Understanding the concepts on Degrees of Freedom

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, have been derived.

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

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

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

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

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

• L_n , M_n and Γ_n for each of the five modes are to be computed from Equations below.
• Also, calculate the effective modal mass M_n* 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?

ANS:

Mass matrix m =

Modified mass matrix

Model property Ln

For the first mode

n=1

Modified mass matrix M1 =

m1=3.861m

Model property L1 =

L1=4.121m

Model participation factor

Γ1=L1/M1

=4.121m/3.861m

=1.067

Effective modal mass

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

M1*=4.398m

For the second mode

n=2

Modified mass matrix M2 =

m2=3.861m

Model property L2 =

L2=-1.293m

Model participation factor

Γ2=L2/M2

=-1.293m/3.861m

=-0.335

Effective modal mass

M2* =(-1.293m)^2/3.861m

M2*=0.436m

For the third mode

n=3

Modified mass matrix M3 =

m3=3.861m

Model property L3 =

L3=0.683m

Model participation factor

Γ3=L3/M3

=0.683m/3.861m

=0.177

Effective modal mass

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

M3*=0.121m

For the fourth mode

n=4

Modified mass matrix M4 =

m4=3.861m

Model property L4 =

L4=-0.381m

Model participation factor

Γ4=L4/M4

=-0.381m/3.861m

=-0.098

Effective modal mass

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

M4*=0.038m

For the fifth mode

n=5

Modified mass matrix M5 =

m5=3.861m

Model property L5 =

L5=0.175m

Model participation factor

Γ5=L5/M5

=0.175m/3.861m

=0.045

Effective modal mass

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

M5*=0.008m

Modal mass participation ratio= Mn*/ Mj

Mj= Total mass of building=1m+1m+1m+1m+1m=5m

Modal mass participation ratio for first node (n=1) = M1*/ Mj

=4.398m/5m =0.88

Modal mass participation ratio for second node (n=2) = M2*/ Mj

=0.436m/5m =0.09

Modal mass participation ratio for third node (n=3) = M3*/ Mj

=0.121m/5m =0.02

Modal mass participation ratio for fourth node (n=4) = M4*/ Mj

=0.038m/5m =0.01

Modal mass participation ratio for fifth node (n=5) = M5*/ Mj

=0.008m/5m =.0002

Adding first two modal mass participation ratios we get, 0.88 + 0.09 = 0.97 i.e. 96%

To achieve 90% first two modes are enough

Base Shear

Vbn = Mn* An

For first node

From graph

A1=.27g

Vb1 = M1* A1

=4.398m/.27g

=1.187mg

For second node

From graph

A2=.75g

Vb2 = M2* A2

=0.436m/.75g

=0.327mg

For third node

From graph

A3=1.05g

Vb3 = M3* A3

=0.121m/1.05g

=0.126mg

For fourth node

From graph

A4=1.02g

Vb4 = M4* A4

=0.038m/1.02g

=0.039mg

For fifth node

From graph

A5=1g

Vb5 = M5* A5

=0.008m/1g

=0.008mg

Average Base shear Vb =square root of (Vb1^2+Vb2^2+Vb3^2+Vb4^2+Vb5^2)

=square root of (1.187^2+0.327^2+0.126^2+0.039^2+0.008^2)

=1.15mg