Forging: open-die, closed-die and forging load

Open-die and impression-die forging, barrelling, flash, forging equipment, and forging load by the slab method for Coulomb and sticking friction.

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Why it matters

Crankshafts, connecting rods, gear blanks, turbine discs and hand tools are forged because forging gives the toughest, most fatigue-resistant parts: the metal's grain flow follows the part's shape and internal porosity is closed. Choosing the press or hammer means estimating the forging load, and that estimate depends on friction and on how thin the part is — which is exactly what GATE tests.

Key ideas

Open-die forging. The work is compressed between flat or simple-shaped dies that do not enclose it, and it is manipulated between blows. Upsetting (reducing height, increasing diameter), drawing out (cogging, reducing cross-section along the length with successive bites), fullering and edging (distributing metal), punching and piercing are the basic operations. It needs cheap dies, suits large parts and small quantities (ship shafts, rotor forgings) and needs skill; accuracy is limited.

Barrelling. Friction at the die faces restrains the metal there, so the middle of an upset cylinder bulges outward. Hot dies also chill the contact faces, which adds to barrelling. With frictionless (well-lubricated) faces the cylinder would stay straight-sided; with sticking friction the contact zone hardly slides at all.

Impression-die (closed-die) forging. The work is pressed into shaped die cavities, usually through several impressions (fullering, edging, blocking, finishing). Extra metal escapes into a thin flash around the parting line. The flash cools quickly and, because it is thin and wide, its friction resistance is very high; this back-pressure forces the metal to fill the cavity corners. Flash is then trimmed. Flashless (true closed-die) and precision forging need very accurate billet volumes. Advantages: complex shapes, good accuracy and properties, high rates. Limits: expensive dies, economic only in quantity; higher loads than open-die for the same area.

Equipment. Hammers (drop/gravity and power hammers) deliver energy in blows; mechanical presses deliver a set stroke; hydraulic presses deliver a set force at slower speed (better for large parts and for strain-rate-sensitive alloys). Hammers are rated by energy, presses by force.

Forging load — the slab method. Taking a thin slice of material between the dies and balancing forces gives a pressure that is lowest (equal to the flow stress) at the free edge and builds up toward the centre because friction resists outward flow. This friction hill means the average die pressure is larger than the flow stress, and grows with the ratio of width (or radius) to height. Thin, wide parts need much higher pressures than thick ones. Two friction models are used:

  • Coulomb (sliding) friction, τ = μp — lubricated, cold or moderate conditions.
  • Sticking friction, τ = k (the shear yield stress) — hot forging with no lubricant; the metal shears internally instead of sliding.

Flow stress. In hot forging use the flow stress at the forging temperature and strain rate (σ_f = C·ε̇ᵐ). In cold forging use the flow stress at the final strain (σ_f = K·εⁿ) for the peak load. For long strips in plane strain use Y′ = 1.155σ_f (von Mises).

Defects. Laps and folds (metal folded over itself when it flows the wrong way), cold shuts, incomplete die filling, internal cracks from tensile secondary stresses, and flow-line exposure at trimmed flash. Proper preform design and generous fillet radii avoid most of them.

Formulas

F = p_avg × A — forging load (N); p_avg = average die pressure (Pa), A = current contact area (m²) at the instant considered (usually the end of the stroke).

Volume constancy: d₀²·h₀ = d²·h — to find the diameter after upsetting.

Cylinder, Coulomb friction: p_avg ≈ σ_f × (1 + 2μr / (3h)) — r = current radius, h = current height, μ = coefficient of friction. Valid when 2μr/h is small (linearised friction hill).

Cylinder, sticking friction (Tresca, k = σ_f/2): p_avg = σ_f × (1 + r / (3h)).

Plane strain strip, Coulomb: p_avg ≈ Y′ × (1 + μb / (2h)); Plane strain strip, sticking: p_avg = Y′ × (1 + b / (4h)) — b = width of the strip in the flow direction, h = thickness, Y′ = 1.155·σ_f.

F = K_f × σ_f × A — empirical closed-die estimate (N); K_f = shape (multiplying) factor from a data book (roughly 1.2 for simple open-die upsetting up to 6–8 for complex parts with flash), A = projected area including flash.

Worked examples

Example 1 (standard — upsetting with friction). A steel cylinder 50 mm diameter × 50 mm high is hot-upset to 25 mm height. Flow stress at the forging temperature = 80 MPa, μ = 0.2. Find the load at the end of the stroke.

  1. Volume constancy: d = 50 × √(50/25) = 70.71 mm, r = 35.36 mm.
  2. p_avg = 80 × (1 + 2 × 0.2 × 35.36 / (3 × 25)) = 80 × (1 + 0.1886) = 95.1 MPa.
  3. A = π × 35.36² = 3927 mm².
  4. F = 95.1 × 3927 = 3.73 × 10⁵ N.

Answer: F ≈ 373 kN. Without friction it would be 80 × 3927 = 314 kN.

Example 2 (GATE level — plane strain, sticking). A long strip 60 mm wide and 300 mm long is hot-forged to 12 mm thickness under sticking friction. Flow stress = 200 MPa; use von Mises. Find the forging load.

  1. Y′ = 1.155 × 200 = 230.9 MPa.
  2. p_avg = Y′ × (1 + b/(4h)) = 230.9 × (1 + 60/48) = 230.9 × 2.25 = 519.6 MPa.
  3. A = 60 × 300 = 18 000 mm².
  4. F = 519.6 × 18 000 = 9.35 × 10⁶ N.

Answer: F ≈ 9.35 MN — more than twice the frictionless value because the strip is thin compared with its width.

