Team H

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Team Members

Initial Brainstorm & Design

When both sections of our team (α & Ω) met together to brainstorm, we initially had different ideas about what kinds of designs would be the most efficient at distributing the force evenly among the arm without allowing for too much displacement.Section α's design ideas were mainly centered around the distribution of multiple circular cuts through the arm to reduce the volume of the arm, while the designs from section Ω focused on hollowing out the arm to reduce volume and adding a lattice structure to support critical stress points in the arm. After running a few simulations of both individual designs, we came to the conclusion that both design types had their merits and we decided to consolidate both into a single overlapping design.

Design Procedure

The next few images show the progression of our model leading up to our finalized design.

Design 2 Half Structured.jpg

This is an example of one of our earliest designs. We tried to cut out as much material as possible while retaining the structural integrity of the arm, but we quickly realized we needed more material to support the remainder of the arm.

Design 3 w Inside Fillet.jpg

We filleted nearly all the flat faces and corners on our model in order to shave off as much material as possible.

Design 4 w Holes near top.jpg

In this design, we tried to optimize hole placement in order to reduce the most material without creating stress zones that increased the displacement of our model.

Design 5 Unfillet Top.jpg

Later on in the design process, we discovered that our design benefited most by having a point or segment of the arm that stayed rigid and we eliminated some of the fillets on the supports closest to the fixture of the arm.

FINAL Design.jpg

After many, many trial and error sessions defined by the excel spreadsheet below, we finally arrived at this design, which is a culmination of all of our design steps.


Arm Attempts.png

Our final design had a final displacement of 0.2072 mm and a volume of 0.88 in^3. Our resulting score was approximately 0.182 mm * in^3.