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= `[[0.334],[0.641],[0.895],[1.078],[1.173]]`

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

ɸ₂= `[[-0.895],[-1.173],[-0.641],[0.334],[1.078]]`

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

ɸ₃= `[[1.173],[0.334],[-1.078],[-0.641],[0.895]]`

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

ɸ₄= `[[-1.078],[0.895],[0.334],[-1.173],[0.641]]`

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

ɸ₅= `[[0.641],[-1.078],[1.173],[-0.895],[0.334]]`

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

Calculations:

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

Mass matrix, m = `[[m,0,0,0,0],[0,m,0,0,0],[0,0,m,0,0],[0,0,0,m,0],[0,0,0,0,m]]`

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

M₁ =`Phi`₁^T m `Phi`₁

i.e, M₁ = `[[0.334,0.641,0.895,1.078,1.173]]*``[[m,0,0,0,0],[0,m,0,0,0],[0,0,m,0,0],[0,0,0,m,0],[0,0,0,0,m]]*``[[0.334],[0.641],[0.895],[1.078],[1.173]]`

M₁ = `[[0.334m,0.641m,0.895m,1.078m,1.173m]]*``[[0.334],[0.641],[0.895],[1.078],[1.173]]`

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, `[[1],[1],[1],[1],[1]]`

L₁ = `[[0.334,0.641,0.895,1.078,1.173]]*``[[m,0,0,0,0],[0,m,0,0,0],[0,0,m,0,0],[0,0,0,m,0],[0,0,0,0,m]]*``[[1],[1],[1],[1],[1]]`

L₁ = `[[0.334m,0.641m,0.895m,1.078m,1.173m]]*``[[1],[1],[1],[1],[1]]`

L₁ = 4.12m.

• The modal participation factor, Γ_n = `(L n)/(Mn)`

Γ₁ = `(4.12m)/(3.86m)`= 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 = `[[m,0,0,0,0],[0,m,0,0,0],[0,0,m,0,0],[0,0,0,m,0],[0,0,0,0,m]]*``[[0.334],[0.641],[0.895],[1.078],[1.173]]``1.067*0.27g`

f₁ = `[[0.334m],[0.641m],[0.895m],[1.078m],[1.173m]]*1.067*0.27g`

f₁ = `[[0.096mg],[0.185mg],[0.258mg],[0.310mg],[0.338mg]]`

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

f₁ = `[[0.096W],[0.185W],[0.258W],[0.310W],[0.338W]]`

• 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^* = `(L n)^2/(Mn)`= `(4.12m)^2/(3.86m)`= 4.397m.

Hence, modal mass participating ratio, M_n^* /m for mode 1 =`( 4.397m)/(5m)`= 0.88 = 88%

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

- For mode 2:

• Mass matrix, m = `[[m,0,0,0,0],[0,m,0,0,0],[0,0,m,0,0],[0,0,0,m,0],[0,0,0,0,m]]`
• Transformed mass matrix, M₂ =`Phi`₂^T m `Phi`₂

i.e, M₂ = `[[-0.895,-1.173,-0.641,0.334,1.078]]*``[[m,0,0,0,0],[0,m,0,0,0],[0,0,m,0,0],[0,0,0,m,0],[0,0,0,0,m]]*``[[-0.895],[-1.173],[-0.641],[0.334],[1.078]]`

M₂ = `[[-0.895m,-1.173m,-0.641m,0.334m,1.078m]]*``[[-0.895],[-1.173],[-0.641],[0.334],[1.078]]`

M₂ = 3.86m.

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

Where, ɩ = Influence matrix which is a unit matrix, `[[1],[1],[1],[1],[1]]`

L₂ = `[[-0.895,-1.173,-0.641,0.334,1.078]]*``[[m,0,0,0,0],[0,m,0,0,0],[0,0,m,0,0],[0,0,0,m,0],[0,0,0,0,m]]*``[[1],[1],[1],[1],[1]]`

L₂ = `[[-0.895m,-1.173m,-0.641m,0.334m,1.078m]]*``[[1],[1],[1],[1],[1]]`

L₂ = -1.297m.

