4. MOMENT OF INERTIA

Moment of Inertia( Second moment area) The product of the elemental area and square of the perpendicular distance between the centroid of area and the axis of reference is the “Moment of Inertia” about the reference axis

y

x

ydA

o

Iox= da1 y12 + da2 y2

2+ da3 y32+ --

= ∑ da y2

Ioy = da1 x12 + da2 x2

2 + da3 x32+ ----

= ∑ da x2

Radius of Gyration

kk

kk

kk

Elemental areaElemental area

rr11

rr22

rr33

AAAA

BB BBRadius of gyration is defined as a constant distance of all Radius of gyration is defined as a constant distance of all elemental areas which have been rearranged with out elemental areas which have been rearranged with out altering the total moment of inertia.altering the total moment of inertia.

IIABAB= da k2 + da k2 + -----------

IIAB AB = ∑ da k2

IIABAB= A k2

k=k=√I√IABAB/A/A

Polar moment of Inertia (Perpendicular Axes theorem)

The moment of inertia of an area about an axis perpendicular to the plane of the area is called “Polar Moment of Inertia” and it is denoted by symbol Izz or J or Ip. The moment of inertia of

an area in xy plane w.r.to z. axis is Izz = Ip = J = ∫r2dA = ∫(x2 +

y2) dA = ∫x2dA + ∫y2dA = Ixx +Iyy 

O

y

x

r

z

x

Y

 Hence polar M.I. for an area w.r.t. an axis perpendicular to its plane of area is equal to the sum of the M.I. about any two mutually perpendicular axes in its plane, passing through the point of intersection of the polar axis and the area.

Polar moment of Inertia (Perpendicular Axes theorem)

Parallel Axis Theorem

d

A B

x x

dA

y*G

Moment of inertia of any area about an axis AB is equal to the M.I. about parallel centroidal axis plus the product of the total area and square of the distance between the two axes.

IAB =∑dA (d +y)2

= ∑dA (d2 + y2 + 2 × d × y) =∑dA. d2 +∑dA y2 + ∑ 2×d×dA y

= ∑dA. d2 +∑dA y2 + 2×d. ∑ y dA In the above term (2×d) is constant & ∑ y dA = 0

IAB = Ixx + A.d2

MOMENT OF INERTIA BY DIRECT INTEGRATION

M.I. about its horizontal centroidal axis :

Find the moment of inertia of rectangular area about centroidal horizontal axis by direct integration  

12

)(

3

2/

2/

2

22/

2/

BD

ydyB

ydAI

D

D

D

Dxx

G

B

x x

dy

yD

D/2

Q 4.1

.

   Find the Moment of Inertia of rectangular area about its base-

(about the line AB) using Parallel Axis Theorem

   IAB = IXX + A(d)2

Where d = D/2, the distance between axes xx and AB

BA

G

B

x x

dy

yD

D/2

  =BD3/12+(BD)(D/2)2

=BD3/12+BD3/4

=BD3/3

Q 4.2

  From similar triangles

Find the moment of inertia of Triangular area about the base by direct integration  

Q 4.3

h

x

Bh/3

x

b

xy

(h-y)dy

AA

  b/h = x/(h-y) x = b . (h-y)/h

  IAB = dA.y2 = (x.dy)y2

h

Ixx = (b . (h-y) y2.dy) /h 0

= b[ h (y3/3) – y4/4 ]/h = bh3/12

Q 4.3

h

x

Bh/3

x

b

xy

(h-y)dy

AA

Using Parallel axis theorem .MI about any line(AB) = MI about cenroidal parallel axis + Ad2

Find the moment of inertia of Triangular area about the centroidal horizontal axis  

Q 4.4

h

x

Bh/3

x

b

xy

(h-y)dy

AA

Centroidal horizontal axisCentroidal horizontal axis

IAB = Ixx + Ad2

Ixx = MI about centroidal axis x xIAB= MI about the Base line AB

Ixx = IAB – Ad2

= bh3/12 – bh/2 . (h/3)2

= bh3/36

Ixx = dA . y2

R 2

= (x.d.dr) r2Sin2 0 0

R 2

= r3.dr Sin2 d 0 0

R 2

= r3 dr {(1- Cos2)/2} d 0

R 0

2

=[r4/4] [/2 – Sin2/4] 0 0

= R4/4[ - 0] = R4/4

IXX = R4/4 = D4/64

BA

xxR

d

y=rSinr

Find the moment of inertia of Circular area about the centroidal horizontal axis  Q 4.5 X

