(^314) A Textbook of Engineering Mechanics
*Solution. Since the truss is supported on rollers at P, threfore the reaction at this end will be
vertical (because of horizontal support). Moreover, it is hinged at Q, therefore the reaction at this end
will be resultant of horizontal and vertical forces and inclined with the vertical.
Taking moments about Q and equating the same,
VP × 6.92 = (20 × 3) + (10 × 6) = 120
∴^120 17.3 kN ( ) and [(10 20 10) sin 60 ] – 17.3 17.3 kN ( )
PQ6.92
VV== ↑ =++ ° = ↑
and HQ = (10 + 20 + 10) cos 60° = 40 × 0.5 = 20 kN (→)
First of all, draw the space diagram and name the members and forces according to Bow’s
notations as shown in Fig. 14.46 (a).
Fig. 14.46.
Now draw the vector diagram as shown in Fig. 14.46 (b). Measuring the various sides of the
vector diagram, the results are tabulated here :
S.No. Member Magnitude of force in kN Nature of force
1 PR (1-7) 20.0 Compression
2 RS (2-8) 17.3 Compression
3 SQ (3-9) 17.3 Compression
4 QT (6-9) 30.0 Tension
5 PT (6-7) 10.0 Tension
6 RT (7-8) 20.0 Tension
7 ST (8-9) 20.0 Compression
Example 14.19. Figure 14.47 shows a truss pin-jointed at one end, and freely supported at
the other. It carries loads as shown in the figure.
Fig. 14.47
Determine forces in all the members of the truss and state their nature.
- We have already solved this example analytically in the last chapter.