`CHAPTER2CENTROIDANDMOMENTOFINERTIAUnder this topic first we will see how to find the areas of given figures and the volumes of given solids. Then the terms centre of gravity and centroids are explained. Though the title of this topic do not indicate the centroid of line segment, that term is also explained, since the centroid of line segment will be useful in finding the surface area and volume of solids using theorems of Pappus and Guldinus. Then the term first moment of area is explained and the method of finding centroid of plane areas and volumes is illustrated. After explaining the term second moment of area, the method of finding moment of inertia of plane figures about x-x or y-y axis is illustrated. The term product moment of inertia is defined and the mehtod of finding principal moment of inertia is presented. At the end the method of finding mass moment of inertia is presented.2.1DETERMINATION OF AREAS AND VOLUMESIn the school education methods of finding areas and volumes of simple cases are taught by many methods. Here we will see the general approach which is common to all cases i.e. by the method of integration. In this method the expression for an elemental area will be written then suitable integrations are carried out so as to take care of entire surface/volume. This method is illustrated with standard cases below, first for finding the areas and latter for finding the volumes:A: Area of Standard Figures(i) Area of a rectangle Let the size of rectangle be b × d as shown in Fig. 2.1. dA is an elemental area of side dx × dy. Area of rectangle, A =d/2 O d/2y dx dy xzb2 d2dA =b2 -b 2-b 2 -d 2z zd2 -d 2dx dyb/2 b/2= x = bd.yFig. 2.170CENTROIDANDMOMENTOF INERTIA71If we take element as shown in Fig. 2.2,d2A=-d 2zd2dA =d2 -d 2-d 2zd/2b  dyd/2 bdy y= b y = bdFig. 2.2(ii) Area of a triangle of base width `b' height `h'. Referring to Fig. 2.3, let the element be selected as shown by hatched lines Theny dA = bdy = b dy hy h b¢ b dyy A = dA = b dy h 0 0=b  y2  bh   = h 2  2  0hzhzhFig. 2.3(iii) Area of a circle Consider the elemental area dA = rddr as shown in Fig. 2.4. Now, dA = rd dr r varies from O to R and  varies from O to 22 RA=2=2= = =zz z LMN z0 0 0 0yrd drr2 22O P QRd0dr r dq q O Rrdq xR d 22 0R2  2R2 2 Fig. 2.4 . 2  = R 2 In the above derivation, if we take variation of  from 0 to , we get the area of semicircle as R 2 R 2 and if the limit is from 0 to  / 2 the area of quarter of a circle is obtained as . 2 472ENGINEERING MECHANICS(iv) Area of a sector of a circle Area of a sector of a circle with included angle 2 shown in Fig. 2.5 is to be determined. The elemental area is as shown in the figure dA = rd . dr y  varies from ­ to and r varies from O to R  A==L R O = M N2 P Q2-z zz z LMN OPQ zdA =r2 2 R - 0  Rr d drRdrrdGd =0-R2 d 20r dGO a=-R2 2 2 = R  2Fig. 2.5a f(v) Area of a parabolic spandrel Two types of parabolic curves are possible (a) y = kx2 (b) y2 = kx Case a: This curve is shown in Fig. 2.6. The area of the element dA = y dx = kx2 dx  A=Lx O = kM P N3Q3zadA =a0za 0 0ykx dxy = kx22ka = 33h x=a dx xxWe know, when x = a, y = h i.e.,  h = ka2 or k = A=h a2Fig. 2.6ka 3 h a3 1 1 = 2 = ha = rd the area of rectangle of size a × h 3 3 3 a 3 Case b: In this case y2 = kx Referring to Fig. 2.7dA = y dx = A=kx dxzay dx =0zakx dx0CENTROIDANDMOMENT2OF INERTIA73=k x3 2L M N2 3O P Qa 0=k2 32 a 3y y = kxWe know that, when x = a, y = h  Hence i.e., h2 = ka A= A=horh2 k= ah x dx x=a2 . . a3 2 a 3Fig. 2.72 2 ha = rd the area of rectangle of