6161103 11.3 principle of virtual work for a system of connected rigid bodies

11.3 Principle of Virtual Work for a System of Connected Rigid Bodies11.3 Principle of Virtual Work for a System of Connected Rigid Bodies

� Method of virtual work most suited for solving equilibrium problems involving a system of several connected rigid bodies

� Before applying the principle of virtual work to � Before applying the principle of virtual work to the systems, specify the number of degrees of freedom for the system and establish the coordinates that define the position of the system


Degrees of Freedom� A system of connected bodies takes on a unique

shape that can be specified provided the position of a number of specific points on the system is knownknown

� Positions are defined using independent coordinates q, measured from fixed reference points

� For every coordinate established, the system will have a degree of freedom for displacement along the coordinate axis such that it is consistent with the constraining action of supports


Degrees of Freedom

� An n degree of freedom system requires n independent coordinates to specify all the locations of all its members

ExampleExample

� Consider one degree of freedom system

where independent coordinate

q = θ is used to specify

location of two connecting

links and the block


Degrees of Freedom

� Coordinate x could be used as the independent coordinate

� Since the block is constrained to move � Since the block is constrained to move within the slot, x is not independent of θ; rather, it can be related to θ using the cosine law b2 = a2 + x2 -2axcos θ


Degrees of Freedom

� Consider double link arrangement as a 2 degrees of freedom system

� To specify the location of

each link, the coordinate each link, the coordinate

angles θ1 and θ2 must be

known since a rotation of

one link is independent of

a rotation of the other


Principles of Virtual Work

� A system of connected rigid bodies is in equilibrium provided that the virtual work done by all the external forces and couples acting on the system is zero for each done by all the external forces and couples acting on the system is zero for each independent displacement of the system

δU = 0

� For a system with n degrees of freedom, it takes n independent coordinates to completely specify the location of the system


Principles of Virtual Work

� For the system, it is possible to write n independent virtual work equations, one for every virtual displacement taken along each of the independent coordinate axes while the every virtual displacement taken along each of the independent coordinate axes while the remaining n-1 independent coordinates are held fixed


Procedure for Analysis

Free Body Diagram

� Draw the FBD of the entire system of connected bodies and sketch the connected bodies and sketch the independent coordinate q

� Sketch the deflected position of the system on the FBD when the system undergoes a positive virtual displacement δq


Procedure for AnalysisVirtual Displacements� Indicate position coordinates si, measured from a

fixed point on the FBD to each i number of active forces and couplesforces and couples

� Each coordinate system should be parallel to line of action of the active force to which it is directed, to calculate the virtual work along the coordinate axis

� Relate each of the position coordinates si to the independent coordinate q, then differentiate for virtual displacements δsi in terms of δq


Procedure for AnalysisVirtual Work Equation� Write the virtual-work equation for the system

assuming that all the position coordinates siundergo positive virtual displacement δsiundergo positive virtual displacement δsi

� Using the relations for δsi, express the work of each active force and couple in the equation in terms of the single independent virtual displacement δq

� Factor out this common displacement from all the terms and solve for unknown force, couple, or equilibrium position, q


Procedure for Analysis

Virtual Work Equation

� If the system contains n degrees of freedom, n independent coordinates qn must be specified

Follow the above procedure and let only one of n

� Follow the above procedure and let only one of the independent coordinate undergo a virtual displacement, while the remaining n – 1 coordinates are held fixed

� n virtual work equations can be written, one for each independent coordinate


Example 11.1

Determine the angle θ for equilibrium of the

two-member linkage. Each member has a

mass of 10 kg.mass of 10 kg.


Solution

FBD

� One degree of freedom since location of both links may be specified by a single independent coordinatemay be specified by a single independent coordinate

� θ undergoes a positive (CW) virtual rotation δθ, only the active forces, F and the 2 9.81N weights do work


SolutionVirtual Displacements� Origin of the coordinates established at the fixed

pin support DLocation of F and W specified by position � Location of F and W specified by position coordinates xB and yw

� To determine work, note these coordinates are parallel to lines of action of their associated forces

� Express position coordinates in terms of independent coordinate θ and taking derivatives


Solution

It can be seen by the signs of these

mymy

mxmx

ww

BB

θδθδθ

θδθδθ

cos5.0)sin1(2

1

sin2)cos1(2

==

−==

� It can be seen by the signs of these equations that an increase in θ causes an increase in xB and an increase in yw


SolutionVirtual Work Equation� If the virtual displacements δxB and δyw were

both positive, then the forces W and F would do positive work since the forces and their positive work since the forces and their corresponding displacements would have the same sense

� For virtual work equation for displacement δθ, δU = 0; Wδyw + Wδyw + FδxB = 0

� Relating virtual displacements to common δθ, 98.1(0.5cosθ δθ) + 9.81(0.5cosθ δθ)

+ 25(-2sinθ δθ) = 0


SolutionVirtual Work Equation� δθ ≠ 0,

(98.1cosθ -50 sinθ) δθ = 0θ = tan-1(9.81/50) = 63.0°θ = tan-1(9.81/50) = 63.0°

� If problem solved using equations of equilibrium, dismember the links and apply 3 scalar equations to each link

� Principle of virtual work, by means of calculus, eliminated this task so that answer is obtained directly


Example 11.2

Determine the angle θ required to maintain

equilibrium of the mechanism. Neglect the weight

of the links. The spring is un-sketched when θ = 0°, of the links. The spring is un-sketched when θ = 0°, and it maintains a horizontal position due to the

roller.


