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SOLID MECHANICS - KTH

Written examination, Material Mechanics (SE2126), 2009-12-17. The results of the examination will be announced at the latest on 2010-01-08. Complaints on the marks must be made within a month thereafter. Observe: The student must show a valid identity paper. Write only on one side of each paper. Write your name and personal identity number on each sheet. All colours except red may be used. Aids: Textbook Material Mechanics by Peter Gudmundson, Handbok och Formelsamling i Hållfasthetslära, mathematical handbooks, complementary formulas (extra sheet available at the exam) and pocket calculator. Note that copied material and lecture notes are not allowed on the exam. Examiner: Martin Kroon Part A 1. a) Describe the general features of the microstructure of polycrystalline metals, and motivate why these metals in general are fairly isotropic on the macroscopic level. (2p) b) Why is plastic deformation in metals insensitive to the hydrostatic stress? (2p) c) The initial yield stress of metals tends to decrease with increasing temperature. Why? (2p) 2. Consider a composite made of epoxy with uniaxially oriented glass fibres, see figure below. The fibres are aligned in a direction that makes an angle 20o with the x-axis. The material is exposed to a stress state σx = 40 MPa, σy = 30 MPa, τxy = 20 MPa (other σij = 0). The elastic properties of the composite are characterised by the entities EL = 90 GPa, ET = 40 GPa, GLT = 20 GPa, and νLT = 0.35. Determine the normal strain in the direction of the fibres. (6p)

3. An elastic-ideal plastic material (obeying von Mises plasticity) is loaded by a positive loading parameter σ0, causing the multi-axial stress state σx = σ0, σy = 2σ0, τxy = σ0. Plane deformation εz = 0 is imposed, such that τxz = τyz = 0, σz ≠ 0. Compute the value of σ0 that will result in initiation of plastic deformation and compute the

Examination in Material Mechanics (SE2126), 2009-12-17.



associated six plastic strain increments at initiation. Express your answer by use of the material constants E, ν, and σs and the equivalent plastic strain increment dλ. (6p) 4. One useful model in viscoelasticity is shown below (it’s also reproduced in Gudmundson’s textbook). Show that for this model, the transformed relaxation modulus can be expressed as

( )2121

11

//11~

EsEEsE

ssE

ηη++

+⋅= .

(6p)

Part B 5. In a manufacturing process, a piece of metal is exposed to a uniaxial pressure pulse

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−=

000 14

tt

tttp σ , , see figure below, where σ0 = 700 MPa and

t0 = 0.12 s. The viscoplastic behaviour of the material follows the Perzyna model, and

the plastic strain increments are defined by

00 tt ≤≤

es

e0

p

23

1σσ

σεε ijij

s−= && , with 0ε& = 0.35 s-1

and σs = 400 MPa, and σe is the effective stress. Compute the plastic deformations , , , caused by the pressure pulse

pxε

pyε

pzε ( )tp .

(6p)

6. In one of its products, a company uses an isotropic polymeric material which has a Young’s modulus E = 30 GPa and a Poisson’s ratio ν = 0.31. The company wants to attain a material with a lower density, and therefore plans to foam the polymer to a relative density of ρ*/ρs = 0.6, where ρ* is the density of the foamed material and ρs is the density of the unfoamed original material. For a low concentration of voids (< 20%), the stiffness can be approximated by use of Eshelby’s solution as described in Gudmundson’s book, and for low density open cell foams (ρ*/ρs < 0.3), the stiffness can be approximated by use of beam theory, which yields E*/Es ≈ (ρ*/ρs)2. Use these two solutions to make an engineering estimate of Young’s modulus E* of the foamed polymer in question for ρ*/ρs = 0.6. (6p)


Written examination, Material Mechanics (SE2126), 2009-12-17. The results of the examination will be announced at the latest on 2010-01-08. Complaints on the marks must be made within a month thereafter. Observe: The student must show a valid identity paper. Write only on one side of each paper. Write your name and personal identity number on each sheet. All colours except red may be used. Aids: Textbook Material Mechanics by Peter Gudmundson, Handbok och Formelsamling i Hållfasthetslära, mathematical handbooks, complementary formulas (extra sheet available at the exam) and pocket calculator. Note that copied material and lecture notes are not allowed on the exam. Examiner: Martin Kroon Part A 1. a) Describe the general features of the microstructure of polycrystalline metals, and motivate why these metals in general are fairly isotropic on the macroscopic level. (2p) b) Why is plastic deformation in metals insensitive to the hydrostatic stress? (2p) c) The initial yield stress of metals tends to decrease with increasing temperature. Why? (2p) 2. Consider a composite made of epoxy with uniaxially oriented glass fibres, see figure below. The fibres are aligned in a direction that makes an angle 20o with the x-axis. The material is exposed to a stress state σx = 40 MPa, σy = 30 MPa, τxy = 20 MPa (other σij = 0). The elastic properties of the composite are characterised by the entities EL = 90 GPa, ET = 40 GPa, GLT = 20 GPa, and νLT = 0.35. Determine the normal strain in the direction of the fibres. (6p)

3. An elastic-ideal plastic material (obeying von Mises plasticity) is loaded by a positive loading parameter σ0, causing the multi-axial stress state σx = σ0, σy = 2σ0, τxy = σ0. Plane deformation εz = 0 is imposed, such that τxz = τyz = 0, σz ≠ 0. Compute the value of σ0 that will result in initiation of plastic deformation and compute the




associated six plastic strain increments at initiation. Express your answer by use of the material constants E, ν, and σs and the equivalent plastic strain increment dλ. (6p) 4. One useful model in viscoelasticity is shown below (it’s also reproduced in Gudmundson’s textbook). Show that for this model, the transformed relaxation modulus can be expressed as

( )2121

11

//11~

EsEEsE

ssE

ηη++

+⋅= .

(6p)

Part B 5. In a manufacturing process, a piece of metal is exposed to a uniaxial pressure pulse

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−=

000 14

tt

tttp σ , , see figure below, where σ0 = 700 MPa and

t0 = 0.12 s. The viscoplastic behaviour of the material follows the Perzyna model, and

the plastic strain increments are defined by

00 tt ≤≤

es

e0

p

23

1σσ

σεε ijij

s−= && , with 0ε& = 0.35 s-1

and σs = 400 MPa, and σe is the effective stress. Compute the plastic deformations , , , caused by the pressure pulse

pxε

pyε

pzε ( )tp .

