Higher Engineering Mathematics - BS Grewal

PG Some applications of integration 385 ll A summary of derived standard results for the second 22 moment of area and radius of gyration of regular sections are listed in Table 38.1. C b Problem 11. Determine the second moment of x area and the radius of gyration about axes AA, BB and CC for the rectangle shown in Fig. 38.18. l 5 12.0 cm A ␦x G C C P B b 5 4.0 cm Figure 38.16 Figure 38.18 B may be determined. In the rectangle shown in Fig. 38.16, A Ipp = bl3 (from above). 3 From the parallel axis theorem 1 2 From Table 38.1, the second moment of area about 2 axis AA, I pp = IGG + (bl) i.e. bl3 = IGG + bl3 IAA = bl3 = (4.0)(12.0)3 = 2304 cm4 3 4 3 3 from which, IGG = bl3 − bl3 = bl3 3 4 12 l 12.0 Perpendicular axis theorem Radius of gyration,kAA = √ = √ = 6.93 cm In Fig. 38.17, axes OX , OY and OZ are mutually per- 33 pendicular. If OX and OY lie in the plane of area A then the perpendicular axis theorem states: Similarly, IBB = lb3 = (12.0)(4.0)3 = 256 cm4 3 3 IOZ = IOX + IOY and b = √4.0 = 2.31 cm kBB = √ 33 Z The second moment of area about the centroid of a bl3 Y rectangle is when the axis through the centroid is 12 parallel with the breadth b. In this case, the axis CC is parallel with the length l. O Hence ICC = lb3 = (12.0)(4.0)3 = 64 cm4 Area A 12 12 Figure 38.17 X b 4.0 and kCC = √ = √ = 1.15 cm 12 12

386 Higher Engineering Mathematics Table 38.1 Summary of standard results of the second moments of areas of regular sections Shape Position of axis Second moment Radius of Rectangle (1) Coinciding with b of area, I gyration, k length l, breadth b (2) Coinciding with l bl3 l (3) Through centroid, parallel to b 3 √ Triangle (4) Through centroid, parallel to l 3 (1) Coinciding with b lb3 3 b bl3 √ 12 lb3 3 12 √l bh3 12 12 √b 12 h √ 6 Perpendicular height h, base b (2) Through centroid, parallel to base bh3 √h 36 18 (3) Through vertex, parallel to base bh3 4 √h Circle (1) Through centre, perpendicular to πr4 2 radius r plane (i.e. polar axis) 2 r (2) Coinciding with diameter πr4 √ 4 2 (3) About a tangent 5π r 4 r 4 2 √ 5 r 2 Semicircle Coinciding with diameter πr4 r radius r 82 Problem 12. Find the second moment of area and lb3 the radius of gyration about axis PP for the IGG = 12 where 1 = 40.0 mm and b = 15.0 mm rectangle shown in Fig. 38.19. Hence IGG = (40.0)(15.0)3 = 11250 mm4 40.0 mm 12 GG From the parallel axis theorem, IPP = IGG + Ad2 , 15.0 mm where A = 40.0 × 15.0 = 600 mm2 and d = 25.0 +7.5 = 32.5 mm, the perpendicular 25.0 mm distance between GG and PP. Hence, P P IPP = 11 250 + (600)(32.5)2 Figure 38.19 = 645000 mm4

IPP = AkP2P , from which, Some applications of integration 387 kPP = IPP = 645000 = 32.79 mm Problem 14. Determine the second moment of area 600 area and radius of gyration of the circle shown in Fig. 38.21 about axis YY . r 5 2.0 cm Problem 13. Determine the second moment of G G area and radius of gyration about axis QQ of the triangle BCD shown in Fig. 38.20. B 3.0 cm 12.0 cm G G Y Y Figure 38.21 CD 8.0 cm 6.0 cm Q Q πr4 π (2.0)4 Figure 38.20 4 4 In Fig. 38.21, IGG = = = 4π cm4. Using the parallel axis theorem: IQQ = IGG + Ad2, Using the parallel axis theorem, IYY = IGG + Ad2, where IGG is the second moment of area about the where d = 3.0 + 2.0 = 5.0 cm. centroid of the triangle, Hence IYY = 4π + [π(2.0)2](5.0)2 i.e. bh3 = (8.0)(12.0)3 = 384 cm4, = 4π + 100π = 104π = 327 cm4 36 36 Radius of gyration, A is the area of the triangle, IY Y = 104π √ area π (2.0)2 = 26 = 5.10 cm kYY = = 1 bh = 1 (8.0)(12.0) = 48 cm2 2 2 and d is the distance between axes GG and QQ, Problem 15. Determine the second moment of area and radius of gyration for the semicircle shown = 6.0 + 1 (12.0) = 10 cm. in Fig. 38.22 about axis XX . 3 Hence the second moment of area about axis QQ, IQQ = 384 + (48)(10)2 = 5184 cm4 G 10.0 mm G B B Radius of gyration, 15.0 mm kQQ = IQQ = 5184 = 10.4 cm X X area 48 Figure 38.22

