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 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
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)
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
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

Project final submission_Coursera_Structure Standing Still: The Statics of Everyday Objects

  • 1.
    Structure Standing Still: TheStatics of EverydayThe Statics of Everyday Objects by Dr. Dan Dickrell Project Final Submission
  • 2.
    Design Scratch: My designlabeling 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.
    Total cost ofmy 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.
    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.
    Stress Analysis: Material Selection: TrussMaterial: 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.
    Final Remarks: Final Remarks:FinalRemarks: AD & CE will fail first due to Compression buckling &Compression buckling & BD & BE will fail first due to tensile i ldiyielding.
  • 7.