Name:
LE TAN LOC
Nim: 23611701
HOMEWORK
Problem 1
Given x1 = X1 + αX 2 , x 2 = X 2 , x 3 = X 3 , α = α ( t )
a.
The deformation gradient F and the deformation tensor C
 ∂x1

∂X1
∂ x  ∂x 2
F ( x ) = ∇f =
=
∂X ∂X1
 ∂x
 3
∂X1
€
∂x1
∂X 2
∂x 2
∂X 2
∂x 3
∂X 2
∂x1 

∂X 3  1 α 0
∂x 2  

= 0 1 0
∂X 3 
∂x 3  0 0 1

∂X 3 
1 0 01 α 0  1
α
0


 

2
C = F .F = α 1 00 1 0 = α α + 1 0
0 0 10 0 1  0
0
1
T
€
€
b.
The displacement gradient H and the strain tensor E
x = x ( X,t )
Displacement
 x1 − X1  αX 2  α

 
  
u = x − X = x 2 − X 2  =  0  =  0 .X 2
x 3 − X 3   0   0 
Strain tensor
€
0
α
2

2
1 T
1
α
α

E = ( F F −1) = [C − I ] = 2
2
2
2

0
0

€
c.
X = X ( x,t )
0

0

0

Assume a small deformation and determine the strain tensor ε and the rotation
tensor R
€
For small deformation reduce to the infinitesimal strain tensor:
€
0 α

2 0

1
ε = ( F + F T − 2I ) = α 2 0 0
2
0
0 0


F = R.U
where R represents a rigid rotation of material point
where U is symmetric tensors
x = F.X = R.(U.X )
€
T
C = F T .F = ( R.U ) ( R.U ) = U T .RT .R.U = U T .U
€
1
α
0


2
We have: ⇒ C = U = α α + 1 0
0
0
1
1


2
1
α
0
 1
1
⇒ U = α 2 (α 2 + 1) 2 0
 0
0
1


€
2
€
Hence,
d.
€
1
3 
  2
2
−

α
+
1
−α 2
(
)
 


1 

 α − (α 2 + 1) 2 

 
1

α 2
R =
1
 α − (α 2 + 1) 2 

 
0




Set α = 2
The right stretch tensor U:
€
1


2
1
α
0  1
 1
1

U = α 2 (α 2 + 1) 2 0 =  2
 0
0
1  0


The rotation tensor R:
€
€


0

1 

α − (α 2 + 1) 2  

 

1
α + (α 2 + 1) 2 0


0
1



1 

−α − α 2 


−2.51 2.48 0


R = −5.99 4.24 0
 0
0
1
2
5
0
0

0
1
Problem 2
A homogeneous deformation is given by:
x1 =
a.
€
2
4
4
X1 − 2X 2 + 2X 3 , x 2 = X1 + X 2 , x 3 = − X1 + X 3
3
3
3
The deformation gradient F and the deformation tensor C
 ∂x1

∂X1
∂x
F = 2
∂X1
 ∂x
 3
∂X1
∂x1   2

 
−2 2
∂X 3 
3

∂x 2   4
=
1 0
∂X 3   3

∂x 3  − 4 0 1


∂X 3   3
4 0
0


C = F T .F = 0 5 −4
0 −4 5 
€
€
∂x1
∂X 2
∂x 2
∂X 2
∂x 3
∂X 2
b.
The principle value λ2k and the principle directions xˆ k of C
The state of strain at a point:
1.5 0 0 
1 € 

E = [C −1] =  0
2 −2
2
 0 −2 2 
€
E − λI = 0
[
2
]
Setting ⇒ (1.5 − λ).(2 − λ) − 0 − 4.(1.5 − λ) = 0
€
⇒ λ1 = 0 ; λ2 = 1.5 ; λ3 = 4
λ1 = 0 ; λ2 = 1.5 ; λ3 = 4 are the principal strain
€
-
The eigenvector components x i associated with ε1 = λ1 = 0 are calculated
from:
€
1.5 0 0  x1 

