SECTION 12.6 Polar Moments of Inertia 15 Polar Moments of Inertia Problem 12.6-1 Determine the polar moment of inertia IP of an isosceles triangle of base b and altitude h with respect to its apex (see Case 5, Appendix D) Solution 12.6-1 Polar moment of inertia y A 2/3 h h C b POINT A (APEX): IP (IP ) c IP bh (4h2 3b 2 ) 144 bh 2 (b 12h2 ) 48 A¢ 2h 2 ≤ 3 bh 2h 2 ¢ ≤ 2 3 POINT C (CENTROID) FROM CASE 5: (IP ) c bh (4h2 144 3b 2 ) Problem 12.6-2 Determine the polar moment of inertia (IP)C with respect to the centroid C for a circular sector (see Case 13, Appendix D). Solution 12.6-2 C y O Polar moment of inertia y A y r x r2 2r sin 3 r2 ¢ ≤ POINT C (CENTROID): (IP ) C (IP ) O r4 (9 18 Ay2 2 r4 2 8 sin2 ) 2r sin 3 2 POINT O (ORIGIN) FROM CASE 13: (IP ) o r4 2 ( radians) Problem 12.6-3 Determine the polar moment of inertia IP for a W 8 wide-flange section with respect to one of its outermost corners. 21 Solution 12.6-3 Polar moment of inertia 2 1 y O 2 C 1 W 8 21 I1 75.3 in.4 I2 9.77 in.4 A 6.16 in.2 Depth d 8.28 in. Width b 5.27 in. Ix I1 A(d 2)2 75.3 6.16(4.14)2 180.9 in.4 Iy I2 Ix A(b 2)2 Iy 9.77 6.16(2.635)2 52.5 in.4 IP 233 in.4 x 16 CHAPTER 12 Review of Centroids and Moments of Inertia Problem 12.6-4 Obtain a formula for the polar moment of inertia IP with respect to the midpoint of the hypotenuse for a right triangle of base b and height h (see Case 6, Appendix D). Solution 12.6-4 Polar moment of inertia d = CP POINT P: IP (IP ) c Ad 2 bh 2 b b 2 h h 2 ¢ ≤ ¢ ≤ 2 3 2 3 b 2 h2 b 2 h2 36 36 36 bh 2 bh b 2 h2 (h b 2) ¢ ≤ 36 2 36 bh 2 (b h2 ) 24 h h/2 C P h/3 b/3 b/2 b A d2 POINT C FROM CASE 6: (IP ) c bh 2 (h 36 b 2) IP Problem 12.6-5 Determine the polar moment of inertia (IP)C with respect to the centroid C for a quarter-circular spandrel (see Case 12, Appendix D). Solution 12.6-5 y Polar moment of inertia yC POINT C (CENTROID): Ix c xC x x r Ix Ay 2 ¢1 C O y 4 POINT O FROM CASE 12: Ix y A 5 ¢1 ≤r 4 16 (10 3 )r 3(4 ) ¢1 COLLECT TERMS AND SIMPLIFY: IxC IyC (IP ) c r4 176 84 9 ¢ 144 4 IxC (by symmetry) 2 IxC 2 ≤ (r 2 ) B ¢1 5 ≤r 4 16 (10 3 )r 2 R 3(4 ) ≤ 4 ≤r 2 r4 176 84 ¢ 72 4 9 2 ≤ SECTION 12.7 Products of Inertia 17 Products of Inertia Problem 12.7-1 Using integration, determine the product of inertia Ixy for the parabolic semisegment shown in Fig. 12-5 (see also Case 17 in Appendix D). Solution 12.7-1 Product of inertia Product of inertia of element dA with respect to axes through its own centroid equals zero. x2 dA y dx h ¢ 1 ≤ dx b2 dIxy product of inertia of element dA with respect to xy axes d 1 x d2 y 2 Parallel-axis theorem applied to element dA: dIxy 0 (dA)(d1d2) (y dx)(x)(y 2) h2x x2 2 ¢1 ≤ dx 2 b2 Ixy dIxy h2 2 b 0 y y h1 ( x2 b2 ) h y/2 O x b dA dx x x ¢1 x2 2 ≤ dx b2 b 2h2 12 Problem 