Mechanical Behaviour of Engineering Materials - Metals, Ceramics, Polymers and Composites 2010 Part 11 pptx

Sincethe stress dependence is linear, Nabarro-Herring creep is most important atlow stresses, whereas dislocation creep is more important at high stresses.Since the creep rate is inverse

applied shear stress, l is the distance between the obstacles, and V∗ is theactivation volume (see section 6.3.2) The strain rate ˙ε is proportional to thevacancy current density j and to the dislocation density % Thus, ˙ε ∝ j% holds.

In the stationary state, the dislocation density % is usually proportional to thesquare of the stress [51].6If we insert the equation for the current density (anduse τ ∝ σ), we finally find

The exponential dependence of the strain rate on the temperature has beenconfirmed experimentally The relation between strain rate and stress is found

to follow a power law, as predicted by the equation (see also section 11.1), but

in reality the creep exponent typically takes values between 3 and 8 Due tothe large variations of the creep exponent in different materials, the value ofthe factor A can differ by several orders of magnitude The activation energy

in equation (11.13) is frequently stated per mole in the units kJ/mol In thiscase, Boltzmann’s constant k has to be replaced by the gas constant R in theequation as explained in appendix C.1

In deriving equation (11.13), we assumed that the vacancy diffusion curs through the undistorted crystal (volume diffusion) However, vacancytransport can also occur mainly along lattice defects like dislocation lines (dis-location pipe diffusion) In this case, the activation energy for site exchange issmaller due to the lattice distortion Because of this difference, vacancy trans-port along dislocation lines dominates at low temperatures, whereas volumediffusion is the faster mechanism at higher temperature Large stresses alsofavour transport along dislocations because the dislocation density increasesdue to the formation of new dislocations (work hardening) We also required

oc-in derivoc-ing equation (11.13) that the product of external stress and activationvolume is small compared to the thermal energy kT This is not the case

at high deformation speeds and stresses In this case, the relation betweenstrain rate and stress is exponential and the power law is not valid anymore(power-law breakdown) [51]

6 A similar relation was already discussed in the context of work hardening, seeequation (6.20)

of the previous section Again, the vacancy current density is calculated andrelated to the strain rate.

As in section 11.2.2, we start by calculation the vacancy concentration

In equilibrium and without externally applied stress, the vacancy centration is n = exp(−QV/kT ), where QVis again the energy for theformation of a vacancy If we now consider a grain boundary whosenormal direction is in the direction of the tensile stress (figure 11.6),the material can elongate in the loading direction if an atom from thecrystal lattice is added at the grain boundary, creating an additionalvacancy in the crystal The external stress does some work because thematerial lengthens (on average) by the quotient of the volume Ω of thevacancy and the cross section of the grain boundary considered Theforce equals the stress multiplied with the cross section of the grain

con-boundary The work done is thus σΩ, independent of the size of thegrain boundary On the other hand, work σΩ must be done to create

a vacancy in a region that is under compressive stress σ

The vacancy concentration in the regions○ and1 ○ from figure 11.62

is thus

n1= exp

„

−QV+ σΩkT

«,

n2= exp

„

−QV− σΩkT

«.The vacancy concentration gradient can be estimated, according toequation (11.10), as (n1− n2)/d, with the grain size d replacing thedislocation distance Thus, a vacancy current density

j = −D0exp

„

−QexkT

Here, ANH is a material parameter, σ the external stress, Ω the volume of

a vacancy, d the grain size, and DV the diffusion coefficient for self-diffusionthrough the bulk material This process is called Nabarro-Herring creep Sincethe stress dependence is linear, Nabarro-Herring creep is most important atlow stresses, whereas dislocation creep is more important at high stresses.Since the creep rate is inversely proportional to the square of the grain size,creep is favoured if the grains are small In contrast to time-independent plas-tic deformation, where small grains are preferred (grain boundary strengthen-ing, see section 6.4.2), large grains are advantageous in materials that creep.Vacancy diffusion needs not to occur through the bulk material in diffusioncreep Instead, vacancies may move directly along the grain boundaries (seefigure 11.7) The activation energy of vacancy diffusion along a grain boundary

