Structures Standing Still : Project final submission

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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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Structures Standing Still : Project final submission

  1. 1. Structure Standing Still: The Statics of EverydayThe Statics of Everyday Objects by Dr. Dan Dickrell Project Final SubmissionProject Final Submission
  2. 2. Design Scratch: My truss design (drawing by AutoCAD) labeling the joint locations withMy truss design (drawing by AutoCAD) labeling the joint locations with letters, and showing important dimensions locating the joints within the design.
  3. 3. Cost Analysis:y My design labeling the joint locations showing important dimensions locating the joints within the design.
  4. 4. Cost Calculation: Given free pin joint at B & two free reaction points, a pinp j p , p 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
  5. 5. 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 KNAB=BC=DE=8.66 KN AD=BD=BE=CE= 10 KN Force Type: AB 8 66 KN (T il i ldi )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)BE 10 KN (Tensile yielding) CE= 10 KN= (Compression buckling)
  6. 6. Stress Analysis: Material Selection: Truss Material: Aluminum Shape: Hollow Pipe Yield strength: 95 MPa = 95000 KN/m² Factor of safety=3 (Assumed) Equation: Yield strength or stress = Load* Factor of safety / Sectional Area After calculation on tensile members: For BD=BE= 10 KN; Outer diameter= 50 mm Inner diameter= 45.8 mm For AB=BC=8.66 KN; Outer diameter= 50 mm Inner diameter= 46.4 mm
  7. 7. Buckling Analysis:g y Euler’s Equation: F i iti l fF= maximum or critical force E= modulus of elasticity=95 Gpa I= area moment of inertia= ∏/2*(R^4-r^4) L= unsupported length of column, K= column effective length factor, For both ends pinned= 1.0. For AD, CE member: Outer diameter( R )= 50 mm & Inner diameter (r)= 48 mm & L= 2.31 m (proposed di i )dimension) Applying Euler’s equation, Maximum or critical force, F= 1.18E+7 GN So our designed load 10 KN in AD & CE member is safe from compressive buckling. F DE bFor DE member: Outer diameter( R )= 50 mm & Inner diameter (r)= 48 mm & L= 4 m (proposed dimension) Applying Euler’s equation Maximum or critical force F 3 93E+6 GNApplying Euler s equation, Maximum or critical force, F= 3.93E+6 GN So our designed load 8.66 KN in DE member is safe from compressive buckling.
  8. 8. Final Remarks: AD & CE will fail first due to Compression buckling &buckling & BD & BE will fail first due to tensile i ldiyielding. Important Notes: I have ignored some parameters such as weight of members, d namic loads ind loaddynamic loads, wind load.
  9. 9. Thanks

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