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Continuum Mechanics Rehearsal Exam AP3 (Example A)

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STATEMENTS

sheet

Observations:

1. Each question may have one or more correct answers. The student must mark them filling with

dark ink the whole space of the corresponding box in the answers sheet.

2. Each correct answer in a same question adds 1/n times the value of the question (where

n={1,2,3,4} is the number of correct answers of the question).

3. Each incorrect answer in a same question subtracts 1/(4-n) times the value of the question (where

n={1,2,3,4} is the number of correct answers of the question).

4. Any unanswered question will be given 0 marks.

Answers must be given exclusively on the answers sheet:

answers given on the statements sheets will be ignored.

Question 1 For a given isotropic linear elastic material which follows Hooke’s law indicate which of

the following situation/s is/are possible for the components of the stress and strain tensors in a Cartesian

coordinate system {x,y,z} :

A σy > 0 ; σx = σz = 0 and εy < 0.

B σm =1

3Tr(σσσ)< 0 and e = Tr(εεε)> 0.

C σx > 0 ; σy = σz = 0 and εy < 0.

D σσσ = σm1 with εεε ′ 6= 0.

Question 2 Indicate which of the following statement/s is/are true for a linear thermoelastic problem:

A If the temperature increment is uniform and

the material is homogeneous, the second

thermal analogy can always be applied.

B The first thermal analogy can only be applied

if the temperature fields are constant or linear.

C The superposition principle can only be ap-

plied if the temperature increment is uniform.

D None of the other answers.

Question 3 Indicate which of the following statement/s is/are correct for the Theory of linear elasticity:

A The elastic potential is always negative or

null.

B σσσ : ε̇εε =1

2εεε : C : εεε .

C Stresses and strains are proportional if and

only if the material is isotropic.

D The material coordinates, X, and spatial

coordinates, x, satisfy x ≈ X for any time

instant t.

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Question 4 The following stress state σσσ(x,y, t) =

σx τxy 0

τxy σy 0

0 0 0

corresponds to:

A A generic stress state with εz = 0.

B A generic plane stress state.

C A plane strain state with ν = 0.

D A generic plane strain state.

Question 5 Indicate which of the following statement/s is/are true for a plane stress state:

A All isostatic lines pass through a singular

point.

B Isobar lines are the locus of points along

which the principal stresses are in the same

direction.

C An isocline line of 0◦ corresponds to an

isostatic line.

D Isostatic lines are the envelopes of the vector

field defined by the principal directions.

Question 6 Indicate which of the yield criteria make/s the following statement true: For a given yield

surface, the hydrostatic stress states never produce plastification:

A Tresca.

B Mohr-Coulomb.

C Von Mises.

D Drucker-Prager.

Question 7

A complete loading-unloading cycle [1]−[2]−[3]−[4] is

applied on a steel bar with Young’s Modulus E, initial yield

stress σe and hardening parameter H ′. It is known that

εmax = 2εe = 2σe/E is the maximum strain reached dur-

ing the whole process. Then, the following is true:

A If H ′ = ∞, the plastic strain at the end of the

process is εp4 = 2εe.

B If H ′ = 0, the plastic strain at the end of the

process is εp4 = εe.

C Necessarily, H ′ ≥−E/2.

D If H ′ = E, the maximum plastic strain

achieved in the process is εpmax = 0,5εe.

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The figure shows a hollow cylinder made of an isotropic elastoplastic material. The outer walls of the

cylinder are fully fixed and a pressure P ≥ 0 is applied uniformly on the inner walls in the radial direction,

as shown in the figure. The inner radius is R1 = 1 and the outer radius is R2 = 2.

HYPOTHESES

• Infinitesimal strain theory.

• Body forces and the effect of

friction are negligible.

• Material properties:

E,ν = 0, c,φ = 30◦.

• The cylinder has radial sym-

metry.

Question 8 The displacement field in the radial direction, ur, is:

A ur =P

5E

(

4

r− r

)

.

