Centroid & Moment of Inertia

UNIT Learning Objectives Centroid & Moment of Inertia After studying this unit, the student will be able to Know what is centre of gravity and centroid Calculate centroid of geometric sections Centre of Gravity Centre of Gravity (or) mass centre of a point in the body where entire mass weight is assumed to be concentrated. In other words, it is a point in the body, through which the resultant of the weights of different parts of the body is assumed to be acting. It is generally written as C.G. Centroid: The plane figure like triangle, rectangle circle etc have only areas and mass is negligible. The centre of area of such plane figures is called Centroid (or) Centre of Area. It is generally denoted by G Centroidal Axis The axis which passes through centre of gravity (or) centroid is known as Centroidal Axis XX 1, YY 1, ZZ 1 are called Centroidal Axis Fig.1

8 Building Construction and Maintenance Technician Axis of Symmetry Axis of Symmetry is the line dividing the figure into two equal parts like mirror images the centroid always lies on the axis of symmetry. Fig. A figure may contain one (or) more axis of symmetry. If there are more axis of symmetry the cntroid lies at the intersection of axis of symmetry Fig. Position of centroids for Standard Geometric Sections. S. No Name Shape of figure Position of centroid At int er sec tions of 1 Rectangle Triangle Diagonals L x B y A L X B At int er sec tions of Diagonals H y 1 A BH

Paper - III Engineering Mechanics 8 Parallelogram 4 Circle At int er sec tions of Diagonals L x B y At int er sec tions of Diagonals D x Radius D y Radius D A 5 Semicircle 6 Trapezium (sloping on both sides) h ab y ( ab ) h ab y 1 ( ab ) h A (a b) 7 Trapezium (One side is vertical and other side is sloping) a ab b x (a b) h A (a b)

84 Building Construction and Maintenance Technician Centroid of Composite sections A composite section is a combination of simple regular shapes as rectangle, triangle circle, semi circle etc. For determining the centroid of composite sections, the entire area is divided into two (or) more regular simple shapes, Then the principle of moments is applied to determine the centroid. Centoid of plane figure having hollow Portion The Centroid of plane figure having hallow portion is determined similar to the composite sections by applying principle of moments, However the negative sign is taken into consideration of hollow positions which are enclosed in a regular shape. Sections Symmetrical about both X and Y axes Fig.4 Sections Symmetrical about horizontal axis (XX) Fig.5 Sections Symmetrical about the vertical axes (YY) Fig.6

Paper - III Engineering Mechanics 85 Sections un symmetric about the both axes (X-X,Y-Y) Methods of determination of centroid The following three methods are available to locate the cntroid of an area. 1. Analytical method. Graphical method. Experimental method Fig.7 Analytical method for location of the centroid Principle: The sum of the moments of a system of a coplanar forces about any point in the plane is equal to the moment of their resultant about the same point. Fig.8 Consider a lamina in area A divided into number of elementary areas A 1, A, A,.etc as shown in fig..8.let the centroids of these elementary areas be at a distance of x 1, x, x.. etc from vertical axis and y 1, y, y from the horizontal axis. Let the centroids of the total area A is at a distance of x and y from vertical and horizontal axis respectively. As per the principle of moments, the sum of moments of all the elementary areas about horizontal axis OX is equal to the moment of the total area about the same horizontal axis i.e OX.

86 Building Construction and Maintenance Technician 1 1 A y y where A=A A 1 +A +A +. Similarly taking moments of areas about vertical axis i.e. OY x 1 1 A x A The terms A1 y 1 & A1 x1 are know as First movement of area about y-axis and x-axis respectively First moment of area: The First moment of area about a line is the product of area and the perpendicular distance of its centroid from the given line. Important Note 1. If the axis passer through the centroid, the moments of areas on one side of the axis will be equal to the moments of areas on the other side of the axis. Example. Locate the centroids if the trapegezium as shown in figure.09 Solution Fig.09 Dived the trapezium into rectangle of size a x h and triangle of base (b-a) and height h Area of rectangle (1) A 1 = a.h Area of rectangle () A = 1 h (b-a)h Total area of trapezium A a b Let the centroid of the trapezium be at a distance y above base and x from last vertical side. Centroidal distance of rectangle from A i. e, a x1

