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Problem 6.1
Determine the axial forces in the mem
bers of the truss and indicate whether they are in tension
(T) or compression (C).
Strategy:
Draw freebody diagram of joint
A
.B
y
writing the equilibrium equations for the joint, you can
determine the axial forces in the two members.
B
A
C
30
°
60
°
800 N
Solution:
30
°
F
AB
F
AC
60
°
A
800 N
y
x
Assume the forces are in the directions shown (both in tension). If a
force turns up negative, that force will be in compression.
Equilibrium Eqns.
X
F
x
:
−
F
AB
cos 30
◦
+
F
AC
cos 60
◦
+800=0
X
F
y
:
−
F
AB
sin 30
◦
−
F
AC
sin 60
◦
=0
B
A
C
30
°
60
°
800 N
Solving: We get
F
AB
= 693
N (tension)
F
AC
=
−
400
N (compression)
Problem 6.2
The truss supports a 10kN load at
C
.
(a) Draw the freebody diagram of the entire truss, and
determine the reactions at its supports.
(b) Determine the axial forces in the members. Indicate
whether they are in tension (T) or compression (C).
10 kN
C
B
A
3 m
4 m
Solution:
(a) The freebody diagram of the system is shown. The
sum of the moments about
B
is:
M
B
=3
A
x
−
4(10) = 0
, from
which
A
x
=13
.
33
kN. The sums of the forces:
X
F
x
=
A
x
+
B
x
,
from which
B
x
=
−
A
x
=
−
13
.
33
kN
.
X
F
y
=
B
y
−
10=0
,
from which
B
y
=10
kN
. (b) The interior angle
ACB
is
α
=
tan
−
1
(0
.
75
)=36
.
87
◦
. (b) Assume that the unknown forces act
away from the joint. Denote the axial force in the member
I
,
K
by
IK
. The axial forces are
F
CB
=
BC
(
−
i
cos
α
+
j
sin
α
)
, and
F
CA
=
−
AC
i
. Summing the forces:
X
F
y
=
sin
α
−
,
from which
=16
.
67
kN
(
T
)
.
X
F
x
=
−
cos
α
−
AC
,
from which
AC
=
−
13
.
33
kN
(
C
)
. For the joint
A
,
10 kN
C
B
A
3 m
4 m
B
X
A
X
B
Y
3 m
4 m
W
X
F
y
=
AB
,
from which
AB
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View Full DocumentProblem 6.3
In Example 6.1, suppose that the 2kN
load is applied at
D
in the horizontal direction, point
ing from
D
toward
B
. What are the axial forces in the
members?
Solution:
First, solve for the support forces and then use the
method of joints at each joint to solve for the forces.
B
X
A
X
A
Y
B
θ
6 m
tan
=
10 m
D
x
y
2 kN
6
10
=
30.96
X
F
x
:
B
x
+
A
x
−
2
kN
=0
X
F
y
:
A
y
X
M
B
:
−
6
A
x
Solving, we get
A
x
=
A
y
,
B
x
=2
kN
Joint A
:
A
X
=
0
A
Y
=
0
AB
AC
=
30.96
θ
=30
.
96
◦
X
F
x
:
AC
cos
θ
X
F
y
:
−
AB
−
AC
sin
θ
Solving, we get
AB
=
AC
Joint C
:
BC
CD
C
AC = 0 from
above
=
30.96
°
X
F
x
:
0
AC
/
cos
θ
−
BC
cos
θ
+
CD
cos
θ
−
+
C
5 m
5 m
A
D
B
2 kN
3 m
3 m
X
F
y
:
0
AC
/
sin
θ
−
sin
θ
−
sin
θ
−
−
Solving, we get
=
Joint B
:
AB
BC
BD
B
X
x
We already know
AB
=
and
B
x
kN
X
F
x
:
0
/
cos
θ
+
BD
+
B
x
+2
kN
=
−
2
kN
(
compression
)
X
F
y
:(
all forces zero
)=0
we have
AB
=
AC
=
=
=
−
2
kN
(
c
)
Note that we did not have to use joint
D
as we had already solved for the forces
there. The
FBD
at
D
is
, with the (
−
) sign, is opposite the direction shown.
CD
=
0
BD
=
−
2 kN (c)
D
2 kN
Problem 6.4
Determine the axial forces in the mem
bers of the truss.
0.6 m
0.4 m
0.3 m
1.2 m
2 kN
C
B
A
Solution:
First, solve for the support reactions at
B
and
C
, and
then use the method of joints to solve for the forces in the members.
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 Winter '08
 ORIENT
 Statics

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