Chapter 6 - The Manhattan Bridge, under construction in...

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Unformatted text preview: The Manhattan Bridge, under construction in 1909, nine months before its opening on December 31, 1909 • Many materials are used to bear and sustain loads. – – – – • buildings bridges cars space shuttle We need to understand… – … how much force a material will withstand without: • deflecting too much. • deforming permanently. • cracking or breaking. – … how the mechanical behavior changes over time due to: • high temperature exposure. • fatigue. • combinations of the above. 2 • What is a “mechanical response”? • If you apply forces to an object it can: – Move – Rotate – Deform • To eliminate the first two you must apply balanced forces. – You can measure deflection and distortion of the material under a load – The actual response depends on the: • material • structure and geometry • A material’s response is correlated by: – stress normalized force – strain normalized displacement 3 What Parameter Correlates A Material’s Behaviour? If it requires 2 baby elephants to break the lecture table how many elephants would it take to break two tables joined together? Why and what is the principle? If a 20 m long bungee cord stretches 6 m when you jump how long will a 30 m cord stretch? • A material’s response is correlated by: – stress normalized force . – strain normalized deflection . 4 • Three ways a load may be applied: – Tension – Compression – Shear • Engineering stress σ : The instantaneous load applied to a specimen divided by its cross‐sectional area before any deformation – Expressed in MPa F A0 1MPa 106N/m2 • Engineering Strain ε : The change in gauge length of a specimen in the direction of an applied force divided by its original gauge length l l l 0 i l0 l0 *NO UNITS • Common mechanical stress‐ strain test • Specimen is deformed to fracture • Increasing tensile load is applied uniaxially along the long axis of the specimen • Circular or rectangular cross section • Several mechanical properties can be determined 0.505 in Titanium Click to Play • Similar to tensile tests • Force is compressive and specimen contracts along the direction of stress • Compressive force is negative, yielding a negative stress and strain • Not as common as tensile tests; harder to perform • Used when a material’s behaviour under large and permanent ie. Plastic strain is desired, as in: – Manufacturing applications – Brittle in tension Compression test using concrete specimen click to play • Shear Stress τ : The instantaneous applied shear load divided by the original cross‐ sectional area across which it is applied F A0 • Shear Strain γ : The tangent of the shear angle that results from an applied shear load • Torsion: – A variation of pure shear; specimen is twisted – Rotational motion is produced about the longitudinal axis of one end of the member relative to the other end. – Example machine axles and drive shafts Aluminum specimen under torsional stress click to play Biaxial Tension Pressurized tank (Courtesy P. M. Anderson) Hydrostatic compression (Courtesy P. M. Anderson) Fish under water > 0 z > 0 9 σh < 0 Nike Bauer Supreme ONE95 hockey stick • Hooke’s Law: Tension Shear E G G= Shear Modulus – Expresses the relationship between engineering stress and strain for elastic deformation tension and compression • Modulus of Elasticity, E Young’s Modulus is the proportionality constant – Ratio of stress to strain – Deformation is completely elastic – Measure of the stiffness of a material • Elastic Deformation: deformation in which stress and strain are proportional linear relationship – Slope of line is the modulus of elasticity, E – Nonpermanent; when applied load is released, piece returns to its original shape – Independent of Time High E stiff material • The elastic modulus: – varies by 106! – depends little on alloying. • same for Fe, steel – depends weakly on temperature in a crystalline solid. Emelting po int 1 E0K 2 1200 10 00 800 600 400 200 10 0 80 60 40 12 Tungsten Molybdenum Steel, Ni Tantalum Platinum Cu alloys Zinc, Ti Silver, Gold Aluminum Magnesium, Tin Si carbide Al oxide Si nitride Carbon fibers only CFRE(|| fibers)* <111> Si crystal Aramid fibers only <100> AFRE(|| fibers)* Glass -soda Glass fibers only G FRE(|| fibers)* Concrete GFRE* 20 10 8 6 4 2 Anelasticity: Time dependent elastic deformation Diamond 1 0.8 0.6 0.4 0.2 CFRE * G FRE( fibers)* G raphite Polyester PET PS PC CFRE( fibers) * AFRE( fibers) * Epoxy only PP HDP E PTF E LDPE E in GPa Wood( grain) • Elasticity E is caused by bonds stretching. • E slope of Interatomic Force vs. separation curve at equilibrium spacing r r0 F E ~ r r r 0 2U ~ 2 r ~S r r r0 0 bonds stretch return to initial 13 F U r0 | r • y x z z • E 2G (1 ) • Permanent or non‐recoverable after release of the applied load; • Permanent atomic displacements; breaking of bonds with original atoms neighbors and reforming bonds with new neighbors • Upon removal of stress, atoms do not return to their original positions – Crystalline solids Accomplished by a process called slip motion of dislocations; discussed in Chapter 7 – Amorphous solids and liquids Viscous flow mechanism • Most metals experience at gradual transition from elastic to plastic region • Yielding: The onset of plastic deformation • Yield Strength σy : The stress required to produce a very specific amount of plastic strain; a strain offset of 0.002 is commonly used “Strength” is used in lieu of “stress” because strength is a property of the metal, whereas stress is related to the magnitude of the applied load Metals/ Alloys Graphite/ Ceramics/ Semicond Composites/ fibers Polymers 20 00 300 Yield strength, 200 Al (6061) ag Steel (1020) hr Ti (pure) a Ta (pure) Cu (71500) hr 100 70 60 50 40 Al (6061) a 30 20 17 10 Tin (pure) Meaning of symbols: ¨ dry PC Nylon 6,6 PET humid PVC PP HDPE LDPE Hard to measure, 700 600 500 400 Ti (5Al-2.5Sn) a W (pure) Cu (71500) cw Mo (pure) Steel (4140) a Steel (1020) cd in ceramic matrix and epoxy matrix composites, since in tension, fracture usually occurs before yield. 