ME51035
Mechanical Vibration
Topic 6: Single Degree of Freedom System
Response to Common Harmonic Excitations
6.1 Objectives
At the end of this lecture, students should be able to:
(1) Understand and solve common harmonic excitations to vibration system
(2) Understand frequency response function and transmissibility for damped system under
rotating unbalance and base excitation
(3) Understand and derive equation of motion for unbalance and base excitation.
(4) Make the sense of unbalance and base excitation and its physical significance in actual
system
6.1 Rotating Unbalance
𝐹 = 𝑚𝜔 𝑒 cos(𝜔𝑡)
𝐾𝑥
𝐶𝑥̇
𝑀 =𝑚 +𝑚 +𝑚
𝑀𝑥̈
Equation of motion,
𝑀𝑥̈ + 𝐶𝑥̇ + 𝐾𝑥 = 𝑚𝜔 𝑒 cos(𝜔𝑡)
where,
𝑚 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑢𝑛𝑏𝑎𝑙𝑎𝑛𝑐𝑒 [𝑘𝑔]
𝑒 = 𝑒𝑐𝑒𝑛𝑡𝑟𝑖𝑐𝑖𝑡𝑦 [𝑚]
𝜔 = 𝑟𝑜𝑡𝑎𝑡𝑖𝑛𝑔 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑢𝑛𝑏𝑎𝑙𝑎𝑛𝑐𝑒 𝑚𝑎𝑠𝑠
Equation of motion to that of SDOF under harmonic excitation except the force magnitude is
𝑚𝜔 𝑒.
For harmonic
𝑋
=
𝐹 ⁄𝑘
1
(1 − 𝑟 ) + (2𝜉𝑟)
Magnitude of unbalance force can be substituted into the force,
𝑋
=
𝑚𝜔 𝑒⁄𝑘
𝑋
=
𝑚𝑒
𝑀𝑋
=
𝑚𝑒
1
(1 − 𝑟 ) + (2𝜉𝑟)
𝜔 ⁄𝑘
(1 − 𝑟 ) + (2𝜉𝑟)
𝑀𝜔 ⁄𝑘
(1 − 𝑟 ) + (2𝜉𝑟 )
𝜔
=
𝑘
𝑀
The response is:
𝑀𝑋
=
𝑚𝑒
𝑟
(1 − 𝑟 ) + (2𝜉𝑟 )
𝑡𝑎𝑛𝜙 =
For maximum
2𝜉𝑟
1−𝑟
,
𝑑 𝑀𝑋
=0
𝑑𝑟 𝑚𝑒
𝑑
𝑟
=0
𝑑𝑟 (1 − 𝑟 ) + (2𝜉𝑟)
1
𝑟 [(1 − 𝑟 ) + (2𝜉𝑟) ] × [2(1 − 𝑟 ) × (2𝑟) + 8𝜉 𝑟] − (1 − 𝑟 ) + (2𝜉𝑟) × 2𝑟
2
=0
(1 − 𝑟 ) + (2𝜉𝑟)
1
𝑟 [−4𝑟 + 4𝑟 + 8𝜉 𝑟] − [(1 − 𝑟 ) + (2𝜉𝑟) ] × 2𝑟 = 0
2
−2𝑟 + 2𝑟 + 4𝜉 𝑟 − [1 − 2𝑟 + 𝑟 + 4𝜉 𝑟 ] × 2𝑟 = 0
−2𝑟 + 2𝑟 + 4𝜉 𝑟 − 2𝑟 + 4𝑟 − 2𝑟 − 8𝜉 𝑟 = 0
[2 − 4𝜉 ]𝑟 − 2𝑟 = 0
𝑟 ≠ 0 𝑜𝑟 [2 − 4𝜉 ]𝑟 − 2 = 0
𝑟 =
𝑟=
2
1
=
2 − 4𝜉
1 − 2𝜉
1
1 − 2𝜉
Maximum of
occurs at 𝑟 =
𝑀𝑋
𝑚𝑒
=
𝑟
𝑀𝑋
𝑚𝑒
=
𝑀𝑋
𝑚𝑒
= (1 − 2𝜉 )
𝑀𝑋
𝑚𝑒
=
𝑀𝑋
𝑚𝑒
=
𝑀𝑋
𝑚𝑒
=
(1 − 𝑟 ) + (2𝜉𝑟)
1
(1 − 2𝜉 ) ×
1−
1
+
(1 − 2𝜉 )
2𝜉
1 − 2𝜉
1
1 − 2𝜉 − 1
1 − 2𝜉
×
+
(2𝜉)
1 − 2𝜉
1
(−2𝜉 ) + (1 − 2𝜉 )(2𝜉 )
1
2𝜉 𝜉 + 1 − 2𝜉
1
2𝜉 1 − 𝜉
Note that condition for maximum amplitudes of harmonic excitation and rotating unbalance is
not the same.
