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centroid
The document defines several terms: 1. Centroid - The point at which the total area of a plane is concentrated. It is where the average position of the total weight of the plane would balance. 2. Radius of gyration - The distance from the axis of rotation to the centroid. It is calculated as the square root of the moment of inertia divided by the total area. 3. Area moment of inertia - The product of the plane area and the square of the perpendicular distance to the axis of reference. It is a measure of the resistance offered by a plane figure to bending or twisting about an axis.
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centroid
- 1. Define following terms 1.Centroid 2.Radius of gyration 3.Area moment of Inertia JNTU – Dec2006 1.Centroid centroid 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 C.G
- 2. x axis Y y axis X Letconsider 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 Area moment of Inertia dA = Y2dA with respect to y axis
- 3. Area moment ofInertia x axis Y y axis centroid C.G X Ixx = Moment of Inertia about x axis Iyy = Moment of Inertia about y axis 𝑿 𝟐 dAIxx = 𝒀 𝟐 dAIyy = 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 Let consider dA is small area on the plane
- 4. Radius of gyration K A=total area of lamina concentrated at this point A Radius of gyration = axis I = Ak2 I = moment of inertia K = I A
- 5. Parallel axis Theorem XX h X axis A B Plane centroid Moment of Inertia about AB axis IAB = IXX + Ah2 Area of plane Moment Inertia about centroid 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
- 6. O b b/2 h/2 h Centroid of rectangle isknown to us by looking at component O Regular rectangle
- 7. Radius = r Diameter= D = 2r O O = centroid of circle is known to us by looking at the component Regular circle
- 8. A hole ona rectangle whose diameter = 10mm 80mm 160mm 120mm 100mm Non Regular component of hole on a rectangle Centroid of the component to be calculated
- 9. 80mm 160mm 100mm 160mm X2=80mm Y2=120mm 100mm X1=50mm Y1=80mm 10mm 0,0 0,0 Y axis Xaxis Y axis X axis Non regular component regular component of rectangle regular component of circle A1= 100X160=16000 mm2 A2= π r2 = π102 = 314 mm2
- 10. Y X X= A1X1 -A2X2 A1 - A2 Y= A1Y1 - A2Y2 A1 - A2 = = A1= 100X160=16000 mm2 A2= π r2 = π102 = 314 mm2 X1=50mm Y1=80mm X2=80mm Y2=120mm Area X cordinate Y cordinate 16000X50 - 314X80 16000 - 314 16000X80 - 314X120 16000 - 314 800000 - 25120 = = 15686 1280000 - 37680 15686 = 49.3mm =79.2mm X = 49.3mm Y = 79.2mm
- 11. 50 mm 80mm 49.3mm 79.2 mm Centroid ofrectangle before making hole Centroid of rectangle after making hole Pl note that shift of centroid
- 12. b h/3 b/3 h For triangle notethe centroid For triangle note centroid distance
- 13. centroid Radius = rRadius= r 4r/3π Radius = r For semicircle note centroid distance
- 14. centroid Radius = rRadius= r 4r/3π Radius = r 4r/3π 4r/3π For quarter circle note the centroid
- 15. R= 2m 6m 3m 4m Determinationof the centroid of the area of composite section is explained in different method developed by me
- 16. R= 2m 6m 3m 4m Drawthe axis as shown
- 17. R= 2m Consider semicircle by splitting the composite section and mark the centroid with respect to axis drawn considering + and – (positive and negative with respect to axis) - 4r/3π = (4x2)/3π = - 0.849 Area = A1= π22 /2= 6.28 m2 X1= - 0.849 Y1= 2.0 Y1= 2.0 - Ve sign is with respect to axis Centroid of semi circle
- 18. 6m 4m X2=3m Y2=2m Area = A2=6X4 24 m2 X2=3 Y2=2 Centroid of rectangle
- 19. 6m 3m 4m 3/3=1 Y3=4/3=1.333 A3=(1/2)x 3x 4=6m2 X3= 7 Y3= 1.333 X3 Centroid of triangle
