Fracture of Brittle Materials & ceramics

Question 1

How does stress and strain vary for metals and ceramics?

Answer 1

Metals can sustain a significant amount of strain for the stress applied linearly to its yield strength then it shall plastically deform to its failure.
Ceramics have brittle fracture at a much smaller strain than metals and will be catastrophic failure. Only linear elastic behaviour. Have a higher youngs modulus.

Question 2

Why do ceramics vary as such against metlas?

Answer 2

Because of the strong atomic bonds in ceramics so they have a more rigid structure of covalent or ionic bonds. And so metals can have lower elastic regions as they have more metallic bonding and is directional.

Question 3

What is super plasticity?

Answer 3

Where a ceramic plastically deforms slightly - not expected to happen and in rare cases. Materials ability to withstand extremely large tensile elongations before fracture

Question 4

what is an ideal brittle ceramic?

Answer 4

Elastic deformation followed immediately by fracture

Question 5

Stress displacement curve

Answer 5

Atoms are pulled apart on the sine curve leading to a displacement from the strength that is applied.

Question 6

What does high strength be dependent on in ceramic behaviour wise?

Answer 6

high modulus(stiffness), high surface energy and small lattice spacing

Question 7

What is the griffith theory?

Answer 7

Where fracture begins at inherent flaws in the material due to stress concentrating effect of cracks. E.g. cut glass by scoring with diamond tipped implement then cracking along the score by applying stress and it should crack down that line.

Question 8

What 2 flaws of the ceramic need to be met to have effective conditions for the griffith theory to propagate?

Answer 8

The local stress must exceed the stress applied.
Must have enough strain energy to form new surfaces .

Question 9

What is the griffith equation ? REMEBR THIS

Question 10

what are the two stresses that must be satisfied to satisfy the 2 flaws for griffiths?

Answer 10

Inglis and Griffith equation however the inglis value can be smaller than the griffith but only one condition would be met so griffith criterion gives the minimum stress for cracks to propagate.

Question 11

What does the strength of individual samples rely on?

Answer 11

A closer look at this equation reveals that for a given series of
similar samples E and γ will be common. Thus the strength of
individual samples will depend on c, the size of the largest flaw
present in the stressed region of the material.
Dependent on largest flaw in the stressed region irrespective of any other flaws - provided the flaws don’t interact.

Question 12

What is the effect of microstructure on flaw depth and flaw shape?

Answer 12

The shape of a pore whether spherical or near spherical will have lower stress concentration effect than an angular pore.

Question 13

How doe the present of cracks adjacent to pore affect propagation?

Answer 13

intersection of the pore with grain
boundaries of the ceramic. If the pore is much larger than the grain size, the
extremities of the pore = critical flaw size. However, if the pore size approaches that of the grains, then the effect of cracks along the grain boundaries will probably predominate leading to an effective larger flaw size.

Question 14

What is the distance between pores and between pores and the surface in propagation?

Answer 14

the physical location of a pore within the ceramic will affect strength. If a pore is close to the surface of the material, the bridge of material separating it from the surface
may break first, resulting in a critical flaw whose dimensions are now the size of the pore plus the length of the bridge.

Question 15

What are pore clusters?

Answer 15

it is known that if a group of pores are close together the material
bridges between them can crack first, linking the pores together and producing a much larger flaw that results in much lower strength.

Question 16

What are the effects of inclusions mechanically?

Answer 16

Contamination during ceramics processing
Effect is similar to that of pores
– Large angular inclusions are more detrimental

Question 17

griffits theorh and rest of 2-4, 5-6

Question 18

What are beneficial properties in ceramics?

Answer 18

high stiffness, high strenghth high corrosion resistance, low density, high hardness, high refractoriness, high wear resistance.

Question 19

what is fracture toughness?

Answer 19

metals are much higher than ceramics (as they have defectts) and ceramic are not good at resisting crack propagation.

Question 20

What is the critical stress intensity factor, Kic?

Answer 20

The stress intensity factor at which a crack propagation will happen and lead to fracture.
If the load is perpendicular to the crack, as is typically the case in a tensile or bend test, the displacement is referred to as mode I and is represented by KI. This is also called the opening mode and is most frequently operational for ceramics. can be applies in 3 mode directions.

