Understanding the concepts on Degrees of Freedom
Aim -
To calculate the base shear of 5 modes as shown
mass matrix `[[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 for first node(m1) = [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 = 3.861m
Model properties = deformation x m x `tau`
For first mode = [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 = 4.121m
Model participation factor for first mode = L1/m1 (4.121/3.861)
= 1.067
For the second mode
m2 = [-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]]`
m2 = -1.297
Model properties = deformation x m x `tau`
For second mode = [-0.895,-1.173,0.334,1.078,] `[[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]]` `[[1],[1],[1],[1],[1]]`
L2 = 3.871m
Model participation factor for first mode = L2/m2 (3.871/-1.297)
= -0.335
For the third mode
m3 = [1.173,0.334,-1.078,-0.641,0.895] `[[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]]` `[[1.173],[0.334],[-1.078],[0.641],[0.895]]`
m3 = 0.683
Model properties = deformation x m x `tau`
For first mode = [1.173,0.334,-1.078,-0.641,0.895] `[[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]]`
L3 = -0.066m
Model participation factor for first mode = L3/m3 (-0.066/.683)
= -0.098
For the fourth mode
m3 = [-1.078,0.895,0.334,-1.173,0.641] `[[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]]` `[[-1.078],[0.895],[0.334],[-1.173],[0.641]]`
m3 = -0.381
Model properties = deformation x m x `tau`
For first mode = [-1.078,0.895,0.334,-1.173,0.641] `[[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]]`
L4 = 0.037m
Model participation factor for first mode = L4/m4 (0.037/-0.381)
= -0.098
For the fifth mode
m5 = [0.641,-1.078,1.173,-0.895,0.334] `[[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]]` `[[0.641],[-1.078],[1.173],[-0.895],[0.334]]`
m5= 0.175
Model properties = deformation x m x `tau`
For first mode = [0.641,-1.078,1.173,-0.895,0.334] `[[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]]`
L5 = 0.007m
Model participation factor for first mode = L5/m5 (0.007/0.175)
= 0.045
Effective model mass (m*) = L`n^2`/Mn
M1* = 4.398m
M2* = 0.436m
M3* = 0.121m
M4* = 0.038m
M5* = 0.008m
Model mass participation ratio = Mn/mj
mj = Total mass of the building = m+m+m+m+m = 5m
Model mass participation ratio of first mode (n=1) = 4.398/5 = 0.88
88% of probability for first mode to happen in the building
Second mode (n=2) = 0.436/5 = 0.09
Third mode (n=3) = 0.121/5 = 0.02
Fourth mode (n =4) = 0.038/5 = 0.01
Fifth mode (n=5) = 0.008/5 = 0.002
`therefore` Adding first two mode participation ratios we get 0.88+0.09 = 0.97
To achieve min 90% of model mass participation ratios, first two are eneough
Finally base shear
Base shear (Vb) = model mass x psedo accleration(from given graph)
Vb1 = 4.398 x 0.27 = 1.187
Vb2 = 0.436 x 0.75 = 0.327
Vb3 = 0.121 x 2.70 = 0.126
Vb4 = 0.038 x 1.02 = 0.039
Vb5 = 0.008 x 1 = 0.008
We have five no of base shear so take average by Square root of the sum of squres
`sqrt(Vb1)^2+(Vb3)^2+(Vb3)^2+(Vb4)^2+(Vb5)^2`
`sqrt(1.187)^2+(0.327)^2+(0.126)^2+(0.039)^2+(0.008)^2`
=
Vb = 1.15