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Upsetting diameter and force calculation

Short answer

Final diameter d₂ = 40√(120/50) = 61.97 mm. True strain ε = ln(120/50) = 0.875. Upsetting force = 100 MPa × (π/4 × 61.97² mm²) ≈ 302 kN. The initial height/diameter ratio is 3, borderline for buckling; use a die or light blows.

A steel billet 40 mm in diameter and 120 mm long is upset hot between flat dies to a height of 50 mm. Assuming volume remains constant and ignoring friction and barrelling, find (a) the final diameter, (b) the true strain, and (c) the upsetting force if the flow stress at the forging temperature is 100 MPa. Comment on whether the billet will buckle during upsetting.

Given: d1=40d_1 = 40 mm, h1=120h_1 = 120 mm, h2=50h_2 = 50 mm, σf=100\sigma_f = 100 MPa.

(a) Final diameter

Volume is constant:

π4d12h1=π4d22h2⇒d2=d1h1h2\frac{\pi}{4}d_1^2 h_1 = \frac{\pi}{4}d_2^2 h_2 \Rightarrow d_2 = d_1\sqrt{\frac{h_1}{h_2}} d2=4012050=40×1.549=61.97 mmd_2 = 40\sqrt{\frac{120}{50}} = 40\times 1.549 = 61.97\ \text{mm}

(b) True strain

ε=ln⁡h1h2=ln⁡12050=0.875\varepsilon = \ln\frac{h_1}{h_2} = \ln\frac{120}{50} = 0.875

(c) Force at the end of the stroke

Final cross-section:

A2=π4(61.97)2=3016 mm2A_2 = \frac{\pi}{4}(61.97)^2 = 3016\ \text{mm}^2 F=σfA2=100×3016=301,600 N≈302 kNF = \sigma_f A_2 = 100\times 3016 = 301{,}600\ \text{N} \approx 302\ \text{kN}

The force is highest at the end of the stroke as the area is largest. With friction the real force would be higher.

Buckling check

h1d1=12040=3\frac{h_1}{d_1} = \frac{120}{40} = 3

This is at the limit of the rule that the unsupported length should not exceed 3 times the diameter in a single blow. The upsetting should be done in a die or by a series of light blows, or the free length should be reduced.

Answer: d₂ = 61.97 mm; true strain = 0.875; F = 302 kN; h/d = 3 is at the safe limit, so use a die or light blows.

✓ Answer checked by Claude Opus 5.5 (high effort)

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