Understanding the concepts on Degrees of Freedom

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 has 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:

• Ln , Mn and Γn for each of the five modes are to be computed from Equations 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 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?

INTRODUCTION:

Degrees of freedom refers to the maximum number of logically independent values, which are values that have the freedom to vary, in the data sample. Degrees of freedom is calculated by subtracting one from the number of items within the data sample.

AIM:

To compute Ln , Mn and Γn for each of the five modes from the given equations.To calculate 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:

Procedure:

Step 1: Mn = ønTmøn

Mode 1:

Mode of vibration, T1 = 2 sec

M1 = ø1T1ø1

Mode 2:

Mode of vibration, T2 = 0.6852 sec

M2 = ø2Tmø2

Mode 3:

Mode of vibration, T3 = 0.4346 sec

M3 = ø3Tmø3

Mode 4:

Mode of vibration, T4 = 0.3383 sec

M4= ø4Tmø4

Mode 5:

Mode of vibration, T5 = 0.2966 sec

M5= ø5Tmø5

Step 2: To find Ln

Ln =ønTmI

Mode 1:

Mode of vibration, T1 = 2 sec

L1 =ø1TmI

Mode 2:

Mode of vibration, T2 = 0.6852 sec

L2 =ø2TmI

Mode 3:

Mode of vibration, T3 = 0.4346 sec

L3 =ø3TmI

Mode 4:

Mode of vibration, T4 = 0.3383 sec

L4 = ø4TmI

Mode 5:

Mode of vibration, T5 = 0.2966 sec

L5 =ø5TmI

Step 3: Finding Tn

Tn = Ln /Mn

T1 =L1/M1 = 1.067

T2 =L2/M2 = -0.33

T3= L3/M3 = 0.178

T4 = L4/M4 = -0.09

T5 = L5/M5 = 0.04

Step 4: Finding Effective modal mass, Mn*

Mn* = Ln2 /Mn

M1* = L12 /M1 = 4.39M

M2* = L22 /M2 = 0.43M

M3* = L32 /M3 = 0.12M

M4* = L42 /M4 = 0.03M

M5* = L52 /M5 = 0.007M

Step 5: Finding Modal mass participating ratio

Modal mass participating ratio = Mn* / Mj

Mj = M+M+M+M+M = 5M

Modal mass participating ratio for Mode 1 = M1* / Mj = 0.88

Modal mass participating ratio for Mode 2 = M2* / Mj = 0.086

Modal mass participating ratio for Mode 3 = M3* / Mj = 0.024

Modal mass participating ratio for Mode 4 = M4* / Mj = 0.006

Modal mass participating ratio for Mode 5 = M5* / Mj = 0.001

Step 6: To obtain the Base shear

Vbn = Mn* * An

For Mode 1

For T1 = 2 sec, A1 = 0.28g

Vb1 = M1* * A1 = 2.612

For Mode 2

For T2 = 0.6852 sec, A2 = 0.77g

Vb2 = M2* * A2 = 0.719

For Mode 3

For T3 = 0.4346 sec, A3 = 1.03g

Vb3 = M3* * A3 = 0.272

For Mode 4

For T4 = 0.3383 sec, A4 = 1.03g

Vb4 = M4* * A4 = 0.067

For Mode 5

For T5 = 0.2966 sec, A5 = 0.77g

Vb5 = M5* * A5 = 0.011

Step 7: Using SRSS rule

Vb = (Vb12+ Vb22+ Vb32+ Vb42+ Vb52)1/2

= 3.2 kips

• The two modes are to be considered such that a modal mass participating ratio of 90% is obtained.
• M1∗Mj">M1∗MjM1∗Mj + M2∗Mj">M2∗MjM2∗Mj = 88% + 8.6% = 96.6%

RESULT:-

• Finally we Calculated Ln , Mn and Γn for each of the five modes from the given equations.
• Calculated 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.