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TSE Engineeiing Job#: ,J�3 <br /> BY: <br /> Page:35 <br /> FLOOR BEAM <br /> INPUT: Uniform Loading Span Length <br /> w(DL) w(LL) L <br /> Roof (psf) 15 25 6 0 <br /> Tribu�ary(n) 8 8 k <br /> Wall (psf) 10 0 <br /> rribinary(n) 9 9 <br /> Floor (psf) 10 40 <br /> 7cibu�ary(n) 1 1 <br /> Other(plf) 10 0 <br /> w(TL) 0 <br /> 230 240 470 0 <br /> plf plf plf <br /> RESULTS: <br /> VI (DL) Vr(DL) VI (LL) Vr(LL) VI (TL) Vr(TL) M (DL) M (LL) M (TL) <br /> 690 690 720 720 1410 1410 1035 1080 2115 <br /> Ibs. Ibs. Ibs. Ibs. Ibs. Ibs. ft.lbs. ft.lbs. k.lbs. <br /> DESIGN: <br /> MATERIAL Fb Fv Fc(perp) E x 10^6 <br /> Manuf.Lbc Co 1.15 <br /> Timber CH 1 <br /> Dimen. Lbc DFL#2 990 180 625 t.6 Cr 1 <br /> psi psi psi psi Ci 1 <br /> b d A S I <br /> 3 9.25 27.75 42.8 198 <br /> In. ln. ln^2 in^3 in.^4 <br /> fv= 57 psi Brg.Lgth: 0.063 ft. CL= 1.000 <br /> fb = 590 psi GL Cv= N/A R = N/A <br /> 0 (Dy= 0.02 in. � (Ly = OA2 in. � Qy = 0.04 in. <br /> RATIOS OF ACTUAL TO ALLOWABLE RATIOS OF SPAN TO DEFLECTION <br /> fv/Fv'= Q27 L/ 3257 for LL <br /> fb/ Fb' = 0.52 L/ 1663 for TL <br /> USE DBL 2x10 OR 4z10 - <br /> �a.a aa <br />