y Area Moment Matrices 1 e y n θ n n y θ x e x The transformation equations for area moments due to a rotation of axes are:


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1 ' t Area Moment Matrices x' t 1 e n 1 t θ θ θ x e x t x n x n n The transformation equations for area moments due to a rotation of axes are: θ θ θ θ 2 2 x x ' xx cos + sin 2 x sin cos 2 2 ' ' xx sin + cos + 2 x sin cos θ θ θ θ x' ' xx x 2 2 sinθ cosθ sinθcosθ + cos θ sin θ We can write these relations in terms of the components of the unit vectors n, t acting along the x' and ' axes, respectivel, where n cosθe + sinθe x t sinθ e + cosθ e x where e x, e are unit vectors along the x and  axes.
2 We find n + n 2 n n 2 2 x' x' xx x x x t + t 2 t t 2 2 ' ' xx x x x n t n t + n t + n t x' ' xx x x x x x These relations can be written in matrix notation as x' x' x' ' nx n xx x nx tx t t n t x ' ' ' ' x x or where [ Q] T [ ' ] [ Q] [ ][ Q] nx tx cos x, x' cos x, ' n t cos x, ' cos, ' cosine of the angle between the x and ' axes, etc. is called the direction cosine matrix
3 Example (in MATLAB) >> M [ ;72 72] M ' xx 288 in 4 72 in 4 x 72 in 4 30 o x' >> angle 30*pi/180; x >> Q [ cos(angle) sin(angle); sin(angle) cos(angle)] Q >> Mr Q'*M*Q Mr [Q].. direction cosine matrix [M r ] [Q] T [M] [Q] x'x' 296 in 4 '' 64 in 4 x'' 58 in 4
4 t is ver eas 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 N x, N and T x,t. Then 0 N N N T 1 x xx x x x 0 T 2 x T x N T where 1, 2 are the principal area moments and the mixed area moment is zero in the principal axis coordinates f we multipl both sides of the above equation b [Q] we find, since [ Q][ Q] [ ] T N T 0 N T x x 1 xx x x x N T 0 2 x N T which gives
5 N x 1 T x 2 xx x Nx Tx N T N T 1 2 x This is equivalent to the two sets of equations: N x 1 xx x Nx N N 1 x T x 2 xx x Tx T T 2 x Thus we see to find either the principal area moment 1 and its principal direction N or the principal area moment 2 and its principal direction T we need to solve the sstem of equations xx x Ux Ux U U x where the unit vector U can be either N or T and can be either 1 or 2
6 The sstem of equations U U xx x x x x U U which can also be written as [ ]{ U} { U} is called an eigenvalue problem, whose solution is a scalar eigenvalue,, and a corresponding eigenvector, U Since this eigenvalue problem can be rewritten as xx 1 x Ux 0 x U 0 to find a solution we must solve the sstem of equations
7 U U 0 xx x x U + U 0 x x But this is a homogeneous set of equations which onl has the solution U 0 unless the determinant of the matrix of coefficients is zero, i.e. 2 xx x 0 Expanding this equation we obtain a quadratic equation for : xx xx x which has the two roots, 1 2 xx + xx ± x which we see are just the principal area moments
8 U U 0 xx x x U + U 0 x x if we place one of the principal values back into the above sstem of equations then we can solve for the corresponding principal direction. However, since we have set the determinant of this sstem equal to zero, the two equations above are not independent. Thus, we can onl solve one of them for a ratio of unit vector components. Thus, for example, from the first equation and using 1 we have U x x U ( ) xx 1 Note: this is equivalent to solving for θ via: ( xx 1 ) tanθ x But since U is a unit vector we have U + U x which we can solve for U in terms of the above ratio as
9 U ± 1 ( U ) 2 x U / + 1 And then we can find U x since the ratio U x /U is known. Note that we onl get the vector solution to within a plus or minus sign since both U and U are principal directions. f we repeat the process for 2 then we can find the second principal direction. However, in MATLAB we can get both principal values and directions out directl b just forming up the area moment matrix M xx x x and then giving that matrix to the builtin 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
10 pdirs ( U ) ( U ) x 1 x 2 ( U) ( U ) 1 2 and the matrix pvals will have the corresponding principal values pvals >> M [ ;72 72]; >> [pdirs, pvals] eig(m) xx x Example: pdirs U e x e U e x e angle (degrees) for value pvals >> atan(pdirs(2,2)/pdirs(1,2))*180/pi ans
11 t can be shown that the eigenvalues 1, 2 of the eigenvalue problem [ ]{ U} { U} are alwas real and the eigenvectors U 1, U 2 are real and orthogonal to each other since the matrix [ ] is a real, smmetrical matrix.
12 One of the reasons for treating the transformation of area moments b a matrix approach is that it easil generalizes to more complex problems. For example, in dnamics the three dimensional angular motion of a bod (such as a spinning satellite, for example) is controlled b the mass moments of inertia defined as m xx m m zz m x m xz m z ( 2 2) ( 2 2) ( 2 2) ρ + z dv ρ x + z dv ρ x + dv ρ ρ ρ x dv xz dv z dv where ρ is the mass densit and dv is a volume element
13 n this case the mass moments transform due to a rotation of axes in just the same manner as we have alread discussed. f we let unit vectors n, t, v be along the x', ', z' axes (which are assumed to be orthogonal to each other), then ' t n x' v x z z' x' x' x' ' x' z ' nx n n z xx x xz nx tx vx t t t n t v x ' ' ' ' z ' ' x z x z zx ' ' z ' ' zz ' ' vx v v z zx z zz nz tz v z
14 x' x' x' ' x' z ' nx n n z xx x xz nx tx vx t t t n t v x ' ' ' ' z ' ' x z x z zx ' ' z ' ' zz ' ' vx v v z zx z zz nz tz v z We see that again we have T [ ' ] [ Q] [ ][ Q] n this case there are three principal mass moments of inertia and three corresponding principal directions. These are again determined b the solution of the eigenvalue problem [ ]{ U} { U}
15 Using MATLAB it is still eas to solve for the principal mass moments of inertia and the principal directions with the same eigenvalue function eig >> [ ; ; ] mass moment of inertia matrix EDU>> [pdirs, pvals]eig() pdirs pvals principal directions (in columns) (x,, z components of a unit vector along the principal axis) principal mass moments of inertia
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