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Solution manual vector mechanics engineers dynamics 8th beer chapter 06

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 W
FBC = 1700 lb C W

Joint C:


FAC Cx 1700 lb
=
=
8
15
17
FAC = 800 lb T W

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
(3.00 kN) −
FAC = 0
5
13
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
FAB = 1080 lb T W

so

FBC = 1170 lb C W

Joint C:
ΣFx = 0:

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

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 FAD 500 lb
=
=
8.4 11.6
8
FAD = 725 lb T W
FCD = 525 lb C W

Joint C:
ΣFx = 0:

FBC − 525 lb = 0
FBC = 525 lb C W

This is apparent by inspection, as is FAC = C y

ΣFx = 0:

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

Joint A:

FAB = 250lb T W
ΣFy = 0:

FAC −

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

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: C y = D y = 6 kN

Joint FBDs:
Joint B:
ΣFy = 0:

− 3 kN +

1
FAB = 0
5

FAB = 3 5 = 6.71 kN T W

Joint C:

ΣFx = 0:

ΣFy = 0:

Joint A:

ΣFx = 0:

ΣFy = 0:

2
FAB − FBC = 0
5

FBC = 6.00 kN C W

3
FAC = 0
5

FAC = 10.00 kN C W

6 kN −

6 kN −

4
FAC + FCD = 0
5

FCD = 2.00 kN T W

 1

3

− 2
3 5 kN  + 2  10 kN  − 6 kN = 0 check
5

 5


By symmetry:

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

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:

ΣM A = 0:

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

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

Joint B:

ΣFX = 0:

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

ΣFy = 0:

FAB = 21.9 kN C W

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 W

Joint D:
ΣFx = 0: −

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

FAD = 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 W


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 W
FDE = 4.00 kN C W

Joint B:
ΣFx = 0:

− FAB +

4
(5 kN) = 0
5
FAB = 4.00 kN T W

ΣFy = 0:

FBD − 6 kN −

3
(5 kN) = 0
5
FBD = 9.00 kN C W

Joint D:

ΣFy = 0:

3
FAD − 9 kN = 0
5
FAD = 15.00 kN T W

ΣFx = 0: FCD −

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

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 W

By inspection:

FBD = 0 W

Joint A:

FAC FAD 12 kips
=
=
5
13
12
FAC = 5.00 kips C W
FAD = 13.00 kips T W

Joint D:
ΣFx = 0: FCD −

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

ΣFy = 0:

5
(13 kips) − FDF = 0
13
FDF = 5.00 kips T W

Joint C:

ΣFx = 0: 30 kips −

12
FCF = 0
13
FCF = 32.5 kips T W

ΣFy = 0: FCE − 5 kips −

5
(32.5 kips)
13
FCE = 17.50 kips C W

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 W


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

FDH = 26.0 kips C !
FAB = FBC = 24.0 kips T !

and, from above:

FGH = FFG = 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:
ΣFx = 0: H x = 0
By symmetry: A y = H y = 4 kips
FAC = FCE and FBC = 0 W

by inspection of joints C and G :

FEG = FGH and FFG = 0 W
also, by symmetry FAB = FFH , FBD = FDF , FCE = FEG and FBE = FEF

Joint FBDs:
Joint A:

FAB FAC 3 kips
=
=
5
4
3
FAB = 5.00 kips C W

so

FAC = 4.00 kips T W
FFH = 5.00 kips C W

and, from above,
and

Joint B:

FCE = FEG = FGH = 4.00 kips T W

4
4
10
(5 kips) − FBE −
FBD = 0
5
5
109

ΣFx = 0:
ΣFy = 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 W

or

Joint E:

FBE = 0.238 kips C W
FDF = 3.98 kips C W

and, from above,

FEF = 0.238 kips C W
ΣFy = 0 :

FDE − 2

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

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:

Ax = 0

ΣM G = 0:

3a Ay − 2a (3 kN) − a (6 kN) = 0

A y = 4 kN
by inspection of joint C,

FAC = FCE and FBC = 0 W

by inspection of joint D,

FBD = FDF and FDE = 6.00 kN C W

Joint FBDs:
Joint A:

FAC FAB 4 kN
=
=
21
29
20
FAB = 5.80 kN C W
FAC = 4.20 kN C W
from above,

Joint B:

ΣFy = 0:

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

ΣFx = 0:

FCE = 4.20 kN C W

29
20

FBE = 1.450 kN T W

21 
29

kN  − FBD = 0
 5.80 kN +
29 
20

FBD = 5.25 kN C W
FDF = 5.25 kN C W

from above,

Joint F:

ΣFx = 0:

5.25 kN −

21
FEF = 0
29
FEF = 7.25 kN T W

ΣFy = 0:

FFG −

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

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 W


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

5
4
FAC − FAB = 0
5
29
2
3
FAC − FAB + 4.9 kN = 0
5
29

ΣFx = 0:

Joint FBDs:
Joint A:

ΣFy = 0:

FAC = 2.8 29 kN

FAC = 15.08 kN T

FAD = 17.50 kN C
Joint B:

ΣFx = 0:
ΣFy = 0:

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

FBC = 1.6 2 kN;
Joint C:

ΣFy = 0:

FBC = 2.26 kN C

4
1
1.6 2 kN −
FCD −
5
2

ΣFx = 0: FCF

2
(2.8 29 kN) = 0
29
FCD = 9.00 kN T
1
3
5
1.6 2 kN + (9 kN) −
(2.8 29 kN) = 0
+
5
2
29
FCF = 7.00 kN T

(

(

)

)

from symmetry,

FEG = 17.50 kN C
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
ΣFy = 0:

Ay − 2.8 kN − 4.2 kN + 2.8 kN = 0

A y = 4.2 kN
Joint FBDs:

5
4
FAC − FAB = 0
5
29
2
3
FAC − FAB + 4.2 kN = 0
5
29

ΣFx = 0:

Joint A:
ΣFy = 0:

FAB = 15.00 kN C !
FAC = 12.92 kN T !

FAC = 2.4 29
Joint B:
ΣFx = 0:
ΣFy = 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
FCD −
5

FBC = 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,

FDE = FEG

and

FEF = 0 !

Joint F:
ΣFy = 0:
ΣFx = 0:

4
FDF −
5

2
FFG = 0
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:

4
FEG −
5

FFG = 8.62 kN T !

5
(1.6 29 kN) = 0
29

FEG = 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.

FDE = 10.00 kN C !


COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 14.
FBD Truss:

ΣFx = 0: A x = 0
ΣM A = 0:

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

ΣFy = 0:

Ay − 1 kN − 2 kN − 2 kN − 1.5 kN − 1 kN
+ 3.625 kN = 0

Joint FBDs:
Joint A:

A y = 3.875 kN

FAB FAC
2.625 kN
=
=
1
29
26

FAB = 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:

FBC = 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:

FDF = 10.0971 kN,

By symmetry:

ΣFy = 0:

FDF = 10.10 kN C !

 2

10.0971 kN  − 2 kN = 0
− FDE + 2 
 29


FDE = 5.50 kN T !

FGH FFH
2.625 kN
=
=
1
26
29

Joint H:

By inspection of joint G:
ΣFx = 0 :

Joint F:

FFH = 14.1361 kN

FFH = 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

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.

FEF = 3.75 kN C !


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
ΣFy = 0:

− 1 kN − 1 kN + 6.7 kN = 0

Joint FBDs:
Joint A:

Ay − 1 kN − 1 kN − 1.4 kN − 4.5 kN − 2.8 kN

ΣFy = 0:

5 kN −

A y = 6.0 kN

3
5 13
FAB = 0, FAB =
kN
3
13

FAB = 6.01 kN C !
ΣFx = 0:

Joint B:

ΣFx = 0:

FAC −

2
13

 5 13

kN  = 0,

 3


FAC = 3.33 kN T !


2  5 13
kN − FBC − FBD  = 0,

13  3


5 13
kN
3

3  5 13
kN + FBC − FBD  − 1 kN = 0,

13  3

4 13
FBD − FBC =
kN
3
3
13 kN
FBD =
FBD = 5.41 kN C !
2
1
13 kN
FBC =
FBC = 0.601 kN C !
6
FBC + FBD =

ΣFy = 0:

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:
ΣFy = 0:

2

1
FDE − 4.5 kN = 0,
17

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 +

Joint D:

FDF = 6.00 kN T !
ΣFy = 0:

3 3
1 9


13 kN  − 1.4 kN − FCD −
17 kN  = 0


13  2
17  4


FCD = 0.850 kN T !

Joint C:

ΣFy = 0:

0.850 kN −

FCG =
ΣFx = 0:

FCI −

3
13

 13
 3
kN  − FCG = 0

 5
 6


1.75
kN
3

FCG = 0.583 kN C !

 10
4  1.75
2  13

kN  −
kN  −
kN = 0


5 3
13  6

 3

FCI = 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
ΣFy = 0:

Ay − 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:

ΣFy = 0:

(6.7 − 1) kN −

3
FHJ = 0,
13

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

ΣFx = 0:

Joint H:

ΣFx = 0:
ΣFy = 0:

2
(1.9 13 kN) − FIJ = 0,
13
2
( FGH + FHI − 1.9 13 kN) = 0
13

3
( FHI − FGH + 1.9 13 kN) − 1 kN = 0
13
26
13 kN,
FGH =
FGH = 6.25 kN C !
15

FHI =
Joint I:

ΣFx = 0:

3.80 kN −

FGI −

13
kN,
6

FHI = 0.601 kN C !


2  13
kN  − FCI = 0

13  6


FCI =
ΣFy = 0:

FIJ = 3.80 kN T !

10.4
kN,
3


3  13
kN  = 0,


13  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:
ΣFy = 0:

 1

FEG  − 4.5 kN = 0,
2
17



FEG =

9
17 kN
4
FEG = 9.28 kN C !

Joint G:

ΣFy = 0:

3  26
3
1 9
 1

13 kN  − kN − FCG −
17 kN 


5
13  15
17  4
 2

− 2.8 kN = 0
FCG = −

ΣFx = 0:

1.75
kN
3

FCG = 0.583 kN C !

