Moment of Inertia of Composite Sections

LESSON 15: MOMENT OF INERTIA OF COMPOSITE SECTIONS E

N G IN E E R IN G M E C

( I 

Introduction We have derived rived moment of inertia nertiaof certain standard sections sections in our last session. Sections ctions of beams and colum columns commonly only used in rea real structures structuresare combination combination of the standard sections. ctions. So, we need to have a method by which which moment of  of  inerti inertia a of such composite posite sections can be determined. H A N IC S

Moment of inertia of composite sections about an axis can be found by thefollo ollowing wing steps steps:

2 =

3. Find the mome oment of ine inertia rtia of each figure about its centroi centroida dal axis. Add theterm  Ay 2  where A is is the area of  the simple figure and y is the distance of the centroid of the simple figu fi gure from from the referenceaxis. This gives moment of  inertia nertia of the the simple figu figure about the refere reference axis. 4. Sumup the moment of inertia inertia of all simple figu figures res to get the moment of inertia of the composite section.

+

 A2 y 22 )

and of semicircl icircle e

( I 

3 =

 I  g 3

+

 A3 y32 )  about axis AB AB are calculate culated.

4. Mome oment of inertia of the composite section about about AB AB is is given by:

 I  AB

1. Divide ivide the given fig figure into a number ber of simple simple fig fi gures ures of  known cross sectional ctional properties properti es.. 2. Locate thecentroid ntroid of each simple simple fig figure by by inspection or or using using standard expression. xpression.

 I  g 2

=

 I 1 +  I 2

=

 I  g 1

+

+

I 3

 A1 y12

+

 I  g 2

+

 A2 y22

+

 I  g 3

+

 A3 y 32

15.1

I n most most engineering ring problem problems, mome moment of inertia nertia about the centroidal ntroidal axis is required quired. I n such cases, first first loca locate the centroidal axis and then find the moment of inertia about this axis.

 The  The procedure give iven above isillu illus strated below low. Refe Referrin ring to the Fig. 15.1, it is required to find out the moment of inertia of the section about axis A – B.

Fig. 15.2 Referring Referring to Fig.15.2, fi first the moment of areaabout any reference axis, say A AB, B, is ta taken and is is divided divided bythe total area of  the section tion to locate locate centroi ntroida dal axis x-x. Then the distancesof  centroid of individual figures  yc1 , y c 2 and  y c3  from the axis x-x are determ determiined. ned. The moment of inertia inerti a of the composite posite section ction about the centroi ntroida dal axis x-x x-x is calculate using the expression:

Fig. 15.1 1. Thesection ction in the figu figure is divided into a rectangle, ngle, a tria triangle and a semicircl micircle. e. The areas of the simple figures  A1 , A2 and  A3 are calculated. 2. Thecentroids ntroids of the recta rectang ngle ( g 1 ) , triangle ( g 2 ) and semicircle ( g 3 )  are located. The distance distances  y1 , y 2 and  y 3 are found found from from the axis AB. AB..

 I  xx

=

( )

=

2

 I  g 1 +  A1 y 1

+

 I  g 2

+

2

 A2 y c2

 I 

+  g 3 +

2

 A3 y c3

15.2

Sometimes, the moment of inert nertia ia is is found found about a convenient nient axis and then using using parall rallel axis theorem theorem, the moment of  inertia inertia about centroida centroidal axis is is found. found. I n the above above example, the moment of inertia  I  AB  is found and  y , the distanceof CG from axis AB AB is is calcula lculated. ted. Then hen from from parallel rallel axis theorem,

3. The moment of inertia inertia of the rectang ctangle about its its centroid ntroid  I  g 1  is calculated using standard expression. To this, this, the term  A1 y12  is added to get the moment of inertia about the axis AB  I 1

2

 I  g 1 +  A1 yc1

 I  AB

=

 I  xx

 I  XX 

=

 I  AB

+

 A y 2

−

 A y 2

Example 1

Determine the moment of inertia inertia of the the symmetric tric I-section I -section shown in Fi F ig.15.3 about its its centroida centroidal axesx-x and y-y.

Similarly, the moment ent of inertia inertia of the tria triangle ngle

© Copy Right: Right: Rai University University 52

7. 1 5 4

E N

Fig. 15.3 Also, determine moment of inertia of the section about a centroidal axis perpendicular to x-x axis and y-y axis. The section is divided into three rectangles  A1 , A2 , A3 . =

200  × 9 = 1800 mm 2

 A2

=

(250 − 9 × 2) × 6.7 = 1554.4 mm

 A3

=

200  × 9 = 1800 mm 2

 A1

Area

Fig. 15.4  The section is divided into three rectangles  A1 , A2 and  A3 .

2

 A1 = 100 ×  13.5 = 1350mm 2 ,  A2 = (400 − 27 ) × 8.1 = 3021.3mm 2  A3 = 100 × 13.5 = 1350.00mm 2

Area,

 Total Area

 Total Area

 A = 5721.3mm 2  A = 5154.4 mm 2

 The given section is symmetric about horizontal axis passing through the centroid  g 2  of the rectangle  A2 .  A referenceaxis (1)-(1) is chosen as shown in Fig. 15.4.

 The section issymmetrical about both x-x and y-yaxes.  Therefore, its centroid will coincide with the centroid of  rectangle  A 2 .

 The distance of the centroid of the section from (1)-(1) 1350× 50 + 3021.3 ×

With respect to the centroidal axesx-x and y-y, the centroid of  rectangle  A1  is  g 1  (0.0,120.5), that of  A2  is  g 2 (0.0,0.0) and that of  A3 is  g 3 (0.0,120.5).

