Solution manual vector mechanics engineers dynamics 8th beer chapter 09

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P.. Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P.. Vector Mechanics for Engineers: Statics

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a Have Iy =∫x dA2

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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

3

13

a

y dyb

3 0

1 5

3 11

b

a yb

=

11 5 6

3

5

33a bb

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b a

y b dyb

y

I = a b

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2 4

4

a a

y

I = a b

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a b b

x b dxa

−( )( ) ( 2 2)

2 1

112

−( ) ( 2 2)

x

I = a b +b b +b

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I = ∫y dA dA= xdy

2 2

2 0

17

x

I = ab

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b b

a

y y b dyb

x

I = ab

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= −

2 44

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sin 1 sin cos

x

I = πab

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I = ab

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a

b

e dxe

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I = πa b

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2

a a

b

x dxa

( )

10 3 1

2

3102

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or 1

1

bk

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1 1

4

2 1 0a x x4

aa

5 4 0

I = ∫y dA dA = x −x dy

1 1

4 0

b

bb

7 4 0

abI

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= ab

1 2

2 2 0

2 a b x a

b x a

y dydx

= ∫ ∫

3 2 3

3 3 0

3

4 3 0

3

593053

abab

=

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4

2 1 0a x x4

aa

5 4 0

I = ∫x dA dA = y − y dxThen Iy 0ax2 b1 x1 b4x4 dx

aa

6 4 0

a x x

aa

7 4 0

a bI

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= ab

Have Iy =∫x dA 2

1 2

2 2 0

2 a b x a

b x a

x dydx

= ∫ ∫

1

2 0

2

2 ax b x bx dx

aa

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kA

3

91453

a bab

=

2770

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a a

b x

a b xa

y dydx

π π

a a

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0.217ab

x IkA

ππ

−

=

−0.642b

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a a

b x

a b xa

x dydx

π π

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1.48228

kA

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112

122

O O

aJ

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2 a bb x a

3 0

= ab

2 20

2 a bb x a

x dydx

−

= ∫ ∫

2 0

= a b

continued

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ab a bab

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a

aa

a y

I = ∫x dA= ∫ x y − y dx

2 0

2

aa

4

5 2 3 2

continued

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P P

ab

Jk

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0

2 r r r y x

4

=

2 6

1 sin 22

x

I = r ∫ππ θ θd

2 4

6

π π

0

2 r r r y y

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Now cos4 cos2 (1 sin2 ) cos2 1sin 22

6

π π

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2 6

6

sin 22

r

π π

ππ

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J = π R −R (b) Now Ix = ( ) ( ) ( )Ix 1+ Ix 2 + Ix 3

x y

I = I = π R −R

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Chapter 9, Solution 26

2 1

38

4

O O

14

m O O

m

tRtR

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 

− +   

 +   

 

− +   

 +   

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O

J = πa

continued

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8a

π

64

O a

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First the circular area is divided into an increasing number of identical circular sectors The sectors can be approximated by isosceles triangles For a large number of sectors the approximate dimensions of one of the isosceles triangles are as shown

For an isosceles triangle (see Problem 9.28)

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( ) 4 cir 2

C

J = π rThe area of the circle is

2 cir

C

J = ∫r dA;

as r decreases, so must J ) CSolution 1

Imagine taking the area A and drawing it into a thin strip of negligible width and of sufficient length so that its area is equal to A To minimize the value of ( )JC A,the area would have to be distributed as closely as possible about C This is accomplished by winding the strip into a tightly wound roll with C as its center; any voids in the roll would place the corresponding area farther from C than is necessary, thus increasing the value of ( )JC A (The process is analogous to rewinding a length of tape back into a roll.) Since the shape of the roll is circular, with the centroid

of its area at C, it follows that

( ) 2 Q.E.D

2

C A AJ

π

where the equality applies when the original area is circular

continued

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Solution 2 Consider an area A, with its centroid at point C, and a circular area of area

