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. 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?
T1 = 2sec `Phi`1 = `((0.334), (0.641), (0.895), (1.078), (1.173))`
T2 = 0.6852 `Phi`2 = `((-0.895), (-1.173), (-0.641), (0.334), (1.078))`
T3 = 0.4346 `Phi`3 = `((1.173), (0.334), (-1.078), (-0.641), (0.895))`
T4 = 0.3383 `Phi`4 = `((-1.078), (0.895), (0.334), (-1.173), (0.641))`
T5 = 0.2966 `Phi`5 = `((0.641), (-1.078), (1.173), (-0.895), (0.334))`
Solution :
L1 = `[[0.334, 0.641, 0.895, 1.078, 1.173]]` m `((1),(1),(1))`
= 4.121m
L2 = `[[-0.895, -1.173, -0.641, 0.334, 1.078]]`m `((1),(1),(1))`
= -1.297m
L3 = `[[1.173, 0.334, -1.078, -0.641, 0.895]]`m `((1),(1),(1))`
= 0.683m
L4 = `[[-1.078, 0.895, 0.334, -1.173, 0.641]]`m`((1),(1),(1))`
= -0.381m
L5 = `[[0.641,-1.078, 1.173,-0.895,0.334]]`m `((1),(1),(1))`
= 0.175m
M1 = `[[0.334, 0.641, 0.895, 1.078, 1.173]]` m `((0.334), (0.641), (0.895), (1.078), (1.173))`
= 3.861m
M2 = `[[-0.895, -1.173, -0.641, 0.334, 1.078]]`m`((-0.895), (-1.173), (-0.641), (0.334), (1.078))`
= 3.861m
M3 = `[[1.173, 0.334, -1.078, -0.641, 0.895]]`m`((1.173), (0.334), (-1.078), (-0.641), (0.895))`
= 3.861m
M4 = `[[-1.078, 0.895, 0.334, -1.173, 0.641]]`m`((-1.078), (0.895), (0.334), (-1.173), (0.641))`
= 3.861m
M5 = `[[0.641,-1.078, 1.173,-0.895,0.334]]`m`((0.641), (-1.078), (1.173), (-0.895), (0.334))`
= 3.861m
`Gamma 1= (L1)/(M1)` = 4.121/ 3.861 = 1.067
`Gamma 2 = (L2)/(M2)` = -1.297/3.861 = -0.335
`Gamma 3 = (L3)/(M3)` = 0.683/3.861 = 0.1768`
`Gamma 4 = (L4)/(M4)` = -0.381/3.861 = -0.0986
`Gamma 5 = (L5)/(M5)` = 0.175/3.861 = 0.045
M1* = `(L1^2)/(M1) = ((4.121^2)m)/3.861 = 4.39m`
M2* = `(L2^2)/(M2) = ((-1.297)^2m)/(3.861) = 0.435m`
M3* = `(L3^2)/(M3) = ((0.683)^2m)/(3.861) = 0.120m`
M4* = `(L4)^2/(M4) = ((-0.381)^2m)/3.861=0.0375m`
M5* = `(L5)^2/(M5) = ((0.175)^2m)/(3.861) = 0.007931m`
From graph
At T1 = 2sec A1 = 0.276g
At T2 = 0.6852sec A2 = 0.76g
At T3 = 0.4346sec A3 = 1.03g
At T4 = 0.3383sec A4 = 1.03g
At T5 = 0.2966sec A5 = 1.03g
`Vb = (L n)^2/(Mn) An = Mn*An`
V1 = M1* X A1 = 4.39m X 0.27g = 1.1853 mg
V2 = M2* X A2 = 0.435m X 0.76g = 0.3306 mg
V3 = M3* X A3 = 0.120m X 1.03g = 0.1236 mg
V4 = M4* X A4 = 0.0375m X 1.03g = 0.0386 mg
V5 = M5* X A5 = 0.007931m X 1.03g = 0.00816 mg
Total base shear = `sqrt(V1^2+V2^2+V3^2+V4^2+V5^2`
= `sqrt(1.1853^2+0.3306^2+0.1236^2+0.0386^2+0.00816^2)mg`
= 1.237mg
Modal mass participating ratio = Mn*/m = M1*/m + M2*/m + M3*/m + M4*/m + M5*/m
= 4.39 + 0.435 + 0.120 + 0.0375 + 0.007931 = 4.99 `approx`5 = Total mass
Modal mass participating ratio of mode 1= 4.39/5 = 0.878 = 87.8%
mode 2 = 0.435/5 = 0.087 = 8.7%
mode 3 = 0.12/5 = 0.024 =2.4%
mode 4 = 0.0375/5= 0.0075 = 0.75%
mode 5 = 0.007931/5 = 0.0014782 = 0.14782%
Thus to obtain 90% in modal mass participating ratio we need to add mode 1, mode 3, mode 4 and mode 5