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© 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
© 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.
366814
© 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.
9–10. 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
Differential Element: The area element parallel to the y axis shown shaded in Fig. a will be considered. The area of the element is
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
© 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.
367816
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* 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
© 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.
368817
© 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.
• 9–13. 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.
Centroid : Applying Eq. 9–4 and performing the integration, we have
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
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369818
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9–14. Determine the area and the centroid of the area.(x, y) y
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
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370819
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9–15. Determine the area and the centroid of the area.(x, y) y
x
a
h y x2h––a2
09a Ch09a 806-861.indd 819 6/17/09 11:35:05 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
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
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371820
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*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
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372
825
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•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
Area and Moment Arm : The area of the differential element is
dA = ydx = 600
xdx and its centroid is x = x.
Centroid : Applying Eq. 9–4 and performing the integration, we have
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
Centroid : Applying Eq. 9–4 and performing the integration, we have
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
© 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.
373826
© 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.
9–23. 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
Area and Moment Arm : The area of the differential element is
dA = ydx = 600
xdx and its centroid is y =
y
2 =
300
x.
Centroid : Applying Eq. 9–4 and performing the integration, we have
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
© 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.
374827
© 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.
•9–25. 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
© 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.
375828
© 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.
9–26. 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
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376839
© 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.
*9–40. 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
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377834
© 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.
9–35. 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,
Centroid: The centroid of the element is located at y = y.
= 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
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378835
© 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.
*9–36. 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
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379836
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•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
Centroid: The centroid of the element is located at y = y.
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
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380837
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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
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381851
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*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
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382852
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•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
Centroid: The centroid of each composite segment is shown in Fig. a,
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
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383853
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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
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384854
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*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
Centroid: The centroid of each composite segment is shown in Fig. a,
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
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385855
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•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
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386861
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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
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387862
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*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
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388868
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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
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389869
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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
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390870
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*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
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391871
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•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
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392298
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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
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393299
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4–143. Replace the distributed loading with an equivalentresultant force, and specify its location on the beammeasured from point A.
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,
If we equate the moments of FR, Fig. b, to the sum of the moment of the forces in Fig. a about point A,
*6–32.
SM_CH06.indd 393 4/8/11 11:52:33 AM
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394300
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*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,
If we equate the moments of FR, Fig. b, to the sum of the moment of the forces in Fig. a about point A,
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
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395301
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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
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396305
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*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
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397306
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4–154. Replace the distributed loading with an equivalentresultant force, and specify its location on the beammeasured from point A.
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
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398307
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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
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399308
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*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
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400916
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•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
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401916
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•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
917
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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
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402918
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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
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403918
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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
919
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*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
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404920
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•10–5. Determine the moment of inertia of the area aboutthe axis.x
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
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405920
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•10–5. Determine the moment of inertia of the area aboutthe axis.x
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
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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
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406922
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10–7. Determine the moment of inertia of the area aboutthe axis.x
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
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407922
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10–7. Determine the moment of inertia of the area aboutthe axis.x
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
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*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
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408924
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•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
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409925
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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.
Moment of Inertia : Applying Eq. 10–1 and performing the integration, we have
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
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410926
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•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.
Moment of Inertia : Applying Eq. 10–1 and performing the integration, we have
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
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411926
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•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.
Moment of Inertia : Applying Eq. 10–1 and performing the integration, we have
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
927
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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
dIx = dI xʹ + dAy 2
= 1
12(dx)y3 + ydx
y
2
2
= 1
3(4 – 4x2)3 dx
= 1
3(–64x6 + 192x4 – 192x2 + 64) dx
Moment of Inertia : Performing the integration, we have
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.
Moment of Inertia : Applying Eq. 10–1 and performing the integration, we have
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
© 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.
412928
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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.
Moment of Inertia : Applying Eq. 10–1 and performing the integration, we have
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
b) Differential Element : Here, x = 1
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
Moment of Inertia : Performing the integration, we have
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
© 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.
413928
© 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.
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.
Moment of Inertia : Applying Eq. 10–1 and performing the integration, we have
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
b) Differential Element : Here, x = 1
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
Moment of Inertia : Performing the integration, we have
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
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*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
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414
936
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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
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415937
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*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
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416938
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10–30. Determine the moment of inertia of the beam’scross-sectional area about the axis.x
y
x
15 mm15 mm60 mm60 mm
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
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417938
© 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.
10–30. Determine the moment of inertia of the beam’scross-sectional area about the axis.x
y
x
15 mm15 mm60 mm60 mm
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
939
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10–31. Determine the moment of inertia of the beam’scross-sectional area about the axis.y
y
x
15 mm15 mm60 mm60 mm
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
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,
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
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418940
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*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
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419940
© 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.
*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
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•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
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420942
© 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.
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
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421942
© 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.
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
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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
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422944
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•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
10–38. Determine the distance to the centroid of thebeam’s cross-sectional area; then find the moment of inertiaabout the axis.x¿
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
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423944
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•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
10–38. Determine the distance to the centroid of thebeam’s cross-sectional area; then find the moment of inertiaabout the axis.x¿
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
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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
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424946
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*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
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425946
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*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
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•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
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426948
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10–42. Determine the moment of inertia of the beam’scross-sectional area about the axis.y
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
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427948
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10–42. Determine the moment of inertia of the beam’scross-sectional area about the axis.y
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
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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
Centroid : The area of each segment and its respective centroid are tabulated below.
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
Centroid : The area of each segment and its respective centroid are tabulated below.
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
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428
913
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*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
Centroid : The area of each segment and its respective centroid are tabulated below.
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
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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
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4291019
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10–118. Determine the moment of inertia of the areaabout the x axis.
y
4y 4 – x2
1 ft
x2 ft
•10–117. Determine the moment of inertia of the areaabout the y axis.
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.
Moment of Inertia : Applying Eq. 10–1 and performing the integration, we have
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
Moment of Inertia : Performing the integration, we have
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
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430312
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•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
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4311016
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*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
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4321020
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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
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4331020
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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
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•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
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•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
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434310
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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