A, with its center (and centroid) at point C Without loss of generality, assume that

It then follows that ( )JC A = ( )JC cir +J AC( )1 −J AC( )2 + J AC( )3 −J AC( )4  Now observe that

( )1 ( )2 0

J A −J A ≥ ( )3 ( )4 0

J A −J A ≥ since as a given area is moved farther away from C its polar moment of inertia with respect to C must increase

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Part : (Same as Part ) I =x 303.3 10 mm× 3 4

Thus for entire area:

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449.33 in

64 in

=2.65 in

=

2.65 in

x

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Part : (Same as Part ) Iy =643.3 10 mm× 3 4

Thus for entire area:

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3 2

8.4375 in 0.9375 in.

9.00 in

xAxΑ

Σ

ΣDetermination of I x:

Part : 1 3 in 0.75 in.( )( ) (3 3 in 0.75 in 3.375 in.)( )( )2 25.734 in4

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A

Σ

=Σ

=

1 84 mm 105 mm 8820 mm 52.5 mm 48.0 mm12

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=Σ

3 2

Σ

=Σ

3 2

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Α

Σ

=Σ

=

i

y Ay

A

Σ

=Σ

=Now Ix =( ) ( )Ix 1+ Ix 2

1 40 mm 90 mm 3600 mm 1 mm12

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( )( )3 ( 2) ( )2

1 48 mm 30 mm 720 mm 25 mm36

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Σ

=Σ

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C 2

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Σ(a) Have Ix =( ) ( ) ( )Ix 1+ Ix 2 − Ix 3

=( ) ( ) ( )1 2 3

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1 12 in2

Α

Σ

=Σ

16 in 72 in 4 in 96 in

72 in 96 in5.8989 in

ππ

( )4 ( )( )3 4

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Shape Data: Fig 9.13A

28.36 5.62 249.38 in 263.36 in

304.8 in22.06 in

263.36 in22.06 in

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i

x Ax

Α

Σ

=Σ

=

i

y Ay

Α

Σ

=Σ

=Now Ix =( ) ( ) ( )Ix 1+ Ix 2 + Ix 3

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1 27 in 0.8in 1.152 in12

y

continued

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AyYA

Σ

=Σ

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1 10 mm 200 mm 6.6667 10 mm12

y

AyYA

Σ

=Σ

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Also Iy = 2( )Iy angle +( )Iy plate

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(a) Using shape data from Fig 9.13A

28.44 3.09 10.8617 in

yAyΑ

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From Sec 9.2: R=γ∫ydA

2 AA

M ′ =γ∫y dALet yP =Distance of center of pressure from AA′

P

y = h

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2 AA

M ′ =γ∫y dALet yP =Distance of center of pressure from AA′

P

aby

πππ

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2 AA

M ′ =γ∫y dALet yP =Distance of center of pressure from AA′:

Iy

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2 AA

M ′=γ∫y dALet yP =Distance of center of pressure from AA′

52

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Then

4 3

1.238216 m 3.001736 m0.4125 m

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FBD of Gate: For gate to open:

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xdV

x x dAx

2

z A A

I

x dAxdA xA

×

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x V =∫x dV = ∫x kx dA = k x dA∫ = kI Where Iz = moment of inertia of area with respect to z-axis

= = which is the same as for center of pressure

For circular area:

54

z v A

aI

x

x A a a

ππ

4

v

x = a

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or, using Equation 9.13, or My′ =γIx y′ ′sinθ

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P =∫dP = ∫γy θdA =γ θ∫ydA( )

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or P Ixy

xAy

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First note

2 2

14

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a a

3 3

0

a b

b x a

12

xy

I = a b

Trang 90

xy

I = b h

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0 0

x a

a

b

xe dxe

2

2 2

0

x a

a

ae

8

a b ee

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Note: Orientation of A corresponding to a 180° rotation of the axes Equation 9.20 then yields 3

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