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CENTROIDS OF SIMPLE SHAPES 469 D 5 0 5x 2 x 3 dx 5 0 5x x 2 dx D 5x 3 3 x 4 4 5 0 5x 2 2 x 3 3 5 0 D 625 3 625 4 125 2 125 3 D 625 12 125 6 D 625 12 6 125 D 5 2 D 2 .5 y D 1 2 5 0 y 2 dx 5 0 y dx D 1 2 5 0 5x x 2 2 dx 5 0 5x x 2 dx D 1 2 5 0 25x 2 10x 3 C x 4 dx 125 6 D 1 2 25x 3 3 10x 4 4 C x 5 5 5 0 125 6 D 1 2 25125 3 6250 4 C 625 125 6 D 2 .5 Figure 57.6 Hence the centroid of the area lies at (2.5, 2.5) (Note from Fig. 57.6 that the curve is symmetrical about x D 2.5 and thus x could have been deter- mined ‘on sight’). Problem 5. Locate the centroid of the area enclosed by the curve y D 2x 2 ,they-axis and ordinates y D 1andy D 4, correct to 3 decimal places From Section 57.4, x D 1 2 4 1 x 2 dy 4 1 x dy D 1 2 4 1 y 2 dy 4 1 y 2 dy D 1 2 y 2 4 4 1 2y 3/2 3 p 2 4 1 D 15 8 14 3 p 2 D 0 .568 and y D 4 1 xy dy 4 1 x dy D 4 1 y 2 y dy 14 3 p 2 D 4 1 y 3/2 p 2 dy 14 3 p 2 D 1 p 2 y 5/2 5 2 4 1 14 3 p 2 D 2 5 p 2 31 14 3 p 2 D 2 .657 Hence the position of the centroid is at (0.568, 2.657) Problem 6. Locate the position of the centroid enclosed by the curves y D x 2 and y 2 D 8x Figure 57.7 shows the two curves intersecting at (0, 0) and (2, 4). These are the same curves as used in Problem 12, Chapter 54, where the shaded area was calculated as 2 2 3 square units. Let the co- ordinates of centroid C be x and y. 470 ENGINEERING MATHEMATICS By integration, x D 2 0 xy dx 2 0 y dx y y = x 2 y 2 = 8 x (or y = √8 x ) y y 4 3 2 1 012 x C x 2 2 Figure 57.7 The value of y is given by the height of the typical strip shown in Fig. 55.7, i.e. y D p 8x x 2 .Hence, x D 2 0 x p 8x x 2 dx 2 2 3 D 2 0 p 8 x 3/2 x 3 2 2 3 D p 8 x 5/2 5 2 x 4 4 2 0 2 2 3 D p 8 p 2 5 5 2 4 2 2 3 D 2 2 5 2 2 3 D 0 .9 Care needs to be taken when finding y in such examples as this. From Fig. 57.7, y D p 8x x 2 and y 2 D 1 2 ( p 8x x 2 ). The perpendicular distance from centroid C of the strip to Ox is 1 2 p 8x x 2 C x 2 . Taking moments about Ox gives: (total area) y D xD2 xD0 (area of strip) (perpendicu- lar distance of centroid of strip to Ox) Hence (area) y D p 8x x 2 1 2 p 8x x 2 Cx 2 dx i.e. 2 2 3 y D 2 0 p 8x x 2 p 8x 2 C x 2 2 dx D 2 0 8x 2 x 4 2 dx D 8x 2 4 x 5 10 2 0 D 8 3 1 5 0 D 4 4 5 Hence y D 4 4 5 2 2 3 D 1 .8 Thus the position of the centroid of the shaded area in Fig. 55.7 is at (0.9, 1.8) Now try the following exercise Exercise 191 Further problems on cen- troids of simple shapes 1. Determine the position of the centroid of a sheet of metal formed by the curve y D 4x x 2 which lies above the x-axis. [(2, 1.6)] 2. Find the coordinates of the centroid of the area that lies between the curve y x D x 2 and the x-axis. [(1, 0.4)] 3. Determine the coordinates of the centroid of the area formed between the curve y D 9 x 2 and the x-axis. [(0, 3.6)] 4. Determine the centroid of the area lying between y D 4x 2 ,they-axis and the ordinates y D 0andy D 4. [(0.375, 2.40] 5. Find the position of the centroid of the area enclosed by the curve y D p 5x,the x-axis and the ordinate x D 5. [(3.0, 1.875)] 6. Sketch the curve y 2 D 9x between the limits x D 0andx D 4. Determine the position of the centroid of this area. [(2.4, 0)] CENTROIDS OF SIMPLE SHAPES 471 7. Calculate the points