Beer & Johnston 9Ed Resolução do Capítulo 6 Análise de Estruturas

Prévia do material em texto

PROBLEM 6.1 
Using the method of joints, determine the force in each member of the 
truss shown. State whether each member is in tension or compression. 
 
 
SOLUTION 
 
 FBD Truss: 
 
 
 Joint FBDs: 
 Joint B: 
 
 N 
 
 
 Joint C: 
N 
 
 
 
( ) ( )( )0: 6.25 m 4 m 315 N 0 240 NB y yM CΣ = − = =C
 
 
0: 315 N 0 75 Ny y y yF B CΣ = − + = =B
 
0: 0x xFΣ = =B
 
 
 
 
 
75 N
5 4 3
AB BCF F
= =
 
125.0 N C ABF = W
 
 
100.0 N T BCF = W
 
 
 
By inspection:
 
260 N CACF = W
 
 
 
 
PROBLEM 6.2 
Using the method of joints, determine the force in each member of the 
truss shown. State whether each member is in tension or compression. 
 
 
SOLUTION 
 FBD Truss: 
 
 
 Joint FBDs: 
 Joint C: 
 
 
 
 Joint A: 
 
 
 
 
 
 
( ) ( )( )0: 14 ft 7.5 ft 5.6 kips 0 3 kipsA x xM CΣ = − = =C
 
 
0: 0 3 kipsx x x xF A CΣ = − + = =A 
 
 
0: 5.6 kips 0 5.6 kipsy y yF AΣ = − = =A
 
 
 
 
 
3 kips
5 4 3
BC ACF F
= =
 
 
5.00 kips C BCF = W
 
 
4.00 kips T ACF = W
 
 
1.6 kips
8.5 4
ABF
=
 
 
3.40 kips T ABF = W
 
 
 
PROBLEM 6.3 
Using the method of joints, determine the force in each member of 
the truss shown. State whether each member is in tension or 
compression. 
 
 
SOLUTION 
 FBD Truss: 
 
 
 Joint FBDs: 
 Joint C: 
 
 Joint B: 
 
 
 
 
 
( )( ) ( )0: 6 ft 6 kips 9 ft 0 4 kipsB y yM CΣ = − = =C
 
 
0: 6 kips 0 10 kipsy y y yF B CΣ = − − = =B 
 
 
0: 0x xFΣ = =C
 
 
 
4 kips
17 15 8
AC BCF F
= =
 
 
8.50 kips T ACF = W
 
 
7.50 kips C BCF = W
 
 
 
 
By inspection:
 
12.50 kips C ABF = W
 
 
 
 
10 kips
5 4
ABF
=
 
 
 
 
PROBLEM 6.4 
Using the method of joints, determine the force in each member of the 
truss shown. State whether each member is in tension or compression. 
 
 
SOLUTION 
 FBD Truss: 
 
 
 Joint FBDs: 
 Joint D: 
 
 
 Joint C: 
 
 
 Joint B: 
 
 
 
( ) ( )( ) ( )0: 1.5 m 2 m 1.8 kN 3.6 m 2.4 kN 0B yM CΣ = + − =
 
3.36 kNy =C
 
 
0: 3.36 kN 2.4 kN 0y yF BΣ = + − =
 
0.96 kNy =B
 
 
 
20: 2.4 kN 0
2.9y AD
F FΣ = − =
 
3.48 kN T ADF = W
 
 
2.10: 0
2.9x CD AD
F F FΣ = − =
 
 
2.1 (3.48 kN)
2.9CD
F =
 
2.52 kN C CDF = W
 
 
By inspection: 3.36 kN C ACF = W 
 
2.52 kN C BCF = W
 
 
 
 
 
40: 0.9 kN 0
5y AB
F FΣ = − =
 
1.200 kN T ABF = W
 
 
 
 
 
PROBLEM 6.5 
Using the method of joints, determine the force in each member of the 
truss shown. State whether each member is in tension or compression. 
 
 
SOLUTION 
 FBD Truss: 
 
 Joint FBDs: 
 Joint B: 
 
 Joint C: 
 
 Joint A: 
 
 
 
0 :xFΣ =
 
0x =C
 
 
By symmetry: 6 kNy y= =C D
 
 
 
 
10: 3 kN 0
5y AB
F FΣ = − + =
 
 
3 5 6.71 kN T ABF = = W
 
 
20: 0
5x AB BC
F F FΣ = − =
 
6.00 kN C BCF = W
 
 
 
30: 6 kN 0
5y AC
F FΣ = − =
 
10.00 kN C ACF = W
 
 
40: 6 kN 0
5x AC CD
F F FΣ = − + =
 
2.00 kN T CDF = W
 
 
 
1 30: 2 3 5 kN 2 10 kN 6 kN 0 check
55y
F    Σ = − + − =      
 
 
 
By symmetry: 6.71 kN TAE ABF F= = W
 
 
10.00 kN C AD ACF F= = W
 
 
6.00 kN C DE BCF F= = W
 
 
 
 
 
 
 
PROBLEM 6.6 
Using the method of joints, determine the force in each member of the 
truss shown. State whether each member is in tension or compression. 
 
 
SOLUTION 
FBD Truss: 
 
 
 
Joint FBDs: 
Joint C: 
 
 
 
 
Joint D: 
 
 
 
 
 
 
 
( ) ( )( ) ( )( )0: 25.5 ft 6 ft 3 kips 8 ft 9.9 kips 0A yM CΣ = + − =
 
2.4 kipsy =C
 
 
0: 2.4 kips 9.9 kips 0y yF AΣ = + − =
 
7.4 kipsy =A
 
 
0: 3 kips 0x xF AΣ = − + =
 
3 kipsx =A
 
 
2.4 kips
12 18.5 18.5
CD BCF F
= =
 
 
3.70 kips T CDF = W
 
 
3.70 kips C BCF = W
 
or: 0:x BC CDF F FΣ = =
 
60: 2.4 kips 2 0
18.5y BC
F FΣ = − =
 
 
same answers
 
 
 
 
 
( )17.5 40: 3 kips 3.70 kips 0
18.5 5x AD
F FΣ = + − =
 
 
8.125 kipsADF =
 
8.13 kips T ADF = W
 
 
( ) ( )6 30: 3.7 kips 8.125 kips 0
18.5 5y BD
F FΣ = + − =
 
 
6.075 kipsBDF =
 
6.08 kips C BDF = W
 
 
 
 
Joint A: 
 
PROBLEM 6.6 CONTINUED 
 
( )4 40: 3 kips 8.125 kips 0
5 5x AB
F FΣ = − + − =
 
4.375 kipsABF =
 
4.38 kips C ABF = W
 
 
 
 
PROBLEM 6.7 
Using the method of joints, determine the force in each member of the 
truss shown. State whether each member is in tension or compression. 
 
SOLUTION 
 FBD Truss: 
 
 Joint FBDs: 
 Joint A: 
 
 Joint B: 
 
 
 
 
 
 
 
 
0: 480 N 0 480 Ny y yF AΣ = − = =A
 
 
( )0: 6 m 0 0A x xM DΣ = = =D
 
 
0: 0 0x x xF AΣ = − = =A
 
 
 
 
 
 
480 N
6 2.5 6.5
AB ACF F
= =
 
200 N C ABF = W
 
 
520 N T ACF = W
 
 
 
 
200 N
2.5 6 6.5
BE BCF F
= =
 
480 N C BEF = W
 
 
520 N T BCF = W 
 
 
 
 
 
 
 
 Joint C: 
 
 Joint D: 
 
PROBLEM 6.7 CONTINUED 
 
 By inspection: 520 N T CD CEF F= = W 
 
 
 
( )2.50: 520 N 0
6.5x DE
F FΣ = − =
 
200 N C DEF = W
 
 
 
 
 
PROBLEM 6.8 
Using the method of joints, determine the force in each member of the 
truss shown. State whether each member is in tension or compression. 
 
SOLUTION 
 
 FBD Truss: 
 
 Joint FBDs: 
 Joint F: 
 
 Joint C: 
 
 
 
 
 
( ) ( )( ) ( )( )0: 9 ft 6.75 ft 4 kips 13.5 ft 4 kips 0E yM FΣ = − − =
 
9 kipsy =F
 
 
0: 9 kips 0 9 kipsy y yF EΣ = − + = =E
 
 
0: 4 kips 4 kips 0 8 kipsx x xF EΣ = − + + = =E
 
By inspection of joint :E
 
9.00 kips T ECF = W
 
 
8.00 kips T EFF = W
 
By inspection of joint :B
 
0ABF = W
 
 
0BDF = W
 
 
40: 8 kips 0
5x CF
F FΣ = − =
 
10.00 kips C CFF = W
 
 
30: (10 kips) 0
5y DF
F FΣ = − =
 
6.00 kips T DFF = W
 
 
 
( )40: 4 kips 10 kips 0
5x CD
F FΣ = − + =
 
 
4.00 kips T CDF = W
 
 
( )30: 9 kips 10 kips 0
5y AC
F FΣ = − + =
 
 
3.00 kips T ACF = W
 
 
 
 
 
 
 Joint A: 
 
PROBLEM 6.8 CONTINUED 
 
 
40: 4 kips 0
5x AD
F FΣ = − =
 
5.00 kips C ADF = W
 
 
 
 
 
PROBLEM 6.9 
Determine the force in each member of the Gambrel roof truss shown. 
State whether each member is in tension or compression. 
 
SOLUTION 
 FBD Truss: 
 
 
 Joint A: 
 
 
 Joint B: 
 
 
 
 Joint D: 
 
 
 
 
0: 0x xFΣ = =H
 
By symmetry: 12 kNy= =A H
 
By inspection of joints and ,C G
 
 and 0CE AC BCF F F= = W
 
 
 
 and 0EG GH FGF F F= = W
 
 
30: 12 kN 3 kN 0
5y AB
F FΣ = − − =
 
15.00 kN C ABF = W
 
 
( )40: 15 kN 0
5x AC
F FΣ = − =
 
12.00 kN T ACF = W
 
 
 
( )4 10 40: 15 kN 0
5 10.44 5x BD BE
F F FΣ = − − =
 
 
 
( )3 3 30: 15 kN 6 kN 0
5 10.44 5y BD BE
F F FΣ = − − + =
 
Solving yields 11.93 kN C BDF = W
 
 
0.714 kN C BEF = W
 
 
0: by symmetryxFΣ =11.93 kN C DFF = W
 
 
( )30: 6 kN 2 11.93 kN 0
10.44y DE
F FΣ = − − + =
 
 
0.856 kN T DEF = W
 
By symmetry:
 
 so 0.714 kN C EF BE EFF F F= = W
 
 
FH ABF F=
 
15.00 kN C FHF = W
 
 
GH ACF F=
 
12.00 kN T GHF = W
 
From above
 
CE ACF F=
 
12.00 kN T CEF = W
 
 
EG GHF F=
 
12.00 kN T EGF = W
 
 
 
PROBLEM 6.10 
Determine the force in each member of the Howe roof truss shown. 
State whether each member is in tension or compression. 
 
SOLUTION 
 FBD Truss: 
 
 
 Joint FBDs: 
 
 
 Joint C: 
 
 
 Joint B: 
 
 
 
0: 0x xFΣ = =A 
By symmetry: 12 kNy y= =A H 
and
 
; ; 
; ; 
FH AB GH AC FG BC
DF BD EF BE EG CE
F F F F F F
F F F F F F
= = =
= = =
 
 
 
 
9 kN
3 4 5
AC ABF F
= =
 
 
12.00 kN T ACF = W
 
 
15.00 kN C ABF = W
 
 
so 15.00 kN C FHF = W
 
 
By inspection: 
 
12.00 kN T GHF = W
 
 
0; 12 kNBC CEF F= =
 
0BC FGF F= = W
 
 
12.00 kN T CEF = W
 
 
12.00 kN T EGF = W
 
1
1
4Note: tan so sin 0.8
3
32 tan so sin 0.96
4
−
−
= =
= =
α α
β β
 
 
1 0: 6 kN sin sin 0y BEF FΣ = − + =α β
 
 
5.00 kN C BEF = W
 
 
so 5.00 kN C EFF = W
 
 
( )3 3 30: 6 kN 15 kN 0
5 5 5y BD BE
F F FΣ = − − + + =
 
 
 
 
 Joint D: 
 
PROBLEM 6.10 CONTINUED 
 
 
( ) ( )3 3 35 kN 15 kN 6 kN
5 5 5BD
F = + −
 
10.00 kN C BDF = W
 
 
so 10.00 kN C DFF = W
 
 
 
30: 6 kN 2 10 kN 0
5y DE
F F Σ = − + − =   6.00 kN T DEF = W 
 
 
 
 
PROBLEM 6.11 
Determine the force in each member of the Fink roof truss shown. State 
whether each member is in tension or compression. 
 
