Solution Manual - Vector Mechanics Engineers Dynamics 8th Beer Chapter 10 PDF

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COSMOS: Complete Online Solutions Manual Organization System Chapter 10, Solution 1. Link ABC: Assume clockwise Then, for point C xC (125 mm ) and for point D xD xC (125 mm ) and for point E 250 mm 2 xD xD 3 375 mm xE Link DEFG: xD ( 375 mm ) Thus (125 mm ) ( 375 mm ) 1 3 ( ) 100 G 100 2 mm 3 Then 2...

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COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 1.

Link ABC: Assume δθ clockwise Then, for point C

δ xC = ( 125 mm) δθ and for point D

δ xD = δ xC = ( 125 mm) δθ and for point E

 250 mm 

2

δ xE =   δ xD = δ xD 3  375 mm  Link DEFG:

δ xD = ( 375 mm ) δφ

Thus

(125 mm ) δθ = (375 mm ) δφ 1 3

δφ = δθ  δ G = 100 2 mm δφ =   3

(

Then

)

100

 2 mm δθ 

 δ y G = δ G cos 45 ° = 

100   100  2 mm δθ cos 45 ° =  mm  δθ  3   3 

Virtual Work:

δU = 0:

Assume P acts downward at G

(9000 N ⋅mm )δθ − (180 N )(δ xE mm ) + P( δ yG mm )

=0

2 100  9000δθ − 180  × 125 δθ  + P  δθ  = 0 3    3  or

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 © 2007 The McGraw-Hill Companies.

P = 180.0 N ⊳

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 2.

FA = 20 lb at A

Have

FD = 30 lb

δ y A = (16 in.) δθ

Link ABC:

Link BF:

at D

δ yF = δ yB

δ yB = (10 in.) δθ δ y F = ( 6 in.) δφ = ( 10 in.) δθ

Link DEFG: or

5 3

δφ = δθ δ y G = (12 in.) δφ = ( 20 in. )δθ dED =

( 5.5 in.)2 + ( 4.8 in.)2 5

= 7.3 in. 

δ D = (7.3 in. )δφ =  × 7.3 in. δθ 3  δ xD = Virtual Work:

4.8 4.8  5  δD = × 7.3 in. δθ = (8 in.)δθ  7.3 7.3  3 

Assume P acts upward at G

δU = 0:

F Aδ y A + FDδ x D + Pδ y G = 0

or

( 20 lb ) (16 in. ) δθ  + (30 lb)  ( 8 in.) δθ 

or

P = − 28.0 lb

+ P ( 20 in.) δθ  = 0

P = 28.0 lb 

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 © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 3.

Link ABC: Assume δθ clockwise Then, for point C

δ xC = (125 mm ) δθ and for point D

δ xD = δ xC = (125 mm ) δθ and for point E

 250 mm  2  δ xD = δ xD 375 mm 3  

δ xE =  Link DEFG:

δ xD = ( 375 mm )δφ

Thus

( 125 mm) δθ = ( 375 mm ) δφ or

Virtual Work:

δ U = 0:

1 3

δφ = δθ Assume M acts clockwise on link DEFG

( 9000 N ⋅ mm) δθ − (180 N )( δ x E mm ) + M δφ = 0 2  1  9000 δθ − 180  ⋅125 δθ  + M  δθ  = 0 3  3  or

M = 18000 N ⋅ mm

or

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 © 2007 The McGraw-Hill Companies.

M = 18.00 N ⋅ m

⊳

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 4.

FA = 20 lb at A

Have

FD = 30 lb

Link ABC:

δ yA = (16 in.) δθ

Link BF:

at D

δ yF = δ yB

δ y B = ( 10 in.) δθ Link DEFG:

δ yF = ( 6 in.) δφ = ( 10 in. )δθ or

d ED =

δφ =

5 δθ 3

( 5.5 in.)2 + ( 4.8 in.)2 5 3

= 7.3 in. 

δ D = (7.3 in. )δφ =  × 7.3 in. δθ δ xD = Virtual Work:



4.8 4.8  5  δD = × 7.3 in.  δθ = (8 in.) δθ 7.3 7.3  3 

Assume M acts

on DEFG

δ U = 0:

FAδ y A + FDδ xD + M δφ = 0

or

(20 lb ) (16 in. )δθ  + (30 lb ) (8 in. )δθ 

or

M = − 336.0 lb ⋅ in.

