solucionario de estatica

CHAPTER 5

30 Dim

30 inmPROBLEM 5.1

Locate the centroid of the plane area shown.

1 300 mil)

-240mm-

SOLUTION

Dimensions in mm

-27T

«E.

' <

U\cfiy~\4

T/TO

4-

JL/or|,/ar j^i

i4, mm2x, mm y, mm x./4, mm3

y.A, mm

1 6300 105 15 0.661 50x1 60.094500x1

6

2 9000 225 150 2.0250x1 61.35000x10*

£ 15300 2.6865xl06

1.44450xl06

Then X £x/5l 2.6865x10*

ZA

ZA 15300

£JM. 1.44450x10*

15300

X = 175.6 mm <

Y = 94.4 mm <«

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541

20 mm 30mm

-~r

—

36 mm

24mm

PROBLEM 5.2


SOLUTION

Dimensions in mm

rZA +K

C,

-A \0

A, mm" x, mm _y, mm xA, mm3yA, mm3

I 1200 10 30 12000 36000

2 540 30 36 16200 19440

1 1740 28200 55440

Then X ZxA 28200

XA 1740

gjvj =55440

XA 1740

X = 16.21 mm <

F = 31.9mm <

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542

-122 in.-*+« 21 in. -

15 in.

PROBLEM 5.3


SOLUTION

Dimensions in in.

zfirt

m +7*

4 in.2

x,in. y> in. x^, in; M in.3

1 -xl2xl5 = 902

8 5 720 450

2 21x15 = 315 22.5 7.5 7087.5 2362.5

I 405.00 7807.5 2812.5

Then-= £x,4 = 7807.5

ZA 405.00

Y^XyA^ 2812.5

1^1 ""405.00

Z = 19.28 in. ^

F = 6.94 in. ^

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543

3 in,

6 in.

6 in. • 6 in.

6 in.

PROBLEM 5.4


SOLUTION

4 WW

A, in.2 x,m. y, in. x^i, in/ _y^, in.

3

1i(12)(6) = 36 4 4 144 144

2 (6X3) = 18 9 7.5 162 135

£ 54 306 279

Then XA = Y,xA

X(54) = 306

YA - XyA

F(54) = 279

X - 5.67 in. -<

r = 5.17in. <

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544

6 in. , 8 in.

PROBLEM 5.5

Locate the eentroid of the plane area shown.

S in.

12 in.

SOLUTION

©,.- .J

c,

|

1 \H*"*

ttiM.

A, in.2

x, in. p, in. X/4, in.3

jM, in.3

] 14x20 = 280 7 10 1960 2800

2 -^(4)2 = "16^ 6 12 -301.59 -603.19

I 229.73 1658.41 2196.8

Then X TsxA 1658.41

1A 229.73

£j^ = 2.196.8

XA ~229.73

X = 7.22 in. ^

F = 9.56 in. <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it withoutpermission.

545

.120 mm

«Ar = 75/mm

ilg \

PROBLEM 5.6


SOLUTION

Z5mw 1

157.5 wwi

•i|

0Omm

©37. 5Vm

3-nv\v»

HS>3».**»***3'K^*W*

J, ram2 x,mm jvmm xA,mm 3yA, mm3

1 !(120)(75) = 4500 80 2.5 360xlO3112.5 xlO

3

2 (75)(75) = 5625 157.5 37.5 885.94 xlO32 10.94 xlO3

3 ™^(75)2 --4417.94

163.169 43.169 -720.86x13 -190.716xl0

3

X 5707.1 525.08 xlO3132.724 xlO

3

Then XA =ZxA ^(5707. 1) = 525.08x 13

YA - ZyA F(5707.1) = 132.724xl03

X = 92.0 mm <

Y = 23.3 mm ^


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546

PROBLEM 5.7


16 in.

1

20 in.

SOLUTION

tM.

ft II*

iA

•*Z

tO 1M.

A, in.2

x, in. J7* in. xA, in; yA, in.

3

1 — (38)2=2268.2

. 16.1277 36581

2 -20x16- -320 -10 8 3200 -2560

Z 1948.23 3200 3402.1

Then X * xA 3200

Y

~ZA 1948.23

SJ^ 3402.1

LA 1948.23

X = 1.643 in. ^

7 = 17.46 in. <«

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547

PROBLEM 5.8


SOLUTION

2.S1N.

©

4-C,

\S l»».

A, in? x, in. ^> 'n * X/4, in? yA, in?

1 30x50 = 1500 15 25 22500 37500

2 -~(15)2 =353.432

23.634 30 -8353.0 -10602.9

£ 1 146.57 14147,0 26.897

Then X

Y =

ZxA_ 14147.0

2^ "1146.57

Xy^ 26897

2,4 1146.57

X = 12.34 in. ^

F = 23.5 in. <

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548

<>() iniii

PROBLEM 5.9


60mm

SOLUTION

^l^-^&.^S <*\**

120-15.445

|

a ^5.43S*w\

X

/4, mm2a-, mm y, mm x,4, mm3

yA, mm3

1 (60)(i 20) = 7200 -30 60 -216xl03432 xlO3

2 -(60)2 = 2827.4

425.465 95.435 72.000 xlO3

269.83 xlO3

3 -—(60)2 --2827.4

4-25.465 25.465 72.000 xlO3

-72.000xlO3

S 7200 -72.000X103629.83 xlO3

Then X4 = ZxA X (7200) = -72.000x 13

YA = Y,yA F(7200) = 629.83xl03

.^ = -10.00 mm A

F = 87.5 mm «

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549

If

Semicllipse

.

70 nun

47 mm 47 mmPROBLEM 5.10

Locate the centroid ofthe plane area shown.

26 mm

SOLUTION

Dimensions in mm

T%

kUi-t)

A, mm2x, mm y, mm jtvf, mm3

jvi, ram3

1 ~x47x26 = 1919.512

11.0347 21181

2 -x 94x70 = 32902

-15.6667 -23.333 -51543 -76766

Z 5209.5 -51543 -55584

Then- S*^ -51543

£/* 5209.5

£v^. -55584

Z.A 5209.5

A' « -9.89 mm ^

F = -10.67 mm <

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550

PROBLEM 5.11


/\A \ /' \

. Ju iiui. 'i • Sin. \id \/

SOLUTION

First note that symmetry implies X = M

,4, in.2

7, in. jv*, in.3

1-*(8)'

=-100.5312

3.3953 -341.33

2*(12)2

=226.192

5.0930 1151.99

I 125.659 810.66

Then r2y^ 8 10.66 in.

3

ZA 125.66 in.2

or Y = 6.45 in. 4

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. M? part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

551

50 nun

15 mull

PROBLEM 5.12

Parabola J-r

Vertex \.-"

*'

-1 80mm *-


X

SOLUTION

»<oOwvw

a 15m«*i

40vav^

A, mm2x, mm _>>, mm X/4, mm3

yA, mm3

1. (15)(80) = 1200 40 . 7.5 48xl03 9xl03

2 i(50)(80)~ 1333.33 60 30 80xI03 40xl03

I 2533.3 128xI03 49xl03

Then

X (2533.3) = 128xlO3

YA^ZyA

7(2533.3) = 49xlO3

^ = 50.5 mm ^

F = 19.34 mm <

PROPRIETARY MATERIAL. & 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,




552

20;

U

1

—

~"\

x

i

mm

1 f x=ky*20 nun

I

*— 30 mm-*X

PROBLEM 5.13


SOLUTION

i .

4*y$+***\

^fflffl2

x, mm ^, mm xA, mm3yA, mm3

1 1x30x20 = 2003

9 15 1800 3000

2 ™(30)2 = 706.864

12.7324 32.7324 9000.0 23137

X 906.86 10800 26137

Then X

Y

ZxA 10800

ZA 906.86

2yA_ 26137

£>4~906.86

X = 1.1.91 mm 4!

F- 28.8 mm ^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, inc. All rights reserved. Ato /w/ ofrftft Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.

553

</ PROBLEM 5.14

"" Locate the centroid of the plane area shown.

^0 =4*2

20 hi. >f/ ^ \

* 20 in.»

X

SOLUTION

Dimensions in in.

*1

©•»+

Aio)

|M»>Btt -no)\o>

/*, in.2

x,in. yM- j/4, in3

yA, in.3

I |x(20X20) = 552 12 7.5 3200 2000

2 ZL(20)(20)^ 15 6.0 -2000 -800

£400

31200 1200

Then X ZxA 1200

Y.A f 400^

I 3 J

XyA_ 1200

XA ( 400^

* = 9.00 in. ^

F = 9.00 in. ^

I 3

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554

y PROBLEM 5.15

r^Vertex

\ ^ParabolaLocate the centroid of the plane area shown.

60 nm

\

60 11m\4xfi§§

*——75 mm *\

SOLUTION

Then

and

iWvn

(fWs.fcA')vwv*\

© \ jb )WM

/4, mm2x, mm j, mm xA, mm3 —— "^

jM, mm"

1 -(75)(120) = 6000 28.125 48 168750 288000

2 —(75)(60) = -2250 25 20 -56250 -45000

I 3750 112500 243000

XT,A=I,xA

X(3750 mm 2) = 1 12500 mm3

YXA = 'LyA

7(3750 mm2) = 243000

or X- 30.0 mm <

or 7 = 64.8 mm "4



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555

y PROBLEM 5,16

Determine thej> coordinate of the centroid of the shaded area in

terms of r\, r2, and a.

«r~*\ z^xk~T«i

SOLUTION

First, determine the location of the centroid.

From Figure 5.8A:_ 2 s *n (f

~

ff)

A,/r

ff

cos a

3'(f-«)

Similarly

Then

cos a

3'(f-«)4

#ff r,

__ . 2 cos aLy/4 =—r

2

/r

3'(f-«)

-(r2

3-r,

3)cosOf

a r2

cos«

. 3'(i-«)

ft I 7

and

Now

Yft

a

YIA = ZyA

-{$ - Oleosa K =2 cos or

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556

« \ 1

PROBLEM 5.17

Show that as r}approaches r2, the location of the centroid

approaches that for an arc ofcircle of radius (/] +r,)/2.

SOLUTION

U®

First, determine the location ofthe centroid.

2 sin(f- or)

From Figure 5.8A:

Similarly

*"3"(f-tf)

2 cos

«

32

(f-«)

_ 2 cos#^i=r'i

3l

(f-<*)

x T

A

«,

71 a

4=l|-ak 2

Jl' Hi

^TTx>H.

Then_„ . 2 cos

#

ZyA-—r3 (f-«).U

j

3)cos

— a IP

2f 3

2 cos or•r,3 -(!-").

#or /;

and

Now

"=if-«W-f-«k>T

ff ll/^-r,2

YZA = ZyA

71

a (ji->n \r2~

}"? 'Cos ff

F=23

f ,.3 „3 "\

'2 -n2 2

2 cos a7c-2a

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557

PROBLEM 5.17 (Continued)

Using Figure 5.8B, Y of an arc of radius -~(t\ +r2 ) is

y =

1 sin(f-a)=— (r, + r7 )2Vl 2;

(f-or)

1 . . COS«f= —(?•, +7%)2' ! 2

'(f-a)(1)

Now3

r2 -n

a

fra-'i)('2 +'i'2 + 'i

2

)

(f2~ r\)if2 +f\)

Let >2=

r, =

= r + A

~r-A

Then /* =

and

.3>2 -I

3

._(r + A)2 + (i- + A)(r - A)(r - A)

2

1r2 -n

2(r + A) + (r~A)

3r2 +A2

2r

In the limit as A -- (i.e., t] - r2 ), then

-I2

3

2

=—x— (r, +r2 )2 2

So that F =2 3, .cosa

=—x— (k +r7 )3 4

V) 2~\~a

- cosa .

or Y={rx+r

2 )-~- A

K~2a

Which agrees with Equation CD-

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558

/':

PROBLEM 5.18

For the area shown, determine the ratio alb for which x - y.

SOLUTION

Then

or

or

Now

X

XZA^XxA

X\ ~ab6

£b12

XJ

a

rLA^yA

Y ahab

l

15

5

X = Y=>1 2.—a~~b2 5

4 X V jM y^

1 loft3

3

8

1*5

a2b

4

lab2

5

22

1

—a3 3 6 3

E — a/;

6 12

£*115

a 4or — -— -^

/? 5

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559

y

'*» = .12 in.

\is

/ Bl _

PROBLEM 5.19

For the semiannular area of Problem 5.11, determine the ratio r2/t'\ so

thaty = 3/]/4.

X

SOLUTION

Then

or

Let

(Q

-vn

/T Y ^

]

2l

3/r 3!

271 o

22

4^3^- 3

2

X %4-t)f('

2W)

YZA^ZyA.

3 7Fi^-^K^-n3

)

9#

16

^'i>

P

—[(/> + l)(p-l)]==(/>~l)(/>2+ /> + !)

16

or 1 6/72 + (1.6- 9^)p + (1 6 - 9/r) =





560


(1 6 - 9/c)± V(l 6 -9nf - 4(1 6)(1 6 - 9n)Then p

2(16)

or p = -0.5726

p = 1 .3397

Taking the positive root — = 1 .340 ^

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561

-300 mm - 12 mm PROBLEM 5.20

W0&60 mm I

""'11!

1.2 mm —*\

12 mm

450 minc

4 I Ij

X

1.2 mm{«) (/>)

A composite beam is constructed by bolting four

plates to four 60 x 60 x 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.

SOLUTION

<e>

22C*w*

From the problem statement: F is proportional to Qx.

Therefore:F„

(Qx )a <Q*)b

£-, or FB J^hLFA

For the first moments: (a). -I 225 + ^1(300x12)

12

831600mnr

(Qxh=(Qx)A + 2\225~ (48xl2)+ 2(225-30)(I.2x60)

= 1364688 mm3

Then F„1364688

831600(280 N) or F„=459N <

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562

</

." ;^k

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 A]and A

2. Determine the first

moment of each component area with respect to the x axis, and explain the

x results obtained.

/ .

1

/"''

7.5 in. .\j

/ ~_-

'_ _ -/£ \

4.5 in. 4.5 in.

SOLUTION

5 ,

M

Sm.

L rk>yn

Note that

Then

and

P* ™ "" <Q \A# ' *l

Qs =*yA

(Az-k

<* ift.

(Q,'h

(Qx%

5.in. x6x5 in."

-x2.5in.lf J-x9x2.5 lin.2

+ | ~-x 2.5 in. x6x2.5 in.

Now Qx =iQx)i+(Qsh=o

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

and Qx = 2,yA = F5^l(J= => g, = 0)

or (fijr),=25.0in.3 ^

or (g^ =-25.0 in.3 ^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AM rights reserved. No part of this Manual may be displayed,




563

1.50 iii.

2.00 in.

4.00 in.

.50 in.

t ,/U!

2.00 in.

0.75 in.

0.75 in.

.1.50 in.

2.00 in.

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 A{and A

2. 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

Then

or

Now

Then

and

A, in.2

Y> in._/ . . 3

y A, in.

I 2(ix2x..5)= 3 0,5 1.5

II 1.5x5.5 = 8.25 2.75 22.6875

III 4.5x2 = 9 6.5 58.5

20.25 82.6875

Y"LA = I,y'A

F'(20.25) = 82.6875

F' = 4.0833 in.

Qx =^y,A

(Or).

(Qx ):

2(5.5-4.0833)in. [(1.5)(5.5-4.0833)]in.

2

<iV*

2L

l! :r *

+[(6.5 - 4.0833) in.][(4.5)(2)]in.2

1

or {Q\ =23.3 in.3 <

2(4.0833 in.) [(1.5)(4.0833)]in/

-[(4.0833 -0.5)in.]x 2 x2xl.5 in. or (fix )2=-23.3in.

3 «

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564


Now Or = (Q,\-K<2,)2=0

This result is expected since x is a centroida! axis (thus F-0)

and a-= XyA == f£i4 (F= 0=>&:-0)

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All lights reserved. M> /wrt <jf rtw Mw«<?/ way Z><? displayed,




565

V

„- -

( 111

}

C X

c;

J

—

/

—*~

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 Oxmaximum, and what is that

maximum value?

SOLUTION

Shaded area:

VA4///97777Z

^(t+y)-yy,

A--=%-^) ca == }H CL

= i(c + ^)[6(c-^)] ^H(fl) Qx ==i%2 "/) «

(b) For2max :

dQ.dy~

-0 or -b(-2y) = Q ;•- = () ^

For y~0: (fir)=

2(&)^fc2 «

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566

30 mm

30 mm

300mm

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.

-240 mm-

SOLUTION

Dimensions in mm

Perimeter of Figure 5.1

xw -J, <=*

ur £.

J-4N;

3oo

fs- h-^<:gti*3/=U

L X y xL t mm2yL, mm"

I 30 15 0.45 xlO3

II 210 105 30 22.05 XlO36.3 xlO3

III 270 210 165 56.7x1.3

44.55 xlO3

IV 30 225 300 6.75x13 9xl03

V 300 240 150 72X.103

45 xlO3

VII 240 120 28.8x13

I 1080 186.3xl03 105.3 xlO 3

XZL^ExL

X(\ 080 mm) = 1 86.3 x 1

3 mm' X = 172.5 mm -4

YZL = ZyL

7(1 080 mm) = 105.3 xlO3 mm 2 r= 97.5 mm A

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AU rights reserved. No pari of this Manual may he displayed,




567

20mm 30 mm

36 nun

,t24 nil!)

PROBLEM 5.25


indicated. Locale the center of gravity of the wire figure thus formed.

SOLUTION

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

corresponding line.

nm A

nm A

1

|\$

/,, mm x, mm y, mm xL, mm2yL, mm2

1 20 10 200

2 24 20 12 480 288

3 30 35 24 1050 720®®4 46.861 35 42 1640.14 1968.16

(l\ V

5 20 1.0 60 200 1200

6 60 30 1800

2 200.86 3570.1 5976.2

Then = £xL 1(200,86) = 3570.1

-ZyL F(200.86) = 5976.2

l = 17.77i

7 = 29.8 i

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed,


distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. Ifyou are a student using this Manual,


568

!/ PROBLEM 5.26

—l2in.-4* 21 in. *

t


indicated. Locate the center of gravity of the wire figure thus formed.

/ 1.5 in.

a-

SOLUTION

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

corresponding line.

