PROBLEM 6.46
A floor truss is loaded as shown. Determine the force in members CF,
EF, and EG.
SOLUTION
FBD Truss:
ΣM K = 0: (1 m ) 1 kN + 2 (1 kN ) + 3 (1.5 kN )
+ 4 ( 2 kN ) + 5 ( 2 kN ) + 6 (1 kN ) − 6 Ay  = 0
A y = 5.25 kN
ΣM E = 0: (1 m ) 1( 2 kN ) + 2 (1 kN − 5.25 kN )  + ( 0.5 m ) FCE = 0
FCF = 13.0 kN
FBD Section:
FCF = 13.00 kN T W
ΣM F = 0: (1 m ) 1( 2 kN ) + 2 ( 2 kN ) + 3 (1 kN − 5.25 kN ) 
− ( 0.5 m ) FEG = 0
FEG = −13.5 kN
FEG = 13.50 kN C W
ΣFy = 0: 5.25 kN − 1 kN − 2 kN − 2 kN −
FEF =
5
= 0.5590 kN
4
1
FEF = 0
5
FEF = 559 N T W
PROBLEM 6.47
A floor truss is loaded as shown. Determine the force in members FI,
HI, and HJ.
SOLUTION
FBD Truss:
ΣFx = 0: K x = 0
(
)
ΣM A = 0: (1m ) 6 K y − 0.5 kN − 5 (1 kN ) − 4 (1 kN ) − 3 (1.5 kN ) − 2 ( 2 kN ) − 1( 2 kN )  = 0


K y = 3.75 kN
FBD Section:
ΣM I = 0:
(1 m )( 3.75 kN − .5 kN ) − (.5 m ) FHJ
=0
FHJ = 6.5 kN
ΣFy = 0:
FHJ = 6.50 kN C W
1
FHI − 1 kN − 0.5 kN + 3.75 kN = 0
5
FHI = −2.25 5 kN
ΣFy = 0: −
FHI = 5.03 kN C W
2
FHI − FFI + FHJ = 0
5
FFI = 2 ( 2.25 kN ) + 6.50 kN
FFI = 11.00 kN T 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:
ΣFx = 0: A x = 0
By load symmetry: A y = I y = 8 kips
FBD Section:
ΣM D = 0: ( 7 ft )( 2 kips − 8 kips ) + ( 3 ft ) ( FCE ) = 0
FCE = 14 kips
FCE = 14.00 kips T W
ΣM E = 0: ( 7 ft ) 1( 4 kips ) + 2 ( 2 kips − 8 kips ) 
( 4.5 ft )
FDF =
7
FDF = 0
51.25
8 51.25
kips
4.5
ΣFy = 0: 8 kips − 2 kips − 4 kips +
FDE = −1.692 kips
FDF = 12.73 kips C W
1.5 8 51.25
kips −
51.25 4.5
3
FDE = 0
58
FDE = 1.692 kips C 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: A y = I y = 8 kips
FBD Section:
ΣM = 0: ( 7 ft )( 8 kips − 2 kips ) − ( 6 ft ) FEG = 0
FEG = 7.00 kips T W
ΣFx = 0:
FHJ =
ΣFy = 0:
7
FHJ − 7.00 kips = 0
51.25
FHJ = 7.16 kips C W
51.25 kips
1.5
51.25
(
)
51.25 kips − FHG + ( 8 − 2 ) kips = 0
FHG = 7.50 kips C 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:
Σ FX = 0: A x = 0
By symmetry: A y = L y = 6 kN
FBD Section:
Notes:
15
m
6
2 5 5
yD = ⋅ = m
3 2 3
2
2
yE = ⋅ 1 = m
3
3
5
yF − yD = m
6
yG = 1 m
yF =
yD − yG =
2
m
3
PROBLEM 6.50 CONTINUED
ΣM D = 0: (1 m )
6
FEG + ( 2 m )( 2 kN ) + ( 4 m )(1 kN − 6 kN ) = 0
37
FEG =
8
37 kN
3
FEG = 16.22 kN T W
 1

 3

ΣM A = 0: ( 2 m )( 2 kN ) + ( 4 m )( 2 kN ) − ( 6 m ) 
FDG  − (1 m ) 
FDG  = 0
 10

 10

FDG =
ΣFx = 0:
4
10 kN
3
6
3
12
FEG −
FDG −
FDF = 0
13
37
10
FDF = 13 kN
FDG = 4.22 kN C W
16 − 4 −
12
FDF = 0
13
FDF = 13.00 kN C 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:
Σ Fx = 0: A x = 0
By symmetry: A y = L y = 6 kN
FBD Section:
Notes:
so
2
3
2
=
3
yI =
m
yH
⋅
5 5
= m
2 3
yH − yI = 1 m
PROBLEM 6.51 CONTINUED
 12

ΣM I = 0: ( 4 m )( 6 kN − 1 kN ) − ( 2 m )( 2 kN ) − (1 m )  FHJ  = 0
13


FHJ =
52
kN
3
FHJ = 17.33 kN C W
 6

ΣM H = 0: ( 4 m )( 6 kN − 1 kN ) − ( 2 m )( 2 kN ) − (1 m ) 
FGI  = 0
 37

FGI =
8
37 kN
3
ΣM L = 0: ( 2 m )( 2 kN ) − ( 4 m ) FHI = 0
60
FGI = 16.22 kN T W
FHI = 1.000 kN T W
PROBLEM 6.52
A Fink roof truss is loaded as shown. Determine the force in members
BD, CD, and CE.
SOLUTION
FBD Truss:
Σ Fx = 0: A x = 0
By symmetry: A y = K y = 3.6 kips
FBD Section:
 16 
ft  FCE + ( 5 ft )(1.2 kips ) + (10 ft )( 0.6 kips ) − (10 ft )( 3.6 kips ) = 0
ΣM D = 0: 
 3 
FCE = 4.50 kips T W
4

ΣM A = 0: ( 6 ft )  FCD  − ( 5 ft )(1.2 kips ) = 0
5

FCD = 1.250 kips T W
ΣFy = 0: ( 3.6 − 0.6 ) kips − 1.2 kips +
FBD = 5.95 kips
4
8
(1.25 kips ) − FBD = 0
5
17
FBD = 5.95 kips C W
PROBLEM 6.53
A Fink roof truss is loaded as shown. Determine the force in members
FH, FG, and EG.
SOLUTION
FBD Truss:
Σ Fx = 0: A x = 0
By symmetry: A y = K y = 3.6 kips
FBD Section:
ΣM F = 0: (15 ft )( 3.6 − .6 ) kips − (10 ft )(1.2 kips ) − ( 5 ft ) (1.2 kips ) − ( 8ft ) FEG = 0
FEG = 3.375 kips
FEG = 3.38 kips T W
 8

ΣM K = 0: ( 5 ft )(1.2 kips ) + (10 ft )(1.2 kips ) − (12 ft ) 
FFG  = 0
 73

FFG =
ΣFy = 0
3
73 kips
16
8  3
 8
73 kips  −
FFH − 1.2 kips − 1.2 kips − 0.6 kip + 3.6 kips = 0

16
73 
 17
FFH = 4.4625 kips
62
FFG = 1.602 kips T W
FFH = 4.46 kips C W