 
SOLUTION 
 FBD Truss: 
 
 
 
 
 Joint FBDs: 
 Joint A: 
 
 
 
 
 Joint B: 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
0 : 0x xFΣ = =A 
 
By symmetry: 24 kipsy y= =A G
 
also; 
 
 ; ; 
; 
FG AB EG AC EF BC
DF BD DE CD
F F F F F F
F F F F
= = =
= = 
 
 
18 kips
;
4 9 97
AC ABF F= =
 
40.5 kips T ACF = W
 
 
44.32 kipsABF =
 
44.3 kips C ABF = W
 
 
so 44.3 kips C FGF = W
 
 
40.5 kips T EGF = W
 
 
( )9 30: 44.32 kips 0
597x BD BC
F F FΣ = − − =
 
 
( )4 40: 44.32 kips 12 0
597y BD BC
F F FΣ = − + − =
 
Solving: 11.25 kips C BCF = W 
 
36.9 kips C BDF = W
 
 
so 11.25 kips C EFF = W
 
 
36.9 kips C DFF = W
 
 
 
 
 
 
 
 
Joint C: 
 
PROBLEM 6.11 CONTINUED 
 
( )4 40: 11.25 kips 0
5 5y CD
F FΣ = − =
 
11.25 kips T CDF = W
 
 
11.25 kips T DEF = W
 
 
( )30: 2 11.25 kips 40.5 kips 0
5x CE
F F  Σ = + − =   
 
27.0 kips T CEF = W
 
 
 
 
PROBLEM 6.12 
Determine the force in each member of the fan roof truss shown. State 
whether each member is in tension or compression. 
 
 
SOLUTION 
 FBD Truss: 
 
 
 Joint FBDs: 
 Joint A: 
 
 
 Joint B: 
 
 
 
 
 Joint D: 
 
 
 
 
 
0 : 0x xFΣ = =A 
 
By symmetry: 12 kNy y= =A I
 
and 
 
; ; 
; ; 
AB HI AC GI BC GH
BD FH DC FG DE EF
CE EG
F F F F F F
F F F F F F
F F
= = =
= = =
=
 
 
 
10 kN
4 9 97
AC ABF F= =
 
22.5 kN T ACF = W
 
 
so 22.5 kN T GIF = W
 
 
2.5 97 kNABF =
 
24.6 kN C ABF = W
 
 
so 24.6 kN C HIF = W
 
 
( )90: 22.5 kN 0
97x BD BC
F F FΣ = − + =
 
 
( )40: 10 kN 4 kN 0
97y BC BD
F F FΣ = − + − =
 
Solving:
 
 19.70 kN C BDF = W
 
 
so 19.70 kN C FHF = W
 
 
and 4.92 kN C BCF = W
 
 
so 4.92 kN C GHF = W
 
By inspection: 19.70 kN C DEF = W 
 
so 19.70 kN C EFF = W
 
 
and 4.00 kN C CDF = W
 
 
so 4.00 kN C FGF = W
 
 
 
 
 
 
 
 Joint C: 
 
 
 
 
PROBLEM 6.12 CONTINUED
 
 
( )9 30: 22.5 kN 4.92 kN 0
597x CE CG
F F FΣ = − + + + =
 
 
( )4 40: 4 kN 4.92 kN 0
597y CE
F FΣ = − − + =
 
Solving:
 
 7.50 kN T CEF = W
 
7.50kN TEGso F = W
 
and 13.50 kN T CGF = W
 
 
 
 
 
 
 
PROBLEM 6.13 
Determine the force in each member of the roof truss shown. State 
whether each member is in tension or compression. 
 
 
SOLUTION 
FBD Truss: 
 
 
( ) ( )( ) ( )( ) ( )( ) ( )( )0: 16 ft 16 ft 200 lb 12 ft 300 lb 8 ft 400 lb 4 ft 400 lb 0A yM HΣ = − − − − =
 
725 lby =H
 
 
0: 200 lb 400 lb 400 lb 300 lb 200 lb 725 lb 0y yF AΣ = − − − − − + =
 
775 lby =A 
 
0 : 0x xFΣ = =A 
Joint FBDs: 
 Joint A: 
 
575 lb
1 26 29
AC ABF F= =
 
 
3096.5 lb; 3.10 kips C AB ABF F= = W
 
 
2931.9 lb; 2.93 kips T AC ACF F= = W
 
 
 
 
 
PROBLEM 6.13 CONTINUED 
 
Joint C: 
 
 By inspection: 0 BCF = W 
2.93 kips T CEF = W
 
 
Joint B: 
 
 
( )2 20: 3097 lb 400 lb 0
29 29y BD
F FΣ = − − =
 
2020.0 lb; 2.02 kips C BD BDF F= = W 
 
( )50: 3097 2020 lb 0
29x BE
F FΣ = − − = 1000.0 lb; 1.000 kip C BE BEF F= = W
 
 
Joint D: 
 
 
( )50: 2.02 kips 0
29x DF
F FΣ = − =
 
2.02 kips C DFF = W
 
 
( )20: 2 2.02 kips 0.4 kips = 0
29y DE
F FΣ = + −
 
1.100 kips C DEF = W
 
Joint H: 
 
 
525 lb
1 29 26
FH GHF F= =
 
2827 lb; 2.83 kips C FH FHF F= = W
 
2677 lb; 2.68 kips T GH GHF F= = W
 
Joint G: 
 
By inspection: 2.68 kips T EGF = W 
 
0 FGF = W
 
 
PROBLEM 6.13 CONTINUED 
Joint F: 
 
 
( )20: 2.83 kips 0.3 kips 0 2.02 kips
29y DF DF
F F FΣ = − − = =
 
 
( )50: 2.02 kips 2.827 kips 0
29x EF
F FΣ = + − = 0.750 kips C EFF = W 
 
 
 
 
PROBLEM 6.14 
Determine the force in each member of the Gambrel roof truss 
shown. State whether each member is in tension or compression. 
 
 
 
SOLUTION 
 FBD Truss: 
 
 
 Joint FBDs: 
 Joint A: 
 
 
 
 Joint B: 
 
 
 
 
 
 
 
 
 
 
 
0 0x xFΣ = =A 
 
By symmetry: 5 kNy y= =A J
 
and
 
 ; ; 
; ; 
AB HJ AC IJ BD GH
CD GI DE EG DF FG
BC HI
F F F F F F
F F F F F F
F F
= = =
= = =
=
 
By inspection of joint F: 0 EFF = W 
 
8 890: 5 kN 1 kN 0 kN
289y AB AB
F F FΣ = − − = =
 
 
4.72 kN C ABF = W
 
 
 
5 890: kN 0
289x AC
F FΣ = − =
 
 
2.50 kN T ACF = W
 
 
so 4.72 kN C HJF = W
 
 
2.50 kN T IJF = W
 
 
5 890: kN 0
289x BD BC
F F F
 Σ = − − =    
 
8 890: kN 1 kN 0
289y BD BC
F F F
 Σ = − + − =    
Solving: 4.127 kNBDF =
 
so 4.13 kN C BDF = W
 
 
0.5896 kNABF =
 
and 0.590 kN C BCF = W
 
 
so 4.13 kN C GHF = W
 
 
and 0.590 kN C HIF = W
 
 
 
 
 
 Joint C: 
 
 
 
 
 
 
 
 Joint D: 
 
 
PROBLEM 6.14 CONTINUED 
 
 
( )50: .59 kN 2.5 kN 0; 2.187 kN
89x CI CI
F F FΣ = + − = =
 
 
2.19 kN T CIF = W
 
 
( )80: .59 kN 0
89y CD
F FΣ = − =
 
0.500 kN T CDF = W
 
 
so 0.500 kN T GIF = W
 
 
( )8 50: 4.127 kN 2.5 kN 0
89 281y DE
F FΣ = − − =
 
 
3.352 kNDEF =
 
 
so 3.35 kN C DEF = W
 
 
and 3.35 kN C EGF = W
 
 
 
( ) ( )5 164.127kN 3.352 kN 0
89 281x DF
F FΣ = − + =
 
 
1.012 kN T DFF = W
 
 
1.012 kN T FGF = W
 
 
 
 
PROBLEM 6.15 
Determine the force in each member of the Pratt bridge truss shown. 
State whether each member is in tension or compression. 
 
 
 
 
SOLUTION 
 FBD Truss: 
 
 
 
 FBDs Joints: 
 
 
 
 
 
 
 
 
0: 0x xFΣ = =A 
By symmetry: 9 kNy y= =A H
 
and
 
 ; 
; 
; 
AB FH AC GH
BC FG BD DF
BE EF CE EG
F F F F
F F F F
F F F F
= =
= =
= =
 
By inspection of joint D: 0 DEF = W 
 
 
40: 9 kN 0
5y AB
F FΣ = − =
 
11.25 kN C ABF = W
 
 
 
30: 0
5x AC AB
F F FΣ = − =
 
6.75 kN T ACF = W
 
 
 
 
0: 6.75 kN 0x CEF FΣ = − =
 
6.75 kN T CEF = W
 
 
0: 6 kN 0y BCF FΣ = − =
 
6.00 kN T BCF = W
 
 
( )4 40: 11.25 kN 6 kN 0
5 5y BE
F FΣ = − + =
 
 
3.75 kN C BEF = W
 
( ) ( )3 30: 11.25 kN 3.75 kN 0
5 5x BD
F FΣ = − − =
 
 
9.00 kN T BDF = W
 
From symmetry conditions above
 
11.25 kN C FHF = W
 
 
6.75 kN T GHF = W
 
 
6.75 kN T EGF = W
 
 
6.00 kN T FGF = W
 
 
3.75 kN C EFF = W
 
 
9.00 kN T DFF = W
 
 
 
 
PROBLEM 6.16 
Determine the force in each member of the Pratt bridge truss shown. 
State whether each member is in tension or compression. Assume 
that the load at G has been removed. 
 
 
SOLUTION 
 FBD Truss: 
 
 
 
 Joint FBDs: 
 Joint H: 
 
 Joint F: 
 
 
 Joint B: 
 
 
 
 
 
 
 
 
0: 0x xFΣ = =A
 
 
( ) ( )( ) ( )( )0: 12 m 6 m 6 kN 3 m 6 kN 0A yM HΣ = − − =
 
 
4.5 kNy =H
 
 
 
0: 6 kN 6 kN 4.5 kN 0y yF AΣ = − − + =
 
 
7.5 kNy =A
 
 
4.5 kN
4 3 5
GH FHF F= =
 
 
3.375 kNGHF =
 
3.38 kN T GHF = W
 
 
5.625 kNFHF =
 
5.63 kN C FHF = W
 
By inspection of joint :G
 
0 FGF = W
 
 
3.38 kN T EG GHF F= = W
 
 
5.625 kN
5 5 6
EF DFF F= =
 
5.63 kN T EFF = W
 
 
6.75 kN C DFF = W
 
By inspection of joint :D
 
0 DEF = W
 
 
6.75 kN C BD DFF F= = W
 
By inspection of joint : AC CEC F F=
 
and 6.00 kN T BCF = W
 
 
( )30: 6.75 kN 0
5x AB BE
F F FΣ = + − =
 
 
 
( )40: 6 kN 0
5y AB BE
F F FΣ = − − =
 
( )40: 6 kN 0
5y AB BE
F F FΣ = − − =
 
 
 
 
 
 
 
 
 Joint A: 
 
 
 
PROBLEM 6.16 CONTINUED 
 
Solving: 
 
9.375 kN so 9.38 kN C AB ABF F= = W
 
 
1.875 kNBEF =
 
1.875 kN T BEF = W
 
7.5 kN 9.375 kN
 5.625 kN
3 4 5
AC
AC
F F= = =
 
 
5.63 kN T ACF = W
 
From above
 
5.63 kN T CEF = W
 
 
 
PROBLEM 6.17 
Determine the force in member DE and in each of the members 
located to the left of DE for the inverted Howe roof truss shown. 
State whether each member is in tension or compression. 
 
