Virtual Displacement Examples

11.1 Virtual Work

11.1 Virtual Work Example 1, page 1 of 5

A

B

L

Smooth surface

P

C

L

W

1. Determine the force P required to keep the two rods in

equilibrium when the angle = 30 and weight W is 50 lb.

The rods are each of length L and of negligible weight.

They are prevented from moving out of the plane of the

figure by supports not shown.


W CP

B

A

C x

C yN

The system has one degree of freedom,

because specifying the value of a single

coordinate, , completely determines the

configuration (shape) of the system.

Consider a free-body diagram and identify

the active forces those forces that would

do work if were increased slightly.

1

2

3

4

5

The reactions Cx and Cy do no

work because point C does not

move. Thus Cx and Cy are not

active forces.

The force W does work because point B

moves up, so W is an active force.

The normal force N does no work

because it is perpendicular to the

displacement of point A. Thus N

is not an active force.

The force P does work as

point A moves to the right,

so P is an active force.

Free-body diagram (The dashed line shows the position

of the system after has been increased a small amount.)

W

L

C

P

L

B

A


W CP

B

A

C x

C yN

y B

x A

y B

x A

Introduce coordinates measured from a fixed point,

point C in the figure, to the point of application of the

active forces.

Compute the work done when the coordinates are

increased positive infinitesimal amounts, xA and xB

(The custom followed by textbook writers is to use the

Greek letter rather than simply writing dxA and dyB

because the infinitesimals represent hypothetical

motions motions that are possible but are not

necessarily motions that actually occur). The principle

of virtual work says that the total work must add to zero

for all possible motions, real or hypothetical that is,

"virtual."

U = 0: P xA W yB = 0 (1)

A negative sign is present because the force P and

displacement xA are in opposite directions. That is, P

does negative work (absorbs work from the system

rather than adding work to the system). Similarly, the

force W does negative work because W points down

and yB is directed up.

6 7


x A

y BL

C

L

B

A

Relate the differentials xA and yB through the change

in the angle, : From the figure, it follows that

yB = L sin (2)

To relate yB to , use the ordinary formula from

calculus for calculating a differential: if y = f( ), then the

differential is

dy = d

Applying this formula to Eq. 2 and using rather than d

gives

yB = L cos (3)

Similarly

xA = 2L cos

xA = 2L sin (4)

Substitute Eqs. 3 and 4 for yB and xA into the

virtual-work equation:

2L sin

P xA W yB = 0 (Eq. 1 repeated)

L cos

8

dfd

or,

(2P sin W cos )(L ) = 0

Because L 0, it follows that

2P sin W cos = 0

Substituting the given values = 30 and W = 50 lb

and solving gives

P = 43.3 lb Ans.

9


CP

B

AC x

C yN

B y

B x

B y

B x

50 lb

Free body of AB Free body of BC plus pin at B

B

Observation: the forces acting between the rods at pin B never

occurred in the virtual-work equation because the work done

by the equal-and-opposite force pairs acting between the parts

of the body cancel out for example the work done by Bx

acting on rod AB has the opposite sign of the work done by Bx

acting on rod BC. Forces such as Bx and By would have had to

be considered if equilibrium equations rather than virtual work

had been used.

Often virtual work is easier to use than equilibrium

equations for problems involving connected rigid

bodies (typically machines and mechanisms), but

this advantage exists only if the relation between

displacements can be found easily. If the geometry

is difficult, then using equilibrium equations is

probably the better approach.

10 11


C

B

A

W

D

2 ft2 ft

M

2. Determine the value of moment M required to

maintain the mechanism in the position shown,

if = 35 and W = 200 lb.


C

B

A

W

D

2 ft 2 ft

D

W

A

B

CA x

A y

C x

C y

M

M

The reactions at A and

C do no work because

points A and C do not

move. Thus the

reactions are not active

forces.

2

The weight W of the

block does work because

the center of gravity of

the block moves

vertically; thus the

weight is an active force.

4

Couple-moment M does

work because member CD

rotates, so M is an active

"force" (better said, "an

active moment" or "active

generalized force")

3

The system has one degree of freedom: specifying the

value of the single coordinate, , completely

determines the configuration of the system. Consider a

free-body diagram and identify the active forces,

1

Free-body diagram (The dashed line shows the position of

the system after has been increased a small amount.)


2 ft

D

W

A

B

CA x

A y

C x

C y

M

y

D

C

y

y

Introduce a coordinate y measured from the fixed

point A to the point of application of the force W.

Compute the work done when y and are increased a

positive infinitesimal amount.

U = M + W y = 0 (1)

Note that the work done by a moment equals moment

times angle of rotation. Here the work is negative

because M and have opposite senses.

Next use geometry to relate the y and :

y = (2 ft) sin

Differentiating gives

y = 2 cos (2)

Substitute Eq. 2 for y into Eq. 1.

M + W y = 0 (Eq. 1 repeated)

2 cos

Thus

( M + 2W cos ) = 0 (3)

5

Substituting the given values W = 200 lb and

= 35 into Eq. 3 and noting 0 gives

M 2(200 lb) cos 35 = 0

Solving gives

M = 328 lb ft Ans.

6


D

A

B C

W

E

P

F

3 m 3 m2 m

3 m

3. Determine the value of the weight W

required to maintain the mechanism in the

position shown, if P = 50 N.


D

A

B C

W

E

P

F

F

P

E

W

C

BA

D

B x

B y

E x

E y

The system has one degree of freedom because if the

displacement of one end of a bar is known, the

displacement of the other bars can be found by similar

triangles (as will be shown below). Consider a free-body

diagram and identify the active forces corresponding to a

small change in configuration of the system.