size a × h 3 3 (vi) Surface area of a cone Consider the cone shown in Fig. 2.8. Now, x R h Surface area of the element,y= dA = 2 y dl = 2 = 2  A=x R dl hdl R a x y dxx dx R h sin x2 2 R h sin  2Rh =  Rl = sin  (vii) Surface area of a sphere Consider the sphere of radius R shown in Fig. 2.9. The element considered is the parallel circle at distance y from the diametral axis of sphere.dS = 2 x Rd = 2 R cos  Rd, since x = R cos 2L O M P N Qhh0Fig. 2.8S = 2 R2 = 2 R = 4 R22- 2zcos  d2 - 2dy y q dqxRdqsin Fig. 2.974ENGINEERING MECHANICSB: Volume of Standard Solids(i) Volume of a parallelpiped. Let the size of the parallelpiped be a × b × c. The volume of the element is dV = dx dy dz V=zzzabc 000 a 0dx dy dz yb 0= x (ii) Volume of a cone: Referring to Fig. 2.8zc 0= abcdV = y2 . dx =  h2 R2x x2 2 R dx , since y = R 2 h h h22V=z0hx dx =2 3R2LM x OP N3Q3h0= (iii) Volume of a sphere Referring to Fig. 2.9 2 h R h = R 2 h 3 3dV =  x2 dy But i.e.,  x2 + y2 = R2 x 2 = R2 ­ y2 dV =  (R 2 ­ y2)dy V=The surface areas and volumes of solids of revolutions like cone, spheres may be easily found using theorems of Pappus and Guldinus. This will be taken up latter in this chapter, since it needs the term centroid of generating lines.j L y OP =  MR y - 3 P M N Q L R a- Rf R | - S- R - =  MR  R - 3 3 M | T N L 1 1O 4  R =  R M1 - + 1 - P = N 3 3Q 3R-Rz R2 - y 2 dy3 Re2-R2333UOP | VP |Q W33CENTROIDANDMOMENTOF INERTIA752.2CENTRE OF GRAVITY AND CENTROIDSConsider the suspended body shown in Fig. 2.10a. The self weight of various parts of this body are acting vertically downward. The only upward force is the force T in the string. To satisfy the equilibrium condition the resultant weight of the body W must act along the line of string 1­1. Now, if the position is changed and the body is suspended again (Fig. 2.10b), it will reach equilibrium condition in a particular position. Let the line of action of the resultant weight be 2­ 2 intersecting 1­1 at G. It is obvious that if the body is suspended in any other position, the line of action of resultant weight W passes through G. This point is called the centre of gravity of the body. Thus centre of gravity can be defined as the point through which the resultant of force of gravity of the body acts.T 1 1T 2w1G W1 2 W = å w1 W = å w1 11(a)(b)Fig. 2.10The above method of locating centre of gravity is the practical method. If one desires to locating centre of gravity of a body analytically, it is to be noted that the resultant of weight of various portions of the body is to be determined. For this Varignon's theorem, which states the moment of resultant force is equal to the sum of moments of component forces, can be used. Referring to Fig. 2.11, let Wi be the weight of an element in the given body. W be the total weight of the body. Let the coordinates of the element be xi, yi, zi and that of centroid G be xc, yc, zc. Since W is the resultant of Wi forces, W = W1 + W2 + W3 . . . = Wi and  Similarly, and Wxc = W1x1 + W2x2 + W3x3 + . . . Wxc = Wixi = Ü xdw Wyc = Wiyi = Ü ydw Wz c = Wizc = Ü zdwG Wi O yi zi xi xc zWyc zcxFig. 2.11U | V | WEqn. (2.1)76ENGINEERING MECHANICSIf M is the mass of the body and mi that of the element, then M=W gandmi =Wi , hence we get gMxc = mixi = Myc = miyi =and Mzc = mizi = zidm If the body is made up of uniform material of unit weight g, then we know Wi = Uig, where U represents volume, then equation 2.1 reduces to Vxc = Vixi = Vyc = Viyi =z z z z z z zxidm yidmU | V | W U | V | WEqn. (2.2)xdV ydVVzc = Vizi= zdV If the body is a flat plate of uniform thickness, in x-y plane, Wi = g Ait (Ref Fig. 2.12). Hence equation 2.1 reduces to Axc = Aixi = Ayc = Aiyi =yzEqn. (2.3)x dA y dAU V WEqn. (2.4)W z xc yc Wi (xi, yc)dL Wi = g A dLFig. 