12.7-2 Using integration, determine the product of inertia Ixy for the quarter-circular spandrel shown in Case 12, Appendix D. Solution 12.7-2 Product of inertia ELEMENT dA: y r x dA dy y d1 (r x)/2 x EQUATION OF CIRCLE: x 2 ( y r)2 r 2 or r 2 x 2 (y r)2 distance to its centroid in x direction (r x) 2 d2 distance to its centroid in y direction y dA area of element (r x) dy Product of inertia of element dA with respect to axes through its own centroid equals zero. Parallel-axis theorem applied to element dA: r x dIxy 0 (dA)(d1d2 ) (r x)(dy) ¢ ≤ (y) 2 1 1 2 (r x 2 ) y dy (y r) 2y dy 2 2 r Ixy 12 0 y(y r) 2 dy r4 24 18 CHAPTER 12 Review of Centroids and Moments of Inertia Problem 12.7-3 Find the relationship between the radius r and the distance b for the composite area shown in the figure in order that the product of inertia Ixy will be zero. y r O b x Solution 12.7-3 yC r C O y Product of inertia C centroid of semicircle xC x b SEMICIRCLE (CASE 10): Ixy Ixcyc Ixy Ixcyc 0 0 ¢ Ad1d2 A r2 2 d1 r d2 2r 4 3 4r 3 TRIANGLE (CASE 7): Ixy b 2h2 24 b 2 (2r) 2 24 b 2r 2 6 COMPOSITE AREA (Ixy Ixy b r 6 2 2 r2 4r ≤ (r) ¢ ≤ 2 3 0) 2r 3 4 0 b 2r Problem 12.7-4 Obtain a formula for the product of inertia Ixy of the symmetrical L-shaped area shown in the figure. b y t t O x b Solution 12.7-4 y t b Product of inertia AREA 2: (Ixy ) 2 Ixc yc A1 A2 b t t x b A2d1d2 (b 0 t2 2 (b 4 t 2) t)(t)(t 2) ¢ t2 (2b2 4 b 2 t ≤ O COMPOSITE AREA: Ixy (Ixy ) 1 (Ixy ) 2 t2 ) AREA 1: (Ixy ) 1 t2b2 4 SECTION 12.7 Products of Inertia 19 Problem 12.7-5 Calculate the product of inertia I12 with respect to the centroidal axes 1-1 and 2-2 for an L 6 6 1 in. angle section (see Table E-4, Appendix E). (Disregard the cross-sectional areas of the fillet and rounded corners.) Solution 12.7-5 y 2 1 in. Product of inertia Coordinates of centroid of aera A1 with respect to 1–2 axes: d1 d2 (x 3.0 0 0.5) 1.3636 in. y 1.1364 in. A1 1 6 in. y O x 2 C 5 in. 1 in. 6 in. x 1 Product of inertia of area A1 with respect to 1-2 axes: I¿ 12 A1d1d2 (6.0 in.2)( 1.3636 in.)(1.1364 in.) 9.2976 in.4 Coordinates of centroid of area A2 with respect to 1–2 axes: d1 3.5 x 1.6364 in. d2 (y 0.5) 1.3636 in. Product of inertia of area A2 with respect to 1-2 axes: I– 12 0 A2d1d 2 (5.0 in.2)(1.6364 in.)( 1.3636 in.) 11.1573 in.4 ANGLE SECTION: I12 I¿ 12 I– 12 20.5 in.4 A2 All dimensions in inches. A1 (6)(1) 6.0 in.2 A2 (5)(1) 5.0 in.2 A A1 A2 11.0 in.2 With respect to the x axis: 6 in. Q 1 (6.0 in.2 ) ¢ ≤ 18.0 in.3 2 Q2 y x (5.0 in.2 ) ¢ 1.0 in. ≤ 2.5 in.3 2 Q 1 Q 2 20.5 in.3 1.8636 in. A 11.0 in.2 y 1.8636 in. Problem 12.7-6 Calculate the product of inertia Ixy for the composite area shown in Prob. 12.3-6. Solution 12.7-6 Product of inertia y d1 AREA A1: (Ixy)1 AREA A2: (Ixy)2 AREA A3: (Ixy)3 0 (By symmetry) 30)(60)(75) A1 0 A2 d1d2 (90 12.15 106 mm4 0 (By symmetry) A2 O d2 x A4 A3 All dimensions in millimeters A2 90 A1 360 30 mm A3 180 30 mm A3 90 d1 60 mm d2 75 mm 30 mm 30 mm AREA A4: (Ixy)4 (Ixy)2 12.15 106 mm4 Ixy (Ixy)1 (Ixy)2 (Ixy)3 (Ixy)4 (2)(12.15 106 mm4) 24.3 106 mm4 20 CHAPTER 12 Review of Centroids and Moments of Inertia Problem 12.7-7 Determine the product of inertia Ixcyc with respect to centroidal axes xc and yc parallel to the x and y axes, respectively, for the L-shaped area shown in Prob. 12.3-7. Solution 12.7-7 Product of inertia y 2 x With respect to the y axis: Q 1 A1x 1 (3.0 in.2 )(0.25 in.) Q2 x xc x 0.75 in.3 3.9375 in.3 A1 6.0 C 3.5 O 4.0 y A2x 2 Q1 Q2 (1.75 in.2 )(2.25 in.) A2 All dimensions in inches. A1 (6.0)(0.5) 3.0 in.2 A2 (3.5)(0.5) 1.75 in.2 A A1 A2 4.75 in.2 With respect to the x axis: Q 1 A1 y1 (3.0 in.2 )(3.0 in.) 9.0 in.3 Q 2 A2 y2 (1.75 in.2 )(0.25 in.) 0.4375 in.3 y Q1 A Q2 9.4375 in.3 4.75 in.2 1.9868 in. 4.6875 in.3 0.98684 in. A 4.75 in.2 Product of inertia of area A1 with respect to xy axes: (Ixy)1 (Ixy)centroid A1 d1 d2 0 (3.0 in.2)(0.25 in.)(3.0 in.) 2.25 in.4 Product of inertia of area A2 with respect to xy axes: (Ixy)2 (Ixy)centroid A2 d1 d2 0 (1.75 in.2)(2.25 in.)(0.25 in.) 0.98438 in.4 ANGLE SECTION Ixy (Ixy)1 (Ixy)2 3.2344 in.4 CENTROIDAL AXES Ixcyc Ixy Ax y 3.2344 in.4 (4.75 in.2)(0.98684 in.)(1.9868 in.) 6.079 in.4 Rotation of Axes The problems for Section 12.8 are to be solved by using the transformation equations for moments and products of inertia. product of inertia Ix y for a square with sides b, as shown in the figure. (Note 1 1 that the x1y1 axes are centroidal axes rotated through an angle with respect to the xy axes.) y1 y x1 b C b x Problem 12.8-1 Determine the moments of inertia Ix1 and Iy1 and the Solution 12.8-1 y1 y Rotation of axes EQ. (12-29): Ix1 x1 Iy1 Ix Iy Iy1 b4 12 b C b x EQ. (12-27): Ix1y1 Ixy 0 Ix 2 Iy sin 2u Ixy cos 2u 0 EQ. (12-25): Ix1 Ix 2 Ix 2 Iy 0 Iy Ix 2 0 Iy FOR A SQUARE: b4 Ix Iy 12 cos 2u b4 12 Ixy sin 2u Since may be any angle, we see that all moments of inertia are the same and the product of inertia is always zero (for axes through the centroid C). SECTION 12.8 Rotation of Axes 21 Problem 12.8-2 Determine the moments and product of inertia with respect to the x1y1 axes for the rectangle shown in the figure. (Note that the x1 axis is a diagonal of the rectangle.) y1 y x1 h C x b Solution 12.8-2 y1 y Rotation of axes (rectangle) ANGLE OF ROTATION: x1 cos u cos 2 sin 2 b b 2 h 2 sin u sin2 h b 2 h2 h C x cos2 2 sin cos b b 2 h2 b 2 h2 2 bh b 2 h2 CASE 1: Ix bh3 12 Iy hb 3 12 Ixy 0 SUBSTITUTE INTO EQS. (12-25), (12-29), AND (12-27) AND SIMPLIFY: Ix1 Ix1y1 b 3h3 6(b 2 h2 ) b 