22-Mec-A4 Design and Manufacture of Machine Elements · December 2017

Question 1 of 6: Sheet-Metal Bending, Springback and Deep-Drawing Radii

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 1: Sheet-Metal Bending, Springback and Deep-Drawing Radii (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Feasibility of substituting 1045 steel

Given. Sheet thickness $T = 5\ \text{mm}$; existing material AISI 1015 steel, bent on a sharp edge (bend radius $R = 0$); proposed material AISI 1045 steel with tensile reduction of area $q_{1045} = 0.45$, i.e. $r = 45\ \%$.

Find. Whether 1045 steel can be bent to the same zero-radius (sharp) corner without cracking, and if not, what bend radius it does require.

Approach. Compare the minimum bend radius that each material's ductility permits, using the empirical Kalpakjian relation between minimum bend radius and tensile reduction of area, and test it against the zero radius the existing part demands.

1. State the ductility criterion for bending. Bending strains the outer fibre in tension, so the limiting bend radius is set by tensile ductility, measured by the reduction of area $r$ (in percent): $$\frac{R_{\min}}{T} = \frac{50}{r} - 1$$ The relation says that a material with $r = 50\ \%$ can just be folded flat on itself ($R_{\min}/T = 0$), and anything less ductile needs a finite radius.
2. Evaluate the incumbent material, 1015 steel. Annealed low-carbon 1015 has a reduction of area of roughly $r \approx 70\ \%$, so $$\frac{R_{\min}}{T} = \frac{50}{70} - 1 = -0.286$$ The negative result simply means the criterion is satisfied with margin: 1015 can be bent flat on a sharp edge, which is exactly what the existing process does. The current process is therefore metallurgically sound — the service failures are a strength problem, not a forming problem.
3. Evaluate the proposed material, 1045 steel. Substituting the value given in the question, $r = 45\ \%$: $$\frac{R_{\min}}{T} = \frac{50}{45} - 1 = 1.111 - 1 = 0.111$$ Because $r$ has fallen below the 50 % threshold, the required radius is now positive rather than negative. Multiplying by the sheet thickness, $$R_{\min} = 0.111 \times 5\ \text{mm}$$ $$\boxed{R_{\min} \approx 0.56\ \text{mm}}$$
4. Make the engineering judgment. The proposal asks for $R = 0$ but the material demands $R \ge 0.56\ \text{mm}$. The requirement is therefore not satisfied, and the substitution is not feasible as a drop-in change: bent on the existing sharp-edge tooling, 1045 sheet will crack on the outer fibre of the bend. The shortfall is a full 5 percentage points of reduction of area (45 % available against the 50 % a sharp bend needs).
5. State what would make the change work. Three routes recover the substitution, in ascending order of cost: (i) put a radius on the punch nose — a nominal 1 mm radius, comfortably above the 0.56 mm minimum, costs nothing in cycle time and only changes the die insert; (ii) bend the 1045 in the annealed or spheroidised condition, where $r$ rises well above 50 %, and heat-treat to the required strength afterwards; or (iii) keep the sharp bend and orient the blank so the bend line runs across the rolling direction, which raises the effective ductility of the outer fibre, while accepting a modest radius. Warm bending is a fourth option but is rarely justified for a 5 mm carbon-steel part.

(b) The two properties governing springback

Springback is the elastic recovery that occurs when the bending moment is released, so it is fixed by the ratio of the elastic strain the material stores to the plastic strain it retains. The two controlling properties are therefore the yield strength $S_y$ and the modulus of elasticity $E$, and they act in opposite senses. The quantitative statement is the Kalpakjian springback expression, in which the recovered curvature depends on the group $S_y/E$ only:

Springback increases with increasing yield strength. A stronger material is carrying a higher stress at every point of the section when the tool is at the bottom of its stroke, so it has stored more elastic strain energy to release; high-strength steels and heat-treated aluminium alloys are notoriously springback-prone for exactly this reason.

Springback decreases with increasing elastic modulus. A stiffer material reaches the same stress at a smaller elastic strain, so there is less elastic strain to recover. This is why steel ($E \approx 207\ \text{GPa}$) springs back roughly a third as much as an aluminium alloy ($E \approx 70\ \text{GPa}$) formed to the same shape and relative strength.

The practical consequence for the part in (a) is worth noting: the very substitution proposed to cure the service failures — going from 1015 to the much stronger 1045 — will simultaneously make the bend angle harder to hold, because it raises $S_y$ while leaving $E$ essentially unchanged. Overbending, bottoming or coining of the die will be needed on top of the radius change.

(c) Critique of "as large as possible" punch and die radii in deep drawing

The suggestion is half right and dangerous as stated. It correctly identifies that small radii are harmful, but it draws the wrong conclusion from that, because the failure mode simply changes character once the radii become large. Deep drawing is bounded on both sides, and the useful design window is a band, not an extreme.

Why small radii are indeed bad. As the sheet is pulled over a sharp die throat it is bent and then unbent through a tight curvature, which consumes ductility, raises the drawing force sharply, and concentrates the wall stress at the punch nose. Because the punch-nose region is the part of the cup that has been work-hardened least — it never passed through the flange — it is the weakest section, and it is where the cup tears. Small radii also gall and wear rapidly. In this much the suggestion is correct.

Why "as large as possible" fails. The blank-holder plate, the die face and the punch face together are what keep the flange flat. Every millimetre added to the die throat radius, and every millimetre added to the punch nose radius, lengthens the span of sheet that is unsupported on both faces, as the right-hand sketch shows. Compressive hoop stress in the flange then has nothing to react against, and the flange buckles out of plane — wrinkling, the second of the two classic deep-drawing failures. On the punch side the same effect produces puckering in the cup base. An oversized punch radius additionally shortens the straight side wall for a given punch travel, so the part geometry itself is compromised.

The production-rate argument does not survive either. Production rate in deep drawing is governed by how much reduction can be achieved per stroke, that is by the limiting drawing ratio, and by how many parts are scrapped. Radii in the correct band maximise the drawing ratio, so they genuinely do reduce the number of redraw and anneal operations. Radii above that band lower the drawing ratio again because wrinkling sets in earlier, and the wrinkled parts are scrap. The rate therefore peaks inside the band and falls off on both sides — the objective is an optimum, not a maximum.

Recommendation. Specify both the punch nose radius and the die throat radius in the range of about four to ten sheet thicknesses, taking the upper end for thin, highly ductile sheet and the lower end for thick or low-ductility sheet, and control the blank-holder force independently. If a still higher rate is required, obtain it from draw beads, better lubrication and a properly matched clearance of roughly $1.1T$ to $1.15T$ — not from ever-larger radii.