---
title: "Truss ILDs and maximum forces under UDL"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:truss-ild-u1u2-u2l2-l2l3-udl"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-ild-u1u2-u2l2-l2l3-udl
site: BCT Engineering
---

# Truss ILDs and maximum forces under UDL

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

- Asked in: 2081 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-u1u2-u2l2-l2l3-udl
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Draw the influence line diagrams for the forces in the members $U_1U_2$, $U_2L_2$ and $L_2L_3$ of the through type truss shown in figure below. Also, determine maximum forces in these members when uniformly distributed load of intensity 10 kN/m and length 4 m passes through the span. [Figure: through truss with bottom joints $L_1$ (hinge), $L_2$, $L_3$, $L_4$ (roller) and top joints $U_1$, $U_2$, $U_3$; 3 panels of 5 m; all inclined members at 60°.]

## Short answer

Maximum forces for 4 m long 10 kN/m UDL: U1U2 = -26.69 kN (compression), U2L2 = +11.29 kN (tension) or -11.29 kN (compression), L2L3 = +23.09 kN (tension).

## Answer

**Geometry:** equilateral triangles of side 5 m (all inclined members at 60°), height $h = 5\sin 60^\circ = 4.33$ m, span 15 m. $L_1$ hinge, $L_4$ roller. The load travels along the bottom chord (through truss), so each ILD is linear between panel points.

Reactions for a unit load at $x$ from $L_1$: $R_{L_1} = (15-x)/15$, $R_{L_4} = x/15$.

#### Method of sections
- **$U_1U_2$:** cut through $U_1U_2$, $U_1L_2$ and $L_1L_2$ and take moments about $L_2$. Top chord is in compression: $F_{U_1U_2} = -M_{L_2}/h$.
- **$U_2L_2$:** pass a section through $U_1U_2$, $U_2L_2$ and $L_2L_3$. The shear $V$ (left forces upward) in this panel is carried by the inclined member: $F_{U_2L_2} = -V/\sin 60^\circ = -1.155\,V$. For a load to the left of the section $V = R_{L_1}-1$; for a load to the right $V = R_{L_1}$.
- **$L_2L_3$:** moments about the joint $U_2$ gives $F_{L_2L_3} = M_{U_2}/h$ (tension).

#### Ordinates (per unit load; + tension, − compression)
| Member | L1 | L2 | L3 | L4 |
|---|---|---|---|---|
| U1U2 | 0.000 | -0.770 | -0.385 | 0.000 |
| U2L2 | 0.000 | 0.385 | -0.385 | 0.000 |
| L2L3 | 0.000 | 0.577 | 0.577 | 0.000 |

Shapes:
- $U_1U_2$: always compressive, triangle with peak $-0.770$ under $L_2$.
- $U_2L_2$: tension for load left of $L_2$ (peak $+0.385$ at $L_2$), compression for load right of $L_2$ (peak $-0.385$ at $L_3$); sign change between $L_2$ and $L_3$.
- $L_2L_3$: tension, flat $+0.577$ between $L_2$ and $L_3$.

#### Maximum forces for 10 kN/m over 4 m
Force $= w\times(\text{area of ILD under the load})$, with the 4 m load moved to the position giving the greatest area.

| Member | Maximum force | Load position (from L1) |
|---|---|---|
| U1U2 | -26.69 kN (C) | 3.67 m to 7.67 m |
| U2L2 (tension) | 11.29 kN | 2.33 m to 6.33 m |
| U2L2 (compression) | -11.29 kN | 8.67 m to 12.67 m |
| L2L3 (tension) | 23.09 kN | 4 m inside the panel L2L3 (5 m) |

**Answer: $U_1U_2 = 26.69$ kN (compression); $U_2L_2 = +11.29$ kN (tension) or $-11.29$ kN (compression); $L_2L_3 = +23.09$ kN (tension).**

## Check yourself

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

1. How is the force in U1U2 found from the ILD?
   A. Moments about L2 divided by h: F = -M_L2/h
   B. Vertical shear divided by sin 60
   C. Moments about U2
   D. Resolving at joint U1 only
   Answer: A. Cutting U1U2, U1L2 and L1L2 and taking moments about L2 gives the chord force. (Key point: Method U1U2)

2. How does the force in U2L2 relate to the shear?
   A. F = V cos 60
   B. F = -V/sin 60 = -1.155 V
   C. F = M/h
   D. F = V
   Answer: B. The vertical shear in the panel is carried by the inclined member. (Key point: Method U2L2)

3. What is the character of the ILD for U2L2?
   A. Always tension
   B. Always compression
   C. Tension for load left of L2, compression for load right of L2, with a sign change between L2 and L3
   D. Zero everywhere
   Answer: C. Ordinates are +0.385 at L2 and -0.385 at L3, so the sign changes in between. (Key point: Sign of U2L2)

4. What is the maximum force in U1U2 from the 10 kN/m, 4 m UDL?
   A. 26.69 kN tension
   B. 11.29 kN compression
   C. 23.09 kN tension
   D. 26.69 kN compression
   Answer: D. U1U2 is always compressive, and the UDL is placed for the maximum ILD area, giving 26.69 kN. (Key point: Maximum forces)

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## Related in other subjects

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