ME 232 Homework Problems

Problems B/15 Page 675 and B/49 Page 689, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: Uniform rod of length 4b and mass m is bent into the green shape shown in the figure.  The coordinate axes are shown in light blue and labeled.  The origin is labeled O.  The rod consists of three straight sections.  One is parallel to the Y axis, of length 2b, and is centered at the origin.  The second straight section is parallel to the Z axis, is of length b, and runs from the point (0, b, 0) to (0, b, -b).  The third straight section is parallel to the X axis, is of length b, and runs from point (0,-b,0) to (b,-b,0).

Find: The angular mass matrix (inertia tensor) of the rod about the origin of the axes shown.

Parameter Values:
m = Rod mass
b = Length of 1/4 of the rod.

To obtain the angular mass matrix for the entire rod we must sum the angular mass matrices for each of the three straight sections of the rod.  To obtain the angular mass matrix for any one of the straight sections of the rod we must evaluate the angular mass matrix of that section about its own center and then add the appropriate transfer terms.  For a slender rod about its mass center, the angular mass matrix is diagonal.  The diagonal terms about the two axes perpendicular to the rod are given by 1/12 the mass of the rod multiplied by the square of the length of the rod.  The diagonal term about the axis parallel to the rod.  The transfer terms depend upon the mass of the rod and the relative location of the rod center and the point about which the angular mass matrix is to be evaluated.

Beginning with the piece of length 2b along the y axis (Piece 1):
Mass = m/2,  Length=2b, Parallel to Y axis, Center at Origin (no transfer necessary).
1/12 m/2 (2b)2       0        0
IO(Piece 1) =            0                 0       0
0                 0    1/12 m/2 (2b)2
For the piece along the Z axis (Piece 2):
Mass = m/4, Length=b, Parallel to Z axis, Center at (0, b, -b/2)
1/12 m/4 (b)2           0                 0
IG2(Piece 2) =      0               1/12 m/4 (b)2       0
0                      0                 0
m/4 (b2 + (-b/2)2)          -0                      -0
IO(Piece 2) = IG2(Piece 2) +       -0                          m/4 (-b/2)2        -m/4 b (-b/2)
-0                         -m/4 b (-b/2)        m/4 b2
1/3 m b2           0                  0
IO(Piece 2) =           0           1/12 m b2      1/8 m b2
0            1/8 m b2       1/4 m b2
For the piece along the X axis (Piece 3):
Mass = m/4, Length =b, Parallel to X axis, Center at (b/2, -b, 0)
0             0                0
IG3(Piece 3) =         0      1/12 m/4 b2       0
0            0         1/12 m/4 b2
m/4 (-b)2      -m/4 b/2 (-b)     0
IO(Piece 3) = IG3(Piece 3) +   -m/4 b/2 (-b)   m/4 (b/2)2        0
0                    0        m/4( (-b)2 + (b/2)2 )
1/4 m b2     1/8 m b2        0
IO(Piece 3) =      1/8 m b2    1/12 m b2      0
0                0            1/3 m b2