Mechanics_Ramkumar_Chapter_01_Center of Gravity

Definition: Centre of Gravity of a body is a point at which entire weight (or mass) of the body may beassumed to be concentrated. A body is having one C.G. for all positions of the body. It is represented by G.C.G. of a body may lie outside of the body. Let us consider a body of total mass(m). Let it is composed of 3 particles ofmass m1, m2 and m3 as shown in figure. Let mass m1 is acting at a distance x1and y1 from axes oy and ox respectively.Similarly mass m2 and m3 are acting atdistances (x2, y2) and (x3, y3) from oy andox axes respectively. Let mass of body (m) acts at adistance x and y from axes oy and oxrespectively.Prof. Ramkumar'sThen, x = m1x1 + m2x2 + m3x3and y m1 + m2 + m3 = m1y1 + m2y2 + m3y3 m1 + m2 + m3Centroid of Uniform Lamina: Let us consider a plane lamina of mass ‘m’, uniform thickness ‘t’ anddensity ‘γ‘. If A is the surface area of lamina which is divided into three parts whose areas are a1, a2 and a3 andcentre of gravity of these parts lie on straight lines at distances x1, x2 and x3 from the reference axis oy thendistance x of C.G.. of the lamina from reference axis oy is given byx = m1x1 + m2x2 + m3x3 mass = volume ×density m1 + m2 + m3 m1 = (a1 t) × γ = (a1tγ)x1 + (a2tγ)x2 + (a3tγ)x3 (a1 + a2 + a3 )tγ similarly, m2 = a2 t γ tγ(a1x1 + a2x2 + a3x3 ) m3 = a3 t γ (a1 + a2 + a3 )tγ = = a1x1 + a2x2 + a3x3 = ax a1 + a2 + a3 aor x = ax aSimilarly, it can be proved that y = a1y1 + a2y2 + a3y3 + a4y4 a1 + a2 + a3 + a4or y = ay a This shows that C.G. of a body of uniform thickness and made up of same material can bedetermined without knowing its thickness and density. For such bodies, we only need the details about itssurface area. C.G. of such bodies is referred as centroid i.e both are identical. 1

Definition of Centroid: The point at which the entire area of a plane figure (rectangle, triangle, circle etc.)is assumed to be concentrated is known as centroid of that area.Note: There is a clear difference in the meaning of terms C.G. and centroid but in practice, both areconsidered one and same thing.Prof. Ramkumar'sCentroid of standard Figures: 2) Square 3) Isosceles Triangle1) RectangleArea = b × h Area = b × b Area = ½ b × h4) Rightangle Triangle 5) Any Triangle 6) CircleArea = ½ b × h Area = ½ b × h7) Semi-circle 8) Quarter-circleArea = πr 2 Area = πr 2 2 4Important Points Related to Centroid1) Any axis that passes through centroid is called as centroidal axis.2) Horizontal and vertical centroidal axis are commonly denoted as x – x and y – y axes.3) Centroidal distance x is the distance between y – reference axis and y – y centroidal axis.4) Centroidal distance y is the distance between x – reference axis and x – x centroidal axis.5) Axis of symmetry is a line that divides the composite area into two equal parts so that one part is mirror image of other. Axis of symmetry is always a centroidal axis.6) If axis of symmetry coincides with x – reference axis, centroidal distance. y = 07) If axis of symmetry coincides with y – reference axis, centroidal distance x = 0 2

Centroid of Triangular Lamina: Consider the triangular lamina with base ‘b’ and height ‘h’. Consider an elemental strip of thickness ‘dy’ at aProf. Ramkumar's distance of ‘y’ from base. Comparing similar triangles b1 = h − y bh b1 = (h − y) b h y= da .y (= a.y) da (= a) where , da = area of elemental strip and y = centroidal distance of strip from base Numerator h = b1dy.y 0 = b h h (h−y) y.dy o = b h − y2 )dy = b hy2 − y3 h h3 − h3 = b . h3 = bh2 h6 6 h (hy h2 3 =b h2 3 00 Denominator = h = b h (h − y)dy = b hy − y2 h h2 − h2 = 1 b.h h 0 h 2 2 2 b1dy =b h 0 0 ∴ y = Numerator = bh2 6 = h Denominator bh 2 3It shows that centroid of triangle of any shape is always at h from base. 3Centroid of Area of Circular Sector: Consider the sector of a circle which subtends an angle 2 α at centre. Due to symmetry about x axis, y = 0. In order to determine x , consider an elemental area which subtends very small angle d at centre as shown. da = ½ r2 d (area of sector) Now treat the elemental area as triangle ∴ x = 2 r cos θ 3 and y = 2 r sinθ 3 da.x +α 1 r 2dθ× 2 r cos θ da.y da −α 2 3 y = from basic principlesx= = 1 r2dθ +α da −α 2 +α 1 r 2dθ× 2 r sinθ 2 3 r3 +α = −α 1 r2dθ 3 cos θ.dθ +α[ [ ] ]x +α= −α = 2r sin θ −α −α 2 3 r2 +α θ +α −α 1 dθ 3 [− ]θ 2 −α r 3 cos α −α [sin α sin(−α)] = = 2 − 1 r2 +α[dθ] 3 r α − (−α) 2 −α 3

