Moment distribution: frame, BMD and SFD

Assumptions (figure is not fully clear)

• Beam A-B-C-D-E is horizontal: AB = BC = CD = 6 m, overhang DE = 3 m. $I_{AB} = I$, $I_{BC} = 1.5I$, $I_{CD} = I_{DE} = 2I$. A is fixed.
• Column BG ($2I$) and column CF ($I$) are 6 m high with fixed bases. The 150 kN horizontal load (taken as acting to the right) acts on column BG at mid-height (3 m + 3 m); the column below C is also taken as 6 m high.
• D is a roller (vertical support), so the overhang DE can carry the 50 kN load. The beam is axially rigid and A is fixed, so the joints cannot sway: this is a non-sway frame.
• Sign: end moments are clockwise (+) on the member end.

Stiffness and distribution factors

Relative stiffness $K = I/L$ (far end fixed), and $\tfrac34 I/L$ for CD because the far end D is a roller with a known moment.

Fixed-end moments

• BC and CD (20 kN/m): $wL^2/12 = 20\times 6^2/12 = 60$ kN·m. So $M_{BC}=-60$, $M_{CB}=+60$, and for CD $M_{CD}=-60$, $M_{DC}=+60$.
• Overhang DE: $M_D = 50\times 3 = 150$ kN·m (hogging) is statically determinate, so the final $M_{DC} = +150$. Releasing D then gives the modified FEM $M_{CD} = -60 - \tfrac12(60-150) = -15$ kN·m.
• BG: load at mid-height, $PL/8 = 150\times 6/8 = 112.5$ kN·m, so $M_{BG}=+112.5$, $M_{GB}=-112.5$.

Moment distribution

Check at joint B: $-10.11 - 82.18 + 92.28 = 0$. At joint C: $38.39 - 29.03 - 9.35 \approx 0$.

Final end moments (kN·m)

• A: 5.05 hogging; B (in AB): 10.11 sagging (zero moment at about 2 m from A).
• B (beam, right side): 82.18 hogging; C (left): 38.39 hogging; C (right): 29.03 hogging; D: 150.00 hogging.
• Column BG: top 92.28, at the load point 117.55, base G 122.61. Column CF: top 9.35, base F 4.68.
• Maximum sagging: in BC 31.05 kN·m at 3.36 m from B; in CD 10.65 kN·m at 1.99 m from C.

Shear forces (kN) and reactions

Reactions: A: 2.53 kN up, 67.61 kN left, M = 5.05 kN·m. D: 130.16 kN up. G: 64.77 kN up, 80.05 kN left, M = 122.61 kN·m. F: 92.54 kN up, 2.34 kN left, M = 4.68 kN·m. Check: $\Sigma V = 2.53+130.16+64.77+92.54 = 290.0 = 20\times12+50$; $\Sigma H = 67.61+80.05+2.34 = 150$.

BMD and SFD shape

• BMD (tension side): AB is a straight line from 5.05 (top tension at A) to 10.11 (bottom tension at B). BC is a parabola: hogging 82.18 at B, hogging 38.39 at C, sagging 31.05 near the middle. CD is a parabola from 29.03 hogging at C, through a small sagging 10.65, to 150 hogging at D. DE is a straight line falling from 150 at D to 0 at E (tension on top). Column BG: 92.28 at the top, 117.55 at the load point on the opposite face, 122.61 at the base; CF: a straight line from 9.35 to 4.68 (double curvature).
• SFD: rectangular blocks in AB, DE (+50, jumping from -80.16 by the 130.16 kN reaction at D) and CF; straight sloping lines in BC and CD (slope 20 kN/m); BG steps from 69.95 to 80.05 at the 150 kN load.

Answer (magnitudes): $M_{AB}=5.05$, $M_{BA}=10.11$, $M_{BC}=82.18$, $M_{CB}=38.39$, $M_{CD}=29.03$, $M_{DC}=150$, $M_{BG}=92.28$, $M_{GB}=122.61$, $M_{CF}=9.35$, $M_{FC}=4.68$ kN·m.