Given. Continuous beam $1\!-\!2\!-\!3\!-\!4$ at $y=4\text{ m}$: free tip $1(0,4)$ with a $16.5\text{ kN}$ load on a $1\text{ m}$ overhang, $12\text{ kN/m}$ over $2\!-\!3$ ($8\text{ m}$), then $3\!-\!4$ ($4\text{ m}$) into a fixed wall at $4$. Columns $2\!-\!5$ ($4\text{ m}$, pinned base $5$) and $3\!-\!6$ ($4\text{ m}$, fixed base $6$). Equal $EI$, inextensible.
Find. SFD and BMD with max/min ordinates for every member.
Q6 Rigid frame; node 4 fixed, base 5 pinned, base 6 fixed.
Approach. Slope-deflection with unknowns $\theta_2,\theta_3$ (node 4 fixed and base 4 restrain sidesway; the overhang applies a known $16.5\text{ kN}\cdot\text{m}$ at 2). Solving the two joint-equilibrium equations and back-substituting gives the member-end moments.
Overhang. The $1\text{ m}$ cantilever delivers $M_{2}=16.5(1)=16.5\text{ kN}\cdot\text{m}$ (hogging) and a $16.5\text{ kN}$ shear into joint 2.
Member-end moments (kN·m, sagging +): span $2\!-\!3$: $-49.5$ at 2, $-60.0$ at 3, with a sagging peak $+41.3$ near mid-span; span $3\!-\!4$: $-30.0$ at 3, $+15.0$ at the fixed end 4; column $2\!-\!5$: $+33.0$ at top, $0$ at the pinned base; column $3\!-\!6$: $-30.0$ at top, $+15.0$ at the fixed base.
Joint checks. At 2: $-16.5-33.0+49.5=0$; at 3: $-60.0=-30.0-30.0$ — both balance.
Reactions. Base 5 (pin): $V=63.2\text{ kN}$, $H=8.25\text{ kN}$; base 6 (fixed): $V=60.6\text{ kN}$, $H=-11.25\text{ kN}$, $M=15.0\text{ kN}\cdot\text{m}$; support 4 (fixed): $V=-11.25\text{ kN}$, $H=3.0\text{ kN}$, $M=15.0\text{ kN}\cdot\text{m}$. Total vertical $=112.5=16.5+12(8)$.
Q6 Bending-moment diagram along the beam (columns: 33 at top of 2-5; −30/+15 on 3-6).
