Ch 4 - Mechanical Properties of Biomaterials

Question 1

Dynamic mechanical analysis (DMA)

Answer 1

• “Dog-bone” geometry: tensile stress at middle of sample → more reproducible fracture
• Instrumentation
1 . Grips/activator (holds sample, moves grip)
2. Load cell (records instantaneous load)
3. Extensometer (records instantaneous length)
4. Computer (converts signals to σ-ε curve)

• Tensile, compressive, creep and stress relaxation tests

Question 2

General σ-ε curves (∀ material)

Answer 2

I. Ceramic: linear, steep slope (\elastic deformation, stronger bonds)
II. Metal (linear, then inverted parabola) = better for loads b/c ductile and more \plastic deformation
III. Semi-crystalline polymer (linear, then upward curl)
IV. Elastomer (logarithmic, then upward curl)

Question 3

Engineering stress

Answer 3

σ = F⊥ / A_o
where A_o = init. CSA
* Meas. in Pa
** Engr. assumes neglig. change in size during testing

Question 4

Engineering strain

Answer 4

ε = (L_i − L_o) / L_o

• Dimensionless
• • Engr. assumes neglig. change in size during testing

Question 5

Shear testing

Answer 5

Produces forces // to top and bottom faces of sample

Question 6

Shear stress

Answer 6

τ = F_(||) / A_o

* F_|| = shear force, can cause sample defor. of angle θ

Question 7

Shear strain

Answer 7

γ=tan⁡(θ)

Question 8

Torsion forces

Answer 8

Twists cylindrical specimen to cause deformation of angle φ

Question 9

Hooke’s law

Answer 9

σ = Eε

where E = \elastic modulus

Question 10

Elastic modulus

Answer 10

“E” = linear slope of stress-strain curve

• Meas. in MPa
• • Stronger interatomic bonds = ↓ defor. = ↑ E (∴ more energy to separate)

Question 11

Elastic deformation

Answer 11

Linear stress/strain relationship

Question 12

Shear modulus (modulus of rigidity)

Answer 12

“G” = slope of τ-γ curve in elastic region

s.t. τ = Gγ

Question 13

Transverse strain

Answer 13

ε_t = Δd /d_o

* Subsequential contraction ⊥ to axial tension

Question 14

Poisson’s ratio

Answer 14

ν = −ε_t / ε_a

* Dimensionless, must be [+]!

Question 15

Elastic & Rigid Moduli

Answer 15

E = 2G (1 + ν)

Question 16

direction dependence

Answer 16

[polymers]
• along axis: 1 ° covalent bonds, mech prop’s similar to metals/ceramics
• else: 2 ° forces dominate, reduces mech prop’s

Question 17

Plastic deformation

Answer 17

• Permanent, sample never completely returns to original shape
• Non-linear portion of stress/strain curve
* Does not follow Hooke’s law

Question 18

Yield strength

Answer 18

“σ_y” = stress end of elastic region (start plastic defor.)

• Draw 0.2% offset line (0.002) on x-axis, // to elastic portion
• • Key design param. b/c end of elasticity

Question 19

Yield point strain

Answer 19

“ε_yp” = strain at end of elastic region

* Draw 0.2% offset line (0.002) on x-axis, // to elastic portion

Question 20

Ultimate Tensile Strength (UTS)

Answer 20

[crystalline e.g. metals/ceramics]
“σ_UTS” = necking of specimen
∴ no longer uniform strain across entire specimen, only defor. at necking area

Question 21

Fracturing strength

Answer 21

“σ_f” = stress at fracture
* after onset of necking, ↓ stress to cause further plastic defor.
∴ σ_frac < σ_UTS

