Mechanical 3 Flashcards
How to increase strength
make it difficult for dislocations to move
Microstructure features
Grains
Precipitates
Grains
- Small crystals
- Within each grain lattice of atoms same, but orientation changes from one grain to another
- typical grain size 10-100 microm
Precipitates
- Small particles, typically 1microm or less in size
- lattice structure of atoms in precipitate diff from that in rest of material
Grains and dislocations
- Dislocation can’t cross grain boundary
- refine grain size - yield strength goes up
Hall-Petch equation
effect of grain size
τᵧ = τₒ + k/√d
τᵧ = yield strength in shear τₒ = yield strength of lattice k = constant d = grain size
Precipitates and dislocations
- Dislocation can’t enter particle bc has diff lattice structure
- Dislocations can overcome precipitates by breaking them or looping around
- happens within each grain, think of it as changing value of τₒ
- reduce spacing between precipitate particles, yield strength goes up
Equation describing changing the value of τₒv
τₒ = Gb/L
G = shear modulus b = spacing between atoms L = spacing between precipitates
Work Hardening
-in stress/strain curve, after σᵧ line continues to rise, need more stress to get more strain
Why Work Hardening
- σᵧ stress at which dislocations able to move
- need lot of dislocations moving to get lots of plastic strain
- more dislocations formed as test goes on
Annealing/Tempering
Heating the metal, making grains larger
Microstructures - Simple single-element metals
eg. aluminium, titanium, have simple microstructures
Cold working
Deforming metal at room temp
- produce distorted grains
- inc sctrength
Recrystallisation
Heating cold-worked metal causes new grains to form & grow
Way to get small grain size
Annealing after cold working (recrystallisation)
Precipitation Strengthening
- To get precipitates need metal with more than one element in it
- Bc chem composition of precipitate particles must be diff to that of rest of material (called the matrix)
- atype of strengthening only possible w/ alloys
eg of alloys
- Steel (Fe + C + other elements)
- Titanium Alloys (eg. Ti + 6% Al + 4%V)
- Brasses (eg. Cu + 40%Zn)
- Aluminium
Amount of carbon in steel
-less than 1%, often less than 0.1%
Tricks steel uses
- Sometimes C “dissolves” in the Fe
- Sometimes C combines w/ some Fe to form new material
- Fe atoms have diff lattice structure at high temp & low temp
C dissolves in Fe
- a solid solution
- C can fit into gaps between larger atoms (intersitial)
- In other alloys one atom can substitute for another
A solid solution
when atoms of one element dissolve in the other
Fe atoms and temp
- High temp: Fe has fcc lattice
- Room temp: bcc
- Change is reversible, happens instantaneously on heating and cooling
Carbon dissolving and temp
- Can dissolve in high-temp lattice (holes bigger)
- Cant dissolve in low-temp lattice
- At high temp: material like single-element metal, has grains but no precip, yield strength low as all C in solid soln
- Cooler: C precipitates out of soln, forming Fe3C particles + raising yield strength
High-temp form of Fe
Austenite fcc
Low temp form of Fe
Ferrite bcc
Alloys vs Single-element metals
Alloys better bc they can use precipitation strengthening, creating lots of small particles to impede dislocation motion
At low temps, C
Instead of dissolving, forms separate phase, Fe3C “cementite”
Pearlite
- high yield strength, v high if layers are v thin
- consists of thin, flat layers of ferrite alternating w/ layers of cementite
- 0.76% Carbon content
Ferrite and pearlite
soft, lower σᵧ than pearlite alone
Normalising
Natural cooling, when you take steel out of furnace and let it cool naturally in air
Cooling rate
- cool slowly: grains and pearlite layers larger, strength lower
- Quenching
- Fe atoms change from fcc to bcc instantaneously
Quenching
Taking steel from furnace and dropping into water at room temp
Matensite
- Distorted bcc material
- Sudden quench leaves C atoms stuck where they are as diffusion doesn’t happen instantaneously
- No fe3C forms
- C atoms distort bcc lattice, not enough room for them in bcc holes
Martensite grains
- Lenticular (lens-shaped
- very small grains
Yield strength of martensite
highest yield strength for any heat treatment of steel
Why anneal/temper
- Everything gets bigger and rounder
- Reduce energy, interfaces have energy associated w/ them
- Heating allows atoms to move around (diffuse), reducing no. of interaces + turning long, thin shapes into round, spherical shapes
Interfaces
(eg. grain boundaries and interfaces between layers in pearlite)
