Mechanical Behaviour of Engineering Materials - Metals, Ceramics, Polymers and Composites 2010 Part 12 potx

The calculation could be improved by using the elastic stress field at some distance from the crack tip and bycalculating the displacements of all atoms inside this region using the forc

The result is: σI= 210 MPa, σII= 100 MPa, σIII= 0 MPa.

b) σeq,T= σI− σIII= 210 MPa > Rp0.2 The material yields

d) It is not possible to decide because both yield criteria are only approximatelytrue

e) τ = σeq,M/M = 58.7 MPa < τF No significant activation of dislocation ment

move-f) The deviator can be calculated from σ0 = σ − 1 σhyd, using σhyd = tr σ/3 =(155 + 155 + 0)/3 MPa = 103.¯3 MPa This results in

The parameter m from equation (3.36) is m = 50 MPa/40 MPa = 1.25

a) The hydrostatic stress state is characterised by σ11 = σ22 = σ33 = σhyd and

–2+ 0

=m − 1

m · 3 σeq,pM,F=3

5σeq,pM,F.This results in a ‘hydrostatic yield strength’ of σeq,pM,F= 5/3 · Rp= 66.7 MPa.Conical: According to equation (3.39), we find at yielding

Rp= 1

2m· [(m − 1) · 3 σeq,cM,F+ 0] = 3

10σeq,cM,F.This results in a ‘hydrostatic yield strength’ of σeq,cM,F= 10/3·Rp= 133.3 MPa.b) Parabolic:

σeq,pM=0.25

2.5 σ11+

s

» 0.252.5 σ11

–2+ 82.5σ

2 11

= 1

10σ11+

r321

100σ

2

11= 1.89 σ11= 1.058 Rp.The material yields

q4σ2 11

–

= 1.9 σ11= 1.064 Rp.The material yields

–2+ 82.5σ

2

11= 1.69 σ11= 0.947 Rp.The material does not yield

Conical:

σeq,cM = 1

2.5

»0.25 (−σ11) + 2.25 ·

q4σ2 11

–

= 1.7 σ11= 0.952 Rp.The material does not yield

According to equation (4.1), we find σmax= Ktσnss= 332 MPa σmax is above

Rp and Rm Thus, the component could not be used

c) Using equation (4.5) yields

σmaxεmax=σ

2 nss

E K

2

t = 1.623 MPa The corresponding Neuber’s hyperbola is shown in figure 13.4

d) The values can be read off the diagram: σmax= 210 MPa, εmax= 0.008 = 0.8%

F = aNaCl

ZaNaCl0

KIc

√2π

0 = √KIc

2π· 2a3/2NaCl.b) The force is F = kx if the bond is strained by a distance x The force at a strain

of aNaCl/10 must, according to the assumption, equal the force from subtask a).The fracture toughness can thus be calculated as follows:

k aNaCl

10 =

KIc

√2π· 2a3/2NaCl,k

10=

KIc

√2π· 2a1/2NaCl,

KIc= k

10

√2π2

1

a1/2NaCl = 0.634 MPa

√

m

This value is of the correct order of magnitude for a ceramic crystal

c) Because we simply used the stress field calculated from continuum mechanics

to find the force at one atom, the calculation is incorrect since there can be

no stresses in between the atomic positions The calculation could be improved

by using the elastic stress field at some distance from the crack tip and bycalculating the displacements of all atoms inside this region using the force law.Furthermore, it would be necessary to quantify the fracture strain of a bondmore precisely Assuming a simple spring force is also a severe approximationbecause the potential curve is not parabolic if the displacements are large (see,for example, figure 2.6) Calculations accounting for all this can yield realisticvalues for the fracture toughness of a material The calculation as presentedhere can be accepted as a very coarse approximation that mainly serves to showwhy a stress singularity does not imply a force singularity at the position of thecrack tip

Solution 14:

We start by adding the elastic line and the 95% line to the diagram (figure 13.5).Reading off the forces yields F5= 13.5 kN, Fmax= 14.5 kN F5is to the left of Fmax,corresponding to the case from figure 5.16(b), yielding FQ= F5 The condition (5.32)has to be met: Fmax/FQ= 1.07 ≤ 1.1 (true)

The geometry factor for the initial crack length is f = 9.66 Using equation (5.30)yields KQ= 23.3 MPa√

m We finally have to check the inequality (5.33) The hand side is 2.5(KQ/Rp)2 = 5.2 mm All required dimensions (B, a, W − a) fulfilthis condition

Fig 13.5 Determination of the forces F5 and Fmax in the load-displacement gram

dia-Thus, the fracture toughness is KIc= 23.3 MPa√

m

Solution 15:

The solution is based on section 5.2.3

a) Design against yielding: The stress state is uniaxial with σ = pD/(2t) Thus, thecondition σ = pD/(2t) < Rp0.2 must be met: Since σ = 1200 MPa < 1420 MPa,there is no yielding

Design against cleavage fracture: σI< σC: 1200 MPa < 2200 MPa Cleavagefracture is not to be expected

Design against crack propagation: σI< KIc/√

πa, where a = 1.5 mm is themaximum half crack length to be expected: 1200 MPa < 1311 MPa There will

be no crack propagation

The tube can be used

b) Since Rpwas stated for uniaxial loading, the result is independent of the yieldcriterion because the service load is also uniaxial.1

c) Yielding occurs at a pressure p = 2tRp0.2/D = 14.2 MPa

Cleavage fracture will be observed at a pressure p = 2tσC/D = 22 MPa.From the fracture toughness, the stress can be calculated using σ =

KIc/√

πa The resulting failure pressure is p = 2tKIc/(√

πa D) = 13.1 MPa.Thus, the tube will fail by crack propagation at a pressure p = 13.1 MPa if

a crack of length a = 1.5 mm is present

d) The yield strength and the fracture toughness are reached simultaneously at avalue ac from equation (5.28): ac= (90 MPa√

m/1420 MPa)2/π = 1.28 mm.e) See figure 13.6

f) According to equation (5.3), the crack opening is

Fig 13.6 Failure-Assessment diagram for exercise 15

where the additional factor 2 is necessary because equation (5.3) uses the placement of one crack surface which is half of the crack opening

corre-b) The equilibrium position at x = a/2 is unstable because the atoms of one layerare situated between those of the other layer, resulting in a maximum strain ofthe bonds An infinitesimal displacement from this position would result in theatoms moving to either x = 0 or x = a This is due to the fact that the stiffness

C = dτ /dx is negative at this point:

b) The dislocation density is the dislocation length per volume To completely shearthe crystal over its width, the dislocation has to extend throughout the crystaland thus have a length of 10 mm The resulting dislocation density is thus

% =aN

a3 = N

a2 = 7 × 109m−2.Not all of the dislocations can contribute because of their orientation Assumethat the shear is in the x direction If we consider screw dislocations, all dis-locations with a line vector in the y direction can contribute (one third of allscrew dislocations) Of the edge dislocations, only those with line vector in the

y direction can contribute that have the additional half-plane in the z tion (one sixth of all edge dislocations) As we are interested in an estimateonly, we can assume that about one fifth of all dislocations contribute to thedeformation This results in a final estimate for the dislocation density of about3.5 × 1010m−2

direc-c) If the grain size is d, the dislocations do not move throughout the crystal, butare limited to one grain because, due to the small amount of deformation, thestresses can be expected to be too small to allow dislocations to pass grainboundaries The number of dislocations thus increases by a factor s/d = 100,resulting in a dislocation density of % = 3.5 × 1012m−2

d) The total length of all dislocations is L = %a3= 3.5 × 106m = 3500 km

Solution 18:

The energy per length of a dislocation line is T ≈ Gb2/2 according to equation (6.3).Strictly speaking, this is only valid for a straight segment, but we will see that therequired energy is so large that this is irrelevant The energy E of a dislocationloop of length 6b is E = 6Gb3/2 = 1.8 × 10−18J The probability to form such adislocation loop is P = exp`−E/(kT )´, resulting in P300 ≈ 1.5 × 10−189

at 300 Kand P900≈ 1.1 × 10−63at 900 K The thermally activated generation of dislocationloops is thus practically impossible

Solution 19:

To calculate the increase in strength, we can use equation (6.20), with the increasebeing the difference of the strengthening contribution in both states:

dcoarsein the coarse-grained and ∆σfine= k/√

dfine in the fine-grained material.Reducing the grain size strengthens by the difference of these two contributions:

∆σ = √k

dfine

−√ k

dcoarse.Solving for the new grain size, we find for ∆σ = 80 MPa

2λ = GbM

∆Rp0.2

= 39 nm b) According to equation (6.28), we find

in figure 13.7 From this diagram, we can read off the Weibull modulus m = 1 whichequals the slope The intersection with the axis is −m ln(σ0/MPa) = −5.6, yielding

