10 3. Determine the moment of inertia of the area about the x axis.

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Bertram Eaton
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1 10 3. Determine the moment of inertia of the area about the ais. m m

2 10 4. Determine the moment of inertia of the area about the ais. m m

3 10 3. Determine the moment of inertia of the shaded area about the ais. SOLUTION Differential Element: The area of the differential element shown shaded in Fig. a is da = (rdu) dr. Moment of Inertia: Appling Eq. 10 1, we have I = LA da = L a> = L a> - a> L0 = L a> - a> = L a> - a> r 0 - a> L0 r4 4 ` r 0 sin udu 0 r sin udu r 0 r 3 sin udrdu r sin u(rdu)dr a a r 0 r 0 However, sin u = 1 (1 - cos u). Thus, I = L a> - a> r 0 4 (1 - cos u)du 8 = r Bu - 1 sin ur ` a> - a> = r 0 4 (a - sin a) 8

4 10 4. Determine the moment of inertia of the shaded area about the ais. a r 0 SOLUTION Differential Element: The area of the differential element shown shaded in Fig. a is da = (rdu)dr. Moment of Inertia: Appling Eq. 10 1, we have I = LA da = L a> = L a> - a> L0 = L a> - a> = L a> - a> r 0 r4 4 ` r 0 cos udu 0 r a> L0 r 3 cos udrdu 4 cos udu r 0 r cos u(rdu)dr a r 0 However, cos u = 1 ( cos u + 1). Thus, a> 4 r 0 I = ( cos u + 1)du L 8 - a> = r B 1 sin u + ur ` a> - a> = r 0 4 ( sin a + a) 8

5 Determine the distance to the centroid of the beam s cross-sectional area; then determine the moment of inertia about the ais. _ 5 mm 50 mm 75 mm 75 mm 50 mm C 5 mm 5 mm SOLUTION Centroid: The area of each segment and its respective centroid are tabulated below. 5 mm Segment A (mm ) (mm) A(mm 3 ) 1 50(100) (10 3 ) 35(5) l.565(10 3 ) 3 5(100) 50 15(10 3 ) 15.65(10 3 ) (10 3 ) Thus, = A A = (103 ) 15.65(10 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, (100) (50) (1003 ) (10 6 ) (10 6 ) 35(5) (35) (53 ) 0.815(10 6 ) 1.36(10 6 ) 3 5(100) (5) (1003 ) (10 6 ) 15.4(10 6 ) I = (I ) i = 34.41A10 6 B mm 4 = 34.4A10 6 B mm 4

6 Determine the moment of inertia of the beam s crosssectional area about the ais. 5 mm 5 mm _ 50 mm 75 mm 75 mm 50 mm C 5 mm SOLUTION 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 1 A i (mm ) [100(5)] Ad B i (mm) 100 AI B i (mm 4 ) C 1 1 (100) (53 )D AAd B i (mm 4 ) 50.0(10 6 ) AI B i (mm 4 ) 50.60(10 6 ) Thus, 1 5(35) 0 1 (5) (353 ) (10 6 ) 3 100(5) (100) (53 ) (10 6 ) I = (I ) i = 11.91A10 6 B mm 4 = 1A10 6 B mm 4

7 Determine the moments of inertia I u and I v of the shaded area. v 0 mm u SOLUTION 00 mm 45 0 mm 0 mm Moment and Product of Inertia about and Aes: Since the shaded area is smmetrical about the ais, I = 0. I = = mm 4 I = = mm 4 40 mm 00 mm Moment of Inertia about the Inclined u and u = 45, we have v Aes: Appling Eq with I u = I + I + I - I cos u - I sin u = a + cos 90-01sin 90 b110 6 = mm 4 I v = I + I - I - I cos u + I sin u = a - cos 90-01sin 90 b110 6 = mm 4

8 Determine the moments of inertia and the product of inertia of the beam s cross sectional area with respect to the u and v aes. v u SOLUTION Moments and product of Inertia with Respect to the and Aes: The perpendicular distances measured from the centroid of the triangular segment to the ais are indicated in Fig. a. C mm 150 mm I = 1 36 (400)(4503 ) = 101.5(10 6 ) mm 4 00 mm 00 mm I = B 1 36 (450)(003 ) + 1 (450)(00)(66.67 )R = 600(10 6 ) mm 4 Since the cross-sectional area is smmetrical about the ais, I = 0. Moment and product of Inertia with Respect to the u and v Aes: Appling Eq with u = 30, we have I u = I + I + I - I cos u - I sin u = B + cos 60-0 sin 60 R(10 6 ) = (10 6 ) mm 4 = 909(10 6 ) mm 4 I v = I + I - I - I cos u + I sin u = B - cos sin 60R(10 6 ) = (10 6 ) mm 4 = 703(10 6 ) mm 4 I uv = I - I = B sin u + I cos u sin cos 60 R(10 6 ) = 178.6(10 6 ) mm 4 = 179(10 6 ) mm 4

