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  1. Home/
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  3. Understanding the concepts on Degrees of Freedom

Understanding the concepts on Degrees of Freedom

AIM:- Understanding the concepts on Degrees of Freedom INTRODUCTION:- A degree of freedom is the number of possible movements of a structural system. The degrees of freedom can be used to describe displacements and rotations at a nodal point. Thus, each degree of freedom allows for a displacement or a rotation in…

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  • Vishal Mahalle

    updated on 19 Aug 2022

AIM:- Understanding the concepts on Degrees of Freedom

INTRODUCTION:-

  • A degree of freedom is the number of possible movements of a structural system. The degrees of freedom can be used to describe displacements and rotations at a nodal point. Thus, each degree of freedom allows for a displacement or a rotation in a certain direction.
  • In this project we have study the details conceptual degree of freedom in structural analysis

Question 1:

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:

 

Φ1=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩0.3340.6410.8951.0781.173⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭Φ1={0.3340.6410.8951.0781.173}

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

 

Φ2=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩−0.895−1.173−0.6410.3341.078⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭Φ2={-0.895-1.173-0.6410.3341.078}

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

 

Φ3=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩1.1730.334−1.078−0.6410.895⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭Φ3={1.1730.334-1.078-0.6410.895}

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

 

Φ4=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩−1.0780.8950.334−1.1730.641⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭Φ4={-1.0780.8950.334-1.1730.641}

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

 

Φ5=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩0.641−1.0781.173−0.8950.334⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭Φ5={0.641-1.0781.173-0.8950.334}

  • Ln , Mn and Γn for each of the five modes are to be computed from Equations below.
  • Also, calculate the effective modal mass Mn* and the modal mass participating ratio for the 5 modes. The mass matrix will be required to evaluate these parameters.

  • Finally, 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.

  • How many modes are to be considered such that a modal mass participating ratio of 90% is obtained?

 

INTRODUCTION:

  • Modal effective mass is a modal dynamic property of a structure associated with the modal characteristics; natural frequencies, mode shapes, generalised masses, and participation factors.
  • Modal participation factors - Mass participation ratio specifies the percentage of how much the structural mass of the model is participating for a given direction and mode.modal participation factors are scalars that measure the interaction between the modes and the directional excitation in a given reference frame. Larger values indicate a stronger contribution to the dynamic response.
  • Effective mass factors - the effective mass factors associated with each mode represents the amount of system mass participating in that mode in a given excitation direction. This value is given as a percentage of the total system mass. Therefore, a mode with a large effective mass will be a significant contributor to the system’s response in the given excitation direction.  A common rule of thumb for linear dynamic analysis is that a mode should be included if it contributes more than 1-2% of the total effective mass.
  • Base shear is an estimate of the maximum expected lateral force on the base of the structure due to seismic activity.

PROCEDURE:-

 a) Modified mass matrix;

Mn">Mn= ΦTnmΦnMn= ΦnTmΦnΦnTmΦn">

Mode 1:

 

M1=(0.334,0.641,0.895,1.078,1.173)(M)⋅⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩0.3340.6410.8951.0781.173⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭M1=(0.334,0.641,0.895,1.078,1.173)(M)⋅{0.3340.6410.8951.0781.173}))

    =Mx(0.3342+0.6412+0.8952+1.0782+1.1732)=Mx(0.3342+0.6412+0.8952+1.0782+1.1732)(0.3342+0.6412+0.8952+1.0782+1.1732)">=(0.334+0.641+0.895+1.078+1.173)M

M1=3.891MM1=3.891M

Mode 2:

M2=(-0.895,-1.173,-0.641,0.334,1.078)(M) *{(-0.895),(-1.173),(-0.641),(0.334),(1.078):}}

      =Mx ((−0.8952)+(−1.1732)+(−0.6912)+(−0.3342)+(−1.0782))=Mx ((−0.8952)+(−1.1732)+(−0.6912)+(−0.3342)+(−1.0782))(-(0.8952)+(-1.1732)+(-0.6912)+(-0.3342)+(-1.0782))">L2=1.298M    

