--- title: "Cold upsetting force with friction" subject: "Manufacturing and Production Processes" chapter: 5 key: "semester_4/manufacturing_and_production_processes:open-die-forging-friction-numerical" source: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/open-die-forging-friction-numerical site: BCT Engineering --- # Cold upsetting force with friction Manufacturing and Production Processes · Chapter 5: Bulk Deformation Process - Asked in: year not given, 6 marks - Video: [Numerical Problem of Forging](https://www.youtube.com/watch?v=L212eomQtT8) — Somnath Chattopadhyaya - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/open-die-forging-friction-numerical - Whole chapter: https://www.bctengineer.com/md/semester_4/manufacturing_and_production_processes/chapter-5 ## Question A solid cylindrical billet of 40 mm diameter and 60 mm height is cold upset by open-die forging to a height of 30 mm. The flow curve of the metal is $\sigma = 500\varepsilon^{0.25}$ MPa. The coefficient of friction at the die interface is 0.2. Determine (a) the final diameter, (b) the true strain, (c) the force at the end of the stroke including the effect of friction, and (d) the ideal work of deformation. Use the forging shape factor $K_f = 1 + 0.4\mu D/h$. ## Short answer Final diameter = 40√(60/30) = 56.6 mm, true strain = ln(60/30) = 0.693. Flow stress = 500(0.693)^0.25 = 456.2 MPa, K_f = 1+0.4·0.2·56.57/30 = 1.151, force = 1.151×456.2×2513 = 1.32 MN. Ideal work = (456.2/1.25)×75398×0.693 = 19.1 kJ. ## Answer Open-die forging (upsetting) reduces the height and increases the cross-section; volume is constant. #### (a) Final diameter $$ \frac{\pi}{4}d_0^2 h_0 = \frac{\pi}{4}d^2 h \quad\Rightarrow\quad d = d_0\sqrt{\frac{h_0}{h}} = 40\sqrt{\frac{60}{30}} = 56.57\ \text{mm} $$ #### (b) True strain $$ \varepsilon = \ln\frac{h_0}{h} = \ln 2 = 0.693 $$ #### (c) Force at end of stroke Flow stress at the final strain: $$ Y_f = K\varepsilon^{n} = 500\,(0.693)^{0.25} = 456.2\ \text{MPa} $$ Final contact area: $$ A = \frac{\pi}{4}(56.57)^2 = 2513\ \text{mm}^2 $$ Shape factor for friction: $$ K_f = 1 + \frac{0.4\,\mu\,D}{h} = 1 + \frac{0.4\times0.2\times56.57}{30} = 1.151 $$ Force: $$ F = K_f\,Y_f\,A = 1.151\times 456.2\times 2513 = 1.32\times10^{6}\ \text{N} $$ Without friction the force would be $456.2\times2513 = 1.147$ MN. Friction therefore adds about 15 percent. The average die pressure is $K_f Y_f = 525$ MPa. #### (d) Ideal work of deformation Initial volume: $V = \frac{\pi}{4}(40)^2(60) = 75\,398\ \text{mm}^3$. Average flow stress: $\bar{Y}_f = \dfrac{K\varepsilon^{n}}{1+n} = \dfrac{456.2}{1.25} = 365.0$ MPa. $$ W = \bar{Y}_f\, V\,\varepsilon = 365.0\times 75\,398\times 0.693 = 1.907\times10^{7}\ \text{N\,mm} = 19.1\ \text{kJ} $$ (Actual work is larger because of friction and redundant work.) **Answer:** final diameter $= 56.6$ mm, true strain $= 0.693$, forging force $= 1.32$ MN, ideal work $= 19.1$ kJ. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A 40 mm dia, 60 mm high billet is upset to 30 mm. The final diameter follows from A. Constant diameter B. Constant volume: d = d0 sqrt(h0/h) = 56.6 mm C. d = d0 x h0/h = 80 mm D. d = d0 + 30 mm Answer: B. Volume is constant, so d = 40 sqrt(2) = 56.57 mm. (Key point: Final diameter) 2. True strain in upsetting from 60 to 30 mm is A. 0.5 B. 0.693 C. 1.0 D. 2.0 Answer: B. epsilon = ln(h0/h) = ln 2 = 0.693. (Key point: True strain) 3. With K_f = 1 + 0.4 mu D/h, what does K_f do? A. Reduces force by friction B. Multiplies the frictionless force to account for friction, here about 1.15 C. Gives the ideal work D. Gives the true strain Answer: B. F = K_f Y_f A; with K_f = 1.151, friction adds about 15% to the force. (Key point: Friction shape factor) 4. Ideal work uses which flow stress? A. Final flow stress K epsilon^n B. Average flow stress K epsilon^n/(1+n) C. Yield stress only D. Flow stress at zero strain Answer: B. W = average Y_f x V x epsilon, with the average Y_f = 456.2/1.25 = 365 MPa, since the metal hardens during deformation. 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