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Problem Set

Module by: Serhat Beyenir. E-mail the author

Summary: Problem Set for MATLAB Essentials

Determine the value of each of the following.

Exercise 1

6×7+4224 6 7 4 2 2 4

Solution

>> (6*7)+4^2-2^4 (ans = 42)

Exercise 2

32+234554+640.55245+56+78 3 2 2 3 4 5 5 4 64 0.5 5 2 4 5 5 6 7 8

Solution

>> ((3^2+2^3)/(4^5-5^4))+((sqrt(64)-5^2)/(4^5+5^6+7^8)) (ans = 0.0426)

Exercise 3

log102+105 10 2 10 5

Solution

>> log10(10^2)+10^5 (ans = 100002)

Exercise 4

e2+23lne2 2 2 3 2

Solution

>> exp(2)+2^3-log(exp(2)) (ans = 13.3891)

Exercise 5

sin2π+cosπ4 2 4

Solution

>> sin(2*pi)+cos(pi/4) (ans = 0.7071)

Exercise 6

tanπ3+cos270+sin270+cosπ3 3 270 270 3

Solution

>> tan(pi/3)+cos(270*pi/180)+sin(270*pi/180)+cos(pi/3) (ans = 1.2321)

Exercise 7

Solve the following system of equations:
2x+4y=1 2 x 4 y 1
x+5y=2 x 5 y 2

Solution


>> A=[2 4; 1 5]

A =

     2     4
     1     5

>> B=[1; 2]

B =

     1
     2

>> Solution=A\B

Solution =

   -0.5000
    0.5000

Exercise 8

Evaluate y at 5.
y=4x4+3x2x y 4 x 4 3 x 2 x

Solution


>> p=[4 0 3 -1 0]

p =

     4     0     3    -1     0

>> polyval(p,5)

ans =

        2570

>> 

Exercise 9

Given below is Load-Gage Length data for a type 304 stainless steel that underwent a tensile test. Original specimen diameter is 12.7 mm. 1

Table 1
Load [kN] Gage Length [mm]
0.000 50.8000
4.890 50.8102
9.779 50.8203
14.670 50.8305
19.560 50.8406
24.450 50.8508
27.620 50.8610
29.390 50.8711
32.680 50.9016
33.950 50.9270
34.580 50.9524
35.220 50.9778
35.720 51.0032
40.540 51.816
48.390 53.340
59.030 55.880
65.870 58.420
69.420 60.960
69.670 (maximum) 61.468
68.150 63.500
60.810 (fracture) 66.040 (after fracture)
The engineering stress is defined as σ=PA σ P A , where P is the load [N] on the sample with an original cross-sectional area A [ m2 m 2 ] and the engineering strain is defined as ε=Δll ε Δl l , where Δl Δl is the change in length and l l is the initial length.

Compute the stress and strain values for each of the measurements obtained in the tensile test. Data available for download.

Solution

First, we need to enter the data sets. Because it is rather a large table, using Variable Editor is more convenient. See the figures below:

Figure 1: Load in Newtons
Load
Figure 2: Extension length in mm.
Length

Next, we will calculate the cross-sectional area.


Area=pi/4*(0.0127^2)

Area =

  1.2668e-004

Now, we can find the Stress values with the following, note that we are obtaining results in MPa:

Sigma=(Load_N./Area)*10^(-6)

Sigma =

         0
   38.6022
   77.1964
  115.8065
  154.4086
  193.0108
  218.0351
  232.0076
  257.9792
  268.0047
  272.9780
  278.0302
  281.9773
  320.0269
  381.9955
  465.9888
  519.9844
  548.0085
  549.9820
  537.9830
  480.0403

For strain calculation, we will first find the change in length:


Delta_L=Length_mm-50.800

Delta_L =

         0
    0.0102
    0.0203
    0.0305
    0.0406
    0.0508
    0.0610
    0.0711
    0.1016
    0.1270
    0.1524
    0.1778
    0.2032
    1.0160
    2.5400
    5.0800
    7.6200
   10.1600
   10.6680
   12.7000
   15.2400

Now we can determine Strain with the following:


Epsilon=Delta_L./50.800

Epsilon =

         0
    0.0002
    0.0004
    0.0006
    0.0008
    0.0010
    0.0012
    0.0014
    0.0020
    0.0025
    0.0030
    0.0035
    0.0040
    0.0200
    0.0500
    0.1000
    0.1500
    0.2000
    0.2100
    0.2500
    0.3000

The final results can be tabulated as foolows:


[Sigma Epsilon]

ans =

         0         0
   38.6022    0.0002
   77.1964    0.0004
  115.8065    0.0006
  154.4086    0.0008
  193.0108    0.0010
  218.0351    0.0012
  232.0076    0.0014
  257.9792    0.0020
  268.0047    0.0025
  272.9780    0.0030
  278.0302    0.0035
  281.9773    0.0040
  320.0269    0.0200
  381.9955    0.0500
  465.9888    0.1000
  519.9844    0.1500
  548.0085    0.2000
  549.9820    0.2100
  537.9830    0.2500
  480.0403    0.3000

Footnotes

  1. Introduction to Materials Science for Engineers by J. F. Shackelford, Macmillan Publishing Company. © 1985, (p.304)

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