1. MOMENT OF INERTIA Moment of Inertia:
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.
Ixx = ∫dA. y2
Iyy = ∫dA. x2 y dA x y x

2. T-2
It is also called second moment of area because first moment of elemental area is dA.y and dA.x; and if it is again multiplied by the distance,we get second moment of elemental area as (dA.y)y and (dA.x)x.

3. 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  Y x r y O x z

4. PERPENDICULAR AXIS THEOREM
 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.

5. Parallel Axis Theorem
T-5 dA y´ x0 x0 * G _ d y x x

6. ( since Ist moment of area about centroidal axis = 0) _
Ixx = ∫dA. y2 _
= ∫dA (d +y')2
_ _
= ∫dA (d2+ y'2 + 2dy')
= ∫dA. d2 + ∫dAy΄2 + ∫ 2d.dAy'
d2 ∫dA = A.(d)2
∫dA. y'2 = Ix0x0
2d ∫ dAy’ = 0
( since Ist moment of area about centroidal axis = 0) _
Ix x = Ix0 x0 +Ad2

7. T-7
Hence, moment of inertia of any area about an axis xx 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.
Radius of Gyration
It is the perpendicular distance at which the whole area may be assumed to be concentrated, yielding the same second moment of the area above the axis under consideration.

8. rxx and ryy are called the radii of gyration
Iyy = A.ryy2
Ixx = A.rxx2
ryy = √ Iyy/A
And rxx = √ Ixx /A A A
rxx
ryy y x x
rxx and ryy are called the radii of gyration

9. MOMENT OF INERTIA BY DIRECT INTEGRATION
M.I. about its horizontal centroidal axis :
RECTANGLE :
  IXoXo = -d/2 ∫ +d/2 dAy2
=-d/2∫+d/2 (b.dy)y2
= bd3/12
 About its base
IXX=IXoXo +A(d)2
Where d = d/2, the distance between axes xx and xoxo
=bd3/12+(bd)(d/2)2
=bd3/12+bd3/4=bd3/3 . dy
d/2 y d x0 x0 G x x b

10. From similar triangles b/h = x/(h-y)  x = b . (h-y)/h h
 (a) M.I. about its base :
  Ixx =  dA.y2 =  (x.dy)y2
From similar triangles
b/h = x/(h-y)
 x = b . (h-y)/h h
Ixx =  (b . (h-y)y2.dy)/h
= b[ h (y3/3) – y4/4 ]/h
= bh3/12
(h-y) dy h x x0 y x0
h/3 x b

11. (b) Moment of inertia about its centroidal axis: _ Ixx = Ix0x0 + Ad2
= bh3/12 – bh/2 . (h/3)2 = bh3/36

12. 3. CIRCULAR AREA: Ix0x0 =  dA . y2 R 2 =   (x.d.dr) r2Sin2 0 0
T-12
3. CIRCULAR AREA:
Ix0x0 =  dA . y2
R 2
=   (x.d.dr) r2Sin2
0 0
R 2
=  r3.dr Sin2 d
0 0
R 
= r3 dr  {(1- Cos2)/2} d
R 
=[r4/4] [/2 – Sin2/4]
= R4/4[ - 0] = R4/4
IXoXo =  R4/4 = D4/64 d r
y=rSin  x0 x0 R x x

13. 4. SEMI CIRCULAR AREA: Ixx =  dA . y2 R  =   (r.d.dr) r2Sin2
T-13
4. SEMI CIRCULAR AREA:
Ixx =  dA . y2
R 
=   (r.d.dr) r2Sin2
0R 
= r3.dr  Sin2 d
R 
=  r3 dr (1- Cos2)/2) d 
=[R4/4] [/2 – Sin2/4]
= R4/4[/2 - 0] = R4/8 y0 x0 R x0
4R/3 x x y0

14. About horizontal centroidal axis: Ixx = Ix0x0 + A(d)2
=  R4/8 R2/2 . (4R/3)2
Ix0x0 = 0.11R4

15. QUARTER CIRCLE: Ixx = Iyy R /2 Ixx =   (r.d.dr). r2Sin2 0 0 R /2
0 0
R /2
= r3.dr  Sin2 d
R /2
= r3 dr  (1- Cos2)/2) d
/2
=[R4/4] [/2 – (Sin2 )/4]
= R4 (/16 – 0) = R4/16 y0 y x0 x0
4R/3π x x y y0
4R/3π

16. Moment of inertia about Centroidal axis, _ Ix0x0 = Ixx - Ad2
= R4/16 - R2. (0. 424R)2
= 0.055R4
The following table indicates the final values of M.I. about X and Y axes for different geometrical figures.

