Chapter 13 : Analysis of Perfect Frames (Analytical Method) 277
Solution. Since the truss is supported on rollers at the right end D, therefore the reaction at
this support will be normal to the support i.e., inclined at 45° with the horizontal. The reaction at A
will be the resultant of horizontal and vertical forces. It will be interesting to know that as the reaction
at D is inclined at 45° with the horizontal, therefore horizontal component (RDH) and vertical compo-
nent (RDV) of this reaction will be equal. Mathematically RDH = RDV.
Taking moments about A and equating the same,
(RDV × 9) – (RDH × 4) = (5 × 3) + (2 × 6)
5 RDH=27 [Q RDH = RDV]
27
5.4 kN ( )
DH 5
R == ←and RDV = 5.4 kN ( ↑ )
The example may be solved either by the method of joints or method of sections. But we shall
solve it by the method of sections, as one section line can cut the members BC, FE and FC and in
which forces are required to be found out.
Now let us pass section (1-1) cutting the truss into two parts as shown in Fig. 13.53.
Fig. 13.53.
Now consider equilibrium of right part of the truss. Let the directions of the forces PBC and
PFE be assumed as shown in Fig 13.53. Taking moments about the joint F and equating the same,
PBC × 4 = (5.4 × 6) – (2 × 3) = 26.4∴
26.4
6.6 kN (Compression)
BC 4
P ==Similarly, taking moments about the joint C and equating the same,
PFE × 4 = (5.4 × 4) – (5.4 × 3) = 5.4∴5.4
1.35 kN (Compression)
4PFE==and now taking moments about the joint B and equating the same,
PFC × 2.4 = (PFE × 4) – (2 × 3) + (5.4 × 6) – (5.4 × 4)
= (1.35 × 4) – 6 + 32.4 – 21.6 = 10.2∴10.2
4.25 kN (Tension)
2.4PFC==