Engineering Mechanics

(^358) „„„„„ A Textbook of Engineering Mechanics
Example 16.12. A square pin-jointed frame ABCD with 500 mm side is subjected to a
force of 100 N at D and 50 N at C as shown in Fig. 16.22. All the pins are smooth and the bar
weights are neglected.
Fig. 16.22.
The force in the bar BD can be adjusted by means of a turnbuckle. Using the method of virtual
work, determine the force in the turnbuckle, when the bar AC carries no load.
Solution. Given : Side of frame = 500 mm; force at D = 100 N and force at C = 50 N
Since the bar AC carries no load, therefore it may
be assumed to be removed. Now let us assume the member
BD (in which the force is required to be found out) to be
removed as shown in Fig. 16.23.
Let T = Magnitude of force in the turnbuckle
(i.e. in bar BD)
y = Virtual vertical displacement of the 100
N load, and
x = Virtual shortening of bar BD.
From the geometry of the figure, we find that
xy= 2
We know that virtual work done by the force in the turbuckle
=+ × =+ ×Tx Ty 2 ...(i)
...(Plus sign due to tension)
and virtual work done by the 100 N load at D
= – 100 × y = – 100 y ...(ii)
...(Minus sign due to downward movement of the load)
We know that from the principle of virtual work, that algebraic sum of the virtual works
done is zero. Therefore
Ty 2 – 100y= 0
or
100
50 2 70.7 N
2
y
T
y
=== Ans.
Note. During the displacement of the members, the forces at A and B do no work. Similarly,
the horizontal force at C does not work in the vertical direction.
Fig. 16.23.