Area ∫ 1m Ans - Faculty Personal Homepage- KFUPMfaculty.kfupm.edu.sa/.../CE202/S_Manual/SM_CH06.pdf · SM_CH06.indd 365 4/8/11 11:52:25 AM © 2011 Pearson Education, Inc., Upper

© 2011 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.

365813

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

•9–9. Determine the area and the centroid of the area.(x, y) y

1 m

1 m

y2 x3

x

09a Ch09a 806-861.indd 813 6/17/09 11:35:01 AM

Differential Element: The area element parallel to the y axis shown shaded in Fig. a will be considered. The area of the element is

Ans

Ans

Ans

dA = y dx = x dx 3/2

Centroid: The centroid of the element is located at x = x and y = y/2 =

Area: Integrating,

x

2

3/2

5 52dA ＝ x dx ＝＝ m2 ＝ 0.4 m2

A∫

0 0

1m 1m

∫ A = 3/2 x5/2 2

x =

x dAA

∫dA

A∫= =

0

1m

∫0

1m

∫

72

1m

0= m = 0.714 m

2/5 2/5 2/5＝

x x dx3/2

x dx5/2

75

x7/2

y =

y dAA

∫dA

A∫= =

0

1m

∫0

1m

∫ 8x

1m

0= m = 0.3125 m

2/5 2/5 2/5＝

x dx

165

3/2x2

3/2x dx3 4x

2

•6–1.

SM_CH06.indd 365 4/8/11 11:52:25 AM


366814


9–10. Determine the area and the centroid of the area.(x, y) y

x

3 ft

3 ft y x31––91 m

1 m

y = x3

1 m

1 m

y = x3


dA = y dx = x3 dx

Centroid: The centroid of the element is located at x = x and y = y / 2 = 1

2x3.

Area: Integrating,

A = dAA∫ = x3

0

1 m

∫ dx = 1

4x4 dx

1 m

0

= 0.25 m2 Ans

x = x dA

dA

A

A

∫∫

= x x dx( )

0.25

3

0

1 m

∫ =

x dx4

0

1 m

0.25

∫ =

1

5

0.25

1 m

0

x5

= 0.8 m Ans

y = y dA

dA

A

A

∫∫

=

1

2( )

0.25

3 3

0

1 m

x x dx

∫

=

1

20.25

6

0

1 m

x dx∫ =

1

14

0.25

1 m

0

x7

= 0.2857 m Ans

09a Ch09a 806-861.indd 814 6/17/09 11:35:01 AM

6–2.

SM_CH06.indd 366 4/8/11 11:52:26 AM


367816


* 9–12. Locate the centroid of the area.x y

x

2 ft

x1/2 2x5/3y

1 m

y

x

1 m

(x, y)

(x, y)

dy ••

1 m

Area and Moment Arm : The area of the differential element is

dA = xdy = y4 dy and its centroid is x = x

2 =

1

2y4.

Centroid : Applying Eq. 9–4 and performing the integration, we have

x = x dA

dA

A

A

∫∫

=

1

2( )4 4

0

1 m

4

0

1 m

y y dy

y dy

∫∫

=

1

18

1

5

91 m

0

51 m

0

y

y

= 5

18 m 8772.0 = Ans

09a Ch09a 806-861.indd 816 6/17/09 11:35:03 AM

•6–3.

SM_CH06.indd 367 4/8/11 11:52:26 AM


368817


• 9–13. Locate the centroid of the area.y y

x

2 ft

x1/2 2x5/3y

1 m

Area and Moment Arm : The area of the differential element is dA = xdy = y4 dy = x2 dx and its centroid is y = y.


y = y dA

dA

A

A

∫∫

= y y dy

y dy

( )4

0

1 m

4

0

1 m

∫∫

=

1

6

1

5

61 m

0

51 m

0

y

y

= 5

6 m 3338.0 = Ans

09a Ch09a 806-861.indd 817 6/17/09 11:35:03 AM

*6–4.

SM_CH06.indd 368 4/8/11 11:52:26 AM


369818



x

a

b

xy c2

09a Ch09a 806-861.indd 818 6/17/09 11:35:04 AM

Differential Element: The element parallel to the y axis shown shaded in Fig. a will be considered. The area of the element is

dA = y dx = c2

dxx

Centroid: The centroid of the element is located at x = x and y = =

Area: Integrating,

y

2 2x

c2

A = dAA∫ =

a

b

∫ dx c In x = = b

a

c

xc In2

b

a

22

x = x dA

dA

A

A

∫∫

= x

a

b

∫ =

c dx x2

a

b

∫ =

c

x b – a

b

ac2

2

dx

=ba

c In2 ba

c In2 ba

c In ba

In2

y = y dA

dA

A

A

∫∫

= a

b

∫ =

dxa

b

∫ =

c

2x c (b – a)

b

a

2

c

x

2

2

c

2x

4

=ba

c In2 ba

c In2 ba

c In ba

2 ab In2

dxc

2x

4

2–Ans

Ans

Ans

6–5.

SM_CH06.indd 369 4/8/11 11:52:26 AM


370819



x

a

h y x2h––a2

09a Ch09a 806-861.indd 819 6/17/09 11:35:05 AM


dA = y dx = h

2

2

2 x dxa

Centroid: The centroid of the element is located at x = x and y = y/2 = 1

2

h

a2

h

2a2

2

x = x

Area: Integrating,

A = dAA∫ =

0

a

∫ x dx = = ah 1

3

h

a

h

a

0

2 x

3

32

2 a2

x

x

= x dA

dA

A

A

∫∫

= dx

0

a

∫ = 0

a

∫ =

h

4

a

0

x4

= a

h

x2

a2 dxh

x3

a2 a

1

3

1

3

1

3

3

4

2

ah ah ah

y = y dA

dA

A

A

∫∫

= 0

a

∫ =

h

2a4

2 2

4 40

a

x2x 2x dx dx∫

=

h

2a

5x

5

a

0

h

2a2

h

a2

=3

10h

1

3ah

1

3

1

3ah ah

Ans

Ans

Ans

6–6.

SM_CH06.indd 370 4/8/11 11:52:26 AM


371820


*9–16. Locate the centroid ( , ) of the area.yx y

x2 m

1 m

y 1 – x21–4

09a Ch09a 806-861.indd 820 6/17/09 11:35:05 AM

Area and Moment Arm: The area of the differential element is dA = ydx

1x . 1

2y2

2

x dx and its centroid is y = = 1–4 4

1

1–2=

Centroid: Due to symmetry

x = 0

Applying Eq. 9–4 and performing the integration, we have

y = y dA

dA

A

A

∫∫

=

2 m

– 2 m∫2 m

– 2 m∫

14

1 1 – 1 –2

2

dx

dx

14

2

x

1 –14

2

x

x

2 m

– 2 m – +

x12 160

x2 2

5

3 x5

2 m

– 2 m

= m

x –x12

3

=

Ans

Ans

6–7.

SM_CH06.indd 371 4/8/11 11:52:27 AM


372

825


•9–21. Locate the centroid of the shaded area.x y

x

a

ka

y 2k(x )x2—2a

9–22. Locate the centroid of the area.x y

x

2 in.

2 in.

y 1

0.5 in.

0.5 in.

x

12 mm

50 mm

50 mm

12 mm

600x

12 mm

12 mm 38 mm

50 mm


dA = ydx = 600

xdx and its centroid is x = x.


x = x dA

dA

A

A

∫∫

=

xx

dx

xdx

600

600

12 mm

50 mm

12 mm

50 mm

∫

∫

=

600

600 ln

50 mm

12 mm

50 mm

12 mm

x

x

= 26.627 mm Ans

09a Ch09a 806-861.indd 825 6/17/09 11:35:17 AM

Area and Moment Arm : The area of the differential element is dA = ydx


dx and its centroid is x = = x.2a

x –x

2k2

2

3

x

x 2k

= x dA

dA

A

A

∫∫

=

=

0

a

∫0

a

∫

x

2a

x – dx

2

2kx

2ax – dx

x

3

5a

8

4x

8a2k

–

0

2x

3

3 a

a

x

6a2k

–

0

= Ans

*6–8.

SM_CH06.indd 372 4/8/11 11:52:27 AM


373826


9–23. Locate the centroid of the area.y y

x

2 in.

2 in.

y 1

0.5 in.

0.5 in.

x

*9–24. Locate the centroid ( , ) of the area.yx y

x

9 ft

3 ft

y 9 x2

12 mm

50 mm

50 mm

12 mm

600x

12 mm

12 mm 38 mm

50 mm

1 m

3 m

y = 3 (1 – x2)

3 m

1 m


dA = ydx = 600

xdx and its centroid is y =

y

2 =

300

x.


y = y dA

dA

A

A

∫∫

=

300 600

600

12 mm

50 mm

1

x xdx

xdx

∫

22 mm

50 mm

∫

=

–( )( )300 6001

600 ln

50 mm

12 mm

50 mm

12 mm

x

x

= 13.31 mm Ans

dA = y dx = 3(1 – x2) dx

x = x

y = y

2 =

3

2(1 – x2)

x = x dA

dA

A

A

∫∫

= x x dx

x dx

[3(1 – )]

3(1 – )

2

0

1

2

0

1

∫

∫ = 0.375 m Ans

y = y dA

dA

A

A

∫∫

=

3

2(1 – ) [3(1 – )]

3(1 – )

2 2

0

1

2

0

1

x x dx

x dx

∫

∫= 1.2 m Ans

09a Ch09a 806-861.indd 826 6/17/09 11:35:21 AM

6–9.

6–10.

SM_CH06.indd 373 4/8/11 11:52:27 AM


374827


•9–25. Determine the area and the centroid of thearea.

(x, y) y

x

y

y x

3 ft

3 ft

x3––9

3 m

3 m

09a Ch09a 806-861.indd 827 6/17/09 11:35:22 AM

Differential Element: The element parallel to the y axis shown shaded in Fig. a will beconsidered. The area of the element is

dA ( = ) dx = x3

x – dx9

y1 – y2

Centroid: The centroid of the element is located at x = x and y = (y + y ) =

Area: Integrating,

1

2

1

2

x

91 2

3

x +

= dAAA∫ =

x

90

3 m

dx =x –∫ = 3 3 m

0

2.25 m

x

36 –

4x

2

22

x = x dA

dA

A

A

∫∫

=

3 m

0∫3 m

0∫x9

x –x x –

dx

x9

2.25 2.25 2.25

dx

3

= = = 1.6 m

24 x

3

3 x45

5

–3 m

0

3 m

0

y = y dA

dA

A

A

∫∫

= =

3 m

0∫ x +x9

12

2.25 2.25

dx dx

3 3 m

0∫ x –x81

12

62x –

x9

3

x567

12

2.25

7x3

3

–

= 1.143 m = 1.14 mAns

Ans

Ans

3 m

3 m

•6–11.

SM_CH06.indd 374 4/8/11 11:52:27 AM


375828


9–26. Locate the centroid of the area.x y

x1 m

y x2

1 m

y2 x

9–27. Locate the centroid of the area.y y

x1 m

y x2

1 m

y2 x

09a Ch09a 806-861.indd 828 6/17/09 11:35:22 AM

Area and Moment Arm: Here, y1 = x and y2 = x . The area of the differential element is dA = (y1 – y2 ) dx = x – x dx and its centroid is x = x.

