cap 5, novena edc_text.pdf

8/17/2019 Cap 5, Novena Edc_text.pdf

1/196

CHAPTER

5


2/196


3/196

30

Dim

30 in

m

PROBLEM

5.1

Locate

the centroid

of the

plane

area shown.

1

300 mil)

-240

mm-

SOLUTION

Dimensions

in

mm

-27T

«E.

'

<

U

cfiy~\4

T

/TO

4-

J

L/or|,/ar

j^i

i4,

mm

2

x,

mm

y,

mm

x./4,

mm

3

y.A,

mm

1

6300

105

15

0.661

50x1

6

0.094500x1

6

2

9000

225

150

2.0250x1

6

1.35000x10*

£

15300

2.6865xl0

6

1.44450xl0

6

Then

X

£x/5l

2.6865x10*

ZA

ZA

15300

£JM.

1.44450x10*

15300

X

=

175.6

mm

<

Y

=

94.4

mm


4/196

20

mm

30mm

-~r

36

mm

24

mm

PROBLEM 5.2

Locate the

centroid

of

the

plane area

shown.

SOLUTION

Dimensions

in mm

r

ZA

+

K

C,

-A

\0

A, mm x, mm

_y,

mm

xA, mm

3

yA,

mm

3

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

<

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

limited

distribution

to

teachers and

educatorspermittedby

McGraw-Hill

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their

individual

course

preparation.

If

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are

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Manual,

you are using it

without permission.

542


5/196

-12

in.-*+«

21

in.

-

15

in.

PROBLEM

5.3

Locate

the

centroid

of

the

plane area

shown.

SOLUTION

Dimensions

in

in.

zfirt

m

+

7*

4

in.

2

x,in.

y>

in.

x^, in;

M

in.

3

1

-xl2xl5

=

90

2

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.

^

PROPRIETARY

MATERIAL.

©

2010 The

McGraw-Hill

Companies,

Inc. All

rights

reserved.

Afo ywrc o///?/i-

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

distribution

to teachers

and

educatorspermitted

by McGraw-Hill

for

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course

preparation.

If

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are a

student

using

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you

are

using

if without permission.

543


6/196

3 in,

6 in.

6 in.

•

6 in.

6

in.

PROBLEM 5.4

Locate the

centroid of

the

plane

area

shown.

SOLUTION

4

WW

A, in.

2

x,m.

y,

in.

x^i,

in/

_y^,

in.

3

1

i(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. <

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 limited

distribution to

teachers

and

educators

permitted by

McGraw-Hill

for

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individual course

preparation.

If

you are

a

student

using

this

Manual,

you are using

it without

permission.

544


7/196

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

any

form

or by

any means,

without

the prior

written

permission

of

the

publisher, or

used beyond the

limited

distribution

to

teachers

andeducators

permitted

by McGraw-Hill

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individual

course

preparation.

If

you

are a student

using

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Manual,

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it

without

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545


8/196

.120

mm

«Ar

=

75

/

mm

ilg

\

PROBLEM 5.6

Locate the

centroid

of

the

plane

area

shown.

SOLUTION

Z5mw

1

157.5

wwi

•i

|

0Omm

©

37.

5Vm

3-n

v\v»

HS>3».**»***3'K^*

W

*

J, ram

2

x,mm jvmm

xA,mm

3

yA,

mm

3

1

(120)(75)

=

4500 80

2.5

360

xlO

3

112.5

xlO

3

2

(75)(75)

=

5625

157.5

37.5

885.94

xlO

3

2

10.94

xlO

3

3

™^(75)

2

--4417.9

4

163.169

43.169

-720.86x1

3

-190.716xl0

3

X

5707.1

525.08

xlO

3

132.724

xlO

3

Then

XA

=

ZxA

^(5707.

1)

=

525.08

x

1

3

YA

-

ZyA

F(5707.1)

=

132.724xl0

3

X

=

92.0

mm <

Y

=

23.3

mm

^

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

limited

distribution to

teachers and

educators

permittedby McGraw-Hill

for

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individual

coursepreparation.

If

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are a

student

using

this

Manual,

you

are

using

it

without

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546


9/196

PROBLEM

5.7

Locate

the

centroid

of

the

plane area

shown.

16

in.

1

20 in.

SOLUTION

tM.

ft

II*

i

A

•*Z

tO

1M.

A,

in.

2

x,

in.

J

7

*

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.

•

fee

displayed,

reproduced

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

educatorspermitted

by McGraw-Hill

for

their

individual

course

preparation.

