Area Moment Matrices
y
y'
t
x'
t
ey
n
θ
ex
ty
x
θ
1
tx
1
θ
nx
n
ny
The transformation equations for area moments due to a rotation of
axes are:
I x′x ' = I xx cos 2 θ + I yy sin 2 θ − 2 I xy sin θ cos θ
I y ' y ' = I xx sin 2 θ + I yy cos 2 θ + 2 I xy sin θ cosθ
I x ' y ' = I xx sin θ cos θ − I yy sin θ cosθ + I xy ( cos 2 θ − sin 2 θ )
We can write these relations in terms of the components of the unit
vectors n, t acting along the x' and y' axes, respectively, where
n = cos θ e x + sin θ e y
t = − sin θ e x + cos θ e y
where ex , ey are unit vectors along the x and y- axes.
We find
I x ' x ' = I xx nx2 + I yy n y2 − 2 I xy nx n y
I y ' y ' = I xx t x2 + I yy t y2 − 2 I xy t x t y
I x ' y ' = − I xx nx t x − I yy n y t y + I xy ( nx t y + n y t x )
These relations can be written in matrix notation as
⎡ Ix'x'
⎢−I
⎣ x' y'
or
where
− I x ' y ' ⎤ ⎡ nx
=
I y ' y ' ⎥⎦ ⎢⎣ t x
n y ⎤ ⎡ I xx
t y ⎥⎦ ⎢⎣ − I xy
[I '] = [Q] [I ][Q]
− I xy ⎤ ⎡ nx
I yy ⎥⎦ ⎢⎣ n y
tx ⎤
t y ⎥⎦
T
⎡ nx
Q
=
[ ] ⎢n
⎣ y
t x ⎤ ⎡ cos ( x, x ')
=
t y ⎥⎦ ⎢⎣cos ( y, x ')
is called the direction cosine matrix
cosine of the angle
between the x and y' axes,
cos ( x, y ') ⎤ etc.
⎥
cos ( y, y ') ⎦
Example (in MATLAB)
>> M = [ 288 72;72 72]
M=
y
y'
288 72
72 72
Ixx = 288 in4
Iyy = 72 in4
Ixy = -72 in4
x'
30o
x
>> angle =30*pi/180;
>> Q = [ cos(angle) -sin(angle); sin(angle) cos(angle)]
Q=
[Q] .. direction cosine matrix
0.8660 -0.5000
0.5000 0.8660
>> Mr =Q'*M*Q
[Mr] = [Q]T [M] [Q]
Mr =
296.3538 -57.5307
-57.5307 63.6462
Ix'x' = 296 in4
Iy'y' = 64 in4
Ix'y' = 58 in4
It is very easy to determine the principal values and principal directions
in this matrix form of the relations
Suppose we can find a set of coordinates where the unit vectors along
the principal axes have components Nx, Ny and Tx ,Ty . Then
⎡ I1
⎢0
⎣
0 ⎤ ⎡Nx
=⎢
⎥
I 2 ⎦ ⎣ Tx
N y ⎤ ⎡ I xx
Ty ⎥⎦ ⎢⎣ − I xy
− I xy ⎤ ⎡ N x
I yy ⎥⎦ ⎢⎣ N y
Tx ⎤
Ty ⎥⎦
where I1, I2 are the principal area moments and the mixed area moment is
zero in the principal axis coordinates
If we multiply both sides of the above equation by [Q] we find, since
⎧1 0 ⎫
T
=
=
Q
Q
I
[ ][ ] [ ] ⎨0 1 ⎬
⎩
⎭
⎡ Nx
⎢N
⎣ y
which gives
Tx ⎤ ⎡ I1
Ty ⎥⎦ ⎢⎣ 0
0 ⎤ ⎡ I xx
=⎢
⎥
I 2 ⎦ ⎣ − I xy
− I xy ⎤ ⎡ N x
I yy ⎥⎦ ⎢⎣ N y
Tx ⎤
Ty ⎥⎦
⎡ N x I1 Tx I 2 ⎤ ⎡ I xx
⎢ N I T I ⎥ = ⎢− I
y 2⎦
⎣ y 1
⎣ xy
− I xy ⎤ ⎡ N x
I yy ⎥⎦ ⎢⎣ N y
Tx ⎤
Ty ⎥⎦
This is equivalent to the two sets of equations:
