Matrices Review
Matrix Multiplication : When the number of columns of the first
matrix is the same as the number of rows in the second matrix then
matrix multiplication can be performed.
Here is an example of matrix multiplication for two 2x2 matrices
Here is an example of matrices multiplication for a 3x3 matrix
When A has dimensions mxn, B has dimensions nxp. Then the
product of A and B is the matrix C, which has dimensions mxp.
Transpose of Matrices : The transpose of a matrix is found by
exchanging rows for columns i.e. Matrix A = (aij) and the transpose
of A is: AT=(aij) Where i is the row number and j is the column
number.
For example, The transpose of a matrix would be:
In the case of a square matrix (m=n), the transpose can be used to
check if a matrix is symmetric. For a symmetric matrix A = AT
The Determinant of a Matrix : Determinants play an
important role in finding the inverse of a matrix and also in
solving systems of linear equations.
Determinant of a 2x2 matrix Assuming A is an arbitrary 2x2
matrix A, where the elements are given by:
Determinant of a 3x3 matrix The determinant of a 3x3 matrix
is more difficult
Inverse Matrix
For a 2x2 matrix
Example:
⎡ Cosθ
A=⎢
⎣− Sinθ
Sinθ ⎤
Cosθ ⎥⎦
For a 3x3 matrix
the matrix inverse is
⎡ Cosθ Sinθ ⎤
2
2
θ
θ =1
A=⎢
A
Cos
Sin
⇒
=
+
⎥
⎣− Sinθ Cosθ ⎦
1 ⎡Cosθ − Sinθ ⎤
−1
T
A = ⎢
=
A
1 ⎣ Sinθ Cosθ ⎥⎦
Coordinate Transformations
σ X 1 = σ X cos θ + σ Y sin θ + 2τ XY sin θ cos θ
σ Y 1 = σ X sin 2 θ + σ Y cos 2 θ − 2τ XY sin θ cosθ
2
2
τ x1 y1 = −σ x ⋅ sin θ ⋅ cos θ + σ y ⋅ sin θ ⋅ cos θ + τ xy (cos 2 θ − sin 2 θ )
⎡ cos 2 θ
⎢
[T ] = ⎢ sin 2 θ
⎢− sin θ cos θ
⎣
⎡ σ x1 ⎤
⎡σ x ⎤
⎢
⎥
⎢ ⎥
⎢ σ y1 ⎥ = [T ]⎢σ y ⎥
⎢τ x1 y1 ⎥
⎢τ xy ⎥
⎣
⎦
⎣ ⎦
sin 2 θ
cos 2 θ