of intersection of the curves x 2 D 4y and y 2 4 D x, and deter- mine the position of the centroid of the area enclosed by them. [(0, 0) and (4, 4), (1.8, 1.8)] 8. Determine the position of the centroid of the sector of a circle of radius 3 cm whose angle subtended at the centre is 40 ° . On the centre line, 1.96 cm from the centre 9. Sketch the curves y D 2x 2 C 5and y 8 D x x C 2 on the same axes and determine their points of intersection. Calculate the coordinates of the centroid of the area enclosed by the two curves. [(1, 7) and (3, 23), (1, 10.20)] 57.7 Theorem of Pappus A theorem of Pappus states: ‘If a plane area is rotated about an axis in its own plane but not intersecting it, the volume of the solid formed is given by the product of the area and the distance moved by the centroid of the area’. With reference to Fig. 57.8, when the curve y D fx is rotated one revolution about the x-axis between the limits x D a and x D b, the volume V generated is given by: volume V D A2 y, from which, y = V 2pA x = ax = b y = f ( x ) x y C Area A y Figure 57.8 Problem 7. Determine the position of the centroid of a semicircle of radius r by using the theorem of Pappus. Check the answer by using integration (given that the equation of a circle, centre 0, radius r is x 2 C y 2 D r 2 ) A semicircle is shown in Fig. 57.9 with its diameter lying on the x-axis and its centre at the origin. Area of semicircle D r 2 2 . When the area is rotated about the x-axis one revolution a sphere is generated of volume 4 3 r 3 . y y C 0 − rrx x 2 + y 2 = r 2 Figure 57.9 Let centroid C be at a distance y from the origin as shown in Fig. 57.9. From the theorem of Pappus, volume generated D area ðdistance moved through by centroid i.e. 4 3 r 3 D r 2 2 2 y Hence y D 4 3 r 3 2 r 2 D 4r 3 By integration, y D 1 2 r r y 2 dx area D 1 2 r r r 2 x 2 dx r 2 2 D 1 2 r 2 x x 3 3 r r r 2 2 D 1 2 r 3 r 3 3 r 3 C r 3 3 r 2 2 D 4r 3 Hence the centroid of a semicircle lies on the axis of symmetry, distance 4r 3p (or 0.424 r)fromits diameter. 472 ENGINEERING MATHEMATICS Problem 8. Calculate the area bounded by the curve y D 2x 2 ,thex-axis and ordinates x D 0andx D 3. (b) If this area is revolved (i) about the x-axis and (ii) about the y-axis, find the volumes of the solids produced. (c) Locate the position of the centroid using (i) integration, and (ii) the theorem of Pappus y 18 y = 2 x 2 12 6 0 x y 123 x Figure 57.10 (a) The required area is shown shaded in Fig. 57.10. Area D 3 0 y dx D 3 0 2x 2 dx D 2x 3 3 3 0 D 18 square units (b) (i) When the shaded area of Fig. 57.10 is revolved 360 ° about the x-axis, the volume generated D 3 0 y 2 dx D 3 0 2x 2 2 dx D 3 0 4x 4 dx D 4 x 5 5 3 0 D 4 243 5 D 194.4p cubic units (ii) When the shaded area of Fig. 57.10 is revolved 360 ° about the y-axis, the volume generated D (volume generated by x D 3 (volume generated by y D 2x 2 ) D 18 0 3 2 dy 18 0 y 2 dy D 18 0 9 y 2 dy D 9y y 2 4 18 0 D 81p cubic units (c) If the co-ordinates of the centroid of the shaded area in Fig. 57.10 are ( x, y) then: (i) by integration, x D 3 0 xy dx 3 0 y dx D 3 0 x2x 2 dx 18 D 3 0 2x 3 dx 18 D 2x 4 4 3 0 18 D 81 36 D 2 .25 y D 1 2 3 0 y 2 dx 3 0 y dx D 1 2 3 0 2x 2 2 dx 18 D 1 2 3 0 4x 4 dx 18 D 1 2 4x 5 5 3 0 