11.1 Virtual Work

11.1 Virtual Work Example 1, page 1 of 5 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.

L L W C A P

Smooth surface

11.1 Virtual Work Example 1, page 2 of 5 B 1 The system has one degree of freedom, L L because specifying the value of a single W coordinate, , completely determines the C configuration (shape) of the system. A P Consider a free-body diagram and identify the active forces those forces that would do work if were increased slightly.

Free-body diagram (The dashed line shows the position of the system after has been increased a small amount.)

2 The force W does work because point B 4 The force P does work as B moves up, so W is an active force. point A moves to the right, so P is an active force. P A W C

3 The reactions Cx and Cy do no N Cy work because point C does not 5 The normal force N does no work move. Thus Cx and Cy are not because it is perpendicular to the active forces. displacement of point A. Thus N is not an active force.

11.1 Virtual Work Example 1, page 3 of 5

6 Introduce coordinates measured from a fixed point, 7 Compute the work done when the coordinates are point C in the figure, to the point of application of the increased positive infinitesimal amounts, xA and xB active forces. (The custom followed by textbook writers is to use the Greek letter rather than simply writing dx and dy y A B B B because the infinitesimals represent hypothetical motions motions that are possible but are not y necessarily motions that actually occur). The principle B of virtual work says that the total work must add to zero P A W C for all possible motions, real or hypothetical that is, "virtual." Cx

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

N Cy A negative sign is present because the force P and displacement xA are in opposite directions. That is, P xA xA 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.

11.1 Virtual Work Example 1, page 4 of 5

8 Relate the differentials xA and yB through the change B in the angle, : From the figure, it follows that L y = L sin (2) L B yB

To relate yB to , use the ordinary formula from A C calculus for calculating a differential: if y = f( ), then the differential is x df A dy = d d Applying this formula to Eq. 2 and using rather than d gives

yB = L cos (3) 9 or, Similarly (2P sin W cos )(L ) = 0 xA = 2L cos Because L 0, it follows that xA = 2L sin (4) 2P sin W cos = 0 Substitute Eqs. 3 and 4 for yB and xA into the virtual-work equation: Substituting the given values = 30° and W = 50 lb

and solving gives 2L sin P = 43.3 lb Ans. P xA W yB = 0 (Eq. 1 repeated)

L cos

11.1 Virtual Work Example 1, page 5 of 5 10 Observation: the forces acting between the rods at pin B never 11 Often virtual work is easier to use than equilibrium occurred in the virtual-work equation because the work done equations for problems involving connected rigid by the equal-and-opposite force pairs acting between the parts bodies (typically machines and mechanisms), but of the body cancel out for example the work done by Bx this advantage exists only if the relation between acting on rod AB has the opposite sign of the work done by Bx displacements can be found easily. If the geometry acting on rod BC. Forces such as Bx and By would have had to is difficult, then using equilibrium equations is be considered if equilibrium equations rather than virtual work probably the better approach. had been used.

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

By By

B B Bx Bx

P A 50 lb C Cx

N Cy

11.1 Virtual Work Example 2, page 1 of 3 2. Determine the value of moment M required to maintain the mechanism in the position shown, if = 35° and W = 200 lb.

A C

2 ft 2 ft W B D

11.1 Virtual Work Example 2, page 2 of 3 A C

2 ft 2 ft 1 The system has one degree of freedom: specifying the value of the single coordinate, , completely W determines the configuration of the system. Consider a free-body diagram and identify the active forces, B D

Free-body diagram (The dashed line shows the position of the system after has been increased a small amount.) 2 The reactions at A and C do no work because points A and C do not A C 3 Couple-moment M does Ax Cx move. Thus the work because member CD M reactions are not active rotates, so M is an active forces. W "force" (better said, "an active moment" or "active A y Cy generalized force")

B D 4 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.

11.1 Virtual Work Example 2, page 3 of 3

5 Introduce a coordinate y measured from the fixed Ax A C point A to the point of application of the force W. Cx M Compute the work done when y and are increased a W y positive infinitesimal amount.

Ay Cy U = M + W y = 0 (1) B D Note that the work done by a moment equals moment y times angle of rotation. Here the work is negative because M and have opposite senses. C Next use geometry to relate the y and :

y = (2 ft) sin y 2 ft Differentiating gives D y = 2 cos (2)

Substitute Eq. 2 for y into Eq. 1. 6 Substituting the given values W = 200 lb and = 35° into Eq. 3 and noting 0 gives M + W y = 0 (Eq. 1 repeated) M 2(200 lb) cos 35° = 0 2 cos Solving gives Thus M = 328 lb ft Ans. ( M + 2W cos ) = 0 (3)

11.1 Virtual Work Example 3, page 1 of 4 3. Determine the value of the weight W required to maintain the mechanism in the position shown, if P = 50 N.

