$$ \begin{array}{l} c_{c u}=c^{\prime}=0 \ \phi_{c u}=17^{\circ} \text { and } \phi_{c u}^{\prime}=34^{\circ} \end{array} $$

• If sample of this soil was tested in a $C U$ Test under a cell pressure of $180 \mathrm{kN} / \mathrm{m}^{2}$.

• Determine

(a) deviator stress at failure (b) pore water pressure at failure (c) minor principal effective stress at failure (d) major principal effective stress at failure (e) the magnitude of $A_{f}$

(a) Since: $$ c_{u}=0 $$ Using, or, (b) $\therefore$ Using, $$ \begin{aligned} \sigma_{1} &=\sigma_{3}+\sigma_{d}=180+148.75=328.75 \mathrm{kN} / \mathrm{m}^{2} \\ \bar{\sigma}_{1} &=\bar{\sigma}_{3} \tan ^{2}\left(45+\frac{\phi_{u}^{\prime}}{2}\right)+2 c^{\prime} \tan \left(45+\frac{\phi_{u}^{\prime}}{2}\right) \\ \bar{\sigma}_{1} &=\bar{\sigma}_{3} \tan ^{2}\left(45+\frac{\phi_{u}^{\prime}}{2}\right)+0 \\ \left(\sigma_{1}^{\prime}-u\right) &=\left(\sigma_{3}-u\right) \tan ^{2}\left(45+\frac{\phi_{c u}^{\prime}}{2}\right) \\ (328.75-u) &=(180-u) \times 3.537 \\ 2.537 u &=307.93 \\ u &=121.37 \mathrm{kN} / \mathrm{m}^{2} \end{aligned} $$ (c) Minor principal effective stress at failure, $$ \bar{\sigma}_{3 t}=\sigma_{3 f}-u_{f}=180-121.37=58.63 \mathrm{kN} / \mathrm{m}^{2} $$ (d) Major principal effective stress at failure, $$ \begin{aligned} \bar{\sigma}_{1 f} &=\bar{\sigma}_{d f}+\bar{\sigma}_{3 f} \\ &=\left(\bar{\sigma}_{1 f}-\bar{\sigma}_{3 f}\right)+\bar{\sigma}_{3 f} \\ &=\left[\left(\sigma_{1 f}-u\right)-\left(\sigma_{3}-u\right)\right]+\bar{\sigma}_{3 f} \\ &=\left(\sigma_{1 f}-\sigma_{3}\right)+\bar{\sigma}_{3 f} \end{aligned} $$ $\begin{aligned} &=\sigma_{d}+\bar{\sigma}_{3 f}=148.75+58.63 \\ &=207.38 \mathrm{kN} / \mathrm{m}^{2} \\ A_{f} &=\frac{u_{f}}{\left(\sigma_{1}-\sigma_{3}\right)_{f}}=\frac{121.37}{148.75}=0.815 \end{aligned}$