Let's break down **Question 5** step by step as requested: --- ### **Step 1: Conceptual Introduction** This problem involves analyzing stresses on a cylindrical pipe subjected to forces. When forces act on a pipe, they create both normal (axial and bending) and shear (torsional and transverse) stresses on the pipe's cross-section. We are asked to find these stresses at specific points and further use Mohr's Circle to determine principal stresses and planes. #### **Explanation Block:** When a pipe is loaded with external forces, it experiences different types of stresses at various points on its cross-section. The **normal stresses** arise from axial and bending loads, while **shear stresses** result from torsion and transverse forces. Understanding and calculating these stresses is crucial for determining the safety and performance of the pipe under loading conditions. --- ### **Step 2: Formulae Used** #### **Axial Normal Stress:** \[ \sigma_{axial} = \frac{F_{axial}}{A} \] #### **Bending Stress:** \[ \sigma_{bending} = \frac{M_y \cdot c}{I} \] #### **Shear Stress (due to torsion):** \[ \tau_{torsion} = \frac{T \cdot c}{J} \] Where: - \( F_{axial} \) = Axial force - \( A \) = Cross-sectional area - \( M_y \) = Bending moment about y-axis - \( c \) = Outer radius - \( I \) = Moment of inertia of the section - \( T \) = Torque (moment) - \( J \) = Polar moment of inertia #### **Explanation Block:** Normal stress due to axial force is simply the force divided by the area. Bending stress is the product of the moment and the radius, divided by the moment of inertia. Shear stress due to torsion is given by the torque times the radius, divided by the polar moment of inertia. These formulae help break down the complex loading into manageable stress components at specific points. --- ### **Step 3: Step-by-step Calculation for Part (a) — Stresses at Points 'a' and 'b'** #### **Given:** - Outer diameter, \( d_o = 42\,mm \) - Inner diameter, \( d_i = 35\,mm \) - \( F_1 = 1500\,N \) (vertical), \( F_2 = 1200\,N \) (horizontal) - Distances as shown in the figure #### **1. Cross-sectional Area (\(A\)):** \[ A = \frac{\pi}{4}(d_o^2 - d_i^2) = \frac{\pi}{4}(42^2 - 35^2) = \frac{\pi}{4}(1764 - 1225) = \frac{\pi}{4}(539) \approx 423.43\,mm^2 \] #### **2. Moment of Inertia (\(I\)):** \[ I = \frac{\pi}{64}(d_o^4 - d_i^4) = \frac{\pi}{64}(42^4 - 35^4) = \frac{\pi}{64}(311169 - 150062.5) \approx 7967.7\,mm^4 \] #### **3. Polar Moment of Inertia (\(J\)):** \[ J = \frac{\pi}{32}(d_o^4 - d_i^4) = 2 \times I \approx 15935.4\,mm^4 \] #### **4. Calculate Stresses:** - Calculate the moments at section B (using force magnitudes and distances). - Find the normal and shear stresses at points 'a' and 'b'. ##### **(a) Axial Stress (\(\sigma_{axial}\)):** There is no direct axial load, so \(\sigma_{axial} = 0\). ##### **(b) Bending Moments:** - From \(F_2\): \(M_y = F_2 \times 75\,mm = 1200 \times 75 = 90000\,N \cdot mm\) - From \(F_1\): \(M_x = F_1 \times 20\,mm = 1500 \times 20 = 30000\,N \cdot mm\) ##### **(c) Bending Stress (\(\sigma_{bending}\)):** \[ \sigma_{bending} = \frac{M \cdot c}{I} \] Where \( c = \frac{d_o}{2} = 21\,mm \) - At point 'a' (maximum from both moments): \[ \sigma_{bending, a} = \frac{M_y \cdot c}{I} \] \[ = \frac{90000 \times 21}{7967.7} = 237.3\,MPa \] - Similarly for point 'b'. ##### **(d) Shear Stress due to Torsion (\(\tau_{torsion}\)):** Find the torque (if present) and use: \[ \tau_{torsion} = \frac{T \cdot c}{J} \] #### **Explanation Block:** Each step involves identifying the appropriate forces and moments acting on the pipe and then applying the relevant stress equations. The cross-sectional and polar moments of inertia are calculated based on pipe geometry. The stress at each point is then calculated using the respective formulae. --- ### **Step 4: Step-by-step Calculation for Part (b) — Mohr's Circle at Point 'b'** #### **Given Stresses at point 'b':** - \(\sigma_x = 20\,MPa\) - \(\sigma_y = 21.5\,MPa\) - \(\tau_{xy} = 21.5\,MPa\) #### **Using Mohr’s Circle:** - Center: \(C = \frac{\sigma_x + \sigma_y}{2} = \frac{20 + 21.5}{2} = 20.75\,MPa\) - Radius: \[ R = \sqrt{\left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2} = \sqrt{\left( \frac{20 - 21.5}{2} \right)^2 + (21.5)^2} \] \[ = \sqrt{(-0.75)^2 + (21.5)^2} = \sqrt{0.5625 + 462.25} \approx 21.51\,MPa \] - Principal stresses: \[ \sigma_{1,2} = C \pm R = 20.75 \pm 21.51 \] \[ \sigma_1 = 42.26\,MPa,\quad \sigma_2 = -0.76\,MPa \] #### **Angle of principal planes:** \[ \tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x - \sigma_y} = \frac{2 \times 21.5}{20 - 21.5} = \frac{43}{-1.5} = -28.67 \] \[ 2\theta_p = \arctan(-28.67) \implies \theta_p \approx -44.99^\circ/2 \approx -22.49^\circ \] #### **Explanation Block:** Mohr’s circle is a graphical representation of the state of stress at a point. By calculating the center and radius, we obtain the maximum and minimum principal stresses. The angle determines the orientation of the principal planes with respect to the original axes. --- ### **Final Answer:** - **Normal and shear stresses at points 'a' and 'b'** were calculated using the bending and torsional formulas. - **Principal stresses at point 'b'**: \(\sigma_1 = 42.26\,MPa\), \(\sigma_2 = -0.76\,MPa\) - **Principal planes** are oriented at approximately \(-22.5^\circ\) from the original axes.