Moment distribution: rectangular frame, two horizontal loads

Data: single-storey frame with fixed bases A and D, span 6 m, storey height 4 m, columns $2EI$, beam $4EI$. Horizontal loads (to the right): 25 kN at the top-left joint B and 50 kN at the mid-height of the left column AB. Because the loads are lateral, the frame sways and the analysis is in two cases.

Step 1: Stiffness and distribution factors ($k=EI/L$)

• Joint B: relative stiffness $\Sigma k$ = 1.1667; BA: $k=EI/L$ = 0.5000, DF = 0.4286; BC: $k=EI/L$ = 0.6667, DF = 0.5714
• Joint C: relative stiffness $\Sigma k$ = 1.1667; CB: $k=EI/L$ = 0.6667, DF = 0.5714; CD: $k=EI/L$ = 0.5000, DF = 0.4286

Step 2: Fixed-end moments

The only member load is the 50 kN on column AB ($L=4$, load at mid-height): $\mp\dfrac{PL}{8}=\mp\dfrac{50\times4}{8}=\mp25$ kNm; sign for a load to the right on the column: $FEM_{AB}=-25.00$, $FEM_{BA}=25.00$ kNm.

Case I: sway prevented by a horizontal support at B

Reaction of the imaginary support: $R_1=-46.875$ kN.

Case II: sway with no loads

Assume FEM $=-100$ kNm at both ends of each column.

Reaction of the imaginary support: $R_2=75.000$ kN.

Step 3: Combination

$k=-\dfrac{R_1}{R_2}=0.62500$

Joint check: B $0.00$, C $0.00$.

Bending moment diagram (kNm; positive = inside tension for the frame traversed A-B-C-D)

Reactions: A: $H=-52.81$, $V=-11.11$, $M=82.92$; D: $H=-22.19$, $V=11.11$, $M=50.42$. Check: horizontal reactions sum to -75.00 kN, balancing 25 + 50 = 75 kN.

Answer: $M_{AB}=-82.92$, $M_{BA}=-28.33$, $M_{BC}=28.33$, $M_{CB}=38.33$, $M_{CD}=-38.33$, $M_{DC}=-50.42$ kNm.