---
title: "Moment area: mid-span deflection under UDL"
subject: "Theory of Structures I"
chapter: 4
key: "semester_4/theory_of_structures_i:moment-area-ss-udl-midspan"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/moment-area-ss-udl-midspan
site: BCT Engineering
---

# Moment area: mid-span deflection under UDL

Theory of Structures I · Chapter 4: Deflection of Beams

- Asked in: 2071 Magh, 7 marks
- Video: [Moment-Area Method - Step by Step Explanation and Solved Example - Deflection and Slope](https://www.youtube.com/watch?v=jkwPzvWsqIY) — AF Math & Engineering
- 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/moment-area-ss-udl-midspan
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4

## Question

Determine the deflection at mid span of a simply supported beam subjected to uniformly distributed load w kN/m on the whole span by moment area. [Figure: simply supported beam of span L, flexural rigidity EI, UDL w kN/m over the whole span.]

## Short answer

Using symmetry, the tangent at mid-span is horizontal. The deviation of the support from this tangent equals the mid-span deflection, yielding 5wL⁴/(384EI) downward.

## Answer

#### Given data
Simply supported beam, span $L$, rigidity $EI$, UDL $w$ over the whole span. The deflection at mid-span C is required.

```text
  w kN/m
 A ===================== B
 ^          L            ^
```

#### Bending moment
$$
R_A = R_B = \frac{wL}{2},\qquad M = \frac{wLx}{2} - \frac{wx^2}{2}
$$

Maximum moment at mid-span: $M_C = \dfrac{wL^2}{8}$. The $M/EI$ diagram is a parabola with peak $\dfrac{wL^2}{8EI}$.

#### Using the symmetry
By symmetry the tangent at mid-span C is horizontal. The deflection at C below the support level is therefore equal to the deviation of A from the tangent at C (Theorem II).

#### Area and its moment about A (portion A to C)
$$
\text{Area} = \int_0^{L/2}\frac{M}{EI}dx = \frac{w}{2EI}\left[\frac{Lx^2}{2} - \frac{x^3}{3}\right]_0^{L/2} = \frac{wL^3}{24EI}
$$

Using the parabola property, the area under a parabola from its vertex over base $L/2$ with height $h$ is $\tfrac23 \cdot \tfrac{L}{2}\cdot h = \tfrac23\cdot\tfrac{L}{2}\cdot\dfrac{wL^2}{8EI} = \dfrac{wL^3}{24EI}$ (check). Its centroid is $\dfrac{5L}{16}$ from A.

#### Deflection at mid-span
$$
t_{A/C} = \text{Area}\times\bar{x}_A = \frac{wL^3}{24EI}\times\frac{5L}{16}
$$

$$
\boxed{y_C = \frac{5wL^4}{384\,EI}}\ \text{(downward)}
$$

Check with the slope: $\theta_A = \dfrac{wL^3}{24EI}$ (clockwise), the area between A and C.

## Check yourself

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

1. Why can the deflection at midspan be found from the deviation of A from the tangent at C?
   A. Because by symmetry the tangent at C is horizontal
   B. Because the moment at C is zero
   C. Because A is fixed
   D. Because the slope at A is zero
   Answer: A. With symmetric loading the slope at mid-span is zero, so the tangent there is horizontal and t_A/C equals the deflection. (Key point: Horizontal tangent at midspan)

2. What is the shape and peak of the M/EI diagram?
   A. Triangle, peak wL^2/(4EI)
   B. Parabola, peak wL^2/(8EI)
   C. Rectangle, height wL^2/(8EI)
   D. Parabola, peak wL^2/(2EI)
   Answer: B. M = wLx/2 - wx^2/2 is parabolic with maximum wL^2/8 at midspan. (Key point: M/EI diagram)

3. What are the area from A to C and the centroid distance from A?
   A. wL^3/24EI, 5L/16
   B. wL^3/12EI, L/4
   C. wL^3/24EI, 3L/16
   D. wL^3/48EI, 5L/16
   Answer: A. Area = 2/3 x L/2 x wL^2/8EI = wL^3/24EI; its centroid is 5L/16 from A. (Key point: Area and centroid)

4. What is the mid-span deflection?
   A. wL^4/(8EI)
   B. 5wL^4/(384EI)
   C. wL^4/(48EI)
   D. 5wL^3/(384EI)
   Answer: B. y = (wL^3/24EI)(5L/16) = 5wL^4/384EI. (Key point: Result)

## Related in Theory of Structures I

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## Related in other subjects

- [Mid-span deflection of beam, 10 kN load](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ass-beam-deflection-mid-span-4m) — Strength of Materials, chapter 7
- [Pure bending; deflection of SS beam under UDL](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Apure-bending-udl-deflection) — Strength of Materials, chapter 7
- [Castigliano: central deflection of simply supported beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-central-deflection-ss-beam-udl) — Theory of Structures II, chapter 2
- [Flexural rigidity; T-beam bending stresses (60 kN loads)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aflexural-rigidity-t-beam-60kn) — Strength of Materials, chapter 7