Engineering Mechanics

Chapter 13 : Analysis of Perfect Frames (Analytical Method) „„„„„ 257

Solution. Since the structure is symmetrical in geometry and loading, therefore reaction at A,

RA=RB = 5 kN

From the geometry of the structure, shown in Fig. 13.18 (a). we find that

3

tan 1.0

3

θ= = or θ = 45°

6

tan 2.0

3

α= = or α = 63.4°
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 only.

First of all, consider the joint A. Let the directions of the forces PAC and PAD be assumed as
shown in Fig 13.18 (a). Resolving the forces horizontally and equating the same,

PAC cos 63.4° = PAD cos 45°

∴

cos 45 0.707

1.58

cos 63.4 0.4477

AD AD

AC AD

PP

PP

°×

===

°

and now resolving the forces vertically and equating the same,

PAC sin 63.4° = 5 + PAD sin 45°

1.58 PAD × 0.8941 = 5 + PAD × 0.707 ...(Q PAC = 1.58 PAD)

∴ 0.7056 PAD=5

5

7.08 kN (Tension)

AD 0.7056

P ==

PAC = 1.58 × PAD = 1.58 × 7.08 = 11.19 kN (Compression)
Now consider the joint D. Let the directions of the forces PCD and PBD be assumed as shown in
Fig. 13.18 (b). Resolving the forces vertically and equating the same,

Fig. 13.18.

PCD = PAD sin 45° + PBD sin 45° = 2 PAD sin 45° ...(QPPBD= AD)

= 2 × 7.08 × 0.707 = 10.0 kN (Tension)

Now tabulate these results as given below :

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

1 AD, DB 7.08 Tension

2 AC, CB 11.19 Compression

3 CD 10.0 Tension