Week 1 Stress Concentration on a Plate with hole
Objective: To determine and compare the maximum deformation and stress developed in the models in below 2 cases.
Case 1:
Plate With Single hole
Plate dimensions:
Length=300mm
Height=120mm
Thickness=30mm
Circular Hole at the center:
Diameter=60mm
Boundary Conditions:
Plate fixed at left face.
Load of 500N on the right face.
Solution:
The 2D sketch has been extruded in SpaceClaim to generate 3D and later exported to Mechanical module.
Material Properties assigned are as below:
Meshing Details:
Meshing has been done using Tetrahedron element geometry with element size of 8 mm.
The mesh metric graph has been represented above which confirms that most of the elements are of high quality and thus acceptable.
Support and Load Details:
The simulation has been applied and the results are as below:
Equivalent Stress:
Total Deformation:
Case: 2
Plate with Multiple holes
Given:
Solution:
The 2D sketch has been extruded in SpaceClaim to generate 3D and later exported to Mechanical module.
Material Properties assigned are as below:
Meshing Details:
Meshing has been done using Tetrahedron element geometry with element size of 8 mm.
The mesh metric graph has been represented above which confirms that most of the elements are of high quality and thus acceptable.
Support and Load Details:
The simulation has been applied and the results are as below:
Equivalent Stress:
Total Deformation:
Coclusion:
Comparing the equivalent stresses and Total deformation in both the cases.
| S. No | Case | Maximum Stress (Mpa) | Total Deformation (mm) |
| 1 | Plate with Single hole | 0.58913 | 0.00027486 |
| 2 | Plate with 3 holes | 0.56106 | 0.00029709 |
From an analysis standpoint, which of the design will you select? Explain why?
From the analysis perspective we can see that there is less deformation in case 1. But, if we look at the stress then we can see that the equivalent stress is lower in case 2 and more uniformly distributed. As the stresses in both the cases is lower than the yield strength and we can also see there is sufficient FOS, I would select design case 2 due to its lesser weight and lower von mises stress value.
From a manufacturing standpoint, which of the design will you select? Explain why?
From manufacturing perspective design 2 will have more cost due to additional drilling costs. So, in order to reduce manufacturing cost. design case 1 can be selected from manufacturing standpoint.
However, design case 2 results in reduction of weight which is again one chief consideration in assembly design.
Report writing.
All the details about the analysis has already been mentioned above. Looking at both the cases from the better design perspective, design case 2 looks to be a more efficient design.