HALLFASTHETSLARA, LTH

Examination in computational materials modeling

TID: 2014-10-27, kl 8.00-13.00

Maximalt 60 poang kan erhallas pa denna tenta. For godkant kravs 30 poang.

Tillatet hjalpmedel: raknare

Uppgift nr 1 2 3 4 5 6Besvarad(satt x)Poang

NAMN:

PERSONNUMMER: ARSKURS:

1

PROBLEM 1 (10p.)

For a non-linear elastic material the stress response ij = sij +13ijkk can

be described by the relations

kk = 3Kkk

sij = 2Geij

where the total strain is kl = ekl +13klpp and

G = A exp(BJ2), J2 =1

2eijeij , I1 = kk

a) Assume that a strain energy function W (I1, J2) exists. Derive the mostgeneral stress strain response from

ij =W

ij

b) Based on the result in a) identify the specific format for the strain energyconsidered in this example.

c) Derive the material secant stiffness Dsijkl given from the relation ij =Dsijklkl.

d) Derive the material tangent stiffness Dtijkl defined by the incrementalrelation dij = D

tijkldkl.

PROBLEM 2 (10p.)

Derive the incremental stress strain relation in elasto-plasticity. Make use ofthat the strain can be decomposed as

ij = eij +

pij

where superscipt e denotes the elastic part and superscipt p the plastic partof the strains. The Hookes law is given as

ij = Dijklekl (1)

2

The evolution laws are defined as

pij = f

ij, = k, 0

where f(ij , K) is the yield surface, K = K() is the hardening function andk = k(ij , K) is related to the hardening of the material.

a) Assume plastic loading, use the consistency condition to derive

f

ijij H = 0

as well asf

ijDijklkl A = 0

and identify H and A.

b) Calculate which signs H and A have for hardening plasticity, ideal plas-ticity and softening plasticity.

c) Use the consistency condition in the above relations to derive the stressdriven incremental law

ij = Cepijklkl

Name one situation where the above relation breaks down.

PROBLEM 3 (10p.)

A transversal isotropic material have same properties in any direction in acertain plane, here taken as x1 x2 plane, i.e. any rotation around the x3-axis. The plane x1 x2 is then called plane of isotropy. The transformationdefining the rotation can be written as

x = Ax, A =

cos sin 0 sin cos 00 0 1

where is an arbitrary angle.

3

For the derivations of Hills orthotropic yield surface will be used as astarting point for derivation of the transversal isotropic yield function. Hillsyield function is given as

f = T P 1 = 0where

=

112233122313

and P =

A F G 0 0 0F B H 0 0 0G H C 0 0 00 0 0 2L 0 00 0 0 0 2M 00 0 0 0 0 2N

a) The transformation of the second order tensor is given by

ij = AikklAjl

Derive the explicit format of transformation of the components.

b) Introducing matrix format such that

= L

Identify L

c) The rotation defines a so-called symmetry plane. Derive the expressionin matrix format that restricts the format of P . The explicit formatfor P should not be derived.

PROBLEM 4 (10p.)

For uniaxial experimental tests, i.e. uniaxial loading

[ij ] =

0 00 0 00 0 0

on a metal it was found that a good fitting could be obtained by using apower law format for the stress strain relation, given by

=

{E when yo

E

yo + kyo(p)n when yo

E

4

where , and p are the uniaxial stress, strain and plastic strain, respectively.Moreover, E is Youngs modulus and yo is the initial yield stress, and finallyn and k are material parameters.

Assuming that the material can be described by an isotropic hardeningvon Mises model during plastic loading

f =

(3

2sijsij

)1/2 yo K() = 0

where sij = ij 13ijkk is the deviatoric stress tensor, K() the hardeningfunction and is the internal variable. Associated plasticity is assumed, i.e

pij = f

ij, 0

a) For a general load case and the choice strain hardening, i.e.

= eff =

(2

3pij

pij

)1/2and plastic work hardening

= ij pij

identify the rate of internal variables by use of the flow rule for plasticstrain rates.

b) Considering uniaxial loading identify the two choices of internal variables(i.e. ).

c) For the two choices of internal variables, given that the experimental datacan be fitted to the power law function, identify the format of K().For simplicity assume n = 1 in this case.

PROBLEM 5 (10p.)

Let initial yielding be determined by the von Mises criterion

F (ij) =

3J2 yo = 0 (1)where

J2 =1

2sijsij ; sij = ij 1

3ijkk

and yo is the initial yield stress in tension.

5

a) Illustrate (1) in the deviatoric plane and in the meridian plane, i.e. theI1

J2 plane. (Note that I1 = kk). A principal sketch is sufficient.

b) For uniaxial tension and uniaxial compression draw the stress paths inthe deviator plane and meridian plane.

c) Assume kinematic hardening of a von Mises material, i.e. the currentyield surface is given as f(ij , ij) = F (ij ij) where ij is thebackstress. Based on (1) write the corresponding yield criterion.

d) Illustrate the behavior of the kinematic hardening yield criterion in thedeviatoric plane and the meridian plane. Geometrically, what does theback-stress ij mean?

e) The flow rule states that

pij = f

ij(2)

Based on the yield criterion established in c), determine f/ij andinsert it into (2). The back-stress ij entering f can be considered tobe a purely deviatoric tensor, i.e. kk = 0.

PROBLEM 6 (10p.)

In a certain experiment it is found that yielding of a material occurs underthe following states of principal stresses

(1, 2, 3) = (20, 0, 0)

(1, 2, 3) = (21, 7, 0)

Assuming that the material is isotropic, that the hydrostatic stress does notaffect the yielding and the yield stress is the same in tension and compression.

Plot as many points as you can derive from these observations in the1 2 space, i.e. biaxial loading, where 3 = 0 (plane stress).

6