2.12Fig. 2.13If the body is a wire of uniform cross section in plane x, y (Ref. Fig. 2.13) the equation 2.1 reduces to Lxc =  Lixi = Eqn. (2.5) Lyc =  Liyi = y dL The term centre of gravity is used only when the gravitational forces (weights) are considered. This term is applicable to solids. Equations 2.2 in which only masses are used the point obtained is termed as centre of mass. The central points obtained for volumes, surfaces and line segments (obtained by eqns. 2.3, 2.4 and 2.5) are termed as centroids.z zx dLU V W2.3CENTROID OF A LINEdx O x L G xCentroid of a line can be determined using equation 2.5. Method of finding the centroid of a line for some standard cases is illustrated below: (i) Centroid of a straight line: Selecting the x-coordinate along the line (Fig. 2.14)Fig. 2.14CENTROIDANDMOMENTOF INERTIA77Lxc =  xc =z0Lx dx =LM x OP N2Q2L=0L2 2L 2 Thus the centroid lies at midpoint of a straight line, whatever be the orientation of line (Ref. Fig. 2.15).y yL G O L 2 L O G x L 2 L cos 2 G L sin 2xFig. 2.15(ii) Centroid of an Arc of a Circle Referring to Fig. 2.16, L = Length of arc = R 2 dL = Rd Hence from eqn. 2.5xcL = i.e., xc R 2 =- -  -z zxRdqxdLa O adqqxR cos  . Rd(i)= R2 sin  xc = yc Land-zR sin  R 2 × 2 sin  =  2 Ry dL =-zFig. 2.16R sin  . Rd -= R2 - cos (ii)=0  yc = 0 From equation (i) and (ii) we can get the centroid of semicircle shown in Fig. 2.17 by putting  = /2 and for quarter of a circle shown in Fig. 2.18 by putting  varying from zero to /2.78ENGINEERING MECHANICSG R G RFig. 2.17Fig. 2.182R  yc = 0 For quarter of a circle,For semicircle xc =2R  2R yc =  (iii) Centroid of composite line segments: The results obtained for standard cases may be used for various segments and then the equations 2.5 in the formxc = xcL = Lixi ycL = Liyi may be used to get centroid xc and yc. If the line segments is in space the expression zcL = Lizi may also be used. The method is illustrated with few examples below: Example 2.1 Determine the centroid of the wire shown in Fig. 2.19.y D G3300m m45° CG2 G1 A 600 mm B200 mm kFig. 2.19CENTROIDANDMOMENTOF INERTIA79Solution. The wire is divided into three segments AB, BC and CD. Taking A as origin the coordinates of the centroids of AB, BC and CD are G1(300, 0); G2(600, 100) and G3(600 ­ 150 cos 45°; 200 + 150 sin 45°) i.e. G3(493.93, 306.07) L1 = 600 mm, L2 = 200 mm, L3 = 300 mm L = 600 + 200 + 300 = 1100 mm Total length From the eqn. Lxc = Lixi, we get 1100 xc = L1x1 + L2x2 + L3x3 = 600 × 300 + 200 × 600 + 300 × 493.93  Now, xc = 407.44 mm Lyc = Liyi 1100 yc = 600 × 0 + 200 × 100 + 300 × 306.07 yc = 101.66 mm Example 2.2 Locate the centroid of the uniform wire bent as shown in Fig. 2.20.G2 250 G1 A 400 mm All dimensions in mm B 150 C 30° G3 DAns.Ans.Fig. 2.20Solution. The composite figure is divided into 3 simple figures and taking A as origin coordinates of their centroids noted as shown below: AB--a straight line L1 = 400 mm, G1 (200, 0) BC--a semicircle L2 = 150  = 471.24, CD--a straight line L3 = 250; x3 = 400 + 300 +2 × 150  i.e., G2 (475, 92.49) G 2 475 ,FG HIJ K250 cos 30° = 808.25 mm 2y3 = 125 sin 30 = 62.5 mm  Total length L = L1 + L2 + L3 = 1121.24 mm  Lxc = Lixi xc = 451.20 mm Lyc = Liyi gives gives Ans. 1121.24 xc = 400 × 200 + 471.24 × 475 + 250 × 