Question 22

Ductility

Answer 22

Ability of material to deform plastically before breaking

Question 23

Brittleness

Answer 23

• Mat’ls w/ low ductility

* Fracture w/ little plastic deformation e.g. ceramics

Question 24

% ELongation

Answer 24

% EL = (L_f − L_o) / L_o ∗100

where “o” = initial, “f” = fracture

Question 25

% Area Reduction

Answer 25

% AR = (A_o − A_f) / A_o ∗100

where “o” = initial, “f” = fracture

Question 26

Plastic defor.: polymers

Answer 26

• Chains become oriented w/ direction of load
• Resist defor. along axes b/c 1 ° interxns ∴ specimen elongation along ENTIRE gauge length (not just necked area, like metals)
• Before frac, signif ↑ σ neces. (b/c overcoming 1 ° bonds within aligned chains) → upward curl
* Complex b/c ∀ grain randomly oriented
** Depends on testing envir e.g. ↑ Temp, ↓ strain rate ∴ E ↓, strength ↓, ductility ↑

Question 27

Elasticity

Answer 27

• Stretch bonds, return to normal length after load removed

* Char. only of \elastomers

Question 28

Elastomers

Answer 28

• Allow large recoverable strains at low stress levels
• Subset of polymers e.g. rubber
• Amorphous in unstressed state: coiled chains w/ nearly free bond rotations about backbone
* b/c the chains are crosslinked, chains cannot slide past each other ∴ no plastic defor.
** Above T_g, tensile force allows uncoiling of chains/alignment along axis (but NOT sliding) → increased entropy ∴ favors original state (like springs)
** would NOT strengthen an elastomer, b/c want to maintain elasticity

Question 29

Resolved shear stress

Answer 29

[metals/crys. ceramics]
τ_r = σ cos⁡(ϕ) cos⁡(λ)
where ϕ = Normal ⊥ to slip plane and λ = angle of slip dir. (both w.r.t vertical)

Question 30

Critical Resolved Shear Stress (CRSS)

Answer 30

[metals/crys. ceramics]
“τ_crss” = amount of shear stress in slip plane to initiate slip
* Factor in determining σ_y of material
** Slip occurs if τ_crss achieved in LEAST favorably aligned grain

Question 31

Processing polymers: Heating

Answer 31

• Grain growth favored at high T
∴ faster cooling → more, smaller grains • More, small grains LESS likely to be aligned
→ Reduce likelihood of slip ∴ STRONGER, σ_y ↑

Question 32

Processing polymers: Additives

Answer 32

• Smaller atom can sit in compressive region around edge defect and cancel lattice strain
• Driving force ↓ ∴ slip ↓

Question 33

strain rate

Answer 33

Faster, higher strain = less time to rearrange chains along axis of loading (in neck region) ∴ ↓ defor.

Question 34

Plastic deformation:

Semi-crystalline Polymers/Elastomers (molecular level)

Answer 34

• Semi…polymers = spherulites → plastic defor. = interxns b/w lamellae & amorph. regions
  1 . Tie chains extend, lamellae slide past each other
  2. Lamellae reorient s.t. chain folds align along axis of loading
  3. Blocks of crystalline phases separate (break 2 ° bonds)
  4. Blocks and tie molec’s extend, oriented along axis of applied tensile force (1 ° covalent bonds still intact, until fracture)
• High force necessary to overcome tie molec/covalent bonds
  ** Any changes that inhibit chain motion (need higher energy to overcome): ↑ σ_y, ↓ ductility

Question 35

1 ° v. 2 ° bonds

Answer 35

1 ° = crosslinking (when broken → fracture)
2 ° = % X-talinity
* also ↑ MW

=> ↓ chain sliding, ↑ σ_y, ↓ ductility

Question 36

Ductile fracture

Answer 36

“cone shape” = plastic deformation prior to breaking
• Crack propagation is slow and stable
• Preferred mode of failure (b/c warning signs from shape change)

Question 37

Brittle fracture

Answer 37

• Little/no plastic deformation (less slip)

* Flat fracture surface (little warning)

Question 38

Charpy and Izod impact tests

Answer 38

Measures impact energy and determines D2B transition temp. (when mat’l fractures brittle-y)

Question 39

Polymer crazing

Answer 39

Network of fine cracks on surface (⊥ to tensile stress), but CAN support some load

