---
title: "Slope deflection: continuous beam, B sinks 19 mm"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:slope-deflection-beam-sinks-19mm"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/slope-deflection-beam-sinks-19mm
site: BCT Engineering
---

# Slope deflection: continuous beam, B sinks 19 mm

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2074 Asoj, 12 marks
- Video: [Slope Deflection Method for Beam Analysis - Problem No 13 ( Sinking / Settlement of Support B )](https://www.youtube.com/watch?v=HHJ7qNxRuw4) — Stan Academy
- 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-beam-sinks-19mm
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Analyse the continuous beam loaded as shown in figure below and draw the bending moment diagrams using the slope deflection method. Support B sinks by 19 mm. Take $EI = 10{,}000\ \text{kN-m}^2$. [Figure: beam ABCD, A fixed; AB = 4 m (1.5EI) with UDL (intensity printed as ':0 kN/m', likely 10 kN/m [?]); BC = 4 m (2EI) with a 50 kN point load at 2 m from C; CD = 2 m + 2 m (1.5EI), D fixed.]

## Short answer

Slope-deflection incorporates chord rotations from the 19 mm settlement. Final end moments: M_AB = -114.63, M_BA = -82.39, M_BC = 82.39, M_CB = 74.97, M_CD = -74.97, M_DC = -37.49 kNm.

## Answer

**Assumptions (figure partly illegible):** the UDL printed as ':0 kN/m' is taken as 10 kN/m on AB; BC carries 50 kN at mid-span (2 m from C); CD (2 m + 2 m) carries no load; spans are 4 m each. A and D are fixed. Support B settles by 19 mm. $EI=10\,000$ kNm²; AB = 1.5EI, BC = 2EI, CD = 1.5EI.

#### Chord rotations (clockwise +, a right-hand end that goes down rotates the chord clockwise)
- $\psi_{AB}=\dfrac{0.019}{4}=0.00475$ (B moves down relative to A)
- $\psi_{BC}=-\dfrac{0.019}{4}=-0.00475$ (C is higher than B)
- $\psi_{CD}=0$

#### Fixed-end moments (clockwise +, kNm)
- AB: $\mp\dfrac{10\times4^2}{12}=\mp13.33$
- BC: $\mp\dfrac{50\times4}{8}=\mp25.00$
- CD: 0

#### Slope-deflection equations
(the constants contain the FEM and the term $-6EI\psi/L$ from the settlement)
- $M_{AB}=7500.00\left(\theta_{B}\right) - 120.21$
- $M_{BA}=7500.00\left(2\theta_{B}\right) - 93.54$
- $M_{BC}=10000.00\left(2\theta_{B} + \theta_{C}\right) + 117.50$
- $M_{CB}=10000.00\left(2\theta_{C} + \theta_{B}\right) + 167.50$
- $M_{CD}=7500.00\left(2\theta_{C}\right)$
- $M_{DC}=7500.00\left(\theta_{C}\right)$

#### Equilibrium and solution
Joint B: $M_{BA}+M_{BC}=0$. Joint C: $M_{CB}+M_{CD}=0$.
$$
\begin{bmatrix}35000.0000 & 10000.0000 \\ 10000.0000 & 35000.0000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}-23.958 \\ -167.500\end{Bmatrix}
$$
$$
\theta_{B}=0.000743519,\quad \theta_{C}=-0.00499815
$$

#### Final end moments (kNm)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | -114.63 | BA | -82.39 |
| BC | 82.39 | CB | 74.97 |
| CD | -74.97 | DC | -37.49 |

Checks:
- Joint B: (-82.39) + (82.39) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Joint C: (74.97) + (-74.97) = 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 | -114.63 | 82.39 | - |
| BC | 82.39 | -74.97 | 53.71 at 2.00 m (under the 50 kN load) |
| CD | -74.97 | 37.49 | - |

Reactions (kN, kNm): A: $H=0.00$, $V=69.26$, $M=114.63$; B: $H=0.00$, $V=-43.60$, $M=0.00$; C: $H=0.00$, $V=92.45$, $M=0.00$; D: $H=0.00$, $V=-28.11$, $M=37.49$.

For comparison, if B did not settle the end moments at B would be $M_{BA}=22.11$ and $M_{BC}=-22.11$ kNm; the settlement therefore changes $M_{BA}$ by -104.50 kNm.

**Answer: $M_{AB}=-114.63$, $M_{BA}=-82.39$, $M_{BC}=82.39$, $M_{CB}=74.97$, $M_{CD}=-74.97$, $M_{DC}=-37.49$ kNm.**

## Check yourself

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

1. When support B sinks 19 mm, what is the chord rotation of span AB (L = 4 m)?
   A. +0.00475 (clockwise), as B goes down relative to A
   B. -0.00475
   C. 0.019
   D. 0
   Answer: A. psi = settlement/L = 0.019/4 = 0.00475; the right-hand end moving down rotates the chord clockwise (positive). (Key point: Chord rotation)

2. Why is psi_BC negative?
   A. Because C is fixed
   B. C is higher than B relative to the chord, so the chord rotates anticlockwise
   C. Because BC carries a point load
   D. Because EI of BC is larger
   Answer: B. B has moved down, so C is higher than B; the right end going up turns the chord anticlockwise. (Key point: Span BC chord rotation)

3. Which joint equations determine the unknown rotations theta_B and theta_C?
   A. M_AB + M_BA = 0
   B. theta_B = theta_C
   C. M_BA + M_BC = 0 and M_CB + M_CD = 0
   D. M_AB = 0 and M_DC = 0
   Answer: C. With no external moment at B or C, the sum of member end moments at each joint is zero; A and D are fixed so their rotations are zero. (Key point: Equilibrium equations)

4. What does the settlement do to the moments at B compared with no settlement?
   A. No change
   B. It only affects the fixed support A
   C. It makes M_BA zero
   D. It changes M_BA by about -104.5 kNm (the -6EI psi/L term adds to the constants)
   Answer: D. Settlement adds the term -6EI psi/L to the equations; M_BA changes from 22.11 to -82.39, a difference of -104.5 kNm. (Key point: Effect of settlement)

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