Solutions of system of nonlinear equations:

Newton-Raphson Example 4:The kinematic equations for a Four-Bar mechanism can be written as (5th semester, Mechanisms Course)

s1

L2

L3

L4θ2

θ3

θ4

L2=0.15 mL3=0.45 mL4=0.28 ms1=0.2 m

0sinLsinLsinL

0scosLcosLcosL

443322

1443322

2

3

4

1

Where link 2 is the input member. How do you calculate θ3 and θ4 when θ2=120°.

)sin(28.0)sin(45.0)120sin(15.0f2.0)cos(28.0)cos(45.0)120cos(15.0f

432

431

-0.075

0.13

Solutions of system of nonlinear equations:

33

1 sin45.0f

44

1 sin28.0f

33

2 cos45.0f

44

2 cos28.0f

Following changes are made in the computer program.

clc, clearx=[0.5 1] ; err=[0.01 0.01];niter1=10;niter2=50;err=transpose(abs(err));for n=1:niter2x%Error Equations--------------------------- a(1,1)=-0.45*sin(x(1));a(1,2)=0.28*sin(x(2)); a(2,1)=0.45*cos(x(1));a(2,2)=-0.28*cos(x(2)); b(1)=-(0.45*cos(x(1))-0.28*cos(x(2))-0.275); b(2)=-(0.13+0.45*sin(x(1))-0.28*sin(x(2)));%---------------------------------------------- bb=transpose(b);eps=inv(a)*bb;x=x+transpose(eps); if n>niter1 if abs(eps)<err break else display ('Roots are not found') end endend

ANSWER:

θ3=0.216 rad (12.37°)

θ4=0.942 rad (53.97°)

(Initial angle values must be given in RADIAN)

clc;clear[x,y]=solve('0.45*cos(x)-0.28*cos(y)=0.275','0.13+0.45*sin(x)-0.28*sin(y)=0');vpa(x,6)vpa(y,6)

Alternative solution with MATLAB

Solutions of system of nonlinear equations:

Newton-Raphson Example 5:Kinematic equations for a crank mechanism are given below (5th semester Mechanisms Course)

s

L2 L3θ2

θ3

L2=0.15 mL3=0.6 m

0sinLsinL

0scosLcosL

3322

3322

)sin(6.0)60sin(15.0fs)cos(6.0)60cos(15.0f

32

31

Where link 2 (crank) is the input member. How dou you calculate θ3 and s with computer when θ2=60°.

0.075

0.1299

33

1 sin6.0f

1sf1

33

2 cos6.0f

0sf2

Following changes are made in the computer program.

ANSWER:

θ3=-0.2182 rad (-12.5°)

s=0.6607 m

Solutions of system of nonlinear equations:

clc;clear[x,y]=solve('0.075+0.6*cos(x)-y=0','0.1299+0.6*sin(x)=0');vpa(x,6)vpa(y,6)

Alternative solution with MATLAB

clc, clearx=[-1 0.8] ; err=[0.01 0.01];niter1=10;niter2=50;err=transpose(abs(err));for n=1:niter2x%Error Equations--------------------------- a(1,1)=-0.6*sin(x(1));a(1,2)=-1; a(2,1)=0.6*cos(x(1));a(2,2)=0; b(1)=-(0.075+0.6*cos(x(1))-x(2)); b(2)=-(0.1299+0.6*sin(x(1)));%---------------------------------------------- bb=transpose(b);eps=inv(a)*bb;x=x+transpose(eps); if n>niter1 if abs(eps)<err break else display ('Roots are not found') end endend

Newton-Raphson Example 6:

Solutions of system of nonlinear equations:

The time-dependent locations of two cars denoted by A and B

are given as

3ts

t4ts2

B

3A

At which time t, two cars meet?

BA ss 3t4ttf 23

4t2t3f 2

n1n xx,

ff

0 0.5 1 1.5 2 2.5 3 3.5 4-10

0

10

20

30

40

50

Zaman (s)

Yol

(m

)

A

B

Newton-Raphson Example 6:

Solutions of system of nonlinear equations:

ANSWER

t=0.713 s

t=2.198 s

0 0.5 1 1.5 2 2.5 3 3.5 4-10

0

10

20

30

40

50

Zaman (s)

Yol

(m

)

A

B

Using roots command in MATLAB

a=[ 1 -1 -4 3]; roots(a)

clc;cleart=solve('t^3-t^2-4*t+3=0');vpa(t,6)

Alternative Solutions with MATLAB

clc, clearx=1;err=0.001;niter=20;%----------------------------------------------for n=1:niter%---------------------------------------------- f=x^3-x^2-4*x+3; df=3*x^2-2*x-4;%---------------------------------------------- eps=-f/df; x =x+eps; if abs(f)<err break endenddisplay('Answer is='),x

From a vibration measurement on a machine, the damping ratio and undamped vibration frequency are calculated as 0.36 and 24 Hz, respectively. Vibration magnitude is 1.2 and phase angle is -42o. Write the MATLAB code to plot the graph of the vibration signal.

Graph Plotting:

Graph Plotting Example 7:

)73.0t7.140cos(e2.1)t(y t3.54

Given:

=0.36

ω0=24*2*π (rad/s)

A=1.2

Φ=-42*π/180 (rad)=-0.73 rad

ω0=150.796 rad/sω

-σ

3.54796.150*36.00

s/rad7.14036.01*796.150

12

20

20

20

α

0

cos

s0416.0796.1501415.3*22

T0

0

s002.0200416.0

20T

t 0 s1155.036.0

0416.0Tt 0

s

Graph Plotting:

0 0.02 0.04 0.06 0.08 0.1 0.12-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Zaman (s)

y

clc;cleart=0:0.002:0.1155;yt=1.2*exp(-54.3*t).*cos(140.7*t+0.73);plot(t,yt)xlabel(‘Time (s)');ylabel(‘Displacement (mm)');