Ch.2. Group Work Units

Continuum Mechanics Course (MMC) - ETSECCPB - UPC

Unit 1

Given the following material strain tensor

Compute at t=1 the length of a segment that at the reference time (t=0) was a straight line joining the points A = (1,1,1) and B = (2,2,2).

0 00 0

0 0

tX

tX

tY

tete

te

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

E

Unit 1 - Solution

0t = 1t =

A

B

A

B

The stretch is defined as:

And its expression is

where is a unitary vector in the direction of the stretch. In this case:

dsdS

λ =

λ = + ˆ ˆ1 2 · ·T ETT̂ ( )= 1ˆ 1,1,1

3T

x

y

z

x

z

yy

x

z

x

y

z

xx

z

x

y

z

x

Unit 1 - Solution

120 0 1

2 21 (1 1 1) 0 0 1 1 (2 )3 3

0 0 1λ

⎡ ⎤⎛ ⎞⎛ ⎞⎢ ⎥⎜ ⎟⎜ ⎟= + = + +⎢ ⎥⎜ ⎟⎜ ⎟

⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

tX

tX tX tY

tY

tete te te

te

For the reference configuration:

( )1,1,1A =

( )2,2,2B = ααα

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

dX dd dY d

dZ dX where [ ]1,2α ∈

α= = 3dS d dX

( )α

α αα

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

XYZ

X

Unit 1 - Solution

For the instant of time t=1: 21 (2 ) 1 23

αλ = + + = +X Ye e e

λ=ds dS

Finally:

1 2 3α α= +ds e d

And the length of the segment joining A and B at that instant can be calculated as:

2

1

1 2 3l ds e dα

α

α

α=

=

= = +∫ ∫2

1

3 1 2l e dα

α

α

α=

=

= +∫

Unit 2

6

The tetrahedron in the figure undergoes a uniform deformation in which:

y

z

x

a

aa

O

A

B

C1.  Points O,A and B do not move.

2. The volume of the solid becomes p times its initial volume, being p>0.

3. The segment becomes p times its initial

length, being p>0.

OC! "!!

Under these conditions the following statement/s concerning the deformation gradient tensor is/are true:

a c

b d

11 22 13 231, 0F F F F= = ≠ =

11 22 13 231, 0F F F F= = = =

331Fp

=

33F p=

Unit 2 - Solution

7

Uniform deformation: =( , ) ( )t tF x F

= ⋅ → = ⋅ → = ⋅∫ ∫ ∫ ∫d d d d d dx F X x F X x F X

From the definition of the deformation gradient tensor: = ⋅d dx F X

= +· ( )tx F X φ

Since , F can be taken out of the integral and the integration constant can depend on time.

=( , ) ( )t tF x F

Point O does not move:

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

11 12 13

21 22 23

31 32 33

0 00 0 ( )0 0

F F FF F F tF F F

φ =( ) 0tφ

Unit 2 - Solution

8

Point A does not move:

11 12 13

21 22 23

31 32 33

0 00 0

a F F F aF F FF F F

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

11

21

31

100

FFF

===

Point B does not move:

12 13

22 23

32 33

0 1 00

0 0 0

F Fa F F a

F F

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

12

22

32

010

FFF

===

So up to now we have

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

13

23

33

1 00 10 0

FFF

F

Unit 2 - Solution

9

Volume of the solid becomes p times its initial volume:

The variation of volume can be expressed as: =t odV dVF

= → = → =∫ ∫ ∫ ∫t o t o t odV dV dV dV dV dVF F F Vt = F Vo

As in the previous case, since , F can be taken out of the integral =( , ) ( )t tF x F

= → = 33 0t o tV V V F VF

t oV pV=33F p=

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

13

23

1 00 10 0

FFp

F

t oV pV=

Segment becomes times its initial length:

Unit 2 - Solution

OC! "!!