1

The force W does work

if A moves vertically, so

W is an active force.

2

The force P does work as point F moves

vertically, so P is an active force.

3

4

Free-body diagram (The dashed line shows the

position of the system after the bars have been

displaced a small amount.)

The reaction forces at B and E

do no work because B and E do

not move; thus the reactions are

not active forces.


D

A B C

W

E

P

F

y F

y F

y A

y A

Introduce coordinates measured from a fixed point to

the point of application of the active forces.

Compute the work done when the coordinates

are increased a positive infinitesimal amount.

U = W yA + P yF = 0 (1)

The force W does negative work because it is

directed down, while the displacement is up.

5

6

D

y A

y D

y C3 m

3 m

2 m

3 m

y F

FE

CB

A

Now relate the differentials yA and yF. By

similar triangles

yD/2 = yF/3 (2)

and

yA/3 = yC/3 (3)

Member DC does not change length so ends C

and D move down the same amount, that is,

yC = yD (4)

Eqs. 2, 3, and 4 imply

yA= (2/3) yF (5)

7


Substitute Eq. 5 into the virtual-work equation, Eq. 1:

W yA + P yF = 0 (Eq. 1 repeated)

(2/3) yF

Thus

W(2/3) + P yF = 0

or, since yF 0 and P is given as 50 N,

W(2/3) + 50 N = 0

Solving gives

W = 75 N Ans.

8


D E

300 mm

250 mm

400 mm

150 mm

A

B

C

G

Q

F

P

250 mm

300 mm

4. Determine the force Q necessary to maintain

equilibrium when force P = 400 N.


P

F

Q

G

CB

A

ED E x

E y

D E

300 mm

250 mm

400 mm

150 mm

A

B C

G

Q

F

P

250 mm

300 mm

B x

B y

The system has one degree of freedom because if member ABC is

rotated about point B a small amount, then the position of CD and

DEFG can be determined. Consider a free-body diagram and identify

the active forces corresponding to the displacements shown.

P does work so

is an active

force.

Points B and E do not

move, so the reactions at

these points do no work.

1

2

3

4

Free-body diagram (The dashed line shows the position of the

system after the bars have been displaced a small amount.)

Q does

work so is an

active force.


B y

B x E y

E x

D E

A

B

C

G

Q

F

P

x Ax A

y G

y G

Introduce coordinates measured from a fixed point

to the point of application of the forces.5


increased a positive infinitesimal amount.

U = 0: P xA + Q yG = 0 (1)

6


300

x A

y D

F

y C

300

150

250D E

y G

400

250

A

B C

G

D

C

All dimensions are in mm.

A

Relate the differentials xA and yG. Begin by

noting that because A is a small angle, the tangent

of A can be replaced by the angle itself:

A = (2)

7 8

xA300 mm

300 mm

xA

xA300 mm

Member ABC is a rigid body, and all parts

must rotate the same amount. Thus

C = A

Substituting for A from Eq. 2 then gives

C =

Again using the small angle approximation for

the tangent gives

yC = (250 mm) C

= (250 mm)

= (5/6) xA

Member CD is a rigid body and thus doesn't

shorten or lengthen. It follows that

yD = yC

Thus

yD = (5/6) xA (3)

9

10


300

x A

y D

F

y C

300

150

250D E

y G

400

250

A

B C

G

All dimensions are in mm.

11

300 mm

xA

Using the small angle approximation for the

tangent gives

D =

[(5/6) xA], by Eq. 3

=

Member DEFG is a rigid body and so all parts

must rotate the same amount. Thus

F = D

Thus

F = (4)

yD250 mm

xA

300 mm

12 F

D


G400 mm

E

F

300 mm(500 mm) F

F

The remaining step is to relate yG to F. We can do

this in two ways, by geometry or by calculus. Let's begin

with the geometric approach. First consider a rotation of

line EG by an amount F.

13

15

16 yG is the vertical distance that point G moves up.

So, considering the small triangle gives

| yG| = (500 F) sin

or,

yG = (5)

(Insert a minus sign, because yG was originally

defined as positive down)

xA300

400 mm

500 mm=

45

, by Eq. 4

43

xA

14 Length = (300 mm)2 + (400 mm)2 = 500 mm


G400 mm

E

F

300 mm500 mm y G

Consider an alternative solution for yG,

based on calculus:17

yG = (500 mm) cos

yG = 500 sin

= 500( ) F

=

= (Same as Eq. 5)

= F because both angles

measure the rotation of line EG)

500 mm

400 mm

xA300

, by Eq. 4400

300

xA

3

18

Note that we can't write

yG = 300 mm

and then differentiate to get yG (which would give yG = 0).

The equation for yG must define a continuous and differentiable

function, not a relationship that is only valid at a single value

of .

Substitute for yG from Eq. 5 into the virtual work equation:

P xA + Q yG = 0 (Eq. 1 repeated)

, by Eq. 5

or,

[P + ( 4/3)Q] xA = 0

Dividing through by xA and using the given value P = 400 N

yields

Q = 300 N Ans.

xA4

3

19

20

4 xA


D

A

B C

F

P

EQ

LL/2L/2

5. Link AB is connected to collar A, which can slide

with negligible friction on horizontal rod EF. Determine

the value of force Q necessary to maintain equilibrium

when = 50, L = 300 mm, and P = 100 N.


D

A

B C

F

P

EQ

Active force

P

CB

A

DD x

D y

A y

Forces from pin D are

not active forces.

L

The force from the rod acting on

the collar is not an active force.Q

L/2L/2

Free-body diagram showing active forces

corresponding to a small increase in 1

Active force


Q

CB

A

DD x

D y

A y

P

x A

x A

y C

y C

Introduce coordinates measured from the fixed point D

to the point of application of the active forces.