808.2580ENGINEERING MECHANICS1121.24 yc = 400 × 0 + 471.24 × 95.49 + 250 × 62.5 yc = 54.07 mm Ans. Example 2.3 Locate the centroid of uniform wire shown in Fig. 2.21. Note: portion AB is in x-z plane, BC in y-z plane and CD in x-y plane. AB and BC are semi circular in shape.zr = 140A x Dr=10 0BC45°yFig. 2.21Solution. The length and the centroid of portions AB, BC and CD are as shown in table below: Table 2.1Portion AB BC CD Li100 140 300xi100 0 300 sin 45°yi0 140 280 + 300 cos 45° = 492.13zi2 × 100  2 × 140 0L = 100 + 140 + 300 = 1053.98 mm 1053.98 xc = 100 × 100 + 140 × 0 + 300 × 300 sin 45° xc = 90.19 mm Ans. Ans.From eqn. Lxc = Lixi, we getSimilarly, 1053.98 yc = 100 × 0 + 140 × 140 + 300 × 492.13 yc = 198.50 mm and 1053.98 zc = 100 ×200 2 × 140 + 140 × + 300 × 0   z c = 56.17 mmAns.2.4FIRST MOMENT OF AREA AND CENTROIDFrom equation 2.1, we have xc = W i xi ,Wyc = Wi y iWandzc = W i ziWCENTROIDANDMOMENTOF INERTIA81From the above equation we can make the statement that distance of centre of gravity of a body from an axis is obtained by dividing moment of the gravitational forces acting on the body, about the axis, by the total weight of the body. Similarly from equation 2.4, we have,A A By terming Aix: as the moment of area about the axis, we can say centroid of plane area from any axis is equal to moment of area about the axis divided by the total area. The moment of area Aix: is termed as first moment of area also just to differentiate this from the term Aix E , which will be dealt latter. It may be noted that since the moment of area about an axis divided by total area gives the distance of the centroid from that axis, the moment of area is zero about any centroidal axis.xc = Ai xi, yc = Ai y iDifference between Centre of Gravity and CentroidFrom the above discussion we can draw the following differences between centre of gravity and centroid: (1) The term centre of gravity applies to bodies with weight, and centroid applies to lines, plane areas and volumes. (2) Centre of gravity of a body is a point through which the resultant gravitational force (weight) acts for any orientation of the body whereas centroid is a point in a line plane area volume such that the moment of area about any axis through that point is zero.Use of Axis of SymmetryCentroid of an area lies on the axis of symmetry if it exits. This is useful theorem to locate the Y centroid of an area. This theorem can be proved as follows: Axis of symmetry Consider the area shown in Fig. 2.22. In this figure y-y is the axis of symmetry. From eqn. 2.4, the distance of centroid from this axis is x x given by: Ai xiOXA Consider the two elemental areas shown in Fig. 2.22, which are equal in size and are equidistant from the axis, but on either side. Now the sum of moments of these areas cancel each other since the Fig. 2.22 areas and distances are the same, but signs of distances are opposite. Similarly, we can go on considering an area on one side of symmetric axis and corresponding image area on the other side, and prove that total moments of area (Aixi) about the symmetric axis is zero. Hence the distance of centroid from the symmetric axis is zero, i.e. centroid always lies on symmetric axis. Making use of the symmetry we can conclude that: (1) Centroid of a circle is its centre (Fig. 2.23); b d (2) Centroid of a rectangle of sides b and d is at distance and from the corner as shown 2 2 in Fig. 2.24.82ENGINEERING MECHANICSb b/2GG d/2dFig. 2.23Fig. 2.24Determination of Centroid of Simple