'13 13

'23 23

'

1 0 00 1 00 0

c

c

c

x F F ay F F az p a pa

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦

Since point O does not move, point C after deformation can be calculate as:

' 'o cl pa=

( )22 2 2 2 2 2' ' 13 23 13 23

2 2 2 2 213 23 13 23

( ) ( ) ( ) ( ) ( ) ( )

( ) 0 ( ) 0

= + + = → + + =

→ + = → + =o cl F a F a ap pa F a F a ap pa

F F a F F

2 2 2' ' 13 23( ) ( ) ( )o cl F a F a ap= + +

13 23 0F F= =

So finally we have ⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

1 0 00 1 00 0 p

F

p

11

The tetrahedron in the figure undergoes a uniform deformation in which:

y

z

x

a

aa

O

A

B

C1.  Points O,A and B do not move.

2. The volume of the solid becomes p times its initial volume, being p>0.

3. The segment becomes p times its initial

length, being p>0.

OC! "!!

Under these conditions the following statement/s concerning the deformation gradient tensor is/are true:

a c

b d

11 22 13 231, 0F F F F= = ≠ =

11 22 13 231, 0F F F F= = = =

331Fp

=

33F p=

Unit 2 - Solution

Unit 3 – Course book problem

12

The solid below is subjected to a uniform deformation such that points A, B and C do not move. Assuming infinitesimal strain: a. Determine the displacement field as a function of “generic” strains and rotations.

b. Identify the null components of the strain tensor and express the rotation vector in terms of the strains.

y

z

x

a

a

aA

B

C

D

E

F

θ

Unit 3 – Course book problem

13

In addition, it is known that: 1. The segment becomes (1+p) its initial length. 2. The volume of the solid becomes (1+q) its initial value. 3. The angle θ increments its value in (r) radians.

Under these conditions: c. Obtain the strain tensor, the rotation vector and the displacement field as a function of p, q and r. NOTE: The values of p, q and r are small and the infinitesimal 2nd order terms can be neglected.

AE! "!!

Unit 3 - Solution

a. Uniform deformation: =( , ) ( )t tF x F

= ⋅ → = ⋅ → = ⋅∫ ∫ ∫ ∫d d d d d dx F X x F X x F X

From the definition of the deformation gradient tensor: = ⋅d dx F X

= +· ( )tx F X φ

For infinitesimal deformations: and = + +F 1 ε Ω = +x X u

= + ⎫⎬= + + ⎭

· ( )tx F X φF 1 ε Ω ( ) ( ) ( ) ( )→ = + + ⋅ + → = + + ⋅ +t tx 1 ε Ω X φ x X ε Ω X φ

( )= + ⎫

⎬= + + ⋅ + ⎭

x X ux X ε Ω X φ

( ) ( )= + ⋅ + tu ε Ω X φ

where and

ε ε εε ε εε ε ε

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

xx xy xz

xy yy yz

xz yz zz

ε⎛ ⎞Ω −Ω⎜ ⎟= −Ω Ω⎜ ⎟⎜ ⎟Ω −Ω⎝ ⎠

00

0

xy zx

xy yz

zx yz

Ω

Unit 3 - Solution

dsdS

λ =

λ = + ˆ ˆ1 2 · ·T ET

b. The stretch is defined as:

And for infinitesimal deformations its expression is

where 11

x xx

y yy

λ ελ ε

= +⎧⎨ = +⎩

Points A and B do not move:

1dsdS

=

is parallel to the x axis: AB! "!!

1 1x xxλ ε= + = 0ε =xx

For the segment joining A and B:

Unit 3 - Solution

Points A and C do not move:

1dsdS

=

is parallel to the y axis: AC! "!!

1 1y yyλ ε= + = 0ε =yy

For the segment joining A and C:

Points A, B and C do not move:

is parallel to the x axis: AB! "!!

is parallel to the y axis: AC! "!!

