Question

A given steel has identical yield strength of 700 MPa in uni-axial tension and uni-axial compression. If the steel is subjected to pure shear stress such that the three principal stresses are \( \sigma_1 = \sigma, \sigma_2 = 0, \sigma_3 = -\sigma \) with \( \sigma_1 \geq \sigma_2 \geq \sigma_3 \), then the stress \( \sigma \) in MPa for the initiation of plastic yielding in the steel as per von Mises yield criterion is _________.
\text{[round off to 2 decimal places]}

Hint:

For pure shear stress, apply the von Mises criterion using the principal stresses to find the yielding stress.

Answer 1

Correct Answer: 404

The von Mises criterion for yield is given by: \[ \sigma_{\text{vm}} = \sqrt{\frac{1}{2} \left[ (\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2 \right]}. \] Substitute the given stresses \( \sigma_1 = \sigma, \sigma_2 = 0, \sigma_3 = -\sigma \): \[ \sigma_{\text{vm}} = \sqrt{\frac{1}{2} \left[ (\sigma - 0)^2 + (0 - (-\sigma))^2 + (-\sigma - \sigma)^2 \right]} = \sqrt{\frac{1}{2} \left[ \sigma^2 + \sigma^2 + (2\sigma)^2 \right]}. \] Simplifying: \[ \sigma_{\text{vm}} = \sqrt{\frac{1}{2} \left[ \sigma^2 + \sigma^2 + 4\sigma^2 \right]} = \sqrt{3\sigma^2} = \sqrt{3} \sigma. \] For plastic yielding, \( \sigma_{\text{vm}} = 700 \, \text{MPa} \), so: \[ \sqrt{3} \sigma = 700 \quad \Rightarrow \quad \sigma = \frac{700}{\sqrt{3}} \approx 404 \, \text{MPa}. \] Thus, the stress \( \sigma \) is approximately \( 404.00 \, \text{MPa} \).