$(DF{)}_{AB}=1$

$(DF{)}_{BA}=\frac{3\frac{I_{\text{A}\text{B}\text{C}}}{6}}{3\frac{I_{\text{A}\text{B}\text{C}}}{6}+4\frac{I_{\text{A}\text{B}\text{C}}}{18}+3\frac{0\mathrm{.}75I_{\text{A}\text{B}\text{C}}}{12}}=0\mathrm{.}5496$

$(DF{)}_{BC}=0\mathrm{.}2443$

$(DF{)}_{BD}=0\mathrm{.}2061$

$(DF{)}_{CB}=\frac{4\frac{I_{\text{A}\text{B}\text{C}}}{18}}{4\frac{I_{\text{A}\text{B}\text{C}}}{18}+3\frac{1\mathrm{.}25I_{\text{A}\text{B}\text{C}}}{12}}=0\mathrm{.}4156$

$(DF{)}_{CE}=0\mathrm{.}5844$

$(DF{)}_{DB}=(DF{)}_{EC}=1$

$(FEM{)}_{AB}=-\frac{4(6{)}^2}{12}=-12\mathrm{.}0k*ft$

$(FEM{)}_{BA}=12\mathrm{.}0k*ft$

$(FEM{)}_{BC}=-\frac{4(18{)}^2}{12}=-108k*ft$

$(FEM{)}_{CB}=108k*ft$

$(FEM{)}_{CE}=-\frac{20(12)}{8}=-30\mathrm{.}0k*ft$

$(FEM{)}_{EC}=30\mathrm{.}0k*ft$

$(FEM{)}_{BD}=(FEM{)}_{DB}=0$

Ex$09$:

Determine the moments.

Assume A$,$ D$,$ and E are pinned. $E=29{10}^3$ksi

Already have end moments after Moment Distribution Method. Find internal member forces, support reactions and draw bending moment diagrams