11.2 Principle of Virtual Work for a Particle and a Rigid Body

11.2 Principle of Virtual Work for a Particle and a Rigid Body

Particle� If the particle undergoes an imaginary or

virtual displacement, then virtual work done by the force system becomesδU = ∑F.δrδU = ∑F.δr

= (∑Fxi + ∑Fyj + ∑Fzk).(δxi + δyj + δzk)= ∑Fx δx + ∑Fy δy + ∑Fz δz

� For equilibrium, ∑Fx = 0, ∑Fy = 0, ∑Fz = 0

� Thus, virtual work, δU = 0

11.2 Principle of Virtual Work for a Particle and a Rigid Body

11.2 Principle of Virtual Work for a Particle and a Rigid Body

Example� Consider the FBD of the ball which

rests on the floor� Imagine the ball to be

displacement downwards a virtual displacement downwards a virtual amount δy and weight does positive virtual work W δy and normal force does negative virtual work -N δy

� For equilibrium, δU = Wδy –Nδy = (W-N)δy =0

� Since δy ≠ 0, then N = W

11.2 Principle of Virtual Work for a Particle and a Rigid Body

11.2 Principle of Virtual Work for a Particle and a Rigid Body

Rigid Body� A similar set of virtual work equations can be

written for a rigid body subjected to a coplanar force system

� If these equations involve separate virtual � If these equations involve separate virtual translations in the x and y directions and a virtual rotation about an axis perpendicular to the x-y plane and passing through an arbitrary point O, it can be shown that

∑Fx = 0; ∑Fy = 0; ∑MO=0� Not necessary to include work done by internal

forces acting within the body

11.2 Principle of Virtual Work for a Particle and a Rigid Body

11.2 Principle of Virtual Work for a Particle and a Rigid Body

� Consider simply supported beam, with a given rotation about point B

� Only forces that do work are P and Ay

� Since δy = lδθ and δy’ = (l/2)δθ, virtual work

δU = A (lδθ) – P(l/2)δθ = (A – P/2)l δθ = 0δU = Ay(lδθ) – P(l/2)δθ = (Ay – P/2)l δθ = 0

� Since δθ ≠ 0, Ay = P/2

� Excluding δθ, terms in parentheses represent moment equilibrium about B