Understanding the concepts on Degrees of Freedom
Question 1:
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.
Given:
The fundamental period of vibration (T1 = 2 seconds) and its associated mode shape:
The second mode of vibration (T2 = 0.6852 seconds) and its associated mode shape:
The third mode of vibration (T3 = 0.4346 seconds) and its associated mode shape:
The fourth mode of vibration (T4 = 0.3383 seconds) and its associated mode shape:
The fifth mode of vibration (T4 = 0.2966 seconds) and its associated mode shape:
- Ln , 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?
ANSWER:
First we calculating the mass matrix(m).
Mass matrix,(m)=[m00000m00000m00000m00000m]</math>">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[�00000�00000�00000�00000�]
Now we calculate the transformed mass matrix for all modes-M=ΦT⋅m⋅Φ</math>">M=ΦT⋅m⋅Φ�=Φ�⋅�⋅Φ
1)Modified mass matrixs:(M)
For mode-1
M_1=Phi_1^T*m*Phi_1
i.e, M1=[0.3340.6410.8951.0781.173]⋅[m00000m00000m00000m00000m]⋅[0.3340.6410.8951.0781.173]</math>">M1=[0.3340.6410.8951.0781.173]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.3340.6410.8951.0781.173⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�1=[0.3340.6410.8951.0781.173]⋅[�00000�00000�00000�00000�]⋅[0.3340.6410.8951.0781.173]
M1=[0.334m0.641m0.895m1.078m1.173m]⋅[0.3340.6410.8951.0781.173]</math>">M1=[0.334m0.641m0.895m1.078m1.173m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.3340.6410.8951.0781.173⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�1=[0.334�0.641�0.895�1.078�1.173�]⋅[0.3340.6410.8951.0781.173]
M1=3.86m</math>">M1=3.86m�1=3.86�
For mode-2
M2=Φ2T⋅m⋅Φ2</math>">M2=ΦT2⋅m⋅Φ2�2=Φ2�⋅�⋅Φ2
i.e, M2=[-0.895-1.173-0.6410.3341.078]⋅[m00000m00000m00000m00000m]⋅[-0.895-1.173-0.6410.3341.078]</math>">M2=[−0.895−1.173−0.6410.3341.078]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−0.895−1.173−0.6410.3341.078⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�2=[-0.895-1.173-0.6410.3341.078]⋅[�00000�00000�00000�00000�]⋅[-0.895-1.173-0.6410.3341.078]
M2=[m-0.895m-1.173m-0.6410.334m1.078m]⋅[-0.895-1.173-0.6410.3341.078]</math>">M2=[−0.895m−1.173m−0.641m0.334m1.078m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−0.895−1.173−0.6410.3341.078⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�2=[�-0.895�-1.173�-0.6410.334�1.078�]⋅[-0.895-1.173-0.6410.3341.078]
M2=3.86m</math>">M2=3.86m�2=3.86�
For mode-3
M3=Φ3T⋅m⋅Φ3</math>">M3=ΦT3⋅m⋅Φ3�3=Φ3�⋅�⋅Φ3
i.e,M3=[1.1730.334-1.078-0.6410.895]⋅[m00000m00000m00000m00000m]⋅[1.1730.334-1.078-0.6410.895]</math>">M3=[1.1730.334−1.078−0.6410.895]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1.1730.334−1.078−0.6410.895⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�3=[1.1730.334-1.078-0.6410.895]⋅[�00000�00000�00000�00000�]⋅[1.1730.334-1.078-0.6410.895]
M3=[1.173m0.334mm-1.078m-0.6410.895m]⋅[1.1730.334-1.078-0.6410.895]</math>">M3=[1.173m0.334m−1.078m−0.641m0.895m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1.1730.334−1.078−0.6410.895⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�3=[1.173�0.334��-1.078�-0.6410.895�]⋅[1.1730.334-1.078-0.6410.895]
M3=3.86m</math>">M3=3.86m�3=3.86�
For mode-4
M4=Φ4T⋅m⋅Φ4</math>">M4=ΦT4⋅m⋅Φ4�4=Φ4�⋅�⋅Φ4
