Frame: 4 m left overhang + roller at $x=4$, rigid joint 2 at $x=12$ with a hanging column to pin 4, fixed end 3 at $x=20$; 8 kN/m over 0–12 m and 72 kN at $x=16$.
Given. A horizontal beam with a $4$ m overhang to a roller (S, $x=4$), continuing to a rigid joint 2 ($x=12$) that carries an $8$ m column down to a pin (4), and on to a fixed end 3 ($x=20$); $8$ kN/m over $0\!-\!12$ m and a $72$ kN load at $x=16$. The frame is $3^\circ$ indeterminate. Find. Reactions and SFD/BMD.
Approach. Because both translations of joint 2 are prevented (the axially-rigid beam 2–3 ties it to the fixed end and the axially-rigid column ties it to pin 4), there is no sidesway; only the joint rotations are unknown. Compute fixed-end moments, distribute at joint 2 in proportion to member stiffnesses, carry over, and iterate to convergence.
Fixed-end moments. Left overhang: the $4$ m cantilever under $8$ kN/m applies a known $-8(4)^2/2=-64$ kN·m at S. Span S–2 (8 m UDL): $\pm wL^2/12=\pm42.7$ kN·m. Span 2–3 (72 kN at mid, $L=8$ m): $\pm PL/8=\pm72$ kN·m.
Distribute at joint 2. The three members framing into joint 2 (beam S–2, beam 2–3, column 2–4) share the unbalanced moment in proportion to $4EI/L$ (with the far-pinned column using $3EI/L$); carry-over one-half to the fixed and continuous ends and re-balance.
Converged reactions. Roller $R_S=66.5$ kN ↑; column pin $R_4=62.5$ kN ↑; fixed end $R_3=39.0$ kN ↑ with fixing moment $\boxed{80\ \text{kN}\cdot ext{m (hog)}}$. Equal and opposite horizontal reactions of $1.5$ kN develop at the column pin and the fixed end (the column shear). Check: $66.5+62.5+39.0=168=8(12)+72$ ✓.
Member moments. Over the roller $M_S=-64$; the beam sags to $+76$ kN·m under the $72$ kN load; the fixed end hogs at $-80$ kN·m; the column top carries $+12$ kN·m, grading to zero at the pin.
Bending-moment diagram along the beam axis (kN·m); the column carries a linear 12 → 0 moment.