σ = 270 MPa

Fig 13.7 Graphical determination of m and σ0 in a diagram analogous to ure 7.17

σlimit

σ0

= mr

−V0

Using V /V0 = 1, we thus find σlimit = 0.541 σ0 Relating this to the pressure,

σlimit= 2pL2/d2min, we find for the thickness

dmin=

r

2pL20.541 σ0

c) From equation (13.19), we find, using V1/V0 = LBdmin/Vspec = 12.08, thenew value of the maximum stress σlimit,1 = 0.458 σ0 With the help of equa-tion (13.20), the thickness is calculated to be dmin,1= 16.35 mm

This, however, changes the specimen volume V , changing the allowedstress σlimit from equation (13.19) Using the current thickness value dmin,1 =16.35 mm yields a permitted stress of σlimit,2 = 0.456 σ0 Using again equa-tion (13.20) results in the new thickness dmin,2= 16.40 mm

The change from dmin,1 to dmin,2 is rather small, making further iterations

of the procedure unnecessary

d) Production needs not to be stopped because all of the material volume is mally stressed in tension, but only a small part of it in bending The strength

maxi-in a tensile test is thus smaller than maxi-in a bendmaxi-ing test Thus, the safety of theproduct is increased by this error

e) d =p2pL2/Rp= 11.07 mm

of the inert strength, we find B = B∗/σn−2c = 0.3912 MPa2h.

Note: Due to the large exponents in this calculation, your results may differfrom those stated here by several percent The values here result if the exactvalues are used

b) The failure probability can be calculated from equation (7.10), with V /V0= 1,

m∗ = m/(n − 2) = 1.2222, and t0(σ) = B∗σ−n = 3.1529 × 1045MPa20h ·(100 MPa)−20= 314 016 h:

c) The failure probability can be reduced using a proof test

d) The calculation is analogous to the derivation of equation (7.16):

The result is σp= 324.1 MPa.

e) The fraction of scrapped parts is calculated using equation (7.3):

The constant of integration C0 can be determined by the fact that the strain ε

is zero at time t = 0 because the dashpot element cannot react immediately tothe stress Thus, we find C0= σ and

b) In a relaxation experiment, the strain is to be increased discontinuously by afinite value This causes an infinite stress in the dashpot element in this model

A relaxation experiment can therefore not be modelled with this approach.c) The three elements in the three-parameter model are denoted as follows: Ele-ment 1 is the spring element in series, element 2 is the parallel spring element,and element 3 is the dashpot element This yields the following relations forstresses and strains:

If the strain ε is constant, we find, using σ1= E1ε1= E1(ε − ε3),

The retardation and relaxation times are the inverse of the prefactor of thevariable t in the exponential function Thus, they are tc= η/E2and tr= η/(E1+

E2), respectively The relaxation time is always smaller than the retardationtime

Solution 26:

a) The time-dependence is shown in figure 13.9(a)

The parameter δ describes the time-shift between stress and strain If δ = 0,stress and strain are in phase, and the material is not viscoelastic; if δ = 90°, thestrain is at its minimum or maximum when the stress is zero In real materials,the value of δ depends on the frequency ω In polymers, δ can take values of afew degrees

b) Using the addition theorem sin(a − b) = sin a cos b + cos a sin b and ωt =arcsin(ε/ε0) (where we have to keep in mind for later that values of the arcsine are limited to [−π/2, π/2]) we get

444 13 Solutions

σ = σ0cos δ sin ωt + σ0sin δ cos ωt

= σ0cos δ · sin“arcsin ε

ε0

”+ σ0sin δ · cos“arcsin ε

The first term is in phase with the strain, the second is out of phase

c) The stress-strain diagram is shown in figure 13.9(b) It has to be noted that therelation between stress and strain is not unique because there are two possiblestress values for any strain This is due to the fact that the cosine in equa-tion (13.22) is always positive when arguments in the interval [−π/2 : π/2] areused For a full circle ωt, negative values of the cosine may also result The fullstress-strain diagram results when we replace the +-sign in equation (13.22) by

no dissipation Thus, only the second term has to be considered in calculatingthe energy The enclosed area – and thus the specific work done – is equal totwice the area above the dashed line in figure 13.9(b) It can be calculated asfollows:

ε0

#ε0

−ε0

= σ0sin δ · ε0· π

This is the area of an ellipse with major and minor axis ε0 and σ0sin δ, for if

we shear the area vertically to move the diagonal to the ε-axis, it is this ellipsethat results

Solution 27:

The activation energy can be determined using equation (8.7) If we compare thestrain rates ˙ε at different temperatures T1and T2and stresses σ1and σ2at identicalvalues of σ/T , we can divide the strain rates to find

“

− Q

kT2

”

13 Solutions 445The unknown activation volume cancels It could be determined in the same way

as Q Solving for Q yields

E /GPa parallel connectionserial connection

Fig 13.10 Young’s modulus in a fibre composite for different fibre arrangements

ff = lf/l and fm = 1 − ff = lm/l We thus find the isostress rule of mixtures,equation (9.5):

