Principle of Virtual Work Aristotle Galileo (1594) Bernoulli - PowerPoint PPT Presentation

MEAM 535 Principle of Virtual Work Aristotle Galileo (1594) Bernoulli (1717) Lagrange (1788) 1. Start with static equilibrium of holonomic system of N particles 2. Extend to rigid bodies 3. Incorporate inertial forces for dynamic analysis 4. Apply to nonholonomic systems University of Pennsylvania 1

MEAM 535 Virtual Work Key Ideas F i ( a )  Virtual displacement e 2  Small  Consistent with constraints  Occurring without passage of time r Pi  Applied forces (and moments)  Ignore constraint forces  Static equilibrium e 1 O  Zero acceleration, or  Zero mass Every point, P i , is subject to a virtual displacement: . The virtual work is the work done by e 3 the applied forces. n generalized coordinates, q j N ( a ) ⋅ δ r P i [ ] ∑ δ W = F i i = 1 University of Pennsylvania 2

MEAM 535 Example: Particle in a slot cut into a rotating disk  Angular velocity Ω constant  Particle P constrained to be in a radial slot on the rotating disk P F r How do describe virtual b Ω 2 b displacements of the particle P ? 1 O  No. of degrees of freedom in A ? B  Generalized coordinates?  Velocity of P in A ? a 2 What is the virtual work done by the force a 1 F = F 1 b 1 + F 2 b 2 ? University of Pennsylvania 3

MEAM 535 Example l Applied forces G = τ /2 r B F acting at P Q r φ θ F m G acting at Q P (assume no gravity) x Constraint forces All joint reaction forces Single degree of freedom Generalized coordinate, θ Motion of particles P and Q can be described by the generalized coordinate θ University of Pennsylvania 4

MEAM 535 Static Equilibrium Implies Zero Virtual Work is Done Forces  Forces that do work  Applied Forces or External Forces  Forces that do no work F i ( a )  Constraint forces Static Equilibrium R i Implies sum of all forces on each particle equals zero N ( a ) + R i ∑ [ ] F i . δ r i = 0 i = 1 University of Pennsylvania 5

MEAM 535 The Key Idea Constraint forces do zero virtual work! 0 N ( a ) + R i [ ] ∑ F i . δ r i = 0 i = 1 N ∑ ( a ) δ W = F i . δ r i = 0 i = 1 University of Pennsylvania 6

MEAM 535 Principle of Virtual Work If a system of N particles ( P 1 , P 2 ,…, P N ) is in static equilibrium, the virtual work done by all the applied (active) forces though any (arbitrary) virtual displacement is zero. Converse: If the virtual work done by all the applied (active) forces on a system of N particles though any (arbitrary) virtual displacement is zero, the system is in static equilibrium. Proof : Assume the system is not in static equilibrium. ( a ) + R i [ ] ≠ 0 for some P i F i Can always find virtual displacement so that University of Pennsylvania 7

MEAM 535 Constraints: Two Particles Connected by Rigid Massless Rod ( x 2 , y 2 ) e F 2 F 2 R 2 F 1 F 1 R 1 ( x 1 , y 1 ) ( x 1 – x 2 ) 2 +( y 1 – y 2 ) 2 = r 2 R 1 = - R 2 = α e University of Pennsylvania 8

MEAM 535 Rigid Body: A System of Particles  A rigid body is a system of infinite particles.  The distance between any pair of particles stays constant through its motion.  Each pair of particles can be considered as connected by a massless, rigid rod.  The internal forces associated with this distance constraint are constraint forces.  The internal forces do no virtual work! University of Pennsylvania 9

MEAM 535 Contact Constraints and Normal Contact Forces Rigid body A rolls and/or slides on rigid Contact Forces body B n P 1 T 1 P 2 N 2 r 1 N 1 B T 2 r 2 O contact C N 1 = - N 2 = α n normal T 1 = - T 2 = β t Contact Kinematics A v P 2 . n = B v P 1 . n P 1 and P 2 are in C v P 2 . n = C v P 1 . n contact implies: δ r 2 . n = δ r 1 . n University of Pennsylvania 10

