Engineering Mechanics

Chapter 14 : Analysis of Perfect Frames (Graphical Method) „„„„„ 317

14.10. FRAMES WITH BOTH ENDS FIXED

Sometimes, a frame or a truss is fixed or built-in at its both ends. In such a case, the reactions
at both the supports can not be determined, unless some assumption is made. The assumptions usu-
ally made are :

1. The reactions are parallel to the direction of the loads and


2. In case of inclined loads, the horizontal thrust is equally shared by the two reactions.
   Generally, the first assumption is made and the reactions are determined as usual by taking
   moments about one of the supports.

Example 14.21. Figure 14.51 shows as roof truss with both ends fixed. The truss is subjected
to wind loads normal to the main rafter.

Fig. 14.51.
Find the force in various members of the truss.
Solution. The reactions may be obtained by any one assumption as mentioned. With the help
of first assumption the reactions have been found out as shown in Fig. 14.52 (a).

Equating the anticlockwise moments and the clockwise moments about A,

1

22 14 8

8 sin 60 9.24 kN

cos 30 cos 30 0.866

R

××

×°= + ==

°°

∴ 1

9.24 9.24

1.33 kN

8 sin 60 8 0.866

R == =

°×

and R 2 = (1 + 2 + 1) – 1.33 = 2.67 kN

First of all, draw the space diagram and name the members according to Bow’s notations as
shown in Fig. 14.52 (a).

Fig. 14.52.
Now draw the vector diagram as shown in Fig. 14.52 (b). Measuring the various sides of the
vector diagram, the results are tabulated here :