 
SOLUTION 
 FBD Truss: 
 Joint FBDs: 
 Joint A: 
 
 
 Joint B: 
 
 
 Joint C: 
 
 
 
 
 
0: 0x xFΣ = =A
 
By load symmetry 2400 lby y= =A H
 
Note: 
 
1 10.08tan 16.26
15.81 18.75
θ −= = °+ 
 
90 2 57.48 ; 180 32.52β θ α β= − = ° = − = °
 
 
 
( )0: 2400 lb 600 lb cos sin 0y ACF Fθ θ′Σ = − − =
 
 
( )1800 lb 6171.5 lb;
tan16.26AC
F = =° 6.17 kips T ACF = W 
0: (6171.5 lb)cos 2 cos 0x ABF Fθ θΣ = − =
 
 
cos32.526171.5 5420.7 lb;
cos16.26AB
F °= =° 5.42 kips C ABF = W 
 
( )0: 5420.7 lb 1200 lb sin 0x BDF Fθ′Σ = + − =
 
 
5420.7 1200sin16.26 5756.7 lbBDF = + ° =
 
5.76 kips C BDF = W
 
 
 
( )0: 1200 lb cos 0 1152 lby BC BCF F F′Σ = − = =θ
 
 
1.152 kips C BCF = W
 
( )0: sin 2 1152 lb cos 0y CDF F θ θ′′Σ = − =
 
 
cos16.261152 2057.2 lb
sin 32.52CD
F °= =° 2.06 kips T CDF = W 
 
( ) ( )0: 1152 lb sin 2057.2 lb cos 2 6171.5 lb 0x CEF F θ θ′′Σ = + + − =
 
 
 
 
 
 
 
 
 Joint E: 
 
PROBLEM 6.17 CONTINUED 
6171.5 1152sin16.26 2057.2cos32.52CEF = − ° − °
 
 
4114.3 lb=
 
4.11 kips T CEF = W
 
 
 
 
( )0: 4114.3 lb sin 2 cos 0y DEF Fθ θΣ = − =
 
 
sin 32.524114.3 2304.0 lb
cos16.26DE
F °= =° 2.30 kips C DEF = W 
 
 
 
PROBLEM 6.18 
Determine the force in each of the members located to the right of 
DE for the inverted Howe roof truss shown. State whether each 
member is in tension or compression. 
 
 
 
SOLUTION 
 FBD Truss: 
 Joint FBDs: 
 Joint H: 
 
 
 
 Joint F: 
 
 
 
 
 
 
 
 
 
0: 0x xFΣ = =A
 
By symmetry of loads 2400 lby y= =A H
 
Note: 
 
1 10.08tan 16.26
15.81 18.75
θ −= = °+ 
 
90 2 57.48= − = °β θ
 
 
 
0: 2400 lb 600 lb sin 0y FHF F θΣ = − − =
 
 
1800 lb 6428.7 lb
sin16.26FH
F = =° 6.43 kips C FHF = W 
 
( )0: 6428.7 lb cos 0x GHF FθΣ = − =
 
 
6428.7cos16.32 6171.5 lbGHF = ° =
 
6.17 kips T GHF = W
 
 
 
( )0: 1200 lb cos 0 1152.0 lby FG FGF F Fθ′Σ = − = =
 
 
1.152 kips C FGF = W
 
( )0: 1200 lb sin 6428.7 lb 0x DFF F θ′Σ = + − =
 
 
6428.7 1200sin16.26 6092.7 lbDFF = − ° =
 
6.09 kips C DFF = W
 
 
 
 
 
 
 
 
Joint G: 
 
 
 
PROBLEM 6.18 CONTINUED 
 
 
0: sin 2 (1152 lb)cos 0y DGF F θ θΣ = − =
 
cos16.261152 2057.2 lb
sin 32.52DG
F °= =° 2.06 kips T DGF = W 
 
0: 6171.5 lb 2057.2cos 2 1152 lb sin 0x EGF FΣ = − − − =θ θ 
( )6171.5 2057.2cos32.52 1152lb sin 16.26 4114.3 lbEGF = − °− ° =
 
4.11 kips T EGF = W
 
 
 
 
PROBLEM 6.19 
Determine the force in each of the members located to the left of 
member FG for the scissor roof truss shown. State whether each 
member is in tension or compression. 
 
SOLUTION 
 FBD Truss: 
 
 Joint FBDs: 
 Joint A: 
 
 Joint B: 
 
 
 Joint E: 
 
 
 
 
 
 
 
( ) ( ) ( ) ( )0: 6 ft 6 3 .5 kip 2 1 kip 1 1 kip 0A yM L Σ = − − − =  
 
0.75 kipy =L
 
 
0: 0.5 kip 1 kip 1 kip 0.5 kip 0.75 kip 0y yF AΣ = − − − − + =
 
 
2.25 kipsy =A
 
 
 
0: 0x xFΣ = =A
 
 
1.75 kips
1 5 8
AC ABF F= =
 
4.95 kips C ABF = W
 
 
3.913 kipsACF =
 
3.91 kips T ACF = W
 
By inspection of joint :C
 
0 BCF = W
 
and 
 
 so 3.91 kips T CE AC CEF F F= = W
 
 
4.95 kips0: 1 kip 0
2 2
BD
y
FFΣ = − − =
 
 
3.536 kipsBDF =
 
3.54 kips C BDF = W
 
 
( ) 10: 4.95 kips 3.536 kips 0
2x BE
F FΣ = − − =
 
 
1.000 kipBEF =
 
1.000 kip C BEF = W
 
 [ ]20: 3.913 kips 1 kip 0
5x EG
F FΣ = − + =
 
 
2.795 kipsEGF =
 
2.80 kips T EGF = W
 
 
( )10: 2.795 kips 3.913 kips 0
5y DE
F FΣ = − + =
 
 
0.500 kip T DEF = W
 
 
 
 
 
 Joint D: 
 
PROBLEM 6.19 CONTINUED 
 
( ) ( )1 20: 3.536 kips 0
2 5x DF DG
F F FΣ = − + =
 
 
( ) ( )1 10: 3.536 kips 1.5 kips 0
2 5y DG DF
F F FΣ = − + − =
 
Solving:
 
2.516 kips so 2.52 kips C
0.280 kip 0.280 kip C 
DF DF
DG DG
F F
F F
= =
= =
W
W 
 
 
 
PROBLEM 6.20 
Determine the force in member FG and in each of the members 
located to the right of member FG for the scissor roof truss shown. 
State whether each member is in tension or compression. 
 
 
SOLUTION 
 
 FBD Truss: 
 
 
 Joint FBDs: 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
( ) ( ) ( ) ( )0: 6 ft 6 3 0.5 kip 2 1 kip 1 1 kip 0A yM L Σ = − − − =  
 
0.75 kipy =L
 
Inspection of joints , , and ,in order, shows thatK J I
 
0 JKF = W
 
 
0 IJF = W
 
 
0 HIF = W
 
and that 
 
; and IK KL HJ JL GI IKF F F F F F= = =
 
 
0.75
 2.1213 kips
1 8 5
JL KL
JL
F F F= = =
 
2.12 kips C JLF = W
 
 
1.6771 kipsKLF =
 
1.677 kips T KLF = W
 
and, from above:
 
2.12 kips C HJF = W
 
 
1.677 kips T GI IKF F= = W
 
 
( ) ( )2 10: 2.1213 kips 0
5 2x FH GH
F F FΣ = + − =
 
 
( ) ( )1 10: 2.1213 kips 0
5 2y GH FH
F F FΣ = + + =
 
Solving: 
 
2.516 kipsFHF =
 
 2.52 kips C FHF = W
 
 
0.8383 kipsGHF = −
 
 0.838 kips T GHF = W
 
 
 
 
 
 
 
 
PROBLEM 6.20 CONTINUED 
 
( )20: 2.516 kips 0
5x DF
F FΣ = − =
 
2.52 kips CDFF =
 
( )( )10: 0.5 kip 2 2.516 kips 0
5y FG
F FΣ = − + =
 
 
1.750 kips T FGF = W
 
 
 
 
PROBLEM 6.21 
The portion of truss shown represents the upper part of a power 
transmission line tower. For the given loading, determine the force in 
each of the members located above HJ. State whether each member is 
in tension or compression. 
 
 
SOLUTION 
 Joint FBDs: 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
1.5 kN
2.29 2.29 1.2
DF EFF F= =
 
 
2.8625 kNDF EFF F= =
 
2.86 kN T DFF = W
 
 
2.86 kN C EFF = W
 
2.8625 kN 1.25 kN
2.21 0.6 2.29
BD DEF F= = =
 
 
2.7625 kNBDF =
 
2.76 kN T BDF = W
 
 
0.750 kN C DEF = W
 
By symmetry of joint vs. joint A F
 
2.86 kN T ABF = W
 
 
2.86 kN T ACF = W
 
 
( )2.21 40: 2.7625 kN 2.8625 kN 0
2.29 5x BE
F FΣ = − + =
 
 
0 BEF = W
 
 
 
( )0.60: 2.8625 kN 0;
2.29y BC
F FΣ = − =
 
 
0.750 kN C BCF = W
 
 
 
( )2.210: 2.8625 kN 0 2.7625 kN
2.29x CE CE
F F FΣ = − = =
 
 
2.76 kN C CEF = W
 
 
( )0.60: 0.75 kN 2.8625 kN 0
2.21y CH
F FΣ = − − =
 
 
1.500 kN C CHF = W
 
 
 
 
 
PROBLEM 6.21 CONTINUED 
 
( )2.21 40: 2.7625 kN 2.8625 kN 0
2.29 5x HE
F FΣ = − − =
 
 
0 HEF = W
 
 
( )0.60: 0.75 kN 2.8625 kN 0
2.29y EJ
F FΣ = − − =
 
 
1.500 kN EJF = W
 
 
 
 
PROBLEM 6.22 
For the tower and loading of Prob. 6.21 and knowing that 
1.5 kNCH EJF F C= = and 0EHF = , determine the force in member 
HJ and in each of the members located between HJ and NO. State 
whether each member is in tension or compression. 
 
SOLUTION 
 
 
 Joint FBDs: 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
1.5 kN
1.2 3.03 3.03
JL KLF F= =
 
 
3.7875 kNJL KLF F= =
 
3.79 kN T JLF = W
 
 
3.79 kN C KLF = W
 
By symmetry:
 
3.79 kN T GHF = W
 
 
3.79 kN C GIF = W
 
 
( )2.970: 3.7875 kN 0
3.03x HJ
F FΣ = − =
 
 
3.7125 kNHJF =
 
3.71 kN T HJF = W
 
 
( )0.60: 3.7875 kN 1.5 kN 0
3.03y JK
F FΣ = − − =
 
 
2.25 kN C JKF = W
 
Knowing 0; by symmetryHEF = 
 
0 HKF = W
 
 
2.25 kN C HIF = W
 
 
( )2.970: 3.7875 kN 0 3.7125
3.03x IK IK
F F FΣ = − = =
 
 
3.71 kN C IKF = W
 
 
( )0.60: 2.25 kN 3.7875 kN 0
3.03y IN
F FΣ = − − =
 
 
3.00 kN C INF = W
 
Knowing that 0, by symmetryHKF =
 
3.00 kN C KOF = W
 
 
0 KNF = W
 
 
 
 
PROBLEM 6.23 
Determine the force in each member of the truss shown. State whether 
each member is in tension or compression. 
 
SOLUTION 
 FBD Truss: 
 
 
 Joint FBDs: 
 
 
 
 
 
 
( ) ( )( ) ( )( )0: 10 m 7.5 m 80 kN 8 m 30 kN 0F yM GΣ = − − =
 
84 kNy =G
 
 
0: 30 kN 0 30 kNx x xF FΣ = − + = =F
 
 
0: 84 kN 80 kN 0 4 kNy y yF FΣ = + − = =F
 
By inspection of joint :G
 
0 EGF = W
 
 
84 kN C CGF = W
 
 
84 kN 10.5 kN
8 61 29
CE ACF F= = =
 
82.0 kN T CEF = W
 
 
56.5 kN C ACF = W
 
 
 
( )2 60: 82.0 kN 80 kN 0
5 61y AE
F FΣ = + − =
 
 
19.01312AEF = 19.01 kN T AEF = W
 
 
( ) ( )1 50: 19.013 kN 82.0 kN 0
5 61x DE
F FΣ = − − + =
 
 
43.99 kNDEF =
 
44.0 kN T DEF = W
 
 
 
10: 30 kN 0
5x DF
F FΣ = − =
 
 
67.082 kNDFF =
 
67.1 kN T DFF = W
 
 
 
 
 
 
 
 
 
 
PROBLEM 6.23 CONTINUED 
 
( )20: 67.082 kN 4 kN 0
5y BF
F FΣ = − − =
 
 
56.00 kNBFF =
 
56.0 kN C BFF = W
 
 
 
5 50: 30 kN 0
61 29x BD AB
F F FΣ = + − =
 
 
6 30: 56 kN 0
61 29y BD AB
F F FΣ = − − =
 
Solving: 
 
42.956 kNBDF =
 
43.0 kN T BDF = W
 
 
61.929 kNABF =
 
 61.9 kN C ABF = W
 
 
( ) ( )6 20: 42.956 N 67.082 N 0
61 5y AD
F FΣ = + − =
 
 
30.157 kNADF =
 
30.2 N T ADF = W
 
 
 
 
 
PROBLEM 6.24 
Determine the force in each member of the truss shown. State whether 
each member is in tension or compression. 
 
