Problem 8.1 Use the method described in Active Example 8.1 to determine IY and ky for the rectangular area. Solution: The height of the vertical strip of width dx is 0.6 m, so the area is dA D (0.6 m) dx. We can use this expression to determine Iy . y x 2 dx D (0.6 m) Iy D 0.4 m x 2 dx 0.2m A D (0.6 m) x3 3 0.4 m D 0.0416 m4 0.2 m The radius of gyration about the y axis is 0.6 m ky D Iy D A 0.0416 m4 D 0.416 m (0.4 m) (0.6 m) x 0.2 m Iy D 0.0416 m4 , ky D 0.416 m 0.4 m Problem 8.2 Use the method described in Active Example 8.1 to determine Ix and kx for the rectangular area. y Solution: It was shown in Active Example 8.1 that the moment of inertia about the x axis of a vertical strip of width dx and height f(x) is 1 Ix strip D [fx]3 dx. 3 For the rectangular strip, fx D 0.6 m. Integrating to determine Ix for the rectangular area., 0.4m Ix D 0.2m 1 1 4 0.6 m3 dx D 0.6 m3 [x]0.4m 0.2m D 0.0288 m 3 3 The radius of gyration about the x axis is 0.6 m kx D Ix D A 0.0288 m4 D 0.346 m (0.4 m)(0.6 m) x 0.2 m Ix D 0.0288 m4 , ky D 0.346 m 0.4 m Problem 8.3 In Active Example 8.1, suppose that the triangular area is reoriented as shown. Use integration to determine Iy and ky . Solution: The height of a vertical strip of width dx is h h/bx, so the area dA D h h x b y h dx. We can use this expression to determine Iy : b x 2 dA D Iy D x2 h 0 A h x b dx D h x3 3 x4 4b x b D 0 1 3 hb 12 b The radius of gyration about the y axis is ky D Iy D Iy D A 1 3 hb b 12 D p 1 6 hb 2 hb3 b , ky D p . 12 6 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 601 Problem 8.4 (a) Determine the moment of inertia Iy of the beam’s rectangular cross section about the y axis. Solution: (b) Determine the moment of inertia Iy 0 of the beam’s cross section about the y 0 axis. Using your numerical values, show that Iy D Iy 0 C d2x A, where A is the area of the cross section. (a) 40 mm 0 x 2 dydx D 1.28 ð 106 mm4 0 (b) 60 mm Iy D 20 mm 30 mm Iy 0 D 20 mm x 2 dydx D 3.2 ð 105 mm4 30 mm y Iy D Iy 0 C dx 2 A y dx 1.28 ð 106 mm4 D 3.2 ð 105 mm4 C 20 mm2 [40 mm60 mm] 60 mm x O dy x O 40 mm Problem 8.5 (a) Determine the polar moment of inertia JO of the beam’s rectangular cross section about the origin O. (b) Determine the polar moment of inertia JO0 of the beam’s cross section about the origin O0 . Using your numerical values, show that JO D JO0 C d2x C d2y A, where A is the area of the cross section. Solution: (a) 40 mm 0 (b) 60 mm JO D x 2 C y 2 dydx D 4.16 ð 106 mm4 0 20 mm 30 mm JO 0 D 20 mm x 2 C y 2 dydx D 1.04 ð 106 mm4 30 mm JO D JO0 C dx 2 C dy 2 A (c) 4.16 ð 106 mm4 D 1.04 ð 106 mm4 C [20 mm2 C 30 mm2 ][40 mm60 mm] Problem 8.6 Determine Iy and ky . y Solution: A D 0.3 m1 m C 1 m 0.3 mC0.3x Iy D 0 0.6 m 0.3 m ky D x 1 0.3 m1 m D 0.45 m2 2 0 Iy D A x 2 dydx D 0.175 m4 0.175 m4 D 0.624 m 0.45 m2 1m 602 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.7 Determine JO and kO . Solution: A D 0.3 m1 m C 1m 0.3 mC0.3x JO D 0 1 0.3 m1 m D 0.45 m2 2 x 2 C y 2 dydx D 0.209 m4 0 0.209 m4 D 0.681 m 0.45 m2 kO D Problem 8.8 Determine Ixy . Solution: 1m 0.3 mC0.3x Ixy D 0 xydydx D 0.0638 m4 0 Problem 8.9 Determine Iy . y Solution: The height of a vertical strip of width dx is 2 x2 , so y 2 x2 the area is dA D 2 x2 dx. We can use this expression to determine Iy : x 2 dA D Iy D 1 x 2 2 x2 dx D 0 A 2x3 x5 3 5 1 0 D 0.467. Iy D 0.467. 1 Problem 8.10 Determine Ix . Solution: It was shown in Active Example 8.1 that the moment of inertia about the x axis of a vertical strip of width dx and height fx is 1 Ix strip D [fx]3 dx. 3 x y y 2 x2 In this problem fx D 2 x3 . Integrating to determine Ix for the area, 1 Ix D 0 D D 1 3 1 3 1 2 x2 3 dx 3 1 8 12x2 C 6x4 x 6 dx 0 1 12x3 6x5 x7 8x C D 1.69. 3 5 7 0 1 x Ix D 1.69. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 603 Problem 8.11 Determine JO . y y 2 x2 Solution: See the solutions to Problems 8.9 and 8.10. The polar moment of inertia is JO D Ix C Iy D 1.69 C 0.467 D 2.15. JO D 2.15. 1 Problem 8.12 Determine Ixy . x y y 2 x2 Solution: It was shown in Active Example 8.1 that the product of inertia of a vertical strip of width dx and height f(x) is Ixy strip D 1 [fx]2 xdx. 2 In this problem, fx D 2 x2 . Integrating to determine Ixy for the area, 1 1 2 x2 2 xdx 2 Ixy D 0 D D 1 2 1 4x 4x3 C x 5 x 0 1 4x4 x6 1 4x2 C D 0.583. 2 2 4 6 0 1 x Ixy D 0.583. Problem 8.13 Determine Iy and ky . y y 1 2 x 4x 7 4 Solution: First we need to locate the points where the curve intersects the x axis. 4 š 1 x 2 C 4x 7 D 0 ) x D 4 Now 14 x x2 /4C4x7 dydx D 72 AD 2 0 14 x2 /4C4x7 Iy D 2 ky D 604 p 16 41/47 D 2,14 21/4 0 Iy D A x 2 dydx D 5126 5126 D 8.44 72 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.14 Determine Ix and kx . Solution: See Solution to Problem 8.13 14 x2 /4C4x7 Ix D 2 0 Ix D A kx D y 2 dydx D 1333 1333 D 4.30 72 Problem 8.15 Determine JO and kO . Solution: See Solution to 8.13 and 8.14 JO D Ix C Iy D 1333 C 5126 D 6459 JO 6459 D D 9.47 kO D A 72 Problem 8.16 Determine Ixy . Solution: 14 x2 /4C4x7 xydydx D 2074 Ixy D 2 0 Problem 8.17 Determine Iy and ky . y y 1 2 x 4x 7 4 y5 Solution: First we need to locate the points where the curve intersects the line. 4 š 1 2 x C 4x 7 D 5 ) x D 4 12 p 16 41/412 D 4,12 21/4 x x2 /4C4x7 dydx D 21.33 AD 4 5 12 x2 /4C4x7 Iy D 4 ky D 5 Iy D A x 2 dydx D 1434 1434 D 8.20 21.33 Problem 8.18 Determine Ix and kx . Solution: See Solution to Problem 8.17 12 x2 /4C4x7 Ix D 4 kx D 5 Ix D A y 2 dydx D 953 953 D 6.68 21.33 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 605 Problem 8.19 (a) Determine Iy and ky by letting dA be a vertical strip of width dx. (b) The polar moment of inertia of a circular area with its center at the origin is JO D 12 R4 . Explain how you can use this information to confirm your answer to (a). y x R Solution: The equation of the circle is x2 C y 2 D R2 , from which p y D š pR2 x 2 . The strip dx wide and y long has the elemental area dA D 2 R2 x 2 dx. The area of the semicircle is AD R2 Iy D 2 R x 2 dA D 2 x2 R2 x 2 dx 0 A R R2 xR2 x 2 1/2 R4 x xR2 x 2 3/2 C C sin1 D2 4 8 8 R 0 (b) If the integration were done for a circular area with the center at the origin, the limits of integration for the variable x would be from R to R, doubling the result. Hence, doubling the answer above, Iy D R4 . 4 By symmetry, Ix D Iy , and the polar moment would be JO D 2Iy D D R4 8 ky D R4 , 2 which is indeed the case. Also, since kx D ky by symmetry for the full circular area, R Iy D A 2 kO D Iy Ix C D A A 2 Iy D A JO A as required by the definition. Thus the result checks. Problem 8.20 (a) Determine Ix and kx for the area in Problem 8.19 by letting dA be a horizontal strip of height dy. (b) The polar moment of inertia of a circular area with its center at the origin is JO D 12 R4 . Explain how you can use this information to confirm your answer to (a). Solution: Use the results of the solution to Problem 8.19, A D R2 . The equation for the circle is x2 C y 2 D R2 , from which x D 2 š R2 y 2 . The horizontal strip is from 0 to R, hence the element of area is CR y 2 dA D Ix D A y2 D kx D 606 and JO D 2Ix D R2 y 2 dy R R4 R4 C D 8 2 8 2 R4 . 4 R y R2 yR2 y 2 1/2 R4 yR2 y 2 3/2 C C sin1 D 4 8 8 R R Ix D By symmetry Iy D Ix , R2 y 2 dy. dA D (b) If the area were circular, the strip would be twice as long, and the moment of inertia would be doubled: R4 R4 , 2 which is indeed the result. Since kx D ky by symmetry for the full circular area, the kO D Iy Ix C D A A 2 Ix D A JO A 8 as required by the definition. This checks the answer. R Ix D . A 2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.21 Use the procedure described in Example 8.2 to determine the moment of inertia Ix and Iy for the annular ring. y Ro x Ri Solution: We first determine the polar moment of inertia JO by integrating in terms of polar coordinates. Because of symmetry and the relation JO D Ix C Iy , we know that Ix and Iy each equal 12 JO . Integrating as in Example 8.2, the polar moment of inertia for the annular ring is r 2 dA D JO D A Therefore Ix D Iy D Ro r 2 2rdr D Ri 1 Ro4 Ri4 2 1 Ro4 Ri4 4 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 607 y Problem 8.22 What are the values of Iy and ky for the elliptical area of the airplane’s wing? y2 x2 21 b a2 Solution: a x 2 dA D Iy D A 0 a y y x 2 dy dx y 0 a Iy D 2 0 5m x2 b1 2 1/2 a [x2 y]0 y dx 0 1/2 x2 x2 b 1 2 dx a a x 2m x 2 dy dx Iy D 2 0 a Iy D 2 x2 + y2 = 1 a2 b2 2b x x2 Iy D 2b 1 0 x2 a2 dx a Rewriting Iy D 2b a a x2 y = b 1– x2 a2 a2 x 2 dx 0 p xa2 x 2 3/2 a2 x a2 x 2 2b Iy D C a 4 8 x a4 C sin1 8 a a 0 D a D 0 p 0 2b aa2 a2 3/2 a3 a2 a2 Iy D C a 4 8 a a4 sin1 8 a x2 b 1 dx a 2b a a a2 x 2 1/2 dx 0 p a x x a2 x 2 a2 C sin1 2 2 a 0 (from the integral tables) C AD2 2b a 0 p a2 a 2b a 0 C sin1 D a 2 2 a 0 0a2 3/2 p a2 0 0 a C sin1 2 2 a 4 0p a2 Ð 0 a2 a4 01 0 C C sin 8 8 a AD ab 2b a2 D a 2 2 2 Evaluating, we get A D 7.85 m2 Iy D a4 2b a 8 2 Finally Iy D A 2a3 b Iy D 8 ky D Evaluating, we get ky D 2.5 m 49.09 7.85 Iy D 49.09 m4 The area of the ellipse (half ellipse) is a x2 b 1 a AD2 0 608 1/2 dy dx 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.23 What are the values of Ix and kx for the elliptical area of the airplane’s wing in Problem 8.22? Solution: y y 2 dA D 2 Ix D a Ix D 2 0 b 3 a y 3 p b yD a y2 x2 — + —2 = 1 a2 b p a2 x2 y 2 dy dx 0 0 A 2b a y = b a2 – x2 a dx 0 b3 2 a x 2 3/2 dx 3a3 Ix D 2 0 2b3 3a3 p 3a2 x a2 x 2 xa2 x 2 3/2 C 4 8 3 4 1 x a sin 8 a C 2b3 Ix D 3 3a 3 a4 8 2 Ix D 2b3 Ð 3a3 Ix D 3ab3 ab3 D 3.8 8 a Evaluating (a D 5, b D 1) 0 p 3a3 0 3 a0 C C a4 4 8 8 2 Ix D p 3a2 Ð 0 a2 0a2 C0 4 8 5 D 1.96 m4 8 From Problem 8.22, the area of the wing is A D 7.85 m2 x a2 x2 a Ix D a kx D Ix D A 1.96 7.85 Problem 8.24 Determine Iy and ky . kx D 0.500 m y Solution: The straight line and curve intersect where x D x2 20. Solving this equation for x, we obtain xD 1š y = x 2 – 20 p 1 C 80 D 4, 5. 2 y=x If we use a vertical strip: the area x dA D [x x2 20] dx. Therefore A D 5 x 2 dA D Iy D x 2 x x 2 C 20 dx 4 x4 x5 20x3 C 4 5 3 5 D 522. The area is y = x2 – 20 A ky D y=x x x 2 C 20 dx x 4 x2 x3 C 20x 2 3 So 5 dA D AD D y 4 Iy D A dA 5 D 122. 4 dx 522 D 2.07. 122 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 609 Problem 8.25 Determine Ix and kx for the area in Problem 8.24. Solution: Let us determine the moment of inertia about the x axis of a vertical strip holding x and dx fixed: y 2 dAs D Ix strip D As x y y 2 dx dy D dx x2 20 y3 3 x y=x y= x2 20 x2 – 20 dAs x dx x 6 C 60x4 C x 3 1200x2 C 8000. 3 D Integrating this value from x D 4 to x D 5 (see the solution to Problem 8.24), we obtain Ix for the entire area: 5 x 1 x 6 C 60x4 C x 3 1200x2 C 8000 dx 3 Ix D 4 dx 7 5 x x4 400x3 8000x D D 10,900. C 4x5 C C 21 12 3 3 4 From the solution to Problem 8.24, A D 122 so kx D Ix D A 10,900 D 9.45. 122 Problem 8.26 A vertical plate of area A is beneath the surface of a stationary body of water. The pressure of the water subjects each element dA of the surface of the plate to a force p0 C y dA, where p0 is the pressure at the surface of the water and is the weight density of the water. Show that the magnitude of the moment about the x axis due to the pressure on the front face of the plate is Mx x A y D p0 yA C Ix , axis where y is the y coordinate of the centroid of A and Ix is the moment of inertia of A about the x axis. Solution: The moment about the x axis is dM D yp0 C y dA integrating over the surface of the plate: p0 C yy dA. MD A Noting that p0 and are constants over the area, y dA C M D p0 y 2 dA. A By definition, y dA A yD A y 2 dA, and Ix D A then M D p0 yA C IX , which demonstrates the result. 610 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.27 Using the procedure described in Active Example 8.3, determine Ix and kx for the composite area by dividing it into rectangles 1 and 2 as shown. Solution: Using results from Appendix B and applying the parallel-axis theorem, the moment on inertia about the x axis for area 1 is Ix 1 D Ix C d2y A D y 1m 1 4m 2 1m 1 (1 m)(3 m)3 C (2.5 m)[(1 m)(3 m)] 12 x 3m D 21.0 m4 The moment of inertia about the x axis for area 2 is Ix 2 D 1 (3 m)(1 m)3 D 1 m4 . 