5  + M  δθ  = 0 3  M = 28.0 lb ⋅ ft

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 © 2007 The McGraw-Hill Companies.



COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 5.

Assume δθ

δ xA = 10δθ in.

δ yC = 4 δθ in. δ yD = δ yC = 4δθ in.

δφ =

δ yD 6

2 = δθ 3 2 

 

δ xG = 15 δφ = 15  δθ  = 10 δθ in. 3 Virtual Work:

δU = 0:

Assume that force P is applied at A.

δ U = − Pδ xA + 30δ yC + 60δ yD + 240δφ + 80δ xG = 0 2  − P ( 10 δ θ in.) + ( 30 lb)( 4δθ in.) + ( 60 lb)( 4 δθ ) + ( 240 lb⋅ in.)  δθ   3  + ( 80 lb)( 10 δθ in.) = 0 −10P + 120 + 240 + 160 + 800 = 0 10P = 1320

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 © 2007 The McGraw-Hill Companies.

P = 132.0 lb



COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 6.

Note:

(a)

x E = 2 xD

δ xE = 2δ xD

x G = 3x D

δ xG = 3 δ xD

x H = 4x D

δ xH = 4 δ xD

x I = 5x D

δ xI = 5 δ xD

Virtual Work:

δ U = 0: FG δ xG − FSPδ xI = 0

(90 N ) (3 δ x D ) − FSP (5 δ x D ) = 0 or Now

FSP = 54.0 N 

FSP = k ∆x I 54 N = (720 N/m ) ∆xI ∆xI = 0.075 m

and

1 4

1 5

δ x D = δx H = δx I ∆xH =

4 4 ∆xI = ( 0.075 m) = 0.06 m 5 5 or

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 © 2007 The McGraw-Hill Companies.

∆xH = 60.0 mm



COSMOS: Complete Online Solutions Manual Organization System

(b)

Virtual Work:

δ U = 0: FG δ xG + FH δ xH − FSP ( δ xI ) = 0

(90 N )(3 δ xD ) + (90 N )(4 δ xD ) − FSP (5 δ xD ) = 0 or Now

FSP = 126.0 N 

FSP = k ∆xI

126.0 N = ( 720 N/m) ∆xI ∆xI = 0.175 m From Part (a)

∆x H =

4 4 ∆x = ( 0.175 m) = 0.140 m 5 I 5 or

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 © 2007 The McGraw-Hill Companies.

∆xH = 140.0 mm



COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 7.

Note:

(a)

x E = 2x D

δ xE = 2 δ xD

xG = 3xD

δ xG = 3δ xD

xH = 4 x D

δ xH = 4δ xD

x I = 5x D

δ xI = 5 δ xD

Virtual Work:

δ U = 0: FE δ xE − FSP δ xI = 0

(90 N )(2 δ xD ) − FSP (5δ xD ) = 0 or Now

FSP = 36.0 N ⊳

FSP = k ∆xI

36 N = (720 N/m ) ∆xI ∆xI = 0.050 m and

1 4

1 5

δ x D = δ xH = δ x I ∆x H =

4 4 ∆x I = ( 0.050 m) = 0.04 m 5 5 or

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 © 2007 The McGraw-Hill Companies.

∆xH = 40.0 mm

⊳

COSMOS: Complete Online Solutions Manual Organization System

(b)

Virtual Work:

δ U = 0: FD δ xD + FE δ xE − FSPδ xI = 0

(90 N )δ xD + (90 N )(2 δ xD ) − FSP (5 δ x D ) = 0 or Now

FSP = 54.0 N ⊳

FSP = k ∆xI

54 N = ( 720 N/m) ∆xI ∆xI = 0.075 m

From Part (a)

∆ xH =

4 4 ∆xI = ( 0.075 ) = 0.06 m 5 5 or

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 © 2007 The McGraw-Hill Companies.

∆xH = 60.0 mm

⊳

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 8.

Assume δ yA

δ yA 16 in.

=

δ yC 8 in.

1 2

δ yC = δ yA

;

Bar CFDE moves in translation 1 2

δ y E = δ yF = δ yC = δ yA Virtual Work:

δ U = 0: − P ( δ y A in. ) + ( 100 lb)( δ y E in.) + (150 lb) (δ y F in.) = 0 1  1  − P δ y A + 100  δ y A  + 150  δ y A  = 0 2  2  P = 125 lb

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 © 2007 The McGraw-Hill Companies.

