Question:

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:

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According to the von Mises yield criterion, the yield stress in pure shear is \( \frac{\sigma_t}{\sqrt{3}} \), where \( \sigma_t \) is the yield stress in uniaxial tension.
Updated On: Apr 15, 2025
  • 115.5
  • 100.0
  • 66.7
  • 141.4
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The Correct Option is A

Solution and Explanation

The von Mises yield criterion relates the yield stress in uniaxial tension (\( \sigma_t \)) and the yield stress in pure shear (\( \sigma_{sh} \)) using the following formula: \[ \sigma_t = \sqrt{3} \cdot \sigma_{sh} \] where:
\( \sigma_t = 200 \, {MPa} \) (yield stress in uniaxial tension),
\( \sigma_{sh} \) is the yield stress in pure shear.
To find the yield stress in pure shear, we solve for \( \sigma_{sh} \): \[ \sigma_{sh} = \frac{\sigma_t}{\sqrt{3}} = \frac{200}{\sqrt{3}} \approx 115.5 \, {MPa} \] Thus, the correct answer is (A) 115.5 MPa.
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