IAB = dA . y2

R

= (r.d.dr) r2Sin2

0

R0

= r3.dr Sin2 d 0 0

R

= r3 dr (1- Cos2)/2) d 0 0

=[R4/4] [/2 – Sin2/4] 0

= R4/4[/2 - 0] = R4/8 = (πD4/32)

4R/3

y0

y0

BAxx R

Find the moment of inertia of Semi-circular area about the Base  & centroidal horizontal axis

Q 4.6

using parallel axis theorem: IAB = Ixx + A(d)2

Ixx = IAB – A(d)2

= R4/8 R2/2 . (4R/3)2

Ixx = 0.11R4

moment of inertia of Semi-circular area about the centriodal horizontal axis  

IAB = ICD R /2

IAB = (r.d.dr). r2Sin2 0 0

R /2

= r3.dr Sin2 d 0 0

R /2

= r3 dr (1- Cos2)/2) d 0 0

/2

=[R4/4] [/2 – (Sin2 )/4] 0

= R4 (/16 – 0) = R4/16

A B

x x

C

D y

y

4R/3π

4R/3π

Find the moment of inertia of Quarter-circular area about the base & centroidal horizontal axis  Q 4.7X

Moment of inertia about Centroidal axis, Ixx = IAB - Ad2

= R4/16 - R2. (0. 424R)2

= 0.055R4

Sl.No Figure I x0

-x0

I y0

-y0

I xx I yy

1

bd3/12 - bd3/3 -

2

bh3/36 - bh3/12 -

3

R4/4 R4/4 - -

4

0.11R4 R4/8 R4/8 -

5

0.055R4 0.055R4 R4/16 R4/16

b

dx0

x

x0

xd/2

b

h

xx

x0x0

h/3

x0x0

y0

y0

O

R

y0

y0xxx0

x0

4R/3π

x0

y y0

4R/3π

4R/3π

Y

Y Xo

Find the moment of Inertia of the shaded area shown in fig.about its base.

20mm5 5

5

5 5

25

15

1010

5

20mm

30mm

X X

Numerical Problems & SolutionsQ 4.8

20mm5 5

5

5 5

25

15

1010

5

20mm

30mm

Ixx = Ixx1+ Ixx2 +Ixx3

4mm310*297.5I

]20*10*1012

10*10[

30*)30*20(2

1

36

30*20

10)20*20(12

20*20

xx

23

23

23

X X

1

3

2

Solution:-

Q 4.8

Compute the M.I. about the base(bottom) for the area given in fig.

100mm

200mm

20mm20mm

80mm

30mm

x x

25mm

Q 4.9

100mm

200mm

20mm20mm

80mm

30mm

x x1

2

3 4

5

SOLUTION :-

25

Q 4.9

Ix x = 200*203/3+[25*1003/12+(25*100)702]

+2[87.5*203/36+0.5*87.5*20*(26.67)2]

+[100*303/12+100*30*1352]

Ix x=71.05*106mm4

Q 4.9

Find M.I. about the horizontal centroidal axis for the area shown and also find the radius of gyration.

xo

463.5mm

250mm250mm

100

100

200mm200mm

100mm

400mm

400mm

1100mm

y=436.5mm

xo

Q 4.10

Solution:-

xo

463.5mm

250mm250mm

100

100

200mm200mm

100mm

400mm

400mm

1100mm

y=436.5mm

xo

1

2

3 45

6

Q 4.10

∑A=1100*100+400*100+400*400+2(1/2*100*400)-π*502

=3,42,150 mm2

∑AY=1100*100*50+100*400*300+400*400*700+[(1/2)*100*

400*633.3]*2 - π *502*700 =14,93,38,200mm3

Y= ∑AY / ∑A=436.5mm

Q 4.10

Moment of Inertia about horizontal centroidal

Axis:-

IXoXo =[1100*1003/12 +1100*100(386.5)2]+[100*4003/12

+(100*400)*(136.5)2]+[400*4003/12+400*400*(263.5)2]+2

[100*4003/36+(1/2*400*100)*(196.8)2]-

[π*(50)4/4+π*502*(263.5)2]

IXoXo =32.36*109mm4

rXoXo=√(IXoXo/A)=307.536mm.

