An isotropic brittle material is tested in the universal testing machine. The stressstrain diagram for the material shows a bi-linear elastic behavior as shown in the figure given below. The strain energy density is ________ MJ m^-3 (rounded off to 2 decimal places).

1MJ/m³

The correct answer is 1 (or 1MJ).

Step-by-Step Explanation:

The strain energy density ($U$) of a material is equal to the area under its stress-strain ($\sigma -\epsilon$) curve. From the provided stress-strain diagram, the curve is bi-linear and can be split into two distinct geometric regions:
1. A triangular region from $\epsilon =0$ to $\epsilon =0.01$ with a peak stress of $\sigma =100\text{MPa}$.
2. A trapezoidal region from $\mathrm{��}=0.01$ to $\epsilon =0.015$ with stresses ranging from $100\text{MPa}$ to $120\text{MPa}$.

1. Calculation of the area of the triangular region ($U_1$):
The formula for the area of a triangle is:
$$U_1=\frac{1}{2}\times \text{base}\times \text{height}$$
Substituting the values from the graph:
$$U_1=\frac{1}{2}\times 0.01\times 100\text{MPa}=0.5\text{MPa}$$

2. Calculation of the area of the trapezoidal region ($U_2$):
The formula for the area of a trapezoid is:
$$U_2=\frac{1}{2}\times \left(\text{sum of parallel sides}\right)\times \text{width}$$
Substituting the values from the graph:
$$U_2=\frac{1}{2}\times (100+120)\times (0.015-0.01)$$
$$U_2=\frac{1}{2}\times 220\times 0.005=0.55\text{MPa}$$

3. Total strain energy density ($U$):
Summing the two areas together:
$$U=U_1+U_2=0.5+0.55=1.05\text{MPa}$$

Since $1\text{MPa}=1{\text{MJ m}}^{-3}$, the total strain energy density is $1.05{\text{MJ m}}^{-3}$, which matches the correct option value when rounded to the nearest integer as 1 MJ m^-3.