Truss ILDs for FG and BC; max forces from 100 kN

Data: bottom chord $A$–$B$–$C$–$D$–$E$, 4 panels of 4 m (span 16 m); apex $G$ 5 m above the bottom chord. Assumed: $F$ and $H$ at 2.5 m (so $AFG$ and $GHE$ are straight rafters); verticals $FB$, $GC$, $HD$; diagonals $FC$ and $CH$. $A$ hinge, $E$ roller. Reactions: $R_A = \dfrac{16-x}{16}$, $R_E = \dfrac{x}{16}$. The unit load moves along the bottom chord.

Member BC (bottom chord, panel $BC$)

Section through $FG$, $FC$, $BC$; moments about $F$ (4, 2.5): lever arm 2.5 m.

• Load right of $C$: $F_{BC} = \dfrac{4R_A}{2.5} = 1.6R_A$.
• Load left of $B$: $F_{BC} = \dfrac{12R_E}{2.5} = 4.8R_E$.

Member FG (top chord)

Same section; moments about $C$. The lever arm of $FG$ (line $y = 0.625x$) about $C$ is $\dfrac{5}{\sqrt{1+0.625^2}} = 4.24$ m.

• Load right of $C$: $F_{FG} = -\dfrac{8R_A}{4.24} = -1.887R_A$ (compression).
• Load left of $B$: $F_{FG} = -\dfrac{8R_E}{4.24} = -1.887R_E$.

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

Between $B$ and $C$ both ILDs are straight lines.

 ILD BC: 0 / 1.2 at B \ 0.8 at C ... 0.4 at D ... 0 at E
 ILD FG: 0 / -0.47 at B / -0.94 at C \ -0.47 at D \ 0

Maximum forces for a single 100 kN rolling load

A single load gives its maximum effect at the largest ordinate:

• FG: largest ordinate $-0.943$ at $C$ (load at the centre): $F_{FG} = 100\times(-0.943) = -94.3$ kN (compression).
• BC: largest ordinate $+1.2$ at $B$: $F_{BC} = 100\times 1.2 = +120$ kN (tension).

Answer: $FG_{max} = 94.3$ kN (C) with the load at $C$; $BC_{max} = 120$ kN (T) with the load at $B$.