Unit 1

A displacement is imposed at the right end of the three bars forming thestructure depicted in the figure below. The elastic limit for the three bars is___and the hardening modulus H for each bar is indicated in the figure. Determinethe plastic strain and total stress at each bar. The length of the three bars is l.

des

2where = e lEsd

H =¥

0H =

/ 2H E= -

d

d

d

1

2

3

l

Unit 1 – Solution

epHE E

E H=

+

The evolution of the stress-strain curve in an elastoplastic material can berepresented as:

s

eE E

epE

Elastic regime:

Elastoplastic regime – Loading:

Elastoplastic regime – Unloading:

d Eds e=

epd E ds e=

d Eds e=

Where the elastoplastic tangent modulus can be calculated as: epE

Since H is different for each bar, will also be different for each bar.Þ epE

Unit 1 – Solution

εT = δ

l=

2σ e

E

BAR 1 H =¥

epH E E=¥ Þ =

2 0ep T e

p

E E s s

e

= Þ =

Þ =

s

e

es

2 es

/e Es 2 /e Es

E

epE E=

The total strain in each of the three bars is: 2= e lEsd

Then, for each bar, once the elastoplastic tangent modulus has beencalculated, the total stress and the plastic strain can be determined:

epEs pe

1

Unit 1 – Solution

BAR 2 0H =

0 0epH E= Þ =

BAR 3 / 2H E= -

s

e

es

2 es

/e Es

E

2 /e Es

s

e

es

2 es

/e Es

E

2 /e Es

0 /ep T e

p e

EE

s s

e s

= Þ =

Þ =

/ 2 epH E E E= - Þ = -

0 2 /ep T

p e

E EE

s

e s

= - Þ =

Þ =

2 3

Eep = E H

E + H

Unit 2

The truss structure OA, OB and OC is composed of concrete, which is assumedto behave as a perfectly elastoplastic material with a tensile elastic limit anda compressive elastic limit . An increasing vertical load P is applied atpoint O, starting at until a vertical displacement is reached atsaid point. Then, the load decreases back to .

a) Draw the diagram of the process, indicating the most significantvalues and the state of plastification of the bars at each instant.

b) Calculate the displacement value at point O at the end of the process.

es10 es

0P = 20 e LEsd =

0P =

P d-

Unit 2

45° 45°

P

L

L L

d

A

CB

s

ees

−10σ e

Unit 2 – Solution

45° 45°

P

L

L L

d

A

CB

1

2 2

Due to symmetry of thestructure and loads, the twodiagonal bars will behave in thesame way.

While bar 1 will be under compression , bar 2will be under tension.

s

ees

−

10σ e

E

e

Es

s

ees

−10σ e

Unit 2 – Solution

Applying equilibrium of vertical forces in the node that joins the threebars, the following equality is obtained:

2N 2N

1N

2 2

1

P2N 2N

1N

FV = 0 → ∑ 2 2

2N2 − (P + N1) = 0

P = 2N2 − N1

ΔP = 2ΔN2 − ΔN1

Next step is to apply compatibility of displacements and calculates the strain at each of the bars:

u1 = −δ

2 2

δ = −u1

δ2= u2

1 u2 = δ cos 45°( ) = δ

2 →

u2 =

δ2

1) EQUILIBRIUM

s

ees

−

10σ e

E

e

Es

s

ees

−10σ e

45° 45°

P

L

L L

d

A

CB

1

2 2

2) COMPATIBILITY

Unit 2 – Solution

u1 = −δ → ε = − δ

L u2 =

δ2

→ ε2 =δ2L

The displacement value at which the elastic limit is reached can be obtained as:

110 10 < e e L

L E Ed s se d= < ® Bar will be in elastic regime.

22 <

2e e LE EL

d s se d= < ® Bars will be in elastic regime.2

Then, concerning the whole structure, three stages must be considered:

1

45° 45°

P

L

L L

d

A

CB

1

2 2

s

ees

−

10σ e

E

e

Es

s

ees

−10σ e

The following sign criteria for strains and stresses are established:

+ Traction

Compression

Þ

- Þ

Unit 2 – Solution

1

2 10 < e eL LE Es s d< Þ

Bar is in elastic regime.

Bars are in plastic regime.2

1

2 10 e eL LE Es sd< < Þ

Bar is in plastic regime.

Bars are in plastic regime.2

Now, considering the possible sates above, four deformation stages are studied.

