Cap 06 - Solucionario estatica beer octaba edicion capitulo 6 PDF

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Solucionario estatica beer octaba edicion capitulo 6...

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COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 1. \

Joint FBDs:

Joint B: FAB 800 lb FBC = = 15 8 17 so

FAB = 1500 lb T  FBC = 1700 lb C 

Joint C:

FAC C 1700 lb = x = 8 15 17 F AC = 800 lb T 

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 6, Solution 2.

Joint FBDs: Joint B: ΣFx = 0:

1 4 FAB − FBC = 0 5 2

ΣFy = 0:

1 3 FAB + FBC − 4.2 kN = 0 5 2

7 FBC = 4.2 kN 5

so

Joint C:

FAB =

ΣFx = 0:

12 2 kN 5

FBC = 3.00 kN C !

FAB = 3.39 kN C !

4 12 F =0 (3.00 kN) − 5 13 AC FAC =

13 kN 5

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.

FAC = 2.60 kN T !

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 3.

Joint FBDs:

Joint B: FAB FBC 450 lb = = 12 13 5 F AB = 1080 lb T 

so

FBC = 1170 lb C 

Joint C: ΣFx = 0:

3 12 FAC − ( 1170 lb) = 0 5 13 FAC = 1800 lb C 

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 6, Solution 4.

Joint FBDs:

Joint D:

FCD F 500 lb = AD = 8.4 11.6 8 FAD = 725 lb T  FCD = 525 lb C 

Joint C: ΣFx = 0:

FBC − 525 lb = 0 F BC = 525 lb C 

This is apparent by inspection, as is F AC= C y

ΣFx = 0:

8.4 3 (725 lb) − FAB − 375 lb = 0 11.6 5

Joint A:

F AB = 250 lb T  ΣFy = 0:

FAC −

4 8 (250 lb) − (725 lb) = 0 5 11.6 FAC = 700 lb C 

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 6, Solution 5. FBD Truss: ΣFx = 0 :

Cx = 0

By symmetry: Cy = Dy = 6 kN

Joint FBDs: Joint B: ΣFy = 0:

− 3 kN +

1 FAB = 0 5

FAB = 3 5 = 6.71 kN T 

Joint C:

ΣFx = 0:

ΣFy = 0:

Joint A:

ΣFx = 0:

ΣFy = 0:

2 FAB − FBC = 0 5

FBC = 6.00 kN C 

3 FAC = 0 5

FAC = 10.00 kN C 

6 kN −

6 kN −

4 FAC + FCD = 0 5

FCD = 2.00 kN T 

 1  3  3 5 kN + 2 10 kN  − 6 kN = 0 check − 2 5 5    

By symmetry:

FAE = FAB = 6.71 kN T  FAD = FAC = 10.00 kN C  FDE = FBC = 6.00 kN C 

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 6, Solution 6. FBD Truss:

Σ M A = 0:

(10.2 m ) Cy + ( 2.4 m ) (15 kN ) − ( 3.2 m)( 49.5 kN) = 0 C y = 12.0 kN

Joint FBDs: Joint C: FBC F 12 kN = CD = 7.4 7.4 8 FBC = 18.50 kN C  FCD = 18.50 kN T 

Joint B:

ΣFX = 0:

4 7 FAB − (18.5 kN) = 0 5 7.4 F AB = 21.875 kN;

ΣFy = 0:

FAB = 21.9 kN C 

3 2.4 (21.875 kN) − 49.5 kN + (18.5 kN) + FBD = 0 5 7.4 FBD = 30.375 kN;

FBD = 30.4 kN C 

Joint D: ΣFx = 0: −

4 7 FAD + (18.5 kN ) + 15 kN = 0 5 7.4

F AD = 40.625 kN;

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.

FAD = 40.6 kN T 

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 7. Joint FBDs:

Joint E: FBE FDE 3 kN = = 5 4 3 FBE = 5.00 kN T  FDE = 4.00 kN C 

Joint B: ΣFx = 0:

− FAB +

4 (5 kN) = 0 5 FAB = 4.00 kN T 

ΣF y = 0:

FBD − 6 kN −

3 (5 kN) = 0 5 FBD = 9.00 kN C 

Joint D:

ΣF y = 0:

3 F AD − 9 kN = 0 5 FAD = 15.00 kN T 

ΣFx = 0: FCD −

4 (15 kN) − 4 kN = 0 5 FCD = 16.00 kN C 

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 6, Solution 8. Joint FBDs:

