Civil Engineering Reference
In-Depth Information
We see from the derivation of Eqs (4.3) and (4.4) that the units of a tension coefficient
are force/unit length, in this case kN/m. Generally, however, we shall omit the units.
We can now proceed to joint B at which, since
t
BA
(
=
t
AB
) has been calculated, there
are two unknowns
x
direction:
t
BA
(
x
A
−
x
B
)
+
t
BC
(
x
C
−
x
B
)
+
t
BD
(
x
D
−
x
B
)
+
3
=
0
(iii)
y
direction:
t
BA
(
y
A
−
y
B
)
+
t
BC
(
y
C
−
y
B
)
+
t
BD
(
y
D
−
y
B
)
=
0
(iv)
Substituting the values of the joint coordinates and
t
BA
in Eqs (iii) and (iv) we have,
from Eq. (iii)
−
0
.
67(0
−
0)
+
t
BC
(1
.
5
−
0)
+
t
BD
(1
.
5
−
0)
+
3
=
0
which simplifies to
1
.
5
t
BC
+
1
.
5
t
BD
+
3
=
0
(v)
and from Eq. (iv)
−
0
.
67(0
−
1
.
5)
+
t
BC
(0
−
1
.
5)
+
t
BD
(1
.
5
−
1
.
5)
=
0
whence
t
BC
=+
0
.
67
Hence, from Eq. (v)
t
BD
=−
2
.
67
There are now just two unknown member forces at joint D. Hence, at D
x
direction:
t
DB
(
x
B
−
x
D
)
+
t
DF
(
x
F
−
x
D
)
+
t
DC
(
x
C
−
x
D
)
=
0
(vi)
y
direction:
t
DB
(
y
B
−
y
D
)
+
t
DF
(
y
F
−
y
D
)
+
t
DC
(
y
C
−
y
D
)
−
5
=
0
(vii)
Substituting values of joint coordinates and the previously calculated value of
t
DB
(
=
t
BD
) in Eqs (vi) and (vii) we obtain, from Eq. (vi)
−
2
.
67(0
−
1
.
5)
+
t
DF
(3
.
0
−
1
.
5)
+
t
DC
(1
.
5
−
1
.
5)
=
0
so that
t
DF
=−
2
.
67
and from Eq. (vii)
−
2
.
67(1
.
5
−
1
.
5)
+
t
DF
(1
.
5
−
1
.
5)
+
t
DC
(0
−
1
.
5)
−
5
=
0
from which
t
DC
=−
3
.
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