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 T_n and their corresponding mode shapes, φ_n has been derived.

• 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) Model mass (M)

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

The model mass ratio for the first mode is

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

The model mass ratio for the second mode is

where M2 = 7.72275m

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

The model mass ratio for the third mode is

where M3 = 11.584425m

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

The model mass ratio for the fourth mode is

where M4 = 15.4459m

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

The model mass for the fifth mode is

where M5 = 15.198565m

Model Property:(L)

As the model property equation given

Where l is influence vector ( Assuming 100% utilization of sesimic forces) =

This are constant for all the modes. where we alreay know the mass matrix for the model we apply them in the equation.

i) model Property for the first mode (n=1)

where we get L1 = 4.121m.

ii) Model propert for the second mode (n=2)

L2= -2.594m.

iii) model property for the third mode is (n=3)

where L3= 2.049m.

iv) Model proerty for the fourth mode is (n=4)

L4= -1.524m.

v) Model property for the fifth mode is (n=5)

L5= 0.875.

Model participation factor:

where L is the model property and M is the model mass. Now substitute them seperatly from the each mode.

1) Model participation factor for the first mode = 4.121m/3.861m

ɼ₁ = 1.067.

2) Model participation factor for the second mode = -2.594m/7.72295m

ɼ₂ = -0.335

3) Model participation factor for the third mode = 2.094m/11.584425m

ɼ₃ = 0.177

4) Model participation factor for the fourth mode = -1.524m/15.4459m

ɼ₄ = -0.098.

5) Model participation factor for the fifth mode = 0.875m/15.198565m

ɼ₅ = 0.045.

Effective model mass:

Where L is the model property and m is the model mass. Now substitute them in the equation and calculate seperatly fot the each mode.

i) Effective model mass for the first mode

m*₁ = (4.121m)² /3.861m = 4.398m.

ii) Effective model mass for second mode

m*₂ = (-2.594m)²/7.72295m = 0.436m.

iii) Effective model mass for the third mode

m*₃ = (2.094m)²/11.584425m = 0.121m.

iv) Effective model mass for the fourth mode

m*₄ = (-1.524m)²/15.4459m = 0.038m.

v) Effective model mass for the fifth mode

m*₅ = (0.875m)²/15.198565m = 0.008m.

Model mass Participation ratio:

As we know Model mass participation ratio = M_n*/m_j

m_j = Total mass of the buildings = m+m+m+m+m = 5m.

Where

1) model mass participation ratio for the first mode (n=1)= 4.398m/5m = 0.88.

2) Model mass participation ratio for the second mode (n=2)= 0.436m/5m = 0.09.

3) Model mass participation ratio for the third mode (n=3)= 0.121m/5m = 0.02.

4) Model mass participation ratio for the fourth mode (n=4)= 0.038m/5m = 0.01.

5) Model mass participation ratio for the fifth mode (n=5)= 0.008m/5m = 0.002.

To achive the 90% of the model mass participation ratio only the first two modes are enough where

= 0.88+0.09=0.97.

Hence the forst two modes are enough to achive 90% model mass participation.

Base shear:

As we know Base shear (V_B) = M_n* . A_n

We will substitute valude in the equation seperatly for each mode.

• Base shear for the first mode V_B1 = 4.398m X 0.27g = 1.187mg.
• Base shear for the first mode V_B2 = 0.436m X 0.76g = 0.327mg.
• Base shear for the first mode V_B3 = 0.121m X 1.03g = 0.126mg.
• Base shear for the first mode V_B4 = 0.038m X 1.03g = 0.039mg.
• Base shear for the first mode V_B5 = 0.008mg.

To get the average base shear we need to use the SRSS method square root of some of squares.

Hence we get The average base shear V_B = 1.15mg.