k2_2000Hz 0.5 10⋅ 8 kN ⋅m

ORIGIN:= 1

STEADY STATE ANALYSIS OF MDOF SYSTEM WITH PROPORTIONAL DAMPING AND FDL

(performed by: Dott. Ing. Simone Caffè - Date: 15/06/2015)

The following example is facing the Stady State analysis of a MDOF system, in which the masses are concentrated in the nodes 2,3,4 and 5 bonded to each other through links whose stiffnes and viscous linear damping assume different intensity values depending on the frequency. The harmonic forces (applied to nodes 2 and 4) depend also on the frequency according to a magnification law described as follows:

Masses and Mass Matrix definitions:

m2:= 20000⋅kg m3:= 10000⋅kg m4:= 5000⋅kg m5:= 2500⋅kg

Ms m2

0 0 0

0 m3

0 0

0 0 m4

0 0 0 0 m5





















1

⋅kg

20000 0 0 0

0 10000

0 0

0 0 5000

0 0 0 0 2500

















= :=

Link stiffness and damping and definitions:

Link 1: k1_0Hz 2 10⋅ ⁸ kN

⋅m

:= k1_2000Hz 1 10⋅ ⁸ kN

⋅m :=

Link 2: k2_0Hz 1 10⋅ ⁸ kN

⋅m

:= k2_2000Hz 0.5 10⋅ ⁸ kN

⋅m :=

Link 3: k3_0Hz 4 10⋅ ⁸ kN

⋅m

:= k3_2000Hz 2 10⋅ ⁸ kN

⋅m :=

Link 4: k4_0Hz 0.5 10⋅ ⁸ kN

⋅m

:= k4_2000Hz 0.8 10⋅ ⁸ kN

⋅m :=

Link 1: c1_0Hz 10000 kN s⋅

⋅ m

:= c1_2000Hz 5000 kN s⋅

⋅ m :=

Link 2: c2_0Hz 5000 kN s⋅

⋅ m

:= c2_2000Hz 1000 kN s⋅

⋅ m :=

Link 3: c3_0Hz 100 kN s⋅

⋅ m

:= c3_2000Hz 3000 kN s⋅

⋅ m :=

Link 4: c4_0Hz 500 kN s⋅

⋅ m

:= c4_2000Hz 4000 kN s⋅

⋅ m :=

Stiffness functions:

It has been considered, for example, a linear stiffness variation depending on frequency:

fmax:= 2000 Hz

k1 f() k1_2000Hz k1_0Hz−

fmax ⋅f+ k1_0Hz :=

k2 f() k2_2000Hz k2_0Hz−

fmax ⋅f+ k2_0Hz :=

k3 f() k3_2000Hz k3_0Hz−

fmax ⋅f+ k3_0Hz :=

k4 f() k4_2000Hz k4_0Hz−

fmax ⋅f+ k4_0Hz :=

0 500 1 10× ³ 1.5 10× ³ 2 10× ³

0 1 10× ¹¹ 2 10× ¹¹ 3 10× ¹¹ 4 10× ¹¹

Frequency dependent stiffness

[Hz]

[kN/m]

k1 f() k2 f() k3 f() k4 f()

f

Linear Viscous Damping functions:

It has been considered, for example, a linear damping variation depending on frequency:

c1 f() c1_2000Hz c1_0Hz−

fmax ⋅f+ c1_0Hz :=

c2 f() c2_2000Hz c2_0Hz−

fmax ⋅f+ c2_0Hz :=

c3 f() c3_2000Hz c3_0Hz−

fmax ⋅f+ c3_0Hz :=

c4 f() c4_2000Hz c4_0Hz−

fmax ⋅f+ c4_0Hz :=

0 500 1 10× ³ 1.5 10× ³ 2 10× ³

0 2 10× ⁶ 4 10× ⁶ 6 10× ⁶ 8 10× ⁶ 1 10× ⁷

Frequency dependent viscous Damping

[Hz]

[kNs/m]

c1 f() c2 f() c3 f() c4 f()

f

Hysteretic Damping functions:

In order to use correctly the FDL in S AP, it has been necessary to convert the linear viscous damping in hysteretic damping.

ch1 f():= c1 f()⋅2⋅π⋅f ch2 f():= c2 f()⋅2⋅π⋅f ch3 f():= c3 f()⋅2⋅π⋅f ch4 f():= c4 f()⋅2⋅π⋅f

0 200 400 600 800 1 10× ³ 0

1 10× ¹⁰ 2 10× ¹⁰ 3 10× ¹⁰ 4 10× ¹⁰ 5 10× ¹⁰

Frequency dependent hysteretic damping

[Hz]

[kN/m]

ch1 f() ch2 f() ch3 f() ch4 f()

f

Stiffness Matrix and Damping Matrix definitions:

Stiffness and Damping Matricies are depending on frequency:

Ks f()

k1 f() +k2 f()

( )

k2 f()

−

0 0

k2 f()

− k2 f() + k3 f()

( )

k3 f()

− 0

0 k3 f()

− k3 f() + k4 f()

( )

k4 f()

−

0 0 k4 f()

− k4 f()





















m

⋅N :=

Cs f()

c1 f() +c2 f()

