Course F - Biomaterials Flashcards

1
Q

define biomaterials

A

“A biomaterial is any material, natural or man-made that comprises whole or part of a living structure, or biomedical device which performs, augments or replaces a natural function”

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2
Q

do a brief comparison between mad-made an biomaterials

A

Man-made:
- often large energy inputs needed
- can be stronger
- regularly metallic
- wide range of materials

Biomaterials:
- synthesized at RTP from CO2, H2O, O2
- sustainable, recyclable, biodegradable
- generally made from C, H, O, N with some others
- usually organic polymers/minerals

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3
Q

what is the elastic behaviour of a standard linear elastic solid

A
  • springs instantaneously back into shape
  • described by σ = Eε and τ = Gγ
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4
Q

what is the ‘elastic behaviour’ of liquid flow

A
  • does not recover original shape
  • response is time dependent
  • described by τ = η dγ/dt

η = viscosity

  • this is the response of a “Newtonian liquid”
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5
Q

what is the continuous spectrum of the two extremes of elastic behaviours (liquid flow and linear elastic solid) called, what properties do most materials have

A
  • the spectrum is called viscoelasticity
  • most materials have a mixture of the two properties/ lie somewhere on the spectrum
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6
Q

what two ‘components’ can we use to help model the elastic behaviour of (bio)materials, what are their responses

A

1) springs
- instantaneous response
- stress and strain in phase
σ = kε

2) dashpots
- leaky cylinder
- delayed response
- stress and strain in antiphase
σ = η dε/dt

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7
Q

what is the Maxwell model for elastic behaviour (using the two components)

A

we add a dashpot in series with a spring
this gives

dashpot response:
dε/dt = σ/η

spring response:
ε = σ/k

the components are in series so the stresses are equal, we add the strains giving:
ε =εd + εs
dε/dt = σ/η + (1/k)(dσ/dt)
this can be rearranged to a differential equation and solved with oscillating stress and strain terms of
σ = σo exp(iωt) and ε = εo exp(iωt - φ)

E = kω^2 (Tr)^2 / (1+ω^2 (Tr)^2))
Tr = response time
tanφ = 1/ωTr

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8
Q

what are the three main cases that we can consider for the Maxwell model (regarding frequency)

A

1) High freq. ωTr»1
- no time for viscous flow, we only see elastic response of spring, E–>k, φ–>0

2) Low Freq. ωTr«1
- plenty of time for viscous flow, elastic ext. of spring negligible compared to displacement of dashpot
E—> kω^2(Tr)^2 —> small
φ–>π/2

3) intermediate freq. ωTr approx. = 1
- movement of spring and dashpot are similar in magnitude
- Energy dissipated but strain is recovered

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9
Q

what are two alternative models to the maxwell model

A

1) Voigt Model:
- spring and dashpot in parallel
- better at describing response at const. stress

2) Standard Linear solid model
- spring in parallel with a series arrangement of a spring and dashpot
- Better than (1) at high freq.

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10
Q

in every materials selection/merit index question, what are the two types of variable

A
  • Fixed Variables, these are constants for the purposes of the question
  • Free variables, these can be varied accordingly
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11
Q

what is the general method for finding a merit index to select materials

A

1) construct two eq’s for the properties of interest (e.g. density, ρ and failure strength, σf) using fixed and free variables

2) rearrange to eliminate the free variables

3) rearrange for a ratio of the variables of interest e.g. σf/ρ, this is the merit index

4) work out whether the merit index should be maximised or minimised for the best material properties for the purpose of the question, state which

5) it is often useful to take logs (as the materials maps are log-log graphs) to determine if the y-intercept should be greater or lower for best properties

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12
Q

what are the two key properties for strength, what do we want them to be

A

1) failure strength, σf
2) density, ρ

Ideally we want max σf and min ρ

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13
Q

what are the two key properties for stiffness, what do we want them to be

A

1) Young Modulus, E
2) Density, ρ

we want max E, min ρ

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14
Q

what are the 3 ways to select for toughness

A

1) Impact resistance
2) stress-limited failure
3) strain-limited failure

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15
Q

how would you select for toughness by maximising impact resistance

A

simply maximise Gc

  • go to Gc against E map
  • draw horizontal line
  • maximise y-int
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16
Q

how would you select for toughness by maximising stress-limited failure

A

maximise the static tensile load that can be supported without crack propagation

we know
σf = sqrt(EGc / πc)

so maximise
sqrt(EGc)

