Define following terms 1.Centroid

Define following terms 1.Centroid
2. Radius of gyration 3.Area moment of Inertia centroid C.G The point at which the total area of plane to be considered is known as centroid , the centroid is represented by C.G The centroid and centre of gravity are at the same point Where centre of gravity consider to be whole mass of an object act at a point

Let consider dA is small area on the plane
Area moment of Inertia Y y axis X x axis Let consider dA is small area on the plane We know that moment of area is = area multiplied by Perpendicular distance to axis of reference i.e. - area moment = XdA Then moment of moment or second moment is = X2dA with respect to x axis = Y2dA with respect to y axis

Ixx = Moment of Inertia about x axis Ixx = ∫ 𝑿𝟐
centroid Area moment of Inertia C.G The area moment about a point is the product of the plane area and the normal distance of the point from the reference line area. If this is multiplied by the perpendicular distance is called second moment of area Y y axis X x axis Let consider dA is small area on the plane Ixx = Moment of Inertia about x axis Ixx = ∫ 𝑿𝟐 dA Iyy = Moment of Inertia about y axis Iyy = ∫ 𝒀𝟐 dA

Radius of gyration A A= total area of lamina concentrated at this point Radius of gyration = K I = moment of inertia I = Ak2 axis I A K =

X axis Parallel axis Theorem Plane Moment Inertia about centroid X X centroid h A B Area of plane Moment of Inertia about AB axis IAB = IXX + Ah2 Moment of Inertia about AB axis IAB = moment of Inertia about centroid + area Multiplied by square of distance between centroid and axis parallel to centroid

Regular rectangle O h h/2 b/2 b O Centroid of rectangle is known to us by looking at component

Regular circle O = centroid of circle is known to us by looking at the component O Radius = r Diameter = D = 2r

Non Regular component of hole on a rectangle
A hole on a rectangle whose diameter = 10mm 160mm 120mm Centroid of the component to be calculated 80mm 100mm

Y axis 10mm Y axis 160mm 160mm Y2=120mm Y1=80mm X axis 0,0 X axis 0,0 80mm X1=50mm 100mm X2=80mm 100mm A2= π r2 = π102 = 314 mm2 A1= 100X160=16000 mm2 Non regular component regular component of rectangle regular component of circle

Area X cordinate Y cordinate A1= 100X160=16000 mm2 A2= π r2 = π102 = 314 mm2 X1=50mm X2=80mm Y1=80mm Y2=120mm X Y A1X1 - A2X2 16000X X80 X= = = = 49.3mm 15686 A1 - A2 16000X X120 A1Y1 - A2Y2 = =79.2mm Y= = 15686 A1 - A2 X = 49.3mm = 79.2mm Y

Pl note that shift of centroid
80mm 79.2 mm 50 mm 49.3mm Centroid of rectangle after making hole Centroid of rectangle before making hole

For triangle note the centroid
b/3 h/3 b For triangle note centroid distance

centroid Radius = r 4r/3π Radius = r Radius = r For semicircle note centroid distance

centroid Radius = r 4r/3π Radius = r 4r/3π 4r/3π Radius = r For quarter circle note the centroid

Determination of the centroid of the area of composite section is explained in different method developed by me R= 2m 4m 6m 3m

Draw the axis as shown R= 2m 4m 6m 3m

Consider semi circle by splitting the composite section and mark the centroid with respect to axis drawn considering + and – (positive and negative with respect to axis) R= 2m Centroid of semi circle Y1= 2.0 - 4r/3π = (4x2)/3π = - Ve sign is with respect to axis Area = A1= π22 /2= 6.28 m2 X1= Y1= 2.0

Centroid of rectangle 4m Y2=2m X2=3m 6m Area = A2= 6X4 24 m2 X2=3 Y2=2

Centroid of triangle X3 4m Y3=4/3=1.333 3/3=1 6m 3m A3=(1/2)x 3x 4=6 m2 X3= 7 Y3= 1.333