(6p)

6. In one of its products, a company uses an isotropic polymeric material which has a Young’s modulus E = 30 GPa and a Poisson’s ratio ν = 0.31. The company wants to attain a material with a lower density, and therefore plans to foam the polymer to a relative density of ρ*/ρs = 0.6, where ρ* is the density of the foamed material and ρs is the density of the unfoamed original material. For a low concentration of voids (< 20%), the stiffness can be approximated by use of Eshelby’s solution as described in Gudmundson’s book, and for low density open cell foams (ρ*/ρs < 0.3), the stiffness can be approximated by use of beam theory, which yields E*/Es ≈ (ρ*/ρs)2. Use these two solutions to make an engineering estimate of Young’s modulus E* of the foamed polymer in question for ρ*/ρs = 0.6. (6p)

SOLID MECHANICS- KTH

Written examination, Material Mechanics (SE2126, 4C1126, 4C1110)December 21, 2007, 08.00 - 13.00

The results of the examination will be registered at the latest January 18, 2008. Com-plaints on the marks must be made at the latest one month thereafter.

Observe: The student must show a valid identity paper.Write only on one page of each paper. Write down your name and personal identity number on each sheet. All colours except red may be used.

Aids: Material Mechanics by Peter Gudmundson, Handbok och Formelsamling i Håll-fasthetslära, Complementary formulas in Material Mechanics, mathematical handbooksand pocket calculator.

Examiner: Peter Gudmundson

1. A thin sheet of wood is glued to a thick plate. The sheet of wood is not fully dry when itis glued to the plate. If the wood had been free to shrink during drying, the resultingshrinkage strains would have been , where L, T denote direc-tions in the plane of the wood sheet. The total strains are however vanishing due tothe constraint of the thick plate. Determine the state of stress in the sheet of wood afterdrying. It is assumed that the wood can be described as a linear, elastic material withorthotropic properties given by: = 12 GPa, = 0.7 GPa, = 0.38.

2. A thin plate in the x, y- plane ( ) made of an elastic-plastic material (E-modulusE, Poisson’s ratio ν, yield stress ) is loaded by the normal stresses( ) so that plastic deformation is initiated. All other stress componentsvanish. Thereafter the strain increments are enforced. Determine theresulting stress increments ( ). (It can be assumed that plastic deformation devel-ops and that the material can be described by von Mises flow rule).

3. A material is at time t = 0 loaded by linearly increasing normal stresses, where denotes a prescribed stress rate and t time. All other stress

components vanish. The material is isotropic and it shows elastic and creep deformation.The creep is described by Norton-Odqvist’s generalized creep law,

where denote the effective stress according to the von Mises hypothesis and thestress deviator respectively. The elastic properties are defined by the E-modulus E and thePoisson ratio . Determine the time when the strain due to creep and elastic deforma-tion are equally large.

εLM -0.001 εT

M, -0.02= =εL εT,

EL ET νLT

σz 0=σs

σx 2σy= σ0 0>=dεx dεy 0>–=

dσx dσy,

σx σy σ· 0t= = σ· 0

ε· ijc

ε· 0σeσc-----⎝ ⎠⎛ ⎞

n 32---

sijσe-----⋅=

σe2 3

2---sijsij=

σe sij,

ν εz

Written examination, Material Mechanics (SE2126, 4C1126, 4C1110) December 21, 2007, 08.00 - 13.00 1


4. A uniaxial tensile tension test is performed on a viscoelastic material that can bedescribed by the Maxwell model with parameters E and . The test is strain controlledwith a constant strain rate . Determine the resulting stress as a function of strain and make a sketch of the stress-strain curve.

5. A carbon fibre composite laminate is considered. The state of strain isdescribed by = 0.002, = -0.001, = 0, where the x- direction coincides with the

- direction of the laminate. Determine the normal stress in the fibre direction of the layers. The stiffness of the composite is described by = 200 GPa, = 10 GPa, = 7 GPa, = 0.3.

6. A linear elastic material with E- modulus E and Poisson ratio = 0.3 is reinforced by asmall fraction of stiff, spherical, elastic particles (much larger E- modulus than the matrixmaterial). The composite is loaded by a uniaxial stress , where is prescribed. Determine the stress in the particles.

ηε· 0 σ ε

0 45±,[ ]sεx εy γxy

0° σL45° EL ETGLT νLT

ν

σx σ σy, σz 0= = = σσy

Written examination, Material Mechanics (SE2126, 4C1126, 4C1110) December 21, 2007, 08.00 - 13.00 2

Solutions Material Mechanicswritten examination, 2007-12-21

Problem 1. Total strains are zero, so

εtot = εmek + εM = 0 ⇒ εmek = −εM

The stresses are given byσ = Cpsεmek,

where the plane stress stiffness matrix is

[Cps] =

⎡⎢⎢⎣

EL

1−νLTνTL

νLTEL

1−νLTνTL0

νLTEL

1−νLTνTL

ET

1−νLTνTL0

0 0 GLT

⎤⎥⎥⎦ where νTL = νLT

ET

EL

.

So the stresses are

[σ] =[

17.47 14.39 0]T

MPa.

Problem 2. The stress deviator is

[sij ] =σ0

2

⎡⎣ 1 0 0

0 0 00 0 −1

⎤⎦ ⇒ σe =

√3

2sijsij =

√3

2σ0 = σs.

The functionf = σ2

x + σ2y − σxσy − σ2

s = 0

describes the yield surface. The consistency relation ∂f∂σij

dσij = 0 gives

2σxdσx + 2σydσy − σxdσy − σydσx = 0 ⇒ dσx = 0

Elastic strains from Hooke’s law and plastic strains from the flow rule dεpij = dλsij

givesdεx = −νdσy

E+ dλ σs√

3

dεy = dσy

E+ 0 ( since sy = 0),

Sodσx = 0dσy = dεyE = −dεxE

1

Problem 3. Elastic strains from Hooke’s law (3.4 in HB)

εez(t) =

−2ν

Eσ̇0t

The stress deviator gives the effective stress

[sij] =σ̇0t

3

⎡⎣ 1 0 0

0 1 00 0 −2

⎤⎦ ⇒ σe =

√3

2sijsij = σ̇0t

The creep strain increment in z-direction

ε̇cz = −ε̇0

(σ̇0t

σc

)n

leads to a separable differential equation which can be integrated as∫ εcz(t)

εcz(0)

dεcz = −ε̇0

(σ̇0

σc

)n ∫ t

0

λndλ,

where λ is just a dummy variable introduced to allow for the use of t as time.

εcz(0) = 0 ⇒ εc

z(t) = −ε̇0

(σ̇0

σc

)ntn+1

n + 1

At t = t�:

εcz(t

�) = εez(t

�) ⇒ t� =

[2νσ̇0

Eε̇0

(σ̇0

σc

)−n

(n + 1)

]1/n

Problem 4. Governing D.E. is

dε

dt=

1

E

dσ

dt+

σ

η

The strain is ε = ε̇0t, which allows for the substitution dt = dε/ε̇0 so theD.E. can be rewritten as

dσ

dε+

E

ε̇0ησ = E.

