6–6.
Determine the force in each member of the truss, and state
if the members are in tension or compression.
600 N
D
4m
SOLUTION
C
Method of Joints: We will begin by analyzing the equilibrium of joint D, and then
proceed to analyze joints C and E.
Joint D: From the free-body diagram in Fig. a,
+ ©F = 0;
:
x
Ans.
Ans.
Joint C: From the free-body diagram in Fig. b,
+ ©F = 0;
:
x
FCE - 900 = 0
FCE = 900 N (C)
+ c ©Fy = 0;
Ans.
800 - FCB = 0
FCB = 800 N (T)
Ans.
Joint E: From the free-body diagram in Fig. c,
R+ ©Fx ¿ = 0;
- 900 cos 36.87° + FEB sin 73.74° = 0
FEB = 750 N (T)
Q+ ©Fy ¿ = 0;
Ans.
FEA - 1000 - 900 sin 36.87° - 750 cos 73.74° = 0
FEA = 1750 N = 1.75 kN (C)
B
6m
4
1000 a b - FDC = 0
5
FDC = 800 N (T)
4m
A
3
FDE a b - 600 = 0
5
FDE = 1000 N = 1.00 kN (C)
+ c ©Fy = 0;
900 N
E
Ans.
6–7.
Determine the force in each member of the Pratt truss, and
state if the members are in tension or compression.
J
2m
K
2m
SOLUTION
2m
Joint A:
+ c ©Fy = 0;
20 - FAL sin 45° = 0
FAB - 28.28 cos 45° = 0
Joint B:
FBC - 20 = 0
FBC = 20 kN (T)
+ c ©Fy = 0;
FBL = 0
Joint L:
R+ ©Fx = 0;
FLC = 0
+Q©Fy = 0;
28.28 - FLK = 0
FLK = 28.28 kN (C)
Joint C:
+ ©F = 0;
:
x
FCD - 20 = 0
FCD = 20 kN (T)
+ c ©Fy = 0;
FCK - 10 = 0
FCK = 10 kN (T)
Joint K:
R+ ©Fx - 0;
10 sin 45° - FKD cos (45° - 26.57°) = 0
FKD = 7.454 kN (L)
+Q©Fy = 0;
28.28 - 10 cos 45° + 7.454 sin (45° - 26.57°) - FKJ = 0
FKJ = 23.57 kN (C)
Joint J:
+ ©F = 0;
:
x
23.57 sin 45° - FJI sin 45° = 0
FJI = 23.57 kN (L)
+ c ©Fy = 0;
H
A
10 kN
FAB = 20 kN (T)
+ ©F = 0;
:
x
L
B
D
C
E
F
2m 2m
2m 2m 2m 2m
FAL = 28.28 kN (C)
+ ©F = 0;
:
x
I
2 (23.57 cos 45°) - FJD = 0
FJD = 33.3 kN (T)
Ans.
FAL = FGH = FLK = FHI = 28.3 kN (C)
Ans.
FAB = FGF = FBC = FFE = FCD = FED = 20 kN (T)
Ans.
FBL = FFH = FLC = FHE = 0
Ans.
FCK = FEI = 10 kN (T)
Ans.
FKJ = FIJ = 23.6 kN (C)
Ans.
FKD = FID = 7.45 kN (C)
Ans.
Due to Symmetry
20 kN
10 kN
G
6–22.
B
Determine the force in each member of the double scissors
truss in terms of the load P and state if the members are in
tension or compression.