Joint B: FAB = 12.00 kips C 

By inspection:

FBD = 0 

Joint A:

FAC F 12 kips = AD = 5 13 12 FAC = 5.00 kips C  FAD = 13.00 kips T 

Joint D: ΣFx = 0:

FCD −

12 (13 kips) − 18 kips = 0 13 FCD = 30.0 kips C 

Σ F y = 0:

5 (13 kips) − FDF = 0 13 F DF = 5.00 kips T 

Joint C:

Σ Fx = 0: 30 kips −

12 FCF = 0 13 FCF = 32.5 kips T 

ΣFy = 0: FCE − 5 kips −

5 (32.5 kips) 13 FCE = 17.50 kips C 

Joint E:

by inspection:

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.

FCF = 0 

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 9. FCH = 0!

First note that, by inspection of joint H:

and

FCG = 0 !

then, by inspection of joint C: and

Joint D:

FBC = FCD FBG = 0 !

then, by inspection of joint G: and

Joint FBDs:

FDH = FGH

FFG = FGH FBF = 0!

then, by inspection of joint B: and

FAB = FBC

FCD FDH 10 kips = = 12 13 5 Joint A:

FCD = 24.0 kips T !

so

F DH = 26.0 kips C! FAB = FBC = 24.0 kips T !

and, from above:

F GH = F FG = 26.0 kips C ! FAF F 24 kips = AE = 5 4 41

FAF = 30.0 kips C !

Joint F:

FAE = 25.6 kips T !

ΣFx = 0: FEF −

12 (26 kips) = 0 13 FEF = 24.0 kips C !

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 6, Solution 10. FBD Truss: ΣF x = 0: H x = 0 By symmetry: A y = H y = 4 kips F AC = FCE and FBC = 0  FEG = FGH and FFG = 0 

by inspection of joints C and G :

also, by symmetry FAB = FFH , FBD = FDF , FCE = FEG and FBE = FEF

Joint FBDs: Joint A:

FAB F 3 kips = AC = 5 4 3 FAB = 5.00 kips C 

so

F AC = 4.00 kips T  FFH = 5.00 kips C 

and, from above, and

Joint B: ΣF x = 0:

FCE = FEG = FGH = 4.00 kips T 

4 4 10 F BD = 0 (5 kips) − F BE − 5 5 109

Σ Fy = 0:

3 3 3 FBD + FBE = 0 ( 5 kips) − 2 − 5 5 109

so

FBD = 3.9772 kips, FBE = 0.23810 kips FBD = 3.98 kips C 

or

F BE = 0.238 kips C 

Joint E:

FDF = 3.98 kips C

and, from above,

F EF = 0.238 kips C  Σ Fy = 0 :

FDE − 2

3 (0.23810 kips) = 0 5 FDE = 0.286 kips T 

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 6, Solution 11. FBD Truss: ΣFx = 0: ΣM

G

Ax = 0

= 0:

3a A y − 2a (3 kN) − a (6 kN) = 0

A y = 4 kN by inspection of joint C,

F AC = FCE and FBC = 0 

by inspection of joint D,

FBD = FDF and FDE = 6.00 kN C 

Joint FBDs: Joint A:

FAC F 4 kN = AB = 21 29 20 FAB = 5.80 kN C  FAC = 4.20 kN C  from above, ΣF y = 0:

Joint B:

20 20 ( 5.80 kN ) − 3 kN − FBE = 0 29 29 FBE =

Σ F x = 0:

FCE = 4.20 kN C 

29 20

FBE = 1.450 kN T 

21  29  kN  − F BD = 0  5.80 kN + 29  20  FBD = 5.25 kN C  FDF = 5.25 kN C 

from above, ΣFx = 0:

5.25 kN −

Joint F:

21 F = 0 29 EF FEF = 7.25 kN T 

ΣF y = 0:

FFG −

20 (7.25 kN) − 1 kN = 0 29 FFG = 6.00 kN C 

by inspection of joint G,

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.

