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by: Mariela Collier


Mariela Collier
GPA 3.86


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Class Notes
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This 15 page Class Notes was uploaded by Mariela Collier on Sunday September 27, 2015. The Class Notes belongs to E M 274H at Iowa State University taught by Staff in Fall. Since its upload, it has received 23 views. For similar materials see /class/214484/e-m-274h-iowa-state-university in Engineering Mechanics at Iowa State University.




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Date Created: 09/27/15
Area Moment Matrices The transformation equations for area moments clue to a rotation of axes are I xx cos2 6 IW sin2 6 21W sin6cos6 IN 1 sin2 61 cos2 621xy sin6cos6 1W xx sin6cos6 Iyy sin6cos61xy cos2 6 sin2 6 We can write these relations in terms of the components of the unit vectors n tacting along the x39 and y39 axes respectively where n cos 96x sin6ey t sin6ex cos6ey where ex ey are unit vectors along the x and y axes We find 2 2 I Ixxnx Iyyny 21xynxny 2 2 I mix Iyyly 21xylxly I xy Inl 1 nl 1 nl nl x Wyy xy xy yx xx x These relations can be written in matrix notation as I Ixy nx ny xx Ixy nx Ix I I t t I I n t x y y y x y xy yy y y T I cosine of the angle Where between the x and y39 axes Qx txcosx2x39 cosxy39 etc y l y cosyx39 cosyy39 or is called the direction cosine matrix Example in MATLAB gtgt M 288 7272 72 M y yl 288 72 IXX 288 in4 I 72 72 I 72 in4 X W IXy 72 m4 30o gtgt angle 30pi180 X gtgt Q cosange sinange sinangle cosange Q Q direction cosine matrix 08660 05000 05000 08660 gtgt Mr Q39MQ M QTM Q Mr I 4 2963538 575307 39x39x39 296 575307 636462 lyy 64 In 58 in4 XY 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 11 0 Nx Ny xx Ixy Nx TX 0 12 Tx Ty 1xy 1yy Ny Ty where 11 12 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 QllQlTIl1 f V Till 0l l quotquotlle El Ny Ty 0 I2 Ixy Iyy Ny Ty which gives Fug g 5 JWng a NyI1 Ty2 1W 1 Ny Ty yy This is equivalent to the two sets of equations lel xx Ixy Nx Nyll 1Xy 1W Ny TX12 xx Ixy TX Tyll Ixy Iyy Ty Thus we see to find either the principal area moment 11 and its principal direction N or the principal area moment 12 and its principal direction T we need to solve the system of equations LW 1 o Lg where the unit vector U can be either N or T and can be either 11 or 12 The system of equations 1W 1W Uy UyI which can also be written as 1U 1U is called an eigenvalue problem whose solution is a scalar eigenvalue 1 and a corresponding eigenvector U Since this eigenvalue problem can be rewritten as In 11 Ixy 1xy Iyy I Uy o to find a solution we must solve the system of equations 1m IUx IWUy 0 1nyx 1W IUy 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 ie 2 1m 11yy 1 1W 0 Expanding this equation we obtain a quadratic equation for 2 Ixx IWIIxxIW 4 0 which has the two roots 1 1 1 1 2 1712 xx yyi xx yy 12 2 2 W which we see are just the principal area moments Im IUx I U 0 xyy 1ny 1yy IUy 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 11 we have U 1 Note this is equivalent to solving for 6 via U Ixx Il tan6M I 0 But since U is a unit vector we have 1lU Uj 1 which we can solve for Uy in terms of the above ratio as i1 Uy Ux Uy2 1 And then we can find UX since the ratio UXUy 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 12 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 M xx Ixy 1 1 and then giving that matrix to the builtin function eig which solves the eigenvalue problem the MATLAB call is pdirs pvals eigM The matrix pdirs will then have the principal direction components in columns as Ux1 Ux2 Uy 52 pdirs and the matrix pvals will have the corresponding principal values I 11 0 VClS p 0 12 Example 11 5o2 u1 02898 ex 09571ey 12 3098 u2 o9571 ex 028981ey angle degrees for 3098 value xx 288 gtgt M 288 7272 72 I I W 72 gtgt pdirs pvals eigM Ixy 72 pdirs 02898 09571 09571 02898 pvals 502002 0 0 3097998 gtgt atanpdirs22pdirs12180pi ans 168450 It can be shown that the eigenvalues I1 12 of the eigenvalue problem 1U 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 where p 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 x39 y39 z39 axes which are assumed to be orthogonal to each other then Ix x 1x39y39 1x Z nx ny n2 xx Ixy Ixz nx Ix Vx Iy x y y 1y 2 x y t2 Iyx Iyy Iyz y ty Vy I I I v v v sz IZy ZZZ Z IZ v2 x x x y x Z x y Z xx xy x2 x x x Iy x y y y 2 2 ix y 2 Iyx Iyy Iyz y ty Vy I I I v v v I I I t v We see that again we have T I I l Q 1in 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 IU IlU 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 100 20 30 20 300 50 3O 50 200 EDUgtgt pdirs pvalseigl pdirs 09670 00322 O2525 01337 09084 03962 O2166 04169 O8827 pvals 914995 0 O 0 1837468 0 O 0 3247537 gtgt 1OO 20 30 20 300 50 3O 50 200 lt mass moment of inertia matrix principal directions in columns x y 2 components of a unit vector along the principal axis principal mass moments of inertia


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