Understanding the concepts on Degrees of Freedom

ANS:

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]]`

Modified mass matrix (M) = `phi`n^t . m . `phi`n

For the first node (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]]` * `[[0.334],[0.641],[0.895],[1.078],[1.173]]`

M1 = 3.861 m

Model property (L) = `phi`n^t . m . l

For the first node (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]]`

L1 = 4.121 m

Model Participation factor (`lceiling`) = L/M

for the first node (n=1)

`lceiling `1= L1/m`

= 4.121/3.861

=1.067

similary `lceiling `2 = -0.335

`lceiling `3 = 0.177

`lceiling `4 = -0.098

`lceiling `5 = 0.045

Effective Model mass (M) = L^2/M

for first node(n=1)

M1 = L1^2/M1

= (4.121)^2/3.861

= 4.398 m

similarly M2 = 0.436 m

M3 = 0.121 m

M4 = 0.038 m

M5 = 0.008 m

Model mass participation ratio = Mn/mj

mj = total mass of building

= 5

Model mass participation ratio for

(n=1) = 4.398/5 = 0.88

(n=2) = 0.09

(n=3) = 0.02

(n=4) = 0.01

(n=5) = 0.002

Addition of first two nodes

= 0.88 + 0.09

= 0.97

• BASE SHEAR (VB) = Mn*An

for the first node (n=1)

4.398 * 0.27

Vb1 = 1.187 mg

Vb2 = 0.327 mg

Vb3 = 0.126 mg

Vb4 = 0.039 mg

Vb5 = 0.008 mg

• AVERAGE BASE SHEAR

= √`(Vb1^2)+(Vb2^2)+(Vb3^2)+(Vb4^2)`

= 1.15 mg