Example 6.1 (e.g. 3.6 from textbook
GIVE:
𝑋 = 0.15 𝑚 𝑎𝑡 𝑟𝑒𝑠𝑜𝑛𝑎𝑛𝑐𝑒
𝑚 = 0.08 𝑀,
𝑚
= 0.08
𝑀
𝜉 = 0.025
FIND:
(a) 𝑒 =?
(𝑏)𝑋
=?
(𝑐)𝑀 𝑓𝑜𝑟 𝑋 = 0.1 𝑚
SOLUTION:
At resonance, 𝜔 = 𝜔 ,
=
=
(a) 𝑒 =
(b) Max:
𝑋
=
= 20 (study response due to harmonic excitation)
× .
=
.
× .
=
= 0.09375 𝑚
=
× .
×
( × .
)
= 20.0063 𝑚
20.0063 × 𝑚𝑒
= 20 × 0.0063 × 0.08 × 0.09375 = 0.15047 𝑚
𝑀
(c) With addition mass at resonance to reduce X to 0.1 m
(𝑀 + 𝑀 )𝑋
= 20
𝑚𝑒
(𝑀 + 𝑀 ) × 0.1
= 20
0.08𝑀 × 0.09375
𝑀 + 𝑀 = 1.5𝑀
𝑀 = 0.5𝑀
Addition mass to be installed is 50%M.
6.2 Base Excitation
Vibration occurs because of the excitation at the base or support. Examples include vibration of
moving vehicle due to road surface roughness. Free body diagram is:
Equation of motion is:
If
,
,
6.2.1 Displacement Transmissibility
Displacement Transmissibility
6
=0.1
=0.3
=0.5
=0.7
=0.9
=1.1
5
X/Y
4
3
2
1
0
0
20
40
60
80
100
120
velocity [km/hr]
Phase of X/Y
200
=0.1
=0.3
=0.5
=0.7
=0.9
=1.1
150
100
50
0
0
20
40
60
velocity [km/hr]
6.2.2 Relative Displacement Transmissibility
Relative displacement transmissibility can be described as:
80
100
120
Relative Displacement Transmissibility
6
=0.1
=0.3
=0.5
=0.7
=0.9
=1.1
5
Z/Y
4
3
2
1
0
0
20
40
60
80
100
120
velocity [km/hr]
Phase of Z/Y
200
=0.1
=0.3
=0.5
=0.7
=0.9
=1.1
150
100
50
0
0
20
40
60
velocity [km/hr]
6.2.3 Force Transmissibility
If F is force transmitted to the support or base,
Force transmissibility is:
80
100
120
Force Transmissibility
6
=0.1
=0.3
=0.5
=0.7
=0.9
=1.1
FT/(KY)
5
4
3
2
1
0
0
20
40
60
80
100
120
velocity [km/hr]
Phase of FT/(KY)
200
=0.1
=0.3
=0.5
=0.7
=0.9
=1.1
150
100
50
0
0
20
40
60
velocity [km/hr]
6.2.4 MATLAB Code
clear all; clc; close all
%% Vibration System property
m=1200;
k=400e3;
zeta=0.1:0.2:1.1;
wn=sqrt(k/m);
%% Base Excitation Information
vk=10:1:120;
v=vk*1e3/3600;
% road roughness
Y=0.05;
lambda=6;
% excitation frequency
w=2*pi*v/lambda;
r=w/wn;
%% calculate amplification factor
phi=zeros(length(zeta),length(w));
XT=zeros(length(zeta),length(w));
FT=zeros(length(zeta),length(w));
ZT=zeros(length(zeta),length(w));
phiZ=zeros(length(zeta),length(w));
for jj=1:length(w)
80
100
120
for ii=1:length(zeta)
