Let’s approach this step by step to ensure clarity and completeness in the solution. --- ### **Given Data:** - **Hole diameter to be punched:** 100 mm - **Plate thickness:** 5.6 mm - **Penetration for cutting completion:** 40% of plate thickness - **Press capacity:** 30 tonnes = 30,000 kg = 294,300 N (since 1 tonne = 9,810 N) - **Material:** Steel (assume ultimate shear strength, τ_u ≈ 400 N/mm² for calculation unless otherwise specified) - **Normal clearance to be provided** --- ### **Step 1: Determination of Punch and Die Diameters** #### **a) Normal Clearance (C):** Normal clearance (C) is typically about 5% of the plate thickness for steel. \[ C = 0.05 \times \text{thickness} = 0.05 \times 5.6 \text{ mm} = 0.28 \text{ mm} \] #### **b) Punch Diameter (d_p):** The punch diameter is equal to the required hole diameter. \[ d_p = 100 \text{ mm} \] #### **c) Die Diameter (d_d):** The die diameter is the punch diameter plus twice the clearance. \[ d_d = d_p + 2C = 100 + 2 \times 0.28 = 100.56 \text{ mm} \] --- ### **Step 2: Calculation of Shear Force Needed (Without Shear Angle)** The maximum punching force required (F) can be calculated by: \[ F = \text{Perimeter} \times \text{Plate Thickness} \times \text{Shear Strength} \] \[ \text{Perimeter} = \pi \times d_p = \pi \times 100 = 314.16 \text{ mm} \] \[ F = 314.16 \times 5.6 \times 400 = 703,564.8 \text{ N} \] #### **Penetration Consideration:** Cutting is completed at 40% penetration. Thus, the actual force required is: \[ \text{Force at 40% penetration} = 0.4 \times F = 0.4 \times 703,564.8 = 281,425.92 \text{ N} \] --- ### **Step 3: Compare with Press Capacity** The press can exert up to 294,300 N. The calculated force (281,425.92 N) is within the press capacity. However, to reduce force further and ensure safety, shear angle can be provided. --- ### **Step 4: Calculation of Suitable Shear Angle (α)** When a shear angle is provided, the force required is reduced. The reduced force (F_shear) is given by: \[ F_{shear} = \frac{F \times t}{t + d_p \tan \alpha} \] Let the required force be equal to the press capacity: \[ F_{shear} = 294,300 \text{ N} \] Substitute values: \[ 294,300 = \frac{703,564.8 \times 5.6}{5.6 + 100 \tan \alpha} \] Solve for \(\tan \alpha\): \[ 294,300 \times (5.6 + 100 \tan \alpha) = 703,564.8 \times 5.6 \] \[ 1,648,080 + 29,430,000 \tan \alpha = 3,939,962.88 \] \[ 29,430,000 \tan \alpha = 3,939,962.88 - 1,648,080 = 2,291,882.88 \] \[ \tan \alpha = \frac{2,291,882.88}{29,430,000} = 0.0779 \] \[ \alpha = \arctan(0.0779) \approx 4.46^\circ \] --- ### **Step 5: Final Answers and Explanation** - **Diameter of punch:** 100 mm (equal to required hole size) - **Diameter of die:** 100.56 mm (punch diameter + 2 × clearance) - **Shear angle on punch:** ≈ 4.46° (to keep force within 30-tonne press) **Explanation:** By providing a suitable clearance and a shear angle of about 4.46°, the punching force is sufficiently reduced to fall within the safe operating limit of the 30-tonne press. --- ### **Final Short Answer** A punch diameter of 100 mm, a die diameter of 100.56 mm, and a shear angle of approximately 4.46° on the punch are suitable for safely punching the required hole in the steel plate using a 30-tonne press.