2

2

12

12

Centroid: Applying Eq. 9 – 4 and performing the integration, we have

x = x dA

dA

A

A

∫∫

=

1 m

0∫1 m

0∫

– dxx x

� �1

2 x 2

– dxx

12 x 2

– x

32 x 4

1 m

0

– x

12 x3

2

5

2

3

1

4

1

3

9

201 m

0

= m = 0.45 mAns

Area and Moment Arm: Here, y1 = x and y2 = x . The area of the differential element is dA = (y1 – y2 ) dx = x – x dx and its centroid is

2

2

12

12

2

121

2

1

2

y1 – y2

2y = y2 + = (y1 + y2 ) = x + x .

Centroid: Applying Eq. 9 – 4 and performing the integration, we have

y =

=

y dA

dA

A

A

∫∫

=

1 m

0∫1 m

0∫

+ x

�1

2 x 2 – dxx

12 x 2

– dxx

12 x 2

1

2�

–

x5

3

x 2

x x

1

2

1

5 9

20

1

2

2

5

1

3

1 m

0

1 m

0

= m = 0.45 m Ans32 –

*6–12.

6–13.

SM_CH06.indd 375 4/8/11 11:52:27 AM


376839


*9–40. Locate the center of mass of the circular coneformed by revolving the shaded area about the y axis. Thedensity at any point in the cone is defined by ,where is a constant.r0

r = (r0 >h)y

y

y

x

z

h

a

z y aa––h

09a Ch09a 806-861.indd 839 6/17/09 11:35:30 AM

Centroid: The centroid of the element is located at y = y.

y

h

2y

h+ y –

y

h

2y

h+ y –

y

h

2y

h+ y –

y

h

2y

h 3+ y –

y

5h

y

2h+ –

y

2

y

4h

2y

3h+ –

y

2

5= h Ans

Differential Element: The thin disk element shown shaded in Fig. a will be considered. The mass of the element is

� �

� �

p

– a

h

h

pa

hy + a dy =

y

h

2y

hdydm = rdV = rpz dy = y2

2r

r

h0

pa r 2 3

2

2

0

+ y –

y dy dy

= y ydm

dm

m

m

∫∫

= = = 0

h

∫

0

h

∫

3

2

2

dy

3

2

2

4

2

32

5

2

432

h

pa r 0

2

dy

3

2

2

h

pa r 0

2

h

pa r 0

0

2

0

h

∫0

h

∫

h

0

4

2

32 h

0

6–14.

SM_CH06.indd 376 4/8/11 11:52:28 AM


377834


9–35. Locate the centroid of the homogeneous solidformed by revolving the shaded area about the y axis.

y

y

x

z

y2 (z a)2 a2

a

09a Ch09a 806-861.indd 834 6/17/09 11:35:27 AM

Differential Element: The thin disk element shown shaded in Fig. a will be considered. The volume of the element is dV = pz2 dy.

Here, z = a – – y2 . Thus,


= y

y 2a – y – 2a p

p

p

= y dV dy

dV

A

A

∫∫

= 0

2 2a

∫ 0

a

∫p

0

a

∫0

2 m

∫

� �2 2a – y

2a – y – 2a dy2 2

2 2a – y 2a – y – 2a dy

dy

2 2

2 2a – y

2a – y – 2ay 2 3

2 2a – y

a y –p 2 24

a

0

3

2a y –23

2 2a – 4a

3a

y = = =

y

4

2a

3

1

12

y

3 6

+

10 – 3 p

a – y p

2 a sin 22 –1

ya

+

a

0� �– a y

3p)2(10 – a Ans

a2

a2 – y2 2

a –

a2 – y2 dy2a2 – y2 – 2a dV = p dy = p

6–15.

SM_CH06.indd 377 4/8/11 11:52:28 AM


378835


*9–36. Locate the centroid of the solid.z

y

z

x

a

z a1

a a y( )2

09a Ch09a 806-861.indd 835 6/17/09 11:35:27 AM

32

12

The volume of the differential thin-disk element dV = py dz = p(a – az) dz

22

2dV = p a + az – 2a z dz and z = z

z

z a + az – 2a z p

p

= z dV dz

dzdV

V

V

∫∫

= = 0

2a

∫0

a

∫

� �a + az – 2a z 2

32

12

32

12

1

5a Ans

*6–16.

SM_CH06.indd 378 4/8/11 11:52:28 AM


379836


•9–37. Locate the centroid of the homogeneous solidformed by revolving the shaded area about the y axis.

y z

y

x

z2 y31––16

2 m

4 m

09a Ch09a 806-861.indd 836 6/17/09 11:35:28 AM

Differential Element: The thin disk element shown shaded in Fig. a will be considered. The volume of the element is

p

163

1

163dV = pz2 dy = p y dy = y dy


p

16p

16

p

16p

16

p

16

y

5

p

16

y

4

y

y y dy

= y dV

dV

A

A

∫∫

= = 0

4 m

∫0

4 m

∫

3

y dy3

y dy

= 0

4 m

∫ 4

y dy0

4 m

∫ 3

5

= 3.2 m

4 m

0

4

4 m

0

Ans

•6–17.

SM_CH06.indd 379 4/8/11 11:52:28 AM


380837


9–38. Locate the centroid of the homogeneous solidfrustum of the paraboloid formed by revolving the shadedarea about the z axis.

z

a

z (a2 y2)h–a2

h–2

h–2

z

x

y

09a Ch09a 806-861.indd 837 6/17/09 11:35:29 AM

Centroid: The centroid of the element is located at zc = z

� � 2

9Ans

Differential Element: The thin disk element shown shaded in Fig. a will be considered. The volume of the element is

a

h

2

dV = py2 dz = p – z dz 2a

a

hz

z p a – z dz

= z dV

dV

A

A

∫∫

= 0

h/2

∫0

h/2

∫

22

a

h p a – z dz

22 a

h a – z dz2

2

a

h p a z – z dz

22

22

2 3a

2 p z – z

= = 0 0

h/2 h/2

0

h/2

∫ p

0

h/2

∫a

2h a z – z2

22 p

= h

2a

3h

6–18.

SM_CH06.indd 380 4/8/11 11:52:29 AM


381851


*9–52. Locate the centroid of the cross-sectional area ofthe concrete beam.

x

y

30 mm

60 mm

30 mm

270 mm

30 mm

120 mm 120 mm

345 mm

30 mm

195 mm

Centroid: The centroid of each composite segment is shown in Fig. a,

y = ΣΣyA

A =

30(120)(60) + 195(270)(60) + 345(240)(30)

1200(60) + 270(60) + 240(30) = 191 mm Ans

09a Ch09a 806-861.indd 851 6/17/09 11:35:40 AM

6–19.

SM_CH06.indd 381 4/8/11 11:52:29 AM


382852


•9–53. Locate the centroid of the cross-sectional area ofthe built-up beam.

y

x

60 mm10 mm

10 mm10 mm30 mm30 mm

60 mm

10 mm

55 mm

90 mm

30 mm


y = ΣΣyA

A =

30[2(60)(10)] + 55(60)(10) + 90(60)(10)

2(60))(10) + 60(10) + 60(10) = 51.25 mm Ans

09a Ch09a 806-861.indd 852 6/17/09 11:35:42 AM

*6–20.

SM_CH06.indd 382 4/8/11 11:52:29 AM


383853


9–55. Locate the distance to the centroid of themember’s cross-sectional area.

x

y

50 mm

600 mm

50 mm

100 mm

300 mm 300 mm

9–54. Locate the centroid of the channel’s cross-sectional area.

20 mm

40 mm

20 mm120 mm

20 mm

C

y

150 mm

50 mm

475 mm

60 mm

20 mm

20 mm

20 mm

10 mm

30 mm

120 mm

150 mm

Centroid : The area of each segment and its respective centroid are tabulated below.

Segment A (mm2) y (mm) yA (mm3) 1 60(40) 30 72000 2 120(20) 10 24000

Σ 4800 96000

Thus,

y = ΣΣyA

A =

96000

4800 = 20 mm Ans

ΣyA = 50(600)(100) + 2(150)1

2

(250)(150) + 475(750)(100)

= 44.25 (106) mm3

ΣA = 600(100) + (2)1

2

(250)(150) + 750(100)

= 17.25 (104) mm2

y = ΣΣyA

A =

44.25 (10 )

17.25 (10 )

6

4 = 257 mm Ans

09a Ch09a 806-861.indd 853 6/17/09 11:35:43 AM

6–21.

6–22.

SM_CH06.indd 383 4/8/11 11:52:29 AM


384854


*9–56. Locate the centroid of the cross-sectional area ofthe built-up beam.

y

x

15 mm

15 mm

115 mm

15 mm

35 mm

15 mm40 mm 40 mm

122.5 mm112.5 mm

57.5

mm


y = ΣΣyA

A =

57.5(115)(15) + 122.5(80)(15) + 112.5(35)(155) + 112.5(35)(15)

115(15) + 80(15) + 35(15)) + 35(15) = 91.7 mm Ans

09a Ch09a 806-861.indd 854 6/17/09 11:35:44 AM

6–23.

SM_CH06.indd 384 4/8/11 11:52:30 AM


385855


•9–57. The gravity wall is made of concrete. Determine thelocation ( , ) of the center of mass G for the wall.yx

y

1.2 m

x

_x

_y

0.6 m 0.6 m2.4 m

3 mG

0.4 m

09a Ch09a 806-861.indd 855 6/17/09 11:35:45 AM

ΣxA = 1.8(3.6)(0.4) + 2.1(3)(3) – 3.4 (3)(0.6) – 1.2 (1.8)(3)

= 15.192 m3

1

2

1

2

1

2

1

2ΣyA = 0.2(3.6)(0.4) + 1.9(3)(3) – 1.4 (3)(0.6) – 2.4 (1.8)(3)

= 9.648 m3

1

2

1

2ΣA = 3.6(0.4) + 3(3) – (3)(0.6) – (1.8)(3)

= 6.84 m2

x = = 2.22 mm Ans

Ans

= 15.192ΣxA

ΣA 6.84

y = = 1.41 m = 9.648ΣyA

ΣA 6.84

*6–24.

SM_CH06.indd 385 4/8/11 11:52:30 AM


386861


9–63. Locate the centroid of the cross-sectional area ofthe built-up beam.

y y

x

450 mm

150 mm150 mm

200 mm

20 mm

20 mm

09a Ch09a 806-861.indd 861 6/17/09 11:35:51 AM

Centroid: The centroid of each composite segment is shown in Fig. a.

Σ

ΣA

yAy = =

= 293 mm Ans

2[225(450)(20)] + 350(200)(20) + 460(300)(20)

2(450)(20)] + 200(20) + 300(20)

6–25.

SM_CH06.indd 386 4/8/11 11:52:30 AM


387862


*9–64. Locate the centroid of the cross-sectional area ofthe built-up beam.

y

200 mm

20 mm50 mm

150 mm

y

x

200 mm

300 mm

10 mm

20 mm 20 mm

10 mm

09b Ch09b 862-915.indd 862 6/18/09 10:09:54 AM

Centroid: The centroid of each composite segment is shown in Fig. a.

y =

ΣyA =

225(450)(40) + 2[400(200)(10)] + 510 (400)(20) ΣA 450(40) + 2(200)(10) + 400(20)

= 324 mm Ans

6–26.

