---
title: "Three moment theorem: SFD and BMD of beam"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:three-moment-beam-100kn-60knm-20kn"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/three-moment-beam-100kn-60knm-20kn
site: BCT Engineering
---

# Three moment theorem: SFD and BMD of beam

Theory of Structures II · Chapter 2: Force method

- Asked in: 2070 Asar
- Video: [SA60: The Three-Moment Equation for the Analysis of Continuous Beams (Part I)](https://www.youtube.com/watch?v=SHrnZtU11mc) — Dr. Structure
- 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-beam-100kn-60knm-20kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Analyse the continuous beam shown in figure below by using the three moment theorem. Draw shear force and bending moment diagram. [Figure: beam ABCD, A fixed; 100 kN at 4 m from A; B at 8 m; BC = 10 m with 60 kN/m UDL; overhang CD = 2 m with 20 kN at D. Printed as the alternative (OR) to the force-method frame question above.]

## Short answer

Using three-moment theorem, moments: M_A = 96.87 kNm, M_B = -493.75 kNm, M_C = -40 kNm. Reactions: R_A = -23.83 kN (down), R_B = 469.20 kN, R_C = 274.63 kN. Maximum sagging moment in BC = 500.28 kNm at 5.76 m from B.

## Answer

Reading of the figure: A fixed; span AB = 8 m with 100 kN at 4 m from A; span BC = 10 m with 60 kN/m; overhang CD = 2 m with 20 kN at D. B and C are rollers and $EI$ is constant.

#### Moment at C (overhang)
$M_C = -20\times2 = -40$ kNm (hogging).

#### Load terms
- Span AB: free BM peak $\dfrac{100\times4\times4}{8} = 200$ kNm, $a_1 = \tfrac12(8)(200) = 800$, $\bar x_1 = 4$ m: $\dfrac{6a_1\bar x_1}{L_1} = \dfrac{6(800)(4)}{8} = 2400$
- Span BC: $\dfrac{wL_2^3}{4} = \dfrac{60(10)^3}{4} = 15000$

#### Three moment equations
**At A (fixed end):** $2M_A(8) + M_B(8) = -2400$

$$
2M_A + M_B = -300
$$

**At B:** $M_A(8) + 2M_B(8+10) + M_C(10) = -2400 - 15000$

$$
8M_A + 36M_B + 10(-40) = -17400 \ \Rightarrow\ 8M_A + 36M_B = -17000
$$

Substituting $M_A = \dfrac{-300 - M_B}{2}$: $-1200 - 4M_B + 36M_B = -17000$, so $32M_B = -15800$:

$$
M_B = -493.750\ \text{kNm},\qquad M_A = 96.875\ \text{kNm}
$$

($M_A$ is positive: the fixed end is sagging for this loading.)

#### Reactions and shear
Span AB ($M_B = M_A + 8R_A - 400$):

$$
R_A = \frac{400 + M_B - M_A}{8} = \frac{400 - 493.75 - 96.875}{8} = -23.828\ \text{kN}
$$

(negative: A is pulled down). Shear just left of B $= -23.828 - 100 = -123.828$ kN.

Span BC ($M_C = M_B + 10V_{BR} - 60\frac{10^2}{2}$):

$$
V_{BR} = \frac{3000 + M_C - M_B}{10} = \frac{3000 - 40 + 493.75}{10} = 345.375\ \text{kN}
$$

$$
R_B = 123.828 + 345.375 = 469.203\ \text{kN},\qquad R_C = 600 - 345.375 + 20 = 274.625\ \text{kN}
$$

Check: $R_A + R_B + R_C = 720.00$ kN $= 100 + 600 + 20 = 720$ kN.

#### Shear force (kN)
| Section | SF |
|---|---|
| A to load point E | -23.83 |
| E to B | -123.83 |
| B to C (at B / at C) | 345.375 / -254.63 |
| C to D (overhang) | +20.00 |

The SF is zero in BC at 5.76 m from B.

#### Bending moment (kNm)
| Section | BM |
|---|---|
| A | 96.87 |
| E (under 100 kN) | 1.56 |
| B | -493.75 |
| Maximum in BC (5.76 m from B) | 500.28 |
| C | -40.00 |
| D | 0 |

**Answer: $M_A = 96.87$ kNm, $M_B = -493.75$ kNm, $M_C = -40$ kNm; $R_A = -23.828$ kN, $R_B = 469.203$ kN, $R_C = 274.625$ kN; maximum sagging moment in BC $= 500.28$ kNm.**

## Check yourself

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

1. How is the moment at C fixed before using three-moment equations?
   A. M_C = -20 × 2 = -40 kNm by statics of the overhang
   B. M_C = 0
   C. M_C = 60 × 10
   D. Found by the third equation
   Answer: A. The overhang is statically determinate: hogging 40 kNm at C. (Key point: Overhang moment)

2. How is the fixed end A handled?
   A. M_A = 0
   B. Treated as a simple support
   C. With the equation 2M_A(8) + M_B(8) = -6a x̄/L
   D. Ignored
   Answer: C. A fixed end gives an equation with the end-span term only: 2M_A L + M_B L = -6a x̄/L. (Key point: Fixed end)

3. For the UDL span of 10 m, 6ax̄/L equals:
   A. wL^3/4 = 15000
   B. wL^2/8
   C. wL^3/24
   D. wL/2
   Answer: A. For a UDL the term is wL^3/4 = 60×1000/4. (Key point: Load term for UDL)

4. Solving the equations gives M_B equal to:
   A. +493.75 kNm
   B. -493.75 kNm (hogging)
   C. -40 kNm
   D. -96.9 kNm
   Answer: B. 32 M_B = -15800 so M_B = -493.75 kNm. (Key point: Result M_B)

## Related in Theory of Structures II

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- [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
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