The angle 𝐴𝐵𝐶  does not change

2 0xy xyθ εΔ = = 0xyε =

Unit 3 - Solution

So up to now we have: and

εε

ε ε ε

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

0 00 0

xz

yz

xz yz zz

ε⎛ ⎞Ω −Ω⎜ ⎟= −Ω Ω⎜ ⎟⎜ ⎟Ω −Ω⎝ ⎠

00

0

xy zx

xy yz

zx yz

Ω

The infinitesimal strain tensor is a symmetric tensor.

Point A does not move: ==( ) 0

Ax xu x

εε

ε ε

⎛ ⎞Ω −Ω⎡ ⎤ ⎡ ⎤⎜ ⎟⎢ ⎥ ⎢ ⎥= −Ω +Ω +⎜ ⎟⎢ ⎥ ⎢ ⎥⎜ ⎟⎢ ⎥ ⎢ ⎥+Ω −Ω⎣ ⎦ ⎣ ⎦⎝ ⎠

0 0 00 0 0 ( )0 0 0

xy xz zx

xy yz yz

xz zx yz yz

tφ =( ) 0tφ

Point B does not move: ==( ) 0

Bx xu x

0 00 0 00 0 0

xy xz zx

xy yz yz

xz zx yz yz

aεε

ε ε

⎛ ⎞Ω −Ω⎡ ⎤ ⎡ ⎤⎜ ⎟⎢ ⎥ ⎢ ⎥= −Ω +Ω⎜ ⎟⎢ ⎥ ⎢ ⎥⎜ ⎟⎢ ⎥ ⎢ ⎥+Ω −Ω⎣ ⎦ ⎣ ⎦⎝ ⎠

0 = 00 = −Ωxya

0 = (εxz +Ωzx )a

⎧

⎨⎪

⎩⎪

0xyΩ =

ε xz = −Ωzx

Unit 3 - Solution

Point C does not move: ==( ) 0

Cx xu x

0 0 0 00 0 00 0 0

xz zx

yz yz

xz zx yz yz

aεε

ε ε

⎛ ⎞−Ω⎡ ⎤ ⎡ ⎤⎜ ⎟⎢ ⎥ ⎢ ⎥= +Ω⎜ ⎟⎢ ⎥ ⎢ ⎥⎜ ⎟⎢ ⎥ ⎢ ⎥+Ω −Ω⎣ ⎦ ⎣ ⎦⎝ ⎠

0 00 00 ( )yz yz yz yzaε ε

⎧ =⎪ =⎨⎪ = −Ω → = Ω⎩

Finally, we have:

εε

ε ε ε

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

0 00 0

xz

yz

xz yz zz

ε and ⎛ ⎞Ω −Ω⎜ ⎟= −Ω Ω⎜ ⎟⎜ ⎟Ω −Ω⎝ ⎠

00

0

xy zx

xy yz

zx yz

Ωεε−⎛ ⎞

⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠0

yz

xzθ

Ω⎡ ⎤⎢ ⎥= Ω⎢ ⎥⎢ ⎥Ω⎣ ⎦

23

31

12

θ

Unit 3 - Solution

c. Segment becomes (1+p) its initial length:

1 1AEds p pdS

λ= + → = +For the segment joining A and E:

and ( )ε

λ λ ε ε εε ε ε

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟= + → = + = + +⎜ ⎟⎜ ⎟

⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

0 0 11 1ˆ ˆ 1 1 0 1 0 0 0 1 12 2

1

xz

AE yz xz zz

xz yz zz

t·ε·t

1 11 12 2xz zz xz zzp pε ε ε ε+ = + + → = +

11 12

AE

AE xz zz

pλ

λ ε ε

= + ⎫⎪⎬

= + + ⎪⎭

AE! "!!