sin θ cos θ
2 sin θ cos θ ⎤
⎥
− 2 sin θ cos θ ⎥
cos 2 θ − sin 2 θ ⎥⎦
⎡σ x ⎤
⎡ σ x1 ⎤
⎢ ⎥
⎥
−1 ⎢
⎢σ y ⎥ = [T ] ⎢ σ y1 ⎥
⎢τ xy ⎥
⎢τ x1 y1 ⎥
⎣ ⎦
⎣
⎦
Theory of Matrix Method for Stress Calculations in 2-D
From equations of rotational transformation of axis, we obtain the following:
⎡ x'⎤ ⎡ cosθ
⎢ y '⎥ = ⎢− sin θ
⎣ ⎦ ⎣
sin θ ⎤ ⎡ x ⎤
cosθ ⎥⎦ ⎢⎣ y ⎥⎦
Hence
⎡ x'⎤ ⎡
⎤⎡ x⎤
T
=
⎢ y '⎥ ⎢
⎥ ⎢ y ⎥ inversely
⎣ ⎦ ⎣
⎦⎣ ⎦
−1
⎡
⎤
⎡cosθ
=
T
⎢
⎥
⎢ sin θ
⎣
⎦
⎣
⎡σ xx τ yx ⎤
⎢τ
⎥
σ
yy ⎦
⎣ xy
− sin θ ⎤ ⎡
⎤
=⎢ T ⎥
⎥
cos θ ⎦ ⎣
⎦
Y
Y'
σy
⎡ x ⎤ ⎡ cos θ
⎢ y ⎥ = ⎢ sin θ
⎣ ⎦ ⎣
− sin θ ⎤ ⎡ x ' ⎤
cos θ ⎥⎦ ⎢⎣ y '⎥⎦
T
AC = Area A
AB = A cos θ
BC = A sin θ
A
X'
τyx
θ
θ
X
τxy
σx
σx
B
τxy
τyx
σy
C
σx
θ
σ x'
τ xy
σy
Using and force equilibrium equation, we obtain
expressions for stress transformations as follows:
{0} = Σ{F }
⎡ Fx ⎤
⎡ Fx ⎤
⎡ Fx ⎤
{0} = ⎢ ⎥ + ⎢ ⎥ + ⎢ ⎥
⎣ Fy ⎦ AB ⎣ Fy ⎦ BC ⎣ Fy ⎦ AC
σx
σ x'
θ
τ xy
σy
⎡σ x ⎤
⎡τ xy ⎤
⎡cos θ ⎤
⎡− sin θ ⎤
{0} = − ⎢ ⎥ A cosθ − ⎢ ⎥ A sin θ + Aσ x ' ⎢
+ Aτ x ' y ' ⎢
⎥
⎥
τ
σ
sin
cos
θ
θ
⎣
⎦
⎣
⎦
⎣ xy ⎦
⎣ y⎦
⎡σ x τ xy ⎤ ⎡cos θ ⎤
⎡cos θ − sin θ ⎤ ⎡ σ x ' ⎤
{0} = − A⎢
+ A⎢
⎥
⎢
⎥
⎥ ⎢τ ⎥
τ
σ
sin
sin
cos
θ
θ
θ
y ⎦⎣
⎦
⎣
⎦ ⎣ x' y' ⎦
⎣ xy
Canceling area A out and pre-multiplying by transformation T
(where T ⊗ T T = I
the identity matrix. The order of the matrix multiplication does
matter in the final outcome., we have
⎡cos θ
⎢ sin θ
⎣
− sin θ ⎤ ⎡ cos θ
×⎢
⎥
cos θ ⎦ ⎣− sin θ
sin θ ⎤ ⎡1 0⎤
=⎢
=I
⎥
⎥
cos θ ⎦ ⎣0 1⎦
⎡σ x τ xy ⎤ ⎡cos θ ⎤
⎡cos θ
{0} = − A⎢
+ A⎢
⎥
⎢
⎥
⎣ sin θ
⎣τ xy σ y ⎦ ⎣ sin θ ⎦
⎡ σ x ' ⎤ ⎡ cosθ
⎢τ ⎥ = ⎢
⎣ x ' y ' ⎦ ⎣− sin θ
− sin θ ⎤ ⎡ σ x ' ⎤
⎢ ⎥
cos θ ⎥⎦ ⎣τ x ' y ' ⎦
sin θ ⎤ ⎡σ x τ xy ⎤ ⎡cosθ ⎤
⎢τ
⎥
⎥
cosθ ⎦ ⎣ xy σ y ⎦ ⎢⎣ sin θ ⎥⎦
For the forces in the X axis we will use the same procedure.
