Virtual Work Principle: Equilibrium Displacements & Forces in Rigid Bodies, Study notes of Engineering

Download Virtual Work Principle: Equilibrium Displacements & Forces in Rigid Bodies and more Study notes Engineering in PDF only on Docsity! Principle of Virtual Work for A Rigid Body rigid body displacements (in 2-D) θ y D x a general displacement of any point in a rigid body consists of a translation plus a rotation. If the translation and rotation are both small then the displacement at any point in a rigid body can be written as d d dθ= + ×r D k r small translation small rotation Thus N N M⎛ ⎞ ( ) 1 1 1 i i i j i i j W Fδ δ δθ δθ = = = = ⋅ + ⋅ × + ⋅⎜ ⎟ ⎝ ⎠ ∑ ∑ ∑D F k r C k but ( ) ( )i i i iδθ δθ⋅ × = ⋅ ×F k r k r F since ( ) ( ) ( )⋅ × = ⋅ × = ⋅ ×a b c b c a c a b so the virtual work done is 1 1 1 N N M i i i j i i j Wδ δ δθ = = = ⎛ ⎞⎛ ⎞ = ⋅ + ⋅ × +⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ∑ ∑ ∑D F k r F C N N M i i i jWδ δ δθ ⎛ ⎞⎛ ⎞ = ⋅ + ⋅ × +⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ∑ ∑ ∑D F k r F C 1 1 1i i j= = = If equilibrium of the rigid body is satisfied then we have N 1 1 1 0 0 i i N M i i j i j = = = = × + = ∑ ∑ ∑ F r F C and, hence δW =0 Thus, if a rigid body is in equilibrium the virtual work done by all the forces and moments acting on the body is zero, or in other words equilibrium δW =0 1 1 1 N N M i i i j i i j Wδ δ δθ = = = ⎛ ⎞⎛ ⎞ = ⋅ + ⋅ × +⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ∑ ∑ ∑D F k r F C However, we can also turn this relationship around say that If the virtual work done by all the forces and couples on a body is zero, for all possible virtual rigid body displacements, then the body must be in equilibrium, i.e. equilibrium δW =0 Principle of Virtual Work for a Rigid Body virtual work approach: T1 T2 5 50 δθ δθ θ δθ 200 50 θ δu=5δθ θ δu=5δθ 200 50 sin 200 cos 0W u uδ δ θ δ θ= − = tan 4θ = note: tensions do no work in this assumed virtual rotation we can find the tensions if we want to by looking at additional virtual translations or rotations: T1 T2 δu δθ δθ3 3 50 200 θ δu=3δθ θ ( ) ( )1 2 1 2 sin sin 0W T u T u T T T δ θ δ θ δ= − + = = = same result as our previous moment equation 50 T1 T2 θ θδu δu δu 200 50 0W T Tδ δ θδ θ same result as first previous force equation 1 2 1 2 cos cos cos cos 50 u u T Tθ θ = − − = + = T1 T2 δ δ 50 200 θ θ u u δu ( ) ( )sin sin 200 0W T u T u uδ θ δ θ δ δ= + − = same result as1 2 1 2sin sin 200T Tθ θ+ = second previous force equation