• The modal participation factor, Γ_n = `(L n)/(Mn)`

Γ₂ = `(-1.297m)/(3.86m)`= -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 = `[[m,0,0,0,0],[0,m,0,0,0],[0,0,m,0,0],[0,0,0,m,0],[0,0,0,0,m]]*``[[-0.895],[-.1.173],[-0.641],[0.334],[1.078]]``-0.336*0.76g`

f₂ = `[[-0.895m],[-1.173m],[-0.641m],[0.334m],[1.078m]]*-0.336*0.76g`

f₂ = `[[0.228mg],[0.299mg],[0.164mg],[-0.085mg],[-0.275mg]]`

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

f₂ = `[[0.228W],[0.229W],[0.164W],[-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^* = `(L n)^2/(Mn)`= `(-1.297m)^2/(3.86m)`= 0.436m.

Hence, modal mass participating ratio, M_n^* /m for mode 2 =`(0.436m)/(5m)`= 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 = `[[m,0,0,0,0],[0,m,0,0,0],[0,0,m,0,0],[0,0,0,m,0],[0,0,0,0,m]]`
• Transformed mass matrix, M₃ =`Phi`₃^T m `Phi`₃

i.e, M₃ = `[[1.173,0.334,-1.078,-0.641,0.895]]*``[[m,0,0,0,0],[0,m,0,0,0],[0,0,m,0,0],[0,0,0,m,0],[0,0,0,0,m]]*``[[1.173],[0.334],[-1.078],[-0.641],[0.895]]`

M₃ = `[[1.173m,0.334m,-1.078m,-0.641m,0.895m]]*``[[1.173],[0.334],[-1.078],[-0.641],[0.895]]`

M₃ = 3.86m.

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

Where, ɩ = Influence matrix which is a unit matrix, `[[1],[1],[1],[1],[1]]`

L₃ = `[[1.173,0.334,-1.078,-0.641,0.895]]*``[[m,0,0,0,0],[0,m,0,0,0],[0,0,m,0,0],[0,0,0,m,0],[0,0,0,0,m]]*``[[1],[1],[1],[1],[1]]`

L₃ = `[[1.173m,0.334m,-1.078m,-0.641m,0.895m]]*``[[1],[1],[1],[1],[1]]`

L₃ = 0.683m.

• The modal participation factor, Γ_n = `(L n)/(Mn)`

Γ₃ = `(0.683m)/(3.86m)`= 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 = `[[m,0,0,0,0],[0,m,0,0,0],[0,0,m,0,0],[0,0,0,m,0],[0,0,0,0,m]]*``[[1.173],[0.334],[-1.078],[-0.641],[0.895]]``0.177*1.03g`

f₃ = `[[1.173m],[0.334m],[-1.078m],[-0.641m],[0.895m]]*0.177*1.03g`

f₃ = `[[0.214mg],[0.061mg],[-0.196mg],[-0.117mg],[0.163mg]]`

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

f₃ = `[[0.214W],[0.061W],[-0.196W],[-0.117W],[0.163W]]`

• 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^* = `(L n)^2/(Mn)`= `(0.683m)^2/(3.86m)`= 0.121m.

Hence, modal mass participating ratio, M_n^* /m for mode 3 =`(0.121m)/(5m)`= 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 = `[[m,0,0,0,0],[0,m,0,0,0],[0,0,m,0,0],[0,0,0,m,0],[0,0,0,0,m]]`
• Transformed mass matrix, M₄ =`Phi`₄^T m `Phi`₄

i.e, M₄ = `[[-1.078,0.895,0.334,-1.173,0.641]]*``[[m,0,0,0,0],[0,m,0,0,0],[0,0,m,0,0],[0,0,0,m,0],[0,0,0,0,m]]*``[[-1.078],[0.895],[0.334],[-1.173],[0.641]]`

M₄ = `[[-1.078m,0.895m,0.334m,-1.173m,0.641m]]*``[[-1.078],[0.895],[0.334],[-1.173],[0.641]]`

M₄ = 3.86m.