 
SOLUTION 
 Joint FBDs: 
 
 
 
 
 
 
 
 
 
 
 
9 kN
9 40 41
AC ABF F= =
 
 
41.0 kN T ABF = W
 
 
40.0 kN C ACF = W
 
By inspection of joint B:
 
41.0 kN T BDF = W
 
 
3.00 kN C BCF = W
 
( )4.6Note: to determine slope of : 2.25 m 1.035 m
10.0
CD DE = =
 
 
 
10: 3 kN 0
5y CD
F FΣ = − =
 
3 5 kN 6.71 kN T CDF = = W
 
 ( )20: 3 5 kN 40 kN 05x CEF FΣ = + − = 46.0 kN C CEF = W 
 
 
( ) ( )40 20: 41 kN 3 5 kN 041 5x DGF FΣ = − − = 
 
47.15 kNDGF =
 
47.2 kN T DGF = W
 
( ) ( )9 10: 47.15 kN 41 kN 3 5 kN 041 5y DEF FΣ = + − − = 
 
1.650 kN C DEF = W
 
 
50: 1.65 kN 0
13y EG
F FΣ = − =
 
4.29 kN T EGF = W
 
 
( )120: 46 kN 4.29 kN 0 49.96 kN
13x EF EF
F F FΣ = + − = =
 
 
50.0 kN C EFF = W
 
 
 
 
 
 
PROBLEM 6.24 CONTINUED 
 
 
49.96 kN
5 4 3
FG FHF F= =
 
40.0 kN C FGF = W
 
 
30.0 kN C FHF = W
 
 
 
 
 
PROBLEM 6.25 
For the roof truss shown in Fig. P6.25 and P6.26, determine the force 
in each of the members located to the left of member GH. State 
whether each member is in tension or compression. 
 
SOLUTION 
 FBD Truss: 
 
 
 
 Joint FBDs: 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
( )( ) ( )( ) ( )( )
( )( ) ( )( ) ( )( )
0: 3 ft 300 lb 9 ft 300 lb 15 ft 300 lb
18 ft 300 lb 24 ft 300 lb 30 ft 200 lb
MMΣ = + +
+ + + 
 
( )( )30 ft 0 890 lby yA− = =A
 
 
0: 0x xFΣ = =A
 
 
690 lb
5 12 13
AC ABF F= =
 
1794 lb C ABF = W
 
 
1656 lb T ACF = W
 
 
By inspection of joint F: 0 CFF = W
 
 
0 EFF = W
 
 
1656 lb
12 13 5
CE BCF F= =
 
1794 lb T CEF = W
 
 
890 lb C BCF = W
 
 
( )50: 1794 lb 690 lb 300 lb 0
13y BD
F FΣ = − + − =
 
 
2808 lbBDF =
 
2.81 kips C BDF = W
 
 
( )120: 1794 lb 2808 lb 0
13x BE
F FΣ = + − =
 
936 lb T BEF = W
 
 
6 120: 936 lb 1794 lb 0
1337x EH
F FΣ = − − =
 
 
432 37 lbEHF =
 
2.63 kips T EHF = W
 
 
( ) ( )5 10: 1794 lb 432 37 lb 013 37y DEF FΣ = − − =1122 lb T DEF = W
 
 
 
 
 
 
 
 
 
 
 
PROBLEM 6.25 CONTINUED 
 
( )12 10: 2808 lb 0
13 2x DG DH
F F FΣ = + − =
 
( )5 10: 2808 lb 300 lb 1122 lb 0
13 2y DG DH
F F FΣ = + + − − =
 
Solving: 1721 lb T DGF = W 
 
1419 lb C DHF = W
 
 
 
 
PROBLEM 6.26 
Determine the force in member GH and in each of the members located 
to the right of GH for the roof truss shown. State whether each member 
is in tension or compression. 
 
 
SOLUTION 
 FBD Truss: 
 
 
 
 Joint FBDs: 
 
 
 
 
 
 
 
 
 
 
 
 
 
( ) ( )( ) ( )( )
( )( ) ( )( )
( )( ) ( )( )
0: 30 ft 30 ft 200 lb 27 ft 300 lb
21 ft 300 lb 15 ft 300 lb
12 ft 300 lb 6 ft 300 lb 0
A yM MΣ = − −
− −
− − =
 
1010 lby =M
 
810 lb
5 12 13
LM KMF F= =
 
 
2106 lbKMF =
 
2.11 kips C KMF = W
 
 
1944 lbLMF =
 
1.944 kips T LMF = W
 
 
1944 lb
1 637
JL KLF F= =
 
324 lb C KLF = W
 
 
1970.8 lbJLF =
 
1.971 kips T JLF = W
 
 
 
( )12 240: 2106 lb 0
13 577x IK JK
F F FΣ = − + =
 
 
( )5 10: 2106 lb 24 lb 0
13 577y IK JK
F F FΣ = − + − + =
 
Solving: 
 
2162.7 lbIKF =
 
2.16 kips C IKF = W
 
 
52.4 lb T JKF = W
 
 
( ) ( )6 240: 1970.8 lb 52.4 lb 0
37 577x HJ
F FΣ = − + =
 
 
2024 lbHJF =
 
2.02 kips T HJF = W
 
 
( ) ( )1 10: 2024 lb 1970.8 lb 52.4 lb 0
37 577y IJ
F FΣ = − + − =
 
 
10.90 lb C IJF = W
 
 
 
 
 
PROBLEM 6.26 CONTINUED 
 
( )0: sin 38.88 10.9 lb 300 lb cos 22.62 0y HIF F′Σ = ° + − ° =
 
 
425.1 lbHJF =
 
425 lb C HIF = W
 
 
( )0: cos38.88 300 lb 10.9 lb sin 22.62x GI HIF F F′Σ = + ° + − °
 
 
2162.7 lb 0− =
 
1720.5 lb 1.721 kips C GIF = = W
 
By symmetry 1720.5 lbDGF =
 
 
( )50: 2 1720.5 lb 300 lb 0
13y GH
F FΣ = − − =
 
 
1023.46 lbGHF =
 
1.023 kips T GHF = W
 
 
 
PROBLEM 6.27 
Determine whether the trusses of Probs. 6.13, 6.14, and 6.25 are 
simple trusses. 
 
 
 
SOLUTION 
 Truss of 6.13: 
 
 Truss of 6.14: 
 
 Truss of 6.25: 
 
 
Start with and add, in order, joints , , , ,ABC E D F G H∆
 
 
This is a simple truss. W
 
 
 
ABDC, DEGF, and GHJI are all individually simple trusses, but no 
simple extension (one joint, two members at a time) will produce the 
given truss: 
 
 Not a simple truss. ∴ W
 
 
 
Starting with , add, in order, joints , , , , , , , , ,ABC E F D H G I J K L M∆
 
 
This is a simple truss. W
 
 
 
 
 
PROBLEM 6.28 
Determine whether the trusses of Probs. 6.19, 6.21, and 6.23 are 
simple trusses. 
 
 
 
SOLUTION 
 Truss of 6.19: 
 Truss of 6.21: 
 Truss of 6.23: 
 
 
Start with ,ABC∆ and add, in order, E, D, G, F, H, I, J, K 
 
 Simple truss ∴ W
 
 
 
Start with ,ABC∆ and add, in order, E, D, F, H, J, K, I, G, L, N, O, R, Q, 
M, P, S, T, etc. 
 
 Simple truss ∴ W
 
 
 
 
 
 
 
 
Start with ,BDF∆ and add, in order, A, E, C, G. 
 
 Simple truss ∴ W
 
 
 
 
 
 
PROBLEM 6.29 
For the given loading, determine the zero-force members in the 
truss shown. 
 
 
 
SOLUTION 
 
 
 
 
 
 
 
By inspection of joint :B
 
0 BCF = W
 
 
Then by inspection of joint :C
 
0 CDF = W
 
 
By inspection of joint :J
 
0 IJF = W
 
 
Then by inspection of joint :I
 
 0 ILF = W
 
 
By inspection of joint :N
 
0 MNF = W
 
Then by inspection of joint :M
 
0 LMF = W
 
 
 
PROBLEM 6.30 
For the given loading, determine the zero-force members in the truss 
shown. 
 
 
SOLUTION 
 
By inspection of joint :C
 
0 BCF = W
 
Then by inspection of joint :B
 
0 BEF = W
 
By inspection of joint :G
 
0 FGF = W
 
Then by inspection of joint :F
 
0 EFF = W
 
Then by inspection of joint :E
 
0 DEF = W
 
 
By inspection of joint :M
 
0 MNF = W
 
Then by inspection of joint :N
 
0 KNF = W
 
By inspection of joint :I
 
0 IJF = W
 
 
 
 
PROBLEM 6.31 
For the given loading, determine the zero-force members in 
the truss shown. 
 
 
SOLUTION 
 
 
 
 
 
By inspection of joint :D
 
 
0 DIF = W
 
By inspection of joint :E
 
 
0 EIF = W
 
Then by inspection of joint :I
 
 
0 AIF = W
 
By inspection of joint :B
 
 
 0 BJF = W
 
By inspection of joint :F
 
 
0 FKF = W
 
By inspection of joint :G
 
 
0 GKF = W
 
Then by inspection of joint :K
 
 
0 CKF = W
 
 
 
 
 
 
 
 
 
PROBLEM 6.32 
For the given loading, determine the zero-force members 
in the truss shown. 
 
 
SOLUTION 
 
By inspection of joint :C
 
0 BCF = W
 
By inspection of joint :M
 
0 LMF = W
 
 
 
 
 
 
 
PROBLEM 6.33 
For the given loading, determine the zero-force members in the truss 
shown. 
 
 
SOLUTION 
 
 
 
 
 
 
By inspection of joint :F
 
0 BFF = W
 
Then by inspection of joint :B
 
0 BGF = W
 
Then by inspection of joint :G
 
0 GJF = W
 
Then by inspection of joint :J
 
0 HJF = W
 
 
By inspection of joint :D
 
0 DHF = W
 
Then by inspection of joint :H
 
0 HEF = W
 
 
 
 
PROBLEM 6.34 
For the given loading, determine the zero-force members in the truss 
shown. 
 
 
 
SOLUTION 
 
 
By inspection of joint :C
 
0 BCF = W
 
Then by inspection of joint :B
 
0 BEF = W
 
Then by inspection of joint :E
 
0 DEF = W
 
By inspection of joint :H
 
0 FHF = W
 
 
and 0 HIF = W
 
 
By inspection of joint :Q
 
0 OQF = W
 
 
and 0 QRF = W
 
 
By inspection of joint :J
 
0 IJF = W
 
 
 
 
PROBLEM 6.35 
Determine the zero-force members in the truss of (a) Prob. 6.9, (b) 
Prob. 6.19. 
 
 
 
SOLUTION 
 
 
 Truss of 6.9: 
 
 
 Truss of 6.19: 
 
 
 
By inspection of joint :C
 
0 BCF = W
 
 
By inspection of joint :G
 
0 FGF = W
 
 
 
 
By inspection of joint :C
 
0 BCF = W
 
 
By inspection of joint :K
 
0 JKF = W
 
Then by inspection of joint :J
 
0 IJF = W
 
Then by inspection of joint :I
 
0 HIF = W
 
 
 
 
PROBLEM 6.36 
The truss shown consists of six members and is supported by a ball 
and socket at B, a short link at C, and two short links at D. Determine 
the force in each of the members for ( )5670 lb= −P j and 0.=Q 
 
 
 
SOLUTION 
 FBD Truss: 
 
 
 Joint B: 
 
 
 
 
 
 
 
 
 
 
 
 
0: 0z zFΣ = =B
 
 ( )( ) ( ) ( )0: 2.4 ft 5670 lb 8.4ft 0 1620 lbBD y yM CΣ = − = =C j
 
 
( ) ( )0: 6.3 ft 6.3 ft 0 x y y y yM D B B DΣ = − = =
 
 
( )0: 5670 lb 1620 lb 0 2025 lby y y y yF B DΣ = + − + = = =B D j
 
 
( ) ( )0: 6.3 ft 6.3 ft 0y x xM B DΣ = − =
 
0x x= =B D
 
 
 
0: 0x x xF B DΣ = + =
 
Where
 
( )
( )
2.4 14.4 6.3
 
15.9
0.1509 0.9057 0.3962
8.4 6.3 0.8 0.6
10.5
BA BA
BA
BC BC BC
F
F
F F
+ −=
= + −
−= = −
i j kF
i j k
i kF i k
 
 
0: 0.90572025 lb 0y BAF FΣ = + =
 
 
2236 lbBAF = −
 
2.24 kips C BAF = W
 
 
By symmetry 2.24 kips CADF = W
 
 
( )0: 0.1509 2236 lb 0.8 0x BCF FΣ = − + = 
 
422 lb T BCF = W
 
 
By symmetry 422lb TDCF = W
 
 
 
 
 
 
 Joint C: 
 
 
PROBLEM 6.36 CONTINUED 
 
 
( ) ( )0: 0.3962 2236 lb 0.6 422 lb 0z BDF FΣ = − − − − =
 
 
633 lb T BDF = W
 
( )6 14.4 0.3846 0.9231
15.6AC AC AC
F F−= = −i jF i j
 
 
( )0: 1620 lb 0.9231 0y ACF FΣ = − =
 
1755 lb C ACF = W
 
 
 
 
PROBLEM 6.37 
The truss shown consists of six members and is supported by a ball 
and socket at B, a short link at C, and two short links at D. Determine 
the force in each of the members for 0=P and ( )7420 lb=Q i . 
 
