Civil Engineering Reference
In-Depth Information
Generally, although not always, the support reactions must be calculated first. So,
taking moments about D for the truss in Fig. 4.14 we obtain
R
A
×
2
−
2
×
1
.
5
−
1
×
1
−
3
×
0
.
5
=
0
which gives
R
A
=
2
.
75 kN
Then, resolving vertically
R
D
+
R
A
−
2
−
1
−
3
=
0
so that
R
D
=
3
.
25 kN
Note that there will be no horizontal reaction at A (D is a roller support) since no
horizontal loads are applied.
The next step is to assign directions to the forces
acting on each joint
. In one approach
the truss is examined to determine whether the force in a member is tensile or com-
pressive. For some members this is straightforward. For example, in Fig. 4.14, the
vertical reaction at A,
R
A
, can only be equilibrated by the vertical component of the
force in AB which must therefore act downwards, indicating that the member is in
compression (a compressive force in a member will push towards a joint whereas a
tensile force will pull away from a joint). In some cases, where several members meet
at a joint, the nature of the force in a particular member is difficult, if not impossible,
to determine by inspection. Then a direction must be assumed which, if incorrect, will
result in a negative value for the member force. It follows that, in the same truss, both
positive and negative valuesmay be obtained for tensile forces and also for compressive
forces, a situation leading to possible confusion. Therefore, if every member in a truss
is initially assumed to be in tension, negative values will always indicate compression
and the solution will then agree with the sign convention adopted in Section 3.2.
We now assign tensile forces to the members of the truss in Fig. 4.14 using arrows to
indicate the
action of the force in the member on the joint
; then all arrows are shown to
pull away from the adjacent joint.
The analysis, as we have seen, is based on a consideration of the equilibrium of each
pin or hinge under the action of
all
the forces at the joint. Thus for each pin or hinge
we can write down two equations of equilibrium. It follows that a solution can only be
obtained if there are no more than two unknown forces acting at the joint. In Fig. 4.14,
therefore, we can only begin the analysis at the joints A or D, since at each of the joints
B and C there are three unknown forces while at E there are four.
Consider joint A. The forces acting on the pin at A are shown in the free body diagram
in Fig. 4.15.
F
AB
may be determined directly by resolving forces vertically.
