(^236) A Textbook of Engineering Mechanics
12.22.FRAMES WITH ONE END HINGED (OR PIN-JOINTED) AND THE
OTHER SUPPORTED ON ROLLERS AND CARRYING INCLINED LOADS
We have already discussed in Art. 12.20 that the support reaction at the roller supported end
will be normal to the support. And the support reaction at the hinged end will be the resultant of
vertical and horizontal forces. The support reactions for such a frame may be found out by the follow-
ing methods :
- Analytical method. 2. Graphical method
Example 12.15. Fig. 12.30 shows a roof truss hinged at one end and rests on rollers at the
other. It carries wind loads as shown in the figure.
Fig. 12.30.
Determine graphically, or otherwise, the reactions at the two supports.
Solution. Given: Span = 10 m
Let RA= Reaction at A, and
RB= Reaction at B.
We know that as the roof truss is supported on rollers at the right hand support (B), therefore
the reaction at this end will be vertical (because of horizontal support). Moreover, as truss is hinged
at the left support (A) and is also carrying inclined loads, therefore the reaction at this end will be the
resultant of horizontal and vertical forces, and thus will be inclined with the vertical.
The example may be solved either analytically or graphically. But we shall solve it by both
the methods one by one.
Analytical Method
From the geometry of the figure, we find that perpendicular distance between the support A
and the line of action of the load at D.
55
5.8 m
cos 30 0.866===
°
and perpendicular distance between the support A and the line of action of the load at C.
5.8
2.9 m
2==Now equating the anticlockwise moments and clockwise moments about A,
RB × 10 = (2 × 2.9) + (1 × 5.8) = 11.6∴
11.6
1.16 kN
B 10
R == Ans.We know that total wind load
= 1 + 2 + 1 = 4 kN
∴ Horizontal component of the total wind load
= 4 cos 60° = 4 × 0.5 = 2 kN