9 Solve Prob using Mohr s circle. Hint: Once the circle is established, rotate u = 60 counterclockwise from the reference OA, then find the coordinates of the points that define the diameter of the circle. v u 300 mm SOLUTION Moments and product of Inertia with Respect to the and Aes: The perpendicular distances measured from the centroid of the triangular segment to the ais are indicated in Fig. a. C mm I = 1 36 (400)(4503 ) = 101.5(10 6 ) mm 4 00 mm 00 mm I = B 1 36 (450)(003 ) + 1 (450)(00)(66.67 )R = 600(10 6 ) mm 4 Since the cross-sectional area is smmetrical about the ais, I = 0. Construction of Mohr s Circle: The center of C of the circle lies along the I ais at a distance I avg = I + I = a b(10 6 )mm 4 = 806.5(10 6 ) mm 4 The coordinates of the reference point A are [101.5, 0](10 6 ) mm 4. The circle can be constructed as shown in Fig. b. The radius of the circle is R = CA = ( )(10 6 ) = 06.5(10 6 ) mm 4 Moment and Product of Inertia with Respect to the u and v Aes: B referring to the geometr of the circle, we obtain I u = ( cos 60 )(10 6 ) = 909(10 6 ) mm 4 I v = ( cos 60 )(10 6 ) = 703(10 6 ) mm 4 I uv = 06.5 sin 60 = 179(10 6 ) mm 4

10 10 7. Locate the centroid of the beam s cross-sectional area and then determine the moments of inertia and the product of inertia ofthis area with respect to the and aes. 50 mm 450 mm 450 mm u 50 mm 400 mm C mm 800 mm Centroid: The perpendicular distances measured from the centroid of each subdivided segment to the bottom of the beam s cross sectional area are indicated in Fig. a. Thus, = Σ A ΣA C = 15(1000)(50) + [1000(400)(50)] + 600(100)(100) 1000(50) + (400)(50) + 100(100) = 85 mm Moment and Product of Inertia with Respect to the and Aes: The perpendicular distances measured from the centroid of each segment to the and aes are indicated in Fig. b. Using the parallel ais theorem, 1 3 I = (1000)(50 ) (50)(400) (50)(400 ) + 50(400)(175) + (100)(100 ) + 100(100)(5) = (10 8 ) mm 4 1 I = 1 (50)(10003 ) (400)(50 ) + 400(50)(75) + 1 = 45 (10 8 ) mm 4 Since the cross sectional area is smmetrical about the ais, I = (100)(1003 ) Moment and Product of Inertia with Respect to the u and v Aes: With = 60, I u = I + I + I I cos I sin = cos 10 0 sin 10 (10 8 ) = (10 8 ) mm 4 = 109 (10 8 ) mm 4 I v = I + I I I cos + I sin = cos sin 10 (10 8 ) = (10 8 ) mm 4 = 38 (10 8 ) mm 4

11 I uv = I I sin + I cos = sin cos 10 (10 8 ) = (10 8 ) mm 4 = 111 (10 8 ) mm 4 50 mm 500 mm 500 mm 75 mm 75 mm 400 mm 50 mm 1000 mm 600 mm 1000 mm 15 mm 4000 mm = 85 mm 175 mm 5 mm

12 Solve Prob using Mohr s circle. Centroid: The perpendicular distances measured from the centroid of each subdivided segment to the bottom of the beam s cross sectional area are indicated in Fig. a. Thus, = Σ A ΣA = 15(1000)(50) + [1000(400)(50)] + 600(100)(100) 1000(50) + (400)(50) + 100(100) = 85 mm Moment and Product of Inertia with Respect to the and Aes: The perpendicular distances measured from the centroid of each segment to the and aes are indicated in Fig. b. Using the parallel ais theorem, 1 3 I = (1000)(50 ) (50)(400) (50)(400 ) + 50(400)(175) (100)(100 ) + 100(100)(5) = (10 8 ) mm 4 1 I = 1 (50)(10003 ) (400)(50 ) + 400(50)(75) + 1 = 45 (10 8 ) mm 4 I = 60(5)( 14.35)(13.15) + 55(15)(15.65)( 14.35) = (10 4 ) mm 4 Since the cross sectional area is smmetrical about the ais, I = (100)(1003 ) Construction of Mohr s Circle: The center C of the circle lies along the u ais at a distance I avg = I + I = (108 ) = (10 8 ) mm 4 The coordinates of the reference point A are (30.44, 0) (10 8 ) mm 4. The circle can be constructed as shown in Fig. c. The radius of the circle is R = CA = ( ) (10 8 ) = 18.7 (10 8 ) mm 4 Moment and Product of Inertia with Respect to the u and v Aes: B referring to the geometr of the circle, I u = ( cos 60 ) (10 8 ) = 109 (10 8 ) mm 4 I v = ( cos 60 ) (10 8 ) = 38 (10 8 ) mm 4 I uv = (18.7 sin 60 ) (10 8 ) = 111 (10 8 ) mm 4 (10 8 mm 4 ) 50 mm 500 mm 500 mm 75 mm 75 mm 400 mm 50 mm 600 mm 1000 mm 15 mm 400 mm = 85 mm 175 mm 5 mm (10 8 mm 4 )

