Five free practice questions on layout, fitting and fabrication: block B of the Red Seal 456A Welder exam, about 28 of its 125 questions. They are in the exam's format: four options, one correct answer. Answer each one before opening the explanation. No account, no email.
01The questions
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Why are tacks on thin austenitic stainless sheet usually placed closer together than on mild steel of the same thickness?
- The code requires twice the tacks on stainless
- Stainless tacks are weaker and crack easily
- Stainless expands more and holds heat locally
- Stainless conducts heat away faster than steel
Show the answer
C. Correct. Austenitic stainless has about 50 % more thermal expansion and about one-third the thermal conductivity of carbon steel, so heat stays concentrated and the edges move more. Closer, smaller tacks and chill bars hold alignment.
Why not the others
A. No code sets a blanket tack count by material. Tack spacing follows the material's behaviour and the job's needs.
B. Austenitic stainless tacks are not weak or crack-prone when made correctly. The reason is thermal movement.
D. It is the reverse: stainless conducts heat poorly, which keeps it concentrated near the joint and increases distortion.
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A right-triangle gusset has legs of 300 mm and 400 mm and is cut from 10 mm plate. Using a steel density of 7.85 g/cm³, what is its mass?
- 4.71 kg
- 47.1 kg
- 0.47 kg
- 9.42 kg
Show the answer
A. Correct. Area = (h × b)/2 = (300 × 400)/2 = 60 000 mm² = 600 cm². Volume = 600 cm² × 1.0 cm = 600 cm³. Mass = 600 × 7.85 = 4710 g = 4.71 kg, the figure for the bill of materials.
Why not the others
B. Off by a factor of ten (for example 10 mm taken as 10 cm). Keep the units consistent: 600 cm² × 1 cm × 7.85 g/cm³ = 4710 g = 4.71 kg.
C. Off by a factor of ten, usually from taking the 10 mm plate as 0.1 cm instead of 1 cm. 600 cm² × 1 cm = 600 cm³; × 7.85 g/cm³ = 4.71 kg.
D. The triangle was not halved: 300 × 400 is the rectangle the gusset is cut from. (h × b)/2 = 600 cm² gives 4.71 kg.
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A tacked 900 × 1200 mm base frame measures 1505 mm on one diagonal and 1495 mm on the other. The drawing allows at most 3 mm difference between diagonals. What is the correct adjustment?
- Push the short diagonal out until it reads 1505 mm
- Accept it; the frame is within tolerance
- Draw the long diagonal in until both read 1500 mm
- Cut one long side 5 mm shorter
Show the answer
C. Correct. A racked rectangle has one long and one short diagonal; drawing the long diagonal's corners together (clamp, come-along or a tap) squares it, and both read 1500 mm. Re-tack if needed, then weld in a balanced sequence and recheck.
Why not the others
A. 1505 mm is the wrong target: square is when both diagonals equal 1500 mm (a 3-4-5 triangle scaled by 300).
B. The diagonals differ by 10 mm, well over the 3 mm allowed. The frame is racked and must be squared.
D. Racking is an angular error, not a length error; the sides are the right length. Shortening one would make the frame wrong in a new way.
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You are cutting a wrap-around for a 273.1 mm (10 in NPS) OD pipe with a 9.3 mm wall. Ignoring overlap and paper thickness, how long must the wrap be to go exactly once around? (π = 3.1416)
- 799.5 mm
- 429.0 mm
- 1715.9 mm
- 858.0 mm
Show the answer
D. Correct. C = πD = 3.1416 × 273.1 = 857.97 mm. In practice cut the wrap longer so it overlaps, and line the edges up at the overlap to keep the line square.
Why not the others
A. This uses the ID (273.1 − 2 × 9.3 = 254.5 mm). A wrap-around goes on the outside of the pipe, so use the OD: 3.1416 × 273.1 = 858.0 mm.
B. This is πr (3.1416 × 136.55), half the circumference: the radius was used where the diameter belongs. Wrap length = πD = 3.1416 × 273.1 = 858.0 mm.
C. This is 2π × D, putting the diameter into 2πr. The circumference is 2πr = πD = 858.0 mm.
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A cracked grey cast-iron pump housing will be repaired by SMAW with a nickel-based electrode. Before grooving out the crack, what is done at each end of it?
- Tack weld across each end
- Drill a small hole just past each end
- Clamp the housing to close the crack
- Centre-punch each end to mark it
Show the answer
B. Correct. Crack-arrest holes stop the crack from running further when the casting is heated and ground. Then grind a groove along the crack, remove oil-soaked metal, and weld short beads with peening and slow cooling.
Why not the others
A. A tack across a cast-iron crack puts a hard, brittle deposit on a stressed spot and often cracks again. Arrest the crack by drilling first.
C. Forcing the crack closed adds stress to a brittle casting and does not stop it spreading. Drill arrest holes, then groove and weld.
D. A punch mark only shows where the crack ends; it does not stop it, and hammering brittle cast iron can start new cracks. Drill a small hole beyond each end.
02Answer key
| Question | Answer | Sub-task |
|---|---|---|
| Q1 | C | Fits components for welding |
| Q2 | A | Transfers dimensions from drawings to materials |
| Q3 | C | Assembles components |
| Q4 | D | Develops templates |
| Q5 | B | Prepares materials |
03What block B covers
This block covers layout and fabrication before the welding starts. It includes developing templates by parallel-line, radial-line and triangulation methods, pipe wrap-arounds and saddles, mitres, bend and shrinkage allowances, layout math and measuring tools, bevel preparation and cleaning of stainless, aluminum and galvanized steel, tacking and fit-up, hi-lo and root opening checks, distortion control and squaring assemblies.
The Red Seal Occupational Standard names this block “Performs layout and fabrication of components for welding”. It carries 22.4% of the exam, about 28 of the 125 questions, split across two tasks. The share column is each task's weight within the block, as the standard publishes it.
| Task and its sub-tasks | Share of block |
|---|---|
| Performs layout Develops templates; Transfers dimensions from drawings to materials | 44% |
| Fabricates components Prepares materials; Fits components for welding; Assembles components | 56% |
04What the questions turn on
Pick the development method
Cylinders are developed by parallel lines, right cones by radial lines swung at the slant height, and transitions such as square-to-round by triangulation. Triangulation needs true lengths, because sloping lines look shorter than they are in both views.
Roll to the mean diameter
Plate rolled into a shell keeps its length only at mid-thickness, so cut it to π times the mean diameter, not the inside diameter. For 12 mm plate and a 1200 mm ID, that is π times 1212 mm.
Tacks follow the WPS
A tack that becomes part of the weld is made to the WPS, with the specified electrode and the specified preheat. Small tacks on thick, cold plate cool very fast, and with damp rod that sets up hydrogen cracking.
Balancing weld shrinkage
Weld shrinkage pulls parts toward the weld. Balance it by welding diagonally opposite corners of a frame in turn and alternating passes from side to side on a double-V, or preset parts by the distortion measured on the first one.
05The other blocks
The 456A exam has four blocks. Each has its own page of free questions:
The 456A Welder practice exam page describes the full question bank. How these questions are written and checked: how our questions are made. Found a mistake? Tell us and we will fix it.