Question $1$

A statically indeterminate beam is shown. The following distribution factors can be calculated: DF$\_($BA$)$

$=0\mathrm{.}5,$DF$\_($BC$)=0\mathrm{.}5$;DF$\_($CB$)=0\mathrm{.}667,$DF$\_($CD$)=0\mathrm{.}333.$ Use w$=45$k$($N$)/($m$).$ Use E$=200$GPa,I$=100\backslash$times $10^(-6)$

m$^(4).$

aC$.$ The following

fixed end moments can be calculated: FEM$\_($AB$)=60$kNm$,$FEM$\_($BA$)=-60$kNm$,$FEM$\_($BC$)=60$kNm$,$

FEM$\_($CB$)=-60$kNm$.$ After the moment distribution process, the following moments are calculated:

M$\_($AB$)=54\mathrm{.}54$kNm$,$M$\_($BA$)=-70\mathrm{.}92$kNm$,$M$\_($BC$)=70\mathrm{.}92$kNm$,$M$\_($CB$)=-21\mathrm{.}82$kNm$,$M$\_($CD$)=21\mathrm{.}82$kNm$.$

What is the value of R$\_(1),$ the force preventing sway?

bC$.$ Fixed end moments in CD are assumed as follows: FEM$\_($CD$)=$FEM$\_($DC$)=30$kNm$.$ What are the corresponding fixed end moments due to sway in element BC $?$

FEM$\_($BC$)$:$,$kNm

FEM$\_($CB$)$:$,$kNm

Unanswered

cM$\_($AB$)=7\mathrm{.}5$kNm$,$M$\_($BA$)=15$kNm$,$M$\_($BC$)=-15$kNm$,$ M$\_($CB$)=-18\mathrm{.}75$kNm$,$M$\_($CD$)=18\mathrm{.}75$kNm$.$ What is the value of R$\_(2),$ the force causing the sway?

R$\_(2)$:$,$kN