Determine whether the following expressions are valid in index notation. If valid, identify the free indices and the dummy indices. 1)
A
m
s
=
b
m
(
c
r
−
d
r
)
{\displaystyle A_{ms}=b_{m}(c_{r}-d_{r})}
Invalid. The free indices are
m
,
s
{\displaystyle m,s}
m
,
r
{\displaystyle m,r}
2)
A
m
s
=
b
m
(
c
s
−
d
s
)
{\displaystyle A_{ms}=b_{m}(c_{s}-d_{s})}
Valid. Both
m
{\displaystyle m}
s
{\displaystyle s}
3)
t
i
=
σ
j
i
n
j
{\displaystyle t_{i}=\sigma _{ji}n_{j}}
Valid. The free index is
i
{\displaystyle i}
j
{\displaystyle j}
4)
t
i
=
σ
j
i
n
i
{\displaystyle t_{i}=\sigma _{ji}n_{i}}
Invalid. The free index is
i
{\displaystyle i}
j
{\displaystyle j}
5)
x
i
x
i
=
r
3
{\displaystyle x_{i}x_{i}=r^{3}}
Valid. The dummy index is
i
{\displaystyle i}
r
3
{\displaystyle r^{3}}
6)
B
i
j
c
j
=
3
{\displaystyle B_{ij}c_{j}=3}
Invalid. There is one free index on the LHS and no free index on the RHS.
Show the following:
1)
δ
i
i
=
3
{\displaystyle \delta _{ii}=3}
δ
i
i
=
δ
11
+
δ
22
+
δ
33
=
1
+
1
+
1
=
3
⟹
δ
i
i
=
3
{\displaystyle \delta _{ii}=\delta _{11}+\delta _{22}+\delta _{33}=1+1+1=3\qquad \implies \qquad \delta _{ii}=3}
2)
e
i
j
k
e
p
q
k
=
δ
i
p
δ
j
q
−
δ
i
q
δ
j
p
{\displaystyle e_{ijk}e_{pqk}=\delta _{ip}\delta _{jq}-\delta _{iq}\delta _{jp}}
The LHS is
e
i
j
k
e
p
q
k
=
e
i
j
1
e
p
q
1
+
e
i
j
2
e
p
q
2
+
e
i
j
3
e
p
q
3
{\displaystyle e_{ijk}e_{pqk}=e_{ij1}e_{pq1}+e_{ij2}e_{pq2}+e_{ij3}e_{pq3}}
If
i
=
j
{\displaystyle i=j}
L
H
S
=
(
e
i
i
k
)
(
e
p
q
k
)
=
0
;
R
H
S
=
δ
i
p
δ
i
q
−
δ
i
q
δ
i
p
=
0
{\displaystyle LHS=({e_{iik}})~~(e_{pqk})=0~;~~RHS=\delta _{ip}\delta _{iq}-\delta _{iq}\delta _{ip}=0}
We get the same result if
p
=
q
{\displaystyle p=q}
i
≠
j
{\displaystyle i\neq j}
p
≠
q
{\displaystyle p\neq q}
Case 1:
i
=
1
{\displaystyle i=1}
j
=
2
{\displaystyle j=2}
p
=
1
{\displaystyle p=1}
q
=
2
{\displaystyle q=2}
L
H
S
=
(
e
121
e
121
)
+
(
e
122
e
122
)
+
(
e
123
e
123
)
=
1
;
R
H
S
=
(
δ
11
δ
11
)
−
(
δ
12
δ
12
)
=
1
{\displaystyle LHS=({e_{121}e_{121}})~~+({e_{122}e_{122}})~~+({e_{123}e_{123}})~~=1~;~~RHS=({\delta _{11}\delta _{11}})~~-({\delta _{12}\delta _{12}})~~=1}
Case 2:
i
=
1
{\displaystyle i=1}
j
=
2
{\displaystyle j=2}
p
=
2
{\displaystyle p=2}
q
=
1
{\displaystyle q=1}
i
=
2
{\displaystyle i=2}
j
=
1
{\displaystyle j=1}
p
=
1
{\displaystyle p=1}
q
=
2
{\displaystyle q=2}
L
H
S
=
(
e
121
e
211
)
+
(
e
122
e
212
)
+
(
e
123
e
213
)
=
−
1
;
R
H
S
=
(
δ
12
δ
21
)
−
(
δ
11
δ
22
)
=
−
1
{\displaystyle LHS=({e_{121}e_{211}})~~+({e_{122}e_{212}})~~+({e_{123}e_{213}})~~=-1~;~~RHS=({\delta _{12}\delta _{21}})~~-({\delta _{11}\delta _{22}})~~=-1}
Case 3:
i
=
2
{\displaystyle i=2}
j
=
1
{\displaystyle j=1}
p
=
2
{\displaystyle p=2}
q
=
1
{\displaystyle q=1}
L
H
S
=
(
e
211
e
211
)
+
(
e
212
e
212
)
+
(
e
213
e
213
)
=
1
;
R
H
S
=
(
δ