18 D 5 .4 (ii) using the theorem of Pappus: Volume generated when shaded area is revolved about Oy D area2 x) i.e. 81 D 182 x, from which, x D 81 36 D 2 .25 Volume generated when shaded area is revolved about Ox D area2 y i.e. 194.4 D 182 y, from which, y D 194.4 36 D 5 .4 Hence the centroid of the shaded area in Fig. 55.10 is at (2.25, 5.4) Problem 9. A cylindrical pillar of diameter 400 mm has a groove cut round its circumference. The section of the groove is a semicircle of diameter 50 mm. Determine the volume of material removed, in cubic centimetres, correct to 4 significant figures A part of the pillar showing the groove is shown in Fig. 57.11. The distance of the centroid of the semicircle from its base is 4r 3 see Problem 7 D 425 3 D 100 3 mm. The distance of the centroid from the centre of the pillar D 200 100 3 mm. CENTROIDS OF SIMPLE SHAPES 473 400 mm 50 mm 200 mm Figure 57.11 The distance moved by the centroid in one revolu- tion D 2 200 100 3 D 400 200 3 mm. From the theorem of Pappus, volume D area ðdistance moved by centroid D 1 2 25 2 400 200 3 D 1168250 mm 3 Hence the volume of material removed is 1168 cm 3 correct to 4 significant figures. Problem 10. A metal disc has a radius of 5.0 cm and is of thickness 2.0 cm. A semicircular groove of diameter 2.0 cm is machined centrally around the rim to form a pulley. Determine, using Pappus’ theorem, the volume and mass of metal removed and the volume and mass of the pulley if the density of the metal is 8000 kg m 3 A side view of the rim of the disc is shown in Fig. 57.12. When area PQRS is rotated about axis XX the volume generated is that of the pulley. The centroid of the semicircular area removed is at a distance of 4r 3 from its diameter (see Problem 7), i.e. 41.0 3 , i.e. 0.424 cm from PQ. Thus the distance of the centroid from XX is (5.0 0.424), i.e. 4.576 cm. The distance moved through in one revolution by the centroid is 2(4.576) cm. Area of semicircle D r 2 2 D 1.0 2 2 D 2 cm 2 Figure 57.12 By the theorem of Pappus, volume generated D area ðdistance moved by centroid D 2 24.576 i.e. volume of metal removed D 45 .16 cm 3 Mass of metal removed D density ðvolume D 8000 kg m 3 ð 45.16 10 6 m 3 D 0.3613 kg or 361. 3g Volume of pulley D volume of cylindrical disc volume of metal removed D 5.0 2 2.0 45.16 D 111.9cm 3 Mass of pulley D density ð volume D 8000 kg m 3 ð 111.9 10 6 m 3 D 0.8952 kg or 895.2g Now try the following exercise Exercise 192 Further problems on the theorem of Pappus 1. A right angled isosceles triangle having a hypotenuse of 8 cm is revolved one revolution about one of its equal sides as axis. Determine the volume of the solid generated using Pappus’ theorem. [189.6 cm 3 ] 2. A rectangle measuring 10.0 cm by 6.0 cm rotates one revolution about one of its longest sides as axis. Determine the 474 ENGINEERING MATHEMATICS volume of the resulting cylinder by using the theorem of Pappus. [1131 cm 2 ] 3. Using (a) the theorem of Pappus, and (b) integration, determine the position of the centroid of a metal template in the form of a quadrant of a circle of radius 4 cm. (The equation of a circle, centre 0, radius r is x 2 C y 2 D r 2 ). On the centre line, distance 2.40 cm from the centre, i.e. at coordinates (1.70, 1.70) 4. (a) Determine the area bounded by the curve y D 5x 2 ,thex-axis and the ordinates x D 0andx D 3. (b) If this area is revolved 360 ° about (i) the x-axis, and (ii) the y-axis, find the volumes of the solids of revolution produced in each case. (c) Determine the co-ordinates of the cen- troid of the area using (i) integral calculus, and (ii) the theorem of Pappus (a) 45 square units (b) (i) 1215 cubic units (ii) 202.5 cubic units (c) (2.25, 13.5) 5. A metal disc has a radius of 7.0 cm and is of thickness 2.5 cm. A semicircular groove of diameter 2.0 cm is machined centrally around the rim to form a pulley. Determine the volume of metal removed using Pappus’ theorem and express this as a percentage of the original volume of the disc. Find also the mass of metal removed if the density of the metal is 7800 kg m 3 . [64.90 cm 3 , 16.86%, 506.2 g] 58 Second moments of area 58.1 Second moments of area and radius of gyration The first moment of area about a fixed axis of a lamina of area A, perpendicular distance y from the centroid of the lamina is defined as Ay cubic units. The second moment of area ofthesamelamina as above is given by Ay 2 , i.e. the perpendicular distance from the centroid of the area to the fixed axis is squared. Second moments of areas are usually denoted by I and have units of mm 4 ,cm 4 , and so on. Radius of gyration Several areas, a 1 , a 2 , a 3 , at distances y 1 , y 2 , y 3 , from a fixed axis, may be replaced by a single area A,whereA D a 1 C a 2 C a 3 CÐÐÐat distance k from the axis, such that Ak 2 D ay 2 . k is called the radius of gyration of area A about the given axis. Since Ak 2 D ay 2 D I then the radius of gyration, k = I A . The second moment of area is a quantity much used in the theory of bending of beams, in the torsion of shafts, and in calculations involving water planes and centres of pressure. 58.2 Second moment of area of regular sections The procedure to determine the second moment of area of regular sections about a given axis is (i) to find the second moment of area of a typical element and (ii) to sum all such second moments of area by integrating between appropriate limits. For example, the second moment of area of the rectangle shown in Fig. 58.1 about axis PP is found by initially considering an elemental strip of width υx, parallel to and distance x from axis PP.Areaof shaded strip D bυx. Second moment of area of the shaded strip about PP D x 2 bυx. The second moment of area of the whole rectangle about PP is obtained by summing all such strips Figure 58.1 between x D 0andx D l,i.e. xDl xD0 x 2 bυx.Itisa fundamental theorem of integration that limit υx!0 xDl xD0 x 2 bυx D l 0 x 2 b dx Thus the second moment of area of the rectangle about PP D b l 0 x 2 dx D b x 3 3 l 0 D bl 3 3 Since the total area of the rectangle, A D lb,then I pp D lb l 2 3 D Al 2 3 I pp D Ak 2 pp thus k 2 pp D l 2 3 i.e. the radius of gyration about axes PP, k pp D l 2 3 D l p 3 58.3 Parallel axis theorem In Fig. 58.2, axis GG passes through the centroid C of area A.AxesDD and GG are in the same plane, are parallel to each other and distance d apart. The parallel axis theorem states: I DD = I GG Y Ad 2 Using the parallel axis theorem the second moment of area