12 10 e eL LE Es sd < < Þ

Bar is in elastic regime.

Bars are in elastic regime.2

State II: While the vertical bar is in elastic regime, inclined bars are in plastic regime.

State III: The three bars of the structure are in plastic regime.

State I: The three bars of the structure are in elastic regime.

s

ees

10 es

10 e

Es

e

Es

s

ees

10 es

45° 45°

P

L

L L

d

A

CB

1

2 2

Unit 2 – Solution

2 12 2P N N E S E S E SL L Ld d d

= - = + = P = 2ES

Lδ

20 e LEsd£ £ 0 2 2 eP Ss< <

12 0e

Es e- < < 12 0es s- < <

20 e

Ese< < 20 es s< £

STAGE I 2 e LEsd <

1 1 1 1 1 1 E E N SL Ld de s e s s= - Þ = ® = - Þ = 1N E S

Ld

= -

2 2 2 2 2 2 2 2

E E N SL L

d de s e s s= Þ = ® = Þ = 2 2N E S

Ld

=

Substituting the values of and into the equality previously obtained fromthe equilibrium of vertical forces, the value of P, as a function of , is:

1N 2Nd

Variation of P during the first stage

s

ees

10 es

10 e

Es

e

Es

s

ees

10 es

45° 45°

P

L

L L

d

A

CB

1

2 2

Unit 2 – Solution

P - δ

2 2 eSs

2 /eL Es

P

d

Ι

1s

1e

− 2σ e

− 2σ e / E

Ι

2s

2e

es

1 1σ - ε 2 2σ - ε

Ι

e

Es

2E SL

EE

s

ees

10 es

10 e

Es

e

Es

s

ees

10 es

45° 45°

P

L

L L

d

A

CB

1

2 2

Δσ2

Δε2

ΔP

Δδ

Unit 2 – Solution

ΔP = 2ΔN2 − ΔN1 = E Δδ

LS = ES

LΔδ

ΔP = ESL

Δδ

2 10 e eL LE Es sd< < Þ ( )2 2 10 2e eS P Ss s< < +

110 2 e e

E Es se- < < - Þ 110 2e es s s- < < -

210

2e e

E Es se< < Þ 2 es s=

STAGE II 2 10 e eL LE Es sd< <

Δε1 = − Δδ

L Elastic( ) ⇒ Δσ 1 = −E Δδ

L

ΔN1 = −E Δδ

LS

Δε2 =

Δδ2L

Plastic( ) ⇒σ 2 =σ e ⇒ Δσ 2 = 0 ΔN2 = Δσ 2S = 0

45° 45°

P

L

L L

d

A

CB

1

2 2

s

ees

−

10σ e

E

e

Es

s

ees

−10σ e

Unit 2 – Solution

P - δ

2 2 eSs

2 /eL Es

P

d10 /eL Es

( )2 10 eSs+

Ι

ΙΙ

1s

1e

− 2σ e

−10σ e

− 2σ e / E −10σ e / E

Ι

ΙΙ

2s

2e

es

1 1σ - ε 2 2σ - ε

ΙΙΙ

e

Es 10

2e

Es

2E SL

E SL

E

E

45° 45°

P

L

L L

d

A

CB

1

2 2

ΔP

Δδ

ΔP

Δδ

Δσ2

Δε2

Δσ1

Δε1

Δσ2

Δε2

s

ees

−

10σ e

E

e

Es

s

ees

−10σ e

Unit 2 – Solution

STAGE III 10 20e eL LE Es sd< <

Δε1 = − Δδ

L Plastic( ) ⇒ σ 1 = −10σ e ⇒Δσ 1 = 0

Δε2 =

Δδ2L

Plastic( ) ⇒ σ 2 =σ e ⇒Δσ 2 = 0 ΔN2 = Δσ 2S = 0⇒ΔN2 = 0

ΔP = 2ΔN2 − ΔN1 = 0 ΔP = 0

10 20 e eL LE Es sd< < Þ ( )2 10 eP Ss= +

120 10 e e

E Es se- < < - Þ 1 10 es s= -

210 20