is smaller than in the bulk because the lattice is distorted

As before, the vacancy current density j is inversely proportional to thegrain size The derivation of the current density in the bulk materialcan thus be copied exactly, simply replacing the activation energy with

d

Fig 11.7 Movement of vacancies along grain boundaries in diffusion creep

the smaller activation energy along the grain boundary The number

of vacancies moving through the grain per unit time is given by thevacancy current density multiplied by the cross section of the regionthe vacancies are moving through This is dδ, where δ denotes thethickness of the grain boundary

In total, the rate of vacancy diffusion is thus jdδ The growthrate equals this rate, normalised by the cross section of the grain i e.,jdδ/d2 = jδ/d To get the strain rate, we again need to divide by thegrain size, yielding the final result ˙ε ∝ jδ/d2∝ 1/d3

.The strain rate for grain boundary diffusion creep is thus

Because the shape of the grains changes in diffusion creep, neighbouringgrains have to deform in a compatible manner, analogous to the compatibility

of deformation in grain boundary strengthening as discussed in section 6.4.2.This is one cause of grain boundary sliding, described in the next section.Diffusion creep is also important in fibre composites It was shown in sec-tion 9.3.2 that the load transfer to a fibre is determined only by the aspectratio, the quotient of length and diameter That the length is considered to

be important in technical applications is only due to the fact that the fibrediameter cannot be made arbitrarily small, whereas the length can be as large

as desired However, at high temperatures, the fibres can be unloaded by sion processes Atoms of the matrix can move along the fibre-matrix interfaceand relax the stress between fibre and matrix Similar to equation (11.16), theabsolute size of the fibre now becomes important and only sufficiently longfibres can have a strengthening effect.

diffu-11.2.4 Grain boundary sliding

At high temperatures, grains in metals and ceramics can move against eachother This process is called grain boundary sliding

The strain rate of grain boundary sliding cannot be estimated as simply

as for the other processes It is [26]

bound-In metals, grain boundary sliding usually contributes only slightly to theoverall deformation, but it is nevertheless important for two reasons: First, indiffusion creep, grain boundary sliding ensures the compatibility of the grainsduring the deformation (see also section 6.4.2 and the end of the previous sec-tion) as sketched in figure 11.8 Second, at points where three grain boundariesmeet (triple points), movement of the grain boundaries by sliding can cause

a large concentration in local stresses and thus induce damage by rupture ofthe grain boundaries (see also section 11.3) It is thus doubly advantageous toincrease the resistance of the grain boundary against sliding: deformation bydiffusion creep is impeded, and the danger of early damage is reduced Thiswill be discussed further in section 11.4

In ceramics, the strength at high temperatures is often limited by grainboundary sliding The reason for this is the presence of a glassy phase at thegrain boundaries (see also section 7.1) These amorphous regions have a muchlower softening temperature than the grains themselves This ‘lubricating film’eases sliding of the grains, without dislocation movement inside the grains be-ing necessary One important goal in manufacturing ceramic high-temperaturematerials is thus to reduce the amount of glassy phase as much as possible

11.2.5 Deformation mechanism maps

The various creep mechanisms discussed so far differ in their temperature pendence because the activation energy of the mechanisms is different Further-more, they differ in their stress-dependence The creep exponent takes values