SM_CH06.indd 387 4/8/11 11:52:31 AM


388868


9–70. Locate the center of mass for the compressorassembly.The locations of the centers of mass of the variouscomponents and their masses are indicated and tabulated inthe figure.What are the vertical reactions at blocks A and Bneeded to support the platform?

x

y

1

2

34

Instrument panel

Filter system

Piping assembly

Liquid storage

Structural framework

230 kg

183 kg

120 kg

85 kg

468 kg

1

2

34

5

5

2.30 m1.80 m

3.15 m

4.83 m

3.26 m

A B

2.42 m 2.87 m1.64 m1.19m

1.20 m

3.68 m

09b Ch09b 862-915.indd 868 6/18/09 10:10:01 AM

Centroid : The mass of each component of the compressor and its respective centroid are tabulated below.

Component m(kg) x(m) y(m) xm(kg · m) ym(kg · m) 1 230 1.80 1.20 414.00 276.00 2 183 5.91 4.83 1081.53 883.89 3 120 8.78 3.26 1053.60 391.20 4 85 2.30 3.68 195.50 312.80 5 468 4.72 3.15 2208.96 1474.20

Σ 1086 4953.59 3338.09

Thus,

x = Σxm = 4953.59 = 4.561 m = 4.56 m Ans Σm 1086

y = Σym = 3338.09 = 3.074 m = 3.07 m Ans Σm 1086

Equations of Equilibrium :

+ΣMA = 0; By(10.42) – 1086(9.81)(4.561) = 0 By = 4663.60 N = 4.66 kN Ans

+cΣFy = 0; Ay + 4663.60 – 1086(9.81) = 0 Ay = 5990.06 N = 5.99 kN Ans

6–27.

SM_CH06.indd 388 4/8/11 11:52:31 AM


389869


9–71. Major floor loadings in a shop are caused by theweights of the objects shown. Each force acts through itsrespective center of gravity G. Locate the center of gravity( , ) of all these components.

y

G2

G4G3

G1

x

3 kN2.7 m

2.1 m

3.6 m1.8 m

2.4 m

1.2 m 0.9 m1.5 m

7.5 kN

2.25 kN

1.4 kN

0.9 m4.8 m

2.4 m

1.8 m3.6 m

2.4 m1.2 m

2.1 m0.9 m

4.8 m

2.4 m

1.8 m3.6 m

2.4 m1.2 m

2.1 m

Centroid : The floor loadings on the floor and its respective centroid are tabulated below.

Loading W (kN) x (m) y (m) xW (kN · m) yW (kN · m) 1 2.25 1.8 2.1 4.05 4.725 2 7.5 5.4 4.8 4.05 3.6 3 3.0 7.8 0.9 23.4 2.7 4 1.4 9 2.4 12.6 3.36

Σ 14.15 80.55 46.785

Thus,

x = Σ

Σ

xW

W =

80.55

14.15 = 5.69 m Ans

y = Σ

Σ

yW

W =

46.785

14.15 = 3.31 m Ans

09b Ch09b 862-915.indd 869 6/18/09 10:10:02 AM

*6–28.

SM_CH06.indd 389 4/8/11 11:52:31 AM


390870


*9–72. Locate the center of mass of thehomogeneous block assembly.

(x, y, z)

y

z

x 150 mm

250 mm

200 mm

150 mm150 mm100 mm

09b Ch09b 862-915.indd 870 6/18/09 10:10:04 AM

Centroid : Since the block is made of a homogeneous material, the center of mass of the block coincides with the centroid of itsvolume. The centroid of each composite segment is shown in Fig. a,

x = ΣxV = = 2.165625(109) = 120 mm Ans ΣV 18(106)

y = ΣyV = = 5.484375(109) = 305 mm Ans ΣV 18(106)

z = ΣyV = = 1.321875(109) = 73.4 mm Ans ΣV 18(106)

(75)(150)(150)(550) + (225)(150)(150)(200) + (200)(1)(150)(150)(100) 2

(150)(150)(550) + (150)(150)(200) + 1(150)(150)(100) 2

(75)(150)(150)(550) + (75)(150)(150)(200) + (50)(1)(150)(150)(100) 2

(150)(150)(550) + (150)(150)(200) + 1(150)(150)(100) 2

(275)(150)(150)(550) + (450)(150)(150)(200) + (50)(1)(150)(150)(100) 2

(150)(150)(550) + (150)(150)(200) + 1(150)(150)(100) 2

6–29.

SM_CH06.indd 390 4/8/11 11:52:32 AM


391871


•9–73. Locate the center of mass of the assembly. Thehemisphere and the cone are made from materials havingdensities of and , respectively.4 Mg>m38 Mg>m3

z

y

z

x

100 mm 300 mm

09b Ch09b 862-915.indd 871 6/18/09 10:10:04 AM

Centroid: The center of mass of each composite segment is shown in Fig. a.

z = Σzm = Σm

= 1.0333� = 0.1107 m = 111 mm Ans 9.3333�

4000(0.175)[1�(0.12)(0.3)] + 8000(0.1 – 3(0.1))[2

�(0.13)] 3 8 3

4000[1�(0.12)(0.3)] + 8000[2

�(0.13)] 3 3

•6–30.

SM_CH06.indd 391 4/8/11 11:52:32 AM


392298


4–142. Replace the distributed loading with an equivalentresultant force, and specify its location on the beammeasured from point A.

A

B

3 m 3 m

15 kN/m

10 kN/m

3 m

04b Ch04b 254-318.indd 298 6/12/09 8:40:27 AM

Loading: The distributed loading can be divided into four parts as shown in Fig. a. The magnitude and locationof the resultant force of each part acting on the beam are also indicated in Fig. a.Resultants: Equating the sum of the forces along the y axis of Figs. a and b,

If we equate the moments of FR, Fig. b, to the sum of the moment of the forces in Fig. a about point A,

+↓FR = ΣFy ; FR = (15)(3) + (5)(3) + 10(3) + (10)(3) = 75 kN ↓ Ans

+(MR)A = ΣMA ; –75(x) = (15)(3)(1) – (5)(3)(1) – 10(3)(1.5) – (10)(3)(4)

x = 1.20 m Ans

1

2

1

2

1

2

1

2

1

2

1

2

6–31.

SM_CH06.indd 392 4/8/11 11:52:32 AM


393299



B

A

8 kN/m

4 kN/m

3 m 3 m

04b Ch04b 254-318.indd 299 6/12/09 8:40:29 AM

+T (FR)y = ©Fy; FR = (8)(3) + (4)(3) + 4(3) = 30 kN T Ans12

12

+(MR)A = ©MA; –30(x) = – (8)(3)(2) – (4)(3)(4) – 4(3)(4.5)

(x) = 3.4 m Ans

12

12

Loading: The distributed loading can be divided into three parts as shown in Fig. a.Resultants: Equating the sum of the forces along the y axis of Figs. a and b,


*6–32.

SM_CH06.indd 393 4/8/11 11:52:33 AM


394300


*4–144. Replace the distributed loading by an equivalentresultant force and specify its location, measured frompoint A.

3 m2 m

A B

800 N/m

200 N/m

•4–145. Replace the distributed loading with anequivalent resultant force, and specify its location on thebeam measured from point A.

A B

L––2

L––2

w0 w0

04b Ch04b 254-318.indd 300 6/12/09 8:40:30 AM

+T (FR) = ©F; FR = 1600 + 900 + 600 = 3100 N

FR = 3.10 kN T Ans

+MRA = ©MA; x(3100) = 1600(1) + 900(3) + 600(3.5)

x = 2.06 Ans

1

2 +T FR = ©F; FR = w0 + w0 = w0L T Ans

L2

12

12

L2

Loading: The distributed loading can be divided into two parts as shown in Fig. a. The magnitude and location of the resultant force of each part acting on the beam are also shown in Fig. a.Resultants: Equating the sum of the forces along the y axis of Figs. a and b,


12

+(MR)A = ©MA; – w0L(x) = – w0 – w0 L

x = L Ans

L2

L6

L2

23

12

512

12

6–33.

•6–34.

SM_CH06.indd 394 4/8/11 11:52:33 AM


395301


4–146. The distribution of soil loading on the bottom ofa building slab is shown. Replace this loading by anequivalent resultant force and specify its location, measuredfrom point O.

3.6 m 2.7 m

2 kN/m1 kN/m

6 kN/m

O

4–147. Determine the intensities and of thedistributed loading acting on the bottom of the slab so thatthis loading has an equivalent resultant force that is equalbut opposite to the resultant of the distributed loadingacting on the top of the plate.

w2w1

6 kN/m

BA

1 m 2 m0.5 m

w2

w1

316

m

12

(6)(0.5) N12

(6)(1) N

12

(W2 – W1)(3.5)

213

m

23

m

W2 – W1

W1

W1(3.5)

1.75 m

2 m

6(2) N

A

+↑FR = ΣFy; FR = 1 (3.6) + 1

2 (5)(3.6)

+ 1

2 (4)(2.7) + 2(2.7)

= 23.4 kN = 23.4 kN Ans

+ MRO = ΣMO; 23.4 (d) = 1 (3.6) (1.8) + 1

2 (5)(3.6)(2.4)

+ 1

2 (4)(2.7)(4.5) + 2 (2.7)(4.95)

d = 3.38 m Ans

4.5 m2.4 m

1.8 m

4.95 m

2(2.7) N1(3.6) N

12

(4)(2.7) N12

(5)(3.6) N

O

+↑FR = ΣF; 0 = w1 (3.5) + 1

2 (w1 – w2)(3.5)

– 1

2 (6)(1) – 6 (2) –

1

2 (6)(0.5)

+ 1

2 (4)(2.7) + 2(2.7)

w1 – w2 = 9.42857 (1)

+ MRA = ΣMA; 0 = w1(3.5)(1.75) + 1

2 (w2 – w1) (3.5) 2

1

3

⎛

⎝⎜

⎞

⎠⎟ –

1

2 (6)(1)

2

3

⎛

⎝⎜⎞

⎠⎟

– 6(2)(2) – 1

2 (6)(0.5) 3

1

6

⎛

⎝⎜

⎞

⎠⎟

w1 – 2 w2 = 15. 06122 (2)

Solving Eqs. (1) and (2),

w1 = 3.766 kN/m Ans

w2 = 5.633 kN/m Ans

04b Ch04b 254-318.indd 301 6/12/09 8:40:33 AM

6–35.

*6–36.

SM_CH06.indd 395 4/8/11 11:52:34 AM


396305


*4–152. Wind has blown sand over a platform such thatthe intensity of the load can be approximated by thefunction Simplify this distributed loadingto an equivalent resultant force and specify its magnitudeand location measured from A.

w = 10.5x32 N>m.

x

w

A

10 m

500 N/m

w (0.5x3) N/m

•4–153. Wet concrete exerts a pressure distribution alongthe wall of the form. Determine the resultant force of thisdistribution and specify the height h where the bracing strutshould be placed so that it lies through the line of action ofthe resultant force. The wall has a width of 5 m.