MEAM 535 Normal and Tangential Contact Forces A 1. Normal contact forces t P 1  Normal contact forces are constraint forces T 1  Equivalently, normal forces do no virtual work N 2 N 1 T 2 2. Tangential contact forces P 2  If A rolls on B (equivalently B rolls on A ) A v P 2 = B v P 1 , C v P 2 = C v P 1 B then, tangential contact forces are constraint forces  In general (sliding with friction), tangential forces will contribute to virtual work A v P 2 = B v P 1 University of Pennsylvania 11

MEAM 535 Principle of Virtual Work for Holonomic Systems A system of N particles ( P 1 , P 2 ,…, P N ) 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. A holonomic system of N particles is in static equilibrium if and only if all the generalized (active) forces are zero.  Only “applied” or “active” forces contribute to the generalized force  The jth generalized force is given by Why? University of Pennsylvania 12

MEAM 535 Principle of Virtual Work vs. Traditional Approach Traditional Principle of Virtual Work P 2 1. Generalized Coordinates F 1 2. Identify forces that do work F 2 3. Analyze motion of points at P 1 which forces act T Free Body Diagram 4. Calculate generalized forces F 2 F 1 T 2 equations for each particle University of Pennsylvania 13

MEAM 535 Partial Velocities In any frame A n speeds P i r Pi a 1 a 2 Define the jth partial velocity of P i O A a 3 N The jth generalized force is ( a ) ⋅ v j [ ] ∑ P i Q j = F i i = 1 University of Pennsylvania 14

MEAM 535 Example Illustrating Partial Velocities Three Degree-of-Freedom Robot Arm differentiating University of Pennsylvania 15

MEAM 535 Example (continued) Equations relating the joint velocities and the end effector velocities 1 ˙ 1 − l 2 ˙ 1 + ˙ 12 − l 3 ˙ 1 + ˙ 2 + ˙ ( ) s ( ) s ˙ x = − l 1 s θ θ θ θ θ θ 2 3 123 1 ˙ 1 c 1 + l 2 ˙ 1 + ˙ ) c 12 + l 3 ˙ 1 + ˙ 2 + ˙ ( ( ) c 123 P ˙ y = l θ θ θ θ θ θ 2 3 in matrix form   u 1   ( ) ( )  x ˙  = − l 1 s  1 + l 2 s 12 + l 3 s − l 2 s 12 + l 3 s − l 3 s   123 123 123 u 2      ( ) ( ) y ˙ l 1 c 1 + l 2 c 12 + l 3 c 123 l 2 c 12 + l 3 c 123 l 3 c 123       u 3   The three partial velocities of the point P (omitting leading superscript A ) are columns of the “Jacobian” matrix University of Pennsylvania 16

MEAM 535 Example 1 l Generalized speed: G =2 =2 τ / r B  u=d θ /d t Q r φ F θ m Generalized Active Forces P  F = - F a 1 x  No friction, gravity  Kinematic Analysis Need Partial Velocities University of Pennsylvania 17

MEAM 535 Example 1 (continued) l Generalized speed: G =2 =2 τ / r B  u=d θ /d t Q r φ Velocities F θ m  P x  Generalized Active Forces  F = - F a 1  The system is in static equilibrium if and only if Q 1 =0 No friction, gravity  University of Pennsylvania 18

MEAM 535 Example 2 Assumptions  No friction at the wall homogeneous rod,  Gravity (center of mass is at Q length 3 l midpoint, C )  Massless string, OP h C Q O θ φ l 2 l P C P University of Pennsylvania 19

MEAM 535 Example 3 A solid circular cylinder B of mass M rolls down a fixed plane wedge A without slipping. The mass m is connected to a massless string which passes over a pulley C and wraps around a massless spool attached to the cylinder at the other end. (a) Determine the relationship between m and M for static equilibrium. (b) Find the tensions in the string on either side of the pulley C . University of Pennsylvania 20

MEAM 535 Conservative Holonomic Systems All applied forces are conservative Or There exists a scalar potential function such that all applied forces are given by: The virtual work done by the applied forces is: University of Pennsylvania 21

MEAM 535 Statement A conservative, holonomic system of N particles ( P 1 , P 2 ,…, P N ) is in static equilibrium if and only if the change in potential energy though any (arbitrary) virtual displacement is zero. University of Pennsylvania 22

MEAM 535 HW Problem What are the generalized forces? z q 2 a 3 y b 2 a 2 a 1 e 2 C b 3 q 4 C* q 3 q 1 x b 1 P q 5 e 3 e 1 University of Pennsylvania 23