22
δ
11
)
−
(
δ
21
δ
12
)
=
1
{\displaystyle LHS=({e_{211}e_{211}})~~+({e_{212}e_{212}})~~+({e_{213}e_{213}})~~=1~;~~RHS=({\delta _{22}\delta _{11}})~~-({\delta _{21}\delta _{12}})~~=1}
Case 4:
i
=
2
{\displaystyle i=2}
j
=
3
{\displaystyle j=3}
p
=
2
{\displaystyle p=2}
q
=
3
{\displaystyle q=3}
L
H
S
=
(
e
231
e
231
)
+
(
e
232
e
232
)
+
(
e
233
e
233
)
=
1
;
R
H
S
=
(
δ
22
δ
22
)
−
(
δ
23
δ
23
)
=
1
{\displaystyle LHS=({e_{231}e_{231}})~~+({e_{232}e_{232}})~~+({e_{233}e_{233}})~~=1~;~~RHS=({\delta _{22}\delta _{22}})~~-({\delta _{23}\delta _{23}})~~=1}
Case 5:
i
=
2
{\displaystyle i=2}
j
=
3
{\displaystyle j=3}
p
=
3
{\displaystyle p=3}
q
=
2
{\displaystyle q=2}
i
=
3
{\displaystyle i=3}
j
=
2
{\displaystyle j=2}
p
=
2
{\displaystyle p=2}
q
=
3
{\displaystyle q=3}
L
H
S
=
(
e
231
e
321
)
+
(
e
232
e
322
)
+
(
e
233
e
323
)
=
−
1
;
R
H
S
=
(
δ
23
δ
32
)
−
(
δ
22
δ
33
)
=
−
1
{\displaystyle LHS=({e_{231}e_{321}})~~+({e_{232}e_{322}})~~+({e_{233}e_{323}})~~=-1~;~~RHS=({\delta _{23}\delta _{32}})~~-({\delta _{22}\delta _{33}})~~=-1}
Case 6:
i
=
3
{\displaystyle i=3}
j
=
2
{\displaystyle j=2}
p
=
3
{\displaystyle p=3}
q
=
2
{\displaystyle q=2}
L
H
S
=
(
e
321
e
321
)
+
(
e
322
e
322
)
+
(
e
323
e
323
)
=
1
;
R
H
S
=
(
δ
33
δ
22
)
−
(
δ
32
δ
23
)
=
1
{\displaystyle LHS=({e_{321}e_{321}})~~+({e_{322}e_{322}})~~+({e_{323}e_{323}})~~=1~;~~RHS=({\delta _{33}\delta _{22}})~~-({\delta _{32}\delta _{23}})~~=1}
Case 7:
i
=
3
{\displaystyle i=3}
j
=
1
{\displaystyle j=1}
p
=
3
{\displaystyle p=3}
q
=
1
{\displaystyle q=1}
L
H
S
=
(
e
311
e
311
)
+
(
e
312
e
312
)
+
(
e
313
e
313
)
=
1
;
R
H
S
=
(
δ
33
δ
11
)
−
(
δ
31
δ
13
)
=
1
{\displaystyle LHS=({e_{311}e_{311}})~~+({e_{312}e_{312}})~~+({e_{313}e_{313}})~~=1~;~~RHS=({\delta _{33}\delta _{11}})~~-({\delta _{31}\delta _{13}})~~=1}
Case 8:
i
=
3
{\displaystyle i=3}
j
=
1
{\displaystyle j=1}
p
=
1
{\displaystyle p=1}
q
=
3
{\displaystyle q=3}
i
=
1
{\displaystyle i=1}
j
=
3
{\displaystyle j=3}
p
=
3
{\displaystyle p=3}
q
=
1
{\displaystyle q=1}
L
H
S
=
(
e
311
e
131
)
+
(
e
312
e
132
)
+
(
e
313
e
133
)
=
−
1
;
R
H
S
=
(
δ
31
δ
13
)
−
(
δ
33
δ
11
)
=
−
1
{\displaystyle LHS=({e_{311}e_{131}})~~+({e_{312}e_{132}})~~+({e_{313}e_{133}})~~=-1~;~~RHS=({\delta _{31}\delta _{13}})~~-({\delta _{33}\delta _{11}})~~=-1}
Case 9:
i
=
1
{\displaystyle i=1}
j
=
3
{\displaystyle j=3}
p
=
1
{\displaystyle p=1}
q
=
3
{\displaystyle q=3}
L
H
S
=
(
e
131
e
131
)
+
(
e
132
e
132
)
+
(
e
133
e
133
)
=
1
;
R
H
S
=
(
δ
11
δ
33
)
−
(
δ
13
δ
31
)
=
1
{\displaystyle LHS=({e_{131}e_{131}})~~+({e_{132}e_{132}})~~+({e_{133}e_{133}})~~=1~;~~RHS=({\delta _{11}\delta _{33}})~~-({\delta _{13}\delta _{31}})~~=1}
Hence the
e
−
δ
{\displaystyle ~e-\delta }
3)
δ
i
j
e
i
j
k
=
0
{\displaystyle \delta _{ij}e_{ijk}=0}
δ
i
j
e
i
j
k
=
e
i
i
k
=
0
.
