Principle of Virtual Work

MEAM 535

Aristotle Galileo (1594) Bernoulli (1717) Lagrange (1788)

University of Pennsylvania 1 MEAM 535 Virtual Work

Key Ideas (a) Fi z Virtual displacement e2 Small

Consistent with constraints Occurring without passage of time rPi z Applied forces (and moments)

Ignore constraint forces z Static equilibrium e Zero acceleration, or O 1

Zero mass

e3 N (a) Pi n generalized coordinates, qj δW = ∑[]Fi ⋅δr i=1 n N Pi   (a) ∂r  δW = ∑ F ⋅ δq ∑ i ∂q j j=1i=1 j  University of Pennsylvania 2 MEAM 535 Example

l Applied forces G=τ/2r B F acting at P Q r φ θ m F G acting at Q P

Constraint forces ? x

Single degree of freedom Generalized coordinate, θ

Motion of particles P and Q can be described by the generalized coordinate θ

University of Pennsylvania 3 MEAM 535 Static Equilibrium Implies Zero Virtual Work is Done Forces z Forces that do work

Applied Forces

External Forces z Forces that do no work (a) Constraint forces Fi

Static Equilibrium Ri Implies sum of all forces on each particle equals zero

(a) []Fi + Ri = 0

N N (a) (a) ∑[Fi + Ri ]= 0 ∑[]Fi + Ri .δri = 0 i=1 i=1

University of Pennsylvania 4 MEAM 535 The Key Idea

Constraint forces do zero virtual work! 0 N (a) ∑[Fi + Ri ].δri = 0 i=1

Why?

University of Pennsylvania 5 MEAM 535 Constraints: Two Particles Connected by Rigid Massless Rod

(x2 , y2) e F2 F2 R2 F 1 F1 R1 (x1 , y1)

2 2 2 (x1 – x2) +(y1 – y2) = r R1 = -R2 = αe ()x1 − x2 (δx1 − δx2 )+ (y1 − y2 )(δy1 − δy2 )= 0

δx1 δx2  []()x1 − x2 (y1 − y2 )  − []()x1 − x2 (y1 − y2 )  = 0 δy1 δy2 

R1.δr1 + R2.δr2 = αe.δr1 − αe.δr2 e⋅()δr1 − δr2 = 0 = αe.()δr1 − δr2 = 0 University of Pennsylvania 6 MEAM 535 Rigid Body: A System of Particles z A rigid body is a system of infinite particles. z The distance between any pair of particles stays constant through its motion. z Each pair of particles can be considered as connected by a massless, rigid rod. z The internal forces associated with this distance constraint are constraint forces. z The internal forces do no virtual work!

University of Pennsylvania 7 MEAM 535 Contact Constraints and Normal Contact Forces

Rigid body A rolls and slides on rigid body B n Contact Forces

P1 T1 P2 r N2 B 1 N1 r T2 1 O contact C N = -N = αn normal 1 2

A P B P v 2 .n = v 1 .n N .δr + N .δr = αn.δr −αn.δr C P C P 1 1 2 2 1 2 Contact Kinematics v 2 .n = v 1 .n = αn.()δr1 −δr2 δr2 .n = δr1 .n = 0 University of Pennsylvania 8 MEAM 535 Normal and Tangential Contact Forces

1. Normal contact forces t z Normal contact forces are constraint forces T1 z Equivalently, normal forces do no virtual work N2 N1 T2 2. Tangential contact forces z If A rolls on B (equivalently B rolls on A)

A P B P v 2 = v 1

then, tangential contact forces are constraint forces z In general (sliding with friction), tangential forces will contribute to virtual work A P B P T .δr + T .δr = βt.δr −βt.δr v 2 = v 1 1 1 2 2 1 2 = βt.()δr1 − δr2 ≠ 0 University of Pennsylvania 9 MEAM 535 Statement

A system of N particles (P1, P2,…, PN) is in static equilibrium if and only if the virtual work done by all the applied (active) forces though any (arbitrary) virtual displacement is zero. n  N Pi    (a) ∂r  δW = ∑ Fi ⋅ δq j = 0 ∑ ∂q j=1 i=1  j  A holonomic system of N particles is in static equilibrium if and only if all the generalized (active) forces are zero. z Only “applied” or “active” forces contribute to the generalized force z The jth generalized force is given by

