4.2.6 NUMERICAL EXAMPLES
Example 4.1
The following are the principal stress at a point in a stressed material. Taking E
0.3 , calculate the volumetric strain and the Lame’s constants.
and
x
200 N / mm 2 ,
150 N / mm 2 ,
y
120 N / mm 2
z
Solution: We have
1
E
x
x
y
1
200 0.3 150 120
210 10 3
x
y
4
5.67 10
1
E
y
z
1
z
2.57 10
1
E
x
210 10 3
v
y
120 0.3 200 150
5
7.14 10
Volumetric strain =
x
y
z
5.67 10
v
200
4
z
1
z
x
150 0.3 120
210 10 3
y
z
4
8.954 10
2.57 10
3
To find Lame’s constants
We have, G
E
21
210 10 3
G
2 1 0.3
G 80.77 10 3 N / mm 2
G 2G E
E 3G
80.77 10 3 2 80.77 10 3 210 10 3
210 10 3 3 80.77 10 3
4
7.14 10
5
210 kN / mm 2
121 .14 10 3 N / mm 2
Example 4.2
The state of strain at a point is given by
x
0.001,
y
0.003,
z
xy
0,
0.004,
xz
Determine
the
stress
tensor
at
this
Poisson’s ratio = 0.28. Also find Lame’s constant.
yz
point.
0.001
Take
Solution: We have
E
21
G
210 10 6
2 1 0.28
82.03 10 6 kN / m 2
G
G 2G E
E 3G
But
82.03 10 6 2 82.03 16 6 210 10 6
210 10 6 3 82.03 10 6
104 .42 10 6 kN / m 2
Now,
2G
x
y
z
2 82.03 104.42 10 6 0.001 104.42 10 6
44780 kN / m 2
x
or
x
44.78 MPa
x
2G
y
y
z
x
2 82.03 104.42
or
y
701020 kN / m 2
y
701 .02 MPa
z
2G
z
x
10 6
= 2 82.03 104.42 10 0
z
208840 kN / m2
z
208 .84 MPa
xy
G
xy
= 82.03 10 6
0
0.003
104.42 10 6 0 0.001
y
6
or
0.003 0
104.42 10 6 0.001 0.003
E
210 10 6 kN / m 2 ,
xy
0
yz
G
yz
82.03 MPa
xz
G
or
or
82.03 10 6 0.001 82030 kN / m 2
yz
82.03 10 6
xz
0.004
328120 kN / m 2
328.12 MPa
xz
The Stress tensor is given by
ij
x
xy
xz
xy
y
yz
xz
yz
z
44.78
0
328 .12
0
701 .02
82.03
328 .12
82.03
208 .84
Example 4.3
The stress tensor at a point is given as
200
160
160
120
100 kN / m 2
240
120
100
160
Determine the strain tensor at this point. Take E
Solution:
1
E
x
x
y
1
=
x
z
200 0.3
210 10 6
1.067 10
210 10 6 kN / m2 and
240 160
6
1
y
y
E
=
y
=
z
Now, G
xy
1
210 10 6
1.657 10
1
E
z
z
210 10 6
0.82 10
E
21
G xy
G
x
240 0.3 160
200
6
x
1
xy
xy
z
y
160 0.3 200 240
6
210 10 6
2 1 0.3
80.77 10 6
160
80.77 10 6
80.77 10 6 kN / m 2
xy
1.981 10
6
0.3
100
80.77 10 6
120
80.77 10 6
yz
yz
G
zx
zx
G
1.24 10
6
1.486 10
6
Therefore, the strain tensor at that point is given by
xy
x
x
xy
xz
2
xz
2
xy
ij
xy
y
yz
zx
zy
z
yz
y
2
zx
2
1.067
0.9905
ij
z
2
0.743
0.9905
1.657
0.62
0.62
0.82
0.743
2
zy
6
10
Example 4.4
A rectangular strain rosette gives the data as below.
0
45
90
670 micrometre s / m
330 micrometre s / m
150 micrometre s / m
Find the principal stresses
1
and
2
2 105 MPa,
if E
0.3
Solution: We have
x
y
2
xy
670 10
0
45
90
150 10
0
90
160 10
xy
6
6
= 2 330 10
6
670 10
6
150 10
6
6
Now, the principal strains are given by
max
i.e.,
or
x
min
max
or
min
max
or
min
y
2
1
2
2
x
y
2
xy
670 150
1
10 6
670 150 10
2
2
410 10 6 272 .03 10 6
6 2
6
max
1
682 .3 10
min
2
137.97 10
6
The principal stresses are determined by the following relations
1
1
1
2
2
.E
682 .03 0.3 137 .97 10
1
0.3
2
6
2 10 5
160 10
6 2
1
Similarly,
159 MPa
2
2
2
1
1 2
75.3MPa
.E
137 .97 0.3 682 .03 10
1
0.3
2
6
2 10 5