CHAPTER 3CHAPTER 7CHAPTER 5 - · PDF fileCHAPTER 3CHAPTER 7CHAPTER 5. 3 ... If you are a student using this Manual, you are using it without permission. PROBLEM 5.1 Locate the centroid

CHAPTER 3CHAPTER 7CHAPTER 7CHAPTER 3CHAPTER 7CHAPTER 7CHAPTER 3CHAPTER 5CHAPTER 5

3

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PROBLEM 5.1

Locate the centroid of the plane area shown.

SOLutiOn

Dimensions in mm

A, mm2 x , mm y, mm xA, mm3 yA, mm3

1 6300 105 15 0 66150 106. ¥ 0 094500 106. ¥

2 9000 225 150 2 0250 106. ¥ 1 35000 106. ¥

S 15300 2 6865 106. ¥ 1 44450 106. ¥

Then XxA

A= = ¥S

S2 6865 10

15300

6. X = 175 6. mm

YyA

A= = ¥S

S1 44450 10

15300

6. Y = 94 4. mm

4


SOLutiOn

Dimensions in mm


1 1200 10 30 12000 36000

2 540 30 36 16200 19440

S 1740 28200 55440

Then XxA

A= =

SS

28200

1740 X = 16 21. mm

YyA

A= =

SS

55440

1740 Y = 31 9. mm

PROBLEM 5.2


5


SOLutiOn

Dimensions in mm

A, mm2 x , mm y, mm xA,mm 3 yA,mm 3

11

2300 400 60000¥ ¥ = 200 400/3 12 ´ 106 8 ´ 106

2 500 400 200000¥ = 550 200 110 ´ 106 40 ´ 106

S 260000 122 ´ 106 48 ´ 106

Then XxA

A= = ¥

¥=

SS

122 10

260 10469 23

6

3. X = 469 mm

YyA

A= = ¥

¥=

SS

48 10

260 10184 62

6

3. Y = 184 6. mm

PROBLEM 5.3


6


SOLutiOn


11

2300 150 22500( )( ) = 100 100 2.25 ´ 106 2.25 ´ 106

2 ( )( )150 75 11250= 225 187.5 2.53125 ´ 106 2.109375 ´ 106

S 33750 4.78125 ´ 106 4.359375 ´ 106

Then XA xA= S

X ( ) .33750 4 78125 106= ¥ X = 141 7. mm

YA yA

Y

=

= ¥

S

( ) .33750 4 359375 106 Y = 129 2. mm

PROBLEM 5.4


7


SOLutiOn

A, mm2 x , mm y, mm xA,mm3 yA,mm3

1 350 500 175 103¥ = ¥ 175 250 30.625 ´ 106 43.75 ´ 106

2 - = - ¥p ( ) .100 31 4159 102 3 150 300 - 4.7124 ´ 106 - 9.4248 ´ 106

S 143.5841 ´ 103 25.9126 ´ 106 34.3252 ´ 106

Then XxA

A= = ¥

¥SS

25 9126 10

143 5841 10

6

3

.

. X = 180 5. mm

YyA

A= = ¥

¥SS

34 3252 10

143 5841 10

6

3

.

. Y = 239 mm

PROBLEM 5.5


8


SOLutiOn


11

2120 75 4500( )( ) = 80 25 360 103¥ 112 5 103. ¥

2 ( )( )75 75 5625= 157.5 37.5 885 94 103. ¥ 210 94 103. ¥

3 - = -p4

75 4417 92( ) . 163.169 43.169 - ¥720 86 103. - ¥190 716 103.

S 5707.1 525 08 103. ¥ 132 724 103. ¥

Then XA x A X= = ¥S ( . ) .5707 1 525 08 103 X = 92 0. mm

YA y A Y= = ¥S ( . ) .5707 1 132 724 103 Y = 23 3. mm

PROBLEM 5.6


9


SOLutiOn


1p2

1000 1 57080 102 6( ) .= ¥ 0 424.413 0 666.668 ´ 106

2 - ¥ = - ¥500 400 0 2 106. –250 200 50 ´ 106 –40 ´ 106

S 1.3708´106 50 ´ 106 626.668 ´ 106

Then XxA

A= = ¥

¥SS

50 10

1 3708 10

6

6. X = 36 5. mm

YyA

A= = ¥

¥SS

626 668 10

1 3708 10

6

6

.

. Y = 457 mm

PROBLEM 5.7


10


SOLutiOn


1 750 1200 0 9375 106¥ = ¥. 375 625 351.5625 106 585.9375 106

2 - = - ¥p2

375 0 22089 102 6( ) . 590.845 750 –130.512 106 –165.6675 106

S 0.71661 106 221.0505 106 420.27 106

Then XxA

A= = ¥

¥SS

221 0505 10

0 71661 10

6

6

.

. X = 308 mm

YyA

A= = ¥

¥SS

420 27 10

0 71661 10

6

6

.

. Y = 586 mm

PROBLEM 5.8


11


SOLutiOn


1 ( )( )60 120 7200= –30 60 - ¥216 103 432 103¥

2p4

60 2827 42( ) .= 25.465 95.435 72 000 103. ¥ 269 83 103. ¥

3 - = -p4

60 2827 42( ) . –25.465 25.465 72 000 103. ¥ - ¥72 000 103.

S 7200 - ¥72 000 103. 629 83 103. ¥

Then XA x A X= = - ¥S ( ) .7200 72 000 103 X = -10 00. mm

YA y A Y= = ¥S ( ) .7200 629 83 103 Y = 87 5. mm

PROBLEM 5.9


12


SOLutiOn

Dimensions in mm


1p2

47 26 1919 51¥ ¥ = . 0 11.0347 0 21181

21

294 70 3290¥ ¥ = –15.6667 –23.333 –51543 –76766

S 5209.5 –51543 –55584

Then Xx A

A= = -S

S51543

5209 5. X = -9 89. mm

Yy A

A= = -S

S55584

5209 5. Y = -10 67. mm

PROBLEM 5.10


13


SOLutiOn

First note that symmetry implies X = 0

A, mm2 y, mm yA, mm3

1 - = -p p( )200

220000

2 800

3p–5.3333 ´ 106

2 p p( ),

300

245 000

2

= 400/p 18 ´ 106

S 78.5398 103 12.6667 ´ 106

Then Yy A

A= = ¥

¥SS

12 667 10

78 5398 10

6

3

.

. or Y = 161 3. mm

PROBLEM 5.11


14


SOLutiOn


1 ( )( )15 80 1200= 40 7.5 48 103¥ 9 103¥

21

350 80 1333 33( )( ) .= 60 30 80 103¥ 40 103¥

S 2533.3 128 103¥ 49 103¥

Then XA xA= S

X ( . )2533 3 128 103= ¥ X = 50 5. mm

YA yA

Y

=

= ¥

S

( . )2533 3 49 103 Y = 19 34. mm

PROBLEM 5.12


15


SOLutiOn


1 1

330 20 200¥ ¥ = 9 15 1800 3000

2p4

706 86(30)2 = . 12.7324 32.7324 9000.0 23137

S 906.86 10800 26137

Then XxA

A= =

SS

10800

906 86. X = 11 91. mm

YyA

A= =

SS

26137

906 86. Y = 28 8. mm

PROBLEM 5.13


16


SOLutiOn

Dimensions in mm

A, mm2 x , mm y, mm xA, mm 3 yA, mm 3

1 2

3500 500

0 5 10

3

6

¥ = ¥( )( )

. 300 187.5 50 ´ 106 31.25 106

2 - = - ¥1

3500 500

0 25 10

3

6

( )( ). 375 150 –31.25 106 –12.5 106

S 0 25 10

3

6. ¥ 18.75 106 18.75 106

Then XxA

A= = ¥

¥ÊËÁ

ˆ¯

SS

18 75 10

0 25 103

6

6

.

. X = 225 mm

YyA

A= = ¥

¥ÊËÁ

ˆ¯

SS

18 75 10

0 25 103

6

6

.

. Y = 225 mm

PROBLEM 5.14


17


SOLutiOn


1 2

375 120 6000( )( ) = 28.125 48 168750 288000

2 - = -1

275 60 2250( )( ) 25 20 –56250 –45000

S 3750 112500 243000

Then X A xAS S=

X ( )3750 1125002mm mm3= or X = 30 0. mm

and Y A yAS S=

Y ( )3750 243000mm2 = or Y = 64 8. mm

PROBLEM 5.15


18


SOLutiOn

First, determine the location of the centroid.

From Figure 5.8A: y r A r

r

2 22

22 2

2

22

2

3 2

2

3

=-( )

-( ) = -ÊËÁ

ˆ¯

=-( )

sin

cos

p

p

p

a

ap a

aa

Similarly y r A r1 12

1 122

3 2=

-( ) = -ÊËÁ

ˆ¯

cosaa

p ap

Then SyA r r r=-( ) -Ê

ËÁˆ¯

ÈÎÍ

˘˚

--( ) -Ê

Ë2

3 2

2

3 222

22

12

cos cosaa

p aaa

p ap p ÁÁ

ˆ¯

ÈÎÍ

˘˚

= -( )

r

r r

12

23

132

3cos a

and SA r r

r r

= -ÊËÁ

ˆ¯

- -ÊËÁ

ˆ¯

= -ÊËÁ

ˆ¯

-( )

p a p a

p a

2 2

2

22

12

22

12

Now Y A yAS S=

Y r r r rp a a2

2

322

12

23

13-Ê

ËÁˆ¯

-( )ÈÎÍ

˘˚

= -( )cos Yr r

r r=

--

Ê

ËÁˆ

¯ -ÊËÁ

ˆ¯

2

3

2

223

13

22

12

cos ap a

PROBLEM 5.16

Determine the y coordinate of the centroid of the shaded area in terms of r

1, r

2, and a.

19


SOLutiOn

First, determine the location of the centroid.

From Figure 5.8A: y r A r

r

2 22

22 2

2

22

2

3 2

2

3

=-( )

-( ) = -ÊËÁ

ˆ¯

=-( )

sin

cos

p

p

p

a

ap a

aa

Similarly y r A r1 12

1 122

3 2=

-( ) = -ÊËÁ

ˆ¯

cosaa

p ap

Then SyA r r r=-( ) -Ê

ËÁˆ¯

ÈÎÍ

˘˚

--( ) -Ê

Ë2

3 2

2

3 222

22

12

cos cosaa

p aaa

p ap p ÁÁ

ˆ¯

ÈÎÍ

˘˚

= -( )

r

r r

12

23

132

3cos a

and SA r r

r r

= -ÊËÁ

ˆ¯

- -ÊËÁ

ˆ¯

= -ÊËÁ

ˆ¯

-( )

p a p a

p a

2 2

2

22

12

22

12

Now Y A yAS S=

Y r r r r

Yr r

r r

p a a2

2

3

2

3

22

12

23

13

23

13

22

-ÊËÁ

ˆ¯

-( )ÈÎÍ

˘˚

= -( )

=--

cos

112

2

2

Ê

ËÁˆ

¯ -ÊËÁ

ˆ¯

cos ap a

PROBLEM 5.17

Show that as r1 approaches r

2, the location of the centroid approaches

that for an arc of circle of radius ( ) .r r1 2 2+ /

20


PROBLEM 5.17 (continued)

Using Figure 5.8B, Y of an arc of radius 1

2 1 2( )r r+ is

Y r r

r r

= +-

-

= +-

1

2

1

2

1 22

2

1 22

( )sin( )

( )

( )cos

( )

p

p

p

aa

aa

(1)

Now r r

r r

r r r r r r

r r r r

r r r

23

13

22

12

2 1 22

1 2 12

2 1 2 1

22

1

--

=- + +( )

- +

=+

( )

( )( )

22 12

2 1

++

r

r r

Let r r

r r2

1

= += -

DD

Then r r r= +1

2 1 2( )

and r r

r r

r r r r

r r

r

r

23

13

22

12

2 2

2 23

2

--

= + + + - -+ + -

= +

( ) ( )( )( )

( ) ( )

D D D DD D

D

In the limit as D Æ 0 (i.e., r r1 2= ), then

r r

r rr

r r

23

13

22

12

1 2

3

2

3

2

1

2

--

=

= ¥ +( )

So that Y r r= ¥ +-

2

3

3

4 1 22

( )cosa

ap or Y r r= +-

( )cos

1 2 2

ap a

which agrees with Equation (1).

21


SOLutiOn

A x y xA yA

12

3ab

3

8a

3

5b a b2

4

2

5

2ab

2 - 1

2ab

1

3a

2

3b - a b2

6- ab2

3

S1

6ab a b2

12

ab2

15

Then X A xA

X aba b

S S=

ÊËÁ

ˆ¯

=1

6 12

2

or X a

Y A y A

Y abab

=

=

ÊËÁ

ˆ¯

=

1

2

1

6 15

2

S S

or Y b= 2

5

Now X Y a b= fi =1

2

2

5 or

a

b= 4

5

PROBLEM 5.18

For the area shown, determine the ratio a/b for which x y= .

22


SOLutiOn

A Y YA

1 - p2 1

2r4

31r

p- 2

3 13r

2 p2 2

2r4

32r

p2

3 23r

Sp2 2

212r r-( ) 2

3 23

13r r-( )

Then Y A yAS S=

or 3

4 2

2

3

9

161

1 22

12

23

13

2

1

22

1

r r r r r

r

r

r

r

¥ -( ) = -( )ÊËÁ

ˆ¯

-È

ÎÍÍ

˘

˚˙˙

=

p

p ÊÊËÁ

ˆ¯

-3

1

Let pr

r

p p p p p

=

+ - = - + +

2

1

29

161 1 1 1

p[( )( )] ( )( )

or 16 16 9 16 9 02p p+ - + - =( ) ( )p p

PROBLEM 5.19

For the semiannular area of Problem 5.11, determine the ratio r2/r

1 so that

y r= 3 41/ .

23



Then p =- - ± - - -( ) ( ) ( )( )

( )

16 9 16 9 4 16 16 9

2 16

2p p p

or p

p

= -=

0 5726

1 3397

.

.

Taking the positive root r

r2

1

1 340= .

24


SOLutiOn

From the problem statement: F is proportional to Qx .

Therefore: F

Q

F

QF

Q

QFA

x A

B

x BB

x B

x AA( ) ( )

,( )

( )= =or

For the first moments: ( ) ( )

( ) ( )

Q

Q Q A

x A

x B x

= +ÊËÁ

ˆ¯

¥

=

= + -Ê

22512

2300 12

831600

2 22512

2

mm3

ËËÁˆ¯

¥ + - ¥

=

( ) ( )( )48 12 2 225 30 12 60

1364688 3mm

Then FB = 1364688

831600280( )N or FB = 459 N

PROBLEM 5.20

A composite beam is constructed by bolting four plates to four 60 ´ 60 ´ 12-mm angles as shown. The bolts are equally spaced along the beam, and the beam supports a vertical load. As proved in mechanics of materials, the shearing forces exerted on the bolts at A and B are proportional to the first moments with respect to the centroidal x axis of the red shaded areas shown, respectively, in Parts (a) and (b) of the figure. Knowing that the force exerted on the bolt at A is 280 N, determine the force exerted on the bolt at B.

25


SOLutiOn

Note that Q yAx = S

Then ( )Qx 150

3

1

2= Ê

ËÁˆ¯

¥ ¥ÊËÁ

ˆ¯

mm 60 50 mm2 or ( ) mm1Qx = 25000 3

and ( )Qx 222

3

1

22

1

32

1

2

= - ¥ÊËÁ

ˆ¯

¥ ¥ÊËÁ

ˆ¯

+ - ¥ÊËÁ

ˆ¯

¥

25 mm 90 5 mm

5 mm 600 5 mm¥ÊËÁ

ˆ¯

2 2

or ( )Qx 2325000= - mm

Now Q Q Qx x x= + =( ) ( )1 2 0

This result is expected since x is a centroidal axis ( )thus y = 0

and Q yA Y A y Qx x= = = fi =S S ( )0 0

PROBLEM 5.21

The horizontal x axis is drawn through the centroid C of the area shown, and it divides the area into two component areas A1 and A2. Determine the first moment of each component area with respect to the x axis, and explain the results obtained.

26


PROBLEM 5.22

The horizontal x axis is drawn through the centroid C of the area shown, and it divides the area into two component areas A1 and A2. Determine the first moment of each component area with respect to the x axis, and explain the results obtained.

SOLutiOn

First determine the location of the centroid C. We have

A, mm2 ¢y , mm ¢y A, mm3

I 21

240 30 1200¥ ¥Ê

ËÁˆ¯

= 10 12000

II 30 110 3300¥ = 55 181500

III 90 40 3600¥ = 130 468000

S 8100 661500

Then ¢ = ¢¢ =Y A y A

Y

S S( )8100 661500

or ¢ =Y 81 667. mm

Now Q y Ax = S 1

Then ( ) ( . ) [( )( . )]Qx 121

2110 81 667 30 110 81 667= -È

ÎÍ˘˚

-mm mm

+ -[( . ) ][( )( )]130 81 667 90 40 2mm mm or ( )Qx 13186040= mm

and ( ) ( . ) [( )( . )]Qx 221

281 667 30 81 667= - È

ÎÍ˘˚

mm mm

- - ¥ ¥ ¥ÊËÁ

ˆ¯

ÈÎÍ

˘˚

[( . ) ]81 667 10 21

240 30 2mm mm or ( mmQx )2

3186040= -

27



Now Q Q Qx x x= + =( ) ( )1 2 0

This result is expected since x is a centroidal axis ( )thus Y = 0

and Q yA Y A Y Qx x= = = fi =S S ( )0 0

28


PROBLEM 5.23

The first moment of the shaded area with respect to the x axis is denoted by Qx.

(a) Express Qx in terms of b, c, and the distance y from the base of the shaded area

to the x axis. (b) For what value of y is Ox maximum, and what is that maximum value?

SOLutiOn

Shaded area:

A b c y

Q yA

c y b c y

x

= -=

= + -

( )

( )[ ( )]1

2

(a) Q b c yx = -1

22 2( )

(b) For Qmax : dQ

dyb y= - =0

1

22 0or ( ) y = 0

For y = 0: ( )Q bcx = 1

22 ( )Q bcx = 1

22

29


SOLutiOn

Perimeter of Figure 5.1

L x y xL, mm2 yL, mm2

I 30 0 15 0 0 45 103. ¥

II 210 105 30 22 05 103. ¥ 6 3 103. ¥

III 270 210 165 56 7 103. ¥ 44 55 103. ¥

IV 30 225 300 6 75 103. ¥ 9 103¥

V 300 240 150 72 103¥ 45 103¥VII 240 120 0 28 8 103. ¥ 0

S 1080 186 3 103. ¥ 105 3 103. ¥

X L x L

X

S S=

= ¥( .1080 186 3 103 mm) mm2 X = 172 5. mm

Y L y L

Y

S S=

= ¥( .1080 105 3 103 mm) mm2 Y = 97 5. mm

PROBLEM 5.24

A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed.

30


SOLutiOn

First note that because wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line.

L, mm x , mm y, mm xL, mm2 yL, mm2

1 20 10 0 200 0

2 24 20 12 480 288

3 30 35 24 1050 720

4 46.861 35 42 1640.14 1968.16

5 20 10 60 200 1200

6 60 0 30 0 1800

S 200.86 3570.1 5976.2

Then X L xL XS S= =( . ) .200 86 3570 1 X = 17 77. mm

Y L yL YS S= =( . ) .200 86 5976 2 Y = 29 8. mm

PROBLEM 5.25


31


SOLutiOn

First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line.


1 800 400 0 320 ´ 103 0

2 400 800 200 320 ´ 103 80 ´ 103

3 500 550 400 275 ´ 103 200 ´ 103

4 ( ) ( )300 400 5002 2+ = 150 200 75 ´ 103 100 ´ 103

S 2200 990 ´ 103 380 ´ 103

Then X L x LS S=

X (2200 990 103) = ¥ or X = 450 mm

and Y L y LS S=

Y ( )2200 380 103= ¥ or Y = 172 7. mm

PROBLEM 5.26


32


SOLutiOn


Y62

1000=p

( ) mm


1 500 –750 0 –375 × 103 0

2 400 –500 200 –200 × 103 80 × 103

3 500 –250 400 –125 × 103 200 × 103

4 400 0 200 0 80 × 103

5 1000 500 0 500 × 103 0

6 p p( )1000 1000= 02000

p0 2000 × 103

S 5941.59 –200 × 103 2360 ×103

Then Xx L

L= = - ¥S

S200 10

5941 59

3

. X = -33 7. mm

Yy L

L= = ¥S

S2360 10

5941 59

3

. Y = 397 mm

PROBLEM 5.27


33


PROBLEM 5.28

A uniform circular rod of weight 4 kg and radius 250 mm is attached to a pin at C and to the cable AB. Determine (a) the tension in the cable, (b) the reaction at C.

SOLutiOn

For quarter circle rr= 2

p

(a) + SM Wr

TrC = ÊËÁ

ˆ¯

- =02

0:p

T W= ÊËÁ

ˆ¯

= ¥ ÊËÁ

ˆ¯

=24 9 81

224 981

p p( . ) .N N T = 25 0. N

(b) + SF T C Cx x x= - = - =0 0 24 981 0: . N Cx = 25 0. N¨

+ SF C W C Cy y y y= - = - ¥ = fi =0 0 4 9 81 39 24: . .N 0 N Cy = ≠39 2. N

C = 46 5. N 57.5°

34


SOLutiOn

First note that for equilibrium, the center of gravity of the component must lie on a vertical line through C. Further, because the tubing is uniform, the center of gravity of the component will coincide with the centroid of the corresponding line. Thus, X = 0

So that Sx L = 0

Then

- - ∞ÊËÁ

ˆ¯

¥

+ - ¥

+ - -

d

d

d

0 75

255 0 75

0 75 1 5

1 5

.cos ( . )

( . ) ( . )

( . )

m m

m m

m11

22 55 2 0¥ ¥ ∞Ê

ËÁˆ¯

ÈÎÍ

˘˚

¥ = m mcos ( )

or ( . . ) ( . ) cos ( . )( . )0 75 1 5 21

20 75 2 55 0 75 1 5 32+ + = -È

ÎÍ˘˚

∞ + +d or d = 0 739. m

PROBLEM 5.29

Member ABCDE is a component of a mobile and is formed from a single piece of aluminum tubing. Knowing that the member is supported at C and that l = 2 m, determine the distance d so that portion BCD of the member is horizontal.

35


PROBLEM 5.30

Member ABCDE is a component of a mobile and is formed from a single piece of aluminum tubing. Knowing that the member is supported at C and that d is 0.50 m, determine the length l of arm DE so that this portion of the member is horizontal.

SOLutiOn

First note that for equilibrium, the center of gravity of the component must lie on a vertical line through C. Further, because the tubing is uniform, the center of gravity of the component will coincide with the centroid of the corresponding line. Thus,

X = 0

So that Sx L = 0

or - ∞ + ∞ÊËÁ

ˆ¯ ¥

+ ¥ ∞ ¥

0 75

220 0 5 35 0 75

0 25 35 1

.sin . sin ( . )

( . sin ) ( .

m m

m 55

1 0 352

0

m

m m

)

. sin ( )+ ¥ ∞ -ÊËÁ

ˆ¯

¥ =ll

or -+

+ ∞ -( )0 096193 35 2.( ) ( )

sin

( )xL xL

l

xLAB BD

l

DE

� �� = 0

The equation implies that the center of gravity of DE must be to the right of C.