= 2 r 2 sin α − 2 r[cos(α) − cos(−α)] 3 2α =3 [θ]α−α 2 r sin α 2 r[cos α − cos α] 3 α 3 α − (−α)Prof. Ramkumar's∴x = . − 2 r×0 3 2α = = = zeroNote for Application: In circular sector, centroidal distance 2 . r sin α is measured along Axis of 3 αSymmetry (A.O.S.).Centroid of Quarter Circle :α = π = 45° 4OG = 2 r sin α = 2 r sin 45 = 8r . 1 3 α 3 π/4 3π 2but x = y = OG cos 45 = 8r . 1 1 = 4r 3π 2 2 3π ∴ x = y = 4r 3πCentroid of Semi-circle: α= π 2 x = 2r sin π 3 2 π , y=0 2 x = 4r 3π 4

Locate the centroid of plane area shown in !figure. Determine the centroid of the area remaining after a circle of diameter ‘r’ is removed from a circle of radius ‘r’ as shown in figure.Prof. Ramkumar'sx = 32.95 mm and y = 89.77 mm y = 0 and x = r 6 to the left of oy axis Determine the centroid of the plane area Determine the co-ordinates of the centroid ofshown in sketch. the shaded portion shown in figure. (June’03)x = 5.49 cm and y = 1.85 cm x = 19.16 cm and y = 19.08 cm Find the centroid of the area shown in figure. . Locate the centroid of shaded area shown in (Dec’09-5 marks) figure.x = 73.29mm, y = 80.19mm x = 7.74 cm and y = 2.485 cm 5

\" Locate the centroid of the shaded area shown # Determine the depth of web of Tee section,in figure. such that centroid coincides with AB axis. (May’05-5 marks)Prof. Ramkumar'sx = y = +0.073 a h = 55.9 mm$ Locate the centroid of the shaded area shown % Find the co-ordinates of centre of gravity forin figure. the following lamina. Please note that the shaded area is the opening in the lamina.x = 15.09 cm, y = 19.04 cm x = 4.243 m and y = 0.908 m Calculate, numerically the co-ordinates xc and yc of the centroid C of theshaded area shown in figure. Each squareof the grid measures 10 mm × 10 mm. (May’2000 6 marks) 6

&'()*&+ a x Y a.x a.yComponent (mm2) (mm) (mm) (mm3) (mm3) 80 × 60 192000 144000Prof. Ramkumar's= 4800 40 30 60 10 – 48000 – 8000 − 40 × 20 70 50 = − 800 75 25 – 28000 – 20000 40 40 − 20 × 20 7500 2500 = − 400 10 30 – 12566.4 – 12566.4 −10 × 10 = −100 8000 24000 − π × (10)2 = − 314.16 − 40 × 20 = − 800 −10 × 10 5 55 – 500 – 5500 = −100 Σ 87433.6 71433.6 Σ 2285.84xc = a.x = 87433.6 = 38.25 mm, a 2285.84yc = a.y = 71433.6 = 31.25 mm a 2285.84 Determine distance ‘h’ for which the centroid of the shaded area is as high above line AA’ as possible.Show that if the distance y of C.G. is maximum, y = h.&'()*&+ Component a (mm2) y (mm) a.y (mm3) Bigger 200 × 1000 500 100 × 106 Rectangle = 200 × 103 Smaller −120 × h 0.5 h 60 h2 rectangle Σ 200 ×103 − 120 h Σ 100 ×106 − 60h2 y = 100 ×106 − 60 h2 ………(1) 200 ×103 − 120 h 7

