Understanding the concepts on Degrees of Freedom

UNDERSTANDING THE CONCEPTS ON DEGREES OF FREEDOM

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 frequency, T₁ = 2 seconds and its associated mode shape:

ɸ₁= ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.3340.6410.8951.0781.173⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.3340.6410.8951.0781.173]

• The second mode of vibration, T₂ = 0.6852 seconds and its associated mode shape:

ɸ₂= ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−0.895−1.173−0.6410.3341.078⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[-0.895-1.173-0.6410.3341.078]

• The third mode of vibration,T₃ = 0.4346 seconds and its associated mode shape:

ɸ₃= ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1.1730.334−1.078−0.6410.895⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[1.1730.334-1.078-0.6410.895]

• The fourth mode of vibration, T₄ = 0.3383 seconds and its associated mode shape:

ɸ₄= ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−1.0780.8950.334−1.1730.641⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[-1.0780.8950.334-1.1730.641]

• The fifth mode of vibration, T₅ = 0.2966 seconds and its associated mode shape:

ɸ₅= ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.641−1.0781.173−0.8950.334⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.641-1.0781.173-0.8950.334]

AIM:

• To calculate L_n , M_n and Γ_n for each of the five modes and the effective modal mass M_n* and the modal mass participating ratio for the 5 modes using the equation given below:

• To 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

`alculations`

• First we are calculating the entire parameters for mode 1:

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

• Now, we have to obtain the transformed mass matrix for mode 1:

M₁ =ΦΦ₁^T m ΦΦ₁

i.e, M₁ = [0.3340.6410.8951.0781.173]⋅[0.3340.6410.8951.0781.173]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅[m00000m00000m00000m00000m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.3340.6410.8951.0781.173⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.3340.6410.8951.0781.173]

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

M₁ = 3.86m.

• Now, we need to find the modal property, L_{n .}

  L_n = ɸ_n^Tmɩ

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

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

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

L₁ = 4.12m.

• The modal participation factor, Γ_n = LnMnLnMn

Γ₁ = 4.12m3.86m4.12m3.86m= 1.067

• Seismic force for mode 1: f_n = mɸ_nΓ_nA_n

f₁ = mɸ₁Γ₁A₁

A₁ = 0.27g for time period of 2 seconds.

f₁ = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅[m00000m00000m00000m00000m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.3340.6410.8951.0781.173⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.3340.6410.8951.0781.173]1.067⋅0.27g1.067⋅0.27g

f₁ = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.334m0.641m0.895m1.078m1.173m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅1.067⋅0.27g[0.334m0.641m0.895m1.078m1.173m]⋅1.067⋅0.27g

f₁ = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.096mg0.185mg0.258mg0.310mg0.338mg⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.096mg0.185mg0.258mg0.310mg0.338mg]

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

f₁ = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.096W0.185W0.258W0.310W0.338W⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.096W0.185W0.258W0.310W0.338W]

• Now, to find out the base shear for mode 1, all the values of the seismic forces can be added.

i.e,

V_b1 = 0.096W+0.185W+0.258W+0.310W+0.338W = 1.187W.

• Similarly, we need to find out the Base shear for all the modes.
• V_b1 can also be calculated by finding out the Effective modal mass, M_n^*.

M₁^* = (Ln)2Mn(Ln)2Mn= (4.12m)23.86m(4.12m)23.86m= 4.397m.

Hence, modal mass participating ratio, M_n^* /m for mode 1 =4.397m5m4.397m5m= 0.88 = 88%

V_b1 = M_n^* .A_n = 4.397m x 0.27g = 1.187mg = 1.187W.

- For mode 2:

• Mass matrix, m = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[m00000m00000m00000m00000m]
• Transformed mass matrix, M₂ =ΦΦ₂^T m ΦΦ₂

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

M₂ = [−0.895m−1.173m−0.641m0.334m1.078m]⋅[-0.895m-1.173m-0.641m0.334m1.078m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−0.895−1.173−0.6410.3341.078⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[-0.895-1.173-0.6410.3341.078]

M₂ = 3.86m.

• The modal property, L_{n ,} L_n = ɸ_n^Tmɩ

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

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

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

L₂ = -1.297m.

• The modal participation factor, Γ_n = LnMnLnMn

Γ₂ = −1.297m3.86m-1.297m3.86m= -0.336

• Seismic force for mode 2: f_n = mɸ_nΓ_nA_n

f₂ = mɸ₂Γ₂A₂

A₂ = 0.76g for time period of 0.6852 seconds.

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

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

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

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

f₂ = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.228W0.229W0.164W−0.085W−0.275W⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.228W0.229W0.164W-0.085W-0.275W]

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

i.e, V_b2 = 0.228W+0.299W+0.164W-0.085W-0.275W = 0.331W

• V_b2 by finding out the Effective modal mass, M_n^*.

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

Hence, modal mass participating ratio, M_n^* /m for mode 2 =0.436m5m0.436m5m= 0.087 = 8.7%

V_b2 = M_n^* .A_n = 0.436m x 0.76g = 0.331mg = 0.331W.

- For mode 3:

• Mass matrix, m = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[m00000m00000m00000m00000m]
• Transformed mass matrix, M₃ =ΦΦ₃^T m ΦΦ₃

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

M₃ = [1.173m0.334m−1.078m−0.641m0.895m]⋅[1.173m0.334m-1.078m-0.641m0.895m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1.1730.334−1.078−0.6410.895⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[1.1730.334-1.078-0.6410.895]

M₃ = 3.86m.

