--- title: "Wire drawing stress, force and power" subject: "Manufacturing and Production Processes" chapter: 5 key: "semester_4/manufacturing_and_production_processes:wire-drawing-stress-force-numerical" source: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/wire-drawing-stress-force-numerical site: BCT Engineering --- # Wire drawing stress, force and power Manufacturing and Production Processes · Chapter 5: Bulk Deformation Process - Asked in: year not given, 6 marks - Video: [Manufacturing || Lecture 8 || Wire drawing || Analysis || All formulae || Gate ME ||](https://www.youtube.com/watch?v=828yUxLnA74) — Mechanical Engineering Make Easy - 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/wire-drawing-stress-force-numerical - Whole chapter: https://www.bctengineer.com/md/semester_4/manufacturing_and_production_processes/chapter-5 ## Question A 10 mm diameter wire of a metal with flow curve $\sigma = 700\varepsilon^{0.2}$ MPa is drawn to 8 mm diameter through a die of semi-angle $6^\circ$. The die friction coefficient is 0.1. Using $\sigma_d = \bar{Y}_f\left(1+\mu/\tan\alpha\right)\phi\ln(A_0/A_f)$ with $\phi = 0.88 + 0.12\,D/L_c$, calculate (a) the area reduction, (b) the drawing stress, (c) the drawing force and (d) the power for a drawing speed of 1.5 m/s. Check that the wire will not fail. ## Short answer Area reduction = 1-(8/10)² = 0.36 (36%), true strain = 0.446. Average flow stress = 700·0.446^0.2/1.2 = 496.4 MPa, drawing stress = 496.4×(1+0.1/tan6°)×0.993×0.446 = 429 MPa. Drawing force = 429×50.27 = 21.6 kN, power = 21.6×1.5 = 32.4 kW. ## Answer In **drawing**, the wire is pulled through a die. The drawing force is limited by the strength of the wire after it leaves the die. #### (a) Area reduction and true strain $$ r = 1 - \left(\frac{8}{10}\right)^2 = 0.36 \;(36\%),\qquad \varepsilon = \ln\frac{A_0}{A_f} = \ln\left(\frac{10}{8}\right)^2 = 0.4463 $$ #### (b) Drawing stress Average flow stress: $$ \bar{Y}_f = \frac{K\varepsilon^n}{1+n} = \frac{700\,(0.4463)^{0.2}}{1.2} = \frac{596.0}{1.2} = 496.4\ \text{MPa} $$ Die contact length and mean diameter: $$ L_c = \frac{D_0 - D_f}{2\sin\alpha} = \frac{10-8}{2\sin6^\circ} = 9.567\ \text{mm},\qquad D = \frac{10+8}{2} = 9\ \text{mm} $$ Shape factor for redundant work: $$ \phi = 0.88 + 0.12\,\frac{9}{9.567} = 0.993 $$ Friction term: $1 + \dfrac{0.1}{\tan 6^\circ} = 1 + \dfrac{0.1}{0.1051} = 1.9515$ $$ \sigma_d = 496.4\times1.9515\times0.993\times0.4463 = 429.3\ \text{MPa} $$ #### (c) Drawing force Area of the drawn wire: $A_f = \dfrac{\pi}{4}(8)^2 = 50.27\ \text{mm}^2$ $$ F = \sigma_d A_f = 429.3\times 50.27 = 21\,580\ \text{N} = 21.6\ \text{kN} $$ #### (d) Power $$ P = F\,v = 21.58\times 1.5 = 32.4\ \text{kW} $$ #### Check of fracture The wire leaving the die has been strain hardened to a flow stress of $K\varepsilon^n = 596$ MPa. Since the drawing stress (429 MPa) is lower than 596 MPa, the drawn wire will not yield or break. The ratio $429/596 = 0.72$ leaves a reasonable safety margin, so a somewhat larger reduction in this pass would still be possible. **Answer:** area reduction $= 36\%$, drawing stress $= 429$ MPa, drawing force $= 21.6$ kN, power $= 32.4$ kW; the wire is safe. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Drawing 10 mm wire to 8 mm gives area reduction of A. 20% B. 36% C. 64% D. 44.6% Answer: B. r = 1 - (8/10)^2 = 0.36 (36%); the true strain is ln(A0/Af) = 0.446. (Key point: Area reduction) 2. In the drawing stress formula, the term (1 + mu/tan alpha) accounts for A. Redundant work B. Die friction C. Strain hardening D. Wire speed Answer: B. It is the friction term (1.95 here); phi accounts for redundant work. (Key point: Friction term) 3. Drawing force is found from A. Drawing stress x initial wire area B. Drawing stress x final wire area (21.6 kN) C. Flow stress x die contact area D. Power / speed with initial area Answer: B. F = sigma_d A_f = 429.3 x 50.27 = 21.6 kN. (Key point: Drawing force) 4. Why is the wire safe against breaking? A. The drawing stress (429 MPa) is below the strain-hardened flow stress of the exit wire (596 MPa) B. The drawing force is below the die force C. The speed is low D. The area reduction is below 50% Answer: A. The exit wire can carry up to K epsilon^n = 596 MPa, which exceeds the 429 MPa drawing stress. 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