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UNDERSTANDING THE CONCEPTS ON DEGREES OF FREEDOM 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. Given: The fundamental period…
Praveen Ps
updated on 21 Jul 2022
UNDERSTANDING THE CONCEPTS ON DEGREES OF FREEDOM
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.
Given:
ɸ1= [0.3340.6410.8951.0781.173]
ɸ2= [-0.895-1.173-0.6410.3341.078]
ɸ3= [1.1730.334-1.078-0.6410.895]
ɸ4= [-1.0780.8950.334-1.1730.641]
ɸ5= [0.641-1.0781.173-0.8950.334]
Aim:
Calculations:
Mass matrix, m = [m00000m00000m00000m00000m]
M1 =Φ1T m Φ1
i.e, M1 = [0.3340.6410.8951.0781.173]⋅[m00000m00000m00000m00000m]⋅[0.3340.6410.8951.0781.173]
M1 = [0.334m0.641m0.895m1.078m1.173m]⋅[0.3340.6410.8951.0781.173]
M1 = 3.86m.
Ln = ɸnTmɩ
Where, ɩ = Influence matrix which is a unit matrix, [11111]
L1 = [0.3340.6410.8951.0781.173]⋅[m00000m00000m00000m00000m]⋅[11111]
L1 = [0.334m0.641m0.895m1.078m1.173m]⋅[11111]
L1 = 4.12m.
Γ1 = 4.12m3.86m= 1.067
f1 = mɸ1Γ1A1
A1 = 0.27g for time period of 2 seconds.
f1 = [m00000m00000m00000m00000m]⋅[0.3340.6410.8951.0781.173]1.067⋅0.27g
f1 = [0.334m0.641m0.895m1.078m1.173m]⋅1.067⋅0.27g
f1 = [0.096mg0.185mg0.258mg0.310mg0.338mg]
We know that, Weight (newton), W=mg, and this can be substituted in the above equation.
f1 = [0.096W0.185W0.258W0.310W0.338W]
i.e,
Vb1 = 0.096W+0.185W+0.258W+0.310W+0.338W = 1.187W.
M1* = (Ln)2Mn= (4.12m)23.86m= 4.397m.
Hence, modal mass participating ratio, Mn* /m for mode 1 =4.397m5m= 0.88 = 88%
Vb1 = Mn* .An = 4.397m x 0.27g = 1.187mg = 1.187W.
- For mode 2:
i.e, M2 = [-0.895-1.173-0.6410.3341.078]⋅[m00000m00000m00000m00000m]⋅[-0.895-1.173-0.6410.3341.078]
M2 = [-0.895m-1.173m-0.641m0.334m1.078m]⋅[-0.895-1.173-0.6410.3341.078]
M2 = 3.86m.
Where, ɩ = Influence matrix which is a unit matrix, [11111]
L2 = [-0.895-1.173-0.6410.3341.078]⋅[m00000m00000m00000m00000m]⋅[11111]
L2 = [-0.895m-1.173m-0.641m0.334m1.078m]⋅[11111]
L2 = -1.297m.
Γ2 = -1.297m3.86m= -0.336
f2 = mɸ2Γ2A2
A2 = 0.76g for time period of 0.6852 seconds.
f2 = [m00000m00000m00000m00000m]⋅[-0.895-.1.173-0.6410.3341.078]-0.336⋅0.76g
f2 = [-0.895m-1.173m-0.641m0.334m1.078m]⋅-0.336⋅0.76g
f2 = [0.228mg0.299mg0.164mg-0.085mg-0.275mg]
We know that, Weight (newton), W=mg, and this can be substituted in the above equation.
f2 = [0.228W0.229W0.164W-0.085W-0.275W]
i.e, Vb2 = 0.228W+0.299W+0.164W-0.085W-0.275W = 0.331W
M2* = (Ln)2Mn= (-1.297m)23.86m= 0.436m.
Hence, modal mass participating ratio, Mn* /m for mode 2 =0.436m5m= 0.087 = 8.7%
Vb2 = Mn* .An = 0.436m x 0.76g = 0.331mg = 0.331W.
- For mode 3:
i.e, M3 = [1.1730.334-1.078-0.6410.895]⋅[m00000m00000m00000m00000m]⋅[1.1730.334-1.078-0.6410.895]
M3 = [1.173m0.334m-1.078m-0.641m0.895m]⋅[1.1730.334-1.078-0.6410.895]
M3 = 3.86m.
Where, ɩ = Influence matrix which is a unit matrix, [11111]
L3 = [1.1730.334-1.078-0.6410.895]⋅[m00000m00000m00000m00000m]⋅[11111]
L3 = [1.173m0.334m-1.078m-0.641m0.895m]⋅[11111]
L3 = 0.683m.