εm= σm

E = 0.02 , εf =

σf

E = 0.014

13 Solutions 447Thus, the fracture strain of the fibre is reached first The failure stress can beestimated using equation (9.7),

σ = σfff+ σm(1 − ff) ,

inserting the stresses at a strain of 0.014 The fibre stress is σf = 4900 MPa,whereas the matrix stress at strain 0.014 is only 42 MPa Altogether, we find atensile strength of

4900 MPa × 0.55 + 42 MPa × 0.45 = 2714 MPa

c) The compressive strength is, according to equation (9.11),

Rc,in phase=

s

ffσm,FEf3(1 − ff) =

s0.55 × 60 MPa × 350 000 MPa

3(1 − 0.55) = 2925 MPa

(13.24)d) We have to check first whether the fibres are larger than the critical length,

lc = dσf/2τi = 0.65 mm The fibres are much longer than this, ensuring aneffective load transfer onto the fibres However, the strain increases locally nearthe fibres by a factor of about two relative to the global strain (see section 9.3.2).According to subtask b), the fracture strain in the matrix is 2% Therefore, thetotal strain in the structure must not exceed 1% At this strain, the matrix andfibre stresses are 30 MPa and 3500 MPa, respectively Using the isostrain rule ofmixtures, we find a tensile strength of 1939 MPa

m Because themaximum stress intensity factor Kmax is clearly below KIc, the component willnot fail statically

c) At R = −1, equation (10.6) yields

∆Kth= E · 2.75 × 10−5· 20.31√

m = 2.39 MPa√

m Since the current range of the stress intensity factor is ∆K = 2Kmax =6.16 MPa√

m according to subtask b), stable crack propagation must be pected

ex-d) The crack growth per cycle at the current crack length can be calculated fromequation (10.8):

da/dN = C∆Kn= 2 × 10−12MPa−2cycle−1× (6.16 MPa√m)2

= 7.6 × 10−11m/cycle = 7.6 × 10−8mm/cycle

«nZ af

a0

1

`Y√a´nda Inserting n = 2, Y (a) = 1 + ba, and b = 0.1 mm−1 yields

–

= 20 621 The component can be cleared until the next service interval because the number

of cycles to failure is significantly larger than the interval time

The total damage in a week is thus Dweek =P4

i=1Di= 9.77 × 10−2 Fracture

of the component is to be expected after n weeks when a damage value of

D = n · Dweek= 1 has been reached:

c) Since we need an approximation only, we can use the stress and strain values

in the middle of the tube The elastic part of bending is negligible, so we canwrite ε = ˙εt because the strain rate is constant at constant stress, We thus find

Aσl2t4d =

A%gl4t32d2 = 2 × 10−4m

In each year, the middle of the tube is displaced by 0.2 mm

450 13 Solutions

Solution 34:

a) Heating causes thermal strains εth in the rod which are compensated for byelastic strains εel because of the clamping In the course of time, the stresscaused by the elastic strains is relaxed by a creep strain εc We can write

For the case of interest, n = 3, this yields

The constant of integration can be found by using the condition σ0= −Eα∆T =

−2223 MPa at time t = 0 in equation (13.25):

C = − 1

2σ2 = − 1

2E2α2∆T2 = −1.01 × 10−7MPa−2.After 100 s, the stress is only σe= −113 MPa

b) The elastic strain at the end of holding time is only εe= σe/E = −8.70 × 10−4.This strain reduces to zero at a temperature difference of ∆T = εe/α = −49.7 K.The rod will fall from the clamping at a temperature of about 950℃

Using tensors

In this chapter, we discuss the basics of how to calculate with tensors For

a more detailed study, the reader is referred to the technical literature e g.,Holzapfel [67]

A.1 Introduction

In general, a tensor is a physical quantity that is associated with coordinates.Although its numerical representation may change when switching to anotherreference or coordinate system, the tensor itself remains unchanged As anexample, consider the vector a in figure A.1 To state the value of the vector,

a single number is not sufficient Instead, its components have to be stated

in a coordinate system Usually, this is done by writing the coordinate values

as a column vector In the x1-x2 coordinate system shown in the figure, thevector’s representation is

If we use the x1 0-x2 0 coordinate system, rotated by 45°, the components are(ai0) =

The components are not the same in both systems, but the vector a itself isnevertheless the same as can be seen in figure A.1

A.2 The order of a tensor

Tensors can be classified by their order, sometimes also called ‘rank’ As a firstexample, we consider again a vector As already explained, its components are