Structure Standing Still: The Statics of EverydayThe Statics of Everyday Objects

by Dr. Dan Dickrell

Project Final SubmissionIqram Hussainq

Coursera Network,Dhaka, Bangladesh

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.

Cost Analysis: y

My design labeling the joint locations showing important dimensions locating the joints within the design.

Cost Calculation:

Given free pin joint at B & two free reaction points, a pin p j p , pjoint 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 j

Total truss (bridge) cost= 1407+100= $1507

Load Calculation:

Blue member (AD, DE, CE) =Compression bucklingRed member (AB, BC, BD, BE) = Tensile yielding

Forces in members:AB=BC=DE=8 66 KNAB=BC=DE=8.66 KNAD=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)

Stress Analysis:

Material Selection:Truss Material: AluminumShape: Hollow Pipe

Yield strength: 95 MPa = 95000 KN/m²Factor of safety=3 (Assumed)Equation:

Yield strength or stress = Load* Factor of safety / Sectional AreaAfter calculation on tensile members: For BD=BE= 10 KN;

Outer diameter= 50 mmInner diameter= 45.8 mm

For AB=BC=8.66 KN;Outer diameter= 50 mmInner diameter= 46.4 mm

Buckling Analysis: g yEuler’s Equation: F i iti l fF= maximum or critical forceE= modulus of elasticity=95 GpaI= 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.

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.

Thanks

For further information please feel easy toFor further information please feel easy to contact

iqramhussain@gmail.comwww.facebook.com/engr.iqram