ME 232 Classroom Problems

January 6, 1997

Plate Supported by 6 Rods

The figure depicts a uniform rectangular plate supported by 6 rods.  The plate (light blue) is supported in a horizontal plane (XY, axes shown in green).  The center of the plate is at the origin.  The plate is 6' by 8' (6' along the Y, 8' along the X).  The plate has a weight per unit area of 12 lb per square foot.  The plate is supported by 6 rods (dark blue).  Each rod is connected to the ceiling (purple spheres).  The connection points are all 6' above the horizontal plane of the plate.  Ball and socket joints are used for all connections so each of the rods may be considered a two force member.  The connection points of the rods are labeled with a lower case letter at one end and the corresponding upper case letter at the other end.  The coordinates of the connection points are given below.

1. Develop 6 equations that could be solved to yield the forces in each of the supporting rods.

2. Solve the six equations.

Connection Coordinates (all coordinates given in feet)

a (0,3,0)              A (7,9,6)
b (-4,3,0)            B (-4,3,6)
c (-4,-3,0)           C (-7,-9,6)
d (0,-3,0)            D (4,-15,6)
e (4,-3,0)            E (12,-3,6)
f (4,3,0)              F (7,5,6)

Applying the four step procedure:

0) Think : The plate is motionless, thus it is in equilibrium.  The sum of the forces and moments acting on the plate must be zero.

1) Mechanical System : Plate in the position shown.

2) Free Body Diagram : Omitted to save time, would look nearly identical to the above figure except that each of the six rods would be replaced by a force acting along the rod from the plate toward the connection to the ceiling.  Additionally the weight of the plate (12 lb/ft2 * 6 ft * 8 ft or 576 lb would be shown acting through the origin in the negative Z direction.  These forces will be labeled using the letters identifying the rod, thus FA will be the force in the rod connecting points a and A and so on.

3) Equations : To express each of the forces as vectors we need to obtain unit vectors parallel to those forces.  These unit vectors can be obtained from the coordinates of the connection points.  Thus (vectors denoted in bold face):

A - a = (7,6,6)      | A-a | = { 72 + 62 + 62 }1/2 = 11    eA = (7/11,6/11,6/11)   FA = FA(7/11,6/11,6/11)
B - b = (0,0,6)      | B-b | = 6    eB = (0,0,1)   FB = FB(0,0,1)
C - c = (-3,-6,6)      | C-c | = 9    eC = (-1/3,-2/3,2/3)   FC = FC(-1/3,-2/3,2/3)
D - d = (4,-12,6)      | D-d | = 14    eD = (2/7,-6/7,3/7)   FD = FD(2/7,-6/7,3/7)
E - e = (8,0,6)      | E-e | = 10    eE = (4/5,0,3/5)   FE = FE(4/5,0,3/5)
F - f = (3,2,6)      | F-f | = 7    eF = (3/7,2/7,6/7)   FF = FF(3/7,2/7,6/7)

Noting that the weight force is 576 lb in the -Z direction, we can now sum forces.  Separating the sum into its X, Y and Z components yields the following three equations :

7/11 FA + 0 FB -1/3 FC + 2/7 FD + 4/5 FE + 3/7 FF               = 0
6/11 FA + 0 FB -2/3 FC - 6/7 FD +  0    FE + 2/7 FF               = 0
6/11 FA + 1 FB +2/3 FC +3/7 FD + 3/5 FE + 6/7 FF - 576 lb  = 0

Unfortunately this is only 3 equations in the 6 unknown forces.  We must consider the moment equation as well.  The moment of a force about a point is given by the cross product of the position vector from the point to the force with the force vector.  If we sum moments about the origin and take the vectors from the origin to the connection points on the plate, then the moments of the forces can be readily calculated (note that the weight passes through the origin and thus produces no moment about the origin).

MA = (0,3',0) X FA(7/11,6/11,6/11) = (18/11',0,-21/11') FA
MB = (-4',3',0) X FB(0,0,1) = (3',4',0) FB
MC = (-4',-3',0) X FC(-1/3,-2/3,2/3) = (-2',8/3',5/3') FC
MD = (0,-3',0) X FD(2/7,-6/7,3/7) = (-9/7',0,6/7') FD
ME = (4',-3',0) X FE(4/5,0,3/5) = (-9/5',-12/5',12/5') FE
MF = (4',3',0) X FF(3/7,2/7,6/7) = (18/7',-24/7',-1/7') FF

Summing the moments and separating the sum into its X, Y, and Z components yields the following three equations :

18/11'  FA + 3' FB -   2'   FC - 9/7' FD - 9/5'   FE + 18/7' FF = 0
0       FA + 4' FB + 8/3' FC +  0   FD - 12/5' FE - 24/7' FF = 0
-21/11' FA + 0  FB + 5/3' FC + 6/7' FD +12/5' FE - 1/7'  FF = 0

We now have a total of six equations in 6 unknowns.  Dividing the moment equations by 1', and taking the 576 lb to the right side in the Z force equation allows us to formulate the six equations in standard matrix form.

|  7/11      0     -1/3      2/7     4/5     3/7  |   |  FA  |         =     |  0          |
|  6/11      0     -2/3    -6/7       0       2/7  |   |  FB  |        =      |  0          |
|  6/11      1      2/3      3/7      3/5     6/7  |   |  FC  |        =      |  576 lb  |
|  18/11    3      -2      -9/7     -9/5   18/7  |  |  FD  |        =      |   0         |
|     0        4      8/3       0     -12/5  -24/7  |  |  FE  |         =     |   0         |
|  -21/11   0     5/3      6/7     12/5   -1/7   |  |   FF |         =     |   0         |

4. Solve : The equations are most easily solved using a calculator or some other computing device.  MathCad was used to define and invert the coefficient matrix.  The inverse of the coefficient matrix was then multiplied by the right hand side vector to yield the following solution for the forces (note that you will be expected to be able to use your calculator or some other computing device to solve equations such as these) :

FA= 319 lb   FB= -192 lb   FC= 648 lb    FD= -182 lb    FE= -110 lb      FF= 357 lb

0. Think : The forces in the rods are all within an order of magnitude of the weight.  Some are negative (compression) and some are positive (tension).  Some of the magnitudes are larger than the weight.  Does this make sense?  Can you think of a two dimensional situation in which the supporting forces are of magnitudes larger than the supported weight?  A two dimensional situation in which some of the supporting forces are tensile and some are compressive?  Do you believe the results obtained? If we believe the results, what are the implications for our support system?

Make sure that you are comfortable with each and every step in the above problem solving process.  These are skills that will be required on many of the dynamics problems that we will tackle over the course of the quarter.