17. Figure I x0-x0 I y0-y0 I xx I yy
T-17
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 4
0.11R4
R4/8 5
0.055R4
R4/16 b Y d x0 x0
d/2 x x Y Xo h x0 x0
h/3 x x b y0 R x0 x0 O y0 y0
4R/3π x0 x0 x x y0 y y0 x0
4R/3π
4R/3π

18. Problems on Moment of Inertia
Q.1. Find the moment of Inertia of the shaded area shown in fig.about its base. 10 10 25
30mm 5 5
20mm 15 X X 5 5 5 5
20mm

19. Solution:- Ixx = Ixx1+ Ixx2 -Ixx3 P-2 10 10 25 30mm 5 5 20mm 15 5 5 5

20. P-3
Q.2. Compute the M.I. about the base(bottom) for the area given in fig.
100mm
30mm
25mm
80mm
Fgfgy
20mm
20mm x x
200mm

21. P-4
SOLUTION :-
100mm 5
30mm
80mm 25
Fgfgy 2
20mm 3 4
20mm 1 x x
200mm

22. P-5
Ix x = *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

23. P-6
Q.3. Find M.I. about the horizontal centroidal axis for the area fig. No.3, and also find the radius of gyration.
463.5mm
250mm
100
200mm
100mm
400mm
1100mm
y=436.5mm xo xo

24. Solution:- xo xo 463.5mm 400mm P-7 250mm 100 200mm 100mm 1100mm
y=436.5mm xo 6 3 4 5 xo 2 1

25. Solution to prob. No.03
P-8
∑A=1100* * *400+2(1/2*100*400)-π*502 =3,42,150 mm2 mm
∑AY=1100*100*50+100*400* *400*700+[(1/2)*100*
400*633.3]*2 - π *502* =14,93,38,200mm3
Y= ∑AY / ∑A=436.5mm

26. Moment of Inertia about horizontal centroidal Axis:-
P-9
Moment of Inertia about horizontal centroidal
Axis:-
IXoXo =[1100*1003/ *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)= mm.

27. P-10
Q. 4. 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) D
150mm C
 = Sin-1(4/5) A X X B
100mm

28. AF=150-FD=75mm FL=ME=75sin53.13o=60mm
P-11
sin =4/5,  =53.13o
= From geometry of fig ,
BK=ABsin( o)
=100sin( o )=60mm
ND=BK=60mm
FD= /sin53.13o= 75mm
AF=150-FD=75mm FL=ME=75sin53.13o=60mm
Solution:- D M
150mm F C N
 = Sin-1(4/5) A K X X L E B
100mm

29. IXX=IDFC+IFCE+IFEA+IAEB =125 (60)3/ 36+ (1/2)*125*60*(60+60/3)2
P-12
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

30. Q.5. Find the M.I. of the shaded area shown in fig.,about AB.
P-13
Q.5. Find the M.I. of the shaded area shown in fig.,about AB.
80mm
40mm A B
40mm
40mm
40mm

31. IAB =IAB1+IAB2+IAB3 [80*803/36 +(1/2)*80*80(80/3)2
P-14
Solution:-
IAB =IAB1+IAB2+IAB3
[80*803/36 +(1/2)*80*80(80/3)2
+[(0.11*404)+(1/2)π(40)2
(0.424r)2] –[π*204/4]
IAB =429.3*10 4mm4
80mm 1
40mm A B 3
40mm 2
40mm
40mm

32. P-15
Q.6. 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 A B
50mm
50mm
50mm
10mm
250mm
10mm
50mm

33. Solution:- P-16 250mm A B 50mm Y=73.03mm 40mm 40mm 250mm 10mm 50mm 2 5

34. P-17
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

35. Moment of Inertia about centroidal axis
P-18
Y= ∑AiYi / ∑A=482000/6600=73.03mm
Moment of Inertia about centroidal axis
=Sum of M.I. of individual rectangles
= 250*103/ *10*68.032
+ [10*403/ *10*(43.03)2 *2
+40*103/12+40*10 (58.03)2 +10*2502 /12+250*
10( )2 +40*103 /12+40*10( )2
IXoXo =5,03,99395mm4

36. P-19
Q.7. Find the second moment of the shaded area shown in fig.about its centroidal x-axis.
30mm
50mm
20mm
40mm
R=20
20mm
40mm
20mm

