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22-Mec-B5 Product Design and Development · December 2018

Question 7 of 7: Design for Manufacturing and Assembly

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

Notes on this paper

Paper format. National Exams, December 2018 — 16-Mec-B5 Product Design and Development. Three hours; OPEN BOOK; an approved Casio or Sharp calculator is permitted. Question 1 is compulsory and carries 40 marks; four of the six remaining questions are attempted at 15 marks each, for a total of 100 marks. The paper prints 40 + 6 × 15 = 130 marks against the 100 that are attempted. All seven questions are solved here. Most questions call for an essay answer or the use of tables, figures and charts, and clarity and organisation of the answer are explicitly marked.

Reference texts for 22-Mec-B5 Product Design and Development. K. T. Ulrich and S. D. Eppinger, Product Design and Development (the framework text for this syllabus); G. E. Dieter and L. C. Schmidt, Engineering Design; G. Pahl and W. Beitz, Engineering Design: A Systematic Approach; G. Boothroyd, P. Dewhurst and W. Knight, Product Design for Manufacture and Assembly; M. F. Ashby, Materials Selection in Mechanical Design; S. Kalpakjian and S. R. Schmid, Manufacturing Engineering and Technology; R. G. Cooper, Winning at New Products. Canadian context is taken from CSA Z412 Office Ergonomics, CSA B651 Accessible Design for the Built Environment, ANSI/BIFMA X5.1 General-Purpose Office Chairs, the Canadian Intellectual Property Office guides, and the Engineers and Geoscientists BC Code of Ethics.

How this paper is answered. Every question on this sitting is descriptive, so the answers are written as engineering prose. Where a claim can be settled with a number rather than asserted — how many people a chair actually fits, how many stations a line needs, whether a warranty improvement is real, which assembly route is cheapest — the calculation is set out with its Given and Find so the reasoning can be checked. That is a deliberate exam tactic as well as good practice: this paper explicitly rewards "the use of tables, figures and charts", and a quantified assertion is the hardest kind to argue with.

Question 7: Design for Manufacturing and Assembly (15 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.

Part A — Outline of the DFMA process and how it improves a product

DFMA is the Boothroyd and Dewhurst method for reducing the cost of a product by changing its design rather than its factory. It runs in two halves and the order matters: design for assembly first, design for manufacture second. Optimising the cost of a part that should not exist is wasted work, so parts are eliminated before the survivors are costed.

  1. Analyse the assembly as it stands. Take the existing or proposed design and list every part in the order it is installed, including every fastener, washer, label and cable tie.
  2. Apply the three minimum-part criteria to each part in turn. A part is theoretically necessary only if, relative to the parts already assembled, it must move; or it must be of a different material for a fundamental reason such as insulation or thermal isolation; or it must be separable to permit assembly or service of other parts. A part failing all three is a candidate for elimination or combination, and fasteners almost never survive.
  3. Estimate handling and insertion times from the classification tables, which score handling on size, thickness, weight, symmetry, and the tendency to tangle or nest, and insertion on access, visibility, alignment, resistance and the securing operation required.
  4. Compute the design-for-assembly index as the ratio of ideal to actual work content.
  5. Redesign against the analysis and repeat until the index stops improving.
  6. Then apply design for manufacture: cost each surviving part against candidate processes and materials, respecting the process rules — uniform wall thickness, draft, generous radii, achievable tolerances — and choose the route on total cost at the programme volume.
1. List every partin assembly order2. Apply the threeminimum-part tests3. Estimate handlingand insertion times4. Compute the DFAindex, alpha5. Redesign, thencost each part (DFM)iterate until the index stops movingKeep a part only if it- moves relative to parts already assembled, or- must be a different material for a fundamental reason, or- must be separable to allow assembly or service.Pressure-regulating valve module- baseline: 38 parts, 285 s, N(min) = 9 -> alpha = 9.47 pct- redesign: 17 parts, 138 s, N(min) = 9 -> alpha = 19.57 pct- rolled throughput yield 82.7 pct -> 91.8 pct
Figure 7.1 — The DFMA loop. Assembly is analysed and simplified before any part is costed, and the loop is repeated until the index stops moving.