Paper - III Engineering Mechanics 87 b a Centroid distance of triangle from A i. e ; x a a b a a b a 1 a b ah b ah A1x1 Ax x A h 1 A a b h b aa b a a ab b a ab h a b a b a ab b a b Similarly A1y1 Ay h h y y 1, y A A Example. Solution 1 h h h h h ah b a ah b a y h h a b a b ah bh ah ah bh h a b a b a h a b a ab b h a b centroid x, y a b a b Locate the position of centroid of lamina in fig.10 Y.Y Axis as symmetry centroid lies in this axis divide the section in to a square and a triangle A 1 = 100x100 = 10,000mm

88 Building Construction and Maintenance Technician y 1 = 100 50mm 1 A 100 60 000mm 0 60 Y 100 10mmabove base Total area = A 1 +A = 10,000+000 = 1000 mm Let be centroid distance from base y = A 1 y 1 +A y = A 1 +A = Fig.10 10000(50) + 000(10) 1,000 500000 +,60,000 1,000 = = 8,60,000 1,000 66.15 mm The centroid of the lamina is 66.15 mm above the base. Example.4 A trapezoidal lamina has uniform batter on both sides. Its top width is 00 mm. bottom width is 00mm and height is 600 mm. determine position of centroid from base. Solution Top width a = 00 mm Bottom width b = 0 mm Height h = 600 m Fig.11 y = Position of centroid from the base

Paper - III Engineering Mechanics 89 Example.5 Solution y ( ) h = a+b = a+b Find the centroid of the following T section Fig.1 shows T- Sections Y-Y axis as symmetrical axis. In this axis only. Taking base X-X axis as reference line. = = 600 x00 + 00 ( 00 + 00 ) 00 ( ) 80 mm 400 + 00 500 web Fig.1 Dividing the T section in to two rectangles areas. (flange +web) Area of rectangles flange A 1 = 100x10 = 1000 mm y 1 = 140 + Area of rectangle web () A =140x10 = 1400 mm distance of centroid from bottom of web XX i. e., 1 10 y = 140 = 145 mm from the bottom of the base A1y1 Ay 1000 1451400 70 y A A 1000 1400 98000 145000 101.5mm 400 = 70 mm from the bottom of the X.X axis

90 Building Construction and Maintenance Technician Example.6(imp) Solution Find the position of the centroid of an I section given. Top angle : 60 x 0 mm Web : 0 x 100 mm Bottom angle : 100x0 mm Fig.1 shows given I sectrim Y-Y axis as axis of symmetry so centroid lien in this axis only. we can find y X-X axis is base of the bottom taken as reference line. Dividing I section in to three rectangles. Area of rectangle (1) A 1 = 60 x 0 =100 mm y 1 = 0 + 100 + 0 10 mm from the base of the bottom flange Area of rectangle () A = 100 x 0 = 000 mm y = 0 + 100 = 70 mm from the base of the bottom flange Area of rectangle () A = 100 x 0 = 000 mm 0 Y = = 10 mm from the base of the botttom of the flange =

Paper - III Engineering Mechanics 91 Example.7(imp) A masonry dam of the trapezoidal section with one face is vertical. Top width of dam is m, bottom width of dam is 6m and height is 6m. Find the position of centroid. Solution (i) Applying for multa Top width Bottom width a = mt b = 6 mt Height of the dam = 6 mt one face is vertical. Fig.14 and Let centroid of the dam be at a distance turn use vertical face centroid. Y about base x x = a + ab + b (a+b) = + x 6 + 6 (+6) 9+18+6 = 7 =. mt y = h = ( ) a + b a+b 6 x+6 ( +6 ) 1 ( ) = 9 =.67 mt II nd method Trapezium OBCD is divided in to tow simple areas 1. Rectangle OLCD. Triangle CLB

9 Building Construction and Maintenance Technician There is no axis as axis of symmetry line. We can find both x and y Finding x Vertical face OD as resume line. For Rectangle OLCD area. A 1 = x 6 = 18 m X l = = 1.5. from vertical face OD. For triangle CLB Area 1 A 6 9m x 4m A1x1 Ax 181.5 9 4 x.m A A 18 9 1 form vertical face. Finding y Base of the dam OB taken as reference line. Area of rectangle (1) A 1 = 18m y 1 6 mt from the base of the dam Area of triangle () A = 9m Example.8 (imp) A1y1 Ay 18 9 y.67m from the base A1 A 18 9 of the dam Determine the centroid of the channel section 00 x 100 x 10 mm as shown is fig.15 Solution Fig.15 shown the given channel section. X-X axis as axis of symmetry line is this only centroid lies it.