2h2 (h2 b 2 ) 12(b 2 h2 ) Iy1 bh(b 4 12(b 2 h4 ) h2 ) Problem 12.8-3 Calculate the moment of inertia Id for a W 12 50 wide-flange section with respect to a diagonal passing through the centroid and two outside corners of the flanges. (Use the dimensions and properties given in Table E-1.) Solution 12.8-3 Rotation of axes y Tan u d 12.19 1.509 b 8.080 56.46º 2 112.92º d C x EQ. (12-25): Id Ix 2 394 Iy Ix 2 Iy cos 2u Ixy sin 2u 0 b W 12 50 Ix 394 Iy 56.3 in.4 Ixy 0 Depth d 12.19 in. Width b 8.080 in. in.4 56.3 394 56.3 cos (112.92 ) 2 2 225 in.4 66 in.4 159 in.4 22 CHAPTER 12 Review of Centroids and Moments of Inertia Problem 12.8-4 Calculate the moments of inertia Ix1 and Iy1 and the product of inertia Ix y with respect to the x1y1 axes for the L-shaped area shown in the 1 1 figure if a 150 mm, b 100 mm, t 15 mm, and 30°. y y1 t a x1 t Probs. 12.8-4 and 12.9-4 O x b Solution 12.8-4 y y1 Rotation of axes Ixy 1 2 2 t a 4 Ad1d2 A = (b t)(t) b t d1 t 2 t 2 d2 a x1 30° Ixy 1 (15) 2 (150) 2 (85)(15)(57.5)(7.5) 4 1.815 106 mm4 30º: SUBSTITUTE into Eq. (12-25) with O b x All dimensions in millimeters. a 150 mm b 100 mm t 15 mm 30º 1 3 1 Ix ta (b t) t 3 3 3 1 1 (15)(150) 3 (85)(15) 3 3 3 16.971 106 mm4 1 1 3 Iy (a t) t 3 tb 3 3 1 1 (135)(15) 3 (15)(100) 3 3 3 5.152 106 mm4 Ix1 Ix Iy Ix 2 106 Iy 2 12.44 cos 2u Ixy sin 2u mm4 120º: SUBSTITUTE into Eq. (12-25) with Iy1 9.68 106 mm4 SUBSTITUTE into Eq. (12-27) with Ix1y1 Ix 2 6.03 Iy sin 2u Ixy cos 2u 30º: 106 mm4 of inertia Ix y with respect to the x1y1 axes for the Z-section shown in the 1 1 figure if b 3 in., h 4 in., t 0.5 in., and 60°. Problem 12.8-5 Calculate the moments of inertia Ix1 and Iy1 and the product h — 2 h — 2 Probs. 12.8-5, 12.8-6, 12.9-5 and 12.9-6 t y1 b y x1 C t t x b SECTION 12.8 Rotation of Axes 23 Solution 12.8-5 Rotation of axes y1 b y Area A2: Area A3: I– y I‡ y I– y I‡ y 1 (h)(t 3 ) 12 I¿ y 0.0417 in.4 3.4635 in.4 6.9688 in.4 h — 2 h — 2 A1 C x1 Iy I¿ y A2 A3 x PRODUCT OF INERTIA Ixy Area A1: I¿ xy 0 (b b All dimensions in inches. b 3.0 in. h 4.0 in. t 0.5 in. 60º t 2 ≤ 2 Area A2: I– xy Ixy I¿ xy I– xy 1 (bt)(b t)(h 4 Area A3: I‡ 0 xy I‡ xy t)(t) ¢ b h ≤¢ 2 2 t) I¿ xy 4 t ≤ 2 3.2813 in.4 MOMENT OF INERTIA Ix Area A1: I¿ x 6.5625 in. 1 h (b t)(t 3 ) (b t)(t) ¢ 12 2 3.8542 in.4 1 Area A2: I– (t)(h3 ) 2.6667 in.4 x 12 Area A3: I‡ I¿ 3.8542 in.4 x x Ix I¿ x I– x I‡ x 10.3751 in.4 SUBSTITUTE into Eq. (12-25) with Ix1 Ix 2 Iy cos 2u 2 13.50 in.4 Ix Iy 60º: Ixy sin 2u SUBSTITUTE into Eq. (12-25) with Iy1 (b t)(t) ¢ b 2 ≤ 2 3.84 in. 4 150º: MOMENT OF INERTIA Iy Area A1: I¿ y 1 (t)(b t) 3 12 3.4635 in.4 SUBSTITUTE into Eq. (12-27) with 60º: Ix Iy Ix1y1 sin 2u Ixy cos 2u 4.76 in.4 2 Problem 12.8-6 Solve the preceding problem if b t 12 mm, and 30°. 