Moment distribution: portal frame, AFD, SFD, BMD

Data: portal frame with fixed bases A and D, columns AB and CD of height 6 m ($2EI$), beam BC of span 7 m ($4EI$) carrying 15 kN/m and a 60 kN load at 2 m from B. Because the load on the beam is not symmetric about the centre line, the frame sways slightly. The analysis is done in two cases.

Step 1: Distribution factors ($k=EI/L$)

• Joint B: relative stiffness $\Sigma k$ = 0.9048; BA: $k=EI/L$ = 0.3333, DF = 0.3684; BC: $k=EI/L$ = 0.5714, DF = 0.6316
• Joint C: relative stiffness $\Sigma k$ = 0.9048; CB: $k=EI/L$ = 0.5714, DF = 0.6316; CD: $k=EI/L$ = 0.3333, DF = 0.3684

Step 2: Fixed-end moments (clockwise +)

Beam BC: UDL $\mp\dfrac{15\times7^2}{12}=\mp61.25$ kNm; point load ($a=2$, $b=5$): $\mp\dfrac{60\times2\times5^2}{7^2}=-61.22$ at B and $+\dfrac{60\times2^2\times5}{7^2}=+24.49$ at C. Total $FEM_{BC}=-122.47$ kNm, $FEM_{CB}=85.74$ kNm.

Case I: sway prevented (imaginary support at B)

Reaction of the imaginary support: $R_1=-2.571$ kN.

Case II: sway without external load

Assume FEM $=-100$ kNm at the ends of both columns (equal EI, equal height, equal $\Delta$).

Reaction of the imaginary support: $R_2=52.667$ kN.

Step 3: Combination

$k=-\dfrac{R_1}{R_2}=0.04882$ and the final moments are $M=M_I+kM_{II}$:

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

Step 4: BMD, SFD and AFD

Bending moment (sagging/inside tension +, kNm):

Shear force and axial force:

Shear sign: positive when the force on the start end acts along the member's local y axis (90° anticlockwise from the direction A→B, B→C, C→D). Axial: negative = compression. The vertical reactions are 95.82 kN at A and 69.18 kN at D (total 165.00 kN = 15×7+60), the horizontal reactions are 14.01 kN and -14.01 kN (equal and opposite).

SFD of the beam: 95.82 kN at B, falling by 15 kN/m to 65.82 kN just left of the 60 kN load, then 5.82 kN, and reaching -69.18 kN at C.

Answer: $M_{AB}=26.40$, $M_{BA}=57.69$, $M_{BC}=-57.69$, $M_{CB}=54.43$, $M_{CD}=-54.43$, $M_{DC}=-29.66$ kNm (clockwise end moments +).