2 m 3 m 3 m 3 m

D E F

A B C

11.1 Virtual Work Example 3, page 2 of 4 1 The system has one degree of freedom because if the D displacement of one end of a bar is known, the E F 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 P small change in configuration of the system. A B C Free-body diagram (The dashed line shows the position of the system after the bars have been displaced a small amount.) W

D E Ex F 3 The force P does work as point F moves vertically, so P is an active force.

Ey P B A

Bx C

W By 4 The reaction forces at B and E do no work because B and E do 2 The force W does work not move; thus the reactions are if A moves vertically, so not active forces. W is an active force.

11.1 Virtual Work Example 3, page 3 of 4 Introduce coordinates measured from a fixed point to 5 the point of application of the active forces. yF

D yF E F

y A P

yA A B C

7 Now relate the differentials yA and yF. By similar triangles D 2 m yF W yD E 3 m F yD/2 = yF/3 (2)

and

yA/3 = yC/3 (3) yA B 3 m C

A 3 m yC Member DC does not change length so ends C and D move down the same amount, that is, 6 Compute the work done when the coordinates

are increased a positive infinitesimal amount. yC = yD (4)

U = W yA + P yF = 0 (1) Eqs. 2, 3, and 4 imply

The force W does negative work because it is yA= (2/3) yF (5) directed down, while the displacement is up.

11.1 Virtual Work Example 3, page 4 of 4 8 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.

11.1 Virtual Work Example 4, page 1 of 7 4. Determine the force Q necessary to maintain equilibrium when force P = 400 N. A P

300 mm

250 mm C

B 150 mm 250 mm E D Q

300 mm 400 mm G F

11.1 Virtual Work Example 4, page 2 of 7 A P 1 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 300 mm DEFG can be determined. Consider a free-body diagram and identify the active forces corresponding to the displacements shown. B 250 mm C Free-body diagram (The dashed line shows the position of the 150 mm system after the bars have been displaced a small amount.) 250 mm A D E P Q 2 P does work so 300 mm is an active force. 400 mm E B C y Bx F G

E D Ex

By Q

3 Points B and E do not move, so the reactions at F G these points do no work. 4 Q does work so is an active force.

11.1 Virtual Work Example 4, page 3 of 7 5 Introduce coordinates measured from a fixed point to the point of application of the forces.

xA P A

E Bx C y 6 Compute the work done when the coordinates are increased a positive infinitesimal amount. B

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

Ex Q By

F G yG

11.1 Virtual Work Example 4, page 4 of 7

7 Relate the differentials xA and yG. Begin by 8 Member ABC is a rigid body, and all parts noting that because A is a small angle, the tangent must rotate the same amount. Thus of A can be replaced by the angle itself: x C = A = A (2) A 300 mm Substituting for A from Eq. 2 then gives x A A xA C = 300 mm

A 9 Again using the small angle approximation for 300 the tangent gives

yC = (250 mm) C B 250 C y C x C = (250 mm) A 150 300 mm D 250 E = (5/6) xA yD

D 10 Member CD is a rigid body and thus doesn't shorten or lengthen. It follows that 300 All dimensions are in mm. yD = yC

yG Thus F 400 G

yD = (5/6) xA (3)

11.1 Virtual Work Example 4, page 5 of 7

A xA 11 Using the small angle approximation for the tangent gives

yD 300 D = 250 mm

[(5/6) xA], by Eq. 3 B 250 C xA = yC 300 mm

150 Member DEFG is a rigid body and so all parts D 250 E must rotate the same amount. Thus yD

D F = D

300 All dimensions are in mm. F 12 Thus y xA G F = 300 mm (4) F 400 G

11.1 Virtual Work Example 4, page 6 of 7

13 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. 2 2 E 14 Length = (300 mm) + (400 mm) = 500 mm

F 15 (500 mm) 300 mm F

F 400 mm G 16 yG is the vertical distance that point G moves up. So, considering the small triangle gives

400 mm 4 = 500 mm 5

| yG| = (500 F) sin x A , by Eq. 4 300 or, 4 xA yG = 3 (5)

(Insert a minus sign, because yG was originally defined as positive down)

11.1 Virtual Work Example 4, page 7 of 7

17 Consider an alternative solution for yG, based on calculus: 19 Note that we can't write E yG = 300 mm

500 mm yG and then differentiate to get yG (which would give yG = 0). 300 mm The equation for yG must define a continuous and differentiable function, not a relationship that is only valid at a single value F 400 mm G of .

18 yG = (500 mm) cos

yG = 500 sin = F because both angles 20 Substitute for yG from Eq. 5 into the virtual work equation: measure the rotation of line EG) 400 mm P x + Q y = 0 (Eq. 1 repeated) = 500 A G ( 500 mm ) F 4 x x A , by Eq. 5 A , by Eq. 4 3 400 x 300 or, = A 300 [P + ( 4/3)Q] xA = 0

4 xA = (Same as Eq. 5) Dividing through by x and using the given value P = 400 N 3 A yields

Q = 300 N Ans.

