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What is Materials Science and Engineering

What is Materials Science and Engineering 30
MaterialsWhat is Materials Science and Engineering Performance Engineering Design Processing Structure Materials Engineering Materials Science PropertiesThe goal of materials science is to empower scientists and engineers to make informed decisions about the design, selection and use of materials for specific applications. Photograph: Nils Jorgensen/REX ntimplants/healthbasezimmertotalhipreplacementimplantco mponents.jpg Under Armour piezoelectric ad Fundamental Tenets Guide Materials Science 1. The principles governing the behavior of materials are grounded in science and are understandable 2. The properties of a given material are determined by its structure. Processing can alter the structure in specific and predictable ways 3. Properties of all materials change over time with use and exposure to environmental conditions 4. When selecting a material for a specific application, sufficient and appropriate testing must be performed to ensure that the material will remain suitable for its intended application throughout the intended life of the productA materials scientist or engineer must be able to: 1. Understand the properties associated with various classes of materials 2. Know why these properties exist and how they can be altered to make a material more suitable for a given application 3. Be able to measure important properties of materials and how those properties will impact performance 4. Evaluate the economic considerations that ultimately govern most material issues 5. Consider the longterm effects of using a material on the environmentFundamental types of materials important to engineering: 1. Crystals Engineering metals and alloys Systemic, regular pattern, minimize volume 2. Engineering Ceramics (including glass) High viscosity at liquidsolid point prevents crystallization. These materials are usually amorphous 3. Polymers Long chains of simple, molecular structures. Plastics and living things 4. Elastomers Long chain polymers which fold or coil. Natural and artificial rubber. Enormous extensions associated with folding and unfolding of chains.Semiconductors Ceramics Metals Steel reinforced concrete Polymers Concrete Fundamental Material Atom 1. What are the atomic building blocks a) Nucleus – Protons (+) and Neutrons (0) b) Electrons () c) Atoms have a neutral charge (protons =electrons) 2. How are electrons distributed through an atom a) Electrons in organized shells in an electron cloud 2 b) electrons/shell = 2N (N = the shell number) 3. What are valence electrons Why are they important a) Valence electrons are in the outermost shell b) Reactivity of atom depends upon valence electrons 4. Why are noble gases inert a) The noble gases have full shells of electrons Atom = stadium Nucleus = housefly in the center of the field 26TBankStadiumDoD.jpg/300px M26TBankStadiumDoD.jpgAtoms – Can we see them Electron Microscopy and Scanning Probe Microscopy Xe on Ni Au resolutionAu100.JPG/200pxAtomicresolutionAu100.JPGThe Fundamental Material Atom 1. All atoms of a given element are 2. Atoms of different elements have different 3. A compound is a specific combination of atoms of more than one element 4. In a chemical reaction, atoms are neither created nor destroyed – only change partners to produce new substances HCl + NH NH Cl 3 4 wikipedia/commons/a/a0/Hy drochloricacidammonia.jpg holds the atoms in metals/crystals, ceramics, polymers and elastomers together BONDSCovalent Bonds • Two or more atoms share electrons • Strong and rigid Shared electron from Shared electron • Found in organics and sometimes ceramics hydrogen from carbon • Strongly directional • Methane CH 4 Carbon has valence electrons Hydrogen has valence electrons • Elemental solids – diamond • Can be stronger – diamond • Can be weaker Bi Bonds • Bonding between a metal and a nonmetal • Metal gives up valence electron(s) to nonmetal • Results in all atoms having stable electron configuration + • Na Cl • Metal becomes +ly charged (cation); nonmetal becomes –ly charged (anion) • Coulombic attraction • Closepacked YES NO YES Example: Na (+) (small) and Cl ()(large) Packing: as close as possible. chloride3Dionic2.pngMetallic Bonds • Hold metals and alloys together • Allows for dense packing of atoms (why metals are heavy) • Valence electrons are not bound to a particular atom and are free to drift through the entire material = “sea of electrons” • Nonvalence electrons + atomic nuclei = ion core (with a net + charge) • Good electrical conductivity • Good heat conductivityIntermolecular Forces Bonds holding molecules together Hydrogen bonds • Intermolecular attraction in which a H atom bonded to a small, electronegative atom (N, O or F) is attracted to a lone pair of electrons on another N, O or F atom. • Weak interactions • Due to the charge distribution on molecule + + • Often seen in organic