.&

©

L, in. x, in. y, in. xL,m.2

yL, in.2

1 33 16.5 544.5

2 15 33 7.5 495 112.5

3 21 22.5 15 472.5 315

4 Vl22 +152=19.2093 6 7.5 115,256 144.070

X 88.209 1627.26 571.57

Then

and

XZL = LxL

X(88.209) = 1627.26

7(88.209) = 571.57

or X = 18.45 in. <

or y= 6.48 in. 4

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed,


distribution to teachers and educatorspermitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,


569

PROBLEM 5.27

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.

20 in.

SOLUTION

First note that because the wire is homogeneous, its center ofgravity will coincide with the centroid of the

corresponding line.

K6 =—(38in.)71

L, in. x, in. y, in. xL, in.2

yL, in.2

1 18 -29 -522

2 16 -20 8 -320 128

3 20 -10 16 -200 320

4 16 8 128

5 38 19 722

6 ^(38) = 119.381 24.192 2888.1

I 227.38 -320 3464.1

Then XliXL -320

ZL 227.38

£7^,^ 3464.1

£/.~227.38

X- -1.407 in. ^

F = 15.23 in. ^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. tfo port o/7/i/a- AAmua/ /«aj> Ae displayed,




570

PROBLEM 5.28

A uniform circular rod of weight 8 lb and radius 10 in. is attached to a pin at Candto the cable AB. Determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION

For quarter circle

(«) +)XMc = 0: W ~ -rr =2r

__2r7t

T^W (81b)

(b) ±^XFx =0: T~Cx=0 5.09lb-C

r=0

+\XFy=0: C

y-W = Q C

y-&\b =

d^Xof/j,

f = 5.0.9 lb ^

C v=5.09 lb—

C, = 8 lb)

C = 9.48 lb ^57.5°-*


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571

PROBLEM 5.29

Member ABCDE is a component of a mobile and is formed from a single

piece ofaluminum tubing. Knowing that the member is supported at Candthat / = 2m, determine the distance d so that portion BCD of the memberis 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 =H

cSo that

Then

ZxL =

d0.75

2cos55< mx(0.75 m)

+ (0.75-d)mx (1.5 m)

+1

>*

(\.5-d)m~\ —x2 mxcos55

or (0.75 + 1.5 + 2)^/ (0.75)'

x(2m) =

cos 55° + (0.75)(1 .5) + 3 or d = 0.739 m <





S72

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 / ofaim DEso 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,

So that

or

HxL =

—sin 20°+ 0.5 sin 35° Imx(0.75 m)

or

+ (0.25 m x sin 35°)x (1 .5 m)

+ |l.0xsin35°-~]mx(/m) =

-0.096193 +(sin35»-j)/ =

(xL),(xL)AB+(xL)m y^jDE

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

Then /2 - 1 . 1 47 1 5/ + 0. 192386 =

1.14715 ±J(-U4715)2-4(0.192386)

or /

or / = 0.204 m

Note that sin 35°-\l > for both values of / so both values are acceptable.

or / = 0.943 m A

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayedreproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it withoutpermission.

573

PROBLEM 5.31

B

W\ W:»>^ Id

?

^The homogeneous wire ^4^C is bent into a semicircular arc and a straight section

as shown and is attached to a hinge at A. Determine the value of 9 for which, the

wire is in equilibrium for the indicated position.

\yCc:C^

SOLUTION

First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through^. Further,

because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding

line. Thus,

So that

Then

or

X =

ZxL = Q

'

'JV

rto^B

rcosO ](/')+ rcos0 \(xr)-0

cos 64

\A-l7t

0.5492.1

or # = 56.7° <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. A'f> part of this Manual may be displayed,


distribution to teachers and educators permittedby McGraw-Hillfor their individualcoursepreparation. Ifyou are a student using this Manual,


574

:

—kb-

/,-

li'

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,

(/?)* = 0.80.

SOLUTION

kkbJ

A y yA

1 ha2

i.

—a3 6

2 ~(kb)h ~h3

-~kbh2

6

£ ~(a-kh) t{a2~~kh

2)

6

Then

or

and

or

YZA = LyA

~{a-kh)2

r

-(a2-Aft

2)

6

«2 - A/?

2

dY 1 -2*% - Mr) - (a2 - *ft

2)(-*)

(1)

dh 3

2h(a-kh)-a2 +kh2 =0

Simplifying Eq. (2) yields

k/r-2ah + a2 ^0

(a-khy

(2)

PROPRIETARY MATERIAL. ® 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,




575


T ,. 2a±J(-2a)

2-4(k)(a

2)

Then h ~ —:

2k

=f[>^]Note that only the negative root is acceptable since h< a. Then

(a) k = 0.10

O.IOL J

(b) k = 0.80

I-Vl-0.8'0.80

or A = 0.5 13a ^

or // = 0.691a ^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual may be displayed,




576

7 \:'::\ f

h!

~ /,

—

B'

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 ~ 2h/3.

SOLUTION

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

y_a2 -kh2

Xa-kh)(1)

2h(a-kh)~a2 +kh2 =0 (2)

Rearranging Eq. (2) (which defines the value of/? which maximizes /) yields

a2 -kh2 =2h{a~kh)

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

f = -—! x2h(a-kh)3(a~kh)

or F = -/? <3





577

y

rj0f$';-\

PROBLEM 5.34

Determine by direct integration the centroid ofthe area shown. Express your

answer in terms ofa and h.

X

SOLUTION

y h

X a

y--h

a

*a == x

ytx~-

1~-y

dA~-= ydx

"{ h \

K-6v

<K,

~Xa

h ^

dx — -~ah2

dx

A~ \"ydx= rJo Jo

xELdA ~ xydx ~\ x\~x

xA = \xEI dA:

ha'

f\ ,\ 1. 2

-h2a

-0

yA ^\y?ELdA -

ah

f

-ha'

2ah -h2a

6

1*

V

f t

V\

0<

x ——a 43

* 3





578

PROBLEM 5.35

Determine by direct integration the centroid of the area shown. Expressyour answer in terms ofa and h.

SOLUTION

At {a, h)

or

or

Now

and

Then

and

)\ : h =-~kaA

k =h

~~

a

2

jv A =~ ma

m-h

a

yEL

d.A

(^+^2)

(y2 ~yO^

hf 2W——(ax— x )ax

h h ,

—x—~xa a

clx

A=\dA^ VJL{ax-x2)dx

\*ElM —(ax-x )dx-~

J IiJliil

a*

a 2 1 3~X XT2 3

_

— —ah6

a 3 1 4—x x3 4rJo

JjO'i + j2X0'2 -*)*] = Ji(

v

2

2 - yf)dx

=~alh

12

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. A'o part of this Manual may be displayedreproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual course, preparation. Ifyou are a student using this Manual,you are using it without permission.

579


xA~ \xFl dA: x -ah.6 j

=-a2h

1.2

x =—a ^12

yA^pEL d.A\ y —ah1 /2— —-aft15 5



distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,


580

il 7Wtr\

PROBLEM 5.36

Determine by direct integration the centroid of the area shown. Express youranswer in terms ofa and h.

SOLUTION

For the element (EL) shown

At

Then

Now

Then

and

Hence

x-a, y-h: h = ka2

or k

1/3x =i*y

dA^xcfy = -^-ym dy

xEL =-x2 2 h

m y1/3

\xEldA=l

4 hm

m-ah

-.Wf -fL VWdi,] == l^_f 3^

A'

ft

2A»*1^ ^J-I^-ll*

J^-f^y°*l=^TO

X// \xIiLdA\

hm \l

x\ —ah \=—

a

2/?

4 J JO

-ah2

7

yA ~\yEidA '- yi'| --«/?\= :-ah'

—

*

**.u

—

rNX\*\\\

i

mi

*~* -» *

10

X = — tf

7

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. A'o/ra/-/ f>/7//w M»»/a/ iimiv /« displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

581

a- b-

PROBLEM 5.37

Determine by direct integration the cenlroid of the area shown.

SOLUTION


and

Then

To integrate, let

Then.

and

h r~2 i

a

dA^(b-y)dx

=ta

JCe-i =X

\a-4a2 -x2

}

dx

-dx

Mel.

V£L

1

yi:i^-(y,+b)

a + yj

la

2 ..2a ~x

"b

o aa~\a -x" \dx

x = a sin 9: *Ja2 - x2 = a cos 6, dx~a cosOdd

pan f}A= —{a- acos 9){a cos $d$)

Jo a

a sin - a — + sin

—

/r/2

£?6 1

tt

K^ = r a -4

a

2 -x2)dx

—*-+-(« -x )

2\3/2

/r/2

!*>»

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Ail rights reserved. No part of this Maiiual may be displayed,




SS2

PROBLEM 5,37 (Continued)

f^-ri(« +^[£(-^*2 ( 3^

a:

2a

6

2^v3

;

xA = \xELdA: x

yA=jyELcM: y

ah

ab\ 1

7t

71

1 2a'h

1 ,2ab'

— 2a .

or x = ^3(4-^)

2/?or y = ^

3(4 -/r)

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All lights reserved. JVe port o/fAfo Manual may be. displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individualcoursepreparation. Ifyou are a student using this Manual,you are using it withoutpermission.

583

,/\

PROBLEM 5.38

Determine by direct integration the centroid of the area shown.

: ss

SOLUTION

First note that symmetry implies A


2r

Then

and

yE[=— (Figure 5.8B)n

dA - fcrdr

A- \dA~|

rcrdr-K

"2 2/

( 2\r

,2

,

It 2 2r2 ~ r

\

\n,,A-\l^rdr) =2\f

--(2/ 3

h ~n

So yA~\yEiM'- y -(^-'f) = -^-n» or ^^—— ^


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584

\

2"\

a \a

2

*S

f\

- f

... Hit-. ! 2

PROBLEM 5.39


SOLUTION

First note that, symmetry implies

Then

and

'"Icix

dA ~adx

xKL =x

dA~—a(adf)

XvEL -OCOS0

A = \dA- adx- —aJ Jo J-a 2

= a[x]Z-j[<pra ^az(\~a)

C— Ca ta 2 (1-1

\xF/ dA= x(adx')- —acosm —a d<f>J Jo J-a 3 1^2

2 3

31 I 2 .

a sina2 3

xA - \xELdA : x[a2(1 - a)] ~ a3 l__2

2 3-sin a

<

_ 3-4sincr .

or x = a -4

6(1- a)


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distribution to teachers and educators permittedby McGraw-Hill for their individual course preparation. Ifyou are. a student using this Manual,you are using it without permission.

585

y = kix-(ir-

PROBLEM 5.40

Determine by direct integration the centroid of the area shown. Express

your answer in terms ofa and b.

SOLUTION

At

Then

Now

0, y = b

k(0-a) 2or k.

x

b

v = \ix-a?

y hf ^2- =—(x-a) -X- M«L

and cIA = ydx = —— (x — a)'dx

Then

and

A-fr-fa-faMx-af] =~abo 3

K^r —{x-d)2dx

a af x

3 - lax2 + a

2x)dx

a

bix* 2 3 ^ a 2 1 2,

21 4 3 2 12

K clAJo 2a

2

1 ,2— aft

-^(Jc-a)2^

2a'7*-*

-0

Hence xA —

vA

—ah3 J 12

'aIX/^dA: x

\ymdA: y[~abU-^abJ\ A 1 .2

4

' 10

PROPRIETARY MATERIAL. © 20 JO The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual may be displayed,




586

Vi =h^

PROBLEM 5.41



SOLUTION

yl-k

]x2

but b-kxa

2y

l=~x2

y2 -k2x4

but b~k2a4 y2

dA = (y2 -yx)dx

b .2 *4\

dx

XEL ~ X

yEL^-iyi+yi)

h^v '

t4\

X — dx

\-,M=[

4 6X X

4 6a"

12-a

lb



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587


jy^A=[

xA = xF, dA : x !— ha

,

JLL

\ 15 J 12a

lh x =—a

yA=\y£i.dA: y

r 2,\ 2, 2

15ba

45aff y^-b <

3





588

A

^-i +i3

PROBLEM 5.42

Determine by direct integration the centroid ofthe area shown.

SOLUTION

We have

Then

and

/'*/"*""""

(

y*-°V"l.

dA = ydx = a

.2 \X X

I + ^r

f 2\X Xt/x

8

2L f X X^a\ 1 +— dx — a

3-ah

K^ = rf

2>X X

l- + dx ~a^

L 1/j

x2

x3—+—

~

2/. 3.L2

-.2/.

2 3.1

+4/.

2

n2A

«/;

a*

^6L

Hence x/1

ril-2:: + 3-

r 2\. X X

c/x

3 4^

z? /4

,

cfir

(3

T,.2 .3 4 5X X X X

x 4.__ +

n2/.

I 7/ 2Ij St

5

\xELdA: -aL

yA ^\yr-iM- y\-

3

11

x=~L 44

_ 33 .

y =—a ^40

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. Afo /««•/ o/'rfe Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it withoutpermission.

58<)

x = kij*

%-"""".'

b,2

b2

a6. "

X

PROBLEM 5.43

Determine by direct integration the centroid of the area, shown. Express


SOLUTION

For_y2 at

Then

Now

and for 0<x

x~a,y~b: a — kb or k —

si- &[ m"~ -A-

yELZ^^t x

2 2

1/2

For

dA = y2dx — b'—r^dx4a

x^a: v\

( , b '* 1 x»2 ^

a 2

Then

v

( .J/2

dA = (y2 — y{)dx — bx X 1_—+_a a 2

C pel/2 x r« ( V

A= \dA =Jb^dx +

Jb -

J Jo Jn Jall

dx

1/2X 1„ +_

a a 2

\

dx

.3/2

3

all

+ b

2_b_

+ bl-

3/2

.3/22 r"

z xz1+_x

3 V« 2a 2

2 12

2a(*

2)

1124

a/.)

3/2

-a/2

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distribution to teachers andeducatorspermittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,


590


and [xFl dA~ \"l2

x\b~chU f .

J J Jn Ja/2 yfa a 2j

dx

-ia/2

+ b2 x x x

+

2 b

5*fa"

+b\—l

-

240

f ^ 5/2

K^-J

5 V« 3a 4

/ \5/2

+ (a)5/2

003 «V 1

+—) 4

all

(*-if

/>*"-£,/2/, x'/2

x"2

b—f=rdx

ca b\ x 1 x+ + -•>«/2 2 a 2 ^

1/2 \

Ja a 2

b2\\ >'

on/>2 f

—XT +—2a 2 2

V

x2

1 (x

2a 3a \ a 2

dx

,3\

all

b_

4a

' a)

> a

6a{ 2 2

Hence xA

-afr48

^,,:i dA:13

/>!7l

2,— a/> = a b24 J 240

*=—a = 0.546a ^130

yA^ jyELdA: y\~ab11

/2—ao48

y=—6 = 0.423/; ^26

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. A'o />«/•/ o/rAra A/<wn«rf maj' be displayed,



5*)1

PROBLEM 5.44



SOLUTION

For V] at

Then

By observation

Now

and for 0S.v<o:

For a^x^2a:

Then

and

a, y — 2b 2b — ka or k

lb

2b

a

Mi

i

•dx.

A

fes^K

J;2 --(x + 2b) = b\2~a \ a j

x*EI.

ysL

}>EL

— y2' !

a—x and dA — yv

dx ——rx dx

—y2=—\ 2——

I and dA= y2dx — b\ 2—- \dx2 2 { a

}

""

\ a

A =\dA=&d**t bV dx

26

a"+ b 2—

-\2a

-0

ab

2b_

*>

a

I2

M-«?^hD4

+ 6

/;| 2--|rfx

n2«

3a-0

a2b + b\[(2a)

2-(a)

2

] + ~-[(2a2)-(a)

72/~a b

6

'**».





592


jy£L<M = rv ~2b 2"

~x ax_a~ *f!H) cix

2/,2 5

X

5

"b1

+—o

2 31 a J

2d

17 2=—ab30

Hence xA = Jxad* J\~aby~a 2h x = a A

yA == \yiii.dA: y\rab \

17 2=—ab30 - 35

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Alt rights reserved. No part of (his Manual may be displayed,


distribution to teachers and educatorspermitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

593

PROBLEM 5.45

A homogeneous wire is bent into the shape shown. Determine by direct

integration the* 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

Where

Then

xEI- a cos

3and dL — ^jdx

2 + dy2

x~a cos3$: dx = ~3a cos

2sin 9d9

y-asur' 9: dy - 3a sin2 9 cos Od'6

dL = [(-3a cos^ 9 sin 0d6f + (3a sinz 9 cos OdGf ]

= 3a cos 9 sin 0(cos2 + sin

20)md9

2 -,1/2

3a cos 9 sin Odd

L - \d'L - 3a cos 9 sin 9d9 ~ 3a

3

1 9-sin

2

2

nil

-0

andf- C*

n3

xELdL— \ a cos~ (3a cos sin 0d0)

3a2

-\7ll2

cos5

-0

-la*5

Hence

Alternative Solution

xL = \xE/ dL:-I

3x\ —a

2

x = acos~ 9=>cos':

y = asm 9^>sm' V"J

2/3

x =—a ^5

x2/3 / x2/3

^ +U =1 or y~(am -xmf2

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, inc. AH rights reserved. No part of this Manual may be displayed,


distribution to teachers andeducators permitted by McGraw-Hillfor their individualcoursepreparation. Ifyou are a student using this Manual,


594


Then

Now

dx

*s=*

and *-Hidx J\ + [(a

m - xmf\~x-m)]

2r dx

Then p-r1/3

1/3dx~a If"

2

and

Hence5

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595

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

Then

and

xEl

~rcos$ and di~rdd

r el*'4 -, /, ^

Thus

•7/T/4

\xELdL= r cos 0(rdO)

rising nIA

2 i_

= -r2V2

xL- YxdL: x —itr -~r2v2

_ 2V2

3tf

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rigiiis reserved. No part of this Manual may be displayed,




596

y = kx~PROBLEM 5.47*

A homogeneous wire is bent into the shape shown. Determine by

direct integration the a- coordinate of its centroid. Express your

answer in terms of a.

SOLUTION

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

corresponding line.

