Example $8\mathrm{.}1$

Determine the minimum cross section required for a rectangular beam so that no shear reinforcement is required. Follow ACl Code requirements and use a concrete strength of $4,000.V_u=38k.$

Solution:

$V_c=2\sqrt[\phantom{2}]{f_c^{\text{'}}}{b}_{w}d$

$V_c=(0\mathrm{.}75)[2(1\mathrm{.}0)(\sqrt[\phantom{2}]{4000})b_{w}d]=94\mathrm{.}87b_{w}d$

But the ACI Code $11\mathrm{.}4\mathrm{.}6\mathrm{.}1$ states that a minimum area of shear reinforcement is to be provided if $V_u$ exceeds $\frac{1}{2}{V}_c$

$38,000Ib=\frac{1}{2}(94\mathrm{.}87b_{w}d)$

$b_{w}d=801\mathrm{.}1In.{?}^2$

$0401314$ Reinforced Concrete Design $1$

Ch$.8-$ Shear and Diagonal Tension

Can someone explain how we assumed $24$in x $36$in beam & d$=33\mathrm{.}5$in$?$ On what basis and how did we assume?