Understanding the concepts on Degrees of Freedom

• To calculate Ln,Mn��,�� and ΓnΓ� for each of the five modes
• To calculate the effective modal mass Mn��* and the modal mass participating ratio for the 5 modes.
• To find the base shear for each of the 5 modes of vibration and the combined base shear as per the SRSS rule.

Calculations:

Calculating the entire parameters for mode 1:

Mass matrix, m=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢`ROOT`⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�=[�00000�00000�00000�00000�]

Now, we have to obtain the transformed mass matrix for mode 1: M1=ΦT1mΦ1�1=Φ1��Φ1

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�

Now, we need to find the modal property, Ln��.

Ln=ΦTnmι��=Φ����

Where, ι=�= Influence matrix which is a unit matrix, ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]

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�

The modal participation factor, Γn=LnMnΓ�=����

Γ1=4.12m3.86m=1.067Γ1=4.12�3.86�=1.067

Seismic force for mode 1:fn=mΦnΓnAn��=�Φ�Γ���

f1=mΦ1Γ1A1�1=�Φ1Γ1�1

Where A1=0.27g�1=0.27� for a time period of 2 seconds.

f1=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.3340.6410.8951.0781.173⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅1.067⋅0.27g�1=[�00000�00000�00000�00000�]⋅[0.3340.6410.8951.0781.173]⋅1.067⋅0.27�

f1=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.334m0.641m0.895m1.078m1.173m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅1.067⋅0.27g�1=[0.334�0.641�0.895�1.078�1.173�]⋅1.067⋅0.27�

f1=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.096mg0.185mg0.258mg0.310mg0.338mg⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�1=[0.096��0.185��0.258��0.310��0.338��]

We know that Weight(newton), W=mg, and this can be substituted in the above equation.

f1=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.096W0.185W0.258W0.310W0.338W⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�1=[0.096�0.185�0.258�0.310�0.338�]

Now, to find out the base shear for mode1, all the values of the seismic forces can be added.

I.e., Vb1=0.096W+0.185W+0.258W+0.310W+0.338W=1.187W��1=0.096�+0.185�+0.258�+0.310�+0.338�=1.187�

Similarly, we need to find out the base shear for all the modes.

Vb1��1 can also be calculated by finding out the effective modal mass, Mn��*

M1�1*=(Ln)2Mn=(4.12m)23.86m=4.397m(��)2��=(4.12�)23.86�=4.397�

Hence, modal mass participating ratio, Mn∗m��∗� for mode 1 = 4.397m5m=0.88=88%4.397�5�=0.88=88%

Vb1=Mn��1=��*⋅An=4.397m×0.27g=1.187mg=1.187W⋅��=4.397�×0.27�=1.187��=1.187�

For mode 2:

Mass matrix, m=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�=[�00000�00000�00000�00000�]

Now, we have to obtain the transformed mass matrix for mode 2: M2=ΦT2mΦ2�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�-0.641�0.334�1.078�]⋅[-0.895-1.173-0.6410.3341.078]

M2=3.86m�2=3.86�

Now, we need to find the modal property, Ln��.

Ln=ΦTnmι��=Φ����

Where, ι=�= Influence matrix which is a unit matrix, ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]

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�-0.641�0.334�1.078�]⋅[11111]

L2=−1.297m�2=-1.297�

The modal participation factor, Γn=LnMnΓ�=����

Γ2=−1.297m3.86m=−0.336Γ2=-1.297�3.86�=-0.336

Seismic force for mode 2: `f_n=mPhi_nGamma_nA_n`

f2=mΦ2Γ2A2�2=�Φ2Γ2�2

A2=0.76g�2=0.76� for a time period of 0.6852 seconds.

f2=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−0.895−1.173−0.6410.3341.078⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦ ⋅−0.336⋅0.76g�2=[�00000�00000�00000�00000�]⋅[-0.895-1.173-0.6410.3341.078] ⋅-0.336⋅0.76�

f2=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−0.895m−1.173m−0.641m0.334m1.078m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅−0.336⋅0.76g�2=[-0.895�-1.173�-0.641�0.334�1.078�]⋅-0.336⋅0.76�

f2=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.228mg0.299mg0.164mg−0.085mg−0.275mg⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�2=[0.228��0.299��0.164��-0.085��-0.275��]

We know that Weight (newton), W=mg, and this can be substituted in the above equation.

f2=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.228W0.299W0.164W−0.085W−0.275W⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�2=[0.228�0.299�0.164�-0.085�-0.275�]

Now, to find out the base shear for mode 2, all the values of the seismic forces can be added.