Fig 11.8 Grain boundary sliding ensures the compatibility of grains which would

be violated if only diffusion creep would occur

volume diffusion

Herring creep

Nabarro-dislocation creep

diffusion creep

disl pipe diffusion

Fig 11.9 Idealised deformation mechanism map (after [26])

between 1 in diffusion creep and 3 in dislocation creep, with even higher ues occurring in reality Thus, depending on the external conditions, differentcreep mechanisms dominate the behaviour

val-So-called deformation mechanism maps allow to read off the dominantmechanism under different conditions Figure 11.9 shows a schematic defor-mation mechanism map In the diagram, the temperature and the external

yield strength

Coble creep

dislocation creep

Herring creep

yield strength

dislocation creep

Herring creep

At low external stresses and low temperatures, the material deforms tically At higher temperatures, diffusion creep starts, being stronger at smallstresses than dislocation creep because of its lower creep exponent Because

elas-of the lower activation energy for grain boundary diffusion, this mechanism

is more important than bulk diffusion at low temperatures Since the creepexponent is the same in both cases, the two regions are separated by a verticalline

If we move on to larger stresses, dislocation creep with its larger creepexponent becomes dominant Vacancy diffusion along dislocation lines is moreimportant than diffusion through the bulk material at lower temperaturessince its activation energy is smaller Because the creep exponent is larger fordiffusion along the dislocations than through the bulk, the separating line isinclined

"1

"3

"4

".4 > ".2 >"1

.

".3 >

diffusion creepdislocation creep

Fig 11.12 Idealised deformation mechanism map at different strain rates ter [26])

(af-At even higher stresses, time-independent plastic deformation begins Ifthe stress level reaches about one tenth of the shear modulus, the theoreticalstrength of the material is reached

Diagrams like this can be compiled for different materials and materialstates Figure 11.10 shows the deformation mechanism maps of aluminiumand tungsten at a grain size of 32µm Although both maps have the sameoverall structure as the schematic map from figure 11.9, they neverthelessdiffer in the size and exact shape of the different regions

Figure 11.11 shows how the grain size changes the deformation nisms, using the example of silver The regions are shown for three different

grain 2

formergrainboundary

)

(a) Accumulation of cavities at

the grain boundaries

(b) Mechanism of formation

Fig 11.13 Schematic illustration of cavern-type pores at grain boundaries

grain sizes, making it easy to see that creep processes are more important atsmall grain size and start at lower temperatures This is due to the grain sizedependence of diffusion creep

Since creep processes are time-dependent, the dominant mechanism alsodepends on the strain rate This can also be represented in the diagrams asshown in figure 11.12 At high strain rates i e., high stresses, diffusion creepbecomes less important in comparison to dislocation creep

11.3 Creep fracture

After sufficiently long loading times, creeping materials fail by creep fracture.The strain rate, which attained its minimum value during secondary creep,increases again, and tertiary creep starts, ending with the final fracture, so-called creep rupture In most cases, creep fracture is distinguished by materialfailure at the grain boundaries, not inside the grains In contrast to ductilefracture, creep fracture is thus usually intercrystalline Transcrystalline frac-ture usually occurs only at high stresses [40]

That fracture occurs at the grain boundaries indicates a damage processthere It is due to the formation of pores and microcracks Microscopically,two different types of damage are distinguished in a creeping material On theone hand, oval cavern-type pores can be formed at grain boundaries which areloaded under tension, on the other hand, wedge-type pores may be induced

at triple points where three grain boundaries meet

Cavern-type pores form by diffusion processes in which the material creases its length in the direction of the tensile stress by moving atoms from

¾

¾

¾grain 1

grain 2

grain 3)

¾

¾

(a) Wedge-type pores at

triple-points of the grains

(b) Mechanism of formation

Fig 11.14 Schematic illustration of wedge-type pores at triple-points

the region of the forming pore to neighbouring zones (figure 11.13) Completelyanalogous to the precipitation of particles (see section 6.4.4), a nucleation bar-rier has to be overcome to form a cavern-type pore, for the energy required

to form an inner surface is proportional to the square of the pore diameter,whereas the energy gain depends cubically on the diameter For small pores,the surface energy dominates Pores are therefore not formed initially, butonly after a long time Their formation is favoured at high temperatures andlarge testing times because this raises the probability to overcome the nucle-ation barrier by thermal activation Local stress concentrations, for exampledue to precipitates on the grain boundary or dislocation pile-up, increase theenergy gain and thus favour pore formation