{\displaystyle {{\delta _{ij}e_{ijk}=e_{iik}=0}.}}
4)
e
q
r
s
d
q
d
s
=
0
{\displaystyle e_{qrs}d_{q}d_{s}=0}
e
q
r
s
d
q
d
s
=
∑
q
=
1
3
∑
s
=
1
3
e
q
r
s
d
q
d
s
=
(
e
1
r
1
d
1
d
1
)
+
e
1
r
2
d
1
d
2
+
e
1
r
3
d
1
d
3
+
e
2
r
1
d
2
d
1
+
(
e
2
r
2
d
2
d
2
)
+
e
2
r
3
d
2
d
3
+
e
3
r
1
d
3
d
1
+
e
3
r
2
d
3
d
2
+
(
e
3
r
3
d
3
d
3
)
{\displaystyle {\begin{aligned}e_{qrs}d_{q}d_{s}=&\sum _{q=1}^{3}\sum _{s=1}^{3}e_{qrs}d_{q}d_{s}\\=&({e_{1r1}d_{1}d_{1}})~~~+e_{1r2}d_{1}d_{2}+e_{1r3}d_{1}d_{3}+\\&e_{2r1}d_{2}d_{1}+({e_{2r2}d_{2}d_{2}})~~~+e_{2r3}d_{2}d_{3}+\\&e_{3r1}d_{3}d_{1}+e_{3r2}d_{3}d_{2}+({e_{3r3}d_{3}d_{3}})\end{aligned}}}
For
r
=
1
{\displaystyle r=1}
e
q
r
s
d
q
d
s
=
(
e
112
)
d
1
d
2
+
(
e
113
)
d
1
d
3
+
(
e
211
)
d
2
d
1
+
(
e
213
)
d
2
d
3
+
(
e
311
)
d
3
d
1
+
(
e
312
)
d
3
d
2
=
0
{\displaystyle e_{qrs}d_{q}d_{s}=({e_{112}})~~~d_{1}d_{2}+({e_{113}})~~~d_{1}d_{3}+({e_{211}})~~~d_{2}d_{1}+({e_{213}})~~~d_{2}d_{3}+({e_{311}})~~~d_{3}d_{1}+({e_{312}})~~~d_{3}d_{2}=0}
For
r
=
2
{\displaystyle r=2}
e
q
r
s
d
q
d
s
=
(
e
122
)
d
1
d
2
+
(
e
123
)
d
1
d
3
+
(
e
221
)
d
2
d
1
+
(
e
223
)
d
2
d
3
+
(
e
321
)
d
3
d
1
+
(
e
322
)
d
3
d
2
=
0
{\displaystyle e_{qrs}d_{q}d_{s}=({e_{122}})~~~d_{1}d_{2}+({e_{123}})~~~d_{1}d_{3}+({e_{221}})~~~d_{2}d_{1}+({e_{223}})~~~d_{2}d_{3}+({e_{321}})~~~d_{3}d_{1}+({e_{322}})~~~d_{3}d_{2}=0}
For
r
=
3
{\displaystyle r=3}
e
q
r
s
d
q
d
s
=
(
e
132
)
d
1
d
2
+
(
e
133
)
d
1
d
3
+
(
e
231
)
d
2
d
1
+
(
e
233
)
d
2
d
3
+
(
e
331
)
d
3
d
1
+
(
e
332
)
d
3
d
2
=
0
{\displaystyle e_{qrs}d_{q}d_{s}=({e_{132}})~~~d_{1}d_{2}+({e_{133}})~~~d_{1}d_{3}+({e_{231}})~~~d_{2}d_{1}+({e_{233}})~~~d_{2}d_{3}+({e_{331}})~~~d_{3}d_{1}+({e_{332}})~~~d_{3}d_{2}=0}
Hence shown.
The elasticity tensor is given by
C
=
λ
1
⊗
1
+
2
μ
I
{\displaystyle {\boldsymbol {\mathsf {C}}}=\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathsf {I}}}}
where
λ
,
μ
{\displaystyle \lambda ,\mu }
1
{\displaystyle {\boldsymbol {\mathit {1}}}}
I
{\displaystyle {\boldsymbol {\mathsf {I}}}}
1
=
δ
i
j
e
i
⊗
e
j
I
=
1
2
[
δ
i
k
δ
j
l
+
δ
i
l
δ
j
k
]
e
i
⊗
e
j
⊗
e
k
⊗
e
l
{\displaystyle {\begin{aligned}{\boldsymbol {\mathit {1}}}&=\delta _{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\{\boldsymbol {\mathsf {I}}}&={\frac {1}{2}}[\delta _{ik}\delta _{jl}+\delta _{il}\delta _{jk}]~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\otimes \mathbf {e} _{k}\otimes \mathbf {e} _{l}\end{aligned}}}
The stress-strain relation is
σ
=
C
:
ε
{\displaystyle {\boldsymbol {\sigma }}={\boldsymbol {\mathsf {C}}}:{\boldsymbol {\varepsilon }}}
Show that the stress-strain relation can be written in index notation as
σ
i
j
=
2
μ
ε
i
j
+
λ
ε
k
k
δ
i
j
.
{\displaystyle \sigma _{ij}=2\mu \varepsilon _{ij}+\lambda \varepsilon _{kk}\delta _{ij}~.}
Write the stress-strain relations in expanded form.