3 The moment of inertia about the x axis for the composite area is Ix D Ix 1 C Ix 2 D 22.0 m4 . The radius of gyration about the x axis is kx D Ix D A 22.0 m4 D 1.91 m 6 m2 Ix D 22.0 m4 , kx D 1.91 m. Problem 8.28 Determine Iy and ky for the composite area by dividing it into rectangles 1 and 2 as shown. Solution: Using results from Appendix B, the moment of inertia about the y axis for area 1 is Iy 1 D 1m 1 4m 1 (3 m)(1 m)3 D 1 m4 . 3 The moment of inertia about the y axis for area 2 is Iy 2 D y 2 1m 3m x 1 (1 m)(3 m)3 D 9 m4 . 3 The moment of inertia about the y axis for the composite area is Iy D Iy 1 C Iy 2 D 10 m4 . The radius of gyration about the y axis is ky D Iy D A 10 m4 D 1.29 m 6 m2 Iy D 10 m4 , kx D 1.29 m. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 611 Problem 8.29 Determine Ix and kx . y Solution: Break into 3 rectangles Ix D 1 1 0.60.23 C 0.20.63 C 0.20.60.52 3 12 0.8 m 1 0.80.23 C 0.80.20.92 D 0.1653 m4 12 C 0.2 m A D 0.20.6 C 0.60.2 C 0.80.2 D 0.4 m2 0.6 m kx D Ix D A 0.1653 m4 D 0.643 m 0.4 m2 0.2 m x Ix D 0.1653 m4 ) 0.2 m kx D 0.643 m 0.6 m Problem 8.30 In Example 8.4, determine Ix and kx for the composite area. y Solution: The area is divided into a rectangular area without the 20 mm cutout (part 1), a semicircular areas without the cutout (part 2), and the circular cutout (part 3). x Using the results from Appendix B, the moment of inertia of part 1 about the x axis is Ix 1 D 1 120 mm80 mm3 D 5.12 ð 106 mm4 , 12 40 mm 120 mm the moment of inertia of part 2 is Ix 2 D 1 40 mm4 D 1.01 ð 106 mm4 , 8 and the moment of inertia of part 3 is Ix 3 D 1 20 mm4 D 1.26 ð 105 mm4 . 4 The moment of inertia of the composite area is Ix D Ix 1 C Ix 2 C Ix 3 D 6.00 ð 106 mm4 . From Example 8.4, the composite area is A D 1.086 ð 104 mm4 , so the radius of gyration about the x axis is kx D Ix D A 6.00 ð 106 mm4 D 23.5 mm. 1.086 ð 104 mm2 Ix D 6.00 ð 106 mm4 , kx D 23.5 mm. 612 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.31 Determine Ix and kx . y 0.8 m 0.2 m x 0.6 m 0.2 m 0.2 m 0.6 m Solution: Break into 3 rectangles — See 8.29 First locate the centroid dD 0.60.20.1 C 0.20.60.5 C 0.80.20.9 D 0.54 m 0.60.2 C 0.20.6 C 0.80.2 1 0.20.63 C 0.20.6d 0.52 12 1 0.60.23 C 0.60.2d 0.12 C 12 1 0.80.23 C 0.80.20.9 d2 D 0.0487 m4 C 12 Ix 0.0487 m4 D kx D D 0.349 m A 0.4 m2 Ix D d Problem 8.32 Determine Iy and ky . Solution: Break into 3 rectangles — See 8.29 Iy D 1 1 1 0.60.23 C 0.20.63 C 0.20.83 12 12 12 D 0.01253 m4 Iy 0.01253 m4 ky D D 0.1770 m D A 0.4 m2 Problem 8.33 Determine JO and kO . Solution: See 8.29, 8.31 and 8.32 JO D Ix C Iy D 0.0612 m4 JO 0.0612 m4 D KO D D 0.391 m A 0.4 m2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 613 y Problem 8.34 If you design the beam cross section so that Ix D 6.4 ð 105 mm4 , what are the resulting values of Iy and JO ? h Solution: The area moment of inertia for a triangle about the x base is Ix D 1 12 bh3 , h from which Ix D 2 1 12 60h3 D 10h3 mm4 , 30 mm 30 mm Ix D 10h3 D 6.4 ð 105 mm4 , from which h D 40 mm. Iy D 2 1 12 2h303 D from which Iy D 1 h303 3 1 40303 D 3.6 ð 105 mm4 3 and JO D Ix C Iy D 3.6 ð 105 C 6.4 ð 105 D 1 ð 106 mm4 614 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 8.35 Determine Iy and ky . 160 mm Solution: Divide the area into three parts: Part (1): The top rectangle. A1 D 16040 D 6.4 ð 103 mm2 , dx1 160 D 80 mm, D 2 Iyy1 D 1 12 40 mm 200 mm 40 mm 40 mm x 120 mm 401603 D 1.3653 ð 107 mm4 . From which Iy1 D d2x1 A1 C Iyy1 D 5.4613 ð 107 mm4 . Part (2): The middle rectangle: A2 D 200 8040 D 4.8 ð 103 mm2 , dx2 D 20 mm, Iyy2 D 1 12 120403 D 6.4 ð 105 mm4 . From which, Iy2 D d2x2 A2 C Iyy2 D 2.56 ð 106 mm4 . Part (3) The bottom rectangle: A3 D 12040 D 4.8 ð 103 mm2 , dx3 D 120 D 60 mm, 2 Iyy3 D 1 12 401203 D 5.76 ð 106 mm4 From which Iy3 D d2X3 A3 C Iyy3 D 2.304 ð 107 mm4 The composite: Iy D Iy1 C Iy2 C Iy3 D 8.0213 ð 107 mm4 ky D Iy D 70.8 mm. A1 C A2 C A3 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 615 Problem 8.36 Determine Ix and kx . Solution: Use the solution to Problem 8.35. Divide the area into from which three parts: Part (1): The top rectangle. A1 D 6.4 ð 103 mm2 , Ix2 D d2y2 A2 C Ixx2 D 5.376 ð 107 mm4 Part (3) The bottom rectangle: A3 D 4.8 ð 103 mm2 , dy1 D 200 20 D 180 mm, Ixx1 D 1 12 dy3 D 20 mm, 160403 D 8.533 ð 105 mm4 . Ixx3 D From which Ix1 D d2y1 A1 C Ixx1 D 2.082 ð 108 mm4 Part (2): The middle rectangle: A2 D 4.8 ð 103 mm2 , dy2 120 C 40 D 100 mm, D 2 Ixx2 D 1 12 1 12 120403 D 6.4 ð 105 mm4 and Ix3 D d2y3 A3 C Ixx3 D 2.56 ð 106 mm4 . The composite: Ix D Ix1 C Ix2 C Ix3 D 2.645 ð 108 mm4 kx D Ix D 128.6 mm A1 C A2 C A3 401203 D 5.76 ð 106 mm4 Problem 8.37 Determine Ixy . Solution: (See figure in Problem 8.35). Use the solutions in from which Problems 8.35 and 8.36. Divide the area into three parts: Part (1): A1 D 16040 D 6.4 ð 103 mm2 , dx1 160 D 80 mm, D 2 dy1 D 200 20 D 180 mm, Ixy2 D dx2 dy2 A2 D 9.6 ð 106 mm4 . Part (3): A3 D 12040 D 4.8 ð 103 mm2 , dx3 D 120 D 60 mm, 2 dy3 D 20 mm, Ixxyy1 D 0, from which from which Ixy1 D dx1 dy1 A1 C Ixxyy1 D 9.216 ð 107 mm4 . Ixy3 D dx3 dy3 A3 D 5.76 ð 106 . The composite: Part (2) A2 D 200 8040 D 4.8 ð 103 mm2 , Ixy D Ixy1 C Ixy2 C Ixy3 D 1.0752 ð 108 mm4 dx2 D 20 mm, dy2 D 616 120 C 40 D 100 mm, 2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.38 Determine Ix and kx . y Solution: The strategy is to use the relationship Ix D d2 A C Ixc , 160 mm where Ixc is the area moment of inertia about the centroid. From this Ixc D d2 A C Ix . Use the solutions to Problems 8.35, 8.36, and 8.37. Divide the area into three parts and locate the centroid relative to the coordinate system in the Problems 8.35, 8.36, and 8.37. 40 mm x 200 mm 40 mm Part (1) A1 D 6.4 ð 103 mm2 , 40 mm 120 mm dy1 D 200 20 D 180 mm. Part (2) A2 D 200 8040 D 4.8 ð 103 mm2 , dx1 D 160 D 80 mm, 2 dy2 D 120 C 40 D 100 mm, 2 dx2 D 20 mm, Part (3) A3 D 12040 D 4.8 ð 103 mm2 , dx3 D 120 D 60 mm, 2 dy3 D 20 mm. The centroid coordinates are xD A1 dx1 C A2 dx2 C A3 dx3 D 56 mm, A yD A1 dy1 C A2 dy2 C A3 dy3 D 108 mm A from which Ixc D y2 A C Ix D 1.866 ð 108 C 2.645 ð 108 D 7.788 ð 107 mm4 The total area is A D A1 C A2 C A3 D 1.6 ð 104 mm2 . kxc D Ixc D 69.77 mm A Problem 8.39 Determine Iy and ky . Solution: The strategy is to use the relationship Iy D d2 A C Iyc , where Iyc is the area moment of inertia about the centroid. From this Iyc D d2 A C Iy . Use the solution to Problem 8.38. The centroid coordinates are x D 56 mm, y D 108 mm, from which Iyc D x2 A C Iy D 5.0176 ð 107 C 8.0213 ð 107 D 3.0 ð 107 mm4 , kyc D Iyc D 43.33 mm A Problem 8.40 Determine Ixy . Solution: Use the solution to Problem 8.37. The centroid coordinates are x D 56 mm, y D 108 mm, from which Ixyc D xyA C Ixy D 9.6768 ð 107 C 1.0752 ð 108 D 1.0752 ð 107 mm4 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 617 Problem 8.41 Determine Ix and kx . Solution: Divide the area into two parts: y 3 ft 4 ft Part (1): a triangle and Part (2): a rectangle. The area moment of inertia for a triangle about the base is Ix D 1 12 3 ft x bh3 . The area moment of inertia about the base for a rectangle is Ix D 1 bh3 . 3 Part (1) Ix1 D Part (2) Ix2 D 1 12 433 D 9 ft2 . 1 333 D 27. 3 The composite: Ix D Ix1 C Ix2 D 36 ft4 . The area: AD 1 43 C 33 D 15 ft2 . 2 kx D Ix D 1.549 ft. A Problem 8.42 Determine JO and kO . Solution: (See Figure in Problem 8.41.) Use the solution to from which Problem 8.41. Part (1): The area moment of inertia about the centroidal axis parallel to the base for a triangle is Iyc D 1 36 bh3 D 1 36 343 D 5.3333 ft4 , from which Iy1 2 8 D A1 C Iyc D 48 ft4 . 3 Iy2 D 5.52 A2 C Iyc D 279 ft4 , where A2 D 9 ft2 . The composite: Iy D Iy1 C Iy2 D 327 ft4 , from which, using a result from Problem 8.41, JO D Ix C Iy D 327 C 36 D 363 ft4 and kO D JO D 4.92 ft A where A1 D 6 ft2 . Part (2): The area moment of inertia about a centroid parallel to the base for a rectangle is Iyc D 618 1 12 bh3 D 1 12 333 D 6.75 ft4 , c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.43 Determine Ixy . Solution: (See Figure in Problem 8.41.) Use the results of the solutions to Problems 8.41 and 8.42. The area cross product of the moment of inertia about centroidal axes parallel to the bases for a 1 2 2 b h , and for a rectangle it is zero. Therefore: triangle is Ix0 y 0 D 72 Ixy1 D 1 72 42 32 C 8 3 A1 D 18 ft4 3 3 and Ixy2 D 1.55.5A2 D 74.25 ft4 , Ixy D Ix0 y 0 1 C Ixy2 D 92.25 ft4 y Problem 8.44 Determine Ix and kx . Solution: Use the results of Problems 8.41, 8.42, and 8.43. The strategy is to use the parallel axis theorem and solve for the area moment of inertia about the centroidal axis. The centroidal coordinate yD A1 1 C A2 1.5 D 1.3 ft. A 4 ft 3 ft 3 ft x From which Ixc D y2 A C Ix D 10.65 ft4 and kxc D Ixc D 0.843 ft A Problem 8.45 Determine JO and kO . Solution: Use the results of Problems 8.41, 8.42, and 8.43. The strategy is to use the parallel axis theorem and solve for the area moment of inertia about the centroidal axis. The centroidal coordinate: A1 xD 8 C A2 5.5 3 D 4.3667 ft, A from which IYC D x2 A C IY D 40.98 ft4 . Using a result from Problem 8.44, JO D IXC C IYC D 10.65 C 40.98 D 51.63 ft4 and kO D JO D 1.855 ft A Problem 8.46 Determine Ixy . Solution: Use the results of Problems 8.41–8.45. The strategy is to use the parallel axis theorem and solve for the area moment of inertia about the centroidal axis. Using the centroidal coordinates determined in Problems 8.44 and 8.45, Ixy D xyA C Ixy D 85.15 C 92.25 D 7.1 ft4 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 619 Problem 8.47 Determine Ix and kx . y 120 mm 20 mm 80 mm x 40 mm 80 mm Solution: Let Part 1 be the entire rectangular solid without the y hole and let part 2 be the hole. m 20 m Area D hb R2 D 80120 40 mm R2 y′ Ix1 D 13 bh3 where b D 80 mm h D 120 mm Ix1 D 13 801203 D 4.608 ð 107 mm4 40 mm 120 mm Part 2 x′ For Part 2, Ix0 2 D 14 R4 D 14 204 mm4 Ix0 2 D 1.257 ð 105 dy = 80 mm mm4 Ix2 D Ix0 2 C d2y A Part 1 x where A D R2 D 1257 mm2 80 mm d D 80 mm Ix2 D 1.257 ð 105 C 202 802 Ix2 D 0.126 ð 106 C 8.042 ð 106 mm4 D 8.168 ð 106 mm4 D 0.817 ð 107 mm4 Ix D Ix1 Ix2 D 3.79 ð 107 mm4 Area D 8343 mm2 kx D 620 Ix D 67.4 mm Area c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.48 Determine JO and kO . Solution: For the rectangle, y 40 mm JO1 D Ix1 C Iy1 D 13 bh3 C 13 hb3 JO1 D 4.608 ð 107 C 2.048 ð 107 mm4 y′ R = 20 mm JO1 D 6.656 ð 107 mm4 x′ A1 D bh D 9600 mm2 120 mm A2 For the circular cutout about x0 y 0 J0O2 D Ix0 2 C Iy 0 D 14 R4 C 14 R4 2 (h) 80 mm A1 J0O2 D 1.257 ð 105 C 1.257 ð 105 mm4 J0O2 D 2.513 ð 105 mm2 x Using the parallel axis theorem to determine JO2 (about x, y) 80 mm (b) JO2 D J002 C d2x C d2y A2 A2 D R2 D 1257 mm2 JO2 D 1.030 ð 107 mm4 JO D 5.63 ð 107 mm4 kO D JO D JO1 JO2 JO D Area JO A1 A2 kO D 82.1 mm JO D 6.656 ð 107 1.030 ð 107 mm4 Problem 8.49 Determine Ixy . Solution: A1 D 80120 D 9600 y 80 mm mm2 A2 D R2 D 202 D 1257 mm2 R = 20 mm y′ For the rectangle A1 x′ Ixy1 D 14 b2 h2 D 14 802 1202 Ixy1 D 2.304 ð 107 mm2 For the cutout A2 A1 120 mm A2 dy = 80 mm A1 Ix 0 y 0 2 D 0 and by the parallel axis theorem x dx = 40 mm Ixy2 D Ix0 y 0 2 C A2 dx dy Ixy2 D 0 C 12574080 Ixy2 D 4.021 ð 106 mm4 Ixy D Ixy1 Ixy2 Ixy D 2.304 ð 107 0.402 ð 107 mm4 Ixy D 1.90 ð 107 mm4 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 621 y Problem 8.50 Determine Ix and kx . 