Wedge-type pores are formed by grain boundary sliding at triple points(see figure 11.14) In these regions, there is a large stress concentration (seesection 11.2.4) that can cause failure of the grain boundaries if their cleavagestrength is exceeded Accordingly, this type of damage usually occurs at highstresses

Both damage mechanisms cause a decrease of the effective cross section ofthe specimen and stress concentrations by notch effects This is the reason forthe rapid increase in the strain rate observed in tertiary creep

11.4 Increasing the creep resistance

Materials heavily loaded under creep conditions must meet particular ments

require-We can conclude from the discussion of the previous sections on anisms that a large activation energy of vacancy diffusion is advantageousbecause vacancy diffusion is important in almost all of the mechanisms dis-cussed Vacancy diffusion is weak if the formation of a vacancy is difficult and

mech-if the dmech-iffusion of any formed vacancy is impeded The enthalpy of vacancyformation is correlated with the binding forces in the material and thus withthe melting temperature Therefore, the homologous temperature T /Tm can

be used as parameter to characterise the creep properties

Vacancy diffusion occurs by the exchange of a vacancy with its bouring atom The atom has to overcome an energy barrier formed by thesurrounding atoms This barrier is the higher, the closer packed the atomsare; close-packed structures are thus more creep resistant For example, thediffusion coefficient for self-diffusion of iron is

by drawing wires They are one important reason for the long life-time oftungsten filaments in light bulbs Gas turbine blades with strongly elongatedgrains in the loading direction are also highly creep resistant One additionaleffect in these blades is that a crack cannot propagate easily even if single grainboundaries fail because the next grain boundary that is perpendicular to theloading direction is usually far away (see figure 11.15) The grains may in fact

be as long as the component, completely avoiding transversal grain boundaries(see figure 2.12) Single-crystal alloys are obviously especially suitable.Similar to time-independent plastic deformation, creep deformation in met-als is dominated by dislocation movement, especially at higher stresses Mecha-nisms that impede dislocation movement are thus also important in producingcreep-resistant materials However, these mechanisms have to be temperatureresistant

Surveying the strengthening mechanisms discussed in chapter 6, we seethat grain boundary strengthening is not suitable in creep applications be-cause we need large grains as explained above Work hardening can also not

be used since a large initial dislocation density would rapidly reduce by ery processes to a value that is determined by temperature and strain rate,destroying any initial hardening effect.

recov-A suitable mechanism is solid solution hardening, provided the dissolvedelements have a large activation energy for diffusion and are thus diffusingslowly Accordingly, carbon in steel cannot strengthen the material at hightemperatures because the interstitially dissolved carbon atoms diffuse rapidlyand move along with the dislocations.7 High-melting foreign atoms, on theother hand, can contribute significantly to the creep strength of metallic high-temperature materials since their bonds with the matrix atoms are usuallystrong, causing a limited mobility and thus making it difficult for dislocations

to take the atoms with them Examples are molybdenum, tungsten, and nium that are added as solid solution strengtheners to nickel-base superalloys

rhe-at weight fractions of up to 10 %

Another possible mechanism is precipitation hardening of metals Here,however, the following problem may occur: To achieve a fine distribution ofthe precipitates, coherent particles are needed These, however, are usuallyonly metastable and transform to an incoherent equilibrium phase at hightemperatures The interfacial energy strongly increases, causing acceleratedparticle growth and a loss of the hardening effect This is the crucial reasonwhy precipitation-hardened aluminium alloys cannot be used above 0.5 T /Tm

7

Carbon is, nevertheless, an important alloying element in creep-resistant steels,for it can form precipitates in the form of carbides

