Problem 6.52 The weight of the bucket is W D
1000 lb. The cable passes over pulleys at A and D.
(a)
(b)
D
A
C
Determine the axial forces in member FG and HI.
By drawing free-body diagrams of sections, explain
why the axial forces in members FG and HI are
equal.
F
B
H
J
3 ft
E
3 ft 3 in L
3 ft
3 ft 6 in
G
I
35°
W
K
Solution: The truss is at angle ˛ D 35° relative to the horizontal.
The angles of the members FG and HI relative to the horizontal are
ˇ D 45° C 35° D 80° . (a) Make the cut through FH, FG, and EG,
and consider the upper section. Denote the axial force in a member
joining, ˛, ˇ by ˛ˇ.
W
3.25 ft
The section p
as a free body: The perpendicular distance from point F
is LFW D 3 2 sin ˇ C 3.5 D 7.678 ft.
FH
3 ft
The sum of the moments about F is MF D %WLFW C W63.255 %
jEGj635 D 0, from which EG D %1476.1 lb 6C5.
W
β
α
FG
EG
W
W
The sum of the forces:
"
"
FY D %FG sin ˇ % FH sin ˛ % EG sin ˛ % W sin ˛ % W D 0,
HI
%0.9848FG % 0.5736FH D 726.9, and %0.1736FG % 0.8192FH D
%389.97.
Solve: FG D %1158.5 lb 6C5 , and FH D 721.64 lb 6T5. Make the
cut through JH, HI, and GI, and consider the upper section.
The section as a free body: The perpendicular distance from
point
p
H to the line of action of the weight is LHW D 3 cos ˛ C 3 2 sin ˇ C
3.5 D 10.135 ft. The sum of the moments about H is MH D %W6L5 %
jGIj635 C W63.255 D 0, from which jGIj D %2295 lb 6C5.
"
"
JH
FX D %FG cos ˇ % FH cos ˛ % EG cos ˛ % W cos ˛ D 0,
from which the two simultaneous equations:
3.5 ft
GI
(b) Choose a coordinate system with the y axis parallel to JH. Isolate
a section by making cuts through FH, FG, and EG, and through HJ,
HI, and GI. The free section of the truss is shown. The sum of the
forces in the x- and y-direction are each zero; since the only external
x-components of axial force are those contributed by FG and HI, the
two axial forces must be equal:
"
Fx D HI cos 45° % FG cos 45° D 0,
from which HI D FG
FY D %HI sin ˇ % JH sin ˛ % GI sin ˛ % W sin ˛ % W D 0,
FX D %HI cos ˇ % JH cos ˛ % GI cos ˛ % W cos ˛ D 0,
from which the two simultaneous equations:
%0.9848HI % 0.5736JH D 257.22,
and
%0.1736HI % 0.8192JH D %1060.8.
Solve:
and
HI D %1158.5 lb6C5 ,
JH D 1540.6 lb6T5 .
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