Choosing a Geometry for a Given Application Material Selection - - PowerPoint PPT Presentation
Choosing a Geometry for a Given Application Material Selection Selection of Materials and Structures Bristol F.2b Air Tractor AT-802A Photo by Alan Wilson CC BY-SA 2.0 Piper Super Cub Photo by flightlog CC BY 2.0 Photo by Ad Meskens CC BY-SA
Selection of Materials and Structures
Photo by Ad Meskens CC BY-SA 3.0 Photo by flightlog CC BY 2.0 Photo by Alan Wilson CC BY-SA 2.0
Bristol F.2b Air Tractor AT-802A Piper Super Cub
3
(fixed by overall geometry of bike frame)
What geometric properties appear in our characteristic equation?
Beam cross-section
y Axis of bending (neutral axis)
x y x
A1 y1 2 1 n i i i
=
2 A
A2 y2 An
3
4
3
24 mm 24 mm
3 3
1 1 24 24 12 12 I bh mm mm = =
□ 4 4
2.76 10 mm = ×
4 4
2.76 10 mm = ×
d
We want to a circular cross section with an equivalent bending stiffness
4
64 d I π =
○
Equate to moment of inertia for square section
4 4 4 64 2.76 10
d mm π = × 27.4mm =
24 mm 24 mm 27.4 mm
( )
2 2
24 576 A mm mm = =
□
( )
2 2
27.4 589 4 A mm mm π = =
○
589 1.023 576 A A = =
○ □
Area, and thus weight, of circular cross section is 2.3% larger than the square section
Neutral axis
Circular section has more area closer to the neutral axis, thus needs a larger diameter to have the same stiffness as the square Moment of inertia is larger for area located further from the neutral axis
4 4
2.76 10 mm = ×
d
We want to a thin-walled circular cross section with an equivalent bending stiffness
3
8 td I π =
○
Equate to moment of inertia for square section
4 4 4
8 2.76 10 0.1 d mm π = × 29.0mm =
t = 0.1d
24 mm 24 mm
( )
2 2
24 576 A mm mm = =
□
( )( )
2
29 2.9 264 A mm mm mm π = =
○
264 0.46 576 A A = =
○ □
Area, and thus weight, of thin walled tube is 46% of the solid square section
t = 2.9 mm 29 mm
t = 2.9 mm 29 mm 24 mm 24 mm 27.4 mm t = 1.7 mm 34.4 mm
Weight = 100% 102% 46% 32%
Is there more than geometry?
steel Al