Ni Fig 11.16 Unit cell of the γ0 phase Ni3Al

in long-term applications Nickel-base superalloys play a special role in thiscontext because a coherent equilibrium phase exists which only coarsens slowly(see figure 11.4(b)) This is the so-called γ0 phase with stoichiometry Ni3Al.Its unit cell only differs slightly from that of the face-centred cubic matrix: thenickel atoms in the binary system Ni3Al are situated almost exclusively on theface-centred positions, whereas the aluminium atoms occupy the corners of thecell (figure 11.16).8 While the lattice constants of precipitates and matrix inthe binary system Ni-Al differ by more than one percent, adding other alloyingelements can reduce the lattice mismatch to 0.1% to 0.2% This reduces theinterfacial energy to a minimum and causes a very slow coarsening process.For this reason, these alloys can be used at 75% of their melting temperature,justifying the name ‘superalloys’

The volume fraction of the second phase in modern superalloys can be ashigh as 70% At these large values, dislocations are constrained to move in thesmall channels between the cuboid precipitates, forming highly elongated dis-location loops (figure 11.17) This impedes dislocation movement and causesthe high strength of these alloys The dislocations can cut the particles only

at rather high stresses Thus, stresses of 100 MPa can be applied at tures of 1000℃ and service times of several thousand hours without materialfailure

tempera-Dispersion-strengthened materials (see section 6.4.4) also have a highcreep resistance If a power law according to equation (11.2) is used

to describe the relation between strain rate and stress, extremely highcreep exponents with values between 20 and 200 result The strain raterapidly reduces to very small values with decreasing stress, resulting in

a high creep strength

This unusual creep property is caused by the interaction of the locations with the dispersoids The dispersoids are small enough to be

dis-8

In multi-component alloys, titanium and tantalum may take some of the sites ally occupied by aluminium Therefore, the notation Ni3(Al, Ti, Ta) is frequentlyused

usu-Fig 11.17 Elongated dislocation loops in a nickel-base superalloy (transmissionelectron microscopic picture) [127]

easily passed by climbing Nevertheless, they strongly impede tion movement because the line tension of the dislocation is reduced

disloca-at the interface to the dispersoid, resulting in an disloca-attractive interaction

To detach the dislocation from the dispersoid, a high stress is required

If the external stress is not sufficient, part of the required energy has

to be provided by thermal activation Because the required energy pends on the external stress, which thus enters the Boltzmann factor inthe exponent, the dependence on the stress is unusually strong and can

de-be approximately descride-bed by a power law with large creep exponent

To enable this mechanism, the line energy of the dislocation at theparticle interface must be reduced.9 This is also the case if the crystalcontains a cavity instead of a particle Tungsten alloys, for example,can be strengthened by adding potassium particles which evaporate athigh temperatures and thus form gas-filled cavities

Another example of how the different strengthening mechanisms interact isshown in table 11.2 The table shows the creep rupture strength at a loadingtime of 100 000 h for different steels Whereas the plain carbon steel C 35 has nosignificant strength at temperatures above 450℃, adding only 1% chromiumand 0.4% molybdenum in the steel 13 CrMo 4-4 significantly increases thecreep resistance Raising the chromium or molybdenum content further andadding vanadium strengthens the effect Strengthening is, on the one hand,due to carbide particles whose stability increases from Fe3C to Cr23C6 to

VCX.10 On the other hand, molybdenum and chromium not bound in thecarbides cause solid solution strengthening The service temperature can be

9

In contrast to the obstacles discussed in section 6.3, it is now important that theinteraction is attractive, not repulsive, because a repulsive obstacle can be easilyavoided by climbing

10 Vanadium carbide does not precipitate in a unique chemical composition and isthus denoted VC

Table 11.2 Creep rupture strength of several alloys (after [39,125,129]) The creeprupture strength Rm/100 000/T i e., the stress needed to cause fracture in a specimen

at temperature T after 105hours (creep rupture time), is stated The creep resistance

of the ferritic steels with large amounts of vanadium and chromium is significantlylarger than that of simpler steels because vanadium and chrome carbides have abetter temperature stability Due to their close-packed face-centred cubic structure,the creep resistance of austenitic steels is larger The creep strength of the nickel-basesuperalloys IN 738 (polycrystalline) and SC 16 (single crystalline) were estimatedfrom Larson-Miller data