4 m

h

(4 ) kPap1/2z

8 kPa

z

p

04b Ch04b 254-318.indd 305 6/12/09 8:40:41 AM

dA = wdx

FR = ∫ dA = ∫0

10

x3dx1

2

= x4

= 1250 N

FR = 1.25 kN Ans

18

⎡

⎣⎢

⎤

⎦⎥

0

10

∫ xdA = ∫0

10

x4dx1

2

10 0001250

= x3

= 10 000 N · m

x = = 8.00 m Ans

110

⎡

⎣⎢

⎤

⎦⎥

0

10

Thus, h = 4 – z = 4 – 2.40 = 1.60 m Ans

Equivalent Resultant Force:

+→ FR = ΣFz ; –FR = – ∫A

dA = – ∫0

zwdz

FR = ∫0

4m(20z )(103) dz

= 106.67(103) N = 107 kN ← Ans

12

Location of Equivalent Resultant Force:

∫A

zdA

∫A

dA

∫0

zzwdz

∫0

zwdzz = =

12 ∫0

4mz[(20z )(103)] dz

12 ∫0

4m[(20z )(103)] dz=

12 ∫0

4m[(20z )(103)] dz

12 ∫0

4m[(20z )(103)] dz=

= 2.40 m

6–37.

•6–38.

SM_CH06.indd 396 4/8/11 11:52:34 AM


397306



w

xA

B

4 m

8 kN/mw (4 x)21––

2

04b Ch04b 254-318.indd 306 6/12/09 8:40:43 AM

Resultant: The magnitude of the differential force dFR is equal to the areaof the element shown shaded in Fig. a. Thus,

Integrating dFR over the entire length of the beam givens the resultant force FR.

+↓ FR = ∫L

dFR = ∫0

4m – 4x + 8 dx = – 2x2 + 8x 0

4m

= 10.667 kN = 10.7 kN ↓ Ans

dFR = w dx = (4 – x)2 dx = – 4x + 8 dx12

x2

2

x2

2

x3

6

Location. The location of dFR on the beam is xc = x, measured from point A. Thus, the location x of FR measured from point A is

x =

x2

2

∫

L

xcdFR

∫L

dFR

=

∫0

4m x – 4x + 8 dx

10.667

x4

8

4x3

3

= = 1 m Ans

– + 4x2 0

4m

10.667

6–39.

SM_CH06.indd 397 4/8/11 11:52:35 AM


398307


4–155. Replace the loading by an equivalent resultantforce and couple moment at point A.

60

6 ft

50 lb/ft

50 lb/ft

100 lb/ft

4 ft

A

B

1 kN/m

1 kN/m

2 kN/m

1.8 m

FRx = 2.338 kN

FRy = 2.55 kN

F3 = 1.2 kN

F1 = 0.9 kN

F2 = 1.8 kN

0.9 m

0.6 m

(1.8 cos 60° + 0.6) m

1.2 m

F1 = 1

2 (1.8) (1) = 0.9 kN

F2 = (1.8) (1) = 1.8 kN

F3 = (1.2) (1) = 1.2 kN

+→FRx = ΣFx; FRx = 0.9 sin 60° + 1.8 sin 60° = 2.338 kN

+↓FRy = ΣFy; FRy = 0.9 cos 60° + 1.8 cos 60° + 1.2 = 2.55 kN

FR = (2.338) + (2.55)2 2 = 3.460 kN Ans

= tan–1 2.55

2.338

⎛

⎝⎜⎞

⎠⎟ = 47.5° Ans

+ MRA = ΣMA; MRA = 0.9 (0.6) + 1.8 (0.9) + 1.2 (1.8 cos 60° + 0.6)

= 3.96 kN · m Ans

04b Ch04b 254-318.indd 307 6/12/09 8:40:44 AM

*6–40.

SM_CH06.indd 398 4/8/11 11:52:35 AM


399308


*4–156. Replace the loading by an equivalent resultantforce and couple moment acting at point B.

60

6 ft

50 lb/ft

50 lb/ft

100 lb/ft

4 ft

A

B

1 kN/m

1 kN/m

1.8 m

2 kN/m

1.2 m

F1 = 1

2 (1.8) (1) = 0.9 kN

F2 = (1.8) (1) = 1.8 kN

F3 = (1.2) (1) = 1.2 kN

+→FRx = ΣFx; FRx = 0.9 sin 60° + 1.8 sin 60° = 2.338 kN

+↓FRy = ΣFy; FRy = 0.9 cos 60° + 1.8 cos 60° + 1.2 = 2.55 kN

FR = (2.338) + (2.55)2 2 = 3.460 kN Ans

= tan–1 2.55

2.338

⎛

⎝⎜⎞

⎠⎟ = 47.5° Ans

+ MRB = ΣMB; MRB = 0.9 cos 60° (1.2 cos 60° + 1.2) + 0.9 sin 60° (1.2 sin 60°)+ 1.8 cos 60° (0.9 cos 60° + 1.2) + 1.8 sin 60° (0.9 sin 60°) + 1.2 (0.6)

MRB = 5.04 kN · m Ans

04b Ch04b 254-318.indd 308 6/12/09 8:40:46 AM

6–41.

SM_CH06.indd 399 4/8/11 11:52:35 AM


400916


•10–1. Determine the moment of inertia of the area aboutthe axis.x

y

x

2 m

2 m

y 0.25 x3

10a Ch10a 916-969.indd 916 6/19/09 3:01:33 PM

The area of the rectangular differential element in Fig. a is dA = (2 – x) dy. Since x = (4y)1/3 then dA = 2 – (4y)1/3 dy.

Ix = A y2dA

= y2 2 – (4y)1/3 dy

= (2y2 – 41/3y7/3) dy

= – (41/3)y10/3 = 0.533 m4 Ans

� �∫∫

2m

0

∫ 2m

0

� �

� �2y2

3

3

10 2m

0

•6–42.

SM_CH06.indd 400 4/8/11 11:52:36 AM


401916



y

x

2 m

2 m

y 0.25 x3

10a Ch10a 916-969.indd 916 6/19/09 3:01:33 PM

The area of the rectangular differential element in Fig. a is dA = (2 – x) dy. Since x = (4y)1/3 then dA = 2 – (4y)1/3 dy.

Ix = A y2dA

= y2 2 – (4y)1/3 dy

= (2y2 – 41/3y7/3) dy

= – (41/3)y10/3 = 0.533 m4 Ans

� �∫∫

2m

0

∫ 2m

0

� �

� �2y2

3

3

10 2m

0

917


10–2. Determine the moment of inertia of the area aboutthe axis.y

y

x

2 m

2 m

y 0.25 x3

10a Ch10a 916-969.indd 917 6/19/09 3:01:34 PM

The area of the rectangular differential element in Fig. a is dA = y dx = dx.

Iy = A x2dA

= x2 dx

= dx

= = 2.67 m4 Ans

∫∫

2m

0

∫ 2m

0

x3

4

x3

4

x5

4

x6

24

2m

0

6–43.

SM_CH06.indd 401 4/8/11 11:52:36 AM


402918


10–3. Determine the moment of inertia of the area aboutthe axis.x

y

x

y2 x31 m

1 m

10a Ch10a 916-969.indd 918 6/19/09 3:01:34 PM

The area of the rectangular differential element in Fig. a is dA = (1 – x) dy. Since x = y2/3, then dA = (1 – y2/3) dy.

Ix = A y2dA

= y2 1 – y2/3 dy

= (y2 – y8/3) dy

= – y11/3 = 0.0606 m4 Ans

∫∫

1m

0

∫ 1m

0

� �

y3

3

3

11

1m

0

*6–44.

SM_CH06.indd 402 4/8/11 11:52:36 AM


403918



y

x

y2 x31 m

1 m

10a Ch10a 916-969.indd 918 6/19/09 3:01:34 PM

The area of the rectangular differential element in Fig. a is dA = (1 – x) dy. Since x = y2/3, then dA = (1 – y2/3) dy.

Ix = A y2dA

= y2 1 – y2/3 dy

= (y2 – y8/3) dy

= – y11/3 = 0.0606 m4 Ans

∫∫

1m

0

∫ 1m

0

� �

y3

3

3

11

1m

0

919


*10–4. Determine the moment of inertia of the area aboutthe axis.y

y

x

y2 x31 m

1 m

10a Ch10a 916-969.indd 919 6/19/09 3:01:35 PM

The area of the rectangular differential element in Fig. a is dA = y dx = x3/2 dx.

Iy = A x2dA

= x2 x3/2 dx

= x7/2 dx

= x9/2 = 0.222 m4 Ans

∫∫

1m

0

∫ 1m

0

29

1m

0

6–45.

SM_CH06.indd 403 4/8/11 11:52:36 AM


404920



y

x

y2 2x

2 m

2 m

10a Ch10a 916-969.indd 920 6/19/09 3:01:35 PM

The area of the rectangular differential element in Fig. a is dA = (2 – x) dy Since x = , then dA = 2 – dy.

Ix = A y2dA

= y2 2 – dy

= 2y2 – dy

= y3 – = 2.13 m4 Ans

∫∫

2m

0

∫ 2m

0

y2

2

2m

0

y2

2

y2

2

y2

4

23

y5

10

•6–46.

SM_CH06.indd 404 4/8/11 11:52:37 AM


405920



y

x

y2 2x

2 m

2 m

10a Ch10a 916-969.indd 920 6/19/09 3:01:35 PM

The area of the rectangular differential element in Fig. a is dA = (2 – x) dy Since x = , then dA = 2 – dy.

Ix = A y2dA

= y2 2 – dy

= 2y2 – dy

= y3 – = 2.13 m4 Ans

∫∫

2m

0

∫ 2m

0

y2

2

2m

0

y2

2

y2

2

y2

4

23

y5

10

921


10–6. Determine the moment of inertia of the area aboutthe axis.y

y

x

y2 2x

2 m

2 m

10a Ch10a 916-969.indd 921 6/19/09 3:01:36 PM

The area of the rectangular differential element in Fig. a is dA = y dx = (2x)1/2 dx.

Iy = A x2dA

= x2(2x)1/2 dA

= 2x 5/2 dx

= 2 x7/2 = 4.57 m4 Ans

∫∫

2m

0

∫ 2m

0

2m

0

� 27 �

6–47.

SM_CH06.indd 405 4/8/11 11:52:37 AM


406922



y

xO

y 2x42 m

1 m

10a Ch10a 916-969.indd 922 6/19/09 3:01:37 PM

∫ 2m

0

y

2

2m

0

�

� �∫ �

y

2 � ∫ 2m

0

1

2

� y3

3

1

2

4

13

� �

The moment of inertia of the area about the x axis will be determined using the rectangular differentialelement in Fig. a. This area is

dA = (1 – x) dy = 1 – 1/4

dy

Ix = A y2dA =

y2 1 –

1/4

dy = y2 –

1/4

y 9/4 dy

= – 1/4

y13/4 = 0.205 m4 Ans

*6–48.

SM_CH06.indd 406 4/8/11 11:52:37 AM


407922



y

xO

y 2x42 m

1 m

10a Ch10a 916-969.indd 922 6/19/09 3:01:37 PM

∫ 2m

0

y

2

2m

0

�

� �∫ �

y

2 � ∫ 2m

0

1

2

� y3

3

1

2

4

13

� �

The moment of inertia of the area about the x axis will be determined using the rectangular differentialelement in Fig. a. This area is

dA = (1 – x) dy = 1 – 1/4

dy

Ix = A y2dA =

y2 1 –

1/4

dy = y2 –

1/4

y 9/4 dy

= – 1/4

y13/4 = 0.205 m4 Ans

923


*10–8. Determine the moment of inertia of the area aboutthe axis.y

y

xO

y 2x42 m

1 m

10a Ch10a 916-969.indd 923 6/19/09 3:01:37 PM

The moment of inertia of the area about the y axis will be determined using the rectangular differentialelement in Fig. a. This area is

dA = y dx = 2x4 dx

Iy = A x2dA =

x2 2x4dx =

2x6dx = x7 = 0.286 m4 Ans∫

1m

0

1m

0

∫

2

7∫ 1m

0

6–49.