Y'
{0} = Σ{F }
Y
BD = Area A
BC = A cos θ
CD = A sin θ
⎡ Fx ⎤
⎡ Fx ⎤
⎡ Fx ⎤
{0} = ⎢ ⎥ + ⎢ ⎥ + ⎢ ⎥
⎣ Fy ⎦ CD ⎣ Fy ⎦ BC ⎣ Fy ⎦ BD
⎡σ x ⎤
⎡τ xy ⎤
⎡− sin θ ⎤
⎡cos θ ⎤
A
A
A
A
sin
cos
θ
σ
τ
θ
−
+
+
⎥
⎢σ ⎥
y' ⎢
x' y' ⎢
⎥
⎥
⎣ cos θ ⎦
⎣ sin θ ⎦
⎣τ xy ⎦
⎣ y⎦
⎡σ τ ⎤ sin θ ⎤
⎡cos θ − sin θ ⎤ ⎡τ x ' y ' ⎤
{0} = A⎢ x xy ⎥ ⎡⎢
A
+
⎢ sin θ cos θ ⎥ ⎢ σ ⎥
⎥
τ
σ
cos
θ
−
xy
y
⎦⎣ y' ⎦
⎣
⎦
⎣
⎣
⎦
{0} = ⎢
X'
θ
D
θ
B
C
X
σ y'
τ x' y'
τ xy
σx
θ
σy
⎡τ y ' x ' ⎤ ⎡ cosθ
⎢σ ⎥ = ⎢
⎣ y ' ⎦ ⎣− sin θ
sin θ ⎤ ⎡σ x τ yx ⎤ ⎡− sin θ ⎤
⎢τ
⎥⎢
⎥
⎥
σ
cosθ ⎦ ⎣ xy
cos
θ
y ⎦⎣
⎦
⎡ σ x ' ⎤ ⎡ cosθ
⎢τ ⎥ = ⎢
⎣ x ' y ' ⎦ ⎣− sin θ
sin θ ⎤ ⎡σ x τ xy ⎤ ⎡cosθ ⎤
⎥⎢
⎢τ
⎥
⎥
σ
cosθ ⎦ ⎣ xy
y ⎦ ⎣ sin θ ⎦
⎡τ x ' y ' ⎤ ⎡ cosθ
⎢σ ⎥ = ⎢
⎣ y ' ⎦ ⎣− sin θ
sin θ ⎤ ⎡σ x τ xy ⎤ ⎡− sin θ ⎤
⎢
⎥
cosθ ⎥⎦ ⎣τ xy σ y ⎦ ⎢⎣ cosθ ⎥⎦
Combining the above expressions
⎡ σ x ' τ x ' y ' ⎤ ⎡ cosθ
=⎢
⎢τ
⎥
⎣ x ' y ' σ y ' ⎦ ⎣− sin θ
or
sin θ ⎤ ⎡σ x τ xy ⎤ ⎡cosθ
⎢τ
⎥⎢
⎥
σ
cosθ ⎦ ⎣ xy
y ⎦ ⎣ sin θ
⎡ σ x' τ y'x' ⎤ ⎡
⎤ ⎡σ x τ yx ⎤ ⎡
⎤
=⎢ T ⎥ ⎢
T ⎥
⎥
⎢τ
⎥
⎢
⎦ ⎣τ xy σ y ⎦ ⎣
⎦
⎣ x' y' σ y' ⎦ ⎣
T
− sin θ ⎤
⎥
cosθ ⎦
State of Stresses in Three Dimensions
z
The general three dimensional state of stress
consists of three unequal principal stresses acting
at a point (triaxial state of stresses).
τxy
σx
τxz
σ
K
τyx
σy
τyz
τzy
J
-x
The plane JKL is assumed to be
a principal plane and σ is the
principal stress acting normal to
the plane.
