Engineering Mechanics

Chapter 13 : Analysis of Perfect Frames (Analytical Method) „„„„„ 281

From the geometry of this triangle, we find that

BD = AD sin 30° = 1.2 × 0.5 = 0.6 m

and BE = BD sin 30° = 0.6 × 0.5 = 0.3 m

∴ DE = BD cos 30° = 0.6 × 0.866 = 0.52 m

and CE = DC – DE = 0.9 – 0.52 = 0.38 m

∴

0.3

tan 0.7895

0.38

BE

BCE

CE

∠== =

or ∠BCE = 38.3°

Resolving the forces horizontally at C and equating the same,

PBC sin 38.3° = 1.889

∴

1.889 1.889

3.049 kN (Compression)

sin 38.3 0.6196

PBC===

°

Now tabulate the results as given below :

S.No. Member Magnitude of force in kN Nature of force

1 AB 2.828 Compression

2 AD 1.974 Tension

3 BD 1.12 Compression

4 CD 2.384 Tension

5 BC 3.049 Compression

Example 13.19. The truss shown in the Fig. 13.58 is made up of three equilateral triangles
loaded at each of the lower panel pains.

Fig. 13.58.
It is supported at the wall on the right hand side and by a cable on the left as shown. Deter-
mine (a) the tension in the cable (b) the reaction at the wall and (c) the nature and magnitude of
the force in each bar.

Solution. The example may be solved either by the method of joints or method of sections.
But we shall solve it by the method of joints, as we have to find out the forces in all the members of
the truss.