This is an first order inhomogeneous differential equation. The homogeneoussolution is

σ(h)(ε) = σ0 exp

(−E

ε̇0ηε

),

2

and the particular solution is a constant, and turns out to be σ(p) = ε̇0η.The constant σ0 can be found from the initial condition

σ(0) = 0 ⇒ σ0 = −ε̇0η.

So

σ(ε) = ε̇0η

(1 − exp

(−E

ε̇0ηε

))and will approach ε̇0η asymptotically as ε → ∞.

Problem 5. For the stiffness matrix [C0] - see Cps in problem 1. Thetransformation matrix is

[L45] =1

2

⎡⎣ 1 1 2

1 1 −2−1 1 0

⎤⎦ .

The stresses in the LT-system can be found from

σ(LT)45 = C0L

−T45 ε

σ(LT)−45 = C0L

T45ε,

which gives the stresses in the longitudinal direction as σL,45 = σL,−45 = 102 MPa.

Problem 6. Stress state in the particle: σpij = σp

h +spij. Handbook (3.11)

and (3.12) gives

σph = Kpε

pv = KpA

Kp εv =

Kp

KmAK

p σh

spij = 2Gpe

pij = . . . =

Gp

GmAG

p sij

Complementary formulas (22):

Kp

KmAK

p =

{Km

Kp� 1

}=

3(1 − νm)

1 + νm

Gp

Gm

AGp =

{Gm

Gp

� 1

}=

15(1 − νm)

2(4 − 5νm)

So

σpy =

3(1 − νm)

1 + νm

σ

3− 15(1 − νm)

2(4 − 5νm)

σ

3= {νm = 0.3} = −0.161σ

3


Written examination, Material Mechanics (4C1126, 4C1110)May 31, 2007, 08.00 - 13.00

The results of the examination will be registered at the latest June 14, 2007. Complaints onthe marks must be made at the latest one month thereafter.


Aids: Handbok och Formelsamling i Hållfasthetslära, Complementary formulas in Mate-rial Mechanics, mathematical handbooks and pocket calculator.

Examiner: Peter Gudmundson, tel. 790 7548.

1. A transversely isotropic uniaxial fibre composite is loaded in the fibre direction by anormal stress . The strain in the z- direction is zero (plane strain). Determine the result-ing normal strain . The stiffness of the composite is described by EL, ET, νLT, GLT, νTT.

2. A fresh, thin slice of an oak tree stem is con-sidered, see Figure 2. It is assumed that the freshslice is stress free. The slice is then dried, whichcorresponds to a (negative) change in the rela-tive moisture content . The moisture swell-ing coefficients read: = 0.13, = 0.29, = 0.02, where R, T, L denote the radial, circum-ferential and longitudinal directions respec-tively.

a. If cracks would occur at drying, in what directions would you suspect them to be found.A thorough motivation must be provided (3p).

b. Formulate the complete set of governing equations and boundary conditions that mustbe solved in order to determine the stresses after drying. Necessary parameters should bedefined (3p).

3. Uniaxial testing of a certain material revealed that the yield stress was 5% lower thanthe expected value. It was suspected that voids had unintentionally been introduced in the

σxεy

x

y

Figure 1

σxσx

R

T

Figure 2

ΔmβR βT βL

Written examination, Material Mechanics (4C1126, 4C1110) May 31, 2007, 08.00 - 13.00 1


processing of the material. Estimate the volume fraction of voids if it is assumed that thematerial with voids can be described by the Gurson-Tvergaard model.

4. An isotropic, elastic-perfectly plastic material is loaded by stresses = , = , = , where is a loading parameter. All other stress components vanish. When the

yield condition is reached, all strain components are kept constant except for , whichcontinues to increase. Determine the rate of change resulting from a rate of change

, right after the initiation of plastic deformation. It can be assumed that the material canbe described by the von Mises flow rule. The complete tangent stiffness matrix isthen defined as,

where denote stresses and strains in vector notation, the yield stress and G theshear modulus. The variables etc. represent components of the stress deviator.

5. An isotropic, elastic, visco-plastic material is subjected to a pure shear loading where denotes the yield stress for vanishing strain rates, t time and a

prescribed time constant. The visco-plastic strain rate is determined from the Perzynamodel,

where are material parameters and where denote the stress deviator and effec-tive von Mises stress respectively. The elastic properties are defined by the E- modulus Eand Poisson´s ratio . Determine the total shear at time .

6. An originally isotropic elastic material has developed small penny shaped cracks, allwith a normal direction aligned with the x- direction. The radius of the cracks is 4 μm andthere are 106 cracks per mm3. The cracked material is uniaxially loaded in the x- directionby a stress . Determine the Poisson ratio defined as . The uncrackedmaterial has an E- modulus E = 210 GPa and a Poisson´s ratio = 0.3.

σx σ0 σy σ– 0τxy σ0 σ0

γxyσ· y

γ· xyCtan

σ· Ctanε· C Cp–[ ]ε· ,= =

Cp 3Gσs

2-------

sx2 sxsy sxsz sxsxy sxsxz sxsyz

sxsy sy2 sysz sysxy sysxz sysyz

sxsz sysz sz2 szsxy szsxz szsyz

sxsxy sysxy szsxy sxy2 sxysxz sxysyz

sxsxz sysxz szsxz sxysxz sxz2 sxzsyz

sxsyz sysyz szsyz sxysyz sxzsyz syz2

,=

σ ε, σssx sy,

τxy σst t0⁄= σs t0ε· ij

vp

ε· ijvp

ε· 0σeσs----- 1–⟨ ⟩

n 3sij2σe---------⋅=

ε· 0 n, sij σe,

ν γxy t0

σx νxy νxy εy εx⁄–=ν

Written examination, Material Mechanics (4C1126, 4C1110) May 31, 2007, 08.00 - 13.00 2


Written examination, Material Mechanics (4C1126, 4C1110)December 14, 2006, 14.00 - 19.00

The results of the examination will be announced at the department notice-board at the lat-est January 5, 2007. Complaints on the marks must be made at the latest one month there-after.


Aids: Handbok och Formelsamling i Hållfasthetslära, Complementary formulas in Mate-rial Mechanics, mathematical handbooks and pocket calculator.

Examiner: Peter Gudmundson, tel. 790 7548.