80 mm, h 120 mm, Solution 12.8-6 Rotation of axes All dimensions in millimeters. y y1 b h — 2 h — 2 t thickness b t 80 mm 12 mm h 120 mm 30º A1 C x1 MOMENT OF INERTIA Ix Area A1: I¿ x A2 A3 x Area A2: I– x b 1 h t 2 (b t)(t 3 ) (b t)(t) ¢ ≤ 12 2 2 2.3892 106 mm4 1 (t)(h3 ) 1.7280 106 mm4 12 I¿x I‡ x 2.3892 6.5065 106 mm4 106 mm4 Area A3: I‡ x Ix I¿ x I– x 24 CHAPTER 12 Review of Centroids and Moments of Inertia MOMENT OF INERTIA Iy Area A1: I¿ y 1 b 2 (t)(b t) 3 (b t)(t) ¢ ≤ 12 2 1.6200 106 mm4 1 (h)(t 3 ) 0.01728 106 mm4 Area A2: I– y 12 Area A3: I‡ I¿ 1.6200 106 mm4 y y Iy I¿ y I– y I‡ y 3.2573 106 mm4 SUBSTITUTE into Eq. (12-25) with Ix1 Ix 2 8.75 Iy Ix 2 Iy cos 2u 30º: Ixy sin 2u 106 mm4 120º: SUBSTITUTE into Eq. (12-25) with Iy1 1.02 106 mm4 SUBSTITUTE into Eq. (12-27) with PRODUCT OF INERTIA Ixy Area A1: I¿ xy 0 (b t)(t) ¢ b h ≤¢ 2 2 t) I¿xy 106 mm4 t ≤ 2 Ix1y1 Ix 2 Iy sin 2u Ixy cos 2u 30º: 1 (bt)(b 4 Area A2: I– xy Ixy I¿xy I– xy 0 I‡ xy 0.356 106 mm4 t)(h Area A3: I‡ xy 3.5251 Principal Axes, Principal Points, and Principal Moments of Inertia Problem 12.9-1 An ellipse with major axis of length 2a and minor axis of length 2b is shown in the figure. (a) Determine the distance c from the centroid C of the ellipse to the principal points P on the minor axis ( y axis). (b) For what ratio a/b do the principal points lie on the circumference of the ellipse? (c) For what ratios do they lie inside the ellipse? a P C P y c c b x b a Solution 12.9-1 Principal points of an ellipse y From Case 16: Iy P1 C P2 a a c c xp b x b ba 3 4 ab Ix ab 3 4 A Parallal-axis theorem: Ixp Ix Ac2 ab 3 4 abc2 (a) LOCATION OF PRINCIPAL POINTS At a principal point, all moments of inertia are equal. At point P1: Ixp Iy Eq. (1) Substitute into Eq. (1): ab 3 ba 3 abc2 4 4 Solve for c: c 1 2 a2 b2 SECTION 12.9 Principal Axes, Principal Points, and Principal Moments of Inertia 25 (b) PRINCIPAL POINTS ON THE CIRCUMFERENCE c b and b 1 2 a2 a b b2 5 (c) PRINCIPAL POINTS INSIDE THE ELLIPSE 0 c b For c For c 1 a 6 b 5 y b — 6 b — 6 b — 6 b — 2 0: b: a b a b and 5 a b 1 a Solve for ratio : b Problem 12.9-2 Demonstrate that the two points P1 and P2, located as shown in the figure, are the principal points of the isosceles right triangle. b — 2 P2 C P1 b — 2 x Solution 12.9-2 Principal points of an isosceles right triangle y1 y3 x2 y2 2 3 4 P1 yC y4 P2 d C x1 x4 xC y2 x2 x3 1 CONSIDER POINT P2: Ix3 y3 Ix2 y2 0 because y2 is an axis of symmetry. 0 (see above). b2 d 4 b 288 b 6 2 4 CONSIDER POINT P1: 0 because y1 is an axis of symmetry. 