Extracted text: A steel column, high 5m, is required to support the lead shown in figure Q4. The column is effectively held in position at both ends restrained in direction at the base, fixed, but not at the cap (i.e. partially retrained, i.e. pinned). Assume self-weight of the column = 2kN. c) Determine the ultimate axial load applied to the column, Ned. d) Determine the ultimate axial load capacity Ne,D. PL=410KN V.I=270N PL = 260KN PL=320EN V.L=170KN V.L=240KN PL = 300KN VL= 190KN