Figures From First PrincipleFor simple figures like triangle and semicircle, we can write general expression for the elemental area and its distance from an axis. Then equations 2.4--y = x =--A The location of the centroid using the above equations may be considered as finding centroid from first principles. Now, let us find centroid of some standard figures from first principles.z zydA A xdACentroid of a Triangle Consider the triangle ABC of base width b and height h as shown in Fig. 2.25. Let us locate the distance of centroid from the base. Let b1 be the width of elemental strip of thickness dy at a distance y from the base. Since DAEF and DABC are similar triangles, we can write:b1 h-y = b hb1 = A dy E b1 y F hFG h - y IJ b = FG 1 - y IJ b H h K H hK= 1-Area of the element = dA = b1dyFG Hy b dy h 1 A = bh 2IJ KB bCFig. 2.25Area of the triangle  From eqn. 2.4--y =Now, ydA = y 1 - h b dy 0zMovement of area = Total areazydA AhFG HyIJ KCENTROIDANDMOMENTOF INERTIA83=L y y OP = bM - M N 2 3h QP2 3z FGHh 0y-y2 b dy hhI JK0= --y =bh 6 ydA2zA=bh 2 1 × 1 6 bh 2--y =h 3 2h h from the base (or from the apex) of 3 3Thus the centroid of a triangle is at a distance the triangle where h is the height of the triangle. Centroid of a SemicircleConsider the semicircle of radius R as shown in Fig. 2.26. Due to symmetry centroid must lie on y axis. Let its distance from diametral axis be y . To find y , consider an element at a distance r from the centre O of the semicircle, radial width being dr and bound by radii at  and  + d. Area of element = r ddr. Its moment about diametral axis x is given by: rd × dr × r sin  = r2 sin  dr d  Total moment of area about diametral axis,Yzz00Rr2Lr O sin  dr d = z M P N3Q 0 3 3Rsin  d0dq r q O R X drR  - cos  0 3 3 R3 2R 1+ 1 = = 3 3 1 2 Area of semicircle A = R 2=Fig. 2.262R 3 Moment of area -- 3 = y = 1 2 Total area R 2 4R = 3Thus, the centroid of the circle is at a distance4R from the diametral axis. 384ENGINEERING MECHANICSCentroid of Sector of a Circle Consider the sector of a circle of angle 2 as shown in Fig. 2.27. Due to symmetry, centroid lies on x axis. To find its distance from the centre O, consider the elemental area shown. Area of the element Its moment about y axis = rd × dr × r cos  = r cos  drd2= rd drY dr r Total moment of area about y axis R=- 0zz3dq q 2a a G Xr cos  drd2 R  -OLr O = M P N3Q= Total area of the sector Rsin 0RR3 2 sin  3Fig. 2.27=- 0 =-zz z LMN2rdrdr2 2OP QRd0R  2 = R 2= -\ The distance of centroid from centre O =Moment of area about y axis Area of the figure2R 3 sin  2R 3 = sin  = 2 3 R Centroid of Parabolic Spandrel Consider the parabolic spandrel shown in Fig. 2.28. Height of the element at a distance x from O is y = kx2CENTROIDANDMOMENTOF INERTIA85Width of element  Area of the element= dx = kx2dx Total area of spandrel =za 0kx 2dx3 a 0= Moment of area about y axis = =L kx OP M3Q N=ka 3 3Yy = kx2hL kx OP = M N4Q4z za a 0­ y) G(x, ­kx 2 dx × x kx 3 dxaOx a dxX0Fig. 2.280= Moment of area about x axis =kx 2 = kx dx = 2 02z z0 aka 44dAy 2za 0k2x4 dx = 2= x = y =From the Fig. 2.28, at x = a, y  k 2 a5 10 ka 4 ka 3 3a ÷ = 4 3 4 2 5 3 k a ka 3 ka 2 ÷ = 10 3 10 =hLM k x OP N2×5 Q2 5 a 0h = ka2 or k =h a2y =Thus, centroid of spandrel isFG 3a , 3h IJ H 4 10 K3 3h h × a2 = 10 a 2 10Centroids of some common figures are shown in Table 2.2.86ENGINEERING MECHANICSTable 2.2. Centroid of Some Common FiguresShape FigureyN OAreaTrianglehG x--h 3bh 2bySemicircleGr x04R 3R2 2yR2 4Quarter circleG x Ry4R 34R 