Manufacturing · Chapter 8 of 10 · Intermediate
Polymer Processing and Additive Manufacturing
Plastics are shaped by melting and moulding at huge volumes; additive builds parts layer by layer with almost no tooling. Two very different routes, one shared idea: geometry sets the time.
Readiness check
This chapter uses polymer behaviour and simple heat and geometry. Tick only what you can do closed-notes.
- Distinguish thermoplastic from thermoset (Materials Chapter 10).
- Recall thermal diffusivity and conduction cooling.
- Evaluate a natural logarithm.
- Compute a count from a length and a step.
- Multiply a rate by a count for total time.
The core idea
Polymers are shaped by melting and forcing them into a mould, where cooling dominates the cycle; additive builds parts one layer at a time, where the number of layers dominates the build.
tcool ∝ s²/αcycle = inject + pack + cool + ejecttbuild = (H/layer)·tlayerA thermoplastic softens when heated and can be moulded repeatedly; a thermoset cures into a permanent network. Injection moulding melts pellets and injects them into a cooled mould, where the part must solidify before ejection, and that cooling time grows with the square of the wall thickness. Additive manufacturing instead deposits or fuses material layer by layer, so build time scales with the part height divided by the layer thickness, trading speed against resolution and tooling-free freedom.
The skills, taught in order
Two process families share this chapter. Five skills cover polymer behaviour, injection moulding, other polymer processes, additive principles, and build planning.
8.1 Polymers in processing
Thermoplastics (polyethylene, nylon) melt and can be reshaped; thermosets (epoxy, phenolic) cure once and cannot remelt. Polymer melts are viscous and shear-thinning, flowing more easily when sheared hard, which is what lets them fill thin moulds under pressure.
8.2 Injection moulding
The cycle is inject, pack, cool, eject. Cooling dominates, and for a wall of thickness s and thermal diffusivity α it follows tcool = (s²/π²α)·ln[(4/π)(Ti−Tmould)/(Teject−Tmould)]. The thickness-squared dependence is why thin, uniform walls are a core design-for-moulding rule.
8.3 Other polymer processes
Different shapes call for different processes.
| Process | Makes |
|---|---|
| Injection moulding | complex 3D parts at high volume |
| Extrusion | continuous profiles, pipe, film |
| Blow moulding | hollow parts such as bottles |
| Thermoforming | thin shells, trays, panels |
8.4 Additive manufacturing principles
Additive builds a part directly from its 3D model, adding material layer by layer with little or no tooling. This frees the geometry (internal channels, lattices) but leaves layer lines and anisotropy.
| Process | Material | Strength / use |
|---|---|---|
| FDM / FFF | thermoplastic filament | cheap prototypes |
| Stereolithography | photopolymer resin | fine detail, smooth |
| Selective laser sintering | nylon powder | functional, no supports |
| Metal (laser melting) | metal powder | end-use metal parts |
8.5 Build time and design for additive
Build time is roughly the number of layers times the per-layer time, so tbuild ≈ (H/layer)·tlayer. Thinner layers improve finish but multiply the build time. Designing for additive means orienting for strength and surface, minimising supports, and accepting layer-direction anisotropy.
Engineering connection: injection moulding makes nearly every plastic product at scale, while additive serves prototypes, low volumes, and shapes no mould can make, with cost crossing over by volume as in Chapter 1.
Worked example 1: injection-moulding cooling time
A plastic part has a 3 mm wall, moulded from melt at 220 °C in a mould at 30 °C, ejected at 80 °C. The polymer's thermal diffusivity is 1.3×10⁻⁷ m²/s. Find the cooling time, and compare with a 2 mm wall.
- ProblemFind the cooling time for the 3 mm wall in Figure 1, and for a 2 mm wall.
- Given / finds = 3 mm = 0.003 m, Ti = 220 °C, Tmould = 30 °C, Teject = 80 °C, α = 1.3×10⁻⁷ m²/s. Find tcool.
- AssumptionsOne-dimensional conduction through the wall to a constant-temperature mould.
- ModelUse the standard slab cooling-time formula for injection moulding.
- Equationstcool = (s²/π²α)·ln[(4/π)(Ti−Tmould)/(Teject−Tmould)]
- SolveThe bracket = (4/π)(190/50) = 1.273 × 3.8 = 4.84, ln 4.84 = 1.58. tcool = (0.003²/(π² × 1.3×10⁻⁷)) × 1.58 = 7.01 × 1.58 = 11.1 s. At 2 mm, the s² term gives (2/3)² × 11.1 = 4.9 s.