Unit 3 - Solution

Volume of the solid becomes (1+q) its initial value:

(1 )

1 ( ) 1

o

zzo

dV qdVdV trdV

ε ε

⎫= + ⎪⎪⎬⎪= + = +⎪⎭

1 1zz qε+ = + zz qε =

1 12 2xz zz xzp p qε ε ε= + → = + 1

2xz p qε = −

11 22 33( )Tr ε ε ε ε= + +

Angle θ increments its value in (r) radians:

( )(1) (2)

10 0 12ˆ ˆ2 · · 2 1 2 1 20 0 1 0 0 1sin 22 2 2sin 1 02

2

yz yz

yz

p qt t p q

p q q

εθ ε εθ

ε

⎛ ⎞−⎜ ⎟ −⎛ ⎞⎜ ⎟⎜ ⎟Δ = − = − = − −⎜ ⎟⎜ ⎟Π⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠−⎜ ⎟⎝ ⎠⎝ ⎠

rθΔ =

Unit 3 - Solution

2 1 222 2 yzr p q ε= − − 1 2

2 2yz p q rε = − −

So finally we have:

⎛ ⎞−⎜ ⎟⎜ ⎟⎜ ⎟= − −⎜ ⎟⎜ ⎟⎜ ⎟− − −⎜ ⎟⎝ ⎠

10 02

1 20 02 2

1 1 22 2 2

p q

p q r

p q p q r q

ε

⎛ ⎞+ −⎜ ⎟

⎜ ⎟⎜ ⎟= −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

1 22 2

120

q r p

p qθ

−⎛ ⎞⎜ ⎟

= − −⎜ ⎟⎜ ⎟⎝ ⎠

(2 )

(2 2 )

p q Z

p q r ZqZ

u

Unit 4

The differential element in the figure below is stretching in the x direction a total amount . Draw the graphics

Where and represent the component of the material, spatial and infinitesimal strain tensors, respectively. Mark the significant values at the graphs.

( )0tdS tδ = < <∞,xx xxE t e t− − xxε xx

, ,ε− − −xx xx xxE t e t t

Unit 4 - Solution

The stretch in the x direction is defined as:

And it can be expressed as:

xdsdS

λ =

11 2 11 2x XX xx

xx

Ee

λ ε= + = = +−

( )0tdS tδ = < <∞The figure is stretching in the x direction:

dsdS

= (1+ t)dSdS

= 1+ t 1x tλ = +

( )2 21 2 1 1 2 1 1 2 1 2XX xx xxE t E t E t t+ = + → + = + → + = + +2

2xxtE t= +

xxE

t

22 2xxE t t→ = +

Exx t=0 = 0dExxdt

t=0

= 1

dExx

dt= t +1

Unit 4 - Solution

( ) ( )21 1 1 1 1 21 2 xx

xx

t t ee

= + → = + −−

21 112 (1 )xxe t⎛ ⎞= −⎜ ⎟+⎝ ⎠ ( )2

1 12 2 1xxe

t= −

+

xxe

texx t=0 = 0

dexxdt

t=0

= 1

1 1xx tε+ = + xx tε =

εxx t=0 = 0dεxxdt

t=0

= 1

t

xxε

( )31

1=

+xxdedt t

Unit 4 - Solution

xxE xxε

xxe12

t

Unit 5

The sphere with radius R depicted in the figure below undergoes a uniform deformation such that points A, B and C are moved to positions A’, B’, C’ and point O remains unchanged. 1. Obtain the deformation gradient tensor as a

function of p and q

2. Obtain the equation of the deformed exterior surface of the sphere indicating the type of resulting surface and the corresponding sketch

3. Obtain the material and spatial strain tensors

and the expression of p as a function of q assuming an incompressible material.

