Moments of inertia of a cross section

BUDAPEST UNVERSTY OF TECHNOLOGY AND ECONOMCS Moments of inertia of a cross section Made by: David Lehotzky Budapest, March 3, 206.

Figure : The cross section under study a b h v d 0 8 4 0.5 5 Table : Parameters of the cross section in mm Problems Determine the moments of inertia about the horizontal and vertical axes fixed to the centroid of the cross section shown in Figure! 2 Calculate the principal moments of inertia at the centroid of the cross section! 3 Determine the moment of inertia about the zero axis of the cross section for the case when the cross section is loaded by a bending moment M which forms angle β 5 with the horizontal axis! Page/ 0

Figure 2: Division of the cross section onto sub cross sections Solution Calculation of the centroid As a first step, the cross section is divided onto parts for which formulas of the moments of inertia are known. One possible division of the cross section is shown in Figure 2. n the following, this particular division is applied.. Area of sub cross sections The sub cross sections shown in Figure 2, have areas A a 2 00 mm 2, A 2 bh 2 6 mm2, A 3 A 2 6 mm 2 A 4 d2 π 4 9.635 mm2, A 5 h2 2 8 mm2. The sub cross sections to be subtracted are considered with a minus sign..2 Centroids of sub cross sections The centroids of the different sub cross sections are given by position vectors (see Figure 2). The values of these position vectors in coordinate system (y, z ) are ys a/2 5 r S mm, z S a/2 5 ys2 a + h/3.333 r S2 mm, z S2 b/3 2.667 ys3 a v h/3 8.67 r S3 mm, z S3 a v b/3 6.833 ys4 h 4 r S4 mm, z S4 v + d/2 3 ys5 h/3.333 r S5 z S5 a h/3 8.667 Page2/ 0

Figure 3: Sub cross section.3 Calculation of the centroid The position vector in coordinate system (y, z ), pointing to the centroid of the cross section is r S ys z S 5 i A ir Si 5 i A i A r S + A 2 r S2 + A 3 r S3 + A 4 r S4 + A 5 r S5 6.377 A + A 2 + A 3 + A 4 + A 5 4.26 2 Moments of inertia at the centroids of sub cross sections The moments of inertia have to be calculated about axes y and z (see Figure 2) fixed to the centroid S of the cross section. Fist, one needs to calculate the moments of inertia at the centroids of sub cross sections, then apply the Parallel Axes Theorem to calculate their moments of inertia about axes y and z. Finally, the summation of moments of inertia of sub cross sections gives the moments of inertia of the cross section about axes y and z. 2. Moments of inertia of sub cross sections about axes y and z Sub cross section The st sub cross section is depicted in Figure 3. The moments of inertia of this sub cross section about its own y and z axes, fixed to its centroid are y a4 2 833.333 mm4, z y 833.333 mm 4, yz 0. The position vector pointing from centroid S of the sub cross section to centroid S of the cross section in the coordinate system (y, z ) is r SS Y Z ys y S z S z S.377 0.784 According to the Parallel Axes Theorem, the moment of inertia of sub cross section about axis y is while its moment of inertia about axis z is,y y + A Z 2 894.790 mm 4,,z z + A Y 2 022.900 mm 4, Page3/ 0

and its product of inertia about axes y and z is Sub cross section 2,yz yz + A Y Z 07.936 mm 4. The 2 nd sub cross section is depicted in Figure 4. The computations are similar to that of the st sub cross section. The moments of inertia of this sub cross section about its own y 2 and z 2 axes, fixed to its centroid are y2 b3 h 36 56.889 mm4, z2 bh3 36 4.222 mm4, y2z 2 b2 h 2 72 4.222 mm4. The position vector pointing from S 2 to S in coordinate system (y 2, z 2 ) is Y2 ys y r S2S S2 z S z S2 Z 2 4.957.549 Using the Parallel Axes Theorem, the moments of inertia of sub cross section 2 about axes y and z are 2,y y2 + A 2 Z 2 2 95.299 mm 4, 2,z z2 + A 2 Y 2 2 407.292 mm 4, 2,yz y2z 2 + A 2 Y 2 Z 2 37.095 mm 4. Figure 4: Sub cross section 2 Figure 5: Sub cross section 3 Sub cross section 3 The 3 rd sub cross section is depicted in Figure 5. The computations are similar to that of the 2 nd sub cross section. However, the moments of inertia about y 3 and z 3 are considered with a minus sign, since sub cross section 3 is to be subtracted. The moments of inertia of this sub cross section about its own y 3 and z 3 axes, fixed to its centroid are y3 b3 h 36 56.889 mm4, z3 bh3 36 4.222 mm4, y3z 3 b2 h 2 72 4.222 mm4. The position vector pointing from S 3 to S in coordinate system (y 3, z 3 ) is Y3 ys y r S3S S3 z S z S3 Z 3.790 2.67 Page4/ 0

