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Chapter 4 · 6 hours

Solidification Process and Powder Metallurgy

Practice questions

Practice questions and answers

5 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.

  • Practice · 6 marks

Describe the steps of the sand casting process with a neat sketch of a mould showing its parts. State the advantages and limitations of sand casting.

Answer

Sand casting is a process in which molten metal is poured into a cavity formed in compacted sand, allowed to solidify, and the sand mould is broken to take out the casting.

Steps

  1. Pattern making - a pattern slightly larger than the part (allowances added) is made of wood, metal or plastic.
  2. Core making - cores of sand bonded with resin or oil form internal cavities.
  3. Moulding - the pattern is placed in the drag, moulding sand is rammed around it, the cope is rammed on top, the pattern is withdrawn, and the gating system is cut.
  4. Core setting and closing - cores are placed in the cavity on core prints, and cope and drag are clamped.
  5. Melting and pouring - metal is melted (cupola, induction or arc furnace) and poured through the sprue.
  6. Solidification and cooling.
  7. Shakeout and fettling - casting is removed, gates and risers are cut, sand is cleaned, flash is ground.
  8. Heat treatment and inspection.

Mould sketch

         pouring basin
            \  |  /
   +---------\ | /---------+
   |   COPE   sprue  riser |  <- cope
   |    ......|...  |  |   |
   |  runner--+  cavity |   |
   +---------------------- +  <- parting line
   |   DRAG     (core)     |  <- drag
   +-----------------------+

Parts: pouring basin, sprue, runner, gate, riser, cavity, core, cope, drag, parting line, vents.

Advantages

  • Almost any metal can be cast, including high melting point alloys.
  • Size ranges from grams to many tonnes; complex shapes are possible.
  • Low tooling cost, suitable for small and large batches.

Limitations

  • Poor surface finish (Ra about 12-25 micrometre) and wide tolerances (about plus or minus 1 mm).
  • Machining allowance is required.
  • Defects such as porosity and sand inclusion are common; labour intensive and slow for high volume.
  • Practice · 6 marks

Explain the solidification of a pure metal and of an alloy in a mould. Describe the three stages of shrinkage in a casting. List and explain any four casting defects with their remedies.

Answer

Solidification of a pure metal

A pure metal solidifies at a single temperature. Heat flows into the mould wall, a thin chill zone of small equiaxed grains forms at the wall, then grains grow inward against the heat-flow direction as columnar (dendritic) crystals. At the centre, equiaxed grains may form. Total solidification time is given by Chvorinov's rule, t=C(V/A)2t = C(V/A)^2.

Solidification of an alloy

An alloy freezes over a range between liquidus and solidus temperatures. A mushy zone of dendrites and liquid exists, so the casting can be segregated (coring) and the feeding of liquid shrinkage is difficult. A wide freezing range gives more porosity and a longer mushy zone.

Shrinkage in three stages

  1. Liquid contraction - temperature falls from pouring to liquidus; compensated by liquid metal from the riser.
  2. Solidification contraction - volume decrease on freezing (about 3-7 percent for most metals); shows as shrinkage cavities if the riser is inadequate.
  3. Solid contraction - cooling from solidus to room temperature; allowed for by the pattern shrinkage allowance.

Casting defects

DefectCauseRemedy
Shrinkage cavityInsufficient feedingProperly designed riser, directional solidification, chills
Gas porosityDissolved gas, damp sand, poor ventingDegas melt, dry sand, add vents
Misrun / cold shutLow pouring temperature, thin sectionHigher pouring temperature, larger gates
Hot tearRestrained contractionFillets, collapsible core, uniform sections
Sand inclusion / blowEroded sand, high moistureStrong sand, correct gating, lower moisture
Core shiftWeak core printsBetter core support
  • Practice · 5 marks

List the allowances provided on a pattern and state the reason for each. A cast iron casting has a finished length of 450 mm between two faces that are to be machined, with 3 mm machining allowance on each face. The shrinkage allowance for cast iron is 1% (10 mm per metre). Find the length of the pattern. Also find the diameter of the core for a finished bore of 80 mm diameter which is cast and then bored with a radial machining allowance of 2.5 mm.

Answer

Pattern allowances

AllowanceReason
ShrinkageMetal contracts while cooling from solidification to room temperature; pattern is made larger
Machining (finish)Extra metal on surfaces to be machined, removing rough skin
Draft (taper)Taper of 0.5-3 degree on vertical faces so the pattern can be withdrawn without damaging the mould
Shake (rapping)Pattern is rapped before withdrawal, enlarging the cavity slightly; pattern is made smaller
DistortionAllowance for warping of shapes such as U or long thin castings

Pattern length

Machined length including allowance on both faces:

Lc=450+2×3=456 mmL_{c} = 450 + 2\times 3 = 456\ \text{mm}

The pattern must be larger by the shrinkage allowance (1 percent):

Lp=456 (1+0.01)=460.56 mmL_p = 456\,(1+0.01) = 460.56\ \text{mm}

Core diameter

Casting hole diameter before boring, with 2.5 mm removed from the radius (5 mm from the diameter): for an internal surface, the machining allowance reduces the cast hole size:

dcast=80−2×2.5=75 mmd_{cast} = 80 - 2\times 2.5 = 75\ \text{mm}

The core print (core size in the pattern) must be larger to allow for shrinkage of the metal on to the core, as the casting is smaller after cooling:

dcore=75 (1+0.01)=75.75 mmd_{core} = 75\,(1+0.01) = 75.75\ \text{mm}

Answer: pattern length =460.56= 460.56 mm (about 460.6 mm); core diameter =75.75= 75.75 mm.

