--- title: "Deep drawing blank size, hydroforming, spinning" subject: "Manufacturing and Production Processes" chapter: 6 key: "semester_4/manufacturing_and_production_processes:deep-drawing-hydroforming-spinning-formability" source: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/deep-drawing-hydroforming-spinning-formability site: BCT Engineering --- # Deep drawing blank size, hydroforming, spinning Manufacturing and Production Processes · Chapter 6: Sheet Metal Product Manufacturing Process - Asked in: year not given, 3+3+2 marks - Video: [Drawing & Deep Drawing Operation | Lecture Video](https://www.youtube.com/watch?v=Fw9-CPamF_M) — MechTalk Academy Labs - 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/deep-drawing-hydroforming-spinning-formability - Whole chapter: https://www.bctengineer.com/md/semester_4/manufacturing_and_production_processes/chapter-6 ## Question (a) A cylindrical cup of 50 mm inside diameter and 70 mm height is to be deep drawn from 1 mm thick sheet (UTS 350 MPa). Find the blank diameter, the draw ratio, the number of draws if the limiting draw ratio for the first draw is 2.0, and the force of the first draw. (b) Write short notes on hydroforming and metal spinning. (c) How is the formability of a sheet metal assessed? ## Short answer Blank diameter = √(50²+4×50×70) = 128.5 mm, draw ratio = 2.57 (>2.0) so two draws; first-draw force ≈ π×64.2×1×350×(2.0−0.7) = 91.8 kN. Hydroforming uses pressurized fluid to form sheet/tube; spinning uses roller on rotating mandrel. Formability assessed by tensile n and r, cupping tests, and forming limit diagrams. ## Answer #### (a) Deep drawing calculation Blank diameter from equal surface area of blank and cup (thickness constant, sharp corner): $$ D = \sqrt{d^2 + 4dh} = \sqrt{50^2 + 4\times50\times70} = \sqrt{16\,500} = 128.5\ \text{mm} $$ Draw ratio: $$ DR = \frac{D}{d} = \frac{128.5}{50} = 2.57 $$ This is greater than the limiting draw ratio (LDR) of 2.0, so a single draw is not possible. First draw cup diameter $d_1 = D/2.0 = 64.2$ mm. The second draw (redraw) from 64.2 mm to 50 mm is a reduction of $1 - 50/64.2 = 22\%$, which is within the usual redraw limit of 25-30 percent. So **two draws** are needed. Force of the first draw (approximate formula): $$ F = \pi d_1\, t\,(UTS)\left(\frac{D}{d_1} - 0.7\right) = \pi\times64.2\times1\times350\times(2.0-0.7) = 91.8\ \text{kN} $$ Blank-holder force is about one third of the drawing force, about 30.6 kN, to prevent wrinkling. #### (b) Short notes **Hydroforming:** a sheet or tube is pressed into a die by pressurised fluid (oil or water) instead of a rigid punch. In sheet hydroforming, a rubber diaphragm or fluid pressure forces the blank around the punch. In tube hydroforming, high internal pressure plus axial feeding expands the tube into the die (automobile frames, exhaust parts). It gives uniform thickness, complex shapes, less springback, and good finish, but the cycle time is longer and the equipment is costly. **Spinning:** a flat disc or tube is clamped against a rotating mandrel and a roller or tool forces the metal over it. In *conventional spinning* the thickness stays nearly constant (cones, hemispheres, cookware, reflectors); in *shear spinning* the wall thins according to the sine law $t_f = t_0\sin\alpha$. Tooling is cheap and suits small batches of axisymmetric shapes. #### (c) Formability assessment - **Tensile test:** n-value (stretchability) and anisotropy ratio $r$ (deep drawability; larger $r$ is better). - **Limiting dome height / Erichsen cupping test:** a punch stretches a clamped sheet until fracture; the depth indicates stretchability. - **Limiting draw ratio test (Swift cup test)** for drawability. - **Forming limit diagram (FLD):** plot of major strain against minor strain at which necking begins, obtained from grid-marked specimens; strains in the part must lie below the curve. **Answer:** blank diameter $= 128.5$ mm, draw ratio $= 2.57$, two draws, first-draw force $= 91.8$ kN. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the blank diameter of the cup found? A. By equating blank area to cup surface area: D = sqrt(d^2 + 4dh) B. D = d + 2h C. D = sqrt(d^2 + dh) D. D = d x LDR regardless of height Answer: A. Assuming constant thickness, equal areas give D = sqrt(50^2 + 4x50x70) = 128.5 mm. (Key point: Blank diameter) 2. Why does the cup need two draws? A. Draw ratio 2.57 is below the limiting ratio of 2.0 B. Draw ratio 2.57 exceeds the limiting draw ratio of 2.0 C. The sheet is thicker than 1 mm D. The blank-holder force exceeds the drawing force Answer: B. A single draw is only possible if D/d is within the LDR; 2.57 > 2.0 so a redraw is needed. (Key point: Number of draws) 3. In shear spinning, how does the wall thickness change? A. It stays constant as in conventional spinning B. It increases with the cone angle C. It thins by the sine law, t_f = t_0 sin(alpha) D. It follows tf = t0 / sin(alpha) Answer: C. Shear spinning deliberately thins the wall by the sine law, while conventional spinning keeps thickness nearly constant. (Key point: Spinning) 4. What does the forming limit diagram (FLD) show? A. The cup height reached in the Erichsen test B. The n-value against the r-value C. The blank-holder force against punch travel D. The major and minor strains at which necking begins; part strains must lie below the curve Answer: D. The FLD plots major against minor strain at the onset of necking, so safe forming keeps strains below the curve. 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