Truss ILDs for AB, BC, BG, lower chord

Data: bottom chord $A$–$F$–$G$–$H$–$E$, 4 panels of 15 m (span 60 m); $B$, $D$ are 5 m and $C$ is 10 m above the bottom chord; verticals $BF$, $CG$, $DH$; diagonals $BG$, $GD$. $A$ hinge, $E$ roller. The load moves on the lower chord. Reactions: $R_A = \dfrac{60-x}{60}$, $R_E = \dfrac{x}{60}$. $AB$ and $BC$ both have slope $\tfrac13$ (so $A$, $B$, $C$ are collinear).

Member AB

Joint $A$: vertical equilibrium, $F_{AB}\sin\alpha = -R_A$ with $\sin\alpha = \dfrac{5}{15.81} = 0.316$. So $F_{AB} = -3.162\,R_A$ (compression); the load at $A$ goes straight to the support, so the ordinate at $A$ is 0.

Member BC

Section through $BC$, $BG$, $FG$; moments about $G$ (lever arm of $BC$: $\dfrac{30}{\sqrt{10}} = 9.487$ m).

• Load right of $G$: $F_{BC} = -\dfrac{30R_A}{9.487} = -3.162R_A$.
• Load left of $F$: $F_{BC} = -3.162R_E$.

Member BG

Moments about $A$ (where $BC$ and $FG$ meet; lever arm of $BG$ = 9.49 m).

• Load right of $G$: $F_{BG} = 0$ (the left part has only $R_A$, which passes through $A$).
• Load left of $F$: $F_{BG} = -\dfrac{60R_E}{9.49} = -6.325R_E$.

Ordinates (per unit load; + tension, − compression)

Shapes

• AB: compression, 0 at $A$, peak $-2.372$ at $F$, falling linearly to 0 at $E$.
• BC: compression, peak $-1.581$ under $G$; $-0.791$ under $F$ and $H$.
• BG: compression, a triangle between $A$ and $G$ with peak $-1.581$ at $F$; zero for load beyond $G$.