388 Higher Engineering Mathematics The centroid of a semicircle lies at 4r from its The polar second moment of area of a circle= πr4 2 diameter. 3π The polar second moment of area of the shaded area is Using the parallel axis theorem: given by the polar second moment of area of the 7.0 cm diameter circle minus the polar second moment of area IBB = IGG + Ad2, of the 6.0 cm diameter circle. where πr4 Hence the polar second moment of area of the cross- IBB = 8 (from Table 38.1) section shown and Hence π 7.0 4 π 6.0 4 = π(10.0)4 = 3927 mm4, =− 8 22 22 A = πr2 = π(10.0)2 = 157.1 mm2 = 235.7 − 127.2 = 108.5 cm4 22 d = 4r = 4(10.0) = 4.244 mm Problem 17. Determine the second moment of 3π 3π area and radius of gyration of a rectangular lamina of length 40 mm and width 15 mm about an axis 3927 = IGG + (157.1)(4.244)2 through one corner, perpendicular to the plane of the lamina. i.e. 3927 = IGG + 2830, The lamina is shown in Fig. 38.24. from which, IGG = 3927 − 2830 = 1097 mm4 Using the parallel axis theorem again: Z Y l 5 40 mm b 5 15 mm IXX = IGG + A(15.0 + 4.244)2 X i.e. IXX = 1097 + (157.1)(19.244)2 X Z Y = 1097 + 58 179 = 59276 mm4 or 59280 mm4, Figure 38.24 correct to 4 significant figures. Radius of gyration, kXX = IXX = 59 276 area 157.1 = 19.42 mm From the perpendicular axis theorem: IZZ = IXX + IYY Problem 16. Determine the polar second moment IXX = lb3 = (40)(15)3 = 45000 mm4 of area of the propeller shaft cross-section shown in 3 3 Fig. 38.23. and IYY = bl3 = (15)(40)3 = 320000 mm4 3 3 Hence IZZ = 45 000 + 320 000 6.0 cm = 365000 mm4 or 36.5 cm4 7.0 cm Radius of gyration, kZZ = IZZ = 365 000 area (40)(15) Figure 38.23 = 24.7 mm or 2.47 cm

Some applications of integration 389 Problem 18. Determine correct to 3 significant Problem 19. Determine the second moment of figures, the second moment of area about axis XX area and the radius of gyration about axis XX for the for the composite area shown in Fig. 38.25. I -section shown in Fig. 38.26. 4.0 cm S 1.0 cm 8.0 cm CD 3.0 cm X X 1.0 cm 2.0 cm 3.0 cm T 8.0 cm CE 7.0 cm 2.0 cm C CT C CF 4.0 cm y 15.0 cm X T 6.0 cm X S Figure 38.26 Figure 38.25 For the semicircle, The I -section is divided into three rectangles, D, E and F and their centroids denoted by CD , CE and CF IXX = πr4 = π (4.0)4 = 100.5 cm4 respectively. 8 8 For rectangle D: For the rectangle, The second moment of area about CD (an axis through CD parallel to XX ) IXX = bl3 = (6.0)(8.0)3 = 1024 cm4 3 3 = bl3 = (8.0)(3.0)3 = 18 cm4 12 12 For the triangle, about axis TT through centroid CT , Using the parallel axis theorem: ITT = bh3 = (10)(6.0)3 = 60 cm4 36 36 IXX = 18 + Ad2 By the parallel axis theorem, the second moment of area where A = (8.0)(3.0) = 24 cm2 and d = 12.5 cm of the triangle about axis XX Hence IXX = 18 + 24(12.5)2 = 3768 cm4. = 60 + 1 (10)(6.0) 8.0 + 1 (6.0) 2 = 3060 cm4. For rectangle E: 2 3 The second moment of area about CE (an axis through CE parallel to XX ) Total second moment of area about XX = bl3 = (3.0)(7.0)3 = 85.75 cm4 = 100.5 + 1024 + 3060 12 12 = 4184.5 = 4180 cm4, correct to 3 significant figures. Using the parallel axis theorem: IXX = 85.75 + (7.0)(3.0)(7.5)2 = 1267 cm4.