We have at

Then

and

Now

and

Then

and

x = a, y = a

a - ka~ " or k

1 3/2

-4a

dy = 3

dx 2^a

1/2

dL~J\ +

1 +

1

\dx j

dx

f3 V-> .1/2

X

1/2

dx

24aa + 9x dx

L= |rfL = r'-^~^a + 9xdxJ Jo l4a

\

24a

a

27

1.4397 k

-x-(4a + 9xf2

3 9V }

[(13)3/2

-8]

M-f yl4a + 9xdx



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597

PROBLEM 5.47* (Continued)

Use integration by parts with

u — x dv~ \}4a + 9x dx

da ~dx v -—(4a + 9x)27

3/2

Then KdL xx—(4a +9xf2

27 -0{

l

'~(4a + 9x)i/2dx]

Jo 27

(4a + 9x)5/2(i3)

3/2

al 1_

27 ° 27>/al45

27j(t3)

3/2~^[(13)

5/2 -32]

= 0.78566a2

xL = \xELdL : x(\ .4397 la) = 0.78566a2

or x = 0.546a A





598

T"XX

l.--'.--.V:0;. \

-« H-< »

L L

PROBLEM 5.48*


SOLUTION

We have

and

Then

and

XEL= X

I a kxy ~—y-— COSLL

2 2 21,

KXdA = ydx = a cos

—

dx21

./4 = IcW = <?cos-

—

dxJ Jo ?/.

21 . TtX—-sin—K 21

112

4iaL

K

jxELdA =J.x| acos-^-dx

2L

^dx-


Then

TtXu — x dv — cos—dx

21,

du = dx2L . kx

v~—sin—it 2L

r„ TVx 7 2L . kx c2L , kx,

x cos

—

dx -— x sin I— sin

—

dx2L K

21

2L J k 21

. KX 21 KXx sin 1 cos—

k \ 2L k 2L

A

Lv£7 <i4 ~aUK

21

. KX 2L KXx sin— H cos

—

21, n 2L

L/2

a-TC

O.I06374«L-

2>/2 K

2L_

K

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599


Also \yKI M-[

xA

112 a Ttx—cos—2 2L{ 21,

....

.

JIXcos— a cos—ax

!.!2

4 In

0.20458^2I

K(

dA : xs

ah

v * j

yA~ jygLdA:4i

71

ah

-0.106374a/:

0.20458A

or x= 0.2361 A

or y - 0.454a ^

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600

PROBLEM 5.49*


SOLUTION

We have2 2

xFL -~rcosO=— ae° cos 6

2 2V =

—

r sin $— ~-ae° sin 6BL3 3

and

Then

dA = ~{r){rd&) = -a2e26d0

A M?n]~a2

eldd0^a2

2 2—e2

2ff

-0

a2{e

2*-\)

and

= 133.623«2

r*2jxELdA^ j*-aed cos&\ ~a 2

e2Od0

rJoia3 few cos&/6>3

Then

Now let

To proceed, use integration by parts, with

u~ew and du = 3ewd0

dv - eos0d6 and v = sin

\ew

cosOdO = ew

sin - (sin 0(3ewd0)

u = ew

then du = 3ewd0

dv = sin &/#, then v = - cos

Then j"e3<?sin&/0 = e

3(?sin 6- 3\-e

wcos#~ |(-cos<9)(3e

3^)

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601


cw

So that Lw cos0d0 = (sin0 + 3cos0)1.0

\xeiM As3

\ew T— (sin0 + 3cos0)10

T

>

= £_(_3e3* -3) = -1239.26a3

30

Also \yEiM -= l^-a/sinAJo 3

(L2ewd&)

,2 J

= -U3 ("V* sin ft/03 Jo

Using integ ation by parts, as above, with

u -

dv--

= e3* and du = 3e

itfd0

- \m\6dd and v = -cos#

Then \ew

sin 0cW --= -e3* cos - |{- cos 0)(3ew

</0)

(?So that \e

3esin0d0--= (~cos# + 3sin#)

10

i J0-i/T

\1eiM '-=-U3

3(-cos0 + 3sin0)

10-0

3

=— (e3*+l) = 413.09a

3

30

Hence xA - \xEl dA: 1 (133.623a2)= -1239.26a

3or x = -921a ^

yA = \yELdA : y{\ 33 .623a2) = 4 13 .09a

3or y = 3.09a ^

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602

!/ = <!-£)

PROBLEM 5.50

Determine the centroid of the area shown when a - 2 in.

SOLUTIONi

We have %/, :~x

Pa'-~y-~\ l—

2^ 2l xj^\^ H

and aW == j>c/x = 1— Lie

Then ^ ==J^.

=f(l~l]| = [-Inxr = (*--In a -1) in.2

and jxELdA =f*HH -r 9 ~|

X"

X 2a +—

2J

in.3

J3 tflM =

Ji 2^ xj[HH4f(1-2

X+±)*

I

"2 x - 2 In x—x_

= — a~2\x\a41

in.3

xA = f-r „ - 4-fl + i-= A/^/j: x- " in.

•» « - In a -

1

y.A =r __ a — 2\na--\yliL dA>. y~ "in.

•* 2(a- In a-!)

Find: xand y when a-= 2 in.

We have__i(2)2 -2 + l

2 - In 2 -

1

or x = 1.629 in. ^

and_ 2-2in2-J-;

2(2-ln2-1)or y = 0.1 853 in. <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual

,


603

.«/

y-o-|>

PROBLEM 5.51

Determine the value ofa for which the ratio xly is 9.

/

.1 iii. —

«

»

X

SOLUTION

We have

and

Then

and

dA = ydx — \\-~\dx

-d* Met-

= («-ln a-l)in.

IUtxj 2

2

[x-lnx]?

M<=r

2

[(4)<s?x =

2-x

_

-a +—2>

in.3

J>e/^ = r

x- 2 In x1

2

' a -Una21 a

in.

xvif= \xELdA:

yA ~ yF, dA :

' y Lh2(a-lna-l)

.9 - « + -J

2 2——.

—

-a-m.a-Ina-1

a - 2 In a - -'

in.

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604


Find: a so that ~ = 9y

We haveX _ JA _ \

XElM

y fA \yELdA

Then\a2

~a-\-\

!(«-2ln«-±)

or «3-]\a

2 +a+ 1 8a In # + 9 =

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

a~\.90 1 in. and a

~

3.74 in. <

PROPRIETARY MATERIAL. CO 2010 The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.

60S

30 mm

30 mm

.300mm

-240mm-

PROBLEM 5.52

Determine the volume and the surface area of the solid obtained

by rotating the area ofProblem 5.1 about (a) the line x = 240 mm,(b) they axis.

PROBLEM 5.1 Locate the centroid of the plane area shown.

•/.- 240 v^w,"

3>Ow\trfl

©

1 <D

i ar

(D

100 yy\*v\

24o **w»

SOLUTION7

From the solution to Problem 5.1 we have

/U 15.3 x10smm2

ZxA = 2.6865 xlO6mm3

Xytf = 1.4445xl06mm3

Applying the theorems of Pappus-Guldinus we have

{a) Rotation about the line x = 240 mm <2)

Volume = 2^(240 ~x)A

=-2x(24QA~I,xA)

= 2^240(1 5.3 x to3) - 2.6865 x 1

6]

Area = 2nXXnJ, = 2f[X(x

liiK)L

- 27r(x, l^ + XjL3+ x4L4 + x

5L

5+ x6L6 )

Where x, ,. . . yx6 are measured with respect to line x = 240 mm.

Area = 2*{(120)(240) + (1 5)(30) + (30)(270)

+(135)(2.10) + (240)(30)]

(b) Rotation about they axis

Volume = 2xXare.AA = 2x(LxA)

= 2^(2.6865x10*mm3

)

Volume = 6. 19xl06mm3 <

Area = 458 x 103mm 2 <

Volume = 1 6.88 x 16mm3 ^

Area = IxX^L = 2nZ(xxinc

)L

— 2x(xlL

l-i-

x

2L2+ x\l^ + x4L4 + x5L$)

= 2/r[(120)(240) + (240)(300)

+ (225)(30) + (2 1 0)(270) + (1 05)(2 1 0)] Area = 1.171x10*mm2 A





606

20 mm 30 mm

36 nmi

24 tut i)

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 Hne>» = 60 mm, (b) they axis.


SOLUTION


4 = 1740 mm2

ZxA = 28200 mm3

ZyA = 55440 mm3


(a) Rotation about the line y - 60 mm

Volume - 2^(60- y)A

= 2jt(6(M -ZyA)

= 27T[60(1 740) -55440]

Area =2 inc

-27tZ{y]mc )L

= 2^(7, l^ + y2L2 + y3Lz +y4L4 +y6L6 )

Where y{,...,y

(iare measured with respect to line y = 60 mm.

20»n*rt -jow^

"34>rv«f*\

24 ww*

Volume = 308x10Jmnr ^3__3

Area - 2/r (60X20) + (48)(24) + (36X30) + (1 8)7(30)2+(36)

2 + (30)(60)

Area = 38.2 x 13mm 2 <


Volume = 2;rXareaA = 2^(Ijetf) = 2tt(28200 mm3

) Volume = 1 77.2 x 16mm3 <

Area = 27t-XVuKL = 2xZ(xMti)L = 2n(xxL\ + x2i2 + *,£, + x4l4 + x5L5 )

(1 0X20) + (20X24) + (35X30) + (35)V(30)2+(36)

2 + (1 0X20)In

Area = 22.4 xlO3mm2 <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

607

y PROBLEM 5.54

Determine the volume and the surface area of the solid obtained by rotating the

r-.-'i.-im

KX

1

20 in.

area of Problem 5.8 about (a) the x axis, (b) the v axis.

30 in.


-— 30 in.

—

-A

SOLUTION


^1146.57 in.2

XE4 = 1.4147.0 in.3

YyA = 26897 in?


(a) Rotation about the x axis:

Volume = 2nYavmA = InTyA

= 2^(26897 in.3

)

Area = 2KYlineA

=2^20WM= 2x(y2L2

+y^ +y4L4+ y5L5

+ y6L6 )

= 24(7.5)(15) + (30)(^x 1 5) + (47.5)(5)

+(50)(30) + (25)(50)]


Volume - 2n:XareaA = 2k'ZxA

= 2^(14147.0 in3)

Area - 2xXlineL = 2/rX(x//W )/,

= 2/r(x,L, + x2l2 + Xj-Lj + x4L4 + x5Z5 )

or Volume = 1 69.0x1Jin.

5 <

3 • 2or Area = 28.4 xlOMn/ ^

or Volume = 88.9x 13in/ 4

2;r (1 5)(30) + (30)(1 5) + [ 30-^^V 71 j

or

(;rxl5) + (30)(5) + (15)(30)

Area = 1 5.48 xlO3in.

2 A

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608

PROBLEM 5.55

Determine the volume of the solid generated by rotating the parabolic

area shown about (a) the x axis, (/;) the axis AA'.

SOLUTION

First, from Figure 5.8a we have A — —ah3

_ 2.

5

Applying the second theorem ofPappus-Guldinus we have

(a) Rotation about the x axis:

Volume = 2iiyA

= 2/r

.5 j

-ah

(b) Rotation about the line AA':

Volume = 2n:{2a)A

1k(2ci) - ah

or Volume =

—

7tah2 A

15

or Volume = -

—

na2h ^

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.

609

PROBLEM 5.56

Determine the volume and the surface area of the chain link

shown, which is made from a 6-mm-diameter bar, ifR ~ 10 mmand L = 30 mm.

SOLUTION

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

A--=—d2and C-nd.

4

Also, the semicircular ends ofthe 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--= 2(Fsi(te ) + 2(K

cncl )

- 2(AL) + 2(tuRA)

= 2{L + 7tR)A

or V--= 2[30mm + jr(10mm)] -(6 mm)2

4

= 3470 mm3or V- 3470 mm3 <

For the area A: A-= 2(4idc) + 2(4„d )

*2(CL) + 2(xRC)

= 2(L + xR)C

or A. -= 2[30 mm + #(10 mm)][^(6 mm)]

= 2320 mm 2or A - 2320 mm2 <

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No par! of this Manual may be displayed,




610

PROBLEM 5.57

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

SOLUTION

Following the second theorem of Pappus-Guldinus, in each case a specific

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 r-W-^gfe 2or V = — fta

3 A

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

We have V = 2nyA = 2n'Aa^

On j

nha or V -—nah A

3

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

We have V = 2nyA - 2lt\ -a)(~ah

(4) Cone: the generating area is a triangle

We have V-TMyA-lJCr a^

v j y

-ha

or V ~—Kc?h2

or V-—tca2h A

3

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

611

PROBLEM 5.58

A| -in.-diameter hole is drilled in a piece of l-in.-thick steel; the hole

is then countersunk as shown. Determine the volume of steel removed

during the countersinking process.

SOLUTION

The required volume can be generated by rotating the area shown about the y axis. Applying the second

theorem of Pappus-Guklinus, we have

|— %«? % irt.

—a- X

V - 27TX.A

= 2713 lflY-+- - in. X8 3UJ J

11. 1

.

—x— m.x— in.

2 4 4V^ 0.0900 in? <





612

PROBLEM 5.59

Determine the capacity, in liters, of the punch bowlshown ifjR = 250 mm.

SOLUTION

The volume can be generated by rotating the triangle and circular sector shown about they axis. Applying thesecond theorem ofPappus-Guldinus and using Figure 5.8a, we have

V - 2kx A = 2tv"LxA

2^"(x,./ij + x2 A.2 )

a,

In

2k

W|""8

8

3 2

F R+

2 2 2

3 \

+2/?sin30 c

3xf J

KIV

16V3 2^3

kR2

7r{Q.25 m)3

Since 1(T/

0.031883 m

lm3

f = 0.03 1883 m3 x10

3/

lm3K = 31.9/'^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

613

rm -y

3 in.

0.625 in.

0.1.25 in.

m

(«)

40°-^, r = 0.25 in.

0.08 in. \L A,

r\i; 0.375 iii.

3 i». 3 in.

KB^Ksa

(W (*)

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

TllA.

i

01> .v o-n;v

\-—~r——/ ^ ca\x*

SOLUTION

Applying the first theorem of Pappus-Guldinus, the contact area Ac of a belt

is given by:

Ac -TtyL-nHyL

where the individual lengths are the lengths ofthe belt cross section that are

in contact with the pulley.

(a) Ac ^n[2(y1L

l) +y2L2 ]

0.125Y7t\2

2 T0.125 in.

cos20 c+ [(3-0.125) in.](0.625 in.)|

rIN..

So'

\ '//

or

(b) A(: =7[[2(yl

Li )]

or

2k 3-0.08-0.375

2in.

0.375 in.

cos 20°

A. = 8. 10 in.2 <

Ac =6.85 in.2 <

T O

(c) Ac ^K[2{yM]

or

K71 J

0(0,25 in.)]

/L. =7.01 in.2 ^

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ft 1

4

PROBLEM 5.61

The aluminum shade for the small high-intensity lamp shown has a uniform thickness of I mm. Knowing that

the density ofaluminum is 2800 kg/m3

, determine the mass of the shade.

56 nun 32 mm 26 mm r 66 mm

32 mm I 28 mm8 mm

SOLUTION

The mass of the lamp shade is given by

m~ pV — pAl

Where A is the surface area and / 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 = 2xyl = 2/rEyL

= 2x(yA + y2Li + J3L3 + yth)

or

Then

or

A = 2x1 3 mm_ (13mm) +

("13 + 16"

2mm x J(32 mm)' + (3 mm)'

+1 —

*

mm x 7(8 mm)2 + (1 2 mm) ;

+28 + 33)mm x y](28 mm)

2 + (5 mm)3

2 ;

= 2^(84.5 + 466.03 + 3 1 7.29 + 867.5 i)

= 10903.4 mm2

m = pAt

- (2800 kg/m3)(1 0.9034 x 1

0"3m2)(0.00 1 m)

m = 0.0305 kg <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Ali rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

615

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 = (density)K, where Kis the volume. Fcan be generated by

rotating the area A about the x-axis.

From the figure: L = V752 ~12.52 = 73.95 1 m

37.5

tan26c76.8864 mm

asLj-i, =2.9324 mm

$ = sin"1— =: 9.594V

?C****

a

75

26° -9.5941'

28.2030° = 0.1 431 68 rad

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

SI*"** fc^L.- i^^Applying the second theorem of Pappus-Guldinus and using Figure 5.8a, we have

V = l7tyA.^l7tXyA

1

1(2.r«^-^

Seg. A, ram2

y, mm yA , mm3

1 —(76.886)(37.5) = 1441 .61 •1(37.5) = 12.5 18020.1

2 ~tf(75)2 =-805.32

2(75)sina . , „ lc -~M——

—

sin (a + 0) = .15.23033a

-12265.3

3 -—(73.95 1)(1 2.5) = -462. 1

9

-(12.5) = 4.16673

-1925.81

4 -(2.9354)(1 2.5) = -36.693 —(12.5) = 6.252

-229.33

I 3599.7

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yon are using it without permission.

616


Then V = InY.yA.

= 2^(3599.7 mm3

)

= 2261 8 mm3

m — (density)V

= (8470 kg/m3)(22.618xi0"6m3

)

0.191574 kg or m = 0.1916 kg <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without (he prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

617

0.50 in.

1.625 iii>|

0.50 in.

r = 0.1875 m.

V

f£i):£jom

-3,00 in.

-

1.00 in.

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 gallon ofpaint covers 100 ft'.

SOLUTION

The number of gallons ofpaint needed is given by

or

Number of gallons - (Number of pegs)(Surface area of 1 peg)

Number of gallons = 400 As{A

s~ ft

2)

1 gallon

100 ft2

(2 coats)

where As

is the surface area ofone peg. Axcan be generated by rotating the line shown about the x axis.

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

We have w

v® v\q

"?> IN.

sin 2a

or

R = 0.875 in.

0.5

0.875

2^ = 34.850° a-17.425<

A^lnYL^lnXyL

r

/>J i

L, in. y, in. yL, in.2

1 0.25 0.125 0.03125

2 0.5 0.25 0.125

3 0.0625°-25 + 03125

=0.281252

0.0175781

4 3 - 0.875(1 - cos 34.850) - 0, 1 875 - 2.6556 0.3125 0.82988

5 —x 0.1 875 = 0.294522

05 _2X0.1875

=03806371

0.112103

6 2a(0.875)0.875 sin 17.425° . ,„ _co—x sin 1 7.425°

a0.137314

2JL = 1.25312 hv



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618


Theniff

2

As-2^(1.25312 in.

2)x144 in.

2

-0.054678 ft2

Finally Number of gallons = 400x 0.054678

= 21.87 gallons Order 22 gallons A



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(.19

0.50 in.

{ 1.025 in>|. \

'.

0.50 id.

r = 0.1875 in.

::

:$\:r = 0,875 m, -<"'*

I ! LOO in.

-3.00 in.

PROBLEM 5.64

The wooden peg shown is turned from a dowel 1 in. in

diameter and 4 in. 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.

