Engineering Materials Vol II (microstructures_ processing_ design) 2nd ed. - M. Ashby_ D. Jones (1999) WW Part 8 ppsx

If the longest microcrack in a given sample has length 2a m then the tensile strength is simply Some engineering ceramics have tensile strengths about half that of steel – around 200 MPa

intensified, the lattice resistance makes slip very difficult It is the crack-tip plasticitywhich gives metals their high toughness: energy is absorbed in the plastic zone, mak-ing the propagation of the crack much more difficult Although some plasticity canoccur at the tip of a crack in a ceramic too, it is very limited; the energy absorbed issmall and the fracture toughness is low.

The result is that ceramics have values of KIC which are roughly one-fiftieth of those

of good, ductile metals In addition, they almost always contain cracks and flaws (seeFig 16.7) The cracks originate in several ways Most commonly the production method(see Chapter 19) leaves small holes: sintered products, for instance, generally containangular pores on the scale of the powder (or grain) size Thermal stresses caused bycooling or thermal cycling can generate small cracks Even if there are no processing orthermal cracks, corrosion (often by water) or abrasion (by dust) is sufficient to createcracks in the surface of any ceramic And if they do not form any other way, cracksappear during the loading of a brittle solid, nucleated by the elastic anisotropy of thegrains, or by easy slip on a single slip system

The design strength of a ceramic, then, is determined by its low fracture toughnessand by the lengths of the microcracks it contains If the longest microcrack in a given

sample has length 2a m then the tensile strength is simply

Some engineering ceramics have tensile strengths about half that of steel – around

200 MPa Taking a typical toughness of 2 MPa m1/2, the largest microcrack has a size of

60 µm, which is of the same order as the original particle size (For reasons givenearlier, particle-size cracks commonly pre-exist in dense ceramics.) Pottery, brickand stone generally have tensile strengths which are much lower than this – around

20 MPa These materials are full of cracks and voids left by the manufacturing cess (their porosity is, typically, 5–20%) Again, it is the size of the longest crack – thistime, several millimetres long – which determines the strength The tensile strength ofcement and concrete is even lower – as low as 2 MPa in large sections – implying thepresence of at least one crack a centimetre or more in length

pro-Fig 17.2 Tests which measure the fracture strengths of ceramics (a) The tensile test measures the tensile

strength, sTS (b) The bend test measures the modulus of rupture, sr, typically 1.7 × sTS (c) The compression

test measures the crushing strength, sc, typically 15 × sTS.

Fig 17.3 (a) In tension the largest flaw propagates unstably (b) In compression, many flaws propagate

stably to give general crushing.

As we shall see, there are two ways of improving the strength of ceramics:

decreas-ing a m by careful quality control, and increasing KIC by alloying, or by making theceramic into a composite But first, we must examine how strength is measured.The common tests are shown in Fig 17.2 The obvious one is the simple tensile test(Fig 17.2a) It measures the stress required to make the longest crack in the samplepropagate unstably in the way shown in Fig 17.3(a) But it is hard to do tensile tests onceramics – they tend to break in the grips It is much easier to measure the forcerequired to break a beam in bending (Fig 17.2b) The maximum tensile stress in thesurface of the beam when it breaks is called the modulus of rupture, σr; for an elastic

beam it is related to the maximum moment in the beam, M, by

axis Fracture is not caused by the rapid unstable propagation of one crack, but the

slow extension of many cracks to form a crushed zone It is not the size of the largest

crack (a m) that counts, but that of the average a The compressive strength is still given

by a formula like eqn (17.1), with

but the constant C is about 15, instead of 1.

Thermal shock resistance

When you pour boiling water into a cold bottle and discover that the bottom drops outwith a smart pop, you have re-invented the standard test for thermal shock resistance.Fracture caused by sudden changes in temperature is a problem with ceramics Butwhile some (like ordinary glass) will only take a temperature “shock” of 80°C beforethey break, others (like silicon nitride) will stand a sudden change of 500°C, and this isenough to fit them for use in environments as violent as an internal combustion engine

One way of measuring thermal shock resistance is to drop a piece of the ceramic,

heated to progressively higher temperatures, into cold water The maximum ature drop ∆T (in K) which it can survive is a measure of its thermal shock resistance.

temper-If its coefficient of expansion is α then the quenched surface layer suffers a shrinkagestrain of α ∆T But it is part of a much larger body which is still hot, and this constrains

it to its original dimensions: it then carries an elastic tensile stress E α ∆T If this tensile

stress exceeds that for tensile fracture, σTS, the surface of the component will crack andultimately spall off So the maximum temperature drop ∆T is given by

Values of ∆T are given in Table 15.7 For ordinary glass, α is large and ∆T is small

– about 80°C, as we have said But for most of the high-performance engineeringceramics, α is small and σTS is large, so they can be quenched suddenly throughseveral hundred degrees without fracturing

Fig 17.4. A creep curve for a ceramic.