4. Repeat question (3) now assuming infinitesimal

deformations. Prove that when p and q are small, the results obtain in questions (3) and (4) coincide.

y

z

x

'

'

'

AA p

BB q

CC q

=

=

=

00

pq>>

B

C

O

'C

'A

'B

A

Unit 5 - Solution

1. Uniform deformation: = ( )tF F

Since : = +F 1 J = ( )tJ J where J is the displacement gradient tensor

∂= → = → = → =∂ ∫ ∫ ∫ ∫d d d d d dUJ U J X U J X U J XX

= +( , ) · ( )t tU X J X φ

Point O does not move: ==( , ) 0t

OX XU X

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟= +⎜ ⎟⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

0 00 0 ( )0 0

xx xy xz

yx yy yz

zx zy zz

J J JJ J J tJ J J

φ =( ) 0tφ

Unit 5 - Solution

Point A deformation:

0 0 00 0 0

xx xy xz xx

yx yy yz yx

zx zy zz zx

p J J J R p J RJ J J J RJ J J J R

⎛ ⎞ =⎛ ⎞ ⎛ ⎞ ⎧⎜ ⎟ ⎪⎜ ⎟ ⎜ ⎟= → =⎨⎜ ⎟⎜ ⎟ ⎜ ⎟

⎪⎜ ⎟ ⎜ ⎟⎜ ⎟ =⎝ ⎠ ⎝ ⎠ ⎩⎝ ⎠

00

xx

yx

zx

pJR

J

J

⎧ =⎪⎪

=⎨⎪ =⎪⎩

Point B deformation:

0 0 00

0 0 0 0

xy xzxy

yy yz yy

zy zz zy

p J J J RRq J J R q J R

J J J R

⎛ ⎞⎜ ⎟ ⎧ =⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎪⎜ ⎟ ⎜ ⎟− = → − =⎨⎜ ⎟⎜ ⎟ ⎜ ⎟

⎪⎜ ⎟ ⎜ ⎟⎜ ⎟ =⎝ ⎠ ⎝ ⎠ ⎩⎜ ⎟⎝ ⎠

0

0

xy

yy

zy

J

qJR

J

=⎧⎪⎪ = −⎨⎪

=⎪⎩

Unit 5 - Solution

00 0 00 0 0 0

0 0

xz

xz

yz yz

zzzz

p JR J R

q J J RR

q R q J RJ

⎛ ⎞⎜ ⎟

=⎛ ⎞ ⎛ ⎞ ⎧⎜ ⎟⎪⎜ ⎟ ⎜ ⎟⎜ ⎟= − → =⎨⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎪⎜ ⎟ ⎜ ⎟− − =⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎩

⎜ ⎟⎜ ⎟⎝ ⎠

Point C deformation:

00

xz

yz

zz

JJ

qJR

⎧⎪ =⎪

=⎨⎪⎪ = −⎩

Since 1F J= +

and

0 0

0 0

0 0

pR

qJR

qR

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟= −⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟⎝ ⎠

1 0 0

0 1 0

0 0 1

pR

qFR

qR

⎛ ⎞+⎜ ⎟⎜ ⎟⎜ ⎟= −⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟⎝ ⎠

Unit 5 - Solution

2. The equation of the original non deformed sphere is 2 2 2 2+ + =X Y Z R

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟= → = − = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠

⎜ ⎟ ⎜ ⎟− −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

0 0

· 0 0

0 0

p p XR RX

q qY YR R

Zq q ZR R

U J X UIt is known:

⎛ ⎞⎛ ⎞⎛ ⎞ +⎜⎜ ⎟ ⎟⎜ ⎟ ⎝ ⎠⎜ ⎟⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎛ ⎞⎜ ⎟ ⎜ ⎟= + → = + − = −⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎜ ⎟⎛ ⎞⎜ ⎟−⎜ ⎟ −⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

1

1

1

pp XX RRXq qY Y YR R

X q qZ ZR R

x X U xAnd since:

1

1

1

px XRqy YRqz ZR

⎧ ⎛ ⎞= +⎜ ⎟⎪ ⎝ ⎠⎪⎪ ⎛ ⎞= −⎨ ⎜ ⎟⎝ ⎠⎪⎪ ⎛ ⎞= −⎪ ⎜ ⎟⎝ ⎠⎩

Unit 5 - Solution

Once the equations of motion are obtained, inverse equations can be also obtained.

