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Unformatted text preview: EXAMPLE 5.3 Given: The two rigid bars are hinged at the support and at point A. The springs have linear force-
deformation response with spring constants shown below. The deformations are assumed to be small, P Q Lilies. Required: Use the principle of virtual work to determine the displacement of the bars. Solution: Since the bars are rigid, it should be clear that only two independent degrees of freedom
are required to describe the motion of the two bars. We could choose the rotation of
each bar, the vertical displacements at points A and B, or any other set of independent
quantities. Suppose we take the vertical displacements at A and B to be qA and (13,
respectively. The deformed geometry is shown below with equilibrium forces produced
by the linear springs on the right-hand bar. Undeformed p Q
center line We apply independent virtual displacements SqA and Sq”, as shown below. The virtual work is given by + 8 +
SW = 0 = P(aq,,/2) — Qan — K, (‘M 2‘”) (Mg—B) — (K2 6111) 5413 Or, regrouping terms,
SW = O = [—(K1/4)qA - (Kl/4)qB + P/2] BqA
+ [—(K1/4)‘1A ‘ (Kl/4 + K2)‘IB _ Q1543 Since the virtual displacements are arbitrary and independent, the above can be satisfied if and
only if the terms in brackets are simultaneously zero. Thus, we obtain two equations: (Kl/4)qA + (Kl/4)q3 = P/2
(Kl/4)qA + (Kl/4 + K2)‘IB = _ Q K1/4 K1/4 q, _ P/2 b
[(1/4 (Kl/4+K2) qB _ —Q () These two linear equations can be solved for qA and qB to obtain (a) Or, in matrix notation 2P/KI + (Q + P/2)/1<2 (C) — (Q + P/2)/K2 ...
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This note was uploaded on 05/19/2010 for the course MAE 472 taught by Professor Peters during the Spring '08 term at N.C. State.
- Spring '08