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

Where, ɩ = Influence matrix which is a unit matrix, `[[1],[1],[1],[1],[1]]`

L₄ =`[[-1.078,0.895,0.334,-1.173,0.641]]*``[[m,0,0,0,0],[0,m,0,0,0],[0,0,m,0,0],[0,0,0,m,0],[0,0,0,0,m]]*``[[1],[1],[1],[1],[1]]`

L₄ = `[[-1.078m,0.895m,0.334m,-1.173m,0.641m]]*``[[1],[1],[1],[1],[1]]`

L₄ = -0.381m.

• The modal participation factor, Γ_n = `(L n)/(Mn)`

Γ₄ = `(-0.381m)/(3.86m)`= -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 = `[[m,0,0,0,0],[0,m,0,0,0],[0,0,m,0,0],[0,0,0,m,0],[0,0,0,0,m]]*``[[-1.078],[0.895],[0.334],[-1.173],[0.641]]`.`-0.099*1.03g`

f₄ = `[[-1.078m],[0.895m],[0.334m],[-1.173m],[0.641m]]*-0.099*1.03g`

f₄ = `[[0.109mg],[-0.091mg],[-0.034mg],[0.119mg],[-0.065mg]]`

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

f₄ = `[[0.109W],[-0.091W],[-0.034W],[0.119W],[-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^* = `(L n)^2/(Mn)`= `(-0.381m)^2/(3.86m)`= 0.037m.

Hence, modal mass participating ratio, M₄^* /m for mode 4 =`(0.037m)/(5m)`= 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 = `[[m,0,0,0,0],[0,m,0,0,0],[0,0,m,0,0],[0,0,0,m,0],[0,0,0,0,m]]`
• Transformed mass matrix, M₅ =`Phi`₅^T m `Phi`₅

i.e, M₅ = `[[0.641,-1.078,1.173,-0.895,0.334]]*``[[m,0,0,0,0],[0,m,0,0,0],[0,0,m,0,0],[0,0,0,m,0],[0,0,0,0,m]]*``[[0.641],[-1.078],[1.173],[-0.895],[0.334]]`

M₅ = `[[0.641m,-1.078m,1.173m,-0.895m,0.334m]]*``[[0.641],[-1.078],[1.173],[-0.895],[0.334]]`

M₅ = 3.86m.

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

Where, ɩ = Influence matrix which is a unit matrix, `[[1],[1],[1],[1],[1]]`

L₅ =`[[0.641,-1.078,1.173,-0.895,0.334]]*``[[m,0,0,0,0],[0,m,0,0,0],[0,0,m,0,0],[0,0,0,m,0],[0,0,0,0,m]]*``[[1],[1],[1],[1],[1]]`

L₅ = `[[0.641m,-1.078m,1.173m,-0.895m,0.334m]]*``[[1],[1],[1],[1],[1]]`

L₅ = 0.175m.

• The modal participation factor, Γ_n = `(L n)/(Mn)`

Γ₅ = `(0.175m)/(3.86m)`= 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 = `[[m,0,0,0,0],[0,m,0,0,0],[0,0,m,0,0],[0,0,0,m,0],[0,0,0,0,m]]*``[[0.641],[-1.078],[1.173],[-0.895],[0.334]]*``0.045*1.03g`

f₅ = `[[0.641m],[-1.078m],[1.173m],[-0.895m],[0.334m]]*0.045*1.03g`

f₅ = `[[0.029mg],[-0.049mg],[0.054mg],[-0.041mg],[0.015mg]]`

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

f₅ = `[[0.029W],[-0.049W],[0.054W],[-0.041W],[0.015W]]`

• 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^* = `(L n)^2/(Mn)`= `(0.175m)^2/(3.86m)`= 0.0079m.

Hence, modal mass participating ratio, M₅^* /m for mode 5 =`(0.0079m)/(5m)`= 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 = `sqrt((Vb1)^2+(Vb2)^2+ (Vb3)^2+(Vb4)^2+(Vb5)^2`= `sqrt((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%.