1

1

1

px XRqy YRqz ZR

⎫⎛ ⎞= +⎜ ⎟ ⎪⎝ ⎠ ⎪⎪⎛ ⎞= − ⎬⎜ ⎟⎝ ⎠ ⎪⎪⎛ ⎞= − ⎪⎜ ⎟⎝ ⎠ ⎭

1

1

1

xX pRyY qRzZ qR

⎫⎪

= ⎪⎪+⎪⎪

= ⎬⎪−⎪⎪

= ⎪⎪−⎭

And now the inverse equations of motion can be substituted within the original equation of the sphere:

2 2 22

2 2 2

1 1 1

x y z Rp q qR R R

+ + =⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

( ) ( ) ( )2 2 2

2 2 2 1x y zR p R q R q

+ + =+ − −

y

z

x

Resulting into the equation of a revolution ellipsoid of semi-axis R+p, R-q and R-q

Unit 5 - Solution

c. The material strain tensor can be obtained as:

( )= −12

TE F ·F 1

For the spatial strain tensor:

( )= −12

T -1e 1 F ·F

2

2

2

11 0 01

1 10 1 02

1

10 0 11

pR

eqR

qR

⎛ ⎞⎜ ⎟⎜ ⎟−⎜ ⎟⎛ ⎞+⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟⎜ ⎟

= −⎜ ⎟⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟

⎜ ⎟−⎜ ⎟⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

2

2

2

1 1 0 0

1 0 1 1 02

0 0 1 1

pR

qER

qR

⎛ ⎞⎛ ⎞+ −⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟⎜ ⎟⎛ ⎞⎜ ⎟= − −⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟

⎛ ⎞⎜ ⎟− −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Unit 5 - Solution

If the material is incompressible: e = F −1= 0

1F =

1F =2

1 1p qFR R

⎛ ⎞⎛ ⎞= + −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

2

211 1 1 1

1

p q pR R R q

R

⎛ ⎞⎛ ⎞+ − = → + =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎛ ⎞−⎜ ⎟⎝ ⎠

21 1

1p R

qR

⎡ ⎤⎢ ⎥⎢ ⎥= −⎢ ⎥⎛ ⎞−⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

d. The infinitesimal strain tensor is defined as: ( )12

TJ Jε = +

0 0

0 0

0 0

pR

qR

qR

ε

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟= −⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟⎝ ⎠

Unit 5 - Solution

If the material is incompressible and considering infinitesimal strains:

=( ) 0Tr ε

= − 2( ) p qTrR R

ε

2 0p qR R

− = 2p q=

When p and q are small, for the results obtained in the previous section:

2 2

21 11 1 11 1 1 22 22 1 11 1

⎡ ⎤ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟⎢ ⎥ − +⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥= − = − = − = =⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟− −− − +⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠

qRp R R R R qq qq q q

R RR R r

2p q=

If p and q are small: and 2 0p ; 2 0q ;

e = F −1= Tr(ε) = 0

2 22 2 22 1 1 1 1 2xx xx xxp p p p p pE E ER R R R R R

⎛ ⎞ ⎛ ⎞= + − = + + − = → = → =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2exx = 1− 1

