Let's break down and **solve the problem step-by-step**: --- ## **Given Data** - **Horizontal stress** (in the cross-section plane): \(\sigma_x = 100\) MPa - **Vertical stress** (perpendicular to the cross-section plane, i.e., along tunnel axis): \(\sigma_z = 20\) MPa - **No mention of vertical stress in the cross-section plane:** So, let's assume the vertical stress in the cross-section plane (\(\sigma_y\)) is **** (common assumption if not specified). - **No shear stresses mentioned:** Assume all shear stresses (\(\tau_{xy}, \tau_{xz}, \tau_{yz}\)) are zero. --- ## **(a) Principal Stresses (Magnitudes and Orientations)** ### **Step 1: Write the Stress Tensor** \[ \begin{align*} \text{Let:} \quad &\sigma_x = 100 \text{ MPa (horizontal in section)} \\ &\sigma_y = \text{ MPa (vertical in section, not specified)} \\ &\sigma_z = 20 \text{ MPa (parallel to tunnel axis)} \\ \end{align*} \] \[ \mathbf{\sigma} = \begin{bmatrix} 100 & & \\ & & \\ & & 20 \\ \end{bmatrix} \] ### **Step 2: Principal Stresses** Since the stress tensor is diagonal and all shear stresses are zero, the **principal stresses are simply the diagonal components**: \[ \boxed{ \begin{align*} \sigma_1 &= 100 \ \text{MPa} \\ \sigma_2 &= 20 \ \text{MPa} \\ \sigma_3 &= \ \text{MPa} \\ \end{align*} } \] - Directions: - \(\sigma_1\) (100 MPa): Horizontal direction (in the section plane) - \(\sigma_2\) (20 MPa): Along the tunnel axis (perpendicular to the section) - \(\sigma_3\) ( MPa): Vertical direction (in the section plane) --- ## **(b) Mohr Circle in Cross-Section Plane** ### **Step 1: State-of-Stress in Cross-Section Plane** In the cross-section plane (**ignore the tunnel axis direction**): - \(\sigma_x = 100\) MPa (horizontal) - \(\sigma_y = \) MPa (vertical) - Shear stress \(\tau_{xy} = \) ### **Step 2: Draw Mohr’s Circle** - The center of the circle, \(C\), is at: \[ C = \frac{\sigma_x + \sigma_y}{2} = \frac{100 + }{2} = 50 \ \text{MPa} \] - The radius, \(R\), is: \[ R = \frac{\sigma_x - \sigma_y}{2} = \frac{100 - }{2} = 50 \ \text{MPa} \] ### **Step 3: Mohr Circle Endpoints** - Left endpoint: \((\sigma_y, ) = (, )\) - Right endpoint: \((\sigma_x, ) = (100, )\) ### **Step 4: Sketch** ``` σ ^ | 100 MPa | o | / | / 50 MPa |------o------ (center at 50 MPa) | \ | \ MPa |-----------o +----------------------> ``` - **Horizontal axis:** Normal stress (\(\sigma\)), in MPa - **Vertical axis:** Shear stress (\(\tau\)), in MPa --- ## **Final Answers** ### **(a) Principal Stresses** - **Magnitudes:** \(\boxed{100 \text{ MPa},\ 20 \text{ MPa},\ \text{ MPa}}\) - **Orientations:** - 100 MPa: Horizontal (cross-section plane) - 20 MPa: Along tunnel axis (out of plane) - MPa: Vertical (cross-section plane) ### **(b) Mohr Diagram** - The Mohr circle has: - **Center:** 50 MPa on the \(\sigma\) axis - **Radius:** 50 MPa - **Endpoints:** MPa and 100 MPa on the \(\sigma\) axis - **No shear stress (\(\tau = \))** Let me know if you want a hand-drawn or software-generated Mohr Circle!