13 Locate the centroid of the beam s cross-sectional area and then determine the moments of inertia of this area and the product of inertia with respect to the u and v aes. The aes have their origin at the centroid C. 5 mm 5 mm v 00 mm C 60 5 mm 75 mm 75 mm u

15 Solve Prob using Mohr s circle.

16 Locate the centroid of the beam s cross-sectional area and then determine the moments of inertia and the product of inertia of this area with respect to the u and v aes. The aes have their origin at the centroid C. 0 mm v 00 mm 00 mm 0 mm C 60 0 mm 175 mm u

17 Solve Prob using Mohr s circle.

18 Determine the principal moments of inertia for the angle s cross-sectional area with respect to a set of principal aes that have their origin located at the centroid C. Use the equation developed in Section For the calculation, assume all corners to be square. 0 mm 3. mm C 3. mm SOLUTION 0 mm I = c 1 1 (0)(100) (0)(50-3.) d + c 1 1 (80)(0)3 + 80(0)( ) d = 3.14(10 6 )mm 4 I = c 1 1 (100)(0) (0)( ) d + c 1 1 (0)(80)3 + 80(0)(60-3.) d = 3.14(10 6 )mm 4 I = A = -( )(50-3.)(100)(0) - (60-3.)(3.- 10)(80)(0) = (10 6 )mm 4 I ma/min = I + I ; a I - I b + I C = 3.14(10 6 ) ; 0 + {( )(10 6 )} I ma = 4.9(10 6 )mm 4 I min = 1.36(10 6 )mm 4

19 Solve Prob using Mohr s circle. 0 mm 3. mm SOLUTION Solve Prob I = 3.14(10 6 )mm 4 I = 3.14(10 6 )mm 4 C 3. mm 0 mm I = 1.778(10 6 )mm 4 Center of circle: I + I = 3.14(10 6 )mm 4 R = ( ) + (-1.778) (10 6 ) = 1.778(10 6 )mm 4 I ma = 3.14(10 6 ) (10 6 ) = 4.9(10 6 )mm 4 I min = 3.14(10 6 ) (10 6 ) = 1.36(10 6 )mm 4

20 Determine the orientation of the principal aes, which have their origin at centroid C of the beam s crosssectional area. Also, find the principal moments of inertia. 0 mm 0 mm 150 mm C 150 mm 0 mm

22 Solve Prob using Mohr s circle.

23 10 8. Locate the centroid of the beam s cross-sectional area and then determine the moments of inertia of this area and the product of inertia with respect to the u and v aes. The aes have their origin at the centroid C. 0 mm 0 mm v 00 mm C 60 0 mm 80 mm 80mm u [100(00)(0)] + 10(0)(10) = = 79.3 mm (00)(0) + 0(10) I I = (0)(00 ) + 0(00)(0.77) + (10)(0 ) + 10(0)(69.3) = 41.70(10 ) mm = (00)(0 ) + 00(0)(70) + (0)(10 ) = 4.35(10 ) mm I u I + I I I = + cos θ Isin θ + = cos( 10 ) 0sin( 10 ) (10 6 ) 6 4 = 4.(10 ) mm I v I + I I I = + cos θ + Isin θ + = cos( 10 ) 0sin( 10 ) (10 6 ) 6 4 = 41.9(10 ) mm

24 I uv I I = sin θ + I cosθ sin( 10 ) 0cos( 10 = + ) 6 4 = 0.8(10 ) mm mm 70 mm 0.77 mm mm 69.3 mm 60 60

25 Solve Prob using Mohr s circle. [100(00)(0)] + 10(0)(10) = = 79.3 mm (00)(0) + 0(10) (0)(00 ) 0(00)(0.77) (10)(0 3 ) 10(0)(69.3) I = = 41.70(10 6 ) mm I (00)(0 ) 00(0)(70) (0)(10 3 = + ) = 4.5(10 6 ) mm 4 I + I I (10 6 )mm (10 6 )mm 4 avg = = = R = CA = ( )(10 6 ) = 0.75(10 6 ) mm 4 I ( cos60 )(10 6 ) 41.86(10 6 )mm 4 v = = I ( cos60 )(10 6 ) 4.19(10 6 )mm 4 u = = I 0.35sin60 0.8(10 6 )mm 4 uv = = I u 70 mm 70 mm 0.77 mm mm 69.3 mm R = I v 4.5

