ME 302 THEORY OF MACHINES II

SECTIONS 2 & 3

SPRING 2013

HOMEWORK 1

IMPORTANT NOTE: This homework is composed of supplementary problems. Solutions

will NOT be collected and graded.

Question-1. The elevation of the load of mass m is controlled by the adjusting screw which

connects joints A and B. The change in the distance between A and B for one revolution of the

screw equals to the lead L of the screw. Determine the moment M, which is applied to the

adjusting screw, necessary to raise the load. Assume that the masses of the links are negligible

with respect to load m.

Answer: cotmgL

M θπ

=

Question-2. The mechanical system shown below is on horizontal plane (no gravity). Find the

necessary spring force Fs to hold the system at static equilibrium for the given position. Note

that W acts at the midpoint of |AC|. Use the virtual work method.

Answer: Fs = 3000 N

Question-3. Considering the mechanism shown in the figure, calculate the torque T which is

necessary to keep the mechanism in equilibrium with the force F. Use the virtual work

method and neglect gravitational and frictional effects.

( 1)i

DOF l j fλ= − − +∑

4l = , 4j = , 4i

f =∑

3(4 4 1) 4 1DOF = − − + =

Virtual work equation can be writen as follows:

14 12 0W T F sδ δθ δ= ⋅ + ⋅ =�� ����� �� �����

In the virtual work equation vector multiplication can be performed as follows:

12 12 12( )( )T T k k Tδθ δθ δθ⋅ = ⋅ ⋅ =

�� ���� � �

12 12 12( )( )F s F j s j F sδ δ δ⋅ = ⋅ − ⋅ = −�� ����� � �

Therefore virtual work equation can be written as follows:

14 12 0W T F sδ δθ δ= − =

Loop closure equation can be written as follows:

14 14( /2)3 /2 (3 /2)12 1 34 3 2

i ii i is e a e s e a e a e

θ θ ππ π π++ = + +

Real part of the LCE is:

12 1 34 14 3 14 2cos(3 / 2) cos( ) cos( ) cos( / 2) cos(3 / 2)s a s a aπ π θ θ π π+ = + + +

12sδ

34sδ

14δθ

1 34 14 3 14cos( ) sin( )a s aθ θ− = −

Imaginary part of the LCE is:

12 1 34 14 3 14 2sin(3 / 2) sin( ) sin( ) sin( / 2) sin(3 / 2)s a s a aπ π θ θ π π+ = + + +

12 34 14 3 14 2sin( ) cos( )s s a aθ θ− = + −

The virtual variation of the real and imaginary parts of the LCEs leads to

34 14 34 14 14 3 14 140 cos( ) sin( ) cos( )s s aδ θ θ δθ θ δθ= − +

3 14 34 1434 14

14

( cos sin )

cos

a ss

θ θδ δθ

θ

+=

12 34 14 34 14 14 3 14 14sin cos sin( )s s s aδ δ θ θ δθ θ δθ− = + −

3 14 34 1412 14 14 34 14 14 3 14 14

14

( cos sin )sin cos sin( )

cos

a ss s a

θ θδ δθ θ θ δθ θ δθ

θ

+− = + −

3 14 34 1412 14 34 14 3 14 14

14

( cos sin )sin cos sin

cos

a ss s a

θ θδ θ θ θ δθ

θ

+− = + −

3412 14

14cos

ssδ δθ

θ− =

If we substitute the preceding expressions into the virtual work equation, we obtain:

14 12 0W T F sδ δθ δ= − =

3414 14

14

0cos

sW T Fδ δθ δθ

θ

−= − =

3414

14

0cos

sT F δθ

θ

−− =

34

14cos

sT F

θ= −

Question-4. Consider the 5-bar mechanism shown in the figure. Determine the torques T1 and

T2 in order to maintain the static equilibrium against a given vertical force F using virtual

work. Take L as the unstretched length of the spring. Assume that the kinematics of the

system have been fully solved and any position related variable is known. Neglect friction and

gravity.

Answer:

Question-5. The mechanical system shown in the figure is at vertical plane. The spring is free

when s14 = b. Only link-4 has mass m4, others are assumed massless. Link-2 and link-3 have

lengths ll2 and ll3, respectively. Using virtual work method, derive the equation of motion.

Take θ12 as independent variable.

Solution to Question-5:

Inspecting the geometry of the mechanism, it can easily be realized that ;

The required torque is

Question-6. The double slider operating in the horizontal plane consists of two sliding blocks

each having mass m and a uniform slender bar with mass m and length l. A spring with

stiffness constant k is attached to block 2 and is free when θ =150°. The mechanism is driven

by force F applied on block 4. Employing virtual work method, derive the differential

equation of motion using θ as the independent variable. Neglect friction and gravitational

effects.

Solution to Question-6:

Question-7. The mechanical system shown works in a horizontal plane (no gravity). The

spring is free when s54

=b. The horizontal force F is known. Find the Torque T and the

hydraulic force P which are required to hold the system at static equilibrium. Use the virtual

work method. Indicate clearly the independent generalized coordinates you select.