If

you are

a

student

using

this

Manual,

you are

using

it

without permission.

547


10/196

PROBLEM

5.8

Locate the

centroid of

the

plane

area

shown.

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

=353.43

2

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.

<

PROPRIETARY

MATERIAL.

CO 2010 The


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

limited

distribution to

teachers

and

educators

permitted

by

McGraw-Hill

for

their

individual course

preparation.

If

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are a student

using

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you

are

using

if without

permission.

548


11/196

() iniii

PROBLEM

5.9

Locate

the

centroid

of the

plane

area

shown.

60mm

SOLUTION

^l^-^&.^S

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

distribution

to teachers

and

educators

permitted

by McGraw-Hill

for

their individual

course

preparation.

If

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are

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this

Manual,

you

are

using it

without

permission.

549


12/196

If

Semicllipse

.

70

nun

47 mm

47 mm

PROBLEM

5.10

Locate the

centroid ofthe

plane area

shown.

26

mm

SOLUTION

Dimensions

in mm

T%

kUi-t)

A,

mm

2

x,

mm

y,

mm

jtvf,

mm

3

jvi,

ram

3

1

~x47x26

=

1919.51

2

11.0347

21181

2

-x

94x70

=

3290

2

-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

<

PROPRIETARY

MATERIAL

C©

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

limited

distribution to

teachers

and

educators

permittedby

McGraw-Hill

for

their

individual course

preparation.

If

you

are

a

student using

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Manual,

you

are

using it

withoutpermission.

550


13/196

PROBLEM

5.11

Locate

the

centroid

of the

plane area shown.

/\

A

\

/'

\

.

Ju

iiui.

'i

•

Sin.

\

id

\/

SOLUTION

First

note that

symmetry

implies

X

=

M

,4,

in.

2

7,

in.

jv*, in.

3

1

-*

(8)

'

=-100.531

2

3.3953 -341.33

2

*

(12)2

=226.19

2

5.0930

1151.99

I 125.659

810.66

Then

r

2y^ 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 any

form

or

by any means,

without

the

prior

written

permission

of

the

publisher, or used

beyond the limited

distribution

to

teachers

andeducatorspermitted

by

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for

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If

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551


14/196

50 nun

15

mull

PROBLEM

5.12

Parabola J-

r

Vertex

\.-

*'

-

1

80

mm

*-

Locate

the centroid of the

plane area

shown.

X

SOLUTION

»>,

mm

X/4,

mm

3

yA,

mm

3

1.

(15)(80)

=

1200

40

.

7.5

48xl0

3

9xl0

3

2

i(50)(80)~

1333.33

60 30

80xI0

3

40xl0

3

I

2533.3

128xI0

3

49xl0

3

Then

X

(2533.3)

=

128

xlO

3

YA^ZyA

7(2533.3)

=

49

xlO

3

^

=

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,

reproduced

or distributed in anyform

or

by

any means,

without

the

prior

written

permission

of

the

publisher, or used

beyond

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distribution to

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552


15/196

20;

U

1

~

\

x

i

mm

1

f

x=ky*

20

nun

I

*—

30

mm-*

X

PROBLEM

5.13

Locate

the

centroid ofthe

plane

area

shown.

SOLUTION

i

.

4*y$

+***\

^fflffl

2

x, mm

^,

mm

xA, mm

3

yA,

mm

3

1

1x30x20 =

200

3

9

15

1800

3000

2

™(30)

2

=

706.86

4

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 any

form

or

by

any

means,

without the prior

written

permission

of

the

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course preparation.

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553


16/196

f

/

^

\

*

20

in.

»

X

SOLUTION

Dimensions

in

in.

*1

©

•»+

Aio)

|M»>

tt

-no)

\o>

/*,

in.

2

x,in.

yM-

j/4,

in

3

yA,

in.

3

I

|x(20X20)

=

552

12

7.5

3200

2000

2

ZL

(20

)(20)^

15

6.0

-2000

-800

£

400

3

1200

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

PROPRIETARY

MATERIAL. ©

2010 The


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rights reserved.

No

pan

of

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permission

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554


17/196

y

PROBLEM

5.15

r

^Vertex

\ ^Parabola

Locate

the centroid

of the plane

area

shown.

60

nm

\

60 11

m

\4

fi§§

*—

—

75 mm

*\

SOLUTION

Then

and

iWvn

(fWs.fcA')

vwv*\

©

\

jb

)WM

/4,

mm

2

x,mm

j,

mm

xA, mm

3

—

—

^

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

mm

3

YXA

=

'LyA

7(3750

mm

2

)

=

243000

or

X-

30.0

mm

<

or

7

=

64.8

mm

4

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

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limited

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preparation.

