Consider a one$-$bay frame, subjected to a combination of distributed gravity loads on the beam $($wDL $=2\mathrm{.}2$ kip$/$ft$,$ wLL $=1\mathrm{.}5$ kip$/$ft$).$ The dimensions of the frame are L $=30$ feet and H $=14$ feet. The frame is statically indeterminate $(1$x indeterminate$),$ but the horizontal reactions are given, so you can solve for the internal forces using the equations of equilibrium. Note that the value of the horizontal base reaction force under gravity depends on the ratio of the beam to column moments of inertia $($i$.$e$.,$ Ib$/$Ic$).$ The horizontal reactions are both Fh $=$ wL$^3/(8($Ib$/$Ic$)$H$^2+2$HL$)$ facing inwards. Gravity load analysis under the distributed load, w:

$($i$)$ Neatly sketch the deflected shape of the frame

$($ii$)$ Assuming that the beam and columns have the same moment of inertia $($i$.$e$.,$ Ib$/$Ic $=1),$ calculate the column base reactions and then sketch the diagram of internal member shear forces and moments and calculate the maximum values of moment and shear in each member.

$($iii$)$ Compare the calculated moment at the end of the beam to the end moment that would occur if the beam had fixed$-$fixed boundary conditions $($i$.$e$.,$ Mend $=$ wL$^2/12).$ Does the difference between the values $($fixed$-$fixed vs your frame calculations$)$ make sense, i$.$e$.,$ can you explain the difference?