Andy Gish
14 Sept 2012
Structural Mechanics
2.080 Recitation 2
Semester Yr
Recitation 2: Stress/Strain Transformations and
Mohr’s Circle
General Transformation Rules
2D Vector
2.1
General Transformation Rules
2.1.1
2D Vector
er a vector v
in the
(x1 , x2 ) coordinate system. A new coordinate system (x
0 ) is component
ined by rotating
system
by
Find
Consider athe
vectorold
v in coordinate
the (x1 , x2 ) coordinate
system. A
newangle
coordinate .system
(x01 , xthe
2
obtained by rotating the old coordinate system by angle ✓. Find the components of v in
new coordinate
system.
the new coordinate system.
geometry, we have:
geometry, weFrom
have:
x01 = x1 cos ✓ + x2 sin ✓
⇡
= x1 cos ✓ + x2 cos( - ✓)
2
L
+
x
L
=
x
1 1 11
2 12
1
2
x s = x cos + x sin
0
= x1xcos
+ x2xcos(
)
2 cos ✓ 2
2 = -x1 sin ✓ +
⇡
cos(
==xx11 L
+✓)x+2xL
2 cos
112 +
12✓
(2.1)
(2.2)
= x1 L21 + x2 L22
x s = x sin + x2 cos
0
x121=cos(
) + x 0 cos
L11 = cos(e01 , e=
cos(e10 , e+
1 ), L
2 2 ), L21 = cos(e22 , e1 ), L22 = cos(e2 , e2 )
= x1 L21 + x2 L22
where Lij is the direction cosine
of the basis
2
1 vectors:
(2.3)
Lij is the direction cosine of the basis vectors:
2-1
L
= cos(e s, e ), L
= cos(e s, e ), L
= cos(e s, e ), L
= cos(e s, e )
Structural Mechanics
2.1.2
2.080 Recitation 2
Semester Yr
General Vector Transformation
A vector is a physical quantity, independent of the coordinate system. But its scalar com­
ponents DO depend on the coordinate system.
v = vi ei = vi0 e0i
(2.4)
The scalar components vi0 in the new coordinate system can be expressed in terms of the
scalar components vi in the original coordinate system. Multiplying the above equation by
e0j gives:
vi ei · ej0 = vi0 ei0 · e0j
= vi0
=
i0 j 0
(2.5)
vj0
The indices i, j are arbitrary and may be reversed to give:
vi0 = vj ei0 · ej
(2.6)
Introduce the direction cosine tensor
Lij ⌘ e0i · ej = cos(e0i , ej )
then we can write vi0 in index form
(i, j = 1, 2, 3)
(2.7)
vi0 = Lij vj
(2.8)
v 0 = [L]v
(2.9)
3
2
3
v10
v1
7
6
6
7
v 0 = 4 v20 5 , and v = 4 v2 5
v30
v3
(2.10)
and in matrix form
where
2
Similarly, the inverse transform
vi = Lji vj0
(2.11)
0
(2.12)
T
v = [L] v
2.1.3
Tensor Transformations
Using the same definition for Lij defined above, the components of a tensor can be trans­
formed to a new coordinate system by:
A0ij = Lik Ljl Akl
2-2
(2.13)
Structural Mechanics
2.080 Recitation 2
Semester Yr
[A0 ] = [L][A][L]T
(2.14)
or in matrix notation:
The inverse transformation can be found by:
Akl = Lik Ljl A0ij
(2.15)
[A] = [L]T [A0 ][L]
(2.16)
Note that the transformation matrix [L] is orthogonal, meaning its transpose is equal to its
inverse.
[L][L]T = I
(2.17)
2.2
Eigenvalues and Eigenvecors
If there exists a vector ni and a scalar Ai for an arbitrary tensor [A] such that
[A]ni = Ai ni
(2.18)
then Ai and ni are eigenvalues and eigenvectors of tensor [A], respectively. The three
eigenvectors are mutually perpendicular, and form a new coordinate system.
The ”eigenvalue problem” for tensor [A] can be written as:
(A - Ai I)ni = 0
(2.19)
This equation has solutions ni 6= 0 only if
det(A - Ai I) = 0
(2.20)
This equation can be solved for Ai , then Ai substituted into the eigenvalue problem to solve
for ni .
For the stress (strain) tensor, the eigenvalues represent principal stresses (strains), and
eigenvectors represent principal axes (i.e., faces with zero shear stress (strain)).
Example
Find the principal stresses and principal axes for [i] =
"
80 30
30 40
#
.
Solution: The given stress tensor can be represented graphically by
2-3
Example
Find the principal stresses and principal axes for [-] =
Structural Mechanics
2.080 Recitation 2
=
80 30
.
30 40
8
Semester Yr
Solution: The given stress tensor can be represented graphically by
The principal stresses are the eigenvalues ( i ) of the stress tensor, and are found by solvThe principal stresses are the eigenvalues (Ai ) of the stress tensor, and are found by solving:
ing:
det(i - AI) = 0
#
80 - A
30
det det(I) = 0 = 0
30
40 - A
8
"
=
(2.21a)
(2.21b)
(80 80
- A)(40 - A)30
- 302 = 0
det
(2.21c)
=0
2
A30
- 120A
40+ 2300 = 0
(2.21d)
A1 = 96.05 and A2 =23.95
(80
)(40
(2.21e)
2
)
30 = 0
The principal directions are the eigenvectors, found by substituting the eigenvalues into the
2
original eigenvalue problem:
120 + 2300 = 0
(i - Ai I)ni = 0
(2.22)
For A1 = 96.05 MPa,
1
= 96.05 and
"
2
= 23.95
#(
)
(
)
- 96.05
30
0
The principal directions are80the
eigenvectors,
foundx1by =
substituting
the eigenvalues
into
(2.23)
30
40
96.05
0
y
1
the original eigenvalue problem:
(80 - 96.05)x1 + 30y1 = 0
For
(2.24)
eigenvector.
