Check (reconstructed figure). The support symbols and web layout on the faint page-6 drawing are ambiguous. Method shown; ordinates depend on the final support reading (a simple pin–roller span over the 6 panels @ 4 m = 24 m is assumed for the qualitative shapes).
Approach. For each requested member, place a unit load successively at each bottom panel point, cut a section exposing that member, and evaluate its force by $\sum M$ (chords) or $\sum F_y$ (diagonals). Joining the panel-point values gives the (piece-wise linear) influence line.
Chord $U_1U_2$ (top chord). From a section between $U_1$ and $U_2$ and moments about the lower panel point, the IL is a smooth “triangle” that is entirely compression, greatest when the unit load sits nearest that panel.
Diagonal $U_2L_5$. By $\sum F_y$ across the panel, the IL changes sign across the panel — one lobe tension, one lobe compression — the classic diagonal shape.
Diagonal/vertical $U_1L_3$. Similar $\sum F_y$ construction; the peak coefficient occurs with the load at the near panel point.
Reading the extremes. The maximum tension/compression coefficient for each member is its largest positive/negative IL ordinate at a panel point.
7(b) — Gerber beam: IL for $M$ over support 3, and moving vehicle
Given. Supports at $x=0$ (pin), $6$, $14$ (support 3), $20$ m; internal hinges at $x=8$ and $x=12$ m; vehicle $64,64,16$ kN at $2$ m spacing.
7(b): Gerber beam and the IL for $M$ at support 3 — a triangle, zero at the hinges (8, 14... i.e. hinge 8 and support 3 at 14) with peak −2.0 m at hinge 12.
Determinacy. Four supports and two hinges: $\text{DSI}=r-3-c=5-3-2=0$ — the beam is a determinate Gerber (compound) beam.
IL for $M_3$. A unit load placed between hinge 8 and support 3 (or on the suspended span 8–12) produces negative moment over support 3. The influence coefficient is $0$ at hinge 8, reaches $\boxed{-2.0\text{ m}}$ at hinge 12, and returns to $0$ at support 3 ($x=14$); it is zero elsewhere.
Position the vehicle. The most negative moment occurs with the two 64 kN axles straddling the peak: one at $x=12$ (IL $=-2.0$), the next at $x=10$ (IL $=-1.0$), the 16 kN axle off the effective length.
Maximum negative moment. $M_3=64(-2.0)+64(-1.0)+16(0)=\boxed{-192\text{ kN}\cdot\text{m}}$.