  • on a graph of Gc against E, this means grad of -1 and maximise y-int
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17
Q

how would you select for toughness by maximising strain-limited failure

A

Maximise “Stretchiness” = amount of tensile strain without crack propagation

we know
εf = σf/E = sqrt(Gc / πcE)

so we maximise
sqrt(Gc/E)

  • for a Gc against E graph this means draw a line of grad 1 and maximise the y-int
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18
Q

what is natural rubber

A
  • a polymer of the monomer isoprene
  • extracted from indigenous amazon trees
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19
Q

what are the 2 conditions for a polymer to have large recoverable strains

A

1) enough energy for the chain backbone shape to change
- depends on Ea for bond rotation and the thermal energy present

2) Chains must uncoil but not slide past each other as this would lead to plastic deformation

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20
Q

What are the elastic properties of natural rubber, how can they be improved

A
  • Normally natural rubber has poor elastic qualities, it is brittle when cold and deforms easily when hot
  • this can be improved by Vulcanization = adding sulphur to make cross-links and prevent plastic deformation
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21
Q

what is the driving force for a polymer chain to recoil i.e. undergo elastic not plastic mechanism

A
  • tensile forces = extension = polymer chain extends, lower entropy
  • elastic energy is stored due to the change in ENTROPY not enthalpy
  • there is a driving force for the chain to recoil back to a higher ENTROPY form
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22
Q

how are proteins formed (brief explanation)

A
  • condensation polymerisation of amino acids, forms peptide bonds
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23
Q

give the two examples of protein rubbers, briefly explain their structure and properties

A
  • resilin and elastin
  • both are randomly coiled and contain cross-links

Resilin:
- ideal density of cross-links, very good elastic properties (better than any mad made)
- found in insect wings

Elastin:
- similar to resilin
- found in neck ligaments

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24
Q

what are the two main folded protein structures, give examples of where each are found

A

1) α-helix, found in Keratin
2) β-sheet, found in spider silk

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25
Q

describe the properties/structure of Keratin and some places it is found

A
  • when dry it’s in α-helix form, when wet the secondary bonds are easily broken and it can be stretched to form β-sheets
  • Keratin is tough
  • Keratin forms hair, nails, hoof etc., in this form it is heavily sulphur cross-linked α-helix
  • fairly strong/stiff
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26
Q

what are the two types of spider silk, what are their compositions, what are their properties

A

Dragline silk:
- stiff, E = 10 GPa
- 20-25% β-sheet miscelles
- σf = 1400 MPa = V high

Viscid silk:
- extensible
- E = 0.1GPa
- V little β-sheet
- very low coefficient of restitution, this means it absorbs energy efficiently and dissipates it, it has a slow response

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27
Q

describe what Collagen is and explain its hierarchical structure

A

Collagen is a protein fibre ONLY found in animals

1) polypeptide chains form left-handed coils (different to right handed α-helix)

2) 3 of these chains coil in a right-handed sense to form a collagen “molecule”

3) several collagen “molecules” arrange themselves into ‘fibrils’ with covalent cross-linking bonds

4) these fibrils are then densely packed

  • this results in a stiff/strong material
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28
Q

explain the structure/materials of skin and summarise its properties

A
  • skin is a composite mostly formed of Collagen ‘fibres’ in an elastin ‘matrix’
  • collagen fibres are short and aligned in 3D, not uniaxially but in preferred directions
  • anisotropic properties, good stretchiness
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29
Q

explain the non-linearity in the stress-strain curve for skin and explain what this means for its energy storage and stiffness

A
  • it has a J shaped σ, ε curve
  • this means it gets stiffer as ε increases
  • this prevents failure by fracture
  • energy absorption per unit strain increases with strain
30
Q

explain the non-linearity in the stress-strain curve for tendons and explain what this means for its energy storage and stiffness