390 Higher Engineering Mathematics For rectangle F: IXX = bl3 = (15.0)(4.0)3 = 320 cm4 EE 3 3 9.0 cm Total second moment of area for the I-section about axis XX, IXX = 3768 + 1267 + 320 = 5355 cm4 Total area of I -section D D Figure 38.28 12.0 cm = (8.0)(3.0) + (3.0)(7.0) + (15.0)(4.0) = 105 cm2. 3. For the circle shown in Fig. 38.29, find the second moment of area and radius of gyration Radius of gyration, about (a) axis FF and (b) axis HH . kXX = IXX = 5355 = 7.14 cm (a) 201 cm4, 2.0 cm area 105 (b) 1005 cm4, 4.47 cm Now try the following exercise H Exercise 152 Further problems on second H r 5 4.0 cm moment of areas of regular sections F 1. Determine the second moment of area and F radius of gyration for the rectangle shown in Fig. 38.27 about (a) axis AA (b) axis BB and (c) axis CC. ⎡ ⎤ Figure 38.29 (a) 72 cm4 , 1.73 cm 4. For the semicircle shown in Fig. 38.30, find the second moment of area and radius of gyration ⎣(b) 128 cm4, 2.31 cm⎦ about axis J J . [3927 mm4, 5.0 mm] (c) 512 cm4, 4.62 cm B C 8.0 cm AA r 5 10.0 mm 3.0 cm J J Figure 38.30 BC Figure 38.27 2. Determine the second moment of area and 5. For each of the areas shown in Fig. 38.31 deter- radius of gyration for the triangle shown in mine the second moment of area and radius of Fig. 38.28 about (a) axis DD (b) axis EE and gyration about axis LL, by using the parallel (c) an axis through the centroid of the triangle axis theorem. ⎡⎤ (a) 335 cm4, 4.73 cm parallel to axis DD.⎡ ⎤ (a) ⎢⎣(b) 22030 cm4, 14.3 cm⎦⎥ 729 cm4, 3.67 cm ⎣(b) 2187 cm4, 6.36 cm⎦ (c) 628 cm4, 7.07 cm (c) 243 cm4, 2.12 cm

Some applications of integration 391 3.0 cm Dia 5 4.0 cm 10. Determine the second moments of areas about 5.0 cm 15 cm 15 cm 5.0 cm the given axes for the shapes shown in L Fig. 38.33. (In Fig. 38.33(b), the circular area (c) 18 cm 10 cm is removed.) ⎡ ⎤ (b) I AA 2.0 cm = 4224 cm4, L ⎣IBB = 6718 cm4, ⎦ (a) ICC = 37300 cm4 Figure 38.31 3.0 cm 6. Calculate the radius of gyration of a rectan- B gular door 2.0 m high by 1.5 m wide about a 4.5 cm vertical axis through its hinge. [0.866 m] 16.0 cm 9.0 cm 7. A circular door of a boiler is hinged so that A 9.0 cm Dia 5 7.0 cm it turns about a tangent. If its diameter is (a) 1.0 m, determine its second moment of area 4.0 cm 15.0 cm and radius of gyration about the hinge. Figure 38.33 A [0.245 m4, 0.559 m] C C 8. A circular cover, centre 0, has a radius of 10.0 cm 12.0 cm. A hole of radius 4.0 cm and centre X , where OX = 6.0 cm, is cut in the cover. Deter- B (b) mine the second moment of area and the radius of gyration of the remainder about a diameter 11. Find the second moment of area and radius through 0 perpendicular to OX . [14280 cm4, 5.96 cm] of gyration about the axis XX for the beam 9. For the sections shown in Fig. 38.32, find section shown in Fig. 38.34. the second moment of area and the radius of gyration about axis XX . 1350 cm4, (a) 12190 mm4, 10.9 mm 5.67 cm (b) 549.5 cm4, 4.18 cm 6.0 cm 1.0 cm 18.0 mm 6.0 cm 2.0 cm 8.0 cm 3.0 mm 2.0 cm 12.0 mm 2.5 cm 3.0 cm X 2.0 cm 10.0 cm X X X 2.0 cm 4.0 mm X X Figure 38.34 (a) (b) Figure 38.32