SOLUTION 
 FBD Truss: 
 
 
 Joint FBDs: 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
0: 0z zFΣ = =B
 
 
( ) ( )( )0: 8.4 ft 14.4 ft 7420 lb 0BD yM CΣ = − =
 
( )12720 lby =C j
 
 
( )( )0: 6.3 ft 0 x y y y yM D B D BΣ = − = =
 
0: 0; 2 12720 lb 0y y y y yF B D C DΣ = + + = + =
 
so 
 
( )6360 lby y= = −B D j
 
 
( )( )0: 6.3 ft 0; y x x x xM B D B DΣ = − = =
 
 
( )0: 7420 lb 0; 3710 lbx x x x xF B DΣ = + + = = = −B D i
 
( )
( )
2.4 ft 14.4 ft 6.3 ft 
15.9 ft
0.1509 0.9057 0.3962 
BA BA
BA BD BD
F
F F
+ −=
= + − =
i j k
F
i j k F k
 
( ) ( )8.4 ft 6.3 ft 0.8 0.6
10.5 ftBC BC BC
F F
−= = −i jF i j
 
 
0: 0.9057 6360 lb 0 7022 lby BA BAF F FΣ = − = =
 
 
7.02 kips T BAF = W
 
By symmetry
 
7.02 kips T DAF = W
 
( )0: 0.1509 7022 lb 0.8 3710 lb 0 3313 lbx BC BCF F FΣ = + − = =
 
 
so 3.31 kips T BCF = W
 
By symmetry
 
3.31 kips T DCF = W
 
 
( ) ( )0: 0.3962 7022 lb 0.6 3313 lb 0z BDF FΣ = − + − =
 
 
4770 lbBDF = −
 
4.77 kips C BDF = W
 
 
 
 
 
 
 
 
PROBLEM 6.37 CONTINUED 
( )
( )
6 ft 14.4 ft 
15.6 ft
0.3846 0.9231
AC AC
AC
F
F
−=
= −
i j
F
i j
 
 
12720 lb 0.9231 0; 13780 lby AC ACF F FΣ = − = =
 
 
1.378 kips C ACF = W
 
 
 
PROBLEM 6.38 
The truss shown consists of six members and is supported by a short 
link at A, two short links at B, and a ball and socket at D. Determine 
the force in each of the members for the given loading. 
 
 
 
SOLUTION 
 FBD Truss: 
 
 Joint FBDs: 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
0: 0z zFΣ = =D
 
 
( ) ( )( ) ( )0: 5 ft 12 ft 800 lb 0 1920 lbz x xM AΣ = − = =A i
 
 
( )( )0: 3.5 ft 0;y x x x xM B D B DΣ = − = =
 
 
( )0: 1920 lb 0 so 960 lbx x x x xF B DΣ = + − = = =B D i
 
 
( )( )0: 3.5 ft 0;x y y y yM D B D BΣ = − = =
 
 
( )0: 800 lb 0 so 400 lby y y y yF B DΣ = + − = = =B D j
 
( ) ( )12 ft 5 ft 12 5
13 ft 13
AC
CA AC
FF
− += = − +i jF i j
 
( ) ( )12 ft 3.5 ft 12 3.5
12.5 ft 12.5
CD
CD CD
FF
− −= = − −i kF i k
 
Similarly 
 
( )12 3.5
12.5
CB
CB
FF = − +i k
 
 
( )3.50: 0; 
12.5z CB CD CB CD
F F F F FΣ = − = =
 
 
50: 800 0 2080 lb
13y AC AC
F F F Σ = − = =   
 
2.08 kips T ACF = W
 
( ) ( )
( )
5 ft 3.5 ft 
5 3.5
6.1033 ft 6.1033
12 3.5
12.5
BA
BA BA
BD BD
CB
BC CB
FF
F
F
−= = −
= −
= − = + −
 
j k
F j k
F k
F F i k
 
 
PROBLEM 6.38 CONTINUED 
 
50: 400 lb 0 488 lb
6.1033
BA
y BA
FF FΣ = + = = −
 
 
so 488 lb C BAF = W 
By symmetry:
 
488 lb C ADF = W
 
 
120: 960 lb 0 1000 lb
12.5x BC BC
F F F Σ = + = = −   
 
1.000 kip C BCF = W
 
By symmetry:
 
1.000 kip C CDF = W
 
 
( )3.5 3.50: 488 lb 1000 lb 0
6.1033 12.5z BD
F F  Σ = − − + =   
 
559.9 lbBDF = −
 
560 lb C BDF = W
 
 
 
 
PROBLEM 6.39 
The portion of a power line transmission tower shown consists of nine 
members and is supported by a ball and socket at B and short links at 
C, D, and E. Determine the force in each of the members for the given 
loading. 
 
 
 
 
SOLUTION 
 FBD Truss: 
 
 Joint FBDs: 
 
 
 
 
 
 
 
 
 
 
 
( )( ) ( )0: 1 m 200 N 0 200 NBD z zM EΣ = − = =E k
 
 
( )( )0: 1 m 1200 N 0x z zM E DΣ = − − =
 
( )1200 N 200 N 1000 N 1000 Nz ZD = − = =D k
 
 
( )0: 1000 N 200 N 0 1200 Nz z zF BΣ = − − = =B k
 
 
( )( ) ( ) ( )0: .5 m 1200 N 1 m 0 600 NBz y yM CΣ = − = =C j
 
 
( )0: 200 N 0 200 Nx x xF BΣ = − = =B i
 
 
0: 1200 N 0y y yF B CΣ = + − =
 
( )1200 N 600 N 600 Ny yB = − =B j
 
 
0: 200 N 0
1.5
EA
z
FFΣ = − =
 
300 N T EAF = W
 
 
.50: 0 100 N
1.5x ED EA ED
F F F FΣ = + = = −
 
 
100 N C EDF = W
 
 
0: 0 200 N
1.5
EA
y EC EC
FF F FΣ = − − = = −
 
 
200 N C ECF = W
 
 
0: 1000 N 0
1.5
AD
z
FFΣ = − =
 
1500 N T ADF = W
 
 
0.5 10: 100 N 0
1.5 2x AD CD
F F FΣ = − − =
 
 
1500 N2 100 N 400 2
3CD
F  = − = −   566 N C CDF = W 
 
0: 0
1.52
CD AD
y BD
F FF FΣ = − − − =
 
 
400 1000 600 NBDF = + − = −
 
600 N C BDF = W
 
 
 
 
 
 
 
 
 
 
 
PROBLEM 6.39 CONTINUED
 
 
0: 600 N 200 N 400 N 0yFΣ = − − =
 
 
0: 0
1.25
AC
z
FFΣ = =
 
0 ACF = W
 
 
.50: 400 N
1.25x BC
F FΣ = − + 0 0ACF =
 
 
400 N T BCF = W
 
 
.50: 200 N 400 N 0
1.25x AB
F FΣ = − − − =
 
 
1200 1.25 NABF =
 
1342 N C ABF = W
 
 
 
PROBLEM 6.40 
The truss shown consists of 18 members and is supported by a ball and 
socket at A, two short links at B, and one short link at G. (a) Check that 
this truss is a simple truss, that it is completely constrained, and that the 
reactions at its supports are statically determinate. (b) For the given 
loading, determine the force in each of the six members joined at E. 
 
 
SOLUTION 
(a) To check for simple truss, start with ABDE and add three members at a time which meet 
at a single joint, thus successively adding joints F, G, H, and C, to complete the truss. 
 This is, therefore, a simple truss.W 
 There are six reaction force components, none of which are in-line, so they are determined 
by the six equilibrium equations. Constraints prevent motion. Truss is completely constrained and W 
 statically determinate W 
(b) FBD Truss: 
 
 ( )( ) ( )( ) ( )0: 5.04 m 550 N 5.5 m 0 504 NBC y yM AΣ = + = = −A j 
 ( )( ) ( )( )0: 5.5 m 480 N 4.8 m 550 N 0 0BF z zM AΣ = + − = =A 
 By inspection of joint : C 0DC BC GCF F F= = = 
 By inspection of joint :A 504 N T AE yF A= − = W 
 0AD zF A= = 
 By inspection of joint : H 0DHF = 
 480 N C EHF = W 
 By inspection of joint :F0 EFF = W 
 
PROBLEM 6.40 CONTINUED 
 Then, since ED is the only non-zero member at D, not in the plane BDG, 0 DEF = W 
Joint E: 4.80: 480 N 0
7.3z EG
F FΣ = − = 730 N T EGF = W 
 
5.040: 504 N 0 746 N
7.46y BE BE
F F FΣ = − − = = − 746 N C BEF = W 
 
 
PROBLEM 6.41 
The truss shown consists of 18 members and is supported by a 
ball and socket at A, two short links at B, and one short link at G. 
(a) Check that this truss is a simple truss, that it is completely 
constrained, and that the reactions at its supports are statically 
determinate. (b) For the given loading, determine the force in each 
of the six members joined at G. 
 
 
SOLUTION 
(a) See part (a) solution 6.40 above 
(b) FBD Truss: 
 
 ( ) ( )( )0: 4.8 m 5.04 m 480 N 0 504 NAB y yM GΣ = + = = −G 
 By inspection of joint : C 0 CGF = W 
 By inspection of joint : H 550 N C HGF = W 
 By inspection of joint :F 0 FGF = W 
Joint G: 
 
5.5 5.50: 550 N 0
7.46 7.30x DG EG
F F FΣ = − − =
 
 
PROBLEM 6.41 CONTINUED 
 
5.04 5.040: 504 N 0
7.46 6.96y DG BG
F F FΣ = − − − = 
 
4.8 4.80: 0
7.30 6.96z EG BG
F F FΣ = + = 
 Solving: 696 NBGF = − 696 N C BGF = W 
 0DGF = 0 DGF = W 
 730 NEGF = 730 N T EGF = W 
 
 
 
PROBLEM 6.42 
A Warren bridge truss is loaded as shown. Determine the force in 
members CE, DE, and DF. 
 
 
 
 
SOLUTION 
 FBD Truss: 
 
 
 
 Section ABDC: 
 
 
 
( )( ) ( )( ) ( )0: 15 m 3 kN 20 m 3 kN 25 m 0K yM AΣ = + − =
 
4.2 kNy =A
 
 
 
 
 
( ) ( )( ) ( )( )0: 6 m 2.5 m 3 kN 7.5 m 4.2 kN 0D CEM FΣ = + − =
 
 
4 kNCEF =
 
4.00 kN T CEF = W
 
 
120: 4.2 kN 3 kN 0
13y DE
F FΣ = − − =
 
 
1.3 kNDEF =
 
1.300 kN T DEF = W
 
 
50: 0
13x DF DE CE
F F F FΣ = + + =
 
( ) ( )5 1.3 kN 4 kN 4.5 kN
13DF
F = − − = −
 
 
4.50 kN C DFF = W
 
 
 
 
PROBLEM 6.43 
A Warren bridge truss is loaded as shown. Determine the force in 
members EG, FG, and FH. 
 
