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

MnMn = ΦnTmΦnΦnTmΦn

Mode:-1

The first mode of vibration (T1T1) = 2 seconds

• Φ1Φ1 = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.3340.640.8951.0781.173⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.3340.640.8951.0781.173]
• M1M1 = Φ1TmΦ1Φ1TmΦ1

= [0.3340.6410.8951.0781.173][0.3340.6410.8951.0781.173] ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[M00000M00000M00000M00000M] ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.3340.640.8951.0781.173⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.3340.640.8951.0781.173]

• By using the scientific calculater we calculate M1M1 and we got, M1M1 = 3.861M
• M1M1 = 3.861M

Mode:-2

The second mode of vibration (T2T2) = 0.6852 seconds

• Φ2Φ2 = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−0.895−1.173−0.6410.3341.078⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[-0.895-1.173-0.6410.3341.078]
• M2M2 = Φ2TmΦ2Φ2TmΦ2

= [−0.895−1.173−0.641−0.3341.078][-0.895-1.173-0.641-0.3341.078] ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[M00000M00000M00000M00000M] ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−0.895−1.173−0.6410.3341.078⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[-0.895-1.173-0.6410.3341.078]

• By using the scientific calculater we calculate M2M2 and we got, M2M2 = 3.86M
• M2M2 = 3.861M

Mode:-3

The third mode of vibration (T3T3) = 0.4346 seconds

• Φ3Φ3 = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1.1730.334−1.078−0.6410.895⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[1.1730.334-1.078-0.6410.895]
• M3M3 = Φ3TmΦ3Φ3TmΦ3

= [1.1730.334−1.078−0.6410.895][1.1730.334-1.078-0.6410.895] ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[M00000M00000M00000M00000M] ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1.1730.334−1.078−0.6410.895⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[1.1730.334-1.078-0.6410.895]

• By using the scientific calculater we calculate M3M3 and we got, M3M3 = 3.85M
• M3M3 = 3.85M

Mode:-4

The fourth mode of vibration (T4T4) = 0.3383 seconds

• Φ4Φ4 = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−1.0780.8950.334−1.1730.641⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[-1.0780.8950.334-1.1730.641]
• M4M4 = Φ4TmΦ4Φ4TmΦ4

= [−1.0780.8950.334−1.1730.641][-1.0780.8950.334-1.1730.641] ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[M00000M00000M00000M00000M] ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−1.0780.8950.334−1.1730.641⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[-1.0780.8950.334-1.1730.641]

• By using the scientific calculater we calculate M4M4 and we got, M4M4 = 3.861M
• M4M4 = 3.861M

Mode:-5

The fifth mode of vibration (T5T5) = 0.2966 seconds

• Φ5Φ5 = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.641−1.0781.173−0.8950.334⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.641-1.0781.173-0.8950.334]
• M5M5 = Φ5TmΦ5Φ5TmΦ5

= [0.641−1.0781.173−0.8950.334][0.641-1.0781.173-0.8950.334] ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[M00000M00000M00000M00000M] ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.641−1.0781.173−0.8950.334⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.641-1.0781.173-0.8950.334]

• By using the scientific calculater we calculate M5M5 and we got, M5M5 = 3.861M
• M5M5 = 3.861M

Step:-2

LnLn = ΦnTΦnTm ιm ι

Mode:-1

The first mode of vibration (T1T1) = 2 seconds

• Φ1Φ1 = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.3340.640.8951.0781.173⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.3340.640.8951.0781.173]
• L1L1 = Φ1Tm ιΦ1Tm ι

= [0.3340.6410.8951.0781.173][0.3340.6410.8951.0781.173] ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[M00000M00000M00000M00000M] ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]

• By using the scientific calculater we calculate L1L1 and we got, L1L1 = 4.121M
• L1L1 = 4.121M

Mode:-2

The second mode of vibration (T2T2) = 0.6852 seconds

• Φ2Φ2 = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−0.895−1.173−0.6410.3341.078⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[-0.895-1.173-0.6410.3341.078]
• L2L2 = Φ2Tm ιΦ2Tm ι

= [−0.895−1.173−0.641−0.3341.078][-0.895-1.173-0.641-0.3341.078] ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[M00000M00000M00000M00000M] ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]

• By using the scientific calculater we calculate L2L2 and we got, L2L2 = -1.298M
• L2L2 = -1.298M

Mode:-3

The third mode of vibration (T3T3) = 0.4346 seconds

• Φ3Φ3 = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1.1730.334−1.078−0.6410.895⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[1.1730.334-1.078-0.6410.895]
• L3L3 = Φ3Tm ιΦ3Tm ι