Therefore, we can choose y1 = 1
Any (x1 , y1 ) that satisfies this equation
(- is an
i I)n
i = 0
and solve for x1 , then normalize the vector as follows:
1
= 96.05 MPa:
8
(80 - 96.05)x1 + 30 = 0
80
30=
x30
96.05
1 =
16.05
(
30
40 96.05
30
n1 = q
1
30 2
( 16.05
)
(80
16.05
12
1
(2.25)
x1
=
)y1 (
=
+
96.05)x
1 + 30y1 = 0
0.88
0.47
0
)0
(2.26)
(2.27)
eigenvector. Therefore, we can choose y1 = 1
Any (x1 , y1 ) that satisfies this equation is an2-4
and solve for x1 , then normalize the vector as follows:
Structural Mechanics
2.080 Recitation 2
Semester Yr
Similarly, the second eigenvector is found to be:
(
)
-0.47
n2 =
0.88
(2.28)
As a check, it is clear that the two eigenvectors are perpendicular (n1 · n2 = 0).
2.3
Mohr’s Circle
Mohr’s circle (named after Otto Mohr (1835-1918)) is a graphical technique to transform
stress (strain) from one coordinate system to another, and to find maximum normal and
shear stresses (strains).
Constructing 2D Mohr’s Circle:
1. Establish a rectangular coordinate system with x =normal stress, y =shear stress.
Scales must be identical.
2. Plot stresses for 2 orthogonal adjacent faces (values from the original stress (strain)
tensor).
3. Connect the 2 points to find center of the circle, C.
4. Draw circle through 2 points with center C.
5. Principal stresses (strains) are values where the circle crosses the x-axis.
6. Max shear stress (strain) is max y-value on the circle.
Sign convention for Mohrs circle: Positive shear stress on a face causes clockwise
rotation of the unit square.
The stress state for a face rotated an angle ✓ from an original coordinate axis may be found
by rotating an angle 2✓ on Mohrs circle in the same direction from the original coordinate
axis.
Example1
Solve the above example using Mohr’s circle.
Solution: The given stress tensor is represented graphically by
1
Example 1.3 from Ugural and Fenster, Advanced Strength and Applied Elasticity, 2003
2-5
mple1
the above example using Mohr’s circle.
Structural Mechanics
2.080 Recitation 2
Semester Yr
olution: The given stress tensor is represented graphically by
Example1
Solve the above example using Mohr’s circle.
Solution: The given stress tensor is represented graphically by
Theon
stress
state
ononthe
x-face
=
+80
MPa,
and
0 and
= MPa
M30a(because
MPa
(because
of
stress state
the
positive
x-face
isisisi+==
+80
MPa,
= s30
MPa
(becau
The
stress
state
thepositive
positive
x-face
+80
MPa,
and
⌧ = -30
of the
the sign
convention
defined
above).
sign convention defined above).
gn convention
defined above).
The
stateononthethe
positive
y-face
is +40
+ = MPa,
+40 MPa,
and
= +30 MPa.
The stress
stress state
positive
y-face
is i =
and ⌧ =
+300 MPa.
he stress state
on the
positive
y-face
isis as shown:
=as+40
The Mohr’s
Mohrs circle
forfor
thethe
given
stress
state state
The
circle
given
stress
is
shown:MPa, and a = +30 MPa.
he Mohr’s circle for the given stress state is as shown:
TheThe
center,
C, C,
is is
located
60MPa
the
+ axis.
center,
locatedatat(40
(40 +
+ 80)/2
80)/2 ==60
MPa ononthe
i axis.
The principal
principal stresses
represented
by A
B1 . The
of thoseofpoints
B1 . coordinates
The coordinates
those are
points
The
stressesareare
represented
by
A1 and
1 and
are found from geometry as:
+1,2 = 60 ±
or
p
(802-6
M 60)2 + 302 MPa
Structural Mechanics
2.080 Recitation 2
Semester Yr
found from geometry as:
i1,2 = 60 ±
or
p
(80 - 60)2 + 302 MPa
i1 = 96.05 MPa and i2 = 23.95 MPa
The principal directions are found by:
(2.29)
(2.30)
The principal directions are found by:
2(2✓p ==arctan
arctan
p
30
o
30
= 56.3
°
=
56.3
(80-660)
60)
(80
°
✓(p =
=28.15
28.15o
p
(2.31a)
(2.31b)
The principal stress state is as shown below:
The principal stress state is as shown below:
3D Mohrs Circle
To draw Mohrs circle for a general 3D stress state, the principal stresses and directions
3D Mohr’s
must firstCircle
be evaluated (by solving the eigenvalue problem). Then the Mohrs circle can be
constructed
shownfor
below:
To draw
Mohr’sascircle
a general 3D stress state, the principal stresses and directions
must first be evaluated (by solving the eigenvalue problem). Then the Mohr’s circle can
The stress state for any rotation will be represented by a point either on one of the 3 circles,
be constructed as shown below:
or in the shaded green area between the inner and outer circles.
2-7
p
= 28.15
The principal stress state is as shown below:
Structural Mechanics
2.080 Recitation 2
Semester Yr
3D Mohr’s Circle
To draw Mohr’s circle for a general 3D stress state, the principal stresses and directions
must first be evaluated (by solving the eigenvalue problem). Then the Mohr’s circle can
be constructed as shown below:
The stress state for any rotation will be represented by a point either on one of the 3
circles, or in the shaded green area between the inner and outer circles.
7
2-8
MIT OpenCourseWare
http://ocw.mit.edu
2.080J / 1.573J Structural Mechanics
Fall 2013
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.