AE 321 – Solution of Homework #5 1.
(5×5=25 POINTS)
Given the displacement field u x (a)
x 2 20 104
m, u y
2 yz 10 3
m, u z z
2
xy 10 3 m
Before deformation, the distance between P x, y , z P 2,5,7 and Q x, y, z Q3,8,9
is given by ds
3 2 2 8 52 9 7 2 1 9 4 14 3.741 m .
The position of points P points P and and Q after deformation is determined using the following relation X x u
(1.1)
Using Equation (1.1), the positions after deformations are
X P
x P u 2e x 5e y 7e z 2 2 20 10 4 e x 2 5 7 10 3 e y 7 2 2 5 10 3 e z m
X P
X Q
2.0024e x 5.070e y 7.039e z m
xQ u 3e x 8e y 9e z 32 20 10 4 e x 2 8 9 10 3 e y 9 2 3 810 3 e z m
X Q
3.0029e x 8.144e y 9.057e z m
The distance between the given points, i.e. P i.e. P and and Q, after deformation is dS
3.0029 2.00242 8.144 5.0702 9.057 7.039 2 3.8109 m .
Therefore, the change in distance between P between P and and Q is dS ds 3.8109 3.7417 0.0692m (b)
Lagrangian strain tensor E ij ij
L
1 2
u
i, j
u j, j, i uk ,k ,iuk ,k , j
1
(1.2)
1
Infinitesimal strain tensor ij
2
u
i, j
u j,i
(1.3)
The components of the Lagrangian strain tensor are:
5 10
5
4 x 10 4 x 4
2
100 y 2
7
5 10 xy
4
5 10
4 z 10 4 z x 3
2
2
5 10 4 y1 2 103 z 4 3 5 10 2 y x 1 2 10 z 2 10 3 z 103 y 2 z 2
The components of the infinitesimal strain tensor are 4
2 10
x
0 3
2 10 z
5 10 4 y 4 5 10 2 y x . 3 2 10 z
Note that this is nothing but the above results for E ij, with all second (and higher) order terms neglected.
(c)
The rotation tensor ij
1 2
u
i , j
u j ,i
0 0 0 0 5 10 4 y 5 10 4 2 y x
ij
(d)
(1.4)
4 5 10 2 y x 0 4
5 10 y
The Lagrangian and the infinitesimal strain tensors are each evaluated at
x 2, y 1, z 3 : 400.58 503 1 106 1 6020 2012 503 2012 6020
E ij
2
400 0 500 106 0 6000 2000 500 2000 6000
ij
Clearly, by comparing the components of the Lagrangian and the infinitesimal strain tensors, we
.
can conclude that E ij
ij
This means that in this case one can assume small
deformation gradients .
(e)
Compatibility equations ij ,kl kl ,ij
ik , jl jl ,ik 0
Only 6 independent equations are obtained from (1.5), namely, 2 2 xy 2 xx yy 2 0 y y x x x y
0 0 20 0
2 yy 2 zz 2 yz 2 0 z z y y y z
0 0 20 0
2 xz 2 zz 2 xx 2 0 x x z z x z
0 0 20 0
2 xy 2 xz 2 yz xx 0 x z x y x x y z
0000 0
2 yz 2 yx 2 zx 2 yy 0 y x y z y y z x
0000 0
2 2 2 zx zy xy 2 zz 0 z y z x z z x y
0000 0
Therefore, the given displacement field is compatible.
3
(1.5)
2. (6×1=6 POINTS) In order to obtain a single-valued continuous displacement field, the strain
field must satisfy the compatibility equations. Then:
2 11 2 22 2 12 2 0 x2 x2 x1 x1 x1 x 2
2 23 2 22 2 33 2 0 x3 x3 x2 x2 x2 x3
2 33 2 11 2 13 2 0 x1 x1 x3 x3 x1 x3
2 A 2 A 2 B 0
0 0 20 0
0 0 20 0
A
B 2
satisfied!
satisfied!
2 12 2 13 2 23 11 0 x1 x3 x1 x 2 x1 x1 x2 x3
2 23 2 21 2 31 2 22 0 x2 x1 x 2 x3 x2 x2 x3 x1
0000 0
satisfied!
2 31 2 32 2 12 2 33 0 x3 x 2 x3 x1 x3 x3 x1 x2
0000 0
satisfied!
0000 0
satisfied!
From the results of the compatibility equations we can conclude the strain will be valid if the following relation is satisfied: A
B 2
4
3.
The infinitesimal strain tensors for simple extension and simple shear deformation are presented in the following table: Simple Extension
Simple Shear
u1 Ax1 , u2 u 3 0 (2 POINTS)
u1 Bx2 ,
21
1
13 31
1
22 u2,2
u 2 u 2
u2,1 0
12
u3,1 0
13 31
1,2
1,3
1 u2,3 u3,2 2 33 u3,3 0
0
23 32
1
u 2
u2,1
1,2
1
2 33 u3,3 0
u
2,3
B 2
0
u3,2 0
A 0 0
ij
21
(2 POINTS)
1 u1,3 u3,1 2 22 u2,2 0
0
23 32
0
11 u1,1 0
11 u1,1 A 12
u2 u3
0 0 0
0
0 0
0 B 20
ij
B 2 0 0
0 0 0
The figures below show the original and final shape of an initially rectangular volume element that was subjected to simple extension and simple shear.
(2 POINTS)
(2 POINTS)
5
4.
Direction cosines of AC : Direction cosines of BD:
1/ 1/
2, 0, 1/ 2
(
11
,12 , 13 )
6, 2 / 3, 1/ 6
21
(2 POINTS)
, 22 , 23
(see also handout given in class explaining how to construct the direction cosine tensor, Q ij)
(a)
Relative Elongation
Deformed Length - Undeformed Length Undeformed Length dS ds ds
Normal Strain of the Element parallel to any of the axes
Since we are interested in determining the relative elongation of AC and BD, it is necessary to find the normal strains along these lines. The infinitesimal strains along x1, x2, and x3 axes are known, and since the infinitesimal strain is a tensor of second order, it can be rotated to obtain the components of the infinitesimal strain at any direction using the transformation law. x2
B
A
O
D C x3
6
x1
Therefore, taking x1 and x2 along CA and DB separately, the relative elongations of CA and DB will be given as follows:
Relative Elongation of AC
dS AC ds AC ' 11 ds AC
(1 POINT)
Relative Elongation of DB
dS DB ' ds DB 22' ds DB
(1 POINT)
'
Using the transformation law kl' kilj ij : (2 POINTS)
' 11 11 11 11 12 12 11 13 13 12 11 21 12 12 22 12 13 23 13 11 31 13 12 32 13 13 33 11
212111 2122 12 2123 13 22 21 21 22 22 22 22 23 23 23 21 31 23 22 32 23 23 33
' 22
Since the direction cosines of AC and DB are known, substituting (2 POINTS)
(b)
Relative Elongation of AC
11' 0.015 and Relative Elongation of DB 22'
0.000333
The change of angle of two elements initially at right angle is given by the angle as follows: A’ and B’ are the positions of A and B after certain
x’2 B
(4 POINTS FOR CORRECT SCHEMATIC)
deformation.
B’
For infinitesimal deformation, 2 12 A’ '
In our case, since CA is along x1 and DB is along x2' ,
D
(2 POINTS)
(1)
12'
A
x’1
we have to calculate 12' using the transformation law:
112111 1122 12 1123 13 12 21 21 12 22 22 12 23 23 13 21 31 13 22 32 13 23 33
Substituting the known values for ,
7
12'
1 1 1 2 1 0.02 0.003 0 0 0 0 0 2 6 2 6 2 1 2 1 1 0.02 0.01 2 6 2 6 1.617 102 rad
0.02 6
2
Substituting into (1), we get 2 1.617 10 2 rad 3.233 10 2 rad 1.85
(2 POINTS)
Using an expression as in the case for stress,
(c)
Q1
ii 11 22 33 0.02 0.01 0.01 0.04
Q2
Q3
1 2
tr
2
tr 2
11
12
21
22
11
13
31
33
(1 POINT)
22
23
23
33
(0.02)(0.01) (0.003)(0.003) (0.02)(0.01) (0) (0.01)(0.01) (0.02)(0.02) (1 POINT) 9.1 10 5
det 0.020.012 0.02 2 0.003 0.0030.01 0 0 6.09 10 6
(1 POINT)
The principal strains can be found by solving the equationdet I 0 .
(d)
0.02 det
Change of ADB 1.85
0.003
0.003 0.01
0.02
0
0.02
0.01
0
0 0.02
3 0.04 2 0.000091 0.000006 0 1 0.0304 2 0.0197 3 0.0102 For 1
0.01
0.02
0.02
0.01
0.003
0.003
0.02
0
0.01
(2 POINTS)
1 0.0304 , 0.003 0 0.02 0.0304 n1 0.201 ^ 0.003 n 0 n1 0.700 0.01 0.0304 0.02 2 0.685 0 0.02 0.01 0.0304 n3
8
(1 POINT)
0
For 2
2 0.0197 ,
0.003 0 0.02 0.0197 n1 0.977 ^ 0.003 n 0 n 2 0.093 0.01 0.0197 0.02 2 0.192 0 0.02 0.01 0.0197 n3 For 3
(1 POINT)
3 0.0102 ,
0.003 0 0.02 0.0102 n1 0.070 ^ 0.003 0.01 0.0102 0.02 n 2 0 n 3 0.708 0 0.02 0.01 0.0102 n3 0.703
9
(1 POINT)