SM_CH06.indd 407 4/8/11 11:52:37 AM


408924


•10–9. Determine the polar moment of inertia of the areaabout the axis passing through point .Oz

y

xO

y 2x42 m

1 m

10a Ch10a 916-969.indd 924 6/19/09 3:01:38 PM

The moment of inertia of the area about the x and y axes will be determined using the rectangular differentialelement in Figs. a and b. The area of these two elements are

dA = (1 – x) dy = 1 – 1/4

dy and dA = y dx = 2x4 dx.

Ix = A y2dA =

y2 1 –

1/4

dy = y2 –

1/4

y9/4 dy

= – 1/4

y13/4 = 0.2051 m4

Iy = A x2dA =

x2 2x4dx =

2x6dx = x7 = 0.2857 m4

Thus, the polar moment of inertia of the area about the z axis is

JO = Ix + Iy = 0.2051 + 0.2857 = 0.491 m4 Ans

y

2

2m

0

�

� �

� y3

3

1

2

4

13

∫ 2m

0

∫ �

y

2� ∫

2m

0

1

2� �

∫ 1m

0

1m

0

∫

2

7∫ 1m

0

•6–50.

SM_CH06.indd 408 4/8/11 11:52:38 AM


409925


10–10. Determine the moment of inertia of the area aboutthe x axis.

y

x

2 in.

8 in.

y x3

10–11. Determine the moment of inertia of the area aboutthe y axis.

y

x

2 in.

8 in.

y x3

2 m

8 m

2 m

8 m

2 m

8 m

2 m

8 m

Differential Element : Here, x = y12 . The area of the

differential element parallel to x axis is dA = xdy = y12 dy.

Moment of Inertia : Applying Eq. 10–1 and performing the integration, we have

Ix = A∫ y2 dA =

0

8 m

∫ y2(y12 ) dy

= 3

10

103y

⎡

⎣⎢

⎤

⎦⎥

8 m

0

= 307 m4 Ans

Differential Element : The area of the differential element parallel to y axis is dA = (8 – y) dx = (8 – x3) dx.


Iy = A∫ x2 dA =

0

2 m

∫ x2(8 – x3) dx

= 8

3–

1

63 6x x

2 m

0

= 10.7 m4 Ans

10a Ch10a 916-969.indd 925 6/19/09 3:01:38 PM

6–51.

*6–52.

SM_CH06.indd 409 4/8/11 11:52:38 AM


410926


•10–13. Determine the moment of inertia of the areaabout the y axis.

x

y

1 in.

2 in.y 2 – 2 x 3

*10–12. Determine the moment of inertia of the areaabout the x axis.

x

y

1 in.

2 in.y 2 – 2 x 3

1 m

2 m

1 m

2 m

1 m

2 m

2 m

1 m

Differential Element : The area of the differential element parallel to y axis is dA = ydx = (2 – 2x3) dx.


Iy = A∫ x2 dA =

0

1 m

∫ x2(2 – 2x3) dx

= 2

3–

1

33 6x x

1 m

0

= 0.333 m4 Ans

Differential Element : The area of the differential element parallel to y axis is dA = ydx. The moment of inertia of this element about x axis is

dIx = dI xʹ + dAy 2

= 1

12(dx)y3 + ydx

y

2

2

= 1

3(2 – 2x3)3 dx

= 1

3(–8x9 + 24x6 – 24x3 + 8) dx

Moment of Inertia : Performing the integration, we have

Ix = ∫ dIx = 1

3 0

1 m

∫ (–8x9 + 24x6 – 24x3 + 8) dx

= 1

3–

4

5+

24

5– 6 + 810 7 4x x x x

1 m

0

= 1.54 m4 Ans

y = 2 – 2x3

y = 2 – 2x3

10a Ch10a 916-969.indd 926 6/19/09 3:01:39 PM

6–53.

•6–54.

SM_CH06.indd 410 4/8/11 11:52:38 AM


411926



x

y

1 in.

2 in.y 2 – 2 x 3

*10–12. Determine the moment of inertia of the areaabout the x axis.

x

y

1 in.

2 in.y 2 – 2 x 3

1 m

2 m

1 m

2 m

1 m

2 m

2 m

1 m

Differential Element : The area of the differential element parallel to y axis is dA = ydx = (2 – 2x3) dx.


Iy = A∫ x2 dA =

0

1 m

∫ x2(2 – 2x3) dx

= 2

3–

1

33 6x x

1 m

0

= 0.333 m4 Ans

Differential Element : The area of the differential element parallel to y axis is dA = ydx. The moment of inertia of this element about x axis is


= 1

12(dx)y3 + ydx

y

2

2

= 1

3(2 – 2x3)3 dx

= 1

3(–8x9 + 24x6 – 24x3 + 8) dx


Ix = ∫ dIx = 1

3 0

1 m

∫ (–8x9 + 24x6 – 24x3 + 8) dx

= 1

3–

4

5+

24

5– 6 + 810 7 4x x x x

1 m

0

= 1.54 m4 Ans

y = 2 – 2x3

y = 2 – 2x3

10a Ch10a 916-969.indd 926 6/19/09 3:01:39 PM

927


10–14. Determine the moment of inertia of the area aboutthe x axis. Solve the problem in two ways, using rectangulardifferential elements: (a) having a thickness of dx, and (b) having a thickness of dy.

1 in. 1 in.

4 in.

y 4 – 4x2

x

y

4 m

1 m 1 m

1 m 1 m

4 m

1 m 1 m

4 m

a) Differential Element : The area of the differential element parallel to y axis is dA = ydx. The moment of inertia of this element about x axis is


= 1

12(dx)y3 + ydx

y

2

2

= 1

3(4 – 4x2)3 dx

= 1

3(–64x6 + 192x4 – 192x2 + 64) dx


Ix = ∫ dIx = 1

3 –1 m

1 m

∫ 1

3(–64x6 + 192x4 – 192x2 + 64) dx

= 1

3–

64

7+

192

5–

192

3+ 647 5 3x x x x

1 m

–1 m

= 19.5 m4 Ans

b) Differential Element : Here, x = 1

24 – y . The area of the differential

element parallel to x axis is dA = 2xdy = 4 – y dy.


Ix = A∫ y2 dA =

0

4 m

∫ y2 4 – y dy

= –2

3(4 – ) –

8

15(4 – ) –

16

105(4 –

2 23

23

yy

yy yy)

23

⎡

⎣⎢⎢

⎤

⎦⎥⎥

4 m

0

= 19.5 m4 Ans

10a Ch10a 916-969.indd 927 6/19/09 3:01:40 PM

6–55.

SM_CH06.indd 411 4/8/11 11:52:39 AM


412928


10–15. Determine the moment of inertia of the area aboutthe y axis. Solve the problem in two ways, using rectangulardifferential elements: (a) having a thickness of dx, and (b) having a thickness of dy.

1 in. 1 in.

4 in.

y 4 – 4x2

x

y

1 m 1 m

4 m

1 m 1 m

4 m

1 m 1 m

4 m

a) Differential Element : The area of the differential element parallel to y axis is dA = ydx = (4 – 4x2) dx.


Iy = A∫ x2 dA =

–1 m

1 m

∫ x2(4 – 4x2) dx

= 4

3–

4

53x x5⎡

⎣⎢

⎤

⎦⎥

1 m

–1 m

= 1.07 m4 Ans


24 – y . The moment of inertia of

the differential element about y axis is

dIy = 1

12(dy) (2x)3 =

2

3x3 dy =

1

12(4 – y)

12 dy


Iy = ∫ dIy = 1

12 0

4 m

∫ (4 – y)12 dy

= 1

12–

2

5(4 – )

32y

⎡

⎣⎢

⎤

⎦⎥

4 m

0

= 1.07 m4 Ans

10a Ch10a 916-969.indd 928 6/19/09 3:01:41 PM

*6–56.

SM_CH06.indd 412 4/8/11 11:52:39 AM


413928


10–15. Determine the moment of inertia of the area aboutthe y axis. Solve the problem in two ways, using rectangulardifferential elements: (a) having a thickness of dx, and (b) having a thickness of dy.

1 in. 1 in.

4 in.

y 4 – 4x2

x

y

1 m 1 m

4 m

1 m 1 m

4 m

1 m 1 m

4 m

a) Differential Element : The area of the differential element parallel to y axis is dA = ydx = (4 – 4x2) dx.


Iy = A∫ x2 dA =

–1 m

1 m

∫ x2(4 – 4x2) dx

= 4

3–

4

53x x5⎡

⎣⎢

⎤

⎦⎥

1 m

–1 m

= 1.07 m4 Ans


24 – y . The moment of inertia of

the differential element about y axis is

dIy = 1

12(dy) (2x)3 =

2

3x3 dy =

1

12(4 – y)

12 dy


Iy = ∫ dIy = 1

12 0

4 m

∫ (4 – y)12 dy

= 1

12–

2

5(4 – )

32y

⎡

⎣⎢

⎤

⎦⎥

4 m

0

= 1.07 m4 Ans

10a Ch10a 916-969.indd 928 6/19/09 3:01:41 PM

929


*10–16. Determine the moment of inertia of the triangulararea about the x axis.

y (b x)h––b

y

x

b

h

•10–17. Determine the moment of inertia of the triangulararea about the y axis.

y (b x)h––b

y

x

b

h

10a Ch10a 916-969.indd 929 6/19/09 3:01:42 PM

Area of the differential element (shaded) dA = xdy

where x = b – y, hence, dA = xdy = b – y dy.

Ix = A y2 dA = y2 b – y dy

= by2 – y3 dy

= y3 – y4

= bh3 Ans

b

h

b

h

∫ h

0

∫

b

h

∫ h

0

b

h

b

4h

b

3 h

0

112

Area of the differential element (shaded) dA = ydx

where y = h – x, hence, dA = ydx = h – x dx.

Iy = A x2 dA = x2 h – x dx

= hx2 – x3 dx

= x3 – x4

= hb3 Ans

Area of the differential element (shaded) dA = xdy

where x = b – y, hence, dA = xdy = b – y dy.

Ix = A y2 dA = y2 b – y dy

= by2 – y3 dy

= y3 – y4

= bh3 Ans

b

h

b

h

∫ h

0

∫

b

h

∫ h

0

b

h

b

4h

b

3 h

0

h

b

h

b

∫ h

0

∫

h

b

∫ h

0

h

b

h

4b

h

3 b

0

112

112

6–57.

•6–58.

SM_CH06.indd 413 4/8/11 11:52:40 AM


414

936


10–26. Determine the polar moment of inertia of the areaabout the axis passing through point O.z

y

x

x2 y2 r2

r0

0

10–27. Determine the distance to the centroid of thebeam’s cross-sectional area; then find the moment of inertiaabout the axis.x¿

y

2 in.

4 in.

1 in.1 in.

Cx¿

x

y

y

6 in.

