PROBLEM 10.51 Denoting by sµ the coefficient of static friction between the block attached to rod ACE and the horizontal surface, derive expressions in terms of P, ,sµ and θ for the largest and smallest magnitudes of the force Q for which equilibrium is maintained.

SOLUTION

For the linkage:

0: 0 or2 2A

B Ax PM x PΣ = − + = =A

Then: 12 2s s sPF A Pµ µ µ= = =

Now 2 sinAx l θ=

2 cosAx lδ θ δθ=

and 3 cosFy l θ=

3 sinFy lδ θ δθ= −

Virtual Work:

( )max0: 0A FU Q F x P yδ δ δ= − + =

( ) ( )max1 2 cos 3 sin 02 sQ P l P lµ θ δθ θ δθ − + − =

or max3 1tan2 2 sQ P Pθ µ= +

( )max 3tan2 sPQ θ µ= +

For min,Q motion of A impends to the right and F acts to the left. We change sµ to sµ− and find

( )min 3tan2 sPQ θ µ= −

PROBLEM 10.52 Knowing that the coefficient of static friction between the block attached to rod ACE and the horizontal surface is 0.15, determine the magnitudes of the largest and smallest force Q for which equilibrium is maintained when o30θ = , 8 in.,l = and 160 lb.P =

SOLUTION

Using the results of Problem 10.52 with

30 , 8 in., 160 lb, and 0.15sl Pθ µ= ° = = =

We have ( ) ( )max160 lb

3tan 30 0.15 150.56 lb2

Q = ° + =

max 150.6 lbQ =

and ( ) ( )min160 lb

3tan 30 0.15 126.56 lb2

Q = ° − =

min 126.6 lbQ =

PROBLEM 10.53 Using the method of virtual work, determine separately the force and the couple representing the reaction at A.

SOLUTION

:yA Consider an upward displacement Ayδ of ABC

ABC: A B Cy y yδ δ δ= =

CDE: 1 ft 2.5 ft

C Ey yδ δ=

or 2.5E Ay yδ δ=

EFG: 0.8 ft 1.2 ft

E Gy yδ δ=

or ( )1.2 ft 2.50.8 ftG Ay yδ δ=

3.75 Ayδ=

Virtual Work:

( ) ( )0: 240 lb 60 lb 0y A B GU A y y yδ δ δ δ= + − =

or ( ) ( )240 lb 60 lb 3.75 0y A A AA y y yδ δ δ+ − =

or 15 lby =A

:xA Consider a horizontal displacement Axδ :

Virtual Work: 0: 0x AU A xδ δ= =

or 0xA = 15.00 lb∴ =A

:AM Consider a counterclockwise rotation about A:

ABC: 2 , 3B A C Ay yδ δθ δ δθ= =

CDE: 1 ft 2.5 ft

C Ey yδ δ=

or ( )2.5 3E Ayδ δθ=

7.5 Aδθ=

EFG: 0.8 ft 1.2 ft

E Gy yδ δ=

PROBLEM 10.53 CONTINUED

or ( )( ) ( )1.2 ft

7.50.8 ftG Ayδ δθ=

11.25 Aδθ=

Virtual Work: ( ) ( )0: 240 lb 60 lb 0A A B GU M y yδ δθ δ δ= + − =

or ( )( ) ( )( )240 lb 2 60 lb 11.25 0A A A AM δθ δθ δθ+ − =

or 195.0 lb ftA = ⋅M

PROBLEM 10.54 Using the method of virtual work, determine the reaction at D.

SOLUTION

Consider an upward displacement Eyδ of pin E.

CDE: 1 ft 3.5 ft

D Ey yδ δ=

or 13.5Dy Eδ δ=

EFG: 0.8 ft 1.2 ft

E Gy yδ δ=

or 1.5G Ey yδ δ=

Virtual Work:

0: 60 0D GU D y yδ δ δ= + =

or ( )( )1 60 lb 1.5 03.5 E ED y yδ δ + =

or 315 lb=D

PROBLEM 10.55

Referring to Problem 10.41 and using the value found for the force exerted by the hydraulic cylinder AB, determine the change in the length of AB required to raise the 480-N load 18 mm.

