Chapter 14 : Analysis of Perfect Frames (Graphical Method) 309
- Find the forces in all the members of a cantilever truss shown in Fig. 14.36.
Ans. BF = 8.4 kN (Tension)
FC = 6.7 kN (Compression)
EF = 0
AD = 12.6 kN (Tension)
DE = 4.3 kN (Compression)
EC = 6.7 kN (Tension) - Find graphically or otherwise the forces in the members 2, 5, 9 and 10 of the truss shown in
Fig 14.37. Also state whether they are in tension or compression.
Ans. 2 = 6.0 kN
5 = 3.55 kN
9 = 2.0 kN
10 = 0- Find the forces in the members of the frame given in Fig. 14.38.
Ans. 1-2 = 12.0 kN (Tension)
2-3 = 6.0 kN (Tension)
3-4 = 2.0 kN (Tension)
4-5 = 2.8 kN (Compression)
5-6 = 2.0 kN (Compression)
6-7 = 6.0 kN (Compression)
2-7 = 8.5 kN (Compression)
2-6 = 4.0 kN (Tension)
3-6 = 5.6 kN (Compression)
3-5 = 2.0 kN (Tension)
14.8.STRUCTURES WITH ONE END HINGED (OR PIN-JOINTED) AND THE OTHER
FREELY SUPPORTED ON ROLLERS AND CARRYING HORIZONTAL LOADS
We have already discussed in Art 14.16 that sometimes a structure is hinged or pin-jointed at
one end and freely supported on rollers at the others end. If such a structure carries vertical loads only,
the problem does not present any special features. Such a problem may be solved just as a simply
supported structure.
But, if such a structure carries horizontal loads (with or without vertical loads) the support
reaction at the roller supported end will be normal to the support; whereas the support reaction at the
hinged end will consist of :
- Vertical reaction, which may be found out by subtracting the vertical support reaction at
the roller supported end from the total vertical load. - Horizontal reaction, which may be found out by algebraically adding all the horizontal
loads. After finding out the reactions, the space and vector diagram may be drawn as
usual.
Fig. 14.37.Fig. 14.36.Fig. 14.38.