EWPAA Structural Plywood and LVL Design Manual - Engineered ...

WIND DOWNW u↓GW u↓+G6.8 x3.6/4= 6.13.29.36.1 + 3.8= 9.96.1 / 1.0= 6.13.8 / 0.57= 6.6Total= 12.73 x 6.8 / 2= 10.210 x 10.2= 10.23.2Total= 13.410.2/1.0= 10.25.6Total=15.819 x 2.8 x3600 3 / 384EI9= 6.5 x 10 / EI2. Strength Limit State - Design Load CombinationsM*crit = M*( W u ↓ + G) = 12.7 kNmV*crit = V*( W u ↓ + G) = 15.8 kNTrial Beams:240 x 45, A =10800 mm 2 , I xx = 51.8 x 10 6 mm 4 , Z xx = 432.0 x 10 3 mm300 x 45, A =13500 mm2 , I xx =101.3 x 10 6 mm 4 , Z xx = 675.0 x 10 3 mm 33Determine k 12 based on critical load combination and trial section sizes.k 12 : ρ b = 14.71 (E/f’ b) -0.480 r -0.061 .AS1720.1-1997,Cl8.4.8= 14.71 (12400/45) -0.480 -0.061(12.2/18.8)= 1.01S 1 = 1.25 d/b (Lay/d)0.5 (downwards loads, L ay =0.9 m) AS 1720.1,Cl3.2.3.2300 x 45 section size: = 1.25(300/45)(900/300) 0.5 =14.4ρ b S 1 = 14.6 => k12 = 1.5 – (0.05 x σ b S 1 ) => k AS 1720.1, Cl3.2.412 = 0.77240 x 45 section size: = 1.25(240/45)(900/300) 0.5 =11.55ρ b S 1 = 11.8 => k12 = 1.5 – (0.05 x σ b S 1 )=> k 12= 0.913. Bending criteria - (establish minimum Z p )φM p ≥ M* pand φM p = φ k 1 k 4 k 6 k 9 k 11 k 12 [f’ b Z p ]=> φk 1 k 4 k 6 k 9 k 11 k 12 [f’ b Z p ] ≥ M* pand Z p ≥ M* p / (φk 1 k 4 k 6 k 9 k 11 k 12 ) f’ bTry 240 x 45 section size k 12 = 0.91Minimum required Z p = 12.7 x 10 6 /(0.9x1.0x1.0x0.9x1.0x1.0x0.91) x 453= 413x10 3 mm3 mm 3 for 240 x 45 section size=> OK> Z = 432.0 x 10xxTry 300 x 45 section size k 12 = 0.77Minimum required Z p = 12.7 x 10 6 /(0.9x1.0x1.0x0.9x1.0x1.0x0.77) x 45= 452.6 x10 3 mm3 mm 3 for 300 x 45 section size=> OK< Z = 675.0 x 10xx=>REQUIRE 240x 45 BBF LVL FOR BENDING STRENGTH4. Shear criteria - (establish minimum A s )ΦVp≥ V* pand ΦV p = Φk1 k 4 k 6 k 9 k 11 [f’ s A s ]=> Φk 1 k 4 k 6 k 9 k 11 k 12 [f’ s A s ] ≥ V* p683