Then l l2 1 14715 0 192386 0- + =. .

or l =± - -1 14715 1 14715 4 0 192386

2

2. ( . ) ( . )

or l = 0 204. m or ml = 0 943.

Note that sin35 012∞ - l for both values of l so both values are acceptable.

36


PROBLEM 5.31

The homogeneous wire ABC is bent into a semicircular arc and a straight section as shown and is attached to a hinge at A. Determine the value of u for which the wire is in equilibrium for the indicated position.

SOLutiOn

First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through A. Further, because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. Thus,

X = 0

So that Sx L = 0

Then -ÊËÁ

ˆ¯

+ -ÊËÁ

ˆ¯

=1

2

20r r

rr rcos ( ) cos ( )q

pq p

or cos

.

qp

=+

=

4

1 2

0 54921

or q = ∞56 7.

37


SOLutiOn

A y yA

1 1

2ba

1

3a

1

62a b

2- 1

2( )kb h

1

3h - 1

62kbh

S ba kh

2( )- b

a kh6

2 2( )-

Then Y A yA

Yb

a khb

a kh

S S=

-ÈÎÍ

˘˚

= -2 6

2 2( ) ( )

or Ya kh

a kh= -

-

2 2

3( ) (1)

and dY

dh

kh a kh a kh k

a kh= - - - - -

-=1

3

20

2 2

2

( ) ( )( )

( )

or 2 02 2h a kh a kh( )- - + = (2)

Simplifying Eq. (2) yields

kh ah a2 22 0- + =

PROBLEM 5.32

Determine the distance h for which the centroid of the shaded area is as far above line BB as possible when (a) k = 0.10, (b) k = 0.80.

38



Then ha a k a

ka

kk

=± - -

= ± -ÈÎ ˘

2 2 4

2

1 1

2 2( ) ( )( )

Note that only the negative root is acceptable since h a . Then

(a) k = 0 10.

ha= - -ÈÎ ˘

0 101 1 0 10

.. or h a= 0 513.

(b) k = 0 80.

ha= - -ÈÎ ˘

0 801 1 0 80

.. or h a= 0 691.

39


SOLutiOn

See solution to Problem 5.32 for analysis leading to the following equations:

Ya kh

a kh= -

-

2 2

3( ) (1)

2 02 2h a kh a kh( )- - + = (2)

Rearranging Eq. (2) (which defines the value of h which maximizes Y ) yields

a kh h a kh2 2 2- = -( )

Then substituting into Eq. (1) (which defines Y )

Ya kh

h a kh=-

¥ -1

32

( )( ) or Y h= 2

3

PROBLEM 5.33

Knowing that the distance h has been selected to maximize the distance y from line BB´ to the centroid of the shaded area, show that y h= 2 /3.

40


PROBLEM 5.34

Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h.

SOLutiOn

y

x

h

a

yh

ax

x x

y y

dA ydx

A ydxh

ax dx ah

EL

EL

a a

=

=

=

=

=

= = ÊËÁ

ˆ¯

=Ú Ú

1

2

1

20 0

x dA xydx xh

ax dx

h

a

xhaEL

aa

= = ÊËÁ

ˆ¯

=È

ÎÍ

˘

˚˙ =Ú Ú Ú

3

0

2

0 3

1

3

y dA y ydxh

ax dx

h

b

xEL

aaa

= ÊËÁ

ˆ¯

= ÊËÁ

ˆ¯

=È

ÎÍ

˘

˚˙Ú ÚÚ

1

2

1

2

1

2 3

2 2

200

3

0

== 1

62h a

xA x dA x ah haEL= ÊËÁ

ˆ¯

=Ú :1

2

1

32 x a= 2

3

yA y dA y ah h aEL= ÊËÁ

ˆ¯

=Ú :1

2

1

62 y h= 1

3

41


PROBLEM 5.35


SOLutiOn

At ( , )a h y h ka12: =

or kh

a=

2

y h ma2 : =

or mh

a=

Now x x

y y y

EL

EL

=

= +1

2 1 2( )

and dA y y dxh

ax

h

ax dx

h

aax x dx

= - = -ÈÎÍ

˘˚

= -

( )

( )

2 1 22

22

Then A dAh

aax x dx

h

a

ax x ah

aa

= = - = -ÈÎÍ

˘˚

=Ú Ú 20

22

2 3

02

1

3

1

6( )

and x dA xh

aax x dx

h

a

ax x a h

y

EL

aa

EL

Ú Ú

Ú

= - =ÈÎÍ

-ÈÎÍ

˘˚

=0 2

22

3 4

0

2

3

1

4

1

12( )

ddA y y y y dx y y dx= + - = -( )Ú Ú1

2

1

21 2 2 1 22

12( )[( ) ]

= -Ê

ËÁˆ

¯

= -È

ÎÍ

˘

˚˙

=

Ú1

2

1

2 3

1

5

1

15

2

22

2

44

0

2

4

23 5

0

h

ax

h

ax dx

h

a

ax x

ah

a

a

22

42



xA x dA x ah a hEL= ÊËÁ

ˆ¯

=Ú :1

6

1

122 x a= 1

2

yA y dA y ah ahEL= ÊËÁ

ˆ¯

=Ú :1

6

1

152 y h= 2

5

43


PROBLEM 5.36


SOLutiOn

For the element (EL) shown

At x a y h h ka kh

a= = = =, : 3

3or

Then xa

hy=

1313

//

Now dA xdya

hy dy

x xa

hy

y y

EL

EL

= =

= =

=

1313

1 3131

2

1

2

//

//

Then A dAa

hy dy

a

hy ah

hh

= = = ( ) =Ú Ú 130

1313

4 3

0

3

4

3

4//

//

and x dAa

hy

a

hy dy

a

hyEL

h

Ú Ú= ÊËÁ

ˆ¯

= ÊËÁ

ˆ¯

1

2

1

2

3

5130

1313

132 3

5 3/

//

//

/˜ =

= ÊËÁ

ˆ¯

= ÊËÁ

ˆ¯Ú Ú

0

2

0 1313

137 3

0

3

10

3

7

h

EL

h

a h

y dA ya

hy dy

a

hy

//

//

hh

ah= 3

72

Hence xA x dA x ah a hEL= ÊËÁ

ˆ¯

=Ú :3

2

3

102 x a= 2

5

yA y dA y ah ahEL= ÊËÁ

ˆ¯

=Ú : 3

4

3

72 y h= 4

7

44


PROBLEM 5.37

Determine by direct integration the centroid of the area shown.

SOLutiOn

For the element (EL) shown yb

aa x= -2 2

and dA b y dx

b

aa a x dx

x x

y y b

b

aa a x

EL

EL

= -

= - -( )=

= +

= + -( )

( )

( )

2 2

2 2

1

2

2

Then A dAb

aa a x dx

a= = - -( )Ú Ú 0

2 2

To integrate, let x a a x a dx a d= - = =sin cos , cosq q q q: 2 2

Then Ab

aa a a d

b

aa a

= -

= - +ÊËÁ

ˆ¯

ÈÎÍ

˘

Ú 0

2

2 2

2

2

4

pq q q

q q q

/( cos )( cos )

sin sin˚

= -ÊËÁ

ˆ¯

0

2

14

p

p

/

ab

and x dA xb

aa a x dx

b

a

ax a x

EL

a

Ú Ú= - -( )ÈÎÍ

˘˚

= + -ÊËÁ

ˆ¯

ÈÎ

2 2

0

2 2 2 3 2

2

1

3( ) /

ÍÍ˘˚

=

0

2

31

6

p/

a b

45



y dA

b

aa a x

b

aa a x dx

b

ax dx

EL

a

a

Ú Ú

Ú

= + -( ) - -( )ÈÎÍ

˘˚

= =

2

2

0

2 2 2 2

2

22

0( )

bb

a

x

ab

a2

2

3

0

2

2 3

1

6

Ê

ËÁˆ

¯

=

xA x dA x ab a bEL= -ÊËÁ

ˆ¯

ÈÎÍ

˘˚

=Ú : 14

1

62p

or xa=-

2

3 4( )p

yA y dA y ab abEL= -ÊËÁ

ˆ¯

ÈÎÍ

˘˚

=Ú : 14

1

62p or y

b=-

2

3 4( )p

46


PROBLEM 5.38


SOLutiOn

First note that symmetry implies x = 0

For the element (EL) shown

yr

dA rdr

EL =

=

2

pp

(Figure 5.8B)

Then A dA rdrr

r rr

r

r

r

= = =Ê

ËÁˆ

¯= -( )Ú Ú p p p2

22

12

2 21

2

1

2

and y dAr

rdr r r rEL r

r

r

r

Ú Ú= = ÊËÁ

ˆ¯

= -( )22

1

3

2

33

23

13

1

2

1

2

pp( )

So yA y dA y r r r rEL= -( )ÈÎÍ

˘˚

= -( )Ú :p2

2

322

12

23

13 or y

r r

r r=

--

4

323

13

22

12p

47


SOLutiOn

First note that symmetry implies y = 0

dA adx dA a ad

x x x aEL EL

= =

= =

1

22

3

( )

cos

f

f

Then A dA adx a d

a xa

a

a

a

= = -

= - = -

Ú Ú Ú -0

2

0

22

1

2

21

f

f a

a

a

aa[ ] [ ] ( )

and x dA x adx a a d

ax

a

EL

a

a

Ú ÚÚ= - ÊËÁ

ˆ¯

=È

ÎÍ

˘

˚˙ -

-( ) cos

2

3

1

2

2

1

3

2

0

2

0

3

f fa

a

[[sin ]

sin

f

a

aa-

= -ÊËÁ

ˆ¯

a3 1

2

2

3

xA x dA x a aEL= - = -ÊËÁ

ˆ¯Ú : [ ( )] sin2 31

1

2

2

3a a or x a= -

-3 4

6 1

sin

( )

aa

PROBLEM 5.39


48


SOLutiOn

Determination of the Constant k

k

b

a=

2

oryb

ax x

b

by= =

22

1 21 2

//

Vertical Differential Element

A dA y dxb

ax dx

b

a

x abaa

= = = =È

ÎÍ

˘

˚˙ =Ú Ú Ú 20

22

3

03 3

Q x dA xy dx xb

ax dx

b

a

x a by el

aa

= = = ÊËÁ

ˆ¯

=È

ÎÍ

˘

˚˙ =Ú Ú Ú 2

2

0 2

4

0

2

4 4

Since Q xAy = , we have

= =xA x dA xab a b

elÚ 3 4

2

x a= 3

4

Q y dAy

y dxb

ax dx

b

a

x abx el

aa

= = = ÊËÁ

ˆ¯

=È

ÎÍ

˘

˚˙ =Ú Ú Ú2

1

2 2 522

0

2 2

4

5

0

2

110

Since Q yAx = , we have

yA y dA yab ab

el= =Ú 3 10

2

y b= 3

10

y

a

dA=ydx

y

x

y–el=y2

x–el=x

PROBLEM 5.40

Determine by direct integration the location of the centroid of a parabolic spandrel.

a

b

y

y=kx2

x

49



Alternate Method:

Horizontal Differential Element. The same results can be obtained by considering a horizontal element. The first moments of the area are

Q x dAa x

a x dya x

dy

aa

by dy

a b

y el

b= = + =

=Ê

ËÁˆ

¯=

Ú Ú Ú2 2

1

2 4

2 2

0

22 2

( )- -

-00

b

Ú

Q y dA y a x dy y aa

by dy

aya

by

x el= = = ÊËÁ

ˆ¯

= ÊËÁ

Ú Ú Ú( )/

/

//

- -

-

1 21 2

1 23 2 ˆ

¯=Ú dy

abb 2

0 10

y

b

dA=(a–x)dy

xx

a+x

a2

x–el= y–el=y

50


PROBLEM 5.41

Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b.

SOLutiOn

y k x b k a yb

ax

y k x b k a yb

ax

dA y y

1 12

12

1 22

2 24

24

2 44

2 1

= = =

= = =

= -

but

but

( )ddxb

ax

x

adx

x x

y y y

b

ax

x

a

EL

EL

= -Ê

ËÁˆ

¯=

= +

= +Ê

ËÁˆ

¯

22

4

2

1 2

22

4

2

1

2

2

( )

A dAb

ax

x

adx

b

a

x x

a

ba

x

a

a

EL

= = -Ê

ËÁˆ

¯

= -È

ÎÍ

˘

˚˙

=

Ú Ú

Ú

22

4

20

2

3 5

20

3 5

2

15

ddAb

ax

x

adx

b

ax

x

adx

b

a

x x

a

a

= ¥ -Ê

ËÁˆ

¯

= -Ê

ËÁˆ

¯

= -

Ú

Ú

0 22

4

2

23

5

20

2

4 6

4 66

1

12

20

2

a

a b

aÈ

ÎÍ

˘

˚˙

=

51



y dAb

ax

x

a

b

ax

x

adx

b

ax

x

a

EL

a

Ú Ú= +Ê

ËÁˆ

¯-

Ê

ËÁˆ

¯

= -

2

2

20

24

2 22

4

2

2

44

8

4

ÊÊ

ËÁˆ

¯

= -È

ÎÍ

˘

˚˙ =

Ú 0

2

4

5 9

40

2

2 5 9

2

45

a

a

dx

b

a

x x

aab

xA x dA x ba a bEL= ÊËÁ

ˆ¯

=Ú :2

15

1

122 x a= 5

8

yA y dA y ba abEL= ÊËÁ

ˆ¯

=Ú :2

15

2

452 y b= 1

3

52


PROBLEM 5.42


SOLutiOn

We have x x

y ya x

L

x

L

dA ydx ax

L

x

Ldx

EL

EL

=

= = - +Ê

ËÁˆ

¯

= = - +Ê

ËÁˆ

¯

1

2 21

1

2

2

2

2

Then A dA ax

L

x

Ldx a x

x

L

x

L

aL

LL

= = - +Ê

ËÁˆ

¯= - +

È

ÎÍ

˘

˚˙

=

Ú Ú 12 3

8

3

2

2

2 3

20

2

0

2

and x dA x ax

L

x

Ldx a

x x

L

x

LEL

L

Ú Ú= - +Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

= - +È

ÎÍ

˘0

2 2

2

2 3 4

21

2 3 4 ˚˙

=

0

2

210

3

L

aL

y dAa x

L

x

La

x

L

x

Ldx

a

EL

L

Ú Ú= - +Ê

ËÁˆ

¯- +

Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

=

21 1

2

0

2 2

2

2

2

2

11 2 3 2

2 2

2

2

3

3

4

40

2 2 3

2

4

3

5

- + - +Ê

ËÁˆ

¯

= - + - +

Úx

L

x

L

x

L

x

Ldx

ax

x

L

x

L

x

L

x

EL

55

11

5

40

2

2

L

a L

LÈ

ÎÍ

˘

˚˙

=

Hence xA x dA x aL aLEL= ÊËÁ

ˆ¯

=Ú :8

3

10

32 x L= 5

4

yA y dA y a aEL= ÊËÁ

ˆ¯

=Ú :1

8

11

52 y a= 33

40

53


PROBLEM 5.43


SOLutiOn

For y2 at x a y b a kb k

a

b= = = =, : 2

2or

Then yb

ax2

1 2= /

Now x xEL =

and for 02

xa

: yy b x

a

dA y dx bx

adx

EL = =

= =

21 2

2

1 2

2 2

/

/

For a

x a y y yb x

a

x

aEL2

1

2 2

1

21 2

1 2

: ( )= + = - +Ê

ËÁˆ

¯

/

dA y y dx bx

a

x

adx= - = - +

Ê

ËÁˆ

¯( )2 1

1 2 1

2

/

Then A dA bx

adx b

x

a

x

adx

a

a

a= = + - +

Ê

ËÁˆ

¯Ú Ú Ú0

2 1 2

2

1 2 1

2

/ /

/

/

= ÈÎÍ

˘˚

+ - +È

ÎÍ

˘

˚˙

= ÊËÁ

ˆ

b

ax b

x

a

x

ax

b

a

a

a

a

a2

3

2

3 2

1

2

2

3 2

3 2

0

2 3 2 2

2

// /

/

¯+ - Ê

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

+ - - ÊËÁ

ˆ¯

È

ÎÍÍ

˘

3 23 2

3 2

22

2

1

2 2

//

/

( )

( )

aa

ba

aa

˚˙˙

+ - ÊËÁ

ˆ¯

ÈÎÍ

˘˚

ÏÌÔ

ÓÔ

¸˝Ô

Ô

=

1

2 2

13

24

( )aa

ab

54



and x dA x bx

adx x b

x

a

x

aEL

a

a

a

Ú Ú Ú=Ê

ËÁˆ

¯+ - +

Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚0

/ /

/

/2 1 2

2

1 2 1

2˙˙dx

= È

ÎÍ˘˚

+ - +È

ÎÍ

˘

˚˙

= ÊËÁ

ˆ

b

ax b

x

a

x

a

x

b

a

a

a

a

a2

5

2

5 3 4

2

5 2

5 2

0

2 5 2 3 4

2

// /

/

¯+ - Ê

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

+ - - ÊËÁ

ˆ¯

È

ÎÍÍ

˘

5 25 2

5 2

33

2

1

3 2

//

/

( )

( )

aa

ba

aa

˚˙˙

+ - ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

=

1

4 2

71

240

22

2

( )aa

a b

y dAb x

ab

x

adx

b x

a

x

a

EL

a

a

a

Ú Ú

Ú

=È

ÎÍ

˘

˚˙

+ - +Ê

ËÁˆ

¯

2

2

1

2

2 1 2 1 2

2

1 2

0

/ / /

/

/

bbx

a

x

adx

b

ax

b x

a

a

1 2

22

0

2 2 2

1

2

2

1

2 2 2

1

/

/

- +Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

= ÈÎÍ

˘˚

+ -33

1

2

4 2 2

3

2

22

a

x

a

b

a

aa

a

a

a

-ÊËÁ

ˆ¯

Ê

ËÁ

ˆ

¯˜

È

ÎÍÍ

˘

˚˙˙

= ÊËÁ

ˆ¯

+ - ÊËÁ

ˆ

/

( )¯

È

ÎÍÍ

˘

˚˙˙

- -ÊËÁ

ˆ¯

=

2 2 3

2

6 2

1

2

11

48

b

a

a

ab

Hence xA x dA x ab a bEL= ÊËÁ

ˆ¯

=Ú :13

24

71

2402 x a a= =17

1300 546.

yA y dA y ab abEL= ÊËÁ

ˆ¯

=Ú :13

24

11

482 y b b= =11

260 423.

55


PROBLEM 5.44


SOLutiOn

For y1 at x a y b b ka k

b

a= = = =, 2 2

222

or

Then yb

ax1 2

22=

By observation yb

ax b b

x

a2 2 2= - + = -ÊËÁ

ˆ¯

( )

Now x xEL =

and for 0 x a: y yb

ax dA y dx

b

ax dxEL = = = =1

2

21 2

21 2

2and

For a x a 2 : y yb x

adA y dx b

x

adxEL = = -Ê

ËÁˆ¯

= = -ÊËÁ

ˆ¯

1

2 22 22 2and

Then A dAb

ax dx b

x

adx

b

a

xb

a

a

a

a

a

= = + -ÊËÁ

ˆ¯

=È

ÎÍ

˘

˚˙ + - -

Ú Ú Ú2

2

2

3 22

20

2 2

2

3

0

xx

aab

aÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

=2

0

27

6

and x dA x

b

ax dx x b

x

adx

b

a

x

EL

a

a

a

Ú Ú Ú= ÊËÁ

ˆ¯

+ -ÊËÁ

ˆ¯

ÈÎÍ

˘˚

=

0

22

2

4

22 2

2

4ÈÈ

ÎÍ

˘

˚˙ + -

È

ÎÍ

˘

˚˙

= + -ÈÎ ˘ +

0

23

0

2

2 2 2 2

3

1

22

1

32

a a

b xx

a

a b b a aa

a( ) ( ) ( ) --ÈÎ ˘ÏÌÓ

¸˝˛

=

( )a

a b

3

27

6

56



y dAb

ax

b

ax dx

b x

ab

x

adEL

a a

Ú Ú Ú= ÈÎÍ

˘˚

+ -ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯2

22

2

0

22

22 2

0xx

b

a

x b a x

a

a

a

a

ÈÎÍ

˘˚

=È

ÎÍ

˘

˚˙ + - -Ê

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

=

2

5 2 32

17

3

2

4

5

0

2 3 2

002ab

Hence xA x dA x ab a bEL= ÊËÁ

ˆ¯

=Ú :7

6

7

62 x a=

yA y dA y ab abEL= ÊËÁ

ˆ¯

=Ú :7

6

17

302 y b= 17

35

57


PROBLEM 5.45

A homogeneous wire is bent into the shape shown. Determine by direct integra-tion the x coordinate of its centroid.

SOLutiOn

First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the corresponding line

Now x a dL dx dyEL = = +cos3 2 2q and

where x a dx a d

y a dy a d

= = -

= =

cos : cos sin

sin : sin cos

3 2

3 2

3

3

q q q q

q q q q

Then dL a d a d

a

= - +

=

[( cos sin ) ( sin cos ) ]

cos sin (cos

3 3

3

2 2 2 2 1 2

2

q q q q q q

q q

/

qq q qq q q

q q qp

+=

= = =Ú Ú

sin )

cos sin

cos sin sin

2 1 2

2

3

3 31

2

/

0

/

d

a d

L dL a d a 22

0

2

3

2

qpÈ

ÎÍ˘˚

=

/

a

and x dL a a d

a a

ELÚ Ú=

= -ÈÎÍ

˘˚

=

cos ( cos sin )

cos

32

2 5

0

2

3

31

5

3

5

0

/

/

p

p

q q q q

q 22

Hence xL x dL x a aEL= ÊËÁ

ˆ¯

=Ú :3

2

3

52 x a= 2

5

Alternative Solution

x ax

a

y ay

a

= fi = ÊËÁ

ˆ¯

= fi = ÊËÁ

ˆ¯

cos cos

sin sin

3 22 3

3 22 3

q q

q q

/

/

x

a

y

ay a xÊ

ËÁˆ¯

+ ÊËÁ

ˆ¯

= = -2 3 2 3

2 3 2 3 3 21/ /

/ / /or ( )

58



Then dy

dxa x x= - - -( ) ( )2 3 2 3 1 2 1 3/ / / /

Now x xEL =

and dLdy

dx

dx a x x dx

= + ÊËÁ

ˆ¯

= + - -ÈÎ ˘{ }-

1

1

2

2 3 2 3 1 2 1 3 2 1 2

( ) ( )/ / / //

Then L dLa

xdx a x a

aa

= = = ÈÎÍ

˘˚

=Ú Ú1 3

1 31 3

0

2 3

0

3

2

3

2

/

// /

and x dL xa

xdx a x aEL

aa

Ú Ú=Ê

ËÁˆ

¯= È

ÎÍ˘˚

=0

1 3

1 31 3 5 3

0

23

5

3

5

/

// /

Hence xL x dL x a aEL= ÊËÁ

ˆ¯

=Ú :3

2

3

52 x a= 2

5

59


PROBLEM 5.46

A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid.