Given. A pressure-regulating valve module in its baseline form: 38 parts, 285 s of total assembly work content, and 9 parts surviving the three minimum-part tests. After one DFMA pass: 17 parts and 138 s, with the same 9 necessary parts. Ideal handling and insertion time $t_a = 3$ s; first-pass yield 99.5 per cent per assembly operation. Find. The improvement in the DFA index and in end-to-end yield.

  1. Compute the index for both designs. The index compares the theoretically ideal assembly work with the actual: $$\alpha_{\mathrm{DFA}} = \frac{N_{\min}\,t_a}{t_{\text{total}}}$$ so the baseline scores $\alpha = (9 \times 3)/285 = 0.0947$ and the redesign $\alpha = (9 \times 3)/138 = 0.1957$, that is $\boxed{9.5\ \% \rightarrow 19.6\ \%}$.
  2. Read what actually moved. The numerator is fixed by function and does not change; the entire gain is a 55.3 per cent reduction in part count (38 to 17) and a 51.6 per cent reduction in work content (285 s to 138 s). An index that rises without the part count falling means the parts were merely made easier to handle, which is a smaller and more fragile gain.
  3. Convert part count into end-to-end yield, a benefit the index never claims. Rolled throughput yield multiplies across operations, so at 99.5 per cent per operation $$RTY = f^{\,N}: \quad 0.995^{38} = 0.8266 \quad \text{against} \quad 0.995^{17} = 0.9183$$ a gain of $\boxed{9.2\ \text{percentage points}}$ obtained purely by removing operations.
  4. Price the redesign and find the volume that justifies it. At a burdened rate of CAD 41.00/h the 147 s saved is worth CAD 1.674 per unit; the consolidated moulding raises piece cost by CAD 1.85, while scrap and rework fall by CAD 0.62 and part-number overhead by CAD 0.88, for a net saving of CAD 1.324 per unit. Against CAD 96 000 of new tooling, $$n^{*} = \frac{96\,000}{1.324} = \boxed{72\,500\ \text{units}}$$

The last step contains the point that is easiest to miss: piece-part cost rose. The consolidated moulding is more expensive than the several simple parts it replaced, and a purchasing organisation that measures only piece price will reject the change. What pays for it is labour, handling, inventory, inspection, rework and the part numbers themselves, none of which appear on a piece-price comparison.

Part B — A typical change a product experiences through DFMA

The changes are strikingly consistent across products, because the criteria that drive them are the same everywhere.

The fasteners largely disappear. Threaded fasteners fail all three minimum-part tests and carry a heavy insertion time, so they are replaced by integral snap-fits, press-fits and self-locating features. A typical first pass removes the majority of them.

Several parts become one moulding with integral features. Springs, bosses, guides, cable routing, living hinges and seals migrate into a single component. Part count typically falls by 30 to 60 per cent, which is the dominant term in every number computed above.

A base or chassis part emerges, and everything inserts into it from one direction. The assembly is reorganised so that parts stack downward under gravity onto a part that never moves, because reorientation of a partly built assembly is among the most expensive operations in the classification tables and delivers nothing.

Parts become either fully symmetric or obviously asymmetric. A part that can be inserted any way round is cheap to handle; a part that physically cannot be inserted wrongly is error-proof. A part that is almost symmetric is the worst of both, because it must be oriented and it can be got wrong. Chamfers and lead-ins appear on every insertion.

The tolerance scheme loosens. Fewer stacked interfaces mean a shorter tolerance stack, so individual tolerances can be relaxed for the same functional result — a real manufacturing saving that follows automatically from consolidation.

And the piece-part cost of the consolidated component rises while total cost falls, together with a larger, more expensive tool and a longer lead time to first parts. That trade is the characteristic signature of a DFMA change and the one that must be defended explicitly.

Part C — How the degree of automation impacts the DFMA process

Automation does not simply scale the analysis; it changes which design features are penalised, and by how much. Boothroyd and Dewhurst provide different classification tables for manual assembly, high-speed automatic assembly and robotic assembly, and the same part can score well in one and be impossible in another.

Under manual assembly the dominant penalties are part count, fastener count and reorientation. A human copes with a tangling spring, a flexible gasket or a near-symmetric part at a modest time penalty, because human hands and eyes are extraordinarily general-purpose.

