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Challenge

Understanding the concepts on Degrees of Freedom

GGaurav Pant9 Nov 2021

Given :-

  • 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 Tn and their corresponding mode shapes, φn has been derived.

  • The fundamental period of vibration (T1 = 2 seconds) and its associated mode shape :-

`Phi`1 = `{(0.334),(0.641),(0.895),(1.078),(1.173):}`

  • The second mode of vibration (T2 = 0.6852 seconds) and its associated mode shape :-

`Phi`2 = `{(-0.895),(-1.173),(-0.641),(0.334),(1.078):}`

  • The third mode of vibration (T3 = 0.4346 seconds) and its associated mode shape :-

`Phi`3 = `{(1.173),(0.334),(-1.078),(-0.641),(0.895):}`

  • The fourth mode of vibration (T4 = 0.3383 seconds) and its associated mode shape :-

`Phi`4 = `{(-1.078),(0.895),(0.334),(-1.173),(0.641):}`

  • The fifth mode of vibration (T4 = 0.2966 seconds) and its associated mode shape :-

`Phi`5 = `{(0.641),(-1.078),(1.173),(-0.895),(0.334):}`

AIM :-

  • To find Ln , Mn and Γn for each of the five modes which are to be computed from the Equations below.
  • Also, calculate the effective modal mass (Mn*) and the modal mass participating ratio for the 5 modes.

  • Also, obtain the base shear for each of the 5 modes of vibration and the combined base shear as per SRSS rule.

  • Also, describe how many modes are to be considered such that a modal mass participating ratio of 90% is obtained?
  • Pseudo Acceleration Design Spectrum :-

ANSWER :-

  • 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]]`
  • Since the degree of freedom moves along the direction of the lateral force, the Infuence Vector will be unity.
  • Infuence Vector, `iota` = `[[1],[1],[1],[1],[1]]`

1. Calculation of Ln :-

  • Ln = `Phi^T`n m`iota`

For mode 1,

  • L1 = `Phi^T`1 m`iota`

`=>` 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 = 4.121m

For mode 2,

  • L2 = `Phi^T`2 m`iota`

`=>` 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 = -1.297m

For mode 3,

  • L3 = `Phi^T`3 m`iota`

`=>` 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 = 0.683m

For mode 4,

  • L4 = `Phi^T`4 m`iota`

`=>` 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 = -0.381m

For mode 5,

  • L5 = `Phi^T`5 m`iota`

`=>` 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 = 0.175m

2. Calculation of Modified Mass Matrix, Mn :-

  • Mn = `Phi^T`n m`Phi`n

For mode 1,

  • M1 = `Phi^T`1 m`Phi`1

`=>` 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 = 3.861m

For mode 2,

  • M2 = `Phi^T`2 m`Phi`2

`=>` 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 = 3.861m

For mode 3,

  • M3 = `Phi^T`3 m`Phi`3

`=>` 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 = 3.861m

For mode 4,

  • M4 = `Phi^T`4 m`Phi`4

`=>` 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 = 3.861m

For mode 5,

  • M5 = `Phi^T`5 m`Phi`5

`=>` 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 = 3.861m

3. Calculation of Γn :-

  • Γn = Ln/Mn

For mode 1,

  • Γ1 = L1/M1

`=>` Γ1 = (4.121m/3.861m)

`=>` Γ1 = 1.067

For mode 2,

  • Γ2 = L2/M2

`=>` Γ2 = (-1.297m/3.861m)

`=>` Γ2 = -0.336

For mode 3,

  • Γ3 = L3/M3

`=>` Γ3 = (0.683m/3.861m)

`=>` Γ3 = 0.177

For mode 4,

  • Γ4 = L4/M4

`=>` Γ4 = (-0.381m/3.861m)

`=>` Γ4 = -0.098

For mode 5,

  • Γ5 = L5/M5

`=>` Γ5 = (0.175m/3.861m)

`=>` Γ5 = 0.045

4. Calculation of Effective Modified Mass, Mn* :-

  • Mn* = `L^2`n/Mn

For mode 1,

  • M1* = `L^2`1/M1

`=>` M1* = (`(4.121m)^2`/3.861m)

`=>` M1* = 4.399m

For mode 2,

  • M2* = `L^2`2/M2

`=>` M2* = (`(-1.297m)^2`/3.861m)

`=>` M2* = 0.435m

For mode 3,

  • M3* = `L^2`3/M3

`=>` M3* = (`(0.683m)^2`/3.861m)

`=>` M3* = 0.121m

For mode 4,

  • M4* = `L^2`4/M4

`=>` M4* = (`(-0.381m)^2`/3.861m)

`=>` M4* = 0.037m

For mode 5,

  • M5* = `L^2`5/M5

`=>` M5* = (`(0.175m)^2`/3.861m)

`=>` M5* = 0.008m

5. Calculation of Modal Mass Participating Ratio :-

  • Participating Ratio = Mn* / Mj

where, Mj = M1* + M2* + M3* + M4* + M5* = 5m (sum of mass of each floor)

For mode 1,

  • Participating Ratio = M1* / Mj = 0.8798

For mode 2,

  • Participating Ratio = M2* / Mj = 0.087

For mode 3,

  • Participating Ratio = M3* / Mj = 0.0242

For mode 4,

  • Participating Ratio = M4* / Mj = 0.0074

For mode 5,

  • Participating Ratio = M5* / Mj = 0.0016

Verifying,

The total mass of the buiding = m + m + m + m + m = 5m

Adding all effective modal mass = M1* + M2* + M3* + M4* + M5* = 5m

Hence OK

  • From the above participation ratios of modal mass, we can say that sum of Mode 1 and Mode 2 are enough to obtain a minimum of 90% of the participation.
  • 0.8796 + 0.087 = 0.9666 = 96.66%

6. Calculation of Base Shear (Vbn) :-

  • Vbn = An * Mn*

From the Pseudo Acceleration design Spectrum graph, Design Acceleration for each modes are :-

  • For T1 = 2 seconds, A1 = 0.279
  • For T2 = 0.6852 second, A2 = 0.769
  • For T3 = 0.4346 second, A3 = 1.039
  • For T4 = 0.3383 second, A4 = 1.039
  • For T5 = 0.2966 second, A5 = 1.039

For mode 1,

  • Vb1 = A1 * M1*

`=>` Vb1 = (0.279) * (4.399m)

`=>` Vb1 = 1.227m

For mode 2,

  • Vb2 = A2 * M2*

`=>` Vb2 = (0.769) * (0.435m)

`=>` Vb2 = 0.334m

For mode 3,

  • Vb3 = A3 * M3*

`=>` Vb3 = (1.039) * (0.121m)

`=>` Vb3 = 0.126m

For mode 4,

  • Vb4 = A4 * M4*

`=>` Vb4 = (1.039) * (0.037m)

`=>` Vb4 = 0.038m

For mode 5,

  • Vb5 = A5 * M5*

For T5 = 2 seconds, A5 = 0.27g

`=>` Vb5 = (1.039) * (0.008m)

`=>` Vb5 = 0.008m

7. Calculation of Combined Base Shear as per SRSS rule :-

  • `V^2`B = `V^2`b1 + `V^2`b2 + `V^2`b3 + `V^2`b4 + `V^2`b5

`=>``V^2`B = `(1.227m)^2+(0.334m)^2+(0.126m)^2+(0.038m)^2+(0.008m)^2`

`=>` VB = `sqrt (1.634)` m

`=>` VB = 1.278m

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