MD 2C
MD 2C
MD 2C
of rows in the second matrix then matrix multiplication can be performed. Here is an example of matrix multiplication for two 2x2 matrices
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:
Example:
Cos A= Sin Sin Cos
Cos Sin 2 2 =1 A= A Cos Sin = + Sin Cos 1 Cos Sin 1 T A = = A 1 Sin Cos
Coordinate Transformations
x1 x y1 = [T ] y x1 y1 xy
x x1 1 y = [T ] y1 xy x1 y1
sin x y cos
x cos y = sin
xx yx xy yy
y yx xy x xy yx y
sin = T cos
Y Y'
A X' X
x
B C
x
xy
x'
{0} = {F }
Using and force equilibrium equation, we obtain expressions for stress transformations as follows:
x
xy
x'
Fx Fx Fx {0} = + + Fy AB Fy BC Fy AC
x xy cos sin {0} = A cos A sin + A x ' + A x ' y ' sin cos xy y x xy cos cos sin x ' {0} = A + A sin sin cos 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 sin
sin 1 0 = =I cos 0 1
sin x ' cos x' y'
For the forces in the X axis we will use the same procedure.
Y' Y BD = Area A BC = A cos CD = A sin
{0} = {F } {0} =
Fx F y
Fx Fx + + Fy BC Fy BD CD
X' D B C X
x xy sin cos A A A A sin cos + + y' x' y' cos sin xy y sin cos sin x ' y ' {0} = A x xy A + sin cos cos xy y y'
{0} =
y'
x' y'
xy x
y
sin cos
yx y yz zy J -x
xy xz zx z
k = cos
l = cos
m = cos
k = xx k + yx l + zx m
Under equilibrium conditions
l = xy k + yy l + zy m m = xz k + yz l + zz m
( xx ) k yx l zx m = 0
xy k + ( yy ) l zy
xz k yz l + ( z ) m = 0
k +l + m =1
2 2 2
xx yx zx yy xy zy = 0 zz xz yz
xx yx zx k zy l = 0 yx yy xz yz zz m
k +l + m =1 3 I1 2 + I 2 I 3 = 0 I1 = x + y + z
2 2 2
The eigenvalues of the stress matrix are the principal stresses. The eigenvectors of the stress matrix are the principal directions.
1 > 2 > 3
2 2 2 I 2 = x y + y z + x z xy xz yz 2 2 2 I 3 = x y z + 2 xy xz yz x yz y xz z xy
The three roots are the three principal stresses 1 , 2 , 3. 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 = x + y + z
x xy y yz x xz I2 = + + xy y zy z z zx z x xy xz I 3 = yx y yz zx yz 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
x + y + z = x1 + y1 + z1 = 1 + 2 + 3
3 I1 2 + I 2 I 3 = 0
I1 1 = 2 A cos + 3 I1 = 2 A cos 60 + 3
0
I I A = 1 2 3 3
1 = 2 3
The solution are the eigenvalues of the stress tensor
I1 = 1 + 2 + 3 I3 = 1 2 3
I 2 = 1 2 + 2 3 + 3 1
1 2 2 3 3 1 , , 2 2 2
max = max
Example: determine the principal stresses for the state of stress (in MPa). Solution: The solution are the eigenvalues of the stress tensor; Substituting:
x xy xz yx y yz = 0 zx zy z
200
0
I1 = 80
I 2 = 113,600
I 3 = 16,128,000
A = 196.4
Cos (3 ) = 0.8620
= 10.15
80 = 359.99 3 80 2 = 2 196.4 Cos (10.15 + 60 ) + = 160.0 3 80 3 = 2 196.4 Cos (10.15 60) + = 279.9 3
Example 2: Determine the maximum 20 principal stresses and the maximum = 40 shear stress for the following triaxial stress state. 30 Solution
40 30 30 25 MPa 25 10
xx yx zx 20 40 30 Stress _ Tensor = [ ] = xy yy zy = 40 30 25 xz yz zz 30 25 10
3 I1 2 + I 2 I 3 = 0
I1 = x + y + z
= 20 + 30 10 = 40 MPa
2 xy 2 xz 2 yz
I 2 = x y + x z + y z
= -3025 MPa
2 2 2 I 3 = x y z + 2 xy xz yz x yz y xz z xy = -89500 MPa
Solution to Exam ple 600000 400000 -51.8 MPa 200000 Sigm a (MPa) 0
-100 -80 -60 -40 -20 0 20 40 60 80 100
26.5 MPa
3 = 51.8MPa
A Mohrs 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.
12 =
1 2
2 2 3 23 = 2 1 3 13 = 2
Compare 2-D and 3-D Mohrs Circle. If z is zero, does it have an effect in 3D?
P
2 2
1 + 2
2
2 + 3
2
1 + 32
2
Consider 3=0 then the plane will be an angle from 1, in the direction of 2 (clockwise). Point P Consider 2=0 then the plane will be an angle from 1, in the direction of 3 (clockwise). Point Q The required system of stresses, fall within P and Q. Loci determined by the center in 2 + 3
2
2 2
Consider 1=0 then the plane will be an angle from 2, in the direction of 3 (anticlockwise). Point R Consider 3=0 then the plane will be an angle from 2, in the direction of 1 (clockwise). Point S The required system of stresses, fall within R and S. Loci determined by the center in +
1 3
Example: Use Mohrs Circle to obtain the principal stresses and maximum shear of a component subjected to the following stresses:
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 = +
2 2 n 2 s 2
Sn = sx + sy + sz
If , and are the angles between the vector Sn and the x, y and z axis respectively and
k = cos
l = cos
m = cos
then
xx yx zx = xy yy zy xz yz zz
1 1 xx x1 y1 1 1 xz
y x y y y z
1 1
1 1
1 1
z x z y z z
xx yx zx T T ] xy yy zy [T ] 1 1 = [ xz yz zz 1 1
1 1
k = cos
y
l = cos
m = cos
k 2 + l 2 + m2 = 1
3 x
We would like to find graphically the normal stress and shear stress on the plane.
k = xx k + yx l + zx m l = xy k + yy l + zy m m = xz k + yz l + zz m
1 =n n1 = n2 = n3 = 3
op = 1 n + 2 n + 3 n =
2 1 2 2 2 3
1 + 2 + 3
3 1 = 3
I1 = 3
op = n ( + + )
2 2 1 2 2 2 3
2 op
( 1 2 )2 + ( 2 3 )2 + ( 3 1 )2
9 op = 2 I12 6 I 2
2
M =
Mean stress matrix
1 + 2 + 3
3 0
x + y + z
3
I1 = op 3
M M = 0 0
M
0
0 0 M
Deviatoric stress
x = x M
y = y M
z = z M
zy 2 1 1 zz xx yy 3 3 3
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
P,D
1 M = 0 0
0 2 M 0
3 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 MPa principal stresses, the principal [ ] = 2 2 0 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
5 + 5 + 22 = 0
3 2
( xx ) k yx l zx m = 0
xz k yz l + ( z ) m = 0
xy k + ( yy ) l zy m = 0
1 2 3 MPa 2 2 0 [ ] = 3 0 2
k3 = 0.714 l3 = 0.364
Mean =
9 op = 2 I12 6 I 2
2 2
9 op = 2 52 6 (5) = 80
op
80 = 9