Lecture 2
The Concept of Strain
Problem 2-1: A thin- walled steel pipe of length 60 cm, diameter 6 cm, and wall thickness 0.12
cm is stretched 0.01 cm axially, expanded 0.001 cm in diameter, and twisted through 1o.
Determine the strain components of the pipe. Note that the shell with a diameter to thickness
ratio of 50 is predominately in the membrane state. For plane stress there are only 3 components
of the strain tensor.
Problem 2-1 Solution:
1
Evaluate deformation components at the end cross-section
ur o
u
o
uz o
0.01/ 2cm
0.0005cm
3 (1o
R
180o
)cm
0.0523cm
0.01cm
Evaluate deformation components across the whole pipe
ur is constant
ur
ur o
0.0005cm
u is linearly proportional to
u
u
z, r
u
o
and r , independent of
z r
LR
u z is linearly proportional to z, independent of r and
uz
uz z
uz o
z
L
Substitute the above equations of ur , u and u z into the expression of six components of the
strain tensor in cylindrical coordinate system, we have
2
rr
zz
r
z
zr
ur
ur o
0
r
r
ur 1 u
ur o 1
0.0005
0
1.67 10 4
r r
r r
6/2
u z u z o 0.01
1.67 10 4
z
L
60
u u
u oz u oz
1 ur 1
1
0
0
r
r
r
LR
LR
2
2
1 uz
2 r
u
z
u or
1
0
2
LR
ur
z
uz
r
1
2
1
2
ur o
z
rr
Summary
zz
z
1 u oR
2 LR
uz ( z)
r
r
1 0.0523
2 60
4.36 10
4
0
0
zr
1.67 10
4
1.67 10
4
4.36 10
4
3
Problem 2-2: Derive an expression for the change in volume of a unit volume element subjected
to an arbitrary small strain tensor.
Problem 2-2 Solution:
A unit volume has sides of unit length
Unit Volume 1 1 1 1[length]3
Strain is defined by
L
, then the new length of each side is 1
If each side is deformed by
L
L
1
Volume after deformation is:
V
(1
1
1
)(1
1
2
2
)(1
2
3
)
1 2
3 2
3 1
1 2 3
Because of small strain, we cancel all higher order terms
Volumetric Strain = change in volume per unit volume, the original volume of a unit volume
element is 1
v
V Vo
Vo
1
1
2
1
3
1
1
v
1
2
2
3
3
4
Problem 2-3: A unit square OABC is distorted to OA’B’C’ in three ways, as shown in the figure
below. In each case write down the displacement field (u1, u2) of every point in the square as a
function of the location of that point (x1, x2) and the strain components ij.
x2
A’
A
2
x2
B’
x2
B’
B
A
A’
O
C x1
A’
B
B’
C’
1
C’
A
B
O
(a)
C
(b)
x1
O
C
(c)
x1
C’
Problem 2-3 Solution:
(a)
Only Axial deformation in this case
u1
u2
x
1 1
x
2 x
The strain components are
5
u1
x1
11
u2
x2
22
12
1
1
2
21
2
u1
x2
u1
x2
0
(b)
Only pure shear in this case
tan
u1
x2
tan
u2
x1
u1
x2
u2
x1
Deformation
Because of small angel
, tan
The strain components are
12
21
11
u1
x1
0
22
u2
x2
0
1
2
u1
x2
u1
x2
6
(c)
Only rigid body rotation in this case, there should be 0 strain everywhere
Deformation
u1
x2
u2
x1
The strain components are
12
21
1
2
11
u1
x1
0
22
u2
x2
0
u1
x2
u1
x2
1
2
0
7
Problem 2-4: The strain (plane strain) in a given point of a body is described by the 2x2 matrix
below. Find components of the strain tensor x' y ' in a new coordinate system rotated by the angle
. Consider four cases = 45o, -45o, 60o and -60o.
0
0.05
0.05
0
xy
Problem 2-4 Solution:
Transformation equations for plane strain, from old coordinate system xy to new coordinate
system x’y’
xx
x'x'
yy
2
yy
xx
2
yy
2
xx
x' y'
yy
2
xx
y' y'
xx
yy
2
cos(2 )
xy
sin(2 )
cos(2 )
xy
sin(2 )
sin(2 )
xy
cos(2 )
Given strain components:
x
y
0
0.05
xy
Simplify the transformation equations:
x'x'
xy
y' y'
x' y'
sin(2 )
xy
xy
sin(2 )
cos(2 )
Substitute the stain components into the simplified transformation equations:
45o
45o
0.05
0
x' y '
0
0.05
60o
x' y '
0.05
0
x' y '
0
0.05
60o
0.04
0.025
0.025
0.04
x' y '
0.04
0.025
0.025
0.04
8
Problem 2-5: A cylindrical pipe of 160-mm outside diameter and 10-mm thickness, spirally
welded at an angle of =40o with the axial (x) direction, is subjected to an axial compressive
load of P=150 kN through the rigid end plates (see below). Determine the normal force
shearing stresses
and
acting simultaneously in the plane of the weld.
y
x
P
P
Problem 2-5 Solution:
Calculate stress in given coordinate system
Compressive force in x-direction
P
A
xx
(80
2
150kN
702 )(10 3 )2 cm2
The stress components are
32MPa
xx
yy
0
xy
0
Transformation equations for plane strain
xx
x'x'
yy
2
yy
xx
2
Substitute the stain components and
yy
2
xx
xy
yy
2
xx
y' y'
xx
yy
2
sin(2 )
cos(2 )
xy
sin(2 )
cos(2 )
xy
sin(2 )
xy
cos(2 )
40 o into the simplified transformation equations:
x'x'
18.77MPa
y'y'
13.22MPa
z'z'
15.76MPa
9
Problem 2-6: A displacement field in a body is given by:
u
v
c x 2 10
2cyz
w
c
z2
xy
where c=10-4. Determine the state of strain on an element positioned at (0, 2, 1).
