Engineering Mechanics

Chapter 13 : Analysis of Perfect Frames (Analytical Method) „„„„„ 267

Fig. 13.37.

We know that in similar triangles OPG and OAE,

AOAP

AEPG

= or^84

42

AO

==

∴ AO = 4 × 4 = 16 m

and DO = 16 – 10 = 6 m

Now in triangle CGP, we find that

2

tan 1

2

∠==GCP or ∠=°GCP 45

∴∠COR = 90° – 45° = 45°

and OR = OC cos 45° = 10 × 0.707 m = 7.07 m

From the geometry of the triangle OPG, we find that

2

tan 0.25

8

∠==GOP or ∠GOP = 14°

Similarly, in triangle OCQ, we find that
CQ = CO sin 14° = 10 × 0.2425 = 2.425 m
Now pass section (1-1) cutting the frame through the members CD, CG and FG. Let the direc-
tions of the forces PCD, PCG and PFG be assumed as shown in Fig. 13.36. Taking moments of the
forces acting on right part of the frame only, about the joint G and equating the same,

PCD × 2 = 2 × 2 or PCD = 2 kN (Tension) Ans.
Similarly, taking moments of the forces acting in the right part of the truss only about the
imaginary joint O and equating the same,

PCG × 7.07 = 2 × 6

or^12 1.7 kN (Tension)

7.07

PCG==^ Ans.

and now taking moments of the forces acting in the right part of the truss only about the joint C and
equating the same,
PFG × 2.425 = 2 × 4 = 8

∴

8

3.3 kN

2.425

PFG== (Compression)

EXERCISE 13.2

1. Determine the forces in the various members of a pin-joined frame as shown in Fig. 13.37.
   Tabulate the result stating whether they are in tension or compression.
   Ans. CD = 2.5 kN (Compression)
   BC = 2.0 kN (Tension)
   AB = 2.0 kN (Tension)
   BD = 1.5 kN (Compression)
   AD = 1.25 kN (Tension)
   ED = 3.75 kN (Compression)