20 mm 120 mm x 80 mm 40 mm 80 mm Solution: We must first find the location of the centroid of the total y area. Let us use the coordinates XY to do this. Let A1 be the rectangle and A2 be the circular cutout. Note that by symmetry Xc D 40 mm Y 80 mm Rectangle1 Circle2 Area Xc Yc 9600 mm2 40 mm 60 mm mm2 40 mm 80 mm 1257 R = 20 mm 120 mm A1 D 9600 mm2 A2 D 1257 mm2 x 80 mm For the composite, Xc D A1 Xc1 A2 Xc2 D 40 mm A1 A2 Yc D A1 Yc1 A2 Yc2 D 57.0 mm A1 A2 Now let us determine Ix and kx about the centroid of the composite body. Rectangle about its centroid (40, 60) mm Ix1 D 1 3 1 bh D 801203 12 12 Ix1 D 1.152 ð 107 mm3 , X 40 mm 40 mm Now to C ! dy2 D 80 57 D 23 mm Ixc2 D Ix2 C dy2 2 A2 Ixc2 D 7.91 ð 105 mm4 For the composite about the centroid Ix D Ixc1 Ixc2 Now to C Ix D 1.08 ð 107 mm4 Ixc1 D Ix1 C 60 Yc 2 A1 The composite Area D 9600 1257 mm2 Ixc1 D 1.161 ð 107 mm4 D 8343 mm2 Circular cut out about its centroid A2 D R2 D 202 D 1257 mm2 kx D Ix D 36.0 mm A Ix2 D 14 R4 D 204 /4 Ix2 D 1.26 ð 105 mm4 622 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.51 Determine Iy and ky . Solution: From the solution to Problem 8.50, the centroid of the y composite area is located at (40, 57.0) mm. The area of the rectangle, A1 , is 9600 mm2 . The area of the cutout, A2 , is 1257 mm2 . The area of the composite is 8343 mm2 . (1) Rectangle about its centroid (40, 60) mm. Iy1 D 1 3 1 hb D 120803 12 12 + x 80 mm Iy1 D 5.12 ð 106 mm4 dx1 D 0 (2) 40 mm Circular cutout about its centroid (40, 80) 80 mm Iy2 D R4 /4 D 1.26 ð 105 mm4 dx2 D 0 Since dx1 and dx2 are zero. (no translation of axes in the xdirection), we get Iy D Iy1 Iy2 Iy D 4.99 ð 106 mm4 Finally, ky D Iy D A1 A2 4.99 ð 106 8343 ky D 24.5 mm Problem 8.52 Determine JO and kO . y Solution: From the solutions to Problems 8.51 and 8.52, Ix D 1.07 ð 107 mm4 Iy D 4.99 ð 106 mm4 and A D 8343 mm2 20 mm JO D Ix C Iy D 1.57 ð 107 mm4 kO D JO D 43.4 mm A x 120 mm 80 mm c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 623 y Problem 8.53 Determine Iy and ky . 12 in x 20 in Solution: Treat the area as a circular area with a half-circular y y y cutout: From Appendix B, 1 Iy 1 D 14 204 in4 and Iy 2 D 18 124 in4 , 12 in. 20 in. 2 x x 2 in. x 20 in. so Iy D 14 204 18 124 D 1.18 ð 105 in4 . The area is A D 202 12 122 D 1030 in2 so, ky D Iy D A 1.18 ð 105 1.03 ð 103 D 10.7 in Problem 8.54 Determine JO and kO . Solution: Treating the area as a circular area with a half-circular cutout as shown in the solution of Problem 8.53, from Appendix B, JO 1 D Ix 1 C Iy 1 D 12 204 in4 and JO 2 D Ix 2 C Iy 2 D 14 124 in4 . Therefore JO D 12 204 14 124 D 2.35 ð 105 in4 . From the solution of Problem 8.53, A D 1030 in2 Ro D D 624 JO A 2.35 ð 105 D 15.1 in. 1.03 ð 103 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.55 Determine Iy and ky if h D 3 m. y Solution: Break the composite into two parts, a rectangle and a semi-circle. 1.2 m For the semi-circle Ix 0 c D 9 8 8 R4 h Iy 0 c D 1 4 4R R d D 8 3 y′ x y d x′ 1.2 m AC d = 4R 3π AR To get moments about the x and y axes, the dxc , dyc for the semicircle are dxc D 0, dyc D 3 m C 3m = h 4R 3 x and Ac D R2 /2 D 2.26 m2 Iy 0 c D To get moments of area about the x, y axes, dxR D 0, dyR D 1.5 m 1 4 R 8 and Iyc D Iy 0 c C d2xc A 0 IyR D Iy 0 R C dxR 2 bh dx D 0 1 32, 43 m4 12 Iyc D Iy 0 c D 1.24 /8 IyR D Iy 0 R D Iyc D 0.814 m4 IyR D 3.456 m2 For the Rectangle Ix 0 R D 1 3 bh 12 Iy 0 R D 1 3 hb 12 AR D bh D 7.2 m2 Iy D Iyc C IyR Iy D 4.27 m2 To find ky , we need the total area, A D AR C Ac AR D bh A D 7.20 C 2.26 m2 y′ y A D 9.46 m2 2.4 m ky D h 3m b Iy D 0.672 m A x′ x c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 625 Problem 8.56 Determine Ix and kx if h D 3 m. Solution: Break the composite into two parts, the semi-circle and the rectangle. From the solution to Problem 8.55, Ix 0 c D 9 8 8 dyc D 3C 4R 3 y Ac R4 m Ac D 2.26 m2 R = 1.2 m h=3m b = 2.4 m AR 3m h Ixc D Ix0 c C Ac d2yc Substituting in numbers, we get dyc D 3.509 m x 2.4 m Ix0 c D 0.0717 m4 yc′ and Ixc D Ix0 c C Ac d2y Ixc D 27.928 m2 xc′ R For the Rectangle h D 3 m, b D 2.4 m Area: AR D bh D 7.20 m2 Ix 0 R D 1 3 bh , dyR D 1.5 m 12 4R 3π IxR D Ix0 R C d2yR AR Substituting, we get Ix0 R D 5.40 m4 IxR D 21.6 m4 For the composite, Ix D IxR C Ixc Ix D 49.5 m4 Also kx D Ix D 2.29 m AR C Ac kx D 2.29 m 626 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.57 If Iy D 5 m4 , what is the dimension of h? Solution: From the solution to Problem 8.55, we have: y c′ y xc′ For the semicircle Iy 0 c D Iy D 1.24/8 D 0.814 m2 h For the rectangle Iy 0 R D IyR D 1.2 y ′R m h x ′R b 1 h2.43 m4 12 2.4 m x Also, we know IyR C Iyc D 5 m4 . Hence 0.814 C 1 h2.43 D 5 12 Solving, h D 3.63 m Problem 8.58 Determine Iy and ky . Solution: Let the area be divided into parts as shown. The areas and the coordinates of their centroids are A1 D 4050 D 2000 in2 , x 1 D 25 in, y 1 D 20 in, A2 D 2030 D 600 in2 , x 2 D 10 in, y 2 D 55 in, A3 D 1 302 D 707 in2 , 4 x 3 D 20 C 430 D 32.7 in, 3 y 3 D 40 C 430 D 52.7 in. 3 Using the results from Appendix B, the moments of inertia of the parts about the y axis are Iy 1 D 1 40 in50 in3 D 167 ð 104 in4 , 3 Iy 2 D 1 30 in20 in3 D 8.00 ð 104 in4 , 3 Iy 3 D 4 16 9 " 430 in 2 ! 30 in4 C 20 in C 30 in2 3 4 D 80.2 ð 104 in4 . The moment of inertia of the composite area about the y axis is Iy D Iy 1 C Iy 2 C Iy 3 D 2.55 ð 106 in4 . 2 The composite area is A D A1 C A2 C A3 D 3310 in . Iy D 27.8 in. The radius of gyration about the y axis is ky D A Iy D 2.55 ð 106 in4 , ky D 27.8 in. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 627 Problem 8.59 Determine Ix and kx . Solution: See the solution to Problem 8.58. Let the area be divided into parts as shown. The areas and the coordinates of their centroids are A1 D 4050 D 2000 in2 , x 1 D 25 in, y 1 D 20 in, A2 D 2030 D 600 in2 , x 2 D 10 in, y 2 D 55 in, x 3 D 20 C 430 D 32.7 in, 3 A3 D 1 302 D 707 in2 , 4 y 3 D 40 C 430 D 52.7 in. 3 Using the results from Appendix B, the moments of inertia of the parts about the x axis are Ix 1 D 1 50 in40 in3 D 1.07 ð 106 in4 , 3 Ix 2 D 1 20 in30 in3 C 55 in2 600 in2 D 1.86 ð 106 in4 , 12 Ix 3 D 4 4 9 " 430 in 2 ! 30 in4 C 40 in C 30 in2 D 2.01 ð 106 in4 . 3 4 The moment of inertia of the composite area about the x axis is Ix D Ix 1 C Ix 2 C Ix 3 D 4.94 ð 106 in4 . 2 The composite area is A D A1 C A2 C A3 D 3310 in . Ix D 38.6 in. The radius of gyration about the y axis is kx D A Ix D 4.94 ð 106 in4 , kx D 38.6 in 628 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.60 Determine Ixy . Solution: See the solution to Problem 8.58. Let the area be divided into parts as shown. The areas and the coordinates of their centroids are A1 D 4050 D 2000 in2 , x 1 D 25 in, y 1 D 20 in, A2 D 2030 D 600 in2 , x 2 D 10 in, y 2 D 55 in, x 3 D 20 C 430 D 32.7 in, 3 A3 D 1 302 D 707 in2 , 4 y 3 D 40 C 430 D 52.7 in. 3 Using the results from Appendix B, the products of inertia of the parts about are Ixy 1 D 1 50 in2 40 in2 D 10.0 ð 105 in4 , 4 Ixy 2 D 10 in55 in600 in2 D 3.30 ð 105 in4 , Ixy 3 D 4 1 8 9 430 in 430 in 30 in4 C 20 in C 40 in C [707 in2 ] 3 3 D 12.1 ð 105 in4 . The product of inertia of the composite area is Ixy D Ixy 1 C Ixy 2 C Ixy 3 D 2.54 ð 106 in4 . Ixy D 2.54 ð 106 in4 . y Problem 8.61 Determine Iy and ky . Solution: See the solution to Problem 8.58. In terms of the coordinate system used in Problem 8.58, the areas and the coordinates of their centroids are A1 D 4050 D 2000 in2 , x 1 D 25 in, y 1 D 20 in, A2 D 2030 D 600 in2 , x 2 D 10 in, y 2 D 55 in, x 3 D 20 C 430 D 32.7 in, 3 A3 D 1 302 D 707 in2 , 4 30 in 40 in x 430 D 52.7 in. y 3 D 40 C 3 20 in The composite area is A D A1 C A2 C A3 D 3310 in2 . x 1 A1 C x 2 A2 C x 3 A3 The x coordinate of its centroid is x D D 23.9 in A The moment of inertia about the y axis in terms of the coordinate system used in Problem 8.58 is Iy D 2.55 ð 106 in4 . Applying the parallel axis theorem, the moment of inertia about the y axis through the centroid of the area is Iy D 2.55 ð 106 in4 23.9 in2 3310 in2 D 6.55 ð 105 in4 . The radius of gyration about the y axis is ky D Iy D 14.1 in. A Iy D 6.55 ð 105 in4 , ky D 14.1 in. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 629 y Problem 8.62 Determine Ix and kx . Solution: See the solution to Problem 8.59. In terms of the coordinate system used in Problem 8.59, the areas and the coordinates of their centroids are A1 D 4050 D 2000 in2 , A2 D 2030 D 600 in2 , A3 D 1 302 D 707 in2 , 4 x 1 D 25 in, y 1 D 20 in, x 2 D 10 in, y 2 D 55 in, x 3 D 20 C 430 D 32.7 in, 3 30 in 40 in x 430 D 52.7 in. 3 y 3 D 40 C 20 in 2 The composite area is A D A1 C A2 C A3 D 3310 in . y A1 C y 2 A2 C y 3 A3 The x coordinate of its centroid is y D 1 D 33.3 in A The moment of inertia about the x axis in terms of the coordinate system used in Problem 8.59 is Ix D 4.94 ð 106 in4 . Applying the parallel axis theorem, the moment of inertia about the x axis through the centroid of the area is Ix D 4.94 ð 106 in4 33.3 in2 3310 in2 D 1.26 ð 106 in4 . The radius of gyration about the x axis is kx D Ix D 19.5 in. A Ix D 1.26 ð 106 in4 , kx D 19.5 in. Problem 8.63 Determine Ixy . Solution: See the solution to Problem 8.60. Ixy D [0 C 1.0 m0.8 mdx 0.5 mdy 0.4 m] [0 C 0.2 m2 dx 0.4 mdy 0.3 m] C 1 1 0.8 m2 0.6 m2 0.8 m0.6 m0.2 m 24 2 C Solving: 0.8 m 3 0.8 m 1 0.8 m0.6 mdx 1.2 m dy 2 3 Ixy D 0.0230 m4 Check using the noncentroidal product of inertia from Problem 8.60 we have Ixy D Ixy 0 Adx dy D 0.2185 m4 0.914 m2 0.697 m0.379 m D 0.0230 m4 630 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.64 Determine Iy and ky . y 18 in x 6 in 6 in Solution: Divide the area into three parts: The composite area: Part (1) The rectangle 18 by 18 inches; Part (2) The triangle with base 6 in and altitude 18 in; Part (3) The semicircle of 9 in radius. AD 3 # 6 in Ai D 397.23 in2 . 1 Part (1): A1 D 1818 D 324 in2 , x1 D 9 in, Iy D x21 A1 C Iyy1 x22 A2 Iyy2 C x23 A3 C Iyy3 , y1 D 9 in, Ixx1 D Iyy1 D 1 12 1 12 The area moment of inertia: Iy D 4.347 ð 104 in4 , 18183 D 8748 in4 , ky D Iy D 10.461 in A 18183 D 8748 in4 . 1 186 D 54 in2 , 2 Part (2): A2 D x2 D 9 in, y2 D 1 18 D 6 in, 3 Ixx2 D 1 36 6183 D 972 in4 , Iyy2 D 1/181833 D 27 in4 . Part (3) A3 D 92 D 127.23 in2 , 2 x3 D 9 in, y3 D 18 C Ixx3 D Iyy3 D 49 3 D 21.82 in, 1 49 2 A3 D 720.1 in4 , 94 8 3 1 94 D 2576.5 in4 . 8 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 631 Problem 8.65 Determine Ix and kx . Solution: Use the results of the solution to Problem 8.64. IX D y21 A1 C IXX1 y22 A2 IXX2 C y23 A3 C IXX3 , Ix D 9.338 ð 104 in4 , kx D Ix D 15.33 in A Problem 8.66 Determine Ixy . Solution: Use the results of the solutions to Problems 8.63 and 8.64. Ixy D x1 y1 A1 x2 y2 A2 C x3 y3 A3 Ixy D 4.8313 ð 104 in4 Problem 8.67 Determine Iy and ky . y Solution: We divide the composite area into a triangle (1), rect- 6 in angle (2), half-circle (3), and circular cutout (4): 2 in Triangle: Iy 1 D 14 1283 D 1536 in4 x Rectangle: Iy 2 D 8 in 1 1283 C 122 812 D 14,336 in4 . 12 8 in y Half-Circle: Iy 3 D 8 8 9 2 46 2 1 62 D 19,593 in4 64 C 16 C 3 2 3 4 1 x Circular cutout: Iy 4 D 14 24 C 162 22 D 3230 in4 . y 8 in. y Therefore Iy D Iy 1 C Iy 2 C Iy 3 Iy 4 D 3.224 ð 104 in4 . 12 in. 2 1 12 in. The area is x A D A1 C A2 C A3 A4 D 1 1 128 C 812 C 62 22 D 188 in2 , 2 2 so ky D Iy D A x 8 in. 12 in. y y 3 6 in. 4 3.224 ð 104 D 13.1 in. 188 6 in. x 16 + 4(6) in. 3π 632 2 in. x 16 in. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.68 Determine JO and kO . Solution: Iy is determined in the solution to Problem 8.67. We will determine Ix and use the relation JO D Ix C Iy . Using the figures in the solution to Problem 8.67, Triangle: Ix 1 D 1 8123 D 1152 in4 . 12 Rectangle: Ix 2 D 13 8123 D 4608 in4 . Half Circle: Ix 3 D 18 64 C 62 21 62 D 2545 in4 . Circular Cutout: Ix 4 D 14 24 C 62 22 D 465 in4 . Therefore Ix D Ix 1 C Ix 2 C Ix 3 Ix 4 D 7840 in4 . Using the solution of Problem 8.67, JO D Ix C Iy D 0.784 ð 104 C 3.224 ð 104 D 4.01 ð 104 in4 . From the solution of Problem 8.67, A D 188 in2 , so R0 D JO D A 4.01 ð 104 D 14.6 in. 188 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 633 Problem 8.69 Determine Iy and ky . y 4 in Solution: Divide the area into four parts: Part (1) The rectangle 8 in by 16 in. Part (2): The rectangle 4 in by 8 in. Part (3) The semicircle of radius 4 in, and Part (4) The circle of radius 2 in. Part (1): A1 D 168 D 128 in2 , x1 D 8 in, 2 in 4 in 8 in y1 D 4 in, Ixx1 D Iyy1 D 1 12 1 12 x 1683 D 682.67 in4 , 16 in 8163 D 2730.7 in4 . Part (2): A2 D 48 D 32 in2 , Iyy2 D 1 12 1 12 843 D 42.667 in4 , AD 42 D 25.133 in2 , 2 Ixx3 D 24 D 12.566 in4 , 3 # Ai A4 D 172.566 in2 . 