Download - Ch.2. Group Work Units
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Ch.2. Group Work Units
Continuum Mechanics Course (MMC) - ETSECCPB - UPC
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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
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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
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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
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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=
=
= +∫
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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=
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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
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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
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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=
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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
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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
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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
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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! "!!
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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
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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:
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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 =
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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
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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
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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! "!!
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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 =
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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
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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
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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
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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
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Unit 4 - Solution
xxE xxε
xxe12
t
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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
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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
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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
=⎧⎪⎪ = −⎨⎪
=⎪⎩
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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
⎛ ⎞+⎜ ⎟⎜ ⎟⎜ ⎟= −⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟⎝ ⎠
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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
⎧ ⎛ ⎞= +⎜ ⎟⎪ ⎝ ⎠⎪⎪ ⎛ ⎞= −⎨ ⎜ ⎟⎝ ⎠⎪⎪ ⎛ ⎞= −⎪ ⎜ ⎟⎝ ⎠⎩
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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
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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
⎛ ⎞⎛ ⎞+ −⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟⎜ ⎟⎛ ⎞⎜ ⎟= − −⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟
⎛ ⎞⎜ ⎟− −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
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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
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟= −⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟⎝ ⎠
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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
![Page 35: Ch.2. Group Work Units](https://reader031.vdocuments.us/reader031/viewer/2022012408/616a2f5911a7b741a34fbc40/html5/thumbnails/35.jpg)
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
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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 =
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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
⎧ = +⎪ = +⎨⎪ = +⎩
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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
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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
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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
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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
![Page 42: Ch.2. Group Work Units](https://reader031.vdocuments.us/reader031/viewer/2022012408/616a2f5911a7b741a34fbc40/html5/thumbnails/42.jpg)
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
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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
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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=
![Page 45: Ch.2. Group Work Units](https://reader031.vdocuments.us/reader031/viewer/2022012408/616a2f5911a7b741a34fbc40/html5/thumbnails/45.jpg)
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
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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
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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
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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
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Unit 7 - Solution
aa
O
'A A=
'B B=
C
y
z
x
'C
The deformed tetrahedron is then: