Problem 6.47 The Howe truss helps support a roof.
Model the supports at Aand Gas roller supports.
(a) Use the method of joints to determine the axial
force in member BI.
(b) Use the method of sections to determine the axial
force in member BI.
2 kN
4 m
A
B
C
G
F
E
D
HI JKL
2 kN
2 kN
2 kN2 kN
Solution: The pitch of the roof is
FYDA5⊲2⊳CGD0,
FxDAB cos ˛CAH D0,
Joint B:
AB
BH
BC
2 kN
2 m
426
Problem 6.48 Consider the truss in Problem 6.47. Use
the method of sections to determine the axial force in
6D2.6667 m.
The interior angle KJE is
2.6667 D6kN⊲T⊳.
The sum of the forces:
FxDDE cos ˛EJ cos ˇJK D0.
FyDDE sin ˛EJ sin ˇ2FCGD0,
from which the two simultaneous equations:
0.8321DE C0.6EJ D6,
0.5547DE 0.8EJ D1.
Solve: EJ D2.5kN⊲C⊳ .
α
EJ
JK
G
2 m 2 m
Problem 6.49 Use the method of sections to determine
the axial forces in member CE,DE, and DF.
CE
G
FD
H
A
B
4 f
t
4 f
t
12 kip
428
Problem 6.50 For the bridge truss shown, use the
method of sections to determine the axial forces in
members CE, CF, and DF.DF HJ
I
200 kN 200 kN 200 kN 200 kN 200 kN
B
A
C
E
G
3 m 4 m
7 m
Problem 6.51 The load FD20 kN and the dimension
LD2 m. Use the method of sections to determine the
axial force in member HK.
Strategy: Obtain a section by cutting members HK,
HI,IJ, and JM. You can determine the axial forces in
members HK and JM even though the resulting free-
body diagram is statically indeterminate.
ABC
D
H
K
G
J
M
E
I
F
F
L
L
L
L
L
Solution: The complete structure as a free body: The sum of the
430
Problem 6.52 The weight of the bucket is WD
1000 lb. The cable passes over pulleys at Aand D.
(a) Determine the axial forces in member FG and HI.
(b) By drawing free-body diagrams of sections, explain
why the axial forces in members FG and HI are
equal. 3 ft 6 in
3 ft
3 ft
3 ft 3 in
35°
L
J
H
F
C
K
I
G
E
B
A
D
W
Solution: The truss is at angle ˛D35°relative to the horizontal.
Problem 6.53 Consider the truss in Problem 6.52. The
weight of the bucket is WD1000 lb. The cable passes
over pulleys at Aand D. Determine the axial forces in
members IK and JL.
Wcos ˛JK cos ˇD0,
and FyDJL sin ˛IK sin ˛
Wsin ˛WJK sin ˇD0,
from which two simultaneous equations:
0.8192JL C0.1736JK D1732
and 0.5736JL C0.9848JK D212.75.
Solve: JL D2360 lb⊲T⊳ ,
and JK D1158.5lb⊲C⊳ .
Problem 6.54 The truss supports loads at N,P, and R.
Determine the axial forces in members IL and KM.
2 m
2 m
2 m
1 m
2 m 2 m 2 m 2 m 2 m
K
I
M
L
O
N
Q
PRJ
H
F
G
E
1 kN 2 kN 1 kN
Solution: The strategy is to make a cut through KM,IM, and
IL, and consider only the outer section. Denote the axial force in a
member joining, ˛, ˇ by ˛ˇ.
432
Problem 6.55 Consider the truss in Problem 6.54.
Determine the axial forces in members HJ and GI.
Solution: The strategy is to make a cut through the four members
AJ,HJ,HI, and GI, and consider the upper section. The axial force
36.87°respectively. Make a cut through the four members AJ,HJ,
HJ D8.25 kN ⊲T⊳ . The sums of the forces:
FxDAJ sin ˛CHJ sin ˇHI sin D0,
from which HI DAJ sin ˛HJ sin ˇ
sin D22
sin D0.
FYDAJ cos ˛HJ cos ˇHI cos GI 4D0,
from which GI D16 kN ⊲C⊳
1 m
Problem 6.56 Consider the truss in Problem 6.54. By
drawing free-body diagrams of sections, explain why the
axial forces in members DE,FG, and HI are zero.
Solution: Define ˛,ˇto be the interior angles BAJ and ABJ
respectively. The sum of the forces in the x-direction at the base yields
Problem 6.57 In Active Example 6.5, draw the free-
body diagram of joint Bof the space truss and use it to
determine the axial forces in members AB,BC, and BD.
1200 lb
B
D (10, 0, 0) ft
C (6, 0, 6) ft
A (5, 3, 2) ft
z
y
x
Solution: From Active Example 6.5 we know that the vertical
reaction force at Bis 440 lb.
The free-body diagram of joint Bis shown. We have the following
434
Problem 6.58 The space truss supports a vertical 10-
kN load at D. The reactions at the supports at joints A,
B, and Care shown. What are the axial forces in the
members AD,BD, and CD?
B (5, 0, 3) m
Ay
Ax
Az
Cy
Cz
By
C (6, 0, 0) m
D (4, 3, 1) m
10 kN
z
y
x
A
Solution: Consider the joint Donly. The position vectors parallel
to the members from Dare
10 kN
Problem 6.59 Consider the space truss in Prob-
lem 6.58. The reactions at the supports at joints A,B,
and Care shown. What are the axial forces in members
AB,AC, and AD?
The equilibrium conditions are:
FDTABeAB CTAC CeAC CTADeAD CAD0,
436
Problem 6.60 The space truss supports a vertical load
Fat A. Each member is of length L, and the truss rests on
the horizontal surface on roller supports at B,C, and D.
Determine the axial forces in members AB,AC, and AD.
F
A
B
C
D
Solution: By symmetry, the axial forces in members AB,AC, and
we see that
Problem 6.61 For the truss in Problem 6.60, deter-
mine the axial forces in members AB,BC, and BD.
Solution: See the solution of Problem 6.60. The axial force in
From the equilibrium equation
Problem 6.62 The space truss has roller supports at B,
C, and Dand supports a vertical 800-lb load at A. What
are the axial forces in members AB,AC, and AD?800 lb
A (4, 3, 4) ft
y
438
Problem 6.63 The space truss shown models an
airplane’s landing gear. It has ball and socket supports
at C,D, and E. If the force exerted at Aby the wheel is
FD40j(kN), what are the axial forces in members AB,
AC, and AD?
B
(1, 0, 0) m
A
(1.1, –0.4, 0) m
0.4 m
y
x
z
E (0, 0.8, 0) m
C
D
F
Solution: The important points in this problem are A(1.1, 0.4,
0), B(1, 0, 0), C(0, 0, 6), and D(0, 0, 0.4). We do not need point
The forces can be written as
TAC D5.26 kN (tension), and TAD D7.42 kN (tension).
y
z
E
C
F
A
(1.1, −0.4, 0) m
TAC
Problem 6.64 If the force exerted at point Aof
the truss in Problem 6.63 is FD10iC60jC20k(kN),
what are the axial forces in members BC,BD and BE?
0), B(1, 0, 0), C(0, 0, 0.6), D(0, 0, 0.4), and E(0, 0.8, 0). The
E
440
Problem 6.65 The space truss is supported by roller
supports on the horizontal surface at Cand Dand a ball
1.0, 0.2) m, C: (0.9, 0, 0.9) m, D: (0.9, 0, 0.6) m,
and E: (0, 0.8, 0) m. Determine the axial forces in
members AB,AC, and AD.
x
y
z
A
D
C
Solution: The important points in this problem are A: (1.6, 0.4,
Problem 6.66 The free-body diagram of the part of the
construction crane to the left of the plane is shown. The
coordinates (in meters) of the joints A,B, and Care (1.5,
1.5, 0), (0, 0, 1), and (0, 0, 1), respectively. The axial
forces P1,P2, and P3are parallel to the xaxis. The axial
forces P4,P5, and P6point in the directions of the unit
vectors
rBA D1.5iC1.5jk(m),
and the unit vector that points from Btoward Ais
442
Problem 6.67 In Problem 6.66, what are the axial
forces P1,P4, and P5?
0.64P50.64P50.426P5
The components of the moment equation are
P1D674.7kN,
Problem 6.68 The mirror housing of the telescope is
supported by a 6-bar space truss. The mass of the
housing is 3 Mg (megagrams), and its weight acts at G.
The distance from the axis of the telescope to points A,
B, and Cis 1 m, and the distance from the axis to points
D,E, and Fis 2.5 m. If the telescope axis is vertical
(˛D90°), what are the axial forces in the members of
the truss?
Mirror housing
A
B
C
G
E
D
F
4 m
1 m
AFD
60°60°
G
END VIEW
y
z
y
α
F⊲2.5 cos 60°,2.5 sin 60°,4⊳D⊲1.25,2.165,4⊳.
y
R = 2.5 m
D
444
Problem 6.69 Consider the telescope described in
Problem 6.68. Determine the axial forces in the members
of the truss if the angle ˛between the horizontal and the
telescope axis is 20°.
Solution: The coordinates of the points are,
A⊲cos 60°,sin 60°,0⊳D⊲0.5,0.866,0⊳⊲m⊳,
values are shown below to four significant figures, the calculations
were done with the full precision permitted (15 digits for TK Solver
Plus.)
rAF 1.75 1.299 4eAF 0.3842 0.2852 0.8781
rBD 2 0.866 4eBD 0.4391 0.1901 0.8781
rCE 0.25 2.165 4eCE 0.0549 0.4753 0.8781
rCF 0.25 2.165 4eCF 0.0549 0.4753 0.8781
The equilibrium condition for the forces is
jTABjeAD CjTAFjeAF CjTBDjeBD CjTBEjeBE CjTCEjeCE
CjTCFjeCF CWD0.
This is three equations in six unknowns. The unit vectors are given in
Table I. The weight vector is WDjWj⊲jcos ˛ksin ˛⊳, where ˛is
the angle from the horizontal of the telescope housing. The remaining
three equations in six unknowns are obtained from the moments:
rAð⊲TAD CTAF⊳CrBð⊲TBD CTBE⊳CrCð⊲TCE
CTCF⊳CrGðWD0.
A
F
y
−100 −50 0 50 100
alpha, deg