Question

Von Mises criterion for plastic yielding of a ductile material predicts that the yield stress in uniaxial tension is related to that in pure torsion as

• A. Equal to each other
• B. 2 times
• C. One half
• D. \( \sqrt{3} \) times

Hint:

The Von Mises criterion is based on the concept of distortion energy and is often used in design for ductile materials. For pure torsion, the relationship with uniaxial yield stress becomes \( \sigma = \sqrt{3} \, \tau \).

Answer 1

The Correct Option is D

Step 1: According to the Von Mises yield criterion for isotropic ductile materials under complex stress states, the equivalent or effective stress is: \[ \sigma_{e} = \sqrt{\frac{1}{2} \left[(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2 \right]} \] Step 2: For uniaxial tension, let the principal stresses be: \[ \sigma_1 = \sigma, \quad \sigma_2 = 0, \quad \sigma_3 = 0 \Rightarrow \sigma_e = \sigma \] Step 3: For pure torsion, the only non-zero stress is the shear stress \( \tau \). The principal stresses are: \[ \sigma_1 = \tau, \quad \sigma_2 = -\tau, \quad \sigma_3 = 0 \Rightarrow \sigma_e = \sqrt{3} \, \tau \] Step 4: Equating the effective stress at yield in both cases: \[ \sigma = \sqrt{3} \, \tau \Rightarrow \frac{\sigma}{\tau} = \sqrt{3} \] Therefore, the yield stress in uniaxial tension is \( \sqrt{3} \) times that in pure torsion.