Creep of ceramics

Like metals, ceramics creep when they are hot The creep curve (Fig 17.4) is just likethat for a metal (see Book 1, Chapter 17) During primary creep, the strain-rate de-creases with time, tending towards the steady state creep rate

Here σ is the stress, A and n are creep constants and Q is the activation energy for

creep Most engineering design against creep is based on this equation Finally, the

creep rate accelerates again into tertiary creep and fracture.

But what is “hot”? Creep becomes a problem when the temperature is greater thanabout 1T m The melting point T m of engineering ceramics is high – over 2000°C – socreep is design-limiting only in very high-temperature applications (refractories, forinstance) There is, however, one important ceramic – ice – which has a low meltingpoint and creeps extensively, following eqn (17.6) The sliding of glaciers, and eventhe spreading of the Antarctic ice-cap, are controlled by the creep of the ice; geophysic-ists who model the behaviour of glaciers use eqn (17.6) to do so

Further reading

W E C Creyke, I E J Sainsbury, and R Morrell, Design with Non-ductile Materials, Applied

Science Publishers, 1982.

R W Davidge, Mechanical Behaviour of Ceramics, Cambridge University Press, 1979.

D W Richardson, Modern Ceramic Engineering Marcel Dekker, 1982.

Problems

17.1 Explain why the yield strengths of ceramics can approach the ideal strength ˜σ,whereas the yield strengths of metals are usually much less than ˜σ How wouldyou attempt to measure the yield strength of a ceramic, given that the fracturestrengths of ceramics in tension are usually much less than the yield strengths?

agree with the values given for ∆T in Table 15.7?

[Hints: (a) assume that σTS≈ σr for the purposes of your estimates; (b) where there

is a spread of values for E, σr or α, use the average values for your calculation.]

The failure probability, P f, for this chalk, loaded in bending under my (standard)writing load is 3/10, that is

When you write on a blackboard with chalk, you are not unduly inconvenienced if

3 pieces in 10 break while you are using it; but if 1 in 2 broke, you might seek an

alternative supplier So the failure probability, P f, of 0.3 is acceptable (just barely) If

the component were a ceramic cutting tool, a failure probability of 1 in 100 (P f = 10−2)might be acceptable, because a tool is easily replaced But if it were the window of a

vacuum system, the failure of which can cause injury, one might aim for a P f of 10−6;and for a ceramic protective tile on the re-entry vehicle of a space shuttle, when one

failure in any one of 10,000 tiles could be fatal, you might calculate that a P f of 10−8 wasneeded

When using a brittle solid under load, it is not possible to be certain that a ent will not fail But if an acceptable risk (the failure probability) can be assigned to the

compon-function filled by the component, then it is possible to design so that this acceptable

risk is met This chapter explains why ceramics have this dispersion of strength; andshows how to design components so they have a given probability of survival Themethod is an interesting one, with application beyond ceramics to the malfunctioning

of any complex system in which the breakdown of one component will cause theentire system to fail

The statistics of strength and the Weibull distribution

Chalk is a porous ceramic It has a fracture toughness of 0.9 MPa m1/2 and, beingpoorly consolidated, is full of cracks and angular holes The average tensile strength of

a piece of chalk is 15 MPa, implying an average length for the longest crack of about

1 mm (calculated from eqn 17.1) But the chalk itself contains a distribution of cracklengths Two nominally identical pieces of chalk can have tensile strengths that differgreatly – by a factor of 3 or more This is because one was cut so that, by chance, all thecracks in it are small, whereas the other was cut so that it includes one of the longer

Fig 18.1. If small samples are cut from a large block of a brittle ceramic, they will show a dispersion of strengths because of the dispersion of flaw sizes The average strength of the small samples is greater than that of the large sample.

flaws of the distribution Figure 18.1 illustrates this: if the block of chalk is cut intopieces, piece A will be weaker than piece B because it contains a larger flaw It isinherent in the strength of ceramics that there will be a statistical variation in strength

There is no single “tensile strength”; but there is a certain, definable, probability that a

given sample will have a given strength

The distribution of crack lengths has other consequences A large sample will fail at

a lower stress than a small one, on average, because it is more likely that it will contain

one of the larger flaws (Fig 18.1) So there is a volume dependence of the strength For

the same reason, a ceramic rod is stronger in bending than in simple tension: intension the entire sample carries the tensile stress, while in bending only a thin layerclose to one surface (and thus a relatively smaller volume) carries the peak tensilestress (Fig 18.2) That is why the modulus of rupture (Chapter 17, eqn 17.2) is largerthan the tensile strength