1 The area moment of inertia: Iy D x21 A1 C Iyy1 C x22 A2 C Iyy2 C x23 A3 C Iyy3 x24 A4 Iyy4 44 3 D 13.698 in. The area moments of inertia about the centroid of the semicircle are Iyy3 D 4 Iyy4 D Ixx4 D 12.566 in4 . 483 D 170.667 in4 . x3 D 12 in. $1% The composite area: Part (3): A3 D y3 D 12 C x4 D 12 in, Ixx4 D y2 D 10 in, Part (4): A4 D 22 D 12.566 in2 , y4 D 12 in, x2 D 12 in, Ixx2 D 12 in 1 44 D 100.53 in4 , 8 Iy D 1.76 ð 104 in4 , ky D Iy D 10.1 in A 1 44 2 A3 D 28.1 in4 . 44 8 3 Check: Ixx3 D 0.1098R4 D 28.1 in4 . check. Problem 8.70 Determine Ix and kx . Solution: Use the results in the solution to Problem 8.69. Ix D y21 A1 C Ixx1 C y22 A2 C Ixx2 C y23 A3 C Ixx3 y24 A4 Ixx4 Ix D 8.89 ð 103 in4 kx D 634 Ix D 7.18 in A c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.71 Determine Ixy . Solution: Use the results in the solution to Problem 8.69. Ixy D x1 y1 A1 C x2 y2 A2 C x3 y3 A3 x4 y4 A4 , Ixy D 1.0257 ð 104 in4 Problem 8.72 Determine Iy and ky . y 4 in 2 in 4 in Solution: Use the results in the solutions to Problems 8.69 to 8.71. The centroid is xD x1 A1 C x2 A2 C x3 A3 x4 A4 A 1024 C 384 C 301.6 150.8 D 9.033 in, D 172.567 x 8 in 12 in 16 in from which Iyc D x2 A C Iy D 1.408 ð 104 C 1.7598 ð 104 D 3518.2 in4 kyc D Iyc D 4.52 in A Problem 8.73 Determine Ix and kx . Solution: Use the results in the solutions to Problems 8.69 to 8.71. The centroid is yD y1 A1 C y2 A2 C y3 A3 y4 A4 D 5.942 in, A from which Ixc D y2 A C Ix D 6092.9 C 8894 D 2801 in4 kxc D Ixc D 4.03 in A Problem 8.74 Determine Ixy . Solution: Use the results in the solutions to Problems 8.69–8.71. Ixyc D xyA C Ixy D 9.263 ð 103 C 1.0257 ð 104 D 994.5 in4 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 635 Problem 8.75 Determine Iy and ky . y Solution: We divide the area into parts as shown: Iy 1 D 5 mm 1 50 C 15 C 15303 D 180,000 mm4 12 Iy 2 D Iy 3 D Iy 4 D 15 mm 50 mm 1 30103 C 202 1030 12 5 mm 5 mm D 122,500 mm4 Iy 5 D Iy 6 D Iy 7 D 8 8 9 x 15 mm 15 mm 154 10 15 15 10 mm mm mm mm 415 2 1 152 D 353,274 mm4 C 25 C 3 2 y 1 Iy 8 D Iy 9 D Iy 10 2 5 8 1 D 54 C 252 52 D 49,578 mm4 . 4 Therefore, 50 mm 7 4 10 Iy D Iy 1 C 3Iy 2 C 3Iy 5 3Iy 8 D 1.46 ð 106 3 6 9 mm4 . x 10 15 mm mm The area is A D A1 C 3A2 C 3A5 3A8 D 3080 C 31030 C 3 1 152 352 2 D 4125 mm2 so ky D Iy D A 1.46 ð 106 D 18.8 mm 4125 Problem 8.76 Determine JO and kO . Solution: Iy is determined in the solution to Problem 8.75. We will determine Ix and use the relation JO D Ix C Iy . Dividing the area as shown in the solution to Problem 8.75, we obtain Ix 1 D 1 30803 C 252 3080 D 2, 780, 000 mm4 12 Ix 2 D 1 10303 C 502 1030 D 772, 500 mm4 12 Ix 3 D Ix 4 D Ix 5 D 1 10303 D 22, 500 mm4 12 Therefore Ix D Ix 1 C Ix 2 C 2Ix 3 C Ix 5 C 2Ix 6 Ix 8 2Ix 9 D 4.34 ð 106 mm4 and JO D Ix C Iy D 5.80 ð 106 mm4 . From the solution to Problem 8.75, A D 4125 mm2 so kO D 1 1 154 C 502 152 D 903,453 mm4 8 2 D Ix 6 D Ix 7 D Ix 8 D 5.80 ð 106 4125 D 37.5 mm. 1 54 C 52 502 , 4 Ix 9 D Ix 10 D 636 1 154 D 19,880 mm4 , 8 JO A 1 54 D 491 mm4 . 4 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 8.77 Determine Ix and Iy for the beam’s cross section. 5 in 2 in Solution: Use the symmetry of the object 1 1 Ix D 3 in.8 in3 C 3 in3 in3 C 3 in2 11.5 in2 2 3 12 C 5 in4 5 in2 16 4 C 4[5 in] 3 2 5 in2 4 8 in 4[5 in] 2 8 in C 3 x 3 in in4 2 16 in2 2 4 C 4[2 in] 3 5 in 3 in 5 in 2 2 in2 4 4[2 in] 2 8 in C 3 Solving we find Ix D 7016 in4 Iy 1 1 D 3 in3 in3 C 8 in3 in3 C 8 in3 in6.5 in2 2 3 12 C 5 in4 5 in2 16 4 4[5 in] 3 2 5 in2 C 4 2 in2 2 in4 16 4 C 4[2 in] 3 4[5 in] 3 in C 3 2 2 2 in2 4 3 in C 4[2 in] 3 2 Solving we find Iy D 3122 in4 y Problem 8.78 Determine Ix and Iy for the beam’s cross section. 5 in 2 in x Solution: Use Solution 8.77 and 7.39. From Problem 7.39 we know that y D 7.48 in, A D 98.987 in2 8 in Ix D 7016 in4 Ay 2 D 1471 in4 Iy D 3122 in4 A02 D 3122 in4 3 in 5 in 5 in 3 in c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 637 Problem 8.79 The area A D 2 ð 104 mm2 . Its moment of inertia about the y axis is Iy D 3.2 ð 108 mm4 . Determine its moment of inertia about the yO axis. ŷ y Solution: Use the parallel axis theorem. The moment of inertia A about the centroid of the figure is Iyc D x 2 A C Iy D 1202 2 ð 104 C 3.2 ð 108 D 3.20 ð 107 mm4 . x, x̂ The moment of inertia about the yO axis is 100 mm 120 mm IyO D x2 A C Iyc IyO D 2202 2 ð 104 C 3.2 ð 107 D 1 ð 109 mm4 Problem 8.80 The area A D 100 in2 and it is symmetric about the x 0 axis. The moments of inertia Ix0 D 420 in4 , Iy 0 D 580 in4 , JO D 11000 in4 , and Ixy D 4800 in4 . What are Ix and Iy ? y y' A x' O' O x Solution: The basic relationships: (1) (2) (3) (4) (5) (6) Ix D y 2 A C Ixc , Iy D x2 A C Iyc , JO D Ar 2 C Jc , JO D Ix C Iy , Jc D Ixc C Iyc , and Ixy D Axy C Ixyc , where the subscript c applies to the primed axes, and the others to the unprimed axes. The x, y values are the displacement of the primed axes from the unprimed axes. The steps in the demonstration are: From symmetry about the xc axis, the product of inertia Ixyc D 0. JO J c D 100 in2 , from which r 2 D x2 C From (3): r 2 D A y 2 D 100 in2 Ixy (iii) From (6) and Ixyc D 0, y D , from which x2 r 2 D x4 C Ax 2 Ixy . From which: x4 100x2 C 2304 D 0. A (iv) The roots: x12 D 64, and x22 D 36. The corresponding values of y p are found from y D r 2 x 2 from which x1 , y1 D 8, 6, and x2 , y2 D 6, 8. (v) Substitute these pairs to obtain the possible values of the area moments of inertia: (i) (ii) Ix1 D Ay12 C Ixc D 4020 in4 , Iy1 D Ax12 C Iyc D 6980 in4 Ix2 D Ay22 C Ixc D 6820 in4 , Iy2 D Ax22 C Iyc D 4180 in4 638 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.81 Determine the moment of inertia of the beam cross section about the x axis. Compare your result with the moment of inertia of a solid square cross section of equal area. (See Example 8.5.) y 20 mm x Solution: We first need to find the location of the centroid of the composite. Break the area into two parts. Use X, Y coords. 160 mm y 2 100 20 A1 = 2000 mm2 XC = 0 160 mm 1 A2 = 3200 mm2 XC2 = 0 20 mm YC2 = 80 mm 100 mm 1 YC1 = 170 mm x 100 mm y Y 20 mm Xc D Xc1 A1 C Xc2 A2 D0 A1 C A2 Yc D Yc1 A1 C Yc2 A2 A1 C A2 C1 1 For the composite 20 mm x 160 mm 2 C2 Substituting, we get Xc D 0 mm X Yc D 114.6 mm We now find Ix for each part about its center and use the parallel axis theorem to find Ix about C. 20 mm Part (1): b1 D 100 mm, h1 D 20 mm Finally, Ix D Ix1 C Ix2 1 1 b1 h13 D 100203 mm4 12 12 Ix D 1.686 ð 107 mm4 Ix 0 1 D for our composite shape. Ix0 1 D 6.667 ð 104 mm4 dy1 D Yc1 Yc D 55.38 mm Ix1 D Ix0 1 C dy1 2 A1 Now for the comparison. For the solid square with the same total area A1 C A2 D 5200 mm2 , we get a side of length l2 D 5200: l D 72.11 mm And for this solid section Ix1 D 6.20 ð 106 mm4 Part (2) b2 D 20 mm, h2 D 160 mm Ix 0 2 D 1 1 b2 h2 3 D 201603 mm4 12 12 IxSQ D 1 3 1 4 bh D l 12 12 IxSQ D 2.253 ð 106 mm4 1.686 ð 107 2.253 ð 106 Ix0 2 D 6.827 ð 106 mm4 Ratio D Ix /IxSQ D dy2 D Yc2 Yc D 34.61 mm Ratio D 7.48 Ix2 D Ix0 2 C dy2 A2 This matches the value in Example 8.5. Ix2 D 1.066 ð 107 mm4 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 639 Problem 8.82 The area of the beam cross section is 5200 mm2 . Determine the moment of inertia of the beam cross section about the x axis. Compare your result with the moment of inertia of a solid square cross section of equal area. (See Example 8.5.) y x 20 mm Solution: Let the outside dimension be b mm, then the inside dimension is b 40 mm. The cross section is A D b2 b 402 D 5200 mm2 . Solve: b D 85 mm. Divide the beam cross section into two parts: the inner and outer squares. Part (1) A1 D 852 D 7225 mm2 , Ixx1 D 1 12 85853 D 4.35 ð 106 . Part (2) A2 D 452 D 2025 mm2 . Ixx2 D 1 12 45453 D 3.417 ð 105 . The composite moment of inertia about the centroid is Ix D Ixx1 Ixx2 D 4.008 ð 106 mm4 . For a square cross section of the same area, h D p 5200 D 72.111 mm. The area moment of inertia is Ixb D 1 12 72.11172.1113 D 2.253 ð 106 in4 . The ratio: RD 4.008 ð 106 D 1.7788 D 1.78 2.253 ð 106 which confirms the value given in Example 8.5. 640 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.83 If the beam in Fig. a is subjected to couples of magnitude M about the x axis (Fig. b), the beam’s longitudinal axis bends into a circular are whose radius R is given by y y z x (a) Unloaded. EIx RD , M where Ix is the moment of inertia of the beam’s cross section about the x axis. The value of the term E, which is called the modulus of elasticity, depends on the material of which the beam is constructed. Suppose that a beam with the cross section shown in Fig. c is subjected to couples of magnitude M D 180 N-m. As a result, the beam’s axis bends into a circular arc with radius R D 3 m. What is the modulus of elasticity of the beam’s material? (See Example 8.5.) Solution: The moment of inertia of the beam’s cross section about the x axis is & ' 1 1 mm4 393 C 2 933 C 62 93 Ix D 12 12 D 2170 mm4 D 2.17 ð 109 m4 . The modulus of elasticity is ED RM 3 m180 N-m D D 2.49 ð 1011 N/m2 Ix 2.17 ð 109 m4 M R M (b) Subjected to couples at the ends. y 3 mm x 9 mm 3 mm 3 mm 9 mm (c) Beam cross section. E D 2.49 ð 1011 N/m2 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 641 Problem 8.84 Suppose that you want to design a beam made of material whose density is 8000 kg/m3 . The beam is to be 4 m in length and have a mass of 320 kg. Design a cross section for the beam so that Ix D 3 ð 105 m4 . (See Example 8.5.) Solution: The strategy is to determine the cross sectional area, and then use the ratios given in Figure 8.14 to design a beam. The volume of the beam is V D AL D 4A m3 . The mass of the beam is m D V8000 D 32000A D 320 kg, from which A D 0.01 m2 . The moment of inertia for a beam of square cross section with this area is Ixxb D 1 12 h 0.10.13 D 8.333 ð 106 m4 . The ratio is R D b ð 105 3 D 3.6. 8.333 ð 106 From Figure 8.6, this ratio suggests an I-beam of the form shown in the sketch. Choose an I-beam made up of three equal area rectangles, of dimensions b by hm in section. The moment of inertia about the centroid is Ix D y21 A1 C Ixx1 C y22 A2 C Ixx2 C y23 A3 C Ixx3 . Since all areas are equal, A1 D A2 D A3 D bh, and y1 D 0, and y3 D y1 , this reduces to Ix D bCh , y2 D 2 1 bCh 2 1 bh3 C 2 hb3 . hb C 6 2 12 A , where A is the known total cross section area. These 3 are two equations in two unknowns. Plot the function Note that bh D fb D 1 bCh 2 1 bh C bh3 C 2 hb3 Ix 6 2 12 Problem 8.85 The area in Fig. (a) is a C230ð30 American Standard Channel beam cross section. Its cross sectional area is A D 3790 mm2 and its moments of inertia about the x and y axes are Ix D 25.3 ð 106 mm4 and Iy D 1 ð 106 mm4 . Suppose that two beams with C230ð30 cross sections are riveted together to obtain a composite beam with the cross section shown in Fig. (b). What are the moments of inertia about the x and y axes of the composite beam? f ( b ) * E + 5 I - beam Flange dimension 1 .8 .6 .4 .2 0 –.2 –.4 –.6 –.8 –1 .03 .04 .05 .06 .07 .08 .09 .1 b A .The function was graphed using 3 TK Solver Plus. The graph crosses the zero axis at approximately b D 0.0395 m. and b D 0.09 m. The lower value is an allowable value for h and the greater value corresponds to an allowable value of b. Thus the I beam design has the flange dimensions, b D 90 mm and h D 39.5 mm. subject to the condition that hb D y y x x 14.8 mm (a) (b) Solution: Ix D 225.3 ð 106 mm4 D 50.6 ð 106 mm4 Iy D 2106 mm4 C [3790 mm2 ][14.8 mm]2 D 3.66 ð 106 mm4 642 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.86 The area in Fig. (a) is an L152ð102ð 12.7 Angle beam cross section. Its cross sectional area is A D 3060 mm2 and its moments of inertia about the x and y axes are Ix D 7.24 ð 106 mm4 and Iy D 2.61 ð 106 mm4 . Suppose that four beams with L152ð102ð12.7 cross sections are riveted together to obtain a composite beam with the cross section shown in Fig. (b). What are the moments of inertia about the x and y axes of the composite beam? y y 24.9 mm x Solution: x 50.2 mm (a) Ix D 47.24 ð 106 mm4 C [3060 mm2 ][50.2 mm]2 D 59.8 ð 106 mm4 Iy D 42.61 ð 106 mm4 C [3060 mm2 ][24.9 mm]2 D 18.0 ð 106 mm4 (b) Problem 8.87 In Active Example 8.6, suppose that the vertical 3-m dimension of the triangular area is increased to 4 m. Determine a set of principal axes and the corresponding principal moments of inertia. y Solution: From Appendix B, the moments and products of inertia of the area are Ix D 1 4 m4 m3 D 21.3 m4 , 12 Iy D 1 4 m3 4 m D 64 m4 , 4 Ixy D 1 4 m2 4 m2 D 32 m4 . 