Understanding the concepts on Degrees of Freedom
1) Aim & Introduction :
To A five storey structure with 5 translational DOF is shown here. Each storey has mass ‘m’. The eigenvalue problem has been solved and the 5 periods of vibration Tn and their corresponding mode shapes, φn has been derived.
- n , Mn and Γn for each of the five modes are to be computed from Equations below.
- Also, calculate the effective modal mass Mn* and the modal mass participating ratio for the 5 modes. The mass matrix will be required to evaluate these parameters.
- 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.
- How many modes are to be considered such that a modal mass participating ratio of 90% is obtained.
Procedure :
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) = Φ^T . m . Φ
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
Modal property (L) = Φ^T . m . l
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
Modal Participation Factor τ = L/M
for the First Mode, τ = L/M
τ = 4.121/3.861
τ1 = 1.067
Similarly
τ2 = -.335
τ3 = .177
τ4 = -.098
τ5 = .045
Effective Modal Mass M* = L^2 /M
M*1 = (4.121)^2/3.861
M*1 = 4.398m
similarly
M*2 = .436m
M*3 = .121m
M*4 = .038m
M*5 = .008m
Modal Mass participation ratio = Mn/mj
where mj =m+m+m+m+m = 5m
For (n=1) = 4.398/5m =.88
Similarly
For (n=2) = .09
For (n=3) = .02
For (n=4) = .01
For (n=5) = .002
Adding First two mode mass Participation ratio we get .97.To Achieve 90% of mode mass participation ratio the first two are enough.
Base Shear Vb = Mn* An
For First Mode Vb1 = 4.398 x .27g
Vb1 = 1.187mg
Similarly
Vb2 = .327mg
Vb3 = .126mg
Vb4 = .039mg
Vb5 = .008mg
Average Base Shear is calculated SRSS method
Vb = √(Vb1)^2 +(Vb2)^2+(Vb3)^2+(Vb4)^2+(Vb5)^2
Vb = 1.15mg
Result : The following results are calculated.