`Lambda3 =0.177`
For Mode 4 (n=1)
`Lambda = frac(L4)(M4)`
`Lambda4 = frac(-0.381m)(3.861m)`
`Lambda4 =-0.098`
For Mode 5 (n=1)
`Lambda = frac(L5)(M5)`
`Lambda5 = frac(0.175m)(3.861m)`
`Lambda5 =0.045`
Now Calculating for Effective Modal Mass,
`M^ast = Lundersetn overset2` / `Mundersetn`
For mode 1
`underset(1)(M)^ast` = `frac((4.121m)^2)(3.861)`
`underset(1)(M)^ast`4.398m
For mode 2
`underset(2)(M)^ast` = 0.436m
For Mode 3
`underset(3)(M)^ast` = 0.121m
For mode 4
`underset(4)(M)^ast` = 0.038
For mode 5
`underset(5)(M)^ast` = 0.008`
Hence,
MODAL MASS PARTICIPATION RATIO =`underset(n)(M)^ast/underset(j)(m)`
where ,
`underset(j)(m)` is the Total building mass i.e. 5m
So,Modal Mass Participation ratio of First mode (n=1) 4.398m/5m = 0.88
Similiarly,
For Mode 2 (n=2) =0.09
For Mode 3 (n=3) = 0.02
For Mode 4 (n=4) =0.01
For Mode 5 (n= 5) =0.002
Hence on adding first 2 modal mass ratio we get
= 0.88 + 0.09
= 0.97
Therefore , for achieving of minimum of 90% of modal mass participation ratio , first two modes are enough
Now calculating the Base Shear `Vundersetb`
`Vundersetb` =`underset(n)(M)^ast * Aundersetn`
For Mode 1
`Vunderset(b1)` = 4.398 m x 0.27 g
`Vunderset(b1)` = 1.187 mg
For mode 2
`Vunderset(b2)` =0.327 m`
For mode 3
`Vunderset(b3)` = 0.126mg
For Mode 4
`Vunderset(b4)` = 0.038mg
For mode 5
`Vunderset(b5)` = 0.008mg
Therefore Avg.Base shear is calculated by SRSS ie. SQUARE ROOT OF SUM OF SQUARES
So,
`sqrt((Vb1)^2 + (Vb2)^2 + (Vb3)^2 + (Vb4)^2 + (Vb5)^2`
`sqrt((1.187)^2 + (0.327)^2 + (0.126)^2 + (0.038)^2 + (0.008)^2`
Hence , `Vundersetb` =1.15mg .....ans