8 3m x 4m From Eq. (8.26), tan2p D 2Ixy 232 D 1.50 ) p D 28.2° . D Iy Ix 64 21.3 From Eqs. (8.23) and (8.24), the principal moments of inertia are Ix ‘ D Ix C Iy Ix Iy C cos 2p Ixy sin 2p 2 2 D 21.3 C 64 2 C 21.3 64 2 cos2[28.2° ] 32 sin2[28.2° ] D 4.21 m4 Iy ‘ D Ix Iy Ix C Iy cos 2p C Ixy sin 2p 2 2 D 21.3 C 64 2 21.3 64 2 cos2[28.2° ] C 32 sin2[28.2° ] D 81.8 m4 p D 28.2° , principal moments of inertia are 4.21 m4 , 81.1 m4 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 643 y Problem 8.88 In Example 8.7, suppose that the area is reoriented as shown. Determine the moments of inertia Ix0 , Iy 0 and Ix0 y 0 if D 30o . Solution: Based on Example 8.7, the moments and product of 1 ft 3 ft inertia of the reoriented area are 1 ft Ix D 10 ft4 , Iy D 22 ft4 , Ixy D 6 ft4 . x 4 ft Applying Eqs. (8.23)–(8.25), Ix 0 D D Iy 0 D D Ix 0 y 0 D D Ix Iy Ix C Iy C cos 2 Ixy sin 2 2 2 10 C 22 10 22 C cos 60° 6 sin 60° D 7.80 ft4 , 2 2 Ix Iy Ix C Iy C cos 2 C Ixy sin 2 2 2 10 C 22 10 22 cos 60° C 6 sin 60° D 24.2 ft4 , 2 2 Ix Iy sin 2 C Ixy cos 2 2 12 22 sin 60° C 6 cos 60° D 2.20 ft4 2 Ix0 D 7.80 ft4 , Iy 0 D 24.2 ft4 , Ix0 y 0 D 2.20 ft4 . y Problem 8.89 In Example 8.7, suppose that the area is reoriented as shown. Determine a set of principal axes and the corresponding principal moments of inertia. Based on the result of Example 8.7, can you predict a value of p without using Eq. (8.26)? Solution: Based on Example 8.7, the moments and product of 1 ft 3 ft inertia of the reoriented area are 1 ft Ix D 10 ft4 , Iy D 22 ft4 , Ixy D 6 ft4 . x 2Ixy 26 D D 1 ) p D 22.5° Ix Iy 22 10 This value could have been anticipated from Example 8.7 by reorienting the axes. Substituting the angle into Eqs. (8.23) and (8.24), the principal moments of inertia are From Eq. (8.26), tan 2p D Ix 0 D D Iy 0 D D 4 ft Ix Iy Ix C Iy C cos 2p Ixy sin 2p 2 2 10 C 22 10 22 C cos 45° 6 sin 45° D 7.51 ft4 , 2 2 Ix Iy Ix C Iy cos 2p C Ixy sin 2p 2 2 10 22 10 C 22 cos 45° C 6 sin 45° D 24.5 ft4 , 2 2 p D 22.5° , principal moments of inertia are 7.51 ft4 , 24.5 ft4 . 644 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.90 The moment of inertia of the area are y y Ix D 1.26 ð 106 in4 , Iy D 6.55 ð 105 in4 , Ixy D 1.02 ð 105 in4 Determine the moments of inertia of the area Ix0 , Iy 0 and Ix0 y 0 if D 30° . x u x Solution: Applying Eqs. (8.23)–(8.25), Ix 0 D Ix Iy Ix C Iy C cos 2 Ixy sin 2Ð 2 2 D 1.26 C 0.655 1.26 0.655 C cos 60° 0.102 sin 60° ð 106 in4 2 2 D 1.20 ð 106 in4 Iy 0 D Ix Iy Ix C Iy cos 2 C Ixy sin 2 2 2 D 1.26 0.655 1.26 C 0.655 cos 60° C 0.102 sin 60° ð 106 in4 2 2 D 7.18 ð 105 in4 Ix 0 y 0 D Ix Iy sin 2 C Ixy cos 2 2 D 1.26 0.655 sin 60° C 0.102 cos 60° ð 106 in4 2 D 2.11 ð 105 in4 Ix0 D 1.20 ð 106 in4 , Iy 0 D 7.18 ð 105 in4 , Ix0 y 0 D 2.11 ð 105 in4 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 645 Problem 8.91 The moment of inertia of the area are y y Ix D 1.26 ð 106 in4 , Iy D 6.55 ð 105 in4 , Ixy D 1.02 ð 105 in4 x u Determine a set of principal axes and the corresponding principal moments of inertia. x Solution: From Eq. (8.26), tan 2p D 2Ixy 2.102 D D 0.337 Iy Ix 0.655 1.26 ) p D 9.32° Substituting this angle into Eqs. (8.23) and (8.24), the principal moments of inertia are Ix 0 D Ix Iy Ix C Iy C cos 2p Ixy sin 2p 2 2 D 1.26 0.655 1.26 C 0.655 C cos 18.63° 0.102 sin 18.63° 2 2 ð 106 in4 D 1.28 ð 106 in4 Iy 0 D Ix C Iy Ix Iy cos 2p C Ixy sin 2p 2 2 D 1.26 0.655 1.26 C 0.655 cos 18.63° C 0.102 sin 18.63° 2 2 ð 106 in4 D 6.38 ð 105 in4 p D 9.32° , principal moments of inertia are 1.28 ð 106 in4 , 6.38 ð 105 in4 . 646 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 8.92* Determine a set of principal axes and the corresponding principal moments of inertia. 160 mm Solution: We divide the area into 3 rectangles as shown: In terms of the xO , yO coordinate system, the position of the centroid is x̂ D 40 mm x̂1 A1 C x̂2 A1 C x̂3 A3 A1 C A2 C A3 x D ŷ D 2040200 C 10012040 C 808040 D 56 mm, 40200 C 12040 C 8040 200 mm 40 mm ŷ1 A1 C ŷ2 A1 C ŷ3 A3 A1 C A2 C A3 40 mm 10040200 C 18012040 C 208040 D D 108 mm. 40200 C 12040 C 8040 120 mm The moments and products of inertia in terms of the xO , yO system are y y^ (Ix D (Ix 1 C (Ix 2 C (Ix 3 D 160 mm 2 1 1 402003 C 120403 C 1802 12040 3 12 C 1 80403 D 26.5 ð 107 mm4 , 3 200 mm x 40 mm (Iy D (Iy 1 C (Iy 2 C (Iy 3 40 mm 3 120 mm D 1 1 200403 C 401203 C 1002 12040 3 12 C 40 mm 1 x^ y y′ 1 40803 C 802 8040 D 8.02 ð 107 mm4 , 12 (Ixy D (Ixy 1 C (Ixy 2 C (Ixy 3 x 12.1° x′ D 2010040200 C 10018040120 C 20804080 D 10.75 ð 107 mm. The moments and product of inertia in terms of the xO , yO system are Ix D (Ix yO 2 A D 77.91 ð 106 mm4 , Iy D (Iy xO 2 A D 30.04 ð 106 mm4 , Ixy D (Ixy x̂ŷA D 10.75 ð 106 mm4 , from Equation (8.26), tan 2p D 2Ixy 210.75 ð 106 , D Iy Ix 30.04 ð 106 77.91 ð 106 we obtain p D 12.1° . We can orient the principal axes as shown: Substituting the values of Ix , Iy and Ixy into Equations (8.23) and (8.24) and setting D 12.1° , we obtain Ix1 D 80.2 ð 106 mm4 Iy 1 D 27.7 ð 106 mm4 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 647 Problem 8.93 Solve Problem 8.87 by using Mohr’s Circle. y 3m x 4m Solution: The vertical 3-m dimension is increased to 4 m. From Problem 8.87, the moments and product of inertia for the unrotated system are Ix D 1 4 m4 m3 D 21.3 m4 , 12 Iy D 1 4 m3 4 m D 64 m4 , 4 Ixy D 1 4 m2 4 m2 D 32 m4 . 8 Mohr’s circle (shown) has a center and radius given by CD 21.3 C 64 D 42.7 m4 2 RD 21.3 64 2 2 C 322 D 38.5 m4 The angle and principal moments are now tan2p D 32 ) p D 28.2° , 64 42.7 I1 D C C R D 81.1 m4 , I2 D C R D 4.21 m4 . p D 28.2° , principal moments of inertia are 4.21 m4 , 81.1 m4 648 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 8.94 Solve Problem 8.88 by using Mohr’s Circle. 1 ft 3 ft 1 ft x 4 ft Solution: Based on Example 8.7, the moments and product of inertia of the reoriented area are Ix D 10 ft4 , Iy D 22 ft4 , Ixy D 6 ft4 . For Mohr’s circle we have the center, radius, and angle CD 10 C 22 D 16 ft4 , 2 RD p D 22 10 2 1 tan1 2 2 C 62 D 8.49 ft4 , 6 22 16 D 22.5° Now we can calculate the new inertias Ix D C R cos 15° D 7.80 ft4 Iy D C C R cos 15° D 24.2 ft4 Ixy D R sin 15° D 2.20 ft4 Ix0 D 7.80 ft4 , Iy 0 D 24.2 ft4 , Ix‘y‘ D 2.20 ft4 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 649 y Problem 8.95 Solve Problem 8.89 by using Mohr’s Circle. 1 ft 3 ft 1 ft x 4 ft Solution: Based on Example 8.7, the moments and product of inertia of the reoriented area are Ix D 10 ft4 , Iy D 22 ft4 , Ixy D 6 ft4 . For Mohr’s circle we have the center, radius, and angle CD 10 C 22 D 16 ft4 , 2 RD p D 22 10 2 1 tan1 2 2 C 62 D 8.49 ft4 , 6 22 16 D 22.5° Now we can calculate the principal moments of inertias I1 D C C R D 24.5 ft4 I2 D C R D 7.51 ft4 p D 22.5° , principal moments of inertia are 7.51 ft4 , 24.5 ft4 . 650 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.96 Solve Problem 8.90 by using Mohr’s Circle. y y x u x Solution: For Mohr’s circle we have the center, radius, and angle CD 12.6 C 6.55 2 RD p D 12.6 6.55 2 1 tan1 2 ð 105 D 9.58 ð 105 in4 , 2 C 1.022 ð 105 D 3.19 in4 , 1.02 12.6 9.58 D 9.32° Now we can calculate the new inertias Ix0 D C C R cos 22.7° D 12.0 ð 105 in4 , Iy 0 D C R cos 22.7° D 7.18 ð 105 in4 , Ix0 y 0 D R sin 22.7° D 2.11 ð 105 in4 . Ix0 D 1.20 ð 106 in4 , Iy 0 D 7.18 ð 105 in4 , Ix0 y 0 D 2.11 ð 105 in4 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 651 Problem 8.97 Solve Problem 8.91 by using Mohr’s Circle. y y x u x Solution: For Mohr’s circle we have the center, radius, and angle CD 12.6 C 6.55 2 RD p D 12.6 6.55 2 1 tan1 2 ð 105 D 9.58 ð 105 in4 , 2 C 1.022 ð 105 D 3.19 in4 , 1.02 12.6 9.58 D 9.32° Now we can calculate the principal inertias I1 D C C R D 12.8 ð 105 in4 , Iy 0 D C R D 6.38 ð 105 in4 , p D 9.32° , principal moments of inertia are 1.28 ð 106 in4 , 6.38 ð 105 in4 . 652 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.98* Solve Problem 8.92 by using Mohr’s circle. Solution: The moments and product of inertia are derived in terms of the xy coordinate system in the solution of Problem 8.92: 50 ◊106 (+) Ix D 77.91 ð 106 mm4 Iy D 30.04 ð 106 mm4 1 2θp Ixy D 10.75 ð 106 mm4 . The Mohr’s circle is: Measuring the 2p, angle we estimate that p D 12° , and the principle moments of inertia are approximately 81 ð 106 mm4 and 28 ð 106 mm4 the orientation of the principal axes is shown in the solution of Problem 8.92. 2 (30.0, –10.8) (77.9,10.8) (+) 100 6 ◊10 - 50 ◊106 Problem 8.99 Derive Eq. (8.22) for the product of inertia by using the same procedure we used to derive Eqs. (8.20) and (8.21). Solution: Suppose that the area moments of inertia of the area A x′ are known in the coordinate system x, y, y 2 y dA, Ix D y′ A A 0 x 2 dA, Iy D x A and Ixy D xyA. A The objective is to find the product of inertia in the new coordinate system x 0 , y 0 in terms of the known moments of inertia. The new x 0 , y 0 system is formed from the old x, y system by rotation about the origin through a counterclockwise angle . By definition, Substitute into the definition: Ix0 y 0 D cos2 sin2 xy dA A A 0 0 x y dA. Ix 0 y 0 D x 2 dA , y 2 dA C cos sin A from which A From geometry, x 0 D x cos C y sin , Ix0 y 0 D cos2 sin2 Ixy C Ix Iy sin cos , which is the expression required. and y 0 D x sin C y cos . The product is x0 y 0 D xy cos2 xy sin2 C y 2 cos sin x2 cos sin . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 653 Problem 8.100 The axis LO is perpendicular to both segments of the L-shaped slender bar. The mass of the bar is 6 kg and the material is homogeneous. Use the method described in Example 8.10 to determine is moment of inertia about LO . Solution: Use Example 8.10 as a model for this solution. 1m LO 2m Introduce the coordinate system shown and divide the bar into two parts as shown y=1 y dy 2 r dx 1 0 2 r 2 dm D I0 1 D Ax 2 dx D A 0 I0 1 D x 2 x x3 3 2 0 8 A 3 However m1 D Al1 D 2A. Since part 1 is 2/3 of the length, its mass is 2/36 kg D 4 kg. Part 2 has mass 2 kg. For part 2, dm D A dy and 22 C y 2 rD 1 r 2 dm D I0 2 D I0 2 D A4y C I0TOTAL D A22 C y 2 dy 0 m2 y3 3 1 D A 0 13 3 13 8 21 A C A D A 3 3 3 I0 TOTAL D 7A The total mass D 3A D 6 kg I0TOTAL D 654 7 6 kg Ð m2 D 7 kg m2 6 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.101 Two homogenous slender bars, each of mass m and length l, are welded together to form the T-shaped object. Use integration to determine the moment of inertia of the object about the axis through point O that is perpendicular to the bars. l O l Solution: Divide the object into two pieces, each corresponding to a slender bar of mass m; the first parallel to the y axis, the second to the x axis. By definition l r 2 dm C ID 0 r 2 dm. m For the first bar, the differential mass is dm D A dr. Assume that the second bar is very slender, so that the mass is concentrated at a distance l from O. Thus dm D A dx, where x lies between the limits l l x . 