MOMENT OF INERTIA

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MOMENT OF INERTIAMoment 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

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.

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

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.

PERPENDICULAR AXIS THEOREMT-4

Parallel Axis Theorem

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 = Ix

0 +Ad2

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.

Iyy = A.ryy2

Ixx = A.rxx2

ryy = √ Iyy/A

And rxx = √ Ixx /A

rxx and ryy are called the radii of gyration

MOMENT OF INERTIA BY DIRECT INTEGRATION

M.I. about its horizontal centroidal axis :

RECTANGLE :

IXoXo = -d/2 ∫

=-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

(2) TRIANGLE :

(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 0

= b[ h (y3/3) – y4/4 ]/h = bh3/12

(b) Moment of inertia about its centroidal axis: _ Ixx = Ix

0 + Ad2

0 = Ixx – Ad2

= bh3/12 – bh/2 . (h/3)2 = bh3/36

x0 = dA . y2

= (x.d.dr) r2Sin2 0 0

= r3.dr Sin2 d 0 0

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

=[r4/4] [/2 – Sin2/4] 0 0

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

IXoXo = R4/4 = D4/64

y=rSin

3. CIRCULAR AREA:

Ixx = dA . y2

= (r.d.dr) r2Sin2

= r3.dr Sin2 d 0 0

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

=[R4/4] [/2 – Sin2/4] 0

= R4/4[/2 - 0] = R4/8

4. SEMI CIRCULAR AREA:

About horizontal centroidal axis: Ixx = Ix

0 + A(d)2

= Ixx – A(d)2

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

= 0.11R4

QUARTER CIRCLE:

Ixx = Iyy R /2

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

= r3.dr Sin2 d 0 0

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

=[R4/4] [/2 – (Sin2 )/4] 0

= R4 (/16 – 0) = R4/16

4R/3π

Moment of inertia about Centroidal axis, _ Ix

0 = 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.

Sl.No Figure I x0

I xx I yy

bd3/12 - bd3/3 -

bh3/36 - bh3/12 -

R4/4 R4/4 - -

0.11R4 R4/8 R4/8 -

0.055R4 0.055R4 R4/16 R4/16

y0xxx0

4R/3π

Q.1. Find the moment of Inertia of the shaded area shown in fig.about its base.

20mm5 5

Problems on Moment of Inertia

20mm5 5

Ixx = Ixx1+ Ixx2 -Ixx3

4mm310*297.5I

]20*10*1012

10*10[

30*)30*20(2

10)20*20(12

Solution:-

Q.2. Compute the M.I. about the base(bottom) for the area given in fig.

20mm20mm

SOLUTION :-

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.3. Find M.I. about the horizontal centroidal axis for the area fig. No.3, and also find the radius of gyration.

463.5mm

250mm250mm

200mm200mm

1100mm

y=436.5mm

Solution:-

463.5mm

250mm250mm

200mm200mm

1100mm

y=436.5mm

Solution to prob. No.03

∑A=1100*100+400*100+400*400+2(1/2*100*400)-π*502 =3,42,150

∑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

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. 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)

= Sin-1(4/5)A

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

FCAE 1258.0

Solution:-

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.5. Find the M.I. of the shaded area shown in fig.,about AB.

40mm40mm

40mmA B

40mm40mm

40mmA B

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 44mmmm44

Solution:-

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.

50mm50mm

Solution:-

Y=73.03mm

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

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

20mm 40mm 20mm

30mm 50mm

solution:-

20mm 40mm 20mm

30mm 50mm

31.5mm

∑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.8. Find the M.I. about top of section and about two centroidal axes.

solution:-

41.21mm

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

solution:-

FY=17.5mm

X=12.5mm

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 =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

Q. 10. Determine he moment of inertia about the horizontal centroidal axes for the area in fig.

100 mm

solution:-

100 mm

Y=40.3mm

Xo Xo1 2

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

EXERCISE PROBLEMS ON M.I.

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

[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]

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]

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]

X X2020

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]

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

Cross sectional area=6971mm2

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

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

Lacing

23.6mm

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

MOMENT OF INERTIA

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