MOMENT OF INERTIAMoment 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 Iox= da1 y12 + da2 y22+ da3 y32+ -- = da y2Ioy = da1 x12 + da2 x22 + da3 x32+ ---- = da x2

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 +IyyOyxrzxY

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 TheoremdABxxdAy*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 + 2ddA y

= dA. d2 +dA y2 + 2d. y dA In the above term (2d) 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 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)2Where d = D/2, the distance between axes xx and AB =BD3/12+(BD)(D/2)2 =BD3/12+BD3/4 =BD3/3

Q 4.2

bdx0xx0xd/2bhxxx0x0h/3x0x0y0y0ORy0y0xxx0x04R/3x0yy04R/34R/3YYXo

Sl.NoFigureI x0-x0I y0-y0I xxI yy1bd3/12-bd3/3-2bh3/36-bh3/12

-3R4/4R4/4--40.11R4R4/8R4/8-50.055R40.055R4R4/16R4/16

Find the M.I. about top of section and about two centroidal axes.Q 4.15

solution:-10mm150mm10mm150mmYoYoXoXo41.21mmQ 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 =6342417.808 mm4 IYoYo =( 10*1503 /12) + (140*103/12)=2824166.7mm4

IXoXo+(150*10+140*10)*(41.21)2=11296667mm3Q 4.15

Find the M.I. about centroidal axes and radius of gyration for the area in given fig. Q 4.16

solution:-40mm10mm50mm10mmXoXoYoYoABCDEGFY=17.5mmX=12.5mmQ 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=17.5mmQ 4.16IXoXo = [10*503/12+(50*10)(17.5-5)2]+ [30*103/12+(30*10) (12.5)2 =181666.66mm4IYoYo=[(50*103/12)+(50*10)(7.5)2]+[10*303/12 +(30*10)*(12.5)2]=101666.66mm4

**************