L
τzx
σz
-y
Letα, β and γ are the angles between the vector σ
and the x, y and z axis respectively and
k = cos α
l = cos β
m = cos γ
σ ⋅ k = σ xx ⋅ k + τ yx ⋅ l + τ zx ⋅ m
Under equilibrium conditions
(σ − σ xx ) ⋅ k − τ yx ⋅ l − τ zx ⋅ m = 0
− τ xy ⋅ k + (σ − σ yy ) ⋅ l − τ zy
σ ⋅ l = τ xy ⋅ k + σ yy ⋅ l + τ zy ⋅ m
σ ⋅ m = τ xz ⋅ k + τ yz ⋅ l + σ zz ⋅ m
As k, l and m are different than
⋅ m = 0 zero (non-trivial solution)
− τ xz ⋅ k − τ yz ⋅ l + (σ − σ z ) ⋅ m = 0
k +l + m =1
2
2
2
⎡σ − σ xx
− τ yx
− τ zx ⎤ ⎡ k ⎤
The determinant must be
⎢
⎥⎢ ⎥
− τ zy ⎥ ⎢ l ⎥ = 0 equal to zero
σ − σ yy
⎢ − τ xy
⎢ − τ xz
− τ yz
σ − σ zz ⎥⎦ ⎢⎣m⎥⎦
⎣
σ − σ xx
− τ yx
− τ zx
σ − σ yy
− τ xy
− τ zy = 0
σ − σ zz
− τ xz
− τ yz
Solution of the determinant results
in a cubic equation in σ
⎡σ − σ xx
− τ yx
− τ zx ⎤ ⎡ k ⎤
⎢
⎥⎢ ⎥
−
−
−
τ
σ
σ
τ
yx
yy
zy ⎥ ⎢ l ⎥ = 0
⎢
⎢ − τ xz
− τ yz
σ − σ zz ⎥⎦ ⎢⎣m⎥⎦
⎣
k +l + m =1
σ 3 − I1 ⋅ σ 2 + I 2 ⋅ σ − I 3 = 0
I1 = σ x + σ y + σ z
2
2
2
The eigenvalues of the stress
matrix are the principal stresses.
The eigenvectors of the stress
matrix are the principal
directions.
σ1 > σ 2 > σ 3
I 2 = σ x ⋅ σ y + σ y ⋅ σ z + σ x ⋅ σ z − τ xy2 − τ xz2 − τ yz2
I 3 = σ x ⋅ σ y ⋅ σ z + 2 ⋅τ xy ⋅τ xz ⋅τ yz − σ x ⋅τ yz2 − σ y ⋅τ xz2 − σ z ⋅τ xy2
The three roots are the three principal stresses σ1 , σ2 , σ3.
I1, I2, and I3 are known as stress invariants as they do not change
in value when the axes are rotated to new positions.
I1 = σ x + σ y + σ z
σ x τ xy σ y τ yz
σ x τ xz
I2 =
+
+
τ xy σ y τ zy σ z z τ zx σ z
σ x τ xy τ xz
I 3 = τ yx σ y τ yz
τ zx τ yz σ z
I1 has been seen before for the two dimensional state of stress. It
states the useful relationship that the sum of the normal stresses for
any orientation in the coordinate system is equal to the sum of the
normal stresses for any other orientation
σ x + σ y + σ z = σ x1 + σ y1 + σ z1 = σ 1 + σ 2 + σ 3
σ − I1 ⋅ σ + I 2 ⋅ σ − I 3 = 0
3
2
I1
σ 1 = 2 ⋅ A ⋅ cos α +
3
(
2
)
I1
= −2 ⋅ A ⋅ cos α ± 60 +
3
⎛ I1 ⎞ ⎛ I 2 ⎞
A = ⎜ ⎟ −⎜ ⎟
⎝3⎠ ⎝ 3⎠
⎡ ⎛ I1 ⎞ 3 ⎛ I1 ⎞
⎤
⎢2 ⋅ ⎜ ⎟ − ⎜ ⎟ ⋅ I 2 + I 3 ⎥
⎢⎣ ⎝ 3 ⎠ ⎝ 3 ⎠
⎥⎦
Cos (3α ) =
2 ⋅ A3
Stress Invariants for Principal Stresses
σ 2,3
0
(
⎤
⎡σ 1
⎥
σ = ⎢⎢
σ2
⎥
⎢⎣
σ 3 ⎥⎦
The solution are the eigenvalues
of the stress tensor
)
I1 = σ 1 + σ 2 + σ 3
I 2 = σ1 ⋅σ 2 + σ 2 ⋅σ 3 + σ 3 ⋅σ 1
I3 = σ 1 ⋅σ 2 ⋅σ 3
τ max
⎛ σ1 − σ 2 σ 2 − σ 3 σ 3 − σ1 ⎞
= max⎜
,
,
⎟
2
2 ⎠
⎝ 2
Example: determine the principal stresses for
the state of stress (in MPa).