• The modal property, L_{n ,} L_n = ɸ_n^Tmɩ

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

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

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

L₃ = 0.683m.

• The modal participation factor, Γ_n = LnMnLnMn

Γ₃ = 0.683m3.86m0.683m3.86m= 0.177

• Seismic force for mode 3: f_n = mɸ_nΓ_nA_n

f₃ = mɸ₃Γ₃A₃

A₃ = 1.03g for time period of 0.4346 seconds.

f₃ = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅[m00000m00000m00000m00000m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1.1730.334−1.078−0.6410.895⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[1.1730.334-1.078-0.6410.895]0.177⋅1.03g0.177⋅1.03g

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

f₃ = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.214mg0.061mg−0.196mg−0.117mg0.163mg⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.214mg0.061mg-0.196mg-0.117mg0.163mg]

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

f₃ = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.214W0.061W−0.196W−0.117W0.163W⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.214W0.061W-0.196W-0.117W0.163W]

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

i.e, V_b3 = 0.214W+0.061W-0.196W-0.117W+0.163W = 0.125W.

• V_b3 by finding out the Effective modal mass, M_n^*.

M₃^* = (Ln)2Mn(Ln)2Mn= (0.683m)23.86m(0.683m)23.86m= 0.121m.

Hence, modal mass participating ratio, M_n^* /m for mode 3 =0.121m5m0.121m5m= 0.024 = 2.4%

V_b3 = M_n^* .A_n = 0.121m x 1.03g = 0.331mg = 0.125W.

- For mode 4:

• Mass matrix, m = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[m00000m00000m00000m00000m]
• Transformed mass matrix, M₄ =ΦΦ₄^T m ΦΦ₄

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

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

M₄ = 3.86m.

• The modal property, L_{n ,} L_n = ɸ_n^Tmɩ

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

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

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

L₄ = -0.381m.

• The modal participation factor, Γ_n = LnMnLnMn

Γ₄ = −0.381m3.86m-0.381m3.86m= -0.099

• Seismic force for mode 4: f_n = mɸ_nΓ_nA_n

f₄ = mɸ₄Γ₄A₄

A₄ = 1.03g for time period of 0.3383 seconds.

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

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

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

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⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.109W-0.091W-0.034W0.119W-0.065W]

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

i.e, V_b4 = 0.109W-0.091W-0.034W+0.119W-0.065W = 0.038W.

• V_b4 by finding out the Effective modal mass, M_n^*.

M₄^* = (Ln)2Mn(Ln)2Mn= (−0.381m)23.86m(-0.381m)23.86m= 0.037m.

Hence, modal mass participating ratio, M₄^* /m for mode 4 =0.037m5m0.037m5m= 0.0074 = 0.74%

V_b4 = M_n^* .A_n = 0.037m x 1.03g = 0.038mg = 0.038W.

- For mode 5:

• Mass matrix, m = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[m00000m00000m00000m00000m]
• Transformed mass matrix, M₅ =ΦΦ₅^T m ΦΦ₅

i.e, M₅ = [0.641−1.0781.173−0.8950.334]⋅[0.641-1.0781.173-0.8950.334]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅[m00000m00000m00000m00000m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.641−1.0781.173−0.8950.334⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.641-1.0781.173-0.8950.334]

M₅ = [0.641m−1.078m1.173m−0.895m0.334m]⋅[0.641m-1.078m1.173m-0.895m0.334m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.641−1.0781.173−0.8950.334⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.641-1.0781.173-0.8950.334]

M₅ = 3.86m.

• The modal property, L_{n ,} L_n = ɸ_n^Tmɩ

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

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

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

L₅ = 0.175m.

• The modal participation factor, Γ_n = LnMnLnMn

Γ₅ = 0.175m3.86m0.175m3.86m= 0.045

• Seismic force for mode 5: f_n = mɸ_nΓ_nA_n

f₅ = mɸ₅Γ₅A₅

A₅ = 1.03g for time period of 0.2966 seconds.

f₅ = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣m00000m00000m00000m00000m⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅[m00000m00000m00000m00000m]⋅⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.641−1.0781.173−0.8950.334⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅[0.641-1.0781.173-0.8950.334]⋅0.045⋅1.03g0.045⋅1.03g

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

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

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⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.029W-0.049W0.054W-0.041W0.015W]

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

i.e, V_b5 = 0.029W-0.049W+0.054W-0.041W+0.015W = 0.008

• V_b5 by finding out the Effective modal mass, M_n^*.

M₅^* = (Ln)2Mn(Ln)2Mn= (0.175m)23.86m(0.175m)23.86m= 0.0079m.

Hence, modal mass participating ratio, M₅^* /m for mode 5 =0.0079m5m0.0079m5m= 0.0016 = 0.16%

V_b5 = M_n^* .A_n = 0.0079m x 1.03g = 0.038mg = 0.008W.

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

Now, Combined base shear by SRSS rule,

V_b = √(Vb1)2+(Vb2)2+(Vb3)2+(Vb4)2+(Vb5)2(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(1.187W)2+(0.331W)2+(0.125W)2+(0.038W)2+(0.008W)2= 1.239W.

V_b = 1.239W.

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, Modal mass participation ratio obtained are:

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 sum of mode 1 and mode 2 will exceed 90%

i.e, 88 + 8.7 = 96.7%.