Question

The elastic behavior of a material for linear stress to linear strain is given in the figure. The energy density for a linear strain of \(4\times10^{-4}\) is center

center

• A. \(20000\ \text{J m}^{-3}\)
• B. \(16000\ \text{J m}^{-3}\)
• C. \(12000\ \text{J m}^{-3}\)
• D. \(8000\ \text{J m}^{-3}\)

Hint:

Energy density equals the area under the stress-strain graph. For linear elastic region: \[ u=\frac12 \sigma \epsilon \]

Answer 1

The Correct Option is B

Step 1: Know the energy density rule.
When a material is stretched, the energy stored per unit volume is the area under the stress versus strain line. For a straight line this area is a triangle. \[ u = \tfrac12 \times \text{stress} \times \text{strain} \]

Step 2: Read the values from the graph.
At the strain $4\times10^{-4}$, the matching stress from the straight graph is \[ \text{stress} = 80\times10^{6}\,\text{N/m}^2 \]
Step 3: Write down both numbers clearly.
Strain $= 4\times10^{-4}$ and stress $= 80\times10^{6}$.
Step 4: Plug into the formula.
\[ u = \tfrac12 \times (80\times10^{6}) \times (4\times10^{-4}) \]
Step 5: Multiply the parts.
First multiply the numbers and powers: $80 \times 4 = 320$ and $10^{6}\times10^{-4} = 10^{2}$. \[ u = \tfrac12 \times 320 \times 10^{2} = \tfrac12 \times 32000 \]
Step 6: Final value.
\[ u = 16000\,\text{J/m}^3 \] \[ \boxed{16000\,\text{J m}^{-3}} \]