U = 0: Q xA + P yC = 0 (1)

2

3

x A

L

L/2 L/2 D

A

B C

Relate the differential xA to

the change in angle,

4

xA = L cos + +

xA = L sin + + (2)

2L

2L

2L

2L

(Length L does not change)

0 0

5


L

L/2 L/2 D

A

B C

y B

y By C

Relate the differential yC to the change in

angle From the figure, we have

yB = L sin

yB = L cos

By similar triangles,

L cos

Thus

yC = (3)

yB

L/2 + L/2=

L/2

yC

6

Substitute Eqs. 2 and 3 for xA and yC into the virtual

work equation, Eq. 1:

L sin , by Eq. 2

Q xA + P yC = 0 (Eq. 1 repeated)

, by Eq. 3

Thus

( Q sin + P )(L ) = 0

Because L 0, it follow that

Q sin + P = 0

Substituting = 50 and P = 100 N and solving for Q

gives

Q = 42.0 N Ans.

2L cos

L cos

2

cos 2

7

2 cos


C

W

B

150 mm

A

D

150 mm

6. Rotating the threaded rod AC of the automobile jack causes

joints A and C to move closer together, thus raising the weight W.

Determine the axial force in the rod, if = 30 and W = 2 kN.

150 mm

150 mm


C

W

B

A

D

F rF r

1

2

The system has one degree of freedom because once

is specified, the location of all parts of the jack can be

determined. Consider a free-body diagram of the jack

and identify the active forces corresponding to a small

change in .

To get a virtual-work equation that contains the

axial force in the rod, it is necessary to exclude the

rod from the free-body diagram. The effect of the

rod is then represented by the two forces Fr. W and

the two Fr forces are the active forces.

Free-body diagram


F rF r

D

B

W

CA

y B

x Ax C

Introduce coordinates measured from fixed points to

the points of application of the active forces.

Calculate the work done.

U = 0: W yB Fr xA Fr xC = 0 (1)

3

4

xA

yB

xC


xC

150 mm

150 mm

D

A

B

C

xA = (150 mm) cos

xA = 150 sin (2)

xC = (150 mm) cos

xC = 150 sin (3)

yB = 2(150 mm) sin + a

yB = 300 cos + a (4)

0

yB

a

6

E

150 mm

150 mm

a

Distance "a," from the intersection of the two sloping

members, point E, to point B, does not change as

changes. Thus when "a" is differentiated, the result is

zero: a = 0.

D

A

BW

C

5

7

Relate yB, xA, and xC through the angle

change, d .

xA


Substituting the expressions for yB, xA, and xC and into the

virtual work equation, Eq. 1, gives

300 cos by Eq. 4 150 sin by Eq. 3

W yB Fr xA Fr xC = 0 (Eq. 1 repeated)

150 sin by Eq. 2

or,

[ 300W cos + 2(150)Fr sin ] = 0

Dividing through by , substituting the given values W = 2 kN

and = 30, and solving gives

Fr = 3.46 kN Ans.

8


A B

L

C

P

k

F

ED

L

LL

L L

7. The original length of the spring is L. Determine the

angle for equilibrium if L = 3 m and P = 300 N.

Spring constant,

k = 200 N/m


1

The spring is not part of the

free-body; the effect of the

spring is represented by the

forces Fs.

2

3

4 Forces P and Fs are

active forces because

their points of

application move in

the direction of the

forces.

Reactions Ay and By do

no work because they

are perpendicular to the

displacement of points

A and B.

Free-body diagram

A B

L

C

P

k

F

ED

L

LL

L L DE

F

P

C

BA

F s F s

A y B y

The system can be described by a single coordinate,

Consider a free-body diagram and identify the active

forces corresponding to a small change in


D E

F

P

C

BA

F s F s

A y B y

y F

y F

x D x E x Ex D

O

Introduce coordinates measured from the point O directly

above pin C to the point of application of the active

forces.



U = P yF + Fs xD + Fs xE = 0 (1)

5

6


D E

F

C

BA

y F

O

L L

LL

LL

x D xE

Relate the differentials yF, xD, and xE to the angle

change From the figure, we see that

yF = 3L sin

xE = L cos

xD = L cos


yF = 3L cos (2)

xE = L sin (3)

xD = L sin (4)

We can use the same figure to calculate the length of the

spring, L , say:

L = distance DE

= 2L cos (5)

7


8

Substituting the given values P = 300 N,

k = 200 N/m, and L = 3 m into Eq. 7 and

solving numerically gives

= 69.1 Ans.

The force in the spring is, then,

Fs = k compression of spring

= k (original length final length)

L (given) L = 2L cos , by Eq. 5

= kL(1 2 cos ) (6)

Substitute from Eqs. 2, 3, 4, and 6 into the virtual

work equation, Eq. 1:

kL(1 2 cos ), by Eq. 6

P yF + Fs xD + Fs xE = 0 (Eq. 1 repeated)

3L cos by Eq. 2 L sin by Eq. 3 L sin by Eq. 4

Thus

3P cos 2kL(1 2 cos ) sin (L ) = 0

or, since L 0,

3P cos 2kL(1 2 cos ) sin = 0 (7)

9


E

B

A

x

y

Q

k 9 kN/m

8. Collars A and B can slide freely on rods CD and CE. Determine the

values of x and y, given that forces P = 900 N and Q = 800 N. The

unstretched length of the spring is 0.2 m, and the weight of the collars is

negligible.

