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ME 302 DYNAMICS OF ME 302 DYNAMICS OF MACHINERYMACHINERY
Dynamic Force Analysis
Dr. Sadettin KAPUCU
© 2007 Sadettin Kapucu
2Gaziantep University
IntroductionIntroduction
If the acceleration of moving links in a mechanism is running with considerable amount of linear and/or angular accelerations, inertia forces are generated and these inertia forces also must be overcome by the driving motor as an addition to the forces exerted by the external load or work the mechanism does. So, and are no longer applicable.Governing rules will be:
I and m are inertial (bodily) properties. At this stage we need to know the description of the inertial properties.
0F
0 τ
aFmατI
3Gaziantep University
CentroidCentroid
Centroid is the point where the resultant of distributed force system is assumed to act and generate the same dynamic results.
Centroid
Resultant Force
4Gaziantep University
Mass CentreMass Centre
If the distributed force is gravity force acting on each particle of mass, then concentrated force itself is called the “weight” and the centroid is called the “center of gravity” or “mass center”. Mass times distance, mr, is called as the first mass moment. This concept of first mass moment is normally used in deriving the center of mass of a system of particles or a rigid body. In figure a series of masses are located on a line. The center of mass or centroid is located at
x3
x1
x2
x-m1 m2 m3
G 321
332211
mmm
mxmxmxx
n
ii
n
iii
m
mxx
1
1
5Gaziantep University
Mass CentreMass Centre
The coordinates of the masses located on a plane can be obtained as:
m1
m2
m3
x-
y-
G321
332211
1
1
mmm
mxmxmx
m
mxx
n
ii
n
iii
321
332211
1
1
mmm
mymymy
m
myy
n
ii
n
iii
6Gaziantep University
Mass CentreMass CentreThis procedure can be extended to masses concentrated in a volume by simply writing an equation for the z axis. A more general form of mass center location for three dimensional body can be obtained by using integration instead of summation. The relations then become
n
ii
n
iii
m
mxx
1
1
n
ii
n
iii
m
myy
1
1
n
ii
n
iii
m
mzz
1
1
m
xdmx
m
ydmy
m
zdmz
7Gaziantep University
Mass Moment of InertiaMass Moment of Inertia
Mass moment of inertia is the name given to rotational inertia, the rotational inertia analog of mass for linear motion. It appears in the relationships for the dynamics of rotational motion. The Mass Moment of Inertia of a solid measures the solid's ability to resist changes in rotational speed about a specific axis.
r
o
o
m
The moment of inertia for a point mass is just the mass times the square of perpendicular distance to the rotation axis. The mass moment of inertia for a single particle is given as:
mrI 200
8Gaziantep University
Mass Moment of InertiaMass Moment of Inertia
When calculating the mass moment of inertia for a rigid body, one thinks of the body as a sum of particles, each having a mass of dm. Integration is used to sum the moment of inertia of each dm to get the mass moment of inertia of body. The equation for the mass moment of inertia of the rigid body is
dmrdmrI 2200
r
o
o
dm
9Gaziantep University
Mass Moment of InertiaMass Moment of InertiaThe integration over mass can be replaced by integration over volume, area, or length. For a fully three dimensional body using the density one can relate the element of mass to the element of volume.
x
y
dm
z
dmzydmrI xxx 222
dmzxdmrI yyy 222
dmyxdmrI zzz 222
rz
ry
rx
These three integrals are called the principle mass moment of inertia of the body. Another three similar integrals are
xydmII yxxy
yzdmII zyyz
xzdmII zxxz
} mass products of inertia of the body
10Gaziantep University
Radius of gyrationRadius of gyration
GICG, mICG, m
Original Body
Sometime in place of the mass moment of inertia the radius of gyration k is provided. The mass moment of inertia can be calculated from k using the relation
where m is the total mass of the body. One can interpret the radius of gyration as the distance from the axis that one could put a single particle of mass m equal to the mass of the rigid body and have this particle have the same mass moment of inertia as the original body.
2mkI
k
Model
k
Model
11Gaziantep University
Parallel-axis theoremParallel-axis theorem
The moment of inertia around any axis can be calculated from the moment of inertia around parallel axis which passes through the center of mass. The equation to calculate this is called the parallel axis theorem and is given as
2mdII CG G
ICG, mICG, m
o od
12Gaziantep University
Example 1Example 1
Solution: Mass moments of inertial of a point mass about an axis passing through itself
What are the mass moments of inertial of a point mass about an axis passing through itself and about an axis r distance away from it ?
o
o
m
002 dmdmrI oo
13Gaziantep University
Example 1Example 1
Solution: Mass moments of inertial of a point mass about an axis passing through itself
What are the mass moments of inertial of a point mass about an axis passing through itself and about an axis r distance away from it ?
o
o
m 002 dmdmrI oo
o
o
rm
x
x
mrmrmrII xxoo222 0
Mass moments of inertial of a point mass about an axis r distance away from it. Using parallel axis theorem
14Gaziantep University
Example 2Example 2
Solution:
Find the mass moment of inertia of a slender rod of length L (slender rod means that it has a length, and the remaining dimensions are negligible small) about an axis perpendicular to the rod and passing through its mass center.
L
L/2
o
o
L
L/2
o
o
dxx
2/
0
2*2L
oo dmxI
Let density of the material is in kg/m.
Then, infinitesimal mass dm=dx. Substituting this into above equation,
2/
0
2*2L
oo dxxI 12
2mL2/
3
|3
2 Lo
x3
232
L
15Gaziantep University
Example 3Example 3An uniform steel bar shown in the figure is used as an oscillating cam follower. Drive the equation of mass moment of inertia of the follower about an axis
through O. Use the density of steel =7800 kg/m3.
k=2kN/m
2 cm 25 cm 50 cm 25 cm
5 cm
y
xO
16Gaziantep University
Example 3Example 3
222 yxr
Solution:
x
y
z
xyr
dx
dzdy
t
w
l
dmrI zz2
dxdydzdm
dxdydzyxI zz )( 22
17Gaziantep University
Example 3Example 3Solution:
x
y
z
dx
t
w
l
x
y
z
dy
t
w
l
dyltydxwtxI zz ******** 22
2/
0
22/
0
2 ****2****2wl
dyyltdxxwt
18Gaziantep University
Example 3Example 3Solution:
*** twlm
1212*
12*
2222 wlm
wm
lmI zz
32
****232
****2
33w
lt
l
wtI zz 12****
12****
22 wwlt
llwtI zz
mass moment of inertia about O can be found by parallel axis theorem
48
47
412*
22
2
2222
0
wlm
lm
wlmdmII zz
mkgI .139.148
05.0*41*78.7
22
0