If

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are a

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using it without permission.

555


18/196

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

cosa

3'(f-«)

4

#

ff r,

__

. 2

cos

a

Ly/4

=—

r

2

/r

3'(f-«)

-(r

2

3

-r,

3

)cosOf

a

r

2

cos«

.

3'(i-«)

ft

I

7

and

Now

Y

ft

a

YIA

=

ZyA

-{$

- Oleosa

K

=

2

cos

or

PROPRIETARY

MATERIAL ©

2010 The

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Alt

rights reserved. No part

of

this

Manual

may

be

displayed


any

form

or by any means,

without

the prior

written permission

of

the

publisher, or used

beyond the

limited

distribution to teachers and

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556


19/196

«

\

1

PROBLEM 5.17

Show

that

as r

}

approaches r

2

,

the location

of the centroid

approaches

that

for

an

arc of

circle

of radius

(/]

+r,)/2.

SOLUTION

U®

First,

determine the location

of

the centroid.

2

sin

(f-

or)

From

Figure 5.8

A:

Similarly

* 3

(f-tf)

2

cos

3

2

(f-«)

_ 2

cos

#

^i=r'i

3

l

(f-


20/196

PROBLEM

5.17 (Continued)

Using

Figure

5.8B,

Y ofan

arc of radius -~(t\ +r

2

)

is

y

=

1

sin(f-a)

=

— (r,

+ r

7

)

2

Vl

2;

(f-or)

1

.

. COS«f

= —(?•,

+7%)

2'

2

'(f-a)

(1)

Now

3

r

2 -n

a

fra-'i)('2

+'i'2

+

'i

2

)

(

f

2~

r

\)i

f

2

+f

\)

Let

>2

=

r,

=

=

r + A

~r-A

Then

/*

=

and

.3

>2

-I

3

.

_

(r + A)

2

+

(i-

+

A)(r

-

A)(r

-

A)

2

1

r

2 -n

2

(r

+ A)

+ (r~A)

3r

2

+A

2

2r

In the

limit

as

A

-

-

(i.e., t]

-

r

2

),

then

-I

2

3

2

=—

x—

(r,

+r

2

)

2 2

So that

F

=

2

3,

.cosa

=

—

x—

(k

+r

7 )

3 4

V)

2

~\~a

-

cosa

.

or Y={r

x

+r

2

)

-~-

A

K~2a

Which

agrees with Equation

CD-

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558


21/196

/':

PROBLEM

5.18

For

the

area

shown,

determine

the

ratio

alb

for which

x

-

y.

SOLUTION

Then

or

or

Now

X

XZA^XxA

X\

~ab

6

£b

12

X

J

a

rLA^yA

Y

ah

ab

l

15

5

X

=

Y=>

1

2.

—a~~b

2 5

4 X

V

jM

y^

1

loft

3

3

8

1*

5

a

2

b

4

lab

2

5

2

2

1

—a

3

3

6

3

E

—a/;

6

12

£*1

15

a 4

or

—

-

—

-^

/?

5

PROPRIETARY

MATERIAL. ©

2010

The

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of

(his

Manual may

be

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reproduced

or

distributed in any

form

or by

any

means, without the prior

written permission

of

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559


22/196

y

'*»

=

.12 in.

\

is

/

Bl _

PROBLEM

5.19

For

the

semiannular

area

of Problem 5.11,

determine

the

ratio r

2

/t'\

so

thaty

=

3/]/4.

X

SOLUTION

Then

or

Let

(Q

-vn

/T Y

^

]

2

l

3/r

3

2

71

o

2

2

4^

3^-

3

2

X

%4-t)

f('

2W)

YZA^ZyA.

3

7F

i^-^K^-n

3

)

9#

16

^

'i>

P

—[(/>

+

l)(p-l)]==(/>~l)(/>

2

+

/>

+

)

16

or

1

6/7

2

+

(1.6

-

9^)p

+

(1

6

-

9/r)

=

PROPRIETARY

MATERIAL.

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All

rights reserved. No

part

of

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displayed,

reproduced

or distributed

in

any

form

or by

any means,

without the

prior written

permission

of

the

publisher, or used beyond the

limited

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560


23/196

PROBLEM

5.19 (Continued)

(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

^

PROPRIETARY

MATERIAL.