37. solution:- P-20 30mm 50mm 20mm Xo Xo 40mm 31.5mm R=20 20mm 40mm 20mm 4

38. ∑A=40*80+1/2*30*30+1/2*50*30-1/2*π*(20)2 =3772mm2
P-21
∑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= mm3
Y = /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 ( *20)]
= 970.3*103mm4

39. Q.8. Find the M.I. about top of section and about two centroidal axes.
150mm
10mm
150mm
10mm

40. P-23
solution:-
150mm Yo
10mm
41.21mm Xo Xo
150mm Yo
10mm

41. It is symmetrical about Y axis, X=0
P-24
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= mm4
IXX = IXoXo + A(d)2 ,where A(d)2
IXoXo = mm4
IYoYo =( 10*1503 /12) + (140*103/12)= mm4
IXoXo+(150*10+140*10)*(41.21)2= mm3

42. P-25
Q.9. Find the M.I. about centroidal axes and radius of gyration for the area in given fig.
10mm
50mm
10mm
40mm

43. solution:- P-26 10mm 50mm Y=17.5mm 10mm 40mm X=12.5mm Yo A B Xo Xo E FC D G Yo
40mm
X=12.5mm

44. X=∑ax/∑a= [(50*10)5+(30*10)25]/800=12.5mm
P-27
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
IXoXo = [10*503/12+(50*10)(17.5-5)2]+
[30*103/12+(30*10) (12.5)2 = mm4
IYoYo=[(50*103/12)+(50*10)(7.5)2]+[10*303/12
+(30*10)*(12.5)2]= mm4
rxx=√( /800) = mm
ryy=√( /800)= mm

45. P-28
Q. 10. Determine he moment of inertia about the horizontal centroidal axes for the area in fig.
60 mm
100 mm
100 mm

46. P-29
solution:-
60 mm
100 mm 2 1 Xo Xo
Y=40.3mm
100 mm

47. Y=[(100*100)50-(π/4)602*74.56]/[(100*100)- (π/4)602] =40.3mm
P-30
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) *(60)2*(34.56)2 ]
=83,75,788.74mm4

48. EXERCISE PROBLEMS ON M.I.
Q.1. Determine the moment of inertia about the centroidal axes.
30mm
30mm 20
30mm
100mm
[Ans: Y = 27.69mm Ixx = x 106mm4
Iyy = x 106mm4]

49. EP-2
Q.2. Determine second moment of area about the centroidal horizontal and vertical axes.
300mm
300mm
200
200mm
900mm
[Ans: X = 99.7mm from A, Y = 265 mm
Ixx = x 109mm4, Iyy = x 109mm4]

50. EP-3
Q.3. Determine M.I. Of the built up section about the horizontal and vertical centroidal axes and the radii of gyration.
200mm 20
140mm 60 20
100mm
[Ans: Ixx = x 106mm4, Iyy = x 106mm4
rxx = 62.66mm, ryy = 45.63mm]

51. EP-4
Q.4. Determine the horizontal and vertical centroidal M.I. Of the shaded portion of the figure. 60 X 20 X 20 60 60
[Ans: X = 83.1mm
Ixx = x 104mm4, Iyy = x 104mm4]

52. EP-5
Q.5. Determine the spacing of the symmetrically placed vertical blocks such that Ixx = Iyy for the shaded area.
200mm
400mm d
200mm
200mm
200mm
600mm
[Ans: d/2 = 223.9mm d=447.8mm]

53. Properties of I section are Ixx = 7983.9 x 104mm4
EP-6
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 = x 104mm4
Iyy = x 104mm4
160mm
2500mm
Cross sectional area=6971mm2
160mm
4000mm
[Ans: Ixx = x 107mm4, Iyy = x 107mm4]

54. Cross sectional area = 4400mm2 Ixx = Iyy ;Cxx = Cyy =18.5mm
EP-7
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.
Properties of ISA
Cross sectional area = 4400mm2
Ixx = Iyy ;Cxx = Cyy =18.5mm
18.5mm
200mm
18.5mm
20mm
300mm
[Ans: Ixx = x 107mm4, Iyy = x 107mm4]

55. EP-8
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.
Properties of ISMC300
C/S Area = 4564mm2
Ixx = x 104mm4
Iyy = x 104mm4
Cyy = 23.6mm
Lacing Y
23.6mm
380mm X X d
[Ans: d = 183.1mm] Y

56. [Ans: Ixx = 2870.43 x 104mm4, Iyy = 521.64 x 104mm4]
EP-9
Q9. Determine horizontal and vertical centroidal M.I. for the section shown in figure.
40mm
160mm
40mm
40mm
90mm
[Ans: Ixx = x 104mm4, Iyy = x 104mm4]