1. CE 102: EngineeringCE 102: Engineering
Mechanics
Principle of Virtual WorkPrinciple of Virtual Work
2. Work of a ForceWork of a Force
rdF=dU
= work of the force corresponding toF
p g
the displacement rd
dU = F ds cosα
α= 0, dU =+ F ds α=
π
2
, dU = 0α= π , dU = − F ds
2
dU = Wdy
3. Work of a CoupleWork of a Couple
Small displacement of a rigid body:Small displacement of a rigid body:
• translation to A’B’
t ti f B’ b t A’ t B”
ddFdFW
• rotation of B’ about A’ to B”
rdθF=dsF=rdF
rd+rdF+rdF=W
22
211
dθM
4. Principle of Virtual Workp
• Imagine the small virtual displacement of particle which
is acted upon by several forces.
• The corresponding virtual work,
δR
rδF+F+F=rδF+rδF+rδF=δU
321321
rδR
Principle of Virtual Work:
• If a particle is in equilibrium, the total virtual work of forces
acting on the particle is zero for any virtual displacement.
• If a rigid body is in equilibrium, the total virtual work
of external forces acting on the body is zero for any
virtual displacement of the bodyvirtual displacement of the body.
• If a system of connected rigid bodies remains connected
during the virtual displacement, only the work of theg p y
external forces need be considered.
5. Different types of forcesDifferent types of forces
• Forces that do work are called
active force.
• Reactive and internal forces doReactive and internal forces do
not do any work.
• Virtual displacements are to be• Virtual displacements are to be
given carefully so that the
active forces are only theactive forces are only the
known forces and the forces we
are interested in obtainingare interested in obtaining
• Similar to FBD we draw active
f di (AFD)force diagram (AFD).
7. Work of a ForceWork of a ForceForces which do no work:
• reaction at a frictionless pin due to rotation of a bodyp y
around the pin
• reaction at a frictionless surface due to motion of a
body along the surfacebody along the surface
• weight of a body with cg moving horizontally
• friction force on a wheel moving without slippingfriction force on a wheel moving without slipping
Sum of work done by several forces may be zero:
• bodies connected by a frictionless pin
• bodies connected by an inextensible cord
• internal forces holding together parts of a rigid body
8. Degrees of FreedomDegrees of Freedom
• DOF in this context is the total number• DOF in this context is the total number
of independent coordinates required to
specify the complete location of everyspecify the complete location of every
member of the structure.
F VW h d i hi ill• For VW method in this course we will
use only 1-DOF systems.
1 DOF 2 DOF
9. To summarizeTo summarize
P i i l f i t l k• Principle of virtual work
– The virtual work done by external active
forces on an ideal mechanical system inforces on an ideal mechanical system in
equilibrium is zero for any all virtual
displacements consistent with the
constraints.constraints.
• Ideal system:
All surfaces joints etc are frictionless– All surfaces, joints etc. are frictionless.
– We will deal with ideal system in this course.
• Consistent with constraints:Consistent with constraints:
– The virtual displacement should be such that they
should not do allow the non-active forces to do any
kwork.
10. Why principle of Virtual WorkWhy principle of Virtual Work
F l h i ( ill l• For complex mechanisms (we will solve some
problems) we do not need to dismember the
systemsystem.
• We obtain the active unknown force in one
shot without bothering about the reactiveshot without bothering about the reactive
forces.
• Such type of analysis will be a stepping stone• Such type of analysis will be a stepping stone
to VW analysis using deformations when you
teach Solid Mechanics Structural Mechanicsteach Solid Mechanics, Structural Mechanics
etc. not to mention powerful Approximate
methods like the Finite Element Method.
11.
12.
13.
14. Problem 1Problem 1
• Assuming frictionless contacts,
determine the magnitude of P forg
equilibrium
This problem will beThis problem will be
referred to as the
Ladder problem
15.