  €
2 −2 x 2  = 0
0
 0 −2 2  x 3 
⇒ 1.5x1 = 0 and 2x 2 − 2x 3 = 0
⇒ x1 = 0 ; x 2 = x 3
€
€
x12 + x 22 + x 32 = 1
Using the normalization
⇒ 0 2 + 2x 22 = 1 ⇒ x 2 = ±
2
2

2 2
Hence, the principal direction associated with principal strain ε1 = 0 is xˆ 1 = ± 0,
, 
 2 2 
€
- The eigenvector components x i associated with ε2 = λ2 = 1.5 are calculated
from:
€
0 0
0  x1 

  €
0 0.5 −2  x 2  = 0
0 −2 0.5 x 3 
⇒ x2 = x3 = 0
€
€
x12 + x 22 + x 32 = 1
Using the normalization
⇒ x1 = 1
€
Hence, the principal direction associated with principal strain ε1 = 1.5 is xˆ 2 = ±(1,0,0)
-
The€eigenvector components x i associated with ε3 = λ3 = 4 are calculated
from:
€
€
2.5 0 0  x1 

  €
€
 0 −2 −2 x 2  = 0
 0 −2 −2 x 3 
⇒ 2.5x1 = 0 and 2x 2 + 2x 3 = 0
⇒ x1 = 0 ; x 2 = −x 3
x12 + x 22 + x 32 = 1
Using the normalization
2
2
⇒ 0 2 + 2x 22 = 1 ⇒ x 2 = ±
; x3 = 
€
2
2

2
2
Hence, the principal direction associated with principal strain ε3 = 4 is xˆ 3 = ± 0,
,−

2 
 2
€
c.
The stretch tensor U
-
€
€
The infinitesimal line elements dX and dX’ need to be parallel, namely:
dX'= UdX = λdX
or upon recalling that dX = adS (where M is unit vector)
€
Ua = λa
€
€
This equation represents a linear eigenvalue probem. The eigenvalues λk > 0 are called
the principal stretches and the associated eigenvectors ak are called the principal directions.
When λk are distinct, one may write:
€
Uak = λk ak
Ual = λl al
€
€
where the parentheses around the subscripts signify that the summation
convention is not in use. Upon premultiplying the preceding two equations with λl and
€
λk respectively, one gets
al .(Uak ) = λk al .ak
€
ak .(Ual ) = λk ak .al
€
Subtracting the two equations from each other leads to
(λk − λl ) ak .al = 0
€
therefore, since by assumption λk ≠ λl it follows that
€
ak .al = δkl
that the principal directions are mutually orthogonal and that {a1 a2 a3 } form an
€
orthogonal basis on E3.
€
It turns out that regardless of whether U has distinct or repeated eigenvalues, the
€
classical spectral representation theorem guarantees that there exists an orthonormal basis ak of
E3 consisting entirely of eigenvectors of U and that, if λk are the associated eigenvalues,
3
U = ∑ λk ak ⊗ ak
k=1
€
€
€
 3
 3

C = U 2 = ∑ λk ak ⊗ ak ∑ λl al ⊗ al 
 k=1
 l=1

3
3
= ∑ ∑ λk λl ( ak ⊗ ak )( al ⊗ al )
k=1 l=1
3
3
= ∑ ∑ λk λl ( ak .al )( ak ⊗ al )
k=1 l=1
3
= ∑ λ2k ak ⊗ ak
k=1
More generally,
€
3
U = ∑ λm ak ⊗ ak
m
k=1
The inverse U −1
-
3
1
ak ⊗ ak
λ
k
k=1
€
The rotation tensor R
U −1 = ∑
€
d.
The eigenvalues λ12 , λ22 and λ23 of matrix [C] are determined by setting
€
[C] − λ2 [I ] = 0
 λ12 = 1  λ1 = 1
 €

⇒  λ22 = 4 ⇒  λ2 = 2
 λ2 = 9  λ = 3
 3
 3
The eigenvectors are (in vector component form)



