Understanding the concepts on Degrees of Freedom
Ans 1.
- 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 matrix (M) = `O/n^T`Xm`∅n.
so for first mode(n=1) =`[[.334,.641,.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]] [[.334], [.641], [.895], [1.078], [1.173]]`
M1= 3.861m.
`or 2nd mode (n=2) = [[-.895, -1.173, -.641,.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]] [[-.895], [-1.173], [-.641], [.334], [1.078]]`
M2= 3.861m.
`For 3rd mode (n=3) = [[1.173, 0.334, -1.078, -.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], [-.641], [0.895]]`
`or 4th mode (n=4) = [[-1.078, 0.895, .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], [.334], [-1.173], [0.641]]`
`For 5th mode (n=5) = [[.641, -1.078, 1.173, -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]] [[.641], [-1.078], [1.173], [-895], [0.334]]`
- `Modal Property (L) =`O/n^T`m*ι where (ι) is influence vector. (Assuming 100% utilization of seidmic force)
`So for first mode(n=1) =[[.334,.641,.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
for 2nd mode (n=2) =`[[-.895, -1.173, -.641,.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= -0.667m.
For 3rd mode (n=3) = `[[1.173, 0.334, -1.078, -.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= -0.683m.
`or 4th mode (n=4) = [[-1.078, 0.895, .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= -0.381m.
`or 5th mode (n=5) = [[.641, -1.078, 1.173, -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= 0.175m.
- Modal Participation factor (Γ) = L/M
for first mode(n=1) Γ1 =L1/M1, 4.121m/3.861m =1.067.
for 2nd mode (n=2) = Γ2 =L2/M2, -0.667m
For 3rd mode (n=3) = Γ3 =L3/M3, -0.683m
For 4th mode (n=4) = Γ4 =L4/M4, -0.381m
For 5th mode (n=5) = Γ5 =L5/M5, 0.175m
- Effective Modal Nass (M*) = `(L n) ^2/ (Mn)`
for first mode (n=1) (M1*) = (4.121m)^2/3.861m= 4.398m
for 2nd mode (n=2) (M2*) = (-0.667m)^2/m=
For 3rd mode (n=3) (M3*) = ( -0.683m)^2/m =
For 4th mode (n=4) (M4*) = ( -0.381m)^2/m =
For 5th mode (n=5) (M4*) = ( 0.175m)^2/m =
- Modal Nass Participation Ratio =(Mn*/mj) where (mj) is Total mass of building = m+m+m+m+m= 5m.
for first mode (n=1) = 4.398m/5m =0.88
for 2nd mode (n=2) = 0.436m/5m = 0.09
For 3rd mode (n=3) = 0.121m/5m = 0.02
For 4th mode (n=4) = 0.038m/5m = 0.01
For 5th mode (n=5) = 0.008m/5m = 0.002
Adding first two mode participation ratio we get 0.88+0.09= 0.97 therefore the mode participation ratio is 90% given in the question.
- Base Shear (Vb) = Mn*.An
for first mode (n=1) = 4.398*0.27g = 1.187mg.
for 2nd mode (n=2) = 0.436m*0.75g = 0.327mg.
For 3rd mode (n=3) = 0.121m*1.03g = 0.126mg.
For 4th mode (n=4) = 0.038m*1.03g = 0.039mg.
For 5th mode (n=5) = 0.008m*1.03g = 0.008mg.
- Average Base Shear by SRSS method =( `sqrt ((Vb1^2) + (Vb2^2) + (Vb3^2) + (Vb4^2) + (Vb5^2)`
Vb = `sqrt ((10187mg^2) + (0.327mg^2) + (0.126mg^2) + (0.039mg^2) + (0.008mg^2))`
Vb = 1.15mg....ans