Challenge

Understanding the concepts on Degrees of Freedom

RRamesh Gupta8 Oct 2022

UNDERSTANDING THE CONCEPT ON DEGREE OF FREEDOM

Q1.A FIVE STOREY STRUCTURE WITH 5 TRANSLATION DOF IS SHOWN HERE, EACH STOREY HAS MASS "m" . THE EIGENVALUE PROBLEM HAS BEEN SOLVED AND THE 5 PERIOD OF VIBRATION Tn AND THEIR CORRESPONDING MODE SHAPES , `∅n HAS BEEN DRIVED.

THE FUNDAMENTAL PERIOD OF VIBRATION (T1 = 2 SECOND ) AND ITS ASSOCIATED MODE SHAPE .

THE SECOND MODE OF VIBRARION(T2 =0.6852 SECOND ) AND ITS ASSOCIATED MODE SHAPE :

THE THIRD MODE OF VIBRATION (T3 =0.4346 SECOND ) AND ITS ASSOCIATED MODE SHAPE :

THE FOURTH MODE OF VIBARATION(T4= 0.3383 SECOND ) AND ITS ASSOCIATED MODE SHAPE ":

THE FIFTH MODE OF VIBARATION ( T5 = 0.2966 SECOND ) AND ITS ASSOCIATED MODE SHAPE :

Ln , Mn , AND `Γn FOR EACH OF THE FIVE MODES ARE TO BE COMPUTED FROM EQUATION 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 COMBINED BASE SHEAR AS PER SRSS RULE . THE PSEUDO ACCELERATION DESIGN SPECTRUM TO OBTAIN THE ASE 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 ?

AIM : TO COMPUTE Ln, Mn AND `Γn FOR EACH OF FIVE MODES FROM THE GIVEN EQUATIONS .

TO CALULATE THE EFFECTIVE MODAL MASS Mn* AND THE MODAL MASS PARTICIPATING RATIO FOR THE 5 MODES.

TO OBTAIN THE BASE SHEAR FOR EACH OF THE 5 MODES OF VIBRATION AND THE COMBINED BASE SHEAR AS PER SRSS RULE .

PROCEDURE :

STEP 1 :

MODEL: 1

THE FIRST MODE OF VIBRATION (T1) = 2 SECONDS

`∅1= [ 0.334

0.64

0.895

1.078

1.173 ]

`M1=Φ1T m Φ1 =[0.334 0.641 0.895 1.078 1.173 ]

[ M 0 0 0 0 [ 0.334

0 M 0 0 0 0.64

0 0 M 0 0 0.895

0 0 0 M 0 1.078

0 0 0 0 M ] 1.173 ]

BY USING THE SCIENTIFIC CALCULATOR WE CALCULATE M1 AND WE GOT ,M1 = 3.861M
M1= 3.861M

MODEL 2:

THE SECOND MODE OF VIBRATION (T2) = 0.6852 SECOND

`Φ2= [ -0.895

-1.173

-0.641

0.334

1.078 ]

M2 = Φ2TmΦ2 = [ -0.895 -1.173 -0.641 0.334 1.078 ]

[ M 0 0 0 0 [ -0.895

0 M 0 0 0 -1.173

0 0 M 0 0 -0.641

0 0 0 M 0 0.334

0 0 0 0 M ] 1.078 ]

BY USING THE SCIENTIFIC CALCULATOR WE CALCULATE M2 AND WE GOT , M2 = 3.86M
M2 = 3.86M

MODEL 3:

THE THIRD MODEL OF VIBRATION (T3) = 0.4346 SECONDS

Φ3 = [ 1.173

0.334

-1.078

-0.641

0.895 ]

[ M 0 0 0 0 [ 1.173

0 M 0 0 0 0.334

0 0 M 0 0 -1.078

0 0 0 M 0 -0.641

0 0 0 0 M ] 0.895 ]

M3 = Φ3T m Φ3 = [ 1.173 0.334 -1.078 -0.641 0.895 ]
BY USING THE SCIENTIFIC CALCULATOR WE CALCULATE M3 AND WE GOT M3 = 3.85M
M3= 3.85M

MODEL 4:

THE FOURTH MODE OF VIBRATION ( T4) = 0.3383 SECOND

Φ4 = [ -1.078

0.895

0.334

-1.173

0.641 ]

