Check (reconstructed figure). Page 5 is faint. Adopted model: a continuous beam $\text{1–2–3–4}$ built-in (fixed) at ends 1 and 4, with 4 m vertical columns (pinned bases 5, 6) rigidly framing into the beam at the two interior points; symmetric spans $18\text{–}16\text{–}18$ m; uniform $EI$; a full-length $w=12$ kN/m downward on the beam. (The drawn 2 m offsets of the column feet are treated as drafting dimension marks.) The method is exact for this model; ordinate magnitudes below are illustrative of it.
Approach. With symmetry, only half the frame need be analysed; slope-deflection/moment distribution gives the member-end moments, from which shears and the SFD/BMD follow.
5 (adopted): fixed ends 1,4; column supports at 2,3 (pinned bases 5,6); 12 kN/m over the beam.
Fixed-end moments. Each beam span carries $\text{FEM}=wL^2/12$ (e.g. $12(18)^2/12=324$ kN·m on an 18 m span).
Distribution factors. At each interior joint the beam spans and the 4 m column share the balancing moment in proportion to $4EI/L$ (columns: $3EI/L$ for the pinned base).
Balance (with symmetry). Distributing and carrying over converges to end moments ≈ $317$–$338$ kN·m (hog) at the beam supports, $\approx248$ kN·m at the centre span, and a column moment $\approx46$ kN·m at each interior joint.
Shears / reactions. From the end moments, the vertical reactions come out $\approx109$ kN at each fixed end and $\approx203$ kN at each column, summing to $12(52)=624$ kN (check).