0 because areas 1 and 2 are symmetrical about the y2 axis and areas 3 and 4 are symmetrical about the x2 axis. Two different sets of principal axes exist at point P1. P1 is a principal point Ix1 y1 Ix2 y2 Parallel-axis theorem: Ix2y2 Ixcyc Ixcyc ¢ Ad1d2 b 4 2 A ≤ 2 d1 d2 b 6 2 6 2 Parallel-axis theorem: Ix4y4 Ix4y4 Ixcyc Ad1d2 ≤¢ b d1 d2 b4 b2 b 2 ¢ ≤ 0 288 4 6 2 Two different sets of principal axes (x3y3 and x4y4) exist at point P2. P2 is a principal point 26 CHAPTER 12 Review of Centroids and Moments of Inertia and p defining the orientations 2 of the principal axes through the origin O for the right triangle shown in the figure if b 6 in. and h 8 in. Also, calculate the corresponding principal moments of inertia I1 and I2. p1 Problem 12.9-3 Determine the angles y y1 h x1 O x b Solution 12.9-3 Principal axes y y1 EQ. (12-30): tan 2up 2 p p 2Ixy Ix Iy 120.256º 60.128º 1.71429 59.744º 29.872º and and h x1 O x b SUBSTITUTE into Eq. (12-25) with Ix1 311.1 in.4 29.872º: RIGHT TRIANGLE b 6.0 in. h 8.0 in. SUBSTITUTE into Eq. (12-25) with Ix1 256 in.4 88.9 in.4 311.1 in.4 up1 88.9 in.4 up2 29.87 60.13 60.128º: CASE 7: Ix Iy Ixy bh3 12 THEREFORE, I1 I2 hb 3 144 in.4 12 b 2h2 96 in.4 24 NOTE: The principal moments of inertia can be verified with Eqs. (12-33a and b) and Eq. (12-29). Problem 12.9-4 Determine the angles p1 and p2 defining the orientations of the principal axes through the origin O and the corresponding principal moments of inertia I1 and I2 for the L-shaped area described in Prob. 12.8-4 (a 150 mm, b 100 mm, and t 15 mm). Solution 12.9-4 Principal axes ANGLE SECTION y y1 t thickness a 150 mm b 100 mm t 15 mm FROM PROB. 12.8-4: a x1 Ix Iy 16.971 106 mm4 5.152 106 mm4 tan 2up Ixy Ix 1.815 2 Ixy Iy 106 mm4 0.3071 EQ. (12-30): O b x 2 p p 17.07º and 162.93º 8.54º and 81.46º SECTION 12.9 Principal Axes, Principal Points, and Principal Moments of Inertia 27 SUBSTITUTE into Eq. (12-25) with Ix1 17.24 106 mm4 8.54º: THEREFORE, I1 I2 17.24 106 mm4 up1 4.88 106 mm4 up2 8.54 81.46 SUBSTITUTE into Eq. (12-25) with Ix1 4.88 10 mm 6 4 81.46º: NOTE: The principal moments of inertia I1 and I2 can be verified with Eqs. (12-33a and b) and Eq. (12-29). Problem 12.9-5 Determine the angles p1 and p2 defining the orientations of the principal axes through the centroid C and the corresponding principal centroidal moments of inertia I1 and I2 for the Z-section described in Prob. 12.8-5 (b 3 in., h 4 in., and t 0.5 in.). Solution 12.9-5 y1 Principal axes y EQ. (12-30): 2 x1 p p tan 2up and and 2 Ixy Ix Iy 255.451º 127.726º 3.8538 75.451º 37.726º h — 2 C h — 2 b x SUBSTITUTE into Eq. (12-25) with Ix1 15.452 in. 4 37.726º: SUBSTITUTE into Eq. (12-25) with Ix1 1.892 in.4 15.45 in.4 1.89 in.4 up1 up2 127.726º: Z-SECTION t b thickness 0.5 in. 3.0 in h 4.0 in THEREFORE, I1 I2 37.73 127.73 FROM PROB. 12.8-5: Ix Ixy 10.3751 in.4 6.5625 in.4 Iy 6.9688 in.4 NOTE: The principal moments of inertia I1 and I2 can be