1. A uniaxial fibre composite is supported by two rigid and frictionless walls, see Figure 1.This means that the strain . The composite is subjected to a temperature change

that causes a free thermal expansion according to and ,where T, L denote the directions transverse (T) and along the fibres (L) respectively, seeFigure 1. The parameter represents the thermal expansion coefficient in the transversedirection. Determine the stress that results from the temperature change. The stiffnessof the composite is described by EL, ET, νLT, GLT.

2. A yield criterion can be defined such that the yield surface lies between the von Misesand the Tresca yield surfaces;

,

where denote the principal stresses and the yield stress in tension. (For the yield criterion reduces to the von Mises criterion and for the Tresca cri-

terion results.) The yield stresses in tension and shear are experimentally obtained as MPa and MPa, respectively. Determine the parameter n ( )

εx 0=∆T εT αT∆T= εL γLT 0= =

αTσx

x

yLT

45°

Figure 1

Rigid and frictionless walls

Composite

121 n⁄---------- σ1 σ2– n σ1 σ3– n σ3 σ2– n+ +[ ]

1 n⁄σs– 0=

σ1 σ2 σ3, , σsn 2= n ∞→

σs 100= τs 55= 2 n ∞≤ ≤

Written examination, Material Mechanics (4C1126, 4C1110) December 14, 2006, 14.00 - 19.00 1


so that the yield criterion is exactly fulfilled for both tension and shear. It is sufficient topresent the equation from which n can be determined.

3. An isotropic, elastic-perfectly plastic material is first loaded by the shears so that plastic deformation develops. (All other strain components van-

ish.). In a second load step, the shear is locked ( ) whereas the shear isfurther increased. Determine the tangent shear modulus defined by at ini-tiation of the second load step by evaluation of the general expression for the tangent stiff-ness matrix expressed as,

where C denotes the elastic stiffness matrix and stress and strain increments in vectornotation according to,

.

4. A strain gage is mounted for measurements of on a specimen made of an isotropic,viscoelastic material. Two experiments are performed. In the first experiment, a uniaxialstress is applied, where t denotes time, is a constant and repre-sents the Heaviside step function. (All other stress components are zero). The strain is then determined as , where is experimentally obtained. When thisexperiment is finished, the loading is removed and the specimen is fully relaxed beforeinitiation of the second experiment. In this experiment, a uniaxial stress is instead applied. The resulting strain is then determined as ,where is experimentally measured.

A volume V of the viscoelastic material is at time subjected to a uniform pressure p.Determine the volume change as a function of time.

Hint: The volume change can be expressed as .

γxy γxz γ0= =γxz γxz γ0= γxy

Gt τ· xy Gtγ·xy=

Ctan

σ· Ctanε· C Cp–[ ]ε· ,= =

Cp 3Gσs

2-------







,=

σ· ε·,

σ·T

σ· x σ· y σ· z τ· xy τ· xz τ· yz, , , , ,( ),= ε·T

ε· x ε· y ε· z γ· xy γ· xz γ· yz, , , , ,( )=

εx

σx t( ) σ0H t( )= σ0 H t( )εx t( )

εx t( ) σ0f t( )= f t( )

σy t( ) σ0H t( )=εx t( ) εx t( ) σ– 0g t( )=

g t( )

t 0=∆V

∆V Vεkk=



5. A laminate is built up of two different isotropic materials with the same volume frac-tion. The first material is linear elastic and the second material is elastic-perfectly plastic.The laminate is first subjected to an in-plane shear , where denotes the shearat which plastic deformation is initiated in material 2. In a second step, the laminate isunloaded to a laminate shear of . Determine the laminate shear stress as well asthe shear stresses in each of the materials at this stage. Both materials have the same shearmodulus G. (It is assumed that the laminate lay-up is symmetric so that only in-planedeformations develop.)

6. A matrix material with E- modulus and Poisson´s ratio contains a small volumefraction of softer particles ( , ). The yield stress for the particles is

and it is assumed that they can be described by the Tresca yield criterion. The compos-ite is loaded by a uniaxial stress . At which stress will plastic deformation be initi-ated in the particles? It is assumed that the matrix material is linear elastic and that thePoisson´s ratio for the matrix material and the particles is 0.3.

Good luck!

Peter

γxy 3γs= γs

γxy γs=

Em νmEp Em 5⁄= νp νm=

σsσ0 σ0



Written examination, Material Mechanics (4C1110), January 16, 2006, 08.00 - 13.00

The results of the examination will be announced at the department notice-board at the lat-est January 30. Complaints on the marks must be made at the latest one month thereafter.


Aids: Handbok och Formelsamling i Hållfasthetslära, Complementary formulas in Material Mechanics, mathematical handbooks and pocket calculator. Examiner: Peter Gudmundson, tel. 790 7548.

1. Single crystals of Copper have a FCC lattice structure and they can be described bycubic anisotropy. The stiffness parameters related to a coordinate system (x1, x2, x3) withaxes parallel to the cubic structure are,

E = 67 GPa, G = 75 GPa, ν = 0.42

The material is loaded by a shear stress in a direction defined by the angle according to Figure 1 below. Determine the shear modulus in this direction by form-ing the ratio between the shear stress and the shear .

2. The stress state in a material point is defined by = in vectornotation, where is a loading parameter. The yield property of the material isdescribed by the Drucker-Prager yield criterion,

,

where denotes the effective von Mises stress, the hydrostatic stress, =400 MPa and = 0.2 (radians). Determine the load parameter k that causes initiation ofplastic deformation.

τxy ϕ π 4⁄=Gxy

τxy γxy

x1

ϕ

x2

τxy

τxyτxy

τxyx

y

Figure 1

σT k [3, 2, 2, 1, 0, -1]⋅k 0>

f σij( ) σe σh βtan σs 1 βtan3-----------+⎝ ⎠

⎛ ⎞–+ 0= =

σe σh σkk 3⁄= σsβ

Written examination, Material Mechanics (4C1110), January 16, 2006, 08.00 - 13.00 1


3. A thin plate in the x, y- plane ( ) made of an elastic-plastic material (E-modulusE, Poisson’s ratio ν, yield stress ) is loaded by the normal stresses( ) so that plastic deformation is initiated. All other stress componentsvanish. Thereafter the strain increments , are enforced. Determine theresulting stress increments ( ). (It can be assumed that plastic deformation devel-ops and that the material can be described by von Mises flow rule).

4. A material is at time t = 0 loaded by the normal stresses MPa. All shear stress components vanish. The material is isotropic and it shows elasticand creep deformation. The creep is described by Norton-Odqvist’s generalized creep law,

where denote the effective stress according to the von Mises hypothesis and thestress deviator respectively. The constitutive parameters are given by: E = 210 GPa, =0.3, = 10-6 s-1, n = 4, = 400 MPa. Determine the strain at time t = 105 s.

