Truss ILDs for BC and BG; max force in BC

Data: bottom chord $A$–$H$–$G$–$F$–$E$, 4 panels of 6 m (span 24 m); $B$, $D$ are 4 m and the apex $C$ is 6 m above the bottom chord; verticals $BH$, $DF$, $CG$; diagonals $BG$, $GD$. $A$ hinge, $E$ roller. The live load moves along the bottom chord; ILDs are straight between panel points.

Reactions: $R_A = \dfrac{24-x}{24}$, $R_E = \dfrac{x}{24}$.

Member BC (top chord, panel $HG$)

Section through $BC$, $BG$ and $HG$. The line $BC$ ($x - 3y + 6 = 0$) has slope $\tfrac13$; the lever arm of $BC$ about $G$ (where $BG$ and $HG$ meet) is $\dfrac{|12+6|}{\sqrt{10}} = 5.692$ m.

• Load right of $G$ (left part): $F_{BC} = -\dfrac{12\,R_A}{5.692} = -2.108\,R_A$ (compression).
• Load left of $H$ (right part): $F_{BC} = -\dfrac{12\,R_E}{5.692} = -2.108\,R_E$.

Member BG (diagonal)

Same section; take moments about $O$, the point where $BC$ (slope $\tfrac13$) produced meets the bottom chord produced, at $x = -6$ m. ($AB$ is steeper, with slope $\tfrac23$, so it is not on this line.) Lever arm of $BG$ about $O$ $= 9.985$ m.

• Load right of $G$: $F_{BG} = +\dfrac{6\,R_A}{9.985} = +0.601\,R_A$ (tension).
• Load left of $H$: $F_{BG} = -\dfrac{30\,R_E}{9.985} = -3.005\,R_E$ (compression).

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

Between $H$ and $G$ the ILDs are straight lines; $BG$ changes sign at $x = 6 + 6\times\dfrac{0.751}{0.751+0.300} = 10.3$ m from $A$.

Maximum force in BC for 6 kN/m over 8 m

All ordinates of $BC$ are negative, so the 8 m load is placed to enclose the largest area, symmetrically about the centre ($x = 8$ m to $16$ m):

Ordinates at $x=8$: $-0.703$; at $x=12$: $-1.054$; at $x=16$: $-0.703$.

$$\text{area} = 2\times\frac{(0.703+1.054)}{2}\times 4 = 7.027\ \text{m}, \qquad F_{BC,max} = 6\times 7.027 = 42.16\ \text{kN}$$

Answer: maximum force in $BC$ = 42.16 kN (compression), with the 8 m load placed symmetrically about mid-span.