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LECTURE+22+COE-3001-A

Aerospace 17 selection of a beam based

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Unformatted text preview: ndix F with S 0.375 ⇥ 106 mm3 •  Candidate beam from appendix F: ☞  65mm x 200mm beam ☞  S=0.456x106 mm3 ☞  weight/length (W/L) = 77.11N/m ☞  Weight density of beam: 5.4kN/m3 2/22/13 M. Mello/Georgia Tech Aerospace 19 EXAMPLE 5- 5 (finish) 0 •  Update the uniform load on the beam: q = q + W/L = 4kN/m + 77.11N/m = 4077N/m •  Calculate the new required secTon modulus: ✓ 0 q S =S q 0 ◆ 4.077 = 0.375 ⇥ 106 mm3 ( 4.0 ◆ = 0.382 ⇥ 106 mm3 •  Previous beam selecTon with S=0.456 x 106 mm3 sTll works •  75mm x 200mm wooden beam cross secTon is saTsfactory Wooden beam actual cross- secTon dimensions 72mm “Dressed Dimension” 147mm SIMPLY SUPPORTED WOOD BEAM 2/22/13 M. Mello/Georgia Tech Aerospace 20 EXAMPLE 5- 6 GIVEN: •  VERTICAL POST HEIGHT: h=2.5m •  Lateral load: P = 12kN (applied at upper end) PROBLEM STATEMENT: •  What is minimum required diameter d1 of wood post if σallow = 15MPa? •  What is minimum required outer diameter d2 of aluminum tube if t=d2/8, (t=wall thickness) and allowable stress σallow = 50MPa? 2/22/13 SOLID WOOD POST ALUMINUM TUBE M. Mello/Georgia Tech Aerospace 21 EXAMPLE 5- 6 (cont.) SOLUTION: •  Determine maximum bending moment: Mmax = P h = (12kN )(2.5m) = 30kN · m •  Wood post: required secTon modulus (circular cross- secTon) ⇡ d3 Mmax 30kN · m i S= = = = 2 ⇥ 106 mm3 32 15M P a allow •  Solve for diameter (d1): d1 = 273mm (must be greater than or equal to) •  Aluminum tube: ⇡ I2 = d2 64 64 (0.75d2 )4 = 0.03356d4 2 I2 0.03356d4 2 3 S2 = = = 0.06712d (a) 2 c d2 /2 •  required secTon modulus: S2 = Mmax allow 30kN · m = 600 ⇥ 103 mm 3 (b) SOLID WOOD POST 50M P a ✓ ◆ 600 ⇥ 103 mm3 = 208mm Equate (a) and (b) : d2 = 0.06712 S2 = ALUMINUM TUBE 2/22/13 M. Mello/Georgia Tech Aerospace 22...
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