4 9
4  1.75 

17 kN  − FFG −  −


5 3 
17  4



2  26

13 kN  = 0

13  15

FFG = 6.00 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 17.
FBD Truss:

By load symmetry,

ΣFx = 0:

θ = tan −1

Joint FBDs:

A y = H y = 1600 lb

Ax = 0

6.72 ft
= 16.2602°
23.04 ft

ΣFy′ = 0:

(1600 lb − 400 lb) cosθ − FAC sin θ = 0

FAC = 4114.3 lb

Joint A:

ΣFx = 0:

FAC cos 2θ − FAB cosθ = 0
FAB = 3613.5 lb

ΣFx′ = 0:
Joint B:

Joint C:

FBC = 0.768 kips C !

FCD sin 2θ − (768 lb) cosθ = 0
FCD = 1371.4 lb

ΣFx′′ = 0:

FBD = 3.84 kips C !

FBC − (800 lb) cosθ = 0
FBC = 768.00 lb

ΣFy′′ = 0:

FAB = 3.61 kips C !

3613.5 lb − FBD + (800 lb)sin θ = 0
FBD = 3837.5 lb

ΣFy′ = 0:

FAC = 4.11 kips T !

FCD = 1.371 kips T !

FCE + (1371.4 lb) cos2θ − (768 lb)sin θ − 4114.3 lb = 0
FCE = 2742.9 lb

FCE = 2.74 kips 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

Joint E:
ΣFy = 0:

(2742.9 lb)sin 2θ − FDE cosθ = 0

FDE = 1536.01 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.

FDE = 1.536 kips C !


COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 18.
FBD Truss:
A y = H y = 1600 lb

By load symmetry,
ΣFx = 0:

Ax = 0

θ = tan −1

6.72 ft
= 16.2602°
23.04 ft

Joint FBDs:
Joint H:

ΣFy = 0:

1600 lb − 400 lb − FFH sin θ = 0
FFH = 4285.7 lb,

( 4285.7 lb ) cosθ

ΣFx = 0:

Joint F:

− FGH = 0

FGH = 4114.3 lb

ΣFy′ = 0:

FGH = 4.11 kips T

FFG − ( 800 lb ) cosθ = 0
FFG = 768.0 lb

ΣFx′ = 0:

FFH = 4.29 kips C

FFG = 0.768 kips C

FDF + ( 800 lb ) sin θ − 4285.7 lb = 0
FDF = 4061.7 lb

FDF = 4.06 kips C

Joint G:
ΣFy = 0:

FDG sin 2θ − (768 lb) cosθ = 0
FDG = 1371.4 lb

ΣFx = 0:

FDG = 1.371 kips T

4114.3 lb − (1371.4 lb ) cosθ − FEG − (768 lb) sin θ = 0
FEG = 2742.9 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.

FEG = 2.74 kips T


COSMOS: Complete Online Solutions Manual Organization System

Chapter 6, Solution 19.
FBD Truss:

ΣFx = 0:
ΣM L = 0:

Ax = 0

(1.5 m ) (6.6 kN) + (3.0 m)(2.2 kN)
+ (5.5 m)(6 kN) − (11 m) Ay = 0

A y = 4.5 kN

Joint FBDs:
Joint A:

ΣFy = 0:

4.5 kN − 6 kN − 2.2 kN − 6.6 kN + L = 0

L = 10.3 kN
4.5 kN FAC
F
=
= AB ,
1
2
5

FAC = 9.00 kN T !

FAB = 4.5 5,

Joint C:

FAB = 10.06 kN C !

9 kN FBC
F
=
= CE ,
16
5
281

FBC =

45
kN
16

FBC = 2.81 kN C !
FCE =

Joint B:

ΣFx = 0:
ΣFy = 0:

Solving:

9
281,
16

FCE = 9.43 kN T !

2 
16
FBD + (4.5 5) kN  +
FBE = 0


5
265
1 
FBD + (4.5 5) kN  −

5

3
45
kN = 0
FBE +
16
265

72
5 kN,
11
45
265 kN,
FBE =
176
FBD = −

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.

FBD = 14.64 kN C !
FBE = 4.16 kN T !


COSMOS: Complete Online Solutions Manual Organization System

Joint E:

ΣFx = 0:

16 
 9
 FEG −  16
281 


FEG =
ΣFy = 0:


16  45


281  kN  −
265  kN = 0

265  176




9
281 kN,
11

FEG = 13.72 kN T !

5 9
9

281 kN −
281 kN  +

16
281  11


3  45

265 kN 

265  176


+ FDE = 0,
FDE = −

Joint D:

ΣFx = 0:

ΣFy = 0:
Solving:

45
kN,
22

FDE = 2.05 kN C !

2 
72
10

FDG = 0
5 kN  +
 FDF +
11
5
101


1 
72
1
45

FDG +
5 kN  −
kN = 0
 FDF +
11
22
5
101

7.5
101 kN,
22

FDG = 3.43 kN T !

FDF = − 8.25 5 kN,

FDF = 18.45 kN C !

FDG =

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.


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