- CheckCooling time fell from 11 to 5 s when the wall dropped from 3 to 2 mm, the thickness-squared rule. Since cooling dominates the cycle, thin walls directly raise output, the key design-for-moulding lesson.
- ConclusionThe cycle is paced by cooling, and cooling by wall thickness squared. Thin, uniform walls are not just a material saving, they are the throughput of an injection-moulding line.
Worked example 2: additive build time
A part 60 mm tall is 3D-printed at a 0.2 mm layer thickness, with each layer taking about 15 s. Find the number of layers and the build time, then compare with a 0.1 mm layer for a finer finish.
- ProblemFind the layer count and build time for the part in Figure 2, and for a 0.1 mm layer.
- Given / findH = 60 mm, layer = 0.2 mm, tlayer = 15 s. Find N and tbuild.
- AssumptionsConstant per-layer time, build time dominated by layering (recoat plus scan).
- ModelNumber of layers is height over layer thickness; build time is that times the per-layer time.
- EquationsN = H/layer tbuild = N·tlayer
- SolveN = 60/0.2 = 300 layers. tbuild = 300 × 15 = 4500 s = 75 min. At 0.1 mm, N = 600 layers, so tbuild = 150 min.
- CheckHalving the layer thickness doubled both the layer count and the build time, the direct resolution-versus-speed trade. Orientation matters too: printing the part lying down would lower H and cut the time.
- ConclusionAdditive build time is paced by layer count, so part height and layer thickness dominate it. Choosing layer thickness and orientation is the core planning decision, just as wall thickness is for moulding.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Thick walls mould faster if filled fast | Long cycle, sink marks | "How does cooling scale with wall?" | Cooling time grows as thickness squared; keep walls thin and uniform. |
| Thermoset can be remelted | Reprocessing fails | "Is it thermoplastic or thermoset?" | Thermosets cure permanently; only thermoplastics remelt. |
| Finer layers are always better | Build times balloon | "Is the finish worth the time?" | Layer thickness trades finish against build time; match to need. |
| Additive parts are isotropic | Weak in the build direction | "Which way do the layers run?" | Layer bonding is weaker across layers; orient for the load. |
Practice ladder
A 90 mm tall part prints at 0.3 mm layers, 12 s each. Find the build time.
Show answer
N = 90/0.3 = 300 layers; t = 300 × 12 = 3600 s = 60 min. Layer count times per-layer time.
For the Worked Example 1 part, what wall thickness would halve the cooling time from the 3 mm value?
Show answer
Since t ∝ s², halving t needs s = 3/√2 = 2.12 mm. A small wall reduction gives a large cycle saving because of the square law.
A bracket is needed in quantities of 50 and of 50 000. Which favours additive and which injection moulding, and why?
Show answer
At 50 units, additive wins: no mould tooling cost to amortize. At 50 000, injection moulding wins: the expensive mould pays off over the volume and the per-part cycle is seconds, not hours. This is the Chapter 1 break-even applied to plastics.
Find one injection-moulded product and one 3D-printed part. Estimate the cooling time or build time, and explain why each process suited the product and its volume.
What good work looks like
A cooling-time or build-time estimate, the wall thickness or layer choice identified as the driver, and a volume-based justification for the process.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Plan one polymer part two ways: estimate the injection-moulding cooling time and the additive build time, and recommend a process for a stated volume.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. What are the steps of the injection-moulding cycle?
Inject, pack, cool, eject; cooling usually dominates.
2. How does cooling time scale with wall thickness?
As the square of the thickness (t ∝ s²/α).
3. Difference between thermoplastic and thermoset in processing?
Thermoplastics remelt and reshape; thermosets cure permanently.
4. How is additive build time estimated?
tbuild ≈ (H/layer) × per-layer time.
5. Name an advantage and a limitation of additive.
Advantage: tooling-free geometric freedom. Limitation: layer-direction anisotropy and slow builds.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Kalpakjian and Schmid, Manufacturing Engineering and Technology, Chapters 19 (polymer processing) and 20 (rapid prototyping and additive) |
| Cross-reference | Groover, Ch. 13 and 33 · Gibson, Rosen, Stucker, Additive Manufacturing Technologies |
| Core topics | 8.1 Polymer behaviour · 8.2 Injection moulding · 8.3 Other polymer processes · 8.4 Additive principles · 8.5 Build planning |
| Engineering connection | Mass-produced plastics by moulding; prototypes and complex parts by additive. |
| Read next | Chapter 9: Metrology, Tolerances, and Quality. |