Moment distribution: frame with 100 kN at D

Assumptions: the girder B-C-D has BC = 8 m with 50 kN/m and CD = 5 m with 100 kN at its mid-point (2.5 m + 2.5 m); D is a hinge (the support prevents horizontal movement). Columns BA (4 m) and CE (6 m) are fixed at A and E. The 20 kN horizontal load acts at B (to the right). All members have the same EI. Because the hinge D prevents horizontal movement of the girder, the frame does not sway; the 20 kN load is carried by axial force in the girder and shared by the supports; it does not change the bending moments.

Fixed-end moments (clockwise +)

• BC: $\mp\dfrac{50\times8^2}{12}=\mp266.67$ kNm
• CD: $\mp\dfrac{100\times5}{8}=\mp62.5$ kNm; D is hinged, so the modified FEM at C is $-62.5-\tfrac12(62.5)=-93.75$ kNm and the end moment at D is zero.

Stiffness and distribution factors ($k=EI/L$; CD uses 3/4 because D is hinged)

• Joint B: relative stiffness $\Sigma k$ = 0.3750; BC: $k=EI/L$ = 0.1250, DF = 0.3333; BA: $k=EI/L$ = 0.2500, DF = 0.6667
• Joint C: relative stiffness $\Sigma k$ = 0.4417; CB: $k=EI/L$ = 0.1250, DF = 0.2830; CD: $k=EI/L$ = 0.1500 (far end hinged: 3/4 factor), DF = 0.3396; CE: $k=EI/L$ = 0.1667, DF = 0.3774

Moment distribution (kNm)

Joint B: $-198.78+(198.78)=0.00$ ✓; Joint C: $253.36+(-169.35)+(-84.00)=0.00$ ✓; hinge D: $M_{DC}=0$.

Bending moment diagram

Reactions: A: $H=74.54$, $V=193.18$, $M=-99.39$; E: $H=-21.00$, $V=290.69$, $M=42.00$; D: $H=-73.54$, $V=16.13$, $M=0.00$.

Answer: $M_{BC}=-198.78$, $M_{CB}=253.36$, $M_{CD}=-169.35$, $M_{BA}=198.78$, $M_{AB}=99.39$, $M_{CE}=-84.00$, $M_{EC}=-42.00$ kNm (end moments in the order of the member names; clockwise +).