Understanding the concepts on Degrees of Freedom
Question :
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.
ANS :
Given data
T1 = 2 sec , `pai`= 0.334
0.641
0.895
1.078
1.173
T2 = 0.685 sec `pai` = -0.895
-1.173
-0.641
0.334
1.078
T3 = 0.4346 sec `pai` = -1.173
0.334
-1.078
-0.641
0.845
T4 = 0.3383 sec `pai` = -1.078
0.895
0.334
-1.173
0.641
T5 = 0.2296 sec `pai` = 0.641
-1.078
1.173
-0.895
0.334
1) WKT : Mn = `pai`n T m `pai`n
Mode 1=
M1= ( 0.334 , 0.641, 0.895 , 1.078, 1.173) (M ) x 0.334
0.641
0.895
1.078
1.173
M1 = (M) X (0.334 ^2 X 0.641 ^2 X 0.895 ^ 2 X 1.078^2 x 1.173^2 )
M1 = 3.861M
Mode 2 =
M2 = (-0.895 X -1.173 X -0.641 X 0.334 X 1.078) (M) X -0.895
-1.173
-0.641
0.334
1.078
M2 = (M ) X (0. 895^2 X -1.173^2 X 0.641 ^2 X 0.334 ^2 X 1.078^ )
M2 = 3.861M
Mode 3 =
M3 = (-1.173 x 0.334 x -1.078 x -0.641 x 0.845) x (M ) x -1.173
0.334
-1.078
0.641
0.845
M3 = (-1.173^2 x 0.334^ 2 X -1.078^ 2 x -0.641^2 X 0.845^2 )(M)
M3 = 3.861M
Mode 4 :
M4 = (1.078 X 0.895 X 0.334 X -1.173 X 0.641 ) (M) X 1.078
0.895
0.334
-1.173
0.641
M4 = (1.078^ 2 X 0.895^ 2 X 0.334 ^ 2 X -1.173^ 2 X 0.641^2 ) (M)
M4 = 3.861 M
Mode 5 =
M5 = (0.641 X -1.078 X 1.173 X -0.895 X 0.334) (M) x 0.641
-1.078
1.173
0.895
0.334
M5 = (0.641^2 X -1.078^2 X 1.173^2 X -0.895^2 X 0.334^@2)(M)
M5 = 3.861M
2)
Ln = `pai`nTmu
Mode 1=
L1= (0.334 , 0.641, 0.895 , 1.078, 1.173) (M ) x 1
1
1
1
1
L1 = (0.334+0.641+0.895 +1.078+1.173) (M )
L1 = 4.121M
Mode 2 =
L2 = (-0.895 X -1.173 X -0.641 X 0.334 X 1.078) (M) X 1
1
1
1
1
L2 = (-0.895 -1.173 -0.641 + 0.334+ 1.078) (M)
L2 = -1.298M
Mode 3 :
M3 = (-1.173 x 0.334 x -1.078 x -0.641 x 0.845) x (M ) X 1
1
1
1
1
M3 = (-1.173+ 0.334 -1.078 -0.641+ 0.845) x (M )
M3= 0.687M
Mode 4 :
M4 = (-1.078 X 0.895 X 0.334 X -1.173 X 0.641 ) (M) X 1
1
1
1
1
M4 = (-1.078 + 0.895 + 0.334 -1.173 + 0.641 ) (M)
M4 = -0.381M
Mode 5 =
M5 = (0.641 X -1.078 X 1.173 X -0.895 X 0.334) (M) X 1
1
1
1
1
M5 = (0.641-1.078 + 1.173 -0.895 + 0.334) (M)
M5 = 0.175M
3)
Tn = Ln/Mn
Mode 1 :
Tn= L1/M1
= 4.121M / 3.861
= 1.067
Mode 2 =
T2 = L2/M2
= -1.298M /3.861M
= - 0.33
Mode 3 =
T3 = L3/M3
= 0.687 / 3.861
= 0.178
Mode 4 =
T4 = L4/M4
= -0.381M / 3.861
= 0.09
Mode 5 =
T5 = L5/M5
= 0.175M / 3.861M
= 0.04
4)
Mn = Ln^2 / Mn
Mode 1 =
M1 = L1^2 / M1
= (4.121m)^2 / 3.861
= 4.39M
Mode 2 =
M2 = L2^2 / M2
= (-1.298M)^2 /3.861M
= 0.43M
Mode 3 =
M3 = L3^2 / M3
= (0.687M)^2 / 3.861
= 0.12M
Mode 4 =
M4 = L4^2 / M4
= (-0.381M)^2 / 3.861
= 0.03M
Mode 5 =
M5 = L5^2 / M5
= (0.175M)^2 / 3.861M
= 0.007M
5)
Model mass participating Ratio = Mn / Mj
Mj = Total mass = M1+ M2 + M3 + M4 + M5=
= M+M+M+M+M
Mj= M
Mode 1 =
M1 / M j = 4.39M / 5 m
= 0.88
Mode 2 =
M2 / M j = 0.43M / 5 m
0.086
Mode 3 =
M3 / M j = 0.12M / 5 m
= 0.024
Mode 4 =
M4 / M j = 0.03M / 5 m
= 0.6
Mode 5 =
M5 / M j = 0.007M / 5 m
= 0.001
