After fracture, the total length was 47.42 mm and the diameter was 18.35 mm. Plot the engineering stress strain curve and calculate (a) the 0.2% offset yield strength; (b) the tensile strength; (c) the modulus of elasticity, using a linear fit to the appropriate data; (d) the % elongation; (e) the % reduction in area;

From the complete question the data is provided for various Loads in ductile testing machine for a sample of d0=20 mm and l0=40mm. The plot is drawn between stress and strain whose values are calculated using following formulae. The corresponding values are attached with the solution.

The engineering-stress is given as

[tex]\sigma=\frac{F}{A}\\\sigma=\frac{F}{\pi \frac{d_0^2}{4}}\\\sigma=\frac{F}{\pi \frac{(20 \times 10^-3)^2}{4}}\\\sigma=\frac{F}{3.14 \times 10^{-4}}[/tex]

Here F are different values of the load

Now Strain is given as

[tex]\epsilon=\frac{l-l_0}{l_0}\\\epsilon=\frac{\Delta l}{40}\\[/tex]

So the curve is plotted and is attached.

Part a

The value of yield strength at 0.2% offset is obtained from the engineering stress-strain curve which is given as 274 MPa.

Part b

The value of tensile strength is obtained from the engineering stress-strain curve which is given as 417 MPa

Part c

Young's Modulus is given as

[tex]E=\frac{\sigma}{\epsilon}\\E=\frac{238 /times 10^6}{0.00138}\\E=172,000 MPa\\E=172 GPa[/tex]

The value of Young's modulus at given point is 172 GPa.

Part d

The percentage elongation is given as

[tex]Elongation=\frac{l_f-l_0}{l_0} \times 100\\Elongation=\frac{47.42-40}{40}\times 100\\Elongation=18.55 \%\\[/tex]

So the percentage elongation is 18.55%

Part e

The reduction in area is given as

[tex]Reduction=\frac{A_0-A_n}{A_0} \times 100\\Reduction=\frac{\pi \frac{d_0^2}{4}-\pi \frac{d_n^2}{4}}{\pi \frac{d_0^2}{4}}\times 100\\Reduction=\frac{{d_0^2}-{d_n^2}}{{d_0^2}} \times 100\\Reduction=\frac{{20^2}-{18.35^2}}{{20^2}} \times 100\\Reduction=15.81\%[/tex]

So the reduction in area is 15.81%