From a vibration measurement on a machine, the damping ratio and undamped vibration frequency are...
-
Upload
osborn-marsh -
Category
Documents
-
view
224 -
download
3
Transcript of From a vibration measurement on a machine, the damping ratio and undamped vibration frequency are...
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)
Roots of a polynomial:
020t6t5t3 24 Find the roots of the polynomial.
with Matlab>> p=[3 0 5 6 -20]>> roots(p)
ans =
-1.5495 0.1829 + 1.8977i 0.1829 - 1.8977i 1.1838
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
-20
-10
0
10
20
30
40
50
60
t
3 t4+5 t2+6 t-20
1.1838-1.5495
>>ezplot('3*t^4+5*t^2+6*t-20',-2,2)
Newton-Raphson Example:
Solution 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:
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
Solution of system of nonlinear equations:
22
3
y1yx
1x3y)x2sin(
How do you find x and y values, which satisfy the equations?
1yyxf
1x3y)x2sin(f22
2
31
ff
2
1
2
1
22
11
ff
yf
xf
yf
xf
1y2yf
,x2xf
y3yf
,3)x2cos(2xf
22
211
x=0.6786
y=0.3885
with Matlab:
>>[x,y]=solve('sin(2*x)+y^3=3*x-1','x^2+y=1-y^2') x=0.6786, y=0.3885
clc, clearx=[1 1]; xe=[0.01 0.01];niter1= 5; niter2=50;fe=transpose(abs(fe));kerr=1;for n=1:niter2x%-----Error equations------------------------a(1,1)=2*cos(2*x(1))-3;a(1,2)=3*x(2)^2;a(2,1)=2*x(1);a(2,2)=2*x(2)+1;b(1)=-(sin(2*x(1))+x(2)^3-3*x(1)+1);b(2)=-(x(1)^2+x(2)^2+x(2)-1);%-------------------------------------------------------bb=transpose(b);eps=inv(a)*bb;x=x+transpose(eps); if n>niter1 if abs(eps)<xe kerr=0; break else display ('Roots are not found') end endend
Solution of system of nonlinear equations:
16)1b(a2
11b2)1a(323
22
How do you calculate a and b, which satisfy given equations by computer?
16)1b(a2f
11b2)1a(3f23
2
221
ff n1n xx
2
1
2
1
22
11
ff
bf
af
bf
af
)1b(2bf
,a6af
b4bf
,a6)a2(3af
222
11
With Matlab
>>[a,b]=solve('3*(a^2-1)+2*b^2=11','2*a^3+(b-1)^2=16')
Matlab gives all possible solutions
761.1b611.1a
clc, clearx=[1 1]; xe=[0.01 0.01];niter1= 5; niter2=50;fe=transpose(abs(fe));kerr=1;for n=1:niter2x%-----Error equations------------------------a(1,1)=6*x(1);a(1,2)=4*x(2);a(2,1)=6*x(1)^2;a(2,2)=2*(x(2)-1);b(1)=-(3*(x(1)^2-1)+2*x(2)^2-11);b(2)=-(2*x(1)^3+(x(2)-1)^2-16);%-------------------------------------------------------bb=transpose(b);eps=inv(a)*bb;x=x+transpose(eps); if n>niter1 if abs(eps)<xe kerr=0; break else display ('Roots are not found') end endend
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),8,'spline')
1. Lagrange Interpolation with Matlab
2. Lagrange Interpolation with Matlab
The pressure values of a fluid flowing in a pipe are given in the table for different locations . Find the pressure value for 5 m.
a) Lagrange interpolation (manually)
a) with computer
Location (m) 3 8 10
Pressure (atm) 7 6.2 6
a) with Lagrange interpolation
6*)810)(310(
)8x)(3x(2.6*
)108)(38()10x)(3x(
7*)103)(83()10x)(8x(
)x(p
6*)810)(310(
)85)(35(2.6*
)108)(38()105)(35(
7*)103)(83()105)(85(
)5(p
)atm(6286.6)x(p b) For computer solution, the MATLAB code is given as
clc;clearx=[3 8 10];P=[7 6.2 6];interp1(x,P,5,'spline')
Lagrange Interpolation:
The x and y coordinates of three points on the screen, which were clicked by a CAD user are given in the figure. Find the y value of the curve obtained from these points at x=50.
x y
25 -10
40 20
70 5
b) How do you find the answer manually?
)5()4070)(2570(
)40x)(25x()20(
)7040)(2540()70x)(25x(
)10()7025)(4025(
)70x)(40x()x(y
1108.269259.0222.2296296.2)50(y
Result: 26.111
a)How do you find the answer with computer?
clc;clearx=[25 40 70];y=[-10 20 5];interp1(x,y,50,'spline')
Lagrange Interpolation:
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
Simpson’s Rule:
Calculate the integral with;
a) Trapezoidal rule
b) Simpson’s rule
c) Using MATLAB, take n=4.
?d)cos(1
5.0
a) Trapezoidal rule: Divide into for equal sections between 0.5 and 1. 125.04
5.01h
2f
fff2f
hI 4321
0TR
k θ f
0 0.5 0.6205
1 0.625 0.6411
2 0.75 0.6337
3 0.875 0.5996
4 1 0.5403
25403.0
5996.06337.06411.02
6205.0125.0ITR
3072.0ITR
Simpson’s Rule:
43210S ff4f2f4f3h
I
b) Simpson’s rule:
5403.05996.0*46337.0*26411.0*46205.03125.0
IS
3085.0IS
using Matlab
>>syms tet
>>I=int(sqrt(tet)*cos(tet),0.5,1);vpa(I,5)
I=0.30796
Lagrange Interpolation + Simpson’s Rule:
For a steel plate weighing 10 N and has a thickness 3 mm, the the coordinates of some points shown in the figure were measured (in cm) by a Coordinate Measuring Machine (CMM). How do you calculate the intensity of the steel by fitting a curve, which passes through these points. ?
