Compendio

Equations of Dynamic Equilibrium and solution for Steady State Responses
Equation of Dynamic Equilibrium : mu + ku = p (t ) = po sin (ωt ) Particular Solution :
Let u p (t ) = Ap sin (ωt ) Substitute this expression into the equation of dynamic equilibrium : − mω 2 Ap sin (ωt ) + kAp sin (ωt ) = po sin (ωt ) Ÿ

(Undamped vibration)

for steady State responses A = p

po = 2 k − mω

po kω2 1− 2 ωn

where ωn = k

m

for applied excitations
ωn

Let angular velocity ratio β = ω Ap = po 1 ; 2 k 1− β u p (t ) =

for natural vibration

po 1 sin (ωt ) 2 k 1− β

Equations of Dynamic Equilibrium and solution for Steady State Responses
Viscously damped vibration

po sin (ωt )

u cu

u ku

mu
Inertial force

Viscous damping force

Elastic resistance

mass= m

total column stiffness = k
Equation of dynamic equilibrium :

u

mu + cu + ku = p(t ) = po sin (ωt )

Equations of Dynamic Equilibrium and Steady State Responses Equation of Dynamic Equilibrium : mu + cu + ku = po sin (ωt )
(viscously damped vibration) Solution : Let u p (t ) = Ap sin (ωt ) + B p cos(ωt )

Substitute this expression into the equation of dynamic equilibrium : − mω2 Ap sin (ωt ) + −mω 2 B p cos(ωt ) + 2ξmωn Apω cos(ωt ) − 2ξmωn B pω sin (ωt ) + kAp sin (ωt ) + kB p cos(ωt ) = po sin (ωt ) Dividing all terms by " k" and using ω n2 = k m and β = ω

ωn :

− β 2 Ap sin (ωt ) − β 2 B p cos(ωt ) + 2ξβAp cos(ωt ) − 2ξβB p sin (ωt ) + Ap sin (ωt ) + B p cos(ωt ) = po Ap 1 − β − 2ξβB p = k po sin (ωt ) k Equating the " sin (ωt )" and " cos(ωt )" termsseparately Ÿ

(

2

)

;

2ξβAp + B p 1 − β 2 = 0 Ÿ

(

)

Equations of Dynamic Equilibrium and Steady State Responses
Solution continuing from the previous page (for viscously damped vibration) :
p Ap = o k p - 2ξβ ; Bp = o 2 2 k 1 - β 2 + 4ξ 2 β 2 1 - β 2 + 4ξ 2 β 2 1- β 2

( )

( )

ª º po « 1- β 2 - 2ξβ sin (ωt ) + cos(ωt )» u p (t ) = » k « 22 2 2 22 2 2 + 4ξ β 1- β + 4ξ β «1- β » ¬ ¼

( )
Ap

( )

Bp

u p (t ) = Ap sin (ωt ) + B p cos(ωt )

Equations of Dynamic Equilibrium and Steady State Responses
Solution continuing from the previous page (for viscously damped vibration) :
Expression of the form : u p (t ) = A p sin (ωt ) + B p cos(ωt ) can be re - written in the more compact form : u p (t ) = C p sin ωt − φ p C p = Ap + B p
2 2

(

)

where :−1¨ − B p ¸ = tan

§

·

¨ Ap ¸ © ¹ It can be shown from results presented in the previous slide that : 1

; φp

po Cp = k p u p (t ) = o k

(1 − β ) + 4ξ
22

2 2

· § −1¨ 2ξβ ¸ ; φ p = tan Ÿ 2¸ ¨ © 1− β ¹ sin ωt − φ p

β

1

(1 − β ) + 4ξ β
22

(

)

2 2

Steady State Response Example
u (t ) = po k 1

(1 − β )

2 2

+ 4ξ 2 β 2

sin (ωt − φ p )
§ 2ξβ ·¸ 2 ¸ ©1− β ¹

For u static
0.5

−1 = 0.1m, β = 1.1, ξ = 0.05, Tn = 0.5s, ωn = 12.56 rad/s φ p = tan ¨ ¨

Ustatic

Upmax
u(t) 0
R d = 4.2

u(t) applied p(t)/k
φ p = −0.48 radian

-0.5 0 1 2 time (secs) 3 4

Rd =

U p max U static

Dynamic Response Factor Rd
u p (t ) = u p max = po k po k 1

(1 − β )

2 2

sin (ωt − φ ) + 4ξ 2 β 2 is the max. displacement of theresponse + 4ξ 2 β 2 po k

1

(1 − β )

2 2

Let u static = displacement of system subject to static loading = Dynamic Response Factor Rd = u p max u static = 1

(1 − β )

2 2

+ 4ξ 2 β 2

Rd is dependent on both β and ξ , and becomes critical when β = 1 ie. when ω = ωn . Let Rd (β = 1) = Rd −crit = For example : Rd −crit = 10 for ξ = 0.05 Rd −crit = 5 for ξ = 0.10 1 2ξ

DynamicResponse Factor Rd
14 Dynamic Response Factor 12 10 8 6 4 2 0 0 0.5 1 ratio of angular velocity 1.5 2
5% damping Dynamic Response Factor ζ=0.02 ζ=0.05
1 = 10 for ξ = 0.05 2ξ

ζ=0.10 ζ=0.2

Rd =

1 + 4ξ β
2 2

(1 − β )

2 2

ζ=0.3 ζ=0.5

Low frequency excitation

High frequency excitation

Phase Angle Change
180 Dynamic Response Factor 160 140 120 100 80 60 40 20 0 0 0.2 0.4 0.6...