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ENGINEERING INNOVATION- July 9, 2010 DUE TODAY:  Bending Mini-Report – place in homework box. CLASS TOPICS TODAY  College Ranking Project  Discussion.

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Presentation on theme: "ENGINEERING INNOVATION- July 9, 2010 DUE TODAY:  Bending Mini-Report – place in homework box. CLASS TOPICS TODAY  College Ranking Project  Discussion."— Presentation transcript:

1 ENGINEERING INNOVATION- July 9, 2010 DUE TODAY:  Bending Mini-Report – place in homework box. CLASS TOPICS TODAY  College Ranking Project  Discussion of Materials Properties  Compression Lab DUE MONDAY  Grand Canyon Problem  College Ranking Presentation

2 How materials work Compression Tension Bending Torsion

3 If the orbit of the single electron of hydrogen were the diameter of the Superdome, then the nucleus would be the size of a pea!!

4 The electrons group into shells.  All chemical properties depend on electron configuration.

5

6 http://www.uky.edu/Projects/Chemcomics/

7 Copper 1955 Uncle Scrooge Adventures

8 Barium – 1997 Thunderbolts

9 Uranium 1964

10 Atoms combine into molecules  The way atoms combine gives materials their chemical properties.  Bonding  Ionic - electrostatic attraction  Covalent - 2 atoms share electrons  Metallic bonds - free electrons are glue

11 Example: H 2 O Covalent bonding (angle of 104 o )  “polar molecule” Intermolecular Bonding  Hydrogen Bonds –  Due to an unequal distribution of charge about the molecule N-H; F-H, O-H covalent bonds required  Directional  Weaker than bonds between atoms  Van der Waals forces (London Dispersion forces)  Due to fluctuating electron cloud  Non-directional  Very weak!

12 Forces between molecules determines the states of materials.  Solid  Liquid  Gas

13 Solids have many forms.  There is a great variety in the types of intermolecular bonds in solids:  crystals  ceramics & glasses  metals  polymers  semiconductors

14 Some material properties of solids  Hardness - resistance to scratching & denting.  Malleability - ability to deform under rolling or hammering.  Toughness - ability to absorb energy, like a blow from a hammer.  Ductility - ability to deform under tensile load without rupture.  Brittleness - material failure with little deformation.  Elasticity - ability to return to original shape/size when unloaded.  Plasticity - ability to deform non-elastically without rupture.  Stiffness- ability to resist deformation; proportional to Young’s modulus E (psi) (slope of linear portion of stress/strain curve).  Reactivity - ability to react to chemicals.

15 Kinds of materials  An elastic material returns to its original length (or shape) when any load is removed.  A plastic material distorts easily but does not break.  A strong material is one with a high breaking stress.  A weak material is one with a low breaking stress.  A stiff material needs a large force (tensile stress) to produce a small extension (tensile strain) - it is difficult to change its shape.  A flexible material only needs a small stress to produce a large extension - it is not difficult to change its shape. http://www.matter.org.uk/schools/Content/YoungModulus/stiffnessExercise.html

16 Characterize these materials  Silly Putty  Straw  Paper Clip  Spaghetti  Rubber Band  Popsicle Stick  Piece of rope  Elastic  Plastic  Strong  Weak  Stiff  Flexible  Brittle

17 Why do solid materials fail?  Bonds break.  Think of bonds like little springs holding atoms or molecules.

18 Hooke’s Law - 1679 Hooke's law: an approximation of the relationship between the deformation of molecules and interatomic forces.

