ME 5550 Intermediate Dynamics
Bearing Loads on a Simple Crank Shaft
The figure to the right shows a
simple crank shaft consisting of seven
segments, each considered to be a
slender bar. Each segment of length
has mass m . There are six segments of
length and one segment of length 2
(segment 4). The mass center of the
system is G and is located on the axis
of rotation.
In previous notes, we found H G the angular momentum of the system to be
2
H G  2m 2 i   (10
3 )m  k
The bearing loads at A and B may be found by applying the Newton/Euler equations of
motion to the free-body diagram shown at the right.
F  m
M  I
A
R
G
aG  0


 R B   R B  H G    r G / A  m R aG 

 I G  R B
B  HG 
R
The terms on the left and right sides of the moment
equation may be calculated as follows:
M
A
  r B / A  FB   T k
2
IG  R B  2m 2 i   (10
3 )m  k
 B  HG   k  H G  2m 2 2 j
R
Substituting these results into the equations above gives the following EOM:
AX   BX   0
AY   BY   0
Kamman – ME 5550 – page: 1/2
4 BY   2m 2
4 BX   2m 2 2
T  103 m 2
Solving these equations gives the following results
BX   12 m  2   AX 
BY    12 m    AY 
T  103 m 2
Kamman – ME 5550 – page: 2/2