Understanding the concepts on Degrees of Freedom

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 T_n and their corresponding mode shapes, φ_n 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:

• L_n , M_n and Γ_n for each of the five modes are to be computed from Equations below.
• Also, calculate the effective modal mass M_n* 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⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[�00000�00000�00000�00000�]

Now we calculate the transformed mass matrix for all modes-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⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�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⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�1=[0.334�0.641�0.895�1.078�1.173�]⋅[0.3340.6410.8951.0781.173]

M1=3.86m�1=3.86�

For mode-2

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⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�2=[-0.895-1.173-0.6410.3341.078]⋅[�00000�00000�00000�00000�]⋅[-0.895-1.173-0.6410.3341.078]

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�2=3.86�

For mode-3

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⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�3=[1.1730.334-1.078-0.6410.895]⋅[�00000�00000�00000�00000�]⋅[1.1730.334-1.078-0.6410.895]

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�3=3.86�

For mode-4

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⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�4=[-1.0780.8950.334-1.1730.641]⋅[�00000�00000�00000�00000�]⋅[-1.0780.8950.334-1.1730.641]

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�4=3.86�

For mode-5

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⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�5=[0.641-1.0781.173-0.8950.334]⋅[�00000�00000�00000�00000�]⋅[0.641-1.0781.173-0.8950.334]

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�5=3.86�

Now we have to determine modal proprety.(L)

2)Modal property(Ln��):ϕTn⋅m⋅l���⋅�⋅�

Where,l=�= infulence matrix is aunit matrix,⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]

For mode-1:

(L1�1):ϕT1⋅m⋅l�1�⋅�⋅�

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⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�1=[0.334�0.641�0.895�1.078�1.173�]⋅[11111]

L1=4.12m�1=4.12�

For mode-2:

(L2�2):ϕT2⋅m⋅l�2�⋅�⋅�

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=[−0.895m−1.173m−0.641m0.334m1.078m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�2=[�-0.895�-1.173�-0.6410.334�1.078�]⋅[11111]

L2=−1.297m�2=-1.297�

For mode-3:

(L3�3):ϕT3⋅m⋅l�3�⋅�⋅�

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=[1.173m0.334m−1.078m−0.641m0.895m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�3=[1.173�0.334��-1.078�-0.6410.895�]⋅[11111]

L3=0.683m�3=0.683�

For mode-4

(L4�4):ϕT4⋅m⋅l�4�⋅�⋅�

L4=[−1.0780.8950.334−1.1730.641]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�4=[-1.0780.8950.334-1.1730.641]⋅[�00000�00000�00000�00000�]⋅[11111]

L4=[−1.078m0.895m0.334m−1.173m0.641m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�4=[�-1.0780.895�0.334��-1.1730.641�]⋅[11111]

L4=−0.381m�4=-0.381�

For mode-5

(L5�5):ϕT5⋅m⋅l�5�⋅�⋅�

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=[0.641m−1.078m1.173m−0.895m0.334m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�5=[0.641��-1.0781.173��-0.8950.334�]⋅[11111]

L5=0.175m�5=0.175�

Now, we have to calculate the 3)Modal participation factor;ΓnΓ�

It is defined as the ratio of modal property (Ln��) to modified mass matrix(M).

Γn=LnMnΓ�=����

For mode-1:

Γ1=L1M1Γ1=�1�1

Γ1=4.121m3.861m=1.067Γ1=4.121�3.861�=1.067

Γ1=1.067Γ1=1.067

For mode-2:

Γ2=L2M2Γ2=�2�2

Γ2=−1.297m3.861m=−0.335Γ2=-1.297�3.861�=-0.335

Γ2=−0.335Γ2=-0.335

For mode-3:

Γ3=L3M3Γ3=�3�3

Γ3=0.683m3.861m=0.177Γ3=0.683�3.861�=0.177

Γ3=0.177Γ3=0.177

For mode-4

Γ4=L4M4Γ4=�4�4

Γ4=−0.381m3.861m=−0.099Γ4=-0.381�3.861�=-0.099

Γ4=−0.099Γ4=-0.099

For mode-5

Γ5=L5M5Γ5=�5�5

Γ5=0.175m3.861m=0.045Γ5=0.175�3.861�=0.045

Γ5=−0.045Γ5=-0.045

After finding the modal participation factor now we have to calculate the effective modal mass(M∗=L2nMn�∗=��2��)

4)effective modal mass(M∗=L2nMn�∗=��2��):

For mode-1:

M∗1=L21M1�1∗=�12�1

M∗1=(4.121m)23.861m=4.398m�1∗=(4.121�)23.861�=4.398�

M∗1=4.398m�1∗=4.398�

For mode-2:

M∗2=L22M2�2∗=�22�2

M∗2=(−1.297m)23.861m=−0.335m�2∗=(-1.297�)23.861�=-0.335�

M∗2=0.436m�2∗=0.436�

For mode-3:

M∗3=L23M3�3∗=�32�3

M∗3=(0.683m)23.861m=0.121m�3∗=(0.683�)23.861�=0.121�

M∗3=0.121m�3∗=0.121�

For mode-4

M∗4=L24M4�4∗=�42�4

M∗4=(−0.381m)23.861m=0.038m�4∗=(-0.381�)23.861�=0.038�

M∗4=0.038m�4∗=0.038�

For mode-5

M∗5=L25M5�5∗=�52�5

M∗5=(0.175m)23.861m=0.008m�5∗=(0.175�)23.861�=0.008�

M∗5=0.008m�5∗=0.008�

Now we have to calculate the modal mass participation ratio.

5)Modal mass particiopation ratio=M∗nmj��∗��

Where mj��is total mass of the buuilding=m+m+m+m+m= 5m.

For mode-1:

M∗1mj=4.938m5m=0.88�1∗��=4.938�5�=0.88

For mode-2:

M∗2mj=0.436m5m=0.09�2∗��=0.436�5�=0.09

For mode-3:

M∗3mj=0.121m5m=0.02�3∗��=0.121�5�=0.02

For mode-4:

M∗4mj=0.038m5m=0.01�4∗��=0.038�5�=0.01

For mode-5:

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��)

Base shear(VB=M∗n.An��=��∗.��)

For mode-1:

VB1=4.398m⋅0.27g=1.187mg��1=4.398�⋅0.27�=1.187��

For mode-2:

VB2=0.436m⋅0.76g=0.327mg��2=0.436�⋅0.76�=0.327��

For mode-3:

VB3=0.121m⋅1.03g=0.126mg��3=0.121�⋅1.03�=0.126��

For mode-4:

VB4=0.038m⋅1.03g=0.039mg��4=0.038�⋅1.03�=0.039��

For mode-5:

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)��=((��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)��=((1.187)2+(0.327)2+(0.126)2+(0.039)2+(0.008)2)

VB=1.15mg��=1.15��