Skip to main content
Engineering LibreTexts

5.2 Problem Set

  • Page ID
    9468
  • Determine the value of each of the following.

    \(6 \times 7+4^{2}-2^{4}\)


    Answer:

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

    \(\frac{3^{2}+2^{3}}{4^{5}-5^{4}}+\frac{64^{0.5}-5^{2}}{4^{5}+5^{6}+7^{8}}\)


    Answer:

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

    \(\log 10^{2}+10^{5}\)


    Answer:

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

    \(e^{2}+2^{3}-\ln \left(e^{2}\right)\)


    Answer:

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

    \(\sin (2 \pi)+\cos \left(\frac{\pi}{4}\right)\)


    Answer:

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

    \(\tan \left(\frac{\pi}{3}\right)+\cos (270)+\sin (270)+\cos \left(\frac{\pi}{3}\right)\)

    Answer:

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

    Solve the following system of equations:

    \(2 x+4 y=1\)

    \(x+5 y=2\)


    Answer:

    >> A=[2 4; 1 5] A = 2 4 1 5 >> B=[1; 2] B = 1 2 >> Solution=A\B Solution = -0.5000 0.5000

    Evaluate y at 5.

    \(y=4 x^{4}+3 x^{2}-x\)

    Answer:

    >> p=[4 0 3 -1 0] p = 4 0 3 -1 0 >> polyval(p,5) ans = 2570 >>

    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

    Load [N] Gage Length [mm]
    0.000 50.8000
    4890 50.8102
    9779 50.8203
    14670 50.8305
    19560 50.8406
    24450 50.8508
    27620 50.8610
    29390 50.8711
    32680 50.9016
    33950 50.9270
    34580 50.9524
    35220 50.9778
    35720 51.0032
    40540 51.816
    48390 53.340
    59030 55.880
    65870 58.420

    69420

    60.960
    69670 (maximum) 61.468
    68150 63.500
    60810 (fracture) 66.040 (after fracture)

    \(\sigma=\frac{P}{A}\) where P is the load [N] on the sample with an original cross-sectional area A \(\left[m^{2}\right]\) and the engineering strain is defined as \(\varepsilon=\frac{\Delta l}{l}\) where \(\Delta l\) is the change in length and ll is the initial length.

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


    Answer:

    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:

    屏幕快照 2019-05-25 19.39.49.png

    \(\PageIndex{1}\) Load in Newtons

    屏幕快照 2019-05-25 19.43.05.png

    \(\PageIndex{2}\) Extension length in mm.

    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)