i.e,M4=[-1.0780.8950.334-1.1730.641]⋅[m00000m00000m00000m00000m]⋅[-1.0780.8950.334-1.1730.641]</math>">M4=[−1.0780.8950.334−1.1730.641]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−1.0780.8950.334−1.1730.641⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�4=[-1.0780.8950.334-1.1730.641]⋅[�00000�00000�00000�00000�]⋅[-1.0780.8950.334-1.1730.641]
M4=[m-1.0780.895m0.334mm-1.1730.641m]⋅[-1.0780.8950.334-1.1730.641]</math>">M4=[−1.078m0.895m0.334m−1.173m0.641m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−1.0780.8950.334−1.1730.641⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�4=[�-1.0780.895�0.334��-1.1730.641�]⋅[-1.0780.8950.334-1.1730.641]
M4=3.86m</math>">M4=3.86m�4=3.86�
For mode-5
M5=Φ5T⋅m⋅Φ5</math>">M5=ΦT5⋅m⋅Φ5�5=Φ5�⋅�⋅Φ5
M5=[0.641-1.0781.173-0.8950.334]⋅[m00000m00000m00000m00000m]⋅[0.641-1.0781.173-0.8950.334]</math>">M5=[0.641−1.0781.173−0.8950.334]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.641−1.0781.173−0.8950.334⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�5=[0.641-1.0781.173-0.8950.334]⋅[�00000�00000�00000�00000�]⋅[0.641-1.0781.173-0.8950.334]
M5=[0.641mm-1.0781.173mm-0.8950.334m]⋅[0.641-1.0781.173-0.8950.334]</math>">M5=[0.641m−1.078m1.173m−0.895m0.334m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.641−1.0781.173−0.8950.334⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�5=[0.641��-1.0781.173��-0.8950.334�]⋅[0.641-1.0781.173-0.8950.334]
M5=3.86m</math>">M5=3.86m�5=3.86�
Now we have to determine modal proprety.(L)
2)Modal property(Ln</math>">Ln��):ϕnT⋅m⋅l</math>">ϕTn⋅m⋅l���⋅�⋅�
Where,l=</math>">l=�= infulence matrix is aunit matrix,[11111]</math>">⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]
For mode-1:
(L1</math>">L1�1):ϕ1T⋅m⋅l</math>">ϕT1⋅m⋅l�1�⋅�⋅�
L1=[0.3340.6410.8951.0781.173]⋅[m00000m00000m00000m00000m]⋅[11111]</math>">L1=[0.3340.6410.8951.0781.173]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�1=[0.3340.6410.8951.0781.173]⋅[�00000�00000�00000�00000�]⋅[11111]
L1=[0.334m0.641m0.895m1.078m1.173m]⋅[11111]</math>">L1=[0.334m0.641m0.895m1.078m1.173m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�1=[0.334�0.641�0.895�1.078�1.173�]⋅[11111]
L1=4.12m</math>">L1=4.12m�1=4.12�
For mode-2:
(L2</math>">L2�2):ϕ2T⋅m⋅l</math>">ϕT2⋅m⋅l�2�⋅�⋅�
L2=[-0.895-1.173-0.6410.3341.078]⋅[m00000m00000m00000m00000m]⋅[11111]</math>">L2=[−0.895−1.173−0.6410.3341.078]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�2=[-0.895-1.173-0.6410.3341.078]⋅[�00000�00000�00000�00000�]⋅[11111]
L2=[m-0.895m-1.173m-0.6410.334m1.078m]⋅[11111]</math>">L2=[−0.895m−1.173m−0.641m0.334m1.078m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�2=[�-0.895�-1.173�-0.6410.334�1.078�]⋅[11111]
L2=-1.297m</math>">L2=−1.297m�2=-1.297�
For mode-3:
(L3</math>">L3�3):ϕ3T⋅m⋅l</math>">ϕT3⋅m⋅l�3�⋅�⋅�
L3=[1.1730.334-1.078-0.6410.895]⋅[m00000m00000m00000m00000m]⋅[11111]</math>">L3=[1.1730.334−1.078−0.6410.895]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�3=[1.1730.334-1.078-0.6410.895]⋅[�00000�00000�00000�00000�]⋅[11111]
L3=[1.173m0.334mm-1.078m-0.6410.895m]⋅[11111]</math>">L3=[1.173m0.334m−1.078m−0.641m0.895m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�3=[1.173�0.334��-1.078�-0.6410.895�]⋅[11111]
