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
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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
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reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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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
=
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reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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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,
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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