2 2 The distance to a differential dx is r D becomes l I D A r 2 dr C A l 2 0 D A r3 3 D ml2 l 2 p l2 C x 2 . Thus the definition l2 C x 2 dx I l/2 x3 C A l2 x C 3 l/2 0 l 1 1 C1C 3 12 D 17 2 ml 12 Problem 8.102 The slender bar lies in the x –y plane. Its mass is 6 kg and the material is homogeneous. Use integration to determine its moment of inertia about the z axis. y 2m 50 x 1m Solution: 1 m 2 Iz D 6 kg D 2 kg/m The density is D 3m x 2 dx y 0 C m 2m [1 m C s cos 50° 2 C s sin 50° 2 ]ds 0 50° Iz D 15.14 kg m2 Problem 8.103 Use integration to determine the moment of inertia of the slender bar in Problem 8.102 about the y axis. x 1m Solution: See solution for 8.102 Iy D 0 1 m 2 x 2 dx C m 1 m C s cos 50° 2 ds D 12.01 kg m2 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 655 y Problem 8.104 The homogeneous thin plate has mass m D 12 kg and dimensions b D 2 m and h D 1 m. Use the procedure described in Active Example 8.9 to determine the moments of inertia of the plate about the x and y axes. h x Solution: From Appendix B, the moments of inertia about the x and y axes are 1 3 1 3 bh , Iy D hb . Ix D 36 36 b Therefore the moments of inertia of the plate about the x and y axes are 1 3 m m 1 1 b D mh2 D 12 kg1 m2 D 0.667 kg-m2 Ixaxis D Ix 1 A 36 18 8 bh 2 Iyaxis D m m Iy D 1 A bh 2 1 3 hb 36 D 1 1 mb2 D 12 kg2 m2 D 2.67 kg-m2 18 18 Ixaxis D 0.667 kg-m2 , Iyaxis D 2.67 kg-m2 . y Problem 8.105 The homogenous thin plate is of uniform thickness and mass m. (a) (b) Determine its moments of inertia about the x and z axes. Let Ri D 0, and compare your results with the values given in Appendix C for a thin circular plate. Ro Ri x Solution: (a) (b) The area moments of inertia for a circular area are Ix D Iy D R4 . For the plate with a circular cutout, Ix D Ro4 Ri4 . The 4 4 m area mass density is , thus for the plate with a circular cut, A m m , from which the moments of inertia D 2 A Ro2 Ri Ri Ro m mRo4 Ri4 D Ro2 C Ri2 4 4Ro2 Ri2 Ix axis D Iz axis D 2Ix axis D m 2 R C Ri2 . 2 o Let Ri D 0, to obtain Ix axis D m 2 R , 4 o Iz axis D m 2 R , 2 o which agrees with table entries. 656 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 8.106 The homogenous thin plate is of uniform thickness and weighs 20 lb. Determine its moment of inertia about the y axis. 1 y = 4 – – x 2 ft 4 Solution: y D4 1 2 x ft 4 x y The plate’s area is 4 4 AD 4 1 2 x 4 dx D 21.3 ft2 . y=4– 1 2 x ft 4 The plate’s density per unit area is υ D 20/32.2/21.3 D 0.0291 slug/ft2 . x –4 ft x The moment of inertia about the y axis is Iy axis D 4 ft dx 1 x 2 υ 4 x 2 dx 4 4 4 D 1.99 slug-ft2 . Problem 8.107 Determine the moment of inertia of the plate in Problem 8.106 about the x axis. Solution: See the solution of Problem 8.106. The mass of the strip so the moment of inertia of the plate about the x axis is element is mstrip 1 D υ 4 x2 4 dx. Ix axis 4 D 4 3 1 1 dx D 2.27 slug-ft2 . υ 4 x2 3 4 The moment of inertia of the strip about the x axis is Istrip D D 2 1 1 mstrip 4 x 2 3 4 3 1 1 υ 4 x2 dx, 3 4 Problem 8.108 The mass of the object is 10 kg. Its moment of inertia about L1 is 10 kg-m2 . What is its moment of inertia about L2 ? (The three axes lie in the same plane.) Solution: The strategy is to use the data to find the moment of inertia about L, from which the moment of inertia about L2 can be determined. IL D 0.62 10 C 10 D 6.4 m2 , 0.6 m L from which IL2 D 1.22 10 C 6.4 D 20.8 m2 0.6 m L1 L2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 657 Problem 8.109 An engineer gathering data for the design of a maneuvering unit determines that the astronaut’s center of mass is at x D 1.01 m, y D 0.16 m and that her moment of inertia about the z axis is 105.6 kg-m2 . Her mass is 81.6 kg. What is her moment of inertia about the z0 axis through her center of mass? y y x x Solution: The distance d from the z axis to the z0 axis is 1.012 C 0.162 dD D 1.0226 m. From the parallel-axis theorem, D Iz0 axis C d2 m : 105.6 D Iz0 axis C 1.02262 81.6. Iz axis Solving, we obtain Iz0 D 20.27 kg-m2 . axis Problem 8.110 Two homogenous slender bars, each of mass m and length l, are welded together to form the Tshaped object. Use the parallel axis theorem to determine the moment of inertia of the object about the axis through point O that is perpendicular to the bars. l O l Solution: Divide the object into two pieces, each corresponding to a bar of mass m. By definition l ID r 2 dm. 0 For the first bar, the differential mass is dm D A dr, from which the moment of inertia about one end is l I1 D A r 2 dr D A 0 r3 3 l D 0 ml2 . 3 For the second bar I2 D A l 2 l 2 r3 r dr D A 3 2 2l l 2 D ml2 12 is the moment of inertia about the center of the bar. From the parallel axis theorem, the moment of inertia about O is Io D 658 17 2 ml2 ml2 C l2 m C D ml . 3 12 12 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.111 Use the parallel-axis theorem to determine the moment of inertia of the T-shaped object in Problem 8.110 about the axis through the center of mass of the object that is perpendicular to the two bars. (See Active Example 8.11.) Solution: The location of the center of mass of the object is x D $ % l 2 C lm 3 D l. Use the results of Problem 8.110 for the moment 2m 4 of inertia of a bar about its center. For the first bar, m I1 D 2 l ml2 7 2 mC D ml . 4 12 48 For the second bar, I2 D 2 l ml2 7 2 mC D ml . 4 12 48 The composite: Ic D I1 C I2 D 7 2 ml 24 y Problem 8.112 The mass of the homogenous slender bar is 20 kg. Determine its moment of inertia about the z axis. y' x' Solution: Divide the object into three segments. Part (1) is the 1 m bar on the left, Part (2) is the 1.5 m horizontal segment, and Part (3) is the segment on the far right. The mass density per unit length is 1m x 20 m p D 5.11 kg/m. D D L 1 C 1.5 C 2 1.5 m 1m The moments of inertia about the centers of mass and the distances to the centers of mass from the z axis are: Part (1) I1 D l31 12 D m1 l21 D 0.426 kg-m2 , 12 m1 D 5.11 kg, d1 D 0.5 m, Part (2), I2 D l2 l32 D m2 2 D 1.437 kg- m2 , 12 12 m2 D 7.66 kg, d2 D Part (3) p 0.752 C 12 D 1.25 m I3 D p 22 l33 D m3 D 1.204 kg-m2 , 12 12 m3 D 7.23 kg, d3 D p 22 C 0.52 D 2.062 m. The composite: I D d21 m1 C I1 C d22 m2 C I2 C d23 m3 C I3 D 47.02 kg-m2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 659 Problem 8.113 Determine the moment of inertia of the bar in Problem 8.112 about the z0 axis through its center of mass. Solution: The center of mass: xD D yD D x 1 m 1 C x2 m 2 C x3 m 3 20 0 C 0.757.66 C 27.23 D 1.01 m. 20 0.5m1 C 1m2 C 0.5m3 20 0.55.11 C 17.66 C 0.57.23 D 0.692 m. 20 The distance from the z axis to the center of mass is d D x2 C y2 D 1.224 m. The moment of inertia about the center o mass: Ic D d2 20 C Io D 17.1 kg-m2 Problem 8.114 The homogeneous slender bar weighs 5 lb. Determine its moment of inertia about the z axis. y y' Solution: The Bar’s mass is m D 5/32.2 slugs. Its length is L D L1 C L 2 C L 3 D 8 C 4 in p 82 C 82 C 4 D 31.9 in. The masses of the parts are therefore, L1 m1 D mD L m2 D L2 mD L m3 D L3 mD L 8 31.9 5 32.2 x' D 0.0390 slugs, p 264 5 D 0.0551 slugs, 31.9 32.2 4 31.9 5 32.2 x 8 in D 0.0612 slugs. y′ y The center of mass of part 3 is located to the right of its center C a distance 2R/ D 24/ D 2.55 in. The moment of inertia of part 3 about C is 2 C r 2 dm D m3 r 2 D 0.061242 D 0.979 slug-in2 . m3 1 The moment of inertia of part 3 about the center of mass of part 3 is therefore 4 in x′ 3 x 8 in I3 D 0.979 m3 2.552 D 0.582 slug-in2 . The moment of inertia of the bar about the z axis is Iz axis D 13 m1 L12 C 13 m2 L22 C I3 C m3 [8 C 2.552 C 42 ] D 11.6 slug-in2 D 0.0802 slug-ft2 . 660 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.115 Determine the moment of inertia of the bar in Problem 8.114 about the z0 axis through its center of mass. Solution: In the solution of Problem 8.114, it is shown that the moment of inertia of the bar about the z axis is Iz axis D 11.6 slug-in2 . The x and y coordinates of the center of mass coincide with the centroid of the axis: xD x1 L1 C x2 L2 C x3 L3 L1 C L2 C L3 p 24 4 48 C 4 82 C 82 C 8 C p D D 6.58 in, 8 C 82 C 82 C 4 yD D y1 L1 C y2 L2 C y3 L3 L1 C L2 C L3 p 0 C 4 82 C 82 C 44 p D 3.00 in. 8 C 82 C 82 C 4 The moment of inertia about the z axis is Iz0 axis D Iz axis x2 C y2 5 32.2 D 3.43 slug-in2 . y Problem 8.116 The rocket is used for atmospheric research. Its weight and its moment of inertia about the z axis through its center of mass (including its fuel) are 10 kip and 10,200 slug-ft2 , respectively. The rocket’s fuel weighs 6000 lb, its center of mass is located at x D 3 ft, y D 0, z D 0, and the moment of inertia of the fuel about the axis through the fuel’s center of mass parallel to z is 2200 slug-ft2 . When the fuel is exhausted, what is the rocket’s moment of inertia about the axis through its new center of mass parallel to z? Solution: Denote the moment of inertia of the empty rocket as IE about a center of mass xE , and the moment of inertia of the fuel as IF about a mass center xF . Using the parallel axis theorem, the moment of inertia of the filled rocket is x From which xE D 186.335 124.224 3 D 4.5 ft 2m , IR D IE C xE2 mE C IF C xF F is the new location of the center of mass. Substitute: about a mass center at the origin (xR D 0). Solve: 2m IE D IR xE2 mE IF xF F 2m . IE D IR xE2 mE IF xF F D 10200 2515.5 2200 1677.01 The objective is to determine values for the terms on the right from the data given. Since the filled rocket has a mass center at the origin, the mass center of the empty rocket is found from 0 D mE xE C mF xF , from which D 3807.5 slug-ft2 xE D mF mE xF . Using a value of g D 32.2 ft/s2 , mF D WF 6000 D D 186.34 slug, g 32.2 mE D 10000 6000 WR WF D D 124.23 slug. g 32.2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 661 Problem 8.117 The mass of the homogenous thin plate is 36 kg. Determine its moment of inertia about the x axis. y 0.4 m 0.4 m Solution: Divide the plate into two areas: the rectangle 0.4 m by 0.6 m on the left, and the rectangle 0.4 m by 0.3 m on the right. The m mass density is D . The area is A 0.3 m A D 0.40.6 C 0.40.3 D 0.36 m2 , 0.3 m x from which D 36 D 100 kg/m2 . 0.36 The moment of inertia about the x axis is Ix axis D $1% 3 0.40.63 C $1% 3 0.40.33 D 3.24 kg-m2 Problem 8.118 Determine the moment of inertia of the plate in Problem 8.117 about the z axis. Solution: The basic relation to use is Iz axis D Ix axis C Iy axis . The value of Ix axis is given in the solution of Problem 8.117. The moment of inertia about the y axis using the same divisions as in Problem 8.117 and the parallel axis theorem is Iy axis D 1 1 0.60.43 C 0.30.43 3 12 C 0.62 0.30.4 D 5.76 kg-m2 , from which Iz axis D Ix axis C Iy axis D 3.24 C 5.76 D 9 kg-m2. y Problem 8.119 The homogenous thin plate weighs 10 lb. Determine its moment of inertia about the x axis. 5 in 5 in Solution: Divide the area into two parts: the lower rectangle 5 in by 10 in and the upper triangle 5 in base and 5 in altitude. The mass W . The area is density is D gA A D 510 C $1% 2 Using g D 32 ft/s2 , the mass density is D 10 in 55 D 62.5 in2 . W D 0.005 slug/in2 . gA 5 in x Using the parallel axis theorem, the moment of inertia about the x axis is Ix axis D 1 1 1053 C 553 3 36 5 2 1 C 5C 55 D 4.948 slug-in2 3 2 Ix 662 axis D 0.03436 slug-ft2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.120 Determine the moment of inertia of the plate in Problem 8.119 about the y axis. Solution: Use the results of the solution in Problem 8.119 for the area and the mass density. Iy axis D 1 1 5103 C 553 3 36 10 2 1 C 5C 55 3 2 D 12.76 slug-in2 D 0.0886 slug-ft2 Problem 8.121 The thermal radiator (used to eliminate excess heat from a satellite) can be modeled as a homogenous, thin rectangular plate. Its mass is 5 slugs. Determine its moment of inertia about the x, y, and z axes. y 3 ft 6 ft 3 ft Solution: The area is A D 93 D 27 ft2 . The mass density is D 5 m D D 0.1852 slugs/ft2 . A 27 2 ft x The moment of inertia about the centroid of the rectangle is Ixc D Iyc D 1 12 1 12 933 D 3.75 slug-ft2 , 393 D 33.75 slug-ft2 . Use the parallel axis theorem: Ix axis D A2 C 1.52 C Ixc D 65 slug-ft2 , Iy axis D A4.5 32 C Iyc D 45 slug-ft2 . Iz axis D Ix axis C Iy axis D 110 slug-ft2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 663 Problem 8.122 The homogeneous cylinder has mass m, length l, and radius R. Use integration as described in Example 8.13 to determine its moment of inertia about the x axis. y x Solution: The volume of the disk element is R2 dz and its mass is dm D R2 dz, where is the density of the cylinder. From Appendix C, the moment of inertia of the disk element about the x0 axis is dIx0 axis D 1 1 dmR2 D R2 dzR2 . 