Paper - III Engineering Mechanics 9 Fig.15 Finding only taking vertical outerface AB as reference line dividing the section in to three rectangles Rectangle (1) AreaA 1 = 100 10 = 1000 mm 100 x1 50mm from vertical face AB Area of Rectangle() Area A = 180 10 = 1800 mm Area of rectangle () A =100 10 = 100mm 100 x from vertical 50mm face AB A1x1 Ax Ax x A A A Example.9 Find the position of centroid for an angle of section from base as shown in fig..16 Solution x = 10 1 = 5mm from vertical face AB 100050 1800 5 100050 1000 1800 1000 50000 9000 50000 800 8.68mm From the vertical face AB Fig.16 shows given angle section there is no X-X and Y-Y axis are axis of symmetry.

94 Building Construction and Maintenance Technician We have to find both x and y Finding x Vertical face AC as reference line. 10 0 Dividing the angle section as two rectangular areas. 10 Fi.g.16 Areas of Rectangle 1 A 1 = 10 x 0 = 400 mm 0 x1 10mm from vertical face AC Area of Rectangle A = 100 x 0 = 000 mm 0 x 100 x 0 70mm from vertical from AC A x A x 40010 00070 4000 140000 A A 400 000 4400 1 1 1 y = 7.7mm from vertical face AC. Finding Bottom AB as axis of reference. 10 y1 60mm from bottom base AB 0 y 10mm from the bottom base AB A1y1 Ay 400 60 000 10 y A A 400 000 1 = 7.7 from the bottom base AB. Example.10 Determine the centroids of the selection shown in figure.17

Paper - III Engineering Mechanics 95 Solution Figure.17 shows selection has 0 no X-X and Y-Y axis x as axis y of symmetry. So both and can be determined. x Finding 100 80 00 CD line vertical face taken as axis of reference line. Dividing given Z selection in to three rectangular areas. 0 80 100 Fig..17 0 Area of rectangular A 1 = 80 x 0 = 1600mm 80 x1 40mm (from vertical face CD) Area of rectangle () A = 0x0 = 4400 mm 0 x from vertical face CD 80 90 mm Area of rectangle () A = 80 x 4 = 190 mm 80 x from the vertical face CD 100 140 mm A1x1 Ax Ax 1600 40 440090 190140 x A A A 1600 4400 190 1 64000 96000 68800 790 = 9.0 mm from the vertical face CD Finding y Bottom base AB as axis of reference 0 y1 00 10 mm from bottom base AB 00 y 110mm from bottom base AB y 1mm from bottom base of AB

96 Building Construction and Maintenance Technician A1y1 A y Ay 1600 10 4400 110 190 1 y A A A 1600 4400 190 1 6000 484000 040 790 84040 790 = 106.44 mm from bottom base AB Example.11 Find the cenrodidal distance for the built up section shown in figure.18 Solution Figure.18 shown Y-Y axis of y symmetry centroid lies in it we can find y Finding Bottom most layer AB line a as axis of reference. Built up section has divided in 5 rectangular areas. Fig.18 Rectangular (1) A1= 100 x 10 = 1000mm 10 y1 10 0 150 0 05mm Rectangular () A = 100 x 0 = 000mm 0 y 10 0 150 190 mm Rectangular () A = 150 x 0 = 000 mm 150 y 10 0 105mm Rectangular (4) A 4 = 0 x 00 = 4000mm 0 y4 10 0 mm Rectangle (5) A 5 = 10 x 00 = 000mm 10 y5 5mm

Paper - III Engineering Mechanics 97 = 8.5mm from bottom base AB Review Questions Short Answer Type Questions 1. Define centre of gravity. Determine the center of gravity and centroid. Locate the position of centroid of the following figures with a neat sketch 1) rectangle ) triangle ) circle 4) Semi circle 4. Find the centroid of triangle of base 80 mm and height 10 mm from the base and the apex Essay Answer Type Questions 1. A masonry dam is trapezoidal in section with one face vertical. Top width is m and bottom width is 10 m height is 10 m. Find the position of centroid axis Ans. x.564m and y = 4.60 m. Determine the centre of gravity of I section having the following dimensions Bottom flange = 00x100mm Top flange Web = 150x50mm = 50x400mm Ans. 198.9mm from bottom flange

98 Building Construction and Maintenance Technician. Find out the centroid of an un equal angle section 100mm x 80mm x 0mm Ans x = 5 mm from left face y = 5mm from bottom face 4. Find the centre of gravity of channel section 100 x 50 x 15 mm 5. Find the centroid of the given T section Top flange of Web 50mmx50mm 50mmx00mm Ans: Ans x= 17.8 mm from outer face of web y = 169.44mm from bottom of the web. 6. Find the centroid of the section shown in figure Ans: x = 95.56mm from the left edge y = 85.55mm from the bottom edge Fig.19