 
 
 
SOLUTION 
 FBD Truss: 
 
 
 
 Section FBD: 
 
 
 
0: 0x xFΣ = =K
 
 
( ) ( )( ) ( )( )0: 25 m 10 m 3 kN 5 m 3 kN 0A yM KΣ = − − =
 
1.8 kNy =K
 
 
 
( )( ) ( )0: 10 m 1.8 kN 6 m 0G FHM FΣ = + =
 
 
3 kNFHF = −
 
3.00 kN C FHF = W
 
 
( )( ) ( )( )0: 12.5 m 1.8 kN 6 m 0F EGM FΣ = − =
 
 
3.75 kNEGF =
 
3.75 kN T EGF = W
 
 
120: 1.8 kN 0
13y FG
F FΣ = + =
 
 
1.95 kNFGF = −
 
1.950 kN C FGF = W
 
 
 
 
PROBLEM 6.44 
A parallel chord Howe truss is loaded as shown. Determine the force 
in members CE, DE, and DF. 
 
 
 
 
SOLUTION 
 FBD Truss: 
 
 
 
 FBD Section: 
 
 
 
 
0: 0x xFΣ = =A
 
 
( ) ( ) ( ) ( )
( ) ( ) ( )
0: 4 ft 1 0.3 kip 2 0.6 kip 3 1 kip
4 0.6 kip 5 0.3 kip 6 0.3 kip 6 0
M
y
M
A
Σ = + +
+ + + − =
 
 
1.7 kipsy =A
 
 
( )( ) ( ) 40: 4 ft 3 kips 1.7 kips 2 ft 0
4.1D CE
M F Σ = − + =   
 
2.87 kipsCEF =
 
2.87 kips T CEF = W
 
 
( )( ) ( )( )0: 8 ft 3 kips 1.7 kips 4 ft 0.3 kipEMΣ = − +
 
( ) 42 ft 0
4.1 DF
F − =   
 
5.125 kipsDFF = −
 
5.13 kips C DFF = W
 
 
( )4 40: 0
4.1 17.21x DF CE DE
F F F FΣ = + + =
 
( )17.21 5.125 2.87 kips 2.28 kips
4.1DE DE
F F= − − + =
 
 
2.28 kips T DEF = W
 
 
 
PROBLEM 6.45 
A parallel chord Howe truss is loaded as shown. Determine the force 
in members GI, GJ, and HJ. 
 
 
 
 
SOLUTION 
 FBD Truss: 
 
 
 
 
 
 
 
 
 
 
 
 
 
 FBD Section: 
 
 
 
 
 
 
 
 
 
 
 
( ) ( ) ( ) ( )24 ft 4 ft 0.3 kip 8 ft 0.6 kip 12 ft 1 kipA yM MΣ = − − −
 
( ) ( ) ( )16 ft 0.6 kip 20 ft 0.3 kip 24 ft 0.3 kip 0− − − =
 
1.7 kipsy =M
 
 
 
( ) ( ) 48 ft 1.7 0.3 kips 4 ft 0.3 kip 2 ft 0
4.1J GI
M F Σ = − − − =   
 
5.125 kips 5.3 kips T GI GIF F= = W
 
( ) ( )12 ft 1.7 0.3 kips 8 ft 0.3 kipGMΣ = − −
 
 
( ) 44 ft 0.6 kip 2 ft 0
4.1 HJ
F − + =   
 
6.15 kips 6.15 kips C HJF = − = W
 
 
( )4 46.15 5.125 kips
4.1 4.94x GJ
F FΣ = − −
 
 
1.235 kips T GJF = W
 
 
 
PROBLEM 6.46 
A floor truss is loaded as shown. Determine the force in members CF, 
EF, and EG. 
 
 
 
 
SOLUTION 
 FBD Truss: 
 
 FBD Section: 
 
 
 
 
( ) ( ) ( )
( ) ( ) ( )
0: 1 m 1 kN 2 1 kN 3 1.5 kN
4 2 kN 5 2 kN 6 1 kN 6 0
K
y
M
A
Σ = + +
+ + + − =
 
5.25 kNy =A
 
 
( ) ( ) ( ) ( )0: 1 m 1 2 kN 2 1 kN 5.25 kN 0.5 m 0E CEM F Σ = + − + =  
 
13.0 kNCFF =
 
13.00 kN T CFF = W
 
 
( ) ( ) ( ) ( )
( )
0: 1 m 1 2 kN 2 2 kN 3 1 kN 5.25 kN
0.5 m 0
F
EG
M
F
 Σ = + + − 
− = 
 
13.5 kNEGF = −
 
13.50 kN C EGF = W
 
 
10: 5.25 kN 1 kN 2 kN 2 kN 0
5y EF
F FΣ = − − − − =
 
 
5 0.5590 kN
4EF
F = =
 
559 N T EFF = W
 
 
 
 
PROBLEM 6.47 
A floor truss is loaded as shown. Determine the force in members FI, 
HI, and HJ. 
 
 
 SOLUTION 
FBD Truss: 
 
 
0: 0x xFΣ = =K
 
 
( ) ( ) ( ) ( ) ( ) ( ) ( )0: 1m 6 0.5 kN 5 1 kN 4 1 kN 3 1.5 kN 2 2 kN 1 2 kN 0A yM K Σ = − − − − − − =  
3.75 kNy =K
 
 FBD Section: 
 
 
( )( ) ( )0: 1 m 3.75 kN .5 kN .5 m 0I HJM FΣ = − − =
 
 
6.5 kNHJF =
 
6.50 kN C HJF = W
 
 
10: 1 kN 0.5 kN 3.75 kN 0
5y HI
F FΣ = − − + =
 
 
2.25 5 kNHIF = −
 
5.03 kN C HIF = W
 
 
20: 0
5y HI FI HJ
F F F FΣ = − − + =
 
 
( )2 2.25 kN 6.50 kNFIF = +
 
11.00 kN T FIF = W
 
 
 
 
PROBLEM 6.48 
A pitched flat roof truss is loaded as shown. Determine the force in 
members CE, DE, and DF. 
 
 
 
 
SOLUTION 
 FBD Truss: 
 
 
 FBD Section: 
 
 
 
 
0: 0x xFΣ = =A
 
By load symmetry: 8 kipsy y= =A I
 
 
 
 
( )( ) ( )( )0: 7 ft 2 kips 8 kips 3 ft 0D CEM FΣ = − + =
 
 
14 kipsCEF =
 
14.00 kips T CEF = W
 
 
( ) ( ) ( )0: 7 ft 1 4 kips 2 2 kips 8 kipsEM  Σ = + −  
( ) 74.5 ft 0
51.25 DF
F =
 
 
8 51.25 kips
4.5DF
F =
 
12.73 kips C DFF = W
 
 
1.5 8 51.25 30: 8 kips 2 kips 4 kips kips 0
4.551.25 58y DE
F FΣ = − − + − =
 
 
1.692 kipsDEF = −
 
1.692 kips C DEF = W
 
 
 
PROBLEM 6.49 
A pitched flat roof truss is loaded as shown. Determine the force in 
members EG, GH, and HJ. 
 
 
SOLUTION 
 FBD Truss: 
 
By load symmetry: 8 kipsy y= =A I
 
 FBD Section: 
 
 
( )( ) ( )0: 7 ft 8 kips 2 kips 6 ft 0EGM FΣ = − − =
 
 
7.00 kips T EGF = W 
 
70: 7.00 kips 0
51.25x HJ
F FΣ = − =
 
 
51.25 kipsHJF =
 
7.16 kips C HJF = W
 
 
( ) ( )1.50: 51.25 kips 8 2 kips 051.25y HGF FΣ = − + − = 
 
7.50 kips C HGF = W
 
 
 
 
PROBLEM 6.50 
A Howe scissors roof truss is loaded as shown. Determine the force in 
members DF, DG, and EG. 
 
 
 
SOLUTION 
FBD Truss: 
 
0: 0X xFΣ = =A
 
By symmetry: 6 kNy y= =A L
 
FBD Section: 
 
Notes:
 
15
 m
6
2 5 5
m
3 2 3
2 21 m
3 3
5
 m
6
1 m
2
 m
3
F
DE
F D
G
D G
y
y
y
y y
y
y y
=
= ⋅ =
= ⋅ =
− =
=
− =
 
 
PROBLEM 6.50 CONTINUED 
 
( ) ( )( ) ( )( )60: 1 m 2 m 2 kN 4 m 1 kN 6 kN 0
37D EG
M FΣ = + + − =
 
 
8 37 kN
3EG
F =
 
16.22 kN T EGF = W
 
 
( )( ) ( )( ) ( ) ( )1 30: 2 m 2 kN 4 m 2 kN 6 m 1 m 0
10 10A DG DG
M F F   Σ = + − − =       
 
4 10 kN
3DG
F =
 
4.22 kN C DGF = W
 
 
6 3 12 120: 0 16 4 0
13 1337 10x EG DG DF DF
F F F F FΣ = − − = − − =
 
 13 kNDFF = 13.00 kN C DFF = W 
 
 
PROBLEM 6.51 
A Howe scissors roof truss is loaded as shown. Determine the force in 
members GI, HI, and HJ. 
 
 
 
 SOLUTION 
FBD Truss: 
 
 
0: 0x xFΣ = =A
 
By symmetry: 6 kNy y= =A L
 
FBD Section: 
 
Notes:
 
2
 m
3
2 5 5
m
3 2 3
I
H
y
y
=
= ⋅ =
 
 
so 
 
1 mH Iy y− =
 
 
60 
PROBLEM 6.51 CONTINUED 
 
( )( ) ( )( ) ( ) 120: 4 m 6 kN 1 kN 2 m 2 kN 1 m 0
13I HJ
M F Σ = − − − =   
 
52 kN
3HJ
F =
 
17.33 kN C HJF = W
 
 
( )( ) ( )( ) ( ) 60: 4 m 6 kN 1 kN 2 m 2 kN 1 m 0
37H GI
M F Σ = − − − =   
 
8 37 kN
3GI
F =
 
16.22 kN T GIF = W
 
 
( )( ) ( )0: 2 m 2 kN 4 m 0L HIM FΣ = − =
 
1.000 kN T HIF = W
 
 
 
PROBLEM 6.52 
A Fink roof truss is loaded as shown. Determine the force in members 
BD, CD, and CE. 
 
 
SOLUTION 
FBD Truss: 
 
 
 
0: 0x xFΣ = =A 
 
By symmetry: 3.6 kipsy y= =A K
 
FBD Section: 
 
 
 
( )( ) ( )( ) ( )( )160: ft 5 ft 1.2 kips 10 ft 0.6 kips 10 ft 3.6 kips 0
3D CE
M F Σ = + + − =   
 
4.50 kips T CEF = W
 
 
( ) ( )( )40: 6 ft 5 ft 1.2 kips 0
5A CD
M F Σ = − =   
 
1.250 kips T CDF = W
 
 
( ) ( )4 80: 3.6 0.6 kips 1.2 kips 1.25 kips 0
5 17y BD
F FΣ = − − + − =
 
 
5.95 kipsBDF =
 
5.95 kips C BDF = W
 
62 
 
PROBLEM 6.53 
A Fink roof truss is loaded as shown. Determine the force in members 
FH, FG, and EG. 
 
 
SOLUTION 
 FBD Truss: 
 
 
0: 0x xFΣ = =A 
 
By symmetry: 3.6 kipsy y= =A K 
FBD Section: 
 
 
( )( ) ( )( ) ( ) ( ) ( )0: 15 ft 3.6 .6 kips 10 ft 1.2 kips 5 ft 1.2 kips 8ft 0F EGM FΣ = − − − − =
 
 
3.375 kipsEGF =
 
3.38 kips T EGF = W
 
 
( )( ) ( )( ) ( ) 80: 5 ft 1.2 kips 10 ft 1.2 kips 12 ft 0
73K FG
M F Σ = + − =   
 
3 73 kips
16FG
F =
 
1.602 kips T FGF = W
 
 
8 3 80 73 kips 1.2 kips 1.2 kips 0.6 kip 3.6 kips 0
16 1773y FH
F F Σ = − − − − + =   
 
4.4625 kipsFHF =
 
4.46 kips C FHF = W
 
 
 
PROBLEM 6.54 
A Fink roof truss is loaded as shown. Determine the force in members 
DF, DG, and EG. (Hint: First determine the force in member EK.) 
 
SOLUTION 
 
FBD Truss: 
 
FBD Sections: 
 Forces in 1b 
 
 
 
 
 
 
 
 
 
 
 0: 0x xFΣ = =A 
By symmetry: 2000 lby y= =A O 
 
 
 
 
 
 
( )( ) ( )0: 2 3 500 lb 4 250 lbHM a a a aΣ = + + + 
 ( )4 2000 lb 2 0EKa aF− + = 
2000 lb TEKF =
 
 
( ) ( ) ( ) ( )0: 500 lb 2 250 lb 2 2000 lb 2000 lbDM a a a aΣ = + − + 
 
( )3 412 ft 2 0
5 5EG EG
a F a F   + + − =       
 1000 lbEGF = 1.000 kip T EGF = W 
 
( ) ( )4 10: 1000 lb 2000 250 500 500 lb 0
5 5y DF
F FΣ = − + − − − =
 
 1550 5 lbDFF = 3.47 kips C DFF = W 
 ( ) ( )3 20: 2000 lb 1000 lb 1550 5 lb 05 5x DGF FΣ = + − + = 
 500 lb T DGF = W 
 
PROBLEM 6.55 
A roof truss is loaded as shown. Determine the force in members FH, 
GJ, and GI. 
 