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2010

The


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AH

rights

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pari

of

this

Manual

may

be

displayed,


any

form

or

by any

means, without the prior written permission

of

the

publisher, or used beyond

the

limited

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to teachers

andeducators

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individual

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If

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are a student

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you are


561


24/196

-300 mm

-

12

mm

PROBLEM

5.20

W0&

60 mm

I

'11

1.2 mm

—*\

12 mm

450

min

c

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

22C*w*

From

the

problem statement:

F is

proportional

to

Q

x

.

Therefore:

F„

(Q

x

)a


25/196


26/196

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.

Q

x

=^y,A

(Or).

(Q

x

):

2

(5.5-4.0833)in.

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

2


27/196

PROBLEM

5.22

(Continued)

Now

Or

=

(Q,\-

K&:-0)

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M> /wrt


28/196

V

„-

-

(

1

11

}

C

X

c;

J

—

—

*~

PROBLEM

5.23

The

first moment of the shaded

area

with respect

to

the x

axis is denoted

by

Q

x

.

(a) Express

Q

x

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 O

x

maximum,

and

what is

that

maximum value?

SOLUTION

Shaded area:

VA4///

97777Z

^(t+y)-y

y,

A--

=

%-^)

c

a

=

=

}H

CL

=

i(c

+

^)[6(c-^)]

^H

(fl) Q

x

=

=

i%

2

/) «

(b) For2

max

:

d

Q.

dy~

-0

or

-b(-2y)

=

Q

;•-

=

()

^

For y~0:

(fir)

=

2

(&)^fc

2

«

PROPRIETARY

MATERIAL. ©

2010 The

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Companies,

Inc.

AH

rights

reserved. Wo /wwf o/rfm

Mmim/

j»qv

'6c displayed,

reproduced

or

distributed

in

any

form

or

by

any means, without the prior

written

permission

of

the publisher,

or

used beyond the limited

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566


29/196

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 ofFigure 5.1

xw -J,


30/196

20

mm

30

mm

36

nun

,t

24

nil )

PROBLEM

5.25

A thin,

homogeneous

wire

is

bent to

form the

perimeter of

the

figure

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

ofthe

corresponding line.

nm

A

nm

A

1

|\$

/,,

mm

x,

mm

y,

mm

xL,

mm

2

yL,

mm

2

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

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2010

The

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Companies, Inc. AH rights reserved.

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be

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any

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or by any means, without

the

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568


31/196

/

PROBLEM

5.26

—

l2in.-4*

21 in.

*

t

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.

/

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

Vl2

2

+15

2

=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.

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any

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569


32/196

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 of

gravity

will

coincide

with

the

centroid

of the

corresponding line.

K

6

=—

(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

X

liXL

-320

ZL

227.38

£7^,^

3464.1

£/.

~

227.38

X-

-1.407

in.

^

F

=

15.23 in.

^

PROPRIETARY MATERIAL.

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/«aj>

Ae displayed,

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form

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any

means, without the prior writ

ten

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570


33/196

PROBLEM

5.28

A

uniform

circular

rod

of

weight

8

lb and radius

10

in.

is

attached

to a pin at Cand

to the

cable AB.

Determine

(a)

the

tension

in

the cable,

(b)

the

reaction

at C.

SOLUTION

For quarter

circle

(«)

+)XM

c

=

0:

W

~

-rr

=

2r

__2r

7t

T^W

(81b)

(b)

±^XF

x

=0:

T~C

x

=0

5.09lb-C

r

=0

+\XF

y

=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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farm

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means, without

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571


34/196

PROBLEM 5.29

Member

ABCDE

is

a

component of

a mobile and

is formed

from

a

single

piece

of

aluminum

tubing. Knowing

that

the

member

is supported at

Cand

that

/

=

2m,

determine the distance d so

that

portion

BCD of the member

is horizontal.

SOLUTION

First note

that

for

equilibrium, the

center

of

gravity

of the component must lie on

a

vertical

line through C.

Further,

because

the tubing is

uniform, the

center

of gravity

of

the component

will coincide with

the

centroid

of the corresponding

line. Thus, X

=

H

c

So that

Then

ZxL

=

d

0.75

2

cos55<

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 <

PROPRIETARY

MATERIAL. ©

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S72


35/196

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

/ of

aim

DE

so that this

portion

of

the member is

horizontal.

SOLUTION

First

note that

for equilibrium,

the

center

of

gravity

of

the

component

must

lie

on a vertical line

through C.

Further,

because

the

tubing

is

uniform,

the

center

of gravity

of the component

will coincide

with

the

centroid

of

the

corresponding

line.