5. A three dimensional volume is subjected to a biaxial stress state, = 100 MPa.All other stress components vanish. The volume contains a small spherical inclusion thatis made of a material with much lower stiffness. Determine the normal strain in theinclusion. The stiffness of the three dimensional volume is defined by an E-modulus

GPa and a Poisson ratio = 0.3.

6. A periodic structure is built up of beams with square cross-sections (thickness h) andlengths L, 2L ( ). A representative part of the structure is shown in Figure 6. Theangle between the short and long beams is . Determine the effective E-modulus and the Poisson ratio for the structure. The beams have E- modulus E.

Good luck!

Peter

σz 0=σs

σx σ– y= σ0 0>=dεx 0> dεy 0=

dσx dσy,

σx 600 σy 500 σz 400=,=,=

ε· ijc

ε· 0σeσc-----⎝ ⎠⎛ ⎞

n 32---

sijσe-----⋅=

σe2 3

2---sijsij=

σe sij,ν

ε· 0 σc εz

σx σy=

εz

E 70= ν

L h»45° E1

ν12

x1

x2

2LL

Figure 6



Written examination, Material Mechanics (4C1110), October 20, 2005, 08.00 - 13.00

The results of the examination will be announced at the department notice-board at the lat-est November 4. Complaints on the marks must be made at the latest one month thereafter.



1. Single crystals of Beryllium are transversely isotropic. The stress-strain relationshipreads,

where the stiffnesses are given in units of GPa. Determine the Poisson ratios .

2. The yield stress in pure shear for a metallic material is 100 MPa. (The influence ofeventual voids on the yield stress in shear can here be neglected). The material is subjectedto a pure hydrostatic state of stress, . It is then found that plastic volumechanges appear for hydrostatic stresses larger than MPa. Determine the volumefraction of voids assuming that the Gurson-Tvergaard model describes the yield condition.

3. A thin walled, closed, cylindrical pressure vessel with radius a and wall thickness h isloaded by a constant internal pressure p. The material is isotropic and it has a yield stress

. The elastic deformation is described by the E-modulus E and the Poisson´s ratio .The plastic deformation is governed by the equations,

where denote the effective stress according to the von Mises hypothesis and thestress deviator respectively. The hardening is described by the constant parameter K. Plas-tic deformation will be initiated above a certain pressure . Determine an expression forthe change in radius as a function of pressure for .

σx

σy

σz

τxy

τxz

τyz

292 24 6 0 0 024 292 6 0 0 06 6 349 0 0 00 0 0 134 0 00 0 0 0 163 00 0 0 0 0 163

εx

εy

εz

γxy

γxz

γyz

=

ν12 ν13,

σij σ0δij=σ0 400=

σs ν

ε· ijp 3

2---sijσe-----

σ· eK-----=

σe2 3

2---sijsij=

σe sij,

ps∆a p ps>

Written examination, Material Mechanics (4C1110), October 20, 2005, 08.00 - 13.00 1


4. A small spherical hole in a large body is considered. The body is at time sub-jected to a constant hydrostatic stress state given by . The material is viscoe-lastic. It can be described by a constant bulk modulus K and at pure shear the materialbehaves like a Kelvin model with stiffness G and viscosity . For an elastic material therelative volume change of the hole can be expressed as,

Apply the viscoelastic correspondance principle to derive an expression for the Laplacetransform of the relative volume change of the hole.

5. A [0, 90]s glass fibre/epoxy laminate is biaxially loaded by tensile stresses =100 MPa in the plane of the laminate. Determine the normal stresses in all plies.(The indices L and T denote along and across the fibre direction respectively.) The uniax-ial plies have the following stiffness parameters: EL = 40 GPa, ET = 8 GPa, GLT = 4 GPa,νLT = 0.25.

6. A three dimensional volume is subjected to a uniaxial stress state, , where is a known stress and where all other stress components vanish. The volume contains asmall spherical inclusion that is made of a material with much higher stiffness. Determinethe stress state in the inclusion. The stiffness of the three dimensional volume is definedby an E-modulus GPa and a Poisson ratio = 0.3.

Good luck!

Peter

t 0=σij σ0δij=

η

∆VV-------

9 1 ν–( )2E--------------------σ0=

σx σy=σL σT,

σx σ0= σ0

E 10= ν



Written examination, Material Mechanics (4C1110), August 25, 2005, 08.00 - 13.00

The results of the examination will be announced at the department notice-board at the lat-est September 15. Complaints on the marks must be made at the latest one month thereaf-ter.



Observe that also a Swedish translation of the exam is available.

1. The E- moduli = 220 GPa, = 10 GPa and Poisson’s ratio = 0.30 have beendetermined from uniaxial tensile tests along (L) and transverse (T) to the fibre direction ofa uniaxially carbon fibre reinforced epoxy material. A tensile test specimen is thereafterprepared with the fibre direction 45° in relation to the loading direction. The E- modulusfor this specimen is determined as = 16 GPa. Determine the shear modulus .

2. The yield criterion for a transverselly isotropic material in plane stress is defined by,

,

where , denote the yield stresses in x- and y- direction respectivelyand where denotes the yield stress in shear. The parameter is known. A tensile testspecimen in the - direction is then considered, see Figure below. The tensile yieldstress in this direction is determined as . Determine the yield stress forshear in the x-, y-plane.

3. An elastic, perfectly plastic material (E- modulus E, Poisson’s ratio ν, yield stress ) isinitially loaded by the stresses until plastic deformation is initiated. Allother stress components vanish. The shear is thereafter kept constant while the strain

continues to increase. Determine the tangent stiffness , defined by , at

EL ET νLT

E45 GLT

σx2

σxs2-------

σy2

σys2-------

σxσy

σxs2------------–

τxy2

τxys2--------- 1–+ + 0=

σxs 2σs= σys σs=τxys σs

45°σ45s 3σs 2⁄= τxys

σ45

x

y

σ45

45°

σsσx τxy σ0= =

γxyεx Et dσx Etdεx=

Written examination, Material Mechanics (4C1110), August 25, 2005, 08.00 - 13.00 1


initiation of plastic deformation. It is assumed that the material is well described by vonMises yield criterion and associated flow rule.

4. A thin walled, closed, cylindrical pressure vessel with radius a and wall thickness h is attime t = 0 loaded by a constant internal pressure p. The elastic deformation can beneglected in comparison to creep deformation. Determine the time when the wall thick-ness has decreased by 1%. The creep deformation is described by the Norton-Odqvist gen-eralized creep law,

,

where are material parameters and where denote the effective von Misesstress and the stress deviator respectively.