L3=0.683m</math>">L3=0.683m�3=0.683�
For mode-4
(L4</math>">L4�4):ϕ4T⋅m⋅l</math>">ϕT4⋅m⋅l�4�⋅�⋅�
L4=[-1.0780.8950.334-1.1730.641]⋅[m00000m00000m00000m00000m]⋅[11111]</math>">L4=[−1.0780.8950.334−1.1730.641]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�4=[-1.0780.8950.334-1.1730.641]⋅[�00000�00000�00000�00000�]⋅[11111]
L4=[m-1.0780.895m0.334mm-1.1730.641m]⋅[11111]</math>">L4=[−1.078m0.895m0.334m−1.173m0.641m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�4=[�-1.0780.895�0.334��-1.1730.641�]⋅[11111]
L4=-0.381m</math>">L4=−0.381m�4=-0.381�
For mode-5
(L5</math>">L5�5):ϕ5T⋅m⋅l</math>">ϕT5⋅m⋅l�5�⋅�⋅�
L5=[0.641-1.0781.173-0.8950.334]⋅[m00000m00000m00000m00000m]⋅[11111]</math>">L5=[0.641−1.0781.173−0.8950.334]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�5=[0.641-1.0781.173-0.8950.334]⋅[�00000�00000�00000�00000�]⋅[11111]
L5=[0.641mm-1.0781.173mm-0.8950.334m]⋅[11111]</math>">L5=[0.641m−1.078m1.173m−0.895m0.334m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�5=[0.641��-1.0781.173��-0.8950.334�]⋅[11111]
L5=0.175m</math>">L5=0.175m�5=0.175�
Now, we have to calculate the 3)Modal participation factor;Γn</math>">ΓnΓ�
It is defined as the ratio of modal property (Ln</math>">Ln��) to modified mass matrix(M).
Γn=LnMn</math>">Γn=LnMnΓ�=����
For mode-1:
Γ1=L1M1</math>">Γ1=L1M1Γ1=�1�1
Γ1=4.121m3.861m=1.067</math>">Γ1=4.121m3.861m=1.067Γ1=4.121�3.861�=1.067
Γ1=1.067</math>">Γ1=1.067Γ1=1.067
For mode-2:
Γ2=L2M2</math>">Γ2=L2M2Γ2=�2�2
Γ2=-1.297m3.861m=-0.335</math>">Γ2=−1.297m3.861m=−0.335Γ2=-1.297�3.861�=-0.335
Γ2=-0.335</math>">Γ2=−0.335Γ2=-0.335
For mode-3:
Γ3=L3M3</math>">Γ3=L3M3Γ3=�3�3
Γ3=0.683m3.861m=0.177</math>">Γ3=0.683m3.861m=0.177Γ3=0.683�3.861�=0.177
Γ3=0.177</math>">Γ3=0.177Γ3=0.177
For mode-4
Γ4=L4M4</math>">Γ4=L4M4Γ4=�4�4
Γ4=-0.381m3.861m=-0.099</math>">Γ4=−0.381m3.861m=−0.099Γ4=-0.381�3.861�=-0.099
Γ4=-0.099</math>">Γ4=−0.099Γ4=-0.099
For mode-5
Γ5=L5M5</math>">Γ5=L5M5Γ5=�5�5
Γ5=0.175m3.861m=0.045</math>">Γ5=0.175m3.861m=0.045Γ5=0.175�3.861�=0.045
Γ5=-0.045</math>">Γ5=−0.045Γ5=-0.045
After finding the modal participation factor now we have to calculate the effective modal mass(M∗=Ln2Mn</math>">M∗=L2nMn�∗=��2��)
4)effective modal mass(M∗=Ln2Mn</math>">M∗=L2nMn�∗=��2��):
For mode-1:
M1∗=L12M1</math>">M∗1=L21M1�1∗=�12�1
M1∗=(4.121m)23.861m=4.398m</math>">M∗1=(4.121m)23.861m=4.398m�1∗=(4.121�)23.861�=4.398�
M1∗=4.398m</math>">M∗1=4.398m�1∗=4.398�
For mode-2:
M2∗=L22M2</math>">M∗2=L22M2�2∗=�22�2
M2∗=(-1.297m)23.861m=-0.335m</math>">M∗2=(−1.297m)23.861m=−0.335m�2∗=(-1.297�)23.861�=-0.335�
M2∗=0.436m</math>">M∗2=0.436m�2∗=0.436�
For mode-3:
M3∗=L32M3</math>">M∗3=L23M3�3∗=�32�3
M3∗=(0.683m)23.861m=0.121m</math>">M∗3=(0.683m)23.861m=0.121m�3∗=(0.683�)23.861�=0.121�
M3∗=0.121m</math>">M∗3=0.121m�3∗=0.121�
For mode-4
M4∗=L42M4</math>">M∗4=L24M4�4∗=�42�4
M4∗=(-0.381m)23.861m=0.038m</math>">M∗4=(−0.381m)23.861m=0.038m�4∗=(-0.381�)23.861�=0.038�
M4∗=0.038m</math>">M∗4=0.038m�4∗=0.038�
For mode-5
M5∗=L52M5</math>">M∗5=L25M5�5∗=�52�5
M5∗=(0.175m)23.861m=0.008m</math>">M∗5=(0.175m)23.861m=0.008m�5∗=(0.175�)23.861�=0.008�
M5∗=0.008m</math>">M∗5=0.008m�5∗=0.008�
Now we have to calculate the modal mass participation ratio.
