ME 365 Manufacturing Techniques

1. ME 365 Manufacturing Techniques

2. Materıal concepts Materials Properties - helps to determine how to make things with it - helps to determine the processing conditions - helps and constrains process optimization Processes - forming, cutting, non-traditional, joining, surface treatments, electronics components, … Process Planning CNC programming Process Evaluation and Quality control Process Economics and Optimization Product design and Fabrication

3. Motivation (1) A bottle of water (~HK$6) Four components (bottle, cap, label, water) - How are each of these manufactured? - What does the equipment cost?

4. Motivation (2) Stapler (~HK$ 45) Approx. 15 components - How do we select the best material for each component? - How are each of these manufactured? Car: ~ 15,000 parts; Boeing 747 plane: ~6 million parts Intel core 2 duo processor: 65 nm feature size, 291 million transistors

5. Materials Ferrous metals: carbon-, alloy-, stainless-, tool-and-die steels Non-ferrous metals: aluminum, magnesium, copper, nickel, titanium, superalloys, refractory metals, beryllium, zirconium, low-melting alloys, gold, silver, platinum, … Plastics: thermoplastics (acrylic, nylon, polyethylene, ABS,…) thermosets (epoxies, Polymides, Phenolics, …) elastomers (rubbers, silicones, polyurethanes, …) Ceramics, Glasses, Graphite, Diamond, Cubic Boron Nitride Composites: reinforced plastics, metal-, ceramic matrix composites Nanomaterials, shape-memory alloys, superconductors, …

6. Properties of materials Mechanical properties of materials Strength, Toughness, Hardness, Ductility, Elasticity, Fatigue and Creep Physical properties Density, Specific heat, Melting and boiling point, Thermal expansion and conductivity, Electrical and magnetic properties Chemical properties Oxidation, Corrosion, Flammability, Toxicity, …

9. Failure in Tension, Young’s modulus and Tensile strength Engineering stress = s = P/Ao Engineering strain =e = (L – Lo)/Lo=d/Lo

10. Final Necking Fracture Failure in Tension, Young’s modulus and Tensile strength.. Original

11. Failure in Tension, Young’s modulus and Tensile strength… In the linear elastic range: Hooke’s law: s = E e or, E = s/e E: Young’s modulus

12. Elastic recovery after plastic deformation

13. Final fracture Necking Fracture fracture engg stress P/Ao true stress P/A engg strain d/Lo true strain ln(L/Lo) True Stress, True Strain, and Toughness Engg stress and strain are “gross” measures: s = F/A => s is the average stress ≠ local stress e = d/Lo => e is average strain Toughness = energy used to fracture = area under true stress-strain curve

14. Ductility Measures how much the material can be stretched before fracture Ductility = 100 x (Lf – Lo)/Lo High ductility: platinum, steel, copper Good ductility: aluminum Low ductility (brittle): chalk, glass, graphite - Walkman headphone wires: Al or Cu?

15. Hardness resistance to plastic deformation by indentation

16. T T L g L T C q D g C’ T d Shear stress and Strain: the torsion test Angle of twist: q = TL/GJ Shear stress: t = Tr/J Maximum shear stress = tmax = TR/J Shear strain = g = rq/L T = torque, J = polar moment of inertia J =  r2 dA Cylindrical shell: J = p( D4-d4)/32 t = G g G: Modulus of rigidity

17. Shear strength and Tensile strength [approximate relation between shear and tensile strengths] Ultimate Tensile Strength = Su Ultimate Shear Strength = Ssu Tensile Yield Strength = Syp Shear yield point = Ssyp References: Machine design Theory and Practice .A.D.Deutschman, W.A Michels & C.E. Wilson.. MacMillan Publishing 1975.

18. Fatigue Fracture/failure of a material subjected cyclic stresses S-N curve for compressive loading

19. Failure under impact Application: Drop forging Testing for Impact Strength

20. Strain Hardening - Metals microstructure: crystal-grains - Under plastic strain, grains slipping along boundaries - Locking up of grains => increase in strength - We can see this in the true-stress-strain curve also Applications: - Cold rolling, forging: part is stronger than casting

21. Residual stresses Internal stresses remaining in material after it is processed Causes: - Forging, drawing, …: removal of external forces - Casting: varying rate of solidification, thermal contraction Problem: warping when machined, creep Releasing residual stresses: annealing

22. Physical Properties

23. Summary Materials have different physical, chemical, electrical properties Knowledge of materials’ properties is required to Select appropriate material for design requirement Select appropriate manufacturing process Optimize processing conditions for economic manufacturing …

24. Examples A cylindrical specimen of steel having a diameter of 15.2 mm and length of 250 mm is deformed elastically in tension with a force of 48,900 N. Using the data contained in Table 6.1, determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen. Will the diameter increase or decrease?

25. Solution We are asked, in this portion of the problem, to determine the elongation of a cylindrical specimen of steel. σ = Eε…………………..(1) F/A=E(Δl/l0)…………….(2) A= πd2/4………………(3) solving for Δl (and realizing that E = 207 GPa, Table 6.1), yields Δl = (4Fl0)/(πd2E)…………(4) =0.325mm