P

DC


E

B

A

x

y

Q

k 9 kN/m

B

800 N

900 N

A

N B (Force from rod CE)

N A (Force from rod CD)

F s

F s

The system has two degree of freedom since both x and

y coordinates must be known if the configuration of the

system is to be determined. Consider a free-body

diagram and identify the active forces corresponding to

a small change in x, while y is held fixed.

Because y is fixed,

collar B does not

move, and so none

of the forces acting

on B is an active

force.

1

2

3

The forces acting on

collar A, Fs (the spring

force) and the 900-N

force, are the only active

forces.

Free-body diagram

P

DC


A

x

y

900 N

800 N

y 0

(y held fixed)

F s

F s

N B

N A

L

x

L

The coordinate x locates the position of the

700-N force. Introduce an additional

coordinate, L, that locates the point of

application of the spring force, Fs.

Compute the work done:

U = 0: (900 N) x Fs L = 0 (1)

Relate x and L:

L2 = x2 + y2 (2)


2L L = 2x x + 2y y

0 (Because y is fixed)

Thus

L = x/L x

Introduce the latter equation into Eq. 1:

(900 N) x Fs L = 0 (Eq. 1 repeated)

x/L x

Thus

(900 Fsx/L) x = 0

Dividing through by x and re-arranging gives

Fsx = 900L (3)

4 5

C

B


C

B

A

x(fixed)

y

900 N

800 N

y

F s

F s

N B

N A

L

L

Next, hold x fixed and compute the work done when

collar B moves an amount y. Following the same

steps as were used for x leads to

Fsy = 800L (4)

The spring force, Fs, is related to L:

Fs = k extension of the spring

= (9000 N/m) (L original length)

0.2 m

Thus

Fs = 9000L 1800 (5)

We now have four simultaneous nonlinear equations

to solve:

L2 = x2 + y2 (2)

Fsx = 900L (3)

Fsy = 800L (4)

Fs = 9000L 1800 (5)

6


These equations can be solved directly if a calculator that

is able to handle such systems is available.

Alternatively, proceed as follows: square both sides of

Eqs. 3 and 4 and add the results to get

(Fsx)2 + (Fsy)

2 = (900L)2 + (800L)2.

or

Fs2(x2 + y2) = L2(9002 + 8002)

L2, by Eq. 2

Solving gives

Fs = 1204 N

Using this result in Eq. 5 gives

Fs = 9000L 1800 (Eq. 5 repeated)

1204 N

Solving gives

L = 0.3338 m.

7 Distance x can now be found from Eq. 3:

Fsx = 900L (Eq. 3 repeated)

1204 N 0.3338 m

Solving gives

x = 0.250 m Ans.

Distance y can be found from Eq. 4:

Fsy = 800L (Eq. 4 repeated)

1204 N 0.3338 m

Solving gives

y = 0.222 m Ans.

8


p

M

C

B

A M

A

B

C

Ay

Ax

N9 in.

4 in.

9. Determine the moment M applied to the crankshaft that

will keep the piston motionless when a pressure p = 400 psi

acts on the top of the piston and = 25. The diameter of

the piston is 3 in., and the piston slides with negligible

friction in the cylinder.

The system can be described by a single coordinate, .


forces corresponding a small change in .

Since friction is negligible, only

the normal force N acts on the

side of the piston. The normal

force does no work since it acts

perpendicular to the motion of

the piston.

1

3

4

2 The resultant of the pressure is an active force:

(400 psi)( )(3 in./2)2 = 2827 lb

The reaction forces Ax and Ay

do no work, because point A

does not move. The moment M

does work as link AB rotates.

Free-body diagram


M

A

B

C

Ay

Ax

N

2827 lb

y C

y C

d

Introduce coordinates measured from

a fixed reference at point A.

Compute the work done when the coordinates are increased

a positive infinitesimal amount.

U = (2827 lb) yC M = 0 (1)

5

6


A

C

Ay

Ax

N

2827 lb

A

C

9 in.

4 in.

y C

Distance "a" does not

change as is

changed so a = 0.

B

a

Relate yC to

yC = (4 in.) cos + (9 in.) cos + a

yC = 4 sin 9 sin + a (2)

0

Relate to by the law of sines,

(3)


Thus

(4)

7

sin

4 in.=

sin

9 in.

cos

4 =

cos

9

=4 cos

9 cos

8


4 cos

9 cos

sin

4 in.=

sin

9 in.

Using Eq. 4 in Eq. 2 gives

yC = 4 sin 9 sin (Eq. 2 repeated)

, by Eq. 4

= ( 4 sin 4 tan cos ) (5)

Substituting Eq. 5 in the virtual work equation gives

(2827) yC M = 0 (Eq. 1 repeated)

( 4 sin 4 tan cos ) by Eq. 5

or

[4(2827)(sin + tan cos ) M = 0

Dividing through by and solving gives

M = 4(2827)(sin tan cos ) (6)

Substituting the given value = 25 into Eq. 3 yields

(Eq. 3 repeated)

which can be solved to give = 10.83.

Using = 25 and = 10.83 in Eq. 6 produces

M = 0 lb in Ans.

Observation: this problem may be more easily

solved by using equations of equilibrium.

Virtual work is superior to using equations of

equilibrium provided that the relation between

displacements can be easily obtained. In the

present problem, deriving the relation between

and yC, Eq. 5, was more complicated than

simply writing equations of equilibrium.

9

11

10


L/2

C

EB

F

D

a

L/2

P

QA

10. Pin B is rigidly attached to member AC and moves in the smooth

quarter-circle slot EF. Determine the value of force Q necessary to

keep the system in equilibrium, if = 30, L = 400 mm, a = 120 mm,

and P = 200 N.