compounds Example: H O • 5 to 30 kJ/mole (as compared to about 100kJ/mole for chemical bond) 2 … OH O=Intermolecular Forces Bonds holding molecules together Van der Waal forces: • Shorttime interactions • Arise from surface differences across molecules • Weaker forces (10 kJ/mole) • Gecko feet: Microscopic branched elastic hairs on toes which take advantage of these atomic scale attractive forces to grip and support heavy loads'stoe.jpgConsequences of Structure • Structure is related to the arrangement of the components of a material • This could be on any length scale – atomic, nano, micro, macro • All length scales matter • Types of carbon (literally just carbon) a) Diamond b) Graphite c) Lonsdeleite d) Buckminster Fullerene C60 e) C540 f) C70 g) Amorphous carbon h) Carbon Nanotube /Science/Chemistry/EightAllotropesofCarbon.pngSpaghetti Take a 10 minute break During the break take time to think about spaghetti How does it break How many ways can it break How can it be made strongerMaterial Properties Young’s Modulus We will StressStrain Curve Tensile Strength determine for Spaghetti Yield Strength Compressive Strength Shear Strength Ductility Poisson’s Ratio Specific Weight Specific Modulus HardnessMechanical Properties How easy does spaghetti break in tension (by pulling) Is thicker spaghetti easier or harder to break in tension Theory says that force needed to break in tension increases with crosssectional area. How easily does spaghetti buckle in compression Depends on force, material strength, length and thickness of spaghetti A longer piece buckles easier than a shorter piece A thinner piece buckles easier than a thicker piece How easily does spaghetti bend if you push on it perpendicularly Is it in tension or compression Deflection depends on force, material strength, span length and crosssectional area. A larger force yields a larger deflection For a given force, longer pieces bend easier For a given force, thin pieces bend easierTerms associated with material properties Hardness resistance to scratching and denting. Malleability ability to deform under rolling or hammering without fracture. Toughness ability to absorb energy, e.g., a blow from a hammer. Area under stressstrain curve is a measure of toughness. Ductility ability to deform under tensile load without rupture; high percentage elongation and percent reduction of area indicate ductility Brittleness material failure with little deformation; low percent elongation and percent area reduction. Elasticity ability to return to original shape and size when unloaded Plasticity ability to deform nonelastically without rupture Stiffness ability to resist deformation; proportional to Young’s Modulus E (psi) E = stress/strain (slope of linear portion of stress/strain curve).Stress and Strain Stress is related to the force or load applied to a F material Stress = σ = Force/original area From Figure: σ = F/A 0 2 Force units Newton = kgm/s 0 2 σ = F/A units = N/m = Pascal 0 6 MPa = 10 Pa 0 F 9 GPa = 10 PaStress and Strain Strain Strain = ε = change in length divided by original length F From Figure: ε = Δ / 0 0 Units Strain is unitless May be reported as: m/m, in/in or Elastic Strain vs. Plastic Strain (Rubber Band vs. Silly Putty) Elasticloading and unloading returns material to original lengthcan be done repeatedly, e.g., a watch spring. Plasticlarger deformations are not reversible when "elastic limit" is exceeded. Some materials are almost purely plastic, e.g., putty. FHooke's Law Robert Hooke, 1679 “As the extension, so the force” i.e., stress is proportional to strain Hooke's Law: an approximation of the relationship between the deformation of molecules and interatomic forces. force (tension) interatomic distance neutral position 10/13PortraitofRobertHooke.JPGHooke’s Law Hooke’s Law of Spring Extension F=kx Hooke’s Law applied to Linear Elastic Solids σ = Eε Young’s Modulus • The ratio of stress to strain within the linear (elastic) region of the stressstrain curve • A measure of the “stiffness” of a material • Also known as the Modulus of Elasticity • Units are the same as the units of stress (F/A) Thomas Young (1773 – 1829) E = σ/ε 9/9f/ThomasYoung(scientist).jpgElastic solids – Young’s Modulus Think of E as the stress required to deform a solid by 100. (Most solids will fail at an extension of about 1, so this is usually hypothetical). Range of E in materials is enormous: E(metal) 45400 GPa Load E(Ceramics) 60500 GPa Slope = Young’s Modulus E(Polymers) 0.014 GPa E(Spaghetti) 4.8 Gpa Unload Higher E implies Greater Stiffness Strain Stress (Pa)Material testing – Tensile Strength Usually tested by controlling extension (strain) and measuring resulting load (stressarea), i.e., independent variable is strain, dependent variable is stress Can also be determined by subjecting material to a predetermined load and measuring elongation, i.e., independent variable is stress, dependent variable is strain Solid Behavior Tension After