4 4 R l Applying the parallel-axis theorem, the moment of inertia of the disk element about the x axis is dIxaxis z 1 D dIx0 axis C z2 dm D R2 dzR2 C z2 R2 dz 4 Integrating this expression from z D 0 to z D l gives the moment of inertia of the cylinder about the x axis. l Ixaxis D 0 1 1 1 R4 C R2 z2 dz D R4 l C R2 l3 . 4 4 3 In terms of the mass of the cylinder m D R2 l, Ixaxis D 1 2 1 2 mR C ml 4 3 Problem 8.123 The homogenous cone is of mass m. Determine its moment of inertia about the z axis, and compare your result with the value given in Appendix C. (See Example 8.13.) y x Solution: The differential mass dm D R 3m m r 2 dz D 2 r 2 dz. V R h The moment of inertia of this disk about the z axis is 12 mr 2 . The radius varies with z, rD h z R z, h from which Iz 664 axis D 3mR2 2h5 h 0 z4 dz D 3mR2 2h5 z5 5 h D 0 3mR2 10 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.124 Determine the moments of inertia of the homogenous cone in Problem 8.123 about the x and y axes, and compare your results with the values given in Appendix C. 3m m D . The differential V R2 h element of mass is dm D r 2 dz. The moment of inertia of this elemental disk about an axis through its center of mass, parallel to the x- and y-axes, is Solution: The mass density is D $1% dIx D 4 r 2 dm. Use the parallel axis theorem, $1% Ix D m 4 r 2 dm C z2 dm. m Noting that r D R z, then h r 2 dm D and z2 dm D R4 h4 R2 h2 z4 dz, z4 dz. Substitute: Ix D R4 4h4 h z4 dz C 0 R2 h2 h z4 dz. 0 Integrating and collecting terms: Ix D 3mR2 3m C 3 4h5 h z5 5 h D m 0 3 2 3 2 R C h . 20 5 By symmetry, Iy D Ix c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 665 Problem 8.125 The mass of the homogeneous wedge is m. Use integration as described in Example 8.13 to determine its moment of inertia about the z axis. (Your answer should be in terms of m, a, b, and h.) y x Solution: Consider a triangular element of the wedge of thickness dz. The mass of the element is the product of the density of the material and the volume of the element, dm D 12 bhdz. The moments of inertia of the triangular element about the x’ and y’ axes are given by Eqs. (8.30) and (8.31) in terms of the mass of the element, its triangular area, and the moments of inertia of the triangular area: dIx0 axis 1 bhdz 1 dm 0 1 2 I D rbh3 dz, bh3 D D 1 A x 12 12 bh 2 dIy 0 axis 1 bhdz 1 dm 0 1 I D 2 hb3 D rhb3 dz, D 1 A y 4 4 bh 2 h a z b The moment of inertia of this thin plate about the z axis is dIzaxis D dIx0 axis C dIy 0 axis D 1 1 bh3 dz C hb3 dz. 12 4 Integrating this expression from z D 0 to z D a gives the moment of inertia of the wedge about the z axis: a Izaxis D 0 1 1 1 1 bh3 C hb3 dz D bh3 a C hb3 a. 12 4 12 4 In terms of the mass m D 12 bha, Izaxis D 666 1 2 1 2 mh C mb . 6 2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.126 The mass of the homogeneous wedge is m. Use integration as described in Example 8.13 to determine its moment of inertia about the x axis. (Your answer should be in terms of m, a, b, and h.) y x Solution: Consider a triangular element of the wedge of thickness dz. The mass of the element is the product of the density of the material and the volume of the element, dm D 12 bhdz. The moments of inertia of the triangular element about the x’ axis is given by Eq. (8.30) in terms of the mass of the element, its triangular area, and the moments of inertia of the triangular area: dIx0 axis h a z b 1 bhdz 1 dm 1 Ix 0 D 2 rbh3 dz, bh3 D D 1 A 36 36 bh 2 Applying the parallel-axis theorem, the moment of inertia of the triangular element about the x axis is 1 dIxaxis D dIx0 axis C z2 C h2 dm 3 D 1 1 1 1 1 bh3 dz C [z2 C h2 ] bhdz D bh3 dz C bhz2 dz 36 3 2 12 2 Integrating this expression from z D 0 to z D a gives the moment of inertia of the wedge about the x axis: a Ixaxis D 0 1 1 1 1 bh3 C bhz2 dz D bh3 a C bha3 . 12 2 12 6 In terms of the mass m D 12 bha, Ixaxis D 1 2 1 2 mh C ma . 6 3 Problem 8.127 In Example 8.12, suppose that part of the 3-kg bar is sawed off so that the bar is 0.4 m long and its mass is 2 kg. Determine the moment of inertia of the composite object about the perpendicular axis L through the center of mass of the modified object. Solution: The mass of the disk is 2 kg. Measuring from the left 0.2 m 0.6 m L end of the rod, we locate the center of mass xD 2 kg0.2 m C 2 kg0.6 m D 0.4 m. 2 kg C 2 kg The center of mass is located at the point where the rod and disk are connected. The moment of inertia is ' & 1 1 I D 2 kg0.4 m2 C 2 kg0.2 m2 C 2 kg0.2 m2 3 2 I D 0.227 kg-m2 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 667 y Problem 8.128 The L-shaped machine part is composed of two homogeneous bars. Bar 1 is tungsten alloy with mass density 14,000 kg/m3 , and bar 2 is steel with mass density 7800 kg/m3 . Determine its moment of inertia about the x axis. 240 mm 1 40 mm 2 80 mm Solution: The masses of the bars are m1 D 14,0000.240.080.04 D 10.75 kg 80 mm z 240 mm m2 D 78000.240.080.04 D 5.99 kg. Using Appendix C and the parallel axis theorem the moments of inertia of the parts about the x axis are Ix axis1 D 1 m1 [0.042 C 0.242 ] C m1 0.122 D 0.2079 kg-m2 , 12 Ix axis2 D 1 m2 [0.042 C 0.082 ] C m2 0.042 D 0.0136 kg-m2 . 12 x Therefore Ix 668 axis D Ix axis1 C Ix axis2 D 0.221 kg-m2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.129 The homogeneous object is a cone with a conical hole. The dimensions R1 D 2 in, R2 D 1 in, h1 D 6 in, and h2 D 3 in. It consists of an aluminum alloy with a density of 5 slug/ft3 . Determine its moment of inertia about the x axis. y x R1 Solution: The density of the material is D 5 slug/ft3 1 ft 12 in 3 D 0.00289 slug/in3 . The volume of the conical object without the conical hole is V1 D R2 h1 z h2 1 2 1 R h1 D 2 in2 6 in D 25.1 in3 . 3 1 3 The mass of the conical object without the conical hole is m1 D V1 D 0.0727 slug. From Appendix C, the moment of inertia of the conical object without the conical hole about the x axis is Ix 1 D m1 3 2 3 2 h C R 5 1 20 1 D 0.0727 slug 3 3 6 in2 C 2 in2 D 1.61 slug-in2 5 20 The volume of the conical hole is V2 D 1 2 1 R h2 D 1 in2 3 in D 3.14 in3 . 3 2 3 The mass of the material that would occupy the conical hole is m2 D V2 D 0.00909 slug. The z coordinate of the center of mass of the material that would occupy the conical hole is z D h1 h2 C 3 3 h2 D 6 in 3 in C 3 in D 5.25 in. 4 4 Using Appendix C and applying the parallel-axis theorem, the moment of inertia about the x axis of the material that would occupy the conical hole is Ix 2 D m2 3 2 3 2 h C R C z2 m2 D 0.255 slug-in2 . 80 2 20 2 The moment of inertia of the conical object with the conical hole is Ix D Ix 1 Ix 2 D 1.36 slug-in2 . Ix D 1.36 slug-in2 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 669 Problem 8.130 The circular cylinder is made of aluminum (Al) with density 2700 kg/m3 and iron (Fe) with density 7860 kg/m3 . Determine its moments of inertia about the x 0 and y 0 axes. y y Al z Fe 600 mm 200 mm z Solution: We have Al D 2700 kg/m3 , Fe D 7860 kg/m3 600 mm x, x We first locate the center of mass xD Al [0.1 m]2 [0.6 m]0.3 m C Fe [0.1 m]2 [0.6 m]0.9 m Al [0.1 m]2 [0.6 m] C Fe [0.1 m]2 [0.6 m] D 0.747 m We also have the masses mAl D Al 0.1 m2 0.6 m, mFe D Fe 0.1 m2 0.6 m Now find the moments of inertia 1 1 mAl 0.1 m2 C mFe 0.1 m2 D 0.995 kg m2 2 2 [0.1 m]2 [0.6 m]2 C D mAl C mAl x 0.3 m2 12 4 [0.1 m]2 [0.6 m]2 C mFe C C mFe 0.9 m x2 12 4 D 20.1 kg m2 Ix 0 D Iy 0 Problem 8.131 The homogenous half-cylinder is of mass m. Determine its moment of inertia about the axis L through its center of mass. Solution: The centroid of the half cylinder is located a distance R L T 4R from the edge diameter. The strategy is to use the parallel 3 axis theorem to treat the moment of inertia of a complete cylinder as the sum of the moments of inertia for the two half cylinders. From Problem 8.118, the moment of inertia about the geometric axis for a cylinder is IcL D mR2 , where m is one half the mass of the cylinder. of By the parallel axis theorem, IcL D 2 4R 3 2 m C IhL . Solve IhL D IcL 2 D mR2 D mR2 670 4R 3 2 2 16 mR mR2 m D 2 2 9 16 1 2 92 16 1 2 92 D 0.31987 mR2 D 0.32 mR2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.132 The homogeneous machine part is made of aluminum alloy with density D 2800 kg/m3 . Determine its moment of inertia about the z axis. y y 20 mm x z 40 mm 120 mm Solution: We divide the machine part into the 3 parts shown: (The dimension into the page is 0.04 m). The masses of the parts are y m1 D 28000.120.080.04 D 1.075 kg, m2 D 2800 12 0.042 0.04 D 0.281 kg, 40 mm 0.12 m 1 0.08 m x m3 D 28000.022 0.04 D 0.141 kg. y Iz axis1 0.12 m C 2 0.12 m 1 m1 [0.082 C 0.122 ] C m1 0.062 D 12 The moment of inertia of part 2 about the axis through the center C that is parallel to the z axis is x 0.02 m Iz axis2 D 0.000144 C m2 0.12 C 0.0172 D 0.00543 kg-m2 . The moment of inertia of the material that would occupy the hole 3 about the z axis is D 12 m2 0.042 . The distance along the x axis from C to the center of mass of part 2 is 40.04/3 D 0.0170 m. Iz axis3 Therefore Iz Therefore, the moment of inertia of part 2 about the z axis through its center of mass that is parallel to the axis is 1 2 2 m2 0.04 3 x – 0.04 m Using this result, the moment of inertia of part 2 about the z axis is D 0.00573 kg-m2 . 1 2 2 m2 R y + Using Appendix C and the parallel axis theorem the moment of inertia of part 1 about the z axis is axis D 12 m3 0.022 C m3 0.122 D 0.00205 kg-m2 . D Iz axis1 C Iz axis2 Iz axis3 D 0.00911 kg-m2 . m2 0.01702 D 0.000144 kg-m2 . Problem 8.133 Determine the moment of inertia of the machine part in Problem 8.132 about the x axis. Solution: We divide the machine part into the 3 parts shown in the solution to Problem 8.132. Using Appendix C and the parallel axis theorem, the moments of inertia of the parts about the x axis are: 1 m1 [0.082 C 0.042 ] D 0.0007168 kg-m2 12 Ix axis1 D Ix axis2 D m2 1 1 0.042 C 0.042 12 4 D 0.0001501 kg-m2 Ix axis3 D m3 1 1 0.042 C 0.022 12 4 D 0.0000328 kg-m2 . Therefore, Ix axis D Ix axis1 C Ix axis2 Ix axis3 D 0.000834 kg-m2 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 671 Problem 8.134 The object consists of steel of density D 7800 kg/m3 . Determine its moment of inertia about the axis LO . 20 mm O 100 mm Solution: Divide the object into four parts: Part (1) The semicylinder of radius R D 0.02 m, height h1 D 0.01 m. Part (2): The rectangular solid L D 0.1 m by h2 D 0.01 m by w D 0.04 m. Part (3): The semi-cylinder of radius R D 0.02 m, h1 D 0.01 m. Part (4) The cylinder of radius R D 0.02 m, height h D 0.03 m. 10 mm 30 mm LO Part (1) m1 D I1 D R2 h1 D 0.049 kg, 2 m1 R 2 D 4.9 ð 106 kg-m2 , 4 Part (2): m2 D wLh2 D 0.312 kg, I2 D 1 12 m2 L 2 C w2 C m2 2 L D 0.00108 kg-m2 . 2 Part (3) m3 D m1 D 0.049 kg, I3 D 4R 3 2 4R 2 m2 C I 1 C m 3 L D 0.00041179 kg m2 . 3 Part (4) m4 D R2 h D 0.294 kg, I4 D $1% 2 m4 R2 C m4 L 2 D 0.003 kg m2 . The composite: ILo D I1 C I2 I3 C I4 D 0.003674 kg m2 Problem 8.135 Determine the moment of inertia of the object in Problem 8.134 about the axis through the center of mass of the object parallel to LO . Solution: The center of mass is located relative to LO 4R 4R m1 C m2 0.05 m3 0.1 C m4 0.1 3 3 xD m1 C m 2 m 3 C m 4 D 0.066 m, Ic D x2 m C ILo D 0.00265 C 0.00367 D 0.00102 kg m2 672 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.136 The thick plate consists of steel of density D 15 slug/ft3 . Determine its moment of inertia about the z axis. y y 4 in 2 in 2 in x Solution: Divide the object into three parts: Part (1) the rectangle 8 in by 16 in, Parts (2) & (3) the cylindrical cut outs. Part (1): m1 D 8164 D 4.444 slugs. I1 D 1 12 z 4 in 4 in 8 in 4 in 4 in m1 162 C 82 D 118.52 slug in2 . Part (2): m2 D 22 4 D 0.4363 slug, I2 D m2 22 C m2 42 D 7.854 slug in2 . 