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...
View
Full Document
 Spring '09
 ZHU
 Second moment of area, Modulus, Secton

Click to edit the document details