Download 7.12 and 7.13

Cams
7.12
1
ANALYSIS OF A RIGID ECCENTRIC CAM
Analysis of a rigid eccentric cam involves the determination of the contact
force, the spring force and the cam shaft torque for one revolution of the cam.
In simplified analysis, all the components of the cam system are assumed to be
rigid and the results are applicable to low speed systems. However, if the
speeds are high and the members are elastic, an elastic body analysis must be
made. The elasticity of the members may be due to extreme length of the
follower or due to use of elastic materials in the system. In such cases, noise,
excessive wear, chatter, fatigue failure of some of the parts are the usual things.
A circular disc cam with the cam shaft hole drilled off centre is known as an
eccentric plate cam. Figure 7.38(a) shows a simplified reciprocating eccentric
cam system consisting of a plate cam, a flat face follower and a retaining
spring.
Fig. 7.38
Let,
e = eccentricity, the distance between the centre of the disc and of the
shaft
m = mass of the follower
s = stiffness of the retaining spring
w = angular velocity of the cam rotation
P = preload including the weight of the follower or the force on the cam
at x = 0
x = motion of the follower (zero at the bottom of the stroke)
If the disc is rotated through an angle q [Fig. 7.38(b)], the mass m is
displaced by a distance x, so that
x = e – e cos q
(7.33)
Velocity,
x = dx = dx d q = e w sin q
dt d q dt
(7.34)
2
Theory of Machines
x = dv = dv d q = ew 2 = cos q
(7.35)
dt d q dt
Now, as the follower is displaced in the upward direction, the acceleration
x of the follower is in the upward direction indicating the inertia force to be in
the downward direction. The spring force is also exerted in the downward
direction. The force exerted by the cam on the follower F, however, will be in
the upward direction.
Thus, the various forces acting on the follower mass are,
(downwards)
Inertia force = mx
Spring force = sx
(downwards)
Preload = P
(downwards)
Force exerted by cam = F
(upwards)
Thus, the equilibrium equation becomes,
mx + sx + P – F = 0
F = mx + sx + P = mew 2 cos w t + se – se cos w t + P
= (se + P) + (mw 2 – s) e cos w t
(7.36)
Equation 7.36 shows that the force F exerted by cam on the follower consists of a constant term (se+P) with a cosine wave superimposed on it, the
maximum value of which occurs at q = 0o and minimum at q = 180o. Figure 7.39
shows the change of this force with the angular displacement of the cam. As
the cam shaft velocity increases, the term involving square of the velocity
increases at a faster rate and becomes zero at a speed when
(7.37)
(se + P) + (mw 2 – s) e cos w t = 0
Acceleration,
Fig. 7.39
This can happen at q = 180o. At that speed, there will be some impact
between the cam and the follower, resulting in rattling, clicking and noisy
operation. This is usually known as jump. However, this can be prevented to
some extent by increasing the preload exerted by the spring or the spring
stiffness.
Cams
At jump speed,
(se + P) – (mw 2 – s) e = 0
or
3
2se + P – mew 2 = 0
2se + P
(7.38)
me
and jump will not occur if preload of the spring is increased such that
P > e(mw 2 – 2s)
The torque applied by the shaft to the cam, T = F ◊ e sin w t
= [(se + P) + (mw 2 – s)e cos w t]e sin w t
or
w=
e2
(mw 2 – s)sin2w t
(7.39)
2
Figure 7.40 shows the variation of torque with the cam rotation. It may be
observed from the plots that area of the torque-displacement diagram above
and below the x-axis is the same meaning that the energy required to raise the
follower is recovered during the return. A flywheel may be used to handle this
fluctuation of energy.
= e(se + P) sin w t +
Fig. 7.40
Example 7.12 A circular disc cam of diameter 120 mm with its centre displaced
40 mm from the camshaft is used with a flat surface follower. The line of action of
the follower is vertical and passes through the shaft axis. The mass of the follower is
3 kg and is pressed downwards with a spring of stiffness 5 N/mm. In the lowest
position, the spring force is 60 N.
Derive an expression for the acceleration of the follower as a function of cam
rotation from the lowest position of the follower. Also, find the speed at which the
follower begins to lift from the cam surface.
Solution
e = 40 mm, m = 3 kg, s = 5 N/mm = 5 000 N/m, P = 60 N + mg = (60 + 3 ¥ 9.81) N
Consider the rotation of the cam through angle q (Refer Fig. 7.41),
Now,
x = 0.04 – 0.04 e cos q = 0.04(1 – cos q)
4
Theory of Machines
Fig. 7.41
x = dx = dx dq = 40w sin q
dt
dq dt
x = dv = dv dq = 40w 2 cos q
dt dq dt
which is the required expression for acceleration of the cam follower system.