5. An isotropic, viscoelastic material is loaded by a strain history defined by where denotes the Heaviside step function and where is a

known parameter. All other strain components vanish. The viscoelastic properties aredescribed by the shear relaxation modulus and an elastic bulk modulus K. Determinethe stress as a function of time.

6. A linear elastic, isotropic matrix material is reinforced by a small volume fraction ofspherically shaped particles. The particles are as well linear elastic and isotropic, but theyare so stiff that they can be considered to be rigid in comparison to the matrix material.The matrix material has an E- modulus of Em = 6 GPa and a Poisson ratio of νm = 0.3. Thecomposite is loaded by a pure shear stress . Determine the largest principal stress in theparticles.

Good luck!

Peter

ε· ijc 1

τ---

σeσc-----⎝ ⎠⎛ ⎞

n 3sij2σe---------⋅=

τ σc n, , σe sij,

εx εy ε0H t( )= = H t( ) ε0

G t( )σz

τ




The results of the examination will be announced at the department notice-board at the lat-est February 4. Complaints on the marks must be made at the latest one month thereafter.




1. A uniaxially reinforced fibre composite with E- moduli , shear modulus and Poisson’s ratio is considered. A rectangular tensile test specimen is prepared withfibres in a direction with respect to the loading direction, see Figure 1. The specimen isloaded by a shear stress . Determine the normal strain in the axial direction of thespecimen.

2. A material with kinematic hardening has previously been plastically deformed. Theremaining plastic strains are given by: , where is known. Theother plastic strain components are zero. The material is thereafter uniaxially loaded by anormal stress in the x- direction, see Figure 2. Determine the maximum and minimumstress that can be applied without further plastic deformation. The yield condition reads,

EL ET, GLTνLTϕ

τ ε

LTϕ

τFigure 1

τ

τ

τ

γxyp γ0= γxz

p 2γ0= γ0

σ

f σ˜

( ) 32--- σ

˜b–( )TDσ σ

˜b–( )

12---

σs– 0,= = b cDs1– ε˜

p,=

Ds

1 0 00 1 00 0 1

0 0 00 0 00 0 0

0 0 00 0 00 0 0

2 0 00 2 00 0 2

,= Dσ

13---

2 1– 1–1– 2 1–1– 1– 2

0 0 00 0 00 0 0

0 0 00 0 00 0 0

2 0 00 2 00 0 2

,=



where denotes the yield stress, c a hardening modulus, the stress state and theplastic strain state in vector notation. Vector notation of stresses and strains is defined by,

.

3. A thin plate in the x, y- plane ( ) made of an elastic-plastic material (E-modulusE, Poisson’s ratio ν, yield stress ) is biaxially loaded ( ) so that plasticdeformation is initiated. Thereafter the strain increments , areenforced. Determine the resulting stress increments ( ). (It can be assumed thatplastic deformation develops).

4. An isotropic, viscoelastic material is loaded by a stress history defined by, where denotes the Heaviside step function and

where are known parameters. All other stress components vanish. The viscoelasticproperties are described by the creep compliance in shear and an elastic bulk modu-lus K. Determine the strain as a function of time.

Hint: If the material at time t = 0 is exposed to a shear stress , the resulting shear will be defined by the creep compliance according to: .

5. A plane plate made of a uniaxially reinforced fibre composite is on both sides coated bya thin colour layer. The thickness of the colour layers is much smaller than the compositethickness. The structure is initially stress free. The temperature of the plate is thereafterincreased by . Determine the stresses that develop in the colour layers. The compositeis linear thermo-elastic with stiffness and thermal expansion parameters , , ,

, , . The colour layers are isotropic and the relevant parameters are denoted asE, , .

Observe that unnecessary information may be included in the problem statement.

6. An isotropic, elastic material with Poisson’s ratio = 0.3 and E-modulus E contains asmall volume fraction of spherically shaped particles. The particles are isotropic and theyare very soft in comparison to the matrix. The composite is uniaxially loaded by a normalstress . Determine the largest principle strain in the particles.

Good luck!

Peter

σs σ˜

ε˜

p

σ˜

T σx σy σz τxy τxz τyz, , , , ,( ),= ε˜

p( )T

εxp εy

p εzp γxy

p γxzp γyz

p, , , , ,( )=

x

y

σσ

Figure 2

σz 0=σs σx σy= σ0=

dεx 0> dεy dγxy 0= =dσx dσy,

σx σ0H t( )= σy σ0H t t0–( )= H t( )σ0 t0,

J t( )εx

τ0 γ t( )J t( ) γ t( ) J t( )τ0=

∆TEL ET GLT

νLT αL αTν α

Fibre composite

Colour layer

Figure 5

ν

σ ε1


KTH SOLID MECHANICS

Written examination, Material Mechanics (4C1110), August 19, 2004, 14.00 - 19.00

The results of the examination will be announced at the department notice-board at the lat-est September 9. Complaints on the marks must be made at the latest one month thereafter.




1. A uniaxially reinforced fibre composite with E- moduli , shear modulus and Poisson’s ratio is considered. A rectangular tensile test specimen is manufacturedwith the fibres in a direction in relation to the tensile axis, see Figure 1. The loading bya stress will cause the ends to rotate an angle . Determine this angle.

2. A material with kinematic hardening has in a prior loading been plastically deformed.The remaining plastic strains read: , where is known. The otherplastic strain components vanish. The material is thereafter loaded by a uniaxial stress in the x- direction, see Figure 2. Determine the maximum tensile and compressive stressesthat can be applied without additional plastic deformation. The yield condition reads,

EL ET, GLTνLT

ϕσ θ

LT

ϕ

θσσ

Figure 1

εxp ε0= εy

p ε– 0= ε0σ

f σ˜

( ) 32--- σ

˜b–( )TDσ σ

˜b–( )

12---

σs– 0,= = b cDs1– ε˜

p,=

Ds

1 0 00 1 00 0 1

0 0 00 0 00 0 0

0 0 00 0 00 0 0

2 0 00 2 00 0 2

,= Dσ

13---

2 1– 1–1– 2 1–1– 1– 2

0 0 00 0 00 0 0

0 0 00 0 00 0 0

2 0 00 2 00 0 2

,=


KTH SOLID MECHANICS

where denotes the yield stress, c a hardening modulus, the stress state and theplastic strain state in vector notation. Stresses and plastic strains are in vector notation rep-resented by,

.

3. A thin walled pipe is loaded by an axial tensile stress corresponding to half the yieldstress. The elongation of the pipe is thereafter kept constant. In a second step, the pipe isloaded by a twisting moment corresponding to a shear stress . For small shear stresses,the relationship between increments of shear stress and shear is governed by the shearmodulus G, . When the yield condition is satisfied, the tangent stiffnessbetween and will change to where . Determine the tangentstiffness . The material is elastic-perfectly plastic and it obeys von Mises flow rule. Thematerial parameters are defined by .