19 Stress - Strain Robert Hooke, 1679 "As the extension, so the force", i.e., stress is proportional to strain  Stress (  ) = Force/area = P/A Units = N/m 2 = Pa  Strain (  ) = the amount of stretch under load per unit length

20 Stress – strain curves  In the elastic region, materials deform and return to their original shape.  In the plastic region, materials permanently deform. stress = force/area strain =  shape

21 Stress-strain curves

22 substitutional defects interstitional defects (e.g., hydrogen embrittlement) ( from IMPRESS, esa) Imperfections leading to strength properties

23 Sketch stress-strain curves  Spaghetti  Silly putty  Rubber Band  Spandex (2009 final exam question)  Glass Rod (2009 final exam question)  Sledgehammer (2009 final exam question)  Paperclip (2009 final exam question)

24 Young's modulus (E) Thomas Young in 1807 realized that stress/strain (  /  ) = constant This constant is now called Young’s Modulus (E) E = stress/strain =  /  = constant E describes flexibility and is a property of the material. E is also used to define stiffness. E has units of stress (load/area) (P/A) psi; Pa; MPa; MN/m 2 Range of E in materials is enormous: E(rubber) = 0.1 GPa E(diamond) = 1220 GPa E(spaghetti) = 4 GPa (4x10 9 Pa) Young’s Modulus

25 Stress - strain curves  At the elastic limit, materials deform and don't return to their original shape.

26 Material strength A. Tensile strength How hard a pull required to break material bonds? steel piano wire = 450,000 p.s.i. aluminum = 10,000 p.s.i. concrete = 600 p.s.i. B. Compression strength 1. Difficult to answer, because materials fail in compression in many ways depending on their geometry and support a) buckling--hollow cylinders, e.g., tin can b) bending--long rod or panel c) shattering--heavily loaded glass

27 C. 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 concrete? D. Other strengths 1. Shear strength--rotating axles fail because their shear strengths were exceeded 2. Ultimate tensile strength--maximum possible load without failure 3. Yield strength--load required to cross line from elastic to plastic deformation

28 Materials good in compression stone, concrete Materials good in tension carbon fiber, cotton, fiberglass Materials good in both compression and tension steel, wood

29 G. Material testing 1. Tensile strength a) Usually tested by controlling extension (strain) and measuring resulting load (stress*area), i.e., independent variable is strain, dependent variable is stress b) Can also be determined by subjecting material to a predetermined load and measuring elongation, i.e., independent variable is stress, dependent variable is strain

30 B. Bending

31 3. Compressive strength of material a) 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 (approximately). b) Crushing: atomic bonds begin to fail, inducing increased local stresses, which cause more bonds to fail. c) Buckling: complicated, because there are many modes 1 st, 2 nd, and 3 rd order bending modes. Lowest order is most likely to occur.

32 Euler buckling

33 Euler buckling load The force at which a slender column under compression will fail by bending E = Young’s modulus I = area moment of inertia L = unsupported length K = 1.0 (pinned at both ends) = 0.699 (fixed at one end, pinned at the other = 0.5 (fixed at both ends) = 2.0 (free at one end, pinned at the other)

34 I = area moment of inertia (dim L 4 )—associated with the bending of beams. Sometimes called second moment of area. (Not to be confused with I = mass moment of inertia (dim ML 2 )— associated with the energy of rotation) Area moment of inertia

35 Some area moments of inertia

36 Spaghetti tests.  Breaking under tension.  Bending under perpendicular force.  Buckling under compression.

37 Failure from bending under perpendicular force  The perpendicular force (load) deforms the material.  The top half is compressed.  The bottom half is pulled apart.  Finally the tension breaks bonds  failure.

38 Follow handout directions  Measure  spaghetti diameter  length between supports  force (weight of nuts)  deflection  Repeat

39 Bending Results - Graph

40 Bending Results – Young’s Modulus TrialTypeDiameter (m) Length of Gap (m)α (N/m)E (Pa) 1Thick0.001850.076245.6533.91E+09 2Thick0.001860.097124.3624.02E+09 3Thick0.001880.11662.513.31E+09 4Regular0.001480.076138.185.37E+09 5Thin0.001430.07674.84753.33E+09 6Angel Hair0.001110.07635.82444.40E+09

41 TODAY  Complete the bending data analysis and mini-report  Prepare for 3-5 minute presentation on college ranking (Monday July 12 th )  Complete your compression testing  Grand Canyon problem (Monday July 12 th )  Work on your formal report for the compression testing lab (Wednesday July 14 th )  Select a research paper topic (Friday July 23 rd )


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