Problem 2-6 Solution:
Substitute the displacement into the expression of strain components, we have
xx
yy
zz
xy
yz
xz
u
x
v
y
w
z
1
2
u
y
v
x
1
0 0
2
1
2
v
z
w
y
1
2cy cx
2
1
2
u
z
w
x
1
0 cy
2
4
c 2 x 10
2 0
0
2cz
2 10
4
1 2 10
4
2cz
2 10
4
1 2 10
4
0
1
2 2 0
2
1 10
2 10
4
4
At (0, 2, 1), the state of strain is
0
0
1 10
0
2 10
4
2 10
4
4
1 10 4
2 10 4
2 10
4
10
Problem 2-7: The distribution of stress in an aluminum machine component is given by:
x
d (cy 2z 2 ),
xy
3dz 2
y
d (cx cz),
yz
dx 2
z
d (3cx cy),
xz
2dy 2
where c= 1 mm and d=1MPa/mm2. Calculate the state of strain of a point positioned at (1, 2, 4)
mm. Use E=70 GPa and =0.3.
Problem 2-7 Solution:
Substitute the given value of c, d , E,
into stress component expressions, we have
x
34MPa
yx
48MPa
y
5MPa
yz
1MPa
z
34MPa
xz
8MPa
Use constitutive equations
xx
yy
zz
And G
1
E
1
E
1
E
xx
yy
zz
xy
yy
xx
zz
yz
zz
yy
xx
xz
1
G
1
G
1
G
xy
yz
xz
E
2(1 )
We have the strain components
xx
4.42 10
8.91 10
4
xy
1.86 10
5
yz
1.49 10
4
xz
4
9.57 10
5
yy
9.57 10
5
zz
11
Problem 2-8: An aluminum alloy plate (E=70 GPa, =1/3) of dimensions a=300 mm, b=400 mm,
and thickness t=10 mm is subjected to biaxial stresses as shown below. Calculate the change in
(a) the length AB; (b) the volume of the plate.
y
a
B
b
30 MPa
A
x
90 MPa
Problem 2-8 Solution:
The state of stress is given
xx
30MPa
yy
90MPa
zz
0MPa
Use constitutive equations to calculate strains
1
E
xx
1
E
yy
1
E
zz
Given strain
yy
xx
yy
zz
yy
xx
zz
zz
xx
yy
xx
and
yy
1
1
30
90 0
9
70 10
3
106
0
1
1
90
30 0
9
70 10
3
106
0.00114
1
1
0
30 90
9
70 10
3
0.00057
, calculate the change in length in y
y
yy
y
y
Lo
Lo
yy
0.46mm
12
Using solutions in problem 2-2, calculate the change in volume
v
V
V0
v
V0
xx
yy
V
V
Vo
zz
a b t
xx
yy
zz
685.72mm3
13
Problem 2-9: Consider the general definition of the strain tensor in the 3D continuum. The three
Euler-Bernoulli hypotheses of the elementary beam theory state:
1. Plane remain plane
2. Normal remain normal
3. Transverse deflections are only a function of the length coordinate x.
Proof that under the above assumptions the state in the beam is uni-axial, meaning that the only
surviving component of the strain is in the length, x-direction. The proof is sketched in the lecture
notes, but we want you to redo the step by step derivation.
Problem 2-9 Solution:
1. Plane remains Plane
This Hypothesis is satisfied when the u-component of the displacement vector is a linear function
of z
uz
uo
z
(1)
where u o is the displacement of the beam axis
2. Normal remains normal
This hypothesis is satisfied when the rotation of the deformed cross-section
local slope of the middle axis
is equal to the
14
dw
dx
(2)
Combing (1) and (2), we have
dw
z
dx
u ( x, z ) u o
(3)
3. Deflections are only a function of x
w
w( x )
(4)
0
(5)
Also, because of planar deformation
v
Evaluate all components of strain tensor
xx
u
x
v
y
0
From equation(5): v
0
From equation(4):w
w(x)
From equation(4):w
w(x) and equation(5): v
From equation(3): u
uo
From equation(3):u
u (x, z) and equation(5): v
Only
xx
yy
zz
dw
z
dx
w( x)
z
zx
0
0
1 v
(
2 z
yz
1 u
(
2 z
w
)
x
0
1
(
2
xy
w
x
w( x)
)
y
1
(0 0) 0
2
w
) 0
x
1 u ( x, z)
(
2
y
v
)
x
1
(0 0) 0
2
survives, therefore the state of strain of a classical beam theory is uniaxial
15
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2.080J / 1.573J Structural Mechanics
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