$($a$)$ For the column section shown in Figure $3\mathrm{.}1,$ compute the ultimate moment and axial load capacities for the case when $\frac{x}{h}=0\mathrm{.}7.$ Assume $f_{ck}=40\frac{N}{m}{m}^2,f_{yk}=500\frac{N}{m}{m}^2,\frac{d}{h}=0\mathrm{.}9,\frac{d_2}{h}=0\mathrm{.}1,E_s=200k\frac{N}{m}{m}^2,f_{yd}=435\frac{N}{m}{m}^2$ and ${}_{yd}=0\mathrm{.}00218.$ Assume the column is short and braced. The non$-$dimensional form of the column equations are given below. $($All symbols have their usual meaning$)$

$N_r=\frac{N}{b_h}=\frac{x}{h}+\frac{f_{s2}}{f_{ck}}{}_2-\frac{f_{st}}{f_{ck}}$

$M_r=\frac{M}{b^2{f}_{ck}}=\frac{x}{h}(0\mathrm{.}5-\frac{}{2}\frac{x}{h})+\frac{f_{s2}}{f_{ck}}{}_2(0\mathrm{.}5-\frac{d_2}{h})+\frac{f_{st}}{f_{ck}}(\frac{d}{h}-0\mathrm{.}5)$

$($b$)$ Calculate the maximum ultimate load $($in $k\frac{N}{m})$ that a $6m$ span simply supported reinforced concrete rectangular beam, subject to only uniformly distributed loading, could sustain without compression reinforcement if the beam is $500mm$ deep by $300mm$ wide. Assume that $d=446mm,f_{ck}=35\frac{N}{m}{m}^2,f_{yk}=500\frac{N}{m}{m}^2.$ You should assume that the ultimate distributed load includes the self$-$weight.