Download - Project final submission_Coursera_Structure Standing Still: The Statics of Everyday Objects
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Structure Standing Still: The Statics of EverydayThe Statics of Everyday Objects
by Dr. Dan Dickrell
Project Final Submission
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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.
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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
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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)
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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
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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.
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Thanks