521
6–33.
SOLUTION
a
+©MA=0;
Ey(8) –2(8) –5(6) –5(4) –5(2) =0Ey=9.5kN
The Howe truss is subjected to the loading shown.
Determine the force in members GF,CD, and GC, and
state if the members are in tension or compression.
3 m
2 kN
5 kN
5 kN
2 m 2 m 2 m 2 m
A
BCD
F
G
H
E
2 kN
5 kN
Ans:
FGF =12.5
kN (C)
FCD =6.67
522
6–34.
The Howe truss is subjected to the loading shown.
Determine the force in members GH,BC, and BG of the
truss and state if the members are in tension or compression.
SOLUTION
a
Ans.
FGH =12.5 kN (C)
+©MB=0; –7.5(2) +FGH sin 36.87°(2) =0
3 m
2 kN
5 kN
5 kN
2 m 2 m 2 m 2 m
A
BCD
F
G
H
E
2 kN
5 kN
Ans:
523
6–35.
Determine the force in members EF, CF, and BC, and state
if the members are in tension or compression.
2 m
1.5 m
2 m
F
A
8 kN
4 kN ED
C
B
SOLUTION
Support Reactions. Not required.
Method of Sections.
FBC
and
FEF
can be determined directly by writing the moment
equations of equilibrium about points F and C, respectively, by referring to the FBD
of the upper portion of the truss section through a–a shown in Fig. a.
a+
ΣMF=0;
FBC(1.5) –4(2) =0
FBC =5.333 kN (C) =5.33 kN (C)
Ans.
FEF(1.5) –4(2) =0
FEF =5.333 kN (T) =5.33 kN (T)
FCF =4.00 kN (T)
Ans:
FBC =5.33 kN (C)
FEF =5.33 kN (T)
FCF =4.00 kN (T)
524
*6–36.
Determine the force in members AF, BF, and BC, and state
if the members are in tension or compression.
2 m
1.5 m
2 m
F
A
8 kN
4 kN ED
C
B
SOLUTION
Support Reactions. Not required.
Method of Sections: Referring to the FBD of the upper portion of the truss section
through a–a shown in Fig. a,
FAF
and
FBC
can be determined directly by writing the
moment equations of equilibrium about points B and F, respectively.
a+
ΣMB=0;
FAF(1.5) –8(2) –4(4) =0
FAF =21.33 kN (T) =21.3 kN (T)
Ans.
ΣMF=0;
FBC(1.5) –4(2) =0
FBC =5.333 kN (C) =5.33 kN (C)
FBF
525
6–37.
Determine the force in members EF, BE, BC and BF of the
truss and state if these members are in tension or
compression. Set P1 = 9 kN, P2 = 12 kN, and P3 = 6 kN.
FE
B
AD
C
3 m
3 m 3 m3 m
P1P2
P3
SOLUTION
Support Reactions. Referring to the FBD of the entire truss shown in Fig. a,
ND
can
be determined directly by writing the moment equation of equilibrium about point A.
a+
ΣMA=0;
ND(9) –12(6) –9(3) –6(3) =0
ND=13.0 kN
FEF
ΣMB=0;
ΣME=0;
FBE
Method of Joints. Using the result of
FBE
, the equilibrium of joint B, Fig. c, requires
526
6–38.
Determine the force in members BC, BE, and EF
of the truss and state if these members are in tension
or compression. Set P1 = 6 kN, P2 = 9 kN, and P3 = 12 kN.
F
E
B
AD
C
3
m
3 m 3 m3 m
P1P2
P3
SOLUTION
Support Reactions. Referring to the FBD of the entire truss shown in Fig. a,
ND
can be determined directly by writing the moment equation of equilibrium about
point A.
a+
ΣMA=0;
ND(9) –12(3) –6(3) –9(6) =0
ND=12.0 kN
FEF
ΣMB=0;
ΣME=0;
Also,
FBE
can be obtained directly by writing the force equation of equilibrium
along the y axis
Ans:
FEF =15.0 kN (C)
FBC =12.0 kN (T)
527
6–39.
