PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

1780

PROBLEM 10.106

Two rods AC and CE are connected by a pin at C and by a spring AE. The constant of the spring is 1.5 N/m, and the spring is unstretched when

30 .θ = ° Knowing that l = .254 m and neglecting the weight of the rods,determine the value of θ corresponding to equilibrium when P = 40 N.

SOLUTION

cosEy l θ=

sinEy lδ θδθ= −

Spring: Unstretched length = ( )2 2 sin30 2l l° =

( )2 2 sin 4 sinx l lθ θ= =

4 cosx lδ θ δθ=

( )2SPF k x l= −

( )4 sin 2k l lθ= −

Virtual Work:

0: 0E SPU P y F xδ δ δ= − =

( ) ( )( )sin 4 sin 2 4 cos 0P l k l l lθδθ θ θδθ− − − =

or ( )sin 8 2sin 1 cos 0P klθ θ θ− − − =

or1 2sin

8 tanPkl

θθ

−=

We have 40 lb, = 10 in., and = 1.5 lb/in.P l k=

Thus( )

( )( )40 N 1 2sin

8 1.5 N/in. 10 in. tanθ

θ−=

Solving 24.98θ = °

or 25.0θ = ° t

PROBLEM 10.106

Two rods AC and CE are connected by a pin at C and by a spring AE. The constant of the spring is 27.16 N/m, and the spring is unstretched when

30 .θ = ° Knowing that l = .254 m and neglecting the weight of the rods,determine the value of θ corresponding to equilibrium when P = 40 N.

P = 40 N, l = 0.254 m, and k = 27.16 N/m

( )40 N 1 2sin8(27.16 N/m)(0.254 m) an

θθ

−=

21.1°θ =

or 21.1°θ =

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

1781

PROBLEM 10.107

For the linkage shown, determine the couple M required for equilibrium when l = .548 m, Q = 40 N, and 65 .θ = °

SOLUTION

12Bx lδ δφ=

cos cos 2cosx BC x l

Cδ δ δφδ

θ θ θ= = =

Virtual Work:

0: 0U M Q Cδ δφ δ= − =

2cosl

M Qδφ δφθ

− = 0

or2cos

QlM

θ=

Thus ( )( )40 N 1.8 ft1

25.5 N m2 cos65

M = = ⋅°

or 25.5 N mM = ⋅ t25.9

( )(0.548 m)40 N125.9 N m

2 cos65°M = = ⋅

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

1782

PROBLEM 10.108

Determine the vertical force P which must be applied at G so that the linkage is in equilibrium for the position shown.

SOLUTION

Assuming

Ayδ

it follows

1201.5

80C A Ay y yδ δ δ= =

1.5E C Ay y yδ δ δ= =

( )1803 1.5 4.5

60D E A Ay y y yδ δ δ δ= = =

( )100 1001.5 2.5

60 60G E A Ay y y yδ δ δ δ= = =

Then, by Virtual Work

( ) ( )0: 300 N 100 N 0A D GU y y P yδ δ δ δ= − + =

( ) ( )300 100 4.5 2.5 0A A Ay y P yδ δ δ− + =

300 450 2.5 0P− + =

60 NP = + 60 N=P t

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

1783

PROBLEM 10.109

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 Chap. 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

3

0: 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 Bδr and ,Fδr 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− =

( )575 mm 0

9P D Pδ − =

41.67 mmDδ = +

41.7 mmDδ = t

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

1784

PROBLEM 10.110

Determine the horizontal movement of joint D if the length of member BF is increased by 75 mm. (See the hint for Prob. 10.109.)

SOLUTION

Apply horizontal load P at D.