Tempering and strength
Tempering reduces strength
Phase Diagram
represents what type of material forms, and at what temp and composition in graphical form
Phase Diagram grapha
x-axis: composition (%)
y-axis: temp
Phases of Fe/C system
- Three solid phases: Ferrite, Austenite, Cementite
- Liquid phase
- Other places on diagram: have two phases at same time
Room temp steel
Most steels at room temp: in two-phase region
Ferrite symbol in phase diagram
α
Eutectoid Point
Type of point having one phase above and two phases below
eg. pearlite that forms when you cool through this point is called a eutectoid
phases - pearlite
-not phase, mixture of ferrite and cementite
phases - martensite
a phase, not there bc phase diagram shows what happens at equilibrium
Martensite
- a non-equilibrium phase
- at room temp: “metastable”
metastable
- would like to change into ferrite and cementite, but wont
- will gradually do if given enough energy by tempering
Cracks and stress
Cracks concentrate stress
- Plain tensile specimen in tension: stress same everywhere
- Notch: stress higher at root of notch
- Crack: stress at tip of crack much higher (theoretically infinite)
- Material at crack tip will break, allowing crack to extend
Local stress equation
σᵣ = Qσ√(a/2r)
σ = applied stress (nominal stress) r = distance
Plasticity
- Metals tough bc of plasticity
- Fact they can yield and deform plastically makes them tough
Plasticity and cracks
- Yield strength acts as upper limit to stress near crack tip
- zone of plastically strained material forms, consuming lot of eneryg
- harder for crack to propagate
Crack deflection and twist
Composites (eg. fibreglass, carbon fibre reinforced polymes) contain long fibres, cause crack to turn sideways
Creep and temp
- Most polymers: show creep at room temp
- Metals creep mostly only at elevated temps
- most ceramics dont creep except concrete + glass
- Most ceramics: high melt points, creep never an issue
Creep: why temp matters
- At any temp above absolute 0: atoms moving
- Vibrate in place, sometimes change places (diffusion)
- as temp goes up, yield strength goes down, dislocations can move more easily
- approach melting point: materials starts to take on properties of liquid, atoms flow past each other, creating constantly inc strain
Materials show creep
Most show creep if T>0.3Tₘ
Tₘ = melting point
Creep Rate
-defined as dε/dt in secondary region
Creep rate = Ae⁻ᴱᵃ/ᴿᵀ
A, R = constants
Eₐ = activation energy for diffusion in material
T = degrees Kelvin
Glass
-Shows creep at temps above 500C
Concrete
- creep at room temp
- bc concrete has flow properties similar to those of a liquid
How to prevent creep
- Keep temp down (relative to melt point)
- Keep stress down (relative to yield str)
Why fatigue
After some cycles, cracks form, and gradually grow
Why fatigue cracks form
- Materials heterogeneous
- weaker in some places
- stress will vary (dep on lattice orientation, precipitates, etc)
- repeated movement of dislocations causes cracks in grains
Heterogeneous
not same everywhere
Growth rate of crack
distance per cycle, da/dN
Stress intensity
K = Qσ√(πa)
rate of growth of fatigue cracks
-depends on applied stress range (△σ) and sq root of crack length
△σ = σmax - σmin
brittle fracture
if K > Kc
fatigue crack propagation threshold
if △K>△Kₜₕ
crack will grow at some rate da/dN
designing for fatigue
a. knowing △σ, use stress/life curve to read off predicted no. of cycles o failure
- below fatigue limit: should last forever
b. can predict how long it will take a crack found to grow to dangerous length - need experimental data
why wear happens
- surfaces never completely flat
- two surfaces pressed together only make contact in few places, where stresses will be high
Abrasive wear
-Hard material will plough up softer one
Spalling
-if both materials equally hard, fatigue cracks develop beneath surface, leading to eventual loss of material
Factors involved in wear
- surface contamination
- lubrication
- temperature
- chemical attack
Reducing wear in components
- develop wear-resistant materials (eg. alloy steels)
- coatings and surface treatments (eg. nitriding)
- making better lubricants
How to prevent wear
- use hard, strong material: may have low toughness + get abrasive wear in other material
- use hard coating eg. carburising
- use soft coating, coating will wear away + can be replaced (sacrificial coating)
carburising
adding extra carbon to surface of steel alloy
explain: The Young’s modulus values of polymers are lower than those of metals.