Note that in contrast with the previous sub cross sections there is a ( ) multiplier in the formula of r S3S. This is due to that y 3 and z 3 axes point in the opposite directions as y and z, respectively. Using the Parallel Axes Theorem, the moments of inertia of sub cross section 3 about axes y and z are 3,y y3 + A 3 Z 2 3 66.49 mm 4, 3,z z3 + A 3 Y 2 3 65.478 mm 4, 3,yz y3z 3 + A 3 Y 3 Z 3 60.730 mm 4. t is important to point out that terms due to the Parallel Axes Theorem already consider the minus sign of subtraction within A 3. Sub cross section 4 The 4 th sub cross section is depicted in Figure 6. The computations are similar as before. The moments of inertia of this sub cross section about its own y 4 and z 4 axes, fixed to its centroid are y4 d4 π 64 30.680 mm4, z4 y4 30.680 mm 4, y4z 4 0. The position vector pointing from S 4 to S in coordinate system (y 4, z 4 ) is Y4 ys y r S4S S4 2.377 Z 4 z S z S4.26 Using the Parallel Axes Theorem, the moments of inertia of sub cross section 4 about axes y and z are 4,y y4 + A 4 Z 2 4 59.76 mm 4, 4,z z4 + A 4 Y 2 4 4.605 mm 4, 4,yz y4z 4 + A 4 Y 4 Z 4 56.752 mm 4. Figure 6: Sub cross section 4 Figure 7: Sub cross section 5 Sub cross section 5 The 5 th sub cross section is depicted in Figure 7. The moments of inertia of this sub cross section about its own y 5 and z 5 axes, fixed to its centroid are y5 h4 36 7. mm4, z5 y5 7. mm 4, y5z 5 h4 72 3.555 mm4. The position vector pointing from S 5 to S in coordinate system (y 5, z 5 ) is r Y5 y S5S S y S5 5.044 Z 5 (z S z S5 ) 4.45 Axes z 5 and z point in opposite directions, therefore in contrast with the previous sub cross section, the corresponding coordinate of r S5S is multiplied by ( ). Using the Parallel Axes Theorem, the moments of inertia of sub cross section 5 about axes y and z are 5,y y5 + A 5 Z 2 5 65.575 mm 4, 5,z z5 + A 5 Y 2 5 20.607 mm 4, 5,yz y5z 5 + A 5 Y 5 Z 5 76.08 mm 4. Page5/ 0