MI /W

-W*-r^T;

% tM. ,

^x

The generating area is next divided into six components as indicated

f\s

a^tu.

£^o.$,ts» m.

sin 2a0.5

0.875

or 2^ = 34.850° a = 17.425°

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

VPEG = ItcYA. = 2nZyA

A, in.2

y, in. y^, in.3

1 0.5x0.25x0.125 0.125 0.015625

2[3- 0.875(1 -cos34.850°)- 0.1875]

x(0.3 125) = 0.829870.15625 0.129667

3 0.1875x0.5x0.9375 0.25 0.023438

4 — (0.1875)2 =-0.027612

4

4X0.1875=042(M2

3x-0.011609

5

6

a(0.875)2

2x0.875sinl7.425° . .„ 4„,~—-xsm 17.425°3a

0.04005

~ (0.875 cos 34.850°)(0.5)= -0. 1 795 1

7

- (0.5) = 0.1 66667 -0.029920

"LyL = 0.1 67252 in.3

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Ato part of (his Manual may be displayed,




620


Then V = 2tt(0, 1 67252 in:4

)

1.05088 in;

Now V(hwel = -(diameter)2(length)

= ^(1 in.)2(4 in.)

= 3.14159 in.3

Then % Waste =^^-x 1 00%^dowel

V, ,-Vdowel ' peg

'dowel

1.05088

3.14.159

xJ00%

x!00% or % Waste = 66.5% A

PROPRIETARY MATERIAL. © 20 JO The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,




621

if =kx'~

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 nradians about they axis. Applying the first theorem of Pappus-Guldinus we have

Now at

and

where

Then

We have

Finally

A — jixL

x - 100 mm, y = 250 mm

250 = *(I00)2

or k = 0.025 mm"1

"FL

dL^A\+\~\ dxdy

dxIkx

dL = 4\ + 4k2x2dx

xL = \xEi dL= Jxl \l\ + 4k

2x2dx

xl.2 2 %3/2

3 4k2

-ilOO

-0

1 1

dL

12 (0.025)"

17543.3 mm 2

A = x(\ 7543,3 mm2)

-{[1 + 4(0.025)2(100)

2f2 - (if2

or A = 55.1x10* mm2 <

PROPRIETARY MATERIAL, © 2010 The McGraw-Hill Companies, Inc. Ail rights reserved. No part of this Manual may be displayed,




022

50 UVil

-A

9 ft

120lM't

;tf

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

tatt

>l.3ft iUw ^

< x y * —t^u8

ISOtW/ft-

i

-—- — no

>>

/

^ CI ft. _>_A < n ft > «> a

Rj = -(1 50 lb/ft) (9 ft) = 675 lb

Rn= -(120 lb/ft) (9 ft) = 5401b

R = Rl+ /?„ = 675 + 540 = 1 2 1. 5 lb

XR = ±XR: ^(1215) = (3)675) + (6)(540) X = 4.3333 ft

(a) R = 12I5 1b| X = 4.33 ft -4

(b) Reactions +)IM(=0: B (9 ft)- (12 1 5 lb) (4.3333 ft) =

B = 585.00 lb B = 5851b)^

+\x,Fy=*0: A + 585.00 lb -121 5 lb =

^ = 630,00 lb A = 6301b{<«

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623

ParabolaVertex PROBLEM 5.67

200 N/m

,

A

soo N/u

4 m

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

Then

or

and

or

(b) Reactions

or

or

/?, = (4 m)(200 N/m) - 800 N

2/?„

- (4 m)(600 N/m) = 1 600 N

IF., -* = -*,-*„

7? = 800 + 1600 = 2400 N

£M, : -^(2400) = -2(800) - 2.5 (1 600)

Xrrlm3

R^2400Ni X = 2.33 m 4

-±*.IF=0: A=0

+>^=0: (4m)5F-||m (2400 N) =

By= 1400 N

+f ZFy = : 4, + 1400 N - 2400 N =

A =1000

N

A = 1000 N | B = 1400Nt^





(.24

G kN/iri

6 in -— 4 in

PROBLEM 5.68

i?. Determine the reactions at the beam supports for the given

~'

'

loading.

SOLUTION

/?,=i(4kN/m)(6m)

= 12 kN

/^=(2kN/m)(.IOm)

= 20kN

4* W-t^/vA 1

2*\ o

*1

>"> y /> j / t / rill a

^ , (at* - - >H 4yv»—

w

+\ZFy^0: ii-12kN-20kN =

A = 32.0kNJ^

+)2A/< =0: MA- (.1 2 kN) (2 m) - (20 kN) (5 m) =

Mj =124.0 k.N-m>

)^

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625

ISO HWI't

In ^1"t ><^r ?. ?..

;a._

-3 ft- •6 ft

"bprf£;

2ft

600 li)/l'i

D

PROBLEM 5.69

Determine the reactions at the beam supports for the given

loading.

SOLUTION

We have Ry

Rm -

= i(3ft)(480 lb/ft) = 720 lb

1 *»fc= -(6ft)(6001b/ft) = 18001b

= (2ft)(600 lb/ft) = 1200 lb

-at

*XB/1 * tn

— t*fc

Then ±-XFs = 0: ^. =

+)ZMB := 0: (2 ft)(720 lb) - (4 ft)(1800 lb)+(6 ft)Cy- (7 ft)(1200 lb) =

or Cy=2360 lb O 2360 lb \<

+J ZF¥== : -720 lb + B - 1 800 lb+ 2360 lb - 1200 lb =

y

or £„=13601b B = l3601b)^

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626

200 lb/ft PROBLEM 5.70

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

SOLUTIONq

R ,- POO Ib/ftHl *> ft> .j^turj i) J. •7'JV*"

,i

'

kr7?, =3000 lb [ J fl

/?„ =-(200 lb/ft) (6 ft) M# ^P~ ?# 4-/^-Js#n =6001b

+)S^=0: -(3000 lb) (1 .5 ft) - (600 lb) (9 ft + 2 ft)+ 5(1 5 ft) =

5 = 740 lb B = 7401b(^

+\ZFy=0: A + 740 lb -3000 lb -600 lb =

A = 2860 lb A = 28601b(^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of (his Manual may be displayed,




627

iOOON/in

iF;;z;=^rfT :ni

3.6 m

PROBLEM 5.71


f 200 .N/m

SOLUTION

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

Uoo^

ttooS

We have

Then

or

or

./?! = -(3.6 m)(2200 N/m) = 3960 N

Rn = (3.6 m)(l 200 N/m) = 4320 N

±^ZFX-0: BX =Q

+)Y,MB = 0: -(3.6 m)Ay+ (2.4 m)(3960 N)

-(I.8m)(4320N)^0

^, =480N

+j £FV. = 0: 480 N - 3960 N + 4320+ By=

« = -840 N

A = 480 N f M

B = 840NH

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this .Manual may be displayed,

reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

628

A

21)0 iivf'tParabolas

tmmmm I/j

PROBLEM 5.72

Determine the reactions at the beam supports for the given

loading.

SOLUTION

t— £(\tl.t\»ito ~m

Hi 5? I*

K1

We have tf, = -(1 2 ft)(200 lb/ft) = 800 ib

K„ = -i (6 ft)(l 00 lb/ft) = 200 lb

Then

+11FV = 0: 4 - 800 lb - 200 lb =

or Ay=1000 lb

+)ZMA- 0: M„ - (3 ft)(800 lb) - (1 6.5 ft)(200 lb) -

A = 1000 lb |^

orMA = 5700 lb • ft M^=57001b-ft

>

)^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. M> par/ o///h's Manual may be displayed,




629

Vertex PROBLEM 5.73

A§jj

Parabola

'

900 N/in

8


SOLUTION

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

We have

Then

or

^lWk« jrWr

.*>m

Niloo &

-*>cx> ta

fcl

R{= (6 m)(300 N/m) = 1 800 N

Rn = -(6 m)(l 200 N/m) = 4800 N

±J£F«0: A-0

+] If; = 0: ,4, + 1800 N - 4800 N =

Ay= 3000 N

+)2M^=0: Af,+(3m)(1800N)-[— m (4800N) =15

or M^=12.6kN-m

A = 3000N(«

M;, =12.6 kN-m")^


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630

ISUON/m

A-f | 000 N/m

wmB

4 m

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

^M-o.1 **\

We have

Then

or

Now

Also

/?, = -(a m) (1 800 N/m) = 900a N

Ru = I [(4 - a)m] (600 N/m) = 300(4- a)N

+| £/;; = 0: Ay- 900a - 300(4 - a) + tf

v=

/(,, + #,.. =1200+ 600a

4, = Bv=> A

v= £

v= 600+ 300a (N)

+)ZMB =Q: ~(4m)/l +

(1)

4--|m [(900a)

N

or

Equating Eqs. (1) and (2)

or

Then

or

Now

(4-a)m 300(4-a)N~| =

4. =400 + 700a -50a'

600+ 300a = 400 + 700a - 50a'

a2 -8a + 4 =

(2)

8±V(-8y-4(l)(4)a-——

2

a = 0.53590 m

a < 4 m =>

a = 7.4641m

a = 0.536 m A





631


(b) We have -±.SF, = 0: AX =Q

Eq.(l) Ay =B,

= 600+ 300(0.53590)

-76.TN A = B = 761Nt^

PROPRIETARY MATERIAL. O 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior mitten permission of the publisher, or used beyond the limited

distribution to teachers and educators permittedby McGraw-Hillfor their individualcourse preparation. Ifyou are a student using this Manual,you are using it without permission.

632

1800 N/mrlhw,-

A

• 4 in

000 N/m

B

PROBLEM 5.75

Determine (a) the distance a so that the reaction at support B is

minimum, (h) the corresponding reactions at the supports.

SOLUTION

(a)

We have

Then

or

Then

(*) Eq.(l)

and

or

<**>£

a- ^<A-<xW

/?, = i(a m)(l 800 N/m) = 900a N

/Jn = I[(4 _ fl)m](600 N/m) = 300(4 - a) N

8 + ^r:)XMA =0: - -m (900a N)- -m [300(4 -a)N] + (4m)

5

>;=0

5^ = 50^-1000 + 800

t/5.

= 100* -100 =

By= 50(1)

2 - 1 00(1) + 800 = 750 N

±+.ZFx=0: Ax =0

+|lFv=0: A- 900(1)N - 300(4 - 1)N + 750 N =

(1)

or a - 1.000 m <

B = 750NJ A

Ay= 1050

N

A = 1050Nf M





633

AX) llj/fi

A g#& B

-4 It -*j-

~:&

12 ft

2 ft

PROBLEM 5.76

iv„

Determine the reactions at the beam supports for the given loading

when <y = 150 lb/ft.

SOLUTION

4So|Kit*

1vso <*

1- vtld. ft it _LPU z ;t

We have A, -~ (1 8 ft)(450 lb/ft) = 4050 lb

/?,r-i(18ft)(1501b/ft) = 13501b

Then +,.SFV

= 0: C,=0

A£*/,:= 0: - (44, 1 00 kip • ft) - (2 ft)- (4050 lb)

~(8 ft)(1350 lb) + (12 ft)Cy=

or Cy= 5250 lb C = 5250 lb J A

+J2F,, = 0: By- 4050 lb - 1 350 lb + 5250 ib =

or By= 150 lb B = 150.0 lb t <




634

45(1 IS>/l!

•M.I kin-It V

411- .1211

2ft

PROBLEM 5.77

Determine (a) the distributed load 0)q at the end D ofthe

beam ABCD for which the reaction at 5 is zero, (6) the

corresponding reaction at C.

SOLUTION

(a)

l&*

<**

We have V= -(18 ft)(450 lb/ft) = 4050 lb

*„:= -(I8ft)(6;()

lb/ft) = 9fi>()lb

Then +)LMC --= : - (44, 1 00 ib • ft) + (1 ft)(4050 lb) + (4 ft)(9^ lb) ==

or aj, =100.0 lb/ft •*

(b) ±JLFX= 0: C

v=0

A^y = 0: -4050 lb - (9 x 100) lb + Cy=

or C, = 4950 lb C = 4950 lb | ^





635

24 kN 30 !<N

a — 0.6 tn 0.3 m

1.8 m

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 coA and (Ob corresponding to equilibrium.

SOLUTION

. CX i /*-*, jj*3fc.|W»L

r i •

30kH

ftV -

8.£————

1

. o.6-»>

^•*•— --*- -,

/?„ = ifife(1.8m) = 0.9fifc

+)XMD =0: (24 kN)(l .2 - a) - (30 kN)(0.3 m) - (0.9^ )(0.6 m) = ( 1

)

For a = 0.6 m: 24(1 .2 - 0.6) - (30)(0.3) - 0.54 (oa=

1 4.4 - 9 _ o.54a)A =0 ^ = 1 0.00 kN/m ^

+\lFy=0: -24 kN - 30 kN + 0.9(10 kN/m) + 0.9% =0 % = 50.0 kN/m ««

PROPRIETARY MATERIAL. €3 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,



636

24 kN

a = 0.6 m

3.8 m

PROBLEM 5.79

For the beam and loading of Problem 5.78, determine (a) the distance a

for which &>,, = 20 kN/m, (6) the corresponding value of (On.

!i 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 6)a and 0)B corresponding to equilibrium.

SOLUTION

14 kN JO tN

_-. O.W*\- ,n

We have

(a)

or

(/>)

/?, =- (1 .8 m)(20 kN/m) = 1 8 kN

J?n= -(1 .8 m){G>H kN/m) = 0.9% kN

+)LMC = 0: (1. .2 - fl)m x 24 kN - 0.6 mx 18 kN - 0.3 mx 30 kN =

+|ZFy = : -24 kN + 1 8 kN + (0.9% ) kN - 30 kN =

a = 0.375 m ^

or coB = 40.0 kN/m. -4

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Afo /wrf o//Am Aftuiwa/ /««>• Ac displayed,




637

* 4.8 in -

Vertex v

I-'iuaholny~'

\A :-

:

-- : -

1/ -.

2.4 m

C

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.

i Br.

SOLUTION

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

p = (7.2 m)(13kg/m3

)(9.8 1 m/s2)

W]=-(4.8 m)(7.2 m)(l m)(2.4xl0

3kg/m

3)(9.81 m/s

2)

= 542.5 kN

W2=-(2.4 m)(7.2 m)(l m)(2.4xl03

kg/m3)(9.81 m/s

2)

= 203.4 kN

1 »3i._/„.3^>P

3= _(2.4 m)(7.2 m)(l m)(10

Jkg/m

J)(9.8 1 mV)

= 84.8kN

p = i /j/7 = 1(7,2 m)(l m)(7.2 m)(103kg/m

3)(9.81 m/s

2)

= 254.3 kN

-±~ ZF¥= 0: tf - 254.3 kN =

+| Lf\, = 0: K - 542.5 - 203.4 - 84.8 =

V = 830.7 kN

H = 254kN~— <

V = 831kN( <

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638


(b)8(4.8 m) = 3 m

Jc2= 4.8 + -(2.4) = 5.6 m

x-, = 4.8 +~ (2.4) -6.4 m

+ )l,MA=0: xV-ZxW + P{2Am) = §

(c) Resultant on face BC

Direct computation:

x(830.7 kN) - (3 m)(542.5 kN) -(5.6 m)(203.4 kN)

- (6.4 m)(84.8 kN) + (2.4 m)(254.3 kN) =

4830.7) - 1 627.5 - 1 1 39.0 - 542.7 + 6 1 0.3 -

x(830.7)- 2698.9 =

p = pgh = (13 kg/m3

)(9.81 m/s2)(7.2 m)

P = 70.63 kN/m2

BC = ^(2A)2+(72)

2

= 7.589 m= 18.43°

R=-PA2

-(70.63 kN/m 2)(7.589 m)(J m)

jc = 3.25 m (To right ofA) <

%2»>

l&Si

R = 268kN "^ 18.43° <

Alternate computation: Use free body of water section BCD.

6i

R = 268 kN ^L 1 8.43°

R = 268kN ^18.43° <





639

8ft PROBLEM 5.81

The cross section of a concrete dam is as shown. For a 1-ft-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 I -ft thick section ofthe dam and the quarter circular section of water above

the dam.

h~^thtt.

Note:

For area 3 first note.

t

r -/ n

21-***ft

3x J

= 12.0873 ft

x2=(21 + 4)ft = 25ft

50-4x21

3tv

41.087 ft

ft

a X

I r2

1

—r2

II7t 7

r-

4

Ar

3>7t

Then .?3=29ft +

(21)(21)2

+(21-fJ)(-fx21;

(21)2-f(2iy

(29 + 4.6907)ft = 33.69 lft

ft

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of litis Manual may be displayed,




640


(a) Now

So that

Also

Then

or

(b) We have

or

or

^=(150 lb/ft3)

71

(21 ft)2(1ft) 51,954 1b

W2= (150 lb/ft

3)[(8 ft)(21 ft)(l ft)] = 25, 200 lb

W3= (1 50 lb/ft

3)

fK4 = (62.4 lb/ft3

)

2\2 ~—x2\ z

|lrx(l ft) 14,196 lb

j(21ft)2(lft) 21,613 1b

p ^~Ap = -[(21 ft)(l ft)][(62.4 lb/ft3)(21 ft)] = 13,759 lb

jUSFv= 0: //-13,7591b = or H = 13.76 kips— •*

+JXFV =0: K- 51,954 lb -25,200 lb -14,1 96 lb -21,613 lb =

K = 112,963 lb V = 113.0 kips f^

+)?.MA = 0: jf(l 1 2,963 lb) - (12.0873 ft)(5 1 ,954 lb) - (25 ft)(25,200 lb)

-(33.69 1 ft)( 3 4, 1 96 lb) - (4 1. .087 ft)(2 1 ,6 1 3 lb)

+(7 ft)( 13,759 lb) =

1 1 2,963x-~ 627,980 - 630,000 -478,280- 888,010 + 96,3 13 =

x = 22.4 ft A(c) Consider water section BCD as the free body

We have £F =

t * \-i,1S9 \b

-R

Then -R = 25.6 kips ^!L 57.5C

or R = 25.6 kips ^57.5° <

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641

T C PROBLEM 5.82

The 3 x 4-m side AB of a tank is hinged at its bottom A and is held in place by a

thin rod EC. 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

Now

Where

Then for d„„„.

