Design the interior beam for $-M_u$ at the interior support $($B$).$ Use the load and

member size from part $($b$).$ Assume ends integral with the support for moment

calculations using the ACI approximate moment equations for two spans.

Note. The flange for this beam will be in tension. You are not allowed to

change the member dimensions selected in part $($b$).$ Start the design process as

a singly reinforced beam. If it is not satisfactory, design the member as a

doubly reinforced beam.

All neccessary designs from part $($b$)$ have been attached.

Total factored load $=(2,619\mathrm{.}5+270)=2,889\mathrm{.}5$ lb$/$ft

$M_u=w_u\frac{I_n^2}{14}$

$=2,889\mathrm{.}5{27}^2\frac{12}{14}=1,805,524lb-in$

$Mn=$ Asfy $(d-\frac{a}{2})Mn=1\mathrm{.}8060,000(19\mathrm{.}56-\frac{0\mathrm{.}42}{2})=2,089,800lb-in$

$Mn=0\mathrm{.}92,089,800lb-in=1,880,820lb-in>1,805,524lb-in$

Please use ACI Tables and Equations