Civil Engineering Reference
In-Depth Information
T
ABLE
16.1
F
1
,
j
L
j
Member
L
j
(m)
F
0,
j
F
1,
j
F
0,
j
F
1,
j
L
j
F
a,
j
AB
L
0
−
0.71
0
0.5
L
+
0.40
P
BC
L
0
−
0.71
0
0.5
L
+
0.40
P
−
−
−
CD
L
P
0.71
0.71
PL
0.5
L
0.60
P
BD
1.41
L
-
1.0
-
1.41
L
−
0.56
P
AC
1.41
L
1.41
P
1.0
2.0
PL
1.41
L
+
0.85
P
AD
L
0
−
0.71
0
0.5
L
+
0.40
P
=
=
2.71
PL
4.82
L
actual member forces (
F
1,
j
) and the member forces produced by the unit load (
F
l
,
j
)
are the same. Therefore, from Eq. (i)
n
F
1,
j
L
j
AE
a
BD
=
(iii)
j
=
1
The solution is completed in Table 16.1.
From Table 16.1
2
.
71
PL
AE
4
.
82
L
AE
BD
=
a
BD
=
Substituting these values in Eq. (i) we have
2
.
71
PL
AE
+
X
BD
4
.
82
L
AE
=
0
from which
X
BD
=−
0
.
56
P
(i.e. compression)
The actual forces,
F
a,
j
, in the members of the complete truss of Fig. 16.18(a) are now
calculated using the method of joints and are listed in the final column of Table 16.1.
We note in the above that
BD
is positive, which means that
BD
is in the direction
of the unit loads, i.e. B approaches D and the diagonal BD in the released structure
decreases in length. Therefore in the complete structure the member BD, which pre-
vents this shortening, must be in compression as shown; also
a
BD
will always be positive
since it contains the term
F
1,
j
. Finally, we note that the cut member BD is included
in the calculation of the displacements in the released structure since its deformation,
under a unit load, contributes to
a
BD
.
E
XAMPLE
16.9
Calculate the forces in the members of the truss shown in Fig.
16.19(a). All members have the same cross sectional area,
A
, and Young's modulus,
E
.
By inspection we see that the truss is both internally and externally statically indeter-
minate since it would remain stable and in equilibrium if one of the diagonals, AD
or BD, and the support at C were removed; the degree of indeterminacy is therefore
