Truss ILDs for FG and BG

Data: bottom chord $A$–$F$–$G$–$H$–$E$, 4 panels of 15 m (span 60 m); $B$, $D$ are 5 m and the apex $C$ is 10 m above the bottom chord; verticals $BF$, $CG$, $DH$; diagonals $BG$ and $GD$. $A$ hinge, $E$ roller. The unit load is taken on the bottom chord (the member forces are zero or linear between panel points).

Reactions for a unit load at $x$ from $A$: $R_A = \dfrac{60-x}{60}$, $R_E = \dfrac{x}{60}$.

Member FG (bottom chord, panel $FG$)

Section through $BC$, $BG$ and $FG$ (panel $FG$). Moments about $B$ ($x=15$, $y=5$), where $BC$ and $BG$ meet (lever arm of $FG$ = 5 m):

• Load to the right of $G$ (left part): $F_{FG}\times 5 = R_A\times 15 \Rightarrow F_{FG} = 3R_A$ (tension).
• Load to the left of $F$ (right part): $F_{FG}\times 5 = R_E\times 45 \Rightarrow F_{FG} = 9R_E$.

Member BG (diagonal)

Use the same section. $AB$ and $BC$ lie on one straight line through $A$ (slope 1/3), so $BC$ and $FG$ meet at $A$; take moments about $A$. The lever arm of $BG$ (line from $B(15,5)$ to $G(30,0)$, length 15.81 m) about $A$ is

$$d = \frac{|15\times(-5) - 5\times 15|}{15.81} = 9.49\ \text{m}$$

• Load to the right of $G$ (left part): the only external force on the left part is $R_A$, acting at $A$, which has zero moment about $A$. So $F_{BG} = 0$.
• Load to the left of $F$ (right part): $F_{BG} = -\dfrac{60\,R_E}{9.49} = -6.325\,R_E$ (compression); at $F$, $R_E = 0.25$, so $F_{BG} = -1.581$.
• Between $F$ and $G$ the ILD is the straight line from $-1.581$ to 0.

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

Shapes

• FG: tension, a triangle with peak $+2.25$ under $F$, falling to $+1.5$ at $G$, $+0.75$ at $H$ and 0 at $E$ (and 0 at $A$).
• BG: compression, a triangle between $A$ and $G$: 0 at $A$, peak $-1.581$ at $F$, 0 at $G$; zero for load beyond $G$.