Continuum Mechanics Evaluation A1 Màster en Enginyeria de...

Evaluation A1. ‒ EXAMPLE TEST B ‒ Permutation 1

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 electronic correction paper.

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.

Q - 1. A continuous medium moves as shown in the figure. At time instant t , all its particles have the same density value 1ρ . At time instant t dt+ , all its particles have the same density value 2 1ρ ρ≠ . The following expressions hold true for all the particles in the continuous medium:

A 0=td

d ρ B 0

tρ∂=

∂

C

ddt tρ ρ∂=∂

D 0ρ∇ =

Q - 2. If the velocity field of a continuous medium is stationary:

A A same particle cannot go twice through the same point in space at different time instants.

C

The spatial acceleration field of this medium is also stationary.

B The particles of this medium have no acceleration. D The spatial description of this field is not dependent on the spatial coordinates.

Q - 3 The equations of the streamlines of the movement in a continuous medium are:

1 2 2 3, ,t tx C C e y C e z C eλ λ λ− += + = = ; with 1 2 3, andC C C constants.

Considering 0t = as the reference time instant, the acceleration field is:

A None of the other answers. C ( , ) , ,Tt t tt Xe Ye Ze− ≡ A X

B [ ]( , ) 0, , Tt y z≡a x D ( , ) 0, ,Tt tt Ye Ze ≡ A X

Permutation

1

Continuum Mechanics ‒ COURSE 2014-2015 ‒ Evaluation A1 ‒ 24/10/2014 Màster en Enginyeria de Camins, Canals i Ports. Màster en Enginyeria Geològica i de Mines.

NAME and SURNAMES: ………………EXAMPLE TEST B………………………………

x y

z

t

ρ1

t + dt

ρ2


Q - 4 The motion of a continuous medium is described by the following equations of motion:

2 2(1 ) ; ;x X a t y Y z Z= + = =

The velocity field in the spatial and/or material description is/are:

A 2

2 2

2( , ) , 0, 01

Txa tt

a t

≡ + v x C

2 2( , ) , 0, 01

Txta t− ≡ +

v x

B [ ]( , ) , 0, 0 Tt X≡ −V X D 2( , ) 2 , 0, 0T

t a Xt ≡ V X

The solid in the figure suffers a uniform infinitesimal deformation such that:

1) Points A , B and C do not move.

2) The segment AE becomes (1 p) + times its initial length. 3) The solid volume becomes (1 q) + times its initial value. 4) The angle β increases in a value r (in radians) its initial value.

Q - 5 The components of the infinitesimal rotation vector θ and the infinitesimal strain vector ε verify:

A 1 2θ θ= C 31 02 xyθ γ= =

B

112 yzθ γ=

D 212 xzθ γ=

Q - 6 Assuming that 0xx yy xyε ε γ= = = , the components of the infinitesimal strain vector ε verify:

A

; 2zz xzq p qε γ= = −

C ; 2zz xzq p qε γ= − = − −

B

rqpyz 22 −−=γ

D 122yz p q rγ = − −

D

x,x1 β=EBC

aADACAB∧

===

A

a

a

F

B

C

y,x2

z,x3

E

β

a


Q - 7 The tensors F , E and e are defined for the motion of a continuous medium. Indicate which of the following

expressions are true.

A If component 0yyE = , then the stretch in direction

y is zero (0). C

F can be split into = ⋅F Q U , where T= ⋅U F F is a stretch and Q is a rotation. The determinants of

these tensors are: 1=Q and =U F .

B If component 0zze = , then the extension ratio in direction y is one (1).

D If the motion is a rigid solid motion, the stretch in any direction will be zero (0).

Q - 8 Indicate which steps are erroneous in the following deduction:

( ) ( ) ( ) ( ) ( )1 2 3 4 5, , , , v ,step step step step stepij i i i i k

ki kjj j j k j

dF t x t x t V t t x l Fdt t X X t X x X

∂ ∂ ∂ ∂ ∂∂ ∂= = = = =

∂ ∂ ∂ ∂ ∂ ∂ ∂

X X X X x

A Step 1 C Step 3

B Step 2 D Step 5

Q - 9 Indicate which of the following statement/s is/are true:

A A finite strain tensor can always be obtained from an arbitrary continuous displacement field. C

If the compatibility conditions are satisfied, the continuity of the medium during a deformation process is guaranteed.