Rm/100 000/T/MPatemperature in℃ 420 450 500 550 600 700 800 900

in the face-centred cubic lattice

Grain boundary sliding, discussed in section 11.2.4, can be impeded byadding discrete particles (e g., carbides) at the grain boundaries In this case,sliding of the grain boundaries requires material transport from the side ofthe particle under compression to the side loaded in tension, slowing downthe process

In cast magnesium alloys, which are increasingly used in automotive try, the influence of grain boundary sliding is rather strong, partially due tothe fine-grained structure of these alloys To improve the creep resistance, sili-con can be added because it forms a Mg2Si phase on the grain boundaries andthus impedes sliding A similar effect can be achieved by adding rare-earthmetals which also cause precipitation hardening [111]

In this chapter, we present exercises to elaborate upon the topics discussed

in this book Simple memorising exercises that can be solved by looking upthe topics are not given Complete solutions to the exercises are provided inchapter 13

Frequently, crystal structures are visualised by drawing the atoms as touchingspheres Using this assumption, calculate the packing density of

a) face-centred cubic,

b) body-centred cubic, and

c) hexagonal close-packed crystals!

Consider a polyethylene molecule with a degree of polymerisation of 104.a) Calculate the molar mass of the molecule! The molar mass of carbon is12.01 g/mol, that of hydrogen is 1.01 g/mol

b) Calculate the length of the chain, assuming that it is in a straight formation! The bond length between two carbon atoms is 0.154 nm, thebond angle is 109°

Assume the following functions for the interaction energy between two atoms

of a molecule of common salt (NaCl):

UA= −1.436

r eV nm , UR=

5.86 × 10−6

r9 eV nm9.a) Calculate the bond length of the diatomic molecule!

b) How large is the binding energy?

c) Compare the result of the calculation with the interatomic distance culated for a NaCl crystal with a density of % = 2.165 g/cm3! The molarmass of Na is 23 g/mol, that of chlorine is 35.4 g/mol Avogadro’s constant(the number of molecules in a mol) is NA= 6.022 × 1023mol−1

cal-d) Estimate Young’s modulus of NaCl in the h100i direction! Neglect thebonds between next-nearest neighbours and those even further away! Note:

1 eV = 1.602 × 10−19J

e) The elastic constants of NaCl are C11 = 48.7 GPa, C12 = 12.6 GPa and

C44= 12.75 GPa Use these values to calculate Eh100i! Compare this value

to your estimate!

The bulk modulus K is a measure of the pressure ∆p needed to change amaterial’s volume V0by ∆V :

a) Derive the relation between K and the elastic constants E, G, and ν in

an isotropic material at small deformations!

b) Calculate the bulk modulus of a material with a Poisson’s ratio of ν1= 0,

ν2=1/3, and ν3= 0.5! How does the volume of a tensile specimen changewith the uniaxial stress σ in the three cases?

c) Some rare materials possess a negative Poisson’s ratio What is thetransversal strain for a positive normal strain in this case?

In section 2.4.3, we introduced equation (2.23), C44= (C11−C12)/2, specifyingthe relation between the components C11, C12, and C44of the elasticity matrix(Cαβ) of an isotropic material Check this equation by prescribing a straintensor

l A ¢l A

(a) Booth A The rubber band is

elon-gated by the same distance ∆lAalways

l B ¢l B

(b) Booth B The rubber band is loadedwith the same force FBalways

Fig 12.1 Candy catapults, shown in the loaded state

and calculating the required stress state in the un-rotated xicoordinate systemand in a coordinate system with axes xi0, rotated by 45° To do so, use Hooke’slaw twice for the strain tensor, once in the xi and once in the xi0 coordinatesystem!