SOLutiOn

First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the corresponding line

Now x r dL rdEL = =cosq qand

Then L dL rd r r= = = =Ú Ú q q pp

ppp

/4

/

//7 4

47 4 3

2[ ]

and x dL r rd

r

r

ELÚ Ú=

=

= - -ÊËÁ

ˆ¯

= -

cos ( )

[sin ]

q q

q

p

p

pp

/

/

//

4

7 4

24

7 4

2 1

2

1

2

rr2 2

Thus xL xdL x r r= ÊËÁ

ˆ¯

= -Ú :3

222p x r= - 2 2

3p

60


PROBLEM 5.47*

A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid. Express your answer in terms of a.

SOLutiOn


We have at x a y a

a ka ka

= =

= =

,

3 2 1/ or

Then ya

x= 1 3 2/

and dy

dx ax= 3

21 2/

Now x xEL =

and dLdy

dxdx

ax dx

aa x d

= + ÊËÁ

ˆ¯

= + ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

= +

1

13

2

1

24 9

2

1 22 1 2

/

/

xx

Then L dLa

a x dx

aa x

a

a

a

= = +

= ¥ +ÈÎÍ

˘˚

=

Ú Ú1

24 9

1

2

2

3

1

94 9

2713

0

3 2

0

3 2

( )

[( )

/

/ --

=

8

1 43971

]

. a

and x dL xa

a x dxEL

a

Ú Ú= +ÈÎÍ

˘˚0

1

24 9

61


PROBLEM 5.47* (continued)

Use integration by parts with

u x dv a x dx

du dx v a x

= = +

= = +

4 9

2

274 9 3 2( ) /

Then x dLa

x a x a x dxEL

aa

Ú Ú= ¥ +ÈÎÍ

˘˚

- +ÏÌÔ

ÓÔ

¸1

2

2

274 9

2

274 93 2

00

3 2( ) ( )/ / ˝Ô

Ô

= - +ÈÎÍ

˘˚

= -

( )( )

( )

13 1

27

2

454 9

2713

2

4

3 22 5 2

0

23 2

//

/

27a

aa x

a

a

5513 32

0 78566

5 2

2

[( ) ]

.

/ -ÏÌÓ

¸˝˛

= a

xL x dL x a aEL= =Ú : ( . ) .1 43971 0 78566 2 or x a= 0 546.

62


PROBLEM 5.48*


SOLutiOn

We have x x

y ya x

L

EL

EL

=

= =1

2 2 2cos

p

and dA ydx ax

Ldx= = cos

p2

Then A dA ax

Ldx

aL x

L

aL

L

L

= =

= ÈÎÍ

˘˚

=

Ú Ú 0

/

/

2

0

2

2

2

2

2

cos

sin

p

pp

p

and x dA x ax

LdxELÚ Ú= Ê

ËÁˆ¯

cosp2

Use integration by parts with u x dvx

Ldx

du dx vL x

L

= =

= =

cos

sin

p

pp

2

2

2

Then xx

Ldx

L x

L

L x

Ldx

Lx

x

L

L x

Ú Ú= ¥ -

= +

cos sin sin

sin cos

pp

pp

p

pp

pp

2

2

2

2

2

2

2

2

22LÊËÁ

ˆ¯

63



x dA a

Lx

x

L

L x

L

aL L

L

EL

L

Ú = +ÈÎÍ

˘˚

= +Ê

ËÁˆ

¯

2

2

2

2

2

2 2

2

2

0pp

pp

p p

sin cos/

--È

ÎÍÍ

˘

˚˙˙

=

2

0 106374 2

L

aL

p

.

Also y dAa x

La

x

Ldx

a x

EL

L

xL

L

Ú Ú= ÊËÁ

ˆ¯

= +È

ÎÍÍ

˘

2 2 2

2 2

0

2

2

2

cos cos

sin

p p

p

p

/

˚˙˙

= +ÊËÁ

ˆ¯

=

-

0

2 2

2

2 4 2

0 20458

La L L

a L

/

p

.

xA x dA x aL aLEL=Ê

ËÁˆ

¯=Ú : .

20 106374 2

p or x L= 0 236.

yA y dA y aL a LEL=Ê

ËÁˆ

¯=Ú : .

20 20458 2

p or y a= 0 454.

64


PROBLEM 5.49*


SOLutiOn

We have x r ae

y r ae

EL

EL

= =

= =

2

3

2

32

3

2

3

cos cos

sin sin

q q

q q

q

q

and dA r rd a e d= =1

2

1

22 2( )( )q qq

Then A dA a e d a e

a e

a

= = = ÈÎÍ

˘˚

= -

=

Ú Ú1

2

1

2

1

2

1

41

133 623

2 2 2 2

00

2 2

2

q qpp

p

q

( )

.

and x dA ae a e d

a e d

ELÚ Ú

Ú

= ÊËÁ

ˆ¯

=

2

3

1

2

1

3

2 2

0

3 3

0

q qp

qp

q q

q q

cos

cos

To proceed, use integration by parts, with

u e= 3q and du e d= 3 3q q

dv d= cosq q and v = sinq

Then e d e e d3 3 33q q qq q q q qÚ Ú= -cos sin sin ( )

Now let u e= 3q then du e d= 3 3q q

dv d= sin ,q q then v = -cosq

Then e d e e e d3 3 3 33 3q q q qq q q q q qÚ Ú= - - - -ÈÎ

˘˚sin sin cos ( cos )( )

65



So that e de

x dA ae

EL

33

33

103

1

3 103

qq

q

q q q q

q q

Ú

Ú

= +

= +È

Î

cos (sin cos )

(sin cos )ÍÍ˘

˚˙

= - - = -

0

33 3

303 3 1239 26

p

pae a( ) .

Also y dA ae a e d

a e d

ELÚ Ú

Ú

= ÊËÁ

ˆ¯

=

2

3

1

2

1

3

2 2

0

3 3

0

q qp

qp

q q

q q

sin

sin

Using integration by parts, as above, with

u e= 3q and du e d= 3 3q q

dv d= Ú sinq q and v = -cosq

Then e d e e d3 3 33q q qq q q q qÚ Ú= - - -sin cos ( cos )( )

So that e de

y dA ae

EL

33

33

103

1

3 103

qq

q

q q q q

q q

Ú

Ú

= - +

= - +

sin ( cos sin )

( cos sin )ÈÈ

ÎÍ

˘

˚˙

= + =

0

33 3

301 413 09

p

pae a( ) .

Hence xA x dA x a aEL= = -Ú : ( . ) .133 623 1239 262 3 or x a= -9 27.

yA y dA y a aEL= =Ú : ( . ) .133 623 413 092 3 or y a= 3 09.

66


PROBLEM 5.50

Determine the centroid of the area shown when a = 2mm.

SOLutiOn

We have x x

y yx

EL

EL

=

= = -ÊËÁ

ˆ¯

1

2

1

21

1

and dA ydxx

dx= = -ÊËÁ

ˆ¯1

1

Then A dAx

dxx x a aaa

= = -ÊËÁ

ˆ¯

= - = - -Ú Ú 11

2111

[ ln ] ( ln )

and x dA xx

dxx

xa

aEL

aa

Ú Ú= -ÊËÁ

ˆ¯

ÈÎÍ

˘˚

= -È

ÎÍ

˘

˚˙ = - +

Ê

ËÁˆ

¯1

1

2 2

1

2

2

1

2

1

y dAx x

dxx x

dxEL

a

Ú = -ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯

ÈÎÍ

˘˚

= - +ÊËÁ

ˆ¯

1

21

11

1 1

21

2 121ÚÚÚ

= - -ÈÎÍ

˘˚

= - -ÊËÁ

ˆ¯

1

1

1

22

1 1

22

1

a

a

x xx

a aa

ln ln

xA x dA x

a

a a

yA y dA ya a

a a

EL

a

ELa

= =- +

- -

= =- -

- -

Ú

Ú

:ln

:ln

( ln )

2

212

1

12

2 1

Find: x and y when a = 2

We have x =- +

- -

12

2 122 2

2 2 1

( )

ln or x = 1 629.

and y =- -

- -2 2 2

2 2 2 1

12ln

( ln ) or y = 0 1853.

67


PROBLEM 5.51

Determine the value of a for which the ratio x y/ is 9.

SOLutiOn

We have x x

y yx

EL

EL

=

= = -ÊËÁ

ˆ¯

1

2

1

21

1

and dA ydxx

dx= = -ÊËÁ

ˆ¯1

1

Then A dAx

dxx x

a a

aa= = -Ê

ËÁˆ¯

= -

= - -

Ú Ú 11

2

1

11[ ln ]

( ln )

and x dA xx

dxx

x

aa

EL

aa

Ú Ú= -ÊËÁ

ˆ¯

ÈÎÍ

˘˚

= -È

ÎÍ

˘

˚˙

= - +Ê

ËÁˆ

¯

11

2

2

1

2

2

11

2

y dAx x

dxx x

dxEL

a

Ú = -ÊËÁ

ˆ¯

-ÊËÁ

ˆ¯

ÈÎÍ

˘˚

= - +ÊËÁ

ˆ¯

1

21

11

1 1

21

2 121ÚÚÚ

= - -ÈÎÍ

˘˚

= - -ÊËÁ

ˆ¯

1

1

1

22

1

1

22

1

a

a

x xx

a aa

ln

ln

xA x dA x

a

a a

yA y dA ya a

a a

EL

a

ELa

= =- +

- -

= =- -

- -

Ú

Ú

:ln

:ln

( ln )

2

212

1

12

2 1

68


PROBLEM 5.51 (Continued)

Find: a so that x

y= 9

We have x

y

xA

yA

x dA

y dA

EL

EL

= = ÚÚ

Then 12

2 12

12

129

a a

a a a

- +- -( ) =

ln

or a a a a a3 211 18 9 0- + + + =ln

Using trial and error or numerical methods and ignoring the trivial solution a = 1, find

a a= =1 901 3 74. .and

69


PROBLEM 5.52

Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.1 about (a) the line x = 240 mm, (b) the y axis.

Problem 5.1 Locate the centroid of the plane area shown.

SOLutiOn

From the solution to Problem 5.1 we have

A

xA

yA

= ¥

= ¥

= ¥

15 3 10

2 6865 10

1 4445 10

3 2

6 3

6 3

.

.

.

mm

mm

mm

S

S

Applying the theorems of pappus-Guldinus we have

(a) Rotation about the line x = 240 mm

Volume = -= -

= ¥ - ¥

2 240

2 240

2 240 15 3 10 2 6865 103 6

pp

p

( )

( )

[ ( . ) .

x A

A xAS

]] Volume mm= ¥6 19 106 3.

Area line line= == + + + +

2 2

2 1 1 3 3 4 4 5 5 6 6

p pp

X L x L

x L x L x L x L x L

S( )

( )

where x x1 6, ,… are measured with respect to line x = 240 mm

Area = + ++ +2 120 240 15 30 30 270

135 210 240 30

p[( )( ) ( )( ) ( )( )

( )( ) ( )( )]] Area mm= ¥458 103 2

(b) Rotation about the y axis

Volume

mm

area= =

= ¥

2 2

2 2 6865 106 3

p p

p

X A xA( )

( . )

S

Volume mm= ¥16 88 106 3.

Area line line= =

= + + + +

=

2 2

2

2

1 1 2 2 3 3 4 4 5 5

p p

p

p

X L x L

x L x L x L x L x L

S( )

( )

[(( )( ) ( )( )

( )( ) ( )( ) ( )( )]

120 240 240 300

225 30 210 270 105 210

+

+ + + Area mm= ¥1 171 106 2.

70


PROBLEM 5.53

Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.2 about (a) the line y = 60 mm, (b) the y axis.


SOLutiOn


A

xA

yA

=

=

=

1740

28200

55440

2

3

3

mm

mm

mm

S

S

Applying the theorems of pappus-Guldinus we have

(a) Rotation about the line y = 60 mm

Volume = -= -= -

2 60

2 60

2 60 1740 55440

ppp

( )

( )

[ ( ) ]

y A

A yAS Volume mm= ¥308 103 3

Area line

line

=== + + + +

2

2

2 1 1 2 2 3 3 4 4 6 6

ppp

Y

y L

y L y L y L y L y L

S( )

( )

where y y1 6, ,… are measured with respect to line y = 60 mm

Area = + + + + +2 60 20 48 24 36 30 18 30 36 30 602 2p ( )( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )( )ÈÈÎÍ

˘˚

Area mm= ¥38 2 103 2.


Volume mmarea= = =2 2 2 28200 3p p pX A xA( ) ( )S Volume mm= ¥177 2 106 3.

Area line line= = = + + + +

=

2 2 2

2

1 1 2 2 3 3 4 4 5 5p p p

p

X L x L x L x L x L x L x LS( ) ( )

(110 20 20 24 35 30 35 30 36 10 202 2)( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )( )+ + + + +ÈÎÍ

˘˚

Area mm= ¥22 4 103 2.

71


PROBLEM 5.54

Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.8 about (a) the x axis, (b) the y axis.


SOLutiOn


A

xA

yA

= ¥

= ¥

= ¥

0 71661 10

221 0505 10

420 27 10

6 2

6 3

6 3

.

.

.

mm

mm

mm

S

S

Applying the theorems of Pappus-Guldinus we have

(a) Rotation about the x axis:

Volume

mm

mm

= =

= ¥

= ¥

2 2

2 420 27 10

26406 10

6 3

9 3

p p

p

Y A y Aarea S

( . )

or Volume mm= ¥2 64 109 3.

Area === + + + +=

2

2

2

22 2 3 3 4 4 5 5 6 6

pppp

Y A

y A

y L y L y L y L y L

line

lineS( )

( )

[(( . )( ) ( )( ) ( . )( )

( )( ) ( )

187 5 375 750 375 1187 5 125

1250 750 625

+ ¥ ++ +

p(( )]1250 or Area mm= ¥17 73 106 2.


Volume

0 mm

= =

= ¥

2 2

2 221 0505 1 6 3

p p

p

X A x Aarea S

( . ) or Volume mm= ¥1 389 109 3.

Area = == + + + +

=

2 2

2

2

1 1 2 2 3 3 4 4 5 5

p pp

p

X L x L

x L x L x L x L x Lline lineS( )

( )

(3375 750 750 375 7502 375

375 750 125)( ) ( )( ) ( ) ( )( ) (+ + - ¥ÊËÁ

ˆ¯

¥ + +p

p 3375 750)( )ÈÎÍ

˘˚

or Area mm= ¥9 67 106 2.

72


PROBLEM 5.55

Determine the volume of the solid generated by rotating the parabolic area shown about (a) the x axis, (b) the axis AA.

SOLutiOn

First, from Figure 5.8a we have A ah

y h

=

=

4

32

5

Applying the second theorem of Pappus-Guldinus we have

(a) Rotation about the x axis:

Volume =

= ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

2

22

5

4

3

p

p

yA

h ah or Volume = 16

152pah

(b) Rotation about the line AA¢:

Volume =

= ÊËÁ

ˆ¯

2 2

2 24

3

p

p

( )

( )

a A

a ah or Volume = 16

32pa h

73


PROBLEM 5.56

Determine the volume and the surface area of the chain link shown, which is made from a 6-mm-diameter bar, if R = 10 mm and L = 30 mm.

SOLutiOn

The area A and circumference C of the cross section of the bar are

A d C d= =p p4

2 and .

Also, the semicircular ends of the link can be obtained by rotating the cross section through a horizontal semicircular arc of radius R. Now, applying the theorems of Pappus-Guldinus, we have for the volume V:

V V V

AL RA

L R A

= +

= +

= +

2 2

2 2

2

( ) ( )

( ) ( )

( )

side end

p

p

or V = + ÈÎÍ

˘˚

2 30 104

6 2[ ( )] ( )mm mm mmp p

= 3470 3mm or mmV = 3470 3

For the area A: A A A

CL RC

L R C

= +

= +

= +

2 2

2 2

2

( ) ( )

( ) ( )

( )

side end

p

p

or A = +2 30 10 6[ ( )][ ( )]mm mm mmp p

= 2320 2mm or mmA = 2320 2

74


PROBLEM 5.57

Verify that the expressions for the volumes of the first four shapes in Figure 5.21 on Page 253 are correct.

SOLutiOn

Following the second theorem of Pappus-Guldinus, in each case a spe-cific generating area A will be rotated about the x axis to produce the given shape. Values of y are from Figure 5.8a.

(1) Hemisphere: the generating area is a quarter circle

We have V y Aa

a= = ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

2 24

3 42p p

pp

or V a= 2

33p

(2) Semiellipsoid of revolution: the generating area is a quarter ellipse

We have V y Aa

ha= = ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

2 24

3 4p p

pp

or V a h= 2

32p

(3) Paraboloid of revolution: the generating area is a quarter parabola

We have V y A a ah= = ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

2 23

8

2

3p p or V a h= 1

22p

(4) Cone: the generating area is a triangle

We have V y Aa

ha= = ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

2 23

1

2p p or V a h= 1

32p

75


SOLutiOn

The required volume can be generated by rotating the area shown about the y axis. Applying the second theorem of Pappus-Guldinus, we have

V x A=

= +ÈÎÍ

˘˚

¥ ¥ ¥ÈÎÍ

˘˚

2

2 7 51

3

1

25

p

p . (5) mm mm 5 mm V = 720 3mm

PROBLEM 5.58

A 15 mm-diameter hole is drilled in a piece of 20 mm-thick steel; the hole is then countersunk as shown. Determine the volume of steel removed during the countersinking process.

76


PROBLEM 5.59

Determine the capacity, in liters, of the punch bowl shown if R = 250 mm.

SOLutiOn

The volume can be generated by rotating the triangle and circular sector shown about the y axis. Applying the second theorem of Pappus-Guldinus and using Figure 5.8a, we have

V x A x A

x A x A

R R R

= == +

= ¥ÊËÁ

ˆ¯

¥ ¥Ê

ËÁˆ

¯+

2 2

2

21

3

1

2

1

2

1

2

3

2

1 1 2 2

p pp

p

S( )

22 30

3 6

216 3 2 3

3

6

2

3 3

RR

R R

sin ∞¥

Ê

ËÁˆ

¯ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

= +Ê

ËÁˆ

¯

=

pp

p

33

8

3 3

80 25

0 031883

3

3

3

p

p

R

=

=

( . )

.

m

m

Since 10 13 3l = m

Vl

= ¥0 03188310

13

3

. mm3

V l= 31 9.

77


PROBLEM 5.60

Three different drive belt profiles are to be studied. If at any given time each belt makes contact with one-half of the circumference of its pulley, determine the contact area between the belt and the pulley for each design.

SOLutiOn

Applying the first theorem of Pappus-Guldinus, the contact area AC of a belt is given by:

A yL yLC = =p p S

where the individual lengths are the lengths of the belt cross section that are in contact with the pulley.

(a) A y L y LC = +

-ÊËÁ

ˆ¯

ÈÎÍ

˘˚ ∞ÈÎÍ

˘˚

p

p

[ ( ) ]

. .

cos

2

2 602 5

2

2 5

20

1 1 2 2

mmmm

++ -ÏÌÓ

¸˝˛

[( . ) ]( . )60 2 5 12 5mm mm

or AC = 3240 2mm

(b) A y LC =

= - -ÊËÁ

ˆ¯

ÈÎÍ

˘˚ ∞ÊËÁ

ˆ¯

p

p

[ ( )]

.. .

cos

2

2 60 1 67 5

2

7 5

20

1 1

mmmm

or AC = 2740 2mm

(c) A y LC =

= -ÊËÁ

ˆ¯

ÈÎÍ

˘˚

p

pp

p

[ ( )]

( )[ ( )]

2

602 5

5

1 1

mm mm

or AC = 2800 2mm

78


PROBLEM 5.61

The aluminum shade for the small high-intensity lamp shown has a uniform thickness of 1 mm. Knowing that the density of aluminum is 2800 kg/m3, determine the mass of the shade.

SOLutiOn

The mass of the lamp shade is given by

m V At= =r r

where A is the surface area and t is the thickness of the shade. The area can be generated by rotating the line shown about the x axis. Applying the first theorem of Pappus Guldinus we have

A yL yL

y L y L y L y L

= == + + +

2 2

2 1 1 2 2 3 3 4 4

p pp

S( )

or A = + +ÊËÁ

ˆ¯

¥ +ÈÎÍ

+ +

213

1313 16

23

16 28

p mm

2 mm) mm (32 mm) mm)2 2( (

2212

28 33

228 5

ÊËÁ

ˆ¯

¥ +

+ +ÊËÁ

ˆ¯

¥ +

mm (8 mm) mm)

mm mm) mm)

2 2

2

(

( ( 22

2 mm

˘˚

= + + +

=

2 84 5 466 03 317 29 867 51

10903 4

p ( . . . . )

.

Then m At=

= ¥ -

r

( )( . )( .2800 10 9034 10 0 0013 kg/m m m)3 2

or m = 0 0305. kg

79


PROBLEM 5.62

The escutcheon (a decorative plate placed on a pipe where the pipe exits from a wall) shown is cast from brass. Knowing that the density of brass is 8470 kg/m3, determine the mass of the escutcheon.

SOLutiOn

The mass of the escutcheon is given by m V= ( ) ,density where V is the volume. V can be generated by rotating the area A about the x-axis.

From the figure: L

L

12 2

2

75 12 5 73 9510

37 5

2676 8864

= - =

=∞

=

. .

.

tan.

m

mm

a L L= - =

= = ∞

= ∞ - ∞ =

-

2 1

1

2 9324

12 5

759 5941

26 9 5941

28 203

.

..

..

mm

sinf

a 00 0 143168∞ = . rad

Area A can be obtained by combining the following four areas:

Applying the second theorem of Pappus-Guldinus and using Figure 5.8a, we have

V yA yA= =2 2p p S

Seg. A, mm2 y, mm yA , mm3

11

276 886 37 5 1441 61( . )( . ) .= 1

337 5 12 5( . ) .= 18020 1.

2 - = -a( ) .75 805 322 2 75

315 2303

( )sinsin ( ) .

aa

a f+ = −12265.3

3 - = -1

273 951 12 5 462 19( . )( . ) .

1

312 5 4 1667( . ) .= -1925 81.

4 - = -( . )( . ) .2 9354 12 5 36 6931

212 5 6 25( . ) .= -229 33.

S 3599 7.

80



Then V yA

m V

=

=

==

=

2

2 3599 7

22618

8470

p

p

S

( . )

(

( )(

mm

mm

density)

kg/m

3

3

3 222 618 10 6. )¥ - m3

= 0 191574. kg or m = 0 1916. kg

81


PROBLEM 5.63

A manufacturer is planning to produce 20,000 wooden pegs having the shape shown. Determine how many gallons of paint should be ordered, knowing that each peg will be given two coats of paint and that one liter of paint covers 9 m2.

SOLutiOn

The number of gallons of paint needed is given by

Number of liters Number of pegs)(Surface area of 1 peg)1 = (

lliter

m coats)

92

2

Ê

ËÁˆ

¯(

or Number of liters 4444.44 m= A As s( )∼ 2

where As is the surface area of one peg. As can be generated by rotating the line shown about the x axis.