11.1 Virtual Work Example 5, page 1 of 4 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.

A E F Q

L L/2 L/2

D B C

11.1 Virtual Work Example 5, page 2 of 4 A E F Q

L L/2 L/2

D B C

1 Free-body diagram showing active forces corresponding to a small increase in

Ay The force from the rod acting on the collar is not an active force. Q A

Active force

Active force Dy P Forces from pin D are not active forces. B C D Dx

11.1 Virtual Work Example 5, page 3 of 4 2 Introduce coordinates measured from the fixed point D 4 Relate the differential xA to to the point of application of the active forces. the change in angle,

Ay A

Q A

P Dy B L/2 C L/2 D

Dx B C D xA

yC L L 5 xA = L cos + + yC 2 2 L L xA xA = L sin + + (2) 2 2 xA 0 0 (Length L does not change) 3 Compute the work done when the coordinates are increased a positive infinitesimal amount.

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

11.1 Virtual Work Example 5, page 4 of 4 A 7 Substitute Eqs. 2 and 3 for xA and yC into the virtual work equation, Eq. 1:

L sin , by Eq. 2 yB L Q xA + P yC = 0 (Eq. 1 repeated) L cos B L/2 C L/2 D , by Eq. 3 2

yC Thus yB ( Q sin + P cos )(L ) = 0 2 6 Relate the differential yC to the change in angle From the figure, we have Because L 0, it follow that cos yB = L sin Q sin + P = 0 2

yB = L cos Substituting = 50° and P = 100 N and solving for Q gives By similar triangles, Q = 42.0 N Ans. L cos yB y = C L/2 + L/2 L/2

Thus L cos y = (3) C 2

11.1 Virtual Work Example 6, page 1 of 5 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.

B 150 mm 150 mm

A C

150 mm 150 mm

11.1 Virtual Work Example 6, page 2 of 5 1 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 .

Free-body diagram 2 To get a virtual-work equation that contains the axial force in the rod, it is necessary to exclude the W rod from the free-body diagram. The effect of the rod is then represented by the two forces Fr. W and B the two Fr forces are the active forces.

A Fr C Fr

11.1 Virtual Work Example 6, page 3 of 5 3 Introduce coordinates measured from fixed points to the points of application of the active forces.

yB B 4 Calculate the work done.

U = 0: W yB Fr xA Fr xC = 0 (1) y B A C Fr Fr

D x x A C x xA C

11.1 Virtual Work Example 6, page 4 of 5

5 Relate yB, xA, and xC through the angle change, d . 6 Distance "a," from the intersection of the two sloping B members, point E, to point B, does not change as changes. Thus when "a" is differentiated, the result is a E zero: a = 0. 150 mm 150 mm W B yB A C a

150 mm 150 mm A C D

xA xC

7 xA = (150 mm) cos

xA = 150 sin (2) D

xC = (150 mm) cos

xC = 150 sin (3)

yB = 2(150 mm) sin + a

yB = 300 cos + a (4) 0

11.1 Virtual Work Example 6, page 5 of 5

8 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.

11.1 Virtual Work Example 7, page 1 of 5 7. The original length of the spring is L. Determine the angle for equilibrium if L = 3 m and P = 300 N.

A B

L L

L L Spring constant, k = 200 N/m

D E k

L L

11.1 Virtual Work Example 7, page 2 of 5 A B 1 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 L L A B 2 Reactions Ay and By do Free-body diagram no work because they are perpendicular to the C displacement of points A and B. Ay By L L C

D E k 3 The spring is not part of the free-body; the effect of the F L L D s Fs spring is represented by the E forces Fs.

4 Forces P and Fs are F active forces because their points of P application move in the direction of the forces. F

11.1 Virtual Work Example 7, page 3 of 5 Introduce coordinates measured from the point O directly 5 A B above pin C to the point of application of the active O forces.

6 Compute the work done when the coordinates are Ay By increased a positive infinitesimal amount.

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

D Fs Fs E

F yF

xD xE x xD E

11.1 Virtual Work Example 7, page 4 of 5

7 Relate the differentials yF, xD, and xE to the angle A O B change From the figure, we see that

yF = 3L sin L L

xE = L cos

xD = L cos C Differentiating gives

y = 3L cos (2) F yF L L

xE = L sin (3)

xD = L sin (4) D E

We can use the same figure to calculate the length of the spring, L , say: L L L = distance DE

= 2L cos (5) F

xD xE

11.1 Virtual Work Example 7, page 5 of 5 8 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 9 Substituting the given values P = 300 N, or, since L 0, k = 200 N/m, and L = 3 m into Eq. 7 and solving numerically gives 3P cos 2kL(1 2 cos ) sin = 0 (7) = 69.1° Ans.

