TEST BANK FOR Introduction to Nonlinear Finite Element Analysis By Nam-Ho Kim

1.6. Exercises
P1.1 Using Cartesian bases, show that (u Ä v) ⋅ (w Ä x) = (v ⋅ w)u Ä x where u, v,
w, and x are rank 1 tensor.
Solution: Using the Cartesian basis, ( )( ) ( )( ) i i j j k k l l u Ä v w Ä x = u e Ä v e ⋅ w e Ä x e .
Since the dot product occurs between adjacent bases, we have
( )( )
( )( )
( )
( )
( )
( )( )
i i j j k k l l
i j k l j k i l
i j k l jk i l
i j j l i l
j j i i l l
u v w x
u v w x
u v w x
u v w x
v w u x
d
Ä ⋅ Ä
= ⋅ Ä
= Ä
= Ä
= Ä
= ⋅ Ä
e e e e
e e e e
e e
e e
e e
v w u x
In the above equation, we used the following properties: j k jk e ⋅ e = d , k jk j w d = w , and
j j v w = v ⋅ w .
P1.2 Any rank 2 tensor T can be decomposed by T = S + W, where S is the symmetric
part of T and W is the skew part of T. Let A be a symmetric rank 2 tensor. Show
A : W = 0 and A : T = A : S.
Solution: Since A is symmetric and W is skew, we have
: ij ij ij ji ji ji A W = AW = -AW = -A W
Since in the above equation, the repeated indices i and j are dummy, the above equation
can be rewritten as
0 ij ij ij ij AW = -AW =
In addition, from the relation T = S + W,
A : T = A : (S +W) = A : S + A : W = A : S
P1.3 For a symmetric rank-two tensor E , using the index notation, show that
I : E = E , where 1
2[ ] ik jl il jk I = d d + d d is a symmetric unit tensor of rank-4.
Solution: Using index notation, the contraction operator can be written as
1
2 ( : ) [ ] ij ik jl il jk kl I E = d d +d d E
Since the Kronecker-delta symbol replaces indices, the above equation can be written as
1
(I : E)ij = 2 [Eij +Eji ]= Eij = (E)ij
The symmetric property of E is used.
P1.4 The deviator of a symmetric rank-2 tensor is defined as m
dev A = A-A 1 where
1
3 11 22 33 Am = (A +A +A ). Find the rank-4 deviatoric identity tensor dev I that satisfies
: dev dev A = I A .
Solution: From Problem P1.3, it can be shown that I : A = A . In addition, Am can be
written in the tensor notation as 1
3 Am = 1 : A. Therefore, m
dev A = A-A 1 and it can be
written as
1
3 : : dev dev
= éê - Ä ùú = A ëI 1 1û A I A
The last equality defined the rank-4 deviatoric identity tensor dev I .
P1.5 The norm of a rank-2 tensor is defined as A = A : A . Calculate the following
derivative ¶ A / ¶A. What is the rank of the derivative?
Solution: From the definition
1/2 1 1/2
( : ) ( : ) (2 : )
2
- ¶ ¶ é ù = êë úû = = ¶ ¶
A A
A A A A A I
A A A
The result is a rank-2 tensor. Note that the property that ¶A / ¶A = I is used.
P1.6 A unit rank-2 tensor in the direction of rank-2 tensor A can be defined as
N = A / A . Show that ¶N / ¶A = [I-N Ä N] / A .
Solution: Using chain-rule of differentiation, the unit normal tensor can be differentiated
as
2
1 ¶ ¶ æç ÷ö çæ¶ ¶ ÷ö = çç ÷÷ = çç - Ä ÷÷ ¶ ¶ ç ÷÷÷ ç¶ ¶ ÷÷÷ çè ø çè ø
N A A A
A A
A A A A A A
It is straightforward to show that ¶A / ¶A = I . From Problem 1.5, we have
1/2 1 1/2
( : ) ( : ) (2 )
2
- ¶ ¶ é ù = êë úû = = ¶ ¶
A A
A A A A A
A A A
Therefore, we have
¶ 1 ( )
= - Ä
¶
N
I N N
A A
P1.7 Through direct calculation of a rank-2 tensor, show that the following identity
det[ ] rst ijk ir js kt e A =e A A A is true
Solution: In the index notation, (r, s, t) are real indices, while (i, j, k) are dummy indices.
Since (r, s, t) only appears in the permutation symbol, it is enough to show the cases of
even and odd permutation. Consider the following case of even permutation: (r, s, t) = (1,
2, 3). In such a case, non-zero components of the right-hand side can be written as
1 2 3 123 11 22 33 132 11 32 23
231 21 32 13 213 21 12 33
312 31 12 23 321 31 22 13
ijk i j k e AA A e A A A e A A A
e A A A e A A A
e A A A e A A A
= +
+ +
+ +
In the above equation, we have 123 231 312 e = e = e = 1 and 132 213 321 e = e = e = -1 .
Therefore, the above equation becomes
1 2 3 11 22 33 32 23 21 32 13 12 33 31 12 23 22 13 ( ) ( ) ( ) ijk i j k e A A A = A A A -A A +A A A -A A +A A A -A A
which is the definition of det[A] . By following a similar approach, it can be shown that
the odd permutation of (r, s, t) will yield -det[A] .
P1.8 For a vector 1 1 2 2 3 3 r = x e + x e + x e and its norm r = r , prove ⋅ (rr) = 4r .
Solution: From the product rule,
⋅ (rr) = r ⋅ r + r⋅ r
Now consider
1/2
1/2
1 1 1
( ) ( ) ( ) ( )
2( ) 2
j j i
i j j j j j j ij j
i k k i i i
x x x
r xx x x x x x
x x x x r x x r r
d
¶ ¶ ¶ ¶
 = = = + = =
¶ ¶ ¶ ¶
Therefore,
2
( ) i i 3 4
i
i
x x r
r r r x r r r
r x r
¶
 ⋅ =  ⋅ +  ⋅ = + = + =
¶
r r r
This completes the proof.
P1.9 A velocity gradient is decomposed