SOLUTION Determine the moment of inertia for the shaded area about the x axis. I x = y 2 da = 2 y 2 (xdy) = 2 y y dy

Transcription

1 5. Determine the moment of inertia for the shaded area about the ais. 4 4m 4 4 I = da = (d) 4 = 4 - d I = B (5 + (4)() + 8(4) ) (4 - ) R m m I = 39. m 4

2 6. Determine the moment of inertia for the shaded area about the ais. 4 4m I = da = (4 - ) d A m m = B R I = 8.53 m 4

3 7. Determine the moment of inertia for the shaded area about the ais. m m.5 m di = d ()3 I = di = 3 ( -.5 )3> d = 3 B 5(-.5) ( -.5)5> R =.533 m 4 Also, da = d = ( - ) d I = da = ( - ) d - = B R - =.533 m 4

4 * 8. Determine the moment of inertia for the shaded area about the ais. m m.5 m da = d I = da = ( -.5) / d = B (8 - (-.5) + 5(-.5) )( -.5) 3 5(-.5) 3 R =.4 4 Also, I = di = 3 3 d 8 = 3 ( - ) 3 d = a 8 3 b B R =.4 4

5 9. Determine the moment of inertia of the shaded area about the ais. Here, the area must be divided into two segments as shown in Fig. a. The moment of inertia of segment () about the ais can be determined using (I ) = while the moment of inertia of segment () about the ais bh3 + A h, can be determined b appling Eq.. The area of the rectangular differential element in Fig. a is da = ( - )d. Here, = 4. Thus, da = 4. - d m 4 m Appling Eq. to segment () about the ais (I ) = da = 4 A m - d = m a4 - bd = ` m = 9 m 4 The moment of inertia of segment () about the ais is (I ) = (3)(3 ) + (3)()(.5 ) = m 4 Thus, I = (I ) + (I ) = 9 + = m 4 The area of the rectangular differential element in Fig. b is da = d. The moment inertia of this element about the ais is di = di + da '. Here,. Thus, di = = = 4 d = 64 d3 + d = 3 3 d 3 3d Performing the integration, we have 64 3 I = dl = d = 3 - m 3 3 ` m = m 4

6 *. Determine the moment of inertia of the shaded area about the ais. The area of the rectangular differential element in Fig. a is da = d. Here, = 4. 4 Thus, da = 4. d Appling Eq., I = da = a 4 A m bd = 4 d = a b ` = 3 m 4 m m Here, the area must be divided into two segments as shown in Fig. b. The moment of inertia of segment () about the ais can be determined using (I ) =, while the moment of inertia of segment () about the bh3 + A h ais can be determined b computing the moment of inertia of the element parallel to the ais shown in Fig. b. The area of this element is da = ( - ) d and its moment of inertia about the ais is m m di = di + da ' = (d)( - )3 + ( - )db ( + )R = a d Here, = 4. Thus, dl = B Rd = d Performing the integration, the moment of inertia of segment () about the ais is (I ) = di = m d = ` = 9 m 4 m The moment of inertia of segment () about the ais is Thus, (I ) = ()(33 ) + ()(3)(.5 ) = m 4 I = (I ) + (I ) = 9 + = 3 m 4

7 46. Determine the distance to the centroid of the beam s cross-sectional area; then determine the moment of inertia about the ais. _ 5 mm 5 mm 75 mm 75 mm 5 mm mm C 5 mm mm 5 mm Centroid: The area of each segment and its respective centroid are tabulated below. 5 mm Segment A (mm ) (mm) A(mm 3 ) 5() ( 3 ) 35(5).5 l.565( 3 ) 3 5() 5 5( 3 ) 5.65( 3 ) ( 3 ) Thus, = A A = (3 ) 5.65( 3 =.5 mm ) Moment of Inertia: The moment of inertia about the ais for each segment can be determined using the parallel-ais theorem I + I + Ad. Segment A Ad AI AAd B AI B i (mm 4 i (mm 4 B i (mm 4 i (mm ) B i (mm) ) ) ) Thus, 5() 5.5 (5) (3 ) 3.78( 6 ) 7.948( 6 ) 35(5) (35) (53 ).85( 6 ).36( 6 ) 3 5() 7.5 (5) (3 ) 3.4( 6 ) 5.4( 6 ) I = (I ) i = 34.4A 6 B mm 4 = 34.4A 6 B mm 4

