Draw the stress-strain graph to the following scales: Stress 10 mm = 25 MPa; Strain 20 mm = 200 × 10⁻⁴…

Question

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

ScanSolve AI Tutor · Accepted Answer

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

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 $0, 200 \times 10^{-6}, 400 \times 10^{-6}, \dots, 1400 \times 10^{-6}$ .
Plot the data points.
Plot the following points on your graph:
- $(0.000470, 100)$
- $(0.000710, 150)$
- $(0.000950, 200)$
- $(0.001100, 250)$
- $(0.001340, 300)$
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.
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:
$\sigma_1 = 100 \text{ MPa} = 100 \times 10^6 \text{ Pa}$
$\epsilon_1 = 470 \times 10^{-6}$
$\sigma_2 = 150 \text{ MPa} = 150 \times 10^6 \text{ Pa}$
$\epsilon_2 = 710 \times 10^{-6}$
$E = \frac{(150 - 100) \times 10^6 \text{ Pa}}{(710 - 470) \times 10^{-6}}$ $E = \frac{50 \times 10^6 \text{ Pa}}{240 \times 10^{-6}}$ $E = \frac{50}{240} \times 10^{12} \text{ Pa}$ $E \approx 0.208333 \times 10^{12} \text{ Pa}$ $E \approx 208.33 \times 10^9 \text{ Pa}$ $E \approx 208.33 \text{ GPa}$
The modulus of elasticity is $\text{208.33 GPa}$ .

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.

Answer

Step-by-step solution

Set up the graph axes and scales.

Plot the data points.

Draw the best-fit line.

Calculate Young's modulus ( $E$ ) from the linear portion of the data.

Try one yourself

Still stuck on this one?

More Physics Questions