Raised Bed Design

Summary: 2 stacked 8x12x2 (96 x 11.5 x 1.5 inch) cedar board sides, less than 1/8 inch differential bulge

• 8 foot long, 23 inches high, two boards, 1.522 psi at bottom
  • w = 4.38 lbf (pounds force) per lineal inch top board
  • w = 13.13 lbf per lineal inch bottom board

• cedar modulus: E = 1M psi Table 9

• 3 point, end and center post support - see http://civilengineer.webinfolist.com/str/prob72.htm

  • L = 96 inch, w = 4.4 lbf/in top, 13 lbf/in bottom, EI tbd
    • wL = 420 lbf top, 1260 lbf bottom

  • deflection of two point supported board, distributed load d = 5 w L⁴/384EI

  • deflection of two point supported board, point load R in middle d = R L³/48EI

  • therefore, R = -(5/8)(wL)
    • top . . . : 79 lbf both ends, 262 lbf middle
    • bottom : 236 lbf both ends, 788 lbf middle

• deflection for uniformly loaded beam: y = ( w x / 24 EI ) ( L³ - 2 Lx² + x³ )

• deflection for center point loaded beam: y = ( R x / 48 EI ) ( 3 L² - 4 x² ) ... 0 < x < L/2

  • see http://en.wikipedia.org/wiki/Deflection_(engineering)

• compound force deflection: y = ( w / 384 EI ) ( L³ x - 12 L x³ + 16 x⁴ )

  • test, x=L/2 : ( L⁴/2 - 3/2 L³ + L³ ) = 0 deflection at center

  • max deflection, first derivative = 0 = L³ -36Lx² + 64x³, x=0.210767582704314 L

  • max deflection = 3.3851E-4 ( L³ (wL) / E I )

• I = w t³ / 12 for beam width w, thickness t

• I = 3.23 in⁴ for 11.5 x 1.5 inch side board

  • (3.3851E-4*96³/(3.23*1E6)) ( wL ) = 9.2E-5 in/lb * (W/L)

  • 8 foot bed deflection 0.042 inch top, 0.127 inch bottom

• I = 0.404 in⁴ for 11.5 x 0.75 inch side board

  • (3.3851E-4*96³/(0.404*1E6)) ( wL ) = 7.4E-4 in/lb * (W/L)

  • 8 foot bed deflection 0.311 inch top, 0.934 inch bottom

  • (3.3851E-4*72³/(0.404*1E6)) ( wL ) = 3.13E-4 in/lb * (W/L)

  • 6 foot bed deflection 0.144 inch top, 0.432 inch bottom