The yield stress of a metal in uniaxial tension is 200 MPa. According to von Mises yield criterion, the yield stress (in MPa) of the metal in pure shear is closest to:

The correct option is 115.5.

Step-by-Step Explanation:

The von Mises yield criterion is used to predict the yielding of materials under multiaxial loading conditions. According to this criterion, yielding begins when the second invariant of the stress deviator tensor reaches a critical value. In terms of principal stresses, the criterion is expressed as:

$${({\sigma }_1-{\sigma }_2)}^2+{({\sigma }_2-{\sigma }_3)}^2+{({\sigma }_3-{\sigma }_1)}^2=2{\sigma }_y^2$$

where:
• ${\sigma }_1$, ${\sigma }_2$, and ${\sigma }_3$ are the principal stresses.
• ${\sigma }_y$ is the yield stress of the metal in uniaxial tension (given as 200 MPa).

For a state of pure shear with shear stress magnitude $\tau$, the principal stresses are calculated as:

$${\sigma }_1=\tau ,{\sigma }_2=-\tau ,{\sigma }_3=0$$

Substituting these values into the von Mises yield equation:

$${(\tau -(-\tau \left)\right)}^2+{(-\tau -0)}^2+{(0-\tau )}^2=2{\sigma }_y^2$$

Simplify each squared term:

$${\left(2\tau \right)}^2+{(-\tau )}^2+{(-\tau )}^2=2{\sigma }_y^2$$

$$4{\tau }^2+{\tau }^2+{\tau }^2=2{\sigma }_y^2$$

Combine the terms on the left side:

$$6{\tau }^2=2{\sigma }_y^2$$

Divide both sides by 2 to solve for the shear yield stress $\tau$:

$$3{\tau }^2={\sigma }_y^2\Rightarrow \tau =\frac{{\sigma }_y}{\sqrt{3}}$$

Now, substitute the given uniaxial tension yield stress value (${\sigma }_y=200\text{MPa}$):

$$\tau =\frac{200}{\sqrt{3}}\approx \frac{200}{1.732}\approx 115.47\text{MPa}$$

This calculated value of approximately 115.5 MPa corresponds to the first option.