26 Determine the mass moment of inertia I of the solid formed b revolving the shaded area around the ais. The total mass of the solid is 1500 kg. z 4m z m O SOLUTION Differential Element: The mass of the disk element shown shaded in Fig. a is dm = rdv = rpr. Here,.Thus, dm = rpa > b d d r = z = 1 = rp 16 3 d 4 3> The mass moment of inertia of this element about the ais is di = 1. dmr = 1 Arpr dbr = rp r4 d = rp a 1 4 3> b 4d = rp 51 6 d Mass: The mass of the solid can be determined b integrating dm. Thus, 4m rp m = dm = L L 16 3 d = rp ` 4 m 0 0 = 4pr The mass of the solid is m = 1500 kg. Thus, 1500 = 4pr r = 375 p Mass Moment of Inertia: Integrating di, 4m rp I = di = L L Substituting r = 375 into I, p kg>m3 0 kg>m d = rp ` 4m 0 = 3p 7 r I = 3p 7 a 375 p b = 1.71(103 )kg# m

27 Determine the mass moment of inertia of the assembl about the z ais. The densit of the material is 7.85 Mg> m 3. z SOLUTION Composite Parts: The assembl can be subdivided into two circular cone segments (1) and (3) and a hemispherical segment () as shown in Fig. a. Since segment (3) is a hole, it should be considered as a negative part. From the similar triangles, we obtain z z = Mass: The mass of each segment is calculated as z = 0.5m 300 mm 450 mm 300 mm m 1 = rv 1 = ra 1 3 pr hb = 7.85(10 3 )c 1 3 p(0.3 )(0.675) d = p kg m = rv = ra 3 pr3 b = 7.85(10 3 )c 3 p(0.33 ) d = 141.3p kg m 3 = rv 3 = ra 1 3 pr hb = 7.85(10 3 )c 1 3 p(0.1 )(0.5) d = p kg Mass Moment of Inertia: Since the z ais is parallel to the ais of the cone and the hemisphere and passes through their center of mass, the mass moment of inertia can be 3 computed from (I and. Thus, 10 m 3r3 z ) 1 = 3 10 m 1r1, (I z ) = 5 m r, I z = (I z ) i = 3 10 ( p)(0.3 ) + 5 (141.3p)(0.3 ) (5.8875p)(0.1 ) = 9.4 kg # m

28 The thin plate has a mass per unit area of 10 kg>m. Determine its mass moment of inertia about the ais. z 00 mm 00 mm SOLUTION 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 (1) and () are m and m = p(0.1 1 = 0.4(0.4)(10) = 1.6 kg )(10) = 0.1p 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 1 1 (1.6)(0.4 ) + 1.6(0. ) d - c 1 4 (0.1p)(0.1 ) + 0.1p(0. ) d 00 mm 00 mm 00 mm 00 mm 00 mm 00 mm = kg # m

29 The thin plate has a mass per unit area of 10 kg>m. Determine its mass moment of inertia about the z ais. z 00 mm 00 mm SOLUTION 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 (1), (), (3), and (4) are m and m 3 = m 4 = p(0.1 1 = m = 0.4(0.4)(10) = 1.6 kg )(10) = 0.1p 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 00 mm 00 mm 00 mm 00 mm 00 mm 00 mm = 1 1 (1.6)(0.4 ) + c 1 1 (1.6)( ) + 1.6(0. ) d (0.1p)(0.1 ) - c 1 (0.1p)(0.1 ) + 0.1p(0. ) d = kg # m

30 Determine the mass moment of inertia of the overhung crank about the ais. The material is steel having a densit of r = 7.85 Mg>m 3. 0 mm 90 mm 30 mm 50 mm 180 mm 0 mm 30 mm 0 mm 50 mm 30 mm = kg m = 3.5 g m.

31 Determine the mass moment of inertia of the overhung crank about the ais. The material is steel having a densit of r = 7.85 Mg>m 3. 0 mm 90 mm 30 mm 50 mm 180 mm 0 mm 30 mm 0 mm 50 mm 30 mm kg m

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

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 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 - ) 3-5 4 R m m I = 39. m 4 6. Determine the moment of inertia for the