 I  xx  A2  I  xx

+

12 =

1800 × 120.5 2

5,92,69202 mm

+

6.7 × 2323 12

+

0+

200 × 9 3 12

+

1800 (120.5)

 I  xx

=

9 × 200 3 12

=

2

∴ Moment of inertia of

100 × 13 .5 3 12

Similarly,

 I  xx +

232 × 6 .7 3

+

= 25.73 mm

=

+

 A1 , A2  and  A3 about x-x

1350 × 193.25 2

+

8.1 × 3733 12

+

100 × 13.53 12

+

1350 × 193.25 2

1.359 × 10 8 mm 4 Ans.

Example 3

9 × 200 3

12

1350× 50

With reference to the centroidal axes x-x and y-y, the centroid of  the rectangle  A1  is  g 1 (24.27,19.25), that of  A2  is  g 2  (21.68,0.0) and that of  A3  is  g 3 (24.27,193.25).

=

4

+

5721.3

of inertia of  A1 + Moment of inertia of  + Moment of inertia of  A3  about x-x 200 × 9 3

2

=

=  Moment

=

8.1

Determine the moment of inertia of the built-up section shown in Fig. 15.5 about its centroidal axesx-x and y-y.

12

1, 20,05815mm 4

 The given composite section may be divided into simple rectanglesand triangles as shown in the Fig. 15.5.

Moment of inertia of the section about a centroidal axis perpendicular to x-x and y-y axis is nothing but polar moment of inertia, and is given by:

 I  zz 

=

 Ixx + I  yy = 5,92,69202 + 1,20,05815

 I  zz 

=

7,12,75017 mm 4 Ans. Fig. 15.5

Example 2

Compute the second moment of area of the channel section shown in Fig.15.4 about centroidal axesx-x and y-y. © Copy Right: Rai University 7.154

53

G IN E E R IN G M E C H

A N IC S

E N G IN E E R

2  A1 = 100 × 30 = 3000 mm ,

Area

 A2 = 100 × 25 = 2500mm 2 IN G M E C H A N IC S

 A3 = 200 × 20 = 4000mm2

 A4 =  A5 =

1

×

2

87 .5 × 20 = 875 mm

1 × 87.5 × 20 2

=

2

875 mm 2

 Total Area A= 11250mm 2

Fig. 15.6

Due to symmetry, centroid lieson the axis y-y. A referenceaxis (1)-(1) is choosen as shown in the figure. The distance of the centroidal axis from (1)-(1) =

Sum of   moments of   areas about  (1

−

(1

Total  area =

Sum of   moments of   areas about  (1) − (1) Total  area

= 59.26 mm With reference to the centroidal axes x-x and y-y, the centroid of  the rectangle  A1 is  g 1 (0.0,75.74), that of  A2  is  g 2 (0.0,10.74), that of  A3  is  g 3 (0.0,49.26), the centroid of  triangle  A4  is  g 4 (41.66,32.59) and that of  A5  is  g 5 (41.66,32.59).  I  xx

+

=

100 × 30 3

4000× 49.26 2

 I  xx

=

 I  yy

=

+

+

12

+

87.5 × 20 3 36

30 × 100 3

+

100 × 25 3

12

36 =

+

3,15, 43447 mm 4

21× 87. 5

 I  yy

3000× 75 .74

2

12

+

25 × 1003

+

12 875 × 32.59 2

2500× 10 .74

+

87.5 × 20 3

2

12

+

36

+

200 × 203

875 × 32.59 2

20 × 200 3 12

2. A plate girder is made up of a web plate of size 400mm x 100mm, 4 angles of size 100mm x 100mm x 10mm and covered plates of size 300mm x 10mmas shown in fig. 15.7. Calculate the moment of inertia about horizontal and vertical centroidal axes.

[Ans. Ixx = 5.35786x 10 8mm 4, Iyy = 60850667 mm 4]

Ans. +

Fig. 15.7

+

20 × 87.5 3 36

+

875 × 41.66 2

3. Thec/ s of a plain concrete culvert is as shown in fig. 15.8. Calculate the moment of inertia about horizontal centroidal axis.(Ans. I xx =5.45865 x 10 10 mm 4)

3

+

87 5× 41.662 ,

4

1,97,45122 mm

Ans.

Problems 1. The C/ S of a gantry girder is shown in fig . 15.6. It is made up of an I sec of depth 450 mm, flange width 200 mmand a channel of size 400 mm x 150 mm. Thickness of all members is 10mm. Calculate the moment of inertia about horizontal centroidal axis.(Ans. I xx =4.2198 x 10 8mm 4)

Fig. 15.8

© Copy Right: Rai University 54

7.154

E N

Fig. 15.9 4. Determine thecentroid of built up section shown in fig.15.9 and calculate the moment of inertia and radius of gyration about horizontal centroidal axis.

[Ans. Ixx = 1267942mm 4 ; Kxx = 18.55a mm]

Question 1. How would you find out the moment of inertia of a plane area? 2. What is Routh’s rule for findingout the moment of inertia of an area? 3. Derive an equation for moment of inertia of the following sections about centroidal axis: a. arectangular section b. a hollow rectangular section c. acircular section , and d. a hollow circular section 4. State and prove the theorem of perpendicular axis, as applied to moment of inertia. 5. Prove theparallel axis theorem, in the determination of  moment of inertia of areas with the help of a neat sketch.. 6. Describe the method of finding out the moment of inertia of a composite section. Notes

© Copy Right: Rai University 7.154

55

G IN E E R IN G M E C H

A N IC S