DESIGN OF MACHINERY
!
SOLUTION MANUAL 10-9-1
PROBLEM 10-9
Statement:
Two dampers are connected in series. One has a damping factor of c1 = 12.5 and the other, c2 = 1.2.
Calculate their effective damping constant. Which damper dominates? Repeat with the two
dampers in parallel. Which damper dominates? (Use any unit system.)
Given:
Damping factors:
c1
Solution:
1.
12.5 .
N .sec
mm
c2
1.2 .
N .sec
mm
See Mathcad file P1009.
For dampers in series, use equation 10.16a to find the effective damping factor.
c eff
1
1
c1
c2
1
c eff = 1.095
N .sec
mm
The softer damper dominates.
2.
For dampers in parallel, use equation 10.16b to find the effective damping factor.
c eff
c1
c2
The stiffer damper dominates.
2nd Edition, 1999
c eff = 13.700
N .sec
mm
DESIGN OF MACHINERY
!
SOLUTION MANUAL 10-12-1
PROBLEM 10-12
Statement:
A mass of m = 2.5 and a spring with k = 42 are attached to one end of a lever at a radius of 4.
Calculate the effective mass and effective spring constant at a radius of 12 on the same lever. (Use
any unit system.)
Given:
Mass:
M
Spring:
2.5 .kg
k
42 .
Radius:
N
mm
Solution:
1.
4 .mm
r2
12 .mm
See Figure 10-8 and Mathcad file P1012.
The effective mass and spring must have the same energy as the original. Taking the mass first and using
equation 10.20b,
m eff
2.
r1
r1
2
.M
r2
m eff = 0.278 kg
To find the effective spring rate, use equation 10.21b.
k eff
2nd Edition, 1999
r1
r2
2
.k
k eff = 4.667
N
mm
DESIGN OF MACHINERY
!
SOLUTION MANUAL 10-15-1
PROBLEM 10-15
Statement:
Given:
Refer to Figure 10-9 and Example 10-1. The data for the valve train are given below. Calculate the
effective spring constant and effective mass of a single-DOF equivalent system placed on the cam
side of the rocker arm. (Use ips unit system.)
Tappet (solid cylinder): d tp 0.75 .in L tp 1.25 .in
Pushrod (hollow cylinder):
d pod 0.375 .in
d pid 0.250 .in
Rocker arm:
w 1.00 .in
h 1.50 .in
a 2.00 .in
b
Cam shaft (cam in center):
Valve spring:
k vs
All parts are steel:
d cs
200 .lbf .in
1.00 .in
12 .in
L pr
3.00 .in
3.00 .in
L cs
1
Modulus of elasticity
6
30 .10 .psi
E
γ
Spec. weight
0.3 .
lbf
3
in
Solution:
See Figure 10-9 and Mathcad file P1015.
1.
Break the system into individual elements as shown in Figure 10-9b.
2.
Define the individual spring constants of each of the six elements.
Cam shaft (simply-supported beam with central load):
Moment of inertia
Spring constant
I cs
k cs
π .d cs
4
64
48 .E .I cs
L cs
6 lbf
k cs = 2.618 .10
3
in
Tappet (solid cylinder):
π .d tp
2
Area
Spring constant
A tp
k tp
4
A tp .E
7 lbf
k tp = 1.060 .10
in
L tp
Pushrod (hollow cylinder):
Area
Spring constant
A pr
k pr
π . d pod
2
d pid
2
4
A pr .E
L pr
5 lbf
k pr = 1.534 .10
in
Rocker arm (side A):
Moment of inertia
Spring constant
Ir
k ra
w .h
3
12
3 .E .I r
a
3
6 lbf
k ra = 3.164 .10
in
Rocker arm (side B):
Spring constant
k rb
3 .E .I r
b
2nd Edition, 1999
3
5 lbf
k rb = 9.375 .10
in
DESIGN OF MACHINERY
SOLUTION MANUAL 10-15-2
3.
Damping will be neglected.
4.
Determine the mass of each of the elements.
Tappet (solid cylinder):
Volume
V tp
Mass
m tp
A tp .L tp
γ .V tp
m tp = 4.291 .10
g
4
2
1
lbf .sec .in
Pushrod (hollow cylinder):
Volume
V pr
Mass
m pr
A pr .L pr
γ .V pr
4
2
1
m pr = 5.721 .10 lbf .sec .in
g
Rocker arm (side A):
Volume
V ra
Mass
m ra
w .h .a
γ .V ra
m ra = 2.331 .10
g
3
2
1
lbf .sec .in
Rocker arm (side B):
Volume
V rb
Mass
m rb
w .h .b
γ .V rb
3
2
1
m rb = 3.497 .10 lbf .sec .in
g
Omit the valve and valve spring because no data are available.
5.
Determine the effective mass and spring constant on either side of the rocker arm..
Left side:
mL
kL
Right side:
mR
kR
6.
m L = 3.332 .10
m tp
m pr
m ra
1
1
1
1
k cs
k tp
k pr
k ra
1
2
1
lbf .sec .in
3
5 lbf
k L = 1.368 .10
in
3
2
1
m R = 3.497 .10 lbf .sec .in
m rb
1
1
1
k rb
k vs
k R = 199.957
lbf
in
Reduce the system to a single DOF.
2nd Edition, 1999
m eff
mL
k eff
kL
2
b
.m
a
b
a
2
.k
R
R
2
1
m eff = 0.011 lbf .sec .in
k eff = 1.372 .10
5 lbf
in