P = ~Ap = ~A.(pgd)

A = 3 m

3w\

(3m)(0.2 x200xl03 N)d„

or

or

(4mx</ )x(103kg/m 3

x9.8I m/s2x</ )

1 20 N •m - 6.54d* N/m 2 =

fiL = 2.64 m <


reproduced or distributed In anyform or by any means, without the prior written permission of the publisher, or used beyond the limited



642

T C B

?

3 m

PROBLEM 5.83

The 3 x 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 densityis 1263 kg/m

3. 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.

SOLUTION

Consider the free-body diagram of the side.

We have P^-Ap =~A{pZd)T

[(2.9 m)(4 m)j [(1263 kg/m3)(9.8l m/s

2)(2.9 m)]

Then

= 208.40 kN

+tlF=0: 4=0

/

* A.<v\

i

NA*

2.9

or

or

LM/t- ; (3 m)T - 1

—m (208.4 kN) =

T -67. 151 kN

-±~ IFX= 0: Ax + 208.40 kN - 67. 1 5 1 kN =

Ax =-141.249 kN

T = 67.2kN— 4

A = 14l.2kN— - ^

PROPRIETARY MATERIAL, © 2010 The McGraw-Hill Companies, Inc. Ail rights reserved. No part of (his Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hill for their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

643

5ft

BitI

T

PROBLEM 5.84

The friction force between a 6 x 6-ft 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 1000 lb.

SOLUTION

Consider the free-body diagram of the gate.

"Now

Then

Finally

I\ =-APj =-[(6*6) ft2][(62.4 lb/ft

3)(9 ft)]

= 10,108.8 lb

Pn =-Ap„ =-[(6x6)ft2][(62.4 lb/ft

3)(15 ft)]

= 16848 lb

F = 0.1/> = 0.1(PF + ^)= 0.1(10108.8 + 16848)lb

= 2695.7 lb

+tIF„ =0: T - 2695.7 lb -

1

000 lb = or T -=3.70 kips t <

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644

U.nOi

If (i fi

n

Md-.ui

•V

:il|-

PROBLEM 5.85

A freshwater marsh is drained to the ocean through an automatic tide

gate that is 4 ft wide and 3 ft 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 A = 6 ft, determine the ocean level d for which the gate will

open. (Specific weight of salt water = 64 lb/ft3.)

SOLUTION

Since gate is 4 ft wide

Thus:

y'* a* v/a

w = (4 ft)/? = 4^(depth)

vi', =4y(/7-3)

w2= 4yh

*{ = 4/(</-3)

m>2 = 4/c/

./r-/i=~(3ft)(H{-H.1 )

-(3ft)[4rV-3)-4/l(/?-3)] = 6/(i/-3)-6r(/?-3)

/S-^^-OftX^-Wz)

- -(3 ft)[4/c/ - 4y/?] - 6/rf - 6yA

)£M.(=0: (3 ft)*-(l ft)(/f- />)-(2 ft)(/^ -i|

1) =

5=-(/r-^)--j(^-/ii)

2=~W(d - 3) - 6y(A - 3)3—[6yd - 6yA]

= 2p(<* - 3) - 2^(/? - 3) + 4yd - 4yh

B = 6/(d~\)-6y(h~l)

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,you are using it without permission.

645


With 5 = and h = 6it = 6/(d - 1)- 6y(h - 1)

r

Data: / = 64 lb/ft3

/= 62.4 lb/ft3

</-l = 562.4 lb/ft

3

64 lb/ft3

4.875 ft d = 5.88 ft <





646

PROBLEM 5.86

The dam for a lake is designed to withstand the additional force causedby silt that has settled on the lake bottom. Assuming that silt is

equivalent to a liquid of density ps- 1 .76xl03

kg/m3and 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.

SOLUTION

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

We have Pw =-Apw =- A(pgh)

[(6.6 m)(l m)][(103kg/m

3)(9.81 m/s

2)(6.6 m)]

= 213.66 kN

Next, determine the force on the dam face with silt

We have />t>-[(4.6 m)(l m)][(10

3kg/m3

)(9.8.i m/s2)(4.6 m)]

= 1 03.790 kN

(./>), =[(2.0 m)(i m)][(103kg/m3

)(9.81 m/s2)(4.6 m)]

- 90.252 k'N

(Px )u =~[(2.0 m)(l m)][(1.76xl0

3kg/m 3

)(9.81 m/s2)(2.0 m)]

= 34.53! kN

Then. P' = p;v + (Px ), + (Ps )n

= 228.57 kN

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

%mc.=/> ~ Fw

xH)0%

(228.57-213.66)

213.66

6.9874%

xl00%

T

H.C#,

n\

CPs\ K*^

% inc. = 6.98% A

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducatorspermittedby McGraw-Hill for their individual coursepreparation. Ifyou are a student using this Manual,you are using it without permission.

647

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 ps- 1.76x1

3kg/m

3) 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 Panm on the dam is then:

Pallow = 1 2PW = (1 .5)(2 1 3,66 kN) - 256.39 kN

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

Pw (*-d) *A

«»',/ V (.dim

f-f.WT t?0JL

We have /y = ±[(6.6 - d)m x (1 m)][(l3kg/m

3)(9.8 1 m/s

2)(6.6 - d)m]

= 4,905(6.6 -~cJ)2 kN

(/>), ^\d{\ m)][(103kg/m

3)(9.81m/s

2X6.6-rf)m]

= 9.81(6.6rf-rf2)kN

1 \3 i.„/_3-(Pg )n = -{d(\ m)][(L76xlO

jkg/m")(9.8.i n^){d)m]

= 8.6328d2 kN

P' = I*+{Ps ) l+(P

s )n= [4.905(43.560-13.2000</ + rf

2)

+ 9.8 \(6.6d - d2) + 8.6328tf

2]kN

^[3.7278^2+2i3.66]kN

Now required that/3' = PMlow to determine the maximum value oft/.

(3.7278c/2+213.66)kN = 256.39 kN

or d~ 3.3856m

Finally 3.3856 m = 12xl(r3-^-xNyear

or N = 282 years 4

PROPRIETARY MATERIAL. © 2010 The McGraw-Hiil Companies, Inc. All rights reserved. No part of this Manual may be displayed,




648

0.27 m

T0.45 m

/ -'

J£i t_

/ : w 1WW 0.48 mJf

SSw

|^-0.64 m-*]

PROBLEM 5.88

A 0.5 x 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 whencable BCD is slack.

SOLUTION

First consider the force of the water on the gate.

We have p=—Ap-—A(pgh)I

o 0i4Sm

so that Pj --[(0.5 m)(0.8 m)]x[(103kg/m

3)(9.81 m/s

2)(0.45 m)]

= 882.9 N "^v#^ 1

/>, =-[(0.5 m)(0.8 m)]x[(103kg/m

3)(9.81 m/s

2)(0.93 m)]

= 1824.66 N

Reactions atA and B when 7" =

We have

Jlft^wftyA

1

+) Y,MA = 0: -(0.8 m)(882.9 N) +-(0.8 m)( 1 824.66 N) - (0.8 m)B =

4 ^^P&*"or B -15 10.74 N ^fcV

or B = 15I1N^53.1° <+\ZF = 0: 4 + 1510.74 N- 882.9 N- 1824.66 N ==

or A = 1197N^53.1° ^




649

,D

\f 0.27 m

0.45 mI

A

t

Ki/''

0.4S in

-0.64 m-

PROBLEM 5.89

A 0.5 x 0.8-m gate AB is located at the bottom of a tank filled with

water. The gate is hinged along its top edge/4 and rests on a frictionless

stop at B. Determine the minimum tension required in cable BCD to

open the gate.

SOLUTION

First consider the foi

We have

so that

ce of the water on the gate.

P^-Ap^-A{pgh)

Px

=i[(0.5 m)(0.8 m)]x[(103 kg/m3

)(9.81 m/s2)(0.45 m)]

- 882 9 N

l\ =-[(0.5 m)(0.8 m)]x[(103kg/m

3)(9.81 m/s

2)(0.93 m)]

%

= 1 824.66 N

T to open gate

First note that when the gate begins to open, the reaction at B —'* 0.iiaawrt . i

nm

Then +)lMA= 0: -(0.8 m)(882.9 N)+-(0.8 m)(l 824.66 N)

-(0.45 + 0.27)mxU7 ;

= ^or 235.44+ 973.152 -0.338827* =

or T = 3570N •«

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650

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.

r = 24 in.-

SOLUTION

Consider free body consisting of 20-in. length of the trough and water

/ = 20-in. length of free body

W = yv = y

_ 4r

4

PA =yr

P = -Pa-1 = -{yr)rl = - yr2l

2 * 2 2

+)'LMA =0\ Tr-iVr-p\-r) =

Tr-\y±* 4r

3/r

yr~l r =0

Data:

Then

1 •> 1 -j 1 >

T=z-yr2l+~yr2I = -yr2

!

3 6 2

94 onr = 62.4 lb/ft

3r =— ft = 2ft / = ^ft

12 12

T = 1(62.4 lb/ft3)(2 ft)

2[ —ft

= 208.00 lb r = 208 lb <





(.51

PROBLEM 5.91

3 It

h?

A 4x2-ft gate is hinged at A and is held in position by rod CD,

End D rests against a spring whose constant is 828 lb/ft. 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

Then

and

Assume

We have

Then

sin<9 = ~ 6> = 30°4

xsl>

=(3 ft) tan 30°

F - Ity

= 828Ib/ftx3ftxtan30°

= 1434.14 lb

rf>4ft

p=~AP =~A(rh)

p}=I[(4 ft)(2 ft)]x [(62.4 lb/ft

3)(d - 4)ft]

= 249.6(</-4)ib

pu = I[(4 ft)(2 ft)]x [(62.4 lb/ft3){d ~ 4 + 4 cos 30°)]

i&,

W~4->&

(4«Lt0M3o

= 249.6(J~0.53590°)l.b

For dmin so that gate opens, W —

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

or

or

+)ZM/}=0: -ft [249.6(rf-4)ib] + -ft [249.6(J-0.53590)lb]

-(3ft)(1434.141b) =

(332.8c/ - 133 1 .2) + (665.6J - 356.70) - 4302.4 =

d = 6.00 ft

t/^4ft=> assumption correct rf = 6.00 ft A

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652

1PROBLEM 5.92

Solve Problem 5.91 if the gate weighs 1000 lb.

PROBLEM 5.91 A 4x2-ft gate is hinged, at A and is held in:'•/-

1I,

{ -! 3 (t

4 it( j D position by rod CD. End D rests against a spring whose constant is

j||;J__>--:— ~Sf\ 828 lb/ft. The spring is undeformed when the gate is vertical.

:_:..:,:- :A._. 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 portion2 It

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

Then

and

Assume

We have

Then

sin0 = - <9 = 30°4

jcS7>

=(3ft)tan30°

Fgp = kxSP = 828 lb/ft x 3 ft x tan30c

= 1434.14 lb

</>4ft

P=L Ap =±A(rh)

*sx

[(4 ft)(2 ft)] x [(62.4 lb/ft3){d - 4)ft]

= 249.6(rf-4)lb

pu = I[(4 ft)(2 ft)]x [(62.4 lb/ft3){d - 4 + 4cos30°)]

(d-<n&

Mttkt**?

-249.6(rf-0.53590°)Ib

For dmin so that gate opens, W - 1 000 lb

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

3f)£Af^=0: -ft [249.6(rf -4) lb] + -ft [249.6(^-0.53590) lb]

3

or

or

d > 4 ft

- (3 ft)(1434. 14 lb) - (1 ft)(l 000 Jb) =

(332.&Z- 1 33 1 .2) + (665.6rf - 356.70) - 4302.4 - 1 000 -

d = 7.00 ft

assumption correct J = 7.00 ft «

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653

PROBLEM 5.93

A prismaticaily 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 Cofthe gate. Determine the depth of water d for which the gate will open.

0.40 m

SOLUTION

First note that when the gate is about to open (clockwise rotation is impending), By

—- and the line of

action of the resultant P of the pressure forces passes through the pin aXA. In addition, if it is assumed that the

gate is homogeneous, then its center of gravity C coincides with the centroid ofthe triangular area. Then

a ^— -(0.25 -A)

3 151,3

b" 1.5

and

Now

so that

Simplifying yields

Alternative solution

f- (0.25 -A) ^ 8

-!(0-4)"il(f) ^

289 . iei 70.6

45 12

i(o-^w) -.o.zs,,

(04&-W)r«\

(1)

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

above the gate.

N ow P'^-Ap^-(dx 1 m)(pgd) i {%*)-2 2

/2-pgd1

(N)

W' ^ pgV ~^ pg\-x~dxdx\m

^~pgd2(N)

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654


Then with By- (as explained above), we have

+)2AtA = 0:

3 3U5 j

!>'•)- |-(0.25-A)lM-°Simplifying yields

289, te . 70.6

45 12

as above.

Find d, h = 0.\0m

Substituting into Eq. (1) ^d + 15(QAQ) =~~ or </ = 0.683 m <

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655

: :

n.j

0,75 in

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 friclionless support

at B. Determine the distance h if the gate is to open when d = 0.75 m.

0.40 m

SOLUTION

First note that when the gate is about to open (clockwise rotation is impending), By—* 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 centraid of the triangular area. Then

a =— (0.25 -A)

3 15 1 3and

Now

so that

4 -(0.25 -A) 8

Simplifying yields

Alternative solution

(0.4) -£($) 1:

289 . lc! 70.6

45 12

^{o.nSwO ,o.2,s w

UUS-K)w*

0)

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 P' = - Ap = -{d x 1 m)(pgd) a (ft**)-

= -/>«*/2 (N)

( 1 8W - pgV = pg\ —x—c/xc/xlm^2 1.5

=^P^2(N) -J|lo.+n»)»_

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656


Then with By— (as exphtineid above), we have

+)XMA== 0:

3 31.15 J_{&»*)

- — -(0,25 -A) (V)-°

Simplifying yields289 . 1C ,

70.6

45 12

as above.

Find h, <1 - 0.75 m

Substituting into Eq. (1)289

(0.75) + 15*=70 -6

45 1.2

or ft = 0.0711m <





657

23 hi.;

PROBLEM 5.95

^yim^ A. 55-galIon 23-in.-diameter drum is placed on its side to act as a

,,

' dam in a 30-in.-wide freshwater channel. Knowing that the drum is

. "i 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 drumand 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

Then

Finally

n=~M=~A(yh)

1

2 (SMH>28 6.542 lb

(62.4 lb/ft3)f— ft

^=^p^~A{yh)

3°\xfi^|fl12 12

(62.4 lb/ft3

)

11.5

12

71.6351b

^= rj/= (62.4 lb/ft3)

W2= yV2 =(62A\b/fr)

W3= yV3 =(62.4 lb/ft

3

)

1.5f 2 _x(lL5s2

12

11.5^

12 j

41 12

ft2 +*flU41 12

ft^

fV

30— ft | -30.746 lb12

30,.

12ft = 255.80 lb

41 12

30

12it = 112.525 lb

.:U£/y /ev=(286.542-71.635)lb = 214.911b

+| £/<; : tfv= (-30.746 + 255.80 + 1 1 2.525) lb = 337.58 lb

R = yJR*+R* =400.18 lb

tan<9 = -^-

9 = 51,5" R=4001b ^57.5°^

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658

PROBLEM 5.96

Determine the location of the centroid of the composite body shown

when (a) h = 2b, (b) h = 2.56.

SOLUTION

* ijj dLii*

V X JF

Cylinder I fta2b

2 2

Cone 11

1 2,—itcrh3 4

-Tta h3

f 1 ^

I 4 >

K=/Tfl'| /? + -/2

2 3 12

(a) For A = 26: j/ = ^' /7+-(26)3

na b

XxV - Tta

= jia~b

-b2+~(2b)b +^(2bf

2 3 12V

1 2 1— +—+~2 3 3

2/.2•Tta b

-(5AT - Xx V : ^

I

-Kerb I = -tfo2/)2 X =—b

3 J 2 10

Centroid is -^-6 to left of base of cone A

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may he displayed,




65')


(b) For h = 2.5b: V --= na2b + -(2.5b) = 1.8333^2

/;

ZxV~--Tta -b2+-(2.5b)b +—(2.5b)2

2 3V

12V ;

= 7ua2h2[{).5 + 0.8333 + 0.52083]

= 1.85416jTflV

XV = Z,xV: X(\.&333na2b)*--\.S54\6aa

2b2 X = 1.01 1366

Centroid is 0.01

1

36b to right of base ofcone <4

Note: Centroid is at base ofcone for h -= yf6b=- 2.4496


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660

u PROBLEM 5,97

1

£•- Determine the y coordinate of the centroid of the body shown.

}

X-- -Amli^-f§

M X

SOLUTION

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

We have

Then —a2(4h -3b)

12

YZ.V = XyV

24cr{2fr-3b2

)

V y yV

Cone1 2,

3" 4 12

Cylinder -n\ — % * a,6 =— /rfl b

4 2

1 ?, ?—Kab"8

X —a2(4/? -3/;)

12 24

- 2/?2 -362

or y = ^2(4/2- 3ft)

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661

/

y PROBLEM 5.98

w'^^?^ Determine the z coordinate of the centroid of the body shown. {Hint: Use--^r :.

'".

.

'\ the res u It of Sa raple Prob 1em 5 . 1 3.

)

SOLUTION

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

V z zV

Hall-Cone1 2,

6

a*

71

1 3,—a n6

Half-Cylinder

2'U, 8 371^2)

2a

371

—a3b

12

L ~?~a2(4h-3b) -—a\2h~h)

12 '

From Sample Problem 5.1.3

We have

Then 224

a2(4h-3b) —a3 (2h-b)

12or Z

a ( Ah- 2b

K \4h~3b

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distribution to teachers and educators permitted by McGraw-Hillfor their individual course, preparation. Ifyou are a student using this Manual,you are using it without permission.

{.(.2

'J

(I

2

PROBLEM 5.99

The composite body shown is formed by removing a semieHipsoid of

revolution of semimajor axis h and semiminor axis all from a

hemisphere of radius a. Determine (a) they coordinate of the ccntroid

when h = all, (b) the ratio hia for which y — -0.4c/.

SOLUTION

V y >JK

Hemisphere2

«r 3— 71a3

3—a8

—nal

4

SemieHipsoid2

JZ

3

— h~—7ta~h

K 2) 6 8

' 5,2+

—

xcrh16

Then LV^~a2 (4a~h)6

EyK =-—

a

2(4a

2 -A2)

16

Now

So that

or

yxK^spK

—

a

2(4a -A) = cr(Acr -h")

16

V a,

\ 3—

—

a8 R3] 0)

(tf) /-? when A-.