To find the speed at which the follower begins to lift from the cam surface or the
jump speed,
w=
=
or
or
7.13
2se + P
me
2 ¥ 5000 ¥ 0.04 + 60 + 3 ¥ 9.81
=
3 ¥ 0.04
[Eq. (7.38)]
4078.6 = 63.86 rad/s
2p N
= 61.91
60
N = 609.9 rpm
ANALYSIS OF AN ELASTIC CAM SYSTEM
To illustrate the effect of follower elasticity upon its displacement and velocity,
consider a simplified model of a cam system with a linear motion used for lowspeed cams (Fig. 7.42).
Let m = lumped mass of the follower
s1 = stiffness of the retaining spring
s2 = stiffness of the follower
x = displacement of the lumped mass of the follower
y = motion machined into the cam surface
h = lift of the follower
5
Cams
Fig. 7.42
Assume a cam profile that gives a uniform rise for an angle of rotation j
followed by a dwell. Thus, y = h
wt
q
=h
j
j
As the follower is usually a rod, its stiffness s2 is far greater than the
stiffness s1 of the spring. The spring is assembled in such a way that it exerts
a preload force. The displacement x of the lumped mass is taken from the
equilibrium position after the spring is assembled. In the equilibrium position,
the spring and the follower exert equal and opposite preload forces on the
mass.
Assuming the displacement x to be more than y, the various forces acting on
the follower mass are
(downwards)
Inertia force = mx
Spring force = s1x
(downwards)
(upwards)
Force of elastic follower = s2(x – y)
Thus, mx + s1x1 + s2 (x – y) = 0
x +
s1 + s2
s
x= 2 y
m
m
s2
y
m
During ascent, the displacement of the follower mass is given by the solution of the above equation i.e.
x + wn 2x =
x = A cos wnt + B sin wn t +
s2
mw n
2
y
(i)
6
Theory of Machines
Differentiating with respect to t,
s2
x = –Awn sin wnt + Bwn cos w nt +
When t = 0, x = x = 0 and y = 0
\ from (i), 0 = A + 0 + 0 or A = 0
s2
From (ii),
0 = 0 + Bwn +
y
2
mw n
\ (i) becomes, x = 0 –
The particular integral
s2
mw n
s2
mw n
x=
=
Let
\
3
2
y sin wn t +
or
B=–
s2
mw n
2
mw n
2
s2
y
mw n
2
y
(ii)
y
(7.40)
y is called the follower command.
s2 Ê
y
ˆ
sin w n t ˜
y2 Á
w
Ë
¯
n
w
m n
Ê
h ˆ
ÁË y = j w ˜¯
s2 Ê h
hw
sin w n t ˆ˜
q2 Á
j
j
w
Ë
¯
n
w
m n
wn
= r; wn t = rw t = rq
w
s2 h
(rq – sin rq)
x=
2
rmw n j
=
x =
s2 h Ê w n
ˆ
s2 h Ê
ˆ
1
w t - sin w nq ˜ =
sin w n t ˜
tÁ
2 Á
wn
rmw nj Ë
¯
¯
rmw n j Ë w
s2 h
s2 h Ê w n
ˆ
1cos w nt ˜ =
(1 – cos rq)
rmw nj ÁË w n
rmw nj
¯
At the peak, x = x1, q = j
x1 =
s2 h
2
rmw n j
(rj – sin rj)
s2 h
(1 – cos j)
rmw nj
These become the initial conditions for dwell period.
x1 =
For dwell period, x = A cos wn t + B sin wn t +
x = –A sin wn t + Bwn cos wn t
s2
mw n
2
h
Ê s2
ˆ
is constant
ÁË mw 2 h
˜¯
n
7
Cams
At t = 0, x = x1, y = h, x = x1
x1 = A +
x1 = Bwn
s2 h
mw n
2
\
,
\ B=
A = x1 –
s2 h
mw n
2
,
x1
wn
Therefore, during the dwell period,
Ê
x
s2
s h ˆ
h
x = Á x1 - 2 2 ˜ cos wnt + 1 sin wnt +
2
wn
mw n
mw n ¯
Ë
(7.41)
Figure 7.43 shows the response of the follower x, the cam motion y, and the
follower command. The follower command is different from the cam motion
due to the elasticity of the follower.
Fig. 7.43