Area X Coordinate Y Coordinate A1=6.28 X1= Y1= 2.0 A2=24 X2=3 Y2=2 A3=6 X3= 7 Y3= 1.333 (6.28 X ( )) +( 24X3) +(6X7) X = A1X1 + A2X2 + A3X3 = mm = ( ) A1+ A2 + A3 Y = A1Y1 + A2Y2 + A3Y3 (6.28 X ( 2.0) +( 24X2) +(6X1.333) = 1.890mm = ( ) A1+ A2 + A3

Centroid of composite bar
R= 2m 4m Y = mm X = mm 3m 6m

x x y Moment of Inertia of standard and regular shape component y
y axis X axis h x x bh3 I xx = 12 h/2 hb3 I yy = Centroid 12 b/2 Where I xx = moment of Inertia about xx axis b y Where I yy = moment of Inertia about yy axis

Y Moment of Inertia of rectangle about the axis AB parallel to axis xx is equal to IAB= Ixx + A(Y1) 2 X axis h x x Where Ixx moment of Inertia about xx axis A area of rectangle and Y1 is distance between the axis AB and x axis Pl note that AB axis is parallel to x axis Y1 b bh3 = IAB +{( b X h) (Y1)2} Y B 12 A

Y centroid X axis X X d = Diameter of circle Y axis Y πd4 Circle Ixx = 64 πd4 Circle Iyy = 64 πd4 πd4 πd4 Plolar moment of Inertia Izz = Ixx+Iyy = + = 64 64 32

Moment of Inertia of semicircle
centroid Radius = r Ixx = d4 For semicircle note centroid distance x x Y1 = 4r/3π Y1 A B For circle moment of Inertia about this axis is known that is πd4/64 and for semi circle it is( πd4/64)X ½ = πd4/128 pl note that we do not have moment f Inertia of semi circle about centroid axis which is to be calculated applying parallel axis theorem Radius = r Radius = r moment of Inertia for semi circle it is ( πd4/64)X ½ = πd4/128 Now IAB= Ixx +(AY1)2 Y1 = 4r/3π Ixx = IAB - (AY1)2 Ixx = d4 A=πr2 or πd4/4 Where IAB= πd4/128

IAB = moment of inertia triangle about centroid
moment of inertia of Triangle about this axis or base of triangle is= bh3 Let us know the moment of inertia of Triangle 12 Ixx= moment of Inertia Of triangle about Centroid of triangle is bh3 36 h Y1=h/3 B A Applying parallel axis theorem b moment of inertia of Triangle about centroid= Ixx =IAB-A(y1)2 = A = area of triangle bh3 36 IAB = moment of inertia triangle about centroid

Ixx of circle = πd4 Pl note that shift of centroid and we need to calculate moment Inertia of Composite section about the shifted axis let us do it 64 d=10 mm bh3 80mm Ixx = 0.8 mm 12 40mm 40.8mm 80mm B A 79.2mm 160mm 80mm 49.3mm 50 mm IA B = to be calculated by applying parallel axis theorem Centroid of rectangle after making hole 100 mm Centroid of rectangle before making hole IA B of rectangle = Ixx of rectangle about its centroid + area of rectangle X square axis shift in y axis bh3 100x803 + 100x80x (0.8)2 = = mm I 4 A B = + bxh (y1)2 = 12 12

Moment of Inertia of composite section =
πd4 πx104 = = mm 4 I of Circle = + π r2 (y1)2 + = πx52x (40.8)2 A B 64 64 Moment of Inertia of composite section = IA B of rectangle - IA B of Circle = = mm4

To find the centroids of given object Let h= 50mm ,D=50mm and R=10mm
Hole h B A R D

A1 area of triangle Y1 = centroid distance from base Y axis A1 = (1/2)50*50 = 1250mm2 centroid Y1=50/3=16.667mm A1Y1= mm3 A1 h = 50mm h 50 3 Y1= 3 = X axis