SolutionFBD� One degree of freedom since location of both

links may be specified by a single independent coordinatecoordinate

� θ undergoes a positive (CW) virtual rotation δθ, links AB and EC rotates by the same amount since they have the same length and link BC only translates

� Since a couple moment does work only when it rotates, work done by M2 = 0

� Reactive forces at A and E does no work


Solution

FBD


Solution

Virtual Displacements

� Position coordinates xB and xD are parallel to lines of action of P and F and these lines of action of P and Fs and these coordinates locate the forces with respect to fixed points A and E

mxmx

mxmx

DD

BB

θδθδθ

θδθδθ

cos2.0sin2.0

cos4.0sin4.0

==

==


Solution


� For positive virtual displacements Fs is opposite to δxD and hence does negative work

δU = 0; M δθ + Pδx - F δx = 0 D

δU = 0; M1δθ + PδxB - FsδxD = 0

� Relating virtual displacements to common δθ,

0.5 δθ + 2(0.4cosθ δθ) - Fs(0.2cosθ δθ) = 0

(0.5 + 0.8cosθ – 0.2Fscosθ) δθ = 0

� For arbitrary angle θ, spring is sketched a distance of xD = (0.2sinθ)m


Solution


� Therefore, Fs = 60N/m(0.2sinθ)m = (12sinθ)N

� δθ ≠ 0, � δθ ≠ 0,

0.5 + 0.8cosθ – 0.2Fscosθ = 0

� Since sin2θ = 2sinθcosθ

1 = 2.4sin2θ – 1.6cosθ

� By trial and Error,

θ = 36.3°


Example 11.3

Determine the horizontal force Cx that the pin

at C must exert on BC in order to hold the

mechanism in equilibrium when θ = 45°. Neglect mechanism in equilibrium when θ = 45°. Neglect

the weight of the members.


SolutionFBD� Cx obtained by releasing the pin constraint at

C in the x direction and allowing the frame to be displaced in this directionbe displaced in this direction

� One degree of freedom since location of both links may be specified by a single independent coordinate

� θ undergoes a positive virtual displacement δθ, only Cx and the 200N force do work


Solution

FBD


Solution

Virtual Displacements

� Forces Cx and 200N are located from the fixed origin A using position coordinates yBfixed origin A using position coordinates yB

and xC

� Using cosine rule,

� Thus,

( ) ( ) ( )

θδθδ

θ

θδθθδ

θ

cos6.0

sin6.0

sin2.1cos2.1200

cos6.026.07.0 222

=

=

+−+=

−+=

B

B

CCC

CC

y

y

xxx

xx


SolutionVirtual Work Equation� yB and xC undergo positive virtual displacements

δyB and δxC

� Cs and 200N do negative work since they act in � Cs and 200N do negative work since they act in opposite sense to δyB and δxC

δU = 0; -200δyB + CxδxC = 0 � Relating virtual displacements to common δθ,

-200(0.6cosθδθ) – Cx[(1.2xCsinθ)/(1.2cosθ – 2xC)]δθ = 0

Cx = [-120cosθ(1.2cosθ – 2xC)]/(1.2xCsinθ)� At required equilibrium position, θ = 45°


Solution

(xC)2 – 1.2cos45°xC – 0.13 = 0

� Solving for positive root,

xC = 0.981mxC = 0.981m

� Thus,

Cx = 114N


Example 11.4

Determine the equilibrium position of the

two-bar linkage. Neglect the weight of the two-bar linkage. Neglect the weight of the

links.


Solution

FBD


Solution

� Two degrees of freedom since the independent coordinates θ1 and θ2 must be known to locate the position of both links

Position vector x , measured from the fixed point � Position vector xB, measured from the fixed point O, is used to specify the location of P

� If θ1 is held fixed and θ2 varies by an amount δ θ2, for the virtual work equation,

[δU = 0]θ1 ; P(δxB)θ1 - M δθ2 = 0

where P and M represent magnitudes of applied force and couple moment acting on link AB


Solution� Position coordinate xB related to independent

coordinates θ1 and θ2

xB = lsinθ1 +lsinθ2

To obtain variation of δx with respect to δθ� To obtain variation of δxB with respect to δθ2

(δxB/δθ2) = l cosθ2

(δxB)θ2 = l cosθ2 δθ2

� Therefore, (Pl cosθ2 – M) δθ2 = 0

� δθ2 ≠ 0, θ2 = cos-1(M/Pl)


Solution

� To obtain variation of δxB with respect to θ1

(δxB/δθ1) = l cosθ1

(δxB)θ1 = l cosθ1 δθ1(δxB)θ1 = l cosθ1 δθ1

� Therefore,

(Pl cosθ1 – M) δθ1 = 0

� δθ1 ≠ 0, θ1 = cos-1(M/Pl)

6161103 11.3 principle of virtual work for a system of connected rigid bodies

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