Using the first theorem of Pappus-Guldinus and Figures 5.8b,

We have

R =

=

17 5

210

17 5

.

sin.

mm

mm

mma

or 2 34.850 17.425a ap p

= ∞ = ∞= =A Y L y Ls 2 2 S

L, mm�y, mm yL, mm2

1 5 2 5. 12 5.

2 10 5 50

3 1 25.5 6 25

25 625

+ =.. 7 03125.

4 60 17 5 1 34 850 3 75 53 111- - - =. ( cos . ) . . 6 25. 331 944.

5p2

3 75 5 8905¥ =. . 102 3 75

7 6127- ¥ =..

p44 843.

6 2 17 5a( . )17 5 17 425

17 425. sin .

sin .∞ ¥ ∞

a54 926.

S y L = 501 244. mm2

82



Then As = =

= ¥ -

2 501 244 3149 41

3149 41 10

2 2

6 2

p ( . ) .

.

mm mm

m

Finally Number of liters

liters

= ¥ ¥=

-4444 44 3149 41 10

13 997

6. .

.

Order 14 liters

83


PROBLEM 5.64

The wooden peg shown is turned from a dowel 20 mm in diameter and 80 mm long. Determine the percentage of the initial volume of the dowel that becomes waste.

SOLutiOn

To obtain the solution it is first necessary to determine the volume of the peg. That volume can be generated by rotating the area shown about the x axis.

The generating area is next divided into six components as indicated

sin.

210

17 5a =

or 2 34 850 17 425a a= ∞ = ∞. .

Applying the second theorem of Pappus-Guldinus and then using Figure 5.8a, we have

V YA yAPEG = =2 2p p S

A, mm2

�y, mm yA, mm3

1 10 5 50¥ = 2 5. 125

2 [ . ( cos . ) . ]

( . ) .

60 17 5 1 34 850 3 75

6 25 331 946

- - ∞ -¥ = 3 125. 1037331

3 3 75 10. ¥ 5 187 5.

4 - = -p4

3 75 11 0452( . ) . 104 3 75

38 4085- ¥ =..

p-92 871.

5 a ( . )17 5 2 2 17 5 17 425

317 425

¥ ∞¥ ∞

. sin .sin .

a320 40.

6 - ∞ = -1

217 5 34 850 10 71 8069( . cos . )( ) .

1

310 3 3333( ) .= -239 356.

S yA mm= 1338 004 3.

84



Then Vpeg =

=

2 1338 004

8406 93

p ( . )

.

mm

mm

3

3

Now Vdowel =

=

=

p

p4

420 80

2( ) (

( (

diameter length)

mm) mm)

25132.74 mm

2

3

Then % %

%

Waste = ¥

=-

¥

V

V

V V

V

waste

dowel

dowel peg

dowel

100

100

= -ÊËÁ

ˆ¯

¥18406 93

25132 74100

.

.% or % . %Waste = 66 5

85


PROBLEM 5.65*

The shade for a wall-mounted light is formed from a thin sheet of translucent plastic. Determine the surface area of the outside of the shade, knowing that it has the parabolic cross section shown.

SOLutiOn

First note that the required surface area A can be generated by rotating the parabolic cross section through p radians about the y axis. Applying the first theorem of Pappus-Guldinus we have

A xL= p

Now at x y

k k

= =

= = -

100 250

100 0 0252 1

mm, mm

250 or mm( ) .

and x x

dLdy

dxdx

EL =

= + ÊËÁ

ˆ¯

12

where dy

dxkx= 2

Then dL k x dx= +1 4 2 2

We have xL x dL x k x dxEL= = +( )Ú Ú 0

100 2 21 4

xLk

k x= +ÈÎÍ

˘˚

= +

1

3

1

41 4

1

12

1

0 0251 4 0 025 1

22 2 3 2

0

100

22

( )

( . )[ ( . ) (

/

000 1

17543 3

2 3 2 3 2) ] ( )

.

/ /-{ }= mm2

Finally A = p ( . )17543 3 mm2 or A = ¥55 1 103. mm2

86


PROBLEM 5.66

For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load, (b) the reactions at the beam supports.

SOLutiOn

R

R

R R R

I

II

I II

N/m)(3m) N

N/m)(3m) N

= =

= =

= + =

1

22500 3750

1

22000 3000

(

(

33750 3000 6750

6750 1 3750 2 3000 1 4444

+ =

= = + =

N

:XR XR X XS ( ) ( )( ) ( )( ) . mm

(a) R = Ø6750 N X = 1 444. m

(b) Reactions: + S M BA = - =0 3 6750 0: m) N)(1.4444 m)( (

B = 3249 99. N B = ≠3250 N

+ S F Ay = + - =0 3249 99 6750 0: N N.

A = 3500 01. N A = ≠3500 N

87


PROBLEM 5.67

For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load, (b) the reactions at the beam supports.

SOLutiOn

(a) We have R

R

I

II

m N/m N

m N/m N

= ( ) =

= =

( )

( )( )

4 200 800

2

34 600 1600

Then S F R R Ry : I II- = - -

or R = + =800 1600 2400 N

and S M XA: - = - -( ) ( ) . ( )2400 2 800 2 5 1600

or X = 7

3m

R = 2400 N Ø X = 2 33. m

(b) Reactions + S F Ax x= =0 0:

+ S M BA y= - ÊËÁ

ˆ¯

=07

32400 0: (4 m) m N)(

or By = 1400 N

+ SF Ay y= + - =0 1400 2400 0: N N

or Ay = 1000 N

A = 1000 N ≠ B = 1400 N ≠

88


PROBLEM 5.68

Determine the reactions at the beam supports for the given loading.

SOLutiOn

R

R

I

II

kN/m)(6 m)

kN

kN/m) m)

kN

=

=

=

=

1

24

12

2 10

20

(

( (

+ SF Ay = - - =0 12 20 0: kN kN

A = 32 0. kN

+ SM MA A= - - =0 12 20 0: kN)(2 m) kN)(5 m( ( )

MA = ◊124 0. kN m

89


PROBLEM 5.69


SOLutiOn

We have R

R

R

I

II

III

m) 8kN/m kN

m) kN/m kN

= ( ) =

= ( ) =

=

1

20 9 3 6

1

21 8 10 9 0

0

( . .

( . .

( .66 10 6m) kN/m kN( ) =

Then + SF Bx x= =0 0:

+ SM CB y= - - =0 1 2 2 1 0: (0.6m)(3.6kN) m)(9kN)+(1.8m) m)(6kN)( . ( .

or Cy = 11 8. kN C = 11 8. kN ≠

+ SF By y= - + - + - =0 3 6 9 0 11 8 0: kN kN kN 6kN. . .

or By = 6 8. kN B = 6 8. kN≠

90


PROBLEM 5.70


SOLutiOn

R

R

R

R

I

I

II

II

kN/m) m)

kN

kN/m)(2m)

3kN

==

=

=

( (

(

3 5

15

1

23

+ SM BA = - - + + =0 15 3 5 0: kN)(0.5 m) kN)(3 m 2/3m) m( ( ( )

B = 3 7. kN B = 3 7. kN≠

+ SF Ay = + - - =0 3 7 15 3 0: kN kN kN.

A = 14 3. kN A = 14 3. kN≠

91


PROBLEM 5.71


SOLutiOn

First replace the given loading with the loading shown below. The two loadings are equivalent because both are defined by a linear relation between load and distance and the values at the end points are the same.

We have R

R

I

II

m N/m N

m N/m N

= =

= =

1

23 6 2200 3960

3 6 1200 4320

( . )( )

( . )( )

Then + SF Bx x= =0 0:

+ SM AB y= - +0 3 6 2 4 3960: m m N( . ) ( . )( )

- =( . )( )1 8 4320 0m N

or Ay = 480 N A = 480 N ≠

+ SF By y= - + + =0 480 3960 4320 0: N N

or By = -840 N B = 840 N Ø

92


PROBLEM 5.72


SOLutiOn

We have R

R

I

II

m N/m N

m N/m N

= =

= =

1

34 3000 4000

1

32 1500 1000

( )( )

( )( )

Then + SF Ax x= =0 0:

+ SF Ay y= - - =0 4000 1000 0: N N

or Ay = 5000 N A = 5000 N ≠

+ SM MA A= 0 1 5 5: ( ( .- -m)(4000 N) m)(1000 N) = 0

or M A = ◊9500 N m MA = ◊9500 N m

93


PROBLEM 5.73


SOLutiOn

First replace the given loading with the loading shown below. The two loadings are equivalent because both are defined by a parabolic relation between load and distance and the values at the end points are the same.

We have R

R

I

II

m N/m N

m N/m N

= =

= =

( )( )

( )( )

6 300 1800

2

36 1200 4800

Then + SF Ax x= =0 0:

+ SF Ay y= + - =0 1800 4800 0: N N

or Ay = 3000 N A = 3000 N ≠

+ SM MA A= + - ÊËÁ

ˆ¯

=0 3 180015

44800 0: m N m N( )( ) ( )

or M A = ◊12 6. kN m MA = ◊12 6. kN m

94


PROBLEM 5.74

Determine (a) the distance a so that the vertical reactions at supports A and B are equal, (b) the corresponding reactions at the supports.

SOLutiOn

(a)

We have R a a

R a a

I

II

m N/m N

m N/m N

= ( )( ) =

= -( )ÈÎ ˘ ( ) = -( )

1

21800 900

1

24 600 300 4

Then + SF A a a By y y= - - - + =0 900 300 4 0: ( )

or A B ay y+ = +1200 600

Now A B A B ay y y y= fi = = +600 300 ( )N (1)

Also + SM Aa

aB y= - + -ÊËÁ

ˆ¯

ÈÎÍ

˘˚

( )ÈÎ ˘0 4 43

900: ( )m m N

+ -ÈÎÍ

˘˚

-( )ÈÎ ˘ =1

34 300 4 0( )a am N

or A a ay = + -400 700 50 2 (2)

Equating Eqs. (1) and (2) 600 300 400 700 50 2+ = + -a a a

or a a2 8 4 0- + =

Then a =± - -8 8 4 1 4

2

2( ) ( )( )

or a = 0 53590. m a = 7 4641. m

Now a 4 m fi a = 0 536. m

95



(b) We have + SF Ax x= =0 0:

Eq. (1) A By y=

= +600 300 0 53590( . )

= 761 N A B= = 761 N ≠

96


PROBLEM 5.75

Determine (a) the distance a so that the reaction at support B is minimum, (b) the corresponding reactions at the supports.

SOLutiOn

(a)

We have R a a

R a a

I

II

m N/m N

m N/m N

= =

= - = -

1

21800 900

1

24 600 300 4

( )( )

[( ) ]( ) ( )

Then + SMa

aa

a BA y= -ÊËÁ

ˆ¯

- +ÊËÁ

ˆ¯

-( )ÈÎ ˘ + ( ) =03

8

3300 4 4 0: m (900 N) m N m

or B a ay = - +50 100 8002 (1)

Then dB

daay = - =100 100 0 or a = 1 000. m

(b) Eq. (1) By = - + =50 1 100 1 800 7502( ) ( ) N B = 750 N ≠

and + SF Ax x= =0 0:

SF Ay y= - - - + =0 900 1 300 4 1 750 0: N N N( ) ( )

or Ay = 1050 N A = 1050 N ≠

+

97


PROBLEM 5.76

Determine the reactions at the beam supports for the given loading when w0 = 1.5 kN/m.

SOLutiOn

We have R

R

I

II

m kN m kN

m kN m kN

= ( ) =

= ( ) =

1

29 3 5 15 75

1

29 1 5 6 75

( ) . .

( ) . .

Then + SF Cx x= =0 0:

+ S ◊M

CB

y

= - - -- + =

0 50 15 75

4 6 75 6 0

: kN m) (1m) kN

m kN m

( ( . )

( )( . ) ( )

or Cy = 15 4583. kN C = 15 46. kN ≠

+ SF By y= - - + =0 15 75 6 75 15 4583 0: kN kN kN. . .

or By = 7 0417. kN B = 7 04. kN ≠

98


PROBLEM 5.77

Determine (a) the distributed load w0 at the end D of the beam ABCD for which the reaction at B is zero, (b) the corresponding reaction at C.

SOLutiOn

(a)

We have R

R

I

II

m kN/m kN

m kN/m kN

= =

= =

1

29 3 5 15 75

1

29 4 50 0

( )( . ) .

( )( ) .w w

Then + S ◊MC = - + + == -

0 50 5 15 75 2 4 5 0

3 1940

0

: kN m m kN m kN

kN/m

( ) ( )( . ) ( )( . )

.

ww

or w0 3 194= - ≠. kN/m

(b) + SF Cx x= =0 0:

+ SF Cy y= - + + =0 15 75 4 5 3 194 0: . ( . )( . )

or Cy = 1 377. kN C = 1 377. kN ≠

99


PROBLEM 5.78

The beam AB supports two concentrated loads and rests on soil that exerts a linearly distributed upward load as shown. Determine the values of wA and wB corresponding to equilibrium.

SOLutiOn

R

R

A A

B B

I

II

m

m

= =

= =

1

21 8 0 9

1

21 8 0 9

w w

w w

( . ) .

( . ) .

+ SM aD A= - - - =0 24 1 2 30 0 3 0 9 0 6 0: kN kN m m( )( . ) ( )( . ) ( . )( . )w (1)

For a a= - - - =0 6 24 1 2 0 6 30 0 3 0 54 0. ( . . ) ( )( . ) .m: w

14 4 9 0 54 0. .- - =w A w A = 10 00. kN/m

+ SFy B= - - + + =0 24 30 0 9 10 0 9 0: kN kN kN/m. ( ) . w wB = 50 0. kN/m

100


SOLutiOn

We have R

R B B

I

II

m kN/m kN

m kN/m kN

= =

= =

1

21 8 20 18

1

21 8 0 9

( . )( )

( . )( ) .w w

(a) + SM aC = - ¥ - ¥ - ¥ =0 1 2 24 0 6 18 0 3 30 0: m kN m kN m kN( . ) . .

or a = 0 375. m

(b) + SFy B= - + + - =0 24 18 0 9 30 0: kN kN kN kN( . )w

or w B = 40 0. kN/m

PROBLEM 5.79

For the beam and loading of Problem 5.78, determine (a) the distance a for which wA = 20 kN/m, (b) the corresponding value of wB.

Problem 5.78 The beam AB supports two concentrated loads and rests on soil that exerts a linearly distributed upward load as shown. Determine the values of wA and wB corresponding to equilibrium.

101


PROBLEM 5.80

The cross section of a concrete dam is as shown. For a 1-m-wide dam section determine (a) the resultant of the reaction forces exerted by the ground on the base AB of the dam, (b) the point of application of the resultant of Part a, (c) the resultant of the pressure forces exerted by the water on the face BC of the dam.

SOLutiOn

(a) Consider free body made of dam and triangular section of water shown. (Thickness = 1 m)

p = ( . )( )( . )7 2 10 9 813m kg/m m/s3 2

W

W

13 3 2

2

2

34 8 7 2 1 2 4 10 9 81

542 5

1

2

= ¥

=

=

( . )( . )( )( . )( . )

.

m m m kg/m m/s

kN

(( . )( . )( )( . )( . )

.

( .

2 4 7 2 1 2 4 10 9 81

203 4

1

22 4

3 3 2

3

m m m kg/m m/s

kN

¥

=

=W mm m m kg/m m/s

kN

m m

)( . )( )( )( . )

.

( . )(

7 2 1 10 9 81

84 8

1

2

1

27 2 1

3 3 2

=

= =P Ap ))( . )( )( . )

.

7 2 10 9 81

254 3

3 3 2m kg/m m/s

kN=

+ SF Hx = - =0: 254.3 kN 0 H = 254 kN Æ

+ SF Vy = - - - =0 542 5 203 4 84 8 0: . . .

V = 830 7. kN V = 831 kN ≠

102



(b) x

x

x

1

2

3

5

84 8 3

4 81

32 4 5 6

4 82

32 4 6 4

= =

= + =

= + =

( . )

. ( . ) .

. ( . ) .

m m

m

m

+ S SM xV xW PA = - + =0 2 4 0: m( . )

x( . ) ( )( . ) ( . )( . )

( . )( . ) (

830 7 3 542 5 5 6 203 4

6 4 84 8 2

kN m kN m kN

m kN

- -- + .. )( . )

( . ) . . . .

( .

4 254 3 0

830 7 1627 5 1139 0 542 7 610 3 0

830

m kN =- - - + =x

x 77 2698 9 0) .- =

x = 3 25. m (To right of A)

(c) Resultant on face BC

Direct computation:

P gh

P

BC

= =

=

= +

r ( )( . )( . )

.

( . ) ( .

10 9 81 7 2

70 63

2 4 7 2

3 3 2

2

2

kg/m m/s m

kN/m

))

.

.

( . )( . )( )

2

2

7 589

18 43

1

21

270 63 7 589 1

== ∞

=

=

m

kN/m m m

q

R PA

R = 268 kN 18 43. ∞

Alternate computation: Use free body of water section BCD.

- =R 268 kN 18 43. ∞

R = 268 kN 18 43. ∞

103


PROBLEM 5.81

The cross section of a concrete dam is as shown. For a 0.3 m-wide dam section determine (a) the resultant of the reaction forces exerted by the ground on the base AB of the dam, (b) the point of application of the resultant of Part a, (c) the resultant of the pressure forces exerted by the water on the face BC of the dam.

SOLutiOn

The free body shown consists of a 6.3 m thick section of the dam and the quarter circular section of water above the dam.

Note: x

x

x

1

2

4

64 6

3

3 4535

6 1

144 6

3

= - ¥ÊËÁ

ˆ¯

=

= + =

= - ¥ÊËÁ

ˆ¯

=

p

p

m

m

m 7 m

m

.

( )

111 4535. m

For area 3 first note.

a x

I r2 1

2r

II - p4

2r rr- 4

3p

Then x3

12

2 4 63 4

2

24

28

6 6 6 6

8

= ++ -( ) - ¥( )

-

È

Î

ÍÍ

˘

˚

˙˙

= +

¥

m6 6

m( )( )

( ) ( )

(

pp

p

11 3402 9 3402. ) .m m=

104



(a) Now W V= Â

So that W

W

1

2

46 0 199 707

2400

= ¥( )ÈÎÍ

˘˚

=

= ¥

2400 9.81 N/m m) .3m) N

9.

3 2p( ( ,

( 881 N/m m)(6 m)(0.3m) N

9.81 N/m

3

3

)[( ] , .

( )

2 84 758 4

2400 643

2

=

= ¥ - ¥Wp

66 54 567 9

1004

6

2

4

ÊËÁ

ˆ¯

¥ÈÎÍ

˘˚

=

= ¥

m (0.3m) N

9.81 N/m m

2

3

, .

( ) (Wp

)) m) N2( . .0 3 83211 4ÈÎÍ

˘˚

=

Also P Ap= = ¥ =1

2

1

26 6[( )( m)(0.3m)][(1000 9.81 N/m m)] 52,974 N3

Then + SF Hx = - =0 52 974: N 0, or H = 53 0. kN Æ

+ SF Vy = - - - - =0 0: 199,707 N 84,758.4 N 54567.9 N 83211.4 N

or V = 422 244 7, . N V = 422kN ≠

(b) We have + SM xA = - --

0 422 244 7 3 4535 7: N) m)(199,707 N) m)(84,758.4 N)( , . ( . (

(99 3402 11 4535

0

. ( .m)(54,567.9 N) m)(83211.4 N)

+(2 m)(52,974 N)

-=

or 422 244 7 689 688 593 309 509 675 953 062 105 948 0, . , , , , ,x - - - - + =

or x = 6 25. m

(c) Consider water section BCD as the free body

We have SF = 0

Then - =R 98 6. kN 57 5. ∞ or R = 98 6. kN 57 5. ∞

105


PROBLEM 5.82

The 3 ¥ 4-m side AB of a tank is hinged at its bottom A and is held in place by a thin rod BC. The maximum tensile force the rod can withstand without breaking is 200 kN, and the design specifications require the force in the rod not to exceed 20 percent of this value. If the tank is slowly filled with water, determine the maximum allowable depth of water d in the tank.

SOLutiOn

Consider the free-body diagram of the side.

We have P Ap A gd= =1

2

1

2( )r

Now + SM hTd

PA = - =03

0:

where h = 3 m

Then for dmax .

( ( ) ( .33

1

24 10 9 813 3 m)(0.2 200 10 N) m kg/m m/s3 2¥ ¥ - ¥ ¥ ¥

ddmax

max ¥¥ÈÎÍ

˘˚

=dmax ) 0

or 120 6 54 03 N m N/m2◊ - =. dmax

or dmax = 2 64. m

106


SOLutiOn

Consider the free-body diagram of the side.

We have P Ap A gd= =

=

1

2

1

2

1

22 9 9 81

( )

[( . )( . )(

r

m)(4 m)] [(1263 kg/m m/s3 2 22 9. m)]

= 208.40 kN

Then + SF Ay y= =0 0:

+ SM TA = - ÊËÁ

ˆ¯

=02 9

3208 4: (3 m) m kN) 0

.( .

or T = 67 151. kN T = 67 2. kN ¨

+ SF Ax x= + - =0 208 40 67 151: kN kN 0. .

or Ax = -141 249. kN A = 141 2. kN ¨

PROBLEM 5.83

The 3 ¥ 4-m side of an open tank is hinged at its bottom A and is held in place by a thin rod BC. The tank is to be filled with glycerine, whose density is 1263 kg/m3. Determine the force T in the rod and the reactions at the hinge after the tank is filled to a depth of 2.9 m.

107


PROBLEM 5.84

The friction force between a 2 ¥ 2-m square sluice gate AB and its guides is equal to 10 percent of the resultant of the pressure forces exerted by the water on the face of the gate. Determine the initial force needed to lift the gate if it weighs 5 kN.

SOLutiOn

Consider the free-body diagram of the gate.

Now P ApI I2 3 m N/m m)]

58,860 N

= = ¥ ¥

=

1

2

1

22 2 1000 9 81 3[( ) ][( . )(

P ApII II2 3m N/m m)]

98,100 N

= = ¥ ¥

=

1

2

1

22 2 1000 9 81 5[( ) ][( . )(

Then F P P P= = +

= +

=

0 1 0 1

0 1 58 860 98 100

. . ( )

. ( , , )

I II

N

15696 N

Finally + SF Ty = - - = fi0 15696 5000: N N 0 = 20696 NT or T = 20 7. kN ≠

108


SOLutiOn

Since gate is 1.5 m wide w p g= =( . .1 5 1 5m) (depth)r

Thus: w g h

w gh

w g d

w gd

1

2

1

2

1 5 1

1 5

1 5 1

1 5

= -

=

¢ = ¢ -

¢ = ¢

.

.

. ( )

.

r

r

r

r

( )

¢- = ¢ -

= ¢ - - - =

P P w w

g d g h g

I I 1m)(

m)[ (

1

21

1

21 1 5 1 1 5 1 0 75

1( )

( . ) . ( )] .r r ¢¢ - - -r r( ) . ( )d g h1 0 75 1

¢ - = ¢ -

= ¢ - = ¢ -

P P w w

gd gh g d

II II m)(

m)[ ]

1

21

1

21 1 5 1 5 0 75 0

2 2( )

( . . . .r r r 775g hr

PROBLEM 5.85

A freshwater marsh is drained to the ocean through an automatic tide gate that is 1.5 m wide and 1 m high. The gate is held by hinges located along its top edge at A and bears on a sill at B. If the water level in the marsh is h = 2m, determine the ocean level d for which the gate will open. ( Density of salt water 1025 kg/m .)3

109



+ SM B P P P PA = - ÊËÁ

ˆ¯ ¢- - Ê

ËÁˆ¯ ¢ - =0

1

3

2

30: (1m) m ( m (I I II II) )

B P P P P

g d g h

= ¢- + ¢ -

= ¢ - - - +

1

3

2

3

1

30 75 1 0 75 1

2

3

( ) ( )

[ . ( ) . ( )] [

I I II II

r r 00 75 0 75

0 25

. . ]

.

