Chapter 13 : Analysis of Perfect Frames (Analytical Method) 263
From the geometry of the figure, we find that
1.5
tan 0.3333
4.5∠==CDE or ∠CDE = 18.4°Resolving the forces vertically at D
PCD sin ∠CDE = 500 or PCD sin 18.4° = 500∴500 500
1584 N (Tension)
CD sin 18.4 0.3156
P ===
°and now resolving the forces horizontally at D
PDE = PCD cos ∠CDE = 1584 cos 18.4°
∴ PDE = 1584 × 0.9488 = 1503 N (Compression)
Now consider the joint E. A little consideration will show that the value of the force PFE
will be equal to the force PED i.e., 1503 N (Compression). Since the vertical components of the
forces PFE and PED are zero, therefore the value of the force PCE will also be zero.
Fig. 13.31.
Now consider the joint C. Let the directions of PBC and PFC be assumed as shown in Fig.
13.31 (a). From the geometry of the figure, we find that the members CD, BC and FC make angle of
18.4° with the horizontal. Resolving the forces horizontally and equating the same,
PBC cos 18.4° = 1584 cos 18.4° + PFC cos 18.4°
or PBC = 1584 + PFC ...(i)and now resolving the forces vertically and equating the same,
1000 + 1584 sin 18.4° =PFC sin 18.4° + PBC sin 18.4°
1000 + (1584 × 0.3156) = (PFC × 0.3156) + (PBC × 0.3156)
1000 + (1581 × 0.3156) = 0.3156 PFC + (1584 + PFC) × 0.3156
...(Q PBC = 1584 + PFC)
1000 + (1581 × 0.3156) = 0.3156 PFC + (1584 × 0.3156) + 0.3156 PFC∴1000
1584 N (Compression)
FC 0.6312
P ==Substituting the value of PFC in equation (i)
PBC = 1584 + 1584 = 3168 N (Tension)
Now consider the joint F. Let the directions of the forces PGF and PFB be assumed as shown in
Fig 13.31 (b). Resolving the forces horizontally,
PGF = 1584 + 1584 cos 18.4° = 1584 + (1584 × 0.9488) N
= 1584 + 1503 = 3087 N (Compression)