What is the ILD of U1L1 for bottom-chord loading?

Triangle with apex +1 at L1, zero elsewhere

How is the force in U2U3 found and what is its nature?

Moments about L3, F = -M_L3/h, compression with peak -3a/(2h) at L3

Why is U3L3 zero for bottom-chord loading?

Joint U3 has two collinear chord members and the vertical, and nothing else, so it is unloaded

Describe the ILD of diagonal U2L3.

Negative left of a neutral point between L2 and L3 (peak -d/(3h) at L2), positive to the right (peak +d/(2h) at L3)

2073 Magh · 12 marks

Truss ILDs for six members, load on bottom chord

Short answer

ILD shapes: U1L1 triangle with peak +1 at L1; U2U3 compressive, peak -3a/(2h) at L3; U2L3 diagonal, negative left of L2-L3, positive right, peak +d/(2h) at L3; U2L2 positive peak +1/3 at L2, negative peak -1/2 at L3; L2L3 tension, peak +4a/(3h) at L2; U3L3 zero throughout. All zero at supports.

Draw ILD for forces in member $U_1L_1$ , $U_2U_3$ , $U_2L_3$ , $U_2L_2$ , $L_2L_3$ and $U_3L_3$ for a given truss, when the load is moving on the bottom chord. [Figure: truss with bottom joints $L_0$ to $L_6$ , top joints $U_1$ to $U_5$ , 6 panels of length a (6a), height h; $L_0$ hinge, $L_6$ roller.]

Assumed truss (Pratt type): bottom joints $L_0$ – $L_6$ , top joints $U_1$ – $U_5$ above $L_1$ – $L_5$ ; 6 panels of length $a$ (span $6a$ ), height $h$ . Verticals $U_iL_i$ ; diagonals $U_1L_2$ , $U_2L_3$ , $U_4L_3$ , $U_5L_4$ ; end posts $L_0U_1$ and $L_6U_5$ . $L_0$ hinge, $L_6$ roller. Let $d = \sqrt{a^2+h^2}$ be the diagonal length, so $\sin\theta = h/d$ . The load travels on the bottom chord, so each ILD is straight between panel points and the ordinate at a panel point equals the force for a unit load placed there.

Reactions for a unit load at $L_k$ ( $x = ka$ ): $R_{L_0} = \dfrac{6-k}{6}$ , $R_{L_6} = \dfrac{k}{6}$ .

Formulas (sections)

$U_1L_1$ : $L_1$ has only three members ( $L_0L_1$ , $L_1L_2$ , $U_1L_1$ ), so it is a hanger: $F = +1$ with the load at $L_1$ , 0 otherwise.
$U_2U_3$ : cut $U_2U_3$ , $U_2L_3$ , $L_2L_3$ ; moments about $L_3$ : $F = -\dfrac{M_{L_3}}{h}$ (compression). With the load right of $L_3$ : $F = -\dfrac{3a\,R_{L_0}}{h}$ ; left of $L_3$ : $F = -\dfrac{3a\,R_{L_6}}{h}$ .
$U_2L_3$ (diagonal): same section, $F = \pm\dfrac{V}{\sin\theta} = \pm\dfrac{d}{h}V$ . Load right of $L_3$ : $F = +\dfrac{d}{h}R_{L_0}$ ; left of $L_2$ : $F = -\dfrac{d}{h}R_{L_6}$ .
$L_2L_3$ : same section, moments about $U_2$ ( $x = 2a$ ): $F = \dfrac{M_{U_2}}{h}$ (tension): right of $L_3$ : $F = \dfrac{2a\,R_{L_0}}{h}$ ; left of $L_2$ : $F = \dfrac{4a\,R_{L_6}}{h}$ .
$U_2L_2$ (vertical): at the unloaded joint $U_2$ : $F = -V_{panel\ L_2L_3}$ , giving $+R_{L_6}$ for a load at or left of $L_2$ and $-R_{L_0}$ for a load at or right of $L_3$ .
$U_3L_3$ : no diagonal meets at $U_3$ , so this unloaded joint has only three members: the two collinear top-chord members $U_2U_3$ , $U_3U_4$ and the vertical $U_3L_3$ . Hence $F = 0$ for bottom-chord loading.

ILD ordinates (per unit load; + tension, − compression; $d=\sqrt{a^2+h^2}$ )

Member	L1	L2	L3	L4	L5
U1L1	$1$	$0$	$0$	$0$	$0$
U2U3	$- \frac{a}{2 h}$	$- \frac{a}{h}$	$- \frac{3 a}{2 h}$	$- \frac{a}{h}$	$- \frac{a}{2 h}$
U2L3	$- \frac{d}{6 h}$	$- \frac{d}{3 h}$	$\frac{d}{2 h}$	$\frac{d}{3 h}$	$\frac{d}{6 h}$
U2L2	$\frac{1}{6}$	$\frac{1}{3}$	$- \frac{1}{2}$	$- \frac{1}{3}$	$- \frac{1}{6}$
L2L3	$\frac{2 a}{3 h}$	$\frac{4 a}{3 h}$	$\frac{a}{h}$	$\frac{2 a}{3 h}$	$\frac{a}{3 h}$
U3L3	$0$	$0$	$0$	$0$	$0$

All ILDs are 0 at $L_0$ and $L_6$ .

Shapes

$U_1L_1$ : triangle with apex $+1$ at $L_1$ .
$U_2U_3$ : compression, peak $-\dfrac{3a}{2h}$ at $L_3$ .
$U_2L_3$ : negative to the left of the neutral point between $L_2$ and $L_3$ (peak $-\dfrac{d}{3h}$ at $L_2$ ), positive to the right (peak $+\dfrac{d}{2h}$ at $L_3$ ).
$U_2L_2$ : positive (peak $+\tfrac13$ at $L_2$ ), crossing zero between $L_2$ and $L_3$ , negative to the right (peak $-\tfrac12$ at $L_3$ ).
$L_2L_3$ : tension, peak $+\dfrac{4a}{3h}$ at $L_2$ .
$U_3L_3$ : zero throughout.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Truss ILDs for six members, load on bottom chord

Formulas (sections)

ILD ordinates (per unit load; + tension, − compression; d=a2+h2d=\sqrt{a^2+h^2}d=a2+h2​)

Shapes

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ILD ordinates (per unit load; + tension, − compression; $d=\sqrt{a^2+h^2}$ )