 
SOLUTION 
 
By symmetry: 2.45 kipsy y= =A M 
 
 
0: 0x xFΣ = =A 
 
JMΣ ( ) ( ) 46 ft 0.3 kip 12 ft 2.35 kips 7 ft 017 GIF
 = − + =   
 3.887 kipsGIF = 3.89 kips T GIF = W 
GMΣ ( ) ( )( ) ( )( ) ( )( )6 ft 4 ft 1.0 kips 10 ft 0.3 kips 16 ft 0.1 kipsHJF= − − − 
 ( )( ) 16 ft 2.45 kips 0+ = 5.10 5.10 kips CHJF = − = 
By inspection: 5.10 kips CFH HJF F= = W 
 
( )1 30: 3.887 kips 2.45 kips 1.4 kips 0
17 13y GJ
F FΣ = + − − = 
 2.40 kips T GJF = W 
 
 
 
PROBLEM 6.56 
A stadium roof truss is loaded as shown. Determine the force in 
members AB, AG, and FG. 
 
 
 
 
 
 
 
SOLUTION 
FBD Section: 
 
 
 
Note: p ( )8.4 2.7 m 1.89 m
12.0
BG = =
 
 
( ) ( )( ) ( )( ) ( )( )0: 2.7 m 3.6 m 8 kN 7.8 m 8 kN 12 m 4 kN 0A FGM FΣ = − − − =
 
 51.56 kNFGF = 51.6 kN C FGF = W 
 
( ) ( )( ) ( )( )120: 1.89 m 4.2 m 8 kN 8.4 m 4 kN 0
12.3G AB
M F Σ = − − =   
 36.44 kNABF = 36.4 kN T ABF = W 
 
( )( ) ( )( ) ( ) 30: 4.2 m 8 kN 8.4 m 8 kN 8.4 m 0
5D AG
M F Σ = + − =   20.0 kN T AGF = W 
 
 
 
 
 
PROBLEM 6.57 
A stadium roof truss is loaded as shown. Determine the force in members 
AE, EF, and FJ. 
 
 
 
 
SOLUTION 
FBD Section: 
 
 
( ) ( )( ) ( )( ) ( )( )80: 2.7 m 3.6 m 8 kN 7.8 m 8 kN 12 m 4 kN 0
145F AE
M F Σ = − − − =   
 
17.4 145 kN
2.7AE
F = 77.6 kN T AEF = W 
 
8 17.40: 145 kN 0
2.7145x EF
F F  Σ = − =   
 51.555 kNEFF = 51.6 kN C EFF = W 
 
( )9 17.40: 145 kN 4 8 8 4 kN 0
2.7145y FJ
F F  Σ = − − + + + =   82.0 kN C FJF = W 
 
 
 
 
 
 
PROBLEM 6.58 
A vaulted roof truss is loaded as shown. Determine the force in members 
BE, CE, and DF. 
 
 
 
SOLUTION 
FBD Truss: 
 
 
( )( ) ( )( ) ( )( ) ( )( )0: 1.2 m 0.2 kN 1.8 m 0.5 kN 2.88 m 0.5 kN 4.2 m 1.3 kNNMΣ = + + +
 
 
( )( ) ( )( ) ( )( ) ( )4.8 m 0.6 kN 6.6 m 0.7 kN 8.4 m 0.1 kN 8.4 m 0yA+ + + − =
 
1.95 kNy =A
 
 
0: 0x xFΣ = =A 
FBD Section: 
 
 
( ) ( )( )0: 0.9 m 1.8 m 1.95 kN 0.1 kN 0B CEM FΣ = − − =
 
3.70 kN T CEF = W
 
 
PROBLEM 6.58 CONTINUED
 
 
( ) ( )1 20: 1.8 m .7 kN .9 m 0
5 5A BE BE
M F F
    Σ = − + =         
 0.35 5 kNBEF = 783 N C BEF = W 
 ( )2 20: 3.70 kN 0.35 5 kN 05 5x BDF FΣ = − − = 
 1.5 5 kN 3.35 kN CBDF = = 
Then by inspection of joint D: DF BDF F= so 3.35 kN C DFF = W 
 
 
 
PROBLEM 6.59 
A vaulted roof truss is loaded as shown. Determine the force in 
members HJ, IJ, and GI. 
 
 
 
SOLUTION 
FBD Truss: 
 
 
( )( ) ( )( ) ( )( ) ( )( )0: 1.8 m 0.7 kN 3.6 m 0.6 kN 4.2 m 1.3 kN 5.52 m 0.5 kNAMΣ = + + +
 
 
( )( ) ( )( ) ( )( ) ( )( )6.6 m 0.5 kN 7.2 m 0.2 kN 8.4 m 0.1 kN 8.4 m 0yN+ + + − =
 
2.05 kNy =N 
 
 
FBD Section: 
 
 
( )( ) ( )( ) ( )0: 1.8 m 2.05 0.1 kN 0.6 m 0.2 kN 0.45 m 0J IKM FΣ = − − + =
 
 7.533 kNIKF = 7.53 kN TIKF = 
 
 
PROBLEM 6.59 CONTINUED
 
 
 
( )( ) ( )( ) ( ) ( )2 10: 2.88 m 2.05 0.1 kN 1.68 m 0.2 kN 0.45 m 1.08 m 0
5 5I HJ HJ
M F F   Σ = − − − − =       
 2.3939 5 kNHJF = 5.35 kN C HJF = W 
 
( )1 50: 2.3939 5 kN 2.05 kN 0.5 kN 0.2 kN 0.1 kN 0135y IJF FΣ = − + + − − − = 
 2.974 kNIJF = 2.97 kN CIJF = W 
FBD Joint: 
 
( )120: 2.974 kN 7.533 kN 0
13x GI
F FΣ = − − + = 
 4.788 kNGIF = 4.79 kN T GIF = W 
 
 
 
PROBLEM 6.60 
Determine the force in members AF and EJ of the truss shown when2 kips.P Q= = (Hint: Use section aa.) 
 
 
 
 
SOLUTION 
FBD Truss: 
 
 
( ) ( )( ) ( )( )0: 24 ft 18 ft 2 kips 36 ft 2 kips 0 A xM KΣ = − − = 
4.5 kipsx =K 
FBD Section: 
 
 
( )( ) ( )( ) ( )( ) ( )0: 12 ft 4.5 kips 18 ft 2 kips 36 ft 2 kips 36 ft 0F EJM FΣ = − − + =
 
 1.500 kips T EJF = W 
 
( )( ) ( )( ) ( )0: 18 ft 2 kips 12 ft 4.5 kips 36 ft 0J AFM FΣ = + − = 
 2.50 kips T AFF = W 
 
 
PROBLEM 6.61 
Determine the force in members AF and EJ of the truss shown when 
2 kipsP = and 0.Q = (Hint: Use section aa.) 
 
 
 
 
 
 
SOLUTION 
FBD Truss: 
 
 
( ) ( )( )0: 24 ft 18 ft 2 kips 0 A xM KΣ = − = 
1.5 kipsx =K 
FBD Section: 
 
 
( )( ) ( )( ) ( )0: 12 ft 1.5 kips 18 ft 2 kips 36 ft 0F EJM FΣ = − + = 0.500 kip T EJF = W 
 
( )( ) ( )( ) ( )0: 18 ft 2 kips 12 ft 1.5 kips 36 ft 0J AFM FΣ = + − = 1.500 kips T AFF = W 
 
 
PROBLEM 6.62 
Determine the force in members DG and FH of the truss shown. 
(Hint: Use section aa.) 
 
 
SOLUTION 
FBD Truss: 
 
 
0: 0x xFΣ = =A 
By symmetry: 3.2 kNy y= =A N 
FBD Section: 
 
 
( ) ( )( ) ( )( )0: 2.4 m 3.2 m 3.2 kN 1.6 m 1 kN 0F DGM FΣ = − + =
 
 3.60 kN C DGF = W 
 
( ) ( )( ) ( )( )0: 2.4 m 1.6 m 1 kN 3.2 m 3.2 kN 0D FHM FΣ = + − = 
 3.60 kN T FHF = W 
 
 
 
PROBLEM 6.63 
Determine the force in members IL, GJ, and HK of the truss shown. 
(Hint: Begin with pins I and J and then use section bb.) 
 
 
SOLUTION 
 
 
FBD Truss: 
 
0: 0x xFΣ = =A 
 
 
By symmetry: 0; 3.2 kNx y yA A N= = = 
 
 
 
 
Joints: By inspection of joint I: GI ILF F= and 1.2 kNIJF = C
 
 By inspection of joint J: ( ) ( )C TGJ HJF F= 
 
 
 
PROBLEM 6.63 CONTINUED 
 
 
 FBD Section: 
 
040; 0 so 0
5
HJx IL HK GJ IL HKF F F F F F F
 Σ = − + − = − =   
 
( )( ) ( )( ) ( )( )0: 3.2 m 3.2 kN 1.2 m 1.6 m 1 kN 0 so 7.2 kNJ IL HK IL HKM F F F FΣ = − + − = + =
 
Solving: 3.6 kNIL HKF F= = 3.60 kN C ILF = W 
 3.60 kN T HKF = W 
 
( )30: 3.2 kN 1 kN 1.2 kN 0
5y GJ HJ
F F FΣ = − − − + = 
 
5
 kN
3GJ HJ
F F+ = but 5 kN
6GJ HJ
F F= = .833 kN C GJF = W 
 
 
 
 
 
PROBLEM 6.64 
The diagonal members in the center panels of the truss shown are 
very slender and can act only in tension; such members are known as 
counters. Determine the forces in the counters which are acting under 
the given loading. 
 
 
 
SOLUTION 
 
FBD Truss: 
 
FBD Section: 
 
 
 
 
0: 0x xFΣ = =F 
 
( ) ( )( ) ( )( )0: 5.5 m 2.75 m 24 kN 2.75 m 24 kNF yM HΣ = + − 
 ( )( ) ( )( )5.5 m 12 kN 8.25 m 12 kN 0− − = 
30 kNy =H 
 
 
( ) ( )0: 30 kN 3 24 kN 2 12 kN 0y yF FΣ = + − − = 
 
66 kNy =F 
Assume there is no pretension in any counter so that, as the truss is 
loaded, one of each pair becomes taut while the other becomes slack. 
 
Here tension is required in BG to provide downward force, so CF is 
slack. 
 
2.40: 66 kN 24 kN 24 kN 0
3.65y BG
F FΣ = − − − =
 
 27.375 kNBGF = 27.4 kN T BGF = W
 0 CFF = W 
 
Here tension is required in GD to provide downward force, so CH is 
slack. 
 0 CHF = W 
 
( ) 2.40: 30 kN 2 12 kN 0
3.65y GD
F FΣ = − − =
 
 
9.125 kNGDF =
 
9.13 kN T GDF = W
 
PROBLEM 6.65 
The diagonal members in the center panels of the truss shown are 
very slender and can act only in tension; such members are known as 
counters. Determine the forces in the counters which are acting under 
the given loading. 
 
SOLUTION 
 
 
 
 
FBD Section: 
FBD Truss: 
 
FBD Section: 
 
 
Assume that there is no pretension in any counter. So that, as the truss is 
loaded, one of each crossing pair becomes taut while the other becomes 
slack. 
 
 
Note: Tension is required in CH to provide upward force; thus GD is 
slack 
 
 
2.40: 24 kN 0
3.65y CH
F FΣ = − =
 
36.5 kN T CHF = W
 
 
 
 
 
( )( ) ( )( ) ( )( )0: 5.5 m 24 kN 2.75 m 24 kN 2.75 mG yM FΣ = + −
 
 
( )( ) ( )( )2.75 m 12 kN 5.5 m 12 kN 0− − =
 
36 kNy =F
 
0: 0x xFΣ = =F
 
 
 
 
Note: Tension is required in FC to provide upward force; so BG is 
slack. 
 
( ) 2.40: 36 24 24 kN 0
3.65y FC
F FΣ = − − + =
 
 18.25 kN T FCF = W 
 
 
 
PROBLEM 6.66 
The diagonal members in the center panel of the truss shown are very 
slender and can act only in tension; such members are known as 
counters. Determine the force in members BD and CE and in the 
counter which is acting when P = 3 kips. 
 
