Design the column under the influence of axial force, the properties of which are given in the figure below, using the AISC 360-16.
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Solution:
Local Buckling
a) For flange:
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| body | --uriencoded--$$ \normalsize b/(2*t_f)≤0.56*\sqrt%7B(E/F_y)%7D$$ |
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| body | --uriencoded--$$ \normalsize 300/(2*24)=6.25≤0.56*\sqrt%7B(200000/355)%7D=13.29$$ |
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b) For web:
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| body | --uriencoded--$$ \normalsize h/(t_w)≤1.49*\sqrt%7B(E/F_y)%7D $$ |
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| body | --uriencoded--$$ \normalsize 400/13.5=29.63≤1.49*\sqrt%7B(200000/355)%7D=35.28$$ |
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Flexural Buckling
According to article 6.4.3(a) of the regulation, K= 1.0 is taken.
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| body | $$ \normalsize L_c_f=K_x*L_x=1.0*9000=9000$$ |
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| body | $$ \normalsize L_c_y=[L_c_y_1;L_c_y_2]_m_a_x=[1.0*4500;1.0*4500]_m_a_x=4500 mm$$ |
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According to the article 8.1.1 of the regulation, the slenderness ratio of compression elements cannot exceed 200.
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| body | $$ \normalsize λ_x=9000/170.8=52.7≤200$$ |
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| body | $$ \normalsize λ_y=4500/74=60.8≤200$$ |
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| body | --uriencoded--$$ \normalsize L_c_i_,_m_a_x=60.8≤4.71\sqrt%7B(E/F_y)%7D=4.71\sqrt%7B(200000/355)%7D=111.79$$ |
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| body | --uriencoded--$$ \normalsize F_E=(\pi%5e2*E)/λ_c_r%5e2=(\pi%5e2*200000)/60.8%5e2=533.97$$ |
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| body | --uriencoded--$$ \normalsize F_c_r=0.658(F_y/F_E)*F_y=0.658%5e(%5e3%5e5%5e5%5e/%5e5%5e3%5e3%5e.%5e9%5e7%5e)*355=268.77$$ |
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| body | --uriencoded--$$ \normalsize P_n=F_c_r*A_g=268.77*19780*10%5e-%5e3=5316.27$$ |
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LRFD:
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| body | $$ \normalsize P_d=\phi_c*P_n=0.90*5316.27=4784.64 kN$$ |
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ASD:
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| body | $$ \normalsize P_d=P_n/Ω_c=5316.27/1.67=3183.4 kN$$ |
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Required Compressive Force Strength
LRFD:
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| body | $$ \normalsize P_u=1.2P_G+1.6P_Q=1.2*750+1.6+2000=4100 kN$$ |
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ASD:
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| body | $$ \normalsize P_a=P_G+P_Q=750+2000=2750 kN$$ |
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PMM Ratio
LRFD:
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| body | $$ \normalsize P_u=P_d=4100/4784.64=0.86 $$ |
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ASD:
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| body | $$ \normalsize P_a=P_d=2750/3183.4=0.86 $$ |
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