10 00 Hard to measure , since in tension, fracture usually occurs before yield. y (MPa) Steel (4140) qt (to be discussed later) a annealed ag aged cw cold worked cd cold drawn hr hot rolled qt quenched and tempered • After they yield, metals get harder to deform. work hardening • Tensile strength: The stress at the maximum on the engineering stress‐strain curve; materials maximum load TS • • 18 Fmax A0 At this maximum stress, a small neck begins to form in the specimen, termed “necking” Beyond Fmax we see tensile instability necking which leads to fracture Fracture • Ductility relates to how much the material plastically deforms before it fractures • May be expressed as percent elongation %EL or percent reduction in area %RA from a tensile test • A material that experiences very little or no plastic deformation up fracture is termed brittle % EL l f l0 l0 100 A0 Af % RA 100 A0 • Work done: dW F d d V A0 0 WV WT 0 • Consequences: – Energy is dissipated by fracture – Packaging & protection applications 20 dW ( )d E d V 0 0 • Resilience: the capacity of a material to absorb energy when it is deformed elastically and then, upon unloading, is able to recover this energy – That means there must be energy stored in the solid • Modulus of resilience, Ur: the elastic energy per unit volume J/m3 • The area under the engineering stress‐strain curve up to the point of yielding Incorporating Hooke’s Law to solve for Modulus of Resilience for elastic behavior y U r d 0 Assuming a linear elastic region and using, y E y and U 1 r yy 2 Solve for stored energy Ur 2 y 2E If you want to protect a pole vaulter from stress, would you use materials with low or high elastic moduli? – Why? – Give examples of materials Olympic pole vaulter Tatiana Grigorieva From abc.net.au • Answer: Remember that 12 WV / E 2 – For a given , to maximize WV, you need to minimize E. – Examples: foam low density polymers 22 “When modern man builds large load‐bearing structures, he uses dense solids; steel, concrete, glass. When nature does the same, she generally uses cellular materials; wood, bone, coral. There must be good reasons for it.” Prof. M. F. Ashby, University of Cambridge aluminum foam Sintered bronze powders (form metal sponge material) cellular iron-based material 23 aluminum sponge nickel sponge Metal foams can be made in several ways: bubbling gas through molten alloys using foaming agents forcing molten alloy into a precursor material They have unique properties which make them well‐suited to a variety of applications. 24 • • (click to play) Foams are also light‐weight • Crash test‐ Toyota Yaris Vs. Toyota Camry Foams can absorb large amounts of energy. They also distribute force isotropically These characteristics make them excellent materials for impact zones on vehicles. CRASHWORTHY CARS Crash‐absorbing aluminum foam might soon be integrated into many automobile components seen here in orange to better guard against collisions. Illustration by: Garry Marshall http://www.popsci.com/popsci/auto/article/0,12543,2 20392,00.html 25 •high strength •high cell‐wall heat conductivity •large surface area 26 from Chapter 19 • • – – – 27 • Example: A rod of fixed length L thermal T room L room L (T Troom ) Lroom L T compressive σ keeps ∆L = 0 28 E thermal • – This can cause cracking in weak or brittle materials. • E (T2 T1 ) • f rapid quench tries to contract during cooling doesn’t want to contract 29 T2 T1 • This is an alternative to tensile testing. • Usually the load required to produce a special indent is measured. Pindent H~ Aindent most plastics brasses Al alloys easy to machine steels file hard increasing hardness 30 cutting tools nitrided steels diamond • What are the advantages and disadvantages of a hardness test compared to a tensile test? • Pros: – – – – – cheaper, quicker non‐destructive quality control works on quasi‐brittle materials used on specific parts of inhomogeneous materials • composites and thin films • Cons: – less precise data, not useful for design calculations 31 • Yielding – onset of plastic deformation • Work hardening – typical of metals • Necking – tensile instability • Fracture 32 Engineering stress‐strain curve • For stress, divide by instantaneous area True stress‐strain curve • For strain, use the instantaneous length • Volume is conserved during deformation 33 • • • • • • 34 ‐2 ‐2 ‐4 ‐2 3 3 5 5 ‐5 ‐5 3. A tie‐rod in a BMW M3 needs to support a force of 10.0 kN without yielding. BMW Engineers are considering three different materials for this application; an advanced steel with a yield stress of 600 MPa and density of 7.80 g/cm3, an aluminum alloy with a yield stress of 190 MPa and density of 2.70 g/cm3 and a Mg alloy with a yield stress of 125 MPa and density of 1.74 g/cm3. Which of these materials will result in the lightest tie‐rod that can sustain the above force? Hint: The length of the rod is the same for all materials, but the diameter may be different . a b c d The steel The Aluminum alloy The Mg alloy They all result in the same tie‐rod mass. 4. Use the Engineering Stress Vs. Engineering Strain Curve s to answer the following questions. ...
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This note was uploaded on 10/14/2011 for the course ENGINEER CHEM ENG 3 taught by Professor Ghosh during the Spring '11 term at McMaster University.

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