(1) 
a =



€






1
0 
0 

 
 1 
1 
 ; a( 2) = 0 ; a( 3) = 

2
0
 2 
 
1 
− 1 

 2 
2
Hence, the stretch tensor can be written as
U = λ1aˆ (1) aˆ (1) + λ2 aˆ ( 2) aˆ ( 2) + λ3 aˆ ( 3) aˆ ( 3)
€
= λ1 a1(1)eˆ1 + a(21)eˆ2 + a(31)eˆ3 a1(1)eˆ1 + a(21)eˆ2 + a(31)eˆ3
(
)(
)
(
)(
)
+ λ ( a( )eˆ + a( )eˆ + a( )eˆ )( a( )eˆ + a( )eˆ + a( )eˆ )



=  λ [ a( ) ] + λ [ a( ) ] + λ [ a( ) ] eˆ eˆ +  λ [ a( ) ] + λ [ a( ) ]



+ ( λ a( ) a( ) + λ a( ) a( ) + λ a( ) a( ) )(eˆ eˆ + eˆ eˆ )
+ ( λ a( ) a( ) + λ a( ) a( ) + λ a( ) a( ) )(eˆ eˆ + eˆ eˆ )
+ ( λ a( ) a( ) + λ a( ) a( ) + λ a( ) a( ) )(eˆ eˆ + eˆ eˆ )
+ λ2 a1( 2)eˆ1 + a(22)eˆ2 + a(32)eˆ3 a1( 2)eˆ1 + a(22)eˆ2 + a(32)eˆ3
3
3
1
1
1
1
1
3
2
3
3
2
2
1
2
2 1
1
1 1
1
3
1
1 2
1
3
3
2
3 1
2 1
2
3
2
2 2
2
3
2
Thus, we have
€
3
3
2
2
1
2
2
3
2
1
3
1
1
1 1
1
2
2
2
3
3
1 1
3
1
1
2
3
2
1 2
2 1
3 1
3
3
1 3
3 1
3
3 2
3
3
2 3
3 2
3
3
2
2
2
2
2
2

+ λ3 a(23) eˆ2eˆ2 +  λ1 a(31)


[ ]
[ ]
2
[ ]
+ λ2 a(32)
2
2
+ λ3 a(33) eˆ3eˆ3

[ ]
[ ]
U11 = λ1 a1(1)
2
[ ]
+ λ2 a1( 2)
2
[ ]
+ λ3 a1( 3)
2
= 1× 0 2 + 2 ×12 + 3 × 0 2 = 2
[ ]
U 22 = λ1 a(21)
2
[ ]
+ λ2 a(22)
2
[ ]
+ λ3 a(23)
2
 1 2
 1 2
2
= 1×   + 2 × 0 + 3 ×   = 2
 2
 2
[ ]
U 33 = λ1 a(31)
2
[ ]
+ λ2 a(32)
2
[ ]
+ λ3 a(33)
2
 1 2
 1 2
2
= 1×   + 2 × 0 + 3 × −
 =2
 2
 2
U12 = U 21 = λ1a1(1) a(21) + λ2 a1( 2) a(22) + λ3 a1( 3) a(23)
1
1
= 1× 0 ×
+ 2 ×1× 0 + 3 × 0 ×
=0
2
2
U13 = U 31 = λ1a1(1) a(31) + λ2 a1( 2) a(32) + λ3 a1( 3) a(33)
 1 
1
= 1× 0 ×
+ 2 ×1× 0 + 3 × 0 × −
=0
 2
2
U 23 = U 32 = λ1a(21) a(31) + λ2 a(22) a(32) + λ3 a(23) a(33)
1
1
1  1 
= 1×
×
+ 2 ×0 ×0 + 3×
× −
 = −1
2
2
2  2
In matrix form
2 0 0 


U = 0 2 −1
0 −1 2 
€
Then the rotation tensor R in matrix form is given by
1
2
 3 −3
2
2
R = FU −1 = 
3
3
2
1
−
 3 3
€
e.
2
 1 −2 2
3
1 1 

 =  2 2 1
3 3
−2 1 2
2
3
The left stretch tensor V
Following an analogous procedure for the left polar decomposition, note for the
€
stretch to be pure it is necessary that
dx = Vdx'= λdx'
or recalling that dx'= RadS
€
€
VRa = λRa
The principal directions bi = Rai . Appealing to the spectral representation theorem,
one may write
€
3
3
V = ∑ λk ( Ra
€ k ) ⊗ ( Rak ) = ∑ λk bk ⊗ bk
k=1
f.
k=1
The shear strain γ 23
1.5 0 0 