For y to be maximum dy = 0 dh( ) ( )200 ×103 − 120h (−120h) − 100 ×106 − 60h2 (−120 ) =0Prof. Ramkumar's ( )200 ×103 − 120h 2 −24 ×106h + 14.4 ×103h2 + 1.2 ×1010 − 7200h2 = 0 7200h2 − 24 ×106h + 1.2 ×1010 = 0On solving, h = 612.57 mm (O.K) and 2720.8 mm (Not valid)Putting the value of h = 612.57 mm in equation (1) y = 100 ×106 − 60 ×(612.57)2 = 612.57 mm = h 200 ×103 − 120 × 612.57Hence y =h Determine the centroid of the shaded area shown in figure. &'()*&+ θ = 45 = π = 2α 4Bigger Sectora = 1 (12)2 × π = 56.55 cm2 242 × 12 × sin(22.5) = 7.796 cm3 π 8Smaller Sectora = 1 (10)2 × π = 39.27 cm2 242 × 10 × sin(22.5) = 6.497 cm3 π 8 a (cm2) x (cm) y (cm) a.x (cm3) a.y (cm3) Component 407.3 168.71 56.55 7.796 cos 22.5 7.796 sin22.5 = 7.203 39.27 6.497 cos 22.5 6.497 sin 22.5 – 235.72 – 97.64 a ax ay = 17.28 = 171.58 = 71.07 x = a x = 171.58 = 9.93 cm = 99.3 mm a 17.28 y = a y = 71.07 = 4.113 cm = 41.13 mm a 17.28 8

Calculate numerically centroid of the shaded area shown in figure. (Dec’00 8 marks) &'()*&+ Due to symmetry about x-axis, y = 0Circular Sectorα = 14° = 14 × π = 0.2443 rad 180R = 1002 + 252 = 103.08 cma = 1 (103.08)2 × 28 × π = 2596.3 cm2 2 1802 ×103.08 × sin(14) = 68.04 cm3 0.2443Prof. Ramkumar'sComponent a (cm2) x (cm) a.x (cm3) π(15)2 = 353.43 2249.93 − 4 × 15 = − 6.366 2 3π 25000 25000 1 ×15 ×100 = 750 100 = 33.333 176652.25 2 3 1 ×15 ×100 = 750 166666.7 2 100 = 33.333 a.x = 57735.62 3 2596.3 68.04 − 1 × 50 ×100 2 ×100 2 3 = 66.67 = − 2500 a = 1949.73x = a x = 57735.62 = 29.61 cm a 1949.73 9

! ! Locate the centroid of plane area shown in Locate the centroid of plane area shown infigure. figure.Prof. Ramkumar'sx = 5 mm and y = 55 mm x = 0 and y = 108.4 mm Find the centroid of shaded area of the Find centroid of the shaded area.semicircle of diameter 100 cm. (Dec’12-8 marks) (May’11-5 marks)x = 58.33 cm, y = 24.76 cm x = 98.59 mm, y = 71.18 mm Three plates ABC, BCDE and DEF are welded Determine coordinate of centroid of shadedtogether as shown in figure. Circle of diameter area shown.1.5 m is cut out from the composite plate. (Dec’04)Determine the centroid of the remaining area.Hint: φ = diameter = 1.5 m.x = 4.37 m and y = 1.82 m x = y = 2.707 m 10

\" Determine the centroid of the shaded area # Locate the centroid of the shaded area as shown in figure. shown in figure. (Dec’11-5 marks & June’02)Prof. Ramkumar'sx = 1.126 m and y = - 0.28 m x = 7.5 m and y = 9 m $ Find centroid of plane area. (Dec’10-6 marks) x = 12.46mm,y = 22.044mm % Locate the centroidal axes x-x and y-y for the figure shown. The lamina has one circular cut of 1.6 m diameter and one rectangular cut of 4 m × 3m. (May’09-6 marks) x = 8.17m, y = 3.67m Find the centroid of the area shown in figure. (June’10-6 marks) x = 101.57 mm, y = 90.34 mm 11

Locate the centroid of the shaded area Find distance y so that CG of given area inshown in figure. figure has co-ordinates (25, 20). (Dec’07-5 marks)Prof. Ramkumar'sx = 1.6 cm and y = 2.08 cm y = 25.625 mm Determine the x and y co-ordinates of the Locate the centroid of the shaded area shownshaded area shown in figure. in figure.x = −0.857 cm and y = +1.764 cm x = 26.33 cm and y = 23.64 cm Determine the centroid of the shaded areashown in figure w.r.to axis ox and oy.x = 10 cm and y = 5.92 cm 12