C
L/3
SOLUTION
c + ©MA = 0;
L
2L
Pa b + P a
b - (Dy)(L) = 0
3
3
Dy = P
+ c ©Fy = 0;
Ay = P
L/3
FFD - FFE - FFB a
1
22
(1)
b = 0
FFD - FFE = P
+ ©F = 0;
:
y
1
FFB a
22
b -P = 0
FFB = 22P = 1.41 P (T)
Similarly,
FEC = 22P
Joint C:
+ ©F = 0;
:
x
2
FCA a
2
25
+c ©Fy = 0;
FCA
25
b - 22P a
FCA 1
25
1
22
1
22
b - FCD a
1
22
b = 0
FCD = P
- 22P
1
22
+ FCD
1
22
=0
FCA =
225
P = 1.4907P = 1.49P (C)
3
FCD =
22
P = 0.4714P = 0.471P (C)
3
FAE -
1
2
22
225
Pa
Pa
b b = 0
3
3
22
25
FAE =
5
P = 1.67 P (T)
3
Joint A:
+ ©F = 0;
:
x
Similarly,
FFD=1.67 P (T)
From Eq.(1), and Symmetry,
FFE = 0.667 P (T)
Ans.
FFD = 1.67 P (T)
Ans.
FAB = 0.471 P (C)
Ans.
FAE = 1.67 P (T)
Ans.
FAC = 1.49 P (C)
Ans.
FBF = 1.41 P (T)
Ans.
FBD = 1.49 P (C)
Ans.
FEC = 1.41 P (T)
Ans.
FCD = 0.471 P (C)
Ans.
F
L/3
P
Joint F:
+ ©F = 0;
:
x
E
A
D
L/3
P
6–25.
Determine the force in each member of the truss in terms of
the external loading and state if the members are in tension
or compression.
P
B
L
C
u
L
L
SOLUTION
A
Joint B:
+ c ©Fy = 0;
L
FBA sin 2u - P = 0
FBA = P csc 2u (C)
+ ©F = 0;
:
x
Ans.
P csc 2u(cos 2u) - FBC = 0
FBC = P cot 2 u (C)
Ans.
Joint C:
+ ©F = 0;
:
x
P cot 2 u + P + FCD cos 2 u - FCA cos u = 0
+ c ©Fy = 0;
FCD sin 2 u - FCA sin u = 0
FCA =
cot 2 u + 1
P
cos u - sin u cot 2 u
FCA = (cot u cos u - sin u + 2 cos u) P (T)
Ans.
FCD = (cot 2 u + 1) P (C)
Ans.
Joint D:
+ ©F = 0;
:
x
FDA - (cot 2 u + 1)(cos 2 u) P = 0
FDA = (cot 2 u + 1)(cos 2 u) (P)
(C)
Ans.
D
P
6–26.
The maximum allowable tensile force in the members of the
truss is 1Ft2max = 2 kN, and the maximum allowable
compressive force is 1Fc2max = 1.2 kN. Determine the
maximum magnitude P of the two loads that can be applied
to the truss. Take L = 2 m and u = 30°.
P
B
L
u
L
L
SOLUTION
A
(Tt)max = 2 kN
L
(FC)max = 1.2 kN
Joint B:
+ c ©Fy = 0;
FBA cos 30° - P = 0
FBA =
+ ©F = 0;
:
x
P
= 1.1547 P (C)
cos 30°
FAB sin 30° - FBC = 0
FBC = P tan 30° = 0.57735 P (C)
Joint C:
+ c ©Fy = 0;
-FCA sin 30° + FCD sin 60° = 0
FCA = FCD a
+ ©F = 0;
:
x
sin 60°
b = 1.732 FCD
sin 30°
P tan 30° + P + FCD cos 60° - FCA cos 30° = 0
FCD = a
tan 30° + 1
23 cos 30° - cos 60°
b P = 1.577 P (C)
FCA = 2.732 P (T)
Joint D:
+ ©F = 0;
:
x
FDA - 1.577 P sin 30° = 0
FDA = 0.7887 P (C)
1) Assume FCA = 2 kN = 2.732 P
P = 732.05 N
FCD = 1.577(732.05) = 1154.7 N 6 (Fc)max = 1200 N
Thus,
Pmax = 732 N
C
(O.K.!)
Ans.
D
P
6–31.
Determine the force in members CD, CJ, KJ, and DJ of the
truss which serves to support the deck of a bridge. State if
these members are in tension or compression.
8000 lb
5000 lb
4000 lb
B
A
C
D
E
F
G
12 ft
SOLUTION
a + ©MC = 0;
L
9 ft
- 9500(18) + 4000(9) + FKJ(12) = 0
FKJ = 11 250 lb = 11.2 kip (T)
a + ©MJ = 0;
Joint D:
Ans.
-9500(27) + 4000(18) + 8000(9) + FCD(12) = 0
FCD = 9375 lb = 9.38 kip (C)
+ ©F = 0;
:
x
K
9 ft
- 9375 + 11 250 -
Ans.
3
FCJ = 0
5
FCJ = 3125 lb = 3.12 kip (C)
Ans.
FDJ = 0
Ans.
J
9 ft
I
9 ft
H
9 ft
9 ft
*6–32.
Determine the force in members EI and JI of the truss
which serves to support the deck of a bridge. State if these
members are in tension or compression.
8000 lb
5000 lb
4000 lb
B
A
C
D
E
F
G
12 ft
SOLUTION
a + ©ME = 0;
L
9 ft
- 5000(9) + 7500(18) - FJI(12) = 0
FJI = 7500 lb = 7.50 kip (T)
+ c ©Fy = 0;
K
9 ft
Ans.
7500 - 5000 - FEI = 0
FEI = 2500 lb = 2.50 kip (C)
Ans.
J
9 ft
I
9 ft
H
9 ft
9 ft
6–46.
Determine the force in members CD and CM of the
Baltimore bridge truss and state if the members are in
tension or compression. Also, indicate all zero-force
members.
M
N
O
SOLUTION
C
21122 + 5182 + 3162 + 2142 - Ay 1162 = 0
Ay = 5.625 kN
Ax = 0
Method of Joints: By inspection, members BN, NC, DO, OC, HJ
LE and JG are zero force members.
Ans.
Method of Sections:
a + ©MM = 0;
FCD142 - 5.625142 = 0
FCD = 5.625 kN 1T2
a + ©MA = 0;
D
E
F
5 kN 3 kN
16 m, 8 @ 2 m
Support Reactions:
Ans.
FCM 142 - 2142 = 0
FCM = 2.00 kN T
Ans.
2m
J
P
I
2 kN
+ ©F = 0;
:
x
K
A
B
a + ©MI = 0;
L
G
2 kN
H
2m
6–47.
Determine the force in members EF, EP, and LK of the
Baltimore bridge truss and state if the members are in
tension or compression. Also, indicate all zero-force
members.
M
N
O
SOLUTION
C
Iy 1162 - 21122 - 31102 - 5182 - 2142 = 0
Iy = 6.375 kN
Method of Joints: By inspection, members BN, NC, DO, OC, HJ
LE and JG are zero force members.
Ans.
Method of Sections:
3122 + 6.375142 - FEF142 = 0
FEF = 7.875 = 7.88 kN 1T2
a + ©ME = 0;
Ans.
6.375182 - 2142 - 3122 - FLK 142 = 0
FLK = 9.25 kN 1C2
+ c ©Fy = 0;
D
E
F
5 kN 3 kN
16 m, 8 @ 2 m
Support Reactions:
Ans.
6.375 - 3 - 2 - FED sin 45° = 0
FED = 1.94 kN T
Ans.
2m
J
P
I
2 kN
a + ©MK = 0;
K
A
B
a + ©MA = 0;
L
G
2 kN
H
2m
6–71.
Determine the support reactions at A, C, and E on the
compound beam which is pin connected at B and D.
10 kN
9 kN
10 kN m
B C
E
D
SOLUTION
A
Equations of Equilibrium: First, we will consider the free-body diagram of
segment DE in Fig. c.
1.5 m 1.5 m 1.5 m 1.5 m 1.5 m 1.5 m
+ ©MD = 0;
NE(3) - 10(1.5) = 0
NE = 5 kN
+ ©ME = 0;
Ans.