A2 area of circle A2 = πr2 = π*102= 314mm2 Y axis Y2=0 since centroid of Circle coincides and with axis i.e. it lies in x axis A2Y2=0 centroid A2 X axis A r=10mm

A3 = area of semi circle A3 = (1/2)πr2 = 0.5*π*252=981.25mm2 A3y3= Y axis centroid X axis A Y3 = - (4r/3π) =-(4*25)/3π= mm Pl note –ve sign is due to the centroid of Semi circle fall below the x axis Radius of circle r=25mm

Y= centroid distance Area X centroid distance = AXY Y1 =16.667mm A1Y1
A1 =1250mm2 Y= centroid distance Area X centroid distance = AXY Y1 =16.667mm A1Y1 Y2 = 0 A2Y2 = 0 = mm3 A2 =314mm2 A3 =981.25mm2 Y3 =10.617mm A3Y3 = mm3 A1 y1-A2Y2 +A3Y3 – = = = mm Y A1 -A2 +A3

Centroid of composite section Calculated
h=50mm 5.4327mm A B r= 10mm D= 50mm

Finding moment of inertia of object about the axis AB
r=d/2= 10mm R=D/2=25mm D=b=50mm

1 50 x 503 B A Moment of inertia of Triangle about AB=(1/12)bh3 =
h=50 mm Axis about which moment of inertia To be taken B A 50 mm Moment of inertia of Triangle about AB=(1/12)bh3 = 1 50 x 503 12 = mm4

Moment of inertia of circle about AB=(πd )/64 = =7850mm4 64
Centroid Axis about which moment of inertia To be taken A B r=d/2= 10mm π 204 Moment of inertia of circle about AB=(πd )/64 = 4 =7850mm4 64

Moment of inertia of semi circle about AB=(1/128)*(πD ) = =153320mm
Centroid Axis about which moment of inertia To be taken A B D=b=50mm π 504 Moment of inertia of semi circle about AB=(1/128)*(πD ) = 4 =153320mm 4 128

h=50 Moment of inertia of object about AB = B A Moment of inertia of Triangle about AB (1/12)bh3 r=d/2= 10mm + π 504 Moment of inertia of semi circle about AB( ( R=D/2=25mm 128 - Moment of inertia of circle about AB{(πd4)/64} D=b=50mm 1 12 π 504 π 204 - 7850 = mm4 = x 50 x 50 + - = 3 128 64

To find the moment of Inertia about centroids. of. given object Let
To find the moment of Inertia about centroids of given object Let h= 50mm ,D=50mm and R=10mm Hole h B A R D

Centroid of composite section Calculated
h=50mm 5.4327mm A B r= 10mm D= 50mm

h=50 Moment of inertia of object about AB = B A Moment of inertia of Triangle about AB (1/12)bh3 r=d/2= 10mm + π 504 Moment of inertia of semi circle about AB( ( R=D/2=25mm 128 - Moment of inertia of circle about AB{(πd4)/64} D=b=50mm 1 12 π 504 π 204 - 7850 = mm4 = x 50 x 50 + - = 3 128 64

moment of Inertia about centroid is to be calculated
To find the moment of Inertia about centroid of given object Let h= 50mm ,D=50mm and R=10mm moment of Inertia about centroid is to be calculated Hole h 5.4327mm B A Centroid of composite section Calculated R Moment of inertia of object about AB = mm4 D

Moment of Inertia about centroid calculation
Moment of Inertia about an axis= IAB= mm4 IAB = IXX+ Ah2 Moment of Inertia about centroid to be calculated A = Triangle area + semicircle area with radius 25mm – circle area with radius 10mm = + – 314 = mm2 h = distance between axis to centroid = 5.4327mm = 2 – Ah IXX IAB = – = mm4 x

Define following terms 1.Centroid

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