( )

c2 f()

−

0 0

c2 f()

− c2 f() + c3 f()

( )

c3 f()

− 0

0 c3 f()

− c3 f() + c4 f()

( )

c4 f()

−

0 0 c4 f()

− c4 f()





















m N s⋅

⋅ :=

Modal - superposition Procedure Employing Modes of the undamped structures:

Λ f( ):= Ms⁻¹⋅Ks f() System eigen-values:

λ f( ):= sort eigenvals Λ f( ( ( )))

The undamped natural frequencies are:

ω_n1( )f := λ f( )₁

fn1 f() ωn1 f() 2⋅π :=

ω_n2( )f := λ f( )₂ fn2 f() ωn2 f() 2⋅π :=

ω_n3( )f := λ f( )₃ fn3 f() ωn3 f() 2⋅π :=

ω_n4( )f := λ f( )₄ fn4 f() ωn4 f() 2⋅π :=

0 500 1 10× ³ 1.5 10× ³ 2 10× ³ 0

5 10× ³ 1 10× ⁴ 1.5 10× ⁴

Undamped natural circular frequencies

ω_n1( )f ω_n2( )f ω_n3( )f ω_n4( )f

f

0 500 1 10× ³ 1.5 10× ³ 2 10× ³ 0

500 1 10× ³ 1.5 10× ³ 2 10× ³

Undamped natural frequencies

f_n1( )f f_n2( )f f_n3( )f f_n4( )f

f

Dynamic Stiffness Matrix (DIRECT METHOD) :

KDYN f() Ks f() + −1⋅(2⋅π⋅f)⋅Cs f() −(2⋅π⋅f)²⋅Ms

 

:= 

Harmonic forces vector:

Consider the case of force exitation with frequency equal to 50 Hz.

F2:= 10000⋅kN F3:= 0⋅kN F4:= −5000⋅kN F5:= 0⋅kN

F0 F2 F3 F4 F5





















1

⋅N

10000000 0 5000000

− 0

















= :=

fF:= 50 Hz

Fharmonic f() F0 f fF









2

⋅ :=

Dynamic responses in physical coordinates:

U f( ):= KDYN f()⁻¹⋅Fharmonic f()

U2 f():= U f( )₁ U3 f():= U f( )₂ U4 f():= U f( )₃ U5 f():= U f( )₄

0 500 1000 1500 2000

0 0.015 0.03 0.045 0.06

Joints 2,3,4,5 - Physical Displacements

[Hz]

[m]

U2 f() U3 f() U4 f() U5 f()

f

Dynamic responses in physical coordinates for 35%F:

X f( )^:= KDYN f()^−1⋅(0.35⋅Fharmonic f())

X2 f():= X f( )₁ X3 f():= X f( )₂ X4 f():= X f( )₃ X5 f():= X f( )₄

0 500 1000 1500 2000

0.01 0.02 0.03 0.04

Joint 2 - Physical Displacements

[Hz]

[m] ^{U2 f()}

X2 f()

f

0 500 1000 1500 2000

0.005 0.01 0.015 0.02

Joint 3 - Physical Displacements

[Hz]

[m] ^{U3 f()}

X3 f()

f

0 500 1000 1500 2000 0.01

0.02 0.03 0.04 0.05

Joint 4 - Physical Displacements

[Hz]

[m] ^{U4 f()}

X4 f()

f

0 500 1000 1500 2000

0.01 0.02 0.03 0.04

Joint 5 - Physical Displacements

[Hz]

[m] ^{U5 f()}

X5 f()

f

SAP 2000 Dynamic responses in physical coordinates:

Harmonic Force = F

Harmonic Force = 35%F

Dynamic responses in physical coordinates - velocity:

v2 f() := U f( )₁ ⋅(2⋅π⋅f) v3 f() := U f( )₂ ⋅(2⋅π⋅f) v4 f() := U f( )₃ ⋅(2⋅π⋅f) v5 f() := U f( )₄ ⋅(2⋅π⋅f)

0 500 1000 1500 2000

0 100 200 300 400

Joints 2,3,4,5 - Velocity

[Hz]

[m]

v2 f() v3 f() v4 f() v5 f()

f

Dynamic responses in physical coordinates - acceleration:

a2 f() := v2 f()⋅(2⋅π⋅f) a3 f() := v3 f()⋅(2⋅π⋅f) a4 f() := v4 f()⋅(2⋅π⋅f) a5 f() := v5 f()⋅(2⋅π⋅f)

0 500 1000 1500 2000

0 1000000 2000000 3000000 4000000

Joints 2,3,4,5 - Acceleration

[Hz]

[m]

a2 f() a3 f() a4 f() a5 f()

f

Displacements Ratio Values:

f:= 1525

U2 f() =0.00585 X2 f() =0.002 ρ2 X2 f()

U2 f() =35⋅% :=

U3 f() =0.01633 X3 f() =0.0057 ρ3 X3 f()

U3 f() =35⋅% :=

U4 f() =0.04057 X4 f() =0.0142 ρ4 X4 f()

U4 f() =35⋅% :=

U5 f() =0.02007 X5 f() =0.007 ρ5 X5 f()

U5 f() =35⋅% :=