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\begin{array}{l}\sigma_{3}=276 \mathrm{kPa} \\\sigma_{d}=276 \mathrm{kPa} \\\sigma_{1}=552 \mathrm{kPa}\end{array}\text { ay } \begin{aligned}\text { minimuus } & =276 \mathrm{kPa} \\\max & =552 \mathrm{kla}\end{aligned}\phi=?\begin{array}{l} \sigma_{1}=\sigma_{3} \tan ^{2} \alpha_{f}+2 \psi_{u}^{0} \tan \alpha f \\\frac{552}{276}=\tan ^{2} \alpha f \\\alpha f=54.73^{\circ} \text { failure plane inclination } \\45+\frac{\phi}{2}=54.73^{\circ} \\\phi=19.47^{\circ} \rightarrow \text { internal friction angle }\end{array}failure plane inclinationd)\begin{aligned}\sigma_{1} & =\frac{\sigma_{1}+\sigma_{3}}{2}+\frac{\sigma_{1}-\sigma_{3}}{2} \cos (2 \alpha f) \\& =\frac{552+276}{2}+\frac{552-276}{2} \cos (2 \times 54.73) \\& =368 \cdot \mathrm{kPa} \\\tau & =\frac{\sigma_{1}-\sigma_{3}}{2} \sin 2 \alpha f \\& =130.11 \mathrm{kPa} .\end{aligned} ...