Civil Engineering Reference
In-Depth Information
The solution then proceeds to joint C to obtain
t
CF
and
t
CE
or to joint F to determine
t
FC
and
t
FE
; joint F would be preferable since fewer members meet at F than at C.
Finally, the remaining unknown tension coefficient (
t
EC
or
t
EF
) is found by considering
the equilibrium of joint E. Then
t
FC
=+
2
.
67,
t
FE
=−
2
.
67,
t
EC
=
0
which the reader should verify.
The forces in the truss members are now calculated by multiplying the tension
coefficients by the member lengths, i.e.
T
AB
=
t
AB
L
AB
=−
×
=−
0
.
67
1
.
5
1
.
0 kN (compression)
T
AC
=
t
AC
L
AC
=+
2
.
0
×
1
.
5
=+
3
.
0 kN (tension)
T
BC
=
t
BC
L
BC
in which
(
x
B
−
(0
L
BC
=
x
C
)
2
+
(
y
B
−
y
C
)
2
=
−
1
.
5)
2
+
(1
.
5
−
0)
2
=
2
.
12 m
Then
T
BC
=+
0
.
67
×
2
.
12
=+
1
.
42 kN (tension)
Note that in the calculation of member lengths it is immaterial in which order the joint
coordinates occur in the brackets since the brackets are squared. Also
T
BD
=
t
BD
L
BD
=−
2
.
67
×
1
.
5
=−
4
.
0 kN (compression)
Similarly
T
DF
=−
4
.
0 kN (compression)
T
DC
=−
5
.
0 kN (compression)
T
FC
=+
5
.
67 kN (tension)
T
FE
=−
4
.
0 kN (compression)
T
EC
=
0
4.9 G
RAPHICAL
M
ETHOD OF
S
OLUTION
In some instances, particularly when a rapid solution is required, the member forces
in a truss may be found using a graphical method.
The method is based upon the condition that each joint in a truss is in equilibrium so
that the forces acting at a joint may be represented in magnitude and direction by the
