Truss ILDs for U3L3 and L3L4

Assumed truss (Pratt type): bottom joints $L_1$–$L_7$, top joints $U_2$–$U_6$ above $L_2$–$L_6$, 6 panels of 4 m (span 24 m), height 4 m. End members $L_1U_2$ and $L_7U_6$ (45°); verticals $U_iL_i$; diagonals $U_2L_3$, $U_3L_4$, $U_5L_4$, $U_6L_5$ (sloping down towards the centre). $L_1$ hinge, $L_7$ roller. The load moves along the bottom chord (ILDs are straight between the panel points).

Reactions for a unit load at $x$ from $L_1$: $R_{L_1} = \dfrac{24-x}{24}$, $R_{L_7} = \dfrac{x}{24}$.

Member L3L4 (bottom chord, panel $L_3L_4$)

Section through $U_3U_4$, $U_3L_4$ and $L_3L_4$; moments about $U_3$ ($x = 8$ m, height 4 m, lever arm 4 m): $F_{L_3L_4} = \dfrac{M_{U_3}}{4}$ (tension).

• Load right of $L_4$ (left part): $F = \dfrac{8R_{L_1}}{4} = 2R_{L_1}$.
• Load left of $L_3$ (right part): $F = \dfrac{16R_{L_7}}{4} = 4R_{L_7}$.

Member U3L3 (vertical)

The shear in panel $L_3L_4$ is carried by the diagonal $U_3L_4$: $F_{U_3L_4} = \sqrt2\,V$. At the unloaded top joint $U_3$ the vertical balances the diagonal: $F_{U_3L_3} = -V$.

• Load at or left of $L_3$: $V = R_{L_1} - 1$, so $F_{U_3L_3} = +R_{L_7}$ (tension).
• Load at or right of $L_4$: $V = R_{L_1}$, so $F_{U_3L_3} = -R_{L_1}$ (compression).

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

Shapes

• L3L4: tension throughout; a triangle with its apex $+1.333$ at $L_3$ (directly below the moment centre $U_3$), passing through $+1.0$ at $L_4$ and falling linearly to 0 at $L_7$.
• U3L3: tension for load left of the panel (peak $+0.333$ at $L_3$), compression to the right (peak $-0.5$ at $L_4$). The neutral point is at $x = 8 + 4\times\frac{0.333}{0.333+0.5} = 9.6$ m from $L_1$.