N Pi  (a) ∂r  Q j = ∑ Fi ⋅  i=1 ∂q j  Why? N  Pi  (a) ∂r& = ∑ Fi ⋅  i=1 ∂q& j 

University of Pennsylvania 10 MEAM 535 Velocity Partials

In any frame A

Pi Pi n speeds r = r ()q1,q2,K,qn ,t dr Pi A vPi = dt Pi

Pi Pi Pi Pi ∂r ∂r dq1 ∂r dq2 ∂r dqn A = + + + + Pi K A P dr ∂t ∂q1 dt ∂q2 dt ∂qn dt v i = rPi dt Pi A Pi ∂r Pi Pi Pi v = + v1 q&1 + v2 q&2 + + vn q&n ∂t K

a2 a1 Define the jth velocity partial O Pi Pi Pi ∂r ∂r& v j = = A ∂q j ∂q& j a3

N (a) Pi The jth generalized force is Q j = ∑[Fi ⋅ v j ] i=1

University of Pennsylvania 11 MEAM 535 Example Illustrating Partial Velocities Three Degree-of-Freedom Robot Arm

REFERENCE POINT x = l cosθ + l cos θ + θ + l cos θ + θ + θ (x,y) φ 1 1 2 ( 1 2 ) 3 ( 1 2 3 )

y = l1 sin θ1 + l2 sin()θ1 + θ2 + l3 sin(θ1 + θ2 + θ3 ) l3 θ3 φ = ()θ1 + θ2 + θ3

l2 differentiating y θ2 l1

θ1 x x& = −l1θ&1s1 − l2 (θ&1 + θ& 2 )s12 − l3(θ&1 + θ& 2 + θ& 3 )s123 ()() u1 = θ& 1 y& = l1θ&1c1 + l2 θ&1 + θ& 2 c12 + l3 θ&1 + θ& 2 + θ& 3 c123 u = θ& 2 2 φ& = ()θ&1 + θ& 2 + θ& 3

u3 = θ& 3 University of Pennsylvania 12 MEAM 535 Example (continued)

Equations relating the joint velocities and the end effector velocities P P v v2 1 x&=−l1θ&1s1 −l2(θ&1 +θ&2)s12−l3(θ&1 +θ&2 +θ&3)s123 REFERENCE P y=lθ& c +l ()θ& +θ& c +l (θ& +θ& +θ& )c P POINT & 1 1 1 2 1 2 12 3 1 2 3 123 φ v3 (x,y)

l3 θ3 in matrix form

y u1 θ2 x −()l s +l s +l s −(l s +l s )−l s l1 &  1 1 2 12 3 123 2 12 3 123 3 123   =  u2 ()()  θ1 y&  l1c1 +l2c12+l3c123 l2c12+l3c123 l3c123  x u3 P P P v1 v2 v3

The three partial velocities of the point P (omitting leading superscript A) are columns of the “Jacobian” matrix University of Pennsylvania 13 MEAM 535 Example 1

Generalized speed: l G=2τ/r B z u=dθ/dt Q Velocities r φ θ F Rsin()θ + φ m z A v P = − a P 1 cosφ 1

r x z A vQ = − ()− sin θa + cosθa 1 2 1 2

x = r cosθ + l cosφ; r sin θ = l sin φ Generalized Active Forces x& = −r sin θθ& − l sin φφ&; r cosθθ& = l cosφφ&

z F = -Fa1 r cosθ 2τ φ& = θ& z G = ()− sin θa + cosθa l cosφ r 1 2

P Q FRsin(θ + φ) Q = F ⋅ v + G ⋅ v Q1 = τ + z No friction, gravity 1 1 1 cosφ

University of Pennsylvania 14 MEAM 535 Example 2

Assumptions z No friction at the wall homogeneous rod, z Gravity (center of mass is at length 3l Q midpoint, C) z Massless string, OP

h C O θ Q l φ 2l P C

University of Pennsylvania 15 MEAM 535 Equivalent System of Forces

A system of forces acting on a rigid body can be replaced by Fi z A resultant force F r F = ∑Fi i=1 z A moment about a convenient reference Fj point O i F r MO = ∑ri × Fi i=1 C rj ri MO O O C' A couple is a set of forces whose resultant force is zero, but the resultant moment is non zero.

University of Pennsylvania 16 MEAM 535 Resultant Moment Depends on Reference Point

Resultant force is independent of origin (reference point)

F Resultant moment is dependent on the origin

MO r PO MP = MO + ∑r × F i=1 O

O P

University of Pennsylvania 17 MEAM 535 Generalized Forces for Rigid Bodies

Generalized force Velocity partials

Pi A Pi ∂r Pi Pi Pi v = + v1 q&1 + v2 q&2 +K+ vn q&n N ∂t (a) Pi Q j = ∑[]Fi ⋅ v j i=1 But,