2.2 Moments of inertia of the cross section about axes y and z Since all the moments of inertia of sub cross sections are calculated about axes y and z, the moments of inertia of the cross section about axes y and z can be obtained by the summation of corresponding moments of inertia of sub cross sections, thus yz y z 5 i,y,y + 2,y + 3,y + 4,y + 5,y 598.307 mm 4, i 5 i,z,z + 2,z + 3,z + 4,z + 5,z 02.500 mm 4, i 5 i,yz,yz + 2,yz + 3,yz + 4,yz + 5,yz 538.53 mm 4. i Using these results, the matrix of moments of inertia at centroid S, in coordinate system (y, z) is written as y yz 598.307 538.53 mm 4. () 538.53 02.500 (y,z) yz 3 Principle moments of inertia and principle axes z n case of pure bending, the normal stress of beams can be calculated using the Navier formula. When only bending moment M η is acting (rotating about axis η) at the centroid of the cross section, the normal stress is given by σ x (ζ) M η η ζ. (2) n this formula, ζ is an axis perpendicular to η within the plane of the cross section. t is important to note that this formula is valid only when η is a principal moment of inertia, that is product of inertia ηζ is zero. n this case η and ζ are principal axes. Since yz 0, hence y and z are not principal axes. Consequently, in order to apply the Navier formula (2), one has to find the principle directions corresponding to η and ζ. This can be done by the calculation of eigenvalues and eigenvectors of. During the eigenvalue-eigenvector calculation a direction given by vector n is sought, with which the multiplication of matrix is equivalent to the product of a number λ and vector n. That is, a direction n is sought in which the multiplication with matrix behaves exactly as a multiplication with some number λ. Consequently, one is looking for all the pairs of λ and n, which satisfy n λ n. (3) Number λ is called the eigenvalue, while vector n is called the eigenvector of. After rearranging (3), one can have ( λe ) n 0. (4) f n 0 was valid, then no direction could be obtained. Consequently, it follows from (4) that det ( λe ) ( ) det y yz 0 λ yz z 0 y λ yz yz z λ 0 has to hold. After the expansion of the determinant (for the calculation of determinant see the Appendix), one has y λ yz z λ ( y λ)( z λ) yz 2 λ 2 ( y + z ) λ + y z yz 2 0. yz The solutions of this second order equation are λ,2,2 2 ( ) y + z ± ( y z ) 2 + 4yz 2 { 382.390 mm 4 228.427 mm 4. Page6/ 0

These eigenvalues give the principle moments of inertia and 2. Eigenvectors n and n 2, corresponding to eigenvalues λ and λ 2, respectively, can be calculated by the substitution of eigenvalues in (4). After the substitution of λ, one has y λ yz n 0. (5) yz z λ n 2 0 }{{} n The expansion of the matrix-vector product (for details see the Appendix) gives equations ( y λ ) n yz n 2 0, (6) yz n + ( z λ ) n 2 0. (7) Here coordinates n and n 2 of vector n are unknowns. However, equations (6)-(7) are not independent from each other, since det( λ E) 0. Consequently, one out of the two unknown coordinates of n can be chosen arbitrarily. Let n be chosen. n this case (6) gives which implies n 2 y λ yz n y λ yz, n y λ yz.456 Similarly, the substitution of λ 2 to (4), gives yz z λ 2 n 22 y λ 2 yz n2 which, after setting n 2, implies n 2 y λ 2 yz } {{ } n 2 0.687. 0 0., (8) Note that the length of eigenvectors n and n 2 can be changed by modifying the arbitrarily chosen coordinates n and n 2, respectively. t is useful to give the eigenvectors as unit vectors. This can be done by the norming of n and n 2. The normed eigenvectors are m m 2 m m2 m 2 m 22 n n n 2 + n 2 2 n 2 n 2 n 2 2 + n 2 22 n n2 n 2 n 22 0.566 0.824 0.824 0.566 The principal axes ζ and η are assigned by the normed eigenvectors m and m 2, respectively (see Figure 8). Consequently, the moments of inertia about these axes are ζ and η 2. t is important to note that depending on how n and n 2 were chosen in (5) and (8), vectors m and m 2 could point in the opposite direction as well. These results would be also correct.,. 4 Moment of inertia about the zero axis The bending moment M acting in the cross section can be decomposed onto coordinates M η and M ζ in the coordinate system of the principle axes (see Figure 8). By assuming that only bending moment M η acts in the cross section (see Figure 9), the normal stress can be calculated according to (2). With the assumption that only bending moment M ζ acts in the cross section (see Figure 0) the normal stress is determined by σ x (η) M ζ ζ ( η). (9) Page7/ 0

