LECTURE+22+COE-3001-A

M mellogeorgia tech aerospace 9 e 3

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Unformatted text preview: HE NEUTRAL AXIS 2/22/13 M. Mello/Georgia Tech Aerospace 14 RELATIVE EFFICIENCY OF BEAM SHAPES ²་  EFFICIENCY IN BENDING DEPENDS UPON SHAPE OF THE CROSS SECTION ²་  CONSIDER: RECTANGULAR CROSS SECTION: I = I S= c bh2 S= 6 bh3 12 c= h 2 Ah 6 S = 0.167Ah ²་  Rectangular cross secTon of fixed area becomes more efficient as height (h) is increased (and width (b) decreased to maintain area constant) 2/22/13 S= M. Mello/Georgia Tech Aerospace 15 Compare square secTon with circle secTon of same given area: CONSIDER SOLID CIRCULAR CROSS SECTION: p p h = b = (d/2) ⇡ A = ⇡ d2 /4 b = (d/2) ⇡ Ssquare Scircle b=h p h3 ⇡ ⇡ d3 = bh /6 = = 6 48 2 ⇡ d3 = = 0.098d3 32 Ssquare = 1.18 Scircle 2/22/13 Areas equal Square beam is more efficient in resisTng bending compared to circular beam of same cross- secTonal area M. Mello/Georgia Tech Aerospace 16 IDEA SHAPE FOR A BEAM OF CROSS- SECTIONAL AREA •  PLACE ONE- HALF OF THE AREA AT A DISTANCE h/2 above and below the neutral axis! ⇥ s2 h I = 2 (bs) + (bs) 12 2 2 0 ◆2 ✓ s h I = 2bs + 12 2 s b 2⇤ Consider extreme (pracTcally una=ainable) case where h>>s In which case: ✓ ◆✓ ◆2 h A I=2 2 2 S= I = 0.5Ah h/2 TheoreTcal limit approached in pracTce by wide- flange secTons and I- secTons For wide- flange secTons: S ⇡ 0.35Ah 2/22/13 M. Mello/Georgia Tech Aerospace 17 SELECTION OF A BEAM BASED UPON ALLOWABLE STRESSES EXAMPLE 5- 5 GIVEN: •  Simply supported wood beam •  Max allowable load: σallow = 12MPa •  Supported against sideways buckling •  Span length: L=3m •  uniform load : q=4kN/m •  wood weight density: 5.4kN/m3 PROBLEM STATEMENT: •  SELECT AN APPROPRIATE WOODEN BEAM FROM APPENDIX F (account for the weight of the beam as well) 2/22/13 SIMPLY SUPPORTED WOOD BEAM M. Mello/Georgia Tech Aerospace 18 EXAMPLE 5- 5 (cont.) SOLUTION: •  HAVE TO USE TRAIL AND ERROR STRATEGY BECAUSE THE BEAM WEIGHT WILL DEPEND UPON THE DIMENSIIONS WE SELECT! •  Step 1. Calculate required secTon modulus (based only upon given uniform load (we’ve encountered this beam bending problem before / see introductory notes on shear force and bending moment diagrams) Mmax qL2 = 8 Mmax (4kN/m)(3m)2 = = 4.5kN · m 8 •  REQUIRED SECTION MODULUS: S = Mmax allow = SIMPLY SUPPORTED WOOD BEAM 4.5kN · m = 0.375 ⇥ 106 mm3 12M P a •  IdenTfy the lightest beam cross- secTon from Appe...
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