4. An isotropic, viscoelastic material is loaded by a strain history defined by:, where denotes the Heaviside step function and

where are known parameters. The remaining strain components vanish. The viscoe-lastic properties are described by the relaxation modulus in shear as well as an elas-tic bulk modulus K. Determine the stress as a function of time.

5. A laminate is built up by Aluminium and a ceramic material. The volume fraction ofAluminium is 50%. The laminate is subjected to a uniaxial tensile stress in the x- directioncorresponding to a strain = 0.1%, see Figure 5. Determine the stress in the ceramicmaterial. The stiffness of the Aluminium is defined by = 70 GPa, = 0.3. For theceramic material we have = 300 GPa, = 0.1.

6. An isotropic elastic material with Poisson’s ratio = 0.3 contains a small volume frac-tion of spherically shaped particles. The material in the particles is elastic, isotropic andvery much stiffer than the matrix. The composite is loaded by a shear stress . Determinethe shear stress in the particles.

Good luck!

Peter

σs σ˜

ε˜

p

σ˜


p( )T

εxp εy

p εzp γxy

p γxzp γyz

p, , , , ,( )=

x

y

σσ

Figure 2

τ

dτ Gdγ=dτ dγ dτ Gtdγ= Gt G<Gt

E ν σs, ,

εx ε0H t( )= εy ε0H t t0–( )= H t( )ε0 t0,

G t( )σz

εx σyEa νa

Ec νc

x

y

Figure 5

ν

τ




The results of the examination will be announced at the department notice-board at the lat-est November 12. Complaints on the marks must be made at the latest one month thereaf-ter.




1. A uniaxially reinforced fibre composite is loaded in the x- direction by a normal stress = 100 MPa. The fibres make an angle of with the x- axis, see Figure 1 below.

Determine the normal strain . The stiffness of the composite is described by = 200GPa, = 10 GPa, = 7 GPa, = 0.3.

2. A thin walled, closed cylindrical pressure vessel with radius a and wall thickness h isloaded by an internal pressure p. The yield property of the material is described by theDrucker-Prager yield criterion,

,

where denotes the effective von Mises stress, the hydrostatic stress, the tensile yield stress and an additional material parameter. Determine the internalpressure that causes initiation of plastic deformation.

σx π 4⁄εy EL

ET GLT νLT

T

x

yπ/4L

σxσx

f σij( ) σe σh βtan σs 1 βtan3-----------+⎝ ⎠

⎛ ⎞–+ 0= =

σe σh σkk 3⁄= σsβ

ps



3. An elastic, isotropic, perfectly plastic material is initially loaded by increasing shearstresses . All other stress components vanish. When the yield criterion accord-ing to the von Mises hypothesis is reached, the shear is locked, hence forcontinued loading. Determine the tangent stiffness defined by at initi-ation of plastic deformation. A general expression for the tangent stiffness matrix is givenbelow,

where C denotes the elastic stiffness matrix and , the increments in stress and strainaccording to the vector representation,

.

4. A thin walled spherical shell with radius a and wall thickness h is at time t = 0 exposedto a constant internal pressure p. Both elastic and creep deformation develop in the shell.The creep deformation is described by the primary creep law,

where denotes the effective von Mises stress, the effective creep strain according tothe von Mises hypothesis and the stress deviator. The creep properties are described bythe parameters m, n, A. The elastic stiffness is defined by the E- modulus E and Poisson’sratio . Determine that change in diameter of the shell as a function of time.

Hint: The stress state is given by .

τxy τxz=γxz dγxz 0=

Gt dτxy Gtdγxy=

dσ C Cp–[ ]dε,=

Cp 3GssT

σs2-------,=

Cp 3Gσs

2-------







,=

dσ dε

dσT dσx dσy dσz dτxy dτxz dτyz, , , , ,( ),= dεT dεx dεy dεz dγxy dγxz dγyz, , , , ,( )=

ε· ijc

Aσe( )m

εec( )

n--------------3sij2σe---------⋅=

σe εec

sij

ν

σϑ σϕpa2h------,= = σr 0=



5. A laminate is composed of two isotropic materials (1, 2) with E- moduli ; Pois-son’s ratios and total thicknesses respectively. Determine the in-plane shearmodulus of the laminate.

Hint: The plane stress stiffness matrix for a laminate is given by:

6. An linear elastic, isotropic material with E- modulus E = 70 GPa and Poisson’s ratio = 0.3 contains 5% of spherically shaped particles. The particles are also assumed to be iso-tropic and linear elastic. Estimate a lower and un upper bound for the E- modulus of thecomposite if the stiffness of the particles is unknown. (The stiffness may be extremelysmall or extremely large).

Good luck!

Peter

E1 E2,ν1 ν2, t1 t2,

Cps

σp Cpsεp,= Cps1t--- tiCps

i ,i 1=

N

∑=

σp σx σy τxy, ,( )T,= εp εx εy γxy, ,( )T.=

ν


KTH SOLID MECHANICS


The results of the examination will be announced at the department notice-board at the lat-est February 5. Complaints on the marks must be made at the latest one month thereafter.




1. A transversely isotropic material is considered.From uniaxial tensile tests, the material parame-ters GPa, GPa, have been determined. The E-modulus for a ten-sile test specimen in the direction relative tothe LT- directions (see Figure) has as well beenobtained, GPa. Determine the shearmodulus .

2. A thin walled pipe with radius a, wall thickness h and length L is loaded by a torsionalmoment that thereafter is released. After the unloading, the pipe shows a permanent twist

between the two ends. The pipe is then loaded by an axial force F. At which force Fwill the pipe again plastically deform? The material shows a kinematic hardening and theyield criterion can be expressed as,

where denotes the yield stress, c a hardening modulus, the stress state and theplastic strain state in vector notation. The vector notation of stresses and plastic strains aredefined by,

L

T45°

EL 40= ET 10= νLT 0.3=

45°

E45 13=GLT

∆ϕ

f σ˜

( ) 32--- σ

˜b–( )TDσ σ

˜b–( )

12---

σs– 0,= = b cDs1– ε˜

p,=

Ds

1 0 00 1 00 0 1

0 0 00 0 00 0 0

0 0 00 0 00 0 0

2 0 00 2 00 0 2

,= Dσ

13---

2 1– 1–1– 2 1–1– 1– 2

0 0 00 0 00 0 0

0 0 00 0 00 0 0

2 0 00 2 00 0 2

,=

σs σ˜

ε˜

p


KTH SOLID MECHANICS

.