P = 125 lb 

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 9.

Have

y A = 2 lcos θ;

θ

CD = 2 lsin ; 2

δ yA = −2 l sin θ δθ θ δ ( CD) = l cos δθ 2

Virtual Work:

δ U = 0: − Pδ yA − Qδ ( CD) = 0

θ   − P (− 2l sin θ δθ ) − Q  l cos δθ  = 0 2   Q = 2P

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 © 2007 The McGraw-Hill Companies.

sin θ  θ  cos    2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 10.

Virtual Work: Have

x A = 2l sinθ

δ xA = 2 l cos θ δθ and

y F = 3l cos θ

δ yF = −3 lsin θ δθ Virtual Work:

δU = 0: Q δ xA + Pδ yF = 0 Q ( 2l cos θ δθ ) + P ( −3l sin θ δθ ) = 0

Q=

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 © 2007 The McGraw-Hill Companies.

3 Ptan θ ⊳ 2

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Chapter 10, Solution 11.

Virtual Work: We note that the virtual work of A x , A y and C is zero, since A is fixed and C is ⊥ to δ xC .

Thus:

δ U = 0:

Pδ x D + Qδ y D = 0

xD = 3 l cos θ

δ x D = − 3l sin θ δθ

y D = l sin θ

δ y D = l cos θ δθ

P( − 3l sinθ δθ ) + Q( l cosθ δθ ) = 0

− 3Pl sin θ + Ql cos θ = 0 Q =

3P sin θ = 3P tan θ cosθ

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 © 2007 The McGraw-Hill Companies.

Q = 3P tanθ 

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 12.

x A = ( a + b) cos θ

δ xA = − ( a + b )sin θ δθ

y G = a sinθ

δ yG = a cos θ δθ

Virtual Work: The reactions at A and B are perpendicular to the displacements of A and B hence do no work.

δ U = 0:

Tδ xA + Wδ yG = 0 T − ( a + b) sin θ δθ  + W ( a cosθ δθ ) = 0

−T ( a + b ) sin θ + Wa cosθ = 0 T =

a W cot θ a +b

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 © 2007 The McGraw-Hill Companies.

T =

a W cotθ  a+b

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 13.

y H = 2l sin θ

Note:

l = 600 mm (length of a link )

Where Then

δ y H = 2l cos θ δθ

Also

1 1 1 W = mg = ( 450 kg ) 9.81 m/s2 2 2 2

(

)

= 2207.3 N 2

3  5  d AF =  l cosθ  +  l sin θ  4 4     =

δ d AF =

l 9 + 16sin 2 θ 4 l 16sin θ cos θ δθ 4 9 + 16sin 2 θ

= 4l

Virtual Work:

Fcyl For θ = 30 °

sin θ cosθ 9 + 16sin 2 θ sin θ

9 + 16sin 2 θ

Fcyl

sin θ cosθ 9 + 16sin 2 θ

δθ

1  Fcyl δ d AF −  W  δ y H = 0 2 

δ U = 0:  Fcyl  4l  

2

  

δθ  − ( 2.2073 kN )( 2l cos θ δθ ) = 0

= 1.10365 kN

sin 30° 9 + 16sin 2 30°

= 1.10365 kN

or

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 © 2007 The McGraw-Hill Companies.

Fcyl = 7.96 kN 

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Chapter 10, Solution 14.

From solution of Problem 10.13: Fcyl

sin θ 9 + 16sin 2 θ

= 1.10365 kN

Fcyl = 35 kN

Then for

( 35 kN )

sinθ 9 + 16sin 2θ

(31.713 sin θ )2

= 1.10365 kN

= 9 + 16sin2 θ

sin 2θ =

9 989.71 or

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 © 2007 The McGraw-Hill Companies.

θ = 5.47 ° 

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Chapter 10, Solution 15.

yB = a sin θ ⇒ δ yB = a cos θδθ

ABC:

yC = 2a sin θ ⇒ δ yC = 2a cos θδθ CDE: Note that as ABC rotates counterclockwise, CDE rotates clockwise while it moves to the left.

δ yC = aδφ

Then

2a cos θδθ = a δφ

or

δφ = 2 cos θδθ

or Virtual Work:

δ U = 0: − Pδ y B − Pδ yC + M δφ = 0 −P( a cosθδθ ) − P ( 2 a cosθδθ ) + M (2 cosθδθ ) = 0 or M =

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 © 2007 The McGraw-Hill Companies.