σ
=
C
:
ε
=
(
λ
1
⊗
1
+
2
μ
I
)
:
ε
=
(
λ
1
⊗
1
)
:
ε
+
2
μ
I
:
ε
{\displaystyle {\begin{aligned}{\boldsymbol {\sigma }}&={\boldsymbol {\mathsf {C}}}:{\boldsymbol {\varepsilon }}\\&=\left(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathsf {I}}}\right):{\boldsymbol {\varepsilon }}\\&=\left(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}\right):{\boldsymbol {\varepsilon }}+2\mu ~{\boldsymbol {\mathsf {I}}}:{\boldsymbol {\varepsilon }}\end{aligned}}}
Now, in dyadic notation
1
⊗
1
=
(
δ
i
j
e
i
⊗
e
j
)
⊗
(
δ
k
l
e
k
⊗
e
l
)
=
δ
i
j
δ
k
l
e
i
⊗
e
j
⊗
e
k
⊗
e
l
and
ε
=
ε
k
l
e
k
⊗
e
l
{\displaystyle {\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}=(\delta _{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j})\otimes (\delta _{kl}~\mathbf {e} _{k}\otimes \mathbf {e} _{l})=\delta _{ij}\delta _{kl}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\otimes \mathbf {e} _{k}\otimes \mathbf {e} _{l}~{\text{and}}~{\boldsymbol {\varepsilon }}=\varepsilon _{kl}~\mathbf {e} _{k}\otimes \mathbf {e} _{l}}
Therefore
(
1
⊗
1
)
:
ε
=
δ
i
j
δ
k
l
ε
k
l
e
i
⊗
e
j
=
δ
i
j
ε
k
k
e
i
⊗
e
j
{\displaystyle {\begin{aligned}({\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}):{\boldsymbol {\varepsilon }}&=\delta _{ij}\delta _{kl}\varepsilon _{kl}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\&=\delta _{ij}\varepsilon _{kk}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\end{aligned}}}
Similarly,
I
:
ε
=
(
1
2
[
δ
i
k
δ
j
l
+
δ
i
l
δ
j
k
]
ε
k
l
)
e
i
⊗
e
j
=
1
2
(
δ
i
k
ε
k
j
+
δ
i
l
ε
j
l
)
e
i
⊗
e
j
=
1
2
(
ε
i
j
+
ε
j
i
)
e
i
⊗
e
j
=
ε
i
j
e
i
⊗
e
j
(symmetry)
{\displaystyle {\begin{aligned}{\boldsymbol {\mathsf {I}}}:{\boldsymbol {\varepsilon }}&=\left({\frac {1}{2}}[\delta _{ik}\delta _{jl}+\delta _{il}\delta _{jk}]\varepsilon _{kl}\right)~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\&={\frac {1}{2}}\left(\delta _{ik}\varepsilon _{kj}+\delta _{il}\varepsilon _{jl}\right)~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\&={\frac {1}{2}}\left(\varepsilon _{ij}+\varepsilon _{ji}\right)~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\&=\varepsilon _{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\qquad {\text{(symmetry)}}\end{aligned}}}
The stress-strain law becomes
σ
=
λ
δ
i
j
ε
k
k
e
i
⊗
e
j
+
2
μ
ε
i
j
e
i
⊗
e
j
=
(
λ
δ
i
j
ε
k
k
+
2
μ
ε
i
j
)
e
i
⊗
e
j
{\displaystyle {\begin{aligned}{\boldsymbol {\sigma }}&=\lambda ~\delta _{ij}\varepsilon _{kk}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}+2\mu ~\varepsilon _{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\&=\left(\lambda \delta _{ij}\varepsilon _{kk}+2\mu \varepsilon _{ij}\right)\mathbf {e} _{i}\otimes \mathbf {e} _{j}\end{aligned}}}
Expanding the left hand side, we get
σ
i
j
e
i
⊗
e
j
=
(
λ
δ
i
j
ε
k
k
+
2
μ
ε
i
j
)
e
i
⊗
e
j
{\displaystyle \sigma _{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}=\left(\lambda \delta _{ij}\varepsilon _{kk}+2\mu \varepsilon _{ij}\right)\mathbf {e} _{i}\otimes \mathbf {e} _{j}}
Therefore,
σ
i
j
=
λ
δ
i
j
ε
k
k
+
2
μ
ε
i
j
{\displaystyle \sigma _{ij}=\lambda \delta _{ij}\varepsilon _{kk}+2\mu \varepsilon _{ij}}
Let (
e
1
,
e
2
,
e
3
{\displaystyle \mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}}
A
{\displaystyle {\boldsymbol {A}}}
u
{\displaystyle \mathbf {u} }
A
=
5
e
1
⊗
e
1
−
4
e
2
⊗
e
1
+
2
e
3
⊗
e
3
u
=
−
2
e
1
+
3
e
3
{\displaystyle {\begin{aligned}{\boldsymbol {A}}&=5~\mathbf {e} _{1}\otimes \mathbf {e} _{1}-4~\mathbf {e} _{2}\otimes \mathbf {e} _{1}+2~\mathbf {e} _{3}\otimes \mathbf {e} _{3}\\\mathbf {u} &=-2~\mathbf {e} _{1}+3~\mathbf {e} _{3}\end{aligned}}}
Write out
A
{\displaystyle {\boldsymbol {A}}}
