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. B 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 Cx 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 M 2 ft 2 ft W B D 11.1 Virtual Work Example 2, page 2 of 3 A C M 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 P A B C W 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.