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



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  F cos a Ds

Thus,
U  Fd
Work

Work done along continuous path, dU  F  dr
U   F  dr  (Fxdx Fy dy  Fz dz)
U   F cosa 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,
 U   F   r  ( Fx i   Fy j  Fz k) (rx i  ry j  rz k)
  Fx rx   Fy ry   Fz rz

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 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 Problem 7/1
Each 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.
Problem 10.2 (Beer book)
Determine the expressions for q and for the
tension in the spring which correspond to the
equilibrium postion of the mechanism. The
unstretched length of the spring is h, and the
con
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 mo,
m, and 2m, respectively.
Potential 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,
x
x
0
0
Ve   Fdx   kx dx
Ve 

For torsional spring
Ve 
1 2
kx
2
1 2
K
2
where K is torsional stiffness
Potential 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
Vg  mgh
Potential Energy

Energy equation in equilibrium condition is
 U   U ' ( Ve   Vg )  0
work from external active forces

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   (Ve  Vg )  0

For a system of 1-dof where the potential energy is continuous
function of the single variable (say, x), the equation becomes
dV
0
dx

For systems with several dof’s, the partial derivative of V with respect
to each coordinate must be zero for equilibrium
dV
0
dxi
Stability of Equilibrium


Three possible conditions for
dV
0
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:
Equilibrium
Stable
Unstable
dV
0
dx
d 2V
0
dx 2
d 2V
0
dx 2
Sample Problem 7/4
The 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 7/5
The 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 7/37
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.