SOLUTION
a+ΣME=0; –Ay(20) +2(20) +4(15) +4(10) +5(5) =0
Ay=8.25 kN
Determine the force in members BC, HC, and HG. After
the truss is sectioned use a single equation of equilibrium
for the calculation of each force. State if these members are
in tension or compression.
ACD
H
G
F
4 kN
3 m
2 m
5 m5 m5 m 5 m
BE
4 kN
5 kN
3 kN
2 kN
Ans:
FBC =10.4
kN (C)
FHG =9.16
FHC =2.24
528
*6–40.
SOLUTION
S
+ΣFx=0; Ex=0
a+ΣMA=0; –4(5) –4(10) –5(15) –3(20) +Ey(20) =0
a+ΣMF=0; –3(5) +9.75(5) –FCD(3)=0
Joint G:
S
+ΣFx=0; FGH =9.155 kN (T)
Determine the force in members CD, CF, and CG and state
if these members are in tension or compression.
ACD
H
G
F
4 kN
3 m
2 m
5 m5 m5 m 5 m
BE
4 kN
5 kN
3 kN
2 kN
Ans:
FCD =11.2 kN (C)
FCF =3.21 kN (T)
529
6–41.
Determine the force developed in members FE, EB, and
BC of the truss and state if these members are in tension or
compression.
11 kN
B
AD
C
FE
22 kN
2 m 1.5 m
2 m
2 m
SOLUTION
Support Reactions. Referring to the FBD of the entire truss shown in Fig. a,
ND
can be determined directly by writing the moment equation of equilibrium about
pointA.
FFE
FEB
Ans:
530
6–42.
Determine the force in members BC, HC, and HG. State if
these members are in tension or compression.
6 kN
12 kN
9 kN
4 kN 6 kN
1.5 m 1.5 m
2 m
1
m1
m
1.5 m 1.5 m
A E
B
H
G
J
CD
SOLUTION
Support Reactions. Referring to the FBD of the entire truss shown in Fig. a,
NA
can
be determined directly by writing the moment equation of equilibrium about point E.
ΣME=0;
Method of Sections. Referring to the FBD of the left portion of the truss sectioned
through a–a shown in Fig. b,
FHG
,
FHC
and
FBC
can be determined directly by writing
the moment equations of equilibrium about points C, A, and H, respectively.
a+
ΣMA=0;
FHC
a2
213b
(3) –6(1.5) =0
ΣMH=0;
Ans:
531
6–43.
Determine the force in members CD, CJ, GJ, and CG and
state if these members are in tension or compression.
6 kN
12 kN
9 kN
4 kN 6 kN
1.5 m 1.5 m
2 m
1
m1
m
1.5 m 1.5 m
A E
B
H
G
J
CD
SOLUTION
Support Reactions. Referring to the FBD of the entire truss shown in Fig. a,
E
y can
be determined directly by writing the moment equation of equilibrium about point A.
Method of Sections. Referring to the FBD of the right portion of the truss sectioned
through a–a shown in Fig. b,
FGJ
,
FCJ
and
FCD
can be determined directly by writing
moment equations of equilibrium about point C, E and J, respectively.
a+
ΣMC=0;
20.25(3) –6(3) –9(1.5) –FGJ
a2
213 b
(3) =0
Method of Joints. Using the result of
FGJ
to consider the equilibrium of joint G,
Fig. c,
532
*6–44.
Determine the force in members BE, EF,and CB,and state
if the members are in tension or compression.
SOLUTION
:
+©Fx=0; 5 +10 –FBE cos 45° =0
Ans:
FBE =21.2 kN (T)
FCB =5 kN (T)
533
6–45.
Determ
i
ne t
h
e force
i
n mem
b
ers BF,BG,an
d
AB,an
d
state
if the members are in tension or compression.
