Question

A workpiece of 30 mm diameter and 40 mm height is compressed between two platens in an open-die forging process. Assume a perfectly plastic material with a flow stress of 300 MPa. The ideal forging load (in kN) at 30% reduction is \(\underline{\hspace{2cm}}\) (round off to nearest integer).

Hint:

Forging load increases with true strain; ideal load ≠ $\sigma_f A$. Always apply $\ln(h_0/h)$ correction.

Answer 1

Correct Answer: 300 - 306

To find the ideal forging load for an open-die forging process, we start by understanding the key parameters:

• Initial diameter (\( D \)) = 30 mm
• Initial height (\( H \)) = 40 mm
• Reduction = 30%
• Flow stress (\( \sigma \)) = 300 MPa

A 30% reduction in height means the final height (\( H_f \)) is 70% of the initial height:

\( H_f = H \times (1 - 0.3) = 40 \times 0.7 = 28 \) mm

In open-die forging, the ideal forging load (\( F \)) is calculated using the formula:

\( F = \sigma \times A \)

where \( A \) is the instantaneous cross-sectional area of the workpiece. Assuming no barreling, the volume before and after deformation is constant. Initial volume (\( V \)):

\( V = \frac{\pi D^2}{4} \times H \)

The same volume post forging at height \( H_f \):

\( V = \frac{\pi D_f^2}{4} \times 28 \)

Equating both expressions,

\(\frac{\pi (30)^2}{4} \times 40 = \frac{\pi D_f^2}{4} \times 28\)

Simplifying,

\( D_f^2 = \frac{(30)^2 \times 40}{28} = \frac{900 \times 40}{28} = 1285.71\)

\( D_f = \sqrt{1285.71} \approx 35.85 \) mm

Thus the cross-sectional area (\( A \)) at 30% reduction is:

\( A = \frac{\pi (35.85)^2}{4} \approx 1009.65 \) mm²

Convert the area to m² to match \( \sigma \)'s units:

\( A = 1009.65 \times 10^{-6} \) m²

Calculate the forging load:

\( F = 300 \times 1009.65 \times 10^{-6} \approx 302.9 \) kN

Rounded to nearest integer, \( F = 303 \) kN.