moment of inertia.pdf

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MOMENT OF INERTIAOF BUILT-UP SECTIONS

Construction of the World Trade CenterPerimeter Column PanelsThree Full Columns Connected by Three Spandrels



Moment of Inertia forBuilt-up Sections

Frequently, standard structural sections are welded togetherto form a built-up section. The moments of inertia of eachsection are easily found in a handboo or from the !endor."owe!er, as we saw in the section on moments of inertia forcomposite sections, we cannot algebraically add moments ofinertia.Since #built-up# section is simply another term for compositesection, then finding moment of inertia for a built-up section isno different.$et#s loo at some e%amples of finding the moment of inertiaof built-up sections.



&%ample '(Consider a built-up section comprised of an S)* % +standard section capped with an C) % *'. miscellaneouschannel. /etermine the centroidal moment of inertia.

!


MC"# $ %&!

S" $ '#



Finding any centroidal moment of inertia requires one nowsthe location of the centroid. Finding the centroid of a built-upsection is no different from finding the centroid of a compositegeometric section.First, impose an %-y coordinate a%is at a location of your

choice. 0n doing so, one should note it is appropriate torecogni1e if symmetry e%ists and to use it to your ad!antage.0n the present e%ample, the section is symmetrical about they-a%is but not about the %-a%is. This places the %-coordinateof the centroid at the center of the section2 this is a !ery good

place to locate the y-a%is of our imposed coordinate system.

(




3lthough we can place the %-a%is anywhere we wish, locatingit at the base of the structure is generally most con!enient.The a%is %c is the centroidal %-a%is, the location of which wewish to find.4ow go to the appendi% of the online te%t and determine the

geometry and moments of inertia of each section.

Centroi) of a Built-up Section

'

"

*

$

$ c

* c

5)6 C-Section 5*6 S-SectionWeb thic ness( .7889 /epth( )*. 9

:c( .; 9 3rea( )7.8 in *

0y ( '.*) in 7 0

%( + in 7

3rea ( '. * in *



The location of the centroid for an #0-beam#-type section is note%plicitly stated in the tables. 0t is assumed you now it isequal to half the depth of the section due to symmetry.0n this case, the centroid is located a distance y c) from ourarbitrary %-a%is. That distance is <. 9.


+

"

*

$

$ c"

$ c

* c"* c



The distance d y* is a bit more tric y. The centroid of a channelsection is measured from the bac of the web and gi!en asdistance : c in the steel tables.This distance must be measured from the same %-a%is as theS-section. Thus, y c* equals depth of the S-section plus the

width of the channel web minus distance : c.


,

"

*

$

$ c"

$ c

$ c

* c"* c

* c

This means(yc* = )*9 > .7889 - .; 9yc* = )).+779



3reas are also obtained from the steel tables.?nce the areas and locations of centroids are obtained, wecan calculate the location of the centroid.


%

"

*

$

$ c"

$ c

$ c

* c"* c

* c

Section 3rea y y3

S )7.8 <. ''.*

C '. * )).+77 ;<. +

Total 23.02 ---- 184.2

yc=184.2in 3

23.02in 2 = 8.0



?nce the centroid is located, the moment of inertia is found byapplying the parallel a%is theorem.The distances between the centroidal a%is of the compositesection and the indi!idual sections are easily calculated as(

"

*

$

$ c"

$ c

$ c) *"

* c

)*

The remainder of the calculationsare e%pressed in the table below

d y1= 8− 6= 2 in

d y2= 11.544 − 8= 3.544in

Section 3rea 0% dy), * d *y), * 536 0% > d*y), * 536

S )7.8 + ' - < +'.' <7

C '. * '.*) )).+7 - ' ) 7.*< ))*

Total 23.02 ---- ---- ---- 476

Moment of Inertiaof a Built-up Section



&%ample ;( 3 column will tend to buc le about the a%is with the leastmoment of inertia. For that reason, it is preferred the momentof inertia about the %- and y-a%is of a column section to beroughly equal.

Consider a built-up column comprised of two C)*% +channels. /etermine the distance #d# between the sectionssuch that the centroidal moment of inertia about the %- and y-a%is are equal.From the appendi%, we can loo up the following data.


"#

Area. "#&! in I$. "+ in (

$c. "&#' in I

*. " &, in (

t/. #&(+, in



Since both sections ha!e the same %-a%is, we now thecomposite moment of inertia about the %-a%is is the sum of themoments of inertia of each indi!idual section about its owncentroidal %-a%is. This gi!es us our desired 0 y."owe!er, we will need to apply the parallel a%is theorem to

find distance #d# that will result in 0 y.


""

*" *

$ c0$" 0$)

I x = I x1 I x2

= 216 216 = 432 in 4

Now we know that: I y = I x = 432 in 4



Since the section is symmetrical about the y-a%is, we nowthat d ) = d * = @ d > t w > %c = @ d > ).+)8The moment of inertia of each section as it references thecommon centroidal y-a%is must be half of the desired 0 y.

3pplying the parallel a%is theorem, we can write(


"

* c

*" *

)

I each y = I y2

= 216 in 4 = I y1 d 12⋅ A

= 12.7 0.5d 1.517 2⋅10.3 in 2

0 = 2.575⋅ d 2 15.625 ⋅ d − 179.60

Applying the quad ati! equation:

d = − 15.625 ± 15.625 2 − 4⋅2.575⋅ − 179.602⋅ 2.575

d = 5.85 in

) " )

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