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

# Slope deflection: B sinks, C rises

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2071 Chaitra, 15 marks
- Video: [Sink beam analysis using slope deflection method | Structural Analysis 1](https://www.youtube.com/watch?v=15ewaHL8zhI) — Parag Pal
- 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-settlement-b-c
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

A continuous beam is shown in figure, support 'B' sinks by 10 mm down and 'C' rises by 20 mm up during loads. Analyse the given beam using the slope deflection method and also draw bending moment diagram and show deflected shape. $E = 200\times10^{6}\ \text{kN/m}^2$ and $I = 80\times10^{-6}\ \text{m}^4$. [Figure: beam ABCD, A and D fixed; 12 kN in AB (1.5EI), 10 kN in BC (2EI), 5 kN/m UDL on CD (EI); dimensions 4 m, 2 m, 3 m, 3 m, 4 m as marked (exact load positions unclear).]

## Short answer

Applying the slope-deflection method with chord rotations due to support settlements (B sinks 10 mm, C rises 20 mm), the equilibrium equations give the end moments: M_AB=-112.13 kNm, M_BA=-116.26 kNm, M_BC=116.26 kNm, M_CB=138.02 kNm, M_CD=-138.02 kNm, M_DC=-119.01 kNm.

## Answer

**Data and assumptions:** $EI=200\times10^6\times80\times10^{-6}=16\,000$ kNm². AB = 4 m (1.5EI) with 12 kN at mid-span; BC = 6 m (2EI) with 10 kN at mid-span; CD = 4 m (EI) with 5 kN/m; A and D are fixed. B sinks 10 mm, C rises 20 mm.

#### Chord rotations (clockwise +)
- $\psi_{AB}=\dfrac{0.010}{4}=0.00250$ (B goes down)
- $\psi_{BC}=\dfrac{-(0.020+0.010)}{6}=-0.00500$ (C is 30 mm higher than B, so the chord turns anticlockwise)
- $\psi_{CD}=\dfrac{0.020}{4}=0.00500$ (D is 20 mm lower than C)

#### Fixed-end moments (kNm, clockwise +)
- AB: $\mp\dfrac{12\times4}{8}=\mp6$
- BC: $\mp\dfrac{10\times6}{8}=\mp7.5$
- CD: $\mp\dfrac{5\times4^2}{12}=\mp6.67$

#### Slope-deflection equations
- $M_{AB}=12000.00\left(\theta_{B}\right) - 96.00$
- $M_{BA}=12000.00\left(2\theta_{B}\right) - 84.00$
- $M_{BC}=10666.67\left(2\theta_{B} + \theta_{C}\right) + 152.50$
- $M_{CB}=10666.67\left(2\theta_{C} + \theta_{B}\right) + 167.50$
- $M_{CD}=8000.00\left(2\theta_{C}\right) - 126.67$
- $M_{DC}=8000.00\left(\theta_{C}\right) - 113.33$

#### Equilibrium and solution
$M_{BA}+M_{BC}=0$ and $M_{CB}+M_{CD}=0$:
$$
\begin{bmatrix}45333.3333 & 10666.6667 \\ 10666.6667 & 37333.3333\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}-68.500 \\ -40.833\end{Bmatrix}
$$
$$
\theta_{B}=-0.00134403,\quad \theta_{C}=-0.000709741
$$

#### Final end moments (kNm)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | -112.13 | BA | -116.26 |
| BC | 116.26 | CB | 138.02 |
| CD | -138.02 | DC | -119.01 |

Checks:
- Joint B: (-116.26) + (116.26) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Joint C: (138.02) + (-138.02) = 0.00 kNm (applied clockwise moment 0.00) ✓

#### Bending moment diagram ordinates (sagging +, kNm)
| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | -112.13 | 116.26 | - |
| BC | 116.26 | -138.02 | - |
| CD | -138.02 | 119.01 | - |

#### Deflected shape
- A and D: fixed (zero deflection, zero slope).
- B: moves down 10 mm; rotation $\theta_B=-0.00134$ rad (clockwise +).
- C: moves up 20 mm; rotation $\theta_C=-0.00071$ rad.
- The curve is concave upward (sagging) where the BM is positive and concave downward (hogging) where it is negative; the points of contraflexure (BM = 0) are at: AB at 1.78 m from its first end, BC at 3.09 m from its first end, CD at 1.99 m from its first end.
- The curve falls from A to B, rises across BC from the low point at B to the raised support C, and then falls again to the fixed end D. At every fixed end the tangent is horizontal.

Reactions (kN, kNm): A: $H=0.00$, $V=63.10$, $M=112.13$; B: $H=0.00$, $V=-88.48$, $M=0.00$; C: $H=0.00$, $V=121.64$, $M=0.00$; D: $H=0.00$, $V=-54.26$, $M=119.01$.

**Answer: $M_{AB}=-112.13$, $M_{BA}=-116.26$, $M_{BC}=116.26$, $M_{CB}=138.02$, $M_{CD}=-138.02$, $M_{DC}=-119.01$ kNm.**

## Check yourself

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

1. EI = ?
   A. 1600
   B. 200e6 x 80e-6 = 16000 kNm^2
   C. 2000
   D. 160000
   Answer: B. E x I = 16000 kNm^2. (Key point: EI value)

2. psi_BC = -0.005 because?
   A. B sinks 20 mm
   B. L = 4
   C. C is 30 mm higher than B over 6 m, chord turns anticlockwise
   D. C sinks 30 mm
   Answer: C. B down 10 and C up 20 give a 30 mm relative rise: 0.030/6 = 0.005 anticlockwise. (Key point: Chord rotation BC)

3. psi_CD = +0.005 because?
   A. C rises
   B. D rises
   C. CD is fixed
   D. D is 20 mm lower than C over 4 m
   Answer: D. Relative drop of the right end is clockwise: 0.020/4. (Key point: Chord CD)

4. Where is the curve concave upward?
   A. Where BM is positive (sagging)
   B. Where BM is negative
   C. Only at fixed ends
   D. Nowhere
   Answer: A. Sagging moment curves the beam concave upward; contraflexure marks the change. (Key point: Deflected shape)

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