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gl Design Example Solution (Cont) <br /> SLep 3 <br /> r2 = 3ft(6ft.--35ft) = 1.25 R. <br /> 6ft. <br /> SLep 4 <br /> B2 =2(125 ft.) =2.5ft. <br /> Hct =125 (from figure 2) <br /> B2. <br /> Sten 5 <br /> Hct =2.5 ft. (1.25)=3.13 R. <br /> ®2 =Tan-i(f25fr/3.13f.) = 21.8° <br /> Stea 6 <br /> Torn®1 =50 <br /> Tan02 =.40 <br /> PN =1312 3_(3-3.5(.50))3 +7 (3-3.5(.50))3 <br /> 30 32(50) 32(.40) <br /> P =247 lbs./ft <br /> N <br /> (12) <br />