Question 40

Stress raisers

Answer 40

• Small flaws which ↑ localized stress → crazes
• e.g. notches, sharp corners and pores
* More signif. for brittle mat’l b/c plastic defor. ↓ localized stress

Question 41

Creep

Answer 41

• Plastic deformation of sample under constant (tensile) load (ε) over time
• 1 °, 2 ° (\SS creep rate), 3 °
• Key param: \SS creep rate and \time to rupture “t_r”
* must be greater than T_g to have viscous effects
** if ↑ Temp OR ↑ σ → then rate ε ↑ , t_r ↓

Question 42

Primary creep

Answer 42

↑ ε w.r.t time, while creep rate (slope) ↓

* due to repositioning of defects e.g. dislocations

Question 43

Secondary creep

Answer 43

• Linear relationship b/w creep & ε (typ. longest period)
* SS creep rate (dε/dt = 0)

Question 44

Tertiary creep

Answer 44

• Gross defects inside mat’l

* e.g. GB separ, cracks/voids → rapid elongation to mat’l failure

Question 45

Molecular causes of creep: Metals

Answer 45

• GB sliding
• Migration of vacancies (esp. at high Temp) → \stress-induced vacancy diffusion
• \dislocation climb

Question 46

Stress-induced vacancy diffusion

Answer 46

• Extra vacancies due to loading (⊥ to axis of loading)
• Vacancies migrate // to axis of loading
• Atoms move in opp. direction

Question 47

Nabarro-Herring creep

Answer 47

Atomic diffusion opp. of \stress-induced vacancy diffusion → elongation of grain along axis of applied stress

Question 48

Coble creep

Answer 48

Vacancies migrate along GB (than through bulk of grain)

Question 49

Dislocation climb

Answer 49

Dislocation moves 1 atomic spacing by diffusion of entire row of vacancies to extra partial plane of edge dislocation

Question 50

Molecular causes of creep: Ceramics

Answer 50

• GB sliding = main microstructural rearrangement

* More resistant to creep b/c electroneutrality

Question 51

Molecular causes of creep: Polymers

Answer 51

• Chain sliding in amorphous regions via viscous flow
• % X-talinity and temp (w.r.t T_g)
→ ↑ % X-tal → creep ↓ b/c less amorphous
→ Below T_g, NO CREEP b/c chains cannot rotate/slide

Question 52

Stress relaxation

Answer 52

[polymers]
↓ stress over time under constant strain
* Similar to creep, depends on % X-talinity and temp (above T_g)

Question 53

Fatigue fracture

Answer 53

• Failure due to repeated loading/dislocations, at stresses signif. less than σ_UTS tensile/yield strengths
• Brittle, little plastic deformation (even in ductile materials)
* Fatigue → ↑ # dislocations → \crack initiation (high σ) → \crack propagation (↑ in size ∀ successive loading cycle) → failure (rapid)

Question 54

Fatigue life

Answer 54

• N_f = N_i + N_p
where N_i = # cycles at crack initiation & N_p = # cycles at propagation to critical size for failure
• ↑ σ or \stress raisers → # cycles to failure ↓ (faster)
* Biodegradable mat’ls = more flaws and pores

Question 55

Corrosion fatigue

Answer 55

Failure due to cyclic stress + chem attack

Question 56

How to improve mechanical properties

Answer 56

Reduce mvmt of dislocations/slip OR chain sliding

*Polycrystalline mat’l gen’lly stronger b/c GB discourage dislocation mvmt (smaller grains = more GB/vol = stronger)

Question 57

Methods to improve mech prop’s: Additives

Answer 57

• [metals] \alloys: reduce lattice strain

* [polymers] \fillers: strength ↑ by acting as crosslinks

Question 58

Methods to improve mech prop’s: Processing

Answer 58

• Polycrystalline, quick cooling → more GB/vol (smaller GB) ∴ stronger (higher yield strength) b/c less likely for grains to align (less suscep. to shear stress)
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