)cm(V)kg(m3
a) Manual calculation:
The volume of the part is calculated by using its surface area and 0.3 cm thickness value. The simpson’s rule is used for area calculation. The x axis must be divided in equal segments in this method. Since the points are not equally spaced on the x axis, the necessary y values should be calculated at suitable x values.
x y
0 5
2.5 7.8
3.7 9.3
4 10
If we divide the interval 0-4 into four equal sections using the increment ∆x=1 , we can obtain the y values for x=1, x=2 and x=3.
10*)7.34)(5.24)(04()7.3x)(5.2x)(0x(
3.9*)47.3)(5.27.3)(07.3(
)4x)(5.2x)(0x(
8.7*)45.2)(7.35.2)(05.2(
)4x)(7.3x)(0x(5*
)40)(7.30)(5.20()4x)(7.3x)(5.2x(
y
Example:
10*)7.34)(5.24)(04()7.31)(5.21)(01(
3.9*)47.3)(5.27.3)(07.3(
)41)(5.21)(01(
8.7*)45.2)(7.35.2)(05.2(
)41)(7.31)(01(5*
)40)(7.30)(5.20()41)(7.31)(5.21(
)1(y
Lagrange Interpolation + Simpson’s Rule:
763.6)1(y
10*)7.34)(5.24)(04()7.32)(5.22)(02(
3.9*)47.3)(5.27.3)(07.3(
)42)(5.22)(02(
8.7*)45.2)(7.35.2)(05.2(
)42)(7.32)(02(5*
)40)(7.30)(5.20()42)(7.32)(5.22(
)2(y
4969.7)2(y
10*)7.34)(5.24)(04()7.33)(5.23)(03(
3.9*)47.3)(5.27.3)(07.3(
)43)(5.23)(03(
8.7*)45.2)(7.35.2)(05.2(
)43)(7.33)(03(5*
)40)(7.30)(5.20()43)(7.33)(5.23(
)3(y
2323.8)3(y
x y
0 5
1 6.763
2 7.4969
3 8.2323
4 10
Lagrange Interpolation + Simpson’s Rule:
b) With computer:
For calculation with computer, MATLAB code is arranged to find the y values for x=1, x=2 and x=3 and the code Lagr.I is run.
The area of the plate can be calculated by using Simpson’s rule. Then, the density of the steel can be calculated as mentioned before.
43210 ff4f2f4f3h
IA
102323.8*44969.7*2763.6*4531
IA 2cm9917.29A 3cm9983.52.0*9917.29t*AV
kg0193.181.9
10m 3cm
kg16994.0
9983.50193.1
Vm
14
04h
clc;clearx=[0 2.5 3.7 4];y=[5 7.8 9.3 10];interp1(x,y,1,'spline')
interp1(x,y,2,'spline')
interp1(x,y,3,'spline')
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
4.12.4cose5.2)(M 4.1
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:
A stationary car starts to move with the acceleration given below
1t
)tsin(1a
2
3
Find the speed of the car at the end of 10 seconds
a) Manually
b) With computer
0 2 4 6 8 10 120
0.2
0.4
0.6
0.8
1
1.2
1.4
Zaman (s)
İvm
e m
/s2
a)
10
010t
v
0
t
0
dtavdtadvdtdv
a
1t
)tsin(1a
2
3
5.24
010n
abth
k t f0 0 1
1 2.5 0.40216
2 5 0.0753
3 7.5 0.2358
4 10 0.18178
4n,1t
)tsin(1f
2
3
43210 ff4f2f4f3h
Iv
s/m236.318178.02358.0*40753.0*240216.0*4135.2
Iv
b) With computer
using Matlab>>syms t>>I=int((1+sin(t)^3)/sqrt(t^2+1),0,10);vpa(I,5)
Simpson’s Rule:
Find the intersection area of the curves y=x2+2 ile y=3x .
2xy 2
x3y
2xx3 2
02x3x2
Roots
2x1x
2
1
a2ac4bb
x2
2,1
2
1
2 ?dx2xx3
k x f0 1 0
1 1.25 0.1875
2 1.5 0.25
3 1.75 0.1875
4 2 0
43210 ff4f2f4f3h
IA
2br16667.001875.0*425.0*21875.0*40325.0
IA
using Matlab>>syms x>>I=int(3*x-x^2-2,1,2);vpa(I,5)
>> roots([1 -3 2])
System of linear equations:
05uzwz612w3u9z3w
How do you calculate u,w and z with computer?
5zwu12z6w3
9z3wu
5129
zwu
111630311
5129
111630311
zwu 1
With Matlabclc;cleara=[-1 1 -3;0 3 -6;1 1 1];b=[9;12;5];c=inv(a)*b 3z
10w8u
System of linear equations:
As a result of the equilibrium conditions, the equations given below are obtained for a truss system. How do you calculate the member forces FJD, FFD, FCD and FFC if FCK=6.157 kN and FCB=-3.888 kN are known?
0F707.0FF894.00F365.2F065.1F3
0738.1F447.0FF707.00466.0F894.0F707.0F
CKFCCD
FCCKJD
CBCDFD
CDFDJD
888.3F,157.6F CBCK
353.4557.6
0466.0
FFFF
1894.000365.200301707.000894.0707.01
FC
CD
FD
JD
A bF
b*AF 1b*)A(invF