20 mm

40 mm

60 mm

10 mm10 mm

20 mm

40 mm

60 mm

40 mm

10 mm

10 mm 10 mm

Centroid :

y = ΣΣyA

A =

10(60)(20) + 2[40(40)(10)]

60(20) + 2[40(10)]] = 22.0 mm Ans

Moment inertia :

Ix′ = 1

12(60)(20)3 + 60(20)(22.0 – 10)2

+ 21

12(10)(40) + 10(40)(40 – 22.0)23

= 57.9 (104) mm4 Ans

10a Ch10a 916-969.indd 936 6/19/09 3:01:47 PM

J0 = Ix + Iy = + =pr0

4

8

pr04

8

pr04

4

6–59.

SM_CH06.indd 414 4/8/11 11:52:40 AM


415937


*10–28. Determine the moment of inertia of the beam’scross-sectional area about the x axis.

2 in.

4 in.

1 in.1 in.

Cx¿

x

y

y

6 in.

•10–29. Determine the moment of inertia of the beam’scross-sectional area about the y axis.

2 in.

4 in.

1 in.1 in.

Cx¿

x

y

y

6 in.

20 mm

40 mm

60 mm

40 mm

10 mm

10 mm 10 mm

10 mm 10 mm

40 mm

20 mm

30 mm 30 mm

15 mm 15 mm

20 mm

40 mm

60 mm

10 mm10 mm

20 mm

40 mm

60 mm

10 mm10 mm

Ix = 1

12(60)(20) + (60)(20)(10)23

+ 21

12(10)(40) + (40)(10)(40)23

= 155 (104) mm4 Ans

Iy = 1

12(20)(60)3 + 2

1

12(40)(10) + 10(40)(15)3 2

= 54.7 (104) mm4 Ans

10a Ch10a 916-969.indd 937 6/19/09 3:01:48 PM

*6–60.

•6–61.

SM_CH06.indd 415 4/8/11 11:52:40 AM


416938


10–30. Determine the moment of inertia of the beam’scross-sectional area about the axis.x

y

x


100 mm

100 mm

50 mm

50 mm

15 mm

15 mm

10a Ch10a 916-969.indd 938 6/19/09 3:01:49 PM

Composite Parts: The composite cross – sectional area of the beam can be subdivided into segments as shown in Fig. a. Theperpendicular distance measured from the centroid of each segment to the x axis is also indicated.

Moment of Inertia: The moment of inertia of each segment about the x axis can be determined using the parallel – axis theorem.Thus,

Ix = Ix¿ + A(dy)2

= 2 (15)(3003) + 2(15)(300)(0)2 + 2 (120)(153) + 2(120)(15)(50)2

= 67.5(106) + 9.0675(106) = 76.6(106) mm4 Ans

� ��

�1

12

1

12

6–62.

SM_CH06.indd 416 4/8/11 11:52:41 AM


417938


10–30. Determine the moment of inertia of the beam’scross-sectional area about the axis.x

y

x


100 mm

100 mm

50 mm

50 mm

15 mm

15 mm

10a Ch10a 916-969.indd 938 6/19/09 3:01:49 PM



Ix = Ix¿ + A(dy)2

= 2 (15)(3003) + 2(15)(300)(0)2 + 2 (120)(153) + 2(120)(15)(50)2

= 67.5(106) + 9.0675(106) = 76.6(106) mm4 Ans

� ��

�1

12

1

12

939


10–31. Determine the moment of inertia of the beam’scross-sectional area about the axis.y

y

x


100 mm

100 mm

50 mm

50 mm

15 mm

15 mm

10a Ch10a 916-969.indd 939 6/19/09 3:01:52 PM



Iy = Iy¿ + A(dx)2

= 2 (300)(153) + 2(300)(15)(67.5)2 + 2 (15)(1203) + 2(120)(15)(0)2

= 41.175(106) + 4.32(106) = 45.5(106) mm4 Ans

� ��

�1

12

1

12

6–63.

SM_CH06.indd 417 4/8/11 11:52:42 AM


418940


*10–32. Determine the moment of inertia of thecomposite area about the axis.x

y

x

150 mm

300 mm

150 mm

100 mm

100 mm

75 mm

10a Ch10a 916-969.indd 940 6/19/09 3:01:55 PM

Composite Parts: The composite area can be subdivided into three segments as shown in Fig. a. Since segment (3) is a hole, it contributes a negative moment of inertia. The perpendicular distance measured from the centroid of each segment to the x axis is also indicated.

Moment of Inertia: The moment of inertia of each segment about the x axis can be determined using the parallel – axis theorem. Thus,

Ix = Ix¿ + A(dx)2

= (300)(2003) + (300)(200) + (300)(2003) + 300(200)(100)2 + – (754) + –p(752) (100)2

= 798(106) mm4 Ans

� � �� 1

36

1

12

1

2

p

42003

2

*6–64.

SM_CH06.indd 418 4/8/11 11:52:42 AM


419940


*10–32. Determine the moment of inertia of thecomposite area about the axis.x

y

x

150 mm

300 mm

150 mm

100 mm

100 mm

75 mm

10a Ch10a 916-969.indd 940 6/19/09 3:01:55 PM

Composite Parts: The composite area can be subdivided into three segments as shown in Fig. a. Since segment (3) is a hole, it contributes a negative moment of inertia. The perpendicular distance measured from the centroid of each segment to the x axis is also indicated.

Moment of Inertia: The moment of inertia of each segment about the x axis can be determined using the parallel – axis theorem. Thus,

Ix = Ix¿ + A(dx)2

= (300)(2003) + (300)(200) + (300)(2003) + 300(200)(100)2 + – (754) + –p(752) (100)2

= 798(106) mm4 Ans

� � �� 1

36

1

12

1

2

p

42003

2

941


•10–33. Determine the moment of inertia of thecomposite area about the axis.y

y

x

150 mm

300 mm

150 mm

100 mm

100 mm

75 mm

10a Ch10a 916-969.indd 941 6/19/09 3:01:56 PM

Composite Parts: The composite area can be subdivided into three segments as shown in Fig. a. Since segment (3) is a hole, it contributes a negative moment of inertia. The perpendicular distance measured from the centroid of each segment to the y axis is also indicated.

Moment of Inertia: The moment of inertia of each segment about the y axis can be determined using the parallel – axistheorem. Thus,

Iy = Iy¿ + A(dx)2

= (200)(3003) + (200)(300)(200)2 + (300)(2003) + 200(300)(450)2 + – (754) + –p(752) (450)2

= 10.3(109) mm4 Ans

� � �� 1

36

1

12

1

2

p

4

•6–65.

SM_CH06.indd 419 4/8/11 11:52:42 AM


420942


10–34. Determine the distance to the centroid of thebeam’s cross-sectional area; then determine the moment ofinertia about the axis.x¿

y

x

x¿C

y

mm 05mm 0575 mm

25 mm

25 mm

75 mm

100 mm

_y

25 mm

25 mm

100 mm

10a Ch10a 916-969.indd 942 6/19/09 3:01:57 PM

Centroid: The area of each segment and its respective centroid are tabulated below.

Segment A (mm2) y (mm) yA (mm3) 1 50(100) 7.5 375(103) 2 325(25) 12.5 101.5625(103) 3 25(100) –50 –125(103)

© 15.625(103) 351.5625(103)

Thus,

Thus,

y = = = 22.5 mm Ans©yA

©A

351.5625(103)15.625(103)

Moment of Inertia: The moment of inertia about the x¿ axis for each segment can be determined using the parallel – axis theorem Ix¿ = Ix¿ + Ady

2.

Segment Ai (mm2) (dy)i (mm) (Ix¿)i (mm4) (Ady2)i (mm4) (Ix¿)i (mm4)

1 50(100) 52.5 (50)(1003) 13.781(106) 17.948(106) 2 325(25) 10 (325)(253) 0.8125(106) 1.236(106) 3 25(100) 72.5 (25)(1003) 13.141(106) 15.224(106)

121

121

121

Ix¿ = ©(Ix¿)i = 34.41(106) mm4 = 34.4(106) mm4 Ans

6–66.

SM_CH06.indd 420 4/8/11 11:52:43 AM


421942


10–34. Determine the distance to the centroid of thebeam’s cross-sectional area; then determine the moment ofinertia about the axis.x¿

y

x

x¿C

y

mm 05mm 0575 mm

25 mm

25 mm

75 mm

100 mm

_y

25 mm

25 mm

100 mm

10a Ch10a 916-969.indd 942 6/19/09 3:01:57 PM

Centroid: The area of each segment and its respective centroid are tabulated below.

Segment A (mm2) y (mm) yA (mm3) 1 50(100) 7.5 375(103) 2 325(25) 12.5 101.5625(103) 3 25(100) –50 –125(103)

© 15.625(103) 351.5625(103)

Thus,

Thus,

y = = = 22.5 mm Ans©yA

©A

351.5625(103)15.625(103)

Moment of Inertia: The moment of inertia about the x¿ axis for each segment can be determined using the parallel – axis theorem Ix¿ = Ix¿ + Ady

2.

Segment Ai (mm2) (dy)i (mm) (Ix¿)i (mm4) (Ady2)i (mm4) (Ix¿)i (mm4)

1 50(100) 52.5 (50)(1003) 13.781(106) 17.948(106) 2 325(25) 10 (325)(253) 0.8125(106) 1.236(106) 3 25(100) 72.5 (25)(1003) 13.141(106) 15.224(106)

121

121

121

Ix¿ = ©(Ix¿)i = 34.41(106) mm4 = 34.4(106) mm4 Ans

943


10–35. Determine the moment of inertia of the beam’scross-sectional area about the y axis.

x

x¿C

y

mm 05mm 0575 mm

25 mm

25 mm

75 mm

100 mm

_y

25 mm

25 mm

100 mm

*10–36. Locate the centroid of the composite area, thendetermine the moment of inertia of this area about thecentroidal axis.x¿

y y

1 in.1 in.

2 in.

3 in.

5 in.x¿

xy

3 in.

C20 mm

50 mm

30 mm30 mm

10 mm 10 mm

Composite Parts: The composite area can be subdivided into three segments. The perpendicular distance measured from the centroid of each segment to the x axis is also indicated.

Centroid: The perpendicular distances measured from the centroid of each segment to the x axis are indicated in Fig. a,

y = Σ

Σy A

AC =

(10)(60)(20) + 2[35(30)(10)]

(60)(20) + 2[(300)(10)] = 18.33 mm Ans

Moment of Inertia: The moment of inertia of each segment about the x′ axis can be determined using the parallel – axis theorem. Thus,

Ix′ = I x + A(dy)2

= 1

12(60)(20 ) + (60)(20)(18.33 – 10)23

+ 2

1

12(10)(30 ) + 10(30)(35 – 18.33)23

= 33.5 (104) mm4 Ans

10a Ch10a 916-969.indd 943 6/19/09 3:01:57 PM

Moment of Inertia : The moment of inertia about the y¿ axis for each segment can bedetermined using the parallel – axis theorem Iy¿ = Iy¿ + Adx

2

Thus.

Segment Ai (mm2) (dx)i (mm) (Ix¿)i (mm4) (Adx2)i (mm4) (Iy¿)i (mm4)

1 2[100(25)] 100 2 (100)(253) 50.0(106) 50.130(106) 2 25(325) 0 (25)(3253) 0 71.519(106) 3 100(25) 0 (100)(253) 0 0.130(106)

121

121

121

� �

Iy¿ = ©(Iy¿)i = 121.78(106) mm4 = 122(106) mm4 Ans

6–67.