2 2e e

E Es se< < Þ 2 es s=

45° 45°

P

L

L L

d

A

CB

1

2 2

ΔN1 = Δσ 1S = 0⇒ΔN1 = 0

s

ees

−

10σ e

E

e

Es

s

ees

−10σ e

Unit 2 – Solution

P - δ

2 2 eSs

2 /eL Es

P

d10 /eL Es 20 /eL Es

( )2 10 eSs+

Ι

ΙΙΙΙΙ

1s

1e

− 2σ e

−10σ e

− 2σ e / E −10σ e / E −20σ e / E

Ι

ΙΙ

ΙΙΙ

2s

2e

es

1 1σ - ε 2 2σ - ε

ΙΙΙ ΙΙΙ

e

Es 10

2e

Es 20

2e

Es

2E SL

E SL

E

E

s

ees

10 es

10 e

Es

e

Es

s

ees

10 es

45° 45°

P

L

L L

d

A

CB

1

2 2

ΔP

Δδ

ΔP

Δδ

Δσ1

Δε1

Δσ1

Δε1

Δσ2

Δε2

Δσ2

Δε2

Unit 2 – Solution

STAGE IV Unloading

1 1 1 1 1 1 * *

E EL L Ld d de s s s s sD D D

D = - Þ D = - Þ = +D = -

ΔN1 = −E Δδ

LS

Δε2 =Δδ2L

⇒ Δσ 2 = E Δδ2L

⇒ σ 2 =σ 2 *+Δσ 2 =σ 2 *+E Δδ2L

Perfect plasticity( )

Elastic br

h a c n

e

eEe es e

ì D = DÞ í D = Dî

⇒ N1 = N1 *+ΔN1 = N1 *−E Δδ

LS

⇒ N2 = N2 *+ΔN2 = N2 *+E Δδ

2LS

ΔN2 = E Δδ

2LS ⇒ N2 = N2 *+E Δδ

2LS

10 es= -

10 eSs= -

es=

eSs=

45° 45°

P

L

L L

d

A

CB

1

2 2

s

ees

−

10σ e

E

e

Es

s

ees

−10σ e

Unit 2 – Solution

ΔP = 2Δ2 − ΔN1 = E Δδ

2LS

⎛⎝⎜

⎞⎠⎟− −E Δδ

LS

⎛⎝⎜

⎞⎠⎟= 2E Δδ

LS = 2E S

LΔδ

* 2P P E SLdD

= +

For * 0P P Dd= Þ =

For 2 100 * 2 2

eLP P LES E

sDd += Þ = - = -

* 0P P Dd< Þ <

20 2 10* 2

e eL LE Es sd d Dd +

Þ = + = - δ = 20− 2 +10

2

⎛

⎝⎜

⎞

⎠⎟σ e

EL = 14.29

σ e

EL

1 1 12 10 2 10

2 2e

eEL E

sde s e sD + +ÞD = - = Þ D = D = Þ

2 2 22 10 2 10

2 2 2 2 2e

eEELsde s e sD + +

ÞD = = - Þ D = D = - Þ

σ 1 =

2 −102

σ e = −4.29σ e

σ 2 = 1− 2 +10

2 2

⎛

⎝⎜

⎞

⎠⎟ σ e = −3.03σ e

1

2

14,29 /4,293,04

e

e

e

E Ld ss ss s

=

= -= -

45° 45°

P

L

L L

d

A

CB

1

2 2

s

ees

−

10σ e

E

e

Es

s

ees

−10σ e

Δσ1

Δε1

Unit 2 – Solution

P - δ

2 2 eSs

2 /eL Es

P

d10 /eL Es 20 /eL Es

( )2 10 eSs+

Ι

ΙΙΙΙΙ

1s

1e

− 2σ e

−10σ e

− 2σ e / E −10σ e / E −20σ e / E

Ι

ΙΙΙ

2s

2e

es

1 1σ - ε 2 2σ - ε

ΙΙ ΙΙΙ

e

Es 10

2e

Es 20

2e

Es

14.29σ eL / E

ΙV

−4.29σ eΙV

−3.03σ e

ΙV

2E SL

2E SL

E SL

ΙΙ

Ι

E

E

s

ees

10 es

10 e

Es

e

Es

s

ees

10 es

45° 45°

P

L

L L

d

A

CB

1

2 2

ΔP

Δδ

Δσ2

Δε2

Unit 3

Indicate the shape of the following surfaces in the three dimensional space ofprincipal stresses:

a) σ m = constant

b) τ oct = constant

c)

σ m

τ oct

= constant

Unit 3 – Solution

a) σ m = const.