= [1.1730.334−1.078−0.6410.895][1.1730.334-1.078-0.6410.895] ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[M00000M00000M00000M00000M] ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]

• By using the scientific calculater we calculate L3L3 and we got, L3L3 = 0.687M
• L3L3 = 0.687M

Mode:-4

The fourth mode of vibration (T4T4) = 0.3383 seconds

• Φ4Φ4 = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−1.0780.8950.334−1.1730.641⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[-1.0780.8950.334-1.1730.641]
• L4L4 = Φ4Tm ιΦ4Tm ι

= [−1.0780.8950.334−1.1730.641][-1.0780.8950.334-1.1730.641] ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[M00000M00000M00000M00000M] ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]

• By using the scientific calculater we calculate L4L4 and we got, L4L4 = -0.381M
• L4L4 = -0.381M

Mode:-5

The fifth mode of vibration (T5T5) = 0.2966 seconds

• Φ5Φ5 = ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0.641−1.0781.173−0.8950.334⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[0.641-1.0781.173-0.8950.334]
• L5L5 = Φ5Tm ιΦ5Tm ι

= [0.641−1.0781.173−0.8950.334][0.641-1.0781.173-0.8950.334] ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣M00000M00000M00000M00000M⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[M00000M00000M00000M00000M] ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣11111⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[11111]

• By using the scientific calculater we calculate L5L5 and we got, L5L5 = 0.175M
• L5L5 = 0.175M

Step:-3

ΓnΓn = LnMnLnMn

Mode:-1

• Γ1Γ1 = L1M1L1M1 = 4.121M3.861M4.121M3.861M = 1.067

Mode:-2

• Γ2Γ2 = L2M2L2M2 = −1.298M3.861M-1.298M3.861M = -0.33

Mode:-3

• Γ3Γ3 = L3M3L3M3 = 0.687M3.85M0.687M3.85M = 0.178

Mode:-4

• Γ4Γ4 = L4M4L4M4 = −0.381M3.861M-0.381M3.861M = -0.09

Mode:-5

• Γ5Γ5 = L5M5L5M5 = 0.175M3.861M0.175M3.861M = 0.04

Step:-4

Mn∗Mn∗ = Ln2MnLn2Mn

Mode:-1

• M1∗M1∗ = L12M1L12M1
• L1L1 = 4.121M, M1M1 = 3.861M
• M1∗M1∗ = (4.121M)23.861M(4.121M)23.861M
• M1∗M1∗ = 4.39M

Mode:-2

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

Mode:-3

• M3∗M3∗ = L32M3L32M3
• L3L3 = 0.687M, M3M3 = 3.85M
• M3∗M3∗ = (0.687M)23.85M(0.687M)23.85M
• M3∗M3∗ = 0.12M

Mode:-4

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

Mode:-5

• M5∗M5∗ = L52M5L52M5
• L5L5 = 0.175M, M5M5 = 3.861M
• M5∗M5∗ = (0.175M)23.861M(0.175M)23.861M
• M5∗M5∗ = 0.007M

Step:-5

• Modal mass participating ratio = Mn∗MjMn∗Mj

Mn∗Mn∗ = Effective modal mass.

MjMj = Total mass = M+M+M+M+M = 5M

Mode:-1

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

Mode:-2

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

Mode:-3

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

Mode:-4

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

Mode:-5

• M5∗MjM5∗Mj = 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.

VbnVbn = Ln2MnLn2Mn××AnAn = Mn∗Mn∗××AnAn

(T1T1) = 2 seconds, A1A1 = 0.27g

(T2T2) = 0.6852 seconds, A2A2 = 0.76g

(T3T3) = 0.4346 seconds, A3A3 = 1.03g

(T4T4) = 0.3383 seconds, A4A4 = 1.03g

(T5T5) = 0.2966 seconds, A5A5 = 0.76g

Mode:-1

• Vb1Vb1 = (M1∗M1∗x A1A1) = (4.39M x 0.27g) = 2.619 kips

Mode:-2

• Vb2Vb2 = (M2∗M2∗x A2A2) = (0.43M x 0.76g) = 0.719 kips

Mode:-3

• Vb3Vb3 = (M3∗M3∗x A3A3) = (0.12M x 1.03g) = 0.272 kips

Mode:-4

• Vb4Vb4 = (M4∗M4∗x A4A4) = (0.03M x 1.03g) = 0.067 kips

Mode:-5

• Vb5Vb5 = (M5∗M5∗x A5A5) = (0.007M x 0.76g) = 0.011 kips

Step:-7

Combined base shear as per SRSS rule:-

• VbVb = √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

= 3.20 kips

Step:-8

• The two modes are to be considered such that a modal mass participating ratio of 90% is obtained.
• M1∗MjM1∗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.