A rectangular beam is $15$ in $.$ wide and $30$ in $.$ deep. For $f_c^{\text{'}}=4000$ psi and $f_{yt}=60,000,$

determine the required spacing of No$.4($No$.13)$ closed stirrups for a factored shear of $80$ kips

and factored torsional moment of $50$ ft $-$kips. The stirrup centroid is located $1\mathrm{.}75$ in $.$ from each

concrete face. The effective depth is $26\mathrm{.}5$ in $.$

Given parameters: $$

${}_1$:$=0\mathrm{.}85,E_s$:$=29000000$ psi, $E_c$:$=57000*\sqrt[\phantom{2}]{f_c^{\text{'}}*}=3604997$ psi

$V_u$:$=80$kip,${}_f$:$=0\mathrm{.}90,{}_y$:$=\frac{f_y}{E_s}=0\mathrm{.}002$

$T_u$:$=50ft*kip,{}_v$:$=0\mathrm{.}75,$:$=1\mathrm{.}0$

Tip: To estimate the required spacing, the required steel area due to torsion and due to shear must

be combined. Therefore, $2A_t+A_v=A_{sNo\mathrm{.}4}.$ Using the equations in the textbook, $A_t$ shall be

estimated in function of $s.$ I.e$.,A_t=x.yys.$ In principle, $\frac{A_v}{s}$ must be estimated from shear

analysis, but to save time it will be given as $\frac{A_v}{s}=0\mathrm{.}0355\frac{i{n}^2}{in}.$