C

B

D

Q (active)

A y

A

P (active)

R

Pin B must move in the slot, that is, in a direction tangent to

the quarter circle and thus perpendicular to radius DB .

Thus the force R from the slot does no work because R is

perpendicular to the motion of the pin.

The system configuration can be

defined by the single coordinate, .

Consider a free-body diagram

showing the active forces

corresponding to a small increase in

1

Free-body diagram

L/2

C

EB

F

D

a

L/2

P

QA

3 Point A must move

horizontally. It is difficult

to tell if A moves to the

right or left, but fortunately

it makes no difference. The

important thing is to note

that force Ay from the

rollers does no work so is

not an active force.

2

Motion of pin B

is perpendicular to BD.

Point C moves both

horizontallly and vertically.

It is difficult to tell if C

moves vertically up or

vertically down, but it

makes no difference. All

we need to note is that

force P is an active force.

4


C

B

D

Q

A y

A

P

R

x Ax A

y C

y CIntroduce coordinates measured from

the fixed point D to the point of

application of the active forces P and Q.

5

Compute the work done when the coordinate are


U = 0: Q xA + P yC = 0 (1)

6


B

D

A

L/2

L/2

C

a

x A

y C

Relate the differentials xA and yC to the angles and

Begin with yC.

yC = L sin

yC = L cos (2)

Note that this equation shows yC is positive if is

positive, that is, point C moves up as increases.

xA = (L/2) cos a cos

xA = (L/2) sin + a sin (3)

8

Relate and through the law of sines:

Because sin (180 ) = sin , the last equation can

be written as

a sin = (L/2) sin (4)


a cos = (L/2) cos

so

= (5)L cos

2a cos

9

sin (180 )

L/2

sin a=

2a cos

L cos

Using Eq. 5 in Eq. 3 gives

xA = (L/2) sin + a sin (Eq. 3 repeated)

, by Eq. 5

= ( sin + cos tan )(L/2) (6)

10

7


11 The angle in Eq. 6 can be calculated by substituting the given

values = 30, L = 400 mm, and a = 120 mm into Eq. 4:

a sin = L/2 sin (Eq. 4 repeated)

and solving to get = 56.44.

Although it is not necessary for solving the problem, we can

now determine whether point A moves to the left or to the

right. From Eq. 6 we have

xA = ( sin + cos tan )(L/2) Eq. 6 repeated)

Substituting = 30 and = 56.44 into this equation gives

xA = (0.8054)(L/2)

That is, xA is positive when is positive, so xA increases,

that is, point A moves to the left for the particular values of

a, and L of our problem.

Substituting Eqs. 2 and 7 for yC and xA into the virtual-work equation, Eq. 1, gives

(0.8054)(L/2) , by Eq. 7

Q xA + P yC = 0 (Eq. 1 repeated)

L cos , by Eq. 2

so

Q(0.8054)/2 + P cos (L ) = 0

Because L 0, it follows that

Q(0.8054)/2 + P cos = 0

Substituting = 30 and P = 200 N and solving gives

Q = 430 N Ans.

12


DE

H

F

G

W

11. A scissors lift is used to raise a weight W = 800 lb.

Determine the force exerted on pin F by the hydraulic

cylinder AF when = 35. Each linkage member is 2-ft

long and pin connected at its midpoint and endpoints. The

lift consists of two identical linkages and cylinders the one

shown and one directly behind it.

C

B

JIK

A


BA

F

W/2

A x

A yB y

FFA


forces associated with a small change in

1

3

Each side of the lift carries half of

the load. The W/2 load does work,

so it is an active force.

2

The force FFA of the hydraulic

cylinder acting on pin F does

work as pin F moves.

The force FFA of the hydraulic

cylinder acting on pin A does no

work because pin A does not

move. For the same reason, the

reaction forces Ax and Ay from the

support do no work.

4

5 The reaction force at B does no

work because it is vertical while

the motion of point B is

horizontal.

Free-body diagram


J

A

Fy J

y J

s F

F FA

s F

Introduce coordinates yJ and sF measured from the fixed point A

to the point of application of the active forces.

Compute the work done when the coordinates are increased a

positive infinitesimal amount:

U = (W/2) yJ + FFA sF = 0 (1)

6

W/2


C

B

JK

A

DE

H

F

G

O

A

F

O

C

s F

(3 ft) sin

(1 ft) cos

(1 ft) sin

(1 ft) sin

(1 ft) sin

(1 ft) sin

(1 ft) sin

(1 ft) sin

y J

(1 ft) cos

1 ft

1 ft

1 ft

1 ft

I

1 ft

Relate the coordinate yJ to the angle

yJ = (6 ft) sin

yJ = 6 cos (2)

To relate sF to ,

consider triangle AFCO.

Use the Pythagorean Theorem and then differentiate to get sF.

sF = (3 sin )2 + (1 cos )2

sF =

=

2(3 sin )(3 cos ) + 2(cos )( sin ) 1

2(3 sin )2 + (1 cos )2

8 sin cos

9 sin2 + cos2(3)

7

8

9

1 ft


Substituting for sF and yJ from Eqs. 2 and 3 in the

virtual-work equation, Eq. 1, gives

(W/2) yJ + FFA sF = 0 (Eq. 1 repeated)

Thus

3W + ] cos = 0

This implies, since cos 0, that

3W +

Substituting the given values = 35 and W = 800 lb and

solving gives

FFA = 997 lb Ans.

6 cos , by Eq. 2

8 sin cos

9 sin2 + cos2, by Eq. 3

9 sin2 + cos2

8FFA sin

8FFA sin

9 sin2 + cos2= 0

Observation: Solving this problem by using

the equations of equilibrium would have

required drawing several free-body diagrams

and writing equations for each diagram.

