Engineering Mechanics: Engineering Mechanics: Statics Statics

Chapter 7: Virtual Work

IntroductionIntroduction

Previous chapters-- FBD & zero-force and zero-moment equations

-- Suitable when equilibrium position is known

For bodies composed of interconnected members, various equilibrium position is possible.

Using equilibrium equations is valid and adequate but the method based on concept of work done by a force is more direct.

WorkWork A rigid body is sometimes constrained not to travel in the

direction of the force.

The block is moved a distance s due to the externally applied force F, which inclines at an angle to the horizontal

The work done from (a) and (b) are the same,

Thus,

cosU F s

U F d

WorkWork

dU F dr

( )x y zU F dr Fdx Fdy Fdz

cosU F ds

dU Md

U Md

Work done along continuous path,

Work of a couple,

and

Dimensions: joules, J (=N.m) and lb.ft

For a particle in static equilibrium position, any assumed and arbitrary small displacement away from this position and consistent with the system constraints is called a virtual displacement.

Work done by any force F acting on the particle during the virtual displacement is called “virtual work”.

or

When there are several forces,

For equilibrium, F = 0. Therefore,

Equilibrium - particleEquilibrium - particle

r

r

U F r U M

( ) ( )x y z x y zU F r F i F j Fk r i r j r k

x x y y z zFr F r Fr

0U

Example: write the equation of virtual work for this problem.

Equilibrium – rigid bodyEquilibrium – rigid body

0U

A rigid body consists of infinite numbers of particles.

Virtual work done on each particle in equilibrium is zero.

Only virtual work done by external forces appears in evaluation of

Since internal forces occur in pairs (which are equal, opposite and collinear), the net work done by these forces is also zero.

Equilibrium – systems of rigid Equilibrium – systems of rigid bodiesbodiesThree types of forces used in the analysis of virtual work

Active forces = external forces capable of doing virtual work

Reactive forces = reactions on supports (no displacement – no work done)

Internal forces = compression or tension forces that the members carry (Always appear in pairs – net work done is zero)

- In virtual work method, active force diagram is used instead of free body diagram!

0U

Principle of Virtual WorkPrinciple of Virtual Work

The virtual work done by external active forces on an ideal mechanical system in equilibrium is zero for any and all virtual displacements consistent with the constraints

Number of Degrees of Freedom = number of independent coordinates needed to specify completely the configuration of the system.

Principle of virtual work states that:

Procedure for AnalysisProcedure for Analysis1. Draw the active-force diagram and define the coordinate q (the dof)

2. Sketch the deflected position of the system when the system undergoes a positive virtual displacement q

3. Indicate position coordinate, s measured from a fixed point directed to the forces that do work and parallel to the force.

4. Relate each position coordinate s to the coordinate q and differentiate these expressions to find s in terms of q

5. Write virtual-work equation assuming that (possible or not) all s are positive. If a force is in the same direction as +s, the work is positive. Otherwise, it is negative.

6. Express the work equation in terms of q

7. Factor out the common terms and solve the equation.

Sample ProblemSample ProblemDetermine the angle for equilibrium of the two-member linkage shown. Each member has a mass of 10 kg.

Sample Problem 7/1Sample Problem 7/1Each of the two uniform hinged bars has a mass m and a length l, and is supported and loaded as shown. For a given force P determine the angle for equilibrium.

Sample Problem 7/2Sample Problem 7/2The mass m is brought to an equilibrium position by the application of the couple M to the end of one of the two parallel links which are hinged as shown. The links have negligible mass, and all friction is assumed to be absent. Determine the expression for the equilibrium angle assumed by the links with the vertical for a given value of M.

Problem 7/19Problem 7/19Determine the couple M which must be applied at O in order to support the mechanism in the position = 30 . The masses of the disk at C, bar OA, and bar BC are mo, m, and 2m, respectively.

Potential EnergyPotential Energy

Previous article – members are assumed perfectly rigid.

The concept of potential energy is useful for elastic elements (ex. spring)

Elastic potential energy Ve= work done stored in the member

For a spring of linear stiffness k, the force F required to compress the spring by a distance x, is F = kx

Work done by F during a movement dx is dU = Fdx

Thus,

For torsional spring where K is torsional stiffness

0 0

x x

eV Fdx kx dx

212eV kx

212eV K

Potential EnergyPotential Energy Gravitational potential energy Vg

Vg is work done on the body by a force (equal and opposite to the weight) to bring the body to the position under consideration from some arbitrary datum

positive if the position is higher than the datum

negative if the position is lower than the datum

gV mgh

Energy equation in equilibrium condition is

Therefore,

important: for this method, spring is internal to the system and weight forces are not shown!

The sum of the work done by spring and gravitational forces is equal to the work done by all other active forces.

Potential EnergyPotential Energy

' ( ) 0e gU U V V

'U V

work from external active forces

Stability of EquilibriumStability of Equilibrium When there is no work done by external force , the

requirement for equilibrium is

For a system of 1-dof where the potential energy is continuous function of the single variable (say, x), the equation becomes

For systems with several dof’s, the partial derivative of V with respect to each coordinate must be zero for equilibrium

' 0U

0dVdx

( ) 0e gV V V

0i

dVdx

Stability of EquilibriumStability of Equilibrium Three possible conditions for

Therefore, for a single degree of freedom x, the conditions for equilibrium and stability are:

0dVdx

2

2

2

2

Equilibrium 0

Stable 0

Unstable 0

dVdxd Vdxd Vdx

V = minimum – Stable equilibrium

V = maximum – Unstable equilibrium

V = constant – neutral equilibrium

Sample Problem 7/4Sample Problem 7/4The 10-kg cylinder is suspended by the spring, which has a stiffness of 2 kN/m. Plot the potential energy V of the system and show that it is minimum at the equilibrium position.

Sample Problem 7/5Sample Problem 7/5The two uniform links, each of mass m, are in the vertical plane and are connected an constrained as shown. As the angle between the links increases with the application of the horizontal force P, the light rod, which is connected at A and passes through a pivoted collar at B, compresses the spring of stiffness k. If the spring is uncompressed in the position where = 0, determine the force P which will produce equilibrium at the angle .

Problem 7/37Problem 7/37The bar of mass m with centre of mass at G is hinged about a horizontal axis through O. Prove the stability conditions for the two positions of equilibrium.

Problem 7/36Problem 7/36The uniform bar of mass m and length L is supported in the vertical plane by two identical springs each of stiffness k and compressed a distance in the vertical position = 0. Determine the minimum stiffness k which will ensure a stable equilibrium position with = 0. The springs may be assumed to act in the horizontal direction during small angular motion of the bar.

Sample Problem 7/6Sample Problem 7/6The ends of the uniform bar of mass m slide freely in the horizontal and vertical guides. Examine the stability conditions for the positions of equilibrium. The spring of stiffness k is undeformed when x = 0.