Truss ILDs for BC, CH, HG, bottom chord

Geometry: bottom chord $A$–$B$–$C$–$D$–$E$, 4 panels of 4 m (span 16 m); top joints $H$, $G$, $F$ at 3 m height above $B$, $C$, $D$. $A$ hinge, $E$ roller. Diagonals $AH$, $HC$, $CF$, $FE$; verticals $HB$, $GC$, $FD$. The unit load moves along the bottom chord, so ordinates are needed at the panel points only and the ILD is a straight line between them (floor beam action).

Reactions: for a unit load at $x$ from $A$, $R_A = \dfrac{16-x}{16}$, $R_E = \dfrac{x}{16}$.

Method of sections (section through panel $BC$, cutting $HG$, $HC$, $BC$)

Diagonal $HC$ has $\sin\theta = 3/5$, $\cos\theta = 4/5$.

Member BC (take moments about $H$, where $HG$ and $HC$ meet; lever arm 3 m):

• Unit load right of the panel ($x\ge 8$, left part considered): $F_{BC}\times 3 = R_A\times 4 \Rightarrow F_{BC} = \tfrac43 R_A$
• Unit load left of the panel ($x\le 4$, right part considered): $F_{BC}\times 3 = R_E\times 12 \Rightarrow F_{BC} = 4R_E = x/4$

Member HG (moments about $C$): $F_{HG} = -\dfrac{8R_A}{3}$ for load right of $C$; $F_{HG} = -\dfrac{8R_E}{3}$ for load left of $B$ (top chord in compression).

Member CH (vertical equilibrium of the left part): $F_{CH}\sin\theta = R_A$ for load right of $C$, so $F_{CH} = \tfrac53 R_A$ (tension); for load left of $B$, $F_{CH} = -\tfrac53 R_E$ (compression).

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

Between $B$ and $C$ the ILDs are straight lines joining the ordinates at $B$ and $C$.

Shapes

• BC: tension only; a triangle with peak $+1.00$ at $B$, $0.667$ at $C$, $0$ at $A$ and $E$.
• CH: compression ($-0.417$ at $B$) for load left of the panel, tension ($+0.833$ at $C$) for load right; a neutral point lies between $B$ and $C$ at $x = 4 + 4\times\frac{0.417}{0.417+0.833} = 5.33$ m from $A$.
• HG: compression throughout with peak $-1.333$ under $C$.