E = 0
2 −2
€ 0 −2 2 
€
γ 23 ≈ 2.E 23 = −4
-
€
The volumetric strain εv
εv = det F −1 = 6 −1 = 5
€
-
€
The extremal values of longitudinal strain is
€
ε1 = 0 ; ε2 = 1.5 ; ε3 = 4
Problem 3
€
a.
The following coordinate transformation equation is used to convert the longitudinal
strain from each strain gage into strain expressed in the x-y coordinates,
εx' =
εx + εy εx − εy
+
cos2θ + εxy sin2θ
2
2
This equation can be used to compute the principal strains and the principal axis orientation
€
directly from the rectangular rosette gage readings. Note that there are many different possible
gage numbering arrangements besides the particular A,B,C layout here, and they can lead to
forms for the final results shown above but with A, B and C interchanged.)
εx + εy εx − εy
+
cos2θ + εxy sin2θ
2
2
ε + ε ε −ε
εII = x y + x y cos2(θ + 45) + εxy sin2(θ + 45)
2
2
ε + ε ε −ε
εIII = x y + x y cos2(θ + 90) + εxy sin2(θ + 90)
2
2
εI =

εx + εy εx − εy
+
cos2θ + εxy sin2θ
εI =
2
2


ε + ε ε −ε
⇒ εII = x y − x y sin2θ + εxy cos2θ
2
2

εx + εy εx − εy

εIII = 2 − 2 cos2θ − εxy sin2θ

€
where
€
cos( A + B) = cos Acos B − sin Asin B
sin(A + B) = sin Acos B + cos Asin B
Hence,

εI + εIII
ε −ε
[1+ sin2θ ] + I III cos2θ − εII sin2θ
€ εx =
2
2

εI + εIII
εI − εIII
cos2θ + εII sin2θ
ε y =
[1− sin2θ ] −
2
2

γ xy = [εI − εIII ] sin2θ − [εI + εIII ] cos2θ + 2εII cos2θ

where γ xy = 2εxy
b.
€
Follow the figure, the value of θ = 0 , thus:
€

εx = εI
€

εy = εIII

γ = 2ε − ε − ε
 xy
II
I
III
The basic Mohr’s Circle and it always has its center on the abscissa at a point given by the
€
€
value, (εx + εy ) /2 . The circle diameter is easily computed as:
The strain average:
ε=
εI + εIII
2
D = γ xy2 + (εx − εy )
€
=
(2εII − εI − εIII )
2
2
+ (εI − εIII )
2
2
= 2εI2 + 4εII2 + 2εIII
− 4εI εII − 4εII εIII + 4εI εIII
2
= 2(εI + εIII ) + 4εII (εII − εI − εIII )
Hence, the principal strain is:
ε1
D ε + εIII 1
2
=ε ± = I
±
2(εI + εIII ) + 4εII (εII − εI − εIII )
ε2
2
2
2
€
⇒
ε1 εI + εIII
1
2
=
±
(εI + εIII ) + εII (εII − εI − εIII )
ε2
2
2
and γ max =
D
1
2
=
(εI + εIII ) + εII (εII − εI − εIII )
2
2
From Mohr’s cycle, the value of θ is:
€
 2ε − ε − ε 
θ = 0.5arctan II I III 
− εIII 
 εI €
The angle θ1 is:
€
€
€
 2(εI − εIII ) 
θ1 = arctan

 2εII − εI − εIII 
Problem 4
The state of stress in a material element with respect to the coordinate system Ox is,
 200 −100 −100


T = −100 100
0 
−100
0
100 
a.
Unit normal
nˆ = [1,1,0] / 2 =
€
1
(eˆ1 + eˆ2 )
2
The components of the stress vector are
 t1 
 200 −100 −100
1
 100 
 

1   1 


0 
 t 2  = T.nˆ = −100 100
1 =
 0 
2 
2
t 

0
100 
 3
−100
0
−100
€
or t ( nˆ ) =
€
1
(100eˆ1 −100eˆ2 )
2
The normal component t nn of the stress vector t on the plane is given by
€
€
b.
€
1
t nn = t ( nˆ ).nˆ = .100 = 50
2
The tangential€component is given by
t ns =
2
2
t − t nn
=
1
(1002 + 1002 ) − 502 = 50 3 = 86.6
2
The stress matrix T˜ with respect to a coordinate system Ox˜ when the
transformation matrix is