verified with Eqs. (12-33a and b) and Eq. (12-29). Problem 12.9-6 Solve the preceding problem for the Z-section described in Prob. 12.8-6 (b 80 mm, h 120 mm, and t 12 mm). Solution 12.9-6 y1 Principal axes y Z-SECTION t x1 h — 2 C h — 2 b b h x thickness 12 mm 80 mm 120 mm FROM PROB. 12.8-6: Ix Ixy 6.5065 106 mm4 Iy 3.5251 106 mm4 3.2573 106 mm4 28 CHAPTER 12 Review of Centroids and Moments of Inertia Eq. (12-30): 2 p p tan 2up and and 2 Ixy Ix Iy 245.257º 122.628º 2.1698 THEREFORE, I1 I2 8.76 1.00 106 mm4 up1 106 mm4 up2 32.63 122.63 65.257º 32.628º SUBSTITUTE into EQ. (12-25) with Ix1 8.763 10 mm 6 4 32.628º: NOTE: The principal moments of inertia I1 and I2 can be verified with Eqs. (12-33a and b) and Eq. (12-29). SUBSTITUTE into Eq. (12-25) with Ix1 1.000 10 mm 6 4 122.628º: and p defining the 2 orientations of the principal axes through the centroid C for the right triangle shown in the figure if h 2b. Also, determine the corresponding principal centroidal moments of inertia I1 and I2. p1 Problem 12.9-7 Determine the angles y1 y h C x1 x b Solution 12.9-7 y1 Principal axes y EQ. (12-30): 2 h 2b C x1 x p p tan 2up and and 2 Ixy Ix Iy 213.6901º 106.8450º 2 3 33.6901º 16.8450º SUBSTITUTE into Eq. (12-25) with Ix1 0.23904 b 4 16.8450º: SUBSTITUTE into Eq. (12-25) with b 106.8450º: Ix1 0.03873 b 4 RIGHT TRIANGLE h 2b THEREFORE, I1 I2 0.2390 b4 up1 0.0387 b4 up2 16.85 106.85 CASE 6 Ix Iy Ixy bh3 2b 4 36 9 hb 3 b 4 36 18 2 2 b h b4 72 18 NOTE: The principal moments of inertia I1 and I2 can be verified with Eqs. (12-33a and b) and Eq. (12-29). SECTION 12.9 Principal Axes, Principal Points, and Principal Moments of Inertia 29 Problem 12.9-8 Determine the angles p1 and p2 defining the orientations of the principal centroidal axes and the corresponding principal moments of inertia I1 and I2 for the L-shaped area shown in the figure if a 80 mm, b 150 mm, and t 16 mm. a y1 t yc x1 t C xc b Probs. 12.9-8 and 12.9-9 Solution 12.9-8 y x y1 a C O Principal axes (angle section) yc x1 t thickness Iy A1 xc y b x A2 a A1 A2 A 80 mm b 150 mm at 1280 mm2 (b t)(t) 2144 mm2 A1 A2 t (a b t) t 16 mm 1 t 2 1 (a)(t 3 ) A1 ¢ ≤ (t)(b t 3 ) 12 2 12 b t 2 A2 ¢ ≤ 2 1 1 (80)(16) 3 (1280)(8) 2 (16)(134) 3 12 12 166 2 (2144) ¢ ≤ 2 18.08738 106 mm4 MOMENTS OF INERTIA (xcyc AXES) Use parallel-axis theorem. Ix Ay2 1.54914 Iy Ax2 7.74386 2.91362 106 106 mm4 18.08738 106 mm4 106 (3424)(19.9626) 2 (3424)(54.9626) 2 3424 mm2 IxC IyC LOCATION OF CENTROID C Qx a Ai y2 (at) ¢ y Qy x 68,352 mm3 Qx 68,352 mm3 19.9626 mm A 3,424 mm2 t b t a Ai xi (at) ¢ 2 ≤ (b t)(t) ¢ 2 ≤ 188,192 mm3 Qy 188,192 mm3 54.9626 mm A 3,424 mm2 a ≤ 2 (b t)(t) ¢ t ≤ 2 PRODUCT OF INERTIA Use parallel-axis theorem: Area A1: I¿ CyC x 0 A1 B ¢x Ixy (1280)(8 1.20449 