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Safety against plastic flow acc. TRESCA (shear stress hypothesis)

This exercise shows how the safety against flow is calculated according to the TRESCA shear stress hypothesis. The equivalent stress is calculated from the principal stresses and then compared with the yield stress of the material.

Task

A load case with the stress tensor is given

\[ \newcommand{\myvec}[1]{{\begin{bmatrix}#1\end{bmatrix}}} \]

\[ S = \myvec{-160 & 120 & 0\\120 & 80 & 0\\0 & 0 & 0} \, MPa \]

The yield point for the material is σF = 650 MPa. Is there a double safety against flowing according to TRESCA?

Solution

The yield condition according to TRESCA (shear stress hypothesis) is

\[ \tag{1} |\tau_{max}| \overset{!}{=} \frac{\sigma_0}{2} \]

with

\[ \tag{2} \frac{1}{2}\,Max \left[ |\sigma_1 -- \sigma_2|,|\sigma_2 -- \sigma_3|,|\sigma_1 -- \sigma_3| \right] = \frac{\sigma_0}{2} \]

Since it is a plane stress state, only σ1 and σ2 have to be considered.

The dimensions of the tensions are not noted in the following equations to improve readability. The angle of the principal stresses is

\[ \tag{3} \phi_0 = \frac{arctan\left( \frac{2\tau_{xy}}{\sigma_x -\sigma_y} \right)}{2}\]

\[ \tag{4} \phi_0 = \frac{arctan\left( \frac{2 \cdot 120}{-160 -80} \right)}{2}\]

\[ \tag{5} \phi_0 = -22.5° \]

The principal stresses are

\[ \tag{6} \sigma_1 = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2} \cdot cos 2 \phi_0 + \tau_{xy} \cdot sin 2 \phi_0 \]

\[ \tag{7} \sigma_1 = \frac{-160 + 80}{2} + \frac{-160-80}{2} \cdot cos (-45°) + 120 \cdot sin (-45°) \]

\[ \tag{8} \sigma_1 = -210 \, MPa \]

\[ \tag{9} \sigma_2 = \frac{\sigma_x + \sigma_y}{2} -- \frac{\sigma_x-\sigma_y}{2} \cdot cos 2 \phi_0 -- \tau_{xy} \cdot sin 2 \phi_0 \]

\[ \tag{10} \sigma_2 = \frac{-160 + 80}{2} -- \frac{-160-80}{2} \cdot cos (-45°) -- 120 \cdot sin (-45°) \]

\[ \tag{11} \sigma_2 = 130 \, MPa \]

This results in the decisive tension

\[ \tag{12} \sigma_0 = |-210 -- 130| \, MPa = 340 \, MPa \]

The safety against flow SF is

\[ \tag{13} S_F = \frac{\sigma_F}{\sigma_0} = \frac{650\,MPa}{340\,MPa} = 1.91 \]

This shows that there is no double safety against flow according to the shear stress hypothesis according to TRESCA.