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⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[m00000m00000m00000m00000m]

Now we calculate the transformed mass matrix for all modes-M=ΦT⋅m⋅ΦM=ΦT⋅m⋅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⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦M1=[0.3340.6410.8951.0781.173]⋅[m00000m00000m00000m00000m]⋅[0.3340.6410.8951.0781.173]

M1=[0.334m0.641m0.895m1.078m1.173m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.3340.6410.8951.0781.173⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦M1=[0.334m0.641m0.895m1.078m1.173m]⋅[0.3340.6410.8951.0781.173]

M1=3.86mM1=3.86m

For mode-2

M2=ΦT2⋅m⋅Φ2M2=Φ2T⋅m⋅Φ2

i.e, M2=[−0.895−1.173−0.6410.3341.078]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−0.895−1.173−0.6410.3341.078⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦M2=[-0.895-1.173-0.6410.3341.078]⋅[m00000m00000m00000m00000m]⋅[-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⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦M2=[m-0.895m-1.173m-0.6410.334m1.078m]⋅[-0.895-1.173-0.6410.3341.078]

M2=3.86mM2=3.86m

For mode-3

M3=ΦT3⋅m⋅Φ3M3=Φ3T⋅m⋅Φ3

i.e,M3=[1.1730.334−1.078−0.6410.895]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1.1730.334−1.078−0.6410.895⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦M3=[1.1730.334-1.078-0.6410.895]⋅[m00000m00000m00000m00000m]⋅[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⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦M3=[1.173m0.334mm-1.078m-0.6410.895m]⋅[1.1730.334-1.078-0.6410.895]

M3=3.86mM3=3.86m

For mode-4

M4=ΦT4⋅m⋅Φ4M4=Φ4T⋅m⋅Φ4

i.e,M4=[−1.0780.8950.334−1.1730.641]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−1.0780.8950.334−1.1730.641⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦M4=[-1.0780.8950.334-1.1730.641]⋅[m00000m00000m00000m00000m]⋅[-1.0780.8950.334-1.1730.641]

M4=[−1.078m0.895m0.334m−1.173m0.641m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−1.0780.8950.334−1.1730.641⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦M4=[m-1.0780.895m0.334mm-1.1730.641m]⋅[-1.0780.8950.334-1.1730.641]

M4=3.86mM4=3.86m

For mode-5

M5=ΦT5⋅m⋅Φ5M5=Φ5T⋅m⋅Φ5

M5=[0.641−1.0781.173−0.8950.334]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.641−1.0781.173−0.8950.334⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦M5=[0.641-1.0781.173-0.8950.334]⋅[m00000m00000m00000m00000m]⋅[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⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦M5=[0.641mm-1.0781.173mm-0.8950.334m]⋅[0.641-1.0781.173-0.8950.334]

M5=3.86mM5=3.86m

Now we have to determine modal proprety.(L)

2)Modal property(LnLn):ϕTn⋅m⋅lϕnT⋅m⋅l

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

For mode-1:

(L1L1):ϕT1⋅m⋅lϕ1T⋅m⋅l

L1=[0.3340.6410.8951.0781.173]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦L1=[0.3340.6410.8951.0781.173]⋅[m00000m00000m00000m00000m]⋅[11111]

L1=[0.334m0.641m0.895m1.078m1.173m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦L1=[0.334m0.641m0.895m1.078m1.173m]⋅[11111]

L1=4.12mL1=4.12m

For mode-2:

(L2L2):ϕT2⋅m⋅lϕ2T⋅m⋅l

L2=[−0.895−1.173−0.6410.3341.078]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦L2=[-0.895-1.173-0.6410.3341.078]⋅[m00000m00000m00000m00000m]⋅[11111]

L2=[−0.895m−1.173m−0.641m0.334m1.078m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦L2=[m-0.895m-1.173m-0.6410.334m1.078m]⋅[11111]

L2=−1.297mL2=-1.297m

For mode-3:

(L3L3):ϕT3⋅m⋅lϕ3T⋅m⋅l

L3=[1.1730.334−1.078−0.6410.895]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦L3=[1.1730.334-1.078-0.6410.895]⋅[m00000m00000m00000m00000m]⋅[11111]

L3=[1.173m0.334m−1.078m−0.641m0.895m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦L3=[1.173m0.334mm-1.078m-0.6410.895m]⋅[11111]