- 20. Area X CoordinateY Coordinate A1=6.28 X1= - 0.849 Y1= 2.0 A2=24 X2=3 Y2=2 A3=6 X3= 7 Y3= 1.333 X = A1X1 + A2X2 + A3X3 A1+ A2 + A3 = (6.28 X ( - 0.849)) +( 24X3) +(6X7) (6.28+ 24 + 6) = 2.995 mm Y = A1Y1 + A2Y2 + A3Y3 A1+ A2 + A3 = (6.28 X ( 2.0) +( 24X2) +(6X1.333) (6.28+ 24 + 6) = 1.890mm
- 21. R= 2m 6m 3m 4m X =2.995 mm Y = 1.890 mm Centroid of composite bar
- 22. Moment of Inertiaof standard and regular shape component h/2 h b/2 b xx y y X axis y axis Centroid I xx = bh3 12 I yy = hb3 12 Where I xx = moment of Inertia about xx axis Where I yy = moment of Inertia about yy axis
- 23. x X axis x Y Y Y1 BA Moment ofInertia of rectangle about the axis AB parallel to axis xx is equal to IAB= Ixx + A(Y1) 2 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 IAB = bh3 12 +{( b X h) (Y1)2 b h }
- 24. XX Y Y X axis Y axis d= Diameter of circle Circle Ixx = πd4 64 centroid Circle Iyy = πd4 64 Plolar moment of Inertia Izz = Ixx+Iyy = πd4 64 + πd4 64 = πd4 32
- 25. centroid Radius = rRadius= r 4r/3π Radius = r For semicircle note centroid distance Moment of Inertia of semicircle 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 x x BA moment of Inertia for semi circle it is ( πd4/64)X ½ = πd4/128 Now IAB= Ixx +(AY1)2 Y1 = Ixx =0.0068598d4 Ixx =0.0068598d4 Where IAB= πd4/128 A=πr2 or πd4/4 Ixx = IAB - (AY1)2 Y1 4r/3πY1 =
- 26. Let us knowthe moment of inertia of Triangle A B moment of inertia of Triangle about this axis or base of triangle is= b Y1=h/3 Ixx= moment of Inertia Of triangle about Centroid of triangle is moment of inertia of Triangle about centroid= Ixx =IAB-A(y1)2 = h A = area of triangle IAB = moment of inertia triangle about centroid bh3 12 bh3 36 bh3 36 Applying parallel axis theorem
- 27. A hole ona rectangle whose diameter = 10mm 80mm 160mm 120mm 100mm Non Regular component of hole on a rectangle Centroid of the component to be calculated
- 28. 80mm 160mm 100mm 160mm X2=80mm Y2=120mm 100mm X1=50mm Y1=80mm 10mm 0,0 0,0 Y axis Xaxis Y axis X axis Non regular component regular component of rectangle regular component of circle A1= 100X160=16000 mm2 A2= π r2 = π102 = 314 mm2
- 29. Y X X= A1X1 -A2X2 A1 - A2 Y= A1Y1 - A2Y2 A1 - A2 = = A1= 100X160=16000 mm2 A2= π r2 = π102 = 314 mm2 X1=50mm Y1=80mm X2=80mm Y2=120mm Area X cordinate Y cordinate 16000X50 - 314X80 16000 - 314 16000X80 - 314X120 16000 - 314 800000 - 25120 = = 15686 1280000 - 37680 15686 = 49.3mm =79.2mm X = 49.3mm Y = 79.2mm
- 30. 50 mm 80mm 49.3mm 79.2 mm Centroid ofrectangle before making hole Centroid of rectangle after making hole Pl note that shift of centroid
- 31. A hole ona rectangle whose diameter = 10mm 80mm 160mm 120mm 100mm Non Regular component of hole on a rectangle Centroid of the component to be calculated
- 32. 80mm 160mm 100mm 160mm X2=80mm Y2=120mm 100mm X1=50mm Y1=80mm 10mm 0,0 0,0 Y axis Xaxis Y axis X axis Non regular component regular component of rectangle regular component of circle A1= 100X160=16000 mm2 A2= π r2 = π102 = 314 mm2
- 33. Y X X= A1X1 -A2X2 A1 - A2 Y= A1Y1 - A2Y2 A1 - A2 = = A1= 100X160=16000 mm2 A2= π r2 = π102 = 314 mm2 X1=50mm Y1=80mm X2=80mm Y2=120mm Area X cordinate Y cordinate 16000X50 - 314X80 16000 - 314 16000X80 - 314X120 16000 - 314 800000 - 25120 = = 15686 1280000 - 37680 15686 = 49.3mm =79.2mm X = 49.3mm Y = 79.2mm
- 34. 50 mm 80mm 49.3mm 79.2 mm Centroid ofrectangle before making hole Centroid of rectangle after making hole Pl note that shift of centroid