B An arbitrary continuous displacement field can always be obtained from a finite strain tensor. D The compatibility conditions are equivalent to the

symmetry of the infinitesimal strain tensor.

Q - 10 If a strain rate tensor ( )d x verifies the compatibility equations, it means that:

A There exists a field such that ( )∇× ×∇ =d 0 . C ( )d x must necessarily be lineal in x .

B There exists a velocity field ( )v x such that

( ) ( )s∇ =v x d x . D ( )d x must necessarily be lineal or quadratic in x .

Solutions: Q1: C, D. Q2: C. Q3: B, D. Q4: A, D. Q5: C, D. Q6: A; B. Q7: C. Q8: D. Q9: A, C. Q10: A, B.

Rehearsal Exam AP1: Example B Continuum Mechanics Course (MMC) - ETSECCPB - UPC

Rehearsal Exam AP1: Example B

Question 1

1

At the time instant all particles have the same density t 1ρAt the time instant all particles have the same density t dt+ 2 1ρ ρ≠

⇒ Uniform density Answer D is TRUE. ρ⇒ ∇ = 0

⇒ The density depends on time Answer B is FALSE. 0 tρ∂≠

∂⇒

0 0 d d d

dt t dt t t dt tρ ρ ρ ρ ρ ρ ρρ ρ

=

∂ ∂ ∂ ∂= + ⋅∇ ⇒ = + ⋅∇ = ≠ ⇒ = ≠∂ ∂ ∂ ∂

v v 0

Answer C is TRUE and answer A is FALSE.


Question 2

2

Stationary velocity ( , ) ( ) t

t ∂=

∂⇒ = ⇒

v0v x v x

Trajectories can intersect each other in a given point at different time instants.

( ), ( ( , )) 0 t tt t

∂ ∂= = ≠∂ ∂V vA X x X

( ) 0 t

∂= + ⋅∇ = ⋅∇ ≠∂va x v v v v

Answer A is FALSE

Answer B is FALSE.

Answer C is TRUE.

Answer D is FALSE

A

B

D

Material description of acceleration: ( ) ( ), ,t tt

∂=∂VA X X

C Spatial description of acceleration: ( ) ( )( )ddt t

∂= = + ⋅∇

∂v va x x x v v

0 =

( , ) ( )t =v x v x


Question 3

3

1 2 2 3Given equations of the streamlines are : , , t tx C C e y C e z C eλ λ λ− += + = =

Differential equation of the streamlines:

2

2

3

v v v

v v v

v v v

t tx x x

y y y

tz z z

dx C e yeddy C e yddz C e zd

λ

λ

λ

λ

λ

λ

− −

+

= → = → =

= → = → =

= → = → =

Differential equation of the trajectories:

tye

y

z

− =

v

( ) ( ( ), )d tdλ λλ

=x v x

( , )d tdt

=x v x

Rehearsal Exam AP1: Example B 4

2 2 1

2 2

3 3

1 ln( )

1 ln( )

t t

t

t

dx ye dx ye dt K dt x K t Kdtdy y dy dt y t K y K edt ydz z dz dt z t K z K edt z

− −= → = = → = +

= → = → = + → =

= → = → = + → =

∫ ∫ ∫

∫ ∫

∫ ∫

Applying the consistency condition

2 1

2

3

t

t

x K t K

y K e

z K e

= +

=

=

( ,0)⇒ =x X X

2 1 1

02 2

03 3

0

y

t

t

X K K K X x Yt X

Y K e K Y Ye

Z K e K Z z Ze

= ⋅ + → = → = +

= → = → =

= → = → =

y

t

t

x Yt X

Ye

z Ze

= +

=

=


x x

ty y

tz z

xV V YtyV V YetzV V Zet

∂= → =∂∂

= → =∂∂

= → =∂

t

t

Y

Ye

Ze

=

V

A 0

A

A

xx x

y ty y

tzz z

VAt

VA Ye

tVA Zet

∂= → =

∂∂

= → =∂∂

= → =∂

0

t

t

Ye

Ze

=

A

Material description of velocity:

Material description of acceleration:

( , )( , ) ttt

∂=

∂x XV X

( , )( , ) ttt

∂=

∂V XA X


( )( , ), t t= ⇒a A X x

0xt t

y

t tz

a

a ye e y

A ze e z

−

−

=

= ⋅ =

= =

( )1

( )2

( , ) ( , )( , ) ·d t ttdt t

∂= = + ∇

∂v x v xa x v v

0 0 0 0 0( , ) 0 , , 1 0

0 0 0 1

t

t t

yet ye y z e y y

z z

−

− −

− = + ⋅ = ⇒ =

a x a

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

0 y

z

⇒ =

a


Question 4

7

2 2Given equations of motion are : (1 ) , , x X a t y Y z Z= + = =

( , )( , ) ttt

∂=

∂x XV XMaterial description of the velocity field :

2 2

0

0

x x

y y

z z

xV V Xa ttyV VtzV Vt

∂= → =∂∂

= → =∂∂

= → =∂

22

0

0

Xa t =

V

Answer B is FALSE and answer D is TRUE

( , ) ( , ) ttt

∂= ⇒

∂x XV X


2 22 2(1 )

(1 )

xx X a t Xa t

y Y Y yz Z Z z

= + → =+

= → == → =

2 2 (1 )

xXa t

Y yZ z

=+

==

Inverse equation of motion:

( , ) ( ( , ), ) t t t= ⇒v x V X x

2

2 2

2v1

v 0

v 0

x

y

z

xa ta t

=+

=

=

2

2 2

21

0 0

xa ta t

+

=

v

Answer A is TRUE and answer C is FALSE

( , ) ( ( , ), )t t t=v x V X xSpatial description of the velocity field :


Question 5

9

Uniform deformation ( , ) ( )t t⇒ =F x F

· ( ) d d d d d d t= ⋅ ⇒ = ⋅ → = ⋅ → = +∫ ∫ ∫ ∫x F X x F X x F X x F X φ

( ) ( ) ( ) ( )t t= + + ⇒ = + + ⋅ + → − = + ⋅ +F 1 ε Ω x 1 ε Ω X φ x X ε Ω X φ

( ) ( ) t= + ⇒ = − ⇒ = + ⋅ +x X u u x X u ε Ω X φ

Points A, B and C do not move:

1 1 1 0 x xx xxdsdS

λ λ ε ε= = ⇒ = + = ⇒ =Segment parallel to the x direction: AB

1 1 1 0 y yy yydsdS

λ λ ε ε= = ⇒ = + = ⇒ =Segment parallel to the y direction: AC

( )1


0 2 0 0 xy xy xy xyBAC BAC ε γ ε γ∆ = ⇒ ∆ = − = − = ⇒ = =

( ) =

=Ax x

u x 0

0 0 0

0 0 0

0

0 00 0 ( ) ( )0 0

xz xy zx

yz xy yz

xz yz zz zx yz

t tε

ε

ε ε ε

Ω −Ω

+ −Ω Ω

Ω −Ω

= + ⇒ =

φ φ 0

( ) =

=Bx x

u x 0

0 0 0

0 0 0

0

00 0 0 0

xz xy zx

yz xy yz

xz yz zz zx yz

aε

ε

ε ε ε

Ω −Ω

+ −Ω Ω

Ω −Ω

= ⇒ ( )

0 00

0xy

xz xz

a

aε

== −Ω

= +Ω

30 0 0 xy xya θ= −Ω ⇒ Ω = ⇒ =


( ) =

=Cx x

u x 0

0 0 0

0 0 0

0

0 00 0 0

xz xy zx

yz xy yz

xz yz zz zx yz

aε

ε

ε ε ε

Ω −Ω

+ −Ω Ω

Ω −Ω

= ⇒ ( )