The Swedish engineer, Weibull, invented the following way of handling the

statist-ics of strength He defined the survival probability P s (V0) as the fraction of identical

samples, each of volume V0, which survive loading to a tensile stress σ He thenproposed that

where σ0 and m are constants This equation is plotted in Fig 18.3(a) When σ = 0 all

the samples survive, of course, and P s (V0) = 1 As σ increases, more and more samples

fail, and P s (V0) decreases Large stresses cause virtually all the samples to break, so

P s (V0) → 0 and σ → ∞

If we set σ = σ0 in eqn (18.2) we find that P s (V0) = 1/e (=0.37) So σ0 is simply the

tensile stress that allows 37% of the samples to survive The constant m tells us how

Fig 18.2. Ceramics appear to be stronger in bending than in tension because the largest flaw may not be near the surface.

Fig 18.3 (a) The Weibull distribution function (b) When the modulus, m, changes, the survival probability

changes as shown.

rapidly the strength falls as we approach σ0 (see Fig 18.3b) It is called the Weibull

modulus The lower m, the greater the variability of strength For ordinary chalk, m is

about 5, and the variability is great Brick, pottery and cement are like this too Theengineering ceramics (e.g SiC, Al2O3 and Si3N4) have values of m of about 10; for these,

the strength varies rather less Even steel shows some variation in strength, but it issmall: it can be described by a Weibull modulus of about 100 Figure 18.3(b) shows

that, for m ≈ 100, a material can be treated as having a single, well-defined failurestress

σ0 and m can be found from experiment A batch of samples, each of volume V0, istested at a stress σ1 and the fraction P s1 (V0) that survive is determined Another batch

is tested at σ2 and so on The points are then plotted on Fig 18.3(b) It is easy todetermine σ0 from the graph, but m has to be found by curve-fitting There is a better way of plotting the data which allows m to be determined more easily Taking natural

logs in eqn (18.2) gives

Fig 18.4. Survival probability plotted on “Weibull probability” axes for samples of volume V0 This is just Fig 18.3(b) plotted with axes that straighten out the lines of constant m.

And taking logs again gives

Thus a plot of ln {ln (1/P s (V0) ) } against 1n (σ/σ0) is a straight line of slope m

Weibull-probability graph paper does the conversion for you (see Fig 18.4)

So much for the stress dependence of P s But what of its volume dependence? Wehave already seen that the probability of one sample surviving a stress σ is P s (V0) The

probability that a batch of n such samples all survive the stress is just {P s (V0)}n If these

n samples were stuck together to give a single sample of volume V = nV0 then its

survival probability would still be {P s (V0)}n So

P V s( ) { ( )} { ( )}= P V s 0 n = P V s 0 V V/ 0 (18.5)This is equivalent to

If we insert this result into eqn (18.7) we get

V s

This, then, is our final design equation It shows how the survival probability depends

on both the stress σ and the volume V of the component In using it, the first step is to fix on an acceptable failure probability, P f: 0.3 for chalk, 10−2 for the cutting tool, 10−6

for the vacuum-chamber window The survival probability is then given by P s = 1 − P f

(It is useful to remember that, for small P f , 1n P s = ln (1 − P f) ≈ −P f.) We can thensubstitute suitable values of σ0, m and V/V0 into the equation to calculate the designstress

The time-dependence of ceramic strength

Most people, at some point in their lives, have been startled by the sudden tion, apparently without cause, of a drinking glass (a “toughened” glass, almost always),

disintegra-or the spontaneous failure of an automobile windshield These poltergeist-like

happen-ings are caused by slow crack growth.

To be more specific, if a glass rod at room temperature breaks under a stress σ in a

short time t, then an identical rod stressed at 0.75 σ will break in a time of order 10t.

