Concentric Reducer Laid Out by the Wrong Method
A ductwork job needs four concentric reducers connecting 400 mm diameter round duct to 250 mm diameter round duct over a height of 300 mm. They are made from 1.6 mm galvanised sheet with a 10 mm lap seam along the slant side. The shop has 8 ft × 4 ft sheets (2438 mm × 1219 mm) and a hand shearing machine, but no CNC plasma. The fitter proposes to lay out each reducer by taking the average circumference, π(400 + 250)/2, developing it as a rectangle, rolling it into a cylinder and then pulling the small end in with a chain block. The foreman doubts this will give a round, dimensionally correct reducer and asks for a proper radial-line layout. The first trial piece is inspected: the small end is oval and its circumference is about 12 mm under size.
1. Calculate the slant height, the development radii and the sector angle for the reducer, taking the neutral axis at mid-thickness. State whether the developed blank fits an 8 ft × 4 ft sheet.
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Given: outside base diameter D = 400 mm, outside top diameter d = 250 mm, height h = 300 mm, thickness t = 1.6 mm. Radii R = 200 mm, r = 125 mm.
1. Slant height (true length of the slant side): L = √(h² + (R − r)²) = √(300² + 75²) = √95625 ≈ 309.2 mm.
2. Use mean (neutral-axis) diameters, since the metal neither stretches nor compresses at mid-thickness: D_m = 400 − 1.6 = 398.4 mm; d_m = 250 − 1.6 = 248.4 mm.
3. Development radii from the common apex: ρ_inner = L × r_m / (R_m − r_m) = 309.2 × 124.2 / 75 ≈ 512 mm ρ_outer = ρ_inner + L ≈ 821 mm
4. Sector angle. Outer arc length must equal the mean base circumference: π × 398.4 = 1251.6 mm. θ = arc / radius = 1251.6 / 821.2 = 1.524 rad ≈ 87.3° (add about 1.4° for the two 10 mm seam allowances, giving about 88.7°). Check: inner arc = 1.524 × 512 ≈ 780 mm = π × 248.4, correct.
5. Blank size and sheet fit. Chord of the outer arc = 2 × 821 × sin(44.35°) ≈ 1148 mm; radial width = 309 mm; bounding box about 1150 mm × 455 mm. An 8 ft × 4 ft sheet is 2438 × 1219 mm, so one blank fits across the 1219 mm width with clearance, and two blanks can be nested along the length. The blank fits comfortably.
2. Explain why the fitter's rectangle method produced an oval, undersized small end, and why the average circumference is not the right development length.
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A cone is not a cylinder. In a cylinder every generator is the same length, so a rectangle is the true development. In a cone the generators are of different lengths and all converge to an apex, so the true surface unfolds as an annular sector. Forming a rectangle into a cone means some material must be stretched near the small end and compressed near the large end; the sheet resists that, so the small end wrinkles, goes oval and cannot be pulled down to size.
Using the average circumference π(400 + 250)/2 = 1021 mm as the developed length of a rectangle is also dimensionally wrong: that figure is a circumference, not a developed edge length. The developed outer arc must equal the base circumference (about 1251 mm) and the inner arc must equal the top circumference (about 780 mm), so the two edges differ by roughly 470 mm which must be built into the layout as an included angle.
The 12 mm undersize at the small end is consistent with a rolled-and-pulled cone: the edge has been drawn in by the chain block past its true circumference because the pattern had no material allocated for the reducing shape, and the lap seam allowance was not placed along the slant.
Conclusion: reject the rectangle approach; the radial-line (sector) development is the correct method, and the pattern must be checked by measuring the arc lengths, not the chord.
3. Describe the marking and forming sequence you would use, including seam allowance and joint choice, so that the next reducer is round and dimensionally correct.
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Layout sequence:
- Scribe a centre line on the sheet and mark the apex point.
- Set dividers/beam trammel to ρ_inner = 512 mm and ρ_outer = 821 mm and strike the two arcs.
- Set the sector angle (about 87.3°, or 88.7° with seam allowance) with a protractor or by stepping the arc length, and scribe the two radial edges.
- Divide the arcs into 12 equal parts and number them so the piece can be checked after rolling and so the seam lands on a generator.
- Add 10 mm seam allowance on each radial edge, measured at right angles to the radial line, and add shear allowance of about 2 mm.
- Centre-punch the line intersections lightly, cut on the waste side of the line, and de-burr.
Forming: roll the piece in the slip rolls progressively, small increments, working from the middle outwards, checking the small-end circumference against π × 248.4 ≈ 780 mm with a steel tape before closing the seam. Because the curvature varies along the slant, roll in several passes rather than one heavy pass.
Joint: a 10 mm lap seam along the slant is acceptable for low-pressure ductwork, but the lap must be set in the direction of airflow and fastened with rivets at 40–50 mm pitch plus sealant, or spot welded if the coating is cleaned back. For a smoother bore, a butt seam with a 1.6 mm backing strip is better. The lap seam increases the effective diameter by one thickness, so the finished size is checked after the joint is closed, not before.