Engineering Mechanics

(^356) „„„„„ A Textbook of Engineering Mechanics
∴ Virtual work done by the effort (P)
= + P × y ...(i)
...(Plus sign due to upward movement of the effort)
and virtual work done by the 1000 N weight
= – 1000 × x = – 1000 × 0.707 y
= – 707 y ...(Q x = 0.707 y) ...(ii)
...(Minus sign due to downward movement of the weight)
We know that from the principle of virtual work, that algebraic sum of the virtual works
done is zero. Therefore
Py – 707 y = 0
or P = 707 N Ans.
Example 16.10. A beam AB of 2 m length is held in equilibrium by the application of a
force P as shown in Fig. 16.20.
Fig. 16.20.
Using the principle of virtual work, find the magnitude of the force P when a weight of 2 kN is
hung from the beam AB at its midpoint.
Solution. Given: Length of beam AB = 2 m; Span AC = 1m; Span CB = 1m and
weight (W ) = 2 kN
Let P = force required to keep the body in equilibrium, and
y = Virtual downward displacement of the effort.
From the geometry of the figure, we find that when the virtual downward displacement of
the force (P) is y, then virtual upward displacement of the beam at B (or length of chord released)
= y
and virtual upward displacement of the load
= 0.5 y
∴ Virtual work done by the load
= + (2 × 0.5 y) = + y ...(i)
...(Plus sign due to upward movement of load)
and virtual work done by the effort
= – Py ...(ii)
...(Minus sign due to downward movement of the effort)
We know that from the principle of virtual work, that algebraic sum of the virtual works
done is zero. Therefore
y – Py=0
or P=y/y = 1 kN Ans.