A
  • s shaped curve
  • good at storing and releasing energy
  • stores more energy than a straight line
31
Q

explain the hysteresis loop of the stress-strain curve for protein structures

A
  • on application of stress, α-helix –> β-sheet
  • then if released β-sheet —> α-helix
  • the elastic energy input is NOT completely recovered on release
  • the area under the curve is the work done per unit volume
  • hence the difference in the area under the loading and unloading curves is the energy absorbed
  • better absorption of energy gives it better toughness
32
Q

what is the cellular structure of wood

A
  • 90-95% of cells are elongated and run vertically
  • the rest run radially outwards
  • none run tangentially (in circle)
  • this makes it anisotropic
33
Q

what is the difference between softwood and hardwood

A

Softwood:
- vertical cells are responsible for support and conduction of materials
- they have a thin wall and a large sap channel in the middle
- the radial cells are used for storage

Hardwood:
- the vertical cells are for support only
- they contain a narrow channel so little sap conduction
- there are specialized vessels which are open-ended vertically stacked tubes for material conduction
- the radial cells are still for storage

34
Q

explain the hierarchical microstructure of wood (specifically the basis of the cell walls in wood)

A
  • most of wood is just the cell walls

1) cellulose = basic fibrous constituent of cell walls
- this is a polymer chain made from glucose

2) the cellulose chains pack together in small bunches to form crystalline regions

3) several bunches can pack together to form a microfibril (about 12x30nm)

4) many microfibrils combine
- millions will form the cell wall
- the crystalline microfibril regions are separated by amorphous material (matrix)

  • the cells are then oriented as before
35
Q

explain the layers of the cell wall

A

1) Primary wall
- Very thin
- random microfibril arrangement

2) outer layer of secondary wall
- < 10% of overall thickness
- microfibrils wound in 2 distinct opposite spirals
- 50-70° from vertical axis

3) Middle later of secondary wall
- Major component
- about 85% of thickness
- microfibrils in single spiral
- 10-30° from vertical

4) inner later of secondary wall
- very similar to 1

36
Q

what can we treat wood as, give the properties of its components

A

We can treat it as a composite where

fibres = cellulose
- E = 100Gpa, Vf = 50%

Matrix = lignin (amorphous material)
- E = 1Gpa, Vf = 50%

  • using the normal assumptions and models for composites, we overpredict the strength
  • this is because we don’t account for the fibres not all being aligned or continuous
37
Q

what are the stiffness properties of wood

A
  • fairly good stiffness
  • using any model we know wood is anisotropic
  • in more dense woods the anisotropy is lower because it becomes more like an isotropic solid
38
Q

what are the σf properties of wood in compression and tension

what effect does moisture have

A
  • σf(compression) < σf(tension)
  • depends on direction of stress
  • when in compression the cell walls can buckle, this causes creases which can then act as cracks if put in tension or cycling between the two
  • the anisotropy does NOT change much with density
  • strength affected by moisture, σf decreases as moisture% increases, until it becomes saturated then σf const.
  • this is due to the cell wall structure changing as the water can affect the hydrogen bonds of cellulose, causing the chains to slide past each other
  • this occurs less with denser woods as the water can’t get into the wood as easily
39
Q

what are wood’s properties regarding toughness

A
  • Exceptional toughness
  • this is due to the fibre pull-out mode;
  • there’s an additional energy absorption mechanism in wood where the middle layer of the cell wall breaks away from the outer layer
40
Q

explain the structure of Chitin

A
  • similar to cellulose
  • nitrogen-containing polysaccharide forming packed microfibrils
  • significant secondary bonding between chains making it stiffer/stronger
41
Q

what are the key properties of Chitin

where is Chitin often found

A
  • σf = 100MPa
  • Toughness = 1500 Jm^-2
  • Max ε = 1-2%
  • more extension, tougher and higher E than bone
  • main tension bearing material in many plants/ animals
  • found structurally in some exoskeletons of beetles
42
Q

explain how hard biomaterials (minerals) are used in magnetoactic bacteria

A
  • Magnetic Ferrous minerals form within bacteria
  • shape, size and orientation are very well controlled and controlled by proteins
  • may have the function of aiding orientation control
43
Q

explain how hard biomaterials (minerals) are used in the ear

A
  • the human inner ear has small crystals of calcite
  • these assist balance
44
Q