B ur =5E

P

(

4

r− r

)

.

C ur =P

5E

(

2

r+ r

)

.

D ur =P

E

(

4

r+ r

)

.

Question 9 Consider the displacement field ur =P

E

(

4

r− r

)

. Valid components of the stress tensor are:

A σrr = P(4ln(r)−1).

B σθθ = P

(

4

r2−1

)

.

C σrr =−P

(

4

r2+1

)

.

D σθθ = P(

4− r2)

.

Question 10 Consider the stress tensor:

σσσ =−P

5

1+4

r20 0

0 1− 4

r20

0 0 0

.

The value of P for which plastification begins according to the Mohr-Coulomb criterion is:

A P =5√

3

7c.

B P =7

5√

3c.

C P =4√

3−2

2√

3c.

D P = 5c.

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Rehearsal Exam AP3: Example AContinuum Mechanics Course (MMC) - ETSECCPB - UPC

Rehearsal Exam AP3: Example A

Question 1

1

Isotropic linear elasticityÞ

Hooke’s Law

A

B

0 , 0 and 0y x z ys s s e> = = <

1 1( ) yy y y yTr

E E E E Esn n n ne s s s+ +

= - + = - + =σ

If 0 0 y ys e> Þ > Answer A is FALSE.

( ) ( )1 0 and 03m Tr e Trs = < = >σ ε

The mean stress can be written as , where the bulk modulus m Kes =

( ) ( )1If 0 0 3me Tr Trs= > Þ = >ε σ

0K >

Answer B is FALSE.

1 ( )TrE En n+

Þ = - +ε σ 1 σ

, where Young’s modulus 0E >


2

C

D

1( ) y y xTrE E En n ne s s+

= - + = -σ

0 , 0 and 0x y z ys s s e> = = <

If 0 0 x ys e> Þ < Answer C is TRUE.

with ms ¢= ¹1 ε 0s

1, where 0 and Poisson's ratio fulfills 0 2

E n> £ £

Hooke’s Law1 ( )Tr

E En n+

Þ = - +ε σ 1 σ

The stress tensor can be decomposed into , being the sphericalpart and the deviatoric partsph m= 1s s 2 µ¢ ¢=s e

sph ¢= +s s s

2 m µ ¢Þ = +σ 1s e Answer D is FALSE.


Question 2

3

A

B

C

Linear thermoelastic problemÞ

The Second Thermal Analogy can only be applied if the thermal strain fieldis integrable. Particular cases :

Homogeneous material + linear thermal increment ( cnt)a = ( linear)qD

Homogeneous material + constant thermal increment ( cnt)qD =

The First Thermal Analogy can be applied for any thermal increment ( , )tqD xAnswer B is FALSE.

The superposition principle can be applied for any thermal increment ( , )tqD xAnswer C is FALSE.

( cnt)a =Answer A is TRUE.

1

2

D Answer D is FALSE.


Question 3

4

A

B

C

Theory of linear elasticityÞ

Elastic potential : 1ˆ : : 0 02

u = > " ¹ε ε ε!

The elastic potential is always positive or null. Answer A is FALSE.Þ

In linear elasticity theory, strains and stresses are always proportional, thematerial does not need to be isotropic. Answer C is FALSE.

Generalized Hooke’s law : :Þ =σ ε ε!

D Under the infinitesimal strain theory : x X t"! Answer D is TRUE.

( )ˆ 1: : :2du ddt dt

= =σ ε ε ε! " Answer B is FALSE.


Question 4

5

The form of the given stress state is : ( )00, ,

0 0 0

x xy

xy yx y ts tt sé ùê ú= ê úê úë û

σ

For a generic plane stress state :

00

0 0 0

x xy

xy y

s tt sé ùê ú= ê úê úë û

σ

Where the component ( )x yz En s se += -

Answer A is FALSE and B is TRUE.