of a rectangle about an axis through the 476 ENGINEERING MATHEMATICS Figure 58.2 centroid may be determined. In the rectangle shown in Fig. 58.3, I pp D bl 3 3 (from above). From the parallel axis theorem I pp D I GG C bl l 2 2 i.e. bl 3 3 D I GG C bl 3 4 from which, I GG D bl 3 3 bl 3 4 D bl 3 12 P G C x b G P l 2 l 2 d x Figure 58.3 58.4 Perpendicular axis theorem In Fig. 58.4, axes OX, OY and OZ are mutually perpendicular. If OX and OY lie in the plane of area A then the perpendicular axis theorem states: I OZ = I OX Y I OY 58.5 Summary of derived results A summary of derive standard results for the second moment of area and radius of gyration of regular sections are listed in Table 58.1. Figure 58.4 Table 58.1 Summary of standard results of the second moments of areas of regular sections Shape Position of axis Second Radius of moment gyration, k of area, I Rectangle (1) Coinciding with b bl 3 3 l p 3 length l (2) Coinciding with l lb 3 3 b p 3 breadth b (3) Through centroid, parallel to b bl 3 l2 l p 12 (4) Through centroid, parallel to l lb 3 12 b p 12 Triangle (1) Coinciding with b bh 3 12 h p 6 Perpendicular (2) Through centroid, parallel to base bh 3 36 h p 18 height h (3) Through vertex, parallel to base bh 3 4 h p 2 base b Circle (1) Through centre, perpendicular to plane (i.e. polar axis) r 4 2 r p 2 radius r (2) Coinciding with diameter r 4 4 r 2 (3) About a tangent 5r 4 4 p 5 2 r Semicircle Coinciding with diameter r 4 8 r 2 radius r 58.6 Worked problems on second moments of area of regular sections Problem 1. Determine the second moment of area and the radius of gyration about axes AA, BB and CC for the rectangle shown in Fig. 58.5 SECOND MOMENTS OF AREA 477 A B C b = 4.0 cm A l =12.0 cm B C Figure 58.5 From Table 58.1, the second moment of area about axis AA, I AA D bl 3 3 D 4.012.0 3 3 D 2304 cm 4 Radius of gyration, k AA D l p 3 D 12.0 p 3 D 6.93 cm Similarly, I BB D lb 3 3 D 12.04.0 3 3 D 256 cm 4 and k BB D b p 3 D 4.0 p 3 D 2 .31 cm The second moment of area about the centroid of a rectangle is bl 3 12 when the axis through the centroid is parallel with the breadth b. In this case, the axis CC is parallel with the length l. Hence I CC D lb 3 12 D 12.04.0 3 12 D 64 cm 4 and k CC D b p 12 D 4.0 p 12 D 1 .15 cm Problem 2. Find the second moment of area and the radius of gyration about axis PP for the rectangle shown in Fig. 58.6 40.0 mm 15.0 mm G 25.0 mm G PP Figure 58.6 I GG D lb 3 12 where l D 40.0mmandb D 15.0mm Hence I GG D 40.015.0 3 12 D 11 250 mm 4 From the parallel axis theorem, I PP D I GG C Ad 2 , where A D 40.0 ð15.0 D 600 mm 2 and d D 25.0 C 7.5 D 32.5 mm, the perpendicular distance between GG and PP. Hence, I PP D 11 250 C60032.5 2 D 645 000 mm 4 I PP D Ak 2 PP from which, k PP D I PP area D 645 000 600 D 32 .79 mm Problem 3. Determine the second moment of area and radius of gyration about axis QQ of the triangle BCD shown in Fig. 58.7 B G G C D QQ 12.0 cm 8.0 cm 6.0 cm Figure 58.7 Using the parallel axis theorem: I QQ D I GG C Ad 2 , where I GG is the second moment of area about the centroid of the triangle, i.e. bh 3 36 D 8.012.0 3 36 D 384 cm 4 , A is the area of the triangle D 1 2 bh D 1 2 8.012.0 D 48 cm 2 and d is the distance between axes GG and QQ D 6.0 C 1 3 12.0 D 10 cm. Hence the second moment of area about axis QQ, I QQ D 384 C 4810 2 D 5184 cm 4 Radius of gyration, k QQ D I QQ area D 5184 48 D 10 .4cm 478 ENGINEERING MATHEMATICS Problem 4. Determine the second moment of area and radius of gyration of the circle shown in Fig. 58.8 about axis YY YY 3.0 cm GG r = 2.0 cm Figure 58.8 In Fig. 58.8, I GG D r 4 4 D 4 2.0 4 D 4 cm 4 . Using the parallel axis theorem, I YY D I GG C Ad 2 , where d D 3.0 C2.0 D 5.0cm. Hence I YY D 4 C [2.0 2 ]5.0 2 D 4 C 100 D 104 D 327 cm 4 Radius of gyration, k YY D I YY area D 104 2.0 2 D p 26 D 5.10 cm Problem 5. Determine the second moment of area and radius of gyration for the semicircle shown in Fig. 58.9 about axis XX G G B B XX 15.0 mm 10.0 mm Figure 58.9 The centroid of a semicircle lies at 4r 3 from its diameter. Using the parallel axis theorem: I BB D I GG C Ad 2 , where I BB D r 4 8 (from Table 58.1) D 10.0 4 8 D 3927 mm 4 , A D r 2 2 D 10.0 2 2 D 157.1mm 2 and d D 4r 3 D 410.0 3 D 4.244 mm Hence 3927 D I GG C 157.14.244 2 i.e. 3927 D I GG C 2830, from which, I GG D 3927 2830 D 1097 mm 4 Using the parallel axis theorem again: I XX D I GG C A15.0 C4.244 2 i.e. I XX D 1097 C 157.119.244 2 D 1097 C 58 179 D 59 276 mm 4 or 59 280 mm 4 , correct to 4 significant figures. Radius of gyration, k XX D I XX area D 59 276 157.1 D 19 .42 mm Problem 6. Determine the polar second moment of area of the propeller shaft cross-section shown in Fig. 58.10 7.0 cm 6.0 cm Figure 58.10 The polar second moment of area of a circle D r 4 2 The polar second moment of area of the shaded area is given by the polar second moment of area of the 7.0 cm diameter circle minus the polar second moment of area of the 6.0 cm diameter circle. Hence the polar second moment of area of the cross-section shown D 2 7.0 2 4 2 6.0 2 4 D 235.7 127.2 D 108.5cm 4 [...]... cm4 , 4. 62 cm 4 For the semicircle shown in Fig 58.15, find the second moment of area and radius of gyration about axis JJ [3927 mm4 , 5.0 mm] Figure 58.15 47 9 48 0 ENGINEERING MATHEMATICS 5 For each of the areas shown in Fig 58.16 determine the second moment of area and radius of gyration about axis LL, by using the parallel axis theorem a 335 cm4 , 4. 73 cm b 22 030 cm4 , 14. 3 cm 4 c 628... applied to matrices and the rules of matrices are such that they obey most of those governing the algebra of numbers 3 7 0 4 1C0 4C 4 −1 0 −1 0 (b) Adding the corresponding elements gives: 3 1 4 3 1 4 4 1 3 C D 3C2 4C 2 1C6 D 5 8 2 4 7 7 2 7 2 1 6 3 1C7 3C1 4C3 5 0 4 4C 5 1C0 3C4 −9 1 1 (ii) Subtraction of matrices If A is a matrix and B is another matrix, then (A B) is a single matrix formed by subtracting... 3.67 mm b 2187 mm4 , 6.36 mm c 243 mm4 , 2.12 mm Figure 58.11 E E From the perpendicular axis theorem: 9.0 cm IZZ D IXX C IYY IXX and IYY Hence IZZ lb3 40 15 D D 3 3 D 3 D 45 000 mm 4 bl3 15 40 3 D D D 320 000 mm4 3 3 D 45 000 C 320 000 3 Radius of gyration, IZZ D area D Figure 58 .13 D 365 000 mm4 or 36.5 cm4 kZZ D 12.0 cm For the circle shown in Fig 58. 14, find the second moment of area and radius... 5355 D 7. 14 cm 105 Now try the following exercise Exercise 1 94 Further problems on second moment of areas of composite areas 1 For the sections shown in Fig 58.19, find the second moment of area and the radius of gyration about axis XX a 12 190 mm4 , 10.9 mm b 549 .5 cm4 , 4. 