¢ -

= ¢ - - - + ¢ -

r r

r r r r

gd gh

g d g h gd gh( 1) 0.25 ( 1) 0.5 0.5

B gd g gh g= ¢ - ¢ - +0 75 0 25 0 75. . .r r r r0.25

With B h= =0 2and m: 0 0 25 3 1 0 25 3 1= ¢. ( ) . ( )r rg d g h- - -

( )( )

3 13 1

dh- -=¢

rr

Data: ¢ =

=

r

r

1025

1000

kg/m

kg/m

3

3

3 11000

10255 1 9593d d- = fi =( ) . m d = 1 959. m

110


SOLutiOn

First, determine the force on the dam face without the silt.

We have P Ap A ghw w= =

=

1

2

1

21

26 6 9 81

( )

[( . )( . )(

r

m)(1 m)][(10 kg/m m/s3 3 2 66 6. m)]

213.66 kN=

Next, determine the force on the dam face with silt

We have ¢ =

=

Pw1

24 6 9 81 4 6[( . )( . )( . m)(1 m)][(10 kg/m m/s m)]

103.790 k

3 3 2

NN

m)(1 m)][(10 kg/m m/s m)]

90.252I

3 3 2( ) [( . )( . )( .Ps ==

2 0 9 81 4 6

kN

m)(1 m)][(1.76 10 kg/m m/sII3 3 2( ) [( . )( . )( .Ps = ¥1

22 0 9 81 2 0 m)]

34.531 kN=

Then ¢ = ¢ + + =P P P Pw s s( ) ( ) .I II kN228 57

The percentage increase, % inc., is then given by

% . %

( . . )

.%

. %

inc =¢ -

¥

= - ¥

=

P P

Pw

w

100

228 57 213 66

213 66100

6 9874 % . % inc. = 6 98

PROBLEM 5.86

The dam for a lake is designed to withstand the additional force caused by silt that has settled on the lake bottom. Assuming that silt is equivalent to a liquid of density rs = ¥1.76 10 kg/m3 3 and considering a 1-m-wide section of dam, determine the percentage increase in the force acting on the dam face for a silt accumulation of depth 2 m.

111


PROBLEM 5.87

The base of a dam for a lake is designed to resist up to 120 percent of the horizontal force of the water. After construction, it is found that silt (that is equivalent to a liquid of density rs = ¥1 76 103. ) kg/m3 is settling on the lake bottom at the rate of 12 mm/year. Considering a 1-m-wide section of dam, determine the number of years until the dam becomes unsafe.

SOLutiOn

From Problem 5.86, the force on the dam face before the silt is deposited, is Pw = 213 66. kN. The maximum allowable force Pallow on the dam is then:

P Pwallow kN) kN= = =1 2 1 5 213 66 256 39. ( . )( . .

Next determine the force ¢P on the dam face after a depth d of silt has settled

We have ¢ = - ¥ -

=

P d dw1

26 6 1 10 9 81 6 63[ . ) ( ( )( . )( . )( m m)][ kg/m m/s m]

4.905(

3 2

66.6 kN

m)][ kg/m m/s m]I3 2

-

= -

=

d

P d ds

)

( ) [ ( ( )( . )( . )

.

2

31 10 9 81 6 6

9 81(( . )

( ) [ ( ( . )( . )( )

6 6

1

21 1 76 10 9 81

2

3

d d

P d ds

-

= ¥

kN

m)][ kg/m m/s mII3 2 ]]

kN= 8 6328 2. d

¢ = ¢ + + = - +

+ -

P P P P d d

d d

w s s( ) ( ) [ . ( . . )

. ( .

I II 4 905 43 560 13 2000

9 81 6 6

2

22 2

2

8 6328

3 7278 213 66

) .

[ . ]

+

= +

d

d

]kN

. kN

Now required that ¢ =P Pallow to determine the maximum value of d.

( . . ) .3 7278 213 66 256 392d + =kN kN

or d = 3 3856. m

Finally 3 3856 12 10 3. mm

yearN= ¥ ¥- or N = 282 years

112


PROBLEM 5.88

A 0.5 ¥ 0.8-m gate AB is located at the bottom of a tank filled with water. The gate is hinged along its top edge A and rests on a frictionless stop at B. Determine the reactions at A and B when cable BCD is slack.

SOLutiOn

First consider the force of the water on the gate.

We have P Ap A gh= =1

2

1

2( )r

so that PI3 3 2 m)(0.8 m)] [(10 kg/m m/s m)]

882.

= ¥

=

1

20 5 9 81 0 45[( . )( . )( .

99 N

PII3 3 2 m)(0.8 m)] [(10 kg/m m/s m)]

182

= ¥

=

1

20 5 9 81 0 93[( . )( . )( .

44.66 N

Reactions at A and B when T = 0

We have

+ SM BA = - =0 0 8 0 8 0 8:1

3 m)(882.9 N) +

2

3 m)(1824.66 N) m) 0( . ( . ( .

or B = 1510 74. N

or B = 1511 N 53 1. ∞

+SF A= + - - =0 1510 74 1824 66: N 882.9 N N 0. .

or A = 1197 N 53 1. ∞

113


PROBLEM 5.89

A 0.5 ¥ 0.8-m gate AB is located at the bottom of a tank filled with water. The gate is hinged along its top edge A and rests on a frictionless stop at B. Determine the minimum tension required in cable BCD to open the gate.

SOLutiOn

First consider the force of the water on the gate.


2

1

2( )r

so that PI3 3 2 m)(0.8 m)] [(10 kg/m m/s m)]

882.

= ¥

=

1

20 5 9 81 0 45[( . )( . )( .

99 N

PII3 3 2 m)(0.8 m)] [(10 kg/m m/s m)]

182

= ¥

=

1

20 5 9 81 0 93[( . )( . )( .

44.66 N

T to open gate

First note that when the gate begins to open, the reaction at B Æ 0.

Then + SM A =

- +

0 0 83

0 8

0 45 0 27

:1

3 m)(882.9 N)+

2 m)(1824.66 N)

m

( . ( .

( . . ) ¥¥ ÊËÁ

ˆ¯

=8

170T

or 235 44 973 152 0 33882 0. . .+ - =T

or T = 3570 N

114


PROBLEM 5.90

A long trough is supported by a continuous hinge along its lower edge and by a series of horizontal cables attached to its upper edge. Determine the tension in each of the cables, at a time when the trough is completely full of water.

SOLutiOn

Consider free body consisting of 500 mm length of the trough and water

l = 500 mm length of free body

W gv g r l

P gr

P P rl gr rl gr l

A

A

= = ÈÎÍ

˘˚

=

= = =

r r p

r

r r

4

1

2

1

2

1

2

2

2( )

+ SM Tr Wr P rA = - - Ê

ËÁˆ¯

=01

30:

Tr g r lr

gr l r- ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

=r pp

r4

4

3

1

2

1

302 2

T gr l gr l gr l= + =1

3

1

6

1

22 2 2r r r

Data: r = = = =1000600

120 6 500kg/m mm m mm = 0.5m3 r l.

Then T = ¥ ¥ ¥

=

1

21000 9 81 0 6 0 53 2. ( . ( .N/m m) m)

882.9 N

T = 883 N

115


SOLutiOn

First determine the forces exerted on the gate by the spring and the water when B is at the end of the cylindrical portion of the floor

We have sin.

. .q q= = = ∞0 4

10 4 23 58

Then xSP = ∞( . ) tan( . )0 75 23 58m

and F kxSP SP=

= ¥ ¥ =24 0 75 7 8565

7856 5

kN/m m tan23.58 kN

= N

. .

.

∞

Assume d 1 m

We have P Ap gh

P d

= =

= ¥ ¥ -

=

1

2

1

2

1

21 1000 9 81 1

24

3

( )

[( ] [( . )( ) ]

r

I m)(0.5m) N/m m

552 5 1. ( )d - N

Then

P dII m m N/m= ¥ ¥ - + ∞

=

1

21 0 5 1000 9 81 1 23 58

2452 5

3[( )( . )] [( . )( cos . )]

. (( . )d - 0 0835 N

For dmin so that gate opens, W = 0

Using the above free-body diagrams of the gate, we have

+ Â = ÊËÁ

ˆ¯ ( ) + Ê

ËÁˆ¯ ( )M d dA 0

1

32452 5 1

2

32452 5 0; . ( ) . ( )m m 0.0835- - - ..75m(7856.5) = 0

or ( . . ) ( . ) .817 5 817 5 1635 136 5225 5892 375 0d d- + - - =

or d = 2 7916. m

d 1 m fi assumption correct d = 2 79. m

PROBLEM 5.91

A 1 0 5¥ . m gate is hinged at A and is held in position by rod CD. End D rests against a spring whose constant is 24 kN/m. The spring is undeformed when the gate is vertical. Assuming that the force exerted by rod CD on the gate remains horizontal, determine the minimum depth of water d for which the bottom B of the gate will move to the end of the cylindrical portion of the floor.

116


PROBLEM 5.92

Solve Problem 5.91 if the gate is 500 kg.

Problem 5.91 A 1 0.5m¥ gate is hinged at A and is held in position by rod CD. End D rests against a spring whose constant is 24 kN/m. The spring is undeformed when the gate is vertical. Assuming that the force exerted by rod CD on the gate remains horizontal, determine the minimum depth of water d for which the bottom B of the gate will move to the end of the cylindrical portion of the floor.

SOLutiOn

First determine the forces exerted on the gate by the spring and the water when B is at the end of the cylindrical portion of the floor

We have sin.

. .q q= = = ∞0 4

10 4 23 58

Then xSP = ∞( . ) tan( . )0 75 23 58m

and F kxSP SP=

= ¥ ¥ =24 0 75 7 8565

7856 5

kN/m m tan23.58 kN

= N

. .

.

∞

Assume d 1 m


2

1

2( )r

Then P d

d

I m)(0.5m) N/m m

N

= ¥ ¥ -

= -

1

21 1000 9 81 1

2452 5 1

3[( ] [( . )( ) ]

. ( )

P dII m m N/m= ¥ ¥ - + ∞

=

1

21 0 5 1000 9 81 1 23 58

2452 5

3[( )( . )] [( . )( cos . )]

. (( . )d - 0 0835 N

For dmin so that gate opens, W = ¥500 9 81. N

Using the above free-body diagrams of the gate, we have

+ SM d dA = ÊËÁ

ˆ¯

- + ÊËÁ

ˆ¯

-

-

01

32452 5 1

2

32452 5 0 0835

0

: ( . )( ) ( . )( . )m m

.. ( . ) ( . )( . )75 7856 5 0 2 500 9 81 0m m- ¥ =

or ( . . ) ( . ) .817 5 817 5 1635 136 5225 5892 375 981 0d d- + - - - =

or d = 3 1916. m

d 1 m fi assumption correct d = 3 19. m

117


PROBLEM 5.93

A prismatically shaped gate placed at the end of a freshwater channel is supported by a pin and bracket at A and rests on a frictionless support at B. The pin is located at a distance h 0.10 m= below the center of gravity C of the gate. Determine the depth of water d for which the gate will open.

SOLutiOn

First note that when the gate is about to open (clockwise rotation is impending), By Æ 0 and the line of action of the resultant P of the pressure forces passes through the pin at A. In addition, if it is assumed that the gate is homogeneous, then its center of gravity C coincides with the centroid of the triangular area. Then

ad

h= - -3

0 25( . )

and bd= - Ê

ËÁˆ¯

2

30 4

8

15 3( . )

Now a

b= 8

15

so that d

d

h323

815 3

0 25

0 4

8

15

- -- ( ) =

( . )

( . )

Simplifying yields

289

4515

70 6

12d h+ = .

(1)

Alternative solution

Consider a free body consisting of a 1-m thick section of the gate and the triangular section BDE of water above the gate.

Now ¢ = ¢ = ¥

=

¢ = = ¥ ¥ ¥ÊËÁ

ˆ¯

P Ap d gd

gd

W gV g d d

1

2

1

21

1

21

2

8

151

2

( )( )

( )

m

N

m

r

r

r r ˜

= 4

152rgd ( )N

118



Then with By = 0 (as explained above), we have

+ SM d gdd

hA = - ÊËÁ

ˆ¯

ÈÎÍ

˘˚ÊËÁ

ˆ¯

- - -È02

30 4

1

3

8

15

4

15 30 252: ( . ) ( . )r

ÎÎÍ˘˚ÊËÁ

ˆ¯

=1

202rgd

Simplifying yields 289

4515

70 6

12d h+ = .

as above.

Find d, h = 0.10 m

Substituting into Eq. (1) 289

4515 0 10

70 6

12d + =( . )

. or d = 0 683. m

119


PROBLEM 5.94

A prismatically shaped gate placed at the end of a freshwater channel is supported by a pin and bracket at A and rests on a frictionless support at B. Determine the distance h if the gate is to open when d = 0.75 m.

SOLutiOn

First note that when the gate is about to open (clockwise rotation is impending), By Æ 0 and the line of action of the resultant P of the pressure forces passes through the pin at A. In addition, if it is assumed that the gate is homogeneous, then its center of gravity C coincides with the centroid of the triangular area. Then

ad

h= - -3

0 25( . )

and bd= - Ê

ËÁˆ¯

2

30 4

8

15 3( . )

Now a

b= 8

15

so that d

d

h323

815 3

0 25

0 4

8

15

- -- ( ) =

( . )

( . )


4515

70 6

12d h+ = .

(1)

Alternative solution

Consider a free body consisting of a 1-m thick section of the gate and the triangular section BDE of water above the gate.

Now

¢ = ¢ = ¥

=

¢ = = ¥ ¥ ¥ÊËÁ

ˆ¯

P Ap d gd

gd

W gV g d d

1

2

1

21

1

21

2

8

151

2

( )( )

( )

m

N

m

r

r

r r ˜

= 4

152rgd ( )N

120



Then with By = 0 (as explained above), we have

+ SM d gdd

hA = - ÊËÁ

ˆ¯

ÈÎÍ

˘˚ÊËÁ

ˆ¯

- - -È02

30 4

1

3

8

15

4

15 30 252: ( . ) ( . )r

ÎÎÍ˘˚ÊËÁ

ˆ¯

=1

202rgd


4515

70 6

12d h+ = .

as above.

Find h, d = 0 75. m

Substituting into Eq. (1) 289

450 75 15

70 6

12( . )

.+ =h or h = 0 0711. m

121


PROBLEM 5.95

A 600 mm diameter drum of height 750 mm is placed on its side to act as a dam in a 750 mm-wide freshwater channel. Knowing that the drum is anchored to the sides of the channel, determine the resultant of the pressure forces acting on the drum.

SOLutiOn

Consider the elements of water shown. The resultant of the weights of water above each section of the drum and the resultants of the pressure forces acting on the vertical surfaces of the elements is equal to the resultant hydrostatic force acting on the drum. Then

P Ap A ghI I

m 0.75m N/m

= =

= ¥[ ]¥ ¥ ¥ÈÎ ˘

=

1

2

1

2

1

20 6 1000 9 81 0 6

132

2

( )

. . .

r

44 35. N

P Ap A ghII II

3m 0.75m N/m m

= =

= ¥[ ]¥ ¥ ¥ÈÎ ˘

=

1

2

1

2

1

20 3 1000 9 81 0 3

( )

. . .

r

3331 0875. N

W gV1 13 2 2 2 21000 9 81 0 3

40 3 0 75 14= = ¥ È

ÎÍ˘˚

=r p( . ) ( . ) ( . ) ( .N/m m m m)- 22 104

1000 9 81 0 34

0 3 0 752 22 2 2 2

.

( . ) ( . ) ( . ) ( .

N

N m m mW gV= = ¥ +ÈÎÍ

˘˚

r p)) N

N m m)

=

= = ¥ ÈÎÍ

˘˚

=

1182 246

1000 9 814

0 3 0 75 523 32 2

.

( . ) ( . ) ( .W gVr p00 071. N

Then + SF Rx x: ( . . ) .= - =1324 35 331 0875 993 26N N

+ SF Ry y: ( . . . ) .= - + + =142 104 1182 246 520 071 1560 21N N

Finally R R R

R

R

x y

y

x

= + =

=

2 2 1849 546.

tan

N

q

q = ∞57 5. R = 1850 N 57.5°

122


PROBLEM 5.96

Determine the location of the centroid of the composite body shown when (a) h b= 2 , (b) h b= 2.5 .

SOLutiOn

V x xV

Cylinder I pa b2 1

2b

1

22 2pa b

Cone II1

32pa h b h+ 1

41

3

1

42pa h b h+Ê

ËÁˆ¯

V a b h

xV a b hb h

= +ÊËÁ

ˆ¯

= + +ÊËÁ

ˆ¯

p

p

2

2 2 2

1

3

1

2

1

3

1

12S

(a) For h b= 2 : V a b b a b= +ÈÎÍ

˘˚

=p p2 21

32

5

3( )

SxV a b b b b

a b

= + +ÈÎÍ

˘˚

= + +ÈÎÍ

˘˚

=

p

p p

2 2 2

2 2

1

2

1

32

1

122

1

2

2

3

1

3

3

2

( ) ( )

aa b2 2

XV xV X a b a b X b= ÊËÁ

ˆ¯

= =S :5

3

3

2

9

102 2 2p p

Centroid is 110 b to left of base of cone

123



(b) For h b= 2 5. : V a b b a b= +ÈÎÍ

˘˚

=p p2 21

32 5 1 8333( . ) .

SxV a b b b b

a b

= + +ÈÎÍ

˘˚

= + +

p

p

2 2 2

2 2

1

2

1

32 5

1

122 5

0 5 0 8333 0

( . ) ( . )

[ . . .552083

1 85416 2 2

]

.= pa b

XV xV X a b a b X b= = =S : ( . ) . .1 8333 1 85416 1 011362 2 2p p

Centroid is 0.01136b to right of base of cone

Note: Centroid is at base of cone for h b b= =6 2 449.

124


SOLutiOn

First note that the values of Y will be the same for the given body and the body shown below. Then

V y yV

Cone1

32pa h - 1

4h - 1

122 2pa h

Cylinder - ÊËÁ

ˆ¯

= -p pab a b

2

1

4

22 - 1

2b

1

82 2pa b

S p12

4 32a h b( )- - -p24

2 32 2 2a h b( )

We have Y V yVS S=

Then Y a h b a h bp p12

4 324

2 32 2 2 2( ) ( )-ÈÎÍ

˘˚

= - - or Yh b

h b= - -

-2 3

2 4 3

2 2

( )

PROBLEM 5.97

Determine the y coordinate of the centroid of the body shown.

125


PROBLEM 5.98

Determine the z coordinate of the centroid of the body shown. (Hint: Use the result of Sample Problem 5.13.)

SOLutiOn

First note that the body can be formed by removing a “half-cylinder” from a “half-cone,” as shown.

V z zV

Half-Cone1

62pa h - a

p* - 1

63a h

Half-Cylinder - ÊËÁ

ˆ¯

= -p p2 2 8

22a

b a b - ÊËÁ

ˆ¯

= -4

3 2

2

3p pa a 1

123a b

Sp24

4 32a h b( )- - -1

1223a h b( )

From Sample Problem 5.13

We have Z V zVS S=

Then Z a h b a h bp24

4 31

1222 3( ) ( )-È

ÎÍ˘˚

= - - or Za h b

h b= - -

-ÊËÁ

ˆ¯p

4 2

4 3

126


PROBLEM 5.99

The composite body shown is formed by removing a semiellipsoid of revolution of semimajor axis h and semiminor axis a/2 from a hemisphere of radius a. Determine (a) the y coordinate of the centroid when h = a/2, (b) the ratio h/a for which y = −0.4a.

SOLutiOn

V y yV

Hemisphere2

33pa - 3

8a - 1

44pa

Semiellipsoid - ÊËÁ

ˆ¯

= -2

3 2

1

6

22p pa

h a h - 3

8h + 1

162 2pa h

Then S

S

V a a h

yV a a h

= -

= - -

p

p6

4

164

2

2 2 2

( )

( )

Now Y V yVS S=

So that Y a a h a a hp p6

416

42 2 2 2( ) ( )-ÈÎÍ

˘˚

= - -

or Yh

aa

h

a4

3

84

2

-ÊËÁ

ˆ¯

= - - ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

(1)

(a) Y ha= =? when2

Substituting h

a= 1

2 into Eq. (1)

Y a41

2

3

84

1

2

2

-ÊËÁ

ˆ¯

= - - ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

or Y a= - 45

112 Y a= -0 402.

127



(b) h

aY a= = -? .when 0 4

Substituting into Eq. (1)

( . )- -ÊËÁ

ˆ¯

= - - ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

0 4 43

84

2

ah

aa

h

a

or 3 3 2 0 8 02h

a

h

aÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

+ =. .

Then h

a=

± - -3 2 3 2 4 3 0 8

2 3

2. ( . ) ( )( . )

( )

= ±3 2 0 8

6

. . or

h

a

h

a= =2

5

2

3and

128


SOLutiOn

Assume that the bracket is homogeneous so that its center of gravity coincides with the centroid of the volume.

V , mm3 x , mm xV , mm4

1 ( )( )( )100 88 12 105600= 50 5280000

2 ( )( )( )100 12 88 105600= 50 5280000

31

262 51 10 15810( )( )( ) = 39 616590

4 - = -1

266 45 12 17820( )( )( ) 34

2

366 78+ =( ) -1389960

S 209190 9786600

Then XV

V= =

SSx 9786600

209190mm or mmX = 46 8.

PROBLEM 5.100

For the stop bracket shown, locate the x coordinate of the center of gravity.

129


PROBLEM 5.101

For the stop bracket shown, locate the z coordinate of the center of gravity.

SOLutiOn

Assume that the bracket is homogeneous so that its center of gravity coincides with the centroid of the volume.

V , mm3 z , mm zV , mm4

1 ( )( )( )100 88 12 105600= 6 633600

2 ( )( )( )100 12 88 105600= 121

288 56+ =( ) 5913600

31

262 51 10 15810( )( )( ) = 12

1

351 29+ =( ) 458490

4 - = -1

266 45 12 17820( )( )( ) 55

2

345 85+ =( ) -1514700

S 209190 5491000

Then ZzV

V= =

SS

5491000

209190mm or mmZ = 26 2.

130


PROBLEM 5.102

For the machine element shown, locate the y coordinate of the center of gravity.

SOLutiOn

First assume that the machine element is homogeneous so that its center of gravity will coincide with the centroid of the corresponding volume.