0

0 0

0

xy

yz yz

a

aε

= Ω

=

= +Ω

( ) 11 1 2 2yz yz yz yz yz yz yzaε ε γ θ γ+Ω ⇒ Ω = ⇒ Ω = ⇒ = −

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

( ) 21 1 2 2xz xz xz xz xz xz xzaε ε γ θ γ+Ω ⇒ Ω = − ⇒ Ω = − ⇒ =


Question 6

12

Segment becomes p times its initial length AE

(1 )AE pλ⇒ = +

1 λ = ⋅ ⋅ +t ε t

[ ]10 0

1 1 , 0 , 1 0 0 0 1 12 2

0

xzzz

yz xz

xz yz zz

εελ ε ε

ε ε ε

= ⋅ ⋅ + = + +

1 1 2 2zz zz

xz xzp pε ελ ε ε⇒ = + + = + ⇒ = +

( )2

The solid volume becomes times its initial value 1 q+ 1o

dV qdV

⇒ = +

( ) ( ) ( ) 1 1 1 zzo o

dV dVTr Tr q Tr q qdV dV

ε= + ⇒ = + = + → = ⇒ =ε ε ε

( )3


The angle increases r radians its initial value β rβ⇒ ∆ =

1 12 sin

ββ

⋅ ⋅∆ =

t ε t

where sin 1β =

[ ] ( )10 0

12 0 , 0 , 1 0 0 1 2 22

0

xz

yz xz yz yz

xz yz zz

p qε

β ε ε ε εε ε ε

∆ = ⋅ ⋅ = − = − −

1 1 12 2 2 2 2 2yz yz yzp q r p q r p q rβ ε ε γ ∆ = − − = → = − − ⇒ = − −

2 2 2

2zz

xz xz xz xzq

p pqp p qε

ε ε ε γ= + = +→ → = − ⇒ = −

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

( )4


Question 7

14

A Component 0yyE =

1 2 1 2 1 1 y yy y yy yE Eλ λ λ= + ⇒ = + = ⇒ = Answer A is FALSE

B Component 0zze =1 0 0

1 2y yyye

λ λ= ≠ ⇒ ≠−

C

0 =

Answer B is FALSE

Polar decomposition of F ( Page 61 Chapter 2 ) Answer C is TRUE

D Rigid body motion 1 E e λ⇒ = = ⇒ =0 Answer D is FALSE


Question 8

15

Step 1: ( ) ( ), ,ij i

j

dF t x tdt t X

∂∂=∂ ∂

X X

Step 2: ( ) ( ), ,i i

j j

x t x tt X X t∂ ∂∂ ∂

=∂ ∂ ∂ ∂

X X

Step 3: ( ) ( ), ,i i

j j

x t V tX t X

∂ ∂∂=

∂ ∂ ∂X X

Step 5: ( ),

i kkj

k jik

v t x Fx X

l∂ ∂

=∂ ∂

x

Answer A is FALSE

Answer B is FALSE

Answer C is FALSE

Answer D is TRUE

A

B

C

D Step 5 is incorrect.⇒


Question 9

16

A

B

C

Given a continuous displacement field , the corresponding strain field___ can ALWAYS be obtained.

( , )tu x( , )tε x Answer A is TRUE

Given a strain field , the corresponding continuous displacement field______ CANNOT ALWAYS be obtained. ( , )tu x

( , )tε x

Answer B is FALSE Compatibility conditions must be satisfied ⇒

Compatibility conditions guarantee the continuity of the medium during the deformation process. Answer C is TRUE

D Compatibility conditions are not equivalent to the symmetry of the infinitesimal strain tensor. Answer D is FALSE


Question 10

17

There exists a field such that ( Slide 19 Ch. 3 )

If verifies the compatibility conditions: ( )d x

( )∇× ×∇ =d 0Answer A is TRUE

There exists a velocity field such that ( Slide 30 Ch. 3 ) Answer B is TRUE

( )v x ( ) ( )s∇ =v x d x

( )d x must be always integrable. ( Slide 5 Ch. 3 )

( )d x linear is always compatible and, thus, integrable. ( Slide 30 Ch. 3 )

Answer C and answer D are FALSE

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