Solving the problem by virtual work is much

easier because we don't have to calculate the

forces acting between the various links. In

general, problems involving connected rigid

bodies can be solved more easily by virtual

work than by equilibrium equations provided

that the relations between displacements can

be easily obtained.

10 11


k 1.5 kN/m

3 m

A

1.5 m

C

B y

B xF s

D

P

2 m

4 m

B

12. The unstretched length of the spring is 1 m. Determine

the value of for equilibrium when force P = 2 kN. The system can be described by a single

coordinate, . Consider a free-body

diagram and identify the active forces

corresponding to a small change in .

The spring is not part of

the free-body; the effect of

the spring is represented

by the force Fs, which is an

active force.

Force P is an active force.

1

2

3

Free-body diagram

B

C

P

D


C

B y

B x

F s

D

P

A

y D

y D

B

s C

s C

Introduce coordinates measured from the fixed points A

and B to the point of application of the forces.



U = P yD Fs sC = 0 (1)

4

5


C

D

A

B

y D4 m

3 m

1.5 m

2 m

s C

s B

Relate the differentials sC and yD to the angle

change

yD = (2 m + 4 m) sin

yD = 6 cos (2)

Law of cosines applied to triangle ABC:

sC2 = (2 m)2 + (sB)

2 2(2 m)(sB) cos ( + ) (3)

Here

sB = (3 m)2 + (1.5 m)2 = 3.354 m (4)

and

= tan-1 = 26.565 (5)

To avoid having to write equations containing several four and

five-digit numbers, introduce intermediate variables a and b:

sC2 = (22 + sB

2) 2(2)(sB) cos ( + ) (Eq. 3 repeated)

Thus

sC2 = a b cos ( + ) (6)

where

3.354 m by Eq. 4

a = 22 + (sB)2

= 4 + 3.3542

= 15.249 m (7)

1.5 m3 m

6

7

8

a b


b sin ( + )

2sC

The parameter b can also be evaluated, for later

use:

3.354 m, by Eq. 4

b = 2(2)(sB)

= 4(3.354)

= 13.416 m (8)

sC can be related to by differentiating Eq. 6:

sC2 = a b cos ( + ) (Eq. 6 repeated)

2sC sC = b sin ( + )

so

sC = (9)

Taking the square root of both sides of Eq. 6

gives an equation for sC.

sC = a b cos ( + ) (10)

b sin ( + )

2sC

The spring force Fs can be expressed in terms of sC:

Fs = k (extension of the spring)

= k (stretched length unstretched length)

= k (sC 1 m) (11)

Substituting for yD, sC, and Fs from Eqs 2, 9, and 11 into

the virtual-work equation, Eq. 1, gives

k(sC 1), by Eq. 11

P yD Fs sC = 0 (Eq. 1 repeated)

6 cos , by Eq. 2 , by Eq. 9

or

[(6P) cos k(sC 1) = 0

Since 0, the expression in brackets must equal to

zero.

(6P) cos k(sC 1) = 0 (12)

109

2sC

b sin ( + )

2sC

b sin ( + )


11 Eq. 12 contains the distance sC, which can be calculated by using Eq. 10:

(6P) cos k(sC 1) = 0 (Eq. 12 repeated)

, by Eq. 10

or

(6P) cos k( a b cos ( + ) 1) = 0

Substituting in the latter equation the values

P = 2 kN (Given)

k = 1.5 kN/m (Given)

a = 15.249 m (Eq. 7 repeated)

b = 13.416 m (Eq. 8 repeated)

= 26.565 (Eq. 5 repeated)

and solving numerically gives

= 53.4 Ans.

a b cos ( + )2

b sin ( + )

a b cos ( + )

b sin ( + )

2sC


A CB

P

DE F

Hinges

10 ft5 ft5 ft5 ft5 ft

13. a) Determine the moment reaction at the wall F.

b) Determine the force reaction at the roller D.

In both cases P = 60 lb.

Part a. Replace the wall at F by a moment

couple MF and a pin support.1

A CB

P

DE

FMF


A

C

B

P

D

EF

F x

F y

D y

B y

MF

A CB

P

D

EF

F x

F y

D y

B y

y A

y A

Consider a free-body diagram and identify the active forces

associated with a small rotation of the segments of the beam

Introduce coordinates yA and for calculating the work. U = 0: P yA MF = 0 (1)

2

4

The active "forces" are the force

P and the couple moment MF.

3

Free-body diagram

MF


A

CB D

E F5 ft 5 ft

5 ft 5 ft

y A

y C

y E

10 ft

Relate the differentials yA and 5


yE = yC and yC = yA

so

yE = yA

For small angles, yE is given by

yE = (10 ft)

But this becomes, after using the relation yE = yA,

yA = 10 (2)

Substitute this result in the virtual-work equation to get

P yA MF = 0 (Eq. 1 repeated)

10

Dividing through by , substituting the known value P = 60 lb,

and then solving for MF gives,

MF = 600 lb ft Ans.

6 7


A

C

B

60 lb

D E F

D y

A

C

B

60 lb

D E F

F y

D y

B y

MF

A

C

B

60 lb

D E F

D yy D y D

y Ay A

Part b. Replace the roller at D by a vertical force, Dy.8

Draw a free-body diagram and show a

small rotation of segments AC and CE.9 Segment EF of the beam does

not move because the wall

support prevents both rotation

and vertical displacement.

Thus MF and Fy do no work.

