Engineering Mechanics

Chapter 13 : Analysis of Perfect Frames (Analytical Method) „„„„„ 271

Solution. Since the truss is supported on rollers at the right hand support (E), therefore the
reaction at this support will be vertical (because of horizontal support). The reaction at the left hand
support (A) will be the resultant of vertical and horizontal forces and inclined with vertical.

Taking moments about A and equating same,

VE × 8 = (2 W × 2) + (W × 2) = 6 W

∴

6

0.75 ( )

E 8

W

VW== ↑

and *VA = 2 W – 0.75 W = 1.25 W ( ↑ ) and HA = W (←)

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 CD, FD and FE in which
the forces are required to be found out. Now let us pass section (1-1) cutting the truss into two parts
as shown in Fig. 13.44.

Fig. 13.44.
Now consider equilibrium of the right part of the truss. Let the directions of the forces PCD,
PFD and PFE be assumed as shown in Fig. 13.44. Taking moments about the joint F and equating the
same,

PCD × 4 sin 45° = (0.75 W × 4) – (W × 2) = W

∴ CD 4 sin 45 4 0.707 0.354 (Compression)

WW

PW== =

°×

Similarly, taking moments about the joint E and equating the same,

PFD × 4 cos 45° = W × 2 = 2 W

∴

22

0.707 (Tension)

FD 4 cos 45 4 0.707

WW

PW== =

°×

and now taking moments about the joint D and equating the same,

PFE × 2 = 0.75 W × 2 = 1.5 W

∴

1.5

0.75 (Tension)

FE 2

W

PW==

* There is no need of finding out the vertical and horizontal reaction at A, as we are not considering this

part of the truss.