V , mm3 x , mm y, mm xV, mm4 yV, mm4

I ( )( )( )100 18 90 162000= 50 9 8100000 1458000

II ( )( )( )16 60 50 48000= 92 48 4416000 2304000

III p ( ) ( ) .12 10 4523 92 = 105 54 475010 244290

IV - = -p ( ) ( ) .13 18 9556 72 28 9 –267590 –86010

S 204967.2 12723420 3920280

We have Y V yVS S=

Y ( . )204967 2 39202803 4mm mm= or Y = 19 13. mm

131


PROBLEM 5.103

For the machine element shown, locate the y coordinate of the center of gravity.

SOLutiOn

For half cylindrical hole: (III)r

y

=

= -

=

25

404 25

329 3897

mm

mm

III( )

.p

For half cylindrical plate: (IV) r

z

=

= + =

40

1404 40

3156 977

mm

mmIV( )

.p

V, mm3 y, mm z, mm yV, mm4 zV , mm4

IRectangular

plate (140)(80)(15) = 168 ×103 -7.5 70 -1.26 × 106 11.76 × 106

II Rectangular plate (80)(40)(20) = 64 ×103 20 40 1.28 × 106 2.56 × 106

III - (Half cylinder) - -p2

19 635 102 3(25) (20) = . ¥ 29.3897 40 -0.577 × 106 -0.7854 × 106

IV (Half cylinder)p2

102 3(40) (15) = 37.699 ¥ -7.5 156.977 -0.2827 × 106 5.9179 × 106

V -(Cylinder) - -p (25) (15) = 29.452 102 3¥ -7.5 140 0.2209 ×106 -4.1233 × 106

S 220.612 ×103 -0.6188 ×106 15.3292 × 106

Y V yV

Y

Y

S S=

¥ = - ¥= -

( . ) .

.

220 612 10 0 6188 10

2 8049

3 3 6 4mm mm

mm Y = -2 80. mm

132


PROBLEM 5.104

For the machine element shown, locate the z coordinate of the center of gravity.

SOLutiOn

For half cylindrical hole: (III)r

y

=

= -

=

25

404 25

329 3897

mm

mm

III( )

.p

For half cylindrical plate: (IV) r

z

=

= + =

40

1404 40

3156 977

mm

mmIV( )

.p

V, mm3 y, mm z, mm yV, mm4 zV , mm4

I Rectangular plate

(140)(80)(15) = 168 × 103 -7.5 70 -1.26 × 106 11.76 × 106

II Rectangular plate

(80)(40)(20) = 64 × 103 20 40 1.28 × 106 2.56 × 106

III - (Half cylinder) - -p2

19 635 102 3(25) (20) = . ¥ 29.3897 40 - 0.577 ×106 - 0.7854 ×106

IV (Half cylinder)p2

102 3(40) (15) = 37.699 ¥ -7.5 156.977 - 0.2827 ×106 5.9179 ×106

V - (Cylinder) - -p (25) (15) = 29.452 102 3¥ -7.5 140 0.2209 ×106 - 4.1233 ×106

S 220.612 × 103 - 0.6188 ×106 15.3292 ×106

Now Z V zVS =

Z

Z

( . ) .

.

220 612 10 15 3292 10

69 4849

3 3 6 4¥ = ¥fi =

mm mm

mm Z = 69 5. mm

133


PROBLEM 5.105

For the machine element shown, locate the x coordinate of the center of gravity.

SOLutiOn

First assume that the machine element is homogeneous so that its center of gravity will coincide with the centroid of the corresponding volume.

V , mm3 x , mm y, mm xV, mm4 yV, mm4

I ( )( )( )100 18 90 162000= 50 9 8100000 1458000

II ( )( )( )16 60 50 48000= 92 48 4416000 2304000

III p ( ) ( ) .12 10 4523 92 = 105 54 475010 244290

IV - = -p ( ) ( ) .13 18 9556 72 28 9 –267590 –86010

S 204967.2 12723420 3920280

We have X V xVS S=

X ( . )204967 2 12723420 4mm mm3 = X = 62 1. mm

134


PROBLEM 5.106

Locate the center of gravity of the sheet-metal form shown.

SOLutiOn

First, assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with the centroid of the corresponding area.

y

z

x

I

I

III

m

m

m

= =

= =

= =

- -

- -

1

31 2 0 4

1

33 6 1 2

4 1 8

3

2 4

( . ) .

( . ) .

( . ) .

p p

A, m2 x , m y, m z , m xA, m3 yA, m3 zA, m3

I1

23 6 1 2 2 16( . )( . ) .= 1.5 - 0.4 1.2 3.24 - 0 864. 2.592

II ( . )( . ) .3 6 1 7 6 12= 0.75 0.4 1.8 4.59 2.448 11.016

IIIp2

1 8 5 08942( . ) .= - 2 4.

p0.8 1.8 -3 888. 4.0715 9.1609

S 13.3694 3.942 5.6555 22.769

We have X V xV XS S= =: m m( . ) .13 3694 3 9422 3 or X = 0 295. m

Y V yV YS S= =: m m( . ) .13 3694 5 65552 3 or Y = 0 423. m

Z V zV ZS S= =: m m( . ) .13 3694 22 7692 3 or Z = 1 703. m

135


PROBLEM 5.107


SOLutiOn

First assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with the centroid of the corresponding area. Now note that symmetry implies

X = 125 mm

y

z

y

II

II

III

mm

mm

= + ¥

=

= ¥

=

= + ¥

=

1502 80

200 93

2 80

50 930

2304 125

32

p

p

p

.

.

883 05. mm

A, mm2 y, mm z , mm yA, mm3 zA, mm3

I ( )( )250 170 42500= 75 40 3187500 1700000

IIp2

80 250 31416( )( ) = 200.93 50930 6312400 1600000

IIIp2

125 245442( ) = 283.05 0 6947200 0

S 98460 16447100 3300000

We have Y A y A YS S= =: ( )98460 164471002 3mm mm or Y = 1670 mm

Z A z A ZS S= = ¥: mm mm( ) .98460 3 300 102 6 3 or Z = 33 5. mm

136


PROBLEM 5.108

A wastebasket, designed to fit in the corner of a room, is 400 mm high and has a base in the shape of a quarter circle of radius 250 mm Locate the center of gravity of the wastebasket, knowing that it is made of sheet metal of uniform thickness.

SOLutiOn

By symmetry: X Z=

For III (Cylindrical surface) xr

A rh

= = =

= = = ¥

2 2 250159 155

2 2250 400 157 080 103 2

p pp p

( ).

( )( ) .

mm

mm

For IV (Quarter-circle bottom) xr

A r

= = =

= = ¥

4

3

4 250

3106 103

4 4250 49 0874 102 2 3 2

p pp p

( ).

( ) .

mm

mm


I (250)(400) = 100 × 103 125 200 12.5 × 106 20 × 106

II (250)(400) = 100 × 103 0 200 0 20 × 106

III 157.680 × 103 159.155 200 25 × 106 31.416 × 106

IV 49.0874 × 103 106.103 0 5.2083 × 106 0

S 406.1674 × 103 42.7083 × 106 71.416 × 106

X A x AS S= : X ( . ) .406 1674 10 42 7083 103 2 6 3¥ = ¥mm mm

X = 105 149. mm X Z= = 1051 mm

Y A y AS S= : Y ( . ) .406 1674 10 71 416 103 2 6 3¥ = ¥mm mm

Y = 175 829. mm Y = 175 8. mm

137


PROBLEM 5.109

A mounting bracket for electronic components is formed from sheet metal of uniform thickness. Locate the center of gravity of the bracket.

SOLutiOn

First, assume that the sheet metal is homogeneous so that the center of gravity of the bracket coincides with the centroid of the corresponding area. Then (see diagram)

z

A

V

V

2

mm

mm

= -

=

= -

= -

454 12 5

339 6948

212 5

245 437

2

( . )

.

( . )

.

p

p

A, mm2 x , mm y, mm z , mm xA, mm3 yA, mm3 zA, mm3

I (50)(120) = 6000 25 0 60 150 × 103 0 360 × 103

II (25)(120) = 3000 50 –12.5 60 150 × 103 – 37.5 × 103 180 × 103

III (15)(120) = 1800 57.5 – 25 60 103.5 × 103 – 45 × 103 108 × 103

IV – (25)(60) = –1500 20 0 75 –30 × 103 0 –112.5 × 103

V – 245.437 20 0 39.6948 – 4.909 × 103 0 – 9.7426 × 103

S 9054.563 368.591 × 103 – 82.5 × 103 525.7574 × 103

We have X A x AS S=

X( . ) .9054 563 368 591 102 3 3mm mm= ¥ or X = 40 7. mm

Y A y A

Y

S S=

= - ¥( . ) .9054 563 82 5 102 3mm mm3 or mmY = -9 11.

Z A z A

Z

S S=

= ¥( . ) .9054 563 525 7574 102 3mm mm3 or Z = 58 1. mm

138


PROBLEM 5.110

A thin sheet of plastic of uniform thickness is bent to form a desk organizer. Locate the center of gravity of the organizer.

SOLutiOn

First assume that the plastic is homogeneous so that the center of gravity of the organizer will coincide with the centroid of the corresponding area. Now note that symmetry implies

Z = 30 0. mm

x

x

x

x

2

4

8

1

62 6

2 1803

362 6

39 820

582 6

54 180

= ¥ =

= + ¥ =

= ¥ =

-

-

p

p

p

.

.

.

mm

mm

mm

00

2 4 8 10

6

1332 6

136 820

62 6

2 1803

752 5

= + ¥ =

= = = = ¥ =

= + ¥

p

p

.

.

mm

mmy y y y

y

-

ppp

p

=

= = = = ¥ ¥ =

= ¥ ¥ =

78 183

26 60 565 49

5 60 942 48

2 4 8 102

6

.

.

.

mm

mm

m

A A A A

A mm2

139




1 ( )( )74 60 4440= 0 43 0 190920

2 565.49 2.1803 2.1803 1233 1233

3 ( )( )30 60 1800= 21 0 37800 0

4 565.49 39.820 2.1803 22518 1233

5 ( )( )69 60 4140= 42 40.5 173880 167670

6 942.48 47 78.183 44297 73686

7 ( )( )69 60 4140= 52 40.5 215280 167670

8 565.49 54.180 2.1803 30638 1233

9 ( )( )75 60 4500= 95.5 0 429750 0

10 565.49 136.820 2.1803 77370 1233

S 22224.44 1032766 604878

We have X A x A XS S= =: ( . )22224 44 10327662 3mm mm or X = 46 5. mm

Y A y A YS S= =: mm mm( . )22224 44 6048782 3 or Y = 27 2. mm

140


PROBLEM 5.111

A window awning is fabricated from sheet metal of uniform thickness. Locate the center of gravity of the awning.

SOLutiOn

First, assume that the sheet metal is homogeneous so that the center of gravity of the awning coincides with the centroid of the corresponding area.

y y

z z

II VI

II VI

mm= = + =

= = =

1004 625

3365 258

4 625

3265 258

( )( ).

( )( ).

p

pmmm

mm

mm

IV

IV

II

y

z

A A

= + =

= =

=

1002 625

497 887

2 625397 887

( )( ).

( )( ).

p

p

VVI3 2

IV2

5 06.796 10 mm

25 mm

= = ¥

= = ¥

p

p4

62 3

62 850 834 486 10

2

3

( )

( )( ) .A

A, mm2 y, mm z , mm yA, mm3 zA, mm3

I (100)(625) = 62.5 × 103 50 312.5 3.125 × 106 19.53125 × 106

II 306.796 × 103 365.258 265.258 112.060 × 106 81.380 × 106

III (100)(850) = 85 × 103 50 625 4.25 × 106 53.125 × 106

IV 834.486 × 103 497.887 397.887 415.480 × 106 332.031 × 106

V (100)(625) = 62.5 × 103 50 312.5 3.125 × 106 19.53125 × 106

VI 306.796 × 103 365.258 265.258 112.060 × 106 81.380 × 106

S 1658.078 × 103 650.1 × 106 586.9785 × 106

141



Now, symmetry implies X = 425 mm

and Y A y A YS S= ¥ = ¥: mm mm2 3( . ) .1658 078 10 650 1 103 6 or Y = 392 mm

Z A z A ZS S= ¥ = ¥: mm mm( . ) .1658 078 10 586 9785 103 2 6 3 or Z = 354 mm

142


PROBLEM 5.112

An elbow for the duct of a ventilating system is made of sheet metal of uniform thickness. Locate the center of gravity of the elbow.

SOLutiOn

First, assume that the sheet metal is homogeneous so that the center of gravity of the duct coincides with the centroid of the corresponding area. Also, note that the shape of the duct implies Y = 38 0. mm

Note that mmI I

II

x z

x

= = =

= =

4002

400 145 352

4002

200 272 68

-

-

p

p

( ) .

( ) . mmm

mm

mm

II

IV IV

z

x z

x

= =

= = =

3002

200 172 676

4004

3400 230 23

-

-

p

p

( ) .

( ) .

VV

V

mm

mm

= =

= =

4004

3200 315 12

3004

3200 215 12

-

-

p

p

( ) .

( ) .z

Also note that the corresponding top and bottom areas will contribute equally when determining x zand .

Thus A, mm2 x , mm z , mm xA, mm3 zA, mm3

Ip2

400 76 47752( )( ) = 145.352 145.352 6940850 6940850

IIp2

200 76 23876( )( ) = 272.68 172.676 6510510 4122810

III 100 76 7600( ) = 200 350 1520000 2660000

IV 2p4

400 2513272ÊËÁ

ˆ¯

=( ) 230.23 230.23 57863020 57863020

V - -24

200 628322pÊËÁ

ˆ¯

=( ) 315.12 215.12 –19799620 –13516420

VI - -2 100 200 40000( )( ) = 300 350 –12000000 –14000000

S 227723 41034760 44070260

143



We have X A x A XS S= =: mm mm( )227723 410347602 3 or X = 180 2. mm

Z A z A ZS S= =: mm mm( )227723 440702602 3 or Z = 193 5. mm

144


PROBLEM 5.113

A 200-mm-diameter cylindrical duct and a 100 × 200 mm rectangular duct are to be joined as indicated. Knowing that the ducts were fabricated from the same sheet metal, which is of uniform thickness, locate the center of gravity of the assembly.

SOLutiOn

Assume that the body is homogeneous so that its center of gravity coincides with the centroid of the area.


1 p(200)(300) = 188.496 × 103 0 150 0 28.2744 × 106

2 - = - ¥p2

200 100 31 4159 103( )( ) . 2 100 200( )

p p= 250 –2 × 106 –7.8540 × 106

3 p2

100 15 708 102 3( ) .= ¥ - = -4 100

3

400

3

( )

p p300 –0.66667 × 106 4.7124 × 106

4 (200)(300) = 60 × 103 150 300 9 × 106 18 × 106

5 (200)(300) = 60 × 103 150 200 9 × 106 12× 106

6 - = - ¥p2

100 15 708 102 3( ) .4 100

3

400

3

( )

p p= 200 –0.66667× 106 –3.1416 × 106

7 (100)(300) = 30 × 103 150 250 4.5 × 106 7.5 × 106

8 (100)(300) = 30 × 103 150 250 4.5 × 106 7.5 × 106

S 337.0801× 103 23.6667 × 106 66.9912 × 106

Then Xx A

A= = ¥

¥SS

23 6667 10

337 0801 10

6

3

.

.mm or X = 70 2. mm

Yy A

A= = ¥

¥SS

66 9912 10

337 0801 10

6

3

.

.mm or Y = 198 7. mm

145


PROBLEM 5.114

A thin steel wire of uniform cross section is bent into the shape shown. Locate its center of gravity.

SOLutiOn

First assume that the wire is homogeneous so that its center of gravity will coincide with the centroid of the corresponding line.

x

z

x

1

1

26

0 3 60 0 15 3

0 3 60 0 15

0 6 30

= ∞ == ∞ =

= ∞Ê

ËÁˆ

¯

. sin .

. cos .

. sins

m

m

p iin.

. sincos

.

(

300 9

0 6 3030

0 93

30

26

2

∞ =

= ∞Ê

ËÁˆ

¯∞ =

= ÊËÁ

ˆ¯

p

p

p

p

m

mz

L .. ) ( . )6 0 2= p m

L, m x , m y, m z , m xL, m2 yL, m2 zL, m2

1 1.0 0 15 3. 0.4 0.15 0.25981 0.4 0.15

2 0 2. p 0 9.

p0 0 9 3.

p0.18 0 0.31177

3 0.8 0 0.4 0.6 0 0.32 0.48

4 0.6 0 0.8 0.3 0 0.48 0.18

S 3.0283 0.43981 1.20 1.12177

We have X L x L XS S= =: m m( . ) .3 0283 0 43981 2 or X = 0 1452. m

Y L yL YS S= =: m m( . ) .3 0283 1 20 2 or Y = 0 396. m

Z L z L ZS S= =: m m( . ) .3 0283 1 12177 2 or Z = 0 370. m

146


PROBLEM 5.115

Locate the center of gravity of the figure shown, knowing that it is made of thin brass rods of uniform diameter.

SOLutiOn

Uniform rod

AB2 2 2 21 0 6 1 5= + +( ) ( . ) ( . )m m m

AB = 1 9. m

L, m x , m y, m z , m xL, m2 yL, m2 SL, m2

AB 1.9 0.5 0.75 0.3 0.95 1.425 0.57

BD 0.6 1.0 0 0.3 0.60 0 0.18

DO 1.0 0.5 0 0 0.50 0 0

OA 1.5 0 0.75 0 0 1.125 0

S 5.0 2.05 2.550 0.75

X L x L XS S= =: m m( . ) .5 0 2 05 2 X = 0 410. m

Y L y L YS S= =: m m( . ) .5 0 2 55 2 Y = 0 510. m

Z L z L ZS S= =: m m( . ) .5 0 0 75 2 Z = 0 1500. m

147


PROBLEM 5.116

Locate the center of gravity of the figure shown, knowing that it is made of thin brass rods of uniform diameter.

SOLutiOn

By symmetry: X = 0

L, mm y, mm z , mm yL, mm2 zL, mm2

AB ( ) ( )750 400 8502 2+ = 375 0 318.75 × 103 0

AD ( ) ( )750 400 8502 2+ = 375 200 318.75 × 103 170 × 103

AE ( ) ( )750 400 8502 2+ = 375 0 318.75 × 103 0

BDE p ( ) .400 1256 637= mm 02 400 800( )

p p= 0 320 × 103

S 3806.637 956.25 × 103 490 × 103

Y L y L YS S= = ¥: ( . ) .3806 637 956 25 103 2mm mm

Y = 251 21. mm Y = 251 mm

Z L z L ZS S= = ¥: mm mm( . )3806 637 490 103 2

Z = 128 723. mm Z = 128 7. mm

148


PROBLEM 5.117

The frame of a greenhouse is constructed from uniform aluminum channels. Locate the center of gravity of the portion of the frame shown.

SOLutiOn

First assume that the channels are homogeneous so that the center of gravity of the frame will coincide with the centroid of the corresponding line.

x x

y y

8 9

8 9

2 0 9 1 8

1 52 0 9

2 07296

= = ¥ =

= = + ¥ =

. .

..

.

p p

p

m

m

L, m x , m y, m z , m xL, m2 yL, m2 zL, m2

1 0.6 0.9 0 0.3 0.54 0 0.18

2 0.9 0.45 0 0.6 0.405 0 0.54

3 1.5 0.9 0.75 0 1.35 1.125 0

4 1.5 0.9 0.75 0.6 1.35 1.125 0.9

5 2.4 0 1.2 0.6 0 2.88 1.44

6 0.6 0.9 1.5 0.3 0.54 0.9 0.18

7 0.9 0.45 1.5 0.6 0.405 1.35 0.54

8p2

0 9 1 4137¥ =. .1 8.

p2.07296 0 0.81 2.9305 0

9p2

0 9 1 4137¥ =. .1 8.

p2.07296 0.6 0.81 2.9305 0.84822

10 0.6 0 2.4 0.3 0 1.44 0.18

S 11.8274 6.21 14.681 4.80822

We have X L x L XS S= =: ( . ) .11 8274 6 21 2m m or X = 0 525. m

Y L y L YS S= =: ( . ) .11 8274 14 681m m2 or Y = 1 241. m

Z L z L ZS S= =: ( . ) .11 8274 4 80822m m2 or Z = 0 407. m

149


PROBLEM 5.118

A scratch awl has a plastic handle and a steel blade and shank. Knowing that the density of plastic is 1030 kg/m3 and of steel is 7860 kg/m ,3 locate the center of gravity of the awl.

SOLutiOn

First, note that symmetry implies Y Z= = 0

x

W

I

I3

mm mm

kg/m m

= =

= ÊËÁ

ˆ¯

5

812 5 7 8125

10302

30 0125 3

( . ) .

( ) ( . )p

== ¥=

= ÊËÁ

ˆ¯

4 2133 10

52 5

10304

0 025

3

2

.

.

( ) ( . ) (

- kg

mm

kg/m m

II

II3

x

Wp

00 08

40 448 10

92 5 25 67 5

1030

3

. )

.

. .

(

m

kg

mm mm mmIII

III

= ¥= =

=

-

-

-

x

W kkg/m m m

kg

3) ( . ) (. )

.

p4

0 0035 05

0 49549 10

2

3

ÊËÁ

ˆ¯

= ¥- -

x

W

IV

IV3

mm mm mm

kg/m m

= =

= ÊËÁ

ˆ¯

182 5 70 112 5

78604

0 0035

. .

( ) ( .

-p

)) ( . ) .

. ( )

(

2 2 30 14 10 5871 10

182 51

410 185

m kg

mm mm mmV

V

= ¥

= + =

=

-

x

W 778603

0 00175 0 01 0 25207 102 3 kg/m m m kg3) ( . ) ( . ) .pÊ

ËÁˆ¯

= ¥ -

150



W, kg x, mm xW, kg mm◊

I 4 123 10 3. ¥ - 7.8125 32 916 10 3. ¥ -

II 40 948 10 3. ¥ - 52 5. 2123 5 10 3. ¥ -

III - -0 49549 10 3. ¥ 67.5 - -33 447 10 3. ¥

IV 10 5871 10 3. ¥ - 112.5 1191 05 10 3. ¥ -

V 0 25207 10 3. ¥ - 185 46 633 10 3. ¥ -

S 55 005 10 3. ¥ - 3360 7 10 3. ¥ -

We have X W xW XS S ◊= ¥ = ¥: ( . ) .55 005 10 3360 7 103 3- -kg kg mm

or X = 61 1. mm

(From the end of the handle)

151


PROBLEM 5.119

A bronze bushing is mounted inside a steel sleeve. Knowing that the r

br= 8800 kg/m3 and of steel is r

steel = 7860 kg/m3, determine the location of the

center of gravity of the assembly.

SOLutiOn

First, note that symmetry implies X Z= = 0

Now W g V= ( )r

y WI I2mm N/m mm mm 10 m= = ¥ Ê

ËÁˆ¯

-( )ÈÎ

˘˚ ¥ -4 7860 9 81

436 15 83 2 2 9 3( . )

p//mm N

mm N/mII II

3

3

0 518873

18 7860 9 814

( )ÏÌÓ

¸˝˛

=

= = ¥ ÊËÁ

ˆ¯

.

