project final submission_coursera_structure standing still: the statics of everyday objects

Structure Standing Still: The Statics of EverydayThe Statics of Everyday Objects

by Dr. Dan Dickrell

Project Final Submission

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.

Total cost of my design: y g

Given at no cost a pin joint at B, two free p j ,reaction points, a pin joint at A and a roller at C. AB=BC=DE=4 mCost= 3*(75+4^4) = $993 AD=BD=BE=CE=2.31 m Cost 4*(75+2 31^4) $414Cost= 4*(75+2.31^4) = $414Total member cost= $1407 Joint (A B C) = $0 (Free)Joint (A, B, C) $0 (Free)Pin joint (D, E) = $50Total pin joint cost =$50*2= $100p jTotal truss (bridge) cost= 1407+100= $1507

Load Calculation:

Blue member (AD, DE, CE) =Compression bucklingRed member (AB, BC, BD, BE) = Tensile yielding

Forces in membersAB=BC=DE=8.66 KNAD=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)

Stress Analysis:

Material Selection:Truss Material: Aluminum alloy 6061-T6Shape: Hollow Pipe

Yield strength: 241 MPa = 241000 KN/m²Factor of safety= 2 (Assumed)Equation:

Yield strength = Load* Factor of safety / Sectional AreaAfter calculation: For AD=BD=BE=CE= 10 KN;

Outer diameter= 50 mmInner diameter= 48.9 mm

For AB=BC=DE=8.66 KN;Outer diameter= 50 mmInner diameter= 49 mm

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.

Thanks