Ch.3. Group Work Units

Continuum Mechanics Course (MMC) - ETSECCPB - UPC

Unit 1- Exam question

Given an infinitesimal deformation tensor , which fulfils the compatibility

equations, the corresponding displacement field is determined except for: a A rigid body motion. c Six integration constants. b A rotation and a translation. d Three integration constants

ε

Unit 1 - Solution

( ),tε xThe form of the solution of the first integration is:

θ x,t( ) = !θ x,t( ) +θ! t( )

The form of the solution of the second integration is:

u x,t( ) = !u x,t( ) + u t( )

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

Where consists in 3 Integration constants.

The train tensor determines the movement in any instant of time except

for a uniform rotation and translation .

The displacement field constructed from these rotation and translation

corresponds in fact to a rigid solid movement:

( ),tε x( )ˆ tθ ( )ˆ tu

( )ˆ tθ

( )ˆ tu ( ) ( )( ) ( )ˆ ˆˆ ˆ,t t t= +u x Ω θ x u

The integration of the strain field is performed two steps:

( )ˆ tθ

Where consists in 3 Integration constants. ( )ˆ tu

Unit 1 - Solution

To summarise, given an infinitesimal deformation tensor, the corresponding

displacement field is determined except for 6 integrations constants, 3 of them

corresponding to a rotation and the other 3 corresponding to a translation. A

rigid solid motion can be constructed, in turn, from these rotation and

translation.

Statements A, B and C are correct while statement D is not.

Unit 1- Exam question

Given an infinitesimal deformation tension , which fulfils the compatibility

equations, the corresponding displacement field is determined except for: a A rigid body motion. c Six integration constants. b A rotation and a translation. d Three integration constants

ε

Unit 2 - Exam question

Given the infinitesimal strain tensor:

And knowing that the displacement field and the infinitesimal rotation vector

nullify at point (0,0,0) for any t and , the components

of the displacement vector verify:

a c

b d

0 01 0

0 0

yy

t

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

ε

u 0,t( )( = 0 θ 0,t( ) = 0)

yu y= 2xu y=

xu y= yu y z= +

Unit 2 - Solution

1

1

1

0 0 0

0 0 0

0 0 0

xyxz

yz yy

yzzz

x y z

y y z

z y z

εθ ε

ε εθ

εθ ε

∂ ⎫∂ ∂= − = − = ⎪∂ ∂ ∂ ⎪⎪∂ ∂∂ = − = − = ⎬∂ ∂ ∂ ⎪⎪∂∂ ∂= − = − = ⎪

∂ ∂ ∂ ⎭

2

2

2

0 0 0

0 0 0

0 0 0

xx xz

xy yz

xz zz

x z x

y z x

z z x

θ ε ε

ε εθ

θ ε ε

∂ ∂ ∂ ⎫= − = − = ⎪∂ ∂ ∂ ⎪∂ ∂∂ ⎪= − = − = ⎬∂ ∂ ∂ ⎪

⎪∂ ∂ ∂= − = − = ⎪∂ ∂ ∂ ⎭

First step is to integrate the derivative of , using the following PDE system:

( )1 1C tθ =

( )2 2C tθ =

θ x,t( )

Unit 2 - Solution

( )3 3y C tθ = − +

Since for any time t the rotation vector is null at point (0,0,0): θ 0,t( ) = 0

θ =

C1 t( )C2 t( )

−y +C3 t( )

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

θ x,t( )x= 0,0,0( )

=

C1

C2

C3

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=000

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 2 3 0C C C= = =

θ =00−y

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

3

3

3

0 0 0

0 1 1

0 0 0

xy xx

yy xy

yz xz

x x y

y x y

z x y

εθ ε

ε εθ

εθ ε

∂ ⎫∂ ∂= − = − = ⎪∂ ∂ ∂ ⎪⎪∂ ∂∂ = − = − = − ⎬∂ ∂ ∂ ⎪⎪∂∂ ∂= − = − = ⎪

∂ ∂ ∂ ⎭

Unit 2 - Solution

Once is known, u is integrated also using a PDE system: θ

1

13

12

0

2

0 0 0

xx

xy

xz

uxu y y yyuz

ε

ε θ

ε θ

⎫∂ = = ⎪∂ ⎪∂ ⎪= − = + = ⎬∂ ⎪

⎪∂ = + = + = ⎪∂ ⎭

( )2 '1 1u y C t= +

23

2

21

0

1

0 0 0

xy

yy

yz

u y yxuyuz

ε θ

ε

ε θ

⎫∂ = + = − = ⎪∂ ⎪∂ ⎪= = ⎬∂ ⎪

⎪∂ = − = − = ⎪∂ ⎭

( )'2 2u y C t= +

Unit 2 - Solution

( )'3 3u zt C t= +

Since for any time t the displacement field is null at point (0,0,0): ( ),t =u 0 0

u =

y 2 +C1' t( )

y +C2' t( )

zt +C3' t( )

