ME 302 THEORY OF MACHINES II SECTIONS 2 & 3...
Transcript of ME 302 THEORY OF MACHINES II SECTIONS 2 & 3...
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 13
13 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