tensile testing: a) Brittle b) Ductile c) Completely Ductile Examples: a) Cast Iron b) Aluminum c) PuttyStressStrain Curves 4 5 3 2 Tensile Test 1 1. Proportionality Limit 2. Elastic Limit 3. Yield Strength 4. Ultimate Tensl Strgth 5. Fracture Strength E = Young’s Modulus, E Ds Modulus of Elasticity E Elastic Region De Plastic Region 0.2 Strain, e (m/m) Stress, s (MPa)Range of Tensile Strengths (TS) How hard a pull is required to break material bonds steel piano wire = 450,000 p.s.i. aluminum = 10,000 p.s.i. concrete = 600 p.s.i.Which curve is typical of: A ductile material A brittle materialStress – Strain CurvesYoung’s ModulusMaterial Strength Compression Materials fail in compression in many ways depending on their geometry and support a) bucklinghollow cylinders, e.g., tin can b) bendinglong rod or panel c) shatteringheavily loaded glass No relation between compressive and tensile strength in part because distinction between a material and a structure is often not clear. e.g., what is a brick or concreteCompressive strength of material Under compression a beam will fail either by crushing or buckling, depending on the material and L/d; e.g., wood will crush if L/d 10 and will buckle if L/d 10 (approximately). Crushing: atomic bonds begin to fail, inducing increased local stresses, which cause more bonds to fail. Buckling: complicated, because there are many modes st nd rd 1 , 2 , and 3 order bending modes. Lowest order is most likely to occur.Euler buckling How the specimen ends are supported must be considered. G/1024pxBuckledmodel.JPGmaterials Material testing Euler buckling load, P c 2 P load (MLT ) c 4 I moment of inertia (L ) 1 2 E Young’s modulus (ML T ) L length (L) We have 4 variables, 3 primitive dimensions = 1 dimensionless groupEuler buckling load The force at which a slender column under compression will fail by bending 2  EI F  2 (KL) E = Young’s modulus K = 1.0 (pinned at both ends) I = area moment of inertia = 0.699 (fixed at one end, pinned at the other L = unsupported length = 0.5 (fixed at both ends) = 2.0 (free at one end, pinned at the other)Area moment of inertia 4 I = area moment of inertia (dim L )—associated with the bending of beams. Sometimes called second moment of area. Not to be confused with 2 I = mass moment of inertia (dim ML ) — associated with the energy of rotation)Some area moments of inertia 4 4 4 4 d a  (D d ) I  I  I  64 12 64 3 3 3 2sb ht bd I  I  12 12Material Strength Bending length L deflection y compression: proportional to distance from neutral axis neutral axis load P support shear tension: proportional to load distance from neutral axisRestoring moment = (moment arm about distance to neutral line) x (force) = neutral line ys(y)dA  s(y) dA y But, s is proportional to strain e, and strain varies linearly with distance to the neutral line. Therefore, s = y s , where s is max max the stress at the maximum distance from the 2 neutral line. So, Restoring moment = s y dA  s I  max max where I is the area moment of inertia of the cross section of the beam about the neutral axis. Moment of inertia depends on crosssection geometry and 4 has units L .materials Material testing bending 3point Bend Test• Bending Test – Setup Materials good in compression stone, concrete Materials good in tension carbon fiber, cotton, fiberglass Materials good in both compression and tension steel, woodOther strengths Shear strengthrotating axles fail because their shear strengths were exceeded Ultimate tensile strengthmaximum possible load without failure Yield strengthload required to cross line from elastic to plastic deformationNEXT: How do we put materials together to form structures. . .Beams and loadstension: Beam under tension Failure occurs when ultimate tensile strength is exceeded. Maximum load is tensile strength times crosssectional area. L = T A max cs For regular spaghetti (diameter = 2mm), maximum load is 10 pounds. Load capacity does not depend on length.Beams and loadscompression: d L Beam in compression Failure occurs two ways: 1) When L/d 10, failure is by crushing 2) When L/d 10, failure is by buckling We are almost always concerned with failure by buckling.Beams and loadscompressive buckling: 4 2 Buckling strength F = k d /L To determine constant of proportionality k: 1) Measure length and diameter of a piece of spaghetti 2) Hold spaghetti vertically on postal scale 3) Press down on spaghetti until it begins to bend 4) Read load F on postal scale 5) Calculate kSome consequences of buckling properties: If a beam of length L and diameter d can support a compressive load of F, d F L then a beam of length L/2 and diameter d can support a compressive load of 4F. 4F d L/2ALSO… If a beam of length L and diameter d can support a compressive load of F, d F L then a beam of length L and diameter 2d can support a compressive load of 16F. 2d 16F LBeams and loadsbending: Very little strength. Never design a structure that relies on bending strength to support a load.