Engineering Mechanics

Chapter 13 : Analysis of Perfect Frames (Analytical Method) „„„„„ 285

and

2.598

tan 1.0392

2.5

DK

FK

β= = = or β = 46.1°

Taking moments about B and equating the same,

RA × 6 = (P × 5) + (2 P × 4) + (P × 2) + (2 P × 1) = 17 P

∴ (^17) 2.83.
A 6
P
R ==P
Let the directions of the various forces be assumed as shown in Fig 13.64. Now resolving the
forces vertically at E and equating the same,
Fig. 13.64.
PED sin 36.6° = P
∴ ED sin 36.6 0.5960 1.68 (Tension)
PP
PP===
°
and now resolving the forces vertically at F and equating the same,
PFD sin 46.1° = 2 P
∴
22
2.78 (Tension)
FD sin 46.1 0.7206
PP
PP===
°
Similarly, resolving the forces vertically at G and equating the same,
PCG sin 46.1° = P
∴ CG sin 46.1 0.7206 1.39 (Tension)
PP
PP===
°
and now resolving the forces vertically at H and equating the same,
PCH sin 36.6° = 2 P
∴
22
3.36 (Tension)
CH sin 36.6 0.5960
PP
PP===
°
From the geometry of the figure, we also find that
∠EDB = ∠ACH = 180° – (36.6° + 60°) = 83.4°
2.24
12.7 4