Understanding the concepts on Degrees of Freedom
1) Aim: To understand modes shapes of multistorey building
Manually calculate the base shear for a multi storey building in model analysis
Understand the terminologies involved in modal analysis
a) 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]]`
Modified mass matrix(M)= `underset(n)(phi)^T`. m. `underset(n)(phi)`
The base shear For the first mode (n=1) , = `[[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]]`
`underset(1)(M)`=3.861m
b) The second mode of vibration `underset(2)(T)`= 0.6852 seconds
`underset(2)(phi)`= `[[overset(-)(0.895)],[overset(-)(1.173)],[overset(-)(0.641)],[0.334],[1.078]]`
`underset(2)(M)`= `underset(2)(phi)^T`,m, `underset(2)(phi)`
=`[[overset(-)(0.895), overset(-)(1.173), overset(-)(0.641), overset(-)(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]] [[overset(-)(0.895)],[overset(-)(1.173)],[overset(-)(0.641)],[0.334],[1.078]]`
By using scientific calculator we calculate `underset(2)(M)` and we got `underset(2)(M)`= 3.86M
c) Mode 3
The third mode of vibration`underset(3)T`= 0.4346 seconds
`underset(3)(phi)`= `[[1.173],[0.334], [overset(-)(1.078)], [overset(-)(0.641)], [0.895]]`
`underset(3)(M)`= `underset(3)(phi)^T` m `underset(3)(phi)`
= `[[1.173, 0.334, overset(-)(1.078), overset(-)(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],[overset(-)(1.078)],[overset(-)(0.641)],[0.895]]`
=`underset(3)(M)`= 3.85m
d) Mode 4
`underset(4)(T)`= 0.3383 sec
`underset(4)(phi)`= `[[overset(-)1.078],[0.895],[0.334],[overset(-)(1.173)],[0.641]]`
`underset(4) (M)`= `underset(4)(phi)^T` m `underset(4)(phi)`
= `[overset(-)(1.078), 0.895, 0.334, overset(-)(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]] [[overset(-)(1.078)],[0.895],[0.334],[overset(-)(1.173)],[0.641]]`
=`underset(4) (M)`= 3.861m
e) Mode 5
`underset(5)(T)`= 0.2966seconds
`underset(5)(phi)`=`[[0.641],[overset(-)(1.078)],[1.173],[overset(-)(0.895)],[0.334]]`
`underset(5)(M)`= `underset(5)(phi)^T` m `underset(5)(phi)`
= `[0.641, overset(-)(1.078), 1.173, overset(-)(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],[overset(-)(1.078)],[1.173],[overset(-)(0.895)],[0.334]]`
=`underset(5)(M)`= 3.861m
Step2
Modal property (L)= `underset(n)(phi)^T`. m. `iota`
Where `iota`= Influence vector( Assuming 100% utilization of seismice forces)= `[[1],[1],[1],[1],[1]]`
For the first mode (n=1),= `[[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]]`
=`underset(1)(L)`= 4.121m
Mode 2
The second mode of vibration `underset(2)(T)`= 0.6852 seconds
`underset(2)(phi)`= `[[overset(-)(0.895)],[overset(-)(1.173)],[overset(-)(0.641)],[0.334],[1.078]]`
`underset(2)(L)`= `underset(2)(phi)^T` m `iota`
=`[[overset(-)(0.895), overset(-)(1.173), overset(-)(0.641), overset(-)(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]]`
=`underset(2)L`= -1.298M
Mode 3
The third mode of vibration`underset(3)T`= 0.4346 seconds
`underset(3)(phi)`= `[[1.173],[0.334], [overset(-)(1.078)], [overset(-)(0.641)], [0.895]]`
`underset(3)(L)`= `underset(3)(phi)^T` m `iota`
=`[[1.173, 0.334, overset(-)(1.078), overset(-)(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]]`
=`underset(3)(L)`= 0.687m
Mode 4
`underset(4)(T)`= 0.3383 sec
`underset(4)(phi)`= `[[overset(-)1.078],[0.895],[0.334],[overset(-)(1.173)],[0.641]]`
`underset(4)(L)`=`underset(4)(phi)^T` m `iota`