M4 = Φ4T m Φ4 =[ -1.078 0.895 0.334 -1.173 0.641 ]

[ M 0 0 0 0 [ -1.078

0 M 0 0 0 0.895

0 0 M 0 0 0.334

0 0 0 M 0 -1.173

0 0 0 0 M ] 0.641 ]

BY USING THE SCIENTIFIC CALCULATOR WE CALCULATE M4 AND WE GOT ,M4 = 3.861M
M4 = 3.861M

MODEL 5:

THE FIFTH MODEL OF VIBRATION ( T5) = 0.2966 SECONDS

Φ5 = [ 0.641

-1.078

1.173

-0.895

0.334 ]

M5 = Φ5T m Φ5 =[ 0.641 -1.078 1.173 -0.895 0.334 ]

[ M 0 0 0 0 [ 0.641

0 M 0 0 0 -1.078

0 0 M 0 0 1.173

0 0 0 M 0 -0.895

0 0 0 0 M ] 0.334 ]

BY USING THE SCIENTIFIC CALCULATER WE CALCULATE M5 AND WE GOT ,M5 = 3.861M

M5 = 3.861M

STEP: 2

Ln = `ΦnTmL

MODEL 1:

THE FIRST MODE OF VINRATION (T1) = 2 SECONDS

`∅1= [ 0.334

0.64

0.895

1.078

1.173 ]

L1 = `Φ1TmL = [ 0.334 0.64 0.895 1.078 1.173 ]

[ M 0 0 0 0 [ 1

0 M 0 0 0 1

0 0 M 0 0 1

0 0 0 M 0 1

0 0 0 0 M ] 1 ]

BY USING THE SCIENTIFIC CALCULATOR WE CALCULATE L1 AND WE GOT ,L1 = 4.12M
L1 = 4.121M

MODEL 2:

THE SECOND MODE OF VIBRATION ( T2 ) = 0.6852 SECONDS

`Φ2 = [ -0.895

-1.173

-0.641

0.334

1.078 ]

`L2 = Φ2TmL = [ -0.895 -1.173 -0.641 0.334 1.078 ]

[ M 0 0 0 0 [ 1

0 M 0 0 0 [ 1

0 0 M 0 0 1

0 0 0 M 0 1

0 0 0 0 M ] 1 ]

BY USING THE SCIENTIFIC CALCULATOR WE CALCULATE L2 AND WE GOT , L2 = -1.298M
L2 = -1.298M

MODEL 3:

THE THIRD MODE OF VIBRATION (T3) = 0.4346 SECONDS

- `Φ3=`[ 1.173
0.334

-1.078

-0.641

`0.895`]
`L3 = Φ3TmL = [ 1.173 0.334 -1.078 -0.641 0.895 ]

[ M 0 0 0 0 [ 1

0 M 0 0 0 1

0 0 M 0 0 1

0 0 0 M 0 1

0 0 0 0 M ] 1 ]

BY USING THE SCIENTIFIC CALCULATOR WE CALCULATE L3 AND WE GOT . L3 =0.687M
L3 = 0.687M

MODEL 4:

THE FOURTH MODE OF VIBRATION (T4) = 0.3383 SECONDS

Φ4=`[ -1.078

0.895

0.334

-1.173

`0.641`]

`L4 = Φ4TmL = [ -1.078 0.895 0.334 -1.173 0.641 ]

[ M 0 0 0 0 [ 1

0 M 0 0 0 1

0 0 M 0 0 1

0 0 0 M 0 1

0 0 0 0 M ] 1 ]

BY USING THE SCIENTIFIC CALCULTATER WE CALCULATE L4 = AND WE GOT ,L4 =-0.381M
L4=-0.381M

MODEL 5:

THE FIFTH MODE OF VIBRATION (T5) = 0.2966 SECONDS

- Φ5 =`[ 0.641
-1.078

1.173

-0.895

0.334`]
L5 = Φ5TmL = [ 0.641 -1.078 1.173 -0.895 0.334 ]
[ M 0 0 0 0 [ 1

0 M 0 0 0 1

0 0 M 0 0 1

0 0 0 M 0 1

0 0 0 0 M ] 1 ]
BY USING THE SCIENTIFIC CALCULTATER WE CALCULATE L4 = AND WE GOT ,L5 = 0.175M
L5 = 0.175 M

STEP 3 :

