Simpson’s Rule:

4.12.4cose5.2)(M 4.1

0

?d)(M

-1.4

4.2i

427.42.44.1 220

316.0427.4

4.1

49.4316.0419.1T0

s

419.1427.422

T0

0

49.4

0

?d)(M

3742.012

049.4

12n

Example: Calculate the integral of the given function.

k θ M(θ)0 0 0.4249

1 0.3742 -1.4592

2 0.7484 -0.1476

3 1.1226 0.5119

4 1.4968 0.0512

5 1.8710 -0.1796

6 2.2452 -0.0178

7 2.6194 0.0630

8 2.9936 0.0062

9 3.3678 -0.0221

10 3.7420 -0.0021

11 4.1162 0.0078

12 4.49 0.000744

1211109876543210 ff4f2f4f2f4f2f4f2f4f2f4f3h

I

000744.00078.0*40021.0*20221.0*40062.0*20630.0*40178.0*21796.0*40512.0*25119.0*41476.0*24592.1*44249.0

33742.0

I

5123.0I

Solution with Matlab:

>>clc;clear;

>> syms teta

>>f=2.5*exp(-1.4*teta)*cos(4.2*teta+1.4)

>>y=int(f,0,4.49)

>>vpa(y,5)

I=-0.4966

Simpson’s Rule:

Example:

Simpson’s Rule:

0

2 ?d)(M

3742.012

049.4

12n

k θ M2(θ)

0 0 0.1806

1 0.3742 2.1293

2 0.7484 0.0218

3 1.1226 0.2621

4 1.4968 0.0026

5 1.8710 0.0323

6 2.2452 0.000316

7 2.6194 0.0040

8 2.9936 3.81x10-5

9 3.3678 4.88x10-4

10 3.7420 4.598x10-6

11 4.1162 6.013x10-5

12 4.49 5.541x10-7

75

645

10x541.510x013.6*4

10x59.4*210x88.4*410x81.3*20040.0*4000316.0*2

0323.0*40026.0*22621.0*40218.0*21293.2*41806.0

33742.0

I

2402.1I

Simpson’s Rule:

0

2 ?d)(M

3742.012

049.4

12n

Solution with Matlab:

We must increase the number of sections!

>>clc;clear;

>> syms teta

>>f=(2.5*exp(-1.4*teta)*cos(4.2*teta+1.4))^2

>>y=int(f,0,4.49)

>>vpa(y,5)

I=0.898

Simpson’s Rule:

Example: Calculate the volume of the 3 meter long beam whose cross section is given in the figure.

2.06

02.1x

6n

k x y(x)0 0 0.5

1 0.2 0.597

2 0.4 0.6864

3 0.6 0.7663

4 0.8 0.8356

5 1.0 0.8944

6 1.2 0.9432

2.1

02

dx4x

1xArea

9432089440483560276630468640259704503

20..*.*.*.*.*.

.Area

2m9012.0Area

370362390120 m..LengthAreaVolume

>>syms x; area=int((x+1)/(sqrt(x^2+4)),0,1.2);vpa(area,5)

Solution with Matlab:

Area=0.9012

4

12

x

xy

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;xe=0.001*x;niter=20;%----------------------------------------------for n=1:niter%---------------------------------------------- f=x^3-x^2-4*x+3; df=3*x^2-2*x-4;%---------------------------------------------- x1=x x=x1-f/df if abs(x-x1)<xe kerr=0;break endendkerr,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)');

Time (s)

Dis

plac

emen

t (m

m)

Lagrange Interpolation:

Example:The temperature (T) of a medical cement increases continuously as the solidification time (t) increases. The change in the cement temperature was measured at specific instants and the measured temperature values are given in the table. Find the cement temperature at t=36 (sec).

68*)1555)(555(

)15t)(5t(43*

)5515)(515()55t)(5t(

30*)555)(155()55t)(15t(

)t(T

C51.6168*)1555)(555()1536)(536(

43*)5515)(515()5536)(536(

30*)555)(155(

)5536)(1536()t(T o

Lagrange Interpolation:

Example:The buckling tests were performed in order to find the critical buckling loads of a clamped-pinned steel beams having different thicknesses. The critical buckling loads obtained from the experiments are given in the table. Find the critical buckling load Pcr (N) of a steel beam with 0.8 mm thickness.

Thickness (t) (mm)

Buckling Load Pcr (N)

0.5 30

0.6 35

0.65 37

0.73 46

0.9 580.8 mm

58*)73.09.0)(65.09.0)(6.09.0)(5.09.0(

)73.0t)(65.0t)(6.0t)(5.0t(46*

)9.073.0)(65.073.0)(6.073.0)(5.073.0()9.0t)(65.0t)(6.0t)(5.0t(

37*)9.065.0)(73.065.0)(6.065.0)(5.065.0(

)9.0t)(73.0t)(6.0t)(5.0t(35*

)9.06.0)(73.06.0)(65.06.0)(5.06.0()9.0t)(73.0t)(65.0t)(5.0t(

30*)9.05.0)(73.05.0)(65.05.0)(6.05.0(

)9.0t)(73.0t)(65.0t)(6.0t()t(Pcr

N779.52164.7810.101600.103538.56134.9)8.0(Pcr

Pcr

Lagrange Interpolation:

Thickness (t) (mm)

Buckling Load Pcr (N)

0.5 30

0.6 35

0.65 37

0.73 46

0.9 580.8 mm

clc;cleart=[0.5 0.6 0.65 0.73 0.9];P=[30 35 37 46 58];interp1(t,P,0.8,'spline')

Pcr

0.5,300.6,350.65,370.73,460.9,58

data.txtclc;clearv=load ('c:\saha\data.txt')interp1(v(:,1),v(:,2),0.8,'spline')

1. Lagrange Interpolation with Matlab

2. Lagrange Interpolation with Matlab