Γ3 = 0.683m3.86m= 0.177
f3 = mɸ3Γ3A3
A3 = 1.03g for time period of 0.4346 seconds.
f3 = [m00000m00000m00000m00000m]⋅[1.1730.334-1.078-0.6410.895]0.177⋅1.03g
f3 = [1.173m0.334m-1.078m-0.641m0.895m]⋅0.177⋅1.03g
f3 = [0.214mg0.061mg-0.196mg-0.117mg0.163mg]
We know that, Weight (newton), W=mg, and this can be substituted in the above equation.
f3 = [0.214W0.061W-0.196W-0.117W0.163W]
i.e, Vb3 = 0.214W+0.061W-0.196W-0.117W+0.163W = 0.125W.
M3* = (Ln)2Mn= (0.683m)23.86m= 0.121m.
Hence, modal mass participating ratio, Mn* /m for mode 3 =0.121m5m= 0.024 = 2.4%
Vb3 = Mn* .An = 0.121m x 1.03g = 0.331mg = 0.125W.
- For mode 4:
i.e, M4 = [-1.0780.8950.334-1.1730.641]⋅[m00000m00000m00000m00000m]⋅[-1.0780.8950.334-1.1730.641]
M4 = [-1.078m0.895m0.334m-1.173m0.641m]⋅[-1.0780.8950.334-1.1730.641]
M4 = 3.86m.
Where, ɩ = Influence matrix which is a unit matrix, [11111]
L4 =[-1.0780.8950.334-1.1730.641]⋅[m00000m00000m00000m00000m]⋅[11111]
L4 = [-1.078m0.895m0.334m-1.173m0.641m]⋅[11111]
L4 = -0.381m.
Γ4 = -0.381m3.86m= -0.099
f4 = mɸ4Γ4A4
A4 = 1.03g for time period of 0.3383 seconds.
f4 = [m00000m00000m00000m00000m]⋅[-1.0780.8950.334-1.1730.641].-0.099⋅1.03g
f4 = [-1.078m0.895m0.334m-1.173m0.641m]⋅-0.099⋅1.03g
f4 = [0.109mg-0.091mg-0.034mg0.119mg-0.065mg]
We know that, Weight (newton), W=mg, and this can be substituted in the above equation.
f4 = [0.109W-0.091W-0.034W0.119W-0.065W]
i.e, Vb4 = 0.109W-0.091W-0.034W+0.119W-0.065W = 0.038W.
M4* = (Ln)2Mn= (-0.381m)23.86m= 0.037m.
Hence, modal mass participating ratio, M4* /m for mode 4 =0.037m5m= 0.0074 = 0.74%
Vb4 = Mn* .An = 0.037m x 1.03g = 0.038mg = 0.038W.
- For mode 5:
i.e, M5 = [0.641-1.0781.173-0.8950.334]⋅[m00000m00000m00000m00000m]⋅[0.641-1.0781.173-0.8950.334]
M5 = [0.641m-1.078m1.173m-0.895m0.334m]⋅[0.641-1.0781.173-0.8950.334]
M5 = 3.86m.
Where, ɩ = Influence matrix which is a unit matrix, [11111]
L5 =[0.641-1.0781.173-0.8950.334]⋅[m00000m00000m00000m00000m]⋅[11111]
L5 = [0.641m-1.078m1.173m-0.895m0.334m]⋅[11111]
L5 = 0.175m.
Γ5 = 0.175m3.86m= 0.045
f5 = mɸ5Γ5A5
A5 = 1.03g for time period of 0.2966 seconds.
f5 = [m00000m00000m00000m00000m]⋅[0.641-1.0781.173-0.8950.334]⋅0.045⋅1.03g
f5 = [0.641m-1.078m1.173m-0.895m0.334m]⋅0.045⋅1.03g
f5 = [0.029mg-0.049mg0.054mg-0.041mg0.015mg]
We know that, Weight (newton), W=mg, and this can be substituted in the above equation.
f5 = [0.029W-0.049W0.054W-0.041W0.015W]
i.e, Vb5 = 0.029W-0.049W+0.054W-0.041W+0.015W = 0.008
M5* = (Ln)2Mn= (0.175m)23.86m= 0.0079m.
Hence, modal mass participating ratio, M5* /m for mode 5 =0.0079m5m= 0.0016 = 0.16%
Vb5 = Mn* .An = 0.0079m x 1.03g = 0.038mg = 0.008W.
Hence, base shear has been obtained for 5 modes of vibration.
Now, Combined base shear by SRSS rule,
Vb = √(Vb1)2+(Vb2)2+(Vb3)2+(Vb4)2+(Vb5)2= √(1.187W)2+(0.331W)2+(0.125W)2+(0.038W)2+(0.008W)2= 1.239W.
Vb = 1.239W.
Total mass of building = 5m
Total effective modal mass = 4.397+0.436+0.121+0.037+0.0079= 4.9989m≈5m. Hence, total mass of the building is verified.
Now, Modal mass participation ratio obtained are:
Mode 1 - 88%
Mode 2 - 8.7%
Mode 3 - 2.4%
Mode 4 - 0.74%
Mode 5 - 0.16%
From the above values, it is clear that sum of mode 1 and mode 2 will exceed 90%
i.e, 88 + 8.7 = 96.7%
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