Let's solve the problem step by step: --- ### **Step 1: Find the Cross-sectional Area at A** The pipe has an **outer diameter** of \( 20 \) mm (since section a-a is shown as a circle of diameter \( 20 \) mm). \[ A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (20 \text{ mm})^2 = \frac{\pi}{4} (400) = 314.16 \text{ mm}^2 \] Convert to \(\text{m}^2\): \[ A = 314.16 \times 10^{-6} \text{ m}^2 = 3.1416 \times 10^{-4} \text{ m}^2 \] --- ### **Step 2: Axial Stress at Point A** Given: \(\sigma_{z,A} = 47.7 \) MPa Axial stress: \[ \sigma_{z,A} = \frac{F_z}{A} \] Where \( F_z \) is the force component along the axis of the pipe at A (which is the vertical member). #### **Resolve P into Components at A** - The force \( P \) is applied at the end of the inclined pipe. - The \( x \)-component: \( \frac{3}{2}P \) - The \( z \)-component: \( P \) The vertical pipe at A is loaded **axially** by the \( z \)-component of the force at the end, which is \( P \). So, \[ \sigma_{z,A} = \frac{P}{A} \] Given \(\sigma_{z,A} = 47.7 \) MPa, \[ 47.7 \times 10^6 = \frac{P}{3.1416 \times 10^{-4}} \] \[ P = 47.7 \times 10^6 \times 3.1416 \times 10^{-4} \] \[ P = 15,000 \text{ N} \quad \boxed{(15.0 \text{ kN})} \] --- ### **Step 3: Shear Stress at Point A (Due to \( P \) and \( \frac{3}{2}P \))** The horizontal force (\( \frac{3}{2}P \)) at the end will cause a **moment** about A, creating a **torsional shear stress** in the pipe. #### **Calculate the Torsional Moment** The force \( \frac{3}{2}P \) acts at a distance of \( 400 \) mm from the axis at A: \[ T = \frac{3}{2}P \times 400 \text{ mm} \] \[ T = \frac{3}{2} \times 15000 \times 0.4 = 9000 \text{ Nm} \] #### **Shear Stress Due to Torsion** For a solid shaft: \[ \tau_{max} = \frac{Tc}{J} \] Where: - \( T \) = torque - \( c \) = outer radius = \( 10 \) mm = \( 0.01 \) m - \( J \) = polar moment of inertia for solid shaft: \[ J = \frac{\pi}{32}d^4 = \frac{\pi}{32} (0.02)^4 = \frac{\pi}{32} \times 1.6 \times 10^{-7} = 1.5708 \times 10^{-8} \text{ m}^4 \] So, \[ \tau_{max} = \frac{9000 \times 0.01}{1.5708 \times 10^{-8}} = \frac{90}{1.5708 \times 10^{-8}} = 5.73 \times 10^9 \text{ Pa} = 5,730 \text{ MPa} \] #### **Shear Stress Due to Shear Force** The vertical reaction at the base will have a horizontal component due to the \( P \) and \( \frac{3}{2}P \) forces, but the primary stress state at A is **axial (normal)** and **torsional (shear)**. --- ### **Step 4: State of Stress at A** At point A, the differential element will have: - **Normal stress in the z-direction:** \(\sigma_{z,A} = 47.7 \) MPa (tension) - **Shear stress due to torsion:** \( \tau_{A} = 5,730 \) MPa (acting on the outer surface, max value) --- ### **Step 5: Sketch the Stress Element** **On a small element at A:** - **Vertical face (perpendicular to z):** - Normal stress, \(\sigma_{z,A} = 47.7 \) MPa (tension, up) - **Shear stress tangent to the pipe's circumference:** - \(\tau_{A} = 5,730 \) MPa (direction depends on the sense of the torque) --- ## **Final Answers** **(a) Magnitude of \( P \):** \[ \boxed{P = 15.0 \text{ kN}} \] **(b) State of stress at A:** - Normal stress: \(\boxed{47.7 \text{ MPa}}\) (tension, z-direction) - Maximum shear stress (due to torsion): \(\boxed{5,730 \text{ MPa}}\) (tangential, at the outer surface) **(c) Sketch:** Draw a small square element at the outer surface of the pipe at A: - Upward (z) arrow: \( 47.7 \) MPa (normal) - Tangential arrow: \( 5,730 \) MPa (shear, direction per right-hand rule) If you need the sketch, let me know! --- **Let me know if you need a diagram of the stress element or clarification on any step!**