 M2=3.861MM2=3.861M

Mode 3:

M3=(1.173,0.334,−1.078,−0.641,0.895)(M)⋅⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩1.1730.334−1.078−0.6410.895⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭M3=(1.173,0.334,-1.078,-0.641,0.895)(M)⋅{1.1730.334-1.078-0.6410.895}`

    =M⋅((1.1732)+(0.3342)+(−1.0782)+(−0.6412)+(0.8952))=M⋅((1.1732)+(0.3342)+(−1.0782)+(−0.6412)+(0.8952))((1.1732)+(0.3342)+(-1.0782)+(-0.6412)+(0.8952))">L3=0.687

M3=3.85MM3=3.85M

Mode 4:

M4=(−1.078,0.895,0.334,−1.173,0.641)(M)⋅⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩−1.0780.8950.334−1.1730.641⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭M4=(-1.078,0.895,0.334,-1.173,0.641)(M)⋅{-1.0780.8950.334-1.1730.641}`

=M⋅((−1.0782)+(0.8952)+(0.3342)+(−1.1732)+(0.6412))=M⋅((-1.0782)+(0.8952)+(0.3342)+(−1.1732)+(0.6412))

M4=3.861MM4=3.861M

Mode 5:

M5=(0.641,−1.078,1.173,−0.895,0.334)(M)⋅⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩0.641−1.0781.173−0.8950.334⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭M5=(0.641,-1.078,1.173,-0.895,0.334)(M)⋅{0.641-1.0781.173-0.8950.334}`

        =M⋅((0.6412)+(−1.0782)+(1.1732)+(−0.8952)+(0.3342)=M⋅((0.6412)+(−1.0782)+(1.1732)+(−0.8952)+(0.3342)

M5=3.861MM5=3.861M

 

 

b) Modal property L;

Ln=ϕTnmlLn=ϕnTmlLn=ϕnTmi">Γn=LnMn

 

Where,l= influence vector(assuming 100% utilization of siesmic forces)=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]`

Mode 1:

      L1=(0.334,0.641,0.895,1.078,1.173)(M)⋅ ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦L1=(0.334,0.641,0.895,1.078,1.173)(M)⋅ [11111](11111)">Γ2=L2M2=−1.298M3.861M=−0.33

     L1=4.121ML1=4.121M

Mode 2:

     L2=(−0.895,−1.173,−0.641,0.334,1.078)(M)⋅ ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦L2=(-0.895,-1.173,-0.641,0.334,1.078)(M)⋅ [11111]`(11111)">Γ5=L5M5=0.175M3.861M=0.04

                =(−0.895−1.173−0.641+0.331+1.078)M(-0.895-1.173-0.641+0.331+1.078)M

     

Mode 3:

       L3=(1.173,0.334,−1.078,−0.641,0.895)(M)⋅ ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦L3=(1.173,0.334,-1.078,-0.641,0.895)(M)⋅ [11111](11111)">M3=L23M3=(0.687M)23.85M=0.12M

       =(1.173+0.334−1.078−0.641+0.895)M=(1.173+0.334-1.078-0.641+0.895)M

     `

Mode 4:

   L4=(−1.078,0.895,0.334,−1.173,0.641)(M)⋅ ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦L4=(-1.078,0.895,0.334,-1.173,0.641)(M)⋅ [11111]`(11111)">=MnMj

                =(−1.078+0.895+0.334−1.173+0.641)M=(-1.078+0.895+0.334-1.173+0.641)M

    L4">M1Mj=4.39M5M=0.88%L4=−0.381ML4=-0.381M

Mode 5:

       L5">M2Mj=0.43M5M=0.086%L5=(0.641,−1.078,1.173,−0.895,0.334)(M)⋅ ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦L5=(0.641,-1.078,1.173,-0.895,0.334)(M)⋅ [11111](11111)">M3Mj=0.12M5M=2.4%

                =(0.641−1.078+1.173−0.895+0.334)M=(0.641-1.078+1.173-0.895+0.334)M

     L5=0.175ML5=0.175M

 

c) Modal participation factor ;