3 Pa  2

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 16.

First note l sin θ +

3 l sin φ = l 2

sin φ =

2 (1 − sin θ ) 3

or Then

2 cosφ δφ = − cos θ δθ 3

or

δφ = −

2 cos θ δθ 3 cos φ = 5 + 8sin θ − 4 sin 2 θ

Now

x C = − l cosθ +

Then

δ xC = l sinθ δθ −

3 l cos φ 2 3 l sinφ δφ 2

  − 2 cos θ  3 = l sinθ − sin φ   δθ 2  3 cos φ  

= l ( sinθ + cosθ tanφ )δθ

 = l sinθ + 

2cos θ (1 − sin θ )

  δθ 5 + 8sin θ − 4sin θ  2

Virtual Work:

δ U = 0: M δθ − Pδ xC = 0  M δθ − Pl sin θ + 

2 cos θ (1 − sin θ )

 δθ = 0 5 + 8sinθ − 4sin2 θ   or M = Pl  sinθ + 

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 © 2007 The McGraw-Hill Companies.

2cosθ (1 − sinθ )

  5 + 8sinθ − 4 sin 2θ 

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 17.

Have x B = l sin θ

δ x B = l cos θδθ yA = lcos θ

δ y A = −l sin θδθ Virtual Work:

δ U = 0: M δθ − Pδ x B + Pδ y A = 0 M δθ − P ( l cosθδθ ) + P( − l sin θδθ ) = 0 M = Pl ( sinθ + cosθ ) 

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 © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 18.

xD = l cos θ

Have

δ x D = − l sin θδθ yD = 3l sin θ

δ yD = 3 lcos θδθ Virtual Work:

δU = 0: M δθ − (P cos β ) δx D − (P sin β) δ yD = 0

M δθ − ( P cos β ) ( −l sin θδθ ) − ( P sin β )( 3l cos θδθ ) = 0 M = Pl( 3sin β cos θ − cos β sin θ )

(1)

(a) For P directed along BCD, β = θ Equation (1):

M = Pl ( 3sin θ cos θ − cos θ sin θ ) M = Pl ( 2sin θ cosθ )

M = Pl sin 2θ 

(b) For P directed , β = 90° Equation (1):

M = Pl ( 3sin 90 °cos θ − cos90 ° sin θ ) M = 3Pl cos θ 

(c) For P directed Equation (1):

, β = 180° M = Pl ( 3sin180° cos θ − cos180° sin θ )

M = Pl sin θ 

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 © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 19.

Analysis of the geometry:

Law of Sines

sin φ sinθ = AB BC sin φ =

AB sin θ BC

(1)

Now

xC = AB cosθ + BC cosφ

δ x C = − AB sin θδθ − BC sin φδφ cosφ δφ =

Now, from Equation (1)

δφ =

or

(2)

AB cosθδθ BC

AB cos θ δθ BC cosφ

(3)

From Equation (2)

 AB cos θ  δθ   BC cos φ 

δ x C = − AB sin θδθ − BC sin φ  or

Then

δ xC = −

AB ( sinθ cosφ + sinφ cosθ )δθ cos φ

δ xC = −

AB sin ( θ + φ) δθ cosφ continued

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 © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

δ U = 0: − Pδ xC − Mδθ = 0

Virtual Work:

 AB sin (θ + φ )  δθ  − M δθ = 0 −P  − cos φ  

M = AB

Thus,

sin (θ + φ ) P cosφ

(4)

For the given conditions: P = 1.0 kip = 1000 lb, AB = 2.5 in., and BC = 10 in.: (a) When

θ = 30 °: sin φ = M = ( 2.5 in.)

2.5 sin 30°, 10

sin ( 30 ° + 7.181 °) cos 7.181°

( 1.0 kip)

φ = 7.181 ° = 1.5228 kip⋅ in. = 0.1269 kip ⋅ ft or M = 126.9 lb ⋅ft

(b) When

θ = 150 °: sin φ = M = ( 2.5 in.)

2.5 sin150°, 10

sin (150° + 7.181° ) cos 7.181°

(1.0 kip)

⊳

φ = 7.181 °

= 0.97722 kip ⋅in. or M = 81.4 lb ⋅ ft

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