⎧ N x I1 ⎫ ⎡ I xx − I xy ⎤ ⎧ N x ⎫
⎨
⎬=⎢
⎥⎨ ⎬
N
I
⎩ y 1 ⎭ ⎣ − I xy I yy ⎦ ⎩ N y ⎭
⎧Tx I 2 ⎫ ⎡ I xx − I xy ⎤ ⎧Tx ⎫
⎨
⎬=⎢
⎥ ⎨T ⎬
−
T
I
I
I
y
2
xy
yy
⎩
⎭ ⎣
⎦⎩ y⎭
Thus we see to find either the principal area moment I1 and its
principal direction N or the principal area moment I2 and its principal
direction T we need to solve the system of equations
⎡ I xx
⎢− I
⎣ xy
− I xy ⎤ ⎧U x ⎫ ⎧U x I ⎫
⎨ ⎬=⎨
⎬
I yy ⎦⎥ ⎩U y ⎭ ⎩U y I ⎭
where the unit vector U can be either N or T and I can be either I1 or I2
The system of equations
⎡ I xx
⎢− I
⎣ xy
− I xy ⎤ ⎧U x ⎫ ⎧U x I ⎫
⎨ ⎬=⎨
⎬
⎥
I yy ⎦ ⎩U y ⎭ ⎩U y I ⎭
which can also be written as
[I ]{U} = I {U}
is called an eigenvalue problem, whose solution is a scalar eigenvalue,
I, and a corresponding eigenvector, U
Since this eigenvalue problem can be rewritten as
⎡ I xx − I1
⎢ −I
xy
⎣
− I xy ⎤ ⎧U x ⎫ ⎧0 ⎫
⎨ ⎬=⎨ ⎬
I yy − I ⎥⎦ ⎩U y ⎭ ⎩0 ⎭
to find a solution we must solve the system of equations
( I xx − I )U x − I xyU y = 0
− I xyU x + ( I yy − I )U y = 0
But this is a homogeneous set of equations which only has the solution U = 0
unless the determinant of the matrix of coefficients is zero, i.e.
( I xx − I ) ( I yy − I ) − I xy2 = 0
Expanding this equation we obtain a quadratic equation for I:
I 2 − ( I xx + I yy ) I + ( I xx I yy − I xy2 ) = 0
which has the two roots
I1 , I 2
I
(
=
xx
+ I yy )
2
⎛ I xx − I yy ⎞
2
± ⎜
⎟ + I xy
2 ⎠
⎝
2
which we see are just the principal area moments
( I xx − I )U x − I xyU y = 0
− I xyU x + ( I yy − I )U y = 0
if we place one of the principal values back into the above system of
equations then we can solve for the corresponding principal direction.
However, since we have set the determinant of this system equal to zero,
the two equations above are not independent. Thus, we can only solve
one of them for a ratio of unit vector components. Thus, for example,
from the first equation and using I = I1 we have
I xy
Ux
=
U y ( I xx − I1 )
Note: this is equivalent to solving for θ via:
I xx − I1 )
(
tan θ =
I xy
But since U is a unit vector we have
U x2 + U y2 = 1
which we can solve for Uy in terms of the above ratio as
Uy =
±1
(U x / U y ) + 1
2
And then we can find Ux since the ratio Ux/Uy is known. Note that we only
get the vector solution to within a plus or minus sign since both U and –U
are principal directions. If we repeat the process for I2 then we can find the
second principal direction.
