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 has been derived. 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?
SOLUTION:
Given
T1 = 2 sec; `phi 1`= `((0.334),(0.641), (0.895), (1.078), (1.173))`
T2 = 0.6852 sec; `phi 2` = `((-0.895), (-1.173), (-0.641),(0.334),(1.078))`
T3 = 0.4346 sec; `phi3` = `((1.173), (0.334), (-1.078), (-0.641), (0.895))`
T4 = 0.3383 sec; `phi 4` = `((-1.078), (0.895), (0.334), (-1.173), (0.641))`
T5 = 0.2966 sec; `phi5` = `((0.641), (-1.078), (1.173), (-0.895), (0.334))`
The mass matrix and Influence vector of the 5 storey structure is
m = `((m,0,0,0,0), (0 , m , 0 , 0 , 0), (0 , 0 , m , 0 , 0), (0 , 0 , 0 , m , 0), (0 , 0 , 0 , 0 , m))`
`l` = `((1),(1),(1),(1),(1))`
Calculation of Mn
Mn = `phi underset(n) overset (T)` m `phi underset (n)`
`M underset (1)` = (0.334 0.641 0.895 1.078 1.173) (m) `((0.334),(0.641),(0.895), (1.078), (1.173))`
`M underset (1)` = ((0.334 x 0.334) + (0.641 x 0.641) + (0.895 x 0.895) + (1.078 x 1.078) + (1.173 x 1.173)) m
`M underset (1)` = 3.8614 m
`M underset(2)`= (-0.895 -1.173 -0.641 0.334 1.078) (m) `((-0.895), (-1.173), (-0.641), (0.334), (1.078))`
`M underset (2)` = 3.8614 m
`M underset(3)` = (1.173 0.334 -1.078 -0.641 0.895) (m) `((1.173), (0.334), (-1.078), (-0.641), (0.895))`
`M underset(3)` = 3.8614 m
`M underset(4)` = (-1.078 0.895 0.334 -1.173 0.641) (m) `((-1.078), (0.895), (0.334), (-1.173), (0.641))`
`M underset(4)` = 3.8614 m
`M underset(5)` = (0.641 -1.078 1.173 -0.895 0.334) (m) `((0.641), (-1.078), (1.173), (-0.895), (0.334))`
`M underset(5)` = 3.8614m
Calculation of `L underset (n)`
`L underset(n)` = `phi underset(n) overset(T)`m `l`
`L underset(1)` = (0.334 0.641 0.895 1.078 1.173) (m) `((1),(1),(1),(1),(1))`
`L underset(1)` = ( (0.334 x 1) + (0.641 x1) + (0.895 x 1) + (1.078 x1) + (1.173 x 1) ) m
`L underset(1)` = 4.121 m
`L underset(2)` = (-0.895 -1.173 -0.641 0.334 1.078) (m) `((1), (1), (1), (1),(1))`
`L underset(2)` = -1.297 m
`L underset(3)` = (1.173 0.334 -1.078 -0.641 0.895) (m) `((1),(1),(1),(1),(1))`
`Lunderset (3)` = 0.683 m
`L underset(4)`=(-1.078 0.895 0.334 -1.173 0.641) (m) `((1),(1),(1),(1),(1))`
`L underset(4)` = 0.381 m
`L underset (5)` = (0.641 -1.078 1.173 -0.895 0.334) (m) `((1),(1),(1),(1),(1))`
`L underset(5)` = 0.175 m
Calculation of Modal participation factor `Gamma underset(n)`
`Gamma underset(n)` = `Lunderset(n)`/ `M underset(n)`
`Gamma underset(1)` = `4.121/3.861` = 1.067
`Gamma underset (2)` = `(-)1.297/3.861` = - 0.336
`Gamma underset(3)` = `0.683/ 3.861` = 0.177
`Gamma underset(4)` = `0.381/3.861` = 0.099
`Gamma underset(5)` = `0.175/ 3.861` = 0.045
Calculation of Effective modal mass `M underset(n) overset(ast)` and participating ratio = `M underset(n) overset(ast)`/`Munderset(i)`
`M underset(n) overset(ast)` = `Lunderset(n) overset(2)` /`Munderset(n)`
`Munderset(i)` = Total mass = 5m
`M underset(1) overset(ast)` = `(4.121 m)overset(2)`/`3.861m` = 4.3980 m
Participating ratio = 4.3980m / 5m = 0.8796
`M underset(2) overset(ast)` = `(1.6820m)overset(2)`/ `3.861m` = 0.4356 m
Participating ratio = 0.4356m / 5m = 0.087
`Munderset(3) overset(ast)` = `(0.466m)overset(2)` / `3.861m` = 0.120 m
Participating ratio = 0.120m / 5m = 0.024
`Munderset(4) overset(ast)` = `(0.145m)overset(2)` / 3.861m = 0.037 m
Participating ratio = 0.037m / 5m = 0.0074
`Munderset(5) overset(ast)` = `(0.0306m) overset(2)` / 3.861m = 0.008m
Participating ratio = 0.008m / 5m = 0.0016
Calculation of design acceleration
From the given pseudo acceleration design spectrum graph. The design acceleration of each mode is
`Tunderset(1)` = 2sec; `Aunderset(1)` = 0.279
`T underset(2)` = 0.6852 sec; `Aunderset (2)`= 0.76
`T underset(3)` = 0.4346 sec; `A underset(3)` = 1.039
`T underset(4)` = 0.3383 sec; `A underset(4)` = 1.039
`Tunderset(5)` = 0.2966 sec; `Aunderset(5)` = 1.039
Calculation of Base shear force `V underset(b)`n
`V underset (b)`n = `Munderset(n) overset(ast)` X `Aunderset (n)`
`Vunderset(b)`1 = `Munderset(1) overset(ast)` X `Aunderset(1)` = `4.3980 X 0.279` = 1.227m
Similarly
`V underset(b)` 2 = 0.4356 X 0.76 = 0.331m
`Vunderset(b)`3 = 0.120 X 1.039 = 0.126 m
`Vunderset(b)`4 = 0.037 X 1.039 = 0.0038 m
`V underset(b)`5 = 0.008 X 1.039 = 0.008 m
Combined Base shear force `V underset(bT)` by Sum of squares (SRSS) method is
`Vunderset(b1) overset(2)` = `(1.227m)overset(2)` = `1.50 m overset(2)`
`Vunderset(b2) overset(2)` = `(0.331m)overset(2)` = `0.1115m overset(2)`
`Vunderset(b3) overset(2)` = `(0.126m) overset(2)` = `0.015m overset(2)`
`V underset(b4) overset(2)` = `(0.038m) overset(2)` = `0.0014m overset(2)`
`V underset(b5) overset(2)` = `(0.008m) overset(2)` = `0.000064m overset(2)`
`Vunderset(bT)` = `sqrt((1.50+0.1115+0.015+0.0014+0.000064))moverset(2)`
`Vunderset(bT)` = 1.275 m