Please activate JavaScript!
Please install Adobe Flash Player, click here for download

TW Catalogue EN Int. 03-2015

3703/2015 11 WT / WR Dowelled purlins 3/5 selected: WT-T-8,2 x 245 F1,Rk = 8,3 kN (according to table on page 4) hmin = 110 mm < 160 mm = h (required minimum height maintained) eerf = = = 112 mm selected: e = 100 mm 2 · Fv,Rd · (h1+h2) 3 · Vd 2 · 5,7 · (160+160) 3 · 10,9 in the two middle quarter zones: eerf = 2 · 100 = 200 mm Fv,Rd = = 5,7 kN 8,3 · 0,9 1,3 Sample Given NKL1;KLED short; kmod = 0,9 span L = 5,00 m cross-section 2-part, C24 b = 100 mm h1 = h2 = 160 mm load gk = 1,00 kN/m pk = 2,00 kN/m qd = 1,35 · 1,00 + 1,5 · 2,00 = 4,35 kN/m Stress resultants max. shearing force max. moment Vd = = = 10,9 kN qd · L 2 1122hh bb hh yy yy yy yy g = 1,00 kN/m, p = 2,00 kN/mg = 1,00 kN/m, p = 2,00 kN/m +V+V -V-V SchubdeckungSchubdeckung QuerkraftverlaufQuerkraftverlauf L = 5,00 mL = 5,00 m L/4L/4 L/4L/4 L/4L/4 L/4L/4 kk kk 4,35 · 5,00 2 Md = = = 13,6 kNm qd · L2 8 4,35 · 5,002 8 Cross-section moment of resistance moment of inertia Wy,ef =  · = 0,85 · = 1,45 · 106 mm3 b · (h1+h2)2 6 100 · (160 +160)2 6 y,ef =  · = 0,65 · = 177 · 106 mm4 b · (h1+h2)3 12 100 · (160 +160)3 12 Proofs bending stress deflection ef,d = = = 9,4 N/mm2 < = 16,6 N/mm2 Md Wy,ef 13,6 · 106 1,45 · 106 ƒef = · = · = 13,8 mm < 16,7 = 3,0 · 50004 10000 · 177 · 106 L 300 5 384 qk · L4 E · y,ef 5 384 0,9 · 24 1,3 Shear overlap Shearing force curve Choice of fastener Arrangement 160160160160 122,5 122,5 122,5 122,5 4040 100100 100100 50505050 Technical documentation and data sheets 11.11. 4040100100100100

Pages Overview