SOLUTION

From the solution to Problem 10.41

cyl 397.08 NF =

And, Virtual Work: cyl0: 0AB DU F S P yδ δ δ= − =

where 0ABSδ < for 0Dyδ >

Then ( ) ( )( )397.08 N 480 N 18 mm 0ABSδ − =

( )or 21.8 mm shortenedABSδ =

PROBLEM 10.56

Referring to Problem 10.45 and using the value found for the force exerted by the hydraulic cylinder BD, determine the change in the length of BD required to raise the platform attached at C by 50 mm.

SOLUTION

Virtual Work: Assume that both Cyδ and BDδ increase

( )0: 2000 N 0C BD BDU y Fδ δ δ= − + =

( )0.05 m and 9473 N from Problem 10.45C BDy Fδ = =

( )2000 0.05 m 9473 0BDδ− + =

0.010556 mBDδ =

10.556 mm=

The positive sign indicates that BD gets longer.

10.56 mmBDδ =

PROBLEM 10.57 Determine the vertical movement of joint D if the length of member BF is increased by 75 mm. (Hint: Apply a vertical load at joint D, and, using the methods of Chapter 6, compute the force exerted by member BF on joints B and F. Then apply the method of virtual work for a virtual displacement resulting in the specified increase in length of member BF. This method should be used only for small changes in the lengths of members.)

SOLUTION

Apply vertical load P at D.

( ) ( )0: 12 m 36 m 0HM P EΣ = − + =

3P

=E

30: 05 3y BF

PF FΣ = − =

59BFF P=

Virtual Work:

We remove member BF and replace it with forces BFF and BF−F at pins F and B, respectively. Denoting the virtual displacements of points B and F as Bδ r and ,Fδ r respectively, and noting that P and Dδ have the same direction, we have

Virtual Work: ( )0: 0BF F BF BU P Dδ δ δ δ= + ⋅ + − ⋅ =F r F r

cos cos 0BF F F BF B BP D F r F rδ δ θ δ θ+ − =

( )cos cos 0BF B B F FP D F r rδ δ θ δ θ− − =

where ( )cos cos ,B B F F BFr rδ θ δ θ δ− = which is the change in length of member BF. Thus,

0BF BFP D Fδ δ− =

( )5 75 mm 09

P D Pδ − =

41.67 mmDδ = +

41.7 mmDδ =

PROBLEM 10.58 Determine the horizontal movement of joint D if the length of member BF is increased by 75 mm. (See the hint for Problem 10.57.)

SOLUTION

Apply horizontal load P at D.

( ) ( )0: 9 m 36 m 0H yM P EΣ = − =

4yPE =

30: 05 4y BF

PF FΣ = − =

512BFF P=

We remove member BF and replace it with forces BFF and BF−F at pins F and B, respectively. Denoting the virtual displacements of points B and F as Bδ r and ,Fδ r respectively, and noting that P and Dδ have the same direction, we have

Virtual Work: ( )0: 0BF F BF BU P Dδ δ δ δ= + ⋅ + − ⋅ =F r F r

cos cos 0BF F F BF B BP D F r F rδ δ θ δ θ+ − =

( )cos cos 0BF B B F FP D F r rδ δ θ δ θ− − =

where ( )cos cos ,B B F F BFr rδ θ δ θ δ− = which is the change in length of member BF. Thus,

0BF BFP D Fδ δ− =

( )5 75 mm 09

P D Pδ − =

31.25 mmDδ = 31.3 mmDδ =

PROBLEM 10.59

Using the method of Section 10.8, solve Problem 10.29.

SOLUTION

Spring:

( )2 2 sin 4 sinAE x l lθ θ= = =

Unstretched length:

0 4 sin 30 2x l l= ° =

Deflection of spring

0s x x= −

( )2 2sin 1s l θ= −

212 EV ks Py= +

( ) ( )21 2 2sin 1 cos2

V k l P lθ θ = − + −

( )24 2sin 1 2cos sin 0dV kl Pld

θ θ θθ

= − + =

( ) cos2sin 1 0sin 8

Pkl

θθθ

− + =

1 2sin8 tanPkl

θθ

−=

PROBLEM 10.60

Using the method of Section 10.8, solve Problem 10.30.

SOLUTION

Using the result of Problem 10.59, with

160 N, 200 mm, and 300 N/mP l k= = =

1 2sin8 tanPkl

θθ

−=

or ( )( )

1 2sin 160 Ntan 8 300 N/m 0.2 m

θθ

−=

13

=

Solving numerically, 25.0θ = °