∣n=LnMn">M5Mj=0.007M5M=0.1%

 

Mode 1:

      Γ1=L1M1=4.121M3.861M=1.067Γ1=L1M1=4.121M3.861M=1.067

Mode 2:

       ∣2=L2M2">A1=0.27g

Mode 3:

     Γ3=L3M3=0.687M3.85M=0.178Γ3=L3M3=0.687M3.85M=0.178

Mode 4:

     Γ4=L4M4=−0.381M3.861M=0.09Γ4=L4M4=-0.381M3.861M=0.09

Mode 5:

     

 

d) Effective modal mass;

 

Mode 1:

       M1⋅=L12M1">T4=0.3383secM1=L21M1=(4.121M)23.861M=4.39MM1=L12M1=(4.121M)23.861M=4.39M

Mode 2:

     M2=L22M2=(−0.298M)23.861M=0.43MM2=L22M2=(-0.298M)23.861M=0.43M

Mode 3:

   

Mode 4:

       M4⋅=L42M4">A5=0.76gM4=L24M4=(−0.381M)23.861M=0.03MM4=L42M4=(-0.381M)23.861M=0.03M

Mode 5:

       M5⋅=L52M5">Vb1=M1A1=1.39M⋅0.27g=2.619kipsM5=L25M5=(0.175M)23.861M=0.007MM5=L52M5=(0.175M)23.861M=0.007M

 

e) Modal mass participating ratio

Mn⋅Mj">Vb2=M2A2=0.43M⋅0.76g=0.719kips

    Mj">MjVb3=M3A3=0.12M⋅1.03g=0.272kips=Total mass of the building = M+M+M+M+M = 5M

 

Mode 1:

   

Mode 2:

      M2⋅Mj">Vb5=M5A5=0.007M⋅0.76g=0.011kips

Mode 3:

      M3⋅Mj">Vb=√V2b1+V2b2+V2b1+V2b1+V2b1

Mode 4:

      M4⋅Mj">=3.2kipsM4Mj=0.03M5M=0.6%M4Mj=0.03M5M=0.6%

Mode 5:

      M5⋅Mj">

  •  Adding first 2 mode mass participation ratio we get,

         = 0.88+0.086= 96.6=96.6%

  • To achieve minimum 90% of modal mass participation ratio,first 2 modes are enough

f) WKT, Vbn=Ln2MnAn=Mn⋅An">

Vbn=Ln2MnAn=Mn⋅An">`V_(bn)=L_n^2/(Mn)An=M_nA_n`

here,

T1">T1=2secT1=2sec, 

 

T2">T2=0.6352secT2=0.6352sec,   A2">A2=0.76gA2=0.76g

 

T3">T3=0.4346secT3=0.4346sec,  A3">A3=1.03gA3=1.03g

 

T4">,  A4">A4=1.03gA4=1.03g

 

T5">T5=0.2966secT5=0.2966sec,  A5">

 

g) Base shear Vb

Mode 1:

Mode 2:

      Vb2=M2⋅A2">

Mode 3:

      Vb3=M3⋅A3">

Mode 4:

      Vb4=M4⋅A4">Vb4=M4A4V=0.03M⋅1.03g=0.067kipsVb4=M4A4V=0.03M⋅1.03g=0.067kips

Mode 5:

      Vb5=M5⋅A5">

 

h) Average base shear is calculated by the method of square root of sum of square (srss)

   

`V_b=2.6132+0.7192+0.2722+0.0672+0.0112"`>   

 

i) Model mass ration of 90% in obtained by 2 model

    `((M_1)/(M_j))+ ((M_2)/(M_j)) =88% +8.6%=96.6%`

 

RESULT:-

  • Calculated Ln , Mn and Γn for each of the five modes from the given equations.
  • Calculated the effective modal mass Mn* and the modal mass participating ratio for the 5 modes.
  • The base shear for each of the 5 modes of vibration are obtained.
  • As per SRSS rule the combined base shear is obtained.
  • The two modes are to be considered such that a modal mass participating ratio of  obove 90% is obtained.

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