A tied compression member has been designed for the given load conditions. However, the original design did not take into account the fact that under a reversal in the direction of lateral load $($wind$),$ the axial load, due to the combined effects of gravity and lateral loads, becomes $85$ kN$,$ with essentially no change in the values of Mu and Vu$.$ See Figure SAM$13\mathrm{.}22.$ Use Figure CODE $542.$Along the positive x$-$axis:Mu $=-120$ kN$-$m$,$ Pu $=750$ kN$,$ Vu $=210$ kNAlong the negative x$-$axis:M$,=+120$ kN$-$m$,$ Pu $=85$ kN$,$ V$4=210$ kNc $=27\mathrm{.}5$ MPa, fyt $=275$Reduction factor for shear $=0\mathrm{.}75$b $=350$ mm$,$ h $=450$ mm$28.$ Calculate the nominal shear strength $($kN$)$ of concrete along the positive x$-$axis using the simplified calculation.A$.95$C$.161$B$.120$D$.24129.$ Calculate the nominal shear strength $($kN$)$ of concrete along the negative x$-$axis using the simplified calculation.A$.78$C$.94$B$.126$D$.23630.$ Calculate the required spacing $($mm$)$ of shear reinforcement.A$.195$C$.97$B$.232$D$.109$