( . )y Wp

222 5 15 20 0 340647

1

2 2 2 9 3 3. .-( )ÈÎ

˘˚ ¥( )Ï

ÌÓ

¸˝˛

=

=

-mm mm 10 m /mm N

IIIy 44 8800 9 814

15 10 283 2 2 2 9mm N/m mm mm 10 mIIIW = ¥ ÊËÁ

ˆ¯

-( )ÈÎ

˘˚ ¥ -( . )

p 33 3 0 237306/mm N( )ÏÌÓ

¸˝˛

= .

We have Y W yW

Y

S S=

=+ +( )( . ) ( )( . ) ( )( .4 0 518873 18 0 340647 14 0 23730mm N mm N mm 66

0 518873 0 340647 0 23730610 512

N

Nmm

)

( . . . ).

+ +=

or Y = 10 51. mm

(above base)

152


PROBLEM 5.120

A brass collar, of length 62.5 mm, is mounted on an aluminum rod of length 100 mm. Locate the center of gravity of the composite body.(Densities: brass = 8470 kg/m3, aluminum = 2800 kg/m3.)

SOLutiOn

Aluminum rod: W gV=

= ¥ ¥ÈÎÍ

˘˚

¥

=

-

rp

Al

3 3 3N/m mm) mm) m /mm2800 9 814

40 100 10

3

2 9. ( (

.44517 N

Brass collar: W gV=

= ¥ ÈÎ ˘ ¥ ¥ -

rp

brass

N/m mm) mm) mm m8470 9 814

75 40 62 5 103 2 2 9. ( ( .- 33 3

16 4168

/mm

N= .

Component W ( )N y(mm) yW (N mm)◊Rod 3.4517 50 172.585

Collar 16.4168 31.25 513.025

S 19.8685 685.61

Y W yW YS S ◊= : ( .19 8685N) = 685.61N mm

Y = 34 507. mm Y = 34 5. mm

153


PROBLEM 5.121

The three legs of a small glass-topped table are equally spaced and are made of steel tubing, which has an outside diameter of 24 mm and a cross-sectional area of 150 mm .2 The diameter and the thickness of the table top are 600 mm and 10 mm, respectively. Knowing that the density of steel is 7860 kg/m3 and of glass is 2190 kg/m ,3 locate the center of gravity of the table.

SOLutiOn

First note that symmetry implies X Z= = 0

Also, to account for the three legs, the masses of components I and II will each bex multiplied by three

yI

mm

= + ¥

=

12 1802 180

77 408

-p

.

m VSTI I3kg/m m m

kg

= = ¥ ¥ ¥

=

r p7860 150 10

20 180

0 33335

6 2( ) ( . )

.

-

yII

mm

= + + ¥

=

12 1802 280

370 25p

.

m VSTII II2kg/m m m

kg

= = ¥ ¥ ¥

=

r p7860 150 10

20 280

0 51855

3 6( ) ( . )

.

-

yIII

mm

= + + +=

24 180 280 5

489

m VGLIII III kg/m m m

kg

= = ¥ ¥

=

r p2190

40 6 0 010

6 1921

3 2( . ) ( . )

.

m, kg y, mm ym, kg mm◊I 3(0.33335) 77.408 77.412

II 3(0.51855) 370.25 515.98

III 6.1921 489 3027.9

S 8.7478 3681.3

We have Y m ym YS S ◊= =: ( . ) .8 7478 3681 3kg kg mm

or Y = 420 8. mm

The center of gravity is 421 mm

(above the floor)

154


PROBLEM 5.122

Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height.

A hemisphere

SOLutiOn

Choose as the element of volume a disk of radius r and thickness dx. Then

dV r dx x xEL= =p 2 ,

The equation of the generating curve is x y a2 2 2+ = so that r a x2 2 2= - and then

dV a x dx= p ( )2 2-

Component 1 V a x dx a xx

a

aa

12 2 2

3

0

2

0

2

3

3

11

24

= =È

ÎÍ

˘

˚˙

=

Ú p p

p

( )/

- -/

and x dV x a x dx

ax x

a

EL

a

a

= ÈÎ ˘

=È

ÎÍ

˘

˚˙

=

ÚÚ p

p

p

( )/

/

2 2

0

2

1

22 4

0

2

4

2 4

7

64

-

-

Now x V x dV x a aEL1 1 1 13 411

24

7

64= Ê

ËÁˆ¯

=Ú : p p or x a121

88=

Component 2

V a x dx a xx

a aa

aa

a

a

a

a

2 2

2 2 23

2

23

2

3

3

= =È

ÎÍ

˘

˚˙

=È

ÎÍ

˘

˚˙

Ú p p

p

//

( )

( )

- -

- -22 3

5

24

2

3

3

ÊËÁ

ˆ¯

( )È

Î

ÍÍ

˘

˚

˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

=

-a

ap

155



and x dV x a x dx ax x

aa

EL a

a

a

a

2

2 2

2

22 4

2

22

2 4

2

Ú Ú= ÈÎ ˘ =È

ÎÍ

˘

˚˙

=

p p

p

( )

( )

//

- -

-- - -( )aa

a

a a42 2

2

2

4

4

4 2 4

9

64

È

ÎÍ

˘

˚˙

( ) ( )È

Î

ÍÍ

˘

˚

˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

= p

Now x V x dV x a aEL2 2 2 23 45

24

9

64= Ê

ËÁˆ¯

=Ú : p p or x a227

40=

156


PROBLEM 5.123


A semiellipsoid of revolution

SOLutiOn



The equation of the generating curve is x

h

y

a

2

2

2

21+ = so that

ra

hh x2

2

22 2= ( )-

and then dVa

hh x dx= p

2

22 2( )-

Component 1 Va

hh x dx

a

hh x

x

a h

hh

1

2

20

2 2 22

22

3

0

2

2

3

11

24

= =È

ÎÍ

˘

˚˙

=

Ú p p

p

//

( )- -

and x dV xa

hh x dx

a

hh

x x

EL

h

h

1

2

22 2

0

2

2

22

2 4

02 4

Ú Ú=È

ÎÍ

˘

˚˙

=È

ÎÍ

˘

˚˙

p

p

( )/

/

-

-22

2 27

64= pa h

Now x V x dV x a h a hEL1 1 1 12 2 211

24

7

64= Ê

ËÁˆ¯

=Ú : p p or x h121

88=

Component 2 Va

hh x dx

a

hh x

x

a

hh h

h

h

h

h

2 2

2

22 2

2

22

3

2

2

22

3= =

È

ÎÍ

˘

˚˙

= ( )

Ú p p

p

//

( )

(

- -

- hhh

h

a h

h)32 2

3

2

3 2 3

5

24

È

ÎÍ

˘

˚˙

ÊËÁ

ˆ¯

( )È

Î

ÍÍ

˘

˚

˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

=

- -

p

157



and x dV xa

hh x dx

a

hh

x x

EL h

h

h

2

2

22 2

2

2

22

2 4

22 4

Ú Ú=È

ÎÍ

˘

˚˙

=È

ÎÍ

˘

˚˙

p

p

( )/

/

-

-hh

=È

ÎÍ

˘

˚˙

( ) ( )È

Î

ÍÍ

˘

˚

˙˙

ÏÌÔ

ÓÔ

¸˝Ô

˛p a

hh

h hh

h h2

22

2 42 2

2

2

4

2 4 2 4

( ) ( )- - -ÔÔ

= 9

642 2pa h


24

9

64= Ê

ËÁˆ¯

=Ú : p p or x h227

40=

158


PROBLEM 5.124


A paraboloid of revolution

SOLutiOn



The equation of the generating curve is x hh

ay= -

22 so that r

a

hh x2

2

= ( )-

and then dVa

hh x dx= p

2

( )-

Component 1 Va

hh x dx

a

hhx

x

a h

h

h

1 0

2 2

2 2

0

2

2

2

3

8

=

=È

ÎÍ

˘

˚˙

=

Ú p

p

p

/

/

( )-

-

and x dV xa

hh x dx

a

hh

x x

EL

h

h

1

2

0

2

2 2 3

0

2

2 3

1

Ú Ú= =È

ÎÍ

˘

˚˙

=È

ÎÍ

˘

˚˙ =

p

p

( )/

/

-

-112

2 2pa h

Now x V x dV x a h a hEL1 1 12 2 2

1

3

8

1

12= Ê

ËÁˆ¯

=Ú : p p or x h12

9=

Component 2

Va

hh x dx

a

hhx

x

a

hh h

h

h

h

h

h

2 2

2 2 2

2

2 2

2

2

= =È

ÎÍ

˘

˚˙

=È

ÎÍ

Ú p p

p

//

( )

( )( )

- -

-˘

˚˙

ÊËÁ

ˆ¯

( )È

Î

ÍÍ

˘

˚

˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

=

- -hh

a h

h

2 2

1

8

2

2

2p

159



and x dV xa

hh x dx

a

hh

x x

a

EL

h

h

h

h

2

2 2 2 3

22

2

2 3Ú Ú=È

ÎÍ

˘

˚˙ =

È

ÎÍ

˘

˚˙

=

p p

p

( )/

/- -

hhh

h hh

h h( ) ( )2 32

2

2

3

2 3 2 3

1

12

- - -È

ÎÍ

˘

˚˙

( ) ( )È

Î

ÍÍ

˘

˚

˙˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

= paa h2 2


8

1

12= Ê

ËÁˆ¯

=Ú : p p or x h22

3=

160


PROBLEM 5.125

Locate the centroid of the volume obtained by rotating the shaded area about the x axis.

SOLutiOn


and z = 0

We have y k X h= ( )- 2

at x y a a k h= = =0 2, ( ): -

or ka

h=

2


dV r dx X xEL= =p 2 ,

Now ra

hx h=

22( )-

so that dVa

hx h dx= p

2

44( )-

Then Va

hx h dx

a

hx h

a h

h h= = ( )È

Î˘˚

=

Ú p p

p

0

2

44

2

45

0

2

51

5

( )- -

and x dV xa

hx h dx

a

hx hx h x h x h

EL

h

Ú Ú=È

ÎÍ

˘

˚˙

= + +

0

2

44

2

45 4 2 3 3 24 6 4

p

p

( )

(

-

- - 44

0

2

46 5 2 4 3 3 4 2

0

1

6

4

5

3

2

4

3

1

2

1

30

x dx

a

hx hx h x h x h x

h

h

)Ú

= + +ÈÎÍ

˘˚

=

p - -

ppa h2 2

Now xV x dV x a h a hEL= ÊËÁ

ˆ¯

=Ú :p p5 30

2 2 2 or x h= 1

6

161


PROBLEM 5.126


SOLutiOn


z = 0



Now r kx= 1 3/

so that dV k x dx= p 2 2 3/

at x h y a= =, : a kh= 1 3/

or ka

h=

1 3/

Then dVa

hx dx= p

2

2 32 3

//

and Va

hx dx

a

hx

a h

h

h

=

= ÈÎÍ

˘˚

=

Ú p

p

p

2

2 32 3

0

2

2 35 3

0

2

3

5

3

5

//

//

Also x dV xa

hx dx

a

hx

a

EL

hh

Ú Ú= =Ê

ËÁˆ

¯= È

ÎÍ˘˚

=

0

2

2 32 3

2

2 38 3

0

3

8

3

8

p p

p

//

//

22 2h

Now xV xdV= Ú : x a h a h3

5

3

82 2 2p pÊ

ËÁˆ¯

= or x h= 5

8

162


PROBLEM 5.127

Locate the centroid of the volume obtained by rotating the shaded area about the line x h= .

SOLutiOn

First, note that symmetry implies x h=

z = 0


dV r dy y yEL= =p 2 ,

Now xh

aa y2

2

22 2= ( )- so that r h

h

aa y= - -2 2

Then dVh

aa a y dy= ( )p

2

22 2

2

- -

and Vh

aa a y dy

a= ( )Ú p

2

22 2

2

0- -

Let y a dy a d= fi =sin cosq q q

Then Vh

aa a a a d

h

aa a a a a

= ( )= +

Úp q q q

p q

p2

22 2 2

2

0

2

2

22 22

- - sin cos

( cos ) (

/

- - 22 2

0

2

2 2 2

0

22 2

sin ) cos

( cos cos sin cos )

/

/

q q q

p q q q q

p

p

ÈÎ ˘

=

Ú

Ú

a d

ah - - ddq

= +ÊËÁ

ˆ¯

ÈÎÍ

˘˚

=Ê

Ë

p q q q q

p

p

p

ah

ah

2 3

0

2

2 2

2 22

2

4

1

3

2 22

sinsin

sin/

- -

- ÁÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

=

- 1

3

0 095870 2. pah

163



and y dV yh

aa a y dy

h

aa y ay a y y

EL

a

Ú Ú= ( )È

ÎÍ

˘

˚˙

= ( )0

2

22 2

2

2

22 2 2 32 2

p

p

- -

- - - ddya

0Ú

= + - -ÈÎÍ

˘˚

= -ÈÎÍ

˘

p

p

h

aa y a a y y

h

aa a a

a2

22 2 2 2 3 2 4

0

2

22 2 4

2

3

1

4

1

4

( )

( )

/

˚- È

ÎÍ˘˚

ÏÌÓ

¸˝˛

=

2

3

1

12

2 3 2

2 2

a a

a h

( ) /

p

Now yV y dVEL= Ú : y ah a h( . )0 0958701

122 2 2p p= or y a= 0 869.

164


PROBLEM 5.128*

Locate the centroid of the volume generated by revolving the portion of the sine curve shown about the x axis.

SOLutiOn

First, note that symmetry implies y = 0

z = 0

Choose as the element of volume a disk of radius r and thickness dx.

Then dV r dx x xEL= =p 2 ,

Now r bx

a= sin

p2

so that dV bx

adx= p p2 2

2sin

Then V bx

adx

bx

b

a

a

xa

a a

a

a a

=

= -È

ÎÍÍ

˘

˚˙˙

= ( ) -

Ú p p

p

p

p

p

22 2

2 2

2

2 22

2

2 2

sin

sin

22

21

2

( )ÈÎ ˘

= pab

and x dV x bx

adxEL a

a

Ú Ú= ÊËÁ

ˆ¯

2 2 2

2p p

sin


u x dVx

a

du dx Vx x

a

a

= =

= = -

sin

sin

2

2

2

2

p

p

p

165



Then x dV b xx x

EL

xa

a a

ax

a

aÚ = -

Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

- -Ê

ËÁˆ

¯p

p

p

p

p2

2

2

22 2

sin sin˜

ÏÌÔ

ÓÔ

¸˝Ô

Ô

= ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

ÈÎÍ

˘˚

- +

Ú dx

b aa

aa

xa

a

a2

2 22

22

2 2

1

4 2p

ppp

2

2

cosx

aa

aÈ

ÎÍ

˘

˚˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

= ÊËÁ

ˆ¯

- + - +È

ÎÍ

˘

˚˙

ÏÌÔ

ÓÔ

¸˝ÔÔ

=

pp p

b a aa

aa2 2 2

2

22

2

2

3

2

1

42

2

1

4 2( ) ( )

ppp

p

a b

a b

2 22

2 2

3

4

1

0 64868

-ÊËÁ

ˆ¯

= .

Now xV x dV x ab a bEL= ÊËÁ

ˆ¯

=Ú :1

20 648682 2 2p p. or x a= 1 297.

166


PROBLEM 5.129*

Locate the centroid of the volume generated by revolving the portion of the sine curve shown about the y axis. (Hint: Use a thin cylindrical shell of radius r and thickness dr as the element of volume.)

SOLutiOn


z = 0

Choose as the element of volume a cylindrical shell of radius r and thickness dr.

Then dV r y dr y yEL= =( )( )( ),21

2p

Now y br

a= sin

p2

so that dV brr

adr= 2

2p p

sin

Then V brr

adr

a

a

= Ú 22

2

p psin


u rd dvr

adr

du dr va r

a

= =

= = -

sin

cos

p

pp

2

2

2

Then V b ra r

a

a r

adr

a

a

a

a

= -ÊËÁ

ˆ¯

ÈÎÍ

˘˚

- ÊËÁ

ˆ¯

ÏÚ2

2

2

2

2

2 2

pp

pp

p( ) cos cosÌÌ

Ô

ÓÔ

¸˝Ô

Ô

= - -[ ]+È

ÎÍ

˘

˚˙

ÏÌÔ

ÓÔ

¸˝Ô

22

2 14

2

2

2

2

pp p

pb

aa

a r

aa

a

( )( ) sin

˛Ô

V ba a

a b

a b

= -Ê

ËÁˆ

¯

= -ÊËÁ

ˆ¯

=

24 4

8 11

5 4535

2 2

2

2

2

pp p

p

.

167



Also y dV br

abr

r

adr

b r

ELa

a

a

a

Ú Ú= ÊËÁ

ˆ¯

ÊËÁ

ˆ¯

=

12

2

2 22

22

2sin sin

sin

p p p

p ÚÚpr

adr

2


u r dvr

adr

du dr vr r

a

a

= =

= = -

sin

sin

2

2

2

2

p

p

p

Then y dV b rr r

EL

ra

a a

ar

a

aÚ = -

Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

- -Ê

ËÁp

p

p

p

p2

2

2

22 2( )

sin sin ˆ

¯

ÏÌÔ

ÓÔ

¸˝Ô

Ô

= ÊËÁ

ˆ¯

- ÊËÁ

ˆ¯

ÈÎÍ

˘˚

-

Ú dr

b aa

aa r

a

a2

22

22

2 2p ( ) ( )

44 2

2

2

2

+È

ÎÍ

˘

˚˙

ÏÌÔ

ÓÔ

¸˝Ô

Ô

a r

aa

a

pp

cos

= - + - +È

ÎÍ

˘

˚˙

ÏÌÔ

ÓÔ

¸˝ÔÔ

= -

pp p

p

b aa a a a

a b

2 22 2

2

2 2

2

2 2

3

2

2

4 2 4 2

3

4

( ) ( )

11

2 0379

2

2 2

pÊËÁ

ˆ¯

= . a b

Now yV y dV y a b a bEL= =Ú : ( . ) .5 4535 2 03792 2 2 or y b= 0 374.

168


PROBLEM 5.130*

Show that for a regular pyramid of height h and n sides ( 3, 4, )n = … the centroid of the volume of the pyramid is located at a distance h/4 above the base.

SOLutiOn

Choose as the element of a horizontal slice of thickness dy. For any number N of sides, the area of the base of the pyramid is given by

A kbbase = 2

where k k N= ( ); see note below. Using similar triangles, have

s

b

h y

h= -

or sb

hh y= -( )

Then dV A dy ks dy kb

hh y dy= = = -slice

22

22( )

and V kb

hh y dy k

b

hh y

kb h

hh

= - = - -ÈÎÍ

˘˚

=

Ú2

20

22

23

0

2

1

3

1

3

( ) ( )

Also y yEL =

So then y dV y kb

hh y dy k

b

hh y hy y dy

kb

EL

h h

Ú Ú Ú= -È

ÎÍ

˘

˚˙ = - +

=

0

2

22

2

22 2 3

0

2

2( ) ( )

hhh y hy y kb h

h

22 2 3 4

0

2 21

2

2

3

1

4

1

12- +È

ÎÍ˘˚

=

Now yV y dV y kb h kb hEL= ÊËÁ

ˆ¯

=Ú :1

3

1

122 2 2 or y h= 1

4 Q.E.D.

Note A N b

Nb

k N b

b

N

Nbase = ¥ ¥

Ê

ËÁˆ

¯

=

=

1

2

4

2

2

2

tan

tan

( )

p

p

169


PROBLEM 5.131

Determine by direct integration the location of the centroid of one-half of a thin, uniform hemispherical shell of radius R.

SOLutiOn


The element of area dA of the shell shown is obtained by cutting the shell with two planes parallel to the xy plane. Now

dA r Rd

yr

EL

=

= -

( )( )p q

p2

where r R= sinq

so that dA R d

yR

EL

=

= -

p q q

pq

2

2

sin

sin

Then A R d R

R

= = -

=

Ú p q q p q

p

pp2

0

22

02

2

//sin [ cos ]

and y dAR

R d

R

ELÚ Ú= -ÊËÁ

ˆ¯

= - -ÈÎÍ

˘˚

2

22

2

4

0

22

3

0

pq p q q

q q

p

sin ( sin )

sin

/

pp

p

/2

3

2= - R

Now yA y dA y R REL= = -Ú : ( )p p2 3

2 or y R= - 1

2

Symmetry implies z y= z R= - 1

2

170


PROBLEM 5.132

The sides and the base of a punch bowl are of uniform thickness t. If t R�� and R = 250 mm, determine the location of the center of gravity of (a) the bowl, (b) the punch.

SOLutiOn

(a) Bowl


z = 0

for the coordinate axes shown below. Now assume that the bowl may be treated as a shell; the center of gravity of the bowl will coincide with the centroid of the shell. For the walls of the bowl, an element of area is obtained by rotating the arc ds about the y axis. Then

dA R Rdwall = ( sin )( )2p q q

and ( ) cosy REL wall = - q

Then A R d

R

R

wall =

= -

=

Ú 2

2

3

2

6

2

26

2

2

p q q

p q

p

p

p

pp

/

/

/

/

sin

[ cos ]

and y A y dA

R R d

R

ELwall wall wall=

= -

=

ÚÚ

( )

( cos )( sin )

[co

/

/q p q q

p

p

p2 2

6

2

3 ss ] //262

33

4

q

p

pp

= - R

By observation A R y Rbase base = = -p4

3

22 ,

Now y A yAS S=

or y R R R R Rp p p p3

4

3

4 4

3

22 2 3 2+Ê

ËÁˆ¯

= - + -Ê

ËÁˆ

¯

or y R R= - =0 48763 250. mm y = -121 9. mm

171



(b) Punch


z = 0

and that because the punch is homogeneous, its center of gravity will coincide with the centroid of the corresponding volume. Choose as the element of volume a disk of radius x and thickness dy. Then

dV x dy y yEL= =p 2 ,

Now x y R2 2 2+ =

so that dV R y dy= -p ( )2 2

Then V R y dy

R y y

R R

R

R

= -

= -ÈÎÍ

˘˚

= - -Ê

ËÁˆ

¯-

-

-

Ú p

p

p

( )/

/

2 2

3 2

0

2 3

3 2

0

2

1

3

3

2

1

33

3

2

3

83

33-

Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

=R Rp

and y dV y R y dy

R y y

EL R

R

Ú Ú= -( )ÈÎ

˘˚

= -ÈÎÍ

˘˚

= -

-

-

( )/

/

3 2

0 2 2

2 2 4

3 2

01

2

1

4

p

p

pp p1

2

3

2

1

4

3

2

15

642

2 44R R R R-

Ê

ËÁˆ

¯- -

Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

= -

Now yV y dV y R REL= ÊËÁ

ˆ¯

= -Ú :3

83

15

643 4p p

or y R R= - =5

8 3250 mm y = -90 2. mm

172


PROBLEM 5.133

After grading a lot, a builder places four stakes to designate the corners of the slab for a house. To provide a firm, level base for the slab, the builder places a minimum of 75 mm of gravel beneath the slab. Determine the volume of gravel needed and the x coordinate of the centroid of the volume of the gravel. (Hint: The bottom surface of the gravel is an oblique plane, which can be represented by the equation y = a + bx + cz.)

SOLutiOn

The centroid can be found by integration. The equation for the bottom of the gravel is:

y a bx cz= + + , where the constants a, b, and c can be determined as follows:

For x = 0, and z = 0: y = -75 mm and therefore,

- -75

10000 075m or m= =a a, .