At a child’s fair, two booths present almost identical candy catapults At bothbooths, candies are accelerated on a horizontal plane using rubber bands Atbooth A, the rubber band is stressed for each shot by lengthening it fromthe initial length lA = lB by ∆lA = const (figure 12.1(a)), whereas booth Bstresses the rubber using a rope, a pulley, and a weight loading the band with

a force FB= const

The cross section of both rubber bands is identical (A) Assume thatboth rubber bands are linear-elastic Young’s modulus of the rubber band

of booth B is twice as large as that at booth A: EB= 2EA

At both booths, the take-off velocity of the candies is disappointingly small

We want to find a way to increase the velocity without using additional terial or changing the construction of the catapults

ma-a) Start by deriving equations for the stored elastic energy W(el) when thebands are stretched!

b) How does the stored energy change when Young’s modulus is increased ordecreased?

c) Derive an equation for the take-off velocity of the candies (mass m) atboth booths! Neglect the mass of the rubber bands and any occurringfriction!

d) Suggest a simple method to increase the take-off velocity! By what factordoes the velocity increase?

Exercise 7: True strain

To compare nominal strain and true strain, we investigate two different mation processes of two rods: The first rod is strained in two steps ∆l1 and

defor-∆l2, the second in a single step ∆l1+2= ∆l1+ ∆l2

a) Prove the inequality ε1+ ε26= ε1+2for the strains!

b) Estimate how the strain measures differ when the length change ∆li issmall and large, respectively! To do so, sketch a diagram of the difference

∆ε versus ε1/ε1+2!

c) Show that, for the true strains, the equation ϕ1+ ϕ2= ϕ1+2 holds!

This exercise illustrates that the question of what initial value is used tocalculate a quantity is not only important in calculating strains, but also incalculating interests

The customer of a bank invests an initial amount of G0= 10 000AC at hisbank He wants to double his money within ten years (G10)

a) Calculate the interest rate z0 required if the interest is always calculatedrelative to the initial deposit G0!

b) How large is the required interest rate z if the interest is always paid onthe current deposit Gi?

As in the solution to exercise 4 a), we again consider the deformation of abrick with edge lengths lito the new lengths li+ ∆li This time, however, wewant to account for large deformations

a) Calculate Green’s strain tensor G for this deformation!

b) Compare Green’s strain tensor G with the strain tensor ε for large andsmall deformations!

A component made of a polycrystalline aluminium alloy with yield strength

Rp0.2= 200 MPa is loaded in a plane-stress state The stress components are

σ11= 155 MPa, σ22= 155 MPa, and τ12= 55 MPa

a) Calculate the principal stresses!

b) Use the Tresca yield criterion to decide whether the material yields!

f) Calculate the stress deviator σ0 for the given stress state!

In a thermoplastic, a yield strength of Rp= 40 MPa was measured in tension,whereas the stress in a compressive test is Rc = 50 MPa The conically andparabolically modified yield criteria are to be compared for several load cases

a) At what purely hydrostatic stress does yielding occur according to thetwo criteria?

b) A component made of this polymer is loaded with a stress state σ11 =

σ22 = −σ33 = 0.56 Rp, σ23 = σ13 = σ12 = 0 Use both criteria to checkwhether the material yields!

c) Does the material yield at a state σ11 = −σ22 = −σ33= 0.56 Rp, σ23 =

σ13= σ12= 0?

A shaft with a circumferential round notch (see figure 4.3) with dimensions

D = 20 mm, d = 16 mm, and % = 4 mm is loaded in tension It has to bechecked whether it can be used at a service load of F = 40 kN The shaft ismade of an aluminium alloy with Young’s modulus of E = 68 000 MPa andthe stress-strain curve shown in figure 12.2 (Rp= 202 MPa, Rm= 280 MPa).a) Determine the stress concentration factor K using diagram 4.3!