Question 6.1: Consider an isotropic material, loaded so that the principal......
Consider an isotropic material, loaded so that the principal stresses coincide with the x, y, and z axes of the material. Assume that the Tresca yield criterion applies. Make a plot of the combinations of \sigma_{y} versus \sigma_{x} that will cause yielding with \sigma_{z} = 0.
Divide the \sigma_{y} versus \sigma_{x} stress space into six sectors as shown in Figure 6.2. The following conditions are appropriate:
1. \sigma_{x}>\sigma_{y}> \sigma_{z}=0, so \sigma_{1} = \sigma_{x}, \sigma_{3} = \sigma_{z}=0, so \sigma_{x} = Y
2. \sigma_{y}>\sigma_{x}> \sigma_{z}=0, so \sigma_{1} = \sigma_{y}, \sigma_{3} = \sigma_{z}=0, so \sigma_{y} = Y
3. \sigma_{y}>\sigma_{z}=0> \sigma_{x}, so \sigma_{1} = \sigma_{y}, \sigma_{3} = \sigma_{x}, so \sigma_{y}-\sigma_{x} = Y
4. \sigma_{z}=0>\sigma_{y}> \sigma_{x}, so \sigma_{1} = 0, \sigma_{3} = \sigma_{x}, so – \sigma_{x} = Y
5. \sigma_{z}=0>\sigma_{x}> \sigma_{y}=0, so \sigma_{1} = 0, \sigma_{3} = \sigma_{y}, so – \sigma_{y} = Y
6. \sigma_{x}>\sigma_{z}=0> \sigma_{y}, so \sigma_{1} = \sigma_{x}, \sigma_{3} = \sigma_{y}, so \sigma_{x} – \sigma_{y}= Y
These are plotted in Figure 6.2.
