Question 2 of 7: Designing a copper drawing schedule with an intermediate anneal

Question 2: Designing a copper drawing schedule with an intermediate anneal (20 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.

Given. Round copper bar to be drawn down to a finished rod, with the property targets, the starting condition and the formability limit below. Percent cold work for a round section drawn without change of shape is measured on area, $\%\mathrm{CW} = \dfrac{A_{0}-A_{d}}{A_{0}}\times 100 = \left[1-\left(\dfrac{d_{d}}{d_{0}}\right)^{2}\right]\times 100$, and it is always counted from the last full anneal.

Find. A drawing and annealing procedure that reaches 13.5 mm diameter without ever exceeding 60% cold work, together with the intermediate diameters it requires, and a statement of which of the two property targets that procedure can actually deliver.

Approach. First test whether the finished size can be reached at all from the stock as received, by converting the stock's 35% cold work into the equivalent annealed diameter and measuring the whole reduction from there. Then read the property window off the page-4 curves, express it as a range of permissible final-pass cold work, and size the intermediate diameter from that final pass working backwards.

1. Recover the annealed diameter the stock came from. Cold work is cumulative from the last anneal, so a bar that is 21.0 mm across and carries 35% cold work behaves as though it had been drawn from$$d_{\text{ann}} = \frac{d_{s}}{\sqrt{1-0.35}} = \frac{21.0}{\sqrt{0.65}} = 26.05\ \text{mm}$$This is the reference diameter against which any further reduction must be counted.
2. Test the direct route and reject it. Drawing the bar as received straight down to 13.5 mm would accumulate$$\%\mathrm{CW} = \left[1-\left(\frac{13.5}{26.05}\right)^{2}\right]\times 100 = (1-0.2686)\times 100 = \boxed{73.1\,\%}$$measured from the last anneal, which is far beyond the 60% at which copper cracks. The bar must therefore be annealed before any further drawing — that conclusion is forced by the numbers and is the first mark of the question.
3. Check the one-anneal, one-pass route. A full recrystallisation anneal at 21.0 mm resets the accumulated cold work to zero. A single pass from there to 13.5 mm is$$\%\mathrm{CW} = \left[1-\left(\frac{13.5}{21.0}\right)^{2}\right]\times 100 = 58.7\,\%$$which is just inside the 60% cracking limit, so the rod can be made in one pass after one anneal. Whether it should be depends entirely on the properties that 58.7% cold work leaves behind, which is what the page-4 curves settle.
4. Read the property window off the curves. For copper, tensile strength climbs from 220 MPa annealed to about 390 MPa at 68% cold work, and ductility falls from about 44%EL annealed to under 5%EL at the same point. Reading each requirement across to the copper curve gives$$\mathrm{TS}\ge 350\ \text{MPa} \;\Rightarrow\; \%\mathrm{CW}\ge 40 \qquad\text{and}\qquad \%\mathrm{EL}\ge 15 \;\Rightarrow\; \%\mathrm{CW}\le 20$$The two ranges do not overlap. Cold-worked commercially pure copper cannot simultaneously carry 350 MPa and retain 15% elongation, and the single 58.7% pass of Step 3 would leave roughly 385 MPa with only about 4%EL.
5. Size the schedule that meets the strength target. A minimum tensile strength is normally the safety-critical requirement, and the least damaging way to meet it is to take exactly the cold work it needs and no more. Setting the final pass at 40% and working backwards from the finished size,$$d_{\text{int}} = \frac{d_{f}}{\sqrt{1-0.40}} = \frac{13.5}{\sqrt{0.60}} = \boxed{17.43\ \text{mm}}$$and the first pass, from the annealed 21.0 mm stock down to that intermediate size, is$$\%\mathrm{CW}_{1} = \left[1-\left(\frac{17.43}{21.0}\right)^{2}\right]\times 100 = 31.1\,\% \;\lt\; 60\,\%$$so neither pass approaches the cracking limit. The finished rod carries about 350 MPa with roughly 5.5%EL.
6. Size the alternative schedule that meets the ductility target. If instead the 15%EL is the governing requirement — because the rod must be bent, staked or upset in service — the final pass is capped at 20%:$$d_{\text{int}} = \frac{13.5}{\sqrt{0.80}} = 15.09\ \text{mm}$$and the first pass from the annealed stock is then$$\%\mathrm{CW}_{1} = \left[1-\left(\frac{15.09}{21.0}\right)^{2}\right]\times 100 = 48.3\,\%$$again with both passes inside the cracking limit. That rod holds about 15%EL but only about 294 MPa, some 56 MPa short of the strength target.
7. State the procedure and the verdict. The procedure is therefore: full-anneal the 21.0 mm stock (recrystallise, of the order of 500 °C for copper, in a protective or reducing atmosphere) to erase the 35% cold work; cold draw to the intermediate diameter; process-anneal a second time; then take the finishing pass to 13.5 mm, the size of that last pass being what sets the delivered properties. With copper, the last pass cannot be chosen to satisfy both targets at once, so the specification must be relaxed on one axis or the material must be changed.