FEG = 0 

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 12. FBD Truss:

ΣFx = 0:

Ax = 0

By symmetry: A y = B y = 4.90 kN FAB = FEG , FAC = FFG , FBC = FEF

and

FBD = FDE , FCD = FDF Joint FBDs: Joint A:

ΣFx = 0:

5 4 FAC − FAB = 0 5 29

ΣF y = 0:

2 3 F AC − F AB + 4.9 kN = 0 5 29

FAC = 2.8 29 kN

FAC = 15.08 kN T ⊳

FAD = 17.50 kN C ⊳ Joint B:

ΣFx = 0: ΣFy = 0:

4 (17.5 kN − FBD ) − 1 FBC = 0 5 2 3 ( 17.5 kN − FBD ) + 1 FBC − 2.8 kN = 0 5 2 FBD = 15.50 kN C ⊳

FBC = 1.6 2 kN ; Joint C:

FBC = 2.26 kN C ⊳

4 1 2 1.6 2 kN − (2.8 29 kN) = 0 FCD − 5 2 29 FCD = 9.00 kN T ⊳ 1 3 5 1.6 2 kN + (9 kN) − (2.8 29 kN) = 0 ΣFx = 0: FCF + 5 2 29 FCF = 7.00 kN T ⊳ F from symmetry, EG = 17.50 kN C ⊳

(

ΣFy = 0:

(

)

)

FFG = 15.08 kN T ⊳ FEF = 2.26 kN C ⊳ FDE = 15.50 kN C ⊳ FDF = 9.00 kN T ⊳

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 6, Solution 13. FBD Truss:

ΣFx = 0: A x = 0 ΣM A = 0: (8 m)Gy − (4 m)(4.2 kN) − (2m)(2.8 kN) = 0

G y = 2.80 kN ΣF y = 0:

A y − 2.8 kN − 4.2 kN + 2.8 kN = 0

A y = 4.2 kN Joint FBDs:

5

ΣF x = 0:

29

Joint A:

2

ΣFy = 0:

29

F AC −

FAC −

4 F AB = 0 5

3 FAB + 4.2 kN = 0 5 F AB = 15.00 kN C !

FAC = 12.92 kN T !

FAC = 2.4 29 Joint B: ΣFx = 0: ΣF y = 0:

4 1 FBC = 0 (15.00 kN − FBD ) − 5 2 3 1 FBC − 2.8 kN = 0 (15.00 kN − FBD ) + 5 2 FBD = 13.00 kN C !

FBC = 1.6 2 kN, Joint C: ΣFy = 0:

4 F − 5 CD

F BC = 2.26 kN C !

2 1 2.4 29 kN − (1.6 2 kN) = 0 29 2

(

)

FCD = 8.00 kN T ! ΣFx = 0:

FCF +

3 ( 8.00 kN) − 5 +

5 (2.4 29 kN) 29

1 (1.6 2 kN) = 0 2

FCF = 5.60 kN T !

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

By inspection of joint E,

F DE = F EG

and

FEF = 0!

Joint F: ΣFy = 0: ΣFx = 0:

4 2 FDF − FFG = 0 5 29 3 5 − 5.6 kN − FDF + FFG = 0 5 29 FDF = 4.00 kN T !

FFG = 1.6 29 kN

Joint G:

ΣFx = 0:

FFG = 8.62 kN T !

4 5 FEG − (1.6 29 kN) = 0 5 29

F EG = 10.00 kN C ! from above (joint E)

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.

F DE = 10.00 kN C!

COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 14. FBD Truss:

ΣFx = 0: Ax = 0 ΣM A = 0:

4a H y − 3a (1.5 kN) − 2a (2 kN) − a (2 kN) = 0

ΣF y = 0:

A y − 1 kN − 2 kN − 2 kN − 1.5 kN − 1 kN + 3.625 kN = 0

Joint FBDs: Joint A:

A y = 3.875 kN

F AB F AC 2.625 kN = = 1 29 26

F AB = 15.4823 kN,

FAB = 15.48 kN C !

FAC = 14.6597 kN,

FAC = 14.66 kN T !

By inspection of joint C: Joint B:

H y = 3.625 kN

ΣFy = 0:

FCE = FAC = 14.66 kN T,

2 (15.4823 kN − FBD ) − 2 kN = 0 29

FBD = 10.0971 kN, ΣFx = 0:

F BC = 0 !

FBD = 10.10 kN C !

5 (15.4823 kN − 10.0971 kN ) − FBE = 0 29

FBE = 5.0000 kN,

FBE = 5.00 kN C !

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

Joint D:

By symmetry:

ΣF y = 0:

FDF = 10.0971 kN,

 2  10.0971 kN  − 2 kN = 0 − FDE + 2   29  FDE = 5.50 kN T !

FGH FFH 2.625 kN = = 26 29 1

Joint H:

By inspection of joint G: ΣFx = 0 :

Joint F:

FDF = 10.10 kN C !

