Using the principle of virtual work, find reactions R b , R e and R f at the roller supports of the compound beam shown in the figure, if three equal vertical loads Q act as shown in the figure.
SOLUTION:
The principle of virtual work states that - if for each virtual displacement of an ideal system, the work produced by the active forces is zero, then the system is in a configuration of equilibrium.
As shown in the figure, the supports, B, E and F are replaced by their reactive forces R b , R e and R f respectively.
As shown in the above figure, the position of D is moved upward by a small distance ΔS d and hence, the position of E moved by a small distance ofΔS e and the position of M is moved by a small distance ofΔS m .
Now, from similar triangle,
ΔS c /0.9 = ΔS d /1.5
=>ΔS c = 3ΔS d /5
and,
ΔS d /1.2 = ΔS m /0.6
=> ΔS m = ΔS d /2
Now, from principle of virtual work,
Re ΔS c - Q ΔS m = 0
=> R e × 3 ΔS d /5 - Q × Î”S d /2 = 0
=> R e = 5 Q/6
As shown in the figure, Position of C is moved upward by a small distanceΔS c , hence the position of M is moved by ΔS` m and position of B is moved by ΔS b .
Now, from similar triangles,
ΔS’ m /0.6 = ΔS c /1.2
=> ΔS` m = ΔS c /2
and
ΔS b /0.9 = ΔS c /1.5
=> ΔS b = 3 ΔS c /5
From principle of virtual work,
Q× Î”S` m + Q × Î”Sc + Q ×ΔS b = R e ΔS b
=> Q × Î”S c /2 + Q × Î”S c + Q × 3 ΔS c /5 = 3ΔS c /5 ×R e
=> Q/2 + Q + 3Q/5 = 3 R e /5
=> R e = 7 Q/ 2
As shown in the figure, the position of F is moved by a small distance ΔS f , hence the position of E is moved byΔS e.
Now, from similar triangles,
ΔS e /0.6 = ΔS f / 1.5
=> ΔS e = 2 ΔS f /5
From principle of virtual work,
R f × Î”S f =ΔS e × R e
=> R f × Î”Sf = 2 ΔS f /5 × R e
=> R f = 2/5 R e
=> R f = 2/5 × 5/6 Q
=> Rf = Q/3
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