3. A 18-foot long column is pinned at the bottom and fixed againstrotation but free to translate at the top. It must support a service deload of 100 kips and a service live load of 100 kips.a. Select a W12 of A992 steel. Use the column load tables.1. Use LRFD.2. Use ASD.b. Select a W16 of A992 steel. Use the trial-and-error approachcovered in Section 4.6.1. Use LRFD.2. Use ASD.

We are given Length of column = 18 ftService Dead load = 100 KipsService Live Load = 100 KipsWe need to calculate the required load strength for LRFD and ASD. Select the section according to meet Required strength based on Column Load table or trial and error approach.Calculations are shown belowCalculation of required Load strengthFor LRFDPu = 1.2*Service Dead Load +1.6*Service Live loadPu = 1.2*100+1.6*100Pu = 280 KipsFor ASDPa = Service Dead Load + Service Live LoadPa = 100+100Pa = 200 Kips(a) Selection of W12 based on Column Load TableTo select a W12 section from Column Load Table 4-1 in AISC Manual we need to calculate Effective length (KL) and obtain fy for A992 steelfy = 50 ksi for A992 AISC Table 2-4Slenderness factor K = 2 from AISC Table C-A -7.1 for pinned and rotation fixed translation free support conditionNow Effective Length KL = 2*18Effective Length KL = 36 ft(1) LRFDEnter Table 4-1 with KL = 36 Ksi and fy = 50 ksi. Look for a section with with greater than Required strength Pu = 280 Kips We get W 12 x 79 section with =261 kips < Pu not OKWe get W 12 x 87 section with =292 kips > Pu hence OK W 12 x 96 section with =326 kips > Pu Hence OKSelect W 12 x 87 section with =326 kips (2) ASDEnter Table 4-1 with KL = 36 Ksi and fy = 50 ksi. Look for a section with with greater than Required strength Pa = 200 Kips We get W 12 x 87 section with =194 Kips < Pa Not OK W 12 x 96 section with =217 Kips > Pa hence OKSelect W 12 x 96 section with =217 Kips (b) Selection of W16 based on Trial and Approach of section 4.6(1) LRFDTo select a section we need to calculate the required gross area(A) of the section using the required strength (Pu)A = Pu / (.fcr)Assume slenderness ration of KL/r = 150 and find .fcr from table 4-22For KL/r = 150 we get .fcr = 10 ksiNow Required gross area A = Pu / (.fcr)A = 280/10A = 28 in2 Select a W16 section from AISC Table 1-1 having A = 12.9 in2 or more Try W16 x 100 (A = 29.4 in2 , rx = 7.10 in, ry = 2.51 in)Now calculate Fcr using below equationFcr = When KL/r ≤ 4.71Fe = critical elastic stress Fe = Fcr =0.877Fe When KL/r > 4.71Check KL/r and 4.71 to use Fcr equation4.71 = 4.71 = 113.43Calculate slenderness ratio (KL/r)For X axis (KL/r)x = (36*12)/7.10(KL/r)x = 60.84For Y axis (KL/r)y= (36*12)/2.51(KL/r)y = 172.11 Governs designKL/r = 172.11 means KL/r > 4.71Use Fcr =0.877*FeFe = Fe = 9.66 MpaFcr =0.877*9.66Fcr = 8.47 Ksiϕc. Fcr =0.9*8.47 (ϕ=0.9)ϕFcr =7.62 KsiDesign Load = 7.62*29.4Design Load = 224.16 Kips < Pu Not OKTry W16 x 100 (A = 29.4 in2 , rx = 7.10 in, ry = 2.51 in)Calculate slenderness ratio (KL/r)For X axis (KL/r)x = (36*12)/7.10(KL/r)x = 60.84For Y axis (KL/r)y= (36*12)/2.51(KL/r)y = 172.11 Governs designHence KL/r = 172.11KL/r > 4.71Use Fcr =0.877*FeFe = Fe = 9.66 MpaFcr =0.877*9.66Fcr = 8.47 Ksiϕc. Fcr =0.9*8.47 (ϕ=0.9)ϕFcr =7.62 KsiDesign Load = 7.62*29.4Design Load = 224.16 Kips < Pu Not OKW16 x 100 is the last section for W 16 shapes with required Area A more than 28 in2 . Remaining W 16 section are smaller in area and will give ess design load than W 16 x 100 section hence we can conclude that " For given effective length(KL and LRFD required strength Pu = 280 Kips , W16 section is not available"(2) ASDWe have calculated above for W 16 x 100 section that Fcr = 8.47 ksi =8.47/1.67 ( = 1.67) = 5.07 ksiAllowable Load = 5.07*29.4Allowable Load = 149 Kips < Pa" For given effective length(KL and ASD required strength Pa = 200 Kips , W16 section is not available All answers are given in above step.