Let's break down the question and solve it step by step. ### **Given Data:** - Diameter of disk (D) = 11 cm = 110 mm - Thickness of sheet (t) = 0.4 cm = 4 mm - Shear strength (τ) = 350 MPa = 350 N/mm² - Clearance allowance (c) = 0.045 --- ### **1. Diameter of Punch (d_p):** The diameter of the punch (d_p) is given by: \[ d_p = D - 2c \cdot t \] Where: - \( D \) = Diameter of blank (in mm) - \( c \) = Clearance allowance - \( t \) = Thickness (in mm) Plug in the values: - \( D = 110 \) mm - \( c = 0.045 \) - \( t = 4 \) mm \[ d_p = 110 - 2 \times 0.045 \times 4 \] \[ d_p = 110 - 2 \times 0.18 \] \[ d_p = 110 - 0.36 \] \[ d_p = 109.64 \text{ mm} \] --- ### **2. Required Blanking Force (F):** Blanking force is calculated as: \[ F = \text{Perimeter} \times t \times \tau \] \[ F = \pi D t \tau \] Where: - \( D = 110 \) mm - \( t = 4 \) mm - \( \tau = 350 \) N/mm² \[ F = \pi \times 110 \times 4 \times 350 \] \[ F = 3.1416 \times 110 \times 4 \times 350 \] \[ F = 3.1416 \times 110 \times 1400 \] \[ F = 3.1416 \times 154000 \] \[ F \approx 483,805 \text{ N} \] \[ F \approx 483.80 \text{ kN} \] --- ### **Summary:** - **Diameter of punch:** **109.64 mm** - **Required blanking force:** **483.80 kN** --- ### **Correct Answer (from options):** - **109.64 mm and 483.80 kN**