Paper - III Engineering Mechanics 99 Moment of Inertia (M.I) Definition The product area (A) and perpendicular distance (x) between the point is know as the first moment of area (Ax) about the point. If this moment is again multipled by the distance x i.e. Ax.x = Ax is called moment of moment of area (or) the second moment of area or simply moment of inertia. Its unit in SI system is mm 4. Moment of inertia for some regular geometrical sections Position of centroids for Standard Geometric Sections. S. No Name Shape of figure MI about MI about XX(Ixx) yy(iyy) 1 Rectangle BD DB 1 1 Hollow Rectangle BD 1 - bd 1 DB 1 - db 1 Solid Circular section D D 64 64 4 4 4 Hollow Circular section 64 64 4 4 4 4 (D d ) (D d ) 5 Triangle b h b h 6 1 about cg about base BC

00 Building Construction and Maintenance Technician Parallel Axis Theorem It states that if the moment of inertia of a plane area about an axis through its centre of gravity (I GG ), is shown in fig.0. Then the moment of inertia of the area about an axis AB parallel to I GG at a distance of h from centre of gravity is given by I AB = I GG + Ah Where Radius of Gyration Fig.0 I AB = M.I of the area about an axis AB I GG = M. I of the area about its C.G A = Area of the section h = distance between C.G of the section and the axis AB. Radius of gyration about a given axis is defined as the effective distance from the given axis at which the whole are may be considered to be located with respect to axis of rotation. It is denoted by k or r Where Perpendicular axis theorem I = Ak (or) Ar I = moment of inertia K(or) r = radius of gyration k (or) r = A= area of cross section Units for k or r in S.I system is mm It states that if I XX and I YY be the moment of inertia of plane section about two perpendicular axes meeting at o shown in figure.1 then, the moment of inertia I ZZ about the axis Z Z which is perpendicular to both XX and YY axises, is given by I ZZ = I XX + I YY I A

Paper - III Engineering Mechanics 01 For symmetrical section like circular I XX = I YY Fig.1 I ZZ = I XX + I xx I ZZ = I XX J = I where J is known as polar moment of Inertia Polar moment of inertia. Definition :- The moment of inertia of an area( I ZZ ) about an axis perpendicular to its plane is called polar moment of Inertia. It is denoted by J BD 400800 10 4 IXX 1.70610 mm 1 1 Solved Problems Problem.1 Find the moment of inertia of a rectangular section 400mm wide and 800mm deep about its base. Solution Breadth of bearn B = 400mm Depth M.I. about C.G i.e. of beam D = 800 mm M. I about its base I I Ah AB I AB = 1.706 x 10 10 + (400x 800) (400) = 1.706 x 10 10 + 5.1 x 10 10 = 6.86 x 10 10 mm 4 Fig.

0 Building Construction and Maintenance Technician Problem.1 Find the M.I of hollow circular sections whose external diameter is 60mm and internal diameter is 50mm about Centroidal axis Solutions External dia D = 60 mm Internal dia d = 50mm IXX IYY D d 64 4 4 60 50 64 4 9. mm 4 4 Fig. Moment of inertia about Centroidal axis is = 9. mm 4 Problem.14 Find the moment of inertia of a rectangle 60mm wide and 10mm deep about Centroidal axis. Find also least radius of gyration. Solutions B = 60mm D = 10mm M. I about Centroidal axis Fig.4 Area of rectangle A = BD = 60 x 10 = 700mm Least radius of gyrations k (or) r = IcG A = 8.64 x 10 6 700

Paper - III Engineering Mechanics 0 Least radius of gyration = 4.64mm Problem.15 Find the radius of gyration of hollow circular sectors of external diameter 00mm and internal dia 00mm. Solution External dia D = 00mm d = 00mm 4 Area D d = 4 00 00.97 10 mm 4 Fig.5 8 I.19110 Radius of gyration K 90.14mm 4 A.97 10 4 4 D d Alternate method 64 D d 00 00 K 4 4 D d 64 = 90.14mm Problem.16 Find the radius of gyration of a triangle whose base is 40mm and height is 60mm about an axis passing through C.G and parallel to base. Base H = 60mm b = 40mm Area = 1 bh M.I of triangle about Centroidal axis Fig.6