€
1
Q = (e˜i .ek ) = 0

0

0
3
−1
2
2
The stress transformation T˜ is:
€
€

0 
1 
2 
3 
2
€

1
T
T˜ = Q.T.Q = 0

0



0
0   200 −100 −100 1

3
1 .−100 100
0 .0
2
2 

0
100  0
3  −100
−1

2
2
 200
−136.6 −36.6


= −136.6
100
0 
 −36.6
0
100 
T
0
3
−1
2
2

0 
1 
2 
3 
2
€
Problem 5
The state of stress in a material element on a free surface solid is,
180 80 0


T =  80 60 0
 0
0 0
a.
The principal stress and the principal direction
Setting T − λI = 0 , we obtain
€
(180 − λ)(60 − λ)(−λ) + λ × 6400 = 0
€
⇒ λ1 = 220 ; λ2 = 20 ; λ3 = 0
Thus, the principal stress are
σ1 = λ1 = 220

 σ 2 = λ2 = 20
 σ =λ =0
 3
3
€
-
The plane associated with the maximum principal stress λ1 = 220 can be
calculated from:
€
180 − 220
80

60 − 220
 80

0
0
⇒ −n1 + 2n 2 = 0 and
0 n1 
 
0 n 2  = 0
0 n 3 
n1 − 2n 2 = 0
€
⇒ n1 = 2n 2 ; n 3 = 0
€
n12 + n 22 + n 32 = 1
Using the normalization
1
2
⇒ n2 =
; n1 =
5
5
€
Hence, the principal direction associated with the maximum principal stress
λ1 = 220 is
n (1) =
€
€
-
2
1
eˆ1 +
eˆ2
5
5
The plane associated with the maximum principal stress λ2 = 20 can be
calculated from:
180 − 20
80
0 n1 

 
80
60
−
20
0

 n 2  = 0
 0
0
0 n 3 
⇒ 2n1 + n 2 = 0 and 2n1 + n 2 = 0
1
2
⇒ n1 = −
; n2 =
; n3 = 0
5
5
€
Hence, the principal direction associated with the maximum principal stress
λ2 = 20 is
€
n ( 2) = −
€
€
-
1
2
eˆ1 +
eˆ2
5
5
The plane associated with the maximum principal stress λ3 = 0 can be
calculated from:
180 80 0 n1 

 
 80 60 0 n 2  = 0
 0
0 0 n 3 
⇒ 18n1 + 8n 2 = 0 and 8n1 + 6n 2 = 0
⇒ n1 = n 2 = 0 ; n 3 = 1
€
Hence, the principal direction associated with the maximum principal stress
λ3 = 0 is
€
n ( 3) = eˆ3
b.
€
€
The normal stress with the x3 – axis:
n ( 3) = eˆ3 ⇒ nˆ = [0 0 1]
The components of the stress vector are
€
 t1 
180 80 00 0
 

   
 t 2  = T.nˆ =  80 60 00 = 0
t 
 0
0 01 0
 3
or t ( nˆ ) = [0 0 0]
The normal component t nn of the stress vector t on the plane is given by
€
€
€
t nn = t ( nˆ ).nˆ = 0
The maximum shear stress given by:
€
1
1
(t ns ) max = (λmax − λmin ) = ( λ1 − λ3 ) = 110
2
2
The plane of the maximum shear stress is given by the vector:
€
(
)
n s = n (1) − n ( 3) =
or nˆ s =
2
1
eˆ1 +
eˆ2 − eˆ3
5
5

∇n s
1  2
1
=
eˆ1 +
eˆ2 − eˆ3 

∇n s

2 5
5
€
The plane of the maximum shear stress lies between the planes of the maximum and minimum
principal stresses (i.e., oriented at ±45◦ to both planes).
c.€
The maximum shear stress
Following part b)
1
2
1
2
(t ns ) max = (λmax − λmin ) = ( λ1 − λ3 ) = 110
€