Truss: max forces in CF and BC under live load

Data: bottom chord $A$–$B$–$C$–$D$–$E$, 4 panels of 6 m (span 24 m); top joints $F$, $H$ 3 m and apex $G$ 4 m above the bottom chord; members $AF$, $FG$, $GH$, $HE$ (top chord), verticals $FB$, $GC$, $HD$ and diagonals $FC$, $CH$. $A$ hinge, $E$ roller. The live load of 28 kN/m is longer than the span, so it can be placed on any part of the span (the loaded length is chosen to suit the ILD). Reactions for a unit load at $x$: $R_A = \dfrac{24-x}{24}$, $R_E = \dfrac{x}{24}$.

Member BC (bottom chord, panel $BC$)

Section through $FG$, $FC$ and $BC$; moments about $F$ (lever arm 3 m):

• Load right of $C$ (left part): $F_{BC} = \dfrac{6\,R_A}{3} = 2R_A$.
• Load left of $B$ (right part): $F_{BC} = \dfrac{18\,R_E}{3} = 6R_E$.

Member CF (diagonal)

Same section; $FG$ (slope $\tfrac16$) meets the bottom chord at $O$ ($x = -12$ m). The lever arm of $CF$ about $O$ is $10.73$ m.

• Load right of $C$: $F_{CF} = \dfrac{12\,R_A}{10.73} = +1.118\,R_A$ (tension).
• Load left of $B$: $F_{CF} = -\dfrac{36\,R_E}{10.73} = -3.354\,R_E$ (compression).

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

$CF$ changes sign between $B$ and $C$ at $x = 9.60$ m from $A$.

Maximum forces for w = 28 kN/m (load of any length)

Force $= w\times$ (area of ILD loaded).

Answer: member $BC$: maximum tension = 504.0 kN (whole span loaded; $BC$ is never in compression). Member $CF$: maximum tension = 112.7 kN (load from $x = 9.60$ m to $E$); maximum compression = 112.7 kN (load from $A$ to $x = 9.60$ m).