Draw the stress-strain graph to the following scales: Stress 10 mm = 25 MPa; Strain 20 mm = 200 x 10^-4. Determine the modulus of elasticity.

Here are the solutions to Question 7.2:

Given data from the test on a steel specimen: | Stress ($\sigma$, MPa) | Strain $\times 10^{-4}$ | | :--------------------- | :---------------------- | | 100 | 470 | | 150 | 710 | | 200 | 950 | | 250 | 1100 | | 300 | 1340 |

Assumption for Strain Units: Based on typical values for Young's Modulus of steel (around 200 GPa) and the previous question (7.1), it is highly probable that the "Strain $\times 10^{-4}$" in the table header is a typo and should be "Strain $\times 10^{-6}$". We will proceed with this assumption to obtain a realistic modulus of elasticity for steel. Therefore, the actual strain values are: | Stress ($\sigma$, MPa) | Strain ($\epsilon$) | | :--------------------- | :------------------ | | 100 | $470 \times 10^{-6} = 0.000470$ | | 150 | $710 \times 10^{-6} = 0.000710$ | | 200 | $950 \times 10^{-6} = 0.000950$ | | 250 | $1100 \times 10^{-6} = 0.001100$ | | 300 | $1340 \times 10^{-6} = 0.001340$ |

7.2.1 Draw the stress-strain graph to the following scales:

• Stress: $10 mm = 25 \text{ MPa}$
• Strain: $20 mm = 200 \times 10^{-4}$ (Using the corrected assumption, this becomes $20 mm = 200 \times 10^{-6}$)

Step 1: Set up the graph axes and scales.

• Y-axis (Stress): Label the vertical axis "Stress ($\sigma$, MPa)".
  • Scale: $10 \text{ mm}$ represents $25 \text{ MPa}$.
  • This means $1 \text{ mm}$ represents $2.5 \text{ MPa}$.
  • Mark points at $0, 25, 50, 75, 100, \dots, 300 \text{ MPa}$.
• X-axis (Strain): Label the horizontal axis "Strain ($\epsilon$)".
  • Scale: $20 \text{ mm}$ represents $200 \times 10^{-6}$ (or $0.0002$).
  • This means $1 \text{ mm}$ represents $10 \times 10^{-6}$ (or $0.00001$) strain.
  • Mark points at $0, 200 \times 10^{-6}, 400 \times 10^{-6}, \dots, 1400 \times 10^{-6}$.

Step 2: Plot the data points. Plot the following points on your graph:

• $(0.000470, 100)$
• $(0.000710, 150)$
• $(0.000950, 200)$
• $(0.001100, 250)$
• $(0.001340, 300)$

Step 3: Draw the best-fit line. Draw a straight line through the initial linear portion of the plotted points. For steel, the stress-strain curve is typically linear up to the yield point.

7.2.2 Determine the modulus of elasticity.

Step 4: Calculate Young's modulus ($E$) from the linear portion of the data. Young's modulus is the slope of the stress-strain curve in the elastic region. We will use the first two data points, which typically represent the initial linear elastic behavior. $E = \frac{\Delta \sigma}{\Delta \epsilon}$ Using the first two points: $\sigma_1 = 100 MPa = 100 \times 10^6 \text{ Pa}$ $\epsilon_1 = 470 \times 10^{-6}$ $\sigma_2 = 150 MPa = 150 \times 10^6 \text{ Pa}$ $\epsilon_2 = 710 \times 10^{-6}$

$E = \frac{(150 - 100) \times 10^6 Pa}{(710 - 470) \times 10^{-6}}$ $E = \frac{50 \times 10^6 Pa}{240 \times 10^{-6}}$ $E = \frac{50}{240} \times 10^{12} Pa$ $E \approx 0.208333 \times 10^{12} Pa$ $E \approx 208.33 \times 10^9 Pa$ $E \approx 208.33 GPa$

The modulus of elasticity is $\boxed{208.33 GPa}$.