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TSE Engineering Job #:'�J�j <br /> By KrC <br /> Page:� <br /> ROOF BEAM <br /> INPUT: Uniform Loading Span Length <br /> w (DL) w (LL) L <br /> Roof (psf) 15 25 4 0 <br /> Tributary(fry 16.5 16.5 ft <br /> Wall (psf) 10 0 <br /> Tributary(fp 0 0 <br /> Floor(psf) 10 40 <br /> Tributary�ft) 0 0 <br /> Other(pN) 10 0 <br /> w (TL) 0 <br /> 257.5 4125 670 0 <br /> plf pN pN <br /> RESULTS: <br /> VI (DL) Vr(DL) VI (LL) Vr(LL) VI (TL) Vr(TL) M (DL) M (LL) M (TL) <br /> 515 515 825 825 1340 1340 515 825 1340 <br /> Ibs. Ibs. Ibs. Ibs. Ibs. Ibs. ft.lbs. ftlbs. ft.lbs. <br /> DESIGN: <br /> MATERIAL Fb W Fc(verp) E x 10^6 <br /> ManutLbr. Co 1.15 <br /> Timber CH i <br /> �imen. Lbr. DFL#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 30.7 111 <br /> in. in. in^2 in^3 in^4 <br /> fv= 55 psi Brg.Lgth.= 0.051 ft C�= 1.000 <br /> fb= 520 psi GL Cv= N/A R = N/A <br /> d (oy = 0.01 in. d (��) = 0.01 in. d (7�) = 0.02 in. <br /> RATIOSOFACTUALTOALLOWABLE RATIOSOFSPANTODEFLECTION <br /> fv/Fv'= Q27 L/ 3593 forLL <br /> fb/ Fb'= 0.39 L/ 2212 forTL <br /> USE 4x8 DF#2 <br /> vs.sos <br />