11.1 Virtual Work Example 8, page 1 of 5 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. x

P C A D

y k 9 kN/m

11.1 Virtual Work Example 8, page 2 of 5 x 1 The system has two degree of freedom since both x and y coordinates must be known if the configuration of the P system is to be determined. Consider a free-body C A diagram and identify the active forces corresponding to D a small change in x, while y is held fixed.

Free-body diagram

y k 9 kN/m 2 The forces acting on 900 N collar A, Fs (the spring A force) and the 900-N force, are the only active F forces. s

B NA (Force from rod CD)

Fs 3 Because y is fixed, collar B does not B Q move, and so none NB (Force from rod CE) of the forces acting E on B is an active force. 800 N

11.1 Virtual Work Example 8, page 3 of 5 4 The coordinate x locates the position of the 5 Compute the work done: 700-N force. Introduce an additional coordinate, L, that locates the point of U = 0: (900 N) x Fs L = 0 (1) application of the spring force, Fs. x Relate x and L: x L2 = x2 + y2 (2)

900 N Differentiating gives A C 2L L = 2x x + 2y y L

Fs 0 (Because y is fixed) Thus

NA y L = x/L x

Introduce the latter equation into Eq. 1: Fs L (900 N) x Fs L = 0 (Eq. 1 repeated) B NB y 0 x/L x (y held fixed) Thus

(900 Fsx/L) x = 0

Dividing through by x and re-arranging gives 800 N Fsx = 900L (3)

11.1 Virtual Work Example 8, page 4 of 5 x(fixed) 6 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 A C 900 N Fsy = 800L (4)

The spring force, Fs, is related to L: Fs

Fs = k extension of the spring NA y = (9000 N/m) (L original length)

Fs 0.2 m L Thus B NB Fs = 9000L 1800 (5) y We now have four simultaneous nonlinear equations to solve:

L2 = x2 + y2 (2) L

800 N Fsx = 900L (3)

Fsy = 800L (4)

Fs = 9000L 1800 (5)

11.1 Virtual Work Example 8, page 5 of 5 7 These equations can be solved directly if a calculator that 8 Distance x can now be found from Eq. 3: is able to handle such systems is available. Fsx = 900L (Eq. 3 repeated) Alternatively, proceed as follows: square both sides of Eqs. 3 and 4 and add the results to get 1204 N 0.3338 m

2 2 2 2 (Fsx) + (Fsy) = (900L) + (800L) . Solving gives

or x = 0.250 m Ans.

2 2 2 2 2 2 Fs (x + y ) = L (900 + 800 ) Distance y can be found from Eq. 4:

2 L , by Eq. 2 Fsy = 800L (Eq. 4 repeated)

Solving gives 1204 N 0.3338 m

Fs = 1204 N Solving gives

Using this result in Eq. 5 gives y = 0.222 m Ans.

Fs = 9000L 1800 (Eq. 5 repeated)

1204 N

Solving gives

L = 0.3338 m .

11.1 Virtual Work Example 9, page 1 of 4 9. Determine the moment M applied to the crankshaft that 1 The system can be described by a single coordinate, . will keep the piston motionless when a pressure p = 400 psi Consider a free-body diagram and identify the active acts on the top of the piston and = 25°. The diameter of forces corresponding a small change in . the piston is 3 in., and the piston slides with negligible friction in the cylinder. 2 The resultant of the pressure is an active force:

p (400 psi)( )(3 in./2)2 = 2827 lb C Free-body diagram C

N 9 in. 3 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 B perpendicular to the motion of the piston. M 4 in. B

A M

Ax 4 The reaction forces Ax and A A y do no work, because point A does not move. The moment M does work as link AB rotates. Ay

11.1 Virtual Work Example 9, page 2 of 4 5 Introduce coordinates measured from a fixed reference at point A. 6 Compute the work done when the coordinates are increased 2827 lb a positive infinitesimal amount.

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

C yC

d B

Ax A

11.1 Virtual Work Example 9, page 3 of 4

7 Relate yC to Distance "a" does not change as is 2827 lb yC = (4 in.) cos + (9 in.) cos + a changed so a = 0.

yC = 4 sin 9 sin + a (2) C C

0 a N

8 Relate to by the law of sines,

sin sin = (3) 4 in. 9 in. yC 9 in.

Differentiating gives B cos cos = 4 9 4 in.