04 Building Construction and Maintenance Technician I XX Radius of gyration bh h K 6 bh 18 60 K 14.14mm 18 Problem.17 Find the moment of inertia about Centroidal axis of hollow rectangular sections shown in fig.7 Solution bh 6 B = 00mm D = 400mm b = 100mm d = 00mm K = Ixx A Fig.7 M.I about XX axis for hollow rectangular sections. = 1000x10 6 mm 4 M. I about Y Y Axis for a hollow rectangular section Problem.18 = 50 x 10 6 mm 4 Determine the position of centroid and calculate the moment of inertia about its horizontal centroidal axis of a T beam shown in figure.8 Solution YY Ixx = 1 1 [ 00 x 400-100 x 00 ] DB db 1 1 1 I 400 00 00 100

Paper - III Engineering Mechanics 05 Finding Centroid YY axis is axis of symmetry centroid lies on it. To Finding y Take AB line axis of reference. Dividing T section into two rectangular areas Fig.8 Area of rectangle (1) A 1 = 00 x 100 = 0000mm 100 y1 00 50mm from bottom base AB Area of rectangle () A = 00 x 100 = 000mm 00 y from bottom base AB 100mm A y Centroidal distance y from bottom A 0000 50 0000100 0000 0000 = 190mm from bottom base AB. A y A 1 1 1 M. I of a rectangle 1 about centroidal axis I XX at (1) = I G + Ah 1 00100 00100y y 1 =.5 x 10 7 x108 x 10 6 = 5 x 10 6 +108 x 10 6 = 1 x 10 6 mm 4 7.5 10 00 100 50 190 h1 y1 y.510 00100 60 7

06 Building Construction and Maintenance Technician M. I of a rectangle of about Centroidal axis I @ I Ah G 100 00 100 0090 1 7 6 6.67 10 1610 66.7 10 1610 6 6 8.7 10 mm 6 4 since h = y- y h = 190-100 = 90mm Moment inertia of T beam about its Centroidal axis I at (1)d I at() Problem.19 An un symmetrical I section has top flange 100x0mm web 100 x 10mm and bottom flange 80x0 mm over all depth is 160mm. Solution 110 8.7 10 Calculate centroid Figure.0 Shows given I section. YY-axis is axis of symmetric line so centroid lies on it. 6 6 61.7010 mm 6 4 Finding y Take line AB, passing through the bottom edge as axis of reference Divide the section into three rectangular areas. Area of rectangle (1) A 1 = 100 x 0 = 000 mm 0 y1 0 10 150mm from base

Paper - III Engineering Mechanics 07 Area of rectangle () A = 10 60 x 10 = 100mm 10 y 0 80mm from base Area of rectangle () A = 80 x 0 = 1600mm I xx 0 y 10mm from the base A1y1 Ay Ay 000 150 100 80 1600 10 y A1 A A 000 100 1600 00000 96000 16000 85.8mm 4800 from the base AB. Finding M.I of I section about X-X axis about centroid M.I of rectangular (1) about X-X axis I @I I A h h y y xx G 1 1 1 1 1 M. I of rectangle () about a X-axis I at I A h M. I of rectangle () about X-X axis 100 0 100064.17 1 6 4 8.10 mm xx G 1010 10105.8 1 6 4 1.4810 mm I at I A h xx G 80 0 80075.8 1 6 4 9.510 mm h1 = 150-85.8 = 64.17 h = y - y = 85.8-50 = 5.8mm h = - y = 85.8-10 = 75.8 mm

08 Building Construction and Maintenance Technician Moment of inertia of given I section about X axis Problem.0 Determine the moment of inertia of the un equal angle section of size 150mm x 100mm x 5mm about Centroidal axis. Solution Finding centroid Finding x Vertical face CD has axis of reference, dividing L section has two rectangular areas. Area of rectangle 1 A 1 = 15 x 5 = 15 mm 5 x1 1.5mm from vertical face CD Area of rectangle A = y100 x 5 = 500mm Finding y I at1 I at I at xx xx xx 8.10 1.4810 9.510 19.010 mm 6 6 6 6 4 100 x 50mm from vertical face CD A1x1 Ax 151.5 50050 x A A 15 500 1 Bottom base AB has taken as axis of reference 15 y1 5 87.5mm from base 5 y 1.5 from base 906.5 15000 565 16406.5 565 = 9.17mm from vertical face CD

