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PROBLEM 9.31 Determine the moment of inertia and the radius of gyration of the shaded area with respect to the x axis. SOLUTION First note that A = A1 + A2 + A3 = [(24)(6) + (8)(48) + (48)(6)] mm 2 = (144 + 384 + 288) mm 2 = 816 mm 2 Now I x = ( I x )1 + ( I x )2 + ( I x )3 where 1 (24 mm)(6 mm)3 + (144 mm 2 )(27 mm)2 12 = (432 + 104,976) mm 4 ( I x )1 = = 105, 408 mm 4 1 (8 mm)(48 mm)3 = 73,728 mm 4 12 1 ( I x )3 = (48 mm)(6 mm)3 + (288 mm 2 )(27 mm) 2 12 = (864 + 209,952) mm 4 = 210,816 mm 4 ( I x )2 = Then I x = (105, 408 + 73, 728 + 210,816) mm 4 = 389,952 mm 4 and k x2 = I x 389,952 mm 4 = A 816 mm 2 or I x = 390 × 103 mm 4 or k x = 21.9 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1435 PROBLEM 9.39 Determine the shaded area and its moment of inertia with respect to the centroidal axis parallel to AA′, knowing that d1 = 30 mm and d2 = 10 mm, and that the moments of inertia with respect to AA′ and BB′ are 4.1 × 106 mm4 and 6.9 × 106 mm4, respectively. SOLUTION I AA′ = 4.1 × 106 mm 4 = I + A(30 mm)2 (1) I BB′ = 6.9 × 106 mm 4 = I + A(40 mm) 2 I BB′ − I AA′ = (6.9 − 4.1) × 106 = A(402 − 302 ) 2.8 × 106 = A(700) Eq. (1): 4.1 × 106 = I + (4000)(30)2 A = 4000 mm 2 I = 500 × 103 mm 4 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1444 PROBLEM 9.41 Determine the moments of inertia I x and I y of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB. SOLUTION Dimensions in mm First locate centroid C of the area. Symmetry implies Y = 30 mm. 1 Then A, mm 2 x , mm 108 × 60 = 6480 54 349,920 46 –59,616 2 1 − × 72 × 36 = −1296 2 Σ 5184 290,304 X Σ A = Σ xA : X (5184 mm 2 ) = 290,304 mm3 or X = 56.0 mm Now I x = ( I x )1 − ( I x )2 where xA, mm3 1 (108 mm)(60 mm)3 = 1.944 × 106 mm 4 12 1 1 × 72 mm × 18 mm (6 mm) 2 ( I x )2 = 2 (72 mm)(18 mm)3 + 36 2 ( I x )1 = = 2(11, 664 + 23,328) mm 4 = 69.984 × 103 mm 4 [( I x ) 2 is obtained by dividing A2 into Then ] I x = (1.944 − 0.069984) × 106 mm 4 or I x = 1.874 × 106 mm 4 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1446 PROBLEM 9.41 (Continued) Also where I y = ( I y )1 − ( I y ) 2 1 (60 mm)(108 mm)3 + (6480 mm 2 )[(56.54) mm]2 12 = (6, 298,560 + 25,920) mm 4 = 6.324 × 106 mm 4 ( I y )1 = 1 (36 mm)(72 mm)3 + (1296 mm 2 )[(56 − 46) mm]2 36 = (373, 248 + 129, 600) mm 4 = 0.502 × 106 mm 4 ( I y )2 = Then I y = (6.324 − 0.502)106 mm 4 or I y = 5.82 × 106 mm 4 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1447 PROBLEM 9.43 Determine the moments of inertia I x and I y of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB. SOLUTION First locate centroid C of the area. Then A, in 2 x , in. 1 5 × 8 = 40 2.5 2 −2 × 5 = −10 1.9 Σ 30 where 4 100 160 4.3 –19 –43 81 117 X = 2.70 in. Y Σ A = Σ yA: Y (30 in 2 ) = 117 in 3 Y = 3.90 in. or Now yA, in 3 X Σ A = Σ xA: X (30 in 2 ) = 81 in 3 or and xA, in 3 y , in. I x = ( I x )1 − ( I x )2 1 (5 in.)(8 in.)3 + (40 in 2 )[(4 − 3.9) in.]2 12 = (213.33 + 0.4) in 4 = 213.73 in 4 ( I x )1 = 1 (2 in.)(5 in.)3 + (10 in 2 )[(4.3 − 3.9) in.]2 12 = (20.83 + 1.60) = 22.43 in 4 ( I x )2 = PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1450 PROBLEM 9.43 (Continued) Then I x = (213.73 − 22.43) in 4 Also I y = ( I y )1 − ( I y ) 2 where or I x = 191.3 in 4 1 (8 in.)(5 in.)3 + (40 in 2 )[(2.7 − 2.5) in.]2 12 = (83.333 + 1.6) in 4 = 84.933 in 4 ( I y )1 = 1 (5 in.)(2 in.)3 + (10 in 2 )[(2.7 − 1.9) in.]2 12 = (3.333 + 6.4) in 4 = 9.733 in 4 ( I y )2 = Then I y = (84.933 − 9.733) in 4 or I y = 75.2 in 4 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1451