Q 4.10

Compute the M.I. of 100 mm x 150mm rectangular shown in fig.about x-x axis to which it is inclined at an angle of   = Sin-1(4/5)

150m

m

100mm

X X

= Sin-1(4/5)A

B

C

D

Q 4.11

150m

m

100mm

X X

= Sin-1(4/5)A

B

C

D

MN

KL

sin =4/5, =53.13o

= From geometry of fig ,

BK=ABsin(90-53.13o)

=100sin(90-53.13o )=60mm

ND=BK=60mm

FD= 60/sin53.13o= 75mm

AF=150-FD=75mm FL=ME=75sin53.13o=60mm

mmAB

FCAE 1258.0

100

90cos

E

F

Solution:-

Q 4.11

IXX=IDFC+IFCE+IFEA+IAEB

=125 (60)3/ 36+ (1/2)*125*60*(60+60/3)2

+125(60)3 /36+(1/2)*125*60*402

+125*603 /36 +(1/2)*125*60 *202

+125*603/36 +(1/2)*125*60*202

Ixx = 36,00,000 mm4

Q 4.11

Find the M.I. of the shaded area shown about AB.

40mm40mm

40mmA B

40mm

80mm

Q 4.12

40mm40mm

40mmA B

40mm

80mm

IA = IA1 + IA2 + IA3

= [80*803/36

+(1/2)*80*80(80/3)2

+[(0.11*404)+(1/2)π(40)2

(0.424r)2] –[π*204/4]

IA =429.3*10 44mmmm44

Solution:-

1

2

3

Q 4.12

Calculate the moment of inertia of the built- up section shown in fig.about the centroidal axis parallel to AB. All members are 10mm thick.

250mm

250mm

50mm

A B

50mm

10mm

10mm

50mm50mm

Q 4.13

250mm

250mm

40mm

A B

50mm

10mm

10mm

1

2

3 45

6

50mm

40mm

Solution:-

Y=73.03mm

Q 4.13

It is divided into six rectangles. Distance of centroidal X-

axis from AB=Y=∑Ai Yi /∑A

∑A=2*250*10+40*4*10=6600mm2

∑Ai Yi =

=250*10*5+2*40*10*30+40*40*15+40*10*255

+250*10*135

=4,82,000mm3

Q 4.13

Y= ∑AiYi / ∑A=482000/6600=73.03mm

Moment of Inertia about centroidal axis

=Sum of M.I. of individual rectangles

= 250*103/ 12+250*10*68.032

+ [10*403/12 +40*10*(43.03)2 *2

+40*103/12+40*10 (58.03)2 +10*2502 /12+250*

10(73.05-135)2 +40*103 /12+40*10(73.05-255)2

IXoXo =5,03,99395mm4

Q 4.13

Find the second moment of the shaded area shown in fig.about its centroidal x-axis.

R=20

20mm 40mm 20mm

40mm

20mm

30mm 50mm

Q 4.14

solution:-

R=20

20mm 40mm 20mm

40mm

20mm

30mm 50mm

1

2 3

4

XoXo

31.5mm

Q 4.14

∑A=40*80+1/2*30*30+1/2*50*30-1/2*π*(20)2 =3772mm2

∑AiXi =3200*40+450*2/3*30+750*(30+50/3)

-1/2* π*202 *40 =146880mm3

∑AiYi =3200*20+450*50+750*50-628*4*20/3 π= 118666.67mm3

Y =118666.67/3772= 31.5mm

IXoXo = [80*403/12+(80*40)(11.5)2 ]+[30*303/36+

1/2 *30*30(18.5)2 ]+ [50*303 /36 +1/2+50*30*(18.50)2]-

[0.11*204 )+π/2*(20)2 (31.5-0.424*20)]

= 970.3*103mm4

Q 4.14

Find the M.I. about top of section and about two centroidal axes.

10mm

150mm

10mm

150mm

Q 4.15

solution:-

10mm

150mm

10mm

150mm

Yo

Yo

Xo Xo

41.21mm

Q 4.15

It is symmetrical about Y axis, X=0

Y=∑AY/∑A =[ (10*150*5) +(10*140*80)]/[(10*150)+(10*140)

=41.21mm from top

IXX= 150*103 /12 + ( 150*10)*52 +10*1403 /12+(10*140)

*(80)2 = 11296667mm4

IXX = IXoXo + A(d)2 ,where A(d)2

IXoXo =6371701.10 mm4

IYoYo =( 10*1503 /12) + (140*103/12)=2824166.7mm4

IXoXo+(150*10+140*10)*(41.21)2=11296667mm3

Q 4.15

Find the M.I. about centroidal axes and radius of gyration for the area in given fig.