Solution:
The solution are the eigenvalues of the
stress tensor;
Substituting:
σ
− (−240)
− (−240)
0
σ − 200
0
− 240
0 ⎤
⎡ 0
⎥
⎢− 240 200
0
⎥
⎢
⎢⎣ 0
0
− 280⎥⎦
σ − σ x − τ xy
− τ xz
− τ yx σ − σ y − τ yz = 0
− τ zx
− τ zy σ − σ z
0
0
=0
σ − (−280)
(σ − (−280) ) ⋅ ((σ ⋅ (σ − 200) ) − ((−240) ⋅ (−240)) ) = 0
One solution σ3=-280MPa is a principal stress because τxz and τyz are
zero, then the other two principal stresses are easy to find by solving the
quadratic equation inside the square brackets for σ
σ − 200σ − (240) = 0
2
σ = 100 ± 260
2
− ( −200) ± (−200) 2 + 4 ⋅ (240) 2
σ=
2
σ1 = 360 MPa
σ 2 = -160 MPa
− 240
0 ⎤
⎡ 0
⎢− 240 200
⎥
0
⎢
⎥
⎢⎣ 0
0
− 280⎥⎦
A = 196.4
I1 = −80
Cos (3α ) = 0.8620
I 2 = −113,600
α = 10.15
− 80
σ 1 = 2 ⋅196.4 ⋅ Cos (10.15) +
= 359.99
3
− 80
σ 2 = −2 ⋅196.4 ⋅ Cos (10.15 + 60 ) +
= −160.0
3
− 80
σ 3 = −2 ⋅196.4 ⋅ Cos (10.15 − 60) +
= −279.9
3
0
0 ⎤
⎡360
⎢ 0 − 160
⎥
0
⎢
⎥
⎢⎣ 0
0
− 280⎥⎦
I 3 = 16,128,000
Example 2: Determine the maximum
⎡ 20
principal stresses and the maximum
σ = ⎢⎢ 40
shear stress for the following triaxial
⎢⎣− 30
stress state.
40 − 30⎤
30 25 ⎥⎥ MPa
25 − 10 ⎥⎦
Solution
⎡σ xx τ yx τ zx ⎤ ⎡ 20 40 − 30⎤
⎥
⎢
Stress _ Tensor = [σ ] = ⎢τ xy σ yy τ zy ⎥ = ⎢⎢ 40 30 25 ⎥⎥
⎢τ xz τ yz σ zz ⎥ ⎢⎣− 30 25 − 10 ⎥⎦
⎦
⎣
σ 3 − I1σ 2 + I 2σ − I 3 = 0
I1 = σ x + σ y + σ z
= 20 + 30 –10 = 40 MPa
I 2 = σ xσ y + σ xσ z + σ yσ z − τ xy2 − τ xz2 − τ yz2
= -3025 MPa
I 3 = σ xσ yσ z + 2τ xyτ xzτ yz − σ τ − σ τ − σ τ
2
x yz
2
y xz
2
z xy=
-89500 MPa
Solution to Exam ple
600000
400000
26.5 MPa
-51.8 MPa
Sigm a (MPa)
200000
0
-100
-80
-60
-40
-20
0
20
40
60
80
100
-200000
65.3 MPa
-400000
-600000
-800000
Stress (MPa)
σ 1 = 65.3MPa
σ 2 = 26.5MPa
τ max = 1 / 2(65.3 + 51.8) = 58.5MPa
σ 3 = −51.8MPa
Mohr’s Circles for 3-D Analysis
Mohr’s circles can make visualization of the stress condition clearer to
the designer. Note that the principal stress values are always
ordered by convention so the σ1 is the largest value in the tensile
direction and σ3 is the largest value in the compressive direction.
Note also that there is one dominant peak shear stress in this diagram.
