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 Process Economics and Optimization Product design and Fabrication CNC programming Process Evaluation and Quality control

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

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

5. Materials Nanomaterials, shape-memory alloys, superconductors, … 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

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

9. Failure in Tension, Young’s modulus and Tensile strength Engineering stress =  = P/A o Engineering strain =  e = (L – L o )/L o =  /L o

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

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

12. Elastic recovery after plastic deformation

13. True Stress, True Strain, and Toughness Engg stress and strain are “gross” measures:  = F/A =>  is the average stress ≠ local stress e =  /L o => e is average strain Final Necking Fracture engg strain d/L o true strain ln(L/L o ) true stress P/A engg stress P/A o fracture 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 (L f – L o )/L o 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. Shear stress and Strain: the torsion test L  C C’  T T L  T T D d T = torque, J = polar moment of inertia J =  r 2 dA Cylindrical shell: J =  D 4  d 4 )/32 Angle of twist:  = TL/GJ Shear stress:  = Tr/J Maximum shear stress =  max = TR/J Shear strain =  = r  /L G: Modulus of rigidity  = G 

17. Shear strength and Tensile strength [approximate relation between shear and tensile strengths] MaterialTensile-RelationYield-Relation Wrought Steel & alloy steelS su ≈ 0.75 x S u S syp = Approx 0,58 x S yp Ductile IronS su ≈ 0.90 x S u S syp = Approx 0,75 x S yp Cast IronS su ≈ 1.3 x S u - Copper & alloysS su ≈ [0.6-0.9] x S u - Aluminum & alloysS su ≈ 0.65 xS u S syp = Approx 0,55 x S yp References: Machine design Theory and Practice.A.D.Deutschman, W.A Michels & C.E. Wilson.. MacMillan Publishing 1975. Ultimate Tensile Strength = S u Ultimate Shear Strength = S su Tensile Yield Strength = S yp Shear yield point = S syp

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

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

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 PropertyApplication (e.g.) Density,  = mass/volume Drop forging, hammering Specific heatCoolant in machining Thermal conductivityCutting titanium Coeff of linear thermal expansion,  =  L/(L  T) Compensation in Casting, … Melting pointBrazing, Casting, … Electrical conductivityEDM, ECM, Plating Magnetic propertiesMagnetic chucking

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

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/l 0 )…………….(2) A= πd 2 /4………………(3) solving for Δl (and realizing that E = 207 GPa, Table 6.1), yields Δl = (4Fl 0 )/(πd 2 E)…………(4) =0.325mm