Thus,

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


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573


36/196

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.

\y

Cc

: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

'

'J

V

rto^B

rcosO

](/')+

rcos0

\(xr)-0

cos

6

4

\A-l7t

0.5492.1

or

#

=

56.7°

<

PROPRIETARY

MATERIAL. ©

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All rights

reserved.

A'f>

part

of

this Manual

may

be

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form

or by

any

means, without

the

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permission

of

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publisher,

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If

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574


37/196

:

—

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

ha

2

i.

—a

3 6

2

~(kb)h

~h

3

-~kbh

2

6

£

~(a-kh)

t{a

2

~~kh

2

)

6

Then

or

and

or

YZA

=

LyA

~{a-kh)

2

r

-(a

2

-Aft

2

)

6

«

2

-

A/?

2

dY

1 -2*%

-

Mr)

-

(a

2

-

*ft

2

)(-*)

(1)

dh

3

2h(a-kh)-a

2

+kh

2

=0

Simplifying

Eq.

(2)

yields

k/r-2ah

+

a

2

^0

(a-khy

(2)

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575


38/196

PROBLEM 5.32

(Continued)

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

^

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MATERIAL. ©

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any

form

or by

any

means,

without

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prior written

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of

the

publisher, or

used

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576


39/196

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_a

2

-kh

2

Xa-kh)

(1)

2h(a-kh)~a

2

+kh

2

=0

(2)

Rearranging Eq.

(2)

(which defines the

value

of/?

which

maximizes

/)

yields

a

2

-kh

2

=2h{a~kh)

Then

substituting

into

Eq.

(1)

(which

defines Y)

f

=

-—

x2h(a-kh)

3(a~kh)

or

F

= -/?

<

3

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the

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577


40/196


41/196

PROBLEM

5.35

Determine

by

direct

integration

the

centroid

of the

area

shown. Express

your

answer

in terms

of

a and

h.

SOLUTION

At {a,

h)

or

or

Now

and

Then

and

)\

: h

=

-~ka

A

k

=

h

~~

2

jv

A

=

~

ma

m-

h

a

yEL

d.A

(^+^2)

(y

2 ~yO^

h

f

2

W

—(ax

—

x

)ax

h

h ,

—x

—

~x

a

a

clx

A=\dA^

VJL{ax-x

2

)dx

\*ElM

—(ax-x

)dx-~

J

I

iJliil

a*

a

2

1

3

~X

XT

2

3

_

—

—ah

6

a

3

1

4

—x

x

3

4

o

JjO'i

+

j

2

X0'

2

-*)*]

=

Ji(

2

2

-

yf)dx

=

~a

l

h

12

PROPRIETARY

MATERIAL

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2010

The

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Inc. All

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reserved.

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may

be

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reproduced

or distributed

in

any

form

or

by

any

means, without

the prior

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permission

of

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used

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579


42/196

PROBLEM 5.35

(Continued)

xA~ \x

Fl

dA:

x -ah

.6

j

=

-a

2

h

1.2

x =—a

^1

2

yA^p

EL

d.A\

y

—ah

1

/2

—-aft

15 5

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or distributed in any form

or by

any

means,

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permission

of

the

publisher, or

used

beyond

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580


43/196

il

7

W

t

\

PROBLEM

5.36

Determine

by

direct

integration

the

centroid

ofthe

area

shown. Express

your

answer

in terms

of

a and h.

SOLUTION

For

the

element

(EL)

shown

At

Then

Now

Then

and

Hence

x-a,

y-h: h

=

ka

2

or k

1/3

x

=i*y

dA^xcfy

=

-^-y

m

dy

x

EL

=-x

2 2

h

m

y

1/3

\x

El

dA=l

4

h

m

m

-ah

-.Wf

-fL

V

W

di

,]

==

l^_f

3

^

A'

ft

2A»*1^

^J-I^-ll*

J^-f^y°*l=^

TO

X//

\x

IiL

dA\

h

m

\l

x\ —ah

\

=

—

2

/?

4

J

JO

-ah

2

7

y

A

~\yEi

dA

'-

y

'|

--«/?

\

=

:

-ah'

—

**.

u

r

NX\*\\\

i

m

i

*~*

-»

*

10

X

=

—

tf

7

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MATERIAL.

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f

>/7//w M»»/a/

iimiv

/« displayed,

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or

distributed

in

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form

or

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the prior

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581


44/196

a-

b-

PROBLEM 5.37

Determine

by

direct

integration

the

cenlroid

of

the

area

shown.