8 47. Determine the moment of inertia of the beam s crosssectional area about the ais. 5 mm 5 mm _ 5 mm 75 mm 75 mm 5 mm mm C mm 5 mm 5 mm Moment of Inertia: The moment of inertia about the ais for each segment can be determined using the parallel-ais theorem I = I + Ad. Segment A i (mm ) [(5)] Ad B i (mm) AI B i (mm 4 ) C () (53 )D AAd B i (mm 4 ) 5.( 6 ) AI B i (mm 4 ) 5.6( 6 ) Thus, 5(35) (5) (353 ) 7.57( 6 ) 3 (5) () (53 ).3( 6 ) I = (I ) i =.9A 6 B mm 4 = A 6 B mm 4

9 63. Determine the product of inertia for the beam s crosssectional area with respect to the u and v aes. 5 mm v 5 mm u Moments of inertia I and I I = (3)(4)3 - (8)(36)3 = 5.36() 6 mm 4 mm C mm mm I = c ()(3)3 d + (36)()3 = 9.4() 6 mm 4 The section is smmetric about both and aes; therefore I =. I uv = I - I sin u + I cos u = a sin 4 + cos 4 b 6 = 35() 6 mm 4

10 * 64. Determine the moments of inertia for the shaded area with respect to the u and v aes. v.5 in. u Moment and Product of Inertia about and Aes: Since the shaded area is smmetrical about the ais, I =. 5in. 3.5 in..5 in. I = ()(53 ) + (4)(3 ) =.75 in 4 I = ()(43 )+(4)(.5) + (5)(3 ) = 3.75 in 4 Moment of Inertia about the Inclined u and v Aes: Appling Eq. -9 with u = 3, we have in. 4in. I u = I + I + I - I cos u - I sin u = cos 6 - (sin 6 ) = 5.75 in 4 I v = I + I - I - I cos u + I sin u = cos 6 + (sin 6 ) = 5.75 in 4

11 74. Determine the principal moments of inertia for the beam s cross-sectional area about the principal aes that have their origin located at the centroid C. Use the equations developed in Section.7. For the calculation, assume all corners to be square. 3 8 in. 4in. 3 8 in. C 4in. I = c (4)a 3 8 b 3 + 4a 3 8 ba4-3 6 b d + a 3 8 ba8-6 8 b 3 4in. = in 4 I = c a 3 8 ba4-3 8 b 3 + (8)a 3 8 b a4-3 8 bea4-3 8 b f d 4in. = 3.89 in 4 I = A = -[( )(3.83)(3.65)(.375)] + = -.73 in 4 I ma>min = I + I ; a I - I b + I C = ; C a b + (-.73) I ma = 64. in 4 I min = 5.33 in 4

12 75. Solve Prob. 74 using Mohr s circle. 4in. 3 8 in. 3 8 in. C 4 in. I = c (4)a 3 8 b 3 + 4a 3 8 ba4-3 6 b d + a 3 8 ba8-6 8 b 3 = in 4 4 in. I = c a 3 8 ba4-3 8 b a4-3 8 bea4-3 8 b f d 4in. + (8)a 3 8 b 3 = 3.89 in 4 I = A = -[( )(3.83)(3.65)(.375)] + = -.73 in 4 Center of circle: I + I = 34.7 in 4 R = ( ) + (-.73) = 9.39 in 4 I ma = = 64. in 4 I min = = 5.33 in 4

13 6. The thin plate has a mass per unit area of kg>m. Determine its mass moment of inertia about the ais. z mm mm mm Composite Parts: The thin plate can be subdivided into segments as shown in Fig. a. Since the segments labeled () are both holes, the should be considered as negative parts. Mass moment of Inertia: The mass of segments () and () are m and m = p(. =.4(.4)() =.6 kg )() =.p kg. The perpendicular distances measured from the centroid of each segment to the ais are indicated in Fig. a. The mass moment of inertia of each segment about the ais can be determined using the parallel-ais theorem. I = AI B G + md = c (.6)(.4 ) +.6(. ) d - c 4 (.p)(. ) +.p(. ) d mm mm mm mm mm mm mm =.44 kg # m

14 7. The thin plate has a mass per unit area of kg>m. Determine its mass moment of inertia about the z ais. z mm mm mm Composite Parts: The thin plate can be subdivided into four segments as shown in Fig. a. Since segments (3) and (4) are both holes, the should be considered as negative parts. Mass moment of Inertia: Here, the mass for segments (), (), (3), and (4) are m and m 3 = m 4 = p(. = m =.4(.4)() =.6 kg )() =.p kg. The mass moment of inertia of each segment about the z ais can be determined using the parallel-ais theorem. I z = AI z B G + md mm mm mm mm mm mm mm = (.6)(.4 ) + c (.6)( ) +.6(. ) d - 4 (.p)(. ) - c (.p)(. ) +.p(. ) d =.3 kg # m