1+ pR

⎛⎝⎜

⎞⎠⎟

2 = 1− 1

1+ 2pR

+ pR

⎛⎝⎜

⎞⎠⎟

2 =1+ 2p

R+ p

R⎛⎝⎜

⎞⎠⎟

2

−1

1+ 2pR

=

2pR

+ pR

⎛⎝⎜

⎞⎠⎟

2

1+ 2pR

= 2pR

→ 2exx =2pR

→ exx =pR

xx xx xxpE eR

ε= = =

2 22 22 2 1 1 1 1

22 2

yy zz

yy zz yy zz

q q q qE ER R R Rq qE E E ER R

⎛ ⎞ ⎛ ⎞= = − − = − + − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

→ = = − → = = −

2eyy = 2ezz = 1− 1

1− qR

⎛⎝⎜

⎞⎠⎟

2 = 1− 1

1− 2qR

+ qR

⎛⎝⎜

⎞⎠⎟

2 =− 2q

R+ q

R⎛⎝⎜

⎞⎠⎟

2

1= − 2q

R

→ 2eyy = 2ezz = − 2qR

→ eyy = ezz = − qR

yy yy yyqE eR

ε= = = −

zz zz zzqE eR

ε= = = −

Unit 5 - Solution

Unit 6

A bar (considered as an unidimensional body) undergoes a uniform deformation characterized by : where Obtain the strain rate tensor and the equations of motion:

, 1, 1atx y zeλ λ λ= = = .a const=

x

ateλ =

Unit 6 - Solution

Because the bar suffers a stretch in the x-direction, the following equation holds: Which once integrated, leads to the following result:

λ

λ

λ

⎧ = = = → =⎪⎪⎪ = = = → =⎨⎪⎪ = = = → =⎪⎩

1

1

at atx

y

z

ds dx e dx e dXdS dXds dy dy dYdS dYds dz dz dZdS dZ

1

2

3

at at atdx e dX dx e dX x e X C

dy dY dy dY y Y C

dz dZ dz dZ z Z C

⎧ = → = → = +⎪⎪ = → = → = +⎨⎪

= → = → = +⎪⎩

∫ ∫∫ ∫∫ ∫

1

2

3

atx e X Cy Y Cz Z C

⎧ = +⎪ = +⎨⎪ = +⎩

Unit 6 - Solution

Considering t = 0 as the reference time, at t = 0 is imposed: The material description of velocity can be obtained through derivation of the equations of motion:

=x X

( ,0) =x X X 1 2 3 0C C C= = = ( ), Tatt e X Y Z⎡ ⎤= ⎣ ⎦x X

( ) ( , ), ttt

∂= →∂x XV X 0

0

atx

y

z

xV ae XtyVtzVt

∂⎧ = =⎪ ∂⎪∂⎪ = =⎨ ∂⎪∂⎪ = =⎪ ∂⎩

( , ) 0 0Tatt ae X⎡ ⎤= ⎣ ⎦V X

Unit 6 - Solution

atx e Xy Yz Z

⎧ =⎪ =⎨⎪ =⎩

atX e xY yZ z

−⎧ =⎪ =⎨⎪ =⎩

From the equations of motion, inverse equations of motion can be obtained: The spatial description of velocity can be obtained by replacing the inverse equations of motion within the material description of velocity:

( ) ( )( ), , ,t t t=v x V X x v x,t( ) = ax 0 0⎡

⎣⎤⎦

T

Unit 6 - Solution

The strain rate tensor is defined, component by component, as: Therefore, the its only non-zero component is:

⎛ ⎞∂∂= +⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠

12

jiij

j i

vvdx x

{ }, 1,2,3i j ∈

111

1

vd ax

∂= =∂

0 00 0 00 0 0

a⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

d

Unit 7

A uniform deformation takes place, at a certain time t*, in the tetrahedron shown in the figure, with the following consequences:

1.  Points O, A and B do not move.

2.  The volume of the solid becomes p times its initial volume.

3.  The length of the segment becomes its initial length.

4.  The final angles has a value of

Then,

a)  Justify why the infinitesimal strain theory cannot be used.

b)  Determine the deformation gradient tensor, the possible

values of p and the displacement field in its material

and spatial descriptions.

d) Draw the deformed solid

a

aa

O

A

B

C

AC! "!!