I.e., Vb2=0.228W+0.299W+0.164W−0.085W−0.275W=0.331W��2=0.228�+0.299�+0.164�-0.085�-0.275�=0.331�

Vb2��2 can also be calculated by finding out the effective modal mass, Mn��*

M2�2*=(Ln)2Mn=(−1.297m)23.86m=0.436m(��)2��=(-1.297�)23.86�=0.436�

Hence, modal mass participating ratio, Mn∗m��∗� for mode 2 = 0.436m5m=0.087=8.7%0.436�5�=0.087=8.7%

Vb2=Mn��2=��*⋅An=0.436m×0.76g=0.331mg=0.331W⋅��=0.436�×0.76�=0.331��=0.331�

For mode 3:

Mass matrix, m=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�=[�00000�00000�00000�00000�]

Now, we have to obtain the transformed mass matrix for mode 3: M3=ΦT3mΦ3�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�-0.641�0.895�]⋅[1.1730.334-1.078-0.6410.895]

M3=3.86m�3=3.86�

Now, we need to find the modal property, Ln��.

Ln=ΦTnmι��=Φ����

Where, ι=�= Influence matrix which is a unit matrix, ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]

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�-0.641�0.895�]⋅[11111]

L3=0.683m�3=0.683�

The modal participation factor, Γn=LnMnΓ�=����

Γ3=0.683m3.86m=0.177Γ3=0.683�3.86�=0.177

Seismic force for mode 3: fn=mΦnΓnAn��=�Φ�Γ���

f3=mΦ3Γ3A3�3=�Φ3Γ3�3

A3=1.03g�3=1.03� for a time period of 0.4346 seconds.

f3=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1.1730.334−1.078−0.6410.895⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅0.177⋅1.03g�3=[�00000�00000�00000�00000�]⋅[1.1730.334-1.078-0.6410.895]⋅0.177⋅1.03�

f3=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1.173m0.334m−1.078m−0.641m0.895m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅0.177⋅1.03g�3=[1.173�0.334�-1.078�-0.641�0.895�]⋅0.177⋅1.03�

f3=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.214mg0.061mg−0.196mg−0.177mg0.163mg⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�3=[0.214��0.061��-0.196��-0.177��0.163��]

We know that Weight (newton), W=mg, and this can be substituted in the above equation.

f3=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.214W0.061W−0.196W−0.177W0.163W⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�3=[0.214�0.061�-0.196�-0.177�0.163�]

Now, to find out the base shear for mode 3, all the values of the seismic forces can be added.

I.e., Vb3=0.214W+0.061W−0.196W−0.177W+0.163W=0.125W��3=0.214�+0.061�-0.196�-0.177�+0.163�=0.125�

Vb3��3 can also be calculated by finding out the effective modal mass, Mn��*

M3�3*=(Ln)2Mn=(0.683m)23.86m=0.121m(��)2��=(0.683�)23.86�=0.121�

Hence, modal mass participating ratio, Mn∗m��∗� for mode 3 = 0.121m5m=0.024=2.4%0.121�5�=0.024=2.4%

Vb3=Mn��3=��*⋅An=0.121m×1.03g=0.125mg=0.125W⋅��=0.121�×1.03�=0.125��=0.125�

For mode 4:

Mass matrix, m=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�=[�00000�00000�00000�00000�]

Now, we have to obtain the transformed mass matrix for mode 4: M4=ΦT4mΦ4�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=[-1.078�0.895�0.334�-1.173�0.641�]⋅[-1.0780.8950.334-1.1730.641]

M4=3.86m�4=3.86�

Now, we need to find the modal property, Ln��.

Ln=ΦTnmι��=Φ����

Where, ι=�= Influence matrix which is a unit matrix, ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]

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=[-1.078�0.895�0.334�-1.173�0.641�]⋅[11111]

L4=−0.381m�4=-0.381�

The modal participation factor, Γn=LnMnΓ�=����

Γ4=−0.381m3.86m=−0.099Γ4=-0.381�3.86�=-0.099

Seismic force for mode 4: fn=mΦnΓnAn��=�Φ�Γ���

f4=mΦ4Γ4A4�4=�Φ4Γ4�4

A4=1.03g�4=1.03� for a time period of 0.3383 seconds.

f4=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−1.0780.8950.334−1.1730.641⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅−0.099⋅1.03g�4=[�00000�00000�00000�00000�]⋅[-1.0780.8950.334-1.1730.641]⋅-0.099⋅1.03�

f4=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−1.078m0.895m0.334m−1.173m0.641m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅−0.099⋅1.03g�4=[-1.078�0.895�0.334�-1.173�0.641�]⋅-0.099⋅1.03�

f4=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.109mg−0.091mg−0.034mg0.119mg−0.065mg⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�4=[0.109��-0.091��-0.034��0.119��-0.065��]

We know that Weight (newton), W=mg, and this can be substituted in the above equation.

f4=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.109W−0.091W−0.034W0.119W−0.065W⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�4=[0.109�-0.091�-0.034�0.119�-0.065�]

Now, to find out the base shear for mode 4, all the values of the seismic forces can be added.