XT(ii,jj)=sqrt((1+(2*zeta(ii)*r(jj)).^2)./((1r(jj).^2).^2+(2*zeta(ii)*r(jj)).^2));
FT(ii,jj)=r(jj).^2.*sqrt((1+(2*zeta(ii)*r(jj)).^2)./((1
-r(jj).^2).^2+(2*zeta(ii)*r(jj)).^2));
ZT(ii,jj)=r(jj).^2./sqrt((1r(jj).^2).^2+(2*zeta(ii)*r(jj)).^2);
phi(ii,jj)=atan((2*zeta(ii)*r(jj).^3)./((1r(jj).^2)+(2*zeta(ii)*r(jj)).^2));
phi(ii,jj)=phi(ii,jj)*180/pi;
phiZ(ii,jj)=atan(2*zeta(ii)*r(jj)./(1-r(jj).^2));
phiZ(ii,jj)=phiZ(ii,jj)*180/pi;
if phi(ii,jj)<0
phi(ii,jj)=180+phi(ii,jj);
end
if phiZ(ii,jj)<0
phiZ(ii,jj)=180+phiZ(ii,jj);
end
end
end
%% plotting
% Displacement transmissibility
figure(1)
subplot(2,1,1)
plot(vk,abs(XT),'LineWidth',2); hold on; grid on;
legend('\zeta=0.1','\zeta=0.3','\zeta=0.5','\zeta=0.7',
'\zeta=0.9','\zeta=1.1')
xlabel('velocity [km/hr]','FontSize',16)
ylabel('X/Y','FontSize',16)
title('Displacement Transmissibility','Fontsize',16)
subplot(2,1,2)
plot(vk,phi,'LineWidth',2); hold on; grid on
legend('\zeta=0.1','\zeta=0.3','\zeta=0.5','\zeta=0.7',
'\zeta=0.9','\zeta=1.1')
xlabel('velocity [km/hr]','FontSize',16)
ylabel('Phase of X/Y','FontSize',16)
% Force transmissibility
figure(2)
subplot(2,1,1)
plot(vk,abs(FT),'LineWidth',2); hold on; grid on;
legend('\zeta=0.1','\zeta=0.3','\zeta=0.5','\zeta=0.7',
'\zeta=0.9','\zeta=1.1')
xlabel('velocity [km/hr]','FontSize',16)
ylabel('F_T/(KY)','FontSize',16)
title('Force Transmissibility','Fontsize',16)
subplot(2,1,2)
plot(vk,phi,'LineWidth',2); hold on; grid on
legend('\zeta=0.1','\zeta=0.3','\zeta=0.5','\zeta=0.7',
'\zeta=0.9','\zeta=1.1')
xlabel('velocity [km/hr]','FontSize',16)
ylabel('Phase of F_T/(KY)','FontSize',16)
% Relative displacement transmissibility
figure(3)
subplot(2,1,1)
plot(vk,abs(ZT),'LineWidth',2); hold on; grid on;
legend('\zeta=0.1','\zeta=0.3','\zeta=0.5','\zeta=0.7',
'\zeta=0.9','\zeta=1.1')
xlabel('velocity [km/hr]','FontSize',16)
ylabel('Z/Y','FontSize',16)
title('Relative Displacement
Transmissibility','Fontsize',16)
subplot(2,1,2)
plot(vk,phiZ,'LineWidth',2); hold on; grid on
legend('\zeta=0.1','\zeta=0.3','\zeta=0.5','\zeta=0.7',
'\zeta=0.9','\zeta=1.1')
xlabel('velocity [km/hr]','FontSize',16)
ylabel('Phase of Z/Y','FontSize',16)
Fig.MenuBar='none';
Calculate Example 3.7 from your textbook by yourself.