-their molecules are held together w/ weak atomic
bonds (hydrogen bonds + Van der Waals bonds)
-These types of bonds have much lower
stiffness than metallic bonds which we find in metals.
explain: There are many different types of steels which have different mechanical
properties but their Young’s modulus values are almost all the same.
- Steel consists largely of iron with small amount of carbon + other elements.
- Young’s modulus determined by characteristics of the atomic bond, in this case, is always the Fe-Fe atomic bond, so it is always the same.
explain: The Young’s modulus of a composite material made from carbon fibre and
epoxy resin can be changed by changing the percentage of carbon fibre.
- Carbon (being a ceramic material) has a higher Young’s modulus than epoxy resin (a polymer).
- If some of epoxy in the composite is replaced by carbon, then overall Young’s modulus will increase.
The potential energy W of a pair of atoms forming an atomic bond can be
written in terms of their separation a and four positive constants A, B, n and m:
check sample paper
derive an equation for the Young’s modulus of the material.
check sample paper
Define the term “wear”. Describe a test to measure wear. Define the
wear factor k and explain how it can be calculated from the results of this test.
- removal of material from one or both surfaces when two bodies rub together.
- requires some compressive force acting between the bodies, and some relative shear motion between them.
Describe the mechanism of
wear which was operating at each stage, and suggest how this wear could be
avoided: A company makes a knee joint implant which consists of a cobalt
alloy component which rubs against a component made from polyethylene. The
polyethylene experiences high levels of wear, visible as parallel grooves in the
surface of the component.
- abrasive wear: metal is much harder than the polymer, so high points of surface of metal dig into polymer + create grooves as they move across it.
- Abrasive wear can be avoided by making both materials equally hard, which is what was actually done in this case
Describe the mechanism of
wear which was operating at each stage, and suggest how this wear could be
avoided: The company makes a material change, replacing the polyethylene
with the cobalt alloy. All goes well for about 10 years but then massive wear
occurs in the form of large pieces of metal falling out from the surfaces of both
components.
- fatigue-wear: both surfaces are equally hard so no abrasion occurs
- every time one component moves over the other, creates stresses near surfaces.
-Cyclic stresses repeated many times can cause fatigue, in form of small cracks which initiate grow
inside metal, just below surface. Eventually cracks become long enough to cause pieces to fall out.
-Fatigue wear can be avoided by using material which has more resistance to fatigue (as shown by the stress/life curve), by reducing applied force between the components, or perhaps by using special coating which is more fatigue resistant.
describe wear test
-wear test involves applying known compressive force, carrying out a known relative motion + measuring loss of material.
sample of wear test
sample of steel in form of straight length with circular cross section
- held stationary + loaded by placing a weight on it to give a downwards force F.
- It presses on a disc of abrasive material (e.g. sandpaper) which is made to rotate.
- After some period of time relative motion d can be calculated knowing the disc radius and rotation speed
- loss of volume △V of the steel can be calculated by measuring its length or by weighing it before and after and knowing its density.
stress definition
the applied force divided by the area over which its acts
strain definition
the change in length divided by the original length
tutorials to redo
tutorial 1 q3
tutorial 2 q1 (a) (c), q2 (c)
tutorial 3
how to conduct cyclic loading test
- subject samples of material to cyclic stress
- eg. apply tensile stress to a sample but stop before failure + cycle the stress on and off until sample fails
- this gives one point on graph, use multiple samples
bcc, fcc, sc, etc formulas
check chemistry report
