Moment distribution: frame with cantilever

Assumptions: the girder F-E-D-C has a 1.5 m cantilever FE (3EI) with 3 kN at F, a 4 m span ED (3EI) with 15 kN/m, and a 1.5 m span DC (3EI) ending on a roller at C. Columns EA (2EI) and DB (EI) are each 4 m high and fixed at their bases. The roller at C allows horizontal movement, so the frame can sway; the moment distribution is done in two cases (sway prevented, then sway induced).

Step 1: Overhang FE

$M_{EF}=3\times1.5=4.5$ kNm (hogging), fixed moment on joint E.

Step 2: Distribution factors

• Joint E: relative stiffness $\Sigma k$ = 1.2500; EF: overhang, DF = 0 (its moment is a known fixed value); ED: $k=EI/L$ = 0.7500, DF = 0.6000; EA: $k=EI/L$ = 0.5000, DF = 0.4000
• Joint D: relative stiffness $\Sigma k$ = 2.5000; DE: $k=EI/L$ = 0.7500, DF = 0.3000; DC: $k=EI/L$ = 1.5000 (far end hinged: 3/4 factor), DF = 0.6000; DB: $k=EI/L$ = 0.2500, DF = 0.1000 (Roller end C: member DC is stiffened by 3/4 because its far end is hinged and the end moment at C is zero.)

Step 3: Case I, sway prevented by an imaginary horizontal support at E

Fixed-end moments from the loads: $\mp15\times4^2/12=\mp20$ kNm on ED.

Reaction of the imaginary support (from the column shears): $R_1=-1.94$ kN (acting to the left).

Step 4: Case II, sway $\Delta_0$ without external loads

Give the girder a trial sway to the right and choose its size so that the fixed-end moment in column EA is $-100$ kNm at both ends. Since $FEM=-6EI\Delta/L^2$ is proportional to $EI/L^2$ and both columns have the same height and the same $\Delta$, the column DB (EI) gets half of this: $-50$ kNm at both ends. All other members have zero FEM.

Reaction of the imaginary support: $R_2=59.69$ kN.

Step 5: Correction for sway

The imaginary support does not exist, so the restraint force must vanish: $R_1+k\,R_2=0$

$$k=-\frac{R_1}{R_2}=-\frac{-1.94}{59.69}=0.0325$$

Final moments $M=M_I+kM_{II}$ (kNm):

Support moments: $M_{AE}=1.26$, $M_{BD}=-2.88$ kNm. Joint E check: $M_{ED}+M_{EA}+M_{EF}=0.00$ ✓. Joint D check: $M_{DE}+M_{DB}+M_{DC}=0.00$ ✓. $M_{CD}=0$.

Bending moment ordinates (kNm)

Answer: $M_{EA}=5.76$, $M_{AE}=1.26$, $M_{ED}=-10.26$, $M_{DE}=19.22$, $M_{DB}=-4.14$, $M_{BD}=-2.88$, $M_{DC}=-15.08$ kNm (clockwise end moments +); $M_{EF}=4.50$ kNm.