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# Project final submission_Coursera_Structure Standing Still: The Statics of Everyday Objects

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The final design project will require students to design a truss structure to span a distance while carrying a load. The necessary skills required to design the truss structure will be acquired throughout the course, no previous skills will be required. Designs will be evaluated based on a set of constraints provided at the beginning of the project and how well the individual design satisfied the constraints.

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### Transcript of "Project final submission_Coursera_Structure Standing Still: The Statics of Everyday Objects "

1. 1. Structure Standing Still: The Statics of EverydayThe Statics of Everyday Objects by Dr. Dan Dickrell Project Final Submission
2. 2. Design Scratch: My design labeling the joint locations with letters, and showingMy design labeling the joint locations with letters, and showing important dimensions locating the joints within the design.
3. 3. Total cost of my design:y g Given at no cost a pin joint at B, two freep j , reaction points, a pin joint at A and a roller at C. AB=BC=DE=4 m Cost= 3*(75+4^4) = \$993 AD=BD=BE=CE=2.31 m Cost 4*(75+2 31^4) \$414Cost= 4*(75+2.31^4) = \$414 Total member cost= \$1407 Joint (A B C) = \$0 (Free)Joint (A, B, C) \$0 (Free) Pin joint (D, E) = \$50 Total pin joint cost =\$50*2= \$100p j Total truss (bridge) cost= 1407+100= \$1507
4. 4. Load Calculation: Blue member (AD, DE, CE) =Compression buckling Red member (AB, BC, BD, BE) = Tensile yielding Forces in members AB=BC=DE=8.66 KN AD=BD=BE=CE= 10 KN AB=8.66 KN= (Tensile yielding) BC=8.66 KN= (Tensile yielding) DE=8.66 KN= (Compression buckling) AD=10 KN= (Compression buckling) BD=10 KN= (Tensile yielding) BE=10 KN= (Tensile yielding) CE= 10 KN= (Compression buckling)
5. 5. Stress Analysis: Material Selection: Truss Material: Aluminum alloy 6061-T6 Shape: Hollow Pipe Yield strength: 241 MPa = 241000 KN/m² Factor of safety= 2 (Assumed) Equation: Yield strength = Load* Factor of safety / Sectional Area After calculation: For AD=BD=BE=CE= 10 KN; Outer diameter= 50 mm Inner diameter= 48.9 mm For AB=BC=DE=8.66 KN; Outer diameter= 50 mm Inner diameter= 49 mm
6. 6. Final Remarks: Final Remarks:Final Remarks: AD & CE will fail first due to Compression buckling &Compression buckling & BD & BE will fail first due to tensile i ldiyielding.
7. 7. Thanks
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