For x z y= = = = -90 0 5 0 125m and mm m, : , .-12 and therefore

- -0.125m + (9m), or= - =0 0751

180. b b

For x z y= = =0 15 0 15, : . ,and m mm= m-150 - and therefore - - -

- - -

0 15 0 0751

200

0 0751

180

1

200

. .

.

m = m+ (15m), or =

= m

C C

y x zTherefore:

Now xx dV

V

EL= Ú

A volume element can be chosen as:

dV y dxdz= | |

or dV x z dx dz= + +ÈÎÍ

˘˚

0 0751

180

1

200.

and x xEL =

173



Then x dV x x z dx dzELÚ ÚÚ= + +ÊËÁ

ˆ¯0

9

0

150 075

1

180

1

200.

= + +È

ÎÍ

˘

˚˙

= +ÊËÁ

ˆ¯

Ú 0 0752

1

540 400

351

80

81

400

23

2

0

15

0

9

.x

xzx

dz

z dz00

15

2

0

15

4

351

80

81

800

1053

16

729

32

2835

32

88 593

Ú

= +ÈÎÍ

˘˚

= + =

=

z z

. m

The volume is: V dV x z dx dzÚ ÚÚ= + +ÊËÁ

ˆ¯

0 0751

180

1

2000

9

0

15.

= + +È

ÎÍ

˘

˚˙

= +ÊËÁ

ˆ¯

=

Ú

Ú

0 075360 200

9

10

9

200

9

2

0

15

0

9

0

15

. xx z

x dz

zdz

z

110

9

400

297

1618 5625

2

0

15

+ÈÎÍ

˘˚

=

z

. m3

Then xx dV

V

EL= = =Ú 88 593

18 56254 7727

.

..

m

mm

4

3

Therefore: V = 18 56. m3

x = 4 77. m

174


PROBLEM 5.134

Determine by direct integration the location of the centroid of the volume between the xz plane and the portion shown of the surface y = 16h(ax - x2)(bz - z2)/a2b2.

SOLutiOn

First note that symmetry implies xa=2

zb=2

Choose as the element of volume a filament of base dx dz¥ and height y. Then

dV ydxdz y yEL= =,1

2

or dVh

a bax x bz z dx dz= - -16

2 22 2( )( )

Then Vh

a bax x bz z dx dz

ab

= - -ÚÚ162 2

00

2 2( )( )

Vh

a bbz z

a

zx x dz

h

a b

aa a

b a

= - -ÈÎÍ

˘˚

= -

Ú16 1

3

16

2

1

3

2 22

0

2 3

0

2 22

( )

( ) ( )33 2 3

0

22 3

2

1

3

8

3 2

1

3

4

9

ÈÎÍ

˘˚

-ÈÎÍ

˘˚

= -ÈÎÍ

˘˚

=

bz z

ah

b

bb b

abh

b

( ) ( )

175



and y dVh

a bax x bz z

h

a bax x bEL

ab

Ú ÚÚ= - -ÈÎÍ

˘˚

-1

2

16 16

002 2

2 22 2

2( )( ) ( )( zz z dx dz

h

a ba x ax x b z bz z

a

-ÈÎÍ

˘˚

= - + - +Ú

2

2

4 42 2 3 4 2 2 3 4

00

1282 2

)

( )( )bb

b a

dx dz

h

a bb z bz z

ax

ax x d

Ú

Ú= - + - +È

ÎÍ

˘

˚˙

1282

3 2

1

5

2

2 42 2 3 4

0

23 4 5

0

( ) zz

= - +È

ÎÍ

˘

˚˙ - +È

ÎÍ

˘

˚˙

128

3 2

1

5 3

1

5

2

4 4

23 4 5

23 4 5h

a b

aa

aa a

bZ

b

ZZ Z( ) ( ) ( )

00

2

4

33 4 5 264

15 3 2

1

5

32

225

b

ah

b

bb

bb b abh= - +

È

ÎÍ

˘

˚˙ =( ) ( ) ( )

Now yV y dV y abh abhEL= ÊËÁ

ˆ¯

=Ú :4

9

32

2252 or y h= 8

25

176


PROBLEM 5.135

Locate the centroid of the section shown, which was cut from a thin circular pipe by two oblique planes.

SOLutiOn


Assume that the pipe has a uniform wall thickness t and choose as the element of volume A vertical strip of width adq and height ( ).y y2 1- Then

dV y y tad y y y z zEL EL= - = + =( ) , ( )2 1 1 21

2q

Now ya

zhh

13

2 6= + y

az h

h

2

23

2

2

3= - +

= +h

az a

6( ) = - +h

az a

32( )

and z a= cosq

Then ( ) ( cos ) ( cos )

( cos )

y yh

aa a

h

aa a

h

2 1 32

6

21

- = - + - +

= -

q q

q

and ( ) ( cos ) ( cos )

( cos )

( co

y yh

aa a

h

aa a

h

dVaht

1 2 6 32

65

21

+ = + + - +

= -

= -

q q

q

ss ) ( cos ), cosq q q qd yh

z aEL EL= - =12

5

177



Then Vaht

d aht

aht

= - = -

=

Ú22

10

0( cos ) [ sin ]q q q q

p

pp

and y dVh aht

d

ah t

ELÚ Ú= - -ÈÎÍ

˘˚

= - +

212

52

1

125 6

0

2

( cos ) ( cos )

( cos

q q q

q

p

ccos )

sinsin

2

0

2

0

2

125 6

2

2

4

11

24

q q

q q q q

p

p

p

Ú

= - + +ÈÎÍ

˘˚

=

d

ah t

ah t

z dV aaht

d

a ht

ELÚ Ú= -ÈÎÍ

˘˚

= - -ÈÎÍ

22

1

2

2

4

0

2

cos ( cos )

sinsin

q q q

q q q

p

˘˚

= -

0

21

2

p

pa ht

Now yV y dV y aht ah tEL= =Ú : ( )p p11

242 or y h= 11

24

and zV y dV z aht a htEL= = -Ú : ( )p p1

22 or z a= - 1

2

178


PROBLEM 5.136*

Locate the centroid of the section shown, which was cut from an elliptical cylinder by an oblique plane.

SOLutiOn


Choose as the element of volume a vertical slice of width zx, thickness dz, and height y. Then

dV xydz y z zEL EL= = =21

24, ,

Now xa

bb z= -2 2

and yh

bz

h h

bb z= - + = -/2

2 2( )

Then Va

bb z

h

bb z dz

b

b= -Ê

ËÁˆ¯

-ÈÎÍ

˘˚-Ú 2

22 2 ( )

Let z b dz b d= =sin cosq q q

Then Vah

bb b b d

abh

= -

= -

Ú2 2

2

2 2

1( cos )[ ( sin )] cos

(cos sin cos

/

/q q q q

q q q

p

p

))

sincos

/

/

/

/

d

abh

V abh

q

q q q

p

p

p

p

p

2

2

3

2

2

2

2

4

1

3

1

2

Ú

= + +ÈÎÍ

˘˚

=

-

179



and y dVh

bb z

a

bb z

h

bb z dzELÚ = ¥ -È

ÎÍ˘˚

-ÊËÁ

ˆ¯

-ÈÎÍ

˘˚

ÏÌÓ

¸1

2 22

22 2( ) ( ) ˝

˛

= - -

-

-

Ú

Ú

b

b

b

bah

bb z b z dz

1

4

2

32 2 2( )


Then y dVah

bb b b d

abh

ELÚ Ú= - ¥

=

-

1

41

1

4

2

32

2

2

2

[ ( sin )] ( cos ) ( cos )/

/q q q q

p

p

((cos sin cos sin cos )/

/ 2 2 2 2

2

22q q q q q q

p

p- +

-Ú d

Now sin ( cos ) cos ( cos )2 21

21 2

1

21 2q q q q= - = +

So that sin cos ( cos )2 2 21

41 2q q q= -

Then y dV abh dELÚ Ú= - + -ÈÎÍ

˘˚-

1

42

1

41 22 2 2 2

2

2cos sin cos ( cos )

/

/q q q q q

p

p

== +ÊËÁ

ˆ¯

+ + - +ÊËÁ

ˆ¯

ÈÎÍ

˘˚

1

4 2

2

4

1

3

1

4

1

4 2

4

82 3abh

q q q q q qsincos

sin˙

=

-p

p

p

/

/

2

2

25

32abh

Also z dVa

ba z

h

bb z dz

ah

bz b z b

EL b

b

b

b

Ú Ú

Ú

= - -ÈÎÍ

˘˚

ÏÌÓ

¸˝˛

= -

-

-

22

2 2

22

( )

( ) -- z dz2


Then z dVah

bb b b b d

ab

ELÚ Ú= - ¥

=

-2 2

2

2

1( sin )[ ( sin )]( cos ) ( cos )/

/q q q q q

p

p

hh d(sin cos sin cos )/

/q q q q q

p

p 2 2 2

2

2-

-Ú

Using sin cos ( cos )2 2 21

41 2q q q= - from above

z dV ab h dELÚ Ú= - -ÈÎÍ

˘˚-

2 2 2

2

2 1

41 2sin cos ( cos )

/

/q q q q

p

p

180



= - - + +ÊËÁ

ˆ¯

ÈÎÍ

˘˚

= --

ab h ab h2 3

2

221

3

1

4

1

4 2

4

8

1

8cos

sin

/

/

q q q q pp

p

Now yV y dV y abh abhEL= ÊËÁ

ˆ¯

=Ú :1

2

5

322p p or y h= 5

16

and z V z dV z abh ab hEL= ÊËÁ

ˆ¯

= -Ú :1

2

1

82p p or z b= - 1

4

181


SOLutiOn


1 126 54 6804¥ = 9 27 61236 183708

2 1

2126 30 1890¥ ¥ = 30 64 56700 120960

3 1

272 48 1728¥ ¥ = 48 -16 82944 -27648

S 10422 200880 277020

Then X A xAS S=

X ( )10422 2008802 2m mm= or X = 19 27. mm

and Y A yAS S=

Y ( )10422 2700202 3m mm= or Y = 26 6. mm

PROBLEM 5.137


182


SOLutiOn

Dimensions in in.


1 2

3100 200 13333 33( )( ) .= 120 37.5 1.6 ´ 106 0.5 ´ 106

2 - -1

2100 200 10000( )( ) = 133.333 33.333 –1.3333 ´ 106 –0.3333 ´ 106

S 3333.33 0.26667 ´ 106 0.16667 ´ 106

Then X A xAS S=

X ( . ) .3333 33 0 26667 102 6 3mm mm= ¥ or X = 80 mm

and Y A yAS S=

Y ( . ) .3333 33 0 166667 102 6 3mm mm= ¥ or Y = 50 mm

PROBLEM 5.138


183


PROBLEM 5.139

The frame for a sign is fabricated from thin, flat steel bar stock of mass per unit length 4.73 kg/m. The frame is supported by a pin at C and by a cable AB. Determine (a) the tension in the cable, (b) the reaction at C.

SOLutiOn

First note that because the frame is fabricated from uniform bar stock, its center of gravity will coincide with the centroid of the corresponding line.

L, m x m, xL m, 2

1 1.35 0.675 0.91125

2 0.6 0.3 0.18

3 0.75 0 0

4 0.75 0.2 0.15

5p2

0 75 1 17810( . ) .= 1.07746 1.26936

S 4.62810 2.5106

Then X L x L

X

S S==( . ) .4 62810 2 5106

or X = 0 54247. m

The free-body diagram of the frame is then

where W m L g= ¢

= ¥ ¥=

( )

. .

.

S

4 73 9 81

214 75

kg/m 4.62810 m m/s

N

2

184



Equilibrium then requires

(a) SM TC BA= ÊËÁ

ˆ¯

- =0 1 553

50 54247 214 75 0: ( . ) ( . )( . ) m m N

or TBA = 125 264. N or TBA = 125 3. N

(b) SF Cx x= - =03

5125 264 0: ( . )N

or Cx = 75 158. N ®

SF Cy y= + - =04

5125 264 214 75 0: ( . ) ( . )N N

or Cy = 114 539. N ≠

Then C = 137 0. N 56.7°

185


PROBLEM 5.140


SOLutiOn

By observation yh

ax h h

x

a1 1= - + = -ÊËÁ

ˆ¯

For y2: at x y h h k k h= = = -( ) =0 1 0, : or

at x a y h ca Ca

= = = - =, ( )0 0 1 22

: or 1

Then y hx

a2

2

21= -

Ê

ËÁˆ

¯

Now dA y y dx

hx

a

x

adx

hx

a

x

a

= -

= -Ê

ËÁˆ

¯- -Ê

ËÁˆ¯

È

ÎÍÍ

˘

˚˙˙

= -Ê

Ë

( )2 1

2

2

2

2

1 1

ÁÁˆ

¯dx

x x

y y y

h x

a

x

a

h x

EL

EL

=

= -

= -ÊËÁ

ˆ¯

+ -Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

= -

1

2

21 1

22

1 2

2

2

( )

aa

x

a-

Ê

ËÁˆ

¯

2

2

Then A dA hx

a

x

adx h

x

a

x

a

ah

a a

= = -Ê

ËÁˆ

¯= -

È

ÎÍ

˘

˚˙

=

Ú Ú0

2

2

2 3

20

2 3

1

6

186



and x dA x hx

a

x

adx h

x

a

x

aELÚ = -

Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

= -Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

2

2

3 4

23 4000

21

12

aa

a h

Ú

=

y dAh x

a

x

ah

x

a

x

adx

h x

a

EL

a

Ú Ú= - -Ê

ËÁˆ

¯-

Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

=

22

22

2

2

2

20

2

-- +Ê

ËÁˆ

¯

= - +È

ÎÍ

˘

˚˙ =

Ú 3

2 5

1

10

2

2

4

40

2 2 3

2

5

40

2

x

a

x

adx

h x

a

x

a

x

aah

a

a

xA x dA x ah a hEL= ÊËÁ

ˆ¯

=Ú :1

6

1

122 x a= 1

2

yA y dA y ah a hEL= ÊËÁ

ˆ¯

=Ú :1

6

1

102 y h= 3

5

187


PROBLEM 5.141


SOLutiOn

First note that symmetry implies y b=

at x a y b= =,

y b ka12: = or k

b

a=

2

Then yb

ax1 2

2=

y b b ca222: = -

or cb

a=

2

Then y bx

a2

2

22= -

Ê

ËÁˆ

¯

Now dA y y dx bx

a

b

ax dx

bx

a

= - = -Ê

ËÁˆ

¯-

È

ÎÍÍ

˘

˚˙˙

= -Ê

ËÁˆ

¯

( )2 1 2

2

2 22

2

2

2

2 1 ddx

and x xEL =

Then A dA bx

adx b x

x

aab

aa

= -Ê

ËÁˆ

¯= -

È

ÎÍ

˘

˚˙ =Ú Ú 2 1 2

3

4

30

2

2

3

20

and x dA x bx

adx b

x x

aa bEL

a

Ú = -Ê

ËÁˆ

¯

È

ÎÍÍ

˘

˚˙˙

= -È

ÎÍ

˘

˚˙ =2 1 2

2 4

1

2

2

2

2 4

20

2

00

a

Ú

xA x dA x ab a bEL= ÊËÁ

ˆ¯

=Ú4

3

1

22 x a= 3

8

188


PROBLEM 5.142

Knowing that two equal caps have been removed from a 250-mm-diameter wooden sphere, determine the total surface area of the remaining portion.

SOLutiOn

The surface area can be generated by rotating the line shown about the y axis. Applying the first theorem of Pappus-Guldinus, we have

A X L x L

x L x L

= == +

2 2

2 2 1 1 2 2

p pp

S( )

Now tana = 4

3or a = ∞53 130.

Then x2180

125 53 130

53 130

107 84

=¥ ∞

∞ ¥=

∞

mm

mm

sin .

.

.

p

and L

A

2 2 53 130180

125

231 8225

2 275

2

= ∞ ¥∞

ÊËÁ

ˆ¯

=

= ÊËÁ

ˆ¯

. ( )

.

(

p

p

mm

mm

mm 775 107 84 231 8225

192420 9 2

mm mm mm

mm

) ( . )( . )

.

+ÈÎÍ

˘˚

=

or A = ¥192 4 103. mm2

189


PROBLEM 5.143


SOLutiOn

R

R

I

II

m N/m

N

m N/m

N

==

=

=

( )( )

( )( )

3 900

2700

1

21 900

450

Now

+

SF Ax x= =0 0:

+ SM AB y= - + - ÊËÁ

ˆ¯

=0 3 1 5 27001

3450 0: ( ) ( . )( ) ( )m m N m N

or Ay = 1300 N A = 1300 N ≠

+

SF By y= - + - =0 1300 2700 450 0: N N N

or By = 1850 N B = 1850 N ≠

190


SOLutiOn

We have R

R

RBC

I

II

m N/m N

m N/m N

m

= =

= =

=

1

26 600 1800

1

26 1200 3600

0 8

( )( )

( )( )

( . )) ( ) ( . )

( . ) ( ) ( )

W W

R W WBC BC

DE DE DE

N/m N

m N/m N

== =

0 8

1 0

Then + SM WG DE= - - + =0 1 1800 3 3600 4 0: ( )( ) ( )( ) ( )( )m N m N m N

or WDE = 3150 N/m

and

+

SF Wy BC= - - + =0 0 8 1800 3600 3150 0: ( . ) N N N N

or WBC = 2812 5. N/m WBC = 2810 N/m

PROBLEM 5.144

A beam is subjected to a linearly distributed downward load and rests on two wide supports BC and DE, which exert uniformly distributed upward loads as shown. Determine the values of w

BC and w

DE corresponding to

equilibrium when wA = 600 N/m.

191


PROBLEM 5.145

The square gate AB is held in the position shown by hinges along its top edge A and by a shear pin at B. For a depth of water d = 1m , determine the force exerted on the gate by the shear pin.

SOLutiOn

First consider the force of the water on the gate. We have

P Ap

A gh

=

=

1

21

2( )d

Then P

P

I3

II

m N/m m

N

m

= ¥

=

=

1

20 5 1000 9 81 0 5

613 125

1

20 5 10

2

2

( . ) ( . )( . )

.

( . ) ( 000 9 81 0 5 0 5 30

1144 107

¥ ¥ + ∞

=

. ) ( . . cos )

.

N/m m

N

3

Now SM L P L P L FA AB AB AB B= ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

- =01

3

2

30: I II

or 1

3613 125

2

31144 107 0( . ) ( . )N N+ - =FB

or FB = 967 113. N FB = 967 N 30.0∞

192


SOLutiOn

V x xV

Rectangular prism Lab1

2L

1

22L ab

Pyramid1

3 2a

bhÊ

ËÁˆ¯

L h+ 1

41

6

1

4abh L h+Ê

ËÁˆ¯

Then S

S

V ab L h

xV ab L h L h

= +ÊËÁ

ˆ¯

= + +ÊËÁ

ˆ¯

ÈÎÍ

˘˚

1

6

1

63

1

42

Now X V xVS S=

so that X ab L h ab L hL h+ÊËÁ

ˆ¯

ÈÎÍ

˘˚

= + +ÊËÁ

ˆ¯

1

6

1

63

1

42 2

or Xh

LL

h

L

h

L1

1

6

1

63

1

4

2

2+Ê

ËÁˆ¯

= + +Ê

ËÁˆ

¯ (1)

(a) X = ? when h L= 1

2

Substituting h

L= 1

2into Eq. (1)

X L11

6

1

2

1

63

1

2

1

4

1

2

2

+ ÊËÁ

ˆ¯

ÈÎÍ

˘˚

= + ÊËÁ

ˆ¯

+ ÊËÁ

ˆ¯

È

ÎÍÍ

˘

˚˙˙

or X L= 57

104 X L= 0 548.

PROBLEM 5.146

Consider the composite body shown. Determine (a) the value of x when h L= /2, (b) the ratio h/L for which x L= .

193



(b) h

L= ? when X L=

Substituting into Eq. (1) Lh

LL

h

L

h

L1

1

6

1

63

1

4

2

2+Ê

ËÁˆ¯

= + +Ê

ËÁˆ

¯

or 11

6

1

2

1

6

1

24

2

2+ = + +h

L

h

L

h

L

or h

L

2

212=

h

L= 2 3

194


PROBLEM 5.147


SOLutiOn

First assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with the centroid of the corresponding area.

y

z

x y

x

I

I

II II

IV

m

m

m

= + =

=

= = ¥ =

0 181

30 12 0 22

1

30 2

2 0 18 0 36

. ( . ) .

( . )

. .

p p

== - ¥

=

0 344 0 05

30 31878

..

.p

m

A, m2 x , m y, m z , m xA, m3 yA, m3 zA, m3

I 1

20 2 0 12 0 012( . )( . ) .= 0 0.22 0 2

3

. 0 0.00264 0.0008

II p p2

0 18 0 2 0 018( . )( . ) .= 0 36.

p0 36.

p0.1 0.00648 0.00648 0.005655

III ( . )( . ) .0 16 0 2 0 032= 0.26 0 0.1 0.00832 0 0.0032

IV - = -p p2

0 05 0 001252( . ) . 0.31878 0 0.1 –0.001258 0 –0.000393

S 0.096622 0.013542 0.00912 0.009262

195



We have X V xV XS S= =: ( . ) .0 096622 0 013542m m2 3 or X = 0 1402. m

Y V yV YS S= =: ( . ) .0 096622 0 00912m m2 3 or Y = 0 0944. m

Z V zV ZS S= =: ( . ) .0 096622 0 009262m m2 3 or Z = 0 0959. m

196


PROBLEM 5.148


SOLutiOn

First, note that symmetry implies y = 0

z = 0

Choose as the element of volume a disk of radius r and thickness dx.

Then dV r dx x xEL= =p 2 ,

Now rx

= -11

so that dVx

dx

x xdx

= -ÊËÁ

ˆ¯

= - +ÊËÁ

ˆ¯

p

p

11

12 1

2

2

Then Vx x

dx x xx

= - +ÊËÁ

ˆ¯

= - -ÈÎÍ

˘˚

= - -ÊËÁ

ˆ¯

Úp p

p

12

1

3

12 1

21

3 2 31

3

3

ln

ln -- - -ÊËÁ

ˆ¯

ÈÎÍ

˘˚

=

1 2 11

1

0 46944 3

ln

( . )p m

and x dV xx x

dxx

x xELÚ Ú= - +ÊËÁ

ˆ¯

ÈÎÍ

˘˚

= - +È

ÎÍ

˘

˚˙

=

1

3

2

2

1

3

2

12 1

22

3

p p

p

ln

222 3 3

1

22 1 1

1 09861

3

- +È

ÎÍ

˘

˚˙ - - +

È

ÎÍ

˘

˚˙

ÏÌÔ

ÓÔ

¸˝ÔÔ

=

( ) ln ( ) ln

( . )p m

Now xV x dV xEL= =Ú : ( . ) .0 46944 1 09861p pm m3 4 or x = 2 34. m

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CHAPTER 3CHAPTER 7CHAPTER 5 - · PDF fileCHAPTER 3CHAPTER 7CHAPTER 5. 3 ... If you are a student using this Manual, you are using it without permission. PROBLEM 5.1 Locate the centroid