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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 of Materials Properties Compression Lab
DUE MONDAY Grand Canyon Problem College Ranking Presentation
If the orbit of the single electron of hydrogen were the diameter ofthe Superdome, then the nucleus would be the size of a pea!!
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
+ +
-Example: H2O Covalent bonding (angle of 104o) “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!
Forces between molecules determines the states of materials.
Solid Liquid Gas
Solids have many forms.
There is a great variety in the types of intermolecular bonds in solids: crystals ceramics & glasses metals polymers semiconductors
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.
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
Characterize these materials
Silly Putty Straw Paper Clip Spaghetti Rubber Band Popsicle Stick Piece of rope
Elastic Plastic Strong Weak Stiff Flexible Brittle
Why do solid materials fail?
Bonds break. Think of bonds like little springs holding atoms or
molecules.
interatomic distance
force (tension)
neutral position
Hooke’s Law - 1679
Hooke's law: an approximation of the relationship between the deformation of molecules and interatomic forces.
Stress - StrainRobert Hooke, 1679 "As the extension, so the force",
i.e., stress is proportional to strain
Stress () = Force/area = P/A
Units = N/m2 = Pa
Strain ( )= the amount of stretch under load per unit length
quantity unitless a
length original L
lengthin change
L
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
substitutionaldefects interstitional defects
(e.g., hydrogenembrittlement)
( from IMPRESS, esa)
Imperfections leading to strength properties
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)
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/m2
Range of E in materials is enormous:
E(rubber) = 0.1 GPaE(diamond) = 1220 GPaE(spaghetti) = 4 GPa (4x109 Pa)
Young’s Modulus
Stress - strain curves
At the elastic limit, materials deform and don't return to their original shape.
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
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
Materials good in compression
stone, concrete
Materials good in tension
carbon fiber, cotton, fiberglass
Materials good in both compression and tension
steel, wood
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
deflection y
load P
length LB. Bending
compression: proportional
to distance from neutral axis
tension: proportional to
distance from neutral axis
neutral axis
shear
load
support
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 and will buckle 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
1st, 2nd, and 3rd orderbending modes. Lowestorder is most likelyto occur.
Euler buckling load
2
2
)(KL
EIF
The force at which a slender column under compression will failby bending
E = Young’s modulusI = area moment of inertiaL = 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)
I = area moment of inertia (dim L4)—associated withthe bending of beams. Sometimes called secondmoment of area.
(Not to be confused with
I = mass moment of inertia (dim ML2)—associated with the energy of rotation)
Area moment of inertia
Spaghetti tests.
Breaking under tension.
Bending under perpendicular force.
Buckling under compression.
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.
Follow handout directions
Measure spaghetti diameter length between supports force (weight of nuts) deflection
Repeat
Bending Results – Young’s Modulus
Trial Type Diameter (m)Length of Gap (m) α (N/m) E (Pa)
1Thick 0.00185 0.076 245.653 3.91E+09
2Thick 0.00186 0.097 124.362 4.02E+09
3Thick 0.00188 0.116 62.51 3.31E+09
4Regular 0.00148 0.076 138.18 5.37E+09
5Thin 0.00143 0.076 74.8475 3.33E+09
6Angel Hair 0.00111 0.076 35.8244 4.40E+09
TODAY
Complete the bending data analysis and mini-report
Prepare for 3-5 minute presentation on college ranking (Monday July 12th)
Complete your compression testing
Grand Canyon problem (Monday July 12th)
Work on your formal report for the compression testing lab (Wednesday July 14th)
Select a research paper topic (Friday July 23rd)