Area A2: I– CyC x 0 b A2 B 54.9626)(40 106 mm4 t xR B t e ≤R B 2 2 Icentroid 19.9626) ¢y yR A d1d2 2 MOMENTS OF INERTIA (xy AXES) Use parallel-axis theorem. Ix 1 a 2 1 (t)(a 3 ) A1 ¢ ≤ (b 12 2 12 1 (16)(80) 3 (1280)(40) 2 12 (2144)(8)2 2.91362 106 mm4 t 2 A2 ¢ ≤ 2 IxCyC I¿ CyC x (2144)(83 0.71910 I–CyC x 54.9626)(8 106 mm4 19.9626) t ≤R 2 t)(t ) 3 1.92359 106 mm4 1 (134)(16) 3 12 SUMMARY IxC IxCyC 1.54914 106 mm4 IyC 7.74386 106 mm4 1.92359 106 mm4 30 CHAPTER 12 Review of Centroids and Moments of Inertia PRINCIPAL AXES Eq. (12-30): 2 p p SUBSTITUTE into Eq. (12-25) with tan 2up 2Ixy Ix Iy 0.621041 Ix1 8.2926 106 mm4 74.0790° 31.8420° and 148.1580° 15.9210° and 74.0790° 15.9210° THEREFORE, I1 I2 8.29 1.00 106 mm4 up1 106 mm4 up2 74.08 15.92 SUBSTITUTE into Eq. (12-25) with Ix1 1.0004 106 mm4 NOTE: The principal moments of inertia I1 and I2 can be verified with Eqs. (12-33a and b) and Eq. (12-29). Problem 12.9-9 Solve the preceding problem if a and t 5/8 in. 3 in., b 6 in., Solution 12.9-9 y Principal axes (angle section) yc x1 MOMENTS OF INERTIA (xy AXES) Use parallel-axis theorem. Ix 1 a 2 1 t 2 (t)(a 3 ) A1 ¢ ≤ (b t)(t 3 ) A2 ¢ ≤ 12 2 12 2 1 5 1 5 3 ¢ ≤ (3.0) 3 (1.875)(1.5) 2 (5.375) ¢ ≤ 12 8 12 8 2 5 (3.35938) ¢ ≤ 16 6.06242 in.4 1 t 2 1 (a)(t 3 ) A1 ¢ ≤ (t)(b t 3 ) 12 2 12 b t 2 A2 ¢ ≤ 2 1 5 3 5 2 1 5 (3.0) ¢ ≤ (1.875) ¢ ≤ ¢ ≤ (5.375) 3 12 8 16 12 8 6.625 2 (3.35938) ¢ ≤ 2 45.1933 in.4 x y1 a C O A1 xc y b x A2 a b t A1 A2 A 3.0 in. 6.0 in. 5 8 in. at 1.875 in.2 (b t)(t) 3.35938 in.2 A1 A2 t (a b t) (at) ¢ a ≤ 2 Iy 5.23438 in.2 t)(t) ¢ t ≤ 2 LOCATION OF CENTROID C Qx a Aiy2 (b y Qy x 3.86230 in.3 Q x 3.86230 in.3 0.73787 in. A 5.23438 in.2 t b t a Ai xi (at) ¢ 2 ≤ (b t)(t) ¢ 2 ≤ 11.71387 in.3 Qy 11.71387 in.3 2.23787 in. A 5.23438 in.2 MOMENTS OF INERTIA (xcyc AXES) Use parallel-axis theorem. IxC IyC Ix Iy Ay2 Ax2 6.06242 45.1933 (5.23438)(0.73787) 2 (5.23438)(2.23787) 2 3.21255 in.4 18.97923 in.4 SECTION 12.9 Principal Axes, Principal Points, and Principal Moments of Inertia 31 PRODUCT OF INERTIA Use parallel-axis theorem: Area A1: I¿ CyC x 0 A1 B ˇ SUBSTITUTE into Eq. (12-25) with ¢x 14.2687° Area A2: I–CyC x IxCyC I¿ CyC x (3.35938)(1.07463)( 0.42537) 1.53562 in.4 I–CyC 4.28696 in.4 x t a yR ≤R B 2 2 (1.875)( 1.92537)(0.76213) 2.75134 in.4 b t t x R B ¢y ≤ R 0 A2 B 2 2 ˇ Ixy Icentroid A d1d2 Ix1 2.1223 in.4 75.7313º SUBSTITUTE into Eq. (12-25) with Ix1 20.0695 in.4 THEREFORE, I1 I2 20.07 in.4 2.12 in.4 up1 up2 75.73 14.27 SUMMARY IxC IxCyC 3.21255 in.4 IyC 18.97923 in.4 NOTE: The principal moments of inertia I1 and I2 can be verified with Eqs. (12-33a and b) and Eq. (12-29). 4.28696 in.4 PRINCIPAL AXES EQ. (12-30): 2 p p tan 2up 2Ixy Ix Iy 0.54380 28.5374° and 151.4626° 14.2687° and 75.7313°