PROBLEM 9.49 Two channels and two plates are used to form the column section shown. For b = 200 mm, determine the moments of inertia and the radii of gyration of the combined section with respect to the centroidal x and y axes. SOLUTION From Figure 9.13B For C250 × 22.8 A = 2890 mm 2 I x = 28.0 × 106 mm 4 I y = 0.945 × 106 mm 4 Total area A = 2[2890 mm 4 + (10 mm)(375 mm)] = 13.28 × 103 mm 2 Given b = 200 mm: I x = 2[28.0 × 106 mm 4 ] + 2 1 (375 mm)(10 mm)3 + (375 mm)(10 mm)(132 mm) 2 12 = 186.743 × 106 k x2 = I x = 186.7 × 106 mm 4 I x 186.743 × 106 = = 14.0620 × 103 mm 2 A 13.28 × 103 k x = 118.6 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. 1459 PROBLEM 9.49 (Continued) Channel I y = Σ( I y + Ad 2 ) = 2[ I y + Ad 2 ] + 2 Plate 1 (10 mm)(375 mm)3 12 200 mm = 2 0.945 × 10 mm + (2890 mm ) + 16.10 mm 2 6 4 2 2 + 87.891 × 106 mm 4 = 2[0.945 × 106 + 38.955 × 106 ] + 87.891 × 106 I y = 167.7 × 106 mm 4 = 167.691 × 106 mm 4 k y2 = Iy A = 167.691 × 106 = 12.6273 × 103 13.28 × 103 k y = 112.4 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. 1460 PROBLEM 9.50 1 Two L6 × 4 × 2 -in. angles are welded together to form the section shown. Determine the moments of inertia and the radii of gyration of the combined section with respect to the centroidal x and y axes. SOLUTION From Figure 9.13A From Figure 9.13A: Area = A = 4.75 in2 I x′ = 17.3 in 4 I y ′ = 6.22 in 4 I x = 2[ I x′ + AyO2 ] = 2[17.3 in 4 + (4.75 in 2 )(1.02 in.) 2 ] = 44.484 in 4 I x = 44.5 in 4 Total area = 2(4.75) = 9.50 in2 k x2 = Ix 44.484 in 4 = = 4.6825 in 2 2 area 9.50 in k x = 2.16 in. I y = 2[ I y′ + AxO2 ] = 2[6.22 in 4 + (4.75 in 2 )(1.269 in.) 2 ] = 27.738 in 4 I y = 27.7 in 4 27.738 in 4 = 2.9198 9.50 in 2 k y = 1.709 in Total area = 2(4.75) = 9.50 in2 k y2 = Iy area = 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. 1461 PROBLEM 9.54 The strength of the rolled W section shown is increased by welding a channel to its upper flange. Determine the moments of inertia of the combined section with respect to its centroidal x and y axes. SOLUTION A = 14, 400 mm 2 W section: I x = 554 × 106 mm 4 I y = 63.3 × 106 mm 4 A = 2890 mm 2 Channel: I x = 0.945 × 106 mm 4 I y = 28.0 × 106 mm 4 First locate centroid C of the section. W Section Channel Σ Then A, mm2 y , mm yA, mm3 14,400 −231 −33,26,400 2,890 49.9 1,44,211 −31,82,189 17,290 Y Σ A = Σ y A: Y (17, 290 mm 2 ) = −3,182,189 mm3 Y = −184.047 mm or Now where I x = ( I x ) W + ( I x )C ( I x ) W = I x + Ad 2 = 554 × 106 mm 4 + (14, 400 mm 2 )(231 − 184.047) 2 mm 2 = 585.75 × 106 mm 4 ( I x )C = I x − Ad 2 = 0.945 × 106 mm 4 + (2,890 mm 2 )(49.9 + 184.047)2 mm 2 = 159.12 × 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. 1465 PROBLEM 9.54 (Continued) Then also I x = (585.75 + 159.12) × 106 mm 4 or I x = 745 × 106 mm 4 or I y = 91.3 × 106 mm 4 I y = ( I y ) W + ( I y )C = (63.3 + 28.0) × 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. 1466

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