Thus Ax A A 4 cos = (4) 9 cos

11.1 Virtual Work Example 9, page 4 of 4

9 Using Eq. 4 in Eq. 2 gives 10 Using = 25° and = 10.83° in Eq. 6 produces

yC = 4 sin 9 sin (Eq. 2 repeated) M = 0 lb in Ans. 4 cos , by Eq. 4 9 cos 11 Observation: this problem may be more easily = ( 4 sin 4 tan cos ) (5) solved by using equations of equilibrium. Virtual work is superior to using equations of Substituting Eq. 5 in the virtual work equation gives equilibrium provided that the relation between displacements can be easily obtained. In the (2827) yC M = 0 (Eq. 1 repeated) present problem, deriving the relation between and yC, Eq. 5, was more complicated than ( 4 sin 4 tan cos ) by Eq. 5 simply writing equations of equilibrium. 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 sin sin = (Eq. 3 repeated) 4 in. 9 in.

which can be solved to give = 10.83°.

11.1 Virtual Work Example 10, page 1 of 5 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, C and P = 200 N.

L/2 E B P

L/2

a F A Q D

11.1 Virtual Work Example 10, page 2 of 5 C

L/2 E B P 1 The system configuration can be defined by the single coordinate, . Consider a free-body diagram L/2 showing the active forces corresponding to a small increase in a F A C Q D Free-body diagram

2 Pin B must move in the slot, that is, in a direction tangent to P (active) the quarter circle and thus perpendicular to radius DB. B Thus the force R from the slot does no work because R is 4 Point C moves both perpendicular to the motion of the pin. horizontallly and vertically. Motion of pin B It is difficult to tell if C 3 Point A must move R is perpendicular moves vertically up or horizontally. It is difficult to BD. vertically down, but it to tell if A moves to the makes no difference. All Q (active) A right or left, but fortunately we need to note is that it makes no difference. The D force P is an active force. important thing is to note that force Ay from the rollers does no work so is

not an active force. Ay

11.1 Virtual Work Example 10, page 3 of 5

C yC 5 Introduce coordinates measured from the fixed point D to the point of application of the active forces P and Q. P B

Q A D

6 Compute the work done when the coordinate are Ay increased a positive infinitesimal amount.

xA U = 0: Q x + P y = 0 (1) xA A C

11.1 Virtual Work Example 10, page 4 of 5

C 7 Relate the differentials xA and yC to the angles and Begin with yC.

y = L sin L/2 C

yC = L cos (2) B yC Note that this equation shows yC is positive if is positive, that is, point C moves up as increases. L/2 a

A 9 Relate and through the law of sines:

D sin (180° ) sin = L/2 a xA Because sin (180° ) = sin , the last equation can 8 xA = (L/2) cos a cos be written as

xA = (L/2) sin + a sin (3) a sin = (L/2) sin (4)

10 Using Eq. 5 in Eq. 3 gives Differentiating gives

xA = (L/2) sin + a sin (Eq. 3 repeated) a cos = (L/2) cos L cos , by Eq. 5 2a cos so L cos = (5) = ( sin + cos tan )(L/2) (6) 2a cos

11.1 Virtual Work Example 10, page 5 of 5

11 The angle in Eq. 6 can be calculated by substituting the given 12 Substituting Eqs. 2 and 7 for yC and xA into the values = 30°, L = 400 mm, and a = 120 mm into Eq. 4: virtual-work equation, Eq. 1, gives

a sin = L/2 sin (Eq. 4 repeated) (0.8054)(L/2) , by Eq. 7

and solving to get = 56.44°. Q xA + P yC = 0 (Eq. 1 repeated) Although it is not necessary for solving the problem, we can L cos , by Eq. 2 now determine whether point A moves to the left or to the right. From Eq. 6 we have so

xA = ( sin + cos tan )(L/2) Eq. 6 repeated) Q(0.8054)/2 + P cos (L ) = 0

Substituting = 30° and = 56.44° into this equation gives Because L 0, it follows that

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

That is, xA is positive when is positive, so xA increases, Substituting = 30° and P = 200 N and solving gives that is, point A moves to the left for the particular values of a, and L of our problem. Q = 430 N Ans.

11.1 Virtual Work Example 11, page 1 of 5 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.

K I J

H G F

E D C

B A

11.1 Virtual Work Example 11, page 2 of 5 1 Consider a free-body diagram and identify the active forces associated with a small change in 2 Each side of the lift carries half of W/2 the load. The W/2 load does work, Free-body diagram so it is an active force.