u
{\displaystyle \mathbf {u} }
A
=
[
5
0
0
−
4
0
0
0
0
2
]
;
u
=
[
−
2
0
3
]
{\displaystyle {\mathbf {A} ={\begin{bmatrix}5&0&0\\-4&0&0\\0&0&2\end{bmatrix}};\qquad \mathbf {u} ={\begin{bmatrix}-2\\0\\3\end{bmatrix}}}}
Find the components of the vector
v
=
A
∙
u
{\displaystyle \mathbf {v} ={\boldsymbol {A}}\bullet \mathbf {u} }
e
1
,
e
2
,
e
3
{\displaystyle \mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}}
The components of
v
{\displaystyle \mathbf {v} }
v
i
=
A
i
j
u
j
{\displaystyle v_{i}=A_{ij}u_{j}}
Therefore
v
1
=
(
5
)
(
−
2
)
=
−
10
;
v
2
=
(
−
4
)
(
−
2
)
=
8
;
v
3
=
(
2
)
(
3
)
=
6
{\displaystyle v_{1}=(5)(-2)=-10;~~v_{2}=(-4)(-2)=8;~~v_{3}=(2)(3)=6}
v
=
−
10
e
1
+
8
e
2
+
6
e
3
{\displaystyle {\mathbf {v} =-10~\mathbf {e} _{1}+8\mathbf {e} _{2}+6\mathbf {e} _{3}}}
Find the components of the vector
w
=
v
×
u
{\displaystyle \mathbf {w} =\mathbf {v} \times \mathbf {u} }
e
1
,
e
2
,
e
3
{\displaystyle \mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}}
The cross product is given by
w
=
v
×
u
=
|
e
1
e
2
e
3
v
1
v
2
v
3
u
1
u
2
u
3
|
=
|
e
1
e
2
e
3
−
10
8
6
−
2
0
3
|
=
(
8
)
(
3
)
e
1
−
[
(
−
10
)
(
3
)
−
(
6
)
(
−
2
)
]
e
2
−
(
8
)
(
−
2
)
e
3
{\displaystyle \mathbf {w} =\mathbf {v} \times \mathbf {u} ={\begin{vmatrix}\mathbf {e} _{1}&\mathbf {e} _{2}&\mathbf {e} _{3}\\v_{1}&v_{2}&v_{3}\\u_{1}&u_{2}&u_{3}\end{vmatrix}}={\begin{vmatrix}\mathbf {e} _{1}&\mathbf {e} _{2}&\mathbf {e} _{3}\\-10&8&6\\-2&0&3\end{vmatrix}}=(8)(3)\mathbf {e} _{1}-[(-10)(3)-(6)(-2)]\mathbf {e} _{2}-(8)(-2)\mathbf {e} _{3}}
Therefore,
w
=
24
e
1
+
18
e
2
+
16
e
3
{\displaystyle {\mathbf {w} =24~\mathbf {e} _{1}+18\mathbf {e} _{2}+16\mathbf {e} _{3}}}
Find the components of the tensor
C
=
v
⊗
w
{\displaystyle {\boldsymbol {C}}=\mathbf {v} \otimes \mathbf {w} }
The tensor product is given by
v
⊗
w
=
v
i
w
j
e
i
⊗
e
j
{\displaystyle \mathbf {v} \otimes \mathbf {w} =v_{i}w_{j}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}}
Hence, in matrix notation
C
=
[
v
1
v
2
v
3
]
[
w
1
w
2
w
3
]
=
[
−
10
8
6
]
[
24
18
16
]
{\displaystyle \mathbf {C} ={\begin{bmatrix}v_{1}\\v_{2}\\v_{3}\end{bmatrix}}{\begin{bmatrix}w_{1}&w_{2}&w_{3}\end{bmatrix}}={\begin{bmatrix}-10\\8\\6\end{bmatrix}}{\begin{bmatrix}24&18&16\end{bmatrix}}}
C
=
[
−
240
−
180
−
160
192
144
128
144
108
96
]
{\displaystyle {\mathbf {C} ={\begin{bmatrix}-240&-180&-160\\192&144&128\\144&108&96\end{bmatrix}}}}
Rotate the basis clockwise by 30 degrees around the
e
3
{\displaystyle \mathbf {e} _{3}}
u
{\displaystyle \mathbf {u} }
v
{\displaystyle \mathbf {v} }
w
{\displaystyle \mathbf {w} }
C
{\displaystyle {\boldsymbol {C}}}
The vector transformation rule is
v
i
′
=
l
i
j
v
i
{\displaystyle v_{i}^{'}=l_{ij}v_{i}}
where
l
i
j
{\displaystyle l_{ij}}
In this case, the direction cosines are
l
11
=
e
1
′
∙
e
1
=
cos
(
30
)
=
0.866
l
12
=
e
1
′
∙
e
2
=
cos
(
60
)
=
0.5
l
13
=
e
1
′
∙
e
3
=
cos
(
90
)
=
0
l
21
=
e
2
′
∙
e
1
=
cos
(
120
)
=
−
0.5
l
22
=
e
2
′
∙
e
2
=
cos
(
30
)
=
0.866
l
23
=
e
2
′
∙
e
3
=
cos
(
90
)
=
0
l
31
=
e
3
′
∙
e
1
=
cos
(
90
)
=
0
l
32
=
e
3
′
∙
e
2
=
cos
(
90
)
=
0
l
33
=
e
3
′
∙
e
3
=
cos
(
0
)
=
1
{\displaystyle {\begin{aligned}l_{11}&=\mathbf {e} _{1}^{'}\bullet \mathbf {e} _{1}=\cos(30)=0.866&l_{12}&=\mathbf {e} _{1}^{'}\bullet \mathbf {e} _{2}=\cos(60)=0.5&l_{13}&=\mathbf {e} _{1}^{'}\bullet \mathbf {e} _{3}=\cos(90)=0\\l_{21}&=\mathbf {e} _{2}^{'}\bullet \mathbf {e} _{1}=\cos(120)=-0.5&l_{22}&=\mathbf {e} _{2}^{'}\bullet \mathbf {e} _{2}=\cos(30)=0.866&l_{23}&=\mathbf {e} _{2}^{'}\bullet \mathbf {e} _{3}=\cos(90)=0\\l_{31}&=\mathbf {e} _{3}^{'}\bullet \mathbf {e} _{1}=\cos(90)=0&l_{32}&=\mathbf {e} _{3}^{'}\bullet \mathbf {e} _{2}=\cos(90)=0&l_{33}&=\mathbf {e} _{3}^{'}\bullet \mathbf {e} _{3}=\cos(0)=1\end{aligned}}}