18 cm Figure 58.21 48 1 48 2 ENGINEERING MATHEMATICS Assignment 15 500 mm This assignment covers the material in Chapters 54 to 58... cm4 , 2.0 cm b 1005 cm4 , 4. 47 cm 365 000 40 15 D 24. 7 mm or 2 .47 cm Now try the following exercise Figure 58. 14 Exercise 193 Further problems on second moments of area of regular sections 1 Determine the second moment of area and radius of gyration for the rectangle shown in Fig 58.12 about (a) axis AA (b) axis BB, and (c) axis CC (a) 72 cm4 , 1.73 cm (b) 128 cm4 , 2.31 cm (c) 512 cm4... hinge bh3 10 6.0 3 D D 60 cm4 36 36 By the parallel axis theorem, the second moment of area of the triangle about axis XX 2 D 60 C 1 10 6.0 8.0 C 1 6.0 D 3060 cm4 2 3 Total second moment of area about XX D 100.5 C 10 24 C 3060 D 41 84. 5 D 41 80 cm4 , correct to 3 significant figures [0. 245 m4 , 0.559 m] 8 A circular cover, centre 0, has a radius of 12.0 cm A hole of radius 4. 0 cm and centre X, where OX... 24 cm2 and d D 12.5 cm IXX D 18 C 24 12.5 Hence Figure 58.19 D 18 cm4 2 2 D 3768 cm4 Determine the second moments of areas about the given axes for the shapes shown in Fig 58.20 (In Fig 58.20(b), the circular area is removed.) IAA D 42 24 cm4 , IBB D 6718 cm4 , ICC D 37 300 cm4 For rectangle E: The second moment of area about CE (an axis through CE parallel to XX) D bl3 3.0 7.0 D 12 12 3 D 85.75 cm4... , 14. 3 cm 4 c 628 cm , 7.07 cm (a) (b) X cm 1.0 cm 2.0 cm X 1.0 cm 8.0 cm 2.0 cm CT T 6.0 cm T (c) 3.0 cm 15 cm 0 4 15 cm m 0c =4 Dia Figure 58.17 5.0 cm 18 cm 2.0 cm L 10 cm 5 cm L For the semicircle, IXX D Figure 58.16 r4 4. 0 4 D 8 8 D 100.5 cm4 bl3 6.0 8.0 3 D 3 3 D 10 24 cm4 6 Calculate the radius of gyration of a rectangular door 2.0 m high by 1.5 m wide about a vertical axis through its hinge... (i) Addition of matrices Corresponding elements in two matrices may be added to form a single matrix Problem 1 Add the matrices 2 1 3 0 and and (a) 7 4 7 4 3 1 4 2 7 5 (b) 4 3 1 and 2 1 0 1 4 3 6 3 4 (a) Adding the corresponding elements gives: 2 7 1 C 4 2C 3 7C7 D 6 D in matrix form The numbers within a matrix are called an array and the coefficients forming the array are called the elements of the... 1267 cm4 For rectangle F: IXX D bl3 15.0 4. 0 D 3 3 3 D 320 cm4 Total second moment of area for the I -section about axis XX , IXX D 3768 C 1267 C 320 D 5355 cm4 Total area of I-section D 8.0 3.0 C 3.0 7.0 C 15.0 4. 0 D 105 cm2 Radius of gyration, kXX D IXX D area Figure 58.20 3 Find the second moment of area and radius of gyration about the axis XX for the beam section shown in Fig 58.21 [135 1 cm4 , 5.67 . moment of area about axis QQ, I QQ D 3 84 C 48 10 2 D 51 84 cm 4 Radius of gyration, k QQ D I QQ area D 51 84 48 D 10 .4cm 47 8 ENGINEERING MATHEMATICS Problem 4. Determine the second moment of. ordinates y D 1andy D 4, correct to 3 decimal places From Section 57 .4, x D 1 2 4 1 x 2 dy 4 1 x dy D 1 2 4 1 y 2 dy 4 1 y 2 dy D 1 2 y 2 4 4 1 2y 3/2 3 p 2 4 1 D 15 8 14 3 p 2 D 0 .568 and y. I GG D r 4 4 D 4 2.0 4 D 4 cm 4 . Using the parallel axis theorem, I YY D I GG C Ad 2 , where d D 3.0 C2.0 D 5.0cm. Hence I YY D 4 C [2.0 2 ]5.0 2 D 4 C 100 D 1 04 D 327 cm 4 Radius