However, in MATLAB we can get both principal values and directions out
directly by just forming up the area moment matrix
⎡ I xx − I xy ⎤
M =⎢
⎥
I
I
−
xy
yy
⎣
⎦
and then giving that matrix to the built-in function eig which solves the
eigenvalue problem. the MATLAB call is:
[ pdirs, pvals] = eig(M)
The matrix pdirs will then have the principal direction components (in
columns) as
⎡(U x )1
pdirs = ⎢
⎢⎣(U y )1
(U x )2 ⎤
⎥
(U y )2 ⎥⎦
and the matrix pvals will have the corresponding principal values
⎡ I1
pvals = ⎢
⎣0
0⎤
I 2 ⎥⎦
I xx = 288
>> M = [ 288 72;72 72];
>> [pdirs, pvals] = eig(M)
I yy = 72
I xy = −72
pdirs =
0.2898 -0.9571
Example:
-0.9571 -0.2898
pvals =
I1 =50.2
U1 = 0.2898 ex - 0.9571ey
I2 =309.8
U2 = -0.9571 ex - 0.28981ey
angle (degrees) for 309.8 value
50.2002
0
0 309.7998
>> atan(pdirs(2,2)/pdirs(1,2))*180/pi
ans = 16.8450
It can be shown that the eigenvalues I1 , I2 of the eigenvalue problem
[I ]{U} = I {U}
are always real and the eigenvectors U1 , U2 are real
and orthogonal to each other since the matrix [ I ] is a
real, symmetrical matrix.
One of the reasons for treating the transformation of area moments by a
matrix approach is that it easily generalizes to more complex problems. For
example, in dynamics the three dimensional angular motion of a body (such
as a spinning satellite, for example) is controlled by the mass moments of
inertia
defined as
I xxm = ∫ ρ ( y 2 + z 2 ) dV
I yym = ∫ ρ ( x 2 + z 2 ) dV
I zzm = ∫ ρ ( x 2 + y 2 ) dV
I xym = ∫ ρ ( xy ) dV
I xzm = ∫ ρ ( xz ) dV
I yzm = ∫ ρ ( yz ) dV
where ρ is the mass density and dV is a volume element
In this case the mass moments transform due to a rotation of axes in just the
same manner as we have already discussed. If we let unit vectors n, t, v be
along the x', y', z' axes (which are assumed to be orthogonal to each other),
then
y
y'
x'
t
n
x
v
z
z'
⎡ Ix'x'
⎢
⎢− I y ' x '
⎢−I z 'x'
⎣
−I x' y'
I y'y'
−Iz' y'
− I x ' z ' ⎤ ⎡ nx
⎥ ⎢
− I y ' z ' ⎥ = ⎢ tx
I z ' z ' ⎥⎦ ⎢⎣ vx
ny
ty
vy
nz ⎤ ⎡ I xx
⎥⎢
t z ⎥ ⎢ − I yx
vz ⎥⎦ ⎢⎣ − I zx
− I xy
I yy
− I zy
− I xz ⎤ ⎡ nx
⎥
− I yz ⎥ ⎢⎢ n y
I zz ⎥⎦ ⎢⎣ nz
tx
ty
tz
vx ⎤
v y ⎥⎥
vz ⎥⎦
⎡ Ix'x'
⎢
⎢− I y ' x '
⎢−I z 'x'
⎣
−I x' y'
I y'y'
−Iz' y'
− I x ' z ' ⎤ ⎡ nx
⎥ ⎢
− I y ' z ' ⎥ = ⎢ tx
I z ' z ' ⎥⎦ ⎢⎣ vx
ny
ty
vy
nz ⎤ ⎡ I xx
⎥⎢
t z ⎥ ⎢ − I yx
vz ⎥⎦ ⎢⎣ − I zx
− I xy
I yy
− I zy
− I xz ⎤ ⎡ nx
⎥
− I yz ⎥ ⎢⎢ n y
I zz ⎥⎦ ⎢⎣ nz
tx
ty
tz
vx ⎤
v y ⎥⎥
vz ⎥⎦
We see that again we have
[I '] = [Q] [I ][Q]
T
In this case there are three principal mass moments of inertia and three
corresponding principal directions. These are again determined by the
solution of the eigenvalue problem
[I ]{U} = I {U}
Using MATLAB it is still easy to solve for the principal mass moments
of inertia and the principal directions with the same eigenvalue function
eig
>> I=[100 20 30; 20 300 50; 30 50 200]
I=
100
20
30
20 300
50
30
mass moment of inertia
matrix
50 200
EDU>> [pdirs, pvals]=eig(I)
pdirs =
0.9670 -0.2166
0.1337
-0.0322
0.4169
0.9084
-0.2525 -0.8827
0.3962
pvals =
91.4995
0
0
0 183.7468
0
0
0 324.7537
principal directions (in columns)
(x, y, z components of a unit
vector along the principal axis)
principal mass moments
of inertia