Therefore,
u
′
=
L
u
=
[
0.866
0.5
0
−
0.5
0.866
0
0
0
1
]
[
−
2
0
3
]
=
[
−
1.732
1
3
]
{\displaystyle \mathbf {u} ^{'}=\mathbf {L} \mathbf {u} ={\begin{bmatrix}0.866&0.5&0\\-0.5&0.866&0\\0&0&1\end{bmatrix}}{\begin{bmatrix}-2\\0\\3\end{bmatrix}}={\begin{bmatrix}-1.732\\1\\3\end{bmatrix}}}
u
′
=
−
1.732
e
1
+
1
e
2
+
3
e
3
{\displaystyle {\mathbf {u} ^{'}=-1.732\mathbf {e} _{1}+1\mathbf {e} _{2}+3\mathbf {e} _{3}}}
Similarly,
v
′
=
L
v
=
[
0.866
0.5
0
−
0.5
0.866
0
0
0
1
]
[
−
10
8
6
]
=
[
−
4.66
11.93
6
]
{\displaystyle \mathbf {v} ^{'}=\mathbf {L} \mathbf {v} ={\begin{bmatrix}0.866&0.5&0\\-0.5&0.866&0\\0&0&1\end{bmatrix}}{\begin{bmatrix}-10\\8\\6\end{bmatrix}}={\begin{bmatrix}-4.66\\11.93\\6\end{bmatrix}}}
v
′
=
−
4.66
e
1
+
11.93
e
2
+
6
e
3
{\displaystyle {\mathbf {v} ^{'}=-4.66\mathbf {e} _{1}+11.93\mathbf {e} _{2}+6\mathbf {e} _{3}}}
Also,
w
′
=
L
w
=
[
0.866
0.5
0
−
0.5
0.866
0
0
0
1
]
[
24
18
16
]
=
[
29.78
3.59
16
]
{\displaystyle \mathbf {w} ^{'}=\mathbf {L} \mathbf {w} ={\begin{bmatrix}0.866&0.5&0\\-0.5&0.866&0\\0&0&1\end{bmatrix}}{\begin{bmatrix}24\\18\\16\end{bmatrix}}={\begin{bmatrix}29.78\\3.59\\16\end{bmatrix}}}
v
′
=
29.78
e
1
+
3.59
e
2
+
16
e
3
{\displaystyle {\mathbf {v} ^{'}=29.78\mathbf {e} _{1}+3.59\mathbf {e} _{2}+16\mathbf {e} _{3}}}
From the handout from Slaughter's book, the tensor transformation
rule is
T
i
j
′
=
l
i
p
l
j
q
T
p
q
{\displaystyle T_{ij}^{'}=l_{ip}l_{jq}T_{pq}}
where
l
i
j
{\displaystyle l_{ij}}
In matrix form,
T
′
=
L
T
L
T
{\displaystyle \mathbf {T} ^{'}=\mathbf {L} \mathbf {T} \mathbf {L} ^{T}}
Therefore the components of
C
{\displaystyle {\boldsymbol {C}}}
C
′
=
[
0.866
0.5
0
−
0.5
0.866
0
0
0
1
]
[
−
240
−
180
−
160
192
144
128
144
108
96
]
[
0.866
−
0.5
0
0.5
0.866
0
0
0
1
]
=
[
−
138.8
−
16.7
−
74.6
355.3
42.8
190.9
178.7
21.5
96
]
{\displaystyle \mathbf {C} ^{'}={\begin{bmatrix}0.866&0.5&0\\-0.5&0.866&0\\0&0&1\end{bmatrix}}{\begin{bmatrix}-240&-180&-160\\192&144&128\\144&108&96\end{bmatrix}}{\begin{bmatrix}0.866&-0.5&0\\0.5&0.866&0\\0&0&1\end{bmatrix}}={\begin{bmatrix}-138.8&-16.7&-74.6\\355.3&42.8&190.9\\178.7&21.5&96\end{bmatrix}}}
C
′
=
[
−
138.8
−
16.7
−
74.6
355.3
42.8
190.9
178.7
21.5
96
]
{\displaystyle {\mathbf {C} ^{'}={\begin{bmatrix}-138.8&-16.7&-74.6\\355.3&42.8&190.9\\178.7&21.5&96\end{bmatrix}}}}
Consider a beam of length
L
{\displaystyle L}
×
{\displaystyle \times }
q
0
{\displaystyle q_{0}}
Hinged-Hinged Beam
The boundary conditions are
w
0
(
0
)
=
u
0
(
L
/
2
)
=
φ
x
(
L
/
2
)
=
0
.
{\displaystyle w_{0}(0)=u_{0}(L/2)=\varphi _{x}(L/2)=0~.}
Compute a plot similar to that shown in Figure 4.3.4 for this case using Beam188 elements. What do you observe?
The result is shown in Table 1.
Table 1. Deflections of a hinged-hinged beam
Load
U
y
{\displaystyle U_{y}}
x
=
L
/
2
{\displaystyle x=L/2}
1
-0.520746
2
-1.04086
3
-1.55922
4
-2.07510
5
-2.58768
6
-3.09622
7
-3.60000
8
-4.09835
9
-4.59065
10
-5.07636
Clamped-Clamped Beam
The boundary conditions are
u
0
(
0
)
=
w
0
(
0
)
=
φ
x
(
0
)
=
u
0
(
L
/
2
)
=
φ
x
=
L
/
2
=
0.
{\displaystyle u_{0}(0)=w_{0}(0)=\varphi _{x}(0)=u_{0}(L/2)=\varphi _{x=L/2}=0.}
Compute a plot for this case using Beam188 elements. Comment on your plot.
The result is shown in Table 2.
Table 2. Deflections of a clamped-clamped beam
Load
U
y
{\displaystyle U_{y}}
x
=
L
/
2
{\displaystyle x=L/2}
1
-0.103456
2
-0.202476
3
-0.294220
4
-0.377753
5
-0.453387
6
-0.521968
7
-0.584455
8
-0.644185
9
-0.696440
10
-0.745243
Listed below is the ANSYS input code for Problem 3A.1 and 3A.2.