Question 21

How is mode 1 stress intensity factor related to the applied stress and crack length?

Answer 21

[ref equation pg 62]
From the equation we see that the toughness determines the critical flaw size that will lead to catastrophic failure at a given stress, i.e. the higher the fracture
toughness the more difficult it is to initiate and propagate a crack.
A doubling of the material toughness will lead to a 4-fold increase in the tolerable flaw size at a given stress.
This will in turn result in an increase in the reliability of the component since the
larger the critical flaw size the greater the chance we have of ensuring no flaws
exceed it during processing.

Question 22

What is the relationship between failure stress, flaw size and fracture toughness for brittle materials?

Answer 22

As the flaw size increases the stress will decrease or as the flaw size increases, the stress will decrease. [ref graph pg63].

Question 23

Why is achieving reliability in ceramics hard to achieve?

Answer 23

The flaw sensitivity and low fracture toughness provides a challenge in struvtural applications

Question 24

What are the 2 main approaches to making ceramics more reliable?

Answer 24

Reduce flaw size- ultra fine powders or sol gel processing (improves processing)

increase toughness- microstructural design(control of microstructure), zirconia toughening(secondary phase addition) or composite reinforcement(particles/whiskers) which improves inherent toughness.

Question 25

What is zirconia toughening?

Answer 25

Zirconia has 3 polymorphic crystal forms : monoclinic -> tetragonal -> cubic -> liquid. The mono tetragonal transformation is accompanied by a 4-5% volume change causing: -cracking of individual particles -prevents use of pure ZrO2 for high temperature applications and now it can be used in engineering applications

Question 26

What is the transformation toughening?

Answer 26

Similar to Zirconia toughening, from the crack in the zirconia matrix the tetragonal particles transform into monoclinic particles through a widening process zone with a crack closing stress field. IT heats up the tetragonal particles as they absorb the energy from the crack and then the region converts into a monoclinic phase. This both increases the fracture toughness and critical flaw size.

Question 27

How can difference in elastic medullae between the inclusion and matrix affect the degree of strength reduction?

Answer 27

- Elastic modulus differences can result in the formation of cracks when a stress is applied.

Question 28

How can difference in coefficient of thermal expansion between the inclusion and matrix affect the degree of strength reduction?

Answer 28

Thermal expansion differences can result in cracks forming adjacent to the inclusion during cooling from the fabrication temperature.

Question 29

how can the strength of ceramics be categorised?

Answer 29

Intrinsically variable Significantly lower in tension than compression (about 1/10) Highly dependant upon component design Dependant upon dimensions of stressed component (large components weaker)

Question 30

What are the main things that can affect the strength of ceramics?

Answer 30

The effects of processing flaws and design defects on the strength of ceramic materials as ceramics can't exhibit plastic deformation.

Question 31

What is the strength of ceramics directly related to ?

Answer 31

The size, shape and distribution of these flaws. Surface cracks and scratches; Sintering defects and porosity; Design features such as sharp corners and undercuts.

Question 32

What is the best tensile test to use for a ceramic ? in terms of price etc

Answer 32

Of a 3 point, 4 point and tensile test; 3point

Question 33

Compare Hot-Pressed Silicon Nitride (HPSN) sintered with glassy additives and Reaction-Bonded Silicon Nitride (RBSN) for strength against temperature.

Answer 33

HPSN is preferred for room-temperature structural applications requiring high strength. The glassy phase softens at high temperatures which weakens grain boundary and therefore strength. RBSN is more suitable for high-temperature environments, provided mechanical loads are moderate. Has lower strength and density at lower temperature due to residual porosity. retains strength better at higher temperatures.

Question 34

How can the probability be estimated for a material to survive under a given load?

Answer 34

Through Weibull statistics: -the probability of survival (or failure) -the effect of sample volume -The variability (or consistency) of the strength

Question 35

What can be statistically categorised about probability of finding a fracture initiation flaw and the flaw distribution?

Answer 35

Chain is as strong as the weakest link: – No two chains will have the identical weakest link (largest flaw) Flaw distribution is essentially random (except design flaws) Probability of finding a fracture initiating flaw increases with the volume of the material under stress

Question 36

How does the Weibull modulus (m) inform the reliability of strength values?