40mm

10mm

50mm

10mm

Q 4.16

solution:-

40mm

10mm

50mm

10mm

Xo Xo

Yo

Yo

A B

C D

E

G

FY=17.5mm

X=12.5mm

Q 4.16

Centroid

X=∑ax/∑a= [(50*10)5+(30*10)25]/800=12.5mm

Y=∑ay/∑a= [(50*10)25+(30*10)5]/800=17.5mm

Q 4.16

IXoXo = [10*503/12+(50*10)(17.5-5)2]+

[30*103/12+(30*10) (12.5)2 =181666.66mm4

IYoYo=[(50*103/12)+(50*10)(7.5)2]+[10*303/12

+(30*10)*(12.5)2]=101666.66mm4

rxx=√(181666.66/800) = 15.07 mm

ryy=√(101666.66/800)= 11.27 mm

Determine he moment of inertia about the horizontal centroidal axes for the area shown.

100 mm

100 mm

60 mm

Q 4.17

solution:-

100 mm

100 mm

60 mm

Y=40.3mm

Xo Xo1 2

Q 4.17

Y=[(100*100)50-(π/4)602*74.56]/[(100*100)- (π/4)602]

=40.3mm

IXoXo= [(100*1003/12)+100*100*(9.7)2]-

[0.55*(60)4+0.785*(60)2*(34.56)2 ]

=83,75,788.74mm4

Q 4.17

EXERCISE PROBLEMS

Q.1. Determine the moment of inertia about the centroidal axes.

100mm

20

30mm

30mm

30mm

[Ans: Y = 27.69mm Ixx = 1.801 x 106mm4

Iyy = 1.855 x 106mm4]

Q.2. Determine second moment of area about the centroidal horizontal and vertical axes.

[Ans: X = 99.7mm from A, Y = 265 mm

Ixx = 10.29 x 109mm4, Iyy = 16.97 x 109mm4]

200mm

200

300mm

300mm

900mm

EXERCISE PROBLEMS

Q.3. Determine M.I. Of the built up section about the horizontal and vertical centroidal axes and the radii of gyration.

[Ans: Ixx = 45.54 x 106mm4, Iyy = 24.15 x 106mm4

rxx = 62.66mm, ryy = 45.63mm]

60

100mm

200mm

140mm

20

20

EXERCISE PROBLEMS

Q.4. Determine the horizontal and vertical centroidal M.I. Of the shaded portion of the figure.

[Ans: X = 83.1mm

Ixx = 2228.94 x 104mm4, Iyy = 4789.61 x 104mm4]

60

60 60

X X2020

EXERCISE PROBLEMS

Q.5. Determine the spacing of the symmetrically placed vertical blocks such that Ixx = Iyy for the shaded area.

[Ans: d/2 = 223.9mm d=447.8mm]

200mm

600mm

d

400mm

200mm

200mm

200mm

EXERCISE PROBLEMS

Q.6. Find the horizontal and vertical centroidal moment of inertia of the section shown in Fig. built up with R.S.J. (I-Section) 250 x 250 and two plates 400 x 16 mm each attached one to each.

Properties of I section are

Ixx = 7983.9 x 104mm4

Iyy = 2011.7 x 104mm4

[Ans: Ixx = 30.653 x 107mm4, Iyy = 19.078 x 107mm4]

4000mm

2500mm

160mm

160mm

Cross sectional area=6971mm2

EXERCISE PROBLEMS

Q.7. Find the horizontal and vertical centroidal moment of inertia of built up section shown in Figure. The section consists of 4 symmetrically placed ISA 60 x 60 with two plates 300 x 20 mm2.

[Ans: Ixx = 111.078 x 107mm4, Iyy = 39.574 x 107mm4]

300mm

Properties of ISA

Cross sectional area = 4400mm2

Ixx = Iyy ;Cxx = Cyy =18.5mm

18.5mm

18.5mm

20mm

200mm

EXERCISE PROBLEMS

Q.8. The R.S. Channel section ISAIC 300 are placed back to back with required to keep them in place. Determine the clear distance d between them so that Ixx = Iyy for the composite section.

[Ans: d = 183.1mm]

Properties of ISMC300

C/S Area = 4564mm2

Ixx = 6362.6 x 104mm4

Iyy = 310.8 x 104mm4

Cyy = 23.6mm

X X

Y

Y

Lacing

d

380mm

23.6mm

EXERCISE PROBLEMS

Q9. Determine horizontal and vertical centroidal M.I. for the section shown in figure.

[Ans: Ixx = 2870.43 x 104mm4, Iyy = 521.64 x 104mm4]90mm

160mm

40mm

40mm

40mm

EXERCISE PROBLEMS