Conical Hopper Layout and Weld Metal Estimate
You are laying out a conical transition for a hopper from 6 mm ASTM A36 sheet. The finished part is a frustum with a 1200 mm inside diameter at the top, 400 mm inside diameter at the bottom, and a vertical height of 800 mm. It will be built in four equal segments, butt-welded with a single-V groove: 60° included bevel, 1.5 mm root face and 2 mm root gap. Filler is 3.2 mm E6013 covered electrode; deposited metal is 65% of electrode weight once stub loss and spatter are counted. Plate stock is 2400 × 1200 mm and the rolls can form 6 mm sheet. The fitter used the finished diameters, 1200 and 400 mm, directly as the arc lengths when marking out the first segment, and it will not close up when tacked. The lead hand asks you to check the layout method and the consumable estimate before the other three segments are cut.
1. Calculate the slant height, the development radii and the included angle, and give the top and bottom arc length per segment. Explain why the fitter's layout failed.
Read the worked answer
Slant height: the radial difference is (1200 − 400)/2 = 400 mm. L = √(400² + 800²) = √800 000 = 894 mm.
Development radii: by similar triangles the apex distance to the top arc is three times that to the bottom arc, and the difference is the slant height:
- r (bottom) = 894/2 = 447 mm
- R (top) = 894 × 600/400 = 1342 mm
Included angle: θ = 360° × 600/1342 = 161° (check: 2.81 rad × 1342 = 3770 mm of top arc).
Arc lengths per segment: top circumference π × 1200 = 3770 mm, so 942 mm per segment; bottom circumference π × 400 = 1257 mm, so 314 mm per segment.
The fitter's error is using diameters as arc lengths. The flat pattern is bounded by radii of 1342 mm and 447 mm, and the arcs must be struck at those radii with the dividers, not marked as 1200 and 400 mm straight chords. That is why the segment is the wrong shape and will not close.
2. Estimate the deposited weld metal and the electrode weight needed for the four seams. State your assumptions.
Read the worked answer
Groove area (per seam, mm²):
- V portion: tan 30° × (6 − 1.5)² = 0.577 × 20.25 = 11.7
- Root gap portion: 2 × 6 = 12.0
- Total ≈ 23.7 mm², ignoring cap reinforcement.
Weld length: 4 × 894 = 3576 mm.
Volume: 23.7 × 3576 = 84 750 mm³ = 84.8 cm³.
Deposited mass: 84.8 × 7.85 g/cm³ = 666 g.
Electrode mass: 666 ÷ 0.65 = 1025 g, so order about 1.5 kg to cover cap reinforcement and handling losses.
Assumptions worth stating: the 2 mm gap runs the full thickness, no backing, no significant melt-through, and the 65% efficiency already includes stubs and spatter. If the root is later back-gouged and rewelded, add roughly 30–40% to the deposit. Making assumptions explicit matters, because a fitters' tack or a changed gap moves the number more than rounding does.
3. What distortion control and fit-up sequence would you impose on the four seams, and what allowance would you make on the finished diameters?
Read the worked answer
- Weld sequence: backstep each longitudinal seam and stagger the stop/start points around the cone so shrinkage is not concentrated on one side. Weld opposite seams in rotation rather than finishing one seam completely.
- Leave one seam loose: fit and tack three seams, then close the last seam against the actual opening rather than forcing the marked arcs together. A cone with four closed seams has no flexibility left.
- Shrinkage allowance: each 894 mm butt seam contracts 1–2 mm across the joint, so allow 3–5 mm on the top circumference and check the top and bottom diameters with a trammel after welding, before attaching flanges.
- Tack and braces: use bridge tacks or a spider brace across the small end to hold roundness through welding, and remove braces only after the part cools.
- Check the flat pattern first: verify one segment's top and bottom arc lengths (942 and 314 mm) against the rolled curvature before cutting the remaining three, so a layout error is caught on one piece, not four.