*6–68.

SM_CH06.indd 421 4/8/11 11:52:43 AM


422944


•10–37. Determine the moment of inertia of thecomposite area about the centroidal axis.y

y

1 in.1 in.

2 in.

3 in.

5 in.x¿

xy

3 in.

C


y

300 mm

100 mm

200 mm

50 mm 50 mm

y

C

x

y

x¿

20 mm

50 mm

30 mm30 mm

10 mm 10 mm

Moment of Inertia: The moment of inertia of each segment about the y axis can be determined using the parallel – axis theorem. Thus,

Iy = I y + A(dx)2

= 1

12(20)(60 )3

+ 2

1

12(30)(10 ) + 30(10)(25)3 2

= 74 (104) mm4 Ans

10a Ch10a 916-969.indd 944 6/19/09 3:01:58 PM

Centroid:

y = ΣΣyA

A =

50(100(200) + 250(100(300)

100(200) + 100(300) = 170 mm Ans

Moment of Inertia:

1

12(200)(100)3 + 200(100)(170 – 50)2

(100)(300)3 + 100(300)(250 – 170)2

Ix œ =

1

12 +

= 722(106) mm4 Ans

•6–69.

6–70.

SM_CH06.indd 422 4/8/11 11:52:44 AM


423944


•10–37. Determine the moment of inertia of thecomposite area about the centroidal axis.y

y

1 in.1 in.

2 in.

3 in.

5 in.x¿

xy

3 in.

C


y

300 mm

100 mm

200 mm

50 mm 50 mm

y

C

x

y

x¿

20 mm

50 mm

30 mm30 mm

10 mm 10 mm

Moment of Inertia: The moment of inertia of each segment about the y axis can be determined using the parallel – axis theorem. Thus,

Iy = I y + A(dx)2

= 1

12(20)(60 )3

+ 2

1

12(30)(10 ) + 30(10)(25)3 2

= 74 (104) mm4 Ans

10a Ch10a 916-969.indd 944 6/19/09 3:01:58 PM

Centroid:

y = ΣΣyA

A =

50(100(200) + 250(100(300)

100(200) + 100(300) = 170 mm Ans

Moment of Inertia:

1

12(200)(100)3 + 200(100)(170 – 50)2

(100)(300)3 + 100(300)(250 – 170)2

Ix œ =

1

12 +

= 722(106) mm4 Ans

945


10–39. Determine the moment of inertia of the beam’scross-sectional area about the x axis.

300 mm

100 mm

200 mm

50 mm 50 mm

y

C

x

y

x¿

10a Ch10a 916-969.indd 945 6/19/09 3:01:59 PM

Ix = 1

12(0.2)(0.1)3 + (0.2)(0.1)(0.05)2

+1

12(0.1)(0.3)3 + (0.1)(0.3)(0.25)2 = 2.17(10–3) m4 Ans

6–71.

SM_CH06.indd 423 4/8/11 11:52:44 AM


424946


*10–40. Determine the moment of inertia of the beam’scross-sectional area about the y axis.

300 mm

100 mm

200 mm

50 mm 50 mm

y

C

x

y

x¿

10a Ch10a 916-969.indd 946 6/19/09 3:02:00 PM

1

12

1

12(100)(200)3 + (300)(100)3 = 91.7(10)6 mm4 AnsIy =

*6–72.

SM_CH06.indd 424 4/8/11 11:52:44 AM


425946


*10–40. Determine the moment of inertia of the beam’scross-sectional area about the y axis.

300 mm

100 mm

200 mm

50 mm 50 mm

y

C

x

y

x¿

10a Ch10a 916-969.indd 946 6/19/09 3:02:00 PM

1

12

1

12(100)(200)3 + (300)(100)3 = 91.7(10)6 mm4 AnsIy =

947


•10–41. Determine the moment of inertia of the beam’scross-sectional area about the axis.x

y

50 mm 50 mm

15 mm115 mm

115 mm

7.5 mmx

15 mm

10a Ch10a 916-969.indd 947 6/19/09 3:02:00 PM

(Ix)1 + (Ix)2Ix =

1

12

1

12(100)(2603) – (92.5)(2303) =

Composite Parts: The composite cross - sectional area of the beam can be subdivided into two segments as shown in Fig. a. Here,segment (2) is a hole, and so it contributes a negative moment of inertia.

Moment of Inertia: Since the x axis passes through the centroid of both rectangular segments,

= 52.7(106) mm4 Ans

•6–73.

SM_CH06.indd 425 4/8/11 11:52:45 AM


426948



y

50 mm 50 mm

15 mm115 mm

115 mm

7.5 mmx

15 mm

10a Ch10a 916-969.indd 948 6/19/09 3:02:01 PM

Σ(Iy)iIy =

1

12

1

122 (15)(1003) + (230)(7.53) =

= 2.5(10)6 mm4 Ans

Composite Parts: The composite cross - sectional area of the beam can be subdivided into two similar segments (2) and onesegment (1) as shown in Fig. a. The location of the centroid of each segment is also indicated.

Moment of Inertia: Since the y axis passes through the centroid of each segment,

� �

6–74.

SM_CH06.indd 426 4/8/11 11:52:45 AM


427948



y

50 mm 50 mm

15 mm115 mm

115 mm

7.5 mmx

15 mm

10a Ch10a 916-969.indd 948 6/19/09 3:02:01 PM

Σ(Iy)iIy =

1

12

1

122 (15)(1003) + (230)(7.53) =

= 2.5(10)6 mm4 Ans

Composite Parts: The composite cross - sectional area of the beam can be subdivided into two similar segments (2) and onesegment (1) as shown in Fig. a. The location of the centroid of each segment is also indicated.

Moment of Inertia: Since the y axis passes through the centroid of each segment,

� �

949


10–43. Locate the centroid of the cross-sectional areafor the angle. Then find the moment of inertia about the

centroidal axis.x¿Ix¿

y

6 in.2 in.

6 in.

x2 in.

C x¿

y¿y

–x

–y

*10–44. Locate the centroid of the cross-sectional areafor the angle. Then find the moment of inertia about the

centroidal axis.y¿Iy¿

x

6 in.2 in.

6 in.

x2 in.

C x¿

y¿y

–x

–y

60 mm

60 mm20 mm

20 mm

60 mm

60 mm20 mm

20 mm

20 mm

60 mm

20 mm

30 mm

60 mm

10 mm

10 mm

20 mm

20 mm

50 mm

20 mm

60 mm60 mm

20 mm

10 mm

10 mm

30 mm


Segment A (mm2) y (mm) yA (mm3) 1 60(20) 30 36000 2 60(20) 10 12000

Σ 2400 48000

Thus,

y = ΣΣyA

A =

48000

2400 = 20 mm Ans

Moment of Inertia : The moment of inertia about the x′ axis for each segment can be determined using the parallel – axis theorem Ix′ = I x′ + Ady

2.

Segment Ai (mm2) (dy)i (mm) ( I x′)i (mm4) (Ady2)i (mm4) (Ix′)i (mm4)

1 20(60) 10 1

12(20)(603) 12.0 (104) 48.0 (104)

2 60(20) 10 1

12(60)(203) 12.0 (104) 16.0 (104)

Thus,

Ix′ = Σ(Ix′)i = 64.0 (104) mm4 Ans


Segment A (mm2) x (mm) xA (mm3) 1 60(20) 10 12000 2 60(20) 50 60000

Σ 2400 72000

Thus,

x = ΣΣxA

A =

72000

2400 = 30 mm Ans

Moment of Inertia : The moment of inertia about the y′ axis for each segment can be determined using the parallel – axis theorem Iy′ = I y′ + Adx

2.

Segment Ai (mm2) (dx)i (mm) ( I y′)i (mm4) (Adx2)i (mm4) (Iy′)i (mm4)

1 60(20) 20 1

12(60)(203) 48.0 (104) 52.0 (104)

2 20(60) 20 1

12(20)(603) 48.0 (104) 84.0 (104)

Thus,

Iy′ = Σ(Iy′)i = 136 (104) mm4 Ans

10a Ch10a 916-969.indd 949 6/19/09 3:02:02 PM

6–75.

*6–76.

SM_CH06.indd 427 4/8/11 11:52:45 AM


428

913


*9–124. The steel plate is 0.3 m thick and has a density ofDetermine the location of its center of mass.

Also compute the reactions at the pin and roller support.7850 kg m3.

A

B

x

y

y2 2x

y x

2 m

2 m

2 m

•9–125. Locate the centroid ( , ) of the area. y

x

30 mm 10 mm

30 mm60 mm

30 mm

10 mm

30 mm60 mm

30 mm15 mm

70 mm4(30)3π

= 40π

mm

4(10)3π

= 403π

mm


Segment A (mm2) x (mm) y (mm) xA (mm3) yA (mm3)

1 1

2(30)(30) 70 10 31.5 (103) 4.50 (103)

2 60(30) 30 15 54.0 (103) 27.0 (103)

3 π4

(302) –40

π

40

π –9.00 (103) 9.00 (103)

4 –π2

(102) 0 40

3π 0 –0.667 (103)

Σ 27.998 (102) 76.50 (103) 39.833 (103)

Thus,

x = ΣΣxA

A = 76.50 (10 )

27.998 (10 )

3

2 = 27.3 mm Ans

y = ΣΣyA

A = 39.833 (10 )

27.998 (10 )

3

2 = 14.2 mm Ans

09b Ch09b 862-915.indd 913 6/18/09 10:11:02 AM

y1 = –x1

y22 = 2x2

dA = (y2 – y1)dx = ( 2x + x)dx

x– = x

y– = = y2 + y1

22x – x

2

x– = = = = 1.2571 = 1.26 m∫A

x–dA

∫A

dA

∫0

2

x( 2x + x)dx

∫0

2

( 2x + x)dx

x5/2 + x322

513

2

0

x3/2 + x222

312

2

0

y– = = = = 0.143 m Ans∫A

y–dA

∫A

dA

– x3x2

216

2

0

x3/2 + x222

312

2

0

∫0

2

( 2x + x)dx

∫0

2

( 2x + x)dx2x – x

2

A = 4.667 m2

W = 7850(9.81)(4.667)(0.3) = 107.81 kN

+ΣMA = 0; –1.2571(107.81) + NB(2 2) = 0

NB = 47.92 = 47.9 kN Ans

S+ ©Fx = 0; –Ax + 47.92 sin 45° = 0

Ax = 33.9 kN Ans

+c©Fy = 0; Ay + 47.92 cos 45° – 107.81 = 0

Ay = 73.9 kN Ans

914


9–126. Determine the location ( , ) of the centroid forthe structural shape. Neglect the thickness of the member.

10 mm10 mm

30 mm

x

y

9–127. Locate the centroid of the shaded area.

x

y

a—2

a—2

a

aa

15 mm 15 mm 15 mm 15 mm

15 mm

33.54 mm

20 mm

15 mm30 mm

Centroid : The length of each segment and its respective centroid are tabulated below.

Segment L (mm) y (mm) yL (mm2) 1 2 (15) 30 9.00 (102) 2 2 (33.54) 15 10.06 (102) 3 20 0 0

Σ 117.1 19.06 (102)

Due to symmetry about y axis,

x = 0 Ans

y = ΣΣyL

L =

19.06 (10 )

117.1

2

= 16.28 mm Ans

09b Ch09b 862-915.indd 914 6/18/09 10:11:04 AM

©y–A©A

y– = = = –0.262a(a)(a cos 30°) + [a(a)]

12

cos 30° (a)(a cos 30°) – [a(a)]a3

a2

12

•6–77.