A Pi A P A B v = v + ω ×ρi

Velocity partials can be rewritten ρj P ρi Pi n A Pi A Pi ∂r ∂[ v ] Pj v = + ∑ q& j Pi ∂t j=1 ∂q& j Fj ∂rOP ∂()AωB = + ×rOP F r ∂t ∂t i i r j n A P A B ∂[]v + ω ×ρi + ∑ q& j j=1 ∂q& j O University of Pennsylvania 18 MEAM 535 Generalized Forces for Rigid Bodies

Generalized force Velocity partials N A P A B (a) Pi A Pi ∂ v ∂( ω ×ρi ) Q j = ∑[]Fi ⋅ v j v j = + i=1 ∂q& j ∂q& j A P A B = v j + ω j ×ρi Angular velocity partial A B A B ∂( ω ) ω j = ∂q& j ρj P ρi Generalized force can be rewritten P P j i N N Q = [F ⋅ vP ]+ [F ⋅(AωB ×ρ )] Fj j ∑ i j ∑ i j i i=1 i=1

Fi ri N rj A B ∑[()ρi × Fi ⋅ ω j ] i=1

Q = F ⋅ vP + M ⋅ AωB O j j P j University of Pennsylvania 19 MEAM 535 Example

Generalized speed: l r B z u=dθ/dt τ Velocities θ φ m F z A B P ω1 = a3

A P Rsin()θ + φ v1 = − a1 cosφ x

Generalized Active Forces

z -Fa1 FRsin(θ + φ) x = r cosθ + l cosφ; r sin θ = l sin φ Q = τ + z τa 1 3 cosφ x& = −r sin θθ& − l sin φφ&; r cosθθ& = l cosφφ& r cosθ φ& = θ& l cosφ

University of Pennsylvania 20 MEAM 535 Example

Generalized coordinates

z (θ1, θ2)

Mz Generalized speeds dθ j P z (u1, u2) u = j dt Velocity Partials (Fx, Fy) A B1 A B2 A C1 A C2 A P l2 ω j , ω j , v j , v j , v j C2

τ θ2 y 2 Generalized forces

A B1 A B2 A C1 l1 Q j = (τ1a3 − τ2a3 )⋅ ω j + τ2a3 ⋅ ω j − m1ga2 ⋅ v j τ1 C1 A C2 A P A B2 − m2ga2 ⋅ v j + ()Fxa1 + Fya2 ⋅ v j + M za3 ⋅ ω j θ1 x

University of Pennsylvania 21 MEAM 535 Example (continued)

M Generalized forces z A B1 A B2 A C1 Q j = (τ1a3 − τ2a3 )⋅ ω j + τ2a3 ⋅ ω j − m1ga2 ⋅ v j

P A C2 A P A B2 − m2ga2 ⋅ v j + ()Fxa1 + Fya2 ⋅ v j + M za3 ⋅ ω j

(F , F ) l2 x y Velocities C2 A B1 A B2 ω = u1a3, ω = (u1 + u2 )a3, τ2 θ2 y A C1 1 v = 2 l1(− s1a1 + c1a2 )u1 l A C2 1 1 v = l1()− s1a1 + c1a2 u1 + 2 l2 ()− s12a1 + c12a2 (u1 + u2 ) τ C1 1 A P v = l1()− s1a1 + c1a2 u1 + l2 (− s12a1 + c12a2 )(u1 + u2 ) θ1 x Generalized Forces

Q1 = τ1 − Fx (l1s1 + l2s12 )+ Fy (l1c1 + l2c12 ) 1 1 − 2 m1gl1c1 − m2g()l1c1 + 2 l2c12 + M z

1 Q2 = τ2 − Fx (l2s12 )+ Fy (l2c12 )− m2g(2 l2c12 )+ M z

University of Pennsylvania 22 MEAM 535 Conservative Holonomic Systems

All applied forces are conservative  ∂x  Or  i  ∂q j  There exists a scalar potential function Pi (a) ∂r  ∂φ ∂φ ∂φ  ∂yi  such that all applied forces are given by: Fi ⋅ = − ∂q ∂x ∂y ∂z ∂q  j  i i i  j  (a) z Fi =− ∇φ()q1,q2,Kqn ,t  ∂ i    ∂q j 

The virtual work done by the applied forces is: ∂φ Q j = − ∂q j n N Pi   (a) ∂r  δW = ∑ F ⋅ δq ∑ i ∂q j j=1i=1 j  n ∂φ δW = −∑ δq j = − δφ j=1 ∂q j

University of Pennsylvania 23 MEAM 535 Statement

A conservative, holonomic system of N particles (P1, P2,…, PN) is in static equilibrium if and only if the change in potential energy though any (arbitrary) virtual displacement is zero.

δφ = 0

University of Pennsylvania 24