Under high-speed automatic assembly the penalties move to feeding and become severe or fatal. Parts that nest or tangle cannot be bowl-fed at all; flexible parts, wires and gaskets essentially cannot be automatically fed; near-symmetry is worse than gross asymmetry because the feeder must detect and correct orientation; every insertion must be vertical, self-aligning and free of obstruction; and the machine cannot see, so it cannot recover from a part presented wrongly. Dedicated automation also has no tolerance for variety: a second variant may require a second machine.

Robotic assembly sits between the two, trading some speed for the flexibility to be reprogrammed and, with vision, to tolerate imperfect presentation.

The consequence for process is decisive: the intended level of automation must be decided before the DFA redesign, not after it. A design optimised for manual assembly and later automated must be re-analysed from the beginning, because the features that were free under manual assembly are the ones that break a feeder.

Given. The redesigned valve module at 138 s of manual assembly content, burdened labour CAD 41.00/h, so a manual variable cost of CAD 1.572 per unit against CAD 25 000 of fixtures. Robotic assembly: CAD 240 000 of equipment and CAD 0.66 per unit. High-speed dedicated automation: CAD 780 000 and CAD 0.21 per unit. Programme demand is 300 000 units a year over a four-year life. Find. The volume ranges over which each route is cheapest, and the route to select.

0.000.601.201.802.403.001k10k100k1MCumulative production volume n (units)Cost per unit (CAD)Manual: T = 25 000, u = 1.572Robotic: T = 240 000, u = 0.66High-speed dedicated: T = 780 000, u = 0.21Manual below 235 800 units; robotic to 1 200 000; dedicated above that.The programme lands exactly on the second break-even.
Figure 7.2 — Assembly route against cumulative volume. The programme's 1.2 million units falls precisely on the robotic-to-dedicated crossover, so the decision is settled by flexibility rather than by cost.
  1. Express each route in the same two-term form. $$c(n) = \frac{T}{n} + u$$ with $T$ the equipment or fixture investment and $u$ the variable cost per unit. The manual variable cost is the work content priced at the burdened rate, $138 \times 41.00/3600 = 1.572$.
  2. Find the manual-to-robotic crossover. $$n^{*}_{mr} = \frac{240\,000-25\,000}{1.572-0.66} = \frac{215\,000}{0.912} = \boxed{235\,800\ \text{units}}$$ Below this the fixtures are cheaper than the robot cell, and manual assembly wins.
  3. Find the robotic-to-dedicated crossover. $$n^{*}_{rd} = \frac{780\,000-240\,000}{0.66-0.21} = \frac{540\,000}{0.45} = \boxed{1\,200\,000\ \text{units}}$$ Dedicated automation never wins below this, because its variable-cost advantage of CAD 0.45 per unit cannot repay a further CAD 540 000 any sooner.
  4. Place the programme and read the decision. Four years at 300 000 units a year is 1 200 000 units, which sits exactly on the second crossover at a cost of CAD 0.86 per unit either way. When the economics tie, the tie-breaker is the option value: a mid-life design change scraps a dedicated line and merely reprograms a robot cell, so $\boxed{\text{robotic assembly is selected}}$.
  5. Feed the decision back into the design, which is the point of the exercise. Having chosen robotic assembly before the detailed design is frozen, the DFA analysis is now run against the robotic tables: presentation on trays or by vision rather than bowl feeding, one insertion direction, chamfered lead-ins, and no flexible or tangling parts. Deciding this after the design would have required the whole analysis to be repeated.
QuantityResult
DFA index, baseline and after one pass9.5 per cent to 19.6 per cent
Part count and assembly content38 to 17 parts; 285 s to 138 s
Rolled throughput yield at 99.5 per cent per operation82.7 per cent to 91.8 per cent
Net unit saving from the redesignCAD 1.324 per unit
Break-even volume on CAD 96 000 of tooling72 500 units
Manual assembly variable costCAD 1.572 per unit
Manual-to-robotic crossover235 800 units
Robotic-to-dedicated crossover1 200 000 units
Programme volume and selected route1 200 000 units; robotic assembly
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