3Sector of a circle2aGx2R sin a 30R 2ParabolahG 2a x03h 54ah 3Semi parabola3a 8y3h 52ah 3Parabolic spandrelh G a x3a 43h 10ah 3Centroid of Composite Sections So far, the discussion was confined to locating the centroid of simple figures like rectangle, triangle, circle, semicircle, etc. In engineering practice, use of sections which are built up of many simple sections is very common. Such sections may be called as built-up sections or composite sections. To locate the centroid of composite sections, one need not go for the first principle (method of integration). The given composite section can be split into suitable simple figures and then the centroid of each simple figure can be found by inspection or using the standard formulae listed in Table 2.2. Assuming the area of the simple figure as concentrated at its centroid, its moment about an axis can be found by multiplying the area with distance of its centroid from theCENTROIDANDMOMENTOF INERTIA87reference axis. After determining moment of each area about reference axis, the distance of centroid from the axis is obtained by dividing total moment of area by total area of the composite section. Example 2.4 Locate the centroid of the T-section shown in the Fig. 2.29. Solution. Selecting the axis as shown in Fig. 2.29, we can say due to symmetry centroid lies on y axis, i.e. x = 0. Now the given T-section may be divided into two rectangles A1 and A2 each of size 100 × 20 and 20 × 100. The centroid of A1 and A2 are g1(0, 10) and g2(0, 70) respectively.  The distance of centroid from top is given by:100 ­ y A1 O g1 20XGy =100 × 20 × 10 + 20 × 100 × 70 100 × 20 + 20 × 100g2 A2100= 40 mm Hence, centroid of T-section is on the symmetric axis at a distance 40 mm from the top. Ans.20 Y All dimensions in mmExample 2.5 Find the centroid of the unequal angle 200 × Fig. 2.29 150 × 12 mm, shown in Fig. 2.30. Solution. The given composite figure can be divided into two rectangles: A1 = 150 × 12 = 1800 mm2 A2 = (200 ­ 12) × 12 = 2256 mm2 Total area A = A1 + A2 = 4056 mm2 Selecting the reference axis x and y as shown in Fig. 2.30. The centroid of A1 is g1 (75, 6) and that of A2 is:g 2 6 , 12 +i.e.,  g2 (6, 106)L M N1 200 - 12 2afOPQO ­ y­ x150 12 g1 G A1 Xx == =y == =Movement about y axis Total area A1 x1 + A2 x2 A 1800 × 75 + 2256 × 6 = 36.62 mm 4056 Movement about x axis Total area A1 y 1 + A2 y 2 A 1800 × 6 + 2256 × 106 = 61.62 mm 4056-- --200g2 A212 Y All dimensions in mmFig. 2.30Thus, the centroid is at x = 36.62 mm and y = 61.62 mm as shown in the figureAns.88ENGINEERING MECHANICSExample 2.6 Locate the centroid of the I-section shown in Fig. 2.31.Y 100 20 A1 g120 A2 100 g2 G ­ y 30 A3 150 g3 O XAll dimensions in mmFig. 2.31Solution. Selecting the co-ordinate system as shown in Fig. 2.31, due to symmetry centroid must lie on y axis, i.e.,x =0Now, the composite section may be split into three rectangles A1 = 100 × 20 = 2000 mm2 Centroid of A1 from the origin is:Similarly20 = 140 mm 2 A2 = 100 × 20 = 2000 mm2y1 = 30 + 100 +100 = 80 mm 2 A3 = 150 × 30 = 4500 mm2, andy2 = 30 +30 = 15 mm 2 A y + A2 y 2 + A3 y 3 y = 1 1  A 2000 + 140 + 2000 × 80 + 4500 × 15 = 2000 + 2000 + 4500 = 59.71 mm Thus, the centroid is on the symmetric axis at a distance 59.71 mm from the bottom as shown in Fig. 2.31. Ans.y3 =`

19 pages

#### Report File (DMCA)

Our content is added by our users. We aim to remove reported files within 1 working day. Please use this link to notify us:

Report this file as copyright or inappropriate

410846