Understanding the concepts on Degrees of Freedom

Q] 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.

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?

AIM:- To compute L_n , M_n and Γ_n for each of the five modes from the given equations.

            To calculate the effective modal mass M_n* 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:  M_n = ø_nT_mø_n

Mode 1:

Mode of vibration, T₁ = 2 sec

M₁ = ø₁T₁ø

   

Mode 2:

Mode of vibration, T₂ = 0.6852 sec

 M₂ = ø₂T_mø₂

  

Mode 3:

Mode of vibration, T₃ = 0.4346 sec

       M₃ = ø₃T_mø₃

 

Mode 4:

Mode of vibration, T₄ = 0.3383 sec

       M₄= ø₄T_mø₄
  

Mode 5:

Mode of vibration, T₅ = 0.2966 sec

      M₅= ø₅T_mø₅
  

Step 2: L_{n =}ø_nT_mI

Mode 1:

Mode of vibration, T₁ = 2 sec

         L_{1 =}ø₁T_mI

                   

Mode 2:

Mode of vibration, T₂ = 0.6852 sec

     L_{2 =}ø₂T_mI
                   

 

Mode 3:

Mode of vibration, T₃ = 0.4346 sec

                             L_{3 =}ø₃T_mI                  

 

Mode 4:

Mode of vibration, T₄ = 0.3383 sec

           L₄ = ø₄T_mI
                   

Mode 5:

Mode of vibration, T₅ = 0.2966 sec

              L_{5 =}ø₅T_mI
                   

                   

Step:-3

T_n = 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

Mn∗">Mn∗Mn∗ = Ln2Mn">Ln2MnLn2Mn

Mode:-1

• M1∗">M1∗M1∗ = L12M1">L12M1L12M1
• L1">L1L1 = 4.121M, M1">M1M1 = 3.861M
• M1∗">M1∗M1∗ = (4.121M)23.861M">(4.121M)23.861M(4.121M)23.861M
• M1∗">M1∗M1∗ = 4.39M

Mode:-2

• M2∗">M2∗M2∗ = L22M2">L22M2L22M2
• L2">L2L2 = -0.298M, M2">M2M2 = 3.861M
• M2∗">M2∗M2∗ = (-0.298M)23.861M">(−0.298M)23.861M(-0.298M)23.861M
• M2∗">M2∗M2∗ = 0.43M

Mode:-3

• M3∗">M3∗M3∗ = L32M3">L32M3L32M3
• L3">L3L3 = 0.687M, M3">M3M3 = 3.85M
• M3∗">M3∗M3∗ = (0.687M)23.85M">(0.687M)23.85M(0.687M)23.85M
• M3∗">M3∗M3∗ = 0.12M

Mode:-4

• M4∗">M4∗M4∗ = L42M4">L42M4L42M4
• L4">L4L4 = -0.381M, M4">M4M4 = 3.861M
• M4∗">M4∗M4∗ = (-0.381M)23.861M">(−0.381M)23.861M(-0.381M)23.861M
• M4∗">M4∗M4∗ = 0.03M

Mode:-5

• M5∗">M5∗M5∗ = L52M5">L52M5L52M5
• L5">L5L5 = 0.175M, M5">M5M5 = 3.861M
• M5∗">M5∗M5∗ = (0.175M)23.861M">(0.175M)23.861M(0.175M)23.861M
• M5∗">M5∗M5∗ = 0.007M

Step:-5

• Modal mass participating ratio = Mn∗Mj">Mn∗MjMn∗Mj

          Mn∗">Mn∗Mn∗ = Effective modal mass.

          Mj">MjMj = Total mass = M+M+M+M+M = 5M

         

Mode:-1

• M1∗Mj">M1∗MjM1∗Mj = 4.39M5M">4.39M5M4.39M5M = 0.88 = 88%

Mode:-2

• M2∗Mj">M2∗MjM2∗Mj = 0.43M5M">0.43M5M0.43M5M = 0.086 = 8.6%

Mode:-3

• M3∗Mj">M3∗MjM3∗Mj = 0.12M5M">0.12M5M0.12M5M = 0.024 = 2.4%

Mode:-4

• M4∗Mj">M4∗MjM4∗Mj = 0.03M5M">0.03M5M0.03M5M = 0.006 = 0.6%

Mode:-5

• M5∗Mj">M5∗MjM5∗Mj = 0.007M5M">0.007M5M0.007M5M = 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.

Vbn">VbnVbn = Ln2Mn">Ln2MnLn2Mn×">××An">AnAn  =  Mn∗">Mn∗Mn∗×">××An">AnAn

(T1">T1T1) = 2 seconds, A1">A1A1 = 0.27g

(T2">T2T2) = 0.6852 seconds, A2">A2A2 = 0.76g

(T3">T3T3) = 0.4346 seconds, A3">A3A3 = 1.03g

(T4">T4T4) = 0.3383 seconds, A4">A4A4 = 1.03g

(T5">T5T5) = 0.2966 seconds, A5">A5A5 = 0.76g

Mode:-1

• Vb1">Vb1Vb1 = (M1∗">M1∗M1∗x A1">A1A1) = (4.39M x 0.27g) = 2.619 kips

Mode:-2

• Vb2">Vb2Vb2 = (M2∗">M2∗M2∗x A2">A2A2) = (0.43M x 0.76g) = 0.719 kips

Mode:-3

• Vb3">Vb3Vb3 = (M3∗">M3∗M3∗x A3">A3A3) = (0.12M x 1.03g) = 0.272 kips

Mode:-4

• Vb4">Vb4Vb4 = (M4∗">M4∗M4∗x A4">A4A4) = (0.03M x 1.03g) = 0.067 kips

Mode:-5

• Vb5">Vb5Vb5 = (M5∗">M5∗M5∗x A5">A5A5) = (0.007M x 0.76g) = 0.011 kips

Step:-7

Combined base shear as per SRSS rule:-

• Vb">VbVb = Vb12+Vb22+Vb32+Vb42+Vb52">√Vb12+Vb22+Vb32+Vb42+Vb52Vb12+Vb22+Vb32+Vb42+Vb52

               = (2.613)2+(0.719)2+(0.272)2+(0.067)2+(0.011)2">√(2.613)2+(0.719)2+(0.272)2+(0.067)2+(0.011)2(2.613)2+(0.719)2+(0.272)2+(0.067)2+(0.011)2

               = 3.20 kips

Step:-8

• 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:-

• Calculated L_n , M_n and Γ_n for each of the five modes from the given equations.
• Calculated the effective modal mass M_n* 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.