Be forewarned the principal stresses
and this peak shear stress are going
to play a strong role in determining
the factor of safety in mechanical
design.
A Mohr’s circle can be generated for triaxial stress states, but it is
often unnecessary, as it is sufficient to know the values of the
principal stresses.
The principal stresses must be ordered from larger to smaller.
τ 12 =
σ1 − σ 2
2
σ 2 −σ3
τ 23 =
2
σ1 − σ 3
τ 13 =
2
τ
Compare 2-D and 3-D Mohr’s Circle. If σz is zero, does it have an
effect in 3D?
Q
P
2α
σ3
σ2 + σ3
2
2α
σ 1 + σ 3σ2
σ1 + σ 2
2
σ1
2
Consider σ3=0 then the plane will be an angle α from σ1, in
the direction of σ2 (clockwise). Point P
Consider σ2=0 then the plane will be an angle α from σ1, in
the direction of σ3 (clockwise). Point Q
The required system of stresses, fall within P and Q. Loci
determined by the center in σ 2 + σ 3
2
2β
2β
σ3
σ2
σ1
Consider σ1=0 then the plane will be an angle β from σ2, in the
direction of σ3 (anticlockwise). Point R
Consider σ3=0 then the plane will be an angle β from σ2, in the
direction of σ1 (clockwise). Point S
The required system of stresses, fall within R and S. Loci
determined by the center in σ + σ
1
3
2
Example:
Use Mohr’s Circle to obtain the principal stresses and maximum
shear of a component subjected to the following stresses:
σ x = 90tension
σ y = 30tension
σ z = −25compression
τ xy = 40ccw
ccw counterclockwise
Stress on ANY Inclined Plane (3-D)
The stress on a plane (S) can
be decomposed into its normal
component (Sn) and its shear
component (Ss).
S = S + S = σ +τ
2
Sn = sx + sy + sz
2
n
2
s
2
If α, β and γ are the angles between the vector Sn
and the x, y and z axis respectively and
k = cos α
l = cos β
m = cos γ
then
2
It can be proved that
⎡σ 1 1
⎢ xx
⎢τ x1 y1
⎢τ 1 1
⎣ xz
τy x
σy y
τy z
1 1
1 1
1 1
⎡σ xx τ yx τ zx ⎤
⎢
⎥
σ = ⎢τ xy σ yy τ zy ⎥
⎢τ xz τ yz σ zz ⎥
⎣
⎦
⎤
⎡σ xx τ yx τ zx ⎤
⎥
⎥
⎢
T
[
]
[
]
T
×
T
τ
σ
τ
=
×
1 1 ⎥
yy
zy ⎥
⎢ xy
⎥
⎢τ xz τ yz σ zz ⎥
1 1
⎦
⎣
⎦
τz x
τz y
σz z
1 1
Mohr’s Circles for 3-D Analysis
There is no easy Mohr’s circle graphical solution for problems of
triaxial stress state. Solution for maximum principal stresses and
maximum shear stress is analytical.
z
Consider the x, y and z axis to coincide with the
axis of the principal stresses σ1, σ2 and σ3.
If α, β and γ are the angles between the
normal to the plane and the x, y and z
σ1
axis respectively and
σ
σ2
y
σ3
x
k = cos α
l = cos β
m = cos γ
k 2 + l 2 + m2 = 1
We would like to find graphically
the normal stress and shear stress on
the plane.
Octahedral Plane and Stresses
An octahedral plane is a plane that makes
three identical angles with the principal planes.
σ ⋅ k = σ xx ⋅ k + τ yx ⋅ l + τ zx ⋅ m
σ ⋅ l = τ xy ⋅ k + σ yy ⋅ l + τ zy ⋅ m
σ ⋅ m = τ xz ⋅ k + τ yz ⋅ l + σ zz ⋅ m
n1 = n2 = n3 =
1
=n
3
σ op = σ 1 ⋅ n12 + σ 2 ⋅ n22 + σ 3 ⋅ n32 =
σ1 + σ 2 + σ 3
τ op = n 2 ⋅ (σ 12 + σ 22 + σ 32 ) − σ op2 =
9τ op = 2 I12 − 6 I 2
2
3
1
3
=
I1
3
(σ 1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ 1 )2
Mean and Deviatoric Stresses
When describing the materials behavior of metals, one concludes
that in certain cases some stress components play a more important
role than other components. Plastic behavior of metals, is reported
to be independent of the average (mean) normal stress.
