A beam of rectangular cross section has $b=22in.$ and $h=15in.$ It is to carry a total factored

load of $3600l\frac{b}{f}t$ uniformly distributed over its $26$ ft span, and in addition the beam will be

subjected to a uniformly distributed torsion of $1800ft-l\frac{b}{f}t$ at factored loads. Closed stirrup$-$ties

will be used to provide for flexural shear and torsion, placed with the stirrup steel centroid $1\mathrm{.}75$

in$.$ from each concrete face. The corresponding flexural effective depth will be approximately

$12\mathrm{.}5$ in $.$ Design the transverse reinforcement for this beam and calculate the increment of

longitudinal steel area needed to provide for torsion, using $f_c^{\text{'}}=4000$ psi and $f_y=60,000$ psi.

Tip: For transverse reinforcement, use No$.4$ bars. $\frac{A_v}{s}$ must be estimated from shear analysis, but

to save time it will be given as $\frac{A_v}{s}=0\mathrm{.}0313\frac{i{n}^2}{in}.V_u=\frac{w_{u}l}{2}$ and $T_u=\frac{t_{u}l}{2}.$ To make the calculation

simple, just consider the spacing required to carry the maximum $V_u$ and $T_u$ at $d$ from the support

$($increasing spacing through the beam length can be skipped$).$ The longitudinal reinforcement

also needs to be calculated.