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

( ) ( )

'1'20,0,0'3

0, 0

0

Ct C

C=

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

xu x ' ' '

1 2 3 0C C C= = =

2yyzt

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

u

32

31

3

0 0 0

0 0 0

xz

yz

zz

uxuyu tz

ε θ

ε θ

ε

⎫∂ = − = − = ⎪∂ ⎪∂ ⎪= + = + = ⎬∂ ⎪

⎪∂ = = ⎪∂ ⎭

Unit 2 – Exam question

Given the infinitesimal strain tensor:

And knowing that the displacement field and the infinitesimal rotation vector

nullify at point (0,0,0) for any t and , the components

of the displacement vector verify:

a c

b d

0 01 0

0 0

yy

t

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

ε

( )( ,t =u 0 0 ( ) ),t =θ 0 0

yu y= 2xu y=

xu y= yu y z= +

Unit 3

Given the infinitesimal strain tensor: Obtain the displacements field and the rotation vector knowing that: and

ε =

8x − 12

y32

x2z

− 12

y x 0

32

x2z 0 x3

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

( ) ( ) [ ]0,0,0

, 3 0 0 Tt t=

=x

u x θ x,t( )

x= 0,0,0( )= 0 0 0⎡⎣

⎤⎦

T

Unit 3 - Solution

1

1

1

0 0 0

0 0 0

0 0 0

xyxz

yz yy

yzzz

x y z

y y z

z y z

εθ ε

ε εθ

εθ ε

∂ ⎫∂ ∂= − = − = ⎪∂ ∂ ∂ ⎪⎪∂ ∂∂ = − = − = ⎬∂ ∂ ∂ ⎪⎪∂∂ ∂= − = − = ⎪

∂ ∂ ∂ ⎭

( )1 1C tθ =

The derivative of can be integrated using aPDE system: ( ),tθ x

∂θ2

∂x=∂ε xx

∂z−∂ε xz

∂x= 0 − 3xz = −3xz

∂θ2

∂y=∂ε xy

∂z−∂ε yz

∂x= 0 − 0 = 0

∂θ2

∂z=∂ε xz

∂z−∂εzz

∂x= 3

2x2 − 3x2 = − 3

2x2

⎫

⎬

⎪⎪⎪

⎭

⎪⎪⎪

( )22 2

32x z C tθ = − +

Unit 3 - Solution

3

3

3

0 0 0

1 312 2

0 0 0

xy xx

yy xy

yz xz

x x y

y x y

z x y

εθ ε

ε εθ

εθ ε

∂ ⎫∂ ∂= − = − = ⎪∂ ∂ ∂ ⎪⎪∂ ∂∂ = − = + = ⎬∂ ∂ ∂ ⎪⎪∂∂ ∂= − = − = ⎪

∂ ∂ ∂ ⎭

( )3 332y C tθ = +

θ =

C1 t( )− 3

2x2z +C2 t( )

32

y +C3 t( )

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

Applying the condition:

θ x,t( )

x= 0,0,0( )= 0 0 0⎡⎣

⎤⎦

T

1 2 3 0C C C= = =

θ =

0

− 32

x2z

32

y

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

Unit 3 - Solution

Is also obtained integrating a PDE system: ( ),tu x

1

13

2 212

8

1 3 22 23 3 02 2

xx

xy

xz

u xxu y y yyu x z x zz

ε

ε θ

ε θ

⎫∂ = = ⎪∂ ⎪∂ ⎪= − = − − = − ⎬∂ ⎪

⎪∂ = + = − = ⎪∂ ⎭

( )2 2 '1 14u x y C t= − +

23

2

21

1 32 2

0 0 0

xy

yy

yz

u y y yxu xyuz

ε θ

ε

ε θ

⎫∂ = + = − + = ⎪∂ ⎪∂ ⎪= = ⎬∂ ⎪

⎪∂ = − = − = ⎪∂ ⎭

( )'2 2u xy C t= +

Unit 3 - Solution

2 2 232

31

33

3 3 32 2

0 0 0

xz

yz

zz

u x z x z x zxuyu xz

ε θ

ε θ

ε

⎫∂ = − = + = ⎪∂ ⎪∂ ⎪= + = + = ⎬∂ ⎪

⎪∂ = = ⎪∂ ⎭

( )3 '3 3u x z C t= +

( )( )( )

2 2 '1

'2

3 '3

4x y C tu xy C t

x z C t

⎡ ⎤− +⎢ ⎥= +⎢ ⎥⎢ ⎥+⎣ ⎦

Applying the condition:

( ) ( ) [ ]0,0,0

, 3 0 0 Tt t=

=x

u x

'1' '2 3

3

0

C t

C C

=

= =

2 2

3

4 3x y tu xy

x z

⎡ ⎤− +⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