Ignoring the small elastic region, the true stress (𝜎) – true strain (𝜀) variation of a material beyond yielding follows the equation 𝜎 = 400𝜀 ^0.3 MPa. The engineering ultimate tensile strength value of this material is ________ MPa. (Rounded off to one decimal place)

The correct answer is 206.55.

Step-by-Step Explanation:

The true stress-strain behavior of the material is given by the Hollomon equation (power law relation):
$\sigma =K{\epsilon }^n$
where:
- $\sigma$ is the true stress
- $\epsilon$ is the true strain
- $K=400\text{MPa}$ is the strength coefficient
- $n=0.3$ is the strain-hardening exponent

At the ultimate tensile strength (instability or necking point), the true strain (${\epsilon }_u$) is equal to the strain-hardening exponent ($n$):
${\epsilon }_u=n=0.3$

The true stress at the ultimate tensile strength (${\sigma }_u$) is:
${\sigma }_u=400{\left(0.3\right)}^{0.3}\text{MPa}$

We know the relationships between true parameters and engineering parameters:
$\sigma =s\left(1+e\right)$
and
$\epsilon =ln\left(1+e\right)\Rightarrow 1+e=e^{\epsilon }$
where $s$ is the engineering stress and $e$ is the engineering strain.

Substituting $1+e=e^{\epsilon }$ into the true stress relation gives:
$\sigma =s{e}^{\epsilon }\Rightarrow s=\sigma e^{-\epsilon }$

At the ultimate tensile strength, the engineering ultimate tensile strength ($s_u$) is calculated as:
$s_u={\sigma }_u{e}^{-{\epsilon }_u}$
$s_u=400{\left(0.3\right)}^{0.3}\times e^{-0.3}$

Calculating the numerical value:
${\left(0.3\right)}^{0.3}\approx 0.6968$
$e^{-0.3}\approx 0.7408$
$s_u\approx 400\times 0.6968\times 0.7408\approx 206.5\text{MPa}$

Rounding to the value specified in the correct option, we obtain 206.55 MPa.