Example 3 (closed-die estimate). A connecting rod has a projected area (including flash) of 0.03 m², flow stress 120 MPa at forging temperature and a shape factor of 2.0. F = 2.0 × 120 × 10⁶ × 0.03 = 7.2 × 10⁶ N → 7.2 MN.

Common mistakes

  • Using the initial area instead of the current (final) area for the peak load.
  • Forgetting volume constancy when finding the final diameter.
  • Using radius where the formula has diameter (or width b where it has half-width) — check which one your formula uses.
  • Using the room-temperature yield stress for a hot-forging problem.
  • Leaving out the 1.155 factor for plane strain when the question says von Mises.
  • Assuming flash is waste with no function; its resistance is what fills the die.

For GATE PI

Numericals ask for forging load in upsetting with Coulomb or sticking friction, plane-strain strip forging, and the final diameter from volume constancy; some ask for the friction-hill peak pressure at the centre. One-mark questions test flash function, barrelling, open vs closed die, and hammer vs press. Practise the slab-method results for both friction models and both geometries.

Quick check

  1. Why does an upset cylinder barrel?
  2. What does the flash do in closed-die forging?
  3. A 40 mm × 40 mm cylinder is upset to 20 mm height. New diameter?
  4. Where on the die face is pressure highest?
  5. Plane-strain flow stress for σ_f = 300 MPa (von Mises)?

Answers: 1. Friction (and chilling) at the die faces restrains flow there. 2. Its high flow resistance builds back-pressure to fill the cavity. 3. 40√2 ≈ 56.6 mm. 4. At the centre (peak of the friction hill). 5. 1.155 × 300 ≈ 346 MPa.

Try answering each one aloud before you open it.

  1. 1.What is open-die forging and how does it differ from closed-die forging?Concept

    Open-die forging involves deforming a metal workpiece between multiple dies that do not enclose the material entirely. It allows for the production of large and simple shapes. Closed-die forging, also known as impression-die forging, involves shaping the metal within a set of dies that enclose the workpiece, allowing for more complex shapes and better dimensional accuracy. The main difference is that open-die forging is used for simpler shapes and larger parts, while closed-die forging is used for more complex and precise components.

  2. 2.Explain the term 'forging load' and its significance in the forging process.Concept

    Forging load refers to the force required to deform a material during the forging process. It is significant because it determines the capacity of the forging equipment needed and affects the quality of the forged part. A higher forging load may be required for harder materials or more complex shapes, and it influences the energy consumption and wear on the forging equipment.

  3. 3.Why is open-die forging often used for large components?Application

    Open-die forging is often used for large components because it allows for the deformation of large workpieces without the need for custom dies. This process is flexible and can accommodate a wide range of sizes and shapes, making it ideal for producing large parts such as shafts, cylinders, and large rings. Additionally, open-die forging can improve the mechanical properties of the material through grain refinement.

  4. 4.What are the advantages of closed-die forging over open-die forging?Application

    Closed-die forging offers several advantages over open-die forging, including the ability to produce more complex shapes with better dimensional accuracy and surface finish. It also allows for higher production rates and less material waste due to the precise control of the material flow within the dies. Additionally, closed-die forging can improve the mechanical properties of the part through controlled deformation and grain flow.

  5. 5.What happens if the forging load is underestimated during the design of a forging process?Application

    If the forging load is underestimated, the forging equipment may not be able to apply sufficient force to properly deform the material, leading to incomplete filling of the die cavities and defects in the forged part. This can result in poor dimensional accuracy, surface defects, and compromised mechanical properties. Additionally, it may cause excessive wear or damage to the forging equipment.

  6. 6.How does temperature affect the forging load required in a forging process?Application

    Temperature significantly affects the forging load required in a forging process. As the temperature of the workpiece increases, the material becomes more ductile and easier to deform, reducing the forging load required. This is why hot forging is often used for materials that are difficult to deform at room temperature, as it allows for easier shaping and reduces the risk of cracking.

  7. 7.Calculate the load needed to start upsetting a cylindrical steel billet 0.1 m in diameter and 0.2 m high if its flow stress is 250 MPa.Numerical

    At the start of upsetting, ignoring friction, F = σ_f × A = 250 × 10⁶ × (π/4)(0.1)² = 250 × 10⁶ × 7.854 × 10⁻³ ≈ 1.96 MN. As the billet is upset its area grows (volume constancy) and friction adds a friction-hill term, p_avg ≈ σ_f(1 + 2μr/(3h)), so the load at the end of the stroke is higher. In cold forging the flow stress also rises with strain.

  8. 8.What is the role of lubrication in the forging process?Application

    Lubrication plays a crucial role in the forging process by reducing friction between the workpiece and the dies. This helps in lowering the forging load required, improving the surface finish of the forged part, and extending the life of the dies by reducing wear. Lubrication also aids in controlling the temperature of the workpiece and dies, preventing overheating and thermal damage.

  9. 9.Explain how grain flow is affected by the forging process and why it is important.Concept

    Grain flow refers to the alignment of the metal's crystalline structure during the forging process. Proper grain flow is important because it enhances the mechanical properties of the forged part, such as strength, toughness, and fatigue resistance. The forging process can be designed to control the grain flow, aligning it in directions that improve the performance of the part under service conditions.

  10. 10.A closed-die forging process requires a forging load of 500 kN. If the die area is 0.02 m², what is the average pressure applied on the workpiece?Numerical

    The average pressure applied on the workpiece can be calculated using the formula: Pressure = Force / Area. Here, the force is 500 kN = 500,000 N, and the area is 0.02 m². Therefore, the average pressure = 500,000 N / 0.02 m² = 25,000,000 N/m² or 25 MPa.

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