σM =
Mean stress matrix
Deviatoric stress
σ1 + σ 2 + σ 3
⎡σ M
M = ⎢⎢ 0
⎢⎣ 0
σ x′ = σ x − σ M
3
0
σM
0
=
σ x +σ y +σ z
3
I1
= = σ op
3
0 ⎤
0 ⎥⎥
σ M ⎥⎦
σ y′ = σ y − σ M
The shear components do not change
σ z′ = σ z − σ M
Deviatoric Stress Matrix
1
1
⎡2
τ yx
⎢ 3 σ xx − 3 σ yy − 3 σ zz
⎢
2
1
1
τ xy
σ yy − σ xx − σ zz
D=⎢
3
3
3
⎢
⎢
τ xz
τ yz
⎢⎣
⎤
⎥
⎥
τ zy
⎥
⎥
2
1
1
σ zz − σ xx − σ yy ⎥
⎥⎦
3
3
3
τ zx
Deviatoric stresses play an important role in the theory of plasticity.
They influence the yielding of ductile materials.
The principal stresses obtained only from the deviatoric matrix is
σ P,D
⎡σ 1 − σ M
⎢
=⎢ 0
⎢⎣ 0
0
σ 2 −σ M
0
⎤
⎥
⎥
σ 3 − σ M ⎥⎦
0
0
Example
For a given stress matrix representing the state of stress at a certain
point
Find the stress invariant, the
⎡1 2 3 ⎤
stresses, the principal
[σ ] = ⎢⎢2 2 0⎥⎥ MPa principal
directions, the octahedral stress
⎢⎣3 0 2⎥⎦
and the shear stress associated
with the octahedral stress.
Solution:
I1 = 1 + 2 + 2 = 5
I 2 = 1 × 2 + 2 × 2 + 2 × 1 − 22 − 02 − 32 = −5
I 3 = 1 × 2 × 2 − 1 × 02 − 2 × 32 − 2 × 22 + 2 × 2 × 0 × 3 = −22
σ 3 − 5 ⋅ σ 2 + −5 ⋅ σ + 22 = 0
σ 1 = 5.14
σ 2 = 2.00
σ 3 = −2.14
(σ − σ xx ) ⋅ k − τ yx ⋅ l − τ zx ⋅ m = 0
− τ xy ⋅ k + (σ − σ yy ) ⋅ l − τ zy ⋅ m = 0
− τ xz ⋅ k − τ yz ⋅ l + (σ − σ z ) ⋅ m = 0
(5.14 − 1) ⋅ k1 − 2 ⋅ l1 − 3 ⋅ m1 = 0
− 2 ⋅ k1 + (5.14 − 2 ) ⋅ l1 − 0 ⋅ m1 = 0
− 3 ⋅ k1 − 0 ⋅ l1 + (5.14 − 2 ) ⋅ m1 = 0
k1 + l1 + m1 = 1
2
2
2
9τ op = 2 × 52 − 6 × (−5) = 80
2
80
9
σ 1 = 5.14
σ 2 = 2.00
σ 3 = −2.14
k1 = 0.657 k2 = 0
k3 = −0.714
l1 = 0.418 l2 = 0.832
l3 = 0.364
m1 = 0.627 m2 = −0.555 m3 = 0.546
2
9τ op = 2 I12 − 6 I 2
τ op =
⎡1 2 3 ⎤
[σ ] = ⎢⎢2 2 0⎥⎥ MPa
⎢⎣3 0 2⎥⎦
σ Mean
I1 5
= =
3 3