( ) ( )0: 9 m 36 m 0H yM P EΣ = − =

4yP

E =

3

0: 05 4y BF

PF FΣ = − =

5

12BFF 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 Bδr and ,Fδr 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− =

( )575 mm 0

12P D Pδ − =

31.25 mmDδ = 31.3 mmDδ = t

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

1785

PROBLEM 10.111

Two uniform rods, AB and CD, are attached to gears of equal radii as shown. Knowing that 3.5 kgABm = and 1.75 kg,CDm = determine the positions of equilibrium of the system, and state in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION

Potential Energy

( ) ( )2 23.5 kg 9.81m/s sin 1.75 kg 9.81m/s cos2 2l l

V θ θ = × − + ×

( ) ( )8.5838 N 2sin cosl θ θ= − +

( ) ( )8.5838 N 2cos sindV

ld

θ θθ

= − −

( ) ( )2

2 8.5838 N 2sin cosd V

ld

θ θθ

= −

Equilibrium: 0: 2cos sin 0dVd

θ θθ

= − − =

or tan 2θ = −

Thus 63.4 and 116.6θ = − ° °

Stability

At 63.4 :θ = − ° ( ) ( ) ( )2

2 8.5838 N 2sin 63.4 cos 63.4d V

ldθ

= − ° − − °

( ) ( )8.5838 N 1.788 0.448 0l= − − <

63.4 ,θ∴ = − ° Unstablet

At 116.6 :θ = ° ( ) ( ) ( )2

2 8.5838 N 2sin 116.6 cos 116.6d V

ldθ

= ° − °

( ) ( )8.5838 N 1.788 0.447 0l= + >

116.6 ,θ∴ = ° Stablet

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

1786

PROBLEM 10.112

A slender rod AB of mass m is attached to two blocks A and B that can move freely in the guides shown. Knowing that the spring is unstretched when AB is horizontal, determine three values of θ corresponding to equilibrium when m = 125 kg, l = 320 mm, and k = 15 kN/m. State in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION

Elongation of Spring: sin coss l l lθ θ= + −

( )sin cos 1l θ θ= + −

Potential Energy:

21sin

2 2l

V ks W θ= −

( )221sin cos 1 sin

2 2l

kl mgθ θ θ= + − −

( )( )2 1sin cos 1 cos sin cos

2dV

kl mgld

θ θ θ θ θθ

= + − − −

Equilibrium: ( )( )0: sin cos 1 cos sin cos 02

dV mgd kl

θ θ θ θ θθ

= + − − − =

or ( )( )cos sin cos 1 1 tan 02mgkl

θ θ θ θ + − − − =

Now with ( )( )2125 kg 9.81 m/s 1226.25 NW mg= = =

320 mm, and = 15 kN/m,l k=

( )( ) ( )( )1226.25 N

cos sin cos 1 1 tan 02 15000 N/m 0.32 m

θ θ θ θ

+ − − − =

or ( )( )cos sin cos 1 1 tan 0.12773 0θ θ θ θ + − − − =

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

1787

PROBLEM 10.112 CONTINUED

By inspection, one solution is cos 0 or 90.0θ θ= = °

Solving numerically: 0.38338 rad = 9.6883 and = 0.59053 rad = 33.8351θ θ= ° °

Stability

( )( ) ( )( )2

22

1cos sin cos sin sin cos 1 sin cos sin

2d V

kl mgld

θ θ θ θ θ θ θ θ θθ

= − − + + − − − +

2 1 + sin cos 2sin22mg

klkl

θ θ θ = + −

( )( ) ( )( )( )

2 1226.25 N15000 N/m 0.32 m 1 sin cos 2sin2

2 15000 N/m 0.32 mθ θ θ

= + + −

( )[ ]1536 N m 1.12773 sin cos 2sin2θ θ θ= ⋅ + −

Thus

At = 90 :θ ° 2

2 1732.2 0d Vdθ

= > 90.0 , Stableθ∴ = ° t

At = 9.6883 :θ ° 2

2 786.4 0d Vdθ

= > 9.69 , Stableθ∴ = ° t

At 33.8351 :θ = ° 2

2 600.6 0d Vdθ

= − < 33.8 , Unstableθ∴ = ° t