Why an 'equal modulus' riser failed: modulus, feeding distance and yield on a steel bracket
A steel foundry is making brackets in 0.30% carbon steel poured at 1580 °C into green sand moulds. The ruling section is a flat plate 200 mm long × 150 mm wide × 40 mm thick, fed in the length direction from one end. The shop rule is that the riser modulus must be 1.2 times the casting modulus, and the top riser is a plain cylinder with height equal to twice its diameter. Earlier brackets were made with a single end riser whose modulus was made equal to the plate modulus, 'so that both freeze together and the riser lasts as long as the casting'. Those castings showed a shrinkage cavity at the far end of the plate about 5 mm below the surface, exposed after machining. The pattern shop wants to repeat the same layout because the arithmetic was correct. Steel needs about 6% of the fed casting mass as feed metal, and the target yield for the job is at least 50%.
1. Calculate the plate modulus, the riser dimensions, the riser mass and the yield for the existing single end riser.
Read the worked answer
Plate volume V = 200 × 150 × 40 = 1 200 000 mm³. Cooling surface area A = 2(200×150) + 2(200×40) + 2(150×40) = 60 000 + 16 000 + 12 000 = 88 000 mm². Modulus M = V/A = 1 200 000 / 88 000 = 13.6 mm.
Required riser modulus = 1.2 × 13.6 = 16.4 mm. For a cylinder with H = 2D, M = V/A = (πD³/2)/(2.5πD²) = D/5, so D = 5 × 16.4 = 82 mm and H = 164 mm.
Riser volume = π/4 × 82² × 164 ≈ 866 000 mm³ = 866 cm³; at 7.85 g/cm³ the riser mass ≈ 6.8 kg. Casting mass = 1 200 cm³ × 7.85 = 9.4 kg. Yield = 9.4 / (9.4 + 6.8) = 58%, so the single riser meets the 50% target on paper, and 6% feed metal demand (0.57 kg) is far below the 6.8 kg available. The arithmetic is indeed right, which is why the failure must be explained by something the modulus method does not model.
2. The shrinkage cavity appeared at the far end. Explain the real cause and state, with reasoning, how many risers should be used.
Read the worked answer
Two separate errors are present.
- Equal modulus cannot feed. A riser only feeds while it is still liquid. If the riser modulus equals the casting modulus, both freeze at about the same time and the riser solidifies before it can draw metal through the last liquid channels. The riser must have the larger modulus (here 1.2 × M) so that it freezes last, and it must still contain liquid feed metal at that moment.
- Feeding distance was ignored. The modulus fixes the riser size, not how far it can reach. For steel plate the reliable feeding distance is roughly 4–5 times the thickness, i.e. about 160–200 mm here. The plate is 200 mm long and is fed from one end, so the far end is at or beyond the limit. Metal there solidifies isolated from the riser and forms a pipe. The observed cavity 5 mm below the surface at the far end is exactly that signature.
Recommendation: keep a riser of at least 82 mm × 164 mm but place it centrally on the 200 mm length (top riser at mid-length), so each side is fed over about 100 mm ≈ 2.5 t, comfortably inside the limit. Alternatively use two end risers. A chill block at the far end is a second-best option if the layout cannot be changed.
3. If two risers of the calculated size are used, what happens to yield, and what would you propose to recover the 50% target?
Read the worked answer
Two risers double the riser mass: 2 × 6.8 = 13.6 kg against a 9.4 kg casting.
Yield = 9.4 / (9.4 + 13.6) = 41%, below the 50% target, so simply doubling the riser is not acceptable.
Options, in order of preference:
- Exothermic sleeves on the two risers. A sleeve keeps the riser metal liquid longer, so the effective modulus rises and the physical diameter can be reduced by roughly 20–30% while still freezing after the casting. Two sleeved risers around 60–65 mm diameter would restore yield above 50%.
- One central riser plus a chill. A central riser feeds 100 mm each way; a chill at each end increases the local thermal gradient and further shortens the required feeding distance, allowing a smaller riser.
- Change the gating so the casting is fed through the riser (riser as feeder of the in-gate), which keeps the riser hotter and improves its efficiency without extra metal.
Whichever route is chosen, the check must include both conditions: riser modulus ≥ 1.2 × casting modulus and riser available feed metal ≥ 6% of the fed mass. The layout that failed satisfied only the first number and ignored feeding distance.