σ m = 1

3σ 1 +σ 2 +σ 3( ) = const. → σ 1 +σ 2 +σ 3 = const.

The surface is a octahedral plane, which isa plane perpendicular to the hydrostaticstress axis.

1 2 3 1 1 3 3 3oct m

I Ids s ss s + += = = ® =

The distance, d, from the origin to the plane is:

3 octd s=

Where is the normal octahedral stressocts

1s

σ 3

σ 2

Hydrostatic stress axis

1 2 3s s s= =

Octahedral plane

σ 1 +σ 2 +σ 3 = const.

Π3 octd s=

Þ

Unit 3 – Solution

b) τ oct = const.

The surface is a cylinder with constant radius andaxis in the hydrostatic stress axis.

The radius of the cylinder is:

3 octR t=

Where is the tangential octahedralstress:

octt

( )1222 2 2

1 2 3 1 2 31 1

33octt s s s s s sé ù= + + - + + =ê úë û

[ ] [ ]1 12 2

2 22 23J R J¢ ¢= ® =

Þ

1s

σ 3

σ 2

Hydrostatic stress axis

1 2 3s s s= =

3 octR t=

Unit 3 – Solution

c)

σ m

τ oct

= const.⇒σ m

τ oct

= const. In this case, the radius of the cylinder varieslinearly with the distance from the origin.

The surface is a cone with axison the hydrostatic stress axisand vertex in the origin.

1s

σ 3

σ 2

Hydrostatic stress axis

1 2 3s s s= =

θ

3 octR t=

q3 octt

3 octs

→

σ oct

τ oct

= 1tanθ

= cotθ

The angle of the cone is constant and itsvalue is :

θ = a cot

τ oct

σ oct

Þ

d = 3σ oct

Unit 4

An elastoplastic material is subjected to a pure shear stress and anuniaxial tensile stress . Plastification occurs, respectively, at ___ and--------. Determine the values for the cohesion c and internal friction angle---, assuming a Mohr-Coulomb yield criterion. Assume also plane stressconditions.

( )Ι( )ΙΙ at =

bs =f

Ιt t

t

t

ΙΙs s

Unit 4 – Solution Ιt t

t

t

ΙΙs s

: PURE SHEAR STRESSI

( ) ( ) ( )1 3 1 3 sin 2 cos 0F c= s -s + s +s f- f =σThe yield surface takes the form:

Where:

1 1 3

1 33

0

2 a

aas = s +s =ù é

Þú ê s -s =s = - ëû

( ) 2 2 cos 0F a cÞ = - f =σ

cosac =f

1s3s 2 0s =

at =

f

PLASTIC ZONE

t

s

Unit 4 – Solution Ιt t

t

t

ΙΙs s

: UNIAXIAL TENSILE STRESSII

( ) ( ) ( )1 3 1 3 sin 2 cos 0F c= s -s + s +s f- f =σThe yield surface takes the form:

Where:

1 1 3

1 33

0

b bb

s = s +s =ù éÞú ê s -s =s = ëû

( ) ( ) 1 sin 2 cos 0F b cÞ = - f - f =σ

2sin 1ab

f = -

1 bs =3 2 0s s= =

f

PLASTIC ZONE

t

s

2 2sin 1 1 1 1 0 2 0 1a a ab b b

f < ® - < - < ® < < Þ < <

Unit 5

Formulate in terms of the stress invariant the equation of theyield surface that, in the principal stress space, is an ellipsoid of revolutionaround the hydrostatic axis with semi-axes a and b :

1 2 3, , and I J J¢ ¢

a

bb

1s 2s

3s

Intersection with the octahedral plane at (0,0,0)

Unit 5 – Solution a

b

b

1s 2s

3s

Since the surface is a surfaceof revolution with axis on thehydrostatic axis, its equationdoes not depend on ¢3J :

( ) ( )1 2 3 1 2, , ,F I J J F I J¢ ¢ ¢Þ

In the x-y-z axis, theequation of a revolutionellipsoid can be written as:

2 2 2

1x y za b bæ ö æ ö æ ö+ + =ç ÷ ç ÷ ç ÷è ø è ø è ø 1s

2s

3s

3 octt

3 octs

xy

z

( )1 2 3, ,F I J J¢ ¢

Unit 5 – Solution a

b

b

1s 2s

3s

2 2 2 2 2 2

2 21 1x y z x y za b b a b

+æ ö æ ö æ ö+ + = Þ + =ç ÷ ç ÷ ç ÷è ø è ø è ø

The equation of a circumference composing the cross-section of the ellipsoid, in the z-y-z axis, is:

R

z

y 2 2 2 2 22 2 2

2 2 2 2 1 1x y z x Ry z Ra b a b

++ = Þ + = ® + =

In the axis: 1 2 3s s s- -

221 1 13 3 x

3 33octI I Ix t= = = ® =

22 2

23 3 2 23octR J J R Jt ¢ ¢ ¢= = = ® =

212 22 1

3I Ja b

¢+ =

Then, the equation of theyield surface is:

Unit 6

Justify which of the yield functions (Von-Mises, Tresca, Mohr-Coulomb andDrucker-Prager) would be suitable for the following cases:

a) The material does not plastify under any hydrostatic stress state.

b) The material only plastifies when the maximum shear stress reaches acertain value.

c) The material behaves significantly different in tension than incompression.

Unit 6 – Solution

( ) 23 0eF J s¢= - =σ ( ) ( )1 2

max2

0eFt

s s s= - - =σ!"#"$

Von-Misesgeneralization

Trescageneralization

Von-Mises Tresca

Drucker-Prager Mohr-Coulomb

Unit 6 – Solution

a) The materials does not plastify under any hydrostatic stress state.

Von-Mises and Tresca

b) The material only plastifies when the maximum shear stress reaches acertain value.

Von-Mises

c) The material behaves significantly different in tension than incompression.

Mohr-Coulomb and Drucker-Prager

Unit 7

The body delimitated by points A,B,C and D is a perfectly elastoplasticmaterial which is tested in the devise illustrated in Figure 1. The action-response curve is shown in Figure 2. An uniaxial stress-strain state isassumed such that:

Consider a thickness b in the out of plane direction. The following values thatlimit the graph in Figure 2 are requested:

a) The value of the elastic load and the corresponding displacement

b) The value of the ultimate tensile and compressive loads in the plasticregime.

c) The value of the ordinate and abscissa corresponding to points 1 and 2 in thegraph at Figure 2.

P d-

0x y z xy xz yzyhLde e e g g g= = = = = =

0 0x y z xy xz yzs s s t t t¹ = = = = =

eP .ed

pP qP

Unit 7P

qP

eP

pP

ed 2 eLEs

1

2

d

Figure 2

y

x

P

h

A B

CD

d

Infinitely rigid

Figure 1

L

e

s

es

es

E

Figure 3

Unit 7 – Solution y

x

P

h

A B

CD

dL

1. ELASTIC REGIME : eP P<

P

R

21 1210 3 332BE EEM P h bh P bhhbhL LLd dd æ öæ ö= Þ × = × = Þ =ç ÷ ç ÷

è ø è øå

sup e eey h L E

d se ee == = = = Þ ee L

Esd =

1 3e eP bhs=

/ Le d=

h

/E Ls d=

y

es

/e Es

( Strain ) ( Stress)

P

qP

eP

pP

ed 2 eLEs

1

2

d

Unit 7 – Solution y

x

P

h

A B

CD

dL

2. PLASTIC REGIME : eP P>

h

es s=

max ee E

se =

P

R

y

es

/e Es ed e s s®¥ Þ ®¥ Þ ®

In the case where: d ®¥

P

qP

eP

pP

ed 2 eLEs

1

2

d

Unit 7 – Solution y

x

P

h

A B

CD

dL

2. PLASTIC REGIME : eP P>

P

R

( ) 2110 · 22B e eM P h bh bhhs sæ ö= Þ = × = Þç ÷

è øå 1 2p eP bhs=

1 2q eP bhs= -

y

es s=

es

/e Es

q pP P= - Þ

P

qP

eP

pP

ed 2 eLEs

1

2

d

Unit 7 – Solution y

x

P

h

A B

CD

dL

3. POINT 1

max2 2 e eeL

E L Es sdd d e= > Þ = =

21111 3 10 · · · 2432 4 4B ee eM P h bhhbh h bh ss s æ öæ ö æ ö æ ö= Þ = + =ç ÷ ç ÷ ç ÷ ç ÷