what are the two main forms of calcium carbonate, CaCO3, which is more stable

A
  • Calcium carbonate, CaCO3 can be found in calcite or aragonite form
  • calcite is more thermodynamically stable but both are commonly found
  • adding Mg2+ ions makes calcite preffered
45
Q

how can the formation of calcite vs. aragonite be controlled in living organisms

A
  • it’s usually determined by enhanced concentrations of mineral ions on specific parts of an organic matrix through binding on active sites
46
Q

how can the crystallographic direction of biomineral crystals be controlled

A
  • controlling heterogeneous nucleation on proteins and constraining growth geometry
47
Q

how can the size/shape of biomineral crystals be controlled

A
  • controlled by the organic matrix in which they grow
  • they can be made to fill pre-defined vesicles
48
Q

explain the composition and 2 main structures of Nacre

A

Nacre is a composite containing:
- organic protein based component- forms sheets
- inorganic aragonite - forms blocks between sheets, usually hexagonal and around 10μm diameter, 500nm thick

  • successive layers of aragonite can nucleate from each other through small holes in the organic sheets
  • this gives either a brick wall structure or a stack of coins structure
49
Q

what are the key properties of Nacre, how are these properties obtained

A
  • Aragonite is very brittle but nacre has a good toughness
  • this is due to a crack deflection mechanism in nacre
  • any cracks which form either in the organic sheets or the aragonite crystals will deflect parallel to the stress along organic layers and propagate in that direction
  • this absorbs lots of energy through the formation of lots of new surface area
  • it also helps to not propagate the crack in the damaging direction perpendicular to the stress
50
Q

explain the Hierarchical structure of bone and give its approximate composition

A
  • Bone is a composite of collagen (protein) and Calcium Phosphate (roughly) in the form of Hydroxyapatite (HA = Ca10(PO4)6(OH)2)

1) collagen molecules
2) fibrils
3) fibres
4) bunched to form sheets
5) sheets are wrapped in concentric circles to form tubes (osteons)
6) these are then bundled to form bone wall

51
Q

is the mineral in bone exactly HA?
how are the HA crystals aligned, what is their size

A
  • the calcium phosphate in bone is not exactly Hydroxyapatite but it is approximately correct
  • the growth of HA is well controlled:
  • the crystals are small about 2.5nm x 20nm x 40nm
  • their long direction // collagen fibres // c-axis
52
Q

what are the main mechanical properties of bone, why is bone special

A

σf (longitudinal) = 100MPa
E(longitudinal) = 25GPa
Gc = 1500Jm^-2

  • bone is special as it is living/responsive, there are specialist cells for making/dissolving (resorbing) bone
53
Q

what do we mean by the stem when doing hip replacements

explain the materials challenges in making the stem

A
  • the step is the part that is inserted into/replaces the upper femur
  • ceramics are too brittle, polymers have creep, so we must use metals
  • the metal used cannot be toxic, allernegenic, carcinogenic
  • this leaves stainless steel or titanium
  • we DON’T want highest stiffness/toughness, just so that it’s similar to bone otherwise the bone would be resorbed
  • so Titanium is best
  • this can also be helped by making the Ti porous through sintering metal powder (compressed powder and heated)
54
Q

explain what is meant by the ball and cap in an artificial hip replacement

what are the materials challenges associated with it

A
  • this is the part that replaces the ball and socket joint itself
  • it is always in moving contact so we need low friction and low wear
  • metal and polyethylene has too much wear
  • metal and ceramic is better
  • metal on metal is now used
  • this can, however, lead to fine metal particles being distributed around the body
  • maybe ceramic on ceramic in future?
55
Q

what is the bonding in an artificial hip joint

explain the materials challenges associated with this

A
  • it is the ‘glue’ that fixes the stem to the bone
  • one option is to coat the stem with HA, this bonds well but can crack off
  • this is due to different thermal expansivities
  • if it’s unstressed when applied hot then it’ll be under tensile stress at RTP
  • this can be helped by adding Mn to the Ti of the stem
56
Q

what is arterial wall made from

what are necessary properties in any replacement material for arterial walls

A
  • real arterial wall is made from a natural composite of collagen and elastin