/ 2p

AOC! 45°

y

z

x

Unit 7 - Solution

a)  The angle changes from to so, obviously this is not a small

deformation. In fact, the small strain theory implies that , and, in

this problem,

b)  Uniform deformation:

Point O does not move:

AOC! 90° 45°

1ΔΦ <<

0,78544ΠΔΦ = ≈

( , ) ( )t t=F x F

x X,t( ) = F�X + ϕ(t)

11 12 13

21 22 23

31 32 33

0 00 0 ( )0 0

F F FF F F tF F F

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

φ

d d d d d d= → = → =∫ ∫ ∫ ∫x F X x F X x F X

From the definition of the deformation gradient tensor: d d=x F X

ϕ(t) = 0

Unit 7 - Solution

Point A does not move:

11 12 13

21 22 23

31 32 33

0 00 0

a F F F aF F FF F F

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

11

21

31

100

FFF

===

Point B does not move:

12 13

22 23

32 33

0 1 00

0 0 0

F Fa F F a

F F

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

12

22

32

010

FFF

===

So up to now we have

13

23

33

1 00 10 0

FFF

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

F

Unit 7 - Solution

Volume of the solid becomes p times its initial volume:

The variation of volume can be expressed as: t odV dV= F

t o t o t odV dV dV dV dV dV= → = → =∫ ∫ ∫ ∫F F F Vt = FVo

33 0t o tV V V F dV= → =F

t oV pV=33F p=

t oV pV=

Segment becomes times its initial length:

Unit 7 - Solution

AC! "!!

'13 13

'23 23

'

1 0 00 1 00 0

c

c

c

x F F ay F F az p a pa

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦

Since point A does not move, point C after deformation can be calculate as:

' 'o cl pa=

( )22 2 2 2 2 2' ' 13 23 13 23

2 2 2 2 213 23 13 23

( ) ( ) ( ) ( ) ( ) ( )

( ) 0 ( ) 0o cl F a F a ap pa F a F a ap pa

F F a F F

= + + = → + + =

→ + = → + =

2 2 2' ' 13 23( ) ( ) ( )o cl F a F a ap= + +

13 23 0F F= =

So finally we have 1 0 00 1 00 0 p

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

F

2p

A O

C

a

a2a

Unit 7 - Solution

The angle becomes : ∑AOC 45°

∑( ) ( )(1) (2)

(1) (2)

2 ·cos ' ' ' cos 452 ·

d dA O Cd d

= ° = = x xx x

· · cosα=uv u v

Since:

(1)

(2)

00

00

dSd

ddS

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

X

X

(1) (1) (1) (1)

(2) (2) (2) (2)

1 0 1 1 1· 0 1 0 0 0

0 0 0 0

1 0 1 0 1· 0 1 0 0 0

0 0 1

d F d dS dSp

d F d dS dSp p

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟= = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟= = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

x X

x X

Then: and (1) (1)d dS=x (2) 2 (2)1d p dS= +x

Unit 7 - Solution

(1) (2)

(1) (2) 2

· 1· 1

d dd d p

=+

x xx x

(1) (2)

(1) (2)

· 22·

dx dxdx dx

=

2 2

2

2

1 2 1 2 1 221

1 1

p pp

p p

= → + = → + =+

→ = → = ±

However since is a condition of the deformation tensor. The only

possible solution is

0p= >F

1p =

And therefore, the deformation gradient tensor is defined as:

1 0 00 1 00 0 p

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

F

Unit 7 - Solution

The equations of motion can be obtained as follows: And the displacement field yields: Replacing in the previous expression the equations of motion we get the spatial description of the displacement field.

( )*

1 0 1, 0 1 0

0 0 1

x X X Zt y Y Y

z Z Z

+⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= → = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x X FX

( ) ( )* *, , 00

X Z X Zt t Y Y

Z Z

+⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − = − =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

U X x X X

( )*, 00

zt

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

u x

Unit 7 - Solution

aa

O

'A A=

'B B=

C

y

z

x

'C

The deformed tetrahedron is then: