---
title: "Slope deflection: three-span beam with settlement"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:slope-deflection-three-span-fixed-ends"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/slope-deflection-three-span-fixed-ends
site: BCT Engineering
---

# Slope deflection: three-span beam with settlement

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2066 Jestha, 20 marks
- Video: [Slope Deflection Multi Span Beam Example 2 Steps 1 and 2 (Part 1)](https://www.youtube.com/watch?v=VKwJYUaEsqg) — Inigo Montoya Rules
- 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/slope-deflection-three-span-fixed-ends
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

A three-spanned continuous beam is fixed at both extreme ends. The left span is of sectional stiffness EI and is loaded with a uniform distributed load of intensity 2 kN/m. The mid span is of stiffness 2 EI and has a vertical concentrated force of magnitude 5 kN applied at a point 2 m from the right end. The right span is of sectional stiffness EI and is centrally loaded with a vertical concentrated force of magnitude 8 kN. The left span is of length 6 m and the other two are of 5 m. The left middle support settles 2 mm down and right middle support is lifted 3 mm up. Analyse using the slope deflection method and draw bending moment diagram for the beam if all the forces being applied are directed vertically downward. Take $EI = 8\times10^{11}\ \text{N-mm}^2$.

## Short answer

Use slope-deflection equations including support settlement chord rotations, with EI=800 kNm^2. Solve joint equilibrium for theta_B=-0.003295 and theta_C=0.0011275. Final moments: M_AB=-7.15, M_BA=3.98, M_BC=-3.98, M_CB=4.85, M_CD=-4.85, M_DC=4.78 kNm.

## Answer

**Data:** three spans AB = 6 m ($EI$, 2 kN/m), BC = 5 m ($2EI$, 5 kN at 2 m from C, i.e. 3 m from B), CD = 5 m ($EI$, 8 kN at mid-span); A and D are fixed. Support B settles 2 mm and support C is lifted 3 mm. $EI=8\times10^{11}$ N mm² $=800$ kNm². Units: kN, m.

#### Chord rotations (clockwise +)
- $\psi_{AB}=\dfrac{0.002}{6}=0.000333$ (B goes down)
- $\psi_{BC}=\dfrac{-(0.003+0.002)}{5}=-0.001000$ (C is 5 mm above B)
- $\psi_{CD}=\dfrac{0.003}{5}=0.000600$ (D is 3 mm below C)

#### Fixed-end moments (kNm, clockwise +)
- AB: $\mp\dfrac{2\times6^2}{12}=\mp6.000$
- BC ($a=3$, $b=2$): $FEM_{BC}=-\dfrac{5\times3\times2^2}{5^2}=-2.400$, $FEM_{CB}=+\dfrac{5\times3^2\times2}{5^2}=3.600$
- CD: $\mp\dfrac{8\times5}{8}=\mp5.000$

#### Slope-deflection equations ($\theta_A=\theta_D=0$)
- $M_{AB}=266.67\left(\theta_{B}\right) - 6.27$
- $M_{BA}=266.67\left(2\theta_{B}\right) + 5.73$
- $M_{BC}=640.00\left(2\theta_{B} + \theta_{C}\right) - 0.48$
- $M_{CB}=640.00\left(2\theta_{C} + \theta_{B}\right) + 5.52$
- $M_{CD}=320.00\left(2\theta_{C}\right) - 5.58$
- $M_{DC}=320.00\left(\theta_{C}\right) + 4.42$

#### Equilibrium and solution
Joint B: $M_{BA}+M_{BC}=0$; joint C: $M_{CB}+M_{CD}=0$.
$$
\begin{bmatrix}1813.3333 & 640.0000 \\ 640.0000 & 1920.0000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}-5.253 \\ 0.056\end{Bmatrix}
$$
$$
\theta_{B}=-0.003295,\quad \theta_{C}=0.0011275
$$

#### Final end moments (kNm)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | -7.15 | BA | 3.98 |
| BC | -3.98 | CB | 4.85 |
| CD | -4.85 | DC | 4.78 |

Checks:
- Joint B: (3.98) + (-3.98) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Joint C: (4.85) + (-4.85) = 0.00 kNm (applied clockwise moment 0.00) ✓

#### Bending moment diagram (sagging +, kNm)
| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | -7.15 | -3.98 | 3.51 at 3.26 m |
| BC | -3.98 | -4.85 | 1.50 at 3.00 m |
| CD | -4.85 | -4.78 | 5.18 at 2.50 m |
Under the 5 kN load: 1.50 kNm; under the 8 kN load: 5.18 kNm.

Reactions (kN, kNm; support moment M anticlockwise +): A: $H=0.00$, $V=6.53$, $M=7.15$; B: $H=0.00$, $V=7.30$, $M=0.00$; C: $H=0.00$, $V=7.19$, $M=0.00$; D: $H=0.00$, $V=3.99$, $M=-4.78$.

**Answer: $M_{AB}=-7.15$, $M_{BA}=3.98$, $M_{BC}=-3.98$, $M_{CB}=4.85$, $M_{CD}=-4.85$, $M_{DC}=4.78$ kNm.**

## Check yourself

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

1. B settles 2 mm down and C is lifted 3 mm up. What is the chord rotation of span BC (L = 5 m)?
   A. +0.001000 (clockwise)
   B. -0.001000 (anticlockwise)
   C. -0.000200
   D. +0.000600
   Answer: B. C is 5 mm above B, so the chord turns anticlockwise: psi = -(0.003+0.002)/5 = -0.001. (Key point: Chord rotation sign)

2. For the 5 kN load 3 m from B on BC (L = 5 m), which expression gives FEM_BC?
   A. -PL/8
   B. -P a^2 b / L^2 = -3.6
   C. -P a b^2 / L^2 = -2.4
   D. -wL^2/12
   Answer: C. FEM at the near end B is -P a b^2/L^2 with a = 3 and b = 2, giving -2.4 kNm. (Key point: Fixed-end moment for off-centre load)

3. Why is the coefficient in the M_BC equation (640) larger than that in M_AB (266.67)?
   A. BC is shorter and has stiffness 2EI, so 2(2EI)/L is larger than 2EI/L of AB
   B. BC carries a concentrated load
   C. BC has settlement
   D. Joint B is fixed
   Answer: A. The coefficient is 2EI/L: 2(2x800)/5 = 640 for BC, but 2x800/6 = 266.67 for AB. (Key point: Stiffness coefficients)

4. What equations are used to find theta_B and theta_C?
   A. M_AB = 0 and M_DC = 0
   B. M_BA + M_BC = 0 and M_CB + M_CD = 0 (no external joint moment)
   C. Sum of vertical forces equals zero at B and C
   D. theta_B = theta_C
   Answer: B. Joints B and C carry no applied moment, so the sum of member end moments at each joint is zero. (Key point: Joint equilibrium equations)

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