SOLUTION

For

the element

(EL) shown

and

Then

To

integrate, let

Then.

and

h

r~

i

a

dA^(b-y)dx

=

t

a

JCe-i

=X

\a-4a

2

-x

2

}

dx

-dx

Mel.

V

£L

1

yi:i^-(y

,+b

)

a

+ yj

la

2 ..2

a

~x

b

o

a

a~\a -x

\dx

x

=

a

sin

9:

*Ja

2

-

x

2

=

a

cos

6,

dx~a

cosOdd

pan

f}

A=

—

{a-

a

cos

9){acos$d$)

Jo

a

a

sin

-

a

—

+

sin

/r/2

£?6

1

tt

K^

=

r

a -4

2

-x

2

)dx

—

*-+-(«

-x

)

2\3/2

/r/2

*>»

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MATERIAL. © 2010

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Maiiual

may be

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distributed in

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any

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SS2


45/196

PROBLEM

5,37

(Continued)

f^-ri(«

+

^[£(-^*

2

(

3^

a:

2a

6

2^

v

3

;

xA

=

\x

EL

dA:

x

yA=jy

EL

cM:

y

ah

ab\

1

7t

71

1

2

a'h

1

,2

ab'

—

2a

.

or

x

=

^

3(4-^)

2/?

or

y

=

^

3(4

-/r)

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MATERIAL.

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or by

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583


46/196

,/\

PROBLEM 5.38

Determine

by

direct

integration the

centroid of

the

area shown.

:

ss

SOLUTION

First note that

symmetry implies

A

For

the

element (EL)

shown

2r

Then

and

y

E[

=

—

(Figure 5.8B)

n

dA

-

fcrdr

A-

\dA~

|

rcrdr-K

2

2/

( 2\

r

,

2

,

It

2

2

r

2

~

r

\

\n,,A-\l^r

dr)

=

2\f

--(

/

3

h ~n

So

yA~\y

E

iM'-

y -(^-'f)

=

-^-n»

or

^^

—

—

^

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MATERIAL.

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(he prior written

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584


47/196


48/196

y

=

kix-(ir-

PROBLEM

5.40

Determine

by

direct

integration the

centroid

of the

area shown. Express

your answer

in terms of

a

and

b.

SOLUTION

At

Then

Now

0, y

=

b

k(0-a)

2

or

k.

x

b

v

=

\ix-a?

y

h

f

^2

-

=

—

(x-a)

-X-

M«L

and

cIA

=

ydx

=

—

—

(x

—

a)' dx

Then

and

A-fr-fa-faMx-af]

=

~ab

o

3

K^r

—{x-d)

2

dx

a

a

f

x

3

-

lax

2

+

a

2

x)dx

a

bix*

2

3

^

a

2

1

2,

21

4

3 2

12

K

clA

Jo

2a

2

1

,2

—

aft

-^(Jc-a)

2

^

2a'

7*-*

-0

Hence

xA

—

vA

—ah

3

J 12

'a

X/^dA:

x

\ymdA:

y[~abU-^ab

\ A

1

.2

4

'

10

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586


49/196

Vi

=

h^

PROBLEM

5.41

Determine

by

direct

integration the

centroid

of the

area shown.

Express

your answer

in

terms ofa and

b.

SOLUTION

y

l

-k

]

x

2

but

b-k

x

a

2

y

l

=~x

2

y

2

-k

2

x

4

but

b~k

2

a

4

y

2

dA

=

(y

2

-y

x

)dx

b

.2

*

4\

dx

X

EL

~

X

yEL^-iyi+yi)

h^v

'

t

4\

X

—

dx

\-,M=[

4 6

X

X

4

6a

12

-a

l

b


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written

permission

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587


50/196

PROBLEM

5.41

(Continued)

jy^A=[

xA

=

x

F

,

dA

:

x

—

ha

,

J

LL

\

15

J

12

a

l

h x

=—a

yA=\y

£

i.dA:

y

r

2,\

2

,

2

15

ba

45

aff y^-b

<

3

PROPRIETARY

MATERIAL.

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588


51/196

A

^-i

+

i3

PROBLEM

5.42

Determine

by direct

integration

the

centroid

of

the

area

shown.

SOLUTION

We have

Then

and

/'*/

*

(

y

*-°V l.

dA

=

ydx

=

a

.2

\

X X

I

+

^r

f

2\

X

X

t/x

8

2L

f

X

X^

a\

1

+

—

dx

—

a

3

-ah

K^

=

r

f

2>

X X

l-

+ dx

~a

^

L

1/j

x

2

x

3

—

+

—

2/.

