Truss ILDs for U2L2, U1L2, U1L1

Assumed truss (Pratt type): bottom joints $L_0$–$L_6$, top joints $U_1$–$U_5$ above $L_1$–$L_5$, 6 panels of 4 m (span 24 m), height 5 m; verticals $U_iL_i$; diagonals $U_1L_2$, $U_2L_3$, $U_4L_3$, $U_5L_4$; end posts $L_0U_1$, $L_6U_5$. $L_0$ hinge, $L_6$ roller. Diagonal length $\sqrt{4^2+5^2} = 6.403$ m, $\sin\theta = 5/6.403 = 0.781$.

Reactions for a unit load at $x$ from $L_0$: $R_{L_0} = \dfrac{24-x}{24}$, $R_{L_6} = \dfrac{x}{24}$.

Method

$U_1L_1$ (vertical): joint $L_1$ has only three members ($L_0L_1$, $L_1L_2$, $U_1L_1$), so $U_1L_1$ is a hanger: it carries the load only when the unit load is at $L_1$ ($F=+1$ for bottom-chord loading). For top-chord loading it is a zero-force member.

$U_1L_2$ (diagonal, panel $L_1L_2$): cut $U_1U_2$, $U_1L_2$, $L_1L_2$. The panel shear is taken by the diagonal: $F = \pm V/\sin\theta = \pm 1.281\,V$.

• Load right of $L_2$: $F = +\dfrac{R_{L_0}}{\sin\theta}$ (tension). At $L_2$: $\dfrac{0.667}{0.781} = 0.854$.
• Load left of $L_1$: $F = -\dfrac{R_{L_6}}{\sin\theta}$ (compression). At $L_1$: $-\dfrac{0.1667}{0.781} = -0.213$.

$U_2L_2$ (vertical): cut $U_2U_3$, $U_2L_3$, $L_2L_3$ for the shear $V$ in panel $L_2L_3$; at the unloaded joint $U_2$, $F_{U_2L_2} = -V$. Load at $L_2$: $V = \tfrac23 - 1$, $F = +0.333$; load at $L_3$: $V = \tfrac12$, $F = -0.5$.

Ordinates, load on the bottom chord (+ tension, − compression)

Ordinates, load on the top chord

Shapes (bottom-chord loading)

• $U_1L_1$: a triangle, $+1$ at $L_1$ and 0 at $L_0$ and $L_2$ (zero elsewhere).
• $U_1L_2$: negative from $L_0$ to a zero point at $x = 4 + 4\times\frac{0.213}{0.213+0.854} = 4.8$ m, then positive with peak $+0.854$ at $L_2$, falling to 0 at $L_6$.
• $U_2L_2$: positive up to the zero point at $x = 8+4\times\frac{0.333}{0.833} = 9.6$ m (peak $+0.333$ at $L_2$), negative thereafter (peak $-0.5$ at $L_3$).