10(1.5) - Dy(3) = 0
Dy = 5 kN
+ ©F = 0;
:
x
Dx = 0
Ans.
Subsequently, the free-body diagram of segment BD in Fig. b will be considered
using the results of Dx and Dy obtained above.
+ ©MB = 0;
NC(1.5) - 5(3) - 10 = 0
NC = 16.67 kN = 16.7 kN
+ ©MC = 0;
Ans.
By(1.5) - 5(1.5) - 10 = 0
By = 11.67 kN
+ ©F = 0;
:
x
By = 0
Finally, the free-body diagram of segment AB in Fig. a will be considered using the
results of Bx and By obtained above.
+ ©F = 0;
:
x
Ax = 0
+ c ©Fy = 0;
11.67 - 9 - A y = 0
A y = 2.67 kN
+ ©MA = 0;
Ans.
Ans.
11.67(3) - 9(1.5) - MA = 0
MA = 21.5 kN # m
Ans.
6–102.
The tractor boom supports the uniform mass of 500 kg in
the bucket which has a center of mass at G. Determine the
force in each hydraulic cylinder AB and CD and the
resultant force at pins E and F. The load is supported
equally on each side of the tractor by a similar mechanism.
G
B
A
0.25 m
E
C
1.5 m
0.3 m
0.1 m
SOLUTION
a + ©ME = 0;
1.25 m
0.2 m
2452.510.12 - FAB10.252 = 0
FAB = 981 N
F
Ans.
+ ©F = 0;
:
x
-Ex + 981 = 0;
Ex = 981 N
+ c ©Fy = 0;
Ey - 2452.5 = 0;
Ey = 2452.5 N
0.6 m
D
0.4 m 0.3 m
FE = 2198122 + 12452.522 = 2.64 kN
a + ©MF = 0;
2452.512.802 - FCD1cos 12.2°210.72 + FCD1sin 12.2°211.252 = 0
FCD = 16 349 N = 16.3 kN
+ ©F = 0;
:
x
Ans.
Ans.
Fx - 16 349 sin 12.2° = 0
Fx = 3455 N
+ c ©Fy = 0;
-Fy - 2452.5 + 16 349 cos 12.2° = 0
Fy = 13 527 N
FF = 21345522 + 113 52722 = 14.0 kN
Ans.
6–109.
The symmetric coil tong supports the coil which has a mass
of 800 kg and center of mass at G. Determine the horizontal
and vertical components of force the linkage exerts on
plate DEIJH at points D and E. The coil exerts only vertical
reactions at K and L.
H
300 mm
D
J
E
I
400 mm
SOLUTION
100 mm
Free-Body Diagram: The solution for this problem will be simplified if one realizes
that links BD and CF are two-force members.
Equations of Equilibrium : From FBD (a),
78481x2 - FK12x2 = 0
a + ©ML = 0;
FK = 3924 N
FBD cos 45°11002 + FBD sin 45°11002 - 39241502 = 0
FBD = 1387.34 N
+ ©F = 0;
:
x
A x - 1387.34 cos 45° = 0
+ c ©Fy = 0;
A y - 3924 - 1387.34 sin 45° = 0
A x = 981 N
A y = 4905 N
From FBD (c),
a + ©ME = 0;
4905 sin 45°17002 - 981 sin 45°17002
- FCF cos 15°13002 = 0
FCF = 6702.66 N
+ ©F = 0;
:
x
Ex - 981 - 6702.66 cos 30° = 0
Ex = 6785.67 N = 6.79 kN
+ c ©Fy = 0;
A
Ans.
Ey + 6702.66 sin 30° - 4905 = 0
Ey = 1553.67 N = 1.55 kN
Ans.
Dx = FBD cos 45° = 1387.34 cos 45° = 981 N
Ans.
Dy = FBD sin 45° = 1387.34 sin 45° = 981 N
Ans.
At point D,
C
30°
45°
30°
F
50 mm
100 mm
K
From FBD (b),
a + ©MA = 0;
45°
B
G
L