3.L

2

-.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

5

X X X

X

x

4.

_

_

+

n2/.

I

7/ 2I

j

St

5

\x

EL

dA:

-aL

y

A

^\yr-iM-

y\-

3

11

x=~L

4

4

_ 33

.

y

=

—a

^

40

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58


52/196

x

=

kij*

%-

.'

b

,2

b

2

a

6.

X

PROBLEM

5.43

Determine

by

direct

integration the

centroid

of the

area,

shown.

Express

your

answer in terms of

a

and

b.

SOLUTION

For_y2 at

Then

Now

and for 0


53/196

PROBLEM

5,43

(Continued)

and

[x

Fl

dA~

\

l2

x\b~chU

f

.

J

J

Jn

Ja/2

yfa

a

2

j

dx

-ia/2

+

b

2

x x x

+

2 b

5*fa

+b\—

l

-

240

f ^

5/2

K^-J

5

V«

3a

4

/ \

5/2

+ (a)

5/2

00

3

«

V

1

+

—

)

4

all

(*-if

/>* -£

,/2/,

x

'/2

x

2

b—f=rdx

c

a

b\

x

1 x

+

+

-

•>«/2

2

a 2

^

1/2

\

Ja

a 2

b

2

\\

>'

on

/>2

f

—XT +

—

2a

2 2

V

x

2

1

(x

2a

3a

\

a

2

dx

,3\

all

b_

4a

'

a

)

>

a

6a{

2

2

Hence xA

-afr

48

^,,

:i

dA:

13

/>

7l

2,

—

a/>

=

a

b

24

J

240

*=—

a

=

0.546a

^

130

yA^

jy

EL

dA:

y\~ab

11

/2

—ao

48

y=—

6

=

0.423/;

^

26


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/>«/•/

o/rAra

A/


54/196

PROBLEM

5.44

Determine

by

direct

integration the

centroid of

the

area

shown.

Express

your

answer in terms

of

a

and

b.

SOLUTION

For

V]

at

Then

By

observation

Now

and

for

0S.vEL

—

y

2'

a

—x

and dA

—

y

v

dx

—

—rx

dx

—y

2

=—\

2——

I and dA=

y

2

dx

—

b\

2

—

-

\dx

2 2

{

a

\

a

A

=\

dA

=&

d

**t

b

V

dx

26

a

+

b 2

—

-\2a

-0

ab

2b_

*>

a

I

2

M-«?^hD

4

+

6

/;|

2--|rfx

n2«

3a

-0

a

2

b + b\[(2a)

2

-(a)

2

]

+ ~-[(2a

2

)-(a)

7

2/

~a

b

6

'**».

PROPRIETARY

MATERIAL

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592


55/196

PROBLEM 5.44

(Continued)

jy£L


56/196

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

x

EI

-

a cos

3

and

dL

—

^jdx

2

+

dy

2

x~a

cos

3

$: dx

=

~3a cos

2

sin

9d9

y-asur'

9:

dy

-

3a

sin

2

9 cos Od'6

dL

=

[(-3a

cos^

9

sin

0d6f

+

(3a

sin

z

9

cos

OdGf

]

=

3a cos

9

sin

0(cos

2

+ sin

2

0)

m

d9

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

and

f- C*

n

3

x

EL

dL— \

a

cos~

(3acos

sin

0d0)

3a

2

-\7ll2

cos

5

-0

-la*

5

Hence

Alternative

Solution

xL

=

\x

E/

dL:

-I

3

x\

—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~(a

m

-x

m

f

2

PROPRIETARY MATERIAL. © 2010

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or

distributed in

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or

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any means,

without

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permission

of

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publisher, or used

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it

without permission.

594


57/196

PROBLEM

5.45 (Continued)

Then

Now

dx

*s=*

and

*-Hi

dx

J\

+ [(a

m

-

x

m

f\~x-

m

)]

2

r

dx

Then

p-r

1/3

1/3

dx~a

If

2

and

Hence

5

PROPRIETARY

MATERIAL.

£5

2010

The


Inc.

AH rights reserved. No

part

of

this Manual may

be

displayed,

reproduced or

distributed in

any

form

or

by

any

means, without

the

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written

permission

of

the publisher, or used beyond the

limited


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for

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preparation.

If

you

are a

student using

this Manual,

you are

using

it

withoutpermission.