For a generic plane strain state :

00

0 0

x xy

xy y

z

s tt s

s

é ùê ú= ê úê úë û

σ

Answer C is TRUE and D is FALSE.

00

0 0 0

x xy

xy y

e ee eé ùê ú= ê úê úë û

ε

Where the component ( )x yz s ss n +=

Consequently, if 0 0zn s= Þ =

00

0 0

x xy

xy y

z

e ee e

e

é ùê ú= ê úê úë û

ε Then, for the given stress state 0zeÞ ¹


Question 5

6

A All isostatic lines pass through a singular point.

Singular points: Points in which In a singular point, alldirections are principal directions.

and 0 x y xys s t= = Þ

Isostatics: Envelopes of the vector field defined by theprincipal directions.

In singular points, isostatics tend to loose their regularityand can abruptly change direction. They do not necessarilypass through a singular point. Answer A is FALSE.

( The stated property corresponds in fact to isoclines )

B Isobars are the locus of points along which the principal stress is constant._____Answer B is FALSE. ( The given definition correspond to isoclines)

Singular point


7

C

D Isostatics are the envelopes of the vector field defined by the principaldirections. _____ Answer D is TRUE.

Isoclines: Locus of points along which principal stresses are in the same direction.Isostatics: Envelopes of the vector field defined by the principal directions. _____

y

x

1s1s

Isocline 0°

Isostatics

Isostaticsy

x

1s1s

y

x

Isocline

1s1s

qq

An isocline of does not correspond to an isostatic. Answer C is FALSE.0°

An isocline of corresponds to an isostatic.0°


Question 6

8

For a given yield surface, a hydrostatic stress state never produces plastification .

Mohr-CoulombVon-Mises Tresca

A hydrostatic stress state is found in the hydrostatic axis, in which 1 2 3 .s s s= =

It will never produce plastification for those criterions in which the yield surfaceis parallel to the hydrostatic axis. This is the case of Von-Mises and Tresacacriterions. Answers A and C are TRUE and answers B and D are FALSE..

Drucker-Prager


Question 7

9

[ ] [ ] [ ] [ ] Complete loading - unloading cycle - - - 31 2 4Þ

max Maxium strain reached during the process is : 2 2 /e e Ee e sÞ = =

The elastoplastic tangent modulus can be computed as : ep HE EE H

¢=

¢+

Elastic strain Plastic strain

s

e

es

maxe

1

2

3

4

Steel bar properties :

- Young’s Modulus

- Initial yield stress

- Hardening parameter

E

es

H ¢


10

B 40 peH e e¢ = Þ =

0 0epH E¢ = Þ =

( )max maxep ep

e e e e eE Es s s s e s e e s= + D = + D = + - =

maxmax max max max 4 , e p e pe e

eE E Es s se e e e e e= = = - = Þ =

Answer B is TRUE.

s

e

es

/e Es

E

2 /e Es

0epE =

A 4 2peH e e¢ = ¥ Þ =

epH E E¢ =¥ Þ =

( )max max 2ep epe e e e eE Es s s s e s e e s= + D = + D = + - =

maxmax max max max 4

2 , 0 0e p e pe

E Es se e e e e= = = - = Þ =

Answer A is FALSE.

s

e

es

2 es

/e Es 2 /e Es

E

epE E=


11

D max 0.5peH E e e¢ = Þ =

2

ep EH E E¢ = Þ =

( )max max32

ep epe e e e eE Es s s s e s e e s= + D = + D = + - =

maxmax max max max max

3 1 , 0.52 2

e p e pe eeE E E

s s se e e e e e= = = - = Þ =

Answer D is TRUE.

s

e

es

/e Es

E

2 /e Es

C

2

epEH E E¢ = - Þ = -

( )max max 0ep epe e e eE Es s s s e s e e= + D = + D = + - =

Answer C is TRUE.