Paper - III Engineering Mechanics 09 y = A1y1 + Ay A1 + A = 15 x 87.5 + 500 x 1.5 1.5 + 500 y Finding I xx = 747.5 + 150 565 = 04687.5 565 = 54.17mm from the base AB. M. I of rectangle (1) about x-x axis 515 Ixxat1IG A 1 1h1 515. 1 6 6 4.0710.47 10 = 7.54 10 6 mm 4 M I of rectangle () about x-x axis h 1 = y - y 1 =54.17-7.5 =.mm 100 5 I xx @.IG A h 100 5 41.67 1 h = y - y =54.17-1.5 = 41.67mm = 0.1 x 10 6 + 4.4 x 10 6 = 4.47 x 10 6 mm 4 Moment Inertia of given angular section about X-X axis Finding I YY I at I I at xx M I of rectangle (1) about Y-Y axis I YY at 1 = I G1 + A 1 h 1 xx 7.5410 4.47 10 6 6 1.0110 mm 6 4 h 1 = x - x 1 = 9.17-1.5

10 Building Construction and Maintenance Technician DB A1 x x1 1 150 5 159.17 1.5 1 6 6 0.19510 0.86810 1.0610 mm M.I Rectangular () about Y-Y axis I at I A h YY G 5100 5000.8 1 6 6.0810 1.0810.1610 mm 6 4 6 4 M. I of a given angular section about Y-Y axis h = x - x = 50-9.17 = 0.8 I at I I at YY YY 1.0610.1610 6 6 4.10 mm 6 4 Review Questions Short Answer Type Questions 1. Explain a) Parallel axis Theorem b) Perpendicular axis theorem. Define the termsa) Moment of inertia b) Radius of gylation. Find the radius of gyration of circle having diameter d Ans: d 4 4. Find the radius of gyration of hollow circular plate of 60mm inner diameter and 100mm outer diameter (Ans:9.15mm)

Paper - III Engineering Mechanics 11 5. Find M.I of a rectangular section 00mm width and 400mm depth about the base Essay Answer Type Questions (Ans. 4.67 x 10 9 mm 4 ) 1. Find the moment of Inertia of a T Section having flange150mm x 50mm and web 50 x 150mm about xx and yy-axis through the C. G of the section. [ Ans: I xx = 5.15 x 10 6 mm 4 I YY =15.65 x 10 6 mm 4 ]. Determine the moment of Inertia of an unequal angle section of size 100mm x 80mm x 0mm about Centroidal axis [ Ans: I xx =.907 x 10 6 mm 4 I YY =1.67 x 10 6 mm 4 ]. Determine the moment of inertia of an I section about XX axis given that top flange 100mm x 10mm web = 00mm x 10mm different flange 160mm x 10mm [Ans: I xx = 4.8 x 10 6 mm 4 ] 4. A built up section is formed by an I section and to flange plates of size 80 x 0mm are an each flange find the moment of inertia about centrodial X-X axis as shown in below figure Key Concepts [Ans: I xx = 188. x 10 6 mm 4 ] 1. The C.G of a body is the fixed point at which its weight is assumed to be concentrated.. The centroid of a surface is the fixed point at which the area of the surface is assumed to be concentrated.

1 Building Construction and Maintenance Technician. The centroid of a surface is determined from the equations: A x x A 4. The centroid of a composite area is treated by the principle of moments, dividing it into regular simple figures. 5. The M.I of an area about a given axis is the sum of the values of ax where a is the area of each element and x is the distance of the centroid of the element from the given axis I ax 6. Radius of gylation (K xx ) of an area about given axis is the distance from the axis at which the area may be assumed concentrated to given the M. I of the area about the given axis I K 7. Parallel axis theorem :- if XX is an axis is parallel to the centrodal axis C.G of surface of area A and if d is the distance between the two parallel axis. I I Ad 1 1 1 1 CG A y and y A 8. Perpendicular axis theorem: If XX and YY are two perpendicular axis is the plane of the area and ZZ is an axis perpendicular to both of them through their intersection. A I I I zz xx YY 9. The M.I about an axis perpendicular to its plane is known as its polar M.I 10. M.I of a built up section = Sum of M.I of all elements of the section about the same axis. 11.M.I of a rectangle bxd about axis through centroid parallel to bd side b 1 1. M. I of a triangle bxh about axis through centroid parallel to base bh 6 4 d 1. M. I of circle of dia d about any diameter 64

Paper - III Engineering Mechanics 1 14. M.I of hollow circular section of diameters D and d about any dia D 4 d 4 64 4 15. Polar M.I of a solid shaft of dia d about axis d 16. Polar M.I of hollow shaft of dia of diameter D and d = D 4 d 4