I.e., Vb4=0.109W−0.091W−0.034W+0.119W−0.065W=0.038W��4=0.109�-0.091�-0.034�+0.119�-0.065�=0.038�

Vb4��4 can also be calculated by finding out the effective modal mass, Mn��*

M4�4*=(Ln)2Mn=(−0.381m)23.86m=0.037m(��)2��=(-0.381�)23.86�=0.037�

Hence, modal mass participating ratio, Mn∗m��∗� for mode 4 = 0.037m5m=0.0074=0.74%0.037�5�=0.0074=0.74%

Vb4=Mn��4=��*⋅An=0.037m×1.03g=0.038mg=0.038W⋅��=0.037�×1.03�=0.038��=0.038�

For mode 5:

Mass matrix, m=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�=[�00000�00000�00000�00000�]

Now, we have to obtain the transformed mass matrix for mode 5: M5=ΦT5mΦ5�5=Φ5��Φ5

I.e., 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�-1.078�1.173�-0.895�0.334�]⋅[0.641-1.0781.173-0.8950.334]

M5=3.86m�5=3.86�

Now, we need to find the modal property, Ln��.

Ln=ΦTnmι��=Φ����

Where, ι=�= Influence matrix which is a unit matrix, ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]

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�-1.078�1.173�-0.895�0.334�]⋅[11111]

L5=0.175m�5=0.175�

The modal participation factor, Γn=LnMnΓ�=����

Γ5=0.175m3.86m=0.045Γ5=0.175�3.86�=0.045

Seismic force for mode 5: fn=mΦnΓnAn��=�Φ�Γ���

f5=mΦ5Γ5A5�5=�Φ5Γ5�5

A5=1.03g�5=1.03� for a time period of 0.2966 seconds.

f5=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.641−1.0781.173−0.8950.334⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅0.045⋅1.03g�5=[�00000�00000�00000�00000�]⋅[0.641-1.0781.173-0.8950.334]⋅0.045⋅1.03�

f5=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.641m−1.078m1.173m−0.895m0.334m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅0.045⋅1.03g�5=[0.641�-1.078�1.173�-0.895�0.334�]⋅0.045⋅1.03�

f5=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.029mg−0.049mg0.054mg−0.041mg0.015mg⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�5=[0.029��-0.049��0.054��-0.041��0.015��]

We know that Weight (newton), W=mg, and this can be substituted in the above equation.

f5=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.029W−0.049W0.054W−0.041W0.015W⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦�5=[0.029�-0.049�0.054�-0.041�0.015�]

Now, to find out the base shear for mode 5, all the values of the seismic forces can be added.

I.e., Vb5=0.029W−0.049W+0.054W−0.041W+0.015W=0.008W��5=0.029�-0.049�+0.054�-0.041�+0.015�=0.008�

Vb5��5 can also be calculated by finding out the effective modal mass, Mn

M5�5*=(Ln)2Mn=(0.175m)23.86m=0.0079m()2=(0.175)23.86=0.0079

Hence, modal mass participating ratio, Mn∗m��∗� for mode 5 = 0.0079m5m=0.0016=0.16%0.0079�5�=0.0016=0.16%

Vb5=Mn��5=��*⋅An=0.0079m×1.03g=0.008mg=0.008W⋅��=0.0079�×1.03�=0.008��=0.008�

Hence, Base shear has been obtained for 5 modes of vibration.

Now, Combined base shear by SRSS Rule.

Vb=√(Vb1)2+(Vb2)2+(Vb3)2+(Vb4)2+(Vb5)2=√(1.187W)2+(0.331W)2+(0.125W)2+(0.038W)2+(0.008W)2=1.239W��=(��1)2+(��2)2+(��3)2+(��4)2+(��5)2=(1.187�)2+(0.331�)2+(0.125�)2+(0.038�)2+(0.008�)2=1.239�

Vb=1.239W=1.239

Total mass of building= 5m

Total effective modal mass= 4.397+0.436+0.121+0.037+0.0079= 4.9989m ≈≈5m

Hence Total mass of the building is verified.

Now, The Modal mass participation ratios obtained is:

Mode 1- 88%

Mode 2- 8.7%

Mode 3- 2.4%

Mode 4- 0.74%

Mode 5- 0.16%

From the above values, it is clear that the sum of mode 1 & mode 2 will exceed 90% .

I.e., 88% + 8.7% = 96.7%