---
title: "Truss: max forces under long UDL"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:truss-max-forces-members-1-2-3-4"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-max-forces-members-1-2-3-4
site: BCT Engineering
---

# Truss: max forces under long UDL

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

- Asked in: 2076 Bhadra, 8 marks
- Video: [Case 2- ILD for max BM & Sf for UDL greater than Span.](https://www.youtube.com/watch?v=r7V263Bnya0) — Aesthetic Civil
- 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-max-forces-members-1-2-3-4
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Determine the maximum forces in the members 2, 3 and 4 of given truss when uniformly distributed load of 10 kN/m longer than the span traverses along the girder. [Figure: truss with bottom joints $L_0$ to $L_5$ (5 × 4 = 20 m), top joints $U_1$ to $U_4$, end angles 45°; members marked 1 (top chord $U_2U_3$), 2, 3 and 4 (diagonals and a vertical near $U_2L_2$).]

## Short answer

For a 10 kN/m UDL longer than the span, maximum forces are: member 2 (U1L2) 63.64 kN tension, 7.07 kN compression; member 3 (U2L2) 20.00 kN tension, 20.00 kN compression; member 4 (U2L3) 28.28 kN tension, 28.28 kN compression.

## Answer

**Assumed truss (figure not available):** Pratt truss, 5 panels of 4 m (span 20 m), height 4 m (end posts $L_0U_1$ and $L_5U_4$ at 45°). Verticals $U_1L_1\ldots U_4L_4$; diagonals $U_1L_2$, $U_2L_3$ and $U_4L_3$. Members: **2 = $U_1L_2$** (diagonal), **3 = $U_2L_2$** (vertical), **4 = $U_2L_3$** (diagonal). The UDL ($w = 10$ kN/m) is longer than the span and acts on the bottom chord, so it can cover any part of the span.

Reactions: $R_{L_0} = \dfrac{20-x}{20}$, $R_{L_5} = \dfrac{x}{20}$.

#### Method
Member forces are found by sections in the panel:

- **Diagonals:** $F = \pm V_{panel}/\sin 45^\circ$ with the panel shear from the reactions: $F_{U_1L_2}=\sqrt2\,V_{L_1L_2}$ and $F_{U_2L_3} = \sqrt2\,V_{L_2L_3}$.
- **Vertical $U_2L_2$:** at the unloaded joint $U_2$, $F_{U_2L_2} = -F_{U_2L_3}/\sqrt2$.

ILD ordinates for the unit load on the bottom chord (+ tension, − compression):

| Member | L0 | L1 | L2 | L3 | L4 | L5 |
|---|---|---|---|---|---|---|
| U1L2 (2) | 0.000 | -0.283 | 0.849 | 0.566 | 0.283 | 0.000 |
| U2L2 (3) | 0.000 | 0.200 | 0.400 | -0.400 | -0.200 | 0.000 |
| U2L3 (4) | 0.000 | -0.283 | -0.566 | 0.566 | 0.283 | 0.000 |

For a UDL longer than the span, the maximum effect of one kind is obtained by loading only the portions of the ILD of that sign. Force $= w\times$ (area of the ILD of that sign).

| Member | Positive area (m) | Negative area (m) |
|---|---|---|
| U1L2 | 6.364 | -0.707 |
| U2L2 | 2.000 | -2.000 |
| U2L3 | 2.828 | -2.828 |

#### Maximum forces for 10 kN/m
- **Member 2 ($U_1L_2$):** maximum tension $=+63.64$ kN (load from the zero point at $x = 5.00$ m to the right support); maximum compression $=-7.07$ kN (load on $L_0$ to the zero point).
- **Member 3 ($U_2L_2$):** maximum tension $=+20.00$ kN (load from $L_0$ to the zero point at $x = 10$ m); maximum compression $=-20.00$ kN (load from the zero point to $L_5$). Here the sign changes at $x = 10$ m.
- **Member 4 ($U_2L_3$):** maximum tension $=+28.28$ kN (load from the zero point at $x = 10$ m, mid-span, to $L_5$); maximum compression $=-28.28$ kN (load from $L_0$ to mid-span).

**Answer: member 2: 63.64 kN (T), 7.07 kN (C); member 3: 20.00 kN (T), 20.00 kN (C); member 4: 28.28 kN (T), 28.28 kN (C) (for 10 kN/m over the length that gives the maximum).**

## Check yourself

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

1. How is the force in a diagonal such as U2L3 obtained from the panel shear?
   A. F = V x sin 45°
   B. F = V / sin 45° = sqrt(2) V
   C. F = V / cos 60°
   D. F = V x tan 45°
   Answer: B. The vertical component of the diagonal balances the panel shear, so F = V / sin 45° = sqrt(2) V. (Key point: Why panel shear)

2. For a UDL longer than the span, how is the maximum force of one sign found?
   A. Load the whole span always
   B. Load only the portions of the ILD of that sign; force = w x area
   C. Place the load at the largest ordinate only
   D. Load half the span from the support
   Answer: B. The UDL can cover any part of the span, so load only the part of the ILD with the required sign. (Key point: UDL longer than span)

3. What is the maximum tension in member 2 (U1L2) for 10 kN/m?
   A. 7.07 kN
   B. 63.64 kN
   C. 20.00 kN
   D. 28.28 kN
   Answer: B. Positive area 6.364 m times 10 kN/m = 63.64 kN, loading from the zero point at x = 5 m to the right support. (Key point: Member 2 maximum)

4. For vertical U2L2 (member 3), why are maximum tension and compression equal (20 kN)?
   A. Its ILD has equal positive and negative areas (2.0 m), changing sign at x = 10 m
   B. It is a zero-force member
   C. The load covers the whole span in both cases
   D. Its ILD has no sign change
   Answer: A. Areas are +2.000 and -2.000, so 10 x 2 = 20 kN in each sense, with the neutral point at mid-span. (Key point: Member 3 sign change)

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