Most oxides behave like this, which is something which must be taken into account inengineering design Carbides and nitrides (e.g SiC or Si3N4) do not suffer from this

time-dependent failure at room temperature, although at high temperatures they may do

so Its origin is the slow growth of surface microcracks caused by a chemical tion between the ceramic and the water in its environment Water or water vapourreaching the crack tip reacts chemically with molecules there to form a hydroxide,breaking the Si–O–Si or M–O–M bonds (Fig 18.5) When the crack has grown to thecritical length for failure at that stress level (eqn 17.1) the part fails suddenly, oftenafter a long period Because it resembles fatigue failure, but under static load, it issometimes called “static fatigue” Toughened glass is particularly prone to this sort offailure because it contains internal stresses which can drive the slow crack growth, andwhich drive spontaneous fast fracture when the crack grows long enough

interac-Fracture mechanics can be applied to this problem, much as it is to fatigue We useonly the final result, as follows If the standard test which was used to measure σTS

takes a time t(test), then the stress which the sample will support safely for a time t is

where n is the slow crack-growth exponent Its value for oxides is between 10 and 20

at room temperature When n = 10, a factor of 10 in time reduces the strength by 20%

For carbides and nitrides, n can be as large as 100; then a factor of 10 in time reduces the strength by only 2% (Data for n are included in Table 15.7) An example of the use

of this equation is given in the following case study

CASE STUDY: THE DESIGN OF PRESSURE WINDOWS

Glass can support large static loads for long times Aircraft windows support a sure difference of up to 1 atmosphere Windows of tall buildings support wind loads;diving bells have windows which support large water pressures; glass vacuum equip-ment carries stress due to the pressure differences at which it operates In Cambridge(UK) there is a cake shop with glass shelves, simply supported at both ends, which onweekdays are so loaded with cakes that the centre deflects by some centimetres Theowners (the Misses Fitzbillies) say that they have loaded them like this, withoutmishap, for decades But what about cake-induced slow crack growth? In this casestudy, we analyse safe design with glass under load

pres-Consider the design of a glass window for a vacuum chamber (Fig 18.6) It is a

circular glass disc of radius R and thickness t, freely supported in a rubber seal around

its periphery and subjected to a uniform pressure difference ∆p = 0.1 MPa (1

atmo-sphere) The pressure bends the disc We shall simply quote the result of the stressanalysis of such a disc: it is that the peak tensile stress is on the low-pressure face ofthe window and has magnitude

σmax = 3 3( +

8

2 2

Table 18.1 Properties of soda glass

Compressive strength sc (MPa) 1000

Modulus of rupture sr (MPa) 50

a critical component: we choose a failure probability of 10−6 The vacuum system isdesigned for intermittent use and is seldom under vacuum for more than 1 hour, sothe design life under load is 1000 hours

The modulus of rupture (σr= 50 MPa) measures the mean strength of the glass in ashort-time bending test We shall assume that the test sample used to measure σr haddimensions similar to that of the window (otherwise a correction for volume is neces-sary) and that the test time was 10 minutes Then the Weibull equation (eqn 18.9) for

a failure probability of 10−6 requires a strength-reduction factor of 0.25 And the static

fatigue equation (eqn 18.10) for a design life of 1000 hours [t/t(test) ≈ 104] requires

a reduction factor of 0.4 For this critical component, a design stress σ = 50 MPa × 0.25

× 0.4 = 5.0 MPa meets the requirements We apply a further safety factor of S = 1.5

to allow for uncertainties in loading, unforeseen variability and so on

We may now specify the dimensions of the window Inverting eqn (18.12) gives

A window designed to these specifications should withstand a pressure difference of

1 atmosphere for 1000 hours with a failure probability of better than 10−6 – provided,

of course, that it is not subject to thermal stresses, impact loads, stress concentrations

or contact stresses The commonest mistake is to overtighten the clamps holdingthe window in place, generating contact stresses: added to the pressure loading, theycan lead to failure The design shown in Fig 18.6 has a neoprene gasket to distributethe clamping load, and a large number of clamping screws to give an even clampingpressure

If, for reasons of weight, a thinner window is required, two options are open to thedesigner The first is to select a different material Thermally toughened glass (quenched

in such a way as to give compressive surface stress) has a modulus of rupture which is

3 times greater than that of ordinary glass, allowing a window 3times thinner than

before The second is to redesign the window itself If it is made in the shape of a

hemisphere (Fig 18.7) the loading in the glass caused by a pressure difference ispurely compressive (σmax= [∆pR/2t]) Then we can utilise the enormous compressive strength of glass (1000 MPa) to design a window for which t /R is 7 × 10−5 with thesame failure probability and life

There is, of course, a way of cheating the statistics If a batch of components has adistribution of strengths, it is possible to weed out the weak ones by loading them all

up to a proof stress (say σ0); then all those with big flaws will fail, leaving the fractionwhich were stronger than σ0 Statistically speaking, proof testing selects and rejects thelow-strength tail of the distribution The method is widely used to reduce the prob-ability of failure of critical components, but its effectiveness is undermined by slowcrack growth which lets a small, harmless, crack grow with time into a large, dangerous

Fig 18.7. A hemispherical pressure window The shape means that the glass is everywhere in compression.