Question

The elastic behavior of material for linear stress and linear strain, is shown in the figure. The energy density for a linear strain of 5×10^–4 is ____ kJ/m³. Assume that material is elastic up to the linear strain of 5×10^–4

Answer 1

Correct Answer: 25

To find the energy density for a linear strain of \(5 \times 10^{-4}\), we use the formula for elastic energy density:

\( \text{Energy Density} = \frac{1}{2} \sigma \epsilon \)

Where \(\sigma\) is the stress and \(\epsilon\) is the strain.

From the graph, the material follows a linear stress-strain relationship, meaning \(\sigma = E \cdot \epsilon\), where \(E\) is the Young's modulus.

We find \(E\) using the slope of the line, \(\frac{\Delta \sigma}{\Delta \epsilon} = \frac{80 \, \text{N/m}^2}{4 \times 10^{-10}}\). Calculating:

\(E = \frac{80}{4 \times 10^{-10}} = 2 \times 10^{11} \, \text{N/m}^2\)

Substituting \(\sigma = E \cdot \epsilon\) and the given \(\epsilon = 5 \times 10^{-4}\):

\(\sigma = 2 \times 10^{11} \times 5 \times 10^{-4} = 1 \times 10^8 \, \text{N/m}^2\)

Now, calculate the energy density:

\(\text{Energy Density} = \frac{1}{2} \times 1 \times 10^8 \times 5 \times 10^{-4}\)

\(= 2.5 \times 10^4 \, \text{J/m}^3 = 25 \, \text{kJ/m}^3\)

The energy density is \(25 \, \text{kJ/m}^3\), which is within the provided range (25, 25).