Question

350 mm -- 20 mm 20 mm Using the AISC equations, check if a column having the cross section with the dimensions shown, can support an axial force of 500 kN. The column has a length of 10 m, is made from steel ( E200 GPa, Gy=345 MPa), and its ends are pinned. Y-Y. 300 mm 10 mm

Answer 1

203 K 310 K 20 300 Due to symmetry, 175 mm 150 mm V = DETERMINATION OF MOMENT OF INERTIA S.No SHAPE 1/12 x bxh 1/12 x 20 x 300 (mm ) = 45000000.0 (7,), 1 / 12 x bx h 3 1 / 12 x 300 x 20 mm 4 ) 200000.0 A(F'-7) (mm 4 ) 6000 10 - 175.000 ) 2 = 1 Rectangle = 163350000.0 2 Rectangle 1/12 x 310 x 10 = 25833.3 1/ 12 x 10 x 310 = 24825833.3 3100 ( 175 - 175.000 ) 2 = 0 3 Rectangle 1/ 12 x 20 x 300 3 = 45000000.0 1/12 x 300 x 20 3 = 200000.0 6000 ( 340 - 175.000 ) ? = 163350000.0 90025833.3 25225833.3 326700000.0 Therefore, moments of inertia, ī, = (), 90025833.3 mm = E(Īy), + E 4(, - ) = 351925833.3 mm Ī,

Area of cross section, A = 15100.0 mm ? Radii of gyration, rx = VITA 77.214 mm ry = VI,JA = 152.664 mm

GIVEN DATA Properties of Steel section Column Section Gross area of cross section, Radii of gyration, Built-up column Ag = 15100 mm? rx = 77.214 mm ry = 152.66 mm Restraints Along x direction Unsupported Length of column, Bottom End Top end Lx = 10 m pinned : pinned Along y direction Unsupported Length of column, Bottom End Top end Ly = 10 m : pinned : pinned Load Factored axial load, Pu= 500 KN CALCULATIONS Yield stress of steel Modulus of Elasticity of steel 345 Mpa 200 Gpa Effective length factors, Kx = Ky = 1 1 1x ( 10 x 1000 mn)/ 77.2 = 129.51 1x 10 x 1000 mn) / 153 = 65.503 Since KxLx/rx REV K, Ly/ry, KxLx/rx controls. V72E = 1.712 > 1.5 Note: For Te<1.5, Fe = (0.658°)F, 2.>1.5, Fe = 0,877|F,

Therefore, critical stress, For = 103.23 Mpa Hence, Design axial strength of column, bPn = 0,F,A, = 0.9 x = 1403 kN SAFE 103.2 x > 15100 500 kN /1000 (=Pu] Hence,