/prep7
b = 1
h = 1
et,1,188
sectype,1,beam,rect
secdata,b,h
MP,EX,1,30e6
MP,PRXY,1,0.3
K,1,0,0,0
K,2,50,0,0
k,3,0,50,0
L,1,2,50
latt,1,,1,,3,3,1
LMESH,ALL
!change this section to d,1,all,0 for Problem 3A.2
d,1,uy,0
d,1,uz,0
d,2,rotz,0
nsel,all
sfbeam,all,,pres,10
fini
/solu
nlgeom,on
autots,on
nsubst,10,100,10
outres,all,all
solve
finish
Simulate the unrolling of a cantilever beam from Section 4.1.1 of Ibrahimbegovic (1995) and compare your results with the results shown in the paper.
The result is shown in Table 3.
Table 3. Cantilever free-end displacement components
Load
U
x
{\displaystyle U_{x}}
U
y
{\displaystyle U_{y}}
Rotation
M
=
10
π
{\displaystyle M=10\pi }
-0.040666
6.3205
-3.1287
M
=
20
π
{\displaystyle M=20\pi }
9.9578
0.12729
-6.2577
The deformation plots are shown in Figure 4 and 5.
Figure 4. Deformed shape under
M
=
10
π
{\displaystyle M=10\pi }
Figure 5. Deformed shape under
M
=
20
π
{\displaystyle M=20\pi }
The ANSYS input code for this problem is listed below.
/prep7
et,1,beam188
sectype,1,beam,rect
secdata,1,1
mp,ex,1,1200
mp,prxy,1,0
l = 10
pi = 4*atan(1)
r = L/2/pi
K,1,0,-r
K,2,-r,0
K,3,0,r
K,4,r,0
K,5,0,-r
k,6,0,0,10
k,7,0,0,0
larc,1,2,7,r,5
larc,2,3,7,r,5
larc,3,4,7,r,5
larc,4,5,7,r,5
latt,1,,1,,6,6,1
lmesh,all
dk,5,all,0
fk,1,mz,-20*pi
/solu
nlgeom,on
cnvtol,f,5,0.001
outres,all,all
arclen,on
nsubst,100
solve
fini
Simulate the clamped-hinged deep circular arch from Example 7.3 of Simo and Vu Quoc (1986) and compare you results with the results shown in the paper.
The inputs are:
R
=
100
{\displaystyle R=100}
φ
=
145
o
{\displaystyle \varphi =145^{o}}
E
I
=
10
6
{\displaystyle EI=10^{6}}
E
A
=
100
E
I
{\displaystyle EA=100EI}
I
=
1
12
h
4
;
A
=
h
2
E
A
E
I
=
A
I
;
12
h
2
h
4
=
12
h
2
=
100
{\displaystyle {\begin{aligned}I&={\cfrac {1}{12}}h^{4}~;&A&=h^{2}\\{\cfrac {EA}{EI}}&={\cfrac {A}{I}}~;&{\cfrac {12h^{2}}{h^{4}}}={\cfrac {12}{h^{2}}}=100\end{aligned}}}
Therefore,
h
=
0.34641
{\displaystyle h=0.34641}
The deformed shape (unconverged) for a load of 905 is shown in Figure 6.
Figure 6. Deformed shape of arch.
The load-displacement curve (up to the last converged solution) is shown in Figure 7.
Figure 7. Load-displacement plot for circular arch.
The buckling load is 900.925 compared to 905.28 in Simo and Vu Quoc.
The ANSYS file use for the calculations is shown below.
/prep7
!*
!* Total load
!*
load = 905.0
!*
!* Element type
!*
et,1,beam188
keyopt,1,1,0
keyopt,1,2,0
keyopt,1,3,0
keyopt,1,4,0
keyopt,1,6,0
keyopt,1,7,0
keyopt,1,8,0
keyopt,1,9,0
keyopt,1,10,0
keyopt,1,11,0
keyopt,1,12,0
!*
!* Beam cross-section type
!*
sectype, 1, beam, rect, , 0
secoffset, cent
secdata, 0.34641, 0.34641, 0,0,0,0,0,0,0,0
!*
!* Material properties
!*
mptemp,,,,,,,,
mptemp,1,0
mpdata,ex,1,,8.3e8
mpdata,prxy,1,,0.33
!*
!* Keypoints
!*
k, 1, 0.000, 0.000, 0.000
k, 2, -95.372, -30.071, 0.000
k ,3, 95.372, -30.071, 0.000
k ,4, 0.000, 100.000, 0.000
!*
!* Arcs
!*
larc,3,4,1,100,
larc,4,2,1,100,
!*
!* Element size = 20 elements per arc
!*
lesize,all, , ,20, ,1, , ,1,
!*
!* Mesh the arcs
!*
lmesh, all
!*
!* Plot the nodes
!*
nplot
/pnum,node,1
/number,0
/replot
!*
!* Apply displacement BCs
!*
!* Hinged end
!*
d, 22, ux, 0
d, 22, uy, 0
d, 22, uz, 0
!*
!* Clamped end
!*
d, 1, all, 0
!*
!* Apply load
!*
f, 2, fy, -load
finish
!*
!* Solve
!*
/solu
antype, static
nlgeom, on
!autots, on
!solcontrol, on
!*
!* Load step 1
!*
time, 1.0
! f, 2, fy, -load
nsubst,100,0,0
kbc, 0
neqit, 100
outres, ,1
arclen,on,100.0,0.0
lswrite
solve
finish
!*
!* See solution
!*
/post26
!*
!* Save solution in variables 2 and 3
!*
nsol, 2, 2, u, x ! Save ux at node 2
nsol, 3, 2, u, y ! Save uy at node 2
!*
!* Scale solution
!*
prod, 4, 1, , , Load, , ,load ! Scale time to get load
prod, 5, 2, , , , , ,-1 ! Make disp +ve
prod, 6, 3, , , , , ,-1 ! Make disp +ve
prvar, 4, 5, 6 ! Print load, ux, uy
!*
!* Plot solution
!*
/axlab, x, Deflection
/axlab, y, Load
/grid, 1
xvar, 5
plvar, 4 ! plot ux vs load
/noerase
xvar, 6
plvar, 4 ! plot uy vs load
/erase
Here is another version of solution to this problem.
