A steel plate having yield strength of 550 MPa, is subjected to biaxial state of stress as σ_x = σ, σ_y = –2σ. Using distortion energy and considering FOS as 2, the permissible value of σ that can be applied to plate is ______ MPa (rounded off to one decimal place)

The correct answer is 103.94.

Step-by-Step Explanation:

We are given the following parameters for the steel plate:
- Yield strength of the steel plate, $S_{\mathrm{yt}}=550\text{MPa}$
- Biaxial state of stress: ${\sigma }_x=\sigma$ and ${\sigma }_y=-2\sigma$
- Factor of Safety (FOS), $N=2$

According to the Distortion Energy Theory (also known as the Von Mises Yield Criterion), yielding occurs when the Von Mises stress (${\sigma }_v$) reaches the yield strength of the material under uniaxial tension. For design purposes, the permissible stress limit is given by dividing the yield strength by the factor of safety:

${\sigma }_v=\frac{S_{\mathrm{yt}}}{N}$

For a biaxial state of stress (where the third principal stress ${\sigma }_z=0$), the Von Mises stress equation simplifies to:

${\sigma }_v=\sqrt{{\sigma }_x^2-{\sigma }_x{\sigma }_y+{\sigma }_y^2}$

Substituting the given values of ${\sigma }_x$ and ${\sigma }_y$ into the expression:

${\sigma }_v=\sqrt{{\sigma }^2-\left(\sigma \right)(-2\sigma )+{(-2\sigma )}^2}$

${\sigma }_v=\sqrt{{\sigma }^2+2{\sigma }^2+4{\sigma }^2}$

${\sigma }_v=\sqrt{7{\sigma }^2}=\sigma \sqrt{7}$

Now, equating this Von Mises stress to the permissible limit ($\frac{S_{\mathrm{yt}}}{N}$):

$\sigma \sqrt{7}=\frac{550}{2}$

$\sigma \sqrt{7}=275$

$\sigma =\frac{275}{\sqrt{7}}$

Since $\sqrt{7}\approx 2.64575$:

$\sigma \approx \frac{275}{2.645751}\approx 103.94\text{MPa}$

Thus, the permissible value of $\sigma$ that can be applied to the plate is 103.94 MPa.