16. Problem 2Problem 2
• The pressure p driving a piston of diameter
100 mm is 1 N/mm2. At the configuration
shown, what weight W will the system hold if
friction is neglected
This along with the ladderg
problem forms a framework
for many other problems
17.
18. Problem 3
(Shames)
• A hydraulic lift platform for loading trucks supports a weight 2W.
Only one side of the system has been shown; the other side is
id ti l Wh t f F h ld th h d li li d id tidentical. What force F should the hydraulic cylinder provide to
support the weight W. Assume that the hydraulic device is
inclined at a clockwise of α with respect to the horizontal. Neglect
f i i hfriction everywhere.
19.
20. Problem 4 MK5Problem 4 MK5
• A power operated loading platform designed for the back of a truck is• A power-operated loading platform designed for the back of a truck is
shown. The position of the platform is controlled by the hydraulic
cylinder, which applies force at C. The links are pivoted to the truck
frame at A, B, and F. Determine the force P supplied by the cylinder in
d t t th l tf i th iti h Th f thorder to support the platform in the position shown. The mass of the
platform and the links may be neglected compared with that of the 250
kg crate with center of mass at G.
21.
22. Problem 5 BJ8Problem 5 BJ8
• A f P i li d t lid C h Th t t f th• A force P is applied to slider C as shown. The constant of the
spring is 1.6 kN/m, and the spring is un-stretched when member
BD is horizontal. Neglecting friction between the slider and the
guide rod and knowing that BC = BD = 150 mm, determine theg g ,
magnitude of P so that θ = 25 deg when the system is in
equilibrium.
23. A small modifiction:A small modifiction:
I have used Q for the
resisting force and P
for the spring force.p g
24. Problem 6Problem 6
D t i Q f ilib i f th t• Determine Q for equilibrium for the system
shown. The pulleys are frictionless and have
masses W1 and W2. The sliding body has massmasses W1 and W2. The sliding body has mass
W3.
25. Problem 7Problem 7
• Find the force Q required to maintain
the system in equilibrium.y q
26.
27.
28. Problem 8Problem 8
• Determine the couple M which must bep
applied at O in order to support the
mechanism in the position θ = 30o. The
f th di k t C b OA d b BCmasses of the disk at C, bar OA, and bar BC
are mo, m, and 2m respectively.
29.
30. Problem 9Problem 9
• Determine force in member CD by usingDetermine force in member CD by using
the method of virtual work.
33. Problem 1
BJ8
Th b li k h i d b• The two-bar linkage shown is supported by a
pin and bracket at B and a collar at D that
slides freely on a vertical rod Determine theslides freely on a vertical rod. Determine the
force P required to maintain the equilibrium of
the linkage.g
38. Problem 3 MK5Problem 3 MK5
• Th l ti f th l tf f t d b th f• The elevation of the platform of mass m supported by the four
identical links is controlled by the hydraulic cylinders AB and AC
which are pivoted at point A. Determine the compression P in
each of the cylinders required to support the platform for they q pp p
specified angle θ
Meriam and KraigeMeriam and Kraige
39.
40. Problem 4
MK2
• Obtain the clamping force Q developed for
the pliers when the handle force is Pthe pliers when the handle force is P.
41.
42.
43.
44. Problem 5 MK5Problem 5 MK5
• The claw of the remote action actuator• The claw of the remote-action actuator
develops a clamping force C as a result of
the tension P in the control rod Express Cthe tension P in the control rod. Express C
in terms of P for the configuration shown
where the jaws are parallel.where the jaws are parallel.
45.
46. Problem 6 ShamesProblem 6 Shames
• Find the force delivered
at C in the horizontal
direction to crush thedirection to crush the
rock. Pressure p1= 100
M-Pa and p2= 60 M-Pa
( d b(measured above
atmospheric pressure).
The diameters ofThe diameters of
pistons are 100mm
each. Neglect the
weight of the rodsweight of the rods.
47.
48.
49. Problem 7 BJ10Problem 7 BJ10
D i h i l f• Determine the vertical movement of
joint C, if the member FG is lengthened
by 50 mm
