Challenge

Understanding the concepts on Degrees of Freedom

BBablu Paswan31 Dec 2022

AIM:-

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.

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

`phi1={(0.334),(0.641),(0.895),(1.078),(1.173):}`

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

1) To Find out the modal mass (Mn)of a given story.

A) For First Node - T= 2sec

i) We know that,

Modal Mass (Mn) = ø_nT_mø_n

_Where,

_{m = Mass Matrix
øn =Mode shape @ nth Mode
øn}_{T = Transpose of Mode Shape}

_{ii) So, For First node (T1) =2 Sec}

_Since

`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]]`

2. Since the degree of freedom move along the direction of the lateral force, the influence vector will be unity.

3.Infuence Vector, `iota =[[1],[1],[1],[1],[1]]`

1.Calculation of Ln:-

Ln = `phi`_nT_m`iota`

_{For mode.1,}

L1 = `phi`_nT_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.

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

`phi2={(-0.895),(-1.173),(-0.641),(0.334),(1.078):}`

L2 = `phi`_nT_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.}

_{The third mode of vibration (T3 = 0.4346 seconds) and its associated mode shape:}

_{`phi3={(1.173),(0.334),(-1.078),(-0.641),(0.895):}`}

L3 = `phi`_nT_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.}

_{The fourth mode of vibration (T4 = 0.3383 seconds) and its associated mode shape:}

_{`phi4={(-1.078),(0.895),(0.334),(-1.173),(0.641):}`}

_{L4 = `phi`nTm`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.}

_{The fifth mode of vibration (T4 = 0.2966 seconds) and its associated mode shape:}

_{`phi5={(0.641),(-1.078),(1.173),(-0.895),(0.334):}`}

_{L5 = `phi`nTm`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^Tn^mphin`

_{For mode,1.}

_{M1= `phi^T1^mphin`}

`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^T2^mphin`}

_{`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^T2^mphin`}

_{`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.}

_{M3= `phi^T2^mphin`}
1.The fundamental period of vibration (T₁ = 2 seconds) and its associated mode shape:

`phi1={(0.334),(0.641),(0.895),(1.078),(1.173):}`
- To Calculate Ln, Mn,& Tn For each of the five modes.
- To calculate the effective modal mass Mn* and the modal mass participating ratio for the five modes.
- To find the base shear for each of the 5 modes of vibration and the combined base shear as per SRSS rule.
1) To Find out the modal mass (Mn)of a given story.

A) For First Node - T= 2sec

i) We know that,

Modal Mass (Mn) = ø_nT_mø_{n - - - - - - - - - - - - - -(A)}

_Where,

_{m = Mass Matrix
øn =Mode shape @ nth Mode
øn}_{T = Transpose of Mode Shape}

_{ii) So, For First node (T1) =2 Sec}

_Since

`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]]`

2. Since the degree of freedom move along the direction of the lateral force, the influence vector will be unity.

3.Infuence Vector, `iota =[[1],[1],[1],[1],[1]]`

1.Calculation of Ln:-
- Ln = `phi`_nT_m`iota`
_{For mode.1,}
- L1 = `phi`_nT_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.

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

`phi2={(-0.895),(-1.173),(-0.641),(0.334),(1.078):}`
- L2 = `phi`_nT_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.}

_{The third mode of vibration (T3 = 0.4346 seconds) and its associated mode shape:}

_{`phi3={(1.173),(0.334),(-1.078),(-0.641),(0.895):}`}
- L3 = `phi`_nT_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.}

_{The fourth mode of vibration (T4 = 0.3383 seconds) and its associated mode shape:}

_{`phi4={(-1.078),(0.895),(0.334),(-1.173),(0.641):}`}

_{L4 = `phi`nTm`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.}

_{The fifth mode of vibration (T4 = 0.2966 seconds) and its associated mode shape:}

_{`phi5={(0.641),(-1.078),(1.173),(-0.895),(0.334):}`}

_{L5 = `phi`nTm`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^Tn^mphin`
_{For mode,1.}
- _{M1= `phi^T1^mphin`}
`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^T2^mphin`}
_{`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^T2^mphin`}

_{`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^T2^mphin}

_{`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^T2^mphin}

_{`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 =(4.121m/3.861m)}

_{Γ1 = 1.067}

_{For mode 2,}

_{Γ2 =(-1.297m/3.861m)}

_{Γ2 = 0.336}

_{For mode 3,}

_{Γ3 =(0.683m/3.861m)}

_{Γ3 = 0.177}

_{For mode 4,}

_{Γ4=(-0.381m/3.861m)}

_{Γ4= -0.098}

_{For mode 5,}

_{Γ5=(0.175m/3.861m)}

_{Γ5= 0.045}

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

M_n* = `L^2`L2_n/M_n

_{For mode 1,}

M1* = `L^2`L21/M1

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

M1* = 4.399m

_{For mode2,}

M2* = `L^2`L22/M2

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

M2* = 0.435m

_{For mode3,}

M3* = `L^2`L23/M3

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

M3* = 0.121m

_{For mode4,}

M4* = `L^2`L24/M4

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

M4* = 0.037m

_{For mode5,}

M5* = `L^2`L25/M5

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

M5* = 0.008m

_{5.Calculation of model mass participating Ratio:-}

Participating Ratio:- = M_n* /Mj

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

_{For mode1,}

Participating Ratio:- = M1* /Mj = 0.8798

_{For mode2,}

Participating Ratio:- = M2* /Mj = 0.087

_{For mode3,}

_{Participating Ratio:- = M3* /Mj = 0.0242}

_{For mode4,}

_{Participating Ratio:-= M4* /Mj = 0.0074}

_{For mode5,}

_{Participating}_{Ratio:- = M5* /Mj = 0.0016}

_Verifying,

_{The total mass of the building = 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 modal 1 and}

_{modal 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) :-}

V_bn = A_{n *}M_n*

From the pseudo Acceleration dessign spectrum graph Acceleration for each

Modes are:-

For T1 =2 Seconds, A 1= 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 mode1,}

V_b1 = A1_*M1*

V_b1 = (0.279)*(0.279m)

V_b1 = 1.227m

_{For mode2,}

V_b2 = A2_*M2*

V_b2 = (0.769)*(0.435m)

V_b2 = 0.334m

_{For mode3,}

V_b3 = A3_*M3*

V_b3 = (1.039)*(0.121m)

V_b3 = 0.126m

_{For mode4,}

V_b4 = A4_*M4*

V_b4 = (1.039)*(0.037m)

V_b4 = 0.038m

_{For mode5,}

V_b5 = A5_*M5*

For T5 = 2 Seconds, A5 = 0.27g

V_b5 = (1.039)*(0.008m)

V_b5 = 0.008m

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

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

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

VB = `sqrt 1.634m`

Result:- VB = 1.278m

Understanding the concepts on Degrees of Freedom

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