Figure 8. Force-displacement diagram for Problem 3B.2.
Figure 9. Shape deformation at the last load step for Problem 3B.2.
The ANSYS input code for Problem 3B.2 is listed below.
/prep7
A = 1
I = A/100
E = 1e6/I
nu = 0.3
et,1,beam188 ! Element type - BEAM188
sectype,1,beam,asec
secdata,A,I,,I,,2*I
mp,ex,1,E
mp,prxy,1,nu
pi = 4*atan(1)
phi = 35/2/180*pi
x = 100*cos(phi)
y = 100*sin(phi)
k,1,0,0,0
k,2,x,-y,0
k,3,0,100,0
k,4,-x,-y
k,5,0,0,100
larc,2,3,1,100
larc,3,4,1,100
latt,1,,1,,5,5,1
lesize,all,,,40
lmesh,all
dk,2,all,0
dk,4,ux,0
dk,4,uy,0
dk,4,uz,0
fk,3,fy,-900
/solu
nlgeom,on
nsubst,100,0,0
outres,all,all
arclen,on
solve
finish
Simulate the buckling of a hinged right-angle frame under both fixed and follower loads from Example 7.4 of Simo and Vu Quoc (1986) and compare your results with those shown in the paper.
The force-displacement diagram is shown in Figure 10. The deformation is illustrated in Figure 11 and 12.
Figure 10. Force-displacement diagram for Problem 3B.3.
Figure 11. Deformation (fixed load) at the last load step for Problem 3B.3.
Figure 12. Force-displacement diagram (follower load) for Problem 3B.3.
The ANSYS input code for Problem 3B.3 is listed below.
/prep7
et,1,188
E = 7.2e6
I = 2
A = 6
sectype,1,beam,asec
secdata,A,I,,I,,2*I
mp,ex,1,7.2e6
mp,prxy,1,0.3
k,1,0,0,0
k,2,0,120,0
k,3,23,120,0
k,4,26,120,0
k,5,120,120,0
k,6,200,0,0
k,7,0,200,0
l,1,2,5
l,2,3,1
l,3,4,2
l,4,5,4
latt,1,,1,,6,6,1
lmesh,1
latt,1,,1,,7,7,1
lmesh,2,4
d,1,ux,0
d,1,uy,0
d,1,uz,0
d,18,ux,0
d,18,uy,0
d,18,uz,0
esel,s,elem,,7,8
!replace the line below with f,15,fy,-40000 for fixed load
sfbeam,all,,pres,40000/2
esel,all
fini
/solu
nlgeom,on
outres,all,all
arclen,on
nsubst,200
solve
fini
Warning: The arc length method no longer converges with Ansys 13. Try using the stabilization option instead of arclen, on :
stabilize, constant, energy, 0.001, anytime, 0
![{\displaystyle {\begin{aligned}{\boldsymbol {\mathit {1}}}&=\delta _{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\{\boldsymbol {\mathsf {I}}}&={\frac {1}{2}}[\delta _{ik}\delta _{jl}+\delta _{il}\delta _{jk}]~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\otimes \mathbf {e} _{k}\otimes \mathbf {e} _{l}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/37cd793d17b780fbbda34c7f9b8e3801f6b5694b)
![{\displaystyle {\begin{aligned}{\boldsymbol {\mathsf {I}}}:{\boldsymbol {\varepsilon }}&=\left({\frac {1}{2}}[\delta _{ik}\delta _{jl}+\delta _{il}\delta _{jk}]\varepsilon _{kl}\right)~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\&={\frac {1}{2}}\left(\delta _{ik}\varepsilon _{kj}+\delta _{il}\varepsilon _{jl}\right)~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\&={\frac {1}{2}}\left(\varepsilon _{ij}+\varepsilon _{ji}\right)~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\&=\varepsilon _{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\qquad {\text{(symmetry)}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc21ae44af56fa8df2d4d4436f511f47ecddbbb8)
![{\displaystyle \mathbf {w} =\mathbf {v} \times \mathbf {u} ={\begin{vmatrix}\mathbf {e} _{1}&\mathbf {e} _{2}&\mathbf {e} _{3}\\v_{1}&v_{2}&v_{3}\\u_{1}&u_{2}&u_{3}\end{vmatrix}}={\begin{vmatrix}\mathbf {e} _{1}&\mathbf {e} _{2}&\mathbf {e} _{3}\\-10&8&6\\-2&0&3\end{vmatrix}}=(8)(3)\mathbf {e} _{1}-[(-10)(3)-(6)(-2)]\mathbf {e} _{2}-(8)(-2)\mathbf {e} _{3}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c85048cc7097fb8e884df583510ff388da3d649f)
