---
title: "Truss ILDs for U2L2, U1L2, U1L1"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:truss-ild-u2l2-u1l2-u1l1"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-ild-u2l2-u1l2-u1l1
site: BCT Engineering
---

# Truss ILDs for U2L2, U1L2, U1L1

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

- Asked in: 2075 Bhadra, 10 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-u2l2-u1l2-u1l1
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Draw ILD for $U_2L_2$, $U_1L_2$ and $U_1L_1$. [Figure: truss with bottom joints $L_0$ to $L_6$ and top joints $U_1$ to $U_5$, 6 panels of 4 m = 24 m, height 5 m.]

## Short answer

For bottom-chord loading: U1L1 ILD is a triangle with +1 at L1, zero elsewhere. U1L2 ILD is negative from L0 to zero at 4.8 m, then positive peaking at +0.854 at L2. U2L2 ILD is positive up to zero at 9.6 m (peak +0.333 at L2), then negative with peak -0.5 at L3.

## Answer

**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)
| Member | L0 | L1 | L2 | L3 | L4 | L5 | L6 |
|---|---|---|---|---|---|---|---|
| U1L1 | 0.000 | 1.000 | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 |
| U1L2 | 0.000 | -0.213 | 0.854 | 0.640 | 0.427 | 0.213 | 0.000 |
| U2L2 | 0.000 | 0.167 | 0.333 | -0.500 | -0.333 | -0.167 | 0.000 |

#### Ordinates, load on the top chord
| Member | U1 | U2 | U3 | U4 | U5 |
|---|---|---|---|---|---|
| U1L1 | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 |
| U1L2 | -0.213 | 0.854 | 0.640 | 0.427 | 0.213 |
| U2L2 | 0.167 | -0.667 | -0.500 | -0.333 | -0.167 |

#### 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 $L_3$).

## Check yourself

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

1. Why does U1L1 carry force only when the unit load is at L1 (bottom-chord loading)?
   A. Joint L1 has only three members, so U1L1 is a hanger carrying just the load at L1
   B. It is a diagonal
   C. It is in compression always
   D. It is a top-chord member
   Answer: A. At L1 with L0L1, L1L2 collinear and U1L1, the vertical carries only the applied load, F = +1 there, 0 otherwise. (Key point: U1L1 as a hanger)

2. For U1L2, F = +/- V / sin(theta). With height 5 m and panel 4 m, what is 1/sin(theta)?
   A. 0.781
   B. 1.281
   C. 1.25
   D. 0.8
   Answer: B. sin(theta) = 5/6.403 = 0.781, so 1/sin(theta) = 1.281. (Key point: Diagonal force from shear)

3. What is the ILD of U1L2 for bottom-chord loading?
   A. Positive all along
   B. Negative from L0 to the zero point at 4.8 m, then positive with peak +0.854 at L2
   C. Positive then negative with peak -0.854
   D. Zero except at L2
   Answer: B. Load left of L1 gives compression (-0.213 at L1), load right of L2 gives tension (+0.854 at L2); the neutral point is at 4.8 m. (Key point: Sign of U1L2 ILD)

4. How is the force in vertical U2L2 obtained?
   A. F = -V of panel L2L3, from the unloaded joint U2
   B. F = +R_L0 always
   C. F = V / sin(theta)
   D. It is always zero
   Answer: A. At the unloaded joint U2, the vertical balances the diagonal's vertical component, so F = -V of panel L2L3 (+0.333 at L2, -0.5 at L3). (Key point: Vertical U2L2)

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