L3=0.683mL3=0.683m

For mode-4

(L4L4):ϕT4⋅m⋅lϕ4T⋅m⋅l

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

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

L4=−0.381mL4=-0.381m

For mode-5

(L5L5):ϕT5⋅m⋅lϕ5T⋅m⋅l

L5=[0.641−1.0781.173−0.8950.334]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦L5=[0.641-1.0781.173-0.8950.334]⋅[m00000m00000m00000m00000m]⋅[11111]

L5=[0.641m−1.078m1.173m−0.895m0.334m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦L5=[0.641mm-1.0781.173mm-0.8950.334m]⋅[11111]

L5=0.175mL5=0.175m

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

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

Γn=LnMnΓn=LnMn

For mode-1:

Γ1=L1M1Γ1=L1M1

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

Γ1=1.067Γ1=1.067

For mode-2:

Γ2=L2M2Γ2=L2M2

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

Γ2=−0.335Γ2=-0.335

For mode-3:

Γ3=L3M3Γ3=L3M3

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

Γ3=0.177Γ3=0.177

For mode-4

Γ4=L4M4Γ4=L4M4

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

Γ4=−0.099Γ4=-0.099

For mode-5

Γ5=L5M5Γ5=L5M5

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

Γ5=−0.045Γ5=-0.045

After finding the modal participation factor now we have to calculate the effective modal mass(M∗=L2nMnM∗=Ln2Mn)

4)effective modal mass(M∗=L2nMnM∗=Ln2Mn):

For mode-1:

M∗1=L21M1M1∗=L12M1

M∗1=(4.121m)23.861m=4.398mM1∗=(4.121m)23.861m=4.398m

M∗1=4.398mM1∗=4.398m

For mode-2:

M∗2=L22M2M2∗=L22M2

M∗2=(−1.297m)23.861m=−0.335mM2∗=(-1.297m)23.861m=-0.335m

M∗2=0.436mM2∗=0.436m

For mode-3:

M∗3=L23M3M3∗=L32M3

M∗3=(0.683m)23.861m=0.121mM3∗=(0.683m)23.861m=0.121m

M∗3=0.121mM3∗=0.121m

For mode-4

M∗4=L24M4M4∗=L42M4

M∗4=(−0.381m)23.861m=0.038mM4∗=(-0.381m)23.861m=0.038m

M∗4=0.038mM4∗=0.038m

For mode-5

M∗5=L25M5M5∗=L52M5

M∗5=(0.175m)23.861m=0.008mM5∗=(0.175m)23.861m=0.008m

M∗5=0.008mM5∗=0.008m

Now we have to calculate the modal mass participation ratio.

5)Modal mass particiopation ratio=M∗nmjMn∗mj

Where mjmjis total mass of the buuilding=m+m+m+m+m= 5m.

For mode-1:

M∗1mj=4.938m5m=0.88M1∗mj=4.938m5m=0.88

For mode-2:

M∗2mj=0.436m5m=0.09M2∗mj=0.436m5m=0.09

For mode-3:

M∗3mj=0.121m5m=0.02M3∗mj=0.121m5m=0.02

For mode-4:

M∗4mj=0.038m5m=0.01M4∗mj=0.038m5m=0.01

For mode-5:

M∗5mj=0.008m5m=0.002M5∗mj=0.008m5m=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(VBVB)

Base shear(VB=M∗n.AnVB=Mn∗.An)

For mode-1:

VB1=4.398m⋅0.27g=1.187mgVB1=4.398m⋅0.27g=1.187mg

For mode-2:

VB2=0.436m⋅0.76g=0.327mgVB2=0.436m⋅0.76g=0.327mg

For mode-3:

VB3=0.121m⋅1.03g=0.126mgVB3=0.121m⋅1.03g=0.126mg

For mode-4:

VB4=0.038m⋅1.03g=0.039mgVB4=0.038m⋅1.03g=0.039mg

For mode-5:

VB5=0.008m⋅1.03g=0.008mgVB5=0.008m⋅1.03g=0.008mg

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)VB=((VB1)2+(VB2)2+(VB3)2+(VB4)2+(VB5)2)

VB=√((1.187)2+(0.327)2+(0.126)2+(0.039)2+(0.008)2)VB=((1.187)2+(0.327)2+(0.126)2+(0.039)2+(0.008)2)

VB=1.15mgVB=1.15mg