If we are to use a material to replace arterial walls then it must be:
- flexible
- J shaped stress-strain curve
- this is necessary to avoid an aneurysm (localised swelling)

  • if there is a normal stress-strain curve then using the expression for hoop stress
    Pex = σt/r
    we find that there are two possible stable radii i.e. aneurysm
  • a J shaped curve stops this
57
Q

what are the key features of the sucrose-water phase diagram

A
  • Eutectic
  • virtually no solid solubility in either ice or sucrose
58
Q

why does the equilibrium phase diagram for sucrose-water not usually help, what does tend to happen (2 possible things)

A
  • kinetics often prevent the expected phase transitions from occurring

1) Crystallisation of sucrose - this is difficult kinetically so generally a super-saturated solution forms instead on cooling (at high sucrose conc.)

2) if the temperature falls enough or the solution is dehydrated enough without crystallisation then a glass forms

59
Q

what is a glass, what occurs at the glass tranasition

A
  • transition of supercooled liquid to glass has no structural changes or enthalpy changes
  • it also has no volume change
  • it is effectively just having a very viscous liquid
  • something is generally considered to be a glass where
    η > 10^12 PaS
60
Q

what are the two risks of dehydration/ ice formation for cells

A

Disrupting composition of cytosol = bad
rupture of cell walls/membranes = fatal

61
Q

explain how dehydration causes problems for cells

A
  • as cells become dehydrated, the cytosol becomes super-saturated in minerals, this can lead to mineral growth
  • mineral growth can upset the ionic ratios in cytosol making it toxic or break the membrane leading to death
62
Q

explain a solution to dehydration in cells

A
  • have a greater concentration of high molecular weight sugars in the cytosol, this makes glass form instead of minerals
  • this preserves biostructures as no vol change
63
Q

explain the problems of ice formation in living systems

A
  • when ice forms it is almost pure H2O
  • this means as it forms, the salt conc. in the remaining cytosol increases lots, this makes it either toxic or the ion balance can change
  • when the salt conc. increases, water moves into the cell via osmosis, this causes swelling and bursting
64
Q

explain the method of water subdivision in animals/plants to avoid ice formation

A
  • pure water can be supercooled by 40K if there are no heterogeneous nucleation sites
  • this can be achieved by forming many very small water droplets
  • i.e. more droplets than nucleation sites to mean nucleation must be homogeneous
  • this occurs in frost-hardened woods, water outside the cells freezes first then inside
65
Q

explain the method of anti-freeze proteins (AFPs) for avoiding ice formation in plants/animals

A
  • Anti-freeze proteins can be used e.g. in the blood
  • these are where you kinetically prevent freezing
  • it works by the AFPs binding to favourable growth sites on ice crystals to prevent growth
66
Q

what are the two methods in plants/animals for tolerating ice formation

A

1) release of latent heat
2) control where water freezes

67
Q

explain the method of release of latent heat for tolerating ice formation in cells

A
  • certain parts of water are allowed to freeze
    e.g. in certain control vessels
  • this releases latent heat and protects the plant (usually) for short periods
68
Q

explain the method of controlling where water freezes using INAs as a method for tolerating ice formation

A
  • the idea is that you cause water to freeze outside of cells using INAs, this causes the salt concentration to increase outside of cells, this causes water to leave the cells via osmosis and hence the cells dehydrate and glass is formed, protecting the cells
69
Q

explain how Ice nucleating agents, INAs work

A
  • they are large proteins as a template for ice formation
  • they have similar structures to AFPs
  • we consider r* = -2γ/ΔGv
    i.e. heterogeneous nucleation on discs

For a given ΔT:
- if min r<r* at a given disc radius, R then ice stops growing
- if min r (this occurs when it is a hemisphere) > r* then ice keeps growing

rearranging gives
ΔTcrit = -2γ/SvR

i.e. the bigger the radius of the INA, the less supercooling is required for ice formation
- so INAs are very large proteins

70
Q

what can we say about the structural features of resilin that make its elastic properties so good

A
  • the density of cross-links between polymeric chains is OPTIMISED
71
Q

what happens to the stiffness of natural rubber as it is heated

A

Natural rubber acts as a spring because of entropy

  • the entropy difference will become greater with higher temperatures
  • so the ‘spring’ will become stiffer