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Chapter 9 · 4 hours

Forging Practice

Practice questions

Practice questions and answers

3 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

Define forging. Explain the following forging operations: upsetting, drawing down, fullering, bending, punching and cutting (hot cutting). List the common forging tools.

Answer

Forging is shaping metal by hammering or pressing it, generally hot (red or white heat), so that plastic flow produces the required shape. It improves the grain flow and strength.

Operations

  • Upsetting: the length of a bar is reduced and the cross-section increased by compressing along its axis, e.g. making a bolt head or a rivet head. If the unsupported length exceeds about 3 times the diameter, the bar bends (buckles).
  • Drawing down: the cross-section is reduced and the length increased by hammering between flat or round faces; the metal is turned after each blow.
  • Fullering: grooves are made across the bar with a fuller to spread the metal in length (before drawing down) and to form shoulders.
  • Bending: the hot bar is bent over the anvil horn or in a bending fork to make hooks, angles and rings.
  • Punching: a punch drives a hole through the hot metal; the slug is removed through a pritchel hole (hardie hole). Drifting enlarges the hole.
  • Cutting: a hot set or hot chisel cuts the hot bar; a cold set is used for cold work.

Other operations: welding (forge welding), swaging, flattening.

Common forging tools

  • Anvil (with horn, face, hardie and pritchel holes), hammers (hand, sledge, ball peen), tongs of various jaws, flatters, set hammer, swages, fullers, drifts, punches, hot sets and chisels, and a forge or furnace.
 upsetting         drawing down
   | |               ___________
  _| |_             /__________/  length increased
  |___|  head
  • Practice · 6 marks

Differentiate between forging hammers and forging presses. State the advantages and limitations of forging.

Answer

Forging hammers and presses

Forging hammerForging press
Energy is delivered as a high-velocity blow (impact)Squeezing action with slow, steady pressure
Types: board, drop, steam/air, power hammerTypes: mechanical (crank, eccentric), hydraulic, screw
Deformation mostly on the surface layersDeformation reaches the full depth of the job
Noise, vibration; needs heavy foundationQuiet, smooth operation
Dies wear quickly; lower accuracyBetter dimensional accuracy; less draft needed
Capacity in kg of falling weightCapacity in tonnes of force
Cheaper, used for small and medium workCostly; suits large and heavy forgings

Drop hammer: ram falls under gravity (board or steam-lifted) on the heated billet in a die. Power hammer: steam or compressed air drives the ram down faster.

Advantages of forging

  • Grain flow follows the shape, giving high strength, toughness and fatigue resistance.
  • Removes internal voids and blowholes; gives dense structure.
  • Good for small and large parts, produces a good surface and accuracy in closed dies.
  • Material saving over machining from solid.

Limitations

  • High cost of dies and equipment for small batches.
  • Complex shapes with deep cavities and undercuts are difficult.
  • Hot forging gives scale and poor surface and dimensional accuracy.
  • Not suitable for brittle metals; needs heating and skilled operation; some safety risk.
  • Practice · 6 marks

A steel billet 40 mm in diameter and 120 mm long is upset hot between flat dies to a height of 50 mm. Assuming volume remains constant and ignoring friction and barrelling, find (a) the final diameter, (b) the true strain, and (c) the upsetting force if the flow stress at the forging temperature is 100 MPa. Comment on whether the billet will buckle during upsetting.

Answer

Given: d1=40d_1 = 40 mm, h1=120h_1 = 120 mm, h2=50h_2 = 50 mm, σf=100\sigma_f = 100 MPa.

(a) Final diameter

Volume is constant:

π4d12h1=π4d22h2⇒d2=d1h1h2\frac{\pi}{4}d_1^2 h_1 = \frac{\pi}{4}d_2^2 h_2 \Rightarrow d_2 = d_1\sqrt{\frac{h_1}{h_2}} d2=4012050=40×1.549=61.97 mmd_2 = 40\sqrt{\frac{120}{50}} = 40\times 1.549 = 61.97\ \text{mm}

(b) True strain

ε=ln⁡h1h2=ln⁡12050=0.875\varepsilon = \ln\frac{h_1}{h_2} = \ln\frac{120}{50} = 0.875

(c) Force at the end of the stroke

Final cross-section:

A2=π4(61.97)2=3016 mm2A_2 = \frac{\pi}{4}(61.97)^2 = 3016\ \text{mm}^2 F=σfA2=100×3016=301,600 N≈302 kNF = \sigma_f A_2 = 100\times 3016 = 301{,}600\ \text{N} \approx 302\ \text{kN}

The force is highest at the end of the stroke as the area is largest. With friction the real force would be higher.

Buckling check

h1d1=12040=3\frac{h_1}{d_1} = \frac{120}{40} = 3

This is at the limit of the rule that the unsupported length should not exceed 3 times the diameter in a single blow. The upsetting should be done in a die or by a series of light blows, or the free length should be reduced.

Answer: d₂ = 61.97 mm; true strain = 0.875; F = 302 kN; h/d = 3 is at the safe limit, so use a die or light blows.

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

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