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# Coursera: Structure Standing Still: Project Presentation

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### Coursera: Structure Standing Still: Project Presentation

1. 1. Structure Standing Still: The Statics of Everyday Objects by Dr. Dan Dickrell Project Final Submission Iqram Hussain q Coursera Network, Dhaka, Bangladesh
2. 2. Design Scratch: My 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 pin p 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) Cost 4*(75+2.31^4) = \$414 Total member cost= \$1407 Joint (A, B, C) = \$0 (Free) (A B Pin joint (D, E) = \$50 Total pin joint cost =\$50*2= \$100 p 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 AB=BC=DE=8 66 KN AD=BD=BE=CE= 10 KN Force Type: AB=8.66 KN= (Tensile yielding) AB 8 66 KN (T il i ldi ) 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= 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= F maximum or critical force i iti l f 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 dimension) di i ) 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. For F DE member: b 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 GN Euler’s equation, force F= 3.93E+6 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 & BD & BE will fail first due to tensile yielding. i ldi Important Notes: I have ignored some parameters such as weight of members, dynamic loads ind load d namic loads, wind load.
9. 9. Thanks
10. 10. For further information please feel easy to contact iqramhussain@gmail.com www.facebook.com/engr.iqram