3 The force FFA of the hydraulic cylinder acting on pin F does work as pin F moves. F FFA

4 The force FFA of the hydraulic cylinder acting on pin A does no

A work because pin A does not B Ax move. For the same reason, the 5 The reaction force at B does no reaction forces Ax and Ay from the work because it is vertical while support do no work.

the motion of point B is horizontal. By Ay

11.1 Virtual Work Example 11, page 3 of 5

W/2

yJ J

6 Introduce coordinates yJ and sF measured from the fixed point A F to the point of application of the active forces. y J sF Compute the work done when the coordinates are increased a

positive infinitesimal amount: FFA

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

11.1 Virtual Work Example 11, page 4 of 5 J K (1 ft) sin 1 ft I 1 ft (1 ft) sin H G 7 Relate the coordinate yJ to the angle 1 ft (1 ft) sin yJ = (6 ft) sin F yJ 1 ft (1 ft) sin yJ = 6 cos (2) E D 1 ft (1 ft) sin C (1 ft) sin A B O 1 ft

(1 ft) cos F 9 Use the Pythagorean Theorem and then differentiate to get sF.

8 To relate s to , 2 2 F sF = (3 sin ) + (1 cos ) consider triangle AFCO. sF (3 ft) sin 1 2(3 sin )(3 cos ) + 2(cos )( sin ) C sF = 2 (3 sin )2 + (1 cos )2

O A 8 sin cos = (3) 9 sin2 + cos2 (1 ft) cos

11.1 Virtual Work Example 11, page 5 of 5

10 Substituting for sF and yJ from Eqs. 2 and 3 in the 11 Observation: Solving this problem by using virtual-work equation, Eq. 1, gives the equations of equilibrium would have required drawing several free-body diagrams 6 cos , by Eq. 2 and writing equations for each diagram. Solving the problem by virtual work is much (W/2) yJ + FFA sF = 0 (Eq. 1 repeated) easier because we don't have to calculate the forces acting between the various links. In 8 sin cos , by Eq. 3 general, problems involving connected rigid 9 sin2 + cos2 bodies can be solved more easily by virtual Thus work than by equilibrium equations provided

8FFA sin that the relations between displacements can 3W + ] cos = 0 be easily obtained. 9 sin2 + cos2

This implies, since cos 0, that

8FFA sin 3W + = 0 9 sin2 + cos2

Substituting the given values = 35° and W = 800 lb and solving gives

FFA = 997 lb Ans.

11.1 Virtual Work Example 12, page 1 of 5 12. The unstretched length of the spring is 1 m. Determine the value of for equilibrium when force P = 2 kN. 1 The system can be described by a single coordinate, . Consider a free-body 3 m diagram and identify the active forces corresponding to a small change in .

A Free-body diagram 1.5 m By B k 1.5 kN/m B

Fs Bx 2 The spring is not part of 2 m the free-body; the effect of C the spring is represented by the force Fs, which is an C active force.

4 m D D

P 3 Force P is an active force. P

11.1 Virtual Work Example 12, page 2 of 5 4 Introduce coordinates measured from the fixed points A and B to the point of application of the forces.

A By

Bx 5 Compute the work done when the coordinates are

Fs B sC increased a positive infinitesimal amount.

U = P yD Fs sC = 0 (1) C

11.1 Virtual Work Example 12, page 3 of 5

6 Relate the differentials sC and yD to the angle 8 Law of cosines applied to triangle ABC: change 2 2 2 sC = (2 m) + (sB) 2(2 m)(sB) cos ( + ) (3) 3 m A Here

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

B and sC = tan-1 1.5 m = 26.565° (5) 3 m

2 m To avoid having to write equations containing several four and five-digit numbers, introduce intermediate variables a and b: C s 2 = (22 + s 2) 2(2)(s ) cos ( + ) (Eq. 3 repeated) C B B yD 4 m a b Thus

2 sC = a b cos ( + ) (6) D where 3.354 m by Eq. 4 2 2 a = 2 + (sB) 7 yD = (2 m + 4 m) sin = 4 + 3.3542 yD = 6 cos (2) = 15.249 m (7)

11.1 Virtual Work Example 12, page 4 of 5

9 The parameter b can also be evaluated, for later 10 The spring force Fs can be expressed in terms of sC: use: Fs = k (extension of the spring) 3.354 m, by Eq. 4 b = 2(2)(sB) = k (stretched length unstretched length)

= 4(3.354) = k (sC 1 m) (11)

= 13.416 m (8) Substituting for yD, sC, and Fs from Eqs 2, 9, and 11 into the virtual-work equation, Eq. 1, gives

sC can be related to by differentiating Eq. 6: k(sC 1), by Eq. 11 2 sC = a b cos ( + ) (Eq. 6 repeated) P yD Fs sC = 0 (Eq. 1 repeated) 2sC sC = b sin ( + ) b sin ( + ) 6 cos , by Eq. 2 , by Eq. 9 2s so C or b sin ( + ) sC = (9) b sin ( + ) 2sC [(6P) cos k(s 1) = 0 C 2s Taking the square root of both sides of Eq. 6 C gives an equation for sC. Since 0, the expression in brackets must equal to zero. sC = a b cos ( + ) (10) b sin ( + ) (6P) cos k(sC 1) = 0 (12) 2sC

11.1 Virtual Work Example 12, page 5 of 5

11 Eq. 12 contains the distance sC, which can be calculated by using Eq. 10: b sin ( + ) (6P) cos k(sC 1) = 0 (Eq. 12 repeated) 2sC a b cos ( + ) , by Eq. 10

or b sin ( + ) (6P) cos k( a b cos ( + ) 1) = 0 2 a b cos ( + )

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.