è ø è ø è ø è øå

111 24 eP bhs= 1

2 e LEsd =

Displacement at point 1 is:

es

/e Es 2 /e Es Þ

P

qP

eP

pP

ed 2 eLEs

1

2

d

es s=

ye

Ese =

max 2 /e Ee s=P

R2h

Unit 7 – Solution y

x

P

h

A B

CD

dL

4. POINT 2

Unloading: ( ) is linear ( ) ( ) is lineary y E ye s eD Þ D = D

es

es-2 e

Ese

Es

Limit case: 2 0 2e e e es s s s s s s-D = - Þ -D = Þ D =

P

qP

eP

pP

ed 2 eLEs

1

2

d

es s=

y

eD

h

sDes s-D

2ess D

-

2h

Unit 7 – Solution y

x

P

h

A B

CD

dL

4. POINT 2

2 es sD =In the limit case where:

es-P

R

es

es-2 e

Ese

Es

25510 · · 2464B eeM P h bhhbh ss æ öæ ö= Þ = = - Þ-ç ÷ ç ÷

è ø è øå 25

24 eP bhs= -

sup2 22 2 / 0 e e

e e E E Es ss s e s eD = Þ D = Þ = - = Þ 2 0 d =

P

qP

eP

pP

ed 2 eLEs

1

2

d

2h

Unit 8 – Exam question

A complete loading-unloading cycle is applied on a steelbar with Young’s Modulus E, initial yield stress and hardening parameter______. It is known that the maximum total strain achieved through theprocess is .The maximum plastic and elastic strain are denoted by___ and respectively. In this situation, it is fulfilled that:

(1) (2) (3) (4)- - -

es

2EH ¢ =

max32

e

Ese =

maxpe max

pe

a c

b d

max76 es s=

max max27

p ee e=

16

epE E=

None of theother answers.

s

e

es

maxe

1

2

3

4

Unit 8 – Solution

The elastoplastic tangent modulus is:

1 1 3 3ep ep

HE E E E EE H

¢= = Þ =

¢+

epHE E

E H¢

=¢+

Answer C is UNCORRECT

The value of the maximum stress can be computed as:

max max3 7

3 3 2 6ep e e e

e e e e eE EE

E E Es s ss s s s e s e s sæ ö æ ö= + D = + D = + - = + - =ç ÷ ç ÷

è ø è ø

max es s s= + D

max7 6 es sÞ = Answer A is CORRECT and D is UNCORRECT

The relation between the maximum and elastic strains is: max max maxp ee e e= -

maxmax max max max

3 3 7 1 1 2 2 6 3 3

p e pe e e e e

E E E E E Es s s s s se e e e= - = - = - = Þ =

max maxmax27

7 6

p ee e

Ee e

se == Þ Answer B is CORRECT

s

e

es

maxe

1

2

3

4

Unit 8 – Solution

A complete loading-unloading cycle is applied on a steelbar with Young’s Modulus E, initial yield stress and hardening parameter______. It is known that the maximum total strain achieved through theprocess is .The maximum plastic and elastic strain are denoted by___ and respectively. In this situation, it is fulfilled that:

(1) (2) (3) (4)- - -

es

2EH ¢ =

max32

e

Ese =

maxpe max

pe

a c

b d

max76 es s=

max max27

p ee e=

16

epE E=

None of theother answers.

s

e

es

maxe

1

2

3

4

Unit 9 – Exam question

Indicate which of the following statement/s is/are correct concerning differentyield surfaces:

a c

b d

The Von-Mises criterion issuitable for materials that do notplastify upon hydrostatic stressstates.

The Drucker-Prager criterion issuitable for materials that do notplastify upon hydrostatic stressstates.

The Mohr-Coulomb criterion issuitable for materials thatbehave differently in tensionand compression.

The Tresca criterion is suitablefor materials that show adifferent behaviour in tensionand compression.

Unit 9 – Solution

Von-Mises Tresca

Drucker-Prager Mohr-Coulomb

Statements A and B are CORRECT and statement C and D are UNCORRECT

Unit 9 – Solution

Indicate which of the following statement/s is/are correct concerning differentyield surfaces:

a c

b d

The Von-Mises criterion issuitable for materials that do notplastify upon hydrostatic stressstates.

The Drucker-Prager criterion issuitable for materials that do notplastify upon hydrostatic stressstates.

The Mohr-Coulomb criterion issuitable for materials thatbehave differently in tensionand compression.

The Tresca criterion is suitablefor materials that show adifferent behaviour in tensionand compression.