  • Practice · 8 marks

State Chvorinov's rule. A cube-shaped casting of side 80 mm solidifies in 3.2 minutes in a sand mould. Using the same mould material and pouring conditions, (a) find the mould constant, (b) find the solidification time of a rectangular plate casting 200 mm x 100 mm x 30 mm, and (c) design a cylindrical riser with height equal to 1.5 times its diameter so that it solidifies 25 percent longer than the plate. Neglect the area of contact between riser and casting.

Answer

Chvorinov's rule: the solidification time of a casting is proportional to the square of its volume-to-surface-area ratio (modulus):

ts=C(VA)2t_s = C\left(\frac{V}{A}\right)^{2}

where C is the mould constant (min/cm2^2), depending on the mould material, metal properties and pouring temperature.

(a) Mould constant

For the cube, side a=80a = 80 mm =8= 8 cm:

VA=a36a2=a6=86=1.333 cm\frac{V}{A} = \frac{a^3}{6a^2} = \frac{a}{6} = \frac{8}{6} = 1.333\ \text{cm} C=ts(V/A)2=3.2(1.333)2=1.8 min/cm2C = \frac{t_s}{(V/A)^2} = \frac{3.2}{(1.333)^2} = 1.8\ \text{min/cm}^2

(b) Solidification time of the plate

V=200×100×30=600 000 mm3V = 200\times100\times30 = 600\,000\ \text{mm}^3 A=2(200×100+200×30+100×30)=2(29 000)=58 000 mm2A = 2(200\times100 + 200\times30 + 100\times30) = 2(29\,000) = 58\,000\ \text{mm}^2 VA=600 00058 000=10.34 mm=1.034 cm\frac{V}{A} = \frac{600\,000}{58\,000} = 10.34\ \text{mm} = 1.034\ \text{cm} tplate=1.8 (1.034)2=1.93 mint_{plate} = 1.8\,(1.034)^2 = 1.93\ \text{min}

(c) Riser design

Required riser time:

tr=1.25×1.926=2.41 mint_r = 1.25\times 1.926 = 2.41\ \text{min}

Required riser modulus:

(VA)r=2.411.8=1.157 cm\left(\frac{V}{A}\right)_r = \sqrt{\frac{2.41}{1.8}} = 1.157\ \text{cm}

For a cylinder with H=1.5DH = 1.5D:

V=π4D2(1.5D)=0.375πD3,A=πD(1.5D)+2⋅π4D2=2πD2V = \frac{\pi}{4}D^2(1.5D) = 0.375\pi D^3,\qquad A = \pi D(1.5D) + 2\cdot\frac{\pi}{4}D^2 = 2\pi D^2 VA=0.375πD32πD2=0.1875 D\frac{V}{A} = \frac{0.375\pi D^3}{2\pi D^2} = 0.1875\,D D=1.1570.1875=6.17 cm,H=1.5×6.17=9.25 cmD = \frac{1.157}{0.1875} = 6.17\ \text{cm},\qquad H = 1.5\times 6.17 = 9.25\ \text{cm}

Check of feeding

Riser volume =0.375π(6.17)3=276.5 cm3= 0.375\pi(6.17)^3 = 276.5\ \text{cm}^3. The casting volume is 600 cm3600\ \text{cm}^3. Even with 6 percent solidification shrinkage (36 cm336\ \text{cm}^3) the riser holds enough liquid, so the riser also satisfies the volume requirement.

Answer: C=1.8C = 1.8 min/cm2^2; plate solidification time =1.93= 1.93 min; riser diameter =61.7= 61.7 mm and height =92.5= 92.5 mm.

  • Practice · 5 marks

Describe the powder metallurgy process with its main steps. Mention its advantages, limitations and four typical products.

Answer

Powder metallurgy (PM) makes parts by compacting metal powders in a die and then heating (sintering) them below the melting point so that the particles bond.

Steps

  1. Powder production - atomisation (molten metal broken by water or gas jet), chemical reduction of oxides, electrolytic deposition. Particle size is about 10-200 micrometre.
  2. Blending and mixing - different powders, alloying elements and a lubricant (about 1 percent zinc stearate) are mixed.
  3. Compaction - powder is pressed in a hardened die at 150-700 MPa to give a "green compact", with density about 70-90 percent of solid metal and enough strength to be handled.
  4. Sintering - the compact is heated in a protective atmosphere at about 70-90 percent of the melting temperature (about 1100-1150 degree C for iron) for 15-60 min. Diffusion welds the particles and strength rises.
  5. Secondary operations - sizing or coining, impregnation with oil, infiltration, heat treatment and machining.
Powder -> Blend -> Press (die) -> Sinter -> Size/finish
                      |
                  green compact

Advantages

  • Near-net shape, little scrap (over 95 percent material use).
  • Can make parts from materials difficult to cast or machine (tungsten carbide, refractory metals).
  • Controlled porosity (self-lubricating bearings, filters).
  • Good dimensional accuracy and surface finish; economical for large batches.

Limitations

  • High cost of powder and tooling; economical only for large quantities.
  • Size limited by press capacity; complex shapes with undercuts are difficult.
  • Lower strength than wrought metal because of residual porosity.

Typical products

Self-lubricating bronze bearings, gears and cams, cemented carbide tool tips, filters, electrical contacts and sintered magnets.

Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.

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