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.

1. Structure Standing Still:
The Statics of EverydayThe Statics of Everyday
Objects
by Dr. Dan Dickrell
Project Final SubmissionProject Final Submission

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. Cost Analysis:y
My design labeling the joint locations showing important dimensions
locating the joints within the design.

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. 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. 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. 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. 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. Thanks