# Step-by-Step Solution: Calculating Torque in Open-Ended Forging **Given:** - Material: 304 stainless steel - Initial diameter (\(D_\)): 8 in - Initial height (\(h_\)): 4 in - Height reduction: 50% (\(h_1 = .5 h_\)) - Coefficient of friction (\(\mu\)): .3 We are to calculate the **torque required** during the forging operation. --- ## 1. **Calculate Final Height** \[ h_1 = .5 \times h_ = .5 \times 4\,\text{in} = 2\,\text{in} \] --- ## 2. **Volume Conservation (Incompressibility)** Volume before = Volume after \[ \frac{\pi}{4} D_^2 h_ = \frac{\pi}{4} D_1^2 h_1 \] \[ D_^2 h_ = D_1^2 h_1 \] \[ D_1^2 = \frac{D_^2 h_}{h_1} = \frac{8^2 \times 4}{2} = \frac{64 \times 4}{2} = \frac{256}{2} = 128 \] \[ D_1 = \sqrt{128} \approx 11.31\,\text{in} \] --- ## 3. **Mean Diameter Calculation** \[ D_m = \frac{D_ + D_1}{2} = \frac{8 + 11.31}{2} = 9.655\,\text{in} \] --- ## 4. **Mean Contact Pressure (\(p_m\))** For forging: \[ p_m = k \left[1 + \frac{\mu D_m}{2 h_m}\right] \] - \(k\) = flow stress (assume average value for 304 SS at room temp: \(\approx 45,000\,\text{psi}\)) - \(h_m = \frac{h_ + h_1}{2} = \frac{4 + 2}{2} = 3\,\text{in}\) ### **Substitute values:** \[ p_m = 45,000 \left[1 + \frac{.3 \times 9.655}{2 \times 3}\right] \] \[ = 45,000 \left[1 + \frac{2.8965}{6}\right] = 45,000 [1 + .4827] = 45,000 \times 1.4827 \approx 66,721\,\text{psi} \] --- ## 5. **Contact Area Calculation** Average area during forging: \[ A = \frac{\pi}{4} \left(D_^2 + D_1^2\right)/2 \] But for torque (friction at interface), use **average area**: \[ A_m = \frac{\pi}{4} D_m^2 \] \[ A_m = \frac{\pi}{4} \times (9.655)^2 = \frac{\pi}{4} \times 93.25 \approx 73.25\,\text{in}^2 \] --- ## 6. **Friction Force** \[ F_f = \mu \times p_m \times A_m \] \[ F_f = .3 \times 66,721 \times 73.25 \approx .3 \times 4,887,714 \approx 1,466,314\,\text{lbf} \] --- ## 7. **Torque Calculation** Assume friction force acts at mean radius: \[ R_m = \frac{D_m}{2} = \frac{9.655}{2} = 4.83\,\text{in} \] \[ T = F_f \times R_m \] \[ T = 1,466,314 \times 4.83 \approx 7,089,293\,\text{in-lbf} \] --- ## 8. **Final Answer** **The required torque is:** \[ \boxed{7,\!090,\!000~\text{in-lbf}} \] (Rounded to the nearest ten thousand in-lbf.) --- ## **Summary Table** | Quantity | Value | |---------------|------------------| | \(D_\) | 8 in | | \(D_1\) | 11.31 in | | \(D_m\) | 9.655 in | | \(h_\) | 4 in | | \(h_1\) | 2 in | | \(h_m\) | 3 in | | \(k\) | 45,000 psi | | \(\mu\) | .3 | | \(A_m\) | 73.25 in² | | \(p_m\) | 66,721 psi | | \(F_f\) | 1,466,314 lbf | | \(R_m\) | 4.83 in | | \(T\) | 7,090,000 in-lbf | --- ### **Assumptions** - Average flow stress for 304 SS at room temperature is 45,000 psi. - Friction force acts at mean diameter. - All units in inch–pound–second (IPS) system. --- **If you need the torque in ft-lbf:** \[ \frac{7,090,000}{12} \approx 590,800~\text{ft-lbf} \]