6–78.

SM_CH06.indd 428 4/8/11 11:52:46 AM


4291019


10–118. Determine the moment of inertia of the areaabout the x axis.

y

4y 4 – x2

1 ft

x2 ft


y

4y 4 – x2

1 ft

x2 ft2 m

1 m

2 m

1 m

2 m 2 m

1 m

2 m 2 m

1 m

Differential Element : Here, y = 1

4(4 – x2). The area of the differential

element parallel to the y axis is dA = ydx = 1

4(4 – x2) dx.


Iy = A∫ x2 dA =

1

4 –2 m

2 m

∫ x2(4 – x2) dx

= 1

4

4

3–

1

53x x5

2 m

–2 m

= 2.13 m4 Ans

Differential Element : Here, y = 1

4(4 – x2). The area of the differential

element parallel to the y axis is dA = ydx. The moment of inertia of this differential element about the x axis is

dIx = dI x′ + dAy 2

= 1

12(dx)y3 + ydx

y

2

2

= 1

3

1

4(4 – )2

3

x

dx

= 1

192(–x6 + 12x4 – 48x2 + 64) dx


Ix = ∫ dIx = 1

192 –2 m

2 m

∫ (–x6 + 12x4 – 48x2 + 64) dx

= 1

192–

1

7+

12

5– 16 + 647 5 3x x x x

2 m

–2 m

= 0.610 m4 Ans

10b Ch10b 970-1022.indd 1019 6/22/09 1:53:01 PM

•6–79.

*6–80.

SM_CH06.indd 429 4/8/11 11:52:47 AM


430312


•4–161. If the distribution of the ground reaction on thepipe per meter of length can be approximated as shown,determine the magnitude of the resultant force due to thisloading.

0.8 m

1 kN/m

0.5 kN/m

w 0.5 (1 + cos ) kN/m

0.8 m

Resultant Components: The magnitude of the differential force dFR is equal to the area of the element shown shaded in Fig. a.

dFR = wr d = 0.5 (1 + cos ) (0.8 d ) = 0.4 (1 + cos ) d

The horizontal and vertical components of dFR are given by

+→ (dFR)x = dFR sin = 0.4 (1 + cos ) sin d = 0.4 sin +

sin 2

2

⎛

⎝⎜

⎞

⎠⎟ d

+↑ (dFR)y = dFR cos = 0.4 (1 + cos ) cos d = 0.4 cos +cos 2 + 1

2

⎛

⎝⎜

⎞

⎠⎟ d

Integrating (dFR)x and (dFR)y from = –π

2 rad to =

–

2

π rad gives the horizontal and vertical components of the resultant for FR.

+→ (FR)x = 0.4

– /2

/2

π

π

∫ sin +sin 2

2

⎛

⎝⎜

⎞

⎠⎟ d = 0.4 –cos +

cos 2

2– /2

/2⎛

⎝⎜

⎞

⎠⎟

π

π

= 0

+↑(FR)y = 0.4– /2

/2

π

π

∫ cos +cos 2 + 1

2

⎛

⎝⎜

⎞

⎠⎟ d = 0.4 sin +

sin 2

4+

1

2– /2

/2⎛

⎝⎜

⎞

⎠⎟

π

π

= 0.4 2 +2

π⎛

⎝⎜

⎞

⎠⎟ = 1.428 kN ↑

Thus,

FR = (FR)y = 1.428 kN ↑

04b Ch04b 254-318.indd 312 6/12/09 8:40:50 AM

•6–81.

SM_CH06.indd 430 4/8/11 11:52:47 AM


4311016


*10–112. Determine the moment of inertia of the beam’scross-sectional area about the x axis which passes throughthe centroid C.

Cx

y

d2

d2

d2

d2 60

60

•10–113. Determine the moment of inertia of the beam’scross-sectional area about the y axis which passes throughthe centroid C.

Cx

y

d2

d2

d2

d2 60

60

10b Ch10b 970-1022.indd 1016 6/22/09 1:52:59 PM

Moment of Inertia : The moment of inertia about the x axis for the composite beam's cross section can be determined using the parallel – axis theoremIx = © (Ix + Ady

2)i

Iy = [ 1 (d)(d3) + 0] 12

+ 4[ 1 (0.2887d)( d )3 + 1 (0.2887d)( d )( d )2] 36 2 2 2 6

= 0.0954d4 Ans

Moment of Inertia : The moment of inertia about y axis for the compositebeam's cross section can be determined using the parallel – axis theoremIy = © (Iy + Ad 2)i

Iy = [ 1 (d)(d3) + 0] 12

+ 2[ 1 (d)(0.2887d)3 + 1 (d)(0.2887d)(0.5962d)2] 36 2

= 0.187d4 Ans

6–82.

•6–83.

SM_CH06.indd 431 4/8/11 11:52:47 AM


4321020


10–119. Determine the moment of inertia of the areaabout the x axis. Then, using the parallel-axis theorem, findthe moment of inertia about the axis that passes throughthe centroid C of the area. .y = 120 mm

x¿

1–––200

200 mm

200 mm

y

x

x¿–y

Cy x2

10b Ch10b 970-1022.indd 1020 6/22/09 1:53:02 PM

Differential Element: Here, x = 200y . The area of the differential

element parallel to the x axis is dA = 2xdy = 2 200y dy.

12

12

Moment of Inertia: Applying Eq. 10 – 1 and performing the integration, wehave

12Ix =

A∫ y2 dA =

=

0

200 mm

∫ y2 2 200y dy

2 200 y

= 914.29(106) mm4 = 914 (106) mm4 Ans

2

7

12

200 m

0

12A =

A∫ dA = 0

200 mm

∫ 2 200y dy = 53.33(103) mm2

The moment of inertia about the x¿ axis can be determined using the parallel –

axis theorem. The area is

Ix¿ + Ady2Ix =

Ix¿ + 53.33(103)(1202) =

146(106) mm4 AnsIx¿

=

914.29(106)

*6–84.

SM_CH06.indd 432 4/8/11 11:52:48 AM


4331020


10–119. Determine the moment of inertia of the areaabout the x axis. Then, using the parallel-axis theorem, findthe moment of inertia about the axis that passes throughthe centroid C of the area. .y = 120 mm

x¿

1–––200

200 mm

200 mm

y

x

x¿–y

Cy x2

10b Ch10b 970-1022.indd 1020 6/22/09 1:53:02 PM

Differential Element: Here, x = 200y . The area of the differential

element parallel to the x axis is dA = 2xdy = 2 200y dy.

12

12

Moment of Inertia: Applying Eq. 10 – 1 and performing the integration, wehave

12Ix =

A∫ y2 dA =

=

0

200 mm

∫ y2 2 200y dy

2 200 y

= 914.29(106) mm4 = 914 (106) mm4 Ans

2

7

12

200 m

0

12A =

A∫ dA = 0

200 mm

∫ 2 200y dy = 53.33(103) mm2

The moment of inertia about the x¿ axis can be determined using the parallel –

axis theorem. The area is

Ix¿ + Ady2Ix =

Ix¿ + 53.33(103)(1202) =

146(106) mm4 AnsIx¿

=

914.29(106)

309


•4–157. The lifting force along the wing of a jet aircraftconsists of a uniform distribution along AB, and asemiparabolic distribution along BC with origin at B.Replace this loading by a single resultant force and specifyits location measured from point A.

x

w

24 ft12 ft

w (2880 5x2) lb/ft2880 lb/ft

A B

C

4–158. The distributed load acts on the beam as shown.Determine the magnitude of the equivalent resultant forceand specify where it acts, measured from point A. w ( 2x2 4x 16) lb/ft

xB

A

w

4 ft

48 kN/mw = (48 – 0.75x2) kN/m

4 m 8 m

2 m 2 m

48(4) = 192 kN

w = (48 – 0.75x2) kN/m

4 m

w = (–2x2 + 4x + 16) kN/m

w = (–2x2 + 4x + 16) kN/m

Equivalent Resultant Force :

+↑FR = ΣFy; FR = 192 + wdxx

0∫

FR = 192 + (48 – 0.75 )2

0

8

xm

∫ dx

= 448 kN ↑ Ans

Location of Equivalent Resultant Foce :

+ MRA = ΣMA;

448 x = 192 (2) + ( + 4)0

xx

∫ wdx

x = 384 + ( + 4)0

8

xm

∫ (48 – 0.75x2) dx

x = 384 + (–0.75 – 3 + 48 + 192)3 2

0

8

x x xm

∫ dx

x = 4.86 m Ans

FR = w x( )∫ dx = (–2 + 4 + 16)2

0

4

x x∫ dx = 53.333 = 53.3 kN Ans

x = x w x dx

w x dx

( )

( )

∫∫

= x x x dx(–2 + 4 + 16)

53.333

2

0

4

∫ = 1.60 m Ans

04b Ch04b 254-318.indd 309 6/12/09 8:40:47 AM

6–86.

432


•6–85. Determine the moment of inertia of the beam’scross-sectional area with respect to the axis passingthrough the centroid C.

x¿

0.5 in.

0.5 in.

4 in.

2.5 in.C x¿

0.5 in.

_y

6–86. The distributed load acts on the beam as shown.Determine the magnitude of the equivalent resultant forceand specify where it acts, measured from point A. w � (�2x2 � 4x � 16) lb/ft

xB

A

w

4 ft

100 mm

12 mm

62 mm

12 mm12 mm

12 mm

50 mm

12 mm12 mm

100 mm6 mm

6

37

37 mm

y 5 SyASA

5 6(12)(100) + 2 [37(50)(12)]12(100) + 2(50)(12)

5 21.5 mm

Ix 5 112

(100)(12)3 + 100(12)(21.5 – 6)2 + 2 c 112

(12)(50)3 + (12)(50)(37 – 21.5)2d= 841(103) mm4

SM_CH06.indd 433 4/8/11 11:52:49 AM


434310


4–159. The distributed load acts on the beam as shown.Determine the maximum intensity . What is themagnitude of the equivalent resultant force? Specify whereit acts, measured from point B.

wmax

w ( 2x2 4x 16) lb/ft

xB

A

w

4 ft

w = (–2x2 + 4x + 16) kN/m

w = (–2x2 + 4x + 16) kN/m

4 m

dw

dx = –4x + 4 = 0

x = 1

wmax = –2(1)2 + 4(1) + 16 = 18 kN/m Ans

FR = w x( )∫ dx = (–2 + 4 + 16)2

0

4

x x∫ dx = 53.333 = 53.3 kN Ans

x = x w x dx

w x dx

( )

( )

∫∫

= x x x dx(–2 + 4 + 16)

53.333

2

0

4

∫ = 1.60 m Ans

So that from B,

xʹ = 4 – 1.60 = 2.40 m Ans

04b Ch04b 254-318.indd 310 6/12/09 8:40:48 AM

6–87.

SM_CH06.indd 434 4/8/11 11:52:49 AM

Documents

Area ∫ 1m Ans - Faculty Personal Homepage- KFUPMfaculty.kfupm.edu.sa/.../CE202/S_Manual/SM_CH06.pdf · SM_CH06.indd 365 4/8/11 11:52:25 AM © 2011 Pearson Education, Inc., Upper