Answer 36

The Weibull equation can be ln and plotted (slope = m) so that there is either as: -There is a sharp cutoff stress above which all fail - the ideal. -Probability of survival is the same at all stresses - the worst case. The higher the m value the better.

Question 37

What is the cause of time dependent failure?

Answer 37

The failure mechanism is due to sub-critical crack growth occurring under stress, sometimes (often) assisted by environmental factors. The crack grows until it reaches ccritical and then failure results. This can be used to calculate the stress intensity factor.

Question 38

What factors is the crack growth dependent on and how can this be determined?

Answer 38

Determined using the double torsion test. The crack growth velocity can be determined by the load applied and the final length of the crack. So the crack velocity is a function of the stress concentration factor.

Question 39

What is the sequence leading to failure in the material?

Answer 39

1. a stress is applied 2. This fixes the value of KI for a given flaw size, c through eq. The value of KI fixes v, the crack front velocity, from the graph. 3.However, due to the crack growth, c is getting larger, which in turn leads to a higher KI, which in turn leads to a higher v. Eventually this leads to failure.

Question 40

What are the two main ways of presenting the lifetime data of failure ?

Answer 40

1. SPT (Strength Probability Time) diagrams 2. Proof Testing

Question 41

What is the procedure for Strength Probability Time (SPT) diagrams?

Answer 41

i) Measure the lifetime, τ, for a batch of samples at a range of stresses, σ. ii) Using the fact that τ = a/ σ^n where n is known, the data can be adjusted to yield a set of failure stresses at a common lifetime, e.g. 1 second. iii) Plot the results on Weibull paper, producing a line for 1 second. iv) Plots for lifetimes of 10s, 100s, etc, will be parallel and equispaced.

Question 42

What can Strength Probability Time (SPT) diagrams be used to predict?

Answer 42

The survival probability for known stress, σ, and lifetime, τ, requirements. The lifetime, τ, for known survival and stress, σ, requirements. The maximum stress, σ, for known survival and lifetime,τ, requirements. It tells us the survival rate but not which ones will survive - only the probability of how many will survive.

Question 43

What is the procedure and goal of proof testing?

Answer 43

Eliminate weak components with cracks that could grow to critical size during service. Key Idea 1: Test at application stress – survivors assumed to be safe. ⚠️ Problem: Subcritical crack growth during test can weaken the sample. Key Idea 2 (Better Approach): Test at a higher stress(than service that should eliminate samples containing cracks) (σ_proof) to detect cracks > c_init (which can grow to c_crit during lifetime τ) The griffith flaw (largest crack) must be smaller than Ccrit after required period of time to survive. Therefore if no cracks are > cinit it will survive for the required time. If a sample survives this proof test, where the applied stress, σproof is greater than the application stress, σA, it means that it has no cracks as large as cinit and so none can grow to ccrit over the required lifetime. This ‘guarantees’ that the sample will survive for the required period of time.

Question 44

How does thermal shock generate fractures?

Answer 44

With ceramics subjected to temperatures that vary throughout the component and may change rapidly with time, the processing of an initial high temperature that is quenched to a low temperature all lead to stresses = cracks/failure.

Question 45

Is rapid cooling or rapid heating worse for thermal shock?

Answer 45

Rapid cooling (tensile) is much worse than rapid heating (compressive). E.g the inner body is hotter than the outside so the surface wishes to contract but is constrained so tensile stresses can appear.

Question 46

What are the 2 main conflicting tendencies in thermal expansion systems with temperature?

Answer 46

i) Heat transfer at the surface makes Ts, the surface temperature, come close to Tf, the environmental temperature, and ii) Heat transfer within the solid makes the average internal temperature come close to the surface value.

Question 47

What can enhance the degree of thermal shock to be greater?

Answer 47

Temperature gradients (through conduction equation) can cause this when they are larger (k/rh where k is thermal conductivity, h is the surface transfer coefficient and r is the diameter of the component)

Question 48

What are factors that encourage crack growth?

Answer 48

a poor Thermal shock resistance can( propogates quicker in high density, high strength ceramics as failure will come faster), greater elastic strain energy than fracture surface energy.