# 5.2 Problem Set

Determine the value of each of the following.

$$6 \times 7+4^{2}-2^{4}$$

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

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

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

$$\log 10^{2}+10^{5}$$

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

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

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

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

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

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

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

Solve the following system of equations:

$$2 x+4 y=1$$

$$x+5 y=2$$

>> 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.

$$y=4 x^{4}+3 x^{2}-x$$

>> 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.

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: $$\PageIndex{1}$$ Load in Newtons $$\PageIndex{2}$$ 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)