F FH = 14.1361 kN

F FH = 14.14 kN C !

FGH = 13.3849 kN

FGH = 13.38 kN T !

FEG = FGH = 13.38 kN T

FEF +

and

FFG = 0 !

2 (10.0971 kN − 14.1361 kN ) = 0 29

FEF = 3.75 kN C !

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 6, Solution 15. FBD Truss:

ΣFx = 0: A x = 0 ΣM A = 0:

8a (J y − 1 kN) − 7a (1 kN) − 6a (2.8 kN) − 4a (4.5 kN) − 2a (4 kN) − a (1 kN) = 0

J y = 6.7 kN ΣF y = 0:

Ay =

− 1 kN − 1 kN + 6.7 kN = 0 6.0 kN

Joint FBDs: Joint A:

A y − 1 kN − 1 kN − 1.4 kN − 4.5 kN − 2.8 kN

ΣF y = 0:

5 kN −

3 13

F AB = 0, F AB =

5 13 kN 3

FAB = 6.01 kN C ! ΣFx = 0:

ΣFx = 0:

Joint B:

ΣF y = 0:

FAC −

2 13

 5 13  kN  = 0,   3 

FAC = 3.33 kN T !

2  5 13 kN − FBC − FBD  13  3 5 FBC + FBD =

  = 0,  13 kN 3  3  5 13 kN + F BC − F BD − 1 kN = 0,  3 13   4 13 kN F BD − F BC = 3 3 13 kN FBD = 5.41 kN C ! FBD = 2 1 13 kN FBC = 0.601 kN C ! FBC = 6 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

FEF = 0!

By inspection of joint F: Joint E:

FDE = FEG

By symmetry: ΣF y = 0:

2

1 17

F DE − 4.5 kN = 0,

FDE =

9 17 kN 4 FDE = 9.28 kN C !

ΣFx = 0:

2 3 4 9 13 kN  − 17 kN  = 0 13  2 17  4  

FDF +

FDF = 6.00 kN T !

Joint D:

ΣF y = 0:

3 3  13 kN − 1.4 kN − F CD − 13  2 

1 9  17 kN = 0 17  4  FCD = 0.850 kN T !

Joint C:

ΣF y = 0:

0.850 kN −

FCG =

ΣFx = 0:

FCI −

3 13

 13  3 kN − F CG = 0   6  5

1.75 kN 3

4  1.75 2  kN  −  5 3 13 

FCG = 0.583 kN C !  13  10 kN  − kN = 0   6  3

F CI = 3.47 kN !

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 6, Solution 16. \

FBD Truss:

ΣFx = 0: ΣM A = 0:

Ax = 0 8a (J y − 1 kN) − 7a (1 kN) − 6a (2.8 kN) − 4a (4.5 kN) − 2a (4 kN) − a (1 kN) = 0

J y = 6.7 kN ΣF y = 0:

A y − 1 kN − 1 kN − 1.4 kN − 4.5 kN − 2.8 kN −1 kN − 1 kN + 6.7 kN = 0

Joint FBDs: Joint J:

ΣF y = 0:

(6.7 − 1) kN −

3 FHJ = 0, 13

A = 6.0 kN FHJ = 1.9 13 kN FHJ = 6.85 kN C !

Joint H:

ΣFx = 0:

2 (1.9 13 kN) − FIJ = 0, 13

ΣFx = 0:

2 ( FGH + FHI − 1.9 13 kN) = 0 13

ΣF y = 0:

Joint I:

ΣFx = 0:

3 ( FHI − FGH + 1.9 13 kN) − 1 kN = 0 13

F GH =

26 13 kN, 15

FGH = 6.25 kN C !

F HI =

13 kN, 6

FHI = 0.601 kN C !

3.80 kN −

 2  13 kN  − FCI = 0   13  6 

FCI = Σ Fy = 0:

FIJ = 3.80 kN T !

FGI −

3 13

10.4 kN, 3

 13  kN = 0,    6 

FCI = 3.47 kN T ! FGI = 0.500 kN T !

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

FEF = 0 !

By inspection of joint F, By symmetry FDE = FEG Joint E: ΣF y = 0:

 1  FEG  − 4.5 kN = 0, 2  17 

FEG =

9 17 kN 4 FEG = 9.28 kN C !

Joint G:

ΣF y = 0:

3  26 3 1 9  1  13 kN  − kN − FCG − 17 ...