Consider a linear rectangular thin sheet of metal, subjected to uniform uniaxial tensile stress of 100 MPa along the length direction. Assume plane stress conditions in the plane normal to the thickness. The Young’s modulus E = 200 MPa and Poisson’s ratio v = 0.3 are given. The principal strains in the plane of the sheet are

The correct option is (0.5, -0.15).

Let us analyze the problem step-by-step to understand why this option is correct.

We are given a linear rectangular thin sheet of metal subjected to a uniform uniaxial tensile stress along the length direction (let this be the $x$-direction). Therefore, the stresses in the plane of the sheet are:
${\sigma }_x=100\text{ MPa}$
${\sigma }_y=0$

We are also given the material properties:
Young's modulus: $E=200\text{ GPa}$ (Note: The question states $E=200\text{ MPa}$, but to get the numerical values in the options, we use the standard hookean relations. Let's calculate using the given values: $E=200\text{ MPa}$ and Poisson's ratio $\nu =0.3$).
Let's compute the principal strains in the plane of the sheet (${\epsilon }_1$ and ${\epsilon }_2$) using the generalized Hooke's law for plane stress:

The principal strain along the direction of the applied tensile stress (${\epsilon }_x$ or ${\epsilon }_1$) is given by:
${\epsilon }_1=\frac{1}{E}\left({\sigma }_x-\nu {\sigma }_y\right)$

Substituting the given values:
${\epsilon }_1=\frac{1}{200}\left(100-0.3\times 0\right)=\frac{100}{200}=0.5$

The principal strain perpendicular to the loading direction in the plane of the sheet (${\epsilon }_y$ or ${\epsilon }_2$) is given by:
${\epsilon }_2=\frac{1}{E}\left({\sigma }_y-\nu {\sigma }_x\right)$

Substituting the given values:
${\epsilon }_2=\frac{1}{200}\left(0-0.3\times 100\right)=\frac{-30}{200}=-0.15$

Thus, the principal strains in the plane of the sheet are $(0.5,-0.15)$.