max If 2 is not reached 2 eEH e e¢Þ < - Þ =

s

e

es

/e Es

E

2 /e Es

epE E= -

Necessarily 2EH ¢ ³ -


Question 8

12

Because of the radial symmetry of the cylinder andthe pressure field, it can be assumed that the form ofthe displacement field is :

( ) ( )[ ] , 0 , Tr zu ur z=u

Due to the null Poisson’s ratio, the vertical componentof the displacement field is zero :

( )0 0 zu zn = Þ =

Navier equations in cylindrical coordinates are used :

( )1 and 0zr zr

dude rur dr dz q= + w = w = w =

Inner and outer radius are 1 21 and 2R R= =

h

1R2R

p


13

( ) ( ) ( ) ( )11 10 0 2 2r rr rd d d dd d A Arru ruru rudr dr r dr drr dr r dræ ö æ ö= Þ = ® = ® =ç ÷ ç ÷è ø è ø

( ) 2 2 r rrd BAr ru Ar B u Arrudr r

® = ® = + Þ = +

[ ] ( ) ( )11 0 0 01 zrr

de d ddu dd rurudr dr dr r drr dr dz

æ ö æ ö= Þ = Þ =+ ç ÷ç ÷è øè ø

Applying the given displacement boundary conditions :

2

40 2 0 2r r

BA AuB=

= Þ × + = Þ = -

From the equation for the radial component, radial displacement can beintegrated :


14

Finally, the stress field can be obtained from : ( ) 2 Trl µ= +1s e e

Applying stress boundary conditions :

1 2

1rr r

PBp E p A BAE=

æ ö= - Þ = - Þ = - +s -ç ÷è ø

4 1 1 4 4 5 5 5rB P p p pB B A u r

E E E E ræ ö- = - + Þ = Þ = - Þ = -ç ÷è ø

Answer A is TRUE and the rest are FALSE.

Where since 0 0 (1 )(1 2 ) 2(1 ) 2

andvE E E Ev v v

n l µ= Þ = = = = Þ =+ - +

σ ε

The strain field can be obtained from the displacement field as :

2 2 , , and 0r r zrr zz r z zr

u u uB BA A Cr r r r zqq q q

¶ ¶e = = - e = = + e = = e = e = e =

¶ ¶


Question 9

15

For this question, the considered displacement field is : 4rPu rE ræ ö= -ç ÷è ø

The strain field components are obtained as :

2 24 4 1 1andr r

rru uP Pr E r Er rqq

¶ æ ö æ öe = = - e = =+ -ç ÷ ç ÷¶ è ø è ø

Finally, the stress field can be obtained from : ( ) 2 Trl µ= +1s e e

Where since 0 0 (1 )(1 2 ) 2(1 ) 2

andvE E E Ev v v

n l µ= Þ = = = = Þ =+ - +

σ ε

2 2

4 4 1 1 and rrrr rr P Pr r

E E qqqq qqs = - s =+ -æ ö æ ös = e Þ s = e Þç ÷ ç ÷è ø è ø

Answers B and C are TRUE and answers A and D are FALSE.


Question 10

16

For this question, the considered stress tensor is :2

2

41 0 0

40 1 05

0 0 0

rP

r

é ù+ê úê ú= - ê ú-ê úê úë û

σ

1 2 32 24 4 , 0 and 1 1

5 5P Pr r

s s sæ ö æ ö= = = -- +ç ÷ ç ÷è ø è ø

According to Mohr-Coulomb plastification criterion :

( ) ( )1 3 1 31 3sin 2cos 0 where 30 sin and cos2 2

cs s s s f f f f f- + + - ³ = °Þ = =

Replacing into the above expression results into :1 3 and s s2 2

2 22

88 3 0 3 0 5 315 5 8P P r rc c P c

r rr-æ ö - ³ Þ - ³ Þ ³-ç ÷ -è ø

For [ ]1,2rÎmin

5 31 7

r P c= Þ =

max 2 5r P c= Þ = min5 3

7P c= Answers A is TRUE

and the rest are FALSE.

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