11.1 Virtual Work Example 13, page 1 of 6 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. Hinges P

D A B C E F

5 ft 5 ft 5 ft 5 ft 10 ft

P 1 Part a. Replace the wall at F by a moment couple MF and a pin support. D F A B C E MF

11.1 Virtual Work Example 13, page 2 of 6 2 Consider a free-body diagram and identify the active forces associated with a small rotation of the segments of the beam

Free-body diagram

P Dy F A B E MF

Fx C D

3 The active "forces" are the force F By y P and the couple moment MF.

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

P Dy F A B C E MF

Fx y D A yA

By Fy

11.1 Virtual Work Example 13, page 3 of 6

5 Relate the differentials yA and

yC A 5 ft 5 ft E 10 ft F B 5 ft C 5 ft D

yA yE

6 By similar triangles, 7 For small angles, yE is given by

yE = yC and yC = yA yE = (10 ft)

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

yE = yA yA = 10 (2)

Substitute this result in the virtual-work equation to get

P yA MF = 0 (Eq. 1 repeated)

Dividing through by , substituting the known value P = 60 lb, and then solving for MF gives,

MF = 600 lb ft Ans.

11.1 Virtual Work Example 13, page 4 of 6

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

60 lb Dy

A B C D E F

9 Draw a free-body diagram and show a 10 Segment EF of the beam does small rotation of segments AC and CE. not move because the wall support prevents both rotation

60 lb Dy and vertical displacement. Thus MF and Fy do no work. A B M C D E F F 11 The active forces are F y By the 60-lb force and Dy.

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

60 lb Dy

yD yD A B

C D E F yA y A 11.1 Virtual Work Example 13, page 5 of 6

13 Relate the differentials yA and yD. y C yD A 5 ft B 5 ft C 5 ft D 5 ft E F

14 By similar triangles,

yA = yC

and

yC y = D 5 + 5 5

Eliminating yC gives y y = A D 2

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

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

yA 2

Dividing through by yA and solving gives

Dy = 120 lb Ans.

11.1 Virtual Work Example 13, page 6 of 6 15 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.

60 lb D A B C E F

18 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 16 Corresponding displacements the constraints and then accounting for F the work done by the force of constraint. y This procedure is equivalent to converting a rigid structure into a 60 lb mechanism, as was done at the D beginning of the present example. A B C E F

17 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.

11.1 Virtual Work Example 14, page 1 of 5 14. Determine the vertical reaction at support C, if P = 2 kN. 1 Convert the structure into a mechanism with one degree of freedom by replacing the pin Cy support at C by a roller C support and a vertical P force, Cy. 3 m C P B

B 4 m

A 3 m

11.1 Virtual Work Example 14, page 2 of 5

Cy Cy

Cx Cx C yC P C P 2 Cy and P are active forces for the displacements

shown. yB B

B yC

Ay y B Ay

Ax Ax A A

3 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)

11.1 Virtual Work Example 14, page 3 of 5 4 Relate the differentials yB and yC through the angles and C Begin by computing the lengths of bars BC and BA (Note that these lengths do not change, as the angles and change). 3 2 m C B 5 2 2 3 2 m BC = (3 m) + (3 m) 3 m yC

B y 3 m B 5 m

A 6 AB = (3 m)2 + (4 m)2 5 m 4 m 7 From the above figure,

A 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 sin sin = (4) 5 m 3 2 m

11.1 Virtual Work Example 14, page 4 of 5 8 Differentiating Eq. 4 gives 9 Thus

cos cos P sin C (sin + tan cos ) (5 ) = 0 = y 5 3 2 Dividing by 5 gives Thus 5 cos P sin C (sin + tan cos ) = 0 (7) = (5) y 3 2 cos

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 cos 3 2 cos

= ( 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

11.1 Virtual Work Example 14, page 5 of 5 C 10 Evaluating the functions of and in Eq. 7, substituting the given 3 value P = 2 kN, and then solving gives tan = = 1 3 m 3 3 3 4 1 5 5 5 3 sin = P sin C (sin + tan cos ) = 0 (Eq. 7 repeated) 3 m 5 y B 4 Cy = 0.857 kN Ans. cos = 5 4 m 5 m

A 11 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.

11.1 Virtual Work Example 15, page 1 of 3 15. Determine the vertical reaction at support I of the truss, if P = 10 kip = Q.