 
SOLUTION 
 
FBD Truss: 
 
 
FBD Section: 
 
 
 
 
 
 
 
 
0: 0x xFΣ = =A 
 
( ) ( )( ) ( )( )0: 25 ft 17 ft 2 kips 8 ft 3 kips 0A yM FΣ = − − =
 
2.32 kipsy =F
 
Assume there is no pretension in either counter so that, as the truss is 
loaded, one becomes taut while the other becomes slack. 
Here tension is required in CD to provide downward force, so 
 0 BEF = 
 
5.60: 2.32 kips 2 kips 0
10.6y CD
F FΣ = − − =
 
 
0.6057 kipCDF =
 
606 lb T CDF = W
 
 
( )( )0: 8 ft 2.32 kips (5.6 ft) 0D CEM FΣ = − =
 
 
3.314 kips CEF =
 
3.31 kips T CEF = W
 
 
( )90: 0.6057 kip 3.314 kips 0
10.6x BD
F FΣ = − − = 
 3.829 kips BDF = 3.83 kips C BDF = W 
 
 
 
PROBLEM 6.67 
Solve Prob. 6.66 when P = 1.5 kips. 
 
SOLUTION 
 
FBD Truss: 
 
FBD Section: 
 
 
 
 
 
 
 
 
 
( ) ( )( ) ( )( )0: 25 ft 17 ft 2 kips 8 ft 1.5 kips 0A yM FΣ = − − =
 
 
1.84 kipsy =F
 
 
Here tension is needed in BE to provide upward force, so 0CDF = 
 
5.60: 1.84 kips 2 kips 0
10.6y BE
F FΣ = + − =
 
0.03029 kipBEF =
 
 
303 lb T BEF = W
 
 
( )( ) ( )0: 8 ft 1.84 kips 5.6 ft 0E BDM FΣ = − =
 
2.629 kipsBDF = 
 2.63 kips C BDF = W 
 
( )90: 2.629 kips 0.3029 kip 0
10.6x CE
F FΣ = − − = 
2.371 kips CEF = 
 2.37 kips T CEF = W 
 
 
 
PROBLEM 6.68 
The diagonal members CF and DE of the truss shown are very slender 
and can act only in tension; such members are known as counters. 
Determine the force in members CE and DF and in the counter which 
is acting. 
 
 
SOLUTION 
FBD Section: 
 
DE must be in tension to provide leftward force, so CF is slack. 
 
40: 4 kips 0
5x DE
F FΣ = − =
 
 
5.00 kips T DEF = W
 
 
( ) ( )( )0: 8 ft 6 ft 4 kips 0D CEM FΣ = − =
 
 
3.00 kips T CEF = W
 
 
( ) ( )( )0: 8 ft 12 ft 4 kips 0E DFM FΣ = − = 
 6.00 kips C DFF = W 
 
 
 
 
 
PROBLEM 6.69 
The diagonal members EH and FG of the truss shown are very slender 
and can act only in tension; such members are known as counters. 
Determine the force in members EG and FH and in the counter which 
is acting. 
 
 
 
 
SOLUTION 
 
FBD Section: 
 
 
 
 
 
 
 
 
It is not obvious which counter is active, so assume FG is (and thus 
EH is slack) 
 
( ) ( )( )60: 8 ft 12 ft 4 kips 0
6.824F GE
M F Σ = − =   
 
6.824 kips TGEF =
 
 
( ) ( )( )60: 14.5 ft 18 ft 4 kips 0
6.824G FH
M F Σ = − =   
 5.647 kips CFHF = 
 
( )6 60: 5.647 kips 6.824 kips 0
6.824 12.75y FG
F F Σ = − − =   
This gives 0FGF < which is impossible, so the assumption is wrong, 
FG is slack, and EH is in tension. 
Then
 
 
( ) ( )( )60: 8 ft12 ft 4 kips 0
6.824E FH
M F Σ = − =   
6.824 kipsFHF =
 
 
6.83 kips C FHF = W
 
 
( ) ( )( )60: 14.5 ft 18 ft 4 kips 0
6.824H GE
M F Σ = − =   
5.647 kipsGEF = 
 5.65 kips T GEF = W 
 
 
( )6 60: 6.824 kips 5.647 kips 0
6.824 12.75y EH
F FΣ = − − = 
2.198 kipsEHF = 
 2.20 kips T EHF = W 
 
PROBLEM 6.70 
Classify each of the structures shown as completely, partially, or 
improperly constrained; if completely constrained, further classify it as 
determinate or indeterminate. (All members can act both in tension 
and in compression.) 
 
SOLUTION 
 
Structure (a): 
 
Structure (b): 
 
Structure (c): 
Also 
consider 
 
 
 
Simple truss with 4,r = 16,m = 10n = 
So 20 2m r n+ = = so completely constrained and determinate W 
Compound truss with 3,r = 16,m = 10n = 
So 19 2 20m r n+ = < = so partially constrained W 
Non-simple truss with 4,r = 12,m = 8n = 
So 16 2m r n+ = = but must examine further, note that reaction 
forces 
xA and xH are aligned, so no equilibrium equation will resolve them. 
 ∴ Statically indeterminate W
For
 
0: 0,y yF HΣ = = but then 
 
0AMΣ ≠ in FBD Truss, 
 ∴ Improperly constrained W 
 
 
PROBLEM 6.71 
Classify each of the structures shown as completely, partially, or 
improperly constrained; if completely constrained, further classify it as 
determinate or indeterminate. (All members can act both in tension and 
in compression.) 
 
 
SOLUTION 
Structure (a): 
 
FBD Sections: 
 
 
 
 
 
 
 
 
Structure (b): 
 
Structure (c): 
 
Non-simple truss with 4,r = 16,m = 10n = 
so 20 2 ,m r n+ = = but must examine further. 
FBD I: 0AMΣ = ⇒ 1T 
 II: 0xFΣ = ⇒ 2T 
 I: 0xFΣ = ⇒ xA 
 I: 0yFΣ = ⇒ yA 
 II: 0EMΣ = ⇒ yC 
 II: 0yFΣ = ⇒ yE 
Since each section is a simple truss with reactions determined, 
 structure is completely constrained and determinate. W 
Non-simple truss with 3,r = 16,m = 10n = 
 
so 19 2 20m r n+ = < = ∴ structure is partially constrainedW 
 
 
Simple truss with 3,r = 17,m = 10n = 
20 2 ,m r n+ = = but the horizontal reaction forces and x xA E are 
collinear and no equilibrium equation will resolve them, so the 
structure is improperly constrained and indeterminate W 
 
 
 
PROBLEM 6.72 
Classify each of the structures shown as completely, partially, or 
improperly constrained; if completely constrained, further classify it as 
determinate or indeterminate. (All members can act both in tension and 
in compression.) 
 
 
SOLUTION 
 
Structure (a): 
 
Structure (b): 
 
Structure (c): 
 
 
 
 
Non-simple truss with 4,r = 12,m = 8n = so 16 2 ,r m n+ = = check 
for determinacy: 
One can solve joint F for forces in EF, FG and then solve joint E for 
yE and force in DE. 
 
 
 
This leaves a simple truss ABCDGH with 
3,r = 9,m = 6n = so 12 2r m n+ = = 
 Structure is completely constrained and determinate W 
 
 
Simple truss (start with ABC and add joints alphabetically to complete 
truss) with 4,r = 13,m = 8n = 
so 17 2 16r m n+ = > = Constrained but indeterminate W 
 
Non-simple truss with 3,r = 13,m = 8n = so 16 2r m n+ = = . To 
further examine, follow procedure in part (a) above to get truss at left. 
Since 1 0≠F (from solution of joint F), 
1AM aFΣ =
 
0≠ and there is no equilibrium. 
 Structure is improperly constrained W
 
 
 
PROBLEM 6.73 
Classify each of the structures shown as completely, partially, or 
improperly constrained; if completely constrained, further classify it as 
determinate or indeterminate. (All members can act both in tension and 
in compression.) 
 
 
SOLUTION 
Structure (a): 
 
 
Structure (b): 
 
 
 
 
 
 
Structure (c): 
 
 
Simple truss (start with ABC and add joints alphabetical to complete 
truss), with 
4,r = 13,m = 8n = so 17 2 16r m n+ = > = 
 Structure is completely constrained but indeterminate. W 
 
 
 
From FBD II: 0GMΣ = ⇒ yJ 
 0xFΣ = ⇒ xF 
 FBD I: 0AMΣ = ⇒ yF 
 0yFΣ = ⇒ yA 
 0xFΣ = ⇒ xA 
 FBD II: 0yFΣ = ⇒ yG 
Thus have two simple trusses with all reactions known, 
 so structure is completely constrained and determinate. W 
 
Structure has 4,r = 13,m = 9n = 
so 17 2 18,r m n+ = < = structure is partially constrained W 
 
 
 
PROBLEM 6.74 
Classify each of the structures shown as completely, partially, or 
improperly constrained; if completely constrained, further classify it as 
determinate or indeterminate. (All members can act both in tension and 
in compression.) 
 
 
SOLUTION 
Structure (a): 
 
 
 
 
Structure (b): 
 
Structure (c): 
 
 
Rigid truss with 3,r = 14,m = 8n = 
so 17 2 16r m n+ = > = 
 so completely constrained but indeterminate W
 
 
Simple truss (start with ABC and add joints alphabetically), with 
3,r = 13,m = 8n = so 16 2r m n+ = = 
 so completely constrained and determinate W 
 
Simple truss with 3,r = 13,m = 8n = so 16 2 ,r m n+ = = but 
horizontal reactions ( and )x xA D are collinear so cannot be resolved 
by any equilibrium equation. 
 ∴ structure is improperly constrainedW 
 
 
 
PROBLEM 6.75 
Classify each of the structures shown as completely, partially, or 
improperly constrained; if completely constrained, further classify it as 
determinate or indeterminate. (All members can act both in tension and 
in compression.) 
 
 
SOLUTION 
Structure (a): 
 
 
FBD of EH: 
 
 
Structure (b): 
 
 
 
 
 
 
 
 
No. of members 12m = 
No. of joints 8n = 16 2m r n+ = = 
No. of react. comps. 4r = unks eqns= 
 
 
 
 
 
0 ;H DEM FΣ = → 0 ;x GHF FΣ = → 0 y yF HΣ = → 
 
 
 
Then ABCDGF is a simple truss and all forces can be determined. 
 This example is completely constrained and determinate. W 
 
No. of members 12m = 
No. of joints 8n = 15 2 16m r n+ = < = 
No. of react. comps. 3r = unks eqns< 
 partially constrained W 
Note: Quadrilateral DEHG can collapse with joint D moving downward: 
in (a) the roller at F prevents this action. 
 
 
 
 
 
 
 
 
Structure (c): 
 
PROBLEM 6.75 CONTINUED 
 
No. of members 13m = 
No. of joints 8n = 17 2 16m r n+ = > = 
No. of react. comps. 4r = unks eqns> 
 completely constrained but indeterminate W 
 
 
 
PROBLEM 6.76 
For the frame and loading shown, determine the force acting on 
member ABC (a) at B, (b) at C. 
 
 
 
 
 
SOLUTION 
FBD member ABC: 
 
 
Note: BD is a two-force member so BDF is through D. 
 
( )a
 
( ) ( )( )40: 8 in. 12 in. 60 lb 0
5C BD
M F Σ = − =   
 112.5 lbBD =F 36.9° W
 
( )b
 
 
( )40: 60 lb 112.5 lb 0
5x x
F CΣ = + − =
 
30 lbxC = 
 
( )30: 112.5 lb 0
5y y
F CΣ = − = 
67.5 lbyC =
 
 so 73.9 lb=C 66.0 °W 
 
 
 
PROBLEM 6.77 
Determine the force in member AC and the reaction at B when (a) 
o30 ,=θ (b) o60 .=θ 
 
 
 
 
SOLUTION 
FBD member BCD: 
 
 
 
 
 
 
 
 
 
 
 
 
 
Note: AC is two-force member so ACF is through A. 
p ( )8 in. cosBC θ= 
( ) ( ) ( )( )0: 8 in. cos sin 10 in. 20 lbB ACM Fθ θΣ = −
 
 
( )( )14 in. 20 lb 0− =
 
60 lb
sin cosAC
F θ θ= 
 
60 lb0: cos 0 
sinx x AC x
F B F Bθ θΣ = − = = 
 
0: sin 20 lb 20 lb 0y y ACF B F θΣ = + − − = 
60 lb40
cos
yB lb θ= −