Substituting «~2 into Eq. (1)

Y\ 4-- 1 3 Of=— c/ 4- -J 8 ^J

J

or112

y = -0.402c/ ^

PROPRIETARY MATERIAL. © 20)0 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,




663


(b) ? when Y = A)Aa

Substituting into Eq, (1.)

(-0.4a)

or '*Y-3,,

4 .-

—

aa) 8

+ 0.8 =

ThenA

s;3

:2±V(-3.2r-4(3)(0.8)

a 2(3)

3.2±0.8 hii

h 2 .

or — —— and — = -— -^

a 5 a 3

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664

;/

-1.2 mmPROBLEM 5.100

sAFor the stop bracket shown, locate the x coordinate of the center

62 iimi1of gravity.

5.1. mm^Jl

.1.2 nun ,,-'\ /RK

34 nun ^.. 4.5 mm

88

\iin

nun

X

SOLUTION

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

volume.

Then

V, mm3x, mm xV, mm4

1 (]00)(88)(1 2) = 105600 50 5280000

2 (100)02X88) = 105 600 50 5280000

3 ~-(62)(5 1)(10) = 15810 39 616590

4 -i(66)(45)(12) = -17820 34 + -(66) = 783

-1389960

X 209190 9786600

- YxV 9786600X =. = mmZV 209190

or X- 46.8 mm A





665

62

>.l inn

34 nun

-12 nunPROBLEM 5.101

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

gravity.

88 mm

SOLUTION

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

V, mm3z, mm —ir 4

zv, mm

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

2 (100)(12)(88) = 105600 12 +-(88) = 56 5913600

3 i(62)(51)(10) = 15810 12 + -(5.1) = 293

458490

4 ~-(66)(45)(12) = -17820 55 + ~(45) = 853

-1514700

2 209190 5491000

Then- YzV 5491000X = rnm

IF 209190or Z = 26.2 mm <4




666

16 mm

28 mm

40 mm s><

(;= lS.mmfc

00 mm ^>2 20 HUH xJf'Bo mm

"V I ^^25 mm20 mm

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, mm 3x, mm y, mm xV, mm4 -,, 4yv,mm

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

11 (I6)(60)(50) = 48000 92 48 4416000 2304000

III a-(12)2(10) = 4523.9 105 54 475010 244290

IV -*(13)2(l 8) = -9556.7 28 9 -267590 -86010

I 204967.2 12723420 3920280

We have YXV^XyV

F(204967.2 mm3) = 3920280 mm4

or Y = 19.13 mm A

PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of (his Manual may be displayed,



667

0.75 in

PROBLEM 5.103

For the machine element shown, locate the y coordinate of the

center of gravity.

SOLUTION

For half cylindrical hole:

For half cylindrical plate:

r = 1.25 in.

4(1.25)y\w

= 23/r

1.470 in.

r = 2 in.

2lv =7 +^^ = 7.85in.3/r

V, in.3

J> in. 2, in. pK,in.4

z K, in.4

I Rectangular plate (7X4X0.75) = 21.0 -0.375 3.5 -7.875 73.50

II Rectangular plate (4X2X0 = 8.0 1.0 2 8.000 16.00

III -(Half cylinder) -—(1.25)2(1) = 2.454 1.470 2 -3.607 -4.908

IV Half cylinder ~(2)2(0.75) = 4.712 -0.375 -7.85 -1.767 36.99

V -(Cylinder) -/r(1.25)2(0.75) = -3.682 -0.375 7 1.381 -25.77

X 27.58 -3.868 95.81.

7(27.58 in.3) = -3.868 in.

4 7 = -0.1403 in. A





668

2 in. ! in.

1,25 in. -rf -4

r r = .!,25 iii.-.

2 in?^^-^

0.75 in.

PROBLEM 5.104

For the machine element shown, locate the z coordinate of the

center of gravity.

SOLUTION

For half cylindrical hole:

For half cylindrical plate:

/* = 1.25 in.

4(1.25).>'iii

23/r

1.470 in.

r — 2 in.

-iv 7+^ = 7.85 in.

3;r

V, in..3

y, in. z, in. jy.in.4

IV, in.4

1 Rectangular plate (7)(4)(0.75) = 21.0 -0.375 3.5 -7.875 73.50

II Rectangular plate (4X2X0 = 8.0 1.0 2 8.000 16.00

III -(Half cylinder) -—(I.25)2(1) = 2.454 1.470 2 -3.607 -4.908

IV Half cylinder ~(2)2(0.75) = 4.7 12 -0.375 -7.85 -1.767 36.99

V -(Cylinder) -^(1.25)2(0.75) = -3.682 -0,375 7 1.381 --25.77

£ 27.58 -3.868 95.81

Now TLV = zV

Z(27.58in.3) = 95.81in.

4 Z = 3.47 in. <





669

28 nun

40 linn

JUO in

i 6 mm

/ 60 itim

24 nun: 10 mm

J^On

KAI 20 mm

25 mm

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.

Hi

F, mm3x, mm y, mm xK, mm4

yV, mm4

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

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

III ff(l2)2(10) -4523.9 105 54 4750.10 244290

IV -tf(t3)2(18) = -9556.7 28 9 -267590 -86010

£ 204967.2 12723420 3920280

We have XZV = ZxV

JT(204967.2 mm3) = 12723420mm4

AT = 62.1mm <

PROPRIETARY MATERIAL. © 20 K) The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,



yon are using it without permission.

670

2 m

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.

Ji=-i(1.2) = -0.4m

z,=-(3.6) = 1.2m

l\u

4(1.8) = 2.4

3;r Km

A., m2x, m >»,m z,m xA, m3

yA, m" Z/4, m3

I ~(3.6)(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

III —(1.8)2 = 5.0894

2

2.4

it

0.8 1.8 -3.888 4.0715 9.1609

L 13.3694 3.942 5.6555 22.769

We have X1.V =LxK: X(13.3694 m2) = 3.942 m3

F2K = £j7K: 7(13.3694 m2) = 5.6555 m3

ZZV = £z K: 2(13.3694 m2) - 22.769 m3

or 1 = 0.295 m A

or F = 0.423 m <«

or Z = 1.703 m ^



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671

.'/ PROBLEM 5.107

y.-\

125 <irm\Locate the center of gravity of the sheet-metal form shown.

gBk - \ 80 mm

k^ ^^

-1150 mm

>250 mm \.

N. ^\X

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 <

We have

3?„=I50 +2x80

K200.93 mm2x80

n50.930 mm

4x1253?n,=230 +

3x

283.05 mm

A, mm2y, mm z, mm yA t mm zA, mm3

I (250)(1 70) = 42500 75 40 3.187500 1700000

11 — (80)(250) = 314I6 200.93 50930 6312400 1600000

III —(125)2 = 24544 283.05 6947200

£ 98460 16447100 3300000

YY.A = ZyA : 7(98460 mm2) - 1 6447 1 00 mm3

ZZA = TzA : £(98460mm 2) = 3.300x 1

6mm3

or Y = 1670 mm <

or 2 = 33.5 mm -4





672

10 in:10m.

16 in.

PROBLEM 5.108

A wastebasket, designed to fit in the corner of a room, is 1 6 in. high

and has a base in the shape of a quarter circle of radius 10 in. Locate

the center of gravity of the wastebasket, knowing that it is made of

sheet metal of uniform thickness.

SOLUTION

By symmetry:

For III (Cylindrical surface)

X = Z

__ 2r 2(10)6.3662 in.

K K

For IV (Quarter-circle bottom.)

A = —yh =—(10)(16) - 25 1 .33 in/

_ 4r 4(10) „.„„,.x -— =~^Z - 4244 1 in.

3;r in

Az=~r2 = ^~(10)2 = 78.540 in.

2

4 4

\Om.

SS10 »«.

Ifciw,

,4, in.2

x, in. x,in. xA y in.3

yA, in.3

I (10)(I6) = 160 5 8 800 1280

II (10X16) = 160 8 1280

III 251.33 6.3662 8 1600.0 2010.6

IV 78.540 4.2441 333.33

I 649.87 2733.3 4570.6

XZA = ZxA: X(649.87 in.2) = 2733.3 in.

3

f = 4.2059 in. X = 2 = 4.21 in. <

YLA = IlyA: 7(649.87 in.2) = 4570.6 in.

3

7 = 7.0331 in. 7 = 7.03 in. <





673

)

!f

= 0.625 in. /^*^-

PROBLEM 5.109

\ „ ^<fin.2.5 in.

> A mounting bracket for electronic components is formed from

- V-^ sheet metal of uniform thickness. Locate the center of gravity

0.75 in. >< ,-<^:) ,;?f

> x>•-

of the bracket.

1.25 in.l

^:fc?7"' ^ 6 in.

0.75 m>/ \f^

SOLUTION

First, assume that the sheet metal is homogeneous so that the center of gravity

the centroid of the corresponding area. Then (see diagram)

of the bracket coincides with

\X/£ zv =2.25-4^625)

\"/^ ^v =~(0.625)2

^K^y^ =-0.61359 in.2

__ A, in.2

x, in. y, in. z , in. x/if, in; yA, in.3

zA, in.3

(2.5X6) = 15 1.25 3 18.75 45

II (1.25)(6) = 7.5 2.5 -0.625 3 18.75 -4.6875 22.5

III (0.75)(6) = 4.5 2.875 -1.25 3 12.9375 -5.625 13.5

IV Jl1(3) = -3.75\4J

1.0 3.75 3.75 -14.0625

V -0.61359 1.0 1.98474 0.61359 -1.21782

s 22.6364 46.0739 10.3125 65.7197

We have XI,A = Ix/i

X(22.6364 in.2) = 46.0739 in.

3or X = 2.04 in. A

YI.A = 2,yA

7(22.6364 in.2) = -10.3 125 in.

3or F = -0.456 in. <

ZZA^YzA

2(22.6364 in.2) = 65.7197 in.

3or Z = 2.90 in. <

PROPRIETARY MATERIAL. CO 2010 The McGraw-Hill Companies, Inc. Alt rights reserved. No part of this Manual may be displayed,




674

60 mm

, r •> mill

74 mm

30 mm/'=6mm r=6mm

r = 6 mm

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 withthe centroid of the corresponding area. Now note that symmetry implies

Z = 30.0 mm <

CD*H *S

&

rfi

J3>

m£_£_%x2 =6

2x6

x4 — 36 +

x8 =58

n2x6

71

2x6

2.1.803 mm

39.820 mm

54.180 mmit

xw = 1 33+~^ = 1 36.820 mm7t

2x6y2 =y4 =ys

= y ^6 ~ - 2. 1 803 mmit

J?6 = 75 +-^ = 78.183mmn

nA2 ^A4 =A8

= Aw-—x 6x 60 = 565.49 mm'

X = xx 5x 60 = 942.48 mm2

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited


675


We have

A, mm2x, mm y, mm. xA, mm3 yA t

mm3

1 (74)(60) = 4440 43 190920

2 565.49 2.1803 2.1803 1233 1233

3 (30)(60) = 1800 21 37800

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 429750

10 565.49 136.820 2.1803 77370 1233

X 22224.44 1032766 604878

XXA = ZxA\ X(22224.44 mm2) = 1032766 mm3

Y1A = XyA : 7(22224.44 mm2) = 604878 mm3

or X - 46.5 mm -4

or Y - 27.2 mm ^


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676

':Uim

PROBLEM 5.111

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

i -?H in.

; in.

SOLUTION

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

yv =yvl =4+(4)(25) = 14.6103 in.

-H ~-"VJ

3/r

(4)(25) 100.2W ,

= = in.

3# 3/r

K„ (2)(25) 50

.

2|y= : -

l ft>

7t

KAn

= ,4VJ- -(25)2 = 490.87 in.

2

4IV (25)(34) = 1335.18in.2

A, in.2

y, in. z, in. yA, in.3

zA, in.3

I (4)(25) = I00 2 12.5 200 1250

II 490.87 14.6103100

3;r7171.8 5208.3

III (4)(34) = 136 2 25 272 3400

IV 1335.18 19.915550

2659.1 21250

v (4)(25) = 100 2 12.5 200 1250

VI 490.87 .1.4.6103100

3x7171.8 5208.3

I 2652.9 41607 37567

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Ail rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using it william permission.

677


Now, symmetry implies X-= 17.00 in. <

and YZA^ZyA: F(2652.9 in.2) = 41607 in.

3or F == 15.68 in. A

ZZA = XIA: Z(2652.9 in.2) = 37567 or Z == 14.16 in. 4

PROPRIETARY MATERIAL. & 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,




678

>:

76mm

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

F = 38.0 mm <4

Note that x, = 5", = 400 -—(400) = 145.352 mmTV

xn = 400 - ~(200) = 272.68 mmn

z = 300 -— (200) = 172.676 mmTV

xw = JIV = 400 -—(400) = 230.23 mm3iz

xv = 400-—(200) = 3 1 5. 1 2 mm3k

zv = 300 (200) = 215.12 mm3k

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

Thus

I TX

A, mm 2 x, mm z,mm xA, mm3zA, mm3

1 -~(400)(76) = 47752 145.352 145.352 6940850 6940850

II ~(200)(76) = 23876 272.68 1.72.676 6510510 4322810

III 100(76) = 7600 200 350 1520000 2660000

IV 2|-j(400)2 =251327 230.23 230.23 57863020 57863020

V-2|™J(200)

2 = -62832 315.12 215.12 -19799620 -13516420

VI -2(1 00)(200) = -40000 300 350 -12000000 -14000000

2 227723 41034760 44070260

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers andeducatorspermittedby McGraw-Hillfor their individualcourse preparation. Ifyou are a student using this Manual,you are using it without permission.

679


We have XLA^ZxA: X(227723 mm2) = 41034760 mm3

or X = 180.2 mm <

ZY.A = Zz A: 2(227723 mm2) = 44070260 mm3

or Z - 193.5 mm ^





(.80

^ 12i'i

4 in.

PROBLEM 5.113

An 8-in.-diameter cylindrical duct and a 4 x 8-in. 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.

Then X

Y

XxA 15.14.67.

ZA 539.33in.

ZyA _ 4287.4.

1A ~539.33

in.

A, in.2 x, in. y, in. xA, in.

3yA, in.

3

1 ;f(8)(12) = 96* 6 576;r

2 ~|(8)(4) =-Uw2(4) 8

10 -128 - 1 60tt

3 ~(4)2-8^r

2

4(4) 16

3# 3/r12 -42.667 96/r

4 (8)(1 2) = 96 6 12 576 1.152

5 (8X1 2) = 96 6 8 576 768

6 -£(4)2 =-8^2

4(4) 16

In 3x8 -42.667 -64/r

7 (4)(12) = 48 6 10 288 480

8 (4)(1 2) = 48 6 10 288 480

L 539.33 1514.6 4287.4

or X = 2.81in, A

or F = 7.95 in. M

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Atl rights reserved. No part of this Manual may be displayed,




681

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, = 0.3 sin 60° = 0.1 5^3 mz, = 0.3cos60° = 0.15 m

U

^0.6sin30°^

/

0.9S in 30° =— m

K

0.6sin 30

n6

o~\

cos30°=—73 mit

K(0.6) = (0.2/zr) m

L, m x, m >», m 2, m xL, m2 yL,m2zL, m2

1 1.0 0.1573 0.4 0.15 0.25981 0.4 0.15

2 0.2/ZT0.9

K0.973

n0.18 0.31177

3 0.8 0.4 0.6 0.32 0.48

4 0.6 0.8 0.3 0.48 0.18

I 3.0283 0.43981 1.20 1.12177

We have XT.L = XxL: X(3.0283 m) = 0.4398 3 m2

YZL = Y,yL: F(3.0283 m) = 1 .20 m2

ZZL = XzL: Z(3.0283 m) = 1 .12177 m ;

or X = 0.1452 m <

or Y = 0.396 m A

or Z = 0.370m A

PROPRIETARY MATERIAL. ©2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed,




6S2

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

ABl = (1 m)2 + (0.6 m)2 + (1 .5 m)'

AB = \.9m

XZL = XxL: X(5.0 m) = 2.05 m2

YZL = -LyL: F(5.0 m) = 2.55 m2

ZY.L = ZzL: 2(5.0 m) = 0.75 m2

L,m x,m y,m z, m xL, m2yL, m2 YLt m2

AB 1.9 0.5 0.75 0.3 0.95 1.425 0.57

BD 0.6 1.0 0.3 0.60 0.18

DO 1.0 0.5 0.50

OA 1.5 0.75 1.125

2 5.0 2.05 2.550 0.75

X- 0.410 m ^

F = 0.510m A

Z~ 0.1 500 m <

PROPRIETARY MATERIAL, © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Ato/wrt o/rii/j Wamw/ may be displayed,


distribution to teachers and educators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,


683

PROBLEM 5.116

Locate the center of gravity of the figure shown, knowing that it is

made of thin brass rods ofuniform diameter.

SOLUTION

By symmetry:

YZL = ZyL : F(l 52.265 in.) = 1530 in.2

F = 10.048 in.

ZZL = TzL : Z(l 52.265 in.) = 784 in.2

f = 5.I49in.

X = <

L, in. y,in. z,in. yl, in.2

zL, in.2

AB 15 510a/302 +162 = 34

AD 15 8 510 272^302 +162 =34

AE 15 510V302H-16

2 =34

BDE ;r(l 6) = 50.2652(' 6

> = 10.18671

512

I 152.265 1530 784

7=10.05 in. <

Z = 5.15 in. <





684

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.

_ _ 2x3 6

„

Xo -x, = -—it

We have

y%=y9=s+

71 71

2x36.9099 ft

71

XZL = TxL : 1X39.4248 ft) = 69 ft2

YXL^ZyL: F(39.4248 ft) = 163.124 ft2

ZXL = ZzL: Z(39.4248 ft) = 53.4248 ft2

L,ft x, ft y,ft z,ft xL, ft2 yU ft

2zL, it

2

1 2 3 1 6 2

2 3 1.5 2 4.5 6

3 5 3 2.5 15 1.2.5

4 5 3 2.5 2 15 1.2.5 10

5 8 4 2 32 16

6 2 3 5 1 6 10 2

7 3 1.5 5 2 4.5 15 6

8 ™x3 ==4.71242

66.9099 9 32.562

9 —x3 = 4.71242

6

7t

6.9099 2 9 32.562 9.4248

10 2 8 1 16 2

X 39.4248 69 163.124 53.4248

or jf = 1.750 ft ^

or f^ 4. 14 ft <

or Z= 1.355 ft <

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685

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/m

and of steel is 7860 kg/m3, locate the center of gravity of the awl.