595


58/196

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

x

El

~rcos$

and di~rdd

r

el*'

4

-,

/, ^

Thus

•7/T/4

\x

EL

dL= r

cos

0(rdO)

rising

nIA

2

i_

=

-r

2

V2

xL- YxdL:

x

—

itr -~r

2

v2

_

2V2

3tf

PROPRIETARY

MATERIAL

©

2010 The

McGraw-Hill

Companies, Inc. All

rigiiis

reserved.

No

part

of

this Manual may be displayed,

reproduced


form

or by

any means, without the prior

written

permission

of

the

publisher, or

used

beyond the limited

distribution

to

teachers

andeducatorspermitted by McGraw-Hill

for

their individual

coursepreparation.

If

you are a student using

this

Manual,

you

are


596


59/196

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

-4

a

dy

=

3

dx

2^a

1/2

dL~J\

+

1

+

1

\dx

j

dx

f

3

V

->

.1/2

X

1/2

dx

24a

a + 9x

dx

L= |rfL

=

r'-^~^a

+

9xdx

J Jo

l4a

\

24a

a

27

1.4397

k

-x-(4a

+

9xf

2

3 9

V }

[(13)

3/2

-8]

M-f

yl4a +

9xdx


©

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


distribution

to teachers and

educatorspermitted

by

McGraw-Hill

for

their

individual

coursepreparation.

If

you

are a student

using this

Manual,

you are

using it withoutpermission.

597


60/196

PROBLEM

5.47*

(Continued)

Use integration

by

parts

with

u

—

x

dv~

\}4a

+

9x

dx

da

~dx v

-

—

(4a

+

9x)

27

3/2

Then

K

dL

xx—(4a

+

9xf

2

27

-0

{

l

'~(4a

+

9x)

i/2

dx

]

Jo

27

(4a

+

9x)

5/2

i3)

3/2

al

1_

27

°

27>/al45

27j

(t3)

3/2

~^[(13)

5/2

-32]

=

0.78566a

2

xL

=

\x

EL

dL

:

x(\

.4397

la)

=

0.78566a

2

or

x

=

0.546a

A

PROPRIETARY MATERIAL. ©

2010 The

McGraw-Hill

Companies,

Inc.

AH rights reserved. No part

of

this

Manual

may be

displayed,


any

form

or by

any

means, without the prior

written

permission

of

the publisher, or used beyond

the

limited

distribution

to

teachers

andeducators

permitted

by

McGraw-Hill

for

their

individual

course

preparation.

If

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area


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are

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permission.

598


61/196

T

XX

l.--'.--.V:0

;

.

\

-«

H-<

»

L L

PROBLEM

5.48*

Determine

by

direct

integration

the

centroid

of the

area

shown.

SOLUTION

We have

and

Then

and

X

EL=

X

I a

kx

y

~

—

y-—

COS

LL

2

2

21,

KX

dA

=

ydx

=

a

cos

dx

21

./4

=

IcW

=


62/196

PROBLEM

5.48*

(Continued)

Also

\y

KI

M-[

xA

112

a

Ttx

—cos

—

2 2L{

21,

....

.

JIX

cos

— a cos

—

ax

. 2

4

In

0.20458^

2

I

K

(

dA

:

x

s

ah

v

*

j

yA~

jy

gL

dA:

4i

71

ah

-0.106374a/:

0.20458

A

or x=

0.2361

A

or

y

-

0.454a ^

PROPRIETARY

MATERIAL.

©

2010 The McGraw-Hill

Companies,

Inc.

All rights reserved.

AV;

yx/r/ o/ /Mv

M»wa/

»mj> Ae displayed,

reproduced

or distributed

in any

form

or

by

any

means,

without

the prior

written

permission

of

the publisher,

or



and

educators

permitted

by McGraw-Hill

far

their individual coursepreparation.

If

you are

a

student using

this Manual,

you

are using it

without

permission.

600


63/196

PROBLEM

5.49*

Determine

by direct integration

the

centroid

of

the

area shown.

SOLUTION

We have

2 2

x

FL

-~rcosO

=—

ae°

cos

6

2 2

V

=

r sin $

—

~-ae°

sin

6

L

3

3

and

Then

dA

=

~{r){rd&)

=

-a

2

e

26

d0

A

M

?

n]

~a

2

e

ld

d0^a

2

2 2

—e

2

2ff

-0

a

2

{e

2

*-\)

and

=

133.623«

2

r*2

jx

EL

dA^

j*-ae

d

cos&\

~a

2

e

2O

d0

ro

ia

cap 5, novena edc_text.pdf

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