---
title: "Truss ILDs for U3L3 and L3L4"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:truss-ild-u3l3-l3l4"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-ild-u3l3-l3l4
site: BCT Engineering
---

# Truss ILDs for U3L3 and L3L4

Theory of Structures I · Chapter 5: Influence Lines for Simple Structures

- Asked in: 2065 Chaitra, 8 marks
- Video: [Influence Line For Trusses Top / Upper Chord Member Easiest way to Draw  | ILD & Rolling Load](https://www.youtube.com/watch?v=fuJ2rSQrsc4) — Ganesh Shegar
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-ild-u3l3-l3l4
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Draw influence line diagram for members $U_3L_3$ and $L_3L_4$ of the truss shown in fig-8. [Figure: truss with bottom joints $L_1$ to $L_7$, top joints $U_2$ to $U_6$, 6 panels of 4 m = 24 m, height 4 m.]

## Short answer

Member L3L4 has a triangular ILD, peak +1.333 at L3, tension throughout. Member U3L3 has a tensile peak +0.333 at L3 and a compressive peak -0.500 at L4, crossing zero at 9.6 m from L1.

## Answer

**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)
| Member | L1 | L2 | L3 | L4 | L5 | L6 | L7 |
|---|---|---|---|---|---|---|---|
| U3L3 | 0.000 | 0.167 | 0.333 | -0.500 | -0.333 | -0.167 | 0.000 |
| L3L4 | 0.000 | 0.667 | 1.333 | 1.000 | 0.667 | 0.333 | 0.000 |

#### 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 $L_1$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. To find the bottom chord force L3L4, moments are taken about which joint?
   A. L4
   B. U3, where U3U4 and U3L4 meet
   C. L3
   D. U4
   Answer: B. The section cuts U3U4, U3L4 and L3L4; U3U4 and U3L4 meet at U3. (Key point: Moment centre for L3L4)

2. What is the L3L4 ordinate at L3 (x = 8 m, span 24 m, height 4 m)?
   A. 0.667
   B. 1.333
   C. 2.667
   D. 1.000
   Answer: B. For load right of the panel, F = 2 R_L1; at x = 8, R_L1 = 16/24 = 0.667, so F = 1.333 (tension). (Key point: Ordinate at L3)

3. How is the vertical U3L3 obtained?
   A. F = -V, because the diagonal carries sqrt(2) V and its vertical component is V (45 degrees)
   B. F = V x 4
   C. Moments about L4
   D. F = 0 always
   Answer: A. The diagonal at 45 degrees carries sqrt(2) V, and the vertical at the unloaded joint U3 balances its vertical component V. (Key point: Vertical from shear)

4. Where is the neutral point of the U3L3 influence line?
   A. At L3
   B. At L4
   C. At 9.6 m from L1, between L3 and L4
   D. At mid-span
   Answer: C. Between ordinates +0.333 at L3 and -0.5 at L4: x = 8 + 4 x 0.333/0.833 = 9.6 m. (Key point: Sign change of U3L3)

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