( 1)i

DOF l j fλ= − − +∑

5l = , 5j = , 5i

f =∑

3(5 5 1) 5 2DOF = − − + =

Since the DOF of the mechanism is 2, we need two independent variables. They can be

selected as 12θ and 23s .

Virtual work equation can be writen as follows:

12 23 54 0sBW T F x P s F sδ δθ δ δ δ= ⋅ + ⋅ + ⋅ + ⋅ =�� ����� �� ����� �� ����� �� �����

If the spring is assumed to be compressed, 54b s> . So,

54( )sF k b s= −

In the virtual work equation vector multiplication can be performed as follows:

12 12 12( )( )T T k k Tδθ δθ δθ⋅ = ⋅ ⋅ =

�� ���� � �

( )( )B B BF x F i x i F xδ δ δ⋅ = − ⋅ ⋅ = −�� ����� � �

Since hydraulic forces are trying to increase s23, virtual work done by hydraulic forces is

23sPδ . Similarly virtual work done by spring forces is 54sFs δ .

Therefore virtual work equation can be written as follows:

12 23 54 0B sW T F x P s F sδ δθ δ δ δ= − + + =

Note that

xB Bxδ

23sδ

54sδ

Fs

23 12cosBx s θ= ⇒ 23 12 12sinBx sδ θ δθ= −

In order to obtain 54sδ interms of the selected independent variables, loop closure equation

(LCE) can be used.

Loop closure equation can be written as follows:

151223 1 54

iis e a s e

θθ = +

Real part of the LCE is:

23 12 1 54 15cos coss a sθ θ= +

Imaginary part of the LCE is:

23 12 54 15sin sins sθ θ=

The virtual variations of the real and imaginary parts are:

23 12 23 12 12 54 15 54 15 15cos sin cos sins s s sδ θ θ δθ δ θ θ δθ− = −

23 12 23 12 12 54 15 54 15 15sin cos sin coss s s sδ θ θ δθ δ θ θ δθ+ = +

If we multiply the first equation by 15cosθ and the second equation by 15sinθ and add both

expressions we obtain:

54 15 12 23 23 15 12 12cos( ) sin( )s s sδ θ θ δ θ θ δθ= − + −

Finally by substituting the expressions into the virtual work equation we find:

23 15 12 23 12 12 15 12 12 23( sin( ) sin ) ( cos( ) cos ) 0s sW T F s Fs P F F sδ θ θ θ δθ θ θ θ δ= + − + + + − =

Since 23sδ and 12δθ are arbitrary independent virtual variations,

23 15 12 23 12sin( ) sin 0sT F s Fsθ θ θ+ − + =

15 12 12cos( ) cos 0sP F Fθ θ θ+ − =

Hence, we find that

23 15 12 23 12sin( ) sinsT F s Fsθ θ θ= − − −

15 12 12cos( ) cossP F Fθ θ θ= − − +

Question-8. The massless rod shown in the figure has two masses on it, one mass m1 is fixed

at the end, while the other m2 is constrained to move along the radius by a linear spring k.

Derive the equations of motion of the system if a constant torque T is applied.

Answer:

( ) ( )

( ) ( )

2

2 2

2 2

1 2 1 2

/ 2 cos 0

2 sin

m r r k r L m g

m L m rr r m L m r g T

θ θ

θ θ θ θ

− + − − =

+ + + + =

���

�� � ���

Question-9 For the mechanical system shown below, find equilibrium state, using virtual work method. Stiffness coefficient of the spring is k, mass of block is m1 and mass of rod is m2. Rod’s

length is L. (|G1G2| = L/2)

Question-10. Consider the cam mechanism shown in the figure. The dimensions of the

mechanism are given as follows.

|FO| = b1, |FC| = c1, |CG| = d1, |AB| = c2

|OA| = b2, |CE| = b3,

Load P is applied to the slider (link 4). Assume that the joint variables and their 1st & 2

nd

derivatives with respect to time are known. Determine the analytical expression for the motor

torque (T) to maintain constant angular velocity ω12 using virtual work method. Neglect

frictional effects on the system.

Link No Center of Gravity Mass Moment of Inertia about CoG

Link 2 Point A m2 IA

Link 3 Point C m3 IC

Link 4 Point D m4 - Answer:

( ) ( ) ( )34 2 12 13 2 12 13 34 2 12 1313 4 14 2 2 12

32 13 32 32 13

sin sin sincos

sin sinC

s b b s bT P I m s m gb

s s s

θ θ θ θ θ θθ θ

θ θ

− − −= − − + +�� ��

s14

A O

B

C

D

E

F

G

θ12

θ13

θ13

s34

s32

b1

c1

d1

b2

c2

b3

(1) (2)

(3)

(4)

(1)

(3)

ω12

P

g

T

+x

+y