Truss ILDs for bc, hg, df, upper chord

Assumed truss (Pratt type, figure not available): bottom joints $a, b, c, d, e$ and top joints $j, i, h, g, f$ directly above them, 4 panels of 10 m (span 40 m), height 12 m; verticals $aj, ib, hc, gd, fe$; diagonals $jb, ic, gc, fd$ (sloping down towards the centre). $a$ hinge, $e$ roller. The load moves on the upper (deck) chord, so each ILD is straight between the top joints.

Reactions for a unit load at $x$ from $a$: $R_a = \dfrac{40-x}{40}$, $R_e = \dfrac{x}{40}$. Diagonal $df$ has length $\sqrt{10^2+12^2} = 15.62$ m, $\sin\theta = 12/15.62 = 0.768$.

Method of sections

Member bc (bottom chord, panel $bc$): cut $ih$, $ic$, $bc$; take moments about $i$ (where $ih$ and $ic$ meet), lever arm 12 m.

• Load right of $h$: $F_{bc} = \dfrac{R_a\times 10}{12} = 0.833R_a$ (tension).
• Load left of $i$: $F_{bc} = \dfrac{R_e\times 30}{12} = 2.5R_e$.

Member hg (top chord, panel $cd$): cut $hg$, $gc$, $cd$; moments about $c$, lever arm 12 m.

• Load right of $g$: $F_{hg} = -\dfrac{R_a\times 20}{12} = -1.667R_a$ (compression).
• Load left of $h$: $F_{hg} = -\dfrac{R_e\times 20}{12} = -1.667R_e$.

Member df (diagonal, panel $de$): cut $gf$, $fd$, $de$; the panel shear is carried by the diagonal.

• Load left of $g$: $F_{df}\sin\theta = R_e \Rightarrow F_{df} = 1.302R_e$ (tension).
• Load at $f$ (over the support): 0. Between $g$ and $f$ the ILD is the straight line joining the ordinates.

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

Shapes

• bc: tension, peak $+0.625$ at $i$, falling to $+0.417$ at $h$, $+0.208$ at $g$, 0 at $j$ and $f$.
• hg: compression, peak $-0.833$ under $h$, $-0.417$ at $i$ and at $g$.
• df: tension, rising from 0 at $j$ to $+0.976$ at $g$, then falling to 0 at $f$.