SOLUTION
Joint F:
Section:
Ans:
534
6–46.
Determine the force in members BC, CH, GH, and CG of
the truss and state if the members are in tension or
compression.
A
CD
H
G
F
8 kN
3 m
2 m
4 m4 m
4 m 4 m
B
E
4 kN 5 kN
SOLUTION
Support Reactions. Referring to the FBD of the entire truss shown in Fig. a,
A
y can
be determined directly by writing the moment equation of equilibrium about point E.
Method of Sections. Referring to the FBD of the left portion of the truss section
through a–a shown in Fig. a,
FBC
,
FGH
and
FCH
can be determined directly by writing
the moment equations of equilibrium about points H, C, and O, respectively,
Method of Joints. Using the result of
FGH
, equilibrium of joint G, Fig. c, requires
Ans:
FBC =11.0 kN(T)
FGH =11.2 kN(C)
FCH =1.25 kN(C)
FCG =10.0 kN(T)
535
SOLUTION
Support Reactions. Referring to the FBD of the entire truss shown in Fig. a,
A
y
can be determined directly by writing the moment equation of equilibrium about
pointG.
Method of Sections. Referring to the FBD of the left portion of the truss sectioned
through a
–
a shown in Fig. b,
FCD
,
FCJ
and
FKJ
can be determined directly by writing
the moment equations of equilibrium about points J, A and C, respectively.
a+
ΣMJ=0;
FCD(3) +6(2) +6(4) – 15.0(6) =0
FCD =18.0 kN (T)
Ans.
6–47.
Determine the force in members CD, CJ, and KJ and state
if these members are in tension or compression.
6 kN
A
BCDE
G
I
H
F
12 m, 6 @ 2 m
J
K
L
6 kN
6 kN
6 kN
6 kN
3 m
Ans:
FCD =18.0 kN (T)
FCJ =10.8 kN (T)
536
*6–48.
Determine the force in members JK,CJ,and CD of the truss,
and state if the members are in tension or compression.
SOLUTION
Method of Joints: Applying the equations of equilibrium to the free – body diagram
of the truss,Fig. a,
Method of Sections: Using the left portion of the free – body diagram, Fig. a.
a
+©MC=0; FJK(3) +4(2) –10.33(4) =0
:
+©Fx=0; Ax=0
A
BCDF
E
G
H
I
J
L
K
6kN
8kN
5kN
4kN
3m
2m 2m 2 m2 m2 m2 m
Ans:
FJK =11.1 kN (C)
FCD =12 kN (T)
FCJ =1.60 kN (C)
537
6–49.
SOLUTION
Support Reactions: Applying the moment equation of equilibrium about point Ato
the free – body diagram of the truss,Fig. a,
Method of Sections: Using the right portion of the free – body diagram, Fig. b.
a
+©MI=0; 12.67(4) –6(2) –FEF(3) =0
Determine the force in members HI,FI, and EF of the truss,
and state if the members are in tension or compression.
A
BCDF
E
G
H
I
J
L
K
6kN
8kN
5kN
4kN
3m
2m 2m 2 m2 m2 m2 m
Ans:
538
Ans:
FCA =FCB =122
lb (C)
FCD =173
FDA =86.6
6–50.
539
Ans:
FAB =6.46
kN (T)
FBC =FBD =3.70
FBE =4.80
6–51.
540
SOLUTION
Support Reactions. Not required
Method of Joints. Perform the joint equilibrium analysis at joint G first and then
proceed to joint E.
Thus,
FGE =6.00 kN (C)
Joint E. Fig. b
ΣF
z
=0;
FED
a2
3b
– 6.00 =0
FED =9.00 kN (T)
Ans.
*6–52.
Determine the force in each member of the space truss and
state if the members are in tension or compression. The truss
is supported by ball-and-socket joints at A, B, C, and D.
Ans:
FGC =4.47 kN (T)
G
A
6 kN
4 kN
B
C
Ey
z
x
D2 m
4 m
4 m
2 m