Figure 8: The principal axes of the cross section shown by blue color and the zero axis shown by green color Here coordinate η has a ( ) multiplier since normal stress σ x is positive for negative η coordinates. When both M η and M ζ act in the cross section, the normal stress is the sum of (2) and (9), that is σ x (η, ζ) M η η ζ M ζ ζ η. (0) The zero axis consists of points in the cross section where the normal stress is zero, that is where σ x (η, ζ) 0. Based on Figure 8 ( ( ) ) m 22 M η M cos(α + δ + β) M cos atan + β 0.650 M, and using (0), the zeros axis is given by the line m 2 M ζ M sin(α + δ + β) 0.760 M, ζ M ζ η M η ζ η tan(α + β + δ) η ζ 0.93 η. The slope of the line is characterized by the angle (see Figure 8) ( α atan tan(α + β + δ) ) η 0.943. ζ Page8/ 0

Figure 9: Cross section parts under tension (red) and compression (green) in the case when only M η acts in the cross section Figure 0: Cross section parts under tension (red) and compression (green) in the case when only M ζ acts in the cross section The matrix () of moments of inertia is given in coordinate system (y, z). Using matrix M, the matrix of moments of inertia can be transformed in the coordinate system (η, ζ) of principle directions as where matrix M m2 m M (y,z) M, () m2 m 0.824 0.566 m 22 m 2 0.566 0.824 describes the transformation from coordinate system (y, z) to the coordinate system (η, ζ) assigned by m and m 2. Details on the computation of the inverse A of a matrix A can be found in the Appendix. Due to that M is always symmetric, its inversion is simplified as M M T, that is the inverse of M is equivalent to its transpose. Details on the computation of the transpose A T of a matrix A can be found in the Appendix. The substitution to () gives η ηζ ηζ ζ 0.824 0.566 0.566 0.824 598.307 538.53 538.53 02.500 0.824 0.566 0.566 0.824 228.427 0 0 382.390 mm 4, that is ηζ 0 is obtained. t can be shown that this follows from (3) in case of any. n general, the transformation from coordinate system (η, ζ) to coordinate system (η, ζ ) can be described by the matrix e2 e T e2 e, e 22 e 2 where vectors e 2 and e assign the direction of coordinates (η, ζ ) in coordinate system (η, ζ). Consequently, any vector given in (η, ζ) can be transformed to (η, ζ ) as r (η,ζ ) T r. The above transformation can be applied for the matrix of moments of inertia as (η,ζ ) T T. (2) Based on Figure 8, coordinate system (η, ζ ) assigned by the zero axis is given by unit vectors cos(α) 0.982 sin(α) 0.90 e 2, e sin(α) 0.90 cos(α) 0.982 in coordinate system (η, ζ). Since unit vectors e and e 2 are perpendicular to each other T T T. After substitution, (2) gives η η ζ 0.982 0.90 382.390 0 0.982 0.90 270.03 25.08 η ζ mm 4. ζ 0.90 0.982 0 228.47 0.90 0.982 25.08 340.800 (η,ζ ) Page9/ 0

Consequently, the moment of inertia about the zero axis is Matrix (η,ζ ) η 270.03 mm 4. can be also determined by a transformation matrix defined in coordinate system (y, z). Based on Figure 8 ( ) m 22 δ atan α 23.539. n coordinate system (y, z), the coordinate system (η, ζ ) can be assigned by unit vectors cos(δ) 0.97 sin(δ) 0.399 e 2, e sin(δ) 0.399 cos(δ) 0.97 (y,z) (y,z) m 2. These unit vectors are again perpendicular to each other thus T T T cos(δ) sin(δ) y sin(δ) cos(δ) (η,ζ ) (y,z) (y,z) (y,z) yz yz z cos(δ) sin(δ) sin(δ) cos(δ) 270.03 25.08 25.08 340.800 mm 4. Appendix Some basic manipulations of 2 2 matrices a a A 2 b b, B 2 a 2 a 22 b 2 b 22, d d d 2 Transpose A T a a 2 a 2 a 22 Multiplication A d C A B a d + a 2 d 2 a 2 d + a 22 d 2 c c 2 c 2 c 22 c a b + a 2 b 2, c 2 a b 2 + a 2 b 22 c 2 a 2 b + a 22 b 2, c a 2 b 2 + a 22 b 22 Determinant nverse A det(a) adj ( A ) det(a) a a 22 a 2 a 2 a a 22 a 2 a 2 a22 a 2 a 2 a Page0/ 0