Let's tackle the problem step by step for each part as requested. --- ## a) Calculate \( P_U / 2A \) for both fields ### **Step 1: Analyze the Upper-Bound Fields** - **Given:** 2:1 reduction by extrusion (50% reduction), plane-strain, dead-metal zones (ABC and EFG). - **Fields:** Two different velocity fields are shown for the same reduction. - **Required:** Calculate the upper-bound pressure per unit area (\( P_U/2A \)) for both fields. #### **Explanation:** The upper-bound pressure can be calculated using the upper-bound theorem of plasticity. The work rate of external forces must equal the internal energy dissipation for admissible velocity fields. --- ### **Step 2: Apply the Upper-Bound Theorem** #### **Field I (Left):** - The work rate per unit area: \( \frac{P_U}{2A} = \tau \cdot \text{Shear Length} \cdot \text{Shear Rate} \) - For a 2:1 reduction, angle bisectors split the deformation zone. - The shear zones are at 45° to the axis (for simplicity). - The shear length = height \( h \). - Shear rate for 45°: \( \dot{\gamma} = \frac{V_}{h} \), where \( V_ \) is the incoming velocity. #### **Explanation:** By considering the geometry and the direction of deformation, we estimate the energy dissipation based on the shear planes and velocities. #### **Field II (Right):** - Has a different configuration but similar logic. The key is to sum the energy dissipations across all shear planes. --- #### **Final Answer (for part a):** - **Field I:** \( \frac{P_U}{2A} = \tau \sqrt{2} \) *(assuming idealized 45° planes and unit heights)* - **Field II:** \( \frac{P_U}{2A} = 2\tau \) *(as the total shear length is longer due to the second shear plane)* *(Note: The exact value depends on the angles, but for classic 2:1 reduction, these are typical results. Please use the exact geometry if more precision is needed.)* --- ## b) Determine the deformation efficiency, \( \eta \), for both cases. ### **Step 1: Define Deformation Efficiency** - **Deformation efficiency:** \( \eta = \frac{\text{Ideal Work}}{\text{Actual Work}} \) - **Ideal work** for extrusion (for a 2:1 reduction): \( W_{ideal} = \sigma_ \ln(2) \) (for perfectly plastic material, \( \sigma_ \) is yield stress) #### **Explanation:** Deformation efficiency shows how much of the work is actually useful (ideal) versus the total expended (actual). --- ### **Step 2: Calculate \( \eta \) for Both Fields** - **Actual work** = \( P_U/2A \) (from part a, replacing \( \tau = \frac{\sigma_}{2} \) for von Mises) - For both fields, substitute \( \tau \) and ideal work: - **Field I:** \( \eta_1 = \frac{\sigma_ \ln 2}{\sigma_ \sqrt{2}} = \frac{\ln 2}{\sqrt{2}} \approx .49 \) - **Field II:** \( \eta_2 = \frac{\sigma_ \ln 2}{2\tau} = \frac{\ln 2}{1} \) (since \( 2\tau = \sigma_ \)), so \( \eta_2 = \ln 2 \approx .693 \) #### **Explanation:** This step compares the minimum (ideal) work needed to the calculated upper-bound work for each field. --- #### **Final Answer (for part b):** - **Field I:** \( \eta_1 \approx .49 \) - **Field II:** \( \eta_2 \approx .693 \) --- ## c) What is the absolute velocity of a particle in triangle AGH? ### **Step 1: Locate Triangle AGH and Analyze Velocities** - **Triangle AGH** is in the central deformation zone. - Velocity components are based on the imposed extrusion speed (\( V_ \)) and the field geometry. #### **Explanation:** Within the deformation zone, particles have velocities derived from the upper-bound field's geometry. --- ### **Step 2: Calculate Absolute Velocity** - For a 2:1 reduction, the output velocity is half the input velocity due to volume constancy. - In triangle AGH, the velocity will be the mean of input and output velocities across the shear plane. - **If \( V_ \) is the input velocity:** **Absolute velocity in AGH:** \( V_{AGH} = \frac{V_ + V_1}{2} = \frac{V_ + V_/2}{2} = \frac{3V_}{4} \) #### **Explanation:** Due to the symmetry and construction of the velocity field, the velocity in the central triangle is the average of the velocities of the incoming and outgoing material. --- #### **Final Answer (for part c):** - **Absolute velocity in triangle AGH:** \( V_{AGH} = \frac{3}{4} V_ \) --- ## **Summary Table** | Part | Field I | Field II | |-------|----------------|---------------| | a | \( \tau \sqrt{2} \) | \( 2\tau \) | | b | \( \approx .49 \) | \( \approx .693 \) | | c | \( \frac{3}{4} V_ \) for both fields | --- **If you need detailed calculations with actual geometry or angles from the diagram, please provide clear values or dimensions.**