=`[overset(-)(1.078), 0.895, 0.334, overset(-)(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]]`
=`underset(4)(L)`= -0.381m
Mode 5
`underset(5)(T)`= 0.2966seconds
`underset(5)(phi)`=`[[0.641],[overset(-)(1.078)],[1.173],[overset(-)(0.895)],[0.334]]`
`underset(5)(L)`= `underset(5)(phi)^T` m `iota`
=`[0.641, overset(-)(1.078), 1.173, overset(-)(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]]`
= `underset(5)(L)`= 0.175m
Step-3
`underset(n)(Gamma)` = `underset(n)(L)/underset(n)M`
Mode-1:
`underset(1)(Gamma)= underset(1)L/underset(1)M= (4.121M)/(3.861M)= 1.067`
Mode-2
`underset(2)(Gamma)= underset(2)(L)/underset(2)M=(-1.298M)/(3.861M)=-0.33`
Mode-3
`underset(3)(Gamma)=underset(3)(L)/underset(3)(M)= (0.687M)/(3.85M)=0.178`
Mode-4
`underset(4)(Gamma)=underset(4)(L)/underset(4)M= (-0.381M)/(3.861M)=-0.09`
Mode-5
`underset(5)(Gamma)= underset(5)L/underset(5)M=(0.175M)/(3.861M)=0.04`
Step-4
`underset(1)M^*`=`(underset(1)(L)^2)/underset(1)M`
`underset(1)L=4.121M, underset(1)M=3.861M`
`underset(1)M^*= (4.121M)^2/(3.861M)`
`underset(1)(M)^*= 4.39M`
Mode-2
`underset(2)M^*= underset(2)L^2/underset(2)M`
`underset(2)L= -0.298M, underset(2)M= 3.861M`
`underset(2)M^*= (-0.298M)^2/(3.861M) = 0.43M`
Mode-3
`underset(3)M^*= underset(3)L^2/underset(3)M= (0.687M)^2/(3.85M)= 0.12M`
Mode-4
`underset(4)M^*= underset(4)L^2/underset(4)M= (0.381M)^2/(3.861M)=0.03M`
Mode-5
`underset(5)M^*= underset(5)L^2/underset(5)M= (0.175M)^2/(3.861M)= 0.007M`
Step-5
Modal mass participating ratio= `underset(n)(M)^*/underset(j)M`
`underset(n)M^*= Effective modal mass`
`underset(j)M = Total mass = M+M+M+M+M= 5M`
Mode-1
`underset(1)M^*/underset(j)M= (4.39M)/(5M)= 0.88=88%`
Mode-2
`underset(2)M^*/underset(j)M= (0.43M)/(5M)= 0.086=8.6%`
Mode-3
`underset(3)M^*/underset(j)M= (0.12M)/(5M)= 0.024=2.4%`
Mode-4
`underset(4)M^*/underset(j)M= (0.03M)/(5M)= 0.006=0.6%`
Mode-5
`underset(5)M^*/underset(j)M= (0.007M)/(5M)= 0.001=0.1%`
Step-6
Now we need to obtain the base shear for each of the 5 modes of vibration, as shown below
`underset(1)T= 2seconds, underset(1)A= 0.27g`
`underset(2)T= 0.6852seconds, underset(2)A= 0.76g`
`underset(3)T= 0.4346seconds, underset(3)A= 1.03g`
`underset(4)T= 0.3383seconds, underset(4)A= 1.03g`
`underset(5)T= 0.2966seconds, underset(5)A= 0.76g`
Mode-1
`underset(b1)V= (underset(1)M^*Xunderset(1)A)= (4.39MX0.27g)= 2.619kips`
Mode-2
`underset(b2)V= (underset(2)M^*Xunderset(2)A)= (0.43MX0.76g)= 0.719kips`
Mode-3
`underset(b3)V= (underset(3)M^*Xunderset(3)A)= (0.12MX1.03g)= 0.272kips`
Mode-4
`underset(b4)V= (underset(4)M^*Xunderset(4)A)= (0.03MX1.03g)= 0.067kips`
Mode-5
`underset(b5)V= (underset(5)M^*Xunderset(5)A)= (0.007MX0.76g)= 0.011kips`
Step 7:
Combined base shear as per SRSS rule:
`underset(b)V= sqrt(underset(b1)(V)^2+ underset(b2)V^2+underset(b3)V^2+underset(b4)V^2+underset(b5)V^2`
= `sqrt((2.613)^2+(0.719)^2+( 0.272)^2+ (0.067)^2+(0.011)^2`
=3.20kips
Step 8
The two modes are to be considered such that a modal mass participating ratio of 90% is obtained
`underset(1)M^*/underset(j)M+underset(2)M^*/underset(j)M= 88%+8.6% = 96.6%`
Result:
Calculated `underset(n)L, underset(n)M and underset(n)Gamma` for each of the five modes from the given equations
Calculated the effective modal mass `underset(n)M^*`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