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.