Introduce coordinates yA and yD for calculating the work. U = 0: (60 lb) yA Dy yD = 0 (3)

10

12

The active forces are

the 60-lb force and Dy.11


A

CB D

y D

y A

5 ft 5 ft

5 ft

5 ft

y C

E F

Relate the differentials yA and yD.


yA = yC

and

=

Eliminating yC gives

yD =

Use this equation to replace yD in the virtual-work equation:

(60) yA Dy yD = 0 (Eq. 3 repeated)

Dividing through by yA and solving gives

Dy = 120 lb Ans.

yC

5 + 5

yD5

yA

2

yA

2

13

14


A CB

60 lb

DE F

F y

A CB

60 lb

DE F

F y

Comment: Let's extend the discussion. If we were asked to

calculate the vertical reaction force at the wall, we would

replace the wall by a support that prevents rotation but permits

vertical displacement.

15

Corresponding displacements16

Segment EF translates but does not rotate. Thus the reaction

moment at the support does no work. The reaction force, Fy,

however, does work and is thus an active force.

Observation: The method applied in this

beam example can be generalized.

Virtual work can be used to calculate a

force of constraint (a reaction) by

considering displacements which violate

the constraints and then accounting for

the work done by the force of constraint.

This procedure is equivalent to

converting a rigid structure into a

mechanism, as was done at the

beginning of the present example.

18

17


A

C

B

4 m

A

C

B

P

P

C y

3 m

3 m

14. Determine the vertical reaction at

support C, if P = 2 kN. Convert the structure

into a mechanism with

one degree of freedom

by replacing the pin

support at C by a roller

support and a vertical

force, Cy.

1


P

C x

C y

A x

A y

A

C

B

P

C x

C y

A x

A y

y C

y C

y B

y B

Cy and P are active

forces for the

displacements

shown.

Define coordinates yB and yC locating the point of

application of the active forces, and compute the work.

U = 0: P yB + Cy yC = 0 (1)

2

3

A

B

C


A

C

B

3 m

4 m

3 m

A

C

B

y B

y C

5 m

BC = (3 m)2 + (3 m)2 23 m

23 m

AB = (3 m)2 + (4 m)2 5 m

Relate the differentials yB and yC through the angles and

Begin by computing the lengths of bars BC and BA (Note that

these lengths do not change, as the angles and change).

From the above figure,

yB = (5 m) cos

yB = 5 sin (2)

yC = (5 m) cos + (3 2 m) cos

yC = 5 sin (3 2 ) sin (3)

Use the law of sines to relate and

(4)

4

5

6

7

=3 2 m5 m

sin sin


Differentiating Eq. 4 gives

Thus

= (5)

The equation relating and , Eq. 5, can be used in Eq. 3

to express yC in terms of alone:

yC = 5 sin 3 2 sin (Eq. 3 repeated)

= ( 5 sin 5 tan cos ) (6)

Substituting Eqs. 2 and 6 for yB and yC into the virtual

work equation gives

5 sin , by Eq. 2

P yB + Cy yC = 0 (Eq. 1 repeated)

( 5 sin 5 tan cos ) , by Eq. 6

=3 2 5

cos cos

5 cos

2 cos

3

3 2 cos

5 cos

8 Thus

P sin Cy (sin + tan cos ) (5 ) = 0

Dividing by 5 gives

P sin Cy (sin + tan cos ) = 0 (7)

9


3 m

4 m

3 m

A

C

B

5 m

tan = = 1

sin =

cos =

3

3

3

5

4

5

Evaluating the functions of and in Eq. 7, substituting the given

value P = 2 kN, and then solving gives

P sin Cy (sin + tan cos ) = 0 (Eq. 7 repeated)

Cy = 0.857 kN Ans.

3

5

3

5

4

51

10

Observation: This example demonstrates that virtual

work can be used to calculate the reaction forces

from the supports acting on a structure. The example

also demonstrates that just because virtual work can

be used doesn't necessarily mean that it should be

used the reaction at support C could have been

found much more easily by employing equilibrium

equations. The usefulness of virtual work depends

on how easy it is to express relations between

coordinates.

11


5 ft 5 ft 5 ft

P Q

A B C D E F G HI

1 Convert the structure to a mechanism with one degree of

freedom by replacing the pin support at I by a vertical force Iy

and a roller.

P Q

A B C D E F G H I

15. Determine the vertical reaction at

support I of the truss, if P = 10 kip = Q.

Iy

5 ft 5 ft 5 ft 5 ft 5 ft


P Q

AB C D

EF G H I

Identify the active forces corresponding to a set of

displacements compatible with the constraints.

Iy

By Dy Fy Hy

2

P Q

A

B C D E F G H I

By Dy Fy Hy yI

yI

yE

yE

yA

yA

3 Introduce coordinates measured from fixed points to the

points of application of the applied forces.

Calculate the work done.

U = 0: P yA + Q yE Iy yI = 0 (1)

Iy

Ix

Ix


A

B C D

E

F G H

IRelate yA, yE, and yI

by geometry (similar

triangles).

45 ft 5 ft 5 ft

5 ft 5 ft 5 ft 5 ft

5 ft

yA = yC (2)

Similarly,

yC = yE, yE = yG, and yG = yI

These equations imply

yA = yI and yE = yI

Substituting the latter pair of equations into the

virtual-work equation, Eq. 1, gives

P yA + Q yE I yI = 0 (Eq. 1 repeated)

yI yIor

(P + Q Iy) yI = 0

Dividing through by yI, substituting the given values P

= 10 kip = Q, and solving gives

Iy = 20 kip Ans.

yA

yC

yE

yG

yI


D

C

m

BA

D

C

BA

T

16. Determine the tension in the cord. The pulleys are

frictionless and m = 90 kg.

Convert the pulley-cord

system into a mechanism

with one degree of

freedom by replacing the

support B by a tensile

force T acting on the end

of the cord.

