Question

A cylindrical billet of diameter 100 mm and length 100 mm is directly extruded into an L-section. The extrusion pressure is given by \[ p = K_s\, \sigma_m \left[0.8 + 1.5\ln(r) + \frac{2l}{d_0}\right] \] where \( \sigma_m = 50\ \text{MPa} \), \( K_s = 1.05 \), \( d_0 = 100\ \text{mm} \). Find the maximum force at the start of extrusion (in kN).

Hint:

Extrusion force calculations require accurate area ratio computation and correct use of logarithm in the extrusion pressure equation.

Answer 1

Correct Answer: 2426

Step 1: Compute the extrusion ratio.
Initial billet cross-section area: \[ A_0 = \frac{\pi}{4} d_0^2 = \frac{\pi}{4}(100^2) = 7854\ \text{mm}^2 \] Final L-section area: Vertical leg area: \[ A_v = 10 \times 60 = 600\ \text{mm}^2 \] Horizontal leg area: \[ A_h = 50 \times 10 = 500\ \text{mm}^2 \] Total extruded area: \[ A_f = A_v + A_h = 1100\ \text{mm}^2 \] Extrusion ratio: \[ r = \frac{A_0}{A_f} = \frac{7854}{1100} \approx 7.14 \] Step 2: Evaluate extrusion pressure.
At the start of extrusion, the full billet length remains: \[ l = 100\ \text{mm} \] Substitute all values: \[ p = 1.05 \times 50 \left[0.8 + 1.5\ln(7.14) + \frac{2(100)}{100}\right] \] Compute term by term: \[ \ln(7.14) \approx 1.964 \] \[ 1.5\ln(7.14) \approx 2.946 \] Now: \[ 0.8 + 2.946 + 2 = 5.746 \] Thus: \[ p = 1.05 \times 50 \times 5.746 \] \[ p = 52.5 \times 5.746 \approx 301.7\ \text{MPa} \] Step 3: Compute force at the start.
\[ F = p A_0 = 301.7\ \text{MPa} \times 7854\ \text{mm}^2 \] Convert: \[ F = 301.7 \times 7854\ \text{N} \approx 2.427 \times 10^6\ \text{N} \] \[ F \approx 2427\ \text{kN} \] Thus, the maximum required force is: \[ \boxed{2427.0\ \text{kN}} \]