---
title: "Derive three moment theorem"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:three-moment-theorem-derive"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/three-moment-theorem-derive
site: BCT Engineering
---

# Derive three moment theorem

Theory of Structures II · Chapter 2: Force method

- Asked in: 2073 Shrawan, 6 marks; 2069 Chaitra, 7 marks (asked 2 times)
- Video: [Clapeyron's Theorem of Three Moments | Three Moment Theorem](https://www.youtube.com/watch?v=PZLpHIK9BME) — Ganesh Shegar
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/three-moment-theorem-derive
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Derive the expression of the three moment theorem for a continuous beam and explain its physical meaning.

## Short answer

The three-moment equation M_A L_1 + 2M_B(L_1+L_2) + M_C L_2 = –(6 a_1 x̄_1/L_1 + 6 a_2 x̄_2/L_2) is derived by equating the slope at interior support B from the left and right spans using moment‑area theorems. It enforces slope continuity; for n supports it provides n–2 equations to find support moments.

## Answer

#### Statement
For two adjacent spans $L_1$ and $L_2$ of a continuous beam between three supports A, B and C, the support moments are related by the three moment equation (Clapeyron's theorem):

$$
M_A L_1 + 2M_B (L_1 + L_2) + M_C L_2 = -\left(\frac{6 a_1 \bar{x}_1}{L_1} + \frac{6 a_2 \bar{x}_2}{L_2}\right)
$$

where $a_1, a_2$ are the areas of the free (simply supported) BM diagrams on spans 1 and 2, $\bar{x}_1$ is the distance of the centroid of $a_1$ from A, and $\bar{x}_2$ is the distance of the centroid of $a_2$ from C. If the supports settle (downward $\delta_A, \delta_B, \delta_C$), the term $+6EI\left[\dfrac{\delta_B-\delta_A}{L_1} + \dfrac{\delta_B-\delta_C}{L_2}\right]$ is added to the right side (EI constant).

#### Derivation (supports at the same level, EI constant)
Let the support moments be $M_A, M_B, M_C$ (sagging positive, so hogging moments are negative). The BM diagram on each span is the sum of the free BM diagram and a trapezoid of support moments.

**Step 1: Slope at B in span AB.** The deviation of A from the tangent at B equals the moment about A of the $M/EI$ diagram between A and B. The diagram is the free BM diagram (area $a_1$, centroid $\bar{x}_1$ from A) plus two triangles from $M_A$ and $M_B$. With the supports at the same level, this deviation equals $\theta_{B1}L_1$:

$$
\theta_{B1}L_1 = \frac{1}{EI}\left[a_1\bar{x}_1 + \frac{M_A L_1}{2}\cdot\frac{L_1}{3} + \frac{M_B L_1}{2}\cdot\frac{2L_1}{3}\right]
$$

$$
\theta_{B1} = \frac{1}{EI}\left[\frac{a_1\bar{x}_1}{L_1} + \frac{M_A L_1}{6} + \frac{M_B L_1}{3}\right]
$$

**Step 2: Slope at B in span BC.** In the same way, taking moments about C (with $\bar{x}_2$ measured from C), the deviation of C from the tangent at B is $-\theta_{B2}L_2$:

$$
\theta_{B2} = -\frac{1}{EI}\left[\frac{a_2\bar{x}_2}{L_2} + \frac{M_C L_2}{6} + \frac{M_B L_2}{3}\right]
$$

**Step 3: Continuity.** The beam is continuous over B, so the slope on the left of B equals the slope on the right: $\theta_{B1} = \theta_{B2}$.

$$
\frac{a_1\bar{x}_1}{L_1} + \frac{M_A L_1}{6} + \frac{M_B L_1}{3} = -\frac{a_2\bar{x}_2}{L_2} - \frac{M_C L_2}{6} - \frac{M_B L_2}{3}
$$

Multiply by 6 and rearrange:

$$
M_A L_1 + 2M_B (L_1 + L_2) + M_C L_2 = -\frac{6a_1\bar{x}_1}{L_1} - \frac{6a_2\bar{x}_2}{L_2}
$$

#### Special results
- UDL $w$ on a span $L$: $\dfrac{6a\bar{x}}{L} = \dfrac{wL^3}{4}$.
- Point load $W$ at distance $a$ from the left support of a span $L$, with $b = L - a$: $\dfrac{6a\bar{x}}{L} = \dfrac{Wab(L+a)}{L}$ when $\bar{x}$ is measured from the left support, and $\dfrac{Wab(L+b)}{L}$ when measured from the right support.

#### Physical meaning
The equation expresses continuity of slope at the middle support B: the rotation at B caused by the unknown support moments must balance the rotation caused by the loads on the two spans acting as simple beams, so the tangent to the elastic curve is the same on both sides of B. For a beam with $n$ supports it gives $n-2$ equations; the remaining equations come from the end conditions (a fixed end is handled by adding an imaginary span of zero length, that is, an infinitely stiff span, beyond it; a free overhang gives a known support moment). Solving the equations gives the support moments, and the reactions and BM follow from statics.

## Check yourself

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

1. Three moment equation left side is:
   A. M_A L1 + 2M_B(L1+L2) + M_C L2
   B. M_A + M_B + M_C
   C. M_B(L1 - L2)
   D. 2M_A L1 + M_C
   Answer: A. It is Clapeyron's equation. (Key point: Equation form)

2. The equation expresses:
   A. Continuity of slope at the middle support
   B. Equilibrium of vertical forces
   C. Zero deflection at midspan
   D. Equal reactions
   Answer: A. Slope on left of B equals slope on right. (Key point: Physical meaning)

3. If supports settle, the equation:
   A. Is unchanged
   B. Gets an added term 6EI[(d_B-d_A)/L1 + (d_B-d_C)/L2] on the right
   C. Is not applicable
   D. Loses M_B
   Answer: B. Settlement adds a term in EI and the differential settlements. (Key point: Settlement term)

4. For a beam with n supports, the theorem gives:
   A. n-2 equations
   B. n equations
   C. n+2 equations
   D. 1 equation
   Answer: A. One per interior support; end conditions supply the rest. (Key point: Number of equations)

## Related in Theory of Structures II

- [Three moment theorem: SFD and BMD of beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Athree-moment-beam-100kn-60knm-20kn) — Theory of Structures II, chapter 2
- [Three moment equation: derive and apply to fixed beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Athree-moment-derive-fixed-beam-udl) — Theory of Structures II, chapter 2
- [ILD: moment at section x-x, continuous beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-moment-section-xx-two-span) — Theory of Structures II, chapter 4
- [Three moment equation: SFD and BMD of beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Athree-moment-beam-300kn-50knm) — Theory of Structures II, chapter 2
- [Three moment equation: beam with overhang](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Athree-moment-beam-100kn-40knm-20kn) — Theory of Structures II, chapter 2
- [Slope deflection: continuous beam, I = 4x10^7 mm^4](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-i4e7-60kn) — Theory of Structures II, chapter 3

## Related in other subjects

- [Slab three edges continuous, one short edge discontinuous (5x4 m, 250 beam)](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aslab-three-edges-continuous-one-short) — Design of RCC Structure, chapter 9, asked 3×
- [Slab continuous over three edges, 5x4 m (225 mm beam)](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aslab-continuous-over-three-edges-5x4-2) — Design of RCC Structure, chapter 9
- [Physical meaning of equilibrium and its application](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aphysical-meaning-equilibrium-structural) — Applied Mechanics, chapter 2
- [Physical meaning of convolution integral](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aconvolution-integral-meaning) — Signals and Systems, chapter 5