Understanding the concepts on Degrees of Freedom
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
`∅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
- M2 = 3.86M
- M3= 3.85M
- M4 = 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
- L2 = -1.298M
- L3 = 0.687M
- L4=-0.381M
- 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 .