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TSE Engineering Job #:�7� <br /> By ✓n <br /> �`1\ <br /> Page:� <br /> HEADER(UPPER LEVEL) <br /> INPUT: Uniform Loading Span Length <br /> w (DL) w (LL) L <br /> Roof (psf) 15 25 5 0 <br /> TribufaryQQ 12 12 ft <br /> Wall (psf) 10 0 <br /> Tributary(�p 0 0 <br /> Floor (psf) 10 40 <br /> Tnbutary(n) 0 0 <br /> Other (plf) 10 0 <br /> w (TL) 0 <br /> 190 300 490 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 /> 475 475 750 750 1225 1225 594 938 1531 <br /> Ibs. Ibs. Ibs. Ibs. Ibs. Ibs. ftlbs. ft.lbs. ft.lbs. <br /> DESIGN: <br /> MATERIAL Fb W Fc(perp) E x 10^6 <br /> Manuf.Lbr. Co 1.15 <br /> Timber CN 1 <br /> oimen. �br. DR#2 1170 180 625 1.6 Cr 1 <br /> psi psi psi psi Ci 1 <br /> b d A S I <br /> 3.5 725 25.38 307 111 <br /> in. in. in.^2 in.^3 in^4 <br /> fv= 55 psi Brg.Lgth.= 0.047 ft. CL = 1.000 <br /> fb = 600 psi GL Cv= N/A R = N/A <br /> d (o�)= 0.02 in. � 0.�)= OA2 in. 4 (T�) = 0.04 in. <br /> RATIOS OF ACTUAL TO ALLOWABLE RATIOS OF SPAN TO DERECT�ON <br /> fv/Fv' = 0.27 L/ 2529 forLL <br /> fb / Fb' = 0.45 L/ 1549 for TL <br /> USE 4x8 DFL#2 <br /> v 33.03 <br />