25i»ni:r„{:

-50 mm -90 min- 10 mm

80 mm

SOLUTION

First, note that symmetry implies

1

Q TLSL

IPLAS"T«33 + C

X2

y^Z-0

^

x, =—(1.2.5 mm) = 7.8125 mm1

8

W,= (1030 kg/m3)^ '(0.0125 m)3

3 '

4.2133x10"' kg

jc,j = 52.5 mmKWn =(1030 kg/m 3

) - (0.025 m)2(0.08 m)

= 40.448 xl0~3kg

Xji, = 92.5 mm - 25 mm = 67.5 mm

Wm = -(1030 kg/m3)(— 1(0.0035 m)2

(.05 m)

-3-0.49549x10"' kg

xiv = 182.5 mm - 70 mm = 1

1

2.5 mmKWw =(7860 kg/m3

) — (0.0035 m)2(0.14m)2 = 10.5871xI0"

3kg

xv = 1 82.5 mm + -(1 mm) = 1 85 mm

Wv = (7860 kg/m3) - (0.001 75 m)2

(0.0.1 m) = 0.25207 x 1.0"3kg

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686

We have X

or


W, kg x, mm xW\ kg -mm

I 4.l23xl(T37.8125 32.91 6xl0~3

II 40.948xl.O"3

52.5 2123.5xl0'~3

III -0.49549xI0~3

67.5 -33.447 xlO"3

IV 10.5871 x 1

0"3112.5 1191.05 xlO

-3

V 0.25207 x!0~3185 46.633 xlO

-3

X 55.005xlO"3 3360.7xI0~3

ZW = XxW; X(55.005xl0_3

kg) = 3360.7 x 1

0~3kg •mm

X = 61.1 mm <

(From the end of the handle)





687

I..125 in.

-

0.5 in.,

-O.i5 in

1.80 in

I 00 in.

t>4() in.

PROBLEM 5.119

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

specific weight of bronze is 0.318 lb/in.3and of steel is 0.284 lb/in.

3

,

determine the location of the center of gravity ofthe assembly.

SOLUTION

First, note that symmetry implies X = Z = Q <

Now W = {pg)V

p, = 0.20 in. \\\ = (0.284 lb/in.3)

71

4 iL

7„ = 0.90 in. W» = (0.284 lb/in. )

ym = 0.70 in. Wm = (0.3 1 8 lb/in.3)

1 .82 - 0.75

2

)in.

2(0.4 in.) \ = 0.23889 lb

1.1252 -0.752

)in.2](lin.)Uo.l568341b

71

4JL

2 Af2V.,20.75^-0.5' in. (1.4 in.) [= 0.1 09269 lb

We have

or

YZW^ZyW(0.20 in.)(0.23889 lb) + (0.90 tn.)(0.156834 lb) + (0.70 in.)(0. 109269 lb)

Y =0,23889 lb + 0. 1 56834 lb + 0. 1 09269 lb

Y = 0.526 in. <

(above base)



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688

,*^U> in. PROBLEM 5.120

A brass collar, of length 2.5 in., is mounted on an aluminum rod

r of length 4 in. Locate the center of gravity of the composite body.£.:;.

4

(Specific weights: brass = 0.306 lb/in.3, aluminum = 0.101 lb/in.

)

1

as**, 2.5 in.

SOLUTION

/A- m.

L

Aluminum rod;

Brass collar:

l.fc m.

\72.5 m.

Tifl,

^WMlMUh'.*

£ = O.IOl lV»/\v»

—(1.6 in.)2(4 in.)

+

B>«sAv6i

y= 0.304. tfe/m 3

= (0.101 lb/in.3

)

= 0.81229 lb

(0.306 .b/in 3)- [(3i„.)-(1.6^](2.5 in.)

3.8693 lb

Component ^(lb) y(™.) pWOb-in.)

Rod 0.81229 2 1 .62458

Collar 3.8693 1.25 4.8366

£ 4.6816 6.4612

YXW ^-ZyW: 7(4.6816 lb) = 6.4612 lb • in.

F = 1.38013 in. 7 = 1.380 in. <





689

PROBLEM 5.121

;, i 280 mm

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 mm2. 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/m3and of glass is 2 1 90 kg/m3

, locate the

center of gravity of the table.

SOLUTION

First note that symmetry implies

Also, to account for the three legs, the masses ofcomponents I and II will each bex

multiplied by three

pt= 12 + 180-

2Xl8Qm

J-/?,yrF1

=7860kg/m3x(15()xl0"

6m2 )x-(0.180m)it

'

2

77.408 mm 0.33335 kg

yn = 1 2+ 1 80 +^?— m = pSTVu = 7860 kg/m3 x (1 50x 1

0"6 m2) x- (0.280 m)

7V 2

370.25 mm = 0.51855 kg

Z = <

ym =24+ 180+ 280 + 5 mm ^

p

GLVm = 2190 kg/m3 x-(0.6m)2 x (0.010 m)

489 mm6.1921kg

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

£ 8.7478 3681.3

We have

or

Ylm = T,ym : 7(8.7478 kg) = 368 1 .3 kg • mm

Y =420.8 mm

The center of gravity is 421 mm ^

(above the floor)



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690

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 ofvolume a disk of radius r and thickness dx. Then

dV = nr2dx, xFL = x

The equation of the generating curve is x2 + y2 — a2

so that r2 = a2 - x

2and then

dV^n(a2 -x2)dx

Component 1

and

Now

Component 2

fall . _

K, = icier -x )dx~nJo

3

2 Xa x

3

-W2

24

J'a/2

-n

a

\xELdV=\ x\n(a2 ~x2)dx\

7t

2 42 X X

a

a/2

7 4—7ia64

_ f_ _ n 3 /

x,K, = x Pt aV: xA -—Tta -—#731

' h hL H 24 J 64

F2 = I#(#

2 - x2 )dx - 11la/2

K{3

2r n «a (a)~

x == # a a

3

a/2

~€ (!)

\ "]]

3

J

5 3

24

_ 21or x, =

—

a



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691


andJxEtAV - I

x\x(.a2 -x2

)dx\ = n

Ad? {of

Now

n

9 4-

—

no*64

*2*2 = \2 *ELdV '- X

2

2 4

(f)

2

(f)

4

f$ 3

ail

24-ncr

9 4—Kcf64

_ 27or x? =—a -^

240

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692

PROBLEM 5.123



divides the shape into two volumes of equal height.

A semieUipsoid of revolution

SOLUTION


dV — izr dx, xEl = x

2 2X V

The equation of the generating curve is —+~rh . a

2

so that

2 a/ 7 2 2\

r =-T (A -x ) V>

and then dV =x~(h2 -x2)dx

Component I V,= n-T(h2 -x2)dx^x-

Jo h2

11 2,— Tta h24

,2 x-h x

-i/j/2

-0

and n-™{h2 -x2)dx

h

Jt-

Y2 Y4? X X1

2 4

A/2

2r2

Now

Component 2

—Ka h64

%l\^\xELdV\ xA—nJhMM24

•" a2

-,2 .2- a

J/;/2 /»2

&2

J/,/2 ^

*2

//2

2 i.27ta"h64'

*2*-^

-i''

-h/2

h2(h)

(hf ( u\

O-J

(tf

24'a A

or jc,

21A ^

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693


and \xEl dv~ r* xh J/1/2

n^{h2 -x2)dx

h

2 fa

h2 2 4

fc/2

( j

a2

h2 2 4

/i) (!)

2 4>

9 2 7 2

64

Now Wl = \*ElAV'- *2f

5 2,1 ^ 2,2

,24 ) 64or x, ~—~h -^

240

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694

PROBLEM 5.124



divides the shape into two volumes of equal height,

A paraboloid of revolution

SOLUTION

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

dV — Kr2dx, xFI

~ x

h.2 a

The equation of the generating curve is x ~ h—- v so that r =—(h - x)

and then

Component 1

h

dV^7t~(h-x)dxh

(m a2

it-

hhx

h/2

—Tta h8

andr f

/)/2

Jt—(h~x)dxh

K-x1 x3

ha

2 7.2

12na h

Now

Component 2

5tf [xELdV. x, I -na2h I

= —Ttcch1

12

th a2

y = ji— (h - x)dx - 7t2h,ii h

7t- h(h)(hy

1 2,—Ttcth

-\h

hx—hll

h\ (!)

:

2 2

or x, ~—h A9





695


and \xELdV= f* xh Jh'2

#

—

(h-x)dxh

a2

h

r ,-2 3

^* £_2 3

-i/i

- ft/2

a"

hh(hf (A)

3

2 3

- A(i) (D2 3

,-

* 2 i 2

12

Now , *2^ =J2%/.^: ^(-*a2Aj==—*a*A2

or jt, ——A -^2

3

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696

ij=k{x-hf

PROBLEM 5.125

Locate the centroid of the volume obtained by rotating the shaded area

about the x axis.

SOLUTION

First note that symmetry implies y = ^

and F = <«

We have y = k{X-hf

at x = 0, y-a\ a = k{-hf

or k~—h2

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

dV---ttr dx, XEL —x 1 k x —

1

— dx

Now

so that

r~

dV =

n

2

~K—~(x~h) 4dx

h4

1 i

4«fe(x-h?

^^-^-~r

""I V 1 J ^™^ ^

Then V--f*

q1( m4j K q1

n—rix-h) dx-——rh h

4 5h4 >~<h

1 2,-—itah5

and \xELdV-2

7Z~~-r{x-h)4dx

h

s zfL [\x5 - 4hx

4 + 6h2x3 - 4/2V + h

4x)dx

h4 h v

a2

/z4

1 /; 4»S 3,24 4-3^ *,42— -r—hx* +-frx4—AV +—AV_6 5 2 3 2

A

30

Now W'~= Jjc^rfF: x —ah15 J

^ 2r2

30or x=~~h A

6

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697

y = kx** PROBLEM 5,126

Locate the centroid of the volume obtained by rotating the shaded area

about the x axis.

SOLUTION



dV = 7zr2dxt xEL = x

Now r - kxm

so that dV = xk2x2,3dx

at x-h, y = a\ a — kH

or

Then

and

J/3

1/3/?

hm

Vh

£12/3

K-

2

fC-Cl

in

xm

dx

h

7ialh

~x~.5/3

3 2

y=0 <

1 = <

's*\<*

Also hdv -i .2/3

a 2/3 ?

-xa2h2

ft-.2/3

~\H

,.8/3

Now xV^ \xdV: x —Jta2h \=-~7ta

2h2

or x h <

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698

a* + r

PROBLEM 5.127

Locate the centroid of the volume obtained by rotating the shaded

area about the line x — h.

SOLUTION



dV = 7tr2dy, yEl = y

h

x = h <

z = <

X-h

Now x2 = ~r-(a2 -y2

) so that r-h-~Ja2 -y2

a1 a

dV^7i\[a-^a2 -y^ dyThen

and

Let

Then

y-lA[°-^hv- o>

y~asm$=>dy = a cos

h2

r'2/V = x-j P (a-^a2 ~a 2 sm 2

o) acosOdO

- 7t^r f* T«2 - 2a(a cos 0)+ (a

2 - a2 sin20)1 a cos </0

a2 h \- J

= #a/?2 T (2cos#-2cos

2 0-sin20cos0)c/0

Jo

ttah'

= ;rahx

2sin0-2 — +sin 20 \ 1 . 1rt

—sin' 61

f *>2-2

v2y

2 4

I3

-i;r/2

-0

0.095870^/?'





699


and J*^-f n-T\a-Jai -y£

\ dy

*\£fa2y-2ayyp^y -y3

)dy

7t'

= 7t-

a2y2 +~a(a2 ~

}n2 \1Q _£_ 4

4

a\af--aA

4\o^r

2 i.2

12ncfh

Now yV =\yEL dV\ y(0.095S70xah

2) = ^~7ta

2h2or J7

= 0.869a ^

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700

//

//

//

/ \

PROBLEM 5.128*

Locate the centroid of the volume generated by revolving the portion

ofthe sine curve shown about the x axis.

SOLUTION


Choose as the element ofvolume a disk ofradius r and thickness dx.

dV - 7tr2dx, xEl ~ xThen

Now

so that

Then

, . 71Xr = osin—

la

dV - Ttb2sin

2-^-dx2a

V- nb sin —dx

Kb~

*4BfHf)]

—Ttab"2

H — X.

(

1

p = ^

F = ^

dx

and


\x(a dV -\ x\ nb2sin

2—-dx

u ~x

du = dx

2a

dV = sm

V

2a

x sin !

In

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701


Then KciV=/rb2

-

Kb 1-

-\2a

x sin-^

2 2.k.

a J

fHf1 T a" JTX—X" H r COS4 2/r a

^NI-«2

4 2n l4

,,,. 3 1

4 7T2

0.64868;wrV

2/r

Now xV = \xelW-2; 2

x I -jwfr2

I= 0.64868;r<r& or x = 1.297a ^

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702

sMj-fogPROBLEM 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


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

x = <

1 = 0-4

Then

Now

so that

Then


Then

dV = {2m-){y){dr\ yEL =~y

y ~ frsinnr

~2a

dV = 2ffbrs'm—dr2a

V~\ 2xbrsin-— dr*« 2a

u — rd

du = dr

V = 27tb-

= lKb<

ov = sin

—

drla

2a Ttrv ~ cos—

Tt 2a

, . 2<7 7tr(r)\ cos—

k 2a

-—[(2aX-0] +Tt

2a

j:

2a(2a Ttr) .|— cos— \dr;

K 2a) \

4a . Ttr—r-sm-TT" 2a

7a

V = 2xb

= Sa2b

Ua24a

2>

v * * J

5.4535^/;

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70.5


Also Wv =i2t3(

1 L 'nV

If 1 L •nV

J-asm— 27z;br$m— dr2

2aA 2a

Kb2I rsin

2—orJ

rsii2a


u = r

du — dr

dv = sin'Ttr

'ladr

r sinv =

2 2/r

Then j?«dV = ^'

= #//

(0

1'

r sin-^

(2a)

2jr_

a J

2a(a)

2 2r a Ttr

+——cos—-,2a

4 2k1

Now

rf ~a2 (2a)2 a

2(a)

2a'

4 2;r'j

2;r'

^2,2.3 I

fta o ,

4 tz:

2

2.0379a2/;2

pK = [yEldV\ y(5-4535a2a) = 2.0379a

2/? or 7 = 0.374/? 4





704

PROBLEM 5.130*

Show that for a regular pyramid of height h and n sides (« = 3, 4,. ..) the centroid of the volume of the pyramid

is located at a distance hiA above the base.

SOLUTION

Choose as the element of a horizontal slice of thickness dy. For any number iV of sides, the area of the base of

the pyramid is given by

Aase_ Kb

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

s _h~~y

bor s = —(h~y)

h

Then dV^A,ikcdy^kS2dy = k~j(h~y)2

dyh

and

Also

So then

•*. b2

V= Vk~T(h~y)2dy = k

Jo hl

h

~kb2h

3

yEL=y

\y^'t kK(h-y)2dy

h

\(H-y?-0

b2

**= k—j\h2y-2hy 2 +y3

)dy

2 3 4~kb2

h2

12

Now

Note

W^pELdV: y -M/h

4~ = *1

A ^

2 imf

N4 tan-*

k(N)b2

2 i.2

12-«//? or v = ~A Q.E.D. <«

4

/ CENTER. <=>P 3>(\y=.

ATI

\ 1

\ 1

H-k£-l

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705

I

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

First note that symmetry impl es x-=0<

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~--- (nr){Rdd) 1

Where r--RsinO \ ^v

so that dA =

yiiL=

'-7zR2 sm0d0 y\ 8 /^§^y2R . a * \ 6fK iX71 ^^^^^

Then A~~-. r/2

xR2 sm0d0 = xR2[-cos0]%

/2

Jo

-~nR2

and \yEidA ~- smh { 7t )

{7tR2 sm0d0)

-- -2R3 > sin 2^

.2 4 -0

2

Now yA ==

J7MaM: j?(^2) = -|i?

3or 37 = -It? ^

Symmetry implies J =J ? =-!« ^

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706

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, (/;) the punch.

SOLUTION

(a) Bowl

First note that symmetry implies x=0 A

z=0 Afor the coordinate axes shown below. Now assume that the bowl may be treated as a shell; the center ofgravity of the bowl will coincide with the centroid of the shell. For the walls of the bowl, an element ofarea is obtained by rotating the arc ds about the y axis. Then

and(3>/sAvail

:

Then Awai]

=

=^(2xRsm0)(Rd&)H

d®= -/?cos6'

C7T/2

2xR 2 sm0d0 \ /30s"

\X\l1— - R. - H ,-&<>

= kSr 2

and Pwa1l4™ll:= j(^/.)wati^

rnll

(~Rcos0)(2xR2 sm0d0)Ja/6

= xR3[cos

2

0Z%

---7t

R

3

4

By observation. AbafiC

-

X D2 - V3

Now y~ZA -= £J^

or y(xJ3R2 +~R2)

4 4i

2>

or y = -0.48763tf J? = 250 mm J7= -121.9 mm A

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707


(b) Punch

First note that symmetry implies x=0 <

1 = <4

and that because the punch is homogeneous, its center of gravity will coincide with the centroid of

the corresponding volume. Choose as the element ofvolume a disk ofradius* and thickness dy. Then

Now

so that

Then

and

Now

or

dV = nx2dy, yEL ~y

x2 +y2 =R 2

dV = 7t(R2 -

y

2)dy

K=f° 7i{R2 -y2

)dy

It

3

-It

-S/2R

( IT \ .. f IX \ 3

R'2R

2-JZ>/3K

3

J J-JlFlR . \ ' _

-71

2 4 StlR

~K !*>u; 2

1u; 4

2 2 4 2\ / v y

64

yV~\yE! dV: yfyc&R15 ^<*

64

>;

8^3

faM

72 7? = 250 mm J = -90.2 mm ^





708

PROBLEM 5.133

After grading a lot, a builder places four stakes to designate the

comers of the slab for a house. To provide a firm, level base for

the slab, the builder places a minimum of 3 in. 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 j> = 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 — -3 in., and therefore

I3 rIt — a, or a -

12

For x - 30 ft, and z = 0: y = -5 in., and therefore

•ft

— ft =—-ft + 6(30 ft), or b12 4 180

For x = 0, and z = 50 ft : y = -6 in., and therefore

Aft = _Ift + c(50ft) sor c =—

—

12 4 200

Therefore:

Now

K 1 1

-_ft x -.4 1 80 200

-r_K dV

V

A volume element can be chosen as:

dV = \y\ dxdz

or

and

dV =— I +

—

x +—2 Itfr fife

4 1 45 50

*££ = *

PROPRIETARY MATERIAL, © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed,


distribution to teachers and educatorspermitted by McGraw-Hillfor their individual coursepreparation, ifyou are a student using this Manual,


70')


Then•50 r30

JLL h Jo 4^ 45 50

I

dxdz

i f5°

4 Jo

I f

50

4 Jo

n30

jc l__ __ 1 * 2JC" + •—-JT t/z

_0

f (650 + 9z)dJo

The volume is: V

9 o

650z +-z2

2

= 1.0937.5 ft4

50 j.30 |

<Z

-i50

Then

Therefore:

K-n ] -f—x -f— z \dxdzo 41 45 50

I f5°

4 Jo

I 2 ^x-f—X + X

90 50

-i30

dz

1 f5°

4£(40 + ?z dz

687.50 ft3

3 740z+—

z

2

10

n50

-0

_ j*a^_ 10937.5 ft4

F 687.5 ft-'

15.9091ft

F = 688ft3 <

x =15.91 ft ^

PROPRIETARY MATERIAL © 201.0 The McGraw-Hill Companies, Inc. All rights reserved. A'o /wrrt o/'/Mv Manual may be displayed,




710

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/y "7 9 9

surfacey — 1 6h(ax - x )(bz ~ z )la b .