- 35. 50 mm 80mm 49.3mm 79.2mm Centroid ofrectangle before making hole Centroid of rectangle after making hole Pl note that shift of centroid and we need to calculate moment Inertia of Composite section about the shifted axis let us do it Ixx = bh3 12 Ixx of circle = πd4 64 d=10 mm 0.8 mm A B IA B = to be calculated by applying parallel axis theorem IA B of rectangle Ixx of rectangle about its centroid + area of rectangle X square axis shift in y axis 160mm 100 mm = IA B = bh3 12 + bxh (y1)2 = 100x803 12 + 100x80x (0.8)2 = 4266666 +5120=4271787mm4 40.8mm 80mm 80mm 40mm
- 36. IA B ofCircle = πd4 64 + π r2 (y1)2 = πx104 64 + πx52x (40.8)2 =490.625+130674.2=131164.8 mm4 Moment of Inertia of composite section = IA B of rectangle - IA B of Circle 4271787 -= 131164.8 =4140622mm4
- 37. To find thecentroids of given object Let h= 50mm ,D=50mm and R=10mm h D R A B Hole
- 38. A1 = (1/2)50*50= 1250mm2 Y1=50/3=16.667mm A1Y1=20833.75 mm3 A1 Y1= = h = 50mm A1 area of triangle centroid Y1 = centroid distance from base h 3 50 3 X axis Y axis
- 39. r=10mm A2 Y2=0 since centroidof Circle coincides and with axis i.e. it lies in x axis A2 = πr2 = π*102= 314mm2 A2Y2=0 centroid A2 area of circle Y axis X axis A
- 40. X axis Y axis Y3= - (4r/3π) =-(4*25)/3π= - 10.617mm Pl note –ve sign is due to the centroid of Semi circle fall below the x axis centroid Radius of circle r=25mm A3y3= - 10417.9 A3 = (1/2)πr2 = 0.5*π*252=981.25mm2 A3 = area of semi circle A
- 41. A1 y1-A2Y2 +A3Y3 Y= A1 -A2 +A3 = 20833.75 - 0 – 10417.9 1250 - 314 +981.25 = 5.4327mm area Y= centroid distance Area X centroid distance = AXY A1 Y1 A1Y1 A2 Y2 = 0 A2Y2 = 0 A3 Y3 A3Y3 =1250mm2 =16.667mm =20833.75 mm3 =314mm2 =981.25mm2 =10.617mm = - 10417.9 mm3
- 42. h=50mm D= 50mm r= 10mm AB 5.4327mm Centroid of composite section Calculated
- 43. Finding moment ofinertia of object about the axis AB h=50 D=b=50mm R=D/2=25mm r=d/2= 10mm A B
- 44. Moment of inertiaof Triangle about AB=(1/12)bh3 = h=50 mm B 50 mm A 1 12 50 x 503 =520833 mm4 Axis about which moment of inertia To be taken
- 45. r=d/2= 10mm A B Momentof inertia of circle about AB=(πd4)/64 = π 204 64 Centroid =7850mm4 Axis about which moment of inertia To be taken
- 46. Moment of inertiaof semi circle about AB=(1/128)*(πD4) = π 504 128 D=b=50mm A B Centroid =153320mm4 Axis about which moment of inertia To be taken
- 47. Moment of inertiaof object about AB = Moment of inertia of Triangle about AB + Moment of inertia of semi circle about AB( - Moment of inertia of circle about AB{(πd4)/64} 1 12 50 x 503= + π 504 128 - π 204 64 =x 520833 + 153320 - 7850 = 666303mm4 h=50 D=b=50mm R=D/2=25mm r=d/2= 10mm A B π 504 128 (1/12)bh3 (
- 48. To find themoment of Inertia about centroids of given object Let h= 50mm ,D=50mm and R=10mm h D R A B Hole
- 49. h=50mm D= 50mm r= 10mm AB 5.4327mm Centroid of composite section Calculated
- 50. Moment of inertiaof object about AB = Moment of inertia of Triangle about AB + Moment of inertia of semi circle about AB( - Moment of inertia of circle about AB{(πd4)/64} 1 12 50 x 503= + π 504 128 - π 204 64 =x 520833 + 153320 - 7850 = 666303mm4 h=50 D=b=50mm R=D/2=25mm r=d/2= 10mm A B π 504 128 (1/12)bh3 (
- 51. To find themoment of Inertia about centroid of given object Let h= 50mm ,D=50mm and R=10mm h D R A B Hole 5.4327mm Centroid of composite section Calculated Moment of inertia of object about AB = 666303mm4 moment of Inertia about centroid is to be calculated
- 52. Moment of Inertiaabout centroid calculation IAB = IXX+ Ah2 Moment of Inertia about an axis= IAB= 666303mm4 Moment of Inertia about centroid to be calculated A = Triangle area + semicircle area with radius 25mm – circle area with radius 10mm h = distance between axis to centroid = 5.4327mm 1250= + 981.25 – 314 =1917.25mm2 IXX = IAB Ah2 = 1917.25 x666303 – – 5.43272 = 610395.99mm4