Engineering Mechanics: Statics Chapter 7: Virtual Work
Introduction 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.
Work In Physics, a force is said to do work when it acts on a body, and there is a displacement of the point of application in the direction of the force. A rigid body is sometimes constrained not to travel in the direction of the force. The block is moved a distance Ds due to the externally applied force F, which inclines at an angle a to the horizontal The work done from (a) and (b) are the same, U Fcosa Ds Thus, U F d
Work Work done along continuous path, U = du = F dr ò F dr = ò (F x dx + F y dy + F z dz) U Fcosa ds Work of a couple, du Md and U Md Dimensions: joules, J (=N.m) and lb.ft
Equilibrium - particle For a particle in static equilibrium position, any assumed and arbitrary small displacement r 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 r is called virtual work. U F r or U M When there are several forces, du = å F dr =( åf x i + åf y j + åf z k) (r x i + r y j + r z k) = åf x r x + åf y r y + åf z r z For equilibrium, SF = 0. Therefore, U 0
Equilibrium rigid body 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 U 0 Since internal forces occur in pairs (which are equal, opposite and collinear), the net work done by these forces is also zero. Example: write the equation of virtual work for this problem.
Equilibrium systems of rigid bodies Three 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!
Principle of Virtual Work Principle of virtual work states that: 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 U 0 Number of Degrees of Freedom = number of independent coordinates needed to specify completely the configuration of the system.
Procedure for Analysis 1. Draw the active-force diagram and define the dof 2. Sketch the deflected position of the system when the system undergoes a positive virtual displacement 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 and differentiate these expressions to find s in terms of 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 7. Factor out the common terms and solve the equation.
Sample Problem Determine the angle for equilibrium of the twomember linkage shown. Each member has a mass of 10 kg.
Sample Problem 7/2 The 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 7/19 Determine 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 m o, m, and 2m, respectively.
Problem 10.2 (Beer book) Determine the expressions for and for the tension in the spring which correspond to the equilibrium position of the mechanism. The unstretched length of the spring is h, and the constant of the spring is k. Neglect the weight of the mechanism.
Potential Energy Previous article members are assumed perfectly rigid. The concept of potential energy is useful for elastic elements (ex. spring) Elastic potential energy V e = 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 e x V Fdx kx dx x 0 0 Ve Ve 1 2 kx 2 1 2 K 2 where K is torsional stiffness
Potential Energy Gravitational potential energy V g V g 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 Vg mgh
Potential Energy Energy equation in equilibrium condition is U U' ( V V ) 0 work from external active forces e g Therefore, U' V The sum of the work done by spring and gravitational forces is equal to the work done by all other active forces. important: for this method, spring is internal to the system and weight forces are not shown!
Stability of Equilibrium When there is no work done by external force U' 0, the requirement for equilibrium is V ( V V ) 0 e g For a system of 1-dof where the potential energy is continuous function of the single variable (say, x), the equation becomes dv dx 0 For systems with several dof s, the partial derivative of V with respect to each coordinate must be zero for equilibrium dv 0 dx i
Sample Problem 7/5 The two uniform links, each of mass m, are in the vertical plane and are connected and 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 7/37 The bar of mass m with centre of mass at G is hinged about a horizontal axis through O. Determine the two positions of equilibrium for the bar.
Stability of Equilibrium Three possible conditions for dv dx V = minimum Stable equilibrium V = maximum Unstable equilibrium V = constant neutral equilibrium Therefore, for a single degree of freedom x, the conditions for equilibrium and stability are: 0 dv Equilibrium 0 dx 2 dv Stable 0 2 dx 2 dv Unstable 0 2 dx
Problem 7/37 The 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 7/36 The 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 7/6 The 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.