Modified Design of a Precision Planter For a Robotic Assistant Farmer

Solution and results discussion: Now that all the boundary conditions and external 
forces are defined, we can solve the problem to get equivalent stress (Von Mises) and total 
deformation. These results are shown in figure 4-4 and 4-5. 
Figure 4- 4- Equivalent (Von Mises) stress contour for depth control link 
To analyze the stresses in this part, brick (hexahedral) elements are used. The results of 
the analysis have been verified by choosing different element sizes and element types. Changing 
the element size, from 3 mm to 10 mm, showed no significant change in stress distribution and 
also maximum stress. A smaller size element could not be used, because the number of generated 
nodes would exceed the number of allowed for the educational license that is available. 

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As it can be seen in Figure 4-4, the maximum stress is calculated as 22.3 MPa. Since the 
material selected for this part is structural Steel, it has a yield strength of 250 MPa. Based on the 
Distortion-Energy theory, this means, for this specific part, a safety factor of more than 10. 
Figure 4- 5- Total Deformation contour for depth control link 
Figure 4-5 shows the total deformation on the link. The maximum value for this 
parameter is 0.023 mm which is very small. 
Also the reaction forces at supports are shown in table 4-2. 

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Table 4- 2- Reaction forces at cylindrical supports A and B 
Support A 
Support B 
X Axis 
-738.8 N 
-48.9 N 
Y Axis -1514.6 N 
568.0 N 
Z Axis 
-169.5 N 
5.6 N 
Total 
1693.6 N 
570.1 N 
To compare the results form FEM with hand-calculations, a simple 2D version of the 
problem is solved by hand. As it can be seen in Figure 4-6, the magnitudes of the external forces 
in the x-y plane are 306.1 N and 1192.8 N.
Figure 4- 6- Depth control link and Schematic forces and supports in 2D for hand 
calculation 
Writing the moment about point B,
(Eq. 5-1) 
Equation 5-1 will result in 
, downward. And subsequently from the 
equilibrium in the y direction we can conclude that 
, upward. The values for 
reaction forces A
y
and B
y
match the results of the FEM, in y direction. 

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It can reliably assume that the maximum moment happen at point A. At this point the 
moment is highest and also the area of cross section is smallest because of the hole. Moment at 
this point will be 
. The link has a cross section of 15 mm × 38.1 
mm and the diameter of the hole is 14.3 mm. So the moment of inertia of the cross section is, 
(Eq. 5-2) 
In Equation 5-2, b= 15 mm is the thickness of the link, and h=38.1 mm is the width of the 
link, and d is the diameter of the hole. This will result in 
. 
Considering that modulus of elasticity for structural steel is E=200 GPa, we can use 
Equation 5-3 to calculate the bending moment in the link, at point A. 
(Eq. 5-3) 
At the same time, there is compression stress on the link due to axial forces in the x 
direction. So, the compression stress can be easily calculated using Eq. 5-4. 
(Eq. 5-4) 
The maximum total stress on the link will be 16.5+2.2= 18.7 MPa, which is close enough 
to 22.3 MPa that was found from FEM. The main reason for the difference in results is that, the 
problem is solved in 2D. This means that the stress caused by force element in the Z direction in 
the link is not considered for simplicity of hand calculations.