SOLUTION

First note that symmetry implies_ a .

x -— ^2

Choose as the element ofvolume a filament of basetft x dz and height/. Then

dV - ydxdz, yEl= —y

- b *2

or dV\6h

a2b2(ax - x2

)(bz - z )dx dz

ThenJo Jo a

2b2{ax-x2

){bz-z2)dxdz

V16ft cb

a2b2

16/7

\\bz-z2)

Jo

a 2 1 3—-x —Xz 3

2 r.2aLh

%ah

3b2

~(b)2 ~ l

-(b?

dz

b 2 1—z —2 3z2 --V

-0

-abh





711


and,b ,*

1 16/?

Jo Jo 2 [a2b2

128/72

j

b pa

-(ax~x2)(bz~z

2) (ax — x )(bz ~z )dx dz

a2b2

~r f r(a2x2 -2ax*+x4

)(b2z2 -2bz3 +z4

)dxdzIf Jo Jo

I28/?2

f*

a2b

4[\b2

z2-2bz* +zA

)Jo

a \ a a 1 %—r —x +-x3 2 5

dz

128/2'

4r4a b

(Aah2

15//~(/>)

3--(/.)

4 +I(ft)5

3 2 5

3 Z 5-0

32 112

225

Now 3?K=Jj^^: yi^obh32

225a/W oi ->

25

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Afo /«»/ o/rtm- Manual may be displayed,




712

PROBLEM 5.135

Locate the centroid of the section shown, which was cut from a thin circular

pipe by two oblique planes.

SOLUTION

First note that symmetry implies x = A

Assume that the pipe has a uniform wall thickness t and choose as the element of volume A vertical strip of

width ad0 and height Ov-*)• Then

ij M /~-Ml

/H

1

v&srf1 wiN>

h

J •flt-s. v I \a a.

dV--= O^ ~ y\ )tad0, yEl -~(yi+y2 ) zEl = z

Now y\ = ^-Z +— V2 = ~Z+— ft

2a 6 2a 3

-—(z + o) -— (~-z + 2a)6a 3a

and z = acos#

Then (>>2 -yxY--—(-a cos + 2a) -(acos0+ a)3a 6a

= -(l-costf)2

and (v, +y2~) z

dV =

/? /?

=— (a cos 6?+ a) +—(~-acos#-T-2a)6a 3a

= -(5~cos<9)6

a/?/ hq _ cosQ}dQ yEL

_— (5 - cos #)5 Zjpt= a cos 6

Z* \ Zj

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Ail rights reserved. No part of this Manual may be displayed,




713


Then

and

V = 2\ (1 - cos 6)dd = aht[0 - sin 0$rn aht

Jo"5

Ttaht

\yEldV = 2 £*A(5 -cos0)oA/

Q-cos0)d0

ahh

12 JoT (5 ~ 6 cos + cos

20)^0

Jo

a/72/

12

iL24

__ . sin 2050-6sm0 +— +

2 4

7uah2t

\zEldV = 2 acos0

alhl sin0-

1 2/r<r/i*

a/?/(\-cos0)d0

sin 20

-0

Now

and

VK= \yELdV: y{n;aht)-—Kah2t

zV~ \zELdV: z(jtahi)-~—na2ht

_ 11. .or y =

—

h -A24

or z -—-a M2




you are using if withoutpermission.

714

PROBLEM 5.136*

Locate the centroid of the section shown, which was cut from an elliptical

cylinder by an oblique plane.

SOLUTION


i~X«*« *a*xvAK

r-f52

A

Choose as the element ofvolume a vertical slice of width zx, thickness dz, and height j\ Then

Now

and

Then

Let

Then

dV — 2xydz, yEl24

z>,EI,

ax~—\jb -z

b

hi! h h ,_ ,

y — z +— =—ib — z)'b 2 2b

V \\2~4b-~z-M b

h_

2b(b-z) dz

z

~

6 sin dz- bcosOdd

V = -Tr\ (b cos 0)[b{\ - sin 0)]b cos 6 ddb2 imi

= abh f (cos2 6- sin $ cos

20)dB

J-/T/2

abh

I

9 sin 2^ 1 3— + +—cos2 4 3

~\Xtl

--nil

y -—jcabh2

/I1 U J

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of/his Manual may be displayed,




715

and

Let

Then

Now

So that

Then

Also

Let

Then

Using


p^v=[ 1 h n x-x

—

(b-z)2 2b

J

1 ah2

r''

4 b3

(b-z) dz

\ (b-zfW-zldz

i-b

z-b sin dz = b cos BdO

h dV =-^- f'

2

[b(\ - sin 0)f (b cos 0) x (b cos d0)J A b *~x/2

= - abh2 f

2

(cos2 - 2sin cos

2 + sin2

cos20)J0

sin2 = i-(l- cos 26) cos

2 = ^(1 + cos 20)

sin20cos

2 = -(l~cos220)

\yEiAV ^-ctbh1

\J-7T/2

cos20~2sin0cos2 + ~(l -cos2

20) 40

-«/?/?'sin 20—+

2 4

1 In I

+ -COS +—3 4

-i*/2'0 sin 40

v2+

t ^32

-rcabh'

l^-lsWla2 -z2

'

h_

2b(b-z) dz

ah ch

fz(b-z)4b 2 ~z2

dzJ—b

z = bsin0 dz = b cos 0d0

c ah c^fe

\zn dV = -T \(b sin 0)\b(\ - sin 0)](7; cos 0) x (b cos c/0)

J 'jy

l i-nll

= aZ>2/* r (sin cos

2 - sin2

cos2 0W0

sin20cos

2 =— (I - cos220) from above

4

\zELdV^ab2h\

= ab2h

•nil

sin cos2 —(1- cos

220)

1 Zn \ n \(0 sin 40-cos —0+— — +3 4 4 2 8

d0

nil

-Kit

1 }

—Ttabk8

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. A'o part of this Manual may be displayed,



716


Now w--'- \y&<W"- y(i ^—nabhU )

——nabh32

or y =

—

h ^J16

and zV --- \zELdV: z\-~xabh\-—nab2h or z=-~b <

4



distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,


717

>J PROBLEM 5.137

Locate the centroid of the plane area shown.30

54 nm

_-—

—

'"

48

J

nm

X-

X

54 mm 72 mm

SOLUTION

Then

and

m-usv

.cii-4-n

+&•\2.lo

11^ =1^.2-

^(1 0422 mO = 200880 mm 2

Y (1 0422 ra2) = 270020 mm3

A, mm2 x, mm j/, mm xA, mm3yA, mm3

1 126x54 = 6804 9 27 61236 183708

2 -xl26x30 = 18902

30 64 56700 120960

3 ix72x48 = 17282

48 -16 82944 -27648

2 10422 200880 277020

or X = 19.27 mm ^

or Y — 26.6 mm ^

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Jnc. All rights reserved. /Vo part o/Wi/s M»h«/ may be displayed,




718

x = ktf

* a - 8 in.~

b - 4 in.

PROBLEM 5.138


SOLUTION

Dimensions in in.

>*4

r*

A, in.2 x, in. y,'m- xA, in.

3yA, in.

3

l ~-(4)(8) = 21.333 4.8 1,5 102.398 32.000

2 ~-(4)(8) = -16.0000 5.3333 1.33333 85.333 -21.333

z 5.3333 17.0650 10.6670

Then

and

XZA = TxA

^(5.3333 in.2) = 17.0650 in.

3

YZA = ZyA

F(5.3333 in.2) = 10.6670 in.

3

or X = 3.20 in. <

or r = 2.00 in. <





719

0.2 m

-

0.6 m

1.35)ii-

PROBLEM 5.139

The frame for a sign is fabricated from thin, flat steel bar stock of massper 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 fabricates from uniform bar stock, its center of gravity will coincide with

the centroid of the corresponding line.

Fit-,. ^.66

L, in x,m. xL, m2

1 1.35 0.675 0.91125

2 0.6 0.3 0.18

3 0.75

4 0.75 0.2 0.15

5 ~(0.75) = 1.178 10 1.07746 1. .26936

I 4.62810 2.5106

Then XZL - ZxL

^(4.62810) = 2.5 106

or X = 0.54247 m

The free-body diagram of the frame is then

Where W = (mZL)g

= 4.73 kg/m. x 4.628 1 m x 9.8 1 m/s2

= 214.75 N

M

&*.

^NC

|w >.

<-x X

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Ato /mi-/ o//A« Mmm«/ bh;>- te displayedreproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited

distribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,you are using if without permission.

720


Equilibrium then requires

(a) EMc =0: (1.55 m) ~TBA J-(0.54247 m)(2 14.75 N) =

or r^=125.264N or TBA = 125.3 N <«

(b) £FV. = 0: C; --(125.264 N) =

or CT=75.158N—

-

EFV=0: Cj, + -(125.264 N)- (2 14.75 N) =

or Cy=11 4.539 nJ

Then C = 137.0 N ^LS6.T<

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed.




Ill

'/

X

y = mx + b-x \

PROBLEM 5.140


your answer in terms ofa and h.

X

SOLUTION

By observation

For y2 : at

at

Then

Now

y ~x + h-h\\-

0, y = h: /r = A(l-0) or k = h

x = a, y = Q: = A(l-ca2) or C= 4j

Then

j2=/2

2 \

dA = (y2 ~yi)dx

2\x x

v* « J

dx

XEl

Jul (y\-yi)

x x

a a'

*\

A^ YA=i

-ah

x

a j

dx

2\

Mev-^i,

/ .?>X Xdx-~= /?

Va cr

j2a 3a'




Ill


and K^ =C-

h

f X *2>

dx = A{"

(x* x^3a Aa £

La*h12

J Jo 21 a a

2\ f t\X X

a a°

dx

a1

a4

2 a a2*5a

4^ah2

10

xA ~ Xfi/dA : x \—ah

^ 1 2

12alh x = ~-a A

2

yA=jyBl dA: y ah =

—

a n10

y=-h <5



distribution to teachers and educators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,


723

2b

J™

y~2b-cx-

£w

PROBLEM 5.141

Determine by direct integration the centroid of the area shown. Express your

answer in terms ofa and b.

SOLUTION


at

Then

or

Then

Now

and

Then

and

x-a, y-b

y x

: b- ka2

or k

JV h:= 2b - cal

c-b

''

"a2

y2'-= b

f2-

x2

)~«2

)

dA---~(y2--yi)&2 = b

f2-

V

x2 '

-¥f r

2 ^= 26 l

a2

dx

^ 1

dx

XEL=*

A K 2vx2\

dx = 2b3a"

y = b <

\x£iLdA~f. 2b\ 1

= 1xA- \xK,dA\ xEIS

dx

-ah3 J 2

2b

a2b

x2

x4

2 4a'

-ab

1 2,—a b2



distribution to teachers and educators permitted by McGraw-Hillfor their individualcoursepreparation. Ifyou are a student using this Manual,you are using it without permission.

724

-lOin.-

-PROBLEM 5.142

Knowing that two equal caps have been removed from a 10-in.-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 = 2tzXL = 2tzZxL

- 2>r(2x]Lj + x2L2 )

4Now

or

Then

tan a

a = 53.130°

„ 5in.xsin53.130°

53. 130°x T

4.3136 in.

w J^is>.U

and

or

l2 =2 53. 130° x-tt

180c

9.2729 in.

(Sin.)

A~2n 2 1 -in. |(3 in.) + (4.3 136 in.)(9.2729 in.)

,4 = 308 in.2 <





725

900 N/m PROBLEM 5.143

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

SOLUTION

Rf=(3m)(900N/m) SooS, 9t

u-USwwU ri-= 2700 N

Un =i(lm)(900N/m)

= 450N

ft*

A *1 B1lm L-_ ^tv\ •^-

Now J^£/"<; == 0: Ax =

+)XMB -= 0: -(3 m)^, +0.5 m)(2700 N)-|im 1(450 N) ==

or Ay-= 1300N A = 1300NJ^

+1 *F?-= 0: 1300N - 2700 N + B

y-450 N =

or i?j;=1850N B = 1850N} <

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Ato/w/7 o/"/Am Muiho/ may be displayed,




726

20(1 N/t!)

PROBLEM 5.144

A beam is subjected to a linearly distributed downwardload and rests on two wide supports BC and DE}

which

exert uniformly distributed upward loads as shown.

Determine the values of wbc and w.w corresponding to

equilibrium when \vA= 600 N/m.

SOLUTION

WOO' \Zoom

*bE

We have

Then

or

and

or

Rx= -(6 m)(600 N/m) = 1 800 N

Rn= -(6 m)(I200 N/m) = 3600 N

RBC = (0.8 m){WBC N/m) = (0.8WBC)N^=0.0m)(%N/m) =(^)N

+)TMG =0: -(1 m)(1800N)-(3m)(3600N) +(4m)(^N) =

0^=3150 N/m ^

+tz^,=0: (0.8^c)N~I800N~3600N +3l50NM)

lfec =2812.5N/m. /iC =2810N/m <

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Ill

PROBLEM 5,145

The square gate AB is held in the position shown by hinges along its top edge Aand by a shear pin at B. For a depth of water c/ = 3.5ft, determine the force

exerted on the gate by the shear pin.

SOLUTION

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

P = ~Ap2

= ~~A{yh)

mt

Then I] =- (1 .8 ft)2(62.4 )b/ft

3)(1 .7 ft)

= 171.850 lb

/?,=i(1.8ft)2 (62.41b/ft3)x(1.7 + 1.8cos30°)ft

= 329.43 lb

(i.atocc^o

Now LM, =0:1 ~LAB l/i +\-L, tl .. \P«~L AIIFI!

=013

aii 'f i '-ab* n

or -(171.850 lb) +™ (329.43 lb) -/^ =0

or 7^ = 276.90 lb F„ = 277 lb ^d. 30.0° ^

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728

PROBLEM 5.146

Consider the composite body shown. Determine (a) the value

of x when h - 272, (b) the ratio hit for which x~L,

SOLUTION

y X xV

Rectangular prism Lab2

-L2ab

2

Pyramid1 (ft>—a\ —3 \2)

h L +-h4

1 f 1 ^

6 { 4 j

Then XF^a6 L + ~h6

632?+A| L + ~h

Now

so that X

XZV^-LxV

ab\ L + ~~h)1-M

1 ^3L2 +hL + -h

or 11 +6L 6

'

h I A2

3 + ~~ +—=-

L 4L2 (1)

(a) AT = ? when A =-L2

/? 1

Substituting — =— into Eq. (J)

X 1 +in612 6

3 +1 ^ If I

2J+4l2

or104

X = 0.548L <

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729


(b) - = ? when X = LL

Substituting into Eq. (1){ 6 L) 6

r, h \hr"

3 +—+ -

or, I h 1 \ h \ h

1

6L 2 61 24 L2

or f-





730

.'/ PROBLEM 5.147

A/:

to.12 in Locate the center of gravity ofthe sheet-metal form shown.

\

-

s^ 1

V^K5>v

0.1m\0,16 t

0.05 in0.2 tn^

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.

p,= 0.18 + ^(0.12) = 0.22 m

z, = 1(0.2 m)

_ _ 2x0.18 0.36xn =yn

=- =—

m

xIV =0.34

K 71

4x0.05

In

0.31878 m

A, m 2 x,m y,ni z,m Xi4, m3yA, m3

zA, m3

Ii(0.2)(0.12) = 0.012 0.22

0.2

30.00264 0.0008

11 —(0.18)(0.2) = 0.018/r0.36

1X

0.36

n0.1 0.00648 0.00648 0.005655

III (0.16)(0.2) = 0.032 0.26 0.1 0.00832 0.0032

IV -— (0.05)2 =-0.001 25^r

20.3 1 878 0.1 -0.001258 -0.000393

E 0.096622 0.013542 0.00912 0.009262

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may he displayed,




731


We have xzv == YxV: X(0.096622 m2) = 0.013542 m3

or X--= 0.1402 m <

Y~LV~~= ZyV- ?(0.096622m 2) = 0.009 12 m 3

or Y-~= 0.0944m <

Z1V--=LIV : Z(0.096622 m2) = 0.009262 m3

or Z -= 0.0959 m A





732

!HI-|>

-3 m-

PROBLEM 5.148

Locate the centroid of the volume obtained by rotating the shaded area about

the x axis.

SOLUTION


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

dV — xr2dx

t xEl - x

I

Then

Now /

I

so that dV^nW

Then

and

Now

jt\\— +

dx

2 1

dxXX)

„=U, + U=* x - 2 In x-i

K 3-2In3-i|»|l-21nl]

(0.469447T) m-

j%^=r

n-

tc\ 1—+— \dxx x

l

2(3) + In 3

n 2x + In x

-i

2(1) + In.

= (1.0986br)m

W = \xELdV : x (0.46944^-m3) = 1 .0986 1n m

1 = <

or x~234m A

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manna/ may be displayed,



733

solucionario de estatica

Documents

x xcx3

prior written permission

individual course preparation

y problem

ato w ofrftft manual

afo orfe manual

homogeneous wire

aproprietary material