P Q

I A B C D E F G H

5 ft 5 ft 5 ft 5 ft 5 ft 5 ft 5 ft 5 ft

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 Iy

A B C D E F G H I

11.1 Virtual Work Example 15, page 2 of 3 2 Identify the active forces corresponding to a set of displacements compatible with the constraints.

P Q Iy

A Ix B C D F G H I E

By Dy Fy Hy

P Q Iy A Ix C G H I yA B D yE E F yI

yA By Dy yE Fy Hy yI

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)

11.1 Virtual Work Example 15, page 3 of 3

4 Relate y , y , and y yC yG A E I A 5 ft 5 ft E 5 ft 5 ft I by geometry (similar triangles). B 5 ft C 5 ft D F 5 ft G 5 ft H yA yE yI

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 yI or

(P + Q Iy) yI = 0

Dividing through by yI, substituting the given values P = 10 kip = Q, and solving gives

Iy = 20 kip Ans.

11.1 Virtual Work Example 16, page 1 of 4 16. Determine the tension in the cord. The pulleys are frictionless and m = 90 kg.

A B T C

A B C 1 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.

D m

Weight = mg

11.1 Virtual Work Example 16, page 2 of 4 2 The tension T and the weight do work (are active forces) if end B of the cord moves up a small amount.

3 Introduce the coordinates yB and yD. T

A yB

B yB C

4 Calculate the work done:

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

D yD

11.1 Virtual Work Example 16, page 3 of 4

5 Relate y to y by first expressing the length, say L, of the cord Diameter dC B D in terms of yB and yD: y A B L = (y t) y s B D B

C + (yD t) u s

+ (yD t) s

+ dD/2 yD Half circumference of pulleys u + dC/2

Thus

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

11.1 Virtual Work Example 16, page 4 of 4 6 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

2 Dividing through by yD, substituting m = 90 kg, g = 9.81 m/s and solving gives

T = 294 N Ans.

11.1 Virtual Work Example 17, page 1 of 9 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.

500 mm

2 k = 0.2 kN/m B D

L 1

M C

11.1 Virtual Work Example 17, page 2 of 9 1 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.

2 Because point A does not move and is fixed, the reactions Ax and Ay, the weight 2 mg, and the spring force Fs do no work when 1 is varied a small amount. Thus they are not active forces. Fs (spring force) mg B

1 3 The couple moment M and the weight M mg of the lower bar BC do work when 1 C is varied, so M and mg are active forces. mg

11.1 Virtual Work Example 17, page 3 of 9

4 In addition to the coordinate 1, introduce a vertical coordinate yE measured downward from point A.

Ay A A

Ax L/2 L yE 2 2 yE L/2 B mg Fs L/2 B E 1

yE E L/2 L/2 6 Relate the differential yE to the angle change, 1, by writing C M mg yE = L cos + (L/2) cos 1 5 Compute the work done when the coordinates and then differentiating with respect to 1, while holding are increased a positive infinitesimal amount. fixed. That is, take the partial derivative with respect to 1 to obtain dU = 0: M 1 + mg yE = 0 (1) yE = (L/2) sin 1 1 (2)

11.1 Virtual Work Example 17, page 4 of 9

7 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.

11.1 Virtual Work Example 17, page 5 of 9 8 Next identify the active forces corresponding to 9 Because point A does not a small change in 2 while 1 is held fixed. Ay move, Ax and Ay do no work and thus are not active forces. A Ax

10 Because 1 is fixed, link BC does not rotate. That is, the mg dashed line representing the new Fs position of link BC is parallel B to BC. Hence couple moment M does no work, and M is not 1 11 Active forces an active force.

M mg C

11.1 Virtual Work Example 17, page 6 of 9 12 Introduce coordinates measured from a fixed point to the point of application of the active forces.

13 Compute the work done. Ay

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

A Ax

L/2 F yF

yE mg Fs

L/2 B

E yE

C mg

11.1 Virtual Work Example 17, page 7 of 9

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

x x = L sin B A B

yF = (L/2) cos L/2 yF yE = L cos (L/2) cos

F Differentiating each equation with respect to with held fixed, L/2 gives 2 yE xB = L cos (6) B yF = (L/2) sin (7) L/2 1 yE = L sin (8) L/2 E

11.1 Virtual Work Example 17, page 8 of 9 15 Substituting Eqs. 6-8 for the differentials into the virtual-work equation, Eq. 5, gives 500 mm L sin by Eq. 8 A mg yF + mg yE + Fs xB = 0 (Eq. 5 repeated)

L/2) sin by Eq. 7 L cos by Eq. 6 L Thus 500 mm L sin 2 2 [ 3mg/2) sin + Fs cos ] L = 0 B Since L 0, it follows that D 3mg/2) sin + Fs cos = 0 (9) 1

16 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)

11.1 Virtual Work Example 17, page 9 of 9

17 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.