5)Modal mass particiopation ratio=Mn∗mj</math>">M∗nmj��∗��
Where mj</math>">mj��is total mass of the buuilding=m+m+m+m+m= 5m.
For mode-1:
M1∗mj=4.938m5m=0.88</math>">M∗1mj=4.938m5m=0.88�1∗��=4.938�5�=0.88
For mode-2:
M2∗mj=0.436m5m=0.09</math>">M∗2mj=0.436m5m=0.09�2∗��=0.436�5�=0.09
For mode-3:
M3∗mj=0.121m5m=0.02</math>">M∗3mj=0.121m5m=0.02�3∗��=0.121�5�=0.02
For mode-4:
M4∗mj=0.038m5m=0.01</math>">M∗4mj=0.038m5m=0.01�4∗��=0.038�5�=0.01
For mode-5:
M5∗mj=0.008m5m=0.002</math>">M∗5mj=0.008m5m=0.002�5∗��=0.008�5�=0.002
As per the question, there are sum of first two modes are enough to considered such as to obtain the MODAL MASS PARTICIPATING RATIO as 90%. i.e, 0.88+0.09=0.97.
Now we have to caluclate the base shear for each mode(VB</math>">VB��)
Base shear(VB=Mn∗.An</math>">VB=M∗n.An��=��∗.��)
For mode-1:
VB1=4.398m⋅0.27g=1.187mg</math>">VB1=4.398m⋅0.27g=1.187mg��1=4.398�⋅0.27�=1.187��
For mode-2:
VB2=0.436m⋅0.76g=0.327mg</math>">VB2=0.436m⋅0.76g=0.327mg��2=0.436�⋅0.76�=0.327��
For mode-3:
VB3=0.121m⋅1.03g=0.126mg</math>">VB3=0.121m⋅1.03g=0.126mg��3=0.121�⋅1.03�=0.126��
For mode-4:
VB4=0.038m⋅1.03g=0.039mg</math>">VB4=0.038m⋅1.03g=0.039mg��4=0.038�⋅1.03�=0.039��
For mode-5:
VB5=0.008m⋅1.03g=0.008mg</math>">VB5=0.008m⋅1.03g=0.008mg��5=0.008�⋅1.03�=0.008��
Now we have to caluculate the average base shear by the method of Square root of sum of square:
i.e,VB=((VB1)2+(VB2)2+(VB3)2+(VB4)2+(VB5)2)</math>">VB=√((VB1)2+(VB2)2+(VB3)2+(VB4)2+(VB5)2)��=((��1)2+(��2)2+(��3)2+(��4)2+(��5)2)
VB=((1.187)2+(0.327)2+(0.126)2+(0.039)2+(0.008)2)</math>">VB=√((1.187)2+(0.327)2+(0.126)2+(0.039)2+(0.008)2)��=((1.187)2+(0.327)2+(0.126)2+(0.039)2+(0.008)2)
VB=1.15mg</math>">VB=1.15mg��=1.15��
Finally the percentages of Modal mass participation ratio are obtained as:
mode 1- 88%
mode 2-8.7%
mode 3-2.4%
mode 4-0.74%
mode 5-0.16%
Total effective modal mass= 4.397m+0.436m+0.121m+0.037m+0.008m=4.9989m≈</math>">≈≈5m
Hence, total mass of the building is verified.
the average base shear value is 1.15mg is calculated by SRSS Rule.