`Γn = Ln / Mn

MODEL 1 :

Γ1 = L1 / M1 = 4.121M /3.861 M = 1.067

MODEL 2:

Γ2 = L2 / M2 = -1.298M /3.861 M = -0.33

MODEL 3:

Γ3 = L2/ M3 = 0.687M / 3.85 M = 0.178

MODEL 4 :

Γ4 = L4 /M4 = -0.381 M / 3.861 M = -0.09

MODEL 5:

Γ5 = L5 / M5 = 0.175 M / 3.861 M = 0.04

STEP 4 :

Mn * = Ln^2 / Mn

MODEL 1:

M1* = L1^2 /M1
PUT L1 = 4.121M , M1 = 3.861M
M1* = (4.121M)^2 / ( 3.861M)
M1 = 4.391M

MODEL 2 :

M2*= L2^2/ M2

PUT L2= -0.298M, M2 = 3.861M

M2* (-0.298M)^2/3.861M
M2* = 0.43M

MODEL 3 :

M3* = L3^2 / M3

M3* = (0.687M) ^2 /(3.85 M)
M3* =0.12M

MODEL 4:

M4* =( -0.381M) ^2 / 3.861M

M4* = 0.03M

MODEL 5 :

M5* =L5*^2 / M5
PUT L5 = 0.175M , M5 = 3.861M
M5 * = ( 0.175M) ^2 / 3.861M
M5* = 0.007M

STEP 5 :

MODAL MASS PARTICIPATING RATIO = Mn * / Mj
Mn* = EFFECTIVE MODAL MASS.
Mj = TOTAL MASS = M+M+M+M+M =5M

MODEL : 1

M1* / Mj = 4.39m / 5M = 0.88 = 88%

MODEL 2 :

M2* / Mj = 0.43M /5M = 0.086 = 86%

MODEL 3 :

M3* / Mj =0.12M /5M = 0.024 = 2.4%

MODE 4:

M4* / Mj = 0.03M /5M = 0.006 =0.6%

MODEL 5:

M5* / Mj = 0.007M /5M =0.001 =0.1%

STEP 6 :

NOW WE NEED TO OBTAIN THE BASE SHEAR FOR EACH OF THE 5 MODE OF VIBRATION , AS SHOWN BELOW .

( T1) = 2 SECONDS , A1 = 0.27g
( T2) = 0.6852 SECONDS , A2 = 0.76g
( T3 )= 0.4346 SECONDS , A3 = 1.03g
(T4 ) = 0.3383 SECONDS , A4 = 1.03g
( T5) = 0.2966 SECONDS , A5 = 0.76g

MODEL :1

Vb1 = ( M1* x A1 ) = ( 4.39M X 0.27g ) = 2.619kips

MODEL : 2

Vb2 = ( M2* X A2 ) = ( 0.43M X 0.76g ) = 0.719 kips

MODEL : 3

Vb3 = ( M3* X A3 ) = ( 0.12M X 1.03g ) = 0.272kips

MODEL : 4

Vb4 = ( M4* x A4) = ( 0.03M X 1.03g ) = 0.067kips

MODEL: 5

Vb5 = ( M5* X A5 ) = ( 0.007 X 0.76g ) = 0.011kips

STEPS : 7

COMBINED BASE SHEAR AS PER SRSS RULE:

Vb = `√ Vb1^2 +Vb2^2 + Vb3^2 + Vb4^2 + Vb5^2 =√ ( 2.613)^2 +(0.719)^2+ (0.272)^2 + ( 0.067)^2 +(0.011)^2

Vb = 3.20kips

STEP:8

THE TWO MODES ARE TO BE CONSIDERED SUCH THAT A MODAL MASS PARTICIPATING RATIO OF 90% IS OBTAINED .

M1* /Mj + M2* / Mj = 88% + 8.6% = 96.6%

RESULT :

CALCULATED Ln ,Mn AND Γn FOR EACH OF FIVE MODES FROM THE GIVEN EQUATIONS.
CALCULATE 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 90% IS OBTAINED .

Understanding the concepts on Degrees of Freedom

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