2 Part (3): m3 D m2 D 0.4363 slugs, I3 D I2 D 7.854 slug-in2 . The composite: Iz axis D I1 2I2 D 102.81 slug-in2 Iz axis D 0.714 slug-ft2 Problem 8.137 Determine the moment of inertia of the plate in Problem 8.136 about the x axis. Solution: Use Problem 8.136. the same divisions of the object as in Part (1): I1x axis D 1 12 m1 82 C 42 D 29.63 slug-in2 , Part (2): I2x axis D 1 12 m2 322 C 42 D 1.018 slug-in2 . The composite: Ix axis D I1x axis 2I2x axis D 27.59 slug in2 D 0.1916 slug ft2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 673 Problem 8.138 Determine Iy and ky . y (1, 1) Solution: 1 dA D dx dyA D dx Iy D A 1 x 2 dx D . 3 x2 x 2 dx 1 dy D 0 0 A ky D 0 1 x 2 dA D Iy D 1 dy D 0 0 y = x2 x2 x 4 dx D 0 x5 5 1 D 0 1 5 x 3 5 Problem 8.139 Determine Ix and kx . Solution: (See figure in Problem 8.138.) dA D dx dy, kx D [x7 ]10 D y 2 dy D 0 0 1 21 x2 dx A D 1 y 2 dA D Ix D 1 3 1 x 6 dx 0 1 21 Ix 1 D p A 7 Problem 8.140 Determine JO and kO . Solution: (See figure in Problem 8.138.) JO D Ix C Iy D kO D 1 26 1 C D , 5 21 105 kx2 C ky2 D 1 3 C D 5 7 26 35 Problem 8.141 Determine Ixy . Solution: (See figure in Problem 8.138.) dA D dx dy D 674 1 2 0 1 x2 x dx 0 A 1 xy dA D Ixy D x 5 dx D y dy 0 1 6 1 1 [x ]0 D 12 12 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.142 Determine Iy and ky . Solution: By definition, y y = x – 1– x 2 4 x 2 dA. Iy D A The element of area is dA D dx dy. The limits on the variable x are 0 x 4. The area is 4 dy D 0 0 4 Iy D x3 x2 2 12 xx2 /4 x 2 dx 4 0 x5 x4 4 20 4 D 2.6667 0 x dy D 0 D xx2 /4 dx AD x 0 x2 4 x 2 dx 4 D 12.8 0 from which ky D Iy D 2.19 A Problem 8.143 Determine Ix and kx . Solution: By definition, y 2 dA, Ix D A from which 4 0 Ix D xx2 /4 dx Ix D y 2 dy D 0 4 3 1 x2 dx x 3 4 0 4 4 1 3 6 x7 x 3 5 D 0.6095. x C x 3 4 20 96 448 0 From Problem 8.142, A D 2.667, kx D Ix D 0.4781 A Problem 8.144 Determine Ixy . Solution: Ixy D xy dA, A 4 x dx D 0 xx2 /4 y dy 0 D 4 2 1 x2 x dx x 2 4 0 D 4 4 x x5 x6 1 C D 2.1333 2 4 10 96 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 675 Problem 8.145 Determine Iy 0 and ky 0 . Solution: The limits on the variable x are 0 x 4. By definition, y dA D Ay D 4 y dy 0 4 2 1 x2 D dx x 2 4 0 D y' y = x – 1– x 2 4 xx2 /4 dx 0 A y x' x 3 4 x4 x5 1 x C D 1.06667. 2 3 8 80 0 From Problem 8.142 the area is A D 2.667, from which y D 0.3999 D 0.4. Similarly, 4 XX2 /4 x dx Ax D 0 dy 0 4 D 0 3 4 x2 x x4 x x D 5.3333, dx D 4 3 16 0 from which x D 1.9999 D 2. The area moment of inertia is Iyy D x2 A C Iy . Using the result of Problem 8.142, Iy D 12.8, from which the area moment of inertia about the centroid is Iy 0 D 10.6666 C 12.8 D 2.133 and ky 0 D Iy 0 D 0.8944 A Problem 8.146 Determine Ix0 and kx0 . Solution: Using the results of Problems 8.143 and 8.145, Ix D 0.6095 and y D 0.4. The area moment of inertia about the centroid is Ix0 D y2 A C Ix D 0.1828 and k D x0 Ix 0 D 0.2618 A Problem 8.147 Determine Ix0 y 0 . Solution: From Problems 8.143 and 8.144, Ixy D 2.133 and x D 2, y D 0.4. The product of the moment of inertia about the centroid is Ix0 y 0 D xyA C Ixy D 2.133 C 2.133 D 0 676 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 8.148 Determine Iy and ky . Solution: Divide the section into two parts: Part (1) is the upper rectangle 40 mm by 200 mm, Part (2) is the lower rectangle, 160 mm by 40 mm. Part (1) A1 D 0.0400.200 D 0.008 m2 , y1 D 0.180 m 40 mm 160 mm x1 D 0, Iy1 D 1 12 0.040.23 D 2.6667 ð 105 m4 . x 80 mm 40 mm 80 mm Part (2): A2 D 0.040.16 D 0.0064 m2 , y2 D 0.08 m, x2 D 0, Iy2 D 1 12 0.160.043 D 8.5 ð 107 m4 . The composite: A D A1 C A2 D 0.0144 m2 , Iy D Iy1 C Iy2 , Iy D 2.752 ð 105 m4 D 2.752 ð 107 mm4 , and ky D Iy D 0.0437 m D 43.7 mm A Problem 8.149 Determine Ix and kx for the area in Problem 8.148. Solution: Use the results in the solution to Problem 8.148. Part (1) A1 D 0.0400.200 D 0.008 m2 , y1 D 0.180 m, Ix1 D 1 12 0.20.043 C 0.182 A1 D 2.603 ð 104 m4 . Part (2): A2 D 0.040.16 D 0.0064 m2 , y2 D 0.08 m, Ix2 D 1 12 0.040.163 C 0.082 A2 D 5.461 ð 105 m4 . The composite: A D A1 C A2 D 0.0144 m2 , The area moment of inertia about the x axis is Ix D Ix1 C Ix2 D 3.15 ð 104 m4 D 3.15 ð 108 mm4 , and kx D Ix D 0.1479 m D 147.9 mm A c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 677 y Problem 8.150 Determine Ix and kx . Solution: Use the results of the solutions to Problems 8.148– 40 mm 8.149. The centroid is located relative to the base at xc D x1 A1 C x2 A2 D 0, A yc D y1 A1 C y2 A2 D 0.1356 m. A x 160 mm The moment of inertia about the x axis is 80 mm Ixc D y2C A C IX D 5.028 ð 107 mm4 and kxc D 40 mm 80 mm Ixc D 59.1 mm A Problem 8.151 Determine JO and kO for the area in Problem 8.150. Solution: Use the results of the solutions to Problems 8.148– 8.149. The area moments of inertia about the centroid are Ixc D 5.028 ð 105 m4 and Iyc D Iy D 2.752 ð 105 m4 , from which JO D Ixc C Iyc D 7.78 ð 105 m4 D 7.78 ð 107 mm4 and kO D JO D 0.0735 m A D 73.5 mm Problem 8.152 Determine Iy and ky . y Solution: For a semicircle about a diameter: Iyy D Ixx D Iy D 1 1 44 24 D 44 24 D 94.25 ft4 , 8 8 8 ky D 678 1 R4 , 8 2 ft x 4 ft 2Iy D 2.236 ft 42 22 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.153 Determine JO and kO . for the area in Problem 8.152. Solution: For a semicircle: Iyy D Ixx D 1 R4 . 8 4 4 24 D 94.248 ft4 . 8 Ix D kx D 2Ix D 2.236 ft. 42 22 Also use the solution to Problem 8.152. JO D Ix C Iy D 294.248 D 188.5 ft4 kO D 2JO D 3.16 ft 42 22 Problem 8.154 Determine Ix and kx . Solution: Break the area into three parts: Part (1) The rectangle with base 2a and altitude h; Part (2) The triangle on the right with base b a and altitude h, and Part (3) The triangle on the left with base b a and altitude h. Part (1) The area is y 3 ft 3 ft 6 ft A1 D 2ah D 24 ft2 . The centroid is x x1 D 0 2 ft h and y1 D D 3 ft. 2 2 ft y b The area moment of inertia about the centroid is Ixc1 D 1 12 2ah3 D Part (2): A2 D x2 D a C y2 D ba D 2.3333 ft, 3 1 36 h 1 hb a D 3 ft2 , 2 2 h D 4 ft, 3 Ixc2 D 1 ah3 D 72 ft4 . 6 a x The composite moment of inertia Ix D y1 2 A1 C Ixc1 C y2 2 A2 C Ixc2 C y3 2 A3 C Ixc3 , Ix D 396 ft4 b ah3 D 6 ft4 . kx D Ix D A 396 D 3.633 ft 30 Part (3): A3 D A2 , x3 D x2 , y3 D y2 , Ixc3 D Ixc2 . The composite area is A D A1 C A2 C A2 D 30 ft2 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 679 Problem 8.155 Determine Iy and ky for the area in Problem 8.154. Solution: Divide the area as in the solution to Problem 8.154. Part (1) The area is A1 D 2ah D 24 ft2 . The centroid is x1 D 0 and h y1 D D 3 ft. The area moment of inertia about the centroid is 2 Iyc1 D 1 12 h2a3 D Part (2): A2 D x2 D a C y2 D 2 ha3 D 32 ft4 3 1 hb a D 3 ft2 , 2 ba D 2.3333 ft, 3 2 h D 4 ft, 3 Iyc2 D 1 36 hb a3 D 0.1667 ft4 . Part (3): A3 D A2 , x3 D x2 , y3 D y2 , Iyc3 D Iyc2 . The composite area is A D A1 C A2 C A2 D 30 ft2 . The composite moment of inertia, Iy D x21 A1 C Iyc1 C x22 A2 C Iyc2 C x23 A3 C Iyc3 , Iy D 65 ft4 and ky D Iy D 1.472 ft A Problem 8.156 The moments of inertia of the area are Ix D 36 m4 , Iy D 145 m4 , and Ixy D 44.25 m4 . Determine a set of principal axes and the principal moment of inertia. Solution: The principal angle is D 1 2Ixy tan1 D 19.54° . 2 Iy Ix y 3m 4m 3m x The principal moments of inertia are IxP D Ix cos2 2Ixy sin cos C Iy sin2 D 20.298 D 20.3 m4 IyP D Ix sin2 C 2Ixy sin cos C Iy cos2 D 160.70 m4 680 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.157 The moment of inertia of the 31-oz bat about a perpendicular axis through point B is 0.093 slugft2 . What is the bat’s moment of inertia about a perpendicular axis through point A? (Point A is the bat’s “instantaneous center,” or center of rotation, at the instant shown.) 31 D 0.06023 slugs. 1632.17 Use the parallel axis theorem to obtain the moment of inertia about the center of mass C, and then use the parallel axis theorem to translate to the point A. Solution: The mass of the bat is m D IC D IA D 12 12 2 12 C 14 12 m C 0.093 D 0.0328 slug-ft2 2 m C 0.0328 D 0.3155 slug-ft2 C 12 in B 14 in A Problem 8.158 The mass of the thin homogenous plate is 4 kg. Determine its moment of inertia about the y axis. Solution: Divide the object into two parts: Part (1) is the semicircle of radius 100 mm, and Part (2) is the rectangle 200 mm by 280 mm. The area of Part (1) A1 D y 100 mm 140 mm x R2 D 15708 mm2 . 2 140 mm The area of Part (2) is 200 mm A2 D 280200 D 56000 mm2 . The composite area is A D A2 A1 D 40292 mm2 . The area mass density is D 4 D 9.9275 ð 105 kg/mm2 . A For Part (1) x1 D y1 D 0, Iy1 D 1 R4 D 3898.5 kg-mm2 . 8 For Part (2) x2 D 100 mm. Iy2 D x22 A2 C 1 12 2802003 D 74125.5 kg-mm2 . The composite: Iy D Iy2 Iy1 D 70226 kg-mm2 D 0.070226 kg-m2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 681 Problem 8.159 Determine the moment of inertia of the plate in Problem 8.158 about the z axis. Solution: Use the same division of the parts and the results of the solution to Problem 8.158. For Part (1), Ix1 D 1 R4 D 3898.5 kg-mm2 . 8 For Part (2) Ix2 D 1 12 2002803 D 36321.5 kg-mm2 . The composite: Ix D Ix2 Ix1 D 32423 kg-mm2 , from which, using the result of the solution to Problem 8.158 Iz D Ix C Iy D 32422 C 70226 D 102649 kg-mm2 D 0.10265 kg-m2 y Problem 8.160 The homogenous pyramid is of mass m. Determine its moment of inertia about the z axis. Solution: The mass density is D x 3m m D 2 . V w h h The differential mass is dm D ω2 dz. The moment of inertia of this element about the z axis is z 1 dIZ D ω2 dm. 6 Noting that ω D dIz D w4 6h4 z4 dz D y y wz , then h mw2 4 z dz. 2h5 Integrating: Iz axis D x w z w h 2 mw 2h5 h z4 dz D 0 1 mw2 10 Problem 8.161 Determine the moment of inertia of the homogenous pyramid in Problem 8.160 about the x and y axes. Solution: Use the results of the solution of Problem 8.160 for the Noting that ω D mass density. The elemental disk is dm D ω2 dz. The moment of inertia about an axis through its center of mass parallel to the x axis is dIX D 1 12 Ix axis D w4 12h4 ω2 dm. Ix 682 axis D 1 12 h z4 dz C 0 w2 h2 h z4 dz. 0 Integrating and collecting terms Use the parallel axis theorem: w z, the integral is h Ix ω2 dm C m axis Dm 1 2 3 2 w C h . 20 5 z2 dm. m By symmetry, Iy axis D Ix axis c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.162 The homogenous object weighs 400 lb. Determine its moment of inertia about the x axis. y y Solution: The volumes are Vcyl D 4692 z 6 in 9 in 3 D 11,706 in , Vcone D 13 62 36 D 1357in3 , 36 in x 36 in 46 in x 46 in so V D Vcyl Vcone D 10,348 in3 . The masses of the solid cylinder and the material that would occupy the conical hole are mcyl D mcone D Vcyl V Vcone V 400 32.2 400 32.2 D 14.052 slug, D 1.629 slug. Using results from Appendix C, Ix axis D 1 3 mcyl 92 mcone 62 2 10 D 551 slug-in2 D 3.83 slug-ft2 Problem 8.163 Determine the moments of inertia of the object in Problem 8.162 about the y and z axes. Solution: See the solution of Problem 8.162. The position of the center of mass of the material that would occupy the conical hole is 3 36 D 37 in. 4 x D 46 36 C 6 in x , x′ From Appendix C, X Iy 0 axiscone y′ y D mcone 3 3 362 C 62 80 20 36 in D 87.97 slug-in2 . The moment of inertia about the y axis for the composite object is Iy axis D mcyl )1 2 3 46 $ Iy 0 C 14 92 axiscone * C x2 mcone % D 7877 slug-in2 D 54.7 slug-ft2 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 683 Problem 8.164 Determine the moment of inertia of the 14-kg flywheel about the axis L. 50 mm 70 mm 120 mm L 100 mm 440 mm 500 mm 150 mm Solution: The flywheel can be treated as a composite of the objects shown: The volumes are = – + – + 4 V1 D 1502502 D 294.5 ð 105 mm3 , 1 2 – 5 6 3 V2 D 1502202 D 228.08 ð 105 mm3 , V3 D 502202 D 76.03 ð 105 mm3 , V4 D 50602 D 5.65 ð 105 mm3 , V5 D 100602 D 11.31 ð 105 mm3 , V6 D 100352 D 3.85 ð 105 mm3 . The volume V D V1 V2 C V3 V4 C V5 V6 D 144.3 ð 105 mm3 , so the density is υD 14 D 9.704 ð 107 kg/mm3 . V The moment of inertia is IL D 12 υV1 2502 12 υV2 2202 C 12 υV3 2202 12 υV4 602 C 12 υV5 602 12 υV6 352 D 536,800 kg-mm2 D 0.5368 kg-m2 . 684 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.