1

Weight = mg


D

CB

A

T

mg

y D

y By B

y D

The tension T and the weight do work (are active

forces) if end B of the cord moves up a small

amount.

Introduce the coordinates yB and yD.

2

3

Calculate the work done:

U = 0: T yB + mg yD = 0 (1)

4


D

C

BA

y D

y B

t

s

Diameter d C

Diameter d D

5 Relate yB to yD by first expressing the length, say L, of the cord

in terms of yB and yD:

L = (yD t) yB

+ (yD t) u s

+ (yD t) s

+ dD/2

Half circumference of pulleys

+ dC/2

Thus

L = 3yD 3t yB u 2s + dD/2 + dC/2

Now differentiate, taking into account that because L, t, u, s, dD,

and dC do not vary as yD and yB vary, we have L = 0 = t = u = s

= dD = dC; the result of the differentiation is, then,

0 = 3 yD yB

Thus

yB = 3 yD (2)

u


Substituting this result in the virtual work equation, Eq. 1, gives

T yB + mg yD = 0 (Eq. 1 repeated)

3 yD, by Eq. 2

Thus

( 3T + mg) yD = 0

Dividing through by yD, substituting m = 90 kg, g = 9.81 m/s2

and solving gives

T = 294 N Ans.

6


D

17. Determine the equilibrium values of and for the

two-bar linkage. The couple moment M = 5 N m; each bar is

uniform and has a mass m of 5 kg; the length L = 400 mm; and

the unstretched length of the spring is 250 mm.

k = 0.2 kN/m

500 mm

2

1

A

B

L

L

CM


mg

The system has two degrees of freedom because two

coordinates and must be specified to define the position

of the linkage. Consider a free-body diagram, and identify the

active forces corresponding to a small change in while

is held fixed.

1

2 Because point A does not move and is

fixed, the reactions Ax and Ay, the weight

mg, and the spring force Fs do no work

when 1 is varied a small amount. Thus

they are not active forces.

The couple moment M and the weight

mg of the lower bar BC do work when 1

is varied, so M and mg are active forces.

3

1

2

C

M

A

B

A y

A x

(spring force)F s

mg


CM

A

B

A y

A x

F s

mg

mg

A

B

L

L/2

E

L/2

L/2

L/2

y E

4

5

6

In addition to the coordinate 1, introduce a vertical

coordinate yE measured downward from point A.

Compute the work done when the coordinates

are increased a positive infinitesimal amount.

dU = 0: M 1 + mg yE = 0 (1)

E

Relate the differential yE to the angle change, 1, by

writing

yE = L cos + (L/2) cos 1

and then differentiating with respect to 1, while holding

fixed. That is, take the partial derivative with respect to 1

to obtain

yE = (L/2) sin 1 1 (2)

L/2

yE

2

2

yE

1


Substitute Eq. 2 for yE into the virtual-work equation:

M 1 + mg yE = 0 (Eq. 1 repeated)

(L/2) sin 1 1, by Eq. 2

Thus

[M (mgL/2) sin 1] 1 = 0

Since 1 0, it follows that

M (mgL/2) sin 1 = 0 (4)

Substituting the following values into Eq. 4

M = 5 N m = 5 000 N mm

L = 400 mm

m = 5 kg

g = 9.81 kg m/s2

and solving gives

1 = 30.6 Ans.

7


Next identify the active forces corresponding to

a small change in 2 while 1 is held fixed.

8

10

9 Because point A does not

move, Ax and Ay do no work

and thus are not active forces.

Because 1 is fixed, link BC

does not rotate. That is, the

dashed line representing the new

position of link BC is parallel

to BC. Hence couple moment

M does no work, and M is not

an active force.11 Active forces1

2

CM

A

B

A y

A x

F s

mg

mg


C

A

B

A y

A x

F s

mg

mg

x B

x B

L/2

L/2

y F

E

F

y E

y F

y E

12 Introduce coordinates measured from a fixed point to

the point of application of the active forces.

13 Compute the work done.

U = 0: mg yE + mg yF + Fs xB = 0 (5)

1

2


C

A

B

E

F

L/2

L/2

L/2

L/2

y E

x B

Relate the differentials yE, yF, and yF to the change in angle

xB = L sin

yF = (L/2) cos

yE = L cos (L/2) cos

Differentiating each equation with respect to with held fixed,

gives

xB = L cos (6)

yF = (L/2) sin (7)

yE = L sin (8)

14

yF

1

2


The force Fs in the spring can be related to 2:

Fs = k extension of spring

= k (current length unstretched length)

= k [(500 mm L sin 2) 250 mm]

= k(250 L sin 2) (10)

Substituting Eqs. 6-8 for the differentials into the virtual-work

equation, Eq. 5, gives

L sin by Eq. 8

mg yF + mg yE + Fs xB = 0 (Eq. 5 repeated)

L/2) sin by Eq. 7 L cos by Eq. 6

Thus

[ 3mg/2) sin + Fs cos ] L = 0

Since L 0, it follows that

3mg/2) sin + Fs cos = 0 (9)

15

16

A

B

L

C

D

500 mm

500 mm L sin 2

1

2


Substituting for Fs from Eq. 10 into the virtual-work equation, Eq. 9, gives

3mg/2) sin + Fs cos = 0 (Eq. 9 repeated)

k(250 L sin 2), by Eq. 10

or,

(3mg/2) sin + k(250 L sin ) cos = 0 (11)

Substituting the following values into Eq. 11

L = 400 mm

m = 5 kg

k = 0.2 kN/m = 0.2 N/mm

g = 9.81 kg m/s2

and solving numerically gives

= 18.50 Ans.

17

Virtual Displacement Examples

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