3. A thin walled pipe is loaded by a torsional moment that corresponds to half the yieldstress in shear. The twist of the pipe is thereafter kept constant. In a second step, the pipe isloaded by an axial tensile stress . Initially, the relation between stress and strain incre-ments will be given by the E- modulus, . When the yield criterion is satisfied,the tangent stiffness between and is changed to where . Deter-mined the tangent stiffness modulus . The material is elastic-perfectly plastic and itobeys von Mises flow rule. The material parameters are defined by .

4. An isotropic, viscoelastic material is considered. From measuments, the relaxationmodulus in shear has been determined. The material shows an elastic respons tohydrostatic loading. The relation between relative volume change and hydrostatic stress isdefined by the bulk modulus K. The material is loaded by a uniaxial strain in the x- direc-tion, , where denotes the Heaviside step function and is a pre-scribed strain. All other strain components vanish. Determine the stress .

5. An aluminium plate with E- modulus 70 GPa, Poisson’s ratio 0.3 and thickness 2 mm iscoated by ceramic layers on both sides. The thickness of the ceramic layers is 0.3 mm(total thickness 0.6 mm) and the stiffness is defined by the E- modulus 300 GPa, Poisson’sratio 0.1. Determine the effective E- modulus for the laminate. (An exact calculation isrequired. It is not sufficient with a “rule of mixtures” estimate).

Hint: The stiffness matrix in plane stress for a laminate is given by,

6. An isotropic, elastic material with E- modulus E = 70 GPa and Poisson’s ratio = 0.3contains a small volume fraction of spherically shaped voids. The material is loaded sothat the total volume change is 0.1 %. Determine the relative volume change for a repre-sentative void.

The correct solution will be published on the course home page (http://www.hallf.kth.se/kurser/4c1110.htm) at 13.00 today.

Good luck!

Peter

σ˜


p( )T

εxp εy

p εzp γxy

p γxzp γyz

p, , , , ,( )=

σdσ Edε=

dσ dε dσ Etdε= Et E<Et

E ν σs, ,

G t( )

εx t( ) ε0H t( )= H t( ) ε0σx t( )

Cps

σ˜

p Cpsε˜

p,= Cps1t--- tiCps

i ,i 1=

N

∑=

σ˜

p σx σy τxy, ,( )T,= ε˜

p εx εy γxy, ,( )T.=

ν




The results of the examination will be announced at the department notice-board at the lat-est November 13. Complaints on the marks must be made at the latest one month thereaf-ter.




1. The shear modulus G of a particle reinforced Aluminium alloy was experimentallydetermined to 27.0 GPa. The volume fraction of hard particles was 5% and the stiffness ofthe particles are given by E = 200 GPa and = 0.3. The shear modulus of the compositewas the same as that for the Aluminium alloy without particles so it was suspected thatvoids had been introduced in the processing of the composite. Estimate the volume frac-tion of voids in the composite. The Poisson ratio of the Aluminium alloy can be assumedto be 0.3.

2. A thin walled pipe with radius a and wall thickness h is at time t = 0 exposed to a con-stant axial force P and torsional moment , see Figure. The material in the pipeshows both elastic and creep deformation. Determine the time when the total elongation ofthe pipe is twice the initial elastic elongation. The material has E- modulus E and Pois-son’s ratio . The creep deformation is described by the generalized Norton-Odqvistcreep law,

,

where are material parameters and denote the von Mises effective stressand the stress deviator respectively.

ν

Mv Pa=

ν

ε· ijc 1

τ---

σeσc-----⎝ ⎠⎛ ⎞

n 3sij2σe---------⋅=

τ σc n, , σe sij,

Mv=Pa

P2a

h

P

Mv=Pa



3. A uniaxial tensile test specimen was exposed to a constant axial stress at time t = 0.The axial and transverse strain were measured as functions of time. It was foundthat the these strains could be described by,

,

where are experimentally determined functions of time. A thin walled sphericalshell with radius a and wall thickness h made of the same material is exposed to a constantinternal pressure p which is applied at time t = 0. Determine the increase in radius as afunction of time. It can be assumed that the material is linearly viscoelastic. The stresses inthe shell resulting from an internal pressure p will be,

,

in a spherical coordinate system.

Hint: A simple route to the solution is to utilize superposition.

4. An elastic-plastic material that obeys the von Mises yield criterion with isotropic hard-ening is considered (shear modulus G, yield stress ). The elastic-plastic tangent stiffnessmatrix is then defined by,

where H denotes the plastic modulus, the von Mises effective stress and s the stressdeviator in vector notation. The vector notation of stresses and strains is defined by,

.

The material is exposed to a shear stress that causes plastic deformation. Determinethe elastic-plastic tangent shear modulus defined by, or equivalently

, for continued plastic loading.

σ0εx εy

εx f t( )σ0,= εy g– t( )σ0=

f t( ) g t( ),

σϑ σϕpa2h------,= = σr 0=

σsCtan

σ· Ctanε,·= Ctan C Cp–[ ],=

Cp 9G2

3G H+------------------ ssT

σe2-------⋅ 3G

1 H3G-------+

----------------- ssT

σe2-------,⋅= =

Cp 3G

1 H3G-------+⎝ ⎠

⎛ ⎞σe2

-----------------------------







,=

σe

σT σx σy σz τxy τxz τyz, , , , ,( ),= εT εx εy εz γxy γxz γyz, , , , ,( )=

τxyGt dτxy Gtdγxy=

τ· xy Gtγ·xy=



5. The yield criterion for a transversely isotropic material in plane stress is given by,

,

where denote the yield stress in the x- direction, y- direction and in shearrespectively. Consider a uniaxial tensile test for a specimen that is cut out in the -direction according to the Figure below. Determine the yield stress for this specimen.

6. Single crystals of Copper have a FCC lattice structure and they can be described bycubic anisotropy. The stiffness parameters related to a coordinate system (x1, x2, x3) withaxes parallel to the cubic structure are,

E = 67 GPa, G = 75 GPa, ν = 0.42

The material is uniaxially loaded by a normal stress in the - direction, see Figurebelow. Determine the E- modulus in this direction by forming the ratio between nor-mal stress and normal strain in the - direction.

The correct solution will be published on the course home page (http://www.hallf.kth.se/kurser/4c1110.htm) at 13.00 today.

Good luck!

Peter

σx2

σxs2-------

σy2

σys2-------

σxσy

σxs2------------–

τxy2

τxys2--------- 1–+ + 0=

σxs σys τxys, ,45°

σ45

x

y

σ45

45°

σ ϕE ϕ( )ϕ

σ

x1σ

ϕ

x2


written examination, material mechanics (se2126), …/menu...solid mechanics - kth written...

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