Moment distribution: multi-bay frame, AFD, SFD, BMD

Assumptions (figure is not fully clear)

• Beam A-B-C-D-E: four members of 1.5 m each; AB and DE are cantilever overhangs ($EI$ each), BC and CD are $2EI$.
• Columns BF ($3EI$, 2 m), CG ($4EI$, 2 m) and DH ($3EI$, 1 m) have fixed bases at F, G, H.
• Vertical loads (upward, as stated): 2 kN at A, 5 kN at the middle of BC, 5 kN at the middle of CD and 2 kN at E. Horizontal loads: 10 kN to the right at B and at D. The beam is axially rigid, so all joints sway equally ($\Delta$): a sway frame.
• Clockwise end moments are positive (kN·m).

Distribution factors

Relative stiffness $K = EI/L$ (far ends fixed). The overhangs AB and DE do not resist rotation of the joint, so they only transfer fixed moments to B and D.

Stage 1: sway prevented (joints held against side movement)

Fixed-end moments from the vertical loads: for BC and CD, $PL/8 = 5\times1.5/8 = 0.9375$ (upward load, so $M_{BC} = M_{CD} = +0.9375$, $M_{CB} = M_{DC} = -0.9375$). The cantilevers apply $2\times1.5 = 3$ kN·m to joints B ($-3$) and D ($+3$). The 10 kN loads act at the joint levels and appear in the sway equation, not in the FEMs.

Column shears in stage 1: $\Sigma H_1 = \Sigma (M_{top}+M_{bot})/h = -1.344$ kN.

Stage 2: sway with no loads

Impose a sway with $6EI\Delta = 100$ (kN·m units). The fixed-end moments $-6(EI)\Delta/h^2$ are: BF $= 3\times100/4 = -75$, CG $= 4\times100/4 = -100$, DH $= 3\times100/1 = -300$ (both ends).

Column shears in stage 2: $\Sigma H_2 = -429.22$ kN.

Sway correction

The columns must resist the 20 kN of applied horizontal load: $\Sigma H_1 + k\,\Sigma H_2 = -20$, so

$$k = \frac{-20 - (-1.344)}{-429.22} = 0.04347$$

Final moment $= M_{stage\,1} + k\,M_{stage\,2}$.

Final end moments (kN·m, clockwise +)

Joint checks: B: $3.555 - 0.555 - 3 = 0$; C: $0.690 + 3.019 - 3.709 = 0$; D: $2.587 - 5.587 + 3 = 0$.

Bending moment diagram (kN·m; + = sagging, tension at bottom)

Column moments at the top/base: BF 0.555 / 1.907; CG 3.709 / 4.028; DH 5.587 / 9.313. Each column bends in double curvature (moment changes sign along its height).

Shear force diagram (kN)

• Beam: AB +2.00 (constant); BC -5.33 up to the load, then -0.33; CD -6.24, then -1.24; DE -2.00 (constant).
• Columns: BF 1.23, CG 3.87, DH 14.90 (all resisting the horizontal load; $1.23+3.87+14.90 = 20$ kN).

Axial force diagram (kN)

• Beam: AB and DE zero; BC 8.77 compression; CD 4.90 compression (the horizontal load is shed to the columns at each joint: $10 - 1.23 = 8.77$, $8.77 - 3.87 = 4.90$, $4.90 + 10 - 14.90 = 0$).
• Columns: BF 7.33 tension, CG 5.91 tension, DH 0.76 tension (the upward loads of 14 kN